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Learning Objectives
• Learn about different ionization techniques
• Explain common ionization methods
A small amount of sample is injected into the mass spectrometer, which is vaporized upon entering. It is then bombarded with a stream of high-energy electrons. When the molecule of interest is struck by this high-energy electron beam, a valence electron is knocked out and the molecule becomes a positive ion. It has become ionized. There are a variety of ways for the instrument to ionize the molecule, which will be discussed here.
Electron Impact (EI)
In electron impact ionization, a vaporized sample is passed through a beam of electrons. The high energy (typically 70 eV) beam strips electrons from the sample molecules leaving a positively charged radical species. The molecular ion is typically unstable and undergoes decomposition or rearrangement to produce fragment ions. Because of this, electron impact is classified as a “hard” ionization technique. One of the main limitations of EI is that the sample must be volatile and thermally stable.
Chemical Ionization (CI)
In chemical ionization, the sample is introduced to a chamber filled with excess reagent gas (such as methane). The reagent gas is ionized by electrons, forming a plasma with species such as CH5+, which react with the sample to form the pseudomolecular ion [M+H]+. Because CI does not involve radical reactions, fragmentation of the sample is generally much lower than that of EI. CI can also be operated in negative mode (to generate anions) by using different reagent gases. For example, a mixture of CH4 and NO2 will generate hydroxide ions, which can abstract protons to yield the [M-H]- species. A related technique, atmospheric pressure chemical ionization (APCI) delivers the sample as a neutral spray, which is then ionized by corona discharge, producing ions in a similar manner as described above. APCI is particularly suited for low molecular weight, nonpolar species that cannot be easily analyzed by other common techniques such as ESI.
Field Ionization/Desorption
Field ionization and desorption are two closely related techniques which use quantum tunneling of electrons to generate ions. Typically, a highly positive potential is applied to an electrode with a sharp point, resulting in a high potential gradient at the tip (see figure below). As the sample reaches this field, electron tunneling occurs to generate the cation, which is repelled into the mass analyzer. Field ionization utilizes gaseous samples whereas in field desorption the sample is adsorbed directly onto the electrode. Both of these techniques are soft, resulting in low energy ions which do not easily fragment.
Electrospray Ionization (ESI)
Electrospray ionization mass spectrometry is a desorption ionization method. Desorption ionization methods can be performed on solid or liquid samples, and allows for the sample to be nonvolatile or thermally unstable. Electrospray ionization is a soft ionization technique that is typically used to determine the molecular weights of proteins, peptides, and other biological macromolecules. Soft ionization is a useful technique when considering biological molecules of large molecular mass, such as the aformetioned, because this process does not fragment the macromolecules into smaller charged particles, rather it turns the macromolecule being ionized into small droplets. These droplets will then be further desolvated into even smaller droplets, which creates molecules with attached protons. These protonated and desolvated molecular ions will then be passed through the mass analyzer to the detector, and the mass of the sample can be determined. As the droplets shrink due to evaporation, the charge density increases until a coulombic explosion occurs, producing daughter droplets that repeat the process until individualized sample ions are generated (see figure below). One of the limitations of is the requirement that the sample be soluble. ESI is best applied to charged, polar, or basic compounds.
Matrix Assisted Laser Desorption Ionization (MALDI)
Laser desorption ionization generates ions by ablation from a surface using a pulsed laser. This technique is greatly improved by the addition of a matrix co-crystallized with the sample. As the sample is irradiated, a plume of desorbed molecules is generated. It is believed that ionization occurs in this plume due to a variety of chemical and physical interactions between the sample and the matrix (see figure below). One of the major advantages of MALDI is that it produces singly charged ions almost exclusively and can be used to volatilize extremely high molecular weight species such as polymers and proteins. A related technique, desorption ionization on silicon (DIOS) also uses laser desorption, but the sample is immobilized on a porous silicon surface with no matrix. This allows the study of low molecular weight compounds which may be obscured by matrix peaks in conventional MALDI.
Inductively Coupled Plasma Mass Spectrometry (ICP-MS)
A plasma torch generated by electromagnetic induction is used to ionize samples. Because the effective temperature of the plasma is about 10,000 °C, samples are broken down to ions of their constituent elements. Thus, all chemical information is lost, and the technique is best suited for elemental analysis. ICP-MS is typically used for analysis of trace elements.
Fast Atom Bombardment (FAB) and Secondary Ion Mass Spectrometry (SIMS)
Both of these techniques involve sputtering a sample to generate individualized ions; FAB utilizes a stream of inert gas atoms (argon or xenon) whereas SIMS uses ions such as Cs+. Ionization occurs by charge transfer between the ions and the sample or by protonation from the matrix material (Figure \(4\)). Both solid and liquid samples may be analyzed. A unique aspect of these techniques for analysis of solids is the ability to do depth profiling because of the destructive nature of the ionization technique.
Choosing an Ionization Technique
Depending on the information desired from mass spectrometry analysis, different ionization techniques may be desired. For example, a hard ionization method such as electron impact may be used for a complex molecule in order to determine the component parts by fragmentation. On the other hand, a high molecular weight sample of polymer or protein may require an ionization method such as MALDI in order to be volatilized. Often, samples may be easily analyzed using multiple ionization methods, and the choice is simplified to choosing the most convenient method. For example, electrospray ionization may be easily coupled to liquid chromatography systems, as no additional sample preparation is required. Table \(1\) provides a quick guide to ionization techniques typically applied to various types of samples.
Information Desired Ionization Technique
Elemental analysis Inductively coupled plasma
Depth profiling Fast atom bombardment/secondary ion mass spectroscopy
Chemical speciation/component analysis (fragmentation desired) Electron impact
Molecular species identification of compounds soluble in common solvents Electrospray ionization
Molecular species identification of hydrocarbon compounds Field ionization
Molecular species identification of high molecular weight compounds Matrix assisted laser desorption ionization
Molecular species identification of halogen containing compounds Chemical ionization (negative mode)
Table \(1\) Strengths of various ionization techniques | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/02%3A_Mass_Spectrometry/2.03%3A_Ionization_Techniques.txt |
Learning Objectives
• Describe the function of common mass analyzers
Mass spectrometry is an analytic method that employs ionization and mass analysis of compounds to determine the mass, formula and structure of the compound being analyzed. A mass analyzer is the component of the mass spectrometer that takes ionized masses and separates them based on charge to mass ratios and outputs them to the detector where they are detected and later converted to a digital output. There are six general types of mass analyzers that can be used for the separation of ions in a mass spectrometry.
Quadrupole Mass Analyzer
Ions are passed through four parallel rods which apply a varying voltage and radiofrequency (RF) potential (figure below). As the field changes, ions respond by undergoing complex trajectories. Depending on the applied voltage and RF frequencies, only ions of a certain m/z ratio will have stable trajectories and pass through the analyzer. If their course goes off too far they will hit the metal rods or the sides of the container and be absorbed. A great deal of selectivity to molecules charge to mass ratio can be obtained because the rods bias can be tuned to specific charge to mass ratios hitting the detector. Quadrupole analyzers are relatively inexpensive, but have limited resolution and low mass range.
TOF (Time of Flight) Mass Analyzer
The amount of time required for an ion to travel a known distance is measured. A pulse of ions is accelerated through and electric analyzer such that they have identical kinetic energies. As a result, their velocity is directly dependent on their mass. Extremely high vacuum conditions are required to extend the mean free path of ions and avoid collisions. TOF mass analyzers are fastest, have unlimited mass ranges, and allow simultaneous detection of all species, but are best coupled with pulsed ionization sources. TOF Analyzers separate ions by time without the use of an electric or magnetic field. In a crude sense, TOF is similar to chromatography, except there is no stationary/ mobile phase, instead the separation is based on the kinetic energy and velocity of the ions.
Ions of the same charges have equal kinetic energies; kinetic energy of the ion in the flight tube is equal to the kinetic energy of the ion as it leaves the ion source.
Unfortunately, at higher masses, resolution is difficult because flight time is longer. Also at high masses, not all of the ions of the same m/z values reach their ideal TOF velocities. To fix this problem, often a reflectron is added to the analyzer. The reflectron consists of a series of ring electrodes of very high voltage placed at the end of the flight tube. When an ion travels into the reflectron, it is reflected in the opposite direction due to the high voltage. The reflectron increases resolution by narrowing the broadband range of flight times for a single m/z value. Faster ions travel further into the reflectrons, and slower ions travel less into the reflector. This way both slow and fast ions, of the same m/z value, reach the detector at the same time rather then at different times, narrowing the bandwidth for the output signal.
Magnetic Sector Mass Analyzer
Similar to time of flight (TOF) analyzer mentioned earlier, in magnetic sector analyzers ions are accelerated through a flight tube, where the ions are separated by charge to mass ratios. The difference between magnetic sector and TOF is that a magnetic field is used to separate the ions. As moving charges enter a magnetic field, the charge is deflected to a circular motion of a unique radius in a direction perpendicular to the applied magnetic field. Ions in the magnetic field experience two equal forces; force due to the magnetic field and centripetal force.
Basically the ions of a certain m/z value will have a unique path radius which can be determined if both magnetic field magnitude, and voltage difference for region of acceleration are held constant. When similar ions pass through the magnetic field, they all will be deflected to the same degree and will all follow the same trajectory path. Those ions which are not selected by voltage and magnetic field values, will collide with either side of the flight tube wall or will not pass through the slit to the detector. Magnetic sector analyzers are used for mass focusing, they focus angular dispersions.
Electrostatic Sector Mass Analyzer
Again, this is similar to time of flight analyzer in that it separates the ions while in flight, but it separates using an electric field. Electrostatic sector analyzer consists of two curved plates of equal and opposite potential. As the ion travels through the electric field, it is deflected and the force on the ion due to the electric field is equal to the centripetal force on the ion. Here the ions of the same kinetic energy are focused, and ions of different kinetic energies are dispersed.
Electrostatic sector analyzers are energy focusers, where an ion beam is focused for energy. Electrostatic and magnetic sector analyzers when employed individually are single focusing instruments. However when both techniques are used together, it is called a double focusing instrument, because in this instrument both the energies and the angular dispersions are focused.
Quadrupole Ion Trap Mass Analyzers
This analyzer employs similar principles as the quadrupole analyzer mentioned above, it uses an electric field for the separation of the ions by mass to charge ratios. The analyzer is made with a ring electrode of a specific voltage and grounded end cap electrodes. The ions enter the area between the electrodes through one of the end caps. After entry, the electric field in the cavity due to the electrodes causes the ions of certain m/z values to orbit in the space. As the radio frequency voltage increases, heavier mass ion orbits become more stabilized and the light mass ions become less stabilized, causing them to collide with the wall, and eliminating the possibility of traveling to and being detected by the detector.
Ion traps are uniquely suited for repeated cycles of mass spectrometry because of their ability to retain ions of desired m/z ratios. Selected fragments can be further fragmented by collision induced dissociation with helium gas. Ion traps are compact, relatively inexpensive, and can be adapted to many hybrid instruments.
Ion Cyclotron Resonance (ICR)
ICR is an ion trap that uses a magnetic field in order to trap ions into an orbit inside of it. In this analyzer there is no separation that occurs rather all the ions of a particular range are trapped inside, and an applied external electric field helps to generate a signal. As mentioned earlier, when a moving charge enters a magnetic field, it experiences a centripetal force making the ion orbit. Again the force on the ion due to the magnetic field is equal to the centripetal force on the ion.
Frequency of the orbit depends on the charge and mass of the ions, not the velocity. If the magnetic field is held constant, the charge to mass ratio of each ion can be determined by measuring the angular velocity. The relationship is that, at high angular velocity, there is low m/z value, and at low angular velocity, there is a high m/z value. Charges of opposite signs have the same angular velocity, the only difference is that they orbit in the opposite direction.
Exercise \(1\)
How are ions separated?
Answer
By accelerating them through magnetic and electric fields.
References
1. K. Downard, Mass Spectrometry: A Foundation Course, The Royal Society of Chemistry: UK 2004, Chapter 3
2. Skoog, Holler, Grouch, Principles of Instrumental Analysis, Thomson Brooks/Cole 2007, chapter 20
3. C. Herbert, R. Johnstone, Mass Spectrometry Basics, CRC Press LLC, 2003 chapter 25, 26, 39
4. E. De Hoffman, V. Stroobant, Mass Spectrometry: Principles and Applications, 2nd ed.;Wiley: England, 2001, chapter 2 | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/02%3A_Mass_Spectrometry/2.04_Mass_Analyzers.txt |
Learning Objectives
• Learn how mass spectrometry is applied in real-world applications
• Understand what mass spectrometry's use is in science
Mass spectrometry is applicable across diverse fields with specific applications including, but not limited to drug testing and discovery, food contamination detection, pesticide residue analysis, isotope ratio determination, protein identification, and carbon dating. One of the new ways that clinical mass spectrometry is being used is to quantitatively detect small amounts of proteins, biomarkers, or drug molecules, with very low concentrations. Some drugs have been given to patients in microdoses and with the ability to use small samples with low concentrations, it allows researchers to determine pharmacokinetic profiles of drugs that have been given. The reason for this is that it protects the patient from possible adverse effects of the drug while allowing scientists to determine what happens to the drug in the body. This will be especially helpful with pediatric patients. Below is a summary of a study detecting quinolones in animal food using mass spectrometry.
Application of LC/ESI-QTOF-MS in the Detection of Quinolones in Edible Animal Food
Quinolones are a family of common antibacterial veterinary medicine which can inhibit DNA-gyrase in bacterial cells. However, the residues of quinolone in edible animal products may be directly toxic or cause resistant pathogens in humans. Therefore, sensitive methods are required to monitor such residues possibly present in different animal-producing food, such as eggs, chicken, milk and fish. The molecular structures of eight quinolones, ciprofloxacin (CIP), anofloxacin methanesulphonate (DAN), enrofloxacin (ENR), difloxacin (DIF), sarafloxacin (SARA), oxolinic, acid (OXO), flumequine (FLU), ofloxacin (OFL), are shown in Figure \(1\).
LC-MS is a common detection approach in the field of food safety. But because of the complex matrix of the samples, it is always difficult to detect those target molecules of low concentration by using single quadrupole MS. The following gives an example of the application of LC/ESI-QTOF-MS.
Using a quaternary pump system, a Q-TOF-MS system, a C18 column (250 mm × 2.0 mm I.D., 5 µm) with a flow rate of 0.2 mL/min, and a mixture of solvents as the mobile phase comprising of 0.3% formic acid solution and acetonitrile. The gradient phofile for mobile phase is shown in Table \(1\). Since at acidic pH condition, the quinolones carried a positive charge, all mass spectra were acquired in the positive ion mode and summarizing 30,000 single spectra in the mass range of 100-500 Da.
Time (min) Volume % of Formic Acid Solution Volume % of Acetonitrile
0 80 20
12 65 35
15 20 80
20 15 85
30 15 85
30.01 80 20
Table \(1\) The gradient phofile for mobile phase
The optimal ionization source working parameters were as follows: capillary voltage 4.5 kV; ion energy of quadrupole 5 eV/z; dry temperature 200 °C; nebulizer 1.2 bar; dry gas 6.0 L/min. During the experiments, HCO2Na (62 Da) was used to externally calibrate the instrument. Because of the high mass accuracy of the TOF mass spectrometer, it can extremely reduce the matrix effects. Three different chromatographs are shown in Figure \(2\). The top one is the total ion chromatograph at the window range of 400 Da. It’s impossible to distinguish the target molecules in this chromatograph. The middle one is at one Da resolution, which is the resolution of single quadrupole mass spectrometer. In this chromatograph, some of the molecules can be identified. But noise intensity is still very high and there are several peaks of impurities with similar mass-to-charge ratios in the chromatograph. The bottom one is at 0.01 Da resolution. It clearly shows the peaks of eight quinolones with very high signal to noise ratio. In other words, due to the fast acquisition rates and high mass accuracy, LC/TOF-MS can significantly reduce the matrix effects.
The quadrupole MS can be used to further confirm the target molecules. Figure \(3\) shows the chromatograms obtained in the confirmation of CIP (17.1 ng/g) in a positive milk sample and ENR (7.5 ng/g) in a positive fish sample. The chromatographs of parent ions are shown on the left side. On the right side, they are the characteristic daughter ion mass spectra of CIP and ENR. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/02%3A_Mass_Spectrometry/2.05%3A_Applications_of_Mass_Spectrometry.txt |
Learning Objectives
• understand how to determine molecular weight.
• interpret mass spectra.
The first piece of information mass spectra can get is the molecular weight of a molecule. In a mass spectrum of a compound, the x axis is the m/z values and the y axis represents the intensity or relative abundance of a given m/z. In addition, there are a number of other lines at a variety of values of m/z; these correspond to the masses of smaller pieces of those that fall apart during the experiment as well as the unfragmented cation. For these smaller pieces, the more stable an ion is, the more likely it is to form. The more of a particular sort of ion that's formed, the higher its peak height will be. The tallest peak is called the base peak and is assigned 100% intensity. The peak that represents the unfragmented cation radical is called the parent peak or molecular ion (M+). The parent peak is how you determine the molecular weight of a molecule. Often, the molecular ion peak is not the same as the base peak. Usually, whole numbers are used for the molecular weights in mass spectrometry. The atomic masses in the periodic table are average masses including different isotopes and because mass spectrometry examines individual molecules, whole numbers are used. Complications can arise with determining the molecular ion peak because it is not always abundant, especially when molecules fragment easily. This is where the soft ionization techniques come into play.
Exercise \(1\)
Which mass spectrum corresponds to hexane, 1-hexene, and 1-hexyne.
(A)
(B)
(C)
Answer
Each molecule has a different molecular weight. Hexane MW = 86; 1-hexene MW = 84, and 1-hexyne MW = 82
To determine which molecule belongs to which look for the molecular ion peak, which will indicate what the molecular weight is of each molecule.
(A) 1-hexyne
(B) 1-hexene
(C) hexane
Another piece to note is that if you look closely at the mass spectrum of C in exercise 1 above, you will notice a little peak at m/z = 87. This is referred to as the M+1 peak (one greater than the molecular ion), and it arises because of 13C. This compound is referred to as an isotopomer; that means the same compound with a different isotope. The chance that a molecule in a sample contains a 13C atom is related to the number of carbons present. If there is just one carbon atom in the molecule, it has a 1% chance of being a 13C. That means the M+1 peak would be only 1/100th as tall as M+, the peak for the molecular ion.
Example \(1\)
Let's have another look at the mass spectrum for pentane:
What causes the line at m/z = 57?
Solution
How many carbon atoms are there in this ion? There can't be 5 because 5 x 12 = 60. What about 4? 4 x 12 = 48. That leaves 9 to make up a total of 57. How about C4H9+ then?
C4H9+ would be [CH3CH2CH2CH2]+, and this would be produced by the following fragmentation:
The methyl radical produced will simply get lost in the machine.
The line at m/z = 43 can be worked out similarly. If you play around with the numbers, you will find that this corresponds to a break producing a 3-carbon ion:
The line at m/z = 29 is typical of an ethyl ion, [CH3CH2]+:
The other lines in the mass spectrum are more difficult to explain. For example, lines with m/z values 1 or 2 less than one of the easy lines are often due to loss of one or more hydrogen atoms during the fragmentation process.
What happens when the molecular weight is the same for the compounds? The fragmentation pattern should be different for the molecules. While knowing the molecular weight is invaluable, mass spectrometry can lend a hand in structure determination. The mass spectrum of a particular compound acts as a "fingerprint," since each compound will fragment in its own unique way (just like a person's fingerprint). In the next section, the fragmentation of functional groups will be discussed.
Exercise \(2\)
The male sex hormone testosterone contains C, H, and O. It has a mass of 288.2089 amu as determined by mass spectrometry. What is the likely molecular formula for testoterone?
Answer
C19H28O2 | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/02%3A_Mass_Spectrometry/2.06%3A_Interpretation_of_Mass_Spectra.txt |
Learning Objectives
• Learn common fragmentation patterns for functional groups
• Interpret the fragmentations of mass spectra
Some fragment ions are very common in mass spectrometry. These ions are seen frequently for either of two reasons:
• there is not a pathway available to break the ion down.
• the ion is relatively stable, so it forms easily.
There are a number of ions commonly seen in mass spectrometry that tell you a little bit about the structure. Just like with anions, there are a couple of common factors influence cation stability:
• Electronegativity plays a role. More electronegative atoms are less likely to be cations.
• Polarizability also plays a role. More polarizable atoms are more likely to be cations.
• Delocalization stabilizes a cation by spreading out the charge onto two or more different atoms. Resonance is a common way to delocalize charge.
Alcohols
When alcohols are subjected to ionization, two fragmentation pathways can occur - alpha cleavage and dehydration. Alpha cleavage occurs by a C-C bond nearest the hydroxyl group being broken. This yields a neutral radical plus a resonance stabilized, oxygen-containing cation. As an example let's look at the mass spectrum of 2-pentanol below.
The parent peak is m/z = 88. The base peak is m/z = 45, which correlates to a fragmentation that occurs with an alpha cleavage. This fragment is the piece due to the resonance stabilized oxygen-containing cation, which is shown in the pathway below.
In dehydration, water is eliminated. This leaves an alkene radical cation that is 18 units less than the molecular ion peak. Using our example of 2-pentanol, this would lead to a peak at m/z = 70 and there is.
Amines
A general principle when nitrogen is part of a molecule is that if there is an odd number of nitrogens, then the molecular weight will be an odd number. This is also known as the nitrogen rule and stems from the fact that nitrogen is a trivalent atom. It also goes that if the molecule contains an even molecular weight, then there will be zero or two nitrogen atoms.
As with alcohols, primary amines undergo a characteristic alpha cleavage. As an example, the mass spectrum of 2-aminopentane is below.
Source: SDBSWeb : https://sdbs.db.aist.go.jp/sdbs/cgi-..._frame_top.cgi (National Institute of Advanced Industrial Science and Technology of Japan) [Accessed August 16, 2022]
The parent peak is very small, but is at m/z = 87. The base peak (m/z = 44) is due to an alpha cleavage and forming a fragment of a resonance stabilized nitrogen-containing cation (see pathway below).
Halides
Halides have isotopes that give distinct appearances in a mass spectrum. Chlorine has two isotopes 35C and 37C with a 3:1 ratio (roughly). This ratio shows up in the mass spectrum for a chlorine-containing compound. Below is the mass spectrum of 2-chloro-2-methylpropane.
Source: SDBSWeb : https://sdbs.db.aist.go.jp/sdbs/cgi-..._frame_top.cgi (National Institute of Advanced Industrial Science and Technology of Japan) [Accessed August 16, 2022]
Looking at the molecular ion peak (m/z = 77 peak), there is another peak at m/z= 79. The peak at 79 is called the M + 2 peak. The ratio of the relative abundance/intensity of the M:M + 2 is about 3:1, which reflecting the isotopic abundance of 35C:37C.
With bromine, the isotopic distribution of 79Br and 81Br is more like 50:50. Again, the ratio of the relative abundance/intensity of the M:M + 2 is about 50:50. In the example below, the mass spectrum of 1-bromohexane is shown.
Source: SDBSWeb : https://sdbs.db.aist.go.jp/sdbs/cgi-..._frame_top.cgi (National Institute of Advanced Industrial Science and Technology of Japan) [Accessed August 16, 2022]
The roughly 50:50 distribution can be seen in the parent peak and M + 2 (m/z = 164 and 165). It is again showing up at the peaks m/z = 135 and 137. The two peaks in each are nearly the same height. In the fragments that contain the bromine, this ratio will be reflected.
Carbonyl Compounds
The McLafferty rearrangement is a common cleavage that occurs in carbonyl compounds that have a hydrogen three atoms away from carbonyl group. This rearragnement yields a carbonyl-containing radical cation via β-cleavage to produce an enol cation and an alkene. This fragmentation is shown below.
In the mass spectrum of 2-hexanone (below), the m/z peak = 59 represents the enol cation that would form from the McLafferty rearrangement.
Source: SDBSWeb : https://sdbs.db.aist.go.jp/sdbs/cgi-..._frame_top.cgi (National Institute of Advanced Industrial Science and Technology of Japan) [Accessed August 16, 2022]
The carbonyl compounds can also undergo alpha cleavage as was seen with the alcohol and primary amines. The alpha cleavage occurs between the carbonyl carbon and the neighboring carbon yielding an acylium ion and a neutral radical (see below).
The acylium ion has an m/z =43. This fragment shows up at m/z = 43. This cleavage can also be seen as the loss of the acylium ion from the parent ion (100-43 = 57). There is a m/z peak at 57 to represent this loss.
Example \(1\)
The mass spectrum of 2-methyl-3-penatonol is shown below.
Source: SDBSWeb : https://sdbs.db.aist.go.jp/sdbs/cgi-...p;sdbsno=13488 (National Institute of Advanced Industrial Science and Technology of Japan) [Accessed August 16, 2022]
What fragments can you identify?
Solution
First, you want to look up the draw or look up the structure. 2-methyl-3-pentanol:
Next, calculate the mass of the molecular ion and identify the functional groups in the molecule. M+1 = 102 and there is an alcohol present.
Then write down the fragmentation patterns you may expect and calculate the masses for those peaks and compare it to the mass spectrum. With an alcohol, there are two pathways for fragmentation - alpha cleavage and dehydration. From the alpha cleavage, one would expect two peaks m/z = 73 and 59. From dehydration, one would expect an m/z of 84. The dehydration fragment peak is not observed, but both the fragments from the alpha cleavage are observed.
Exercise \(1\)
What are the masses of the charged fragments produced in the following cleavage pathways?
a. alpha cleavage of triethylamine
b. McLafferty rearrangment of 4-methyl-2-pentanone
Answer
a. m/z = 86
b. m/z = 58
Exercise \(2\)
Nicotine is a diamino compound with two rings and a molecular ion peak of 162.1157. Remembering the nitrogen rule, give the molecular formula for nicotine and calculate the number of double bonds.
Answer
C10H14N2 | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/02%3A_Mass_Spectrometry/2.07_Mass_Spectrometry_of_Some_Common_Functional_Groups.txt |
Learning Objectives
• Interpret mass spectra
Exercise \(1\)
The following figure shows the mass spectrum of a saturated hydrocarbon (containing only carbon and hydrogen with only single bonds between carbons, not double bonds).
1. Draw five different structures that would have the molecular weight of this compound.
2. Choose three smaller m/z values from the spectrum and draw one structure for each of them. Note that these fragments will not have complete Lewis structures.
Answer
1. Molecular Weight = 114, which cooresponds to a C8H18 hydrocarbon. There is the possibility of 18 isomers, but here are a few isomers:
2. m/z = 57; [CH3CH2CH2CH2]+
m/z = 43; [CH3CH2CH2]+
m/z = 29; [CH3CH2]+
Exercise \(2\)
Caffeine has a mass of 194.19 amu, determined by mass spectrometry, and contains C, N, H, O. What is a molecular formula for this molecule?
Answer
C8H10N4O2
C = 12 × 8 = 96
N = 14 × 4 = 56
H = 1 × 10 = 10
O = 2 × 16 = 32
96+56+10+32 = 194 g/mol
Exercise \(3\)
The following are the spectra for 2-methyl-2-hexene and 2-heptene, which spectra belongs to the correct molecule. Explain.
A:
B:
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)
Answer
The (A) spectrum is 2-methyl-2-hexene and the (B) spectrum is 2-heptene. Looking at (A) the peak at 68 m/z is the fractioned molecule with just the tri-substituted alkene present. While (B) has a strong peak around the 56 m/z, which in this case is the di-substituted alkene left behind from the linear heptene.
Exercise \(4\)
What are the masses of all the components in the following fragmentations?
Answer
The first undergoes an alpha cleavage. The second undergoes a dehydration. The final one goes througha McLafferty rearrangement.
Exercise \(5\)
5-Chloro-2-pentanone has the mass spectrum shown. Which peak represents the M+? Which is the base peak? Why is there a peak at 122? Explain what the fragment for the base peak would be.
Source: SDBSWeb : https://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_frame_disp.cgi?sdbsno=10178 (National Institute of Advanced Industrial Science and Technology, 16 August 2022)
Answer
M+ = 120
base peak = 43
The m/z peak at 122 is the M + 2 peak. It occurs because chlorine has two isotopes 35C and 37C in a 3:1 ratio.
The m/z = 43 occurs due to the alpha cleavage. The acylium ion has an m/z of 43. This fragment is particularly stable due to resonance.
2.S Summary of Mass Spectrometry
Concepts & Vocabulary
2.1: Chapter Objectives and Preview of Mass Spectrometry
• Mass spectrometry is a way to determine the molecular weight of a structure to begin structure elucidation.
2.2 Instrumentation
• Mass spectrometry breaks apart molecules to detect fragments based on a mass to charge ratio.
• The particles in the sample (atoms or molecules) are bombarded with a stream of electrons, and some of the collisions are energetic enough to knock one or more electrons out of the sample particles to make positive ions.
• Most of the positive ions formed will carry a charge of +1 because it is much more difficult to remove further electrons from an already positive ion.
• Different ions are deflected by the magnetic field by different amounts, which depends on the mass of the ion and the charge of the ion.
• The output from the chart recorder is usually simplified into a "stick diagram". This shows the relative current produced by ions of varying mass/charge ratio.
2.3 Ionization Techniques
• There is a variety of ionization techniques. The ones discussed in this section are electron impact, chemical ionization, field ionization/desorption, electrospray ionization, matrix assisted laser desorption ionization, inductively coupled plasma mass spectrometry, and fast atom bombardment.
• Depending on the information desired from mass spectrometry analysis, different ionization techniques may be desired.
2.4 Mass Analyzers
• A mass analyzer is the component of the mass spectrometer that takes ionized masses and separates them based on charge to mass ratios and outputs them to the detector where they are detected and later converted to a digital output.
• There are six general types of mass analyzers that can be used for the separation of ions in a mass spectrometry (quadrapole, time of flight, magnetic sector, electrostatic sector, quadropole ion trap, and ion cyclotron resonance).
2.5 Applications of Mass Spectrometry
• Mass spectrometry is applicable across diverse fields with specific applications including, but not limited to drug testing and discovery, food contamination detection, pesticide residue analysis, isotope ratio determination, protein identification, and carbon dating.
• One of the new ways that clinical mass spectrometry is being used is to quantitatively detect small amounts of proteins, biomarkers, or drug molecules, with very low concentrations.
2.6 Interpretation of Mass Spectra
• Mass spectrum looks like a bar graph with the x axis as the m/z values and the y axis represents the intensity or relative abundance of a given m/z.
• The lines correspond to the fragments of the different ions with different m/z values.
• The more of a particular sort of ion that's formed, the higher its peak height will be. The tallest peak is called the base peak and is assigned 100% intensity.
• The peak that represents the unfragmented cation radical is called the parent peak or molecular ion (M+).
• There is often an M+1 peak (one greater than the molecular ion), and it arises because of 13C. This compound is referred to as an isotopomer.
2.7 Mass Spectrometry of Common Functional Groups
• Some fragment ions are very common in mass spectrometry often due to having no pathway available to break the ion down or the ion is relatively stable, so it forms easily.
• When alcohols are subjected to ionization, two fragmentation pathways can occur - alpha cleavage and dehydration.
• A general principle when nitrogen is part of a molecule is that if there is an odd number of nitrogens, then the molecular weight will be an odd number. This is also known as the nitrogen rule.
• Primary amines undergo a characteristic alpha cleavage.
• Other isotopomers are common when chlorine and bromine are part of the molecule. For chlorine and bromine there will be an additional M + 2 peak representing one of the isotopomers.
• Chlorine has two isotopes 35C and 37C with a 3:1 ratio (roughly), which appears in the mass spectrum in the fragments containing chlorine.
• The ratio of the relative abundance/intensity of the M:M + 2 is about 3:1, which reflecting the isotopic abundance of 35C:37C.
• With bromine, the isotopic distribution of 79Br and 81Br is more like 50:50. The ratio of the relative abundance/intensity of the M:M + 2 is about 50:50.
• Carbonyl compounds can undergoe an alpha cleavage and the McLafferty rearragnement.
Skills to Master
• Skill 2.1 Know what mass to charge means and measures.
• Skill 2.2 Distinguish between different types of ionization techniques.
• Skill 2.3 Determine the best mass analyzer to use for sample.
• Skill 2.4 Determine the ratio of different types of protons present in an organic compound.
• Skill 2.5 Interpret fragmentation patterns
• Skill 2.6 Interpret mass spectra | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/02%3A_Mass_Spectrometry/2.08%3A_Mass_Spectrometry_Problems.txt |
Learning Objectives
After completing this chapter, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• analyze problems which may require the interpretation of ultraviolet spectroscopy.
• define, and use in context, the key terms introduced in this chapter.
Many of the pigments that are responsible for the beautiful colors in nature are conjugated compounds. Many of the early organic dyes used for pigmenting cloth and art are conjugated compounds. These included the crimson pigment - kermesic acid, the blue dye - indigo, and the yellow saffron pigment - crocetin. A common feature of all these colored compounds, displayed below, is a system of extensive pi bonds. In this chapter, we will look at how ultraviolet (UV) spectroscopy is a technique that is only applicable to conjugated compounds giving information on the nature of the conjugated pi electron system.
3.02: Conjugated Dienes
Objective
• Determine whether or not a molecule contains a conjugated system, given its Kekulé, condensed or shorthand formula.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• conjugated diene
• conjugated double bonds
• diene
• enone
• polyene
A diene is a hydrocarbon chain that has two double bonds that may or may not be adjacent to each other. The arrangements of these double bonds can have varying affects on the compounds reactivity and stability. This section focuses on the delocalization of pi systems by comparing two neighboring double bonds, specifically conjugated molecules. A molecule is defined as conjugated when there is a system of connected p orbitals where electron density can be shared across the system. The arrangement of bonds alternates between single and multiple bonds in conjugated molecules.
Conjugated vs. Nonconjugated vs. Cumulated Dienes
Conjugated dienes are two double bonds separated by a single bond. An example of this is penta-1,3-diene.
Nonconjugated dienes are two double bonds are separated by more than one single bond. The double bonds are considered isolated from each other. An example of this is 2,5-heptadiene.
Cumulated dienes are two double bond connected to a similar atom. These are also known as allenes. An example of this is 2,3-heptadiene.
Another conjugated diene is a enone, which is compound containing both a double bond and carbonyl where the double bond is conjugated to the carbonyl. An example of this is cyclohex-2-enone.
There are many compounds that have more than two double bonds present in the molecule. The term given to describe these molecules is polyene. An example of a polyene is beta-carotene (below). Beta-carotene is a red-orange pigment found in plants and fruits like carrots, which is converted to vitamin A in the human body.
Exercise $1$
Are the following structures conjugated?
(A)
(B)
(C)
Answer
(A) Yes - conjugated
(B) Yes - conjugated
(C) No - not conjugated
3.03: Electronic Transitions
Objectives
• discuss the bonding in 1,3-butadiene in terms of the molecular orbital theory, and draw a molecular orbital for this and similar compounds.
• understand how electronic transitions occur.
• get an understanding of when electronic transitions can be observed with UV spectroscopy
Let’s take as our first example the simple case of molecular hydrogen, H2. The molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as a σ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.
When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength (lower energy) than molecular hydrogen.
The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated pi systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores.
Next, we'll consider the 1,3-butadiene molecule (below). From valence orbital theory alone we might expect that the C2-C3 bond in this molecule, because it is a sigma bond, would be able to rotate freely.
Experimentally, however, it is observed that there is a significant barrier to rotation about the C2-C3 bond (colored in red above), and that the entire molecule is planar. In addition, the C2-C3 bond is 148 pm long, shorter than a typical carbon-carbon single bond (about 154 pm), though longer than a typical double bond (about 134 pm). Molecular orbital theory accounts for these observations with the concept of delocalized π bonds. In this picture, the four p atomic orbitals combine mathematically to form four pi molecular orbitals of increasing energy. Two of these - the bonding pi orbitals - are lower in energy than the p atomic orbitals from which they are formed, while two - the antibonding pi orbitals - are higher in energy.
The lowest energy molecular orbital, pi1, has only constructive interaction and zero nodes. Higher in energy, but still lower than the isolated p orbitals, the pi2 orbital has one node but two constructive interactions - thus it is still a bonding orbital overall. Looking at the two antibonding orbitals, pi3* has two nodes and one constructive interaction, while pi4* has three nodes and zero constructive interactions.
By the aufbau principle, the four electrons from the isolated 2pz atomic orbitals are placed in the bonding pi1 and pi2 MO’s. Because pi1 includes constructive interaction between C2 and C3, there is a degree, in the 1,3-butadiene molecule, of pi-bonding interaction between these two carbons, which accounts for its shorter length and the barrier to rotation. The valence bond picture of 1,3-butadiene shows the two pi bonds as being isolated from one another, with each pair of pi electrons ‘stuck’ in its own pi bond. However, molecular orbital theory predicts (accurately) that the four pi electrons are to some extent delocalized, or ‘spread out’, over the whole pi system.
1,3-butadiene is the simplest example of a system of conjugated pi bonds. Remember to be considered conjugated, two or more pi bonds must be separated by only one single bond – in other words, there cannot be an intervening sp3-hybridized carbon, because this would break up the overlapping system of parallel p orbitals. In the compound below, for example, the C1-C2 and C3-C4 double bonds are conjugated (highlighted in blue), while the C6-C7 double bond (highlighted in red) is isolated from the other two pi bonds by sp3-hybridized C5.
A very important concept to keep in mind is that there is an inherent thermodynamic stability associated with conjugation. This stability can be measured experimentally by comparing the heat of hydrogenation of two different dienes. When the two conjugated double bonds of 1,3-pentadiene are 'hydrogenated' to produce pentane, about 225 kJ is released per mole of pentane formed. Compare that to the approximately 250 kJ/mol released when the two isolated double bonds in 1,4-pentadiene are hydrogenated, also forming pentane.
The conjugated diene is lower in energy: in other words, it is more stable. In general, conjugated pi bonds are more stable than isolated pi bonds. Conjugated pi systems can involve heteroatoms like oxygen and nitrogen as well as carbon. In the metabolism of fat molecules, some of the key reactions involve alkenes that are conjugated to carbonyl groups.
As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.
In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange.
The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO:
This is referred to as an n - π* transition. The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an n - π* transition is smaller that that of a π - π* transition – and thus the n - π* peak is at a longer wavelength. In general, n - π* transitions are weaker (less light absorbed) than those due to π - π* transitions.
Exercise \(1\)
Without calculations, which molecule (2,5-heptadiene or 2,4-heptadiene) would you predict to have a lower heat of hydrogenation?
or
Answer
I would predict 2,4-heptadiene to have a lower heat of hydrogenation than 2,5-heptadiene. This is due to the conjugation between the double bonds in 2,4-heptadiene, which is stabilizing.
Exercise \(2\)
Which of the following molecules would you expect to have a smaller gap in the electronic transition? Explain your answer.
Answer
B. The entire molecule is conjugated, so it has a more extended pi system than A. More extended pi systems typically have smaller absorption gaps. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/03%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/3.01%3A_Chapter_Objectives_and_Preview_of_Ultraviolet_Spectroscopy.txt |
Objectives
• identify the ultraviolet region of the electromagnetic spectrum which is of most use to organic chemists.
The electromagnetic spectrum was discussed in Chapter 1, but as a reminder most of the radiation that surrounds us is not visible. However, this radiation can be detected by instruments. The following chart displays many of the important regions of the electromagnetic spectrum, and demonstrates the inverse relationship between wavelength and frequency.
UV-Visible Absorption Spectra
To understand why some compounds are colored and others are not, and to determine the relationship of conjugation to color, we must make accurate measurements of light absorption at different wavelengths in and near the visible part of the spectrum. Commercial optical spectrometers enable such experiments to be conducted with ease, and usually survey both the near ultraviolet and visible portions of the spectrum. The visible region of the spectrum comprises photon energies of 36 to 72 kcal/mol, and the near ultraviolet region, out to 200 nm, extends this energy range to 143 kcal/mol. Ultraviolet radiation having wavelengths less than 200 nm is difficult to handle, and is seldom used as a routine tool for structural analysis.
The energies noted above are sufficient to promote or excite a molecular electron to a higher energy orbital. Consequently, absorption spectroscopy carried out in this region is sometimes called "electronic spectroscopy". In Section 3.3, the electronic transitions were discussed, but a summary of the various kinds of electronic excitation that may occur in organic molecules is shown in the diagram below. Of the six transitions outlined, only the non-bonding to anti-bonding and bonding to anti-bonding are achieved by the energies available in the 200 to 800 nm spectrum. As a rule, energetically favored electron promotion will be from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO), and the resulting species is called an excited state.
When sample molecules are exposed to light having an energy that matches a possible electronic transition within the molecule, some of the light energy will be absorbed as the electron is promoted to a higher energy orbital. An optical spectrometer records the wavelengths at which absorption occurs, together with the degree of absorption at each wavelength. The resulting spectrum is presented as a graph of absorbance (A) versus wavelength, as in the spectrum shown below of beta-carotene as well as the structure. Beta-carotene absorbs most strongly from 400 to 500 nm. This is the blue/green part of the spectrum and beta-carotene appears orange because it reflects back red/yellow. Absorbance usually ranges from 0 (no absorption) to 2 (99% absorption), and is precisely defined in context with spectrometer operation.
Spectrum from: https://scilearn.sydney.edu.au/organ...The%20Spectrum
Exercise \(1\)
What is the energy range for 400 nm to 500 nm in the ultraviolet spectrum where beta-carotene absorbs?
Answer
E = hc/λ
E = (6.62 × 10−34 Js)(3.00 × 108 m/s)/(4.00 × 10−7 m)
E = 6.62 × 10−19 J
The range of 3.972 × 10-19 to 4.965 × 10-19 joules.
Exercise \(2\)
Would 1,3-cyclohexadiene or 1,4-cyclohexadiene absorb a longer wavelength?
Answer
1,3-cyclohexadiene would absorb a longer wavelength. Typically, the more conjugation in the molecule, the longer the wavelength absorbed. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/03%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/3.04%3A_Ultraviolet_Absorption.txt |
Ultraviolet-visible (UV-vis) spectroscopy is used to obtain the absorbance spectra of a compound in solution or as a solid. What is actually being observed spectroscopically is the absorbance of light energy or electromagnetic radiation, which excites electrons from the ground state to the first singlet excited state of the compound or material. The UV-vis region of energy for the electromagnetic spectrum covers 1.5 - 6.2 eV which relates to a wavelength range of 800 - 200 nm. The Beer-Lambert Law, Equation \ref{1} , is the principle behind absorbance spectroscopy. For a single wavelength, A is absorbance (unitless, usually seen as arb. units or arbitrary units), ε is the molar absorptivity of the compound or molecule in solution (M-1cm-1), b is the path length of the cuvette or sample holder (usually 1 cm), and c is the concentration of the solution (M).
$A\ =\ \varepsilon b c \label{1}$
All of these instruments have a light source (usually a deuterium or tungsten lamp), a sample holder and a detector, but some have a filter for selecting one wavelength at a time. The single beam instrument (Figure $1$) has a filter or a monochromator between the source and the sample to analyze one wavelength at a time. The double beam instrument (Figure $2$) has a single source and a monochromator and then there is a splitter and a series of mirrors to get the beam to a reference sample and the sample to be analyzed, this allows for more accurate readings. In contrast, the simultaneous instrument (Figure $3$) does not have a monochromator between the sample and the source; instead, it has a diode array detector that allows the instrument to simultaneously detect the absorbance at all wavelengths. The simultaneous instrument is usually much faster and more efficient, but all of these types of spectrometers work well.
What Information can be Obtained from UV-vis Spectra?
UV-vis spectroscopic data can give qualitative and quantitative information of a given compound or molecule. Irrespective of whether quantitative or qualitative information is required it is important to use a reference cell to zero the instrument for the solvent the compound is in. For quantitative information on the compound, calibrating the instrument using known concentrations of the compound in question in a solution with the same solvent as the unknown sample would be required. If the information needed is just proof that a compound is in the sample being analyzed, a calibration curve will not be necessary; however, if a degradation study or reaction is being performed, and concentration of the compound in solution is required, a calibration curve is needed.
To make a calibration curve, at least three concentrations of the compound will be needed, but five concentrations would be ideal for a more accurate curve. The concentrations should start at just above the estimated concentration of the unknown sample and should go down to about an order of magnitude lower than the highest concentration. The calibration solutions should be spaced relatively equally apart, and they should be made as accurately as possible using digital pipettes and volumetric flasks instead of graduated cylinders and beakers. An example of absorbance spectra of calibration solutions of Rose Bengal (4,5,6,7-tetrachloro-2',4',5',7'-tetraiodofluorescein, Figure $4$, can be seen in Figure $5$. To make a calibration curve, the value for the absorbances of each of the spectral curves at the highest absorbing wavelength, is plotted in a graph similar to that in Figure $6$ of absorbance versus concentration. The correlation coefficient of an acceptable calibration is 0.9 or better. If the correlation coefficient is lower than that, try making the solutions again as the problem may be human error. However, if after making the solutions a few times the calibration is still poor, something may be wrong with the instrument; for example, the lamps may be going bad.
Limitations of UV-vis Spectroscopy
Sample
UV-vis spectroscopy works well on liquids and solutions, but if the sample is more of a suspension of solid particles in liquid, the sample will scatter the light more than absorb the light and the data will be very skewed. Most UV-vis instruments can analyze solid samples or suspensions with a diffraction apparatus (Figure $7$), but this is not common. UV-vis instruments generally analyze liquids and solutions most efficiently.
Choice of Solvent or Container
Every solvent has a UV-vis absorbance cutoff wavelength. The solvent cutoff is the wavelength below which the solvent itself absorbs all of the light. So when choosing a solvent be aware of its absorbance cutoff and where the compound under investigation is thought to absorb. If they are close, chose a different solvent. Table $1$ provides an example of solvent cutoffs.
Table $1$: UV absorbance cutoffs of various common solvents
Solvent UV Absorbance Cutoff (nm)
Acetone 329
Benzene 278
Dimethylformamide 267
Ethanol 205
Toluene 285
Water 180
The material the cuvette (the sample holder) is made from will also have a UV-vis absorbance cutoff. Glass will absorb all of the light higher in energy starting at about 300 nm, so if the sample absorbs in the UV, a quartz cuvette will be more practical as the absorbance cutoff is around 160 nm for quartz (Table $2$).
Table $2$: Three different types of cuvettes commonly used, with different usable wavelengths.
Material Wavelength Range (nm)
Glass 380-780
Plastic 380-780
Fused Quartz < 380
Concentration of Solution
To obtain reliable data, the peak of absorbance of a given compound needs to be at least three times higher in intensity than the background noise of the instrument. Obviously using higher concentrations of the compound in solution can combat this. Also, if the sample is very small and diluting it would not give an acceptable signal, there are cuvettes that hold smaller sample sizes than the 2.5 mL of a standard cuvettes. Some cuvettes are made to hold only 100 μL, which would allow for a small sample to be analyzed without having to dilute it to a larger volume, lowering the signal to noise ratio.
Exercise $1$
What are the two most common light sources used in a UV-Vis spectrophotometer?
Answer
deuterium or tungsten lamp
Exercise $2$
How does concentration affect UV-Vis?
Answer
The concentration of sample present is directly proportional to the intensity of light absorption, thus influencing the spectrum.
Exercise $3$
When should you use a calibration curve in UV-Vis?
Answer
To determine the concentration of the sample from absorbance. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/03%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/3.05%3A_UV-Visible_Spectrometer.txt |
Objective
• be able to interpret UV-Vis spectra.
• understand the effect of conjugation.
The wavelength necessary to make the transition from π -π* in a conjugated molecule depends on the energy gap between the HOMO and LUMO. This energy gap depends on the conjugated system of the molecule being studied. If you recall from Section 3.3, the energy gap for π -π* transitions is smaller for conjugated systems than for isolated double bonds, and thus the wavelength absorbed is longer. Therefore, by measuring the UV spectrum of a molecule, structural information can be derived about the nature of the conjugated pi electron system present.
We have been talking in general terms about how molecules absorb UV and visible light – now let's look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred. Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD+. This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems.
You’ll notice that this UV spectrum has only one peak, although many molecules have more than one. Notice also that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm). Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height. Typically, there are two things that we look for and record from a UV-Vis spectrum. The first is lambda max (λmax), which is the wavelength at maximal light absorbance. As you can see, NAD+ has λmax, = 260 nm. We also want to record how much light is absorbed at λmax. Here we use a unitless number called absorbance, abbreviated 'A'. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength before it passes through the sample (I0), divides this value by the intensity of the same wavelength after it passes through the sample (I), then takes the log10 of that number:
A = log I0/I
You can see that the absorbance value at 260 nm (A260) is about 1.0 in this spectrum.
Here is the absorbance spectrum of the common food coloring Red #3:
Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the λmax of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes.
Now, take a look at the spectrum of another food coloring, Blue #1:
Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue.
Exercise \(1\)
Which of the following would show UV absorptions in the 200-300 nm range?
Answer
You are looking for conjugated systems, which leads to B and D.
Exercise \(2\)
What is the lambda max for caffeine?
UV-vis spectra of caffeine in water
Belay, Abebe & Beketie, Kassahun & Redi, Mesfin & Asfaw, Araya. (2008). Measurement of Caffeine in Coffee Beans with UV/Vis Spectrometer. Food Chemistry. 108. 310-315. 10.1016/j.foodchem.2007.10.024.
Answer
λmax of 275 nm.
Exercise \(3\)
A colleague has isolated a compound that has the formula C20H32O and a λmax of 275 nm. Your colleague subjected the product to hydrogenation (Pd/C and H2), which resulted in no change in the λmax. They then tried to reduce the molecule with sodium borohydride, which led to no change in λmax. They have proposed four different structures, but cannot figure out what structure they have isolated. Which do you think is most likely?
A)
B)
C)
D)
Answer
With a λmax, there needs to be a conjugated pi system. A is lacking a conjugated pi system, so it can't be molecule A. Ketones can be reduced to alcohols when treated with sodium borohydride, which means the λmax would change for both C and D. Therefore it can't be C or D. In additon, C would react under the hydrogenation conditions, so again its λmax would change. It can definitely not be C. Which leaves us B. It has a conjugated pi system and would not react in either set of conditions, so its λmax would stay the same. The isolated compound is B - | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/03%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/3.06%3A_Interpreting_Ultraviolet_Spectra.txt |
Objectives
• explain why some organic compounds have different colours based on compound structure and our perception of light.
• state the relationship between frequency of light absorbed and the extent of conjugation in an extended pi electron system.
Eyes receive light energy then transfer and passing the energy into neural impulses to brain. This page will show the role of light plays in vision.
Introduction
Light is one of the most important resources for civilization, it provides energy as it pass along by the sun. Light influence our everyday live. Living organisms sense light from the environment by photoreceptors. Light, as waves carry energy, contains energy by different wavelength. In vision, light is the stimulus input. Light energy goes into eyes stimulate photoreceptor in eyes. However, as an energy wave, energy is passed on through light at different wavelength. For example, "white" light from a lamp consists of all the wavelengths in the visible region. When this white light hits beta-carotene, the wavelengths from 400 to 500 nm are absorbed and all the other wavelengths are transmitted to our eyes. The carrot looks orange because the wavelengths of 400 to 500 nm that were absorbed occur in the blue range, when blue is removed our eyes see an orange color for beta-carotene.
Energy converting chemicals
Light energy can convert chemicals to other forms. Vitamin A, also known as retinol, anti-dry eye vitamins, is a required nutrition for human health. The predecessor of vitamin A is present in the variety of plant carotene. Vitamin A is critical for vision because it is needed by the retina of eye. Retinol can be converted to retinal, and retinal is a chemical necessary for rhodopsin. As light enters the eye, the 11-cis-retinal is isomerized to the all-"trans" form.
Mechanism of Vision
We now know in rhodopsin, there is protein and retinal. The large protein is called opsin. Opsin does not absorb visible light, but when it bonded with 11-cis-retinal by its lysine side-chain to from rhodopsin, the new molecule has a very broad absorption band in the visible region of the spectrum.[2][3]
The reaction above shows Lysine side-chain from the opsin react with 11-cis-retinal when stimulated. By removing the oxygen atom form the retinal and two hydrogen atom form the free amino group of the lysine, the linkage show on the picture above is formed, and it is called Schiff base.
Signal transduction pathway
In human eyes, rod and cones react to light stimulation, and a series of chemical reactions happen in cells. These cells receive light, and pass on signals to other receiver cells. This chain of process is class signal transduction pathway. Signal transduction pathway is a mechanism that describes the ways cells react and respond to stimulation.
The molecule cis-retinal can absorb light at a specific wavelength. When visible light hits the cis-retinal, the cis-retinal undergoes an isomerization, or change in molecular arrangement, to all-trans-retinal. The new form of trans-retinal does not fit as well into the protein, and so a series of geometry changes in the protein begins. The resulting complex is referred to a bathrhodopsin (there are other intermediates in this process, but we'll ignore them for now).
As the protein changes its geometry, it initiates a cascade of biochemical reactions that results in changes in charge so that a large potential difference builds up across the plasma membrane. This potential difference is passed along to an adjoining nerve cell as an electrical impulse. The nerve cell carries this impulse to the brain, where the visual information is interpreted.
Exercise \(1\)
Indigo is an organic dye with a distinctive blue color. What wavelength does indigo absorb? What color is absorbed for our eyes to perceive blue?
Answer
Indigo absorbs orange light, which is a wavelength of 375-475 nm. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/03%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/3.07%3A_Conjugation_and_the_Absorption_of_Light_in_the_Real_World.txt |
Concepts & Vocabulary
3.1 Chapter Objectives and Preview of Ultraviolet Spectroscopy
• Ultraviolet (UV) spectroscopy is a technique that is only applicable to conjugated compounds giving information on the nature of the conjugated pi electron system.
3.2 Conjugated Dienes
• A diene is a hydrocarbon chain that has two double bonds that may or may not be adjacent to each other.
• The arrangements of these double bonds can have varying affects on the compounds reactivity and stability.
• A molecule is defined as conjugated when there is a system of connected p orbitals where electron density can be shared across the system.
• Enones are a type of conjugated diene.
• Polyenes are compounds that have more than two double bonds present in the molecule.
3.3 Electronic Transitions
• When both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
• If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO.
• π- π* energy gaps are narrower than σ - σgaps.
• Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated pi systems.
• In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer.
3.4 Ultraviolet Absorption
• To understand why some compounds are colored and others are not, and to determine the relationship of conjugation to color, we must make accurate measurements of light absorption at different wavelengths in and near the visible part of the spectrum.
• The visible region of the spectrum comprises photon energies of 36 to 72 kcal/mol, and the near ultraviolet region, out to 200 nm, extends this energy range to 143 kcal/mol.
• When sample molecules are exposed to light having an energy that matches a possible electronic transition within the molecule, some of the light energy will be absorbed as the electron is promoted to a higher energy orbital.
3.5 UV- Visible Spectrometer
• UV-Vis spectrometers have a light source (usually a deuterium or tungsten lamp), a sample holder and a detector.
• UV-vis spectroscopic data can give qualitative and quantitative information of a given compound or molecule.
• For a sample, it is important to use a reference cell to zero the instrument for the solvent the compound is in.
• For quantitative information on the compound, calibrating the instrument using known concentrations of the compound in question in a solution with the same solvent as the unknown sample would be necessary for accurate results.
• UV-vis spectroscopy works well on liquids and solutions, but if the sample is more of a suspension of solid particles in liquid, the sample will scatter the light more than absorb the light and the data will be very skewed.
• Every solvent has a UV-vis absorbance cutoff wavelength, which is the wavelength below which the solvent itself absorbs all of the light.
3.6 Interpreting Ultraviolet Spectra
• The basic setup is: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred.
• Some UV spectra have only one broad peak, although many molecules have more than one broad peak.
• The convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm).
• There are two things that we look for and record from a UV-Vis spectrum. The first is lambda max (λmax), which is the wavelength at maximal light absorbance and the secon is to record how much light is absorbed at λmax.
3.7 Conjugation and the Absorption of Light in the Real World
• Light influence our everyday lives.
• Eyes receive light energy then transfer and passing the energy into neural impulses to brain.
• As an energy wave, energy is passed on through light at different wavelength.
• Light energy can convert chemicals to other forms.
• In human eyes, rod and cones react to light stimulation, and a series of chemical reactions happen in cells.
Skills to Master
• Skill 3.1 Distinguish between different types of dienes.
• Skill 3.2 Compare molecules to determine if there is a larger or smaller energy gap.
• Skill 3.3 Understand which molecules will absorb longer wavelengths.
• Skill 3.4 Calculate the energy to excite a molecule from HOMO to LUMO.
• Skill 3.5 Understand how the spectrometer works.
• Skill 3.6 Understand the limitations of a spectrometer.
• Skill 3.7 Determine λmax.
• Skill 3.8 Gather information from a UV-vis spectrum.
• Skill 3.9 Understand why humans see different colors. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/03%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/3.S%3A_Summary.txt |
Learning Objectives
After completing this chapter, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• solve problems which may require the interpretation of IR spectra in addition to other spectral data.
• define, and use in context, the key terms introduced in this chapter.
When a molecule absorbs infrared (IR) radiation, the molecule vibrates, which causes the irradiated molecules to heat up. Molecules in the atmosphere, such as carbon dioxide, methane, and water, absorb IR radiation, which in turn creates more heat at the earth's surface. This in turn responsible for the greenhouse effect. In structure determination, IR spectroscopy is an important tool. It provides valuable information on what functional groups are present or absent in the molecule. An IR spectrum can be thought of as a fingerprint for the molecule.
This chapter will focus on IR spectroscopy. To start, some basic theory behind this technique will be discussed, followed by what type of information you can glean from spectra.
4.02: Theory
Objectives
After completing this section, you should be able to
• identify (by wavelength, wavenumber, or both) the region of the electromagnetic spectrum which is used in infrared (IR) spectroscopy.
• discuss, in general terms, the effect that the absorption of infrared radiation can have on a molecule.
Infrared (IR) spectroscopy is one of the most common and widely used spectroscopic techniques employed mainly by inorganic and organic chemists due to its usefulness in determining structures of compounds and identifying them. Chemical compounds have different chemical properties due to the presence of different functional groups.
Introduction
Absorbing groups in the infrared region absorb within a certain wavelength region. The absorption peaks within this region are usually sharper when compared with absorption peaks from the ultraviolet and visible regions. In this way, IR spectroscopy can be very sensitive to determination of functional groups within a sample since different functional group absorbs different particular frequency of IR radiation. Also, each molecule has a characteristic spectrum often referred to as the fingerprint. A molecule can be identified by comparing its absorption peak to a data bank of spectra, which will be discussed in detail in other sections. IR spectroscopy is very useful in the identification and structure analysis of a variety of substances, including both organic and inorganic compounds. It can also be used for both qualitative and quantitative analysis of complex mixtures of similar compounds.
The use of infrared spectroscopy began in the 1950's by Wilbur Kaye. He had designed a machine that tested the near-infrared spectrum and provided the theory to describe the results. Karl Norris started using IR Spectroscopy in the analytical world in the 1960's and as a result IR Spectroscopy became an accepted technique. There have been many advances in the field of IR Spectroscopy, the most notable was the application of Fourier Transformations to this technique thus creating an IR method that had higher resolution and a decrease in noise. The year this method became accepted in the field was in the late 1960's.
Absorption Spectroscopy
There are three main processes by which a molecule can absorb radiation and each of these routes involves an increase of energy that is proportional to the light absorbed. The first route occurs when absorption of radiation leads to a higher rotational energy level in a rotational transition. The second route is a vibrational transition which occurs on absorption of quantized energy. This leads to an increased vibrational energy level. The third route involves electrons of molecules being raised to a higher electron energy, which is the electronic transition. It’s important to state that the energy is quantized and absorption of radiation causes a molecule to move to a higher internal energy level. This is achieved by the alternating electric field of the radiation interacting with the molecule and causing a change in the movement of the molecule. There are multiple possibilities for the different possible energy levels for the various types of transitions.
The energy levels can be rated in the following order: electronic > vibrational > rotational. Each of these transitions differs by an order of magnitude. Rotational transitions occur at lower energies (longer wavelengths) and this energy is insufficient and cannot cause vibrational and electronic transitions but vibrational (near infra-red) and electronic transitions (ultraviolet region of the electromagnetic spectrum) require higher energies.
The energy of IR radiation is weaker than that of visible and ultraviolet radiation, and so the type of radiation produced is different. Absorption of IR radiation is typical of molecular species that have a small energy difference between the rotational and vibrational states. A criterion for IR absorption is a net change in dipole moment in a molecule as it vibrates or rotates. Using the molecule HBr as an example, the charge distribution between hydrogen and bromine is not evenly distributed since bromine is more electronegative than hydrogen and has a higher electron density. $HBr$ thus has a large dipole moment and is thus polar. The dipole moment is determined by the magnitude of the charge difference and the distance between the two centers of charge. As the molecule vibrates, there is a fluctuation in its dipole moment; this causes a field that interacts with the electric field associated with radiation. If there is a match in frequency of the radiation and the natural vibration of the molecule, absorption occurs and this alters the amplitude of the molecular vibration. This also occurs when the rotation of asymmetric molecules around their centers results in a dipole moment change, which permits interaction with the radiation field.
Diatomic Molecular Vibration
The absorption of IR radiation by a molecule can be likened to two atoms attached to each other by a massless spring. Considering simple diatomic molecules, only one vibration is possible. The Hook's law potential on the other hand is based on an ideal spring
\begin{align} F &= -kx \label{1} \[4pt] &= -\dfrac{dV(x)}{dx} \label{2} \end{align}
this results in one dimensional space
$V(r) = \dfrac{1}{2} k(r-r_{eq})^2 \label{3}$
One thing that the Morse and Harmonic oscillator have in common is the small displacements ($x=r-r_{eq}$) from the equilibrium. Solving the Schrödinger equation for the harmonic oscillator potential results in the energy levels results in
$E_v = \left(v+\dfrac{1}{2}\right)hv_e \label{4}$
with $v=0,1,2,3,...,\,infinity$
$v_e = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{\mu}} \label{5}$
When calculating the energy of a diatomic molecule, factors such as anharmonicity (has a similar curve with the harmonic oscillator at low potential energies but deviates at higher energies) are considered. The energy spacing in the harmonic oscillator is equal but not so with the anharmonic oscillator. The anharmonic oscillator is a deviation from the harmonic oscillator. Other considered terms include; centrifugal stretching, vibrational and rotational interactions have to be taken into account. The energy can be expressed mathematically as
$E_v = \underset{\text{Harmonic Oscillator}}{\left(v+\dfrac{1}{2}\right)hv_e} - \underset{\text{anharmonicity}}{\left(v+\dfrac{1}{2}\right)^2 X_e hv_e} + \underset{\text{Rigid Rotor}}{B_e J (J+1)} - \underset{\text{centrifugal stretching}}{D_e J^2 (J+1)^2} -\alpha_e \underset{\text{rovibrational coupling}}{\left(v+\dfrac{1}{2}\right) J(J+1)} \label{6}$
The first and third terms represent the harmonicity and rigid rotor behavior of a diatomic molecule such as HCl. The second term represents anharmonicity and the fourth term represents centrifugal stretching. The fifth term represents the interaction between the vibration and rotational interaction of the molecule.
Polyatomic Molecular Vibration
The bond of a molecule experiences various types of vibrations and rotations. This causes the atom not to be stationary and to fluctuate continuously. Vibrational motions are defined by stretching and bending modes. These movements are easily defined for diatomic or triatomic molecules. This is not the case for large molecules due to several vibrational motions and interactions that will be experienced. When there is a continuous change in the interatomic distance along the axis of the bond between two atoms, this process is known as a stretching vibration. A change in the angle occurring between two bonds is known as a bending vibration. Four bending vibrations exist namely, wagging, twisting, rocking and scissoring. A CH2 group is used as an example to illustrate stretching and bending vibrations below.
Symmetric Stretch Asymmetric Stretch Twisting
Wagging Scissoring Rocking
Types of Vibrational Modes. To ensure that no center of mass motion occurs, the center atom (yellow ball) will also move. Figure from Wikipedia
As stated earlier, molecular vibrations consist of stretching and bending modes. A molecule consisting of (N) number of atoms has a total of 3N degrees of freedom, corresponding to the Cartesian coordinates of each atom in the molecule. In a non-linear molecule, 3 of these degrees of freedom are rotational, 3 are translational and the remainder is fundamental vibrations. In a linear molecule, there are 3 translational degrees of freedom and 2 are rotational. This is because in a linear molecule, all of the atoms lie on a single straight line and hence rotation about the bond axis is not possible. Mathematically the normal modes for a linear and non linear can be expressed as
Linear Molecules: (3N - 5) degrees of freedom
Non-Linear molecules: (3N - 6) degrees of freedom
Example 1: Vibrations of Water
Diagram of Stretching and Bending Modes for H2O.
Solution
H2O molecule is a non-linear molecule due to the uneven distribution of the electron density. O2 is more electronegative than H2 and carries a negative charge, while H has a partial positive charge. The total degrees of freedom for H2O will be 3(3)-6 = 9-6 = 3 degrees of freedom which correspond to the following stretching and bending vibrations. The vibrational modes are illustrated below:
Example Vibrations of $CO_2$
Diagram of Stretching and Bending Modes for CO2.
Solution
CO2 is a linear molecule and thus has the formula (3N-5). It has 4 modes of vibration (3(3)-5). CO2 has 2 stretching modes, symmetric and asymmetric. The CO2 symmetric stretch is not IR active because there is no change in dipole moment because the net dipole moments are in opposite directions and as a result, they cancel each other. In the asymmetric stretch, O atom moves away from the C atom and generates a net change in dipole moments and hence absorbs IR radiation at 2350 cm-1. The other IR absorption occurs at 666 cm-1. CO2 symmetry with $D_{\infty h}$ CO2 has a total of four of stretching and bending modes but only two are seen. Two of its bands are degenerate and one of the vibration modes is symmetric hence it does not cause a dipole moment change because the polar directions cancel each other. The vibrational modes are illustrated below:
Selection Rules of IR
In order for vibrational transitions to occur, they are normally governed by some rules referred to as selection rules.
1. An interaction must occur between the oscillating field of the electromagnetic radiation and the vibrational molecule for a transition to occur. This can be expressed mathematically as
$\left(\dfrac{d\mu}{dr}\right)_{r_{eq}} \not= 0 \label{30}$
$\triangle v = +1$ and $\triangle J = +1 \label{31}$
1. This holds for a harmonic oscillator because the vibrational levels are equally spaced and that accounts for the single peak observed in any given molecular vibration. For gases J changes +1 for the R branch and -1 for the P branch.$\triangle J = 0$ is a forbidden transition and hence a q branch for a diatomic will not be present. For any anharmonic oscillator, the selection rule is not followed and it follows that the change in energy becomes smaller. This results in weaker transitions called overtones, then $\triangle v = +2$ (first overtone) can occur, as well as the 2nd overtone $\triangle v = +3$. The frequencies of the 1st and 2nd overtones provides information about the potential surface and about two to three times that of the fundamental frequency.
2. For a diatomic, since $\mu$ is known, measurement of ue provides a value for k, the force constant.
$k = \left(\dfrac{d^2 V(r)}{dr^2}\right)_{r_{eq}} \label{32}$
where k is the force constant and indicates the strength of a bond.
Influence Factors of IR
• Isotope Effects: It's been observed that the effect on k when an atom is replaced by an isotope is negligible but it does have an effect on $\nu$ due to changes in the new mass. This is because the reduced mass has an effect on the rotational and vibrational behavior.
• Solvent Effects: The polarity of solvent will have an influence on the IR spectra of organic compounds due to the interactions between solvent and compounds, which is called solvent effects. If we place a compound, which contains n, pi and pi* orbitals, into a polar solvent, the solvent will stabilizes these three orbitals in different extent. The stabilization effects of polar solvent on n orbital is the largest one, the next larger one is pi* orbital, and the effects on pi orbital is the smallest one. The spectra of n→pi* transition will shift to blue side, which means it will move to shorter wavelengths and higher energies since the polar solvent causes the energy difference between n orbital and pi* orbital to become bigger. The spectra of pi→pi* transition will shift to red side, which means it will move to longer wavelengths and lower energies since the polar solvent causes the energy difference between n orbital and pi* orbital to become smaller.
In summary, some bonds absorb infrared light more strongly than others, and some bonds do not absorb at all. In order for a vibrational mode to absorb infrared light, it must result in a periodic change in the dipole moment of the molecule. Such vibrations are said to be infrared active. In general, the greater the polarity of the bond, the stronger its IR absorption. The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon triple bond in most alkynes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkyne groups absorb rather weakly compared to carbonyls. Some kinds of vibrations are infrared inactive. The stretching vibrations of completely symmetrical double and triple bonds, for example, do not result in a change in dipole moment, and therefore do not result in any absorption of light (but other bonds and vibrational modes in these molecules do absorb IR light).
Exercise $1$
Which of the three main processes a molecule can absorb radiation leads to IR absorptions? What
Answer
Absorption of IR radiation is typical of molecular species that have a small energy difference between the rotational and vibrational states.
Exercise $2$
What is key for a molecule to be IR active?
Answer
A criterion for IR absorption is a net change in dipole moment in a molecule as it vibrates or rotates.
Exercise $3$
Define a stretching vibration.
Answer
In a stretching vibration, the distance between two atoms increases and decreases in a rhythmic manner, but the atoms remain aligned along the bond axis.
Exercise $4$
Define a bending vibration.
Answer
In a bending vibration, the positions of the atoms change relative to the bond axis.
Exercise $5$
What are the vibrational modes of the methylene group, CH2?
Answer
The stretching and bending vibrations of methylene chloride are:
• symmetric stretching
• asymmetric stretching
• wagging
• twisting
• rocking
• scissoring
Exercise $6$
The intensity of C=O stretching is stronger than that of C=C stretching. The C=O also appears at a higher wavenumber than a C=C. Explain it.
Answer
The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon double bond in most alkenes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkene groups absorb rather weakly compared to carbonyls.
A C=O bond is stronger than a C=C bond. Stronger bonds lead to a higher frequency absorbed due to Hooke's Law. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/04%3A_Infrared_Spectroscopy/4.01%3A_Chapter_Objectives_and_Preview_of_Infrared_Spectroscopy.txt |
Learning Objectives
After completing this section, you should be able to:
• have basic understanding of how the IR spectrometer works
• understand the basic components of the IR spectrometer
There are two types of instruments used to measure IR absorption: Fourier transform (FT) spectrometers and dispersive spectrometers. FTIR spectrometers are the most commonly used instruments for obtaining IR spectra. FTIR spectrometers have several prominent advantages:
1. The signal-to-noise ratio of spectrum is significantly higher than the previous generation infrared spectrometers.
2. The accuracy of wavenumber is high. The error is within the range of ± 0.01 cm-1.
3. The scan time of all frequencies is short (approximately 1 s).
4. The resolution is extremely high (0.1 ~ 0.005 cm-1).
5. The scan range is wide (1000 ~ 10 cm-1).
6. The interference from stray light is reduced.
Due to these advantages, FTIR Spectrometers have replaced dispersive IR spectrometers.
The Infrared Spectrometer
Development of IR Spectrometers
Up till FTIR spectrometers, there have been three generations of IR spectrometers.
1. The first generation IR spectrometer was invented in late 1950s. It utilizes prism optical splitting system. The prisms are made of NaCl. The requirement of the sample’s water content and particle size is extremely strict. Further more, the scan range is narrow. Additionally, the repeatability is fairly poor. As a result, the first generation IR spectrometer is no longer in use.
2. The second generation IR spectrometer was introduced to the world in 1960s. It utilizes gratings as the monochrometer. The performance of the second generation IR spectrometer is much better compared with IR spectrometers with prism monochrometer, But there are still several prominent weaknesses such as low sensitivity, low scan speed and poor wavelength accuracy which rendered it out of date after the invention of the third generation IR spectrometer.
3. The invention of the third generation IR spectrometer, Fourier transform infrared spectrometer, marked the abdication of monochrometer and the prosperity of interferometer. With this replacement, IR spectrometers became exceptionally powerful. Consequently, various applications of IR spectrometer have been realized.
Dispersive IR Spectrometers
To understand the powerfulness and usefulness of FTIR spectrometer, it is essential to have some background information of dispersive IR Spectrometer. The basic components of a dispersive IR spectrometer include a radiation source, monochromator, and detector. The common IR radiation sources are inert solids that are heated electrically to promote thermal emission of radiation in the infrared region of the electromagnetic spectrum. The monochromator is a device used to disperse or separate a broad spectrum of IR radiation into individual narrow IR frequencies.
Generally, dispersive spectrometers have a double-beam design with two equivalent beams from the same source passing through the sample and reference chambers as independent beams. These reference and sample beams are alternately focused on the detector by making use of an optical chopper, such as, a sector mirror. One beam will proceed, traveling through the sample, while the other beam will pass through a reference species for analytical comparison of transmitted photon wavefront information.
After the incident radiation travels through the sample species, the emitted wavefront of radiation is dispersed by a monochromator (gratings and slits) into its component frequencies. A combination of prisms or gratings with variable-slit mechanisms, mirrors, and filters comprise the dispersive system. Narrower slits gives better resolution by distinguishing more closely spaced frequencies of radiation and wider slits allow more light to reach the detector and provide better system sensitivity. The emitted wavefront beam (analog spectral output) hits the detector and generates an electrical signal as a response.
Detectors are devices that convert the analog spectral output into an electrical signal. These electrical signals are further processed by the computer using mathematical algorithm to arrive at the final spectrum. The detectors used in IR spectrometers can be classified as either photon/quantum detectors or thermal detectors.
It is the absorption of IR radiation by the sample, producing a change of IR radiation intensity, which gets detected as an off-null signal (e.g. different from reference signal). This change is translated into the recorder response through the actions of synchronous motors. Each frequency that passes through the sample is measured individually by the detector which consequently slows the process of scanning the entire IR region. A block diagram of a classic dispersive IR spectrometer is shown in Figure 4.3.1.
FTIR Spectrometers
The Components of FTIR Spectrometers
A common FTIR spectrometer consists of a source, interferometer, sample compartment, detector, amplifier, A/D convertor, and a computer. The source generates radiation which passes the sample through the interferometer and reaches the detector. Then the signal is amplified and converted to digital signal by the amplifier and analog-to-digital converter, respectively. Eventually, the signal is transferred to a computer in which Fourier transform is carried out. Figure 4.3.2 is a block diagram of an FTIR spectrometer.
The major difference between an FTIR spectrometer and a dispersive IR spectrometer is the Michelson interferometer.
Michelson Interferometer
The Michelson interferometer, which is the core of FTIR spectrometers, is used to split one beam of light into two so that the paths of the two beams are different. Then the Michelson interferometer recombines the two beams and conducts them into the detector where the difference of the intensity of these two beams are measured as a function of the difference of the paths. Figure 4.3.3 is a schematic of the Michelson Interferometer.
A typical Michelson interferometer consists of two perpendicular mirrors and a beamsplitter. One of the mirror is a stationary mirror and another one is a movable mirror. The beamsplitter is designed to transmit half of the light and reflect half of the light. Subsequently, the transmitted light and the reflected light strike the stationary mirror and the movable mirror, respectively. When reflected back by the mirrors, two beams of light recombine with each other at the beamsplitter.
If the distances travelled by two beams are the same which means the distances between two mirrors and the beamsplitter are the same, the situation is defined as zero path difference (ZPD). But imagine if the movable mirror moves away from the beamsplitter, the light beam which strikes the movable mirror will travel a longer distance than the light beam which strikes the stationary mirror. The distance which the movable mirror is away from the ZPD is defined as the mirror displacement and is represented by ∆. It is obvious that the extra distance travelled by the light which strikes the movable mirror is 2∆. The extra distance is defined as the optical path difference (OPD) and is represented by delta. Therefore,
δ=2Δ(4.3.1.1)(4.3.1.1)δ=2Δ
It is well established that when OPD is the multiples of the wavelength, constructive interference occurs because crests overlap with crests, troughs with troughs. As a result, a maximum intensity signal is observed by the detector. This situation can be described by the following equation:
δ=nλ(4.3.1.2)(4.3.1.2)δ=nλ
with n = 0,1,2,3...
In contrast, when OPD is the half wavelength or half wavelength add multiples of wavelength, destructive interference occurs because crests overlap with troughs. Consequently, a minimum intensity signal is observed by the detector. This situation can be described by the following equation:
δ=(n+12)λ(4.3.1.3)(4.3.1.3)δ=(n+12)λ
with n = 0,1,2,3...
These two situations are two extreme situations. If the OPD is neither n-fold wavelengths nor (n+1/2)-fold wavelengths, the interference should be between constructive and destructive. So the intensity of the signal should be between maximum and minimum. Since the mirror moves back and forth, the intensity of the signal increases and decreases which gives rise to a cosine wave. The plot is defined as an interferogram. When detecting the radiation of a broad band source rather than a single-wavelength source, a peak at ZPD is found in the interferogram. At the other distance scanned, the signal decays quickly since the mirror moves back and forth. Figure 4.3.1.44.3.1.4(a) shows an interferogram of a broad band source.
In an infrared spectrometer (Figure $4$ ) the sample to be analyzed is held in front of an infrared laser beam, in order to do this, the sample must be contained in something, consequently this means that the very container the sample is in will absorb some of the infrared beam.
IR Sample Preparation
IR spectra can be obtained from solid, liquid, or gas samples. Nujol mulls and pressed pellets are typically used for collecting spectra of solids, while thin-film cells are used for solution-phase IR spectroscopy. In these methods, infrared radiation is passed through the pellet or thin-film cells. A newer method for obtaining IR spectra uses attenuated total reflectance (ATR), which is discussed below. Gas samples require a special cell for sampling, but are not often used in organic chemistry, so will not be discussed further here.
When preparing a liquid, a drop of neat sample is pressed between two disks. A thin film of solid can be prepared by placing a drop of concentrated solution of the compound in the center of a disk and then allowing the solvent to evaporate. Since all materials have some sort of vibration associated with them pellets and thin-film cells must be considered carefully. If the sample holder has an optical window made of something that absorbs near where your sample does, the sample might not be distinguishable from the optical window of the sample holder. The range that is not blocked by a strong absorbance is known as a window (not to be confused with the optical materials of the cell). Windows are an important factor to consider when choosing the method to perform an analysis, as seen in Table $1$ there are a number of different materials each with their own characteristic absorption spectra and chemical properties. Keep these factors in mind when performing analyses and precious sample will be saved. For most organic compounds NaCl works well though it is susceptible to attack from moisture.
Material Transparent Ranges (cm -1) Solubility Notes
NaCl 40,000 - 625 H2O Easy to polish, hygroscopic
Silica glass 55,000-3,000 HF Attacked by HF
Quartz 40,000-2,500 HF Attacked by HF
Sapphire 20,000-1,780 - Strong
Diamond 40,000-2,500 and 1,800-200 - Very strong, expensive, hard, useless for pellets
CaF2 70,000-1,110 Acids Attacked by acids, avoid ammonium salts
BaF2 65,000-700 - Avoid ammonium salts
ZnSe 10,000 - 550 Acids Brittle, attacked by acids
AgCl 25,000-400 - Soft, sensitive to light.
KCl 40,000-500 H2O, Et2O, acetone Hygroscopic, soft, easily polished, commonly used in making pellets.
KBr 40,000-400 H2O, EtOH Hygroscopic, soft, easily polished, commonly used in making pellets.
CsBr 10,000-250 H2O, EtOH, acetone Hygroscopic soft
CsI 10,000-200 H2O, EtOH, MeOH, acetone Hygroscopic, soft.
Teflon 5,000-1,200; 1,200-900 - Inert, disposable
Polyethylene 4,000-3,000; 2,800-1,460; 1,380 - 730; 720- 30 - Inert, disposable
Table $1$ Various IR-transparent materials and their solubilities and other notes. M. R. Derrick, D. Stulik, and J. M. Landry, in Scientific Tools in Conservation: Infrared Spectroscopy in Conservation Science. Getty Conservation Institute (1999).
A common method of preparing solid samples for IR analysis is mulling, which is not a true solution, but a fine dispersion of a solid compound in a viscous liquid. The principle here is by grinding the particles to below the wavelength of incident radiation that will be passing through there should be limited scattering. To suspend those tiny particles, an oil, often referred to as Nujol is used. IR-transparent salt plates are used to hold the sample in front of the beam in order to acquire data. To prepare a sample for IR analysis using a salt plate, first decide what segment of the frequency band should be studied, refer to Table $1$ for the materials best suited for the sample.
Attenuated Total Reflectance- Fourier Transform Infrared Spectroscopy
First publicly proposed in 1959 by Jacques Fahrenfort from the Royal Dutch Shell laboratories in Amsterdam, ATR IR spectroscopy was described as a technique to effectively measure weakly absorbing condensed phase materials. In Fahrenfort's first article describing the technique, published in 1961, he used a hemicylindrical ATR crystal (see Experimental Conditions) to produce single-reflection ATR (Figure $5$ ). ATR IR spectroscopy was slow to become accepted as a method of characterization due to concerns about its quantitative effectiveness and reproducibility. The main concern being the sample and ATR crystal contact necessary to achieve decent spectral contrast. In the late 1980’s FTIR spectrometers began improving due to an increased dynamic range, signal to noise ratio, and faster computers. As a result ATR-FTIR also started gaining traction as an efficient spectroscopic technique. These days ATR accessories are often manufactured to work in conjunction with most FTIR spectrometers.
ATR-FTIR is a physical method of compositional analysis that builds upon traditional transmission FTIR spectroscopy to minimize sample preparation and optimize reproducibility. Sample accessories make obtaining IR spectra of solids and liquids easier and it is not necessary to prepare Nujol mulls or disks. With ATR, infrared radiation is passed through an infrared transmitting crystal with a high refractive index. This allows the radiation to reflect within the crystal. This should be a material that is fully transparent to the incident infrared radiation to give a real value for the refractive index, $6$.
We can consider the sample to be absorbing in the infrared. Electromagnetic energy will pass through the crystal/sample interface and propagate into the sample via the evanescent wave. This energy loss must be compensated with the incident IR light. Thus, total reflectance is no longer occurring and the reflection inside the crystal is attenuated. If a sample does not absorb, the reflectance at the interface shows no attenuation. Therefore if the IR light at a particular frequency does not reach the detector, the sample must have absorbed it.
The penetration depth of the evanescent wave within the sample is on the order of 1µm. The expression of the penetration depth is given in \ref{8} and is dependent upon the wavelength and angle of incident light as well as the refractive indices of the ATR crystal and sample. The effective path length is the product of the depth of penetration of the evanescent wave and the number of points that the IR light reflects at the interface between the crystal and sample. This path length is equivalent to the path length of a sample in a traditional transmission FTIR setup.
$d_{p} = \frac{ \lambda }{2 \pi n_{1}} (sin \omega - ( \frac{n_{1}}{n_{2}} )^{2} )^{1/2} \label{8}$
Refractive Indices of ATR Crystal and Sample
Typically an ATR attachment can be used with a traditional FTIR where the beam of incident IR light enters a horizontally positioned crystal with a high refractive index in the range of 1.5 to 4, as can be seen in Table $2$ will consist of organic compounds, inorganic compounds, and polymers which have refractive indices below 2 and can readily be found on a database. The most commonly used crystal is zinc selenide.
Material Refractive Index (RI) Spectral Range (cm-1)
Zinc Selenide (ZnSe) 2.4 20,000 - 650
Germanium (Ge) 4 5,500 - 870
Sapphire (Al2O3) 1.74 50,000 - 2,000
Diamond (C) 2.4 45,000 - 2,500,
1650 - 200
Table $2$ A summary of popular ATR crystals. Data obtained from F. M. Mirabella, Internal reflection spectroscopy: Theory and applications, 15, Marcel Dekker, Inc., New York (1993).
Single and Multiple Reflection Crystals
Multiple reflection ATR was initially more popular than single reflection ATR because of the weak absorbances associated with single reflection ATR. More reflections increased the evanescent wave interaction with the sample, which was believed to increase the signal to noise ratio of the spectrum. When IR spectrometers developed better spectral contrast, single reflection ATR became more popular. The number of reflections and spectral contrast increases with the length of the crystal and decreases with the angle of incidence as well as thickness. Within multiple reflection crystals some of the light is transmitted and some is reflected as the light exits the crystal, resulting in some of the light going back through the crystal for a round trip. Therefore, light exiting the ATR crystal contains components that experienced different number of reflections at the crystal-sample interface.
Angle of Incidence
It was more common in earlier instruments to allow selection of the incident angle. In all cases for total internal reflection to hold, the angle of incidence must exceed the critical angle and ideally complement the angle of the crystal edge so that the light enters at a normal angle of incidence. These days 45° is the standard angle on most ATR-FTIR setups.
ATR Crystal Shape
For the most part ATR crystals will have a trapezoidal shape as shown in Figure $6$. This shape facilitates sample preparation and handling on the crystal surface by enabling the optical setup to be placed below the crystal. However, different crystal shapes (Figure $7$ ) may be used for particular purposes, whether it is to achieve multiple reflections or reduce the spot size. For example, a hemispherical crystal may be used in a microsampling experiment in which the beam diameter can be reduced at no expense to the light intensity. This allows appropriate measurement of a small sample without compromising the quality of the resulting spectral features.
Because the path length of the evanescent wave is confined to the interface between the ATR crystal and sample, the sample should make firm contact with the ATR crystal. The sample sits atop the crystal and intimate contact can be ensured by applying pressure above the sample. However, one must be mindful of the ATR crystal hardness. Too much pressure may distort the crystal and affect the reproducibility of the resulting spectrum.
Exercise $1$
Why did FTIR spectrometers become the standard over dispersive spectrometers?
Answer
The advantages of FTIR are:
• The signal-to-noise ratio of spectrum is significantly higher than the previous generation infrared spectrometers.
• The accuracy of wavenumber is high.
• The error is within the range of ± 0.01 cm-1.
• The scan time of all frequencies is short (approximately 1 s).
• The resolution is extremely high (0.1 ~ 0.005 cm-1).
• The scan range is wide (1000 ~ 10 cm-1).
• The interference from stray light is reduced.
Essentially, FTIR spectrometers produce more reliable data in a better and faster way.
Exercise $2$
In transmission IR samples, how are solids prepared?
Answer
Nujol mulls and pressed pellets are typically used for collecting spectra of solids.
Exercise $3$
What is the standard angle of incidence for an ATR-FTIR spectrometer?
Answer
These days 45° is the standard angle on most ATR-FTIR setups. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/04%3A_Infrared_Spectroscopy/4.03%3A_IR_Instrumentation.txt |
Learning Objectives
After completing this section, you should be able to:
• understand the idea behind why peaks fall at particular wavelengths
• begin to figure out which peaks
The IR spectrum is a graph where the x-axis is frequency and is labeled as wavenumber (cm-1). The y-axis is the amount of light absorbed and labeled as Transmittance (%). This is measuring how much light has been transmitted at a particular frequency. In other words, if all of the light is detected, then the molecule has absorbed no light at that frequency. If you run your finger along the baseline, which is at the top of the spectrum, you can tell when light has been absorbed due to a dip in the line. The dips are actually called peaks. In the theory section of this chapter, it was discussed that different types of bonds will absorb at different frequencies. Below is an IR spectrum of pentane.
Now that we have oriented ourselves with what an IR spectrum looks like, let's think back to the theory of IR spectroscopy and begin to understand why peaks appear as they do.
Origin of Peak Positions, Intensities, and Widths
Peak Positions
The equation below gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light.
There are two variables in the above equation - a chemical bond's force constant and reduced mass. Here, the reduced mass refers to (M1M2)/(M1+M2) where M1 and M2 are the masses of the two atoms in the vibrating bond, respectively. These two molecular properties determine the wavenumber at which a molecule will absorb infrared light. No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique. To understand the effect of atomic masses and force constant on the positions of infrared bands, table 1 and 2 are shown as an example, respectively.
Table 1. An Example of an Mass Effect
Bond C-H Stretch in cm-1
C-1H ~3000
C-2D ~2120
In this example, the force constant remains the same, while the reduced masses of C-1H and C-2D are different. Remember, deuterium is one of the isotopes of hydrogen - 2H (or 2D). Here, we simply doubled the mass of the hydrogen atom, the carbon-hydrogen stretching vibration is reduced by over 800 cm-1. Mass is inversely proportional to the wavenumber.
Table 2. An Example of a Bond Strength Effect
Bond C-H Stretch in cm-1
Csp3-H ~3000
Csp2-H ~3100
Csp-H ~3330
In this example, the reduced mass remains the same, while the force constant changes. Here, we can see that as we increase the strength of the bond (or increase the force constant), the wavenumber increase. Therefore, force constant is proportional to the frequency.
The Origin of Peak Intensities
The different vibrations of the different functional groups in the molecule give rise to bands of differing intensity. This is because $\frac{\partial \mu}{\partial x}$ is different for each of these vibrations. For example, the most intense band in the spectrum of pentane shown above is at 2958 cm-1 and is due to stretching of the C-H bond. One of the weaker bands in the spectrum of pentane is at 727 cm-1, and it is due to long-chain methyl rock of the carbon-carbon bonds in pentane. The change in dipole moment with respect to distance for the C-H stretching is greater than that for the C-C rock vibration, which is why the C-H stretching band is the more intense than C-C rock vibration.
Another factor that determines the peak intensity in infrared spectra is the concentration of molecules in the sample. The equation below that relates concentration to absorbance is Beer's law,
The absorptivity is the proportionality constant between concentration and absorbance. The absorptivity is an absolute measure of infrared absorbance intensity for a specific molecule at a specific wavenumber.
The Origin of Peak Widths
In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. The figure below shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.
In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is common in alcohols and carboxylic acids. When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.
The Origin of Group Frequencies
An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3300 and 2800 cm-1 and carbonyl (C=O) stretching vibrations usually appear between 1800 and 1600 cm-1. This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample.
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
Exercise $1$
Why do carbon-nitrogen triple bonds appear at a higher wavenumber than carbon-nitrogen double bonds?
Answer
If we look at the equation for frequency, there are two variables. The reduced mass and the bond strength. In this case, our reduced mass will remain the same because we are looking at bonds with the same atoms attached - carbon and nitrogen. The difference is bond strength. A triple bond is a stronger bond than a double bond. Bond strength and frequency are proportional to each other, so the stronger the bond, the larger the wavenumber.
Exercise $2$
Would you expect the C=O bond to have a more intense or less intense peak?
Answer
Since there is a greater dipole change in the C=O bond stretch, this will lead to a more intense peak.
Exercise $3$
Alcohols typically show up as a very wide broad (3322 cm-1) as can be seen in the figure below. Why does the peak show up as broad?
Answer
The width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Alcohols have the ability to hydrogen bond, which creates the broad band. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/04%3A_Infrared_Spectroscopy/4.04_The_IR_Spectrum.txt |
Learning Objectives
• Understand where certain group frequencies are
• Be able to read and understand the data table for IR spectroscopy
As stated in the previous section, similar bonds will show up grouped near the same frequency because they tell us about the presence or absence of specific functional groups in a sample. From there, a data table of approximate frequencies for different types of bonds has been created to use to help IR spectrum analysis.
Table of Common IR Absorptions.
Approximate Frequency (cm-1) Description Bond Vibration Notes
3500 - 3200 broad, round O-H much broader, lower frequency (3200-2500)
if next to C=O
3400-3300 weak, triangular N-H stronger if next to C=O
3300 medium-strong =C-H (sp C-H)
3100-3000 weak-medium =C-H (sp2 C-H) can get bigger if lots of bonds present
3000-2900 weak-medium -C-H (sp3 C-H) can get bigger if lots of bonds present
2800 and 2700 medium C-H in O=C-H two peaks; "alligator jaws"
2250 medium C≡N
2250-2100 weak-medium C≡C stronger if near electronegative atoms
1800-1600 strong C=O lower frequency (1650-1550)
if attached to O or N
middle frequency if attached to C, H
higher frequency (1800) if attached to Cl
1650-1450 weak-medium C=C lower frequency (1600-1450) if conjugated
(i.e. C=C-C=C)
often several if benzene present
1450 weak-medium H-C-H bend
1300 - 1000 medium-strong C-O higher frequency (1200-1300) if conjugated
(i.e. O=C-O or C=C-O)
1250-1000 medium C-N
1000-650 strong C=C-H bend often several if benzene present
Note: strong, medium, weak refers to the length of the peak (in the y axis direction).
Note: spectra taken by ATR method (used at CSB/SJU) have weaker peaks between 4000-2500 cm-1 compared to reference spectra taken by transmittance methods (typical on SDBS and other sites).
Exercise \(1\)
What wavenumber range would you predict the triple bond region to be?
Answer
2200 - 2500 cm-1
Exercise \(2\)
How could you determine the presence of an aldehyde and rule out a ketone?
Answer
The aldehyde would have peaks at 2700 and 2800 cm-1, whereas the ketone would lack these. The absorption is due to the the Csp2-H bond.
Exercise \(3\)
How can you tell the difference between and alcohol and a carboxylic acid?
Answer
Here the main point of difference is either the presence of a carbonyl (C=O) or its absence. The carboxylic acid contains a C=O, so you would expect a peak somewhere between 1800-1600 cm-1, whereas an alcohol would not have a peak here.
4.06: Interpretation
Learning Objectives
• After completing this section, you should be able to
• be able to read an IR spectrum
• understand where the different regions for bonds in the IR spectrum are
• determine what type of hydrocarbon you have in your molecule
When analyzing IR spectra, there are a few things to remember:
1. You are looking to determine what functional groups are a part of a molecule. This is not a method for complete elucidation of a structure for the molecule.
2. You do NOT need to analyze every single peak. For example, the fingerprint region (1500 - 600 cm-1) is a forest of peaks, so this region is often ignored in analysis.
3. Remember, similar functional groups will have similar frequencies, which also indicates certain bonds will fall in the same region.
From this, different regions have been determined by previous research. The hydrogen bond region falls from 4000 to 2500 cm-1. This is the area you will find all of your O-H, N-H, and C-H bonds that typically appear in organic molecules. The next region is the triple bond region (2500 - 2000 cm-1), where the C≡C and C≡N bonds commonly found in organic molecules will absorb. The triple bond region leads into the double bond region (2000 - 1500 cm-1). Here you will find the C=C, C=O, and C=N bonds. Finally, the fingerprint region is 1500 - 600 cm-1, where all the single bonds will be found with the exception of some of the bonds to hydrogen that are found in the hydrogen bond region. Below is an image that summarizes what was just described. It is very helpful to have an idea of where these regions are when interpreting an IR spectrum, even with the data table in hand.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• Let's take a deeper look at pentane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum, which appear at 2873 - 2958 cm-1 in this spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense. The peak at 1460 cm-1, is due to C-H scissoring. 1379 cm-1 is a methyl rock. The C-H scissoring and methyl rock an often get lost in the fingerprint region when looking at more complex molecules. Since so many molecules with have similar peaks to an alkane, we can consider these our baseline. Therefore, the regions that will be of interest will be the hydrogen bond region, triple bond region, and double bond region when determining what functional groups are present or if there are other characteristic functional groups present.
In spectrum of 1-hexene, a terminal alkene, is shown below. Here will will look for differences compared to pentane. For alkenes, there will be a Csp2-H peak in the hydrogen bond region (if there are hydrogens attached to the double bond carbons as in this case) and the C=C bond. For 1-hexene, the Csp2-H peak is at 3079 cm-1 and the C=C bond is at 1642 cm-1. At 992 cm-1 and 907 cm-1, there are peaks due to the =C-H bend, but often get lost in the fingerprint region, so will not be a notable peak.
The spectrum of 1-hexyne, a terminal alkyne, is shown below. The notable peaks for the alkyne are the peak greater than 3000 cm-1 and the peak in the triple bond region. In 1-hexyne, there is a peak at 3309 cm-1, which is due to the Csp-H bond. The other peak of note is at 2118 cm-1 for the C≡C. We can ignore the peaks at 2873 cm-1 to 2959 cm-1 because these are your Csp3-H bonds, which are in so many organic molecules that they are not distinctive to the triple bond. This goes for the fingerprint region peaks as well.
Here, we have looked at hydrocarbons, but these are not the only functional groups to consider in organic molecules. In the next section, the focus will be on functional groups containing a heteroatom.
Exercise \(1\)
What regions of the IR would you focus on to determine if you have an alkane or alkene?
Answer
If we use the alkane as our baseline, the regions that would stick out for an alkene are the hydrogen bond region above 3000 cm-1 and the double bond region.
Exercise \(2\)
What two regions do you find the alkyne notable peaks?
Answer
The hydrogen bond region for the Csp-H bond and the triple bond region for the carbon-carbon triple bond.
Exercise \(3\)
Below is a spectrum of toluene. How does it differ from a plain alkene like 1-hexene?
Answer
Both still have a peak around 3000 cm-1 for the Csp2-H bond. However, due to conjugation the C=C shifts to a lower wavenumber in an aromatic molecule. It shifts to about 1600 cm-1 due to conjugation. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/04%3A_Infrared_Spectroscopy/4.05_IR_Data_Table.txt |
Learning Objectives
After completing this section, you should be able to
• describe how the so-called “fingerprint region” of an infrared spectrum can assist in the identification of an unknown compound.
• be able to use an infrared spectrum to determine the presence of functional groups, such as alcohols, amines and carbonyl groups, in an unknown compound, given a list of infrared absorption frequencies.
• identify the broad regions of the infrared spectrum in which occur absorptions caused by
• N−H, C−H, and O−H
• C≡C and C≡N
• C=O and C=C
One of the most common application of infrared spectroscopy is to the identification of organic compounds. In this section, we will focus key IR peaks for common functional groups in organic chemistry. We will focus on functional groups with OH, NH, C-O, C=O, and C≡N. It is possible to identify functional groups such as tertiary amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. This means you may or may not actually be able to determine if they are present as a functional group in your molecule.
Functional Groups Containing the C-O Bond
• Ethers have IR absorptions associated with both the C-O stretching vibrations.
• The figure below shows the spectrum of diethyl ether.
• Notable peak: C-O stretch at 1117 cm-1. Note: Since this falls in the fingerprint region, it can be hard to interpret sometimes.
• Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• The figure below shows the spectrum of ethanol.
• Notable peaks: the very broad, strong band of the O–H stretch at 3322 cm-1 and C-O stretch at 1113 cm-1.
Functional Groups Containing the C=O Bond
• Ketones have IR absorptions associated with the C=O bond.
• Below is a spectrum of 2-butanone.
• Notable peak: the strong band at 1712 cm-1 for the C=O.
• Note: for conjugated ketones, the carbonyl peak will shift 20-30 cm-1 lower.
• Aldehydes have IR absorptions associated with the C=O bond and the aldehyldic proton.
• Below is a spectrum of butanal.
• Notable peaks: the strong band at 1723 cm-1 for the C=O and for the aldehyldic proton there are two peaks at 2822 cm-1 and 2721 cm-1.
• Note: for conjugated aldehydes, the carbonyl peak will shift 20-30 cm-1 lower.
• Carboxylic acids have IR absorptions associated with the C=O bond and the carboxylic acid proton.
• Below is a spectrum of propanoic acid.
• Notable peaks: the strong band at 1706 cm-1 for the C=O and for the carboxylic acid proton is very broad band from 3100 to 2800 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• Note: for conjugated carboxylic acids, the carbonyl peak will shift 20-30 cm-1 lower.
• Esters have IR absorptions associated with the C=O bond and the C-O bond.
• Below is a spectrum of ethyl acetate.
• Notable peaks: the strong band at 1736 cm-1 for the C=O and for the C-O bond is at 1232 cm-1. Depending on the fingerprint region, it may be hard to determine the C-O bond peak.
• Note: for conjugated esters, the carbonyl peak will shift 20-30 cm-1 lower.
Organic Nitrogen Compounds
• Amines have IR absorptions associated with the N-H bond. There is a C-N peak as well, but it is often buried in the fingerprint region and difficult to discern.
• Below is a spectrum of butylamine.
• Notable peaks: the bands at 3368 cm-1 and 3290 cm-1
• Note: 2 peaks in the range 3400 - 3300 cm-1 indicates a primary amine, 1 peak in this range indicates a secondary amine. A tertiary amine will not have a peak in this region due to the lack of a N-H bond.
• Nitriles have IR absorptions associated with the C≡N bond. There is a C-N peak as well, but it is often buried in the fingerprint region and difficult to discern.
• Below is a spectrum of acetonitrile.
• Notable peaks: the band at 2252 cm-1
Organic Compounds Containing Halogens
• Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• The spectrum of propyl chloride is shown below.
• There are stretches due to the C-Cl, but they are located in the fingerprint region and difficult to discern from other peaks.
As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol.
Exercise \(1\)
What functional groups give the following signals in an IR spectrum?
A) 1700 cm-1
B) 2250 cm-1
C) 1700 cm-1 and 2510-3000 cm-1
Answer
A) Carbonyl: C=O bond
B) Nitrile: C≡N bond
C) Carboxylic acid: for the C=O at 1700 cm-1 and the broad OH at 2510-3000 cm-1
Exercise \(2\)
How can you distinguish the following pairs of compounds through IR analysis?
A)
B)
C)
Answer
A) A OH peak will be present around 3300 cm-1 for methanol and will be absent in the ether.
B) 1-hexene will have a alkene peak around 1650 cm-1 for the C=C and there will be another peak around 3100 cm-1 for the sp2 C-H group on the alkene, which will both be absent in cyclohexane
C) Cannot distinguish these two isomers. They both have the same functional groups and therefore would have the same peaks on an IR spectra.
Exercise \(3\)
3-Ethynyltoluene has the following spectrum. What notable peaks can you identify in the spectrum.
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)
Answer
Frequency (cm-1) Functional Group
3200 C≡C-H
3050 Csp2-H
2100 C≡C
1610 C=C
Exercise \(4\)
What absorptions would the molecule below have in an IR spectrum?
Answer
Frequency (cm-1) Functional Group
3000-3100 C=C-H
1710 C=O
1610 C=C
1100 C-O | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/04%3A_Infrared_Spectroscopy/4.07_Identifying_Characteristic_Functional_Groups.txt |
Learning Objectives
After completing this section, you should be able to
• determine functional groups in an IR spectrum
Here are some problems for IR analysis.
Exercise \(1\)
What functional group is present in the spectrum below?
Answer
Alkyne
Exercise \(2\)
What functional group is present in the spectrum below?
Answer
Carboxylic Acid
Exercise \(3\)
Which IR spectrum goes with which molecule below?
Molecule A:
Molecule B:
IR spectrum 1:
IR spectrum 2:
Answer
Molecule A is Spectrum 2.
Molecule B is Spectrum 1.
Exercise \(4\)
What notable peaks in the IR spectrum verify that benzaldehyde is present?
Answer
Frequency (cm-1) Functional Group
3063 Csp2-H
2818 and 2736 aldehyldic proton
1695 C=O
1653-1455 aromatic ring
4.09: Application
Learning Objectives
After completing this section, you should be able to
• who uses IR spectroscopy and why
Infrared spectroscopy, an analytical technique that takes advantage of the vibrational transitions of a molecule, has been of great significance to scientific researchers in many fields such as protein characterization, nanoscale semiconductor analysis and space exploration.
The most common application of IR spectroscopy is determining the functional groups present or absent in a molecule. Chemists often synthesize a molecule and need to determine if they made it or not and the presence or absence of a functional group could indicate whether or not the chemist made the molecule or not. For example, if one is performing a synthesis and oxidizes an alcohol to an aldehyde, the IR spectrum could help deduce quickly if you made the product or not. If the alcohol band disappears and the carbonyl band appears, this would be proof that the alcohol did in fact oxidize to the aldehyde. Since different molecules with different combination of atoms produce their unique spectra, infrared spectroscopy can be used to qualitatively identify substances. In addition, the intensity of the peaks in the spectrum is proportional to the amount of substance present, enabling its application for quantitative analysis.
Infrared spectroscopy is used across a wide variety of industries to monitor substances, since it is an simple and reliable technique. Environmental scientists use IR for detecting industrial pollutants. This can be to monitor air quality in big cities, such as Tokyo or London, to monitoring the methane levels in arctic circle. IR helps the environmental scientists study the pollutants or greenhouse gases and how it effects the everyday life.
In art conservation, IR spectroscopy is used to help identify what pigments, adhesives, fibers, plastics, and binders were used. The ability to know what molecular structures may be present in the artwork allows conservationists to determine the best way possible to preserve or clean the art. This helps the longevity of these important pieces of art and history persevere over time for many to enjoy and learn from.
Reference
1. Settle, F. A. Handbook of instrumental techniques for analytical chemistry; Prentice Hall PTR: Upper Saddle River, NJ, 1997.
2. Heigl, J. J.; Bell, M.; White, J. U. Anal. Chem.1947, 19, 293.
3. Baker, A. W. J. Phys. Chem. 1957, 61, 450.
4. Kamariotis, A.; Boyarkin, O. V.; Mercier, S. R.; Beck, R. D.; Bush, M. F.; Williams, E. R.; Rizzo, T. R. J. Am. Chem. Soc. 2006, 128, 905.
5. Stuart, B. Infrared spectroscopy fundamentals and applications; J. Wiley: Chichester, Eng.; Hoboken, N.J., 2004.
6. Günzler, H.; Heise, H. M. IR spectroscopy: an introduction; Wiley-VCH: Weinheim, 2002.
7. Wartewig, S. IR and Raman spectroscopy: fundamental processin; Wiley-VCH: Weinheim, 2003. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/04%3A_Infrared_Spectroscopy/4.08_Infrared_Spectroscopy_Problems.txt |
Concepts & Vocabulary
4.2 Theory
• Molecules can absorb light energy by three different processes and IR absorptions occur with small differences between the rotational and vibrational modes.
• IR absorption is a net change in dipole moment in a molecule as it vibrates or rotates.
• The absorption of IR radiation by a molecule can be likened to two atoms attached to each other by a massless spring often based using Hooke's Law.
• The bond of a molecule experiences various types of vibrations and rotations since they are not stationary and to fluctuate continuously, which are defined by stretching and bending modes.
• A molecule consisting of (N) number of atoms has a total of 3N degrees of freedom, corresponding to the Cartesian coordinates of each atom in the molecule.
• The larger the molecule, the more complex the vibrational modes.
4.3 Instrumentation
• There are two types of instruments used to measure IR absorption: Fourier transform (FT) spectrometers and dispersive spectrometers.
• The basic components of a dispersive IR spectrometer include a radiation source, monochromator, and detector.
• Dispersive spectrometers have a double-beam design with two equivalent beams from the same source passing through the sample and reference chambers as independent beams.
• A common FTIR spectrometer consists of a source, interferometer, sample compartment, detector, amplifier, A/D convertor, and a computer.
• The major difference between an FTIR spectrometer and a dispersive IR spectrometer is the Michelson interferometer.
• The Michelson interferometer, which is the core of FTIR spectrometers, is used to split one beam of light into two so that the paths of the two beams are different.
• IR spectra can be obtained from solid, liquid, or gas samples.
4.4 The IR Spectrum
• The IR spectrum is a graph where the x-axis is frequency and is labeled as wavenumber (cm-1).
• The y-axis is the amount of light absorbed and labeled as Transmittance (%).
• The spectrum is measuring how much light has been transmitted at a particular frequency.
• Bond strength and reduced mass are the two molecular properties determine the wavenumber at which a molecule will absorb infrared light.
• No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique.
• Peak intensity is determined by dipole moment and the concentration of the sample.
• The width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding.
• An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule.
4.5 IR Data Table
• Since similar functional groups appear at similar frequencies, a data table was curated for scientists to use as a guide for where they would expect certain peaks.
4.6 Interpretation
• The hydrogen bond region falls from 4000 to 2500 cm-1. This is the area you will find all of your O-H, N-H, and C-H bonds that typically appear in organic molecules.
• The next region is the triple bond region (2500 - 2000 cm-1), where the C≡C and C≡N bonds commonly found in organic molecules will absorb.
• The double bond region (2000 - 1500 cm-1). Here you will find the C=C, C=O, and C=N bonds.
• The fingerprint region is 1500 - 600 cm-1, where all the single bonds will be found with the exception of some of the bonds to hydrogen that are found in the hydrogen bond region.
• You are looking to determine what functional groups are a part of a molecule.
• You do NOT need to analyze every single peak.
4.7 Identifying Characteristic Functional Groups
• Ethers have IR absorptions associated with both the C-O stretching vibrations.
• Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• Ketones have IR absorptions associated with the C=O bond.
• Aldehydes have IR absorptions associated with the C=O bond and the aldehydic proton.
• Carboxylic acids have IR absorptions associated with the C=O bond and the carboxylic acid proton.
• Esters have IR absorptions associated with the C=O bond and the C-O bond.
• Amines have IR absorptions associated with the N-H bond. There is a C-N peak as well, but it is often buried in the fingerprint region and difficult to discern.
• Nitriles have IR absorptions associated with the C≡N bond. There is a C-N peak as well, but it is often buried in the fingerprint region and difficult to discern.
• IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups.
4.9 Application
• The most common application of IR spectroscopy is determining the functional groups present or absent in a molecule.
• Environmental scientists use IR for detecting industrial pollutants.
• In art conservation, IR spectroscopy is used to help identify what pigments, adhesives, fibers, plastics, and binders were used.
• IR spectroscopy is used in a wide variety of industries.
Skills to Master
• Skill 4.1 Distinguish between different types of stretching.
• Skill 4.2 Determine the number of vibrational modes molecules may have.
• Skill 4.3 Understand the instrumentation used to obtain IR spectra.
• Skill 4.4 Orient oneself with an IR spectrum.
• Skill 4.5 Be able to read a data table for IR spectroscopy.
• Skill 4.6 Determine functional groups present in the molecule based on an IR spectrum. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/04%3A_Infrared_Spectroscopy/4.S_Summary.txt |
Objectives
After completing this chapter, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• solve road-map problems which may require the interpretation of 1H NMR spectra in addition to other spectral data.
• define, and use in context, the key terms introduced in this chapter.
In the previous chapter, it was discussed that Infrared (IR) Spectroscopy gives information about the functional groups present in a molecule. Nuclear Magnetic Resonance (NMR) is another type of absorption spectroscopy similar to Ultraviolet-Visible spectroscopy (UV) or IR spectroscopy. In the presence of a magnetic field, a sample can absorb electromagnetic radiation, specifically in the radiofrequency (rf) region, based on the function of certain nuclei in the molecule. For organic chemists, NMR spectroscopy is an invaluable resource for determining the structure of molecules and often used first when analyzing a molecule. NMR spectroscopy complements IR spectroscopy because information on the hydrocarbon portion of the molecule can be obtained as well as additional information about the functional groups.
This chapter will focus on proton nuclear magnetic spectroscopy (1H NMR). To start, some basic theory behind this technique will be discussed, followed by what type of information you can glean from spectra and finishing with interpretation of an NMR spectrum for the determination of a structure of a molecule.
5.02: Theory of NMR
Objectives
After reading this section, you should be able to
• understand why NMR works
• understand why different hydrogens do not all show up in the same location on a spectrum
Introduction
Nuclear magnetic resonance (NMR) plays an important role in the fields chemistry, materials science, physics and engineering. In 1946, NMR was co-discovered by Purcell, Pound and Torrey of Harvard University and Bloch, Hansen and Packard of Stanford University. The discovery first came about when it was noticed that magnetic nuclei, such as 1H and 31P (read: proton and Phosphorus 31) were able to absorb radiofrequency energy when placed in a magnetic field of a strength that was specific to the nucleus. Upon absorption, the nuclei begin to resonate and different atoms within a molecule resonated at different frequencies. This observation allowed a detailed analysis of the structure of a molecule. Since then, NMR has been applied to solids, liquids and gasses, kinetic and structural studies, resulting in 6 Nobel prizes being awarded in the field of NMR.
Nuclear magnetic resonance, NMR, is a physical phenomenon of resonance transition between magnetic energy levels, happening when atomic nuclei are immersed in an external magnetic field and an applied electromagnetic radiation with specific frequency. By detecting the absorption signals, one can acquire NMR spectrum. According to the positions, intensities and fine structure of resonance peaks, people can study the structures of molecules quantitatively. The size of molecules of interest varies from small organic molecules, to biological molecules of middle size, and even to some macromolecules such as nucleic acids and proteins. Apart from these commonly utilized applications in organic compounds, NMR also plays an important role in analyzing inorganic molecules, which makes NMR spectroscopy a powerful technique.
But the a major question still remains- Why does NMR work? Some types of atomic nuclei act as though they spin on their axis similar to the Earth. Since they are positively charged they generate an electromagnetic field just as the Earth does. In effect, they will act as tiny bar magnetics. Not all nuclei act this way, but fortunately both 1H and 13C do have nuclear spins and will respond to this technique.
The concept of spin is regularly addressed in subatomic particle physics. However, to most people spin seems like an abstract concept. This is due to the fact there is no macroscopic equivalent of what spin is. However, for those people who have taken an introduction to chemistry course have seen the concept of spin in electrons. Electrons are subatomic particles which have spin intrinsic to them. The nucleus is not much different. Spin is just another form of angular momentum. The nucleus consists of protons and neutrons and neutrons and protons are comprised of subatomic particles known as quarks and gluons. The neutron has 2 quarks with a -e/3 charge and one quark with a +2e/3 charge resulting in a total charge of 0. The proton however, has 2 quarks with +2e/3 charge and only one quark with a -e/3 charge giving it a net positive charge. Both protons and neutrons are spin=1/2.
For any system consisting of n multiple parts, each with an angular momentum the total angular momentum can be described by J where
$J=|J_1+J_2+...+J_n|, |J_1+J_2+...+J_n| -1,...|J_1-J_2-...-J_n|$
$a=mx^2$
Here are some examples using the isotopes of hydrogen
• 1H = 1 proton so J=1/2
• 2H= 1 proton and 1 neutron so $J=1$ or 0.
For larger nuclei, it is not immediately evident what the spin should be as there are a multitude of possible values. For the remainder of the discussion we will attribute the spin of the nucleus, I, to be an intrinsic value. There are some rules that the nuclei do follow with respect to nuclear spin. They are summarized in the table below.
Table $1$: General rules for determination of nuclear spin quantum numbers
Mass Number Number of Protons Number of Neutrons Spin (I) Example
Even Even Even 0 16O
Odd Odd Integer (1,2,...) 2H
Odd Even Odd Half-Integer (1/2, 3/2,...) 13C
Odd Even Half-Integer (1/2, 3/2,...) 15N
Atomic nuclei with even numbers of protons and neutrons have zero spin and all the other atoms with odd numbers have a non-zero spin. Furthermore, all molecules with a non-zero spin have a magnetic moment, $\mu$, given by
$\mu=\gamma I$
where $\gamma$ is the gyromagnetic ratio, a proportionality constant between the magnetic dipole moment and the angular momentum, specific to each nucleus (Table 2). In other words, 1H and 13C are not unique in their ability to undergo NMR. All nuclei with an odd number of protons (1H, 2H, 14N, 19F, 31P) or nuclei with an odd number of neutrons (i.e. 13C) show the magnetic properties required for NMR. Only nuclei with even number of both protons and neutrons (12C and 16O) do not have the required magnetic properties.
Table $2$: The gyromagnetic ratios for several common nuclei
Nuclei Spin Gyromagetic Ratio (MHz/T) Natural Abundance (%)
1H 1/2 42.576 99.9985
13C 1/2 10.705 1.07
31P 1/2 17.235 100
27Al 5/2 11.103 100
23Na 3/2 11.262 100
7Li 3/2 16.546 92.41
29Si
1/2 -8.465 4.68
17O 5/2 5.772 0.038
15N 1/2 -4.361 0.368
The magnetic moment of the nucleus forces the nucleus to behave as a tiny bar magnet. In the absence of an external magnetic field, each magnet is randomly oriented (as can be seen in the figure below on the left). During the NMR experiment the sample is placed in an external magnetic field, $B_0$, which forces the bar magnets to align with (low energy) or against (high energy) the $B_0$. The nuclear spins will adopt specific orientations like a compass needle responds to the Earth’s magnetic field and aligns with it. Two possible orientations are possible, with the external field (i.e. parallel to and in the same direction as the external field) or against the field (i.e. antiparallel to the external field) as can be seen in the figure below on the right.
(Left) Random nuclear spin without an external magnetic field. (Right) Ordered nuclear spin in an external magnetic field
If the ordered nuclei are now subjected to electromagnetic (EM) radiation, specifically radiofrequency, of the proper frequency the nuclei aligned with the field will absorb energy and "spin-flip" to align themselves against the field, a higher energy state. When this spin-flip occurs the nuclei are said to be in "resonance" with the field, hence the name for the technique, Nuclear Magentic Resonance.
In order for the NMR experiment to work, a spin flip between the energy levels must occur. The energy difference between the two states corresponds to the energy of the electromagnetic radiation that causes the nuclei to change their energy levels. For most NMR spectrometers, $B_0$ is on the order of Tesla (T) while $\gamma$ is on the order of $10^7$. Consequently, the electromagnetic radiation required is on the order of 100's of MHz and even GHz. The energy of a photon is represented by
$E=h\nu$
and thus the frequency necessary for absorption to occur is represented as:
$\nu=\dfrac{\gamma B_0}{2\pi}$
For the beginner, the NMR experiment measures the resonant frequency that causes a spin flip. For the more advanced NMR users, the sections on NMR detection and Larmor frequency should be consulted.
The amount of energy, and hence the exact frequency of EM radiation required for resonance to occur is dependent on both the strength of the magnetic field applied and the type of the nuclei being studied. As the strength of the magnetic field increases the energy difference between the two spin states increases and a higher frequency (more energy) EM radiation needs to be applied to achieve a spin-flip (see image below).
Superconducting magnets can be used to produce very strong magnetic field, on the order of 21 tesla (T). Lower field strengths can also be used, in the range of 4 - 7 T. At these levels the energy required to bring the nuclei into resonance is in the MHz range and corresponds to radio wavelength energies, i.e. at a field strength of 4.7 T 200 MHz bring 1H nuclei into resonance and 50 MHz bring 13C into resonance. This is considerably less energy then is required for IR spectroscopy, ~10-4 kJ/mol versus ~5 - ~50 kJ/mol.
If all 1H nuclei have the same gyromagnetic ratio, shouldn't all the 1H absorb the same energy, so would all show up at the same spot? Luckily, no. The power of NMR is based on the concept of nuclear shielding, which allows for structural assignments. Every atom is surrounded by electrons, which orbit the nucleus. Charged particles moving in a loop will create a magnetic field which is felt by the nucleus. Therefore the local electronic environment surrounding the nucleus will slightly change the magnetic field experienced by the nucleus, which in turn will cause slight changes in the energy levels! This is known as shielding. Nuclei that experience different magnetic fields due to the local electronic interactions are known as nonequivalent nuclei (Bexperienced = B0 - Bisigma). The change in the energy levels requires a different frequency to excite the spin flip, which as will be seen below, creates a new peak in the NMR spectrum. The shielding allows for structural determination of molecules.
The shielding of the nucleus allows for chemically nonequivalent environments to be determined by Fourier Transforming the NMR signal. The result is a spectrum, shown below, that consists of a set of peaks in which each peak corresponds to a distinct chemical environment.
In 1H NMR spectrum, hydrogen atoms bound to a carbon consisting of a double bond are typically found in low field of the NMR spectrum and the hydrogens are considered deshielded. The cause for this is due to the movement of the electrons in the pi bond of the carbon-carbon double bond and how the magnetic field effects the protons bound to the pi bond.
The pi system creates an external magnetic field that is perpendicular to the double bond axis and causes the electrons in the pi bond to enter a circular motion (shown in red). The circular motion actually reinforces the external field at the edge of the double bond on both sides of the pi bond but creates a local field (shown in purple and green) that opposes the external field in the center of the double bond. Because of this pulling force within the pi bond across the double bond which reinforces the regions occupied by alkenyl hydrogens, the alkenyl hydrogens are strongly deshielded.
Bexperienced = B0 - Bisigma + Bipi
Unlike alkenyl hydrogens, alkynyl hydrogens give rise to shielded hydrogens, or relatively high field chemical shifts for H-NMR when subjected to an external magnetic field (Bexperienced = B0 - Bisigma - Bipi). The molecules are tumbling all around in solution, but for the alkyne the triple bond has a cylindrical pi cloud around the carbon-carbon triple bond. When the two bonds are subjected to an external magnetic field, these electrons will enter into a cylindrical motion that results in this strong shielding effect with high field chemical shifts. This means that the alkynyl hydrogens will be experiencing less of the external magnetic field. This electron cloud can be seen in the figure below.
Relaxation
Relaxation refers to the phenomenon of nuclei returning to their thermodynamically stable states after being excited to higher energy levels. The energy absorbed when a transition from a lower energy level to a high energy level occurs is released when the opposite happens. This can be a fairly complex process based on different timescales of the relaxation. The two most common types of relaxation are spin lattice relaxation (T1) and spin spin relaxation (T2).
To understand relaxation, the entire sample must be considered. By placing the nuclei in an external magnetic field, the nuclei create a bulk magnetization along the z-axis. The spins of the nuclei are also coherent. The NMR signal may be detected as long as the spins are coherent with one another. The NMR experiment moves the bulk magnetization from the z-axis to the x-y plane, where it is detected.
• Spin-Lattice Relaxation ($T_1$): T1 is the time it takes for the 37% of bulk magnetization to recovery along Z-axis from the x-y plane. The more efficient the relaxation process, the smaller relaxation time (T1) value you will get. In solids, since motions between molecules are limited, the relaxation time (T1) values are large. Spin-lattice relaxation measurements are usually carried out by pulse methods.
• Spin-Spin Relaxation ($T_2$): T2 is the time it takes for the spins to lose coherence with one another. T2 can either be shorter or equal to T1.
In the next sections, the topics to be discussed will build up to actually analyzing NMR spectra.
Exercise $1$
Where is radiofrequency (RF) radiation on the energy scale of the electromagnetic spectrum?
Answer
Radiofrequency (RF) radiation, which includes radio waves and microwaves, is at the low-energy end of the electromagnetic spectrum. It is a type of non-ionizing radiation.
Exercise $2$
If in a field strength of 4.7 T, 1H requires 200 MHz of energy to maintain resonance. If atom X requires 150 MHz, calculate the amount of energy required to spin flip atom X’s nucleus. Is this amount greater than the energy required for hydrogen?
Answer
E = hυ
E = (6.62 × 10−34)(150 MHz)
E = 9.93 × 10−26 J
The energy is equal to 9.93x10-26 J. This value is smaller than the energy required for hydrogen (1.324 × 10−25 J).
Exercise $3$
Calculate the energy required to spin flip at 400 MHz. Does changing the frequency to 500 MHz decrease or increase the energy required? What about 300 MHz.
Answer
E = hυ
E = (6.62 × 10−34)(400 MHz)
E = 2.648 × 10−25 J
The energy would increase if the frequency would increase to 500 MHz, and decrease if the frequency would decrease to 300 MHz.
Exercise $4$
What “things” in a molecule generate magnetic fields that will influence Bo for a particular hydrogen nucleus?
Answer
The electrons in bonds can generate magnetic fields that will influence BO.
Contributors and Attributions
• Derrick Kaseman (UC Davis) and Revathi Srinivasan Ganesh Iyer (UCD) | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.01%3A_Chapter_Objectives_and_Preview_of_Nuclear_Magnetic_Resonance_Spectroscopy.txt |
Objectives
After completing this section, you should be able to:
• have basic understanding of how the NMR works
• understand the basic components of the NMR spectrometer
NMR spectroscopy works by varying the machine’s emitted frequency over a small range while the sample is inside a constant magnetic field. Most of the magnets used in NMR instruments are superconducting to create the magnetic field range from 6 to 24 T. The diagram below is of an NMR spectrometer. In a basic experiment, an organic sample is dissolved in a suitable solvent, which is often deuterated chloroform or another deuterated solvent, and placed in a thin glass tube. A reasonable concentration for your sample is 5 to 25 mg of desired compound dissolved in 0.6 to 1.0 mL of a suitable solvent, which should be free of particulate matter for solution NMR experiments. The sample is inserted through the NMR spinner and pneumatically lowered into the NMR probe, which is between two magnetic poles. The coils around the NMR tube in the diagram below is where the sample must be. NMR experiments require a uniform magnetic field over the whole of the NMR sample volume that sits within the detection coil. Deviation from this ideal introduces various line shape distortions, compromising both sensitivity and resolution.The strong magnetic field causes the 1H nuclei (or other NMR active nuclei in other experiments) in a molecule to align in one of the two possible orientations. The sample is then subjected to a frequency from the radio wave source. A detector then interprets the results and sends it to the main console. Using a data analysis program, the free induction decay (FID) data is transformed into the spectrum typically shown in papers and books.
NMR Instrument
An NMR can be divided into three main components: the workstation computer where one operates the NMR instrument, the NMR spectrometer console, and the NMR magnet, which is shown in the picture below.
The NMR is layered with the superconducting magnet just outside the probe towards the center of the spectrometer (diagram below). The magnet only works if a coil is kept very cool, so it is immersed in the first layer inside the NMR, which is liquid helium (4.2 K). To reduce the boil off rate of the liquid helium, the next layer is filled with liquid nitrogen at 77 K. The liquid nitrogen reservoir space is mostly above the magnet. This way, it can act as a less expensive refrigerant to block infrared radiation from reaching the liquid helium jacket. Aluminum and stainless steel are also used to contain the liquids and block infrared irradiation as well. All the layers are working to keep the magnet coil very cold.
Exercise \(1\)
Consider a sample in an NMR tube.The crosshatched region in the tube is the area over which signal is recorded.Why is it important that BAPPL be homogeneous over this entire region?
Answer
NMR experiments require a uniform magnetic field over the whole of the NMR sample volume that sits within the detection coil, which is represented by the crosshatched region. Deviation from this ideal introduces various line shape distortions, compromising both sensitivity and resolution. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.03%3A_Instrumentation.txt |
Objectives
After completing this section, you should be able to:
1. identify those protons which are equivalent in a given chemical structure.
2. use the 1H NMR spectrum of a simple organic compound to determine the number of equivalent sets of protons present.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• diastereotopic
• enantiotopic
• homotopic
If all protons in all organic molecules had the same resonance frequency in an external magnetic field of a given strength, 1H NMR would not be terribly useful to organic chemists. Fortunately for organic chemists, resonance frequencies are not uniform for all protons in a molecule. In an external magnetic field of a given strength, protons in different locations in a molecule have different resonance frequencies, because they are in non-identical electronic environments. In methyl acetate, below for example, there would be two peaks in the 1H NMR spectrum, which means there are two types of protons. The three protons labeled Ha have a different - and easily distinguishable – resonance frequency than the three Hb protons, because the two sets of protons are in non-identical environments: they are, in other words, chemically nonequivalent.
On the other hand, the three Ha protons are all in the same electronic environment, and are chemically equivalent to one another. They have identical resonance frequencies. The same can be said for the three Hb protons.
A good test to determine if hydrogens are chemically equivalent is by doing a thought exercise. In the thought exercise, you replace hydrogens by X to determine what the "thought molecules'" relationship would be to each other. If the protons in the two "thought molecules" are identical, then the protons are said to be homotopic. Homotopic protons are identical protons and will be chemically equivalent. This means that they will show up at the same location in the NMR spectrum.
Example \(1\)
Are the protons in methane, CH4, homotopic, enantiotopic, or diastereotopic?
Solution
Let's do the thought experiment. Two of the hydrogens in methane have been labeled a and b.
Now exhange replace each with X to form two separate "thought molecules". If we exchange Ha for X, then the molecule would be . Doing the same with Hb, the resulting molecule is . The next step is to determine the relationship between these two molecules. Both "thought molecules" are identical, so the protons are homotopic.
You might expect that the equatorial and axial hydrogens in cyclohexane would be non-equivalent, and would have different resonance frequencies. In fact, an axial hydrogen is in a different electronic environment than an equatorial hydrogen. Remember, though, that the molecule rotates rapidly between its two chair conformations, meaning that any given hydrogen is rapidly moving back and forth between equatorial and axial positions. It turns out that, except at extremely low temperatures, this rotational motion occurs on a time scale that is much faster than the time scale of an NMR experiment.
In this sense, NMR is like a camera that takes photographs of a rapidly moving object with a slow shutter speed - the result is a blurred image. In NMR terms, this means that all 12 protons in cyclohexane are equivalent.
The next example we will consider is bromochloromethane. Are the protons of the CH2 chemically equivalent?
Example \(2\)
Are the protons in bromochloromethane, C2H4BrCl, homotopic, enantiotopic, or diastereotopic?
Solution
Start with the same thought experiment that we did with methane and exchange one of the hydrogens with X to make one "thought molecule" and then repeat with the other hydrogen. The two molecules you get are and . The relationship between the two molecules is that they are enantiomers. These protons are considered enantiotopic.
In Example 5.4.2, the "thought" molecules were enantiomers of each other. The hydrogens are termed enantiotopic and like enantiomers the protons are only different in the presence of something that is chiral. The solvents typically used for NMR spectroscopy are achiral. Therefore, the two methylene protons are equivalent protons and will have the same chemical shift. For bromochloromethane, one would expect there to be one NMR absorption for the CH2 group. In summary, enantiotopic protons will be chemically equivalent. This means that they will show up at the same location in the NMR spectrum.
The final type of protons to discuss is diastereotopic protons.
Example \(3\)
Are the methylene protons in 2-bromo-2-chlorobutane, C4H8BrCl, chemically equivalent?
Solution
Start with the same thought experiment that we did with methane and exchange one of the hydrogens with X to make one "thought molecule" and then repeat with the other hydrogen. The two "thought molecules" are and . The relationship between these two molecules is diastereomers. These protons are considered diastereotopic protons. Diastereomers have different chemical properties, which means these protons are not chemically equivalent.
Overall, 2-bromo-2-chlorobutane will have four different types of hydrogens: the two methyl groups will be different giving two NMR absorptions and the CH2 will give two NMR absorptions. In general, diastereotopic protons occur when there is a chirality center already present in the molecule. In summary, diastereotopic protons will be chemically different. This means that they will show up at different locations in the NMR spectrum.
When stereochemistry is taken into account, the issue of equivalence versus nonequivalence in NMR starts to get a little more complicated. It should be fairly intuitive that hydrogens on different sides of asymmetric ring structures and double bonds are in different electronic environments, and thus are nonequivalent and have different resonance frequencies. In the alkene and cyclohexane structures below, for example, Ha is trans to the chlorine substituent, while Hb is cis to chlorine. Ha and Hb in both the alkene and cyclohexane structures would give different absorptions in the NMR spectrum.
Most organic molecules have several sets of protons in different chemical environments, and each set, in theory, will have a different resonance frequency in 1H-NMR spectroscopy. The ability to recognize chemical equivalency and nonequivalency among atoms in a molecule will be central to understanding NMR.
Exercise
Exercise \(2\)
Are the labeled protons homotopic, enantiotopic, or diastereotopic?
a)
b)
Answer
a) Diastereotopic
b) Enantiotopic
Exercise \(2\)
How many non-equivalent hydrogens are in the following molecules? How many different signals will you see in a 1H NMR spectrum?
a)
b)
c)
Answer
a) 1-chloropropane has three non-equivalent hydrogens and would have 3 signals in an 1H NMR spectrum.
b) Diethylether has two non-equivalent hydrogens and would have 2 signals in an 1H NMR spectrum.
c) Ethylbenzene has five non-equivalent hydrogens and would have 5 signals in an 1H NMR spectrum.
Contributors
• Derrick Kaseman (UC Davis), Sureyya OZCAN, Siyi Du | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.04%3A_Types_of_Protons.txt |
Objectives
After completing this section, you should be able to
• understand why the delta scale is used in NMR spectroscopy.
• explain how chemical environment of the proton is related to the chemical shift.
• understand how to read a table of the approximate chemical shift (δ) for protons
• predict the approximate chemical shifts of each of the protons in an organic compound, given its structure and a table of chemical shift correlations.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• chemical shift
• delta scale
• upfield/downfield
Study Notes
Although the calculations described in this section will help you understand the principles of NMR, it is the actual delta values, not the calculations, which are of greatest importance to the beginning organic chemist. Thus, we shall try to focus on the interpretation of NMR spectra, not the mathematical aspects of the technique.
You should not attempt to memorize the chemical shifts listed in the table of this section, you will refer to it quite frequently throughout your analysis of spectra. However, if you have an approximate idea of the chemical shifts of some of the most common types of protons, you will find the interpretation of 1H NMR spectra less arduous than it might otherwise be.
NMR spectra are displayed on a plot that shows the applied field strength increasing from left to right. The left side of the plot is low-field or downfield and the right side of the plot is high-field or upfield. The different local chemical environments surrounding any particular nuclei causes them to resonate at slightly different frequencies. This is a result of a nucleus being more or less shielded than another. There are structural features of the molecule will have an effect on the exact magnitude of the magnetic field experienced by a particular nucleus causing an upfield or downfield shift. This means that nuclei which have different chemical environments will show up in different regions of the NMR plot or spectrum. This is what makes NMR so useful for structure determination in organic chemistry. There are three main features that will effect the shielding of the nucleus: 1) electronegativity, 2) magnetic anisotropy of π systems (Section 5.2) and 3) hydrogen bonding. The position on the spectrum at which a nucleus absorbs is called its chemical shift (δ).
The chemical shift is dependent on the applied field of the spectrometer and as discussed in Section 5.2, there are a variety of spectrometers. Therefore, if you were to run the same sample in a 300 MHz spectrometer and a 600 MHz spectrometer, the same proton would give a different chemical shift in Hertz (Hz). It would be inconvenient and confusing to always have to convert NMR data according to the field strength of the instrument used. Luckily, chemists report resonance frequencies not as absolute values in Hz, but rather as values relative to a common standard, generally the signal generated by the protons in tetramethylsilane (TMS). TMS (structure below) was chosen as a standard because it is chemically inert, symmetrical, volatile (bp = 27 C), and soluble in most organic solvents. Most importantly, it gives a single absorption peak and its protons are more shielded than almost all organic protons. Remember - more shielded means more toward the right of the spectrum and a lower chemical shift. With a standard in place, the units parts per million (ppm) comes in to play. Regardless of the magnetic field strength of the instrument being used, the resonance frequency of the 12 equivalent protons in TMS is defined as a zero point. The resonance frequencies of protons in the sample molecule are then reported in terms of how much higher they are, in ppm, relative to the TMS signal. Now whether a 300 MHz instrument was used or a 600 MHz instrument, the same peak report out at the same chemical shift in ppm. The equation below shows the conversion from Hertz (Hz) to part per million (ppm). Chemical shifts can be used to identify structural properties in a molecule based on our understanding of different chemical environments.
Example \(1\)
For CHCl3, there is a peak at 1451 Hz in a H1 NMR spectra from a 400 MHz spectrometer. Convert to δ (ppm) units.
Solution
δ = ((1451 Hz - 0 Hz)/400,000,000 Hz) x 106 ppm = 3.627 ppm
Remember: TMS is our standard and is set to 0 Hz (or 0 ppm).
Chemical Shifts in 1H NMR Spectroscopy
The NMR spectra is displayed as a plot of the applied radiofrequency versus the absorption. In 1H NMR spectra, the typical range for protons is 0 to 14 ppm. Let's look at an actual 1H-NMR plot for methyl acetate. Just as in IR and UV-Vis spectroscopy, the vertical axis corresponds to intensity of absorbance (typically not shown in an NMR spectrum) and the horizontal axis to frequency.
We see three absorbance signals: two of these correspond to Ha and Hb, while the peak at the far right of the spectrum corresponds to the 12 chemically equivalent protons in TMS. In the spectrum above, you can see that each methyl group has a different chemical shift, so the 1H nuclei must be in different chemical environments. The two methyl groups are attached to different functionalities, which causes a difference in the shielding for each methyl group, creating the unique 1H environments. The main difference is how close each methyl group is to an electronegative atom in methyl acetate.
Electronegativity
If you remember from Section 5.2, the electrons that surround the nucleus are in motion creating their own electromagnetic field. If this field opposes the the applied magnetic field, it reduces the magnetic field experienced by the nucleus. Thus the electrons are said to shield the nucleus. Since the magnetic field experienced at the nucleus defines the energy difference between spin states it also defines what the chemical shift will be for that nucleus. Electron with-drawing groups can decrease the electron density at the nucleus, deshielding the nucleus, resulting in a larger chemical shift. Let's look at some data in the table below.
Compound, CH3X CH3F CH3OH CH3Cl CH3Br CH3I CH4 (CH3)4Si
Electronegativity of X 4.0 3.5 3.1 2.8 2.5 2.1 1.8
Chemical shift, δ (ppm) 4.26 3.4 3.05 2.68 2.16 0.23 0
As can be seen from the data, as the electronegativity of X increases the chemical shift, δ. This is an effect of the electronegative atom pulling the electron density away from the nuclei of 1H atoms and exposing them more to the magnetic field, which "deshields" the nuclei and shifting the peak downfield. Looking back at the two methyl groups in methyl acetate, the methyl group directly attached to the oxygen atom brings that signal further downfield to about 3.6 ppm, while the methyl group attached to the carbonyl group, putting an additional carbon atom between the electron-withdrawing oxygen and the 1H nuclei and thus there is less of an electron withdrawing effect on the nuclei, so they are more shielded. This leads the signal to be more upfield around 2 ppm. The diagram below shows where protons that are part of or near different organic functional groups generally show up on an NMR spectrum.
The effects of electron withdrawing groups are cumulative so the presence of more electron withdrawing groups will produce greater deshielding and therefore a larger chemical shift. As you can see in the table below, the more electronegative chlorine atoms present, increases the chemical shift. In general, the more electronegative atoms present in the environment, the more exposed the nuclei becomes and thus more deshielded.
Compound CH4 CH3Cl CH2Cl2 CHCl3
δ (ppm) 0.23 3.05 5.30 7.27
These inductive effects are not only felt by the immediately adjacent atoms, but also the deshielding can occur further down the chain. The effect lessens the further away the nuclei is from the electronegative atom and after about three bonds away is not felt much any more. In the piece of a molecule below, you can see this trend. The protons closes to the bromine atom are more downfield at 3.30 ppm and an upfield shift occurs with the protons attached to carbons 2 bonds and 3 bonds away from the bromine.
NMR signal -CH2-CH2-CH2Br
δ (ppm) 1.25, 1.69, 3.30
Another way to view chemical shift data is in a table. You will notice that there are ranges. The range is where the proton peak may generally occur.
Hydrogen type
Chemical shift (ppm)
RCH3
0.9 - 1.0
RCH2R
1.2 - 1.7
R3CH
1.5 – 2.3
1.5 – 1.8
RNH2
1 - 3
ArCH3
2.2 – 2.4
2.3 – 3.0
ROCH3
3.2 – 3.8
3.7 – 3.9
ROH
2 - 5
5.0 – 6.5
5 - 9
ArH
6.0 – 8
9 – 10.0
10 - 13
Example \(2\)
Calculate the chemical shift for the protons in ethanol (CH3CH2OH) using the table above.
Solution
-OH proton = between 1-5 ppm
-CH2- protons = 3.2-3.8 ppm
-CH3 protons = > 0.9-1.0 ppm, since the methyl group is not directly attached to the electron-withdrawing group it will not feel the pull away of electron density as much as the methylene group, but it is still close enough to feel it a bit. This means the methyl group will be slightly higher than expected.
Below is an actual image of the NMR spectrum of ethanol. Our predicted values do align with the actual spectral data. The OH actually shows at 3.71 ppm, which is between 1-5ppm as predicted, the methylene occurs at 3.60 ppm also falls within the predicted rage of 3.2-3.8 ppm, and the methyl group does fall just higher than the predicted 0.9-1.0 ppm at 1.16 ppm.
Magnetic Anisotropy: π Electron Effects
The π electrons in a compound, when placed in a magnetic field, will move and generate their own magnetic field. The new magnetic field will have an effect on the shielding of atoms within the field. The best example of this is benzene (see the figure below). Also, see Section 5.2 for more details on generating magnetic fields.
This effect is common for any atoms near a π bond.
Proton Type Effect Chemical shift (ppm)
C6H5-H highly deshielded 6.5 - 8
C=C-H deshielded 5 - 6.5
C≡C-H shielded* ~2.5
O=C-H very highly deshielded 9 - 10
* the acetylene H is shielded due to its location relative to the π system
Example \(3\)
The chemical shift for the highlighted proton Ha is 8.8 ppm, whereas the chemical shift for each of the other protons explicitly shown is between 7.3-8.0 ppm. Provide an explanation for this downfield shift of Ha.
Solution
Ha feels the benzene pi system as do all the other hydrogens on naphthalene. The difference is that it also feels the pi system from the carbonyl as well, which brings Ha further downfield than the other hydrogens.
Hydrogen Bonding
Protons that are involved in hydrogen bonding (i.e.-OH or -NH) are usually observed over a wide range of chemical shifts. This is due to the deshielding that occurs in the hydrogen bond. Since hydrogen bonds are dynamic, constantly forming, breaking and forming again, there will be a wide range of hydrogen bonds strengths and consequently a wide range of deshielding. This exchange is faster than the time it takes to acquire a 1H NMR spectrum. This, as well as, solvation effects, acidity, concentration and temperature make it very difficult to predict the chemical shifts for these atoms. The exchange process is facilitated by even traces of acid or base in the sample. The functional groups affected are alcohols, carboxylic acids, amines, and amides.
Experimentally -OH and -NH can be identified by carrying out a simple D2O exchange experiment since these protons are exchangeable.
• run the normal H-NMR experiment on your sample
• add a few drops of D2O
• re-run the H-NMR experiment
• compare the two spectra and look for peaks that have "disappeared"
Exercise
Exercise \(1\)
The following peaks were from a 1H NMR spectra from a 300 MHz spectrometer. Convert to δ units (ppm).
a) CH3OH: 693 Hz
b) CH2Cl2: 1060 Hz
Answer
a) δ = ((693 Hz - x 0 Hz)/400,000,000 Hz) x 106 ppm = 1.73 ppm
b) δ = ((1060 Hz - x 0 Hz)/400,000,000 Hz) x 106 ppm = 2.65 ppm
Exercise \(2\)
Butan-2-one shows a chemical shift around 2.1 on a 300 MHz spectrometer in the H1 NMR spectrum.
a) How far downfield is this peak from TMS in Hz?
b) If the spectrum was done with a 400 MHz instrument, would a different chemical shift be seen?
c) On this new 400 MHz spectrum, what would be the difference in Hz from the chemical shift and TMS?
Answer
a) (2.1 ppm x 300000000 Hz)/106 ppm = 630 Hz The peak at 2.1 ppm would show at 630 Hz on a 300 MHz spectrometer. TMS is always at 0 Hz, so the peak would be 630 Hz downfield from TMS.
b) No, a different chemical shift would not be seen in ppm. The point of using the ppm scale is to be able to discuss the same peak at the same chemical shift unlike when using Hertz.
c) (2.1 ppm x 400000000 Hz)/106 ppm = 840 Hz The 2.1 ppm peak would show at 840 Hz on a 400 MHz spectrometer.
Exercise \(3\)
The following have one H1 NMR peak. In each case predict approximately where this peak would be in a spectra.
a)
b)
Answer
a) 1.2-1.7 ppm
b) 6.0-8.0 ppm
Exercise \(4\)
Identify the different equivalent protons in the following molecule and predict their expected chemical shift.
Answer
There are 6 different protons in this molecule.
Proton Predicted Chemical Shift (ppm) from Table Actual Chemical Shift (ppm) Data
Ha 1.5-1.8 1.72
Hb 5.0-6.5 6.03
Hc 5.0-6.5 6.26
Hd 6.5-8 7.33
He 6.5-8 7.19
Hf 6.5-8 7.22
Exercise \(5\)
Which methyl group would you expect to be more downfield on the molecule below.
Answer
The methyl group directly attached to the oxygen will be further downfield (predicted 3.2-3.8 ppm) versus the methyl group directly attached to a double bond (predicted 1.5-1.8 ppm). | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.05%3A_Chemical_Shift.txt |
Objectives
After completing this section, you should be able to
1. explain what information can be obtained from an integrated 1H NMR spectrum, and use this information in the interpretation of such a spectrum.
2. use an integrated 1H NMR spectrum to determine the ratio of the different types of protons present in an organic compound.
Integration
In a previous sections 5.4: Types of Protons and 5.5: Chemical Shift, equivalent and nonequivalent hydrogens were discussed as well as unique nonequivalent hydrogens having different chemical shifts. In this section, integration will let the researcher know how many hydrogens there are for each unique NMR absorption. The area of a peak in a 1H NMR is proportional to the number of hydrogens to which the peak corresponds. This is very helpful when determining how many of each unique hydrogen exist and total number of hydrogens present in a molecule. The computer in an NMR instrument can be instructed to automatically integrate the area under a signal or group of signals, so it can be superimposed on the spectrum. The integration curve appears as a series of steps with the height being proportional to the area of the corresponding absorption peak, and the number of protons responsible for the absorption. In practice since it can be difficult to decide exactly where to start and stop, the ratios may not be exact whole numbers. If we look at methyl acetate (below), there are two types of protons that would give two separate peaks in an NMR spectrum. The peaks would integrate to approximately the same area, since both correspond to a set of three equivalent protons.
Example \(1\)
If we look at a spectrum of para-xylene (common name) or 1,4-dimethylbenzene (IUPAC name), how many types of protons are there and what does the ratio mean?
Solution
There are still two signals like in methyl acetate indicating there are two types of hydrogens, but now instead of ratios being about 1 to 1, the ratio is 1 to 1.5. When we instruct the instrument to integrate the areas under the two signals, we find that the area under the peak at 2.6 ppm is 1.5 times greater than the area under the peak at 7.4 ppm. This molecule has two sets of protons: the six methyl (Ha) protons and the four aromatic (Hb) protons, which matches our 1:1.5 ratio since 6 is 1.5 times 4. This (along with the actual chemical shift values, discussed in section 5.5: Chemical Shift) tells us which set of protons corresponds to which NMR signal. The aromatic protons (Hb) are at 7.4 ppm and the methyl protons, Ha, are at 2.6 ppm.
Integration can also be used to determine the relative amounts of two or more compounds in a mixed sample. If we have a sample that is a 50:50 (mole/mole) mixture of benzene and acetone, for example, the acetone signal should integrate to the same value as the benzene sample, because both signals represent six equivalent protons. If we have a 50:50 mixture of acetone and cyclopentane, on the other hand, the ratio of the acetone peak area to the cylopentane peak area will be 3:5 (or 6:10), because the cyclopentane signal represents ten protons.
Exercises
Exercise \(1\)
Predict how many signals the following molecule would have and the integrations in 1H NMR? Sketch the spectra and estimate the integration of the peaks.
Answer
There would be two signals. One for Ha that would integrate to 6 and another for Hb that would integrate to 4.
Ideal general spectrum shown with integration (below).
Exercise \(2\)
Using integration, how could you determine the difference between these two molecules.
vs.
Answer
The two will have very similar chemical shifts for their protons, but the total number of protons will be different. If you add up all the integrations, then you will get the total number of hydrogens for each molecule. The first molecule has 14 total protons, while the second molecule has 18 total protons.
Exercise \(3\)
You take the 1H-NMR spectrum of a mixed sample of 36% para-xylene and 64% acetone in CDCl3 solvent. How many peaks do you expect to see? What is the expected ratio of integration values for these peaks? (set the acetone peak integration equal to 1.0)
Answer
There are three peaks: two from para-xylene and one from acetone. The acetone peak and the para-xylene methyl peak both represent six protons, so the ratio of their integration values is simply 64 to 36 or 1 (64/64) to 0.56 (36/64). The ratio of the para-xylene methyl peak to the para-xylene aromatic peak is 6 to 4 (1.5: 1 ratio), or 0.56 to 0.37 (still 1.5: 1 ratio). So the final integral ratio of acetone:methyl:aromatic signals should be 1 to 0.56 to 0.37. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.06%3A_Integration_of_Proton_Spectra.txt |
Objectives
After completing this section, you should be able to
1. understand what spin-spin splitting is and what information it tells you
2. explain the spin-spin splitting pattern observed in the 1H NMR spectrum of a simple organic compound.
3. interpret the splitting pattern of a given 1H NMR spectrum.
4. determine the structure of a relatively simple organic compound, given its 1H NMR spectrum and other relevant information.
5. use coupling constants to determine which groups of protons are coupling with one another in a 1H NMR spectrum.
6. predict the splitting pattern which should be observed in the 1H NMR spectrum of a given organic compound.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• coupling constant
• multiplet
• quartet
• triplet
• doublet
From what we have learned about 1H NMR spectra so far, we might predict that the spectrum of 1-chloroethane, CH3CH2Cl, would consist of two peaks—one, at about 0.9 δ, expected for CH3 and one shifted downfield because of the presence of an additional electronegative chlorine atom on the second carbon. However, when we look at the spectrum (below) it appears to be much more complex. True, we see absorptions in the regions we predicted, but instead of the single peak we have seen thus far, these absorptions appear as a group of three peaks (a triplet) and a group of four peaks (a quartet). This complication is in fact very useful to the organic chemist, and adds greatly to the power of NMR spectroscopy as a tool for the elucidation of chemical structures. The split peaks (multiplets) arise because the magnetic field experienced by the protons of one group is influenced by the spin arrangements of the protons in an adjacent group.
The source of spin-spin coupling
The 1H-NMR spectra that we have seen so far (of methyl acetate and para-xylene) are somewhat unusual in the sense that in both of these molecules, each set of protons generates a single NMR signal. In fact, the 1H-NMR spectra of most organic molecules contain proton signals that are split into two or more sub-peaks. This splitting behavior actually provides us with more information about our sample molecule.
Consider the spectrum for 1,1,2-trichloroethane. In this and in many spectra to follow, enlargements of individual signals will be shown so that the signal splitting patterns are recognizable.
The signal at 3.96 ppm, corresponding to the two Ha protons, is split into two subpeaks of equal height (and area) – this is referred to as a doublet. The two Hb protons at 5.76 ppm, on the other hand, is split into three sub-peaks, with the middle peak higher than the two outside peaks - if we were to integrate each subpeak, we would see that the area under the middle peak is twice that of each of the outside peaks. This is called a triplet.
The source of signal splitting is a phenomenon called spin-spin coupling, a term that describes the magnetic interactions between neighboring, non-equivalent NMR-active nuclei. In our 1,1,2--trichloroethane example, the Ha and Hb protons are spin-coupled to each other. Here's how it works, looking first at the Hsignal. The magnetic environment experienced by Hb is influenced by the fields of both neighboring Ha protons, which we will call Ha1 and Ha2. There are four possibilities here, each of which is equally probable. First, the magnetic fields of both Ha1 and Ha2 could be aligned with B0, which would deshield Hb, shifting its NMR signal slightly downfield. Second, both the Ha1 and Ha2 magnetic fields could be aligned opposed to B0, which would shield Hb, shifting its resonance signal slightly upfield due to being more shielded. Third and fourth, Ha1 could be with B0 and Ha2 opposed, or Ha1 opposed to B0 and Ha2 with B0. In each of the last two cases, the shielding effect of one Ha proton would cancel the deshielding effect of the other, and the chemical shift of Hb would be unchanged. These ideas an be illustrated by a splitting diagram, as shown below.
So in the end, the signal for Hb is a triplet, with the middle peak twice as large as the two outer peaks because there are two ways that Ha1 and Ha2 can cancel each other out.
The Hhydrogens give rise to a doublet signal at 3.96 ppm – notice that the two peaks are equal in height to one another. This splitting pattern results from the spin-coupling effect of the one Hb hydrogen next door, and can be explained by an analysis similar to that which we used to explain the triplet pattern above. The proton could be aligned with the magnetic field, which would deshield Ha or it could oppose the magnetic field, which would shield Ha.
Example \(1\)
Now, consider the spectrum for ethyl acetate:
Explain the splitting patterns of Ha, Hb, and Hc in ethyl acetate.
Solution
We see an unsplit singlet peak at 1.83 ppm that corresponds to the acetyl (Ha) hydrogens. This signal is unsplit because there are no hydrogens on the adjacent carbon. The signal at 1.055 ppm for the Hc hydrogens is split into a triplet by the two Hb hydrogens next door. The explanation here is the same as the explanation for the triplet peak we saw previously for 1,1,2-trichloroethane.
The Hhydrogens give rise to a quartet signal at 3.91 ppm – notice that the two middle peaks are taller then the two outside peaks. This splitting pattern results from the spin-coupling effect of the three Hc hydrogens next door, and can be explained by an analysis similar to that which we used to explain the doublet and triplet patterns previously discussed.
Instead of drawing a splitting diagram for every molecule, there is a recognizable pattern which is usually referred to as the n + 1 rule: if a set of hydrogens has n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks. This is very useful information if we are trying to determine the structure of an unknown molecule: if we see a triplet signal, we know that the corresponding hydrogen or set of hydrogens has two neighboring hydrogens. When we begin to determine structures of unknown compounds using 1H-NMR spectral data, it will become more apparent how this kind of information can be used.
Three important points need to be emphasized here. First, signal splitting only occurs between non-equivalent hydrogens – in other words, Ha1 in 1,1,2-trichloroethane is not split by Ha2, and vice-versa.
Second, splitting occurs primarily between hydrogens that are separated by three bonds. This is why the Ha hydrogens in ethyl acetate form a singlet– the nearest hydrogen neighbors are five bonds away (count the bonds between hydrogen to hydrogen), too far for coupling to occur.
Occasionally we will see four-bond and even 5-bond splitting, but in these cases the magnetic influence of one set of hydrogens on the other set is much more subtle than what we typically see in three-bond splitting (more complex coupling interactions is provided in section 5.8).
Finally, splitting is most noticeable with hydrogens bonded to carbon. Hydrogens that are bonded to heteroatoms (alcohol or amino hydrogens, for example) are coupled weakly - or not at all - to their neighbors. This has to do with the fact that these protons exchange rapidly with solvent or other sample molecules, which is greatly enhanced by even traces of acid or base (see mechanism with ethanol below). Remember, NMR is like a camera that takes photographs of a rapidly moving object with a slow shutter speed - the result is a blurred image. This blurred image here creates a broader singlet for exchanging protons, even if there are neighboring hydrogens.
To see the splitting of exchangeable protons, the NMR experiment needs to be carried out as low temperatures.
Multiplicity in Proton NMR
The number of lines in a peak is always one more (n+1) than the number of hydrogens on the neighboring carbon. This table summarizes coupling patterns that arise when protons have different numbers of neighbors.
# of lines
ratio of lines
term for peak
# of neighbors
1
-
singlet
0
2
1:1
doublet
1
3
1:2:1
triplet
2
4
1:3:3:1
quartet
3
5
1:4:6:4:1
quintet
4
6
1:5:10:10:5:1
sextet
5
7
1:6:15:20:15:6:1
septet
6
8
1:7:21:35:35:21:7:1
octet
7
9
1:8:28:56:70:56:28:8:1
nonet
8
Coupling constants
Chemists quantify the spin-spin coupling effect using something called the coupling constant, which is abbreviated with the capital letter J. The coupling constant is simply the difference, expressed in Hz (not ppm), between two adjacent sub-peaks in a split signal. For our triplet in the 1,1,2-trichloroethane spectrum, for example, the three subpeaks are separated by 6.1 Hz, and thus we write 3Ja-b = 6.1 Hz.
The superscript 3 tells us that this is a three-bond coupling interaction, and the a-b subscript tells us that we are talking about coupling between Ha and Hb. Unlike the chemical shift value, the coupling constant, expressed in Hz, is the same regardless of the applied field strength of the NMR magnet. This is because the strength of the magnetic moment of a neighboring proton, which is the source of the spin-spin coupling phenomenon, does not depend on the applied field strength. Also, the coupling constant 3Ja-b quantifies the magnetic interaction between the Ha and Hb hydrogen sets, and this interaction is of the same magnitude in either direction. In other words, Ha influences Hb to the same extent that Hb influences Ha. When looking at more complex NMR spectra (Section 5.8), this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets.
Similar types of bonds lead to similar coupling constants and this has resulted in a range for particular types of hydrogens interacting through a particular bond. For example, coupling constants between proton sets on neighboring sp3-hybridized carbons (3 bonds) is typically in the region of 6-8 Hz. The table below lists typical three-bond coupling constant values for different types of bonds. Remember, chemically equivalent protons do not couple with one another to give spin-spin splitting.
Type of bond Image of type of bond Coupling constant (Hz)
Standard three bond coupling 6-8
Vinylic geminal coupling 0-3
Vinylic cis coupling 6-15
Vinylic trans coupling 11-18
Benzylic ortho coupling 6-10
Benzylic meta coupling* 0-4
Aldehydic proton coupling 2-3
*Actually 4-bond coupling.
Exercise
Exercise \(1\)
Predict the splitting patterns of the following molecules:
a)
b)
Answer
a) There are three types of protons, so there will be three peaks. The methyl directly attached to the oxygen will be a singlet. The methylene will be split into a quartet by the neighboring methyl group. The other methyl group will be split into a triplet by the neighboring -CH2-.
b) This molecule is symmetrical, which leads to two types of hydrogens and therefore two peaks. The methylene will be split into a triplet by the two neighboring hydrogens on either side. The -CH- peak will be split into a triplet as well due to the neighboring -CH2-.
Exercise \(2\)
How many proton signals would you expect to see in the 1H-NMR spectrum of the structure shown below? For each of the proton signals, predict the splitting pattern. Assume that you see only 3-bond coupling.
Answer
Because of the symmetry in the molecule, there are only four proton signals.
OH proton = singlet
Ha = singlet
Hb = doublet
Hc = doublet
Exercise \(3\)
The following spectrum is for C4H10O. Determine the structure. The ratio of integrals from left to right is 2:3.
Answer
The total number of hydrogens is 2+3 = 5. All the hydrogens are not accounted for since we need a total of 10 hydrogens. This means there is symmetry in the molecule. Ten is divisible by five. If we multiply each in the 2:3 ratio, then we get 4:6, which does add to 10.
Typically, integrations of 3 or multiples of 3 are methyl groups. The peak at 1.14 ppm is a methyl group with two neighbors, since the peak is split into a triplet. n +1 = 3, n = 2, n = number of neighbors. The first fragment is a -CH2-CH3. There are two of these fragments, since there are actually 6 total methyl protons.
The peak at 3.43 ppm integrates to 4 and is a quartet. The integration indicates it is a methylene group and since it is a quartet, it must have three neighbors. So far, the fragment would be -CH2-CH3. It is also very far downfield for a proton just bonded to other carbons with hydrogens, which would typically be around 1.2-1.7 ppm (see the table in Section 5.5). This means it must be attached to an electron withdrawing group. The electron withdrawing group in this molecule is the -O-. Our final fragment is CH3-CH2-O-.
If we look at all of our fragments, we can begin to put them together. Let's take the two fragments (-CH2-CH3 and -O-CH2-CH3), each makes one more bond. If you connect them, then you would get CH3-CH2-O-CH2-CH3. The molecule is symmetrical with two types of protons with teh 4:6 (or 2:3) ratio and correct splitting pattern. This spectrum correlates to diethyl ether. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.07%3A_Spin-Spin_Splitting_in_Proton_NMR__Spectra.txt |
Objectives
After completing this section, you should be able to
1. explain how multiple coupling can give rise to complex-looking 1H NMR spectra.
2. predict the splitting pattern expected in the 1H NMR spectrum of an organic compound in which multiple coupling is possible.
3. interpret 1H NMR spectra in which multiple coupling is evident.
Study Notes
Another effect that can complicate a spectrum is the “closeness” of signals. If signals accidently overlap they can be difficult to identify. You can try this yourself by drawing a tree diagram. Keep this point in mind when interpreting real 1H NMR spectra.
Also, when multiplets are well separated, they form patterns. However, when multiplets approach each other in the spectrum they sometimes become distorted. Usually, the inner peaks become larger than the outer peaks. Note the following examples:
Complex coupling
In all of the examples of spin-spin coupling that we have seen so far, the observed splitting has resulted from the coupling of one set of hydrogens to just one neighboring set of hydrogens. What happens if there are hydrogens on both sides?
Similar Coupling Constants
When a proton is coupled to two different neighboring proton sets with identical or very close coupling constants, the splitting pattern that emerges often appears to follow the simple n + 1 rule of non-complex splitting. Looking at 1,1,3-trichloropropane below, Ha and Hc only have Hb as neighbors. Remember, splitting occurs primarily between hydrogens that are separated by three bonds. This means that Ha would be a triplet as would Hc. However, Hb has two sets of neighboring protons (Ha and Hc). Each follows the simple spin-spin coupling patterns, since they each just have one neighboring set of hydrogens (Hb). However with Hb, it has neighboring hydrogens on both sides. There are two Ha neighbors and there is one Hc neighbor. It turns out that since the types of interveneing bonds between Ha and Hb are the same as Hb and Hc, the coupling constants will be very similar. Ha and Hc are not equivalent (their chemical shifts are different), but it turns out that 3Jab is very close to 3Jbc.
If we perform a splitting diagram analysis for Hb, we see that, due to the overlap of sub-peaks, the signal appears to be a quartet, and for all intents and purposes follows the n + 1 rule.
Therefore, we can still use the n + 1 rule when the coupling constants are similar and making it easier for organic chemists.
Different Coupling Constants
When a set of hydrogens is coupled to two or more sets of nonequivalent neighbors, the result is a phenomenon called complex coupling. A good illustration is provided by the 1H-NMR spectrum of methyl acrylate:
First, let's first consider the Hc signal, which is centered at 6.21 ppm. Here is a closer look:
With this enlargement, it becomes evident that the Hc signal is actually composed of four sub-peaks. Why is this? Hc is coupled to both Ha and Hb , but with two different coupling constants. Once again, a splitting diagram (or tree diagram) can help us to understand what we are seeing. Ha is trans to Hc across the double bond, and splits the Hc signal into a doublet with a coupling constant of 3Jac = 17.4 Hz. In addition, each of these Hc doublet sub-peaks is split again by Hb (geminal coupling) into two more doublets, each with a much smaller coupling constant of 2Jbc = 1.5 Hz.
The result of this double splitting is a pattern referred to as a doublet of doublets, abbreviated dd.
The signal for Ha at 5.95 ppm is also a doublet of doublets, with coupling constants 3Jac= 17.4 Hz and 3Jab = 10.5 Hz. We know it is not a quartet because the intensity of the lines is the same for each like a doublet, whereas for a quartet the ratio of the lines is 1:3:3:1 where the outside lines would be shorter.
The signal for Hb at 5.64 ppm is split into a doublet by Ha, a cis coupling with 3Jab = 10.4 Hz. Each of the resulting sub-peaks is split again by Hc, with the same geminal coupling constant 2Jbc = 1.5 Hz that we saw previously when we looked at the Hc signal. The overall result is again a doublet of doublets, this time with the two sub-doublets spaced slightly closer due to the smaller coupling constant for the cis interaction. Here is a blow-up of the actual Hsignal:
You may have noticed in the splitting diagram, the larger of coupling constant came first followed by the smaller or finer coupling constant.While it does not matter which coupling constant you start with, it is usually easier to construct the splitting diagram for analysis of complex coupling patterns if you begin with the larger coupling constant. However, the end result will be the same regardless of how you begin.
Example $1$
Construct a splitting diagram for the Hb signal in the 1H-NMR spectrum of methyl acrylate. Show the chemical shift value for each sub-peak, expressed in Hz (assume that the resonance frequency of TMS is exactly 300 MHz).
Solution
The coupling tree for Hb is:
The complex coupling pattern would be a doublet of doublets (dd).
In many cases, it is difficult to fully analyze a complex splitting pattern. Aromatic ring protons quite commonly have overlapping signals and multiplet distortions. Sometimes you cannot distinguish between individual signals, and one or more messy multiplets often appear in the aromatic region. In the spectrum of toluene, for example, if we consider only 3-bond coupling we would expect the signal for Ha to be a doublet, Hb a triplet, and Hc a triplet.
In practice, however, all three aromatic proton groups have very similar chemical shifts and their signals overlap substantially, making such detailed analysis difficult. In this case, we would refer to the aromatic part of the spectrum (~7.0 ppm) as a multiplet.
When we start trying to analyze complex splitting patterns in larger molecules, we gain an appreciation for why scientists are willing to pay large sums of money (hundreds of thousands of dollars) for higher-field NMR instruments. Quite simply, the stronger our magnet is, the more resolution we get in our spectrum. In a 100 MHz instrument (with a magnet of approximately 2.4 Tesla field strength), the 12 ppm frequency 'window' in which we can observe proton signals is 1200 Hz wide. In a 500 MHz (~12 Tesla) instrument, however, the window is 6000 Hz - five times wider. In this sense, NMR instruments are like digital cameras and HDTVs: better resolution means more information and clearer pictures (and higher price tags!).
It is much easier to rationalize the observed 1H NMR spectrum of a known compound than it is to determine the structure of an unknown compound from its 1H NMR spectrum. However, rationalizations can be a useful learning technique as you try to improve your proficiency in spectral interpretation. Remember that when a chemist tries to interpret the 1H NMR spectrum of an unknown compound, he or she usually has additional information available to make the task easier. For example, the chemist will almost certainly have an infrared spectrum of the compound and possibly a mass spectrum too. Details of how the compound was synthesized may be available, together with some indication of its chemical properties, its physical properties, or both.
Exercises
Exercise $1$
The molecule below is 2-pentanone, what would the splitting pattern be for the two hydrogens explicitly indicated (below) in an 1H NMR spectrum.
Answer
The indicated hydrogens (Hc below) in the spectrum of 2-pentanone appears as a sextet. They would be split by the five combined protons (Hb and Hd below). Since the intervening bonds are the same, Jbc and Jcd would be very similar and the n+1 rule can be followed. Technically, this 'sextet' could be considered to be a 'triplet of quartets' with overlapping sub-peaks.
Exercise $2$
The following spectrum is for C3H8O. Determine the structure. The ratio of integrals from left to right is 2:1:2:3.
Answer
The total number of hydrogens is 2+1+2+3 = 8. All the hydrogens are accounted for.
Typically, integrations of 3 or multiples of 3 are methyl groups. The peak at 0.9 ppm is a methyl group with two neighbors, since the peak is split into a triplet. n +1 = 3, n = 2, n = number of neighbors. The first fragment is a -CH2-CH3.
The peak just above 1.5 ppm integrates to 2 protons and is a sextet (split into 6 peaks). The integration of 2 indicates this is a methylene group (-CH2-) and the spin-spin splitting pattern tells us that there are 5 neighbors. A methylene makes two more bonds, so one side must be the methyl group and the other side must be another methylene in order to have 5 total neighbors. The next fragment is -CH2-CH2-CH3.
The peak just below 3.0 ppm is a broad singlet. Broad singlets typically come from exchangeable protons like an -OH or -NH-. In our molecular formula we have no N, but we do have an O, so this peak must be due to the -OH.
The final peak at 3.5 ppm integrates to 2 and is a triplet. The integration indicates it is another methylene group and since it is a triplet, it must have two neighbors. So far, the fragment would be -CH2-CH2-. It is also very far downfield for a proton just bonded to other carbons with hydrogens, which would typically be around 1.2-1.7 ppm (see the table in Section 5.5). This means it must be attached to an electron withdrawing group. The electron withdrawing group in this molecule is the -OH. Our final fragment is -CH2-CH2-OH.
If we look at all of our fragments, we can see that it is too many carbons and hydrogens if they all represented a separate piece to include. However, spin-spin splitting is the information from neighboring hydrogens. This means that the fragments overlap each other. Let's take the first two fragments (-CH2-CH3 and -CH2-CH2-CH3). The integration information indicates that there is only one peak that integrates to 3, so there is only one methyl group, not two. This means one of the methyl groups is a duplicate and there is actually only one methyl group. In the second fragment, the splitting pattern indicated that there must be a neighboring methyl group, but the information is actually about the central methylene in that fragment. Considering this, the molecule must be HO-CH2-CH2-CH3. It has four different types of hydrogens, with the correct ratios as well as splitting pattern.
Exercise $3$
For the molecule below, what would the coupling pattern be in a 1H NMR spectrum?
Answer
Hb would be a doublet of doublet of triplets. Hc splits Hb by trans coupling into a doublet. Ha splits Hb by cis coupling into a doublet. The two Hd protons split Hb into a triplet. All of the J coupling constants are different, so you cannot use the n+1 rule.
Exercise $4$
Unknown compound D (C15H14O) has the following spectral properties.
Infrared spectrum:
3010 cm−1 (medium)
1715 cm−1 (strong)
1610 cm−1 (strong)
1500 cm−1 (strong)
1H NMR spectrum:
δ (ppm) Relative Area Multiplicity
3.00 2 triplet
3.07 2 triplet
7.1-7.9 10 Multiplets
Answer | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.08%3A_More_Complex_Spin-Spin_Splitting_Patterns.txt |
Objective
After completing this section,
• understand how 1H NMR data can be used to distinguish between two (or more) possible structures for an unknown organic compound
• determine if your reaction went to completion based on how clean the 1H NMR spectrum is
• understand how the ratio of mixtures can be obtained
The two major areas where NMR has proven to be of critical importance is in the fields of medicine and chemistry, with new applications being developed daily.
Nuclear magnetic resonance imaging, better known as magnetic resonance imaging (MRI) is an important medical diagnostic tool used to study the function and structure of the human body. It provides detailed images of any part of the body, especially soft tissue, in all possible planes and has been used in the areas of cardiovascular, neurological, musculoskeletal and oncological imaging. Unlike other alternatives, such as computed tomography (CT), it does not used ionized radiation and hence is very safe to administer.
In many laboratories today, chemists use nuclear magnetic resonance to determine structures of important chemical and biological compounds. In NMR spectra, different peaks give information about different atoms in a molecule according specific chemical environments and bonding between atoms. The most common isotopes used to detect NMR signals are 1H and 13C but there are many others, such as 2H, 3He, 15N, 19F, etc., that are also in use.
NMR has also proven to be very useful in other area such as environmental testing, petroleum industry, process control, earth’s field NMR and magnetometers. Non-destructive testing saves a lot of money for expensive biological samples and can be used again if more trials need to be run. The petroleum industry uses NMR equipment to measure porosity of different rocks and permeability of different underground fluids. Magnetometers are used to measure the various magnetic fields that are relevant to one’s study.
1H NMR spectroscopy provides a lot of information. Each signal in the spectrum is a unique proton(s). Chemical shift indicates the type of environment protons are in. Integration tells how many of the unique protons there are. Spin-spin coupling tells how many neighbors a particular proton has. Each adding a bit more complexity, but how can 1H NMR be used?
1) There will be cases in which you already know what the structure might be. In these cases:
• You should draw attention to pieces of data that most strongly support your expected structure. This approach will demonstrate evaluative understanding of the data; that means you can look at data and decide what parts are more crucial than others.
• You should also draw attention to negative results: that is, peaks that might be there if this spectrum matched another, possible structure, but that are in fact missing.
2) One of the most complicated problems to deal with is the analysis of a mixture. This situation is not uncommon when running a reaction in lab.
• Sometimes the spectra shows a little starting material mixed in with the product.
• Sometimes solvents show up in the spectrum.
• As you might expect, the minor component usually shows up as smaller peaks in the spectrum. If there are fewer molecules present, then there are usually fewer protons to absorb in the spectrum.
• In this case, you should probably make two completely separate sets of data tables for your analysis, one for each compound, or else one for the main compound and one for impurities.
It can be helpful to either take a 1H NMR spectrum of your starting materials or look it up in a database, which can be used as a reference for the product 1H NMR.
3) Remember that integration ratios are really only meaningful within a single compound. If your NMR sample contains some benzene (C6H6) and some acetone (CH3COCH3), and there is a peak at 7.15 that integrates to 1 proton and a peak at 2.10 ppm integrating to 6 protons, you may think it might mean there are 6 protons in acetone and 1 in benzene, but you can tell that isn't true by looking at the structures. Both benzene and acetone have six protons. In benzene, they should all show up near 7 ppm and in acetone they should all show up near 2 ppm. Assuming that small integral of 1H for the benzene is really supposed to be 6H, then the large integral of 6H for the acetone must also represent six times as many hydrogens, too. It would be 36 H. There are only six hydrogens in acetone, so it must represent six times as many acetone molecules as there are benzene molecules.
Similarly, if you have decided that you can identify two sets of peaks in the 1H spectrum, such as starting material and product, analyzing them in different tables makes it easy to keep the integration analysis completely separate too; 1 H in one table will not be the same size integral as 1 H in the other table unless the concentrations of the two compounds in the sample are the same. Comparing the ratio of two integrals for two different compounds can give you the ratio of the two compounds in solution, just as we could determine the ratio of benzene to acetone in the mixture described above.
Example \(1\)
Three students performed a synthesis of a fragrant ester, ethyl propanoate, CH3CH2CO2CH2CH3. During their reactions, they each used a different solvent (acetonitrile, tetrahydrofuran, and dichloromethane). Did any of the students see just ethyl propanoate?
See the first student's spectrum:
See the second student's spectrum:
See the third student's spectrum:
Solution
The students saw peaks in the NMR spectrum for ethyl propanoate, chloroform (CHCl3, in the CDCl3 they used to make their NMR samples), and trace amounts of solvent from running the reaction. They were also able to determine that they had some leftover solvent in their samples by consulting a useful table of solvent impurities in NMR (which they found in Goldberg et. al., Organometallics 2010, 29, 2176-2179).
Exercises
Exercise \(1\)
How could you tell the difference between o-xylene and p-xylene in 1H NMR?
o-xylene:
p-xylene:
Answer
There are a variety of ways to determine the difference between these two molecules. In o-xylene, there are 3 types of protons and in p-xylene there are two types of protons. There will be spin-spin splitting in the o-xylene aromatic protons, since there are two types of aromatic protons and the protons are neighbors. p-Xylene will show now splitting in the aromatic protons, since there is just one type of aromatic proton with no neighbors. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.09%3A_Uses_of_%281H%29_NMR_Spectroscopy.txt |
Objectives
• understand how chemical shift, integration, spin-spin splitting all come together to solve 1H NMR problems.
• solve unknown 1H NMR problems given the molecular formula.
This page is devoted to explaining solved 1H NMR problems to help guide you through different scenarios. The first will be a case in which you know the structure of the product you are synthesizing, but need to prove you did indeed make it. Your strategy should be:
• Draw your attention to pieces of data that most strongly support your expected structure. A skill necessary for this will be your ability to demonstrate evaluative understanding of the data. In other words, you can look at the data and decide what parts are more important than other parts.
• Draw you attention to negative results: what peaks might be there if the spectrum matched another possible structure, but is missing.
Example \(1\)
In the laboratory, you performed an esterification to make ethyl acetate. Based on the 1H NMR spectrum, did you make your desired product?
Expected Product:
1H NMR: The ratio of protons is 2:3:3.
Solution
Yes, ethyl acetate was synthesized. First, the ratio of protons equals the number of protons in our expected product.
The peak at 1.20 ppm is a triplet that integrates to 3 protons. The integration indicates that the peak corresponds to a methyl group. The splitting pattern of a triplet indicates that there are two neighboring. All of this information together gives the piece -CH2CH3
The peak at 1.98 ppm is a singlet that integrates to 3 protons. The integration tells us the peak corresponds to a methyl group. The splitting pattern of a singlet indicates that there are no neighbors. The chemical shift is quite far downfield at 1.98 ppm (a methyl group attached to another carbon would be at about 0.9 ppm). Being further downfield means that the methyl group is attached to an electron-withdrawing group, which could be directly to an oxygen or the carbonyl carbon. All of this information together gives the piece -EWG-CH3. EWG = Electron-Withdrawing Group
The peak at 4.06 ppm is a quartet that integrates to 2 protons. The integration tells us the peak corresponds to a methylene group (-CH2-). The splitting pattern of a quartent indicates that there are 3 neighbors. The chemical shift is quite far downfield at 4.06 ppm (a methylene group attached to another carbon would be at about 1.2 ppm). Being the furthest downfield, the methylene group needs to be attached to the electron-withdrawing group that will be the most deshielding for the protons (Chemical Shift information). This is the proton must be directly attached to the oxygen. All of this information together gives the piece -O-CH2CH3, which means the methyl group at 1.98 ppm is directly attached to the carbonyl carbon.
From this analysis, it can be concluded that yes, ethyl acetate was synthesized as expected.
Next, we will look at a 1H NMR spectrum of an unknown and determine its structure.
Example \(2\)
Using the chemical formula and 1H NMR spectrum, determine the structure of your unknown molecule.
Chemical Formula: C5H9ClO
Unknown molecule 1H NMR spectrum: The ratio of protons is 2:2:2:3.
Solution
Unknown molecule:
First, if the molecular formula is known, then start by calculating the degree of unsaturation (DU), which will tell how many double bond, triple bond, or rings are present in the molecule. A double bond is a DU = 1. A ring is a DU = 1. A triple bond is a DU = 2. The degree of unsaturation for C5H9ClO is 1. Therefore, the molecule either contains a ring or a double bond.
The peak at 1.05 ppm is a triplet that integrates to 3 protons. The integration indicates that the peak corresponds to a methyl group. The splitting pattern of a triplet indicates that there are two neighboring. All of this information together gives the piece -CH2CH3
The peak at 2.40 ppm is a quartet that integrates to 2 protons. The integration tells us the peak corresponds to a methylene group. The splitting pattern of a quartet indicates that there are 3 neighboring protons. The chemical shift is quite far downfield at 2.40 ppm (a methylene group attached to another carbon would be at about 1.2 ppm). Being further downfield means that the methylene group is attached to an electron-withdrawing group, which could be directly to the chlorine or carbonyl carbon. All of this information together gives the piece -EWG-CH2CH3. EWG = Electron-Withdrawing Group
The peak at 2.90 ppm is a triplet that integrates to 2 protons. The integration tells us the peak corresponds to a methylene group. The splitting pattern of a triplet indicates that there are 2 neighboring protons. The chemical shift is quite far downfield at 2.90 ppm. Being further downfield means that the methylene group is attached to an electron-withdrawing group, which could be directly to the chlorine or carbonyl carbon. All of this information together gives the piece -EWG-CH2CH2-.
The peak at 3.77 ppm is a triplet that integrates to 2 protons. The integration tells us the peak corresponds to a methylene group. The splitting pattern of a triplet indicates that there are 2 neighbors. The chemical shift is quite far downfield at 3.77 ppm. Being the furthest downfield, the methylene group needs to be attached to the electron-withdrawing group that will be the most deshielding for the protons (Chemical Shift information). This is the proton must be directly attached to the chlorine. All of this information together gives the piece Cl-CH2CH2-, which means the methylene groups at 2.40 and 2.90 ppm are directly attached to the carbonyl carbon, since the chlorine only makes one bond.
All of this information yields the unknown molecule structure to be .
In the example just discussed, the number of hydrogens in the spectrum was equal the number of hydrogens in the molecular formula. What if this is not the case? Symmetry in a molecule can complicate your ability to determine a structure because the 1H NMR spectrum seems more simple than expected.
Example \(3\)
Using the chemical formula and 1H NMR, determine the structure of the unknown molecule.
Molecular formula: C6H7N
Unknown molecule 1H NMR spectrum: The ratio is 1:1:2.
Solution
Unknown molecule: .
Thus far, we have not needed to use the J-coupling constants from the splitting patterns, because the constants would have all been very similar of about 7 Hz. However, they can be very useful in determining your final product and ruling out other possibilities.
Example \(4\)
Using the chemical formula and the 1H NMR spectrum, determine the structure of the unknown molecule.
Molecular formula: C8H10O.
Unknown molecule 1H NMR spectrum: The proton ratio is 2:1:1:6. The coupling constant for the doublet (6.93 ppm) and triplet (6.71 ppm) is 8 Hz.
Solution
DU = 4. Anytime there is a DU of at least 4, it is often an aromatic ring. Each double bond is 1 DU and the ring is 1 DU, so 3(1) + 1 = 4.
To double check this, look to see if there are any protons in the aromatic proton region (6.5 ppm - 8 ppm). This spectrum does in fact have aromatic protons, so there must be an aromatic ring. There is a total of 3 protons in the aromatic region, which means that there are three hydrogens coming of the aromatic ring and three groups coming off the aromatic ring. The coupling constant of 8 Hz indicates that the aromatic protons are ortho to one anther. The piece from this information is
For the groups, since there are only two types of aromatic protons, this means that the groups ortho to the hydrogens must be the same.
The peak at 4.66 ppm is broad and small and integrates to one proton. Broad peaks typically are from exchangeable protons attached to oxygen or nitrogen. Since our molecular formula contains an O, this must be a hydroxyl group (-OH).
The singlet at 2.20 ppm integrates to 6 protons. The most common way to have 6 equivalent protons is that there are two methyl groups with no neighbors. Two methyl groups would be two of the same group and therefore are the groups ortho to the protons on the aromatic ring.
All of this information yields the unknown molecule structure to be .
The final example brings in a more complex splitting pattern.
Example \(5\)
Using the chemical formula and the 1H NMR spectrum, determine the structure of the unknown molecule.
Molecular formula: C4H6O.
Unknown molecule 1H NMR spectrum: The proton ratio is 1:1:1:1:2. The coupling constant for the doublet of doublet of triplets (5.93 ppm) has three coupling constants which are 17 Hz, 10 Hz, and 7 Hz. and triplet (6.71 ppm) is 8 Hz. The doublet of doublets at 5.20 ppm has coupling constants of 10 Hz and 2 Hz and the doublet of doublets at 5.10 ppm has coupling constants of 17 Hz and 2 Hz. The doublet at 3.12 ppm has a coupling constant of 7 Hz.
Solution
DU = 2. Therefore, there is either a double bond or a ring. If we look at chemical shifts, there are peaks in the double bond region, so the molecule has at least one double bond.
Another thing to note is that the ratio of protons equals our total number of protons of 6. Therefore, all the protons are accounted for and there is not integration of 3, so this molecule has NO methyl groups.
The peak out at 10 ppm is an aldehyde proton. Very few peaks show up that far downfield and using the chemical shift table, it indicates that it is an aldehydic proton. An aldehyde contains a carbonyl (C=O), which is our second degree of unsaturation.
The peak at 3.12 ppm is a methylene (-CH2-) group, since the integration is 2. It is quite far downfield, so it must be near an electron-withdrawing group. So far, there is an aldehyde, which would be an "end" to the molecule much like a methyl group. The fragment piece could be .
This brings us to the double bond region. In total, there are three double bond protons, which means we have a terminal double bond. There is only one way to have a double bond with three protons, but let's consider the splitting. If we combine the double bond with the fragment we have, then our molecule would be . HA is a doublet of doublets. It is split by cis coupling to Hc of 10 Hz and geminal coupling to HB of 2 Hz, which yields a doublet of doublets. HB is a doublet of doublets. It is split by trans coupling to Hc of 17 Hz and geminal coupling to HA of 2 Hz, which yields a doublet of doublets. The doublet of doublet of triplets is Hc. As we just discussed, it couples through cis bonding to HA (10 Hz) and trans to HB (17 Hz). The triplet comes from regular 3-bond coupling to the methylene group, which is J = 7 Hz.
All of this information yields the unknown molecule structure to be .
The next section will have problems for you to solve. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.10%3A_Interpreting_%281H%29_NMR_Spectra.txt |
Objectives
• Solve unknown problems using 1H NMR spectra and molecular formula.
Note
Helpful resources for solving these types of problems:
You may also want to read through some worked problems on how to solve unknown structure determination problems (Section 5.10).
Exercise \(1\)
Determine the structure for the unknown molecule with the molecular formula of C5H10O2.
1H NMR: The ratio of protons is 2:3:2:3. J = 7 Hz for all coupling.
Answer
Exercise \(2\)
Determine the structure for the unknown molecule with the molecular formula of C5H10O2.
1H NMR: The ratio of protons is 1:1:1:1:3:3. The peak at 9.5 ppm is a singlet. The peak at 2.41 ppm is sextet (J = 7 Hz). The peak at 1.72 ppm is a multiplet (J = 7Hz, 25 Hz). The peak at 1.53 ppm is a multiplet (J = 7 Hz, 25 Hz). The peak at 1.20ppm is a doublet (J = 7 Hz). The peak at 0.95 ppm is a triplet (J = 7 Hz).
Answer
Note: The -CH2- protons are diastereotopic, so they show up differently in 1H NMR spectra.
Exercise \(3\)
Determine the structure for the unknown molecule with the molecular formula of C10H12O2.
1H NMR: The ratio of protons is 5:1:3:3. The peak at 7.5 ppm is a multiplet. The peak at 3.70 ppm is quartet (J = 7 Hz). The peak at 3.58 ppm is a singlet. The peak at 1.48 ppm is a doublet (J = 7 Hz).
Answer
Exercise \(4\)
Determine the structure for the unknown molecule with the molecular formula of C10H15N.
1H NMR: The ratio of protons is 1:2:2:4:6. The peak at 7.3 ppm is a triplet (J = 2Hz) and the peak at 6.24 ppm is a doublet (J = 2 Hz).
Answer
Exercise \(5\)
Determine the structure for the unknown molecule with the molecular formula of C7H14O.
1H NMR: The ratio of protons is 1:3:2:3:2:3. The peak at 4.36 ppm is a triplet of quartets (J = 17 Hz and 7Hz). The peak at 3.37 ppm is a singlet. The peak at 2.04 ppm is a doublet of triplets (J = 17 Hz and 7Hz). The peak at 1.56 ppm is a doublet (J = 7 Hz). The peak at 1.48 ppm is a sextet (J = 7 Hz). The peak at 0.85 ppm is a triplet (J = 7 Hz).
Answer
5.S: Summary
Concepts & Vocabulary
5.1: Chapter Objectives and Preview of Nuclear Magnetic Resonance Spectroscopy
• Nuclear magnetic resonance spectroscopy is based on the function of nuclei as they interact with a magnetic field.
5.2 Theory of NMR
• Some types of atomic nuclei act as though they spin on their axis similar to the Earth.
• Atomic nuclei with even numbers of protons and neutrons have zero spin and all the other atoms with odd numbers have a non-zero spin.
• The magnetic moment of the nucleus acts as a tiny bar magnet.
• In the absence of an external magnetic field, each magnet is randomly oriented. However in the presence of an external magnetic field, the nuclear spins will either align with the magnetic field or oppose it.
• In order for the NMR experiment to work, a spin flip between the energy levels must occur.
• The local electronic environment surrounding the nucleus will slightly change the magnetic field experienced by the nucleus, which in turn will cause slight changes in the energy levels.
• Hydrogens attached to single or triple bonds are more shielded than alkenyl hydrogens due to position of induced magnetic fields of the electrons.
• Relaxation refers to the phenomenon of nuclei returning to their thermodynamically stable states after being excited to higher energy levels.
5.3 Instrumentation
• NMR spectroscopy works by varying the machine’s emitted frequency over a small range while the sample is inside a constant magnetic field.
• The magnets used by NMR instruments are superconducting to create the magnetic field range from 6 to 24 T.
• A sample is inserted into the NMR probe, which sits in a uniform magnetic field, and is irradiated by radio wave frequency. A detector interprets the results.
5.4 Types of Protons
• There are homotopic, enantiotopic, and diastereotopic protons.
• Homotopic protons are identical protons and will be chemically equivalent. These will show up at the same location in NMR spectroscopy.
• Enantiotopic protons are chemically equivalent. These will show up at the same location in NMR spectroscopy.
• Diastereoptopic protons are different protons and will be chemically nonequivalent. These will show up at different locations in NMR spectroscopy.
5.5 Chemical Shift
• NMR spectra are displayed on a plot that shows the applied field strength increasing from left to right.
• The left side of the plot is low-field or downfield and the right side of the plot is high-field or upfield.
• The different local chemical environments surrounding any particular nuclei causes them to resonate at slightly different frequencies. This means that nuclei which have different chemical environments will show up in different regions of the NMR plot or spectrum.
• Chemical shift is dependent on the applied field of the spectrometer, so TMS is used as a standard in order to use the units parts per million (ppm). This allows scientists to talk about the same peak in the same units regardless of field strength.
• The main "things" that effect the shielding of a nucleus are electronegative of atoms, magnetic anisotropy of pi systems, and hydrogen bonding.
• Electron with-drawing groups can decrease the electron density at the nucleus, deshielding the nucleus, resulting in a larger chemical shift.
• The π electrons in a compound, when placed in a magnetic field, will move and generate their own magnetic field. The new magnetic field will have an effect on the shielding of atoms within the field.
• Protons that are involved in hydrogen bonding (i.e.-OH or -NH) are usually observed over a wide range of chemical shifts since hydrogen bonds are dynamic
5.6 Integration of Proton Spectra
• The area of a peak in a 1H NMR is proportional to the number of hydrogens to which the peak corresponds.
• The integration curve appears as a series of steps with the height being proportional to the area of the corresponding absorption peak, and the number of protons responsible for the absorption.
• Integration can be used to determine the relative amounts of two or more compounds in a mixed sample.
5.7 Spin-Spin Splitting in Proton NMR spectra
• The split peaks (multiplets) arise because the magnetic field experienced by the protons of one group is influenced by the spin arrangements of the protons in an adjacent group.
• There is a recognizable pattern which is usually referred to as the n + 1 rule: if a set of hydrogens has n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks.
• Splitting occurs primarily between hydrogens that are separated by three bonds.
• The spin-spin coupling effect is quantified by the coupling constant, J. The coupling constant is simply the difference, expressed in Hz (not ppm), between two adjacent sub-peaks in a split signal.
5.8 More Complex Spin-Spin Splitting Patterns
• When a proton is coupled to two different neighboring proton sets with identical or very close coupling constants, the splitting pattern that emerges often appears to follow the simple `n + 1 rule` of non-complex splitting.
• If the set of protons is coupled to two or more sets of nonequivalent neighbors, the result is complex coupling.
• Complex coupling occurs when the coupling constants are different from each other on the neighboring protons.
• The stronger our magnet is, the more resolution we get in our spectrum. This means a clearer picture to gather more information.
• Nuclear magnetic resonance imaging (MRI) is an important medical diagnostic tool developed from NMR spectroscopy.
• Scientists use NMR spectroscopy to determine the structure of a molecule.
Skills to Master
• Skill 5.1 Distinguish between different types of protons in a molecule.
• Skill 5.2 Estimate the chemical shift of protons.
• Skill 5.3 Know which protons will be more downfield.
• Skill 5.4 Determine the ratio of different types of protons present in an organic compound.
• Skill 5.5 Provide a splitting diagram for a proton.
• Skill 5.6 Predict splitting patterns using a splitting diagram or the n + 1 rule.
• Skill 5.7 Interpret complex splitting in a spectrum.
• Skill 5.8 Solve unknown structure determination problems with 1H NMR spectroscopy. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/05%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.11%3A_%281H%29_NMR_problems.txt |
Learning Objectives
After completing this chapter, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• solve road-map problems which will require the interpretation of 13C NMR spectra in addition to other spectral data.
• define, and use in context, the key terms introduced in this chapter.
In the previous chapter, 1H Nuclear Magnetic Resonance (NMR) was discussed. The theory for NMR spectroscopy is the same for all NMR spectroscopy. In the presence of a magnetic field, a sample can absorb electromagnetic radiation, specifically in the radiofrequency (rf) region, based on the function of certain nuclei in the molecule. This chapter will focus on carbon nuclear magnetic spectroscopy (13C NMR) and what type of information you can glean from spectra and how it complements 1H NMR for the determination of a structure of a molecule.
6.02: C-13 NMR Spectroscopy- Signal Averaging and FT-NMR
Learning Objectives
• Learn how 1H NMR and 13C NMR are similar and different.
• Understand the basics of how 13C NMR works.
Why are we talking about 13C NMR when the 12C isotope of carbon - which accounts for up about 99% of the carbons in organic molecules? The 12C isotope does not have a nuclear magnetic moment, and thus is NMR-inactive (cannot be seen by the NMR). Fortunately for organic chemists, however, the 13C isotope, which accounts for most of the remaining 1% of carbon atoms in nature, has a magnetic moment just like protons, which is why we will be discussing 13C NMR spectroscopy and not 12C NMR spectroscopy. Most of what the theory we have learned about 1H-NMR spectroscopy also applies to 13C-NMR, although there are several important differences.
The basics of 13C NMR spectroscopy
The magnetic moment of a 13C nucleus is much weaker than that of a proton. This means that NMR signals from 13C nuclei are inherently much weaker than proton signals. Combining the weaker magnetic moment with the low natural abundance of 13C, means that it is much more difficult to observe carbon signals: more sample is required, and often the data from hundreds of scans must be averaged in order to bring the signal-to-noise ratio down to acceptable levels.
Chemical Shift
The resonance frequencies of 13C nuclei are lower than those of protons in the same applied field - in a 7.05 Tesla instrument, protons resonate at about 300 MHz, while carbons resonate at about 75 MHz. This is fortunate, as it allows us to look at 13C signals using a completely separate 'window' of radio frequencies. Just like in 1H-NMR, the standard used in 13C NMR experiments to define the 0 ppm point is tetramethylsilane (TMS), although of course in 13C NMR it is the signal from the four equivalent carbons in TMS that serves as the standard. Chemical shifts for 13C nuclei in organic molecules are spread out over a much wider range than for protons – up to 200 ppm for 13C compared to 12 ppm for protons (see Table 3 for a list of typical 13C NMR chemical shifts). This is also fortunate, because it means that the signal from each carbon in a compound can almost always be seen as a distinct peak, without the overlapping that often plagues 1H NMR spectra. The chemical shift of a 13C nucleus is influenced by essentially the same factors that influence a proton's chemical shift: bonds to electronegative atoms and diamagnetic anisotropy effects tend to shift signals downfield (higher resonance frequency). In addition, sp2 hybridization results in a large downfield shift. The 13C NMR signals for carbonyl carbons are generally the furthest downfield (170-220 ppm), due to both sp2 hybridization and to the double bond to oxygen. Symmetry will also play a role in how many signals are observed on a 13C NMR spectrum. If there is no symmetry, then each carbon should show up as a signal in the spectrum. If symmetry is present in the molecule, then there will be less than the total number of carbons in the molecule. Only non-equivalent carbons will appear as a signal in 13C NMR.
Example \(1\)
How many sets of non-equivalent carbons are there in ethyl benzene?
Solution
There are 5 different carbons in ethyl benzene. There is symmetry in ethyl benzene in the aromatic ring. On the labeled molecule, there are two carbon 3's and two carbon 4's due to the symmetry of the molecule.
Integration
Unlike 1H NMR signals, the area under a 13C NMR signal cannot be used to determine the number of carbons to which it corresponds. This is because the signals for some types of carbons are inherently weaker than for other types – peaks corresponding to carbonyl carbons, for example, are much smaller than those for methyl or methylene (CH2) peaks. There are some quantitative 13C NMR experiments that when enriched with 13C can be integrated, but typically it is not done.
Spin-spin Splitting
Because of the low natural abundance of 13C nuclei, it is very unlikely to find two 13C atoms near each other in the same molecule, which means that spin-spin coupling is not observed between neighboring carbons in a 13C NMR spectrum. However,there is heteronuclear coupling between 13C carbons and the hydrogens to which they are bound. Carbon-proton coupling constants are very large, on the order of 100 – 250 Hz. Proton-coupled 13C spectra show complex overlapping multiplets, which makes for a very difficult interpretation. For clarity, chemists generally use a technique called broadband decoupling , which essentially 'turns off' C-H coupling, resulting in a spectrum where all carbon signals are singlets. Below is the proton-decoupled 13C NMR spectrum of ethyl acetate, showing the expected four signals, one for each of the carbons with no spin-spin splitting.
While broadband decoupling results in a much simpler spectrum, useful information about the presence of neighboring protons is lost. However, there are other 13C NMR experiments that can give more information, such as Distortionless Enhancement by Polarization Transfer (DEPT) allows us to determine how many hydrogens are bound to each carbon.
Exercise \(1\)
How many sets of non-equivalent carbons are there in:
a. 2-pentanone:
b. para-xylene:
c. serotonin:
Answer
a. 5 non-equivalent carbons
b. 3 non-eqivalent carbons
c. 10 non-equivalent carbons | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/06%3A_Carbon-13_NMR_Spectroscopy/6.01%3A_Chapter_Objectives_and_Preview_of_C-13_Nuclear_Magnetic_Resonance_Spectroscopy.txt |
Learning Objectives
• Understand where different types of carbon appear on the spectrum
Simply, 13C NMR allows you to determine how many different carbons are in a molecule. It will also be seen that information on functional groups present in a molecule can be determined using 13C NMR. In a spectrum, each signal represents a resonance for a different carbon atom. The typical range for the resonance frequencies is 0 to 220 ppm from tetramethylsilane (TMS) reference. Like 1H NMR, the chemical shift of 13C nuclei is influenced by its chemical environment like the 1H nuclei.
One of the greatest advantages of 13C-NMR compared to 1H-NMR is the breadth of the spectrum - carbons resonate from 0-220 ppm relative to the TMS standard, as opposed to only 0-12 ppm for protons. Because of this, 13C signals rarely overlap, meaning we can almost always distinguish separate peaks for each carbon, even in a relatively large compound containing carbons in very similar environments. In a 1H NMR spectrum of 1-heptanol, for example, many of the signals overlap and it becomes difficult to analyze, only the signals for the alcohol proton (Ha) and the two protons on the adjacent carbon (Hb) are easily analyzed.
In the 13C spectrum of 1-heptanol, we can easily distinguish each carbon signal, and we know from this data that our sample has seven non-equivalent carbons. (Notice also that, as we would expect, the chemical shifts of the carbons get progressively smaller as they get farther away from the deshielding oxygen.)
This property of 13C NMR makes it very helpful in the elucidation of larger, more complex structures.
Example \(1\)
Predict the number of carbon resonances expected in a 13C NMR spectrum of ethyl prop-2-enoate, C5H8O2
Solution
There is no symmetry in this molecule, so you would expect 5 resonances - one for each C in the molecule - in a 13C NMR spectrum.
13C NMR spectrum of ethyl prop-2-en-oate:
13C NMR Chemical Shifts
The 13C NMR is used for determining functional groups based on characteristic shift values. 13C chemical shifts are greatly affected by electronegative effects and magnetic anisotropy. If a H atom in an alkane is replaced by substituent X, electronegative atoms (O, N, halogen), 13C signals for nearby carbons shift downfield (left; increase in ppm) with the effect diminishing with distance from the electron withdrawing group just as in 1H NMR. Below, a typical 13C chemical shift table shows the regions of some of the common organic functional groups.
13C Chemical shift range for organic compounds
Example \(2\)
Assign the resonances in a 13C NMR spectrum of ethyl prop-2-enoate, C5H8O2.
Solution
There are 5 carbons in the molecule, which equate to the 5 peaks in the 13C NMR spectrum.
Exercise \(1\)
Using 13C NMR spectrum, how could you tell the difference between the isomers acetone and methoxy ethene?
vs.
Answer
There are a few ways to tell the difference between the two molecules. Acetone would have 2 different resonances in a 13C NMR spectrum due to the symmetry of the molecule. Acetone is a ketone and ketone carbons appear far downfield 180-220 ppm. Methoxy ethene is difunctional molecule with an ether and an alkene. It would show 3 resonances in the 13C NMR spectrum, which all would be lower than the ketone resonance. Alkene resonances are 100-150 ppm and a C-O bond would be 40-85 ppm.
Exercise \(2\)
Assign as many peaks as you can to the 13C NMR spectrum to specific carbons in the N-ethylbenzamide.
13C NMR spectrum:
Answer
While there are 9 total carbons, there are only 7 non-equivalent carbons.
Labeled Carbon Number Chemical Shift (ppm)
1 14.80
2 34.80
3 134.40
4 126.70
5 128.10
6 130.90
7 167.20 | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/06%3A_Carbon-13_NMR_Spectroscopy/6.03%3A_Characteristics_of_C-13_NMR_Spectroscopy.txt |
Learning Objectives
• Understand the difference between 1D 13C NMR and DEPT.
• Determine what information is gained when using DEPT
There are many types of experiments that can be run using the NMR spectrometer, including gathering information from 13C NMR spectra and while broadband decoupling gives a simpler spectrum, it loses information about neighbors. Distortionless enhancement by polarization transfer, DEPT, is one of these techniques to gain this type of information back and making it possible to distinguish between methyl (CH3), methylene (CH2), methine (CH), and quaternary carbons. In other words, the number of hydrogens attached to a particular carbon can be determined. In DEPT, it takes advantage of the 13C to 1H coupling that is removed in broadband-decoupled 13C spectra.
DEPT experiments often start by running an ordinary 13C spectrum (typically broadband-decoupled spectrum). This allows one to know where the chemical shifts for the carbons in a molecule to be known. For a DEPT, there is a final step in the data acquisition that has a final flip angle of 90 or 135. During the experiment, polarization is transferred from one nuclei to another. Typically, the small gyromagnetic nuclei is observed, so the transfer would be from 1H to 13C. This allows for the DEPT to determine which carbons are attached to hydrogens. In a DEPT-135 spectrum, the CH3 and CH resonances are upright or positive, the CH2 resonances are inverted or negative, and quaternary carbons do not show up since these carbons are not directly attached to a hydrogen. Another DEPT experiment is the DEPT-90. While DEPT-135 showed all of the resonances of protonated carbons, DEPT-90 only shows CH peaks (upright/positive).
Example \(1\)
Propose a structure for an alcohol, C6H14O, that has the following 13C NMR spectral data.
Broadband-decoupled 13C NMR spectrum:
DEPT-135:
DEPT-90: No positive peaks.
Solution
From the 13C NMR spectrum, the resonances inform that there are 6 distinct carbons. There are only 6 carbons in our molecular formula, therefore they are all different. Based on chemical shift, there are no multiple bonds in the molecule. You can also begin by calculating the degrees of unsaturation, which indicates know multiple bonds or rings. The DEPT-135 shows 5 negative peaks and 1 positive peak. The negative resonances are methylenes and the positive peak is either a methyl or methine. The DEPT-90 indicates that there are no CH peaks in the molecule. Therefore, the resonance at 14.20 ppm is a methyl group. Our structure is:
There are many more complicated experiments as well, but that is beyond the scope of this chapter.
Exercise \(1\)
Propose a structure for an alcohol, C6H10O2, that has the following 13C NMR spectral data.
Broadband-decoupled 13C NMR spectrum:
DEPT-135:
DEPT-90:
Answer
There are 2 degrees of unsaturation. The 13C NMR spectrum indicates that all 6 carbons are different. The resonance at 196 ppm indicates a C=O based on chemical shift and the resonances at 170 and 104 ppm indicate a C=C bond. This would account for both degrees of unsaturation. The DEPT-135 shows no negative resonances, so there are no CH2s. The loss of the resonances at 196 and 170 ppm means that they are quaternary carbons (not attached to hydrogen). The DEPT-90 shows one resonance at 104 ppm, which means it is a CH and before was indicated as part of a double bond. The peaks at 56, 31, and 20 ppm are all methyl groups. The one at 56 ppm must be attached to an electron withdrawing group, which in this example an O. The fragment is -OCH3. To have a quaternary carbon that is not the carbonyl, there must be a methyl group attached to the one of the carbon of the C=C double bond. 196 ppm indicates a ketone, which would be where the last methyl group goes. The final structure based on chemical shift and DEPT is: | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/06%3A_Carbon-13_NMR_Spectroscopy/6.04%3A_DEPT_C-13_NMR_Spectroscopy.txt |
Learning Objectives
• understand how chemical shift and number of peaks come together to determine functional groups.
• solve unknown 13C NMR problems given the molecular formula.
This section will help identify how to interpret information from 13C NMR spectra.
Starting with the 13C NMR spectrum for ethanol, C2H6O.
Remember that each peak identifies a carbon atom in a different environment within the molecule. In this case there are two peaks because there are two different environments for the carbons. The carbon in the CH3 group is attached to 3 hydrogens and a carbon. The carbon in the CH2 group is attached to 2 hydrogens, a carbon and an oxygen. So which peak is which?
You might remember from Section 6.2 that the external magnetic field experienced by the carbon nuclei is affected by the electronegativity of the atoms attached to them. The effect of this is that the chemical shift of the carbon increases if you attach a more electronegative atom like oxygen to it. That means that the peak at about 60 (the larger chemical shift) is due to the CH2 group because it has a more electronegative atom attached.
In principle, you should be able to work out the fact that the carbon attached to the oxygen will have the larger chemical shift. In practice, you always work from tables of chemical shift values for different groups (see below).
What if you needed to work it out? The electronegative oxygen pulls electrons away from the carbon nucleus leaving it more exposed to any external magnetic field. That means that you will need a smaller external magnetic field to bring the nucleus into the resonance condition than if it was attached to less electronegative things. The smaller the magnetic field needed, the higher the chemical shift.
A table of typical chemical shifts in 13C NMR spectra
carbon environment chemical shift (ppm)
C=O (in ketones) 205 - 220
C=O (in aldehydes) 190 - 200
C=O (in acids and esters) 170 - 185
C in aromatic rings 125 - 150
C=C (in alkenes) 115 - 140
RCH2OH 50 - 65
RCH2Cl 40 - 45
RCH2NH2 37 - 45
R3CH 25 - 35
CH3CO- 20 - 30
R2CH2 16 - 25
RCH3 10 - 15
In the table, the "R" groups will not necessarily be simple alkyl groups. If a substituent is very close to the carbon in question, and very electronegative, that might affect the values given in the table slightly. For example, ethanol has a peak at about 60 because of the -CH2OH group. No problem! It also has a peak due to the RCH3 group. The "R" group this time is -CH2OH. The electron pulling effect of the oxygen atom increases the chemical shift slightly from the one shown in the table to a value of about 18 ppm.
A simplification of the table:
carbon environment chemical shift (ppm)
C-C 0 - 50
C-O 50 - 100
C=C 100 - 150
C=O 150 - 200
Now, we will look at 3-buten-2-one:
The structure for the compound is: . This molecule has carbons and all four are in different chemical environments. Therefore, in the 13C NMR spectrum there should be four signals.
The 13C NMR spectrum for 3-buten-2-one is:
Using the table above, you can assign each peak to each carbon.
• The peak at just under 200 ppm is due to a carbon-oxygen double bond. The two peaks at 137 ppm and 129 ppm are due to the carbons at either end of the carbon-carbon double bond. And the peak at 26 is the methyl group which, of course, is joined to the rest of the molecule by a carbon-carbon single bond. If you want to use the more accurate table, you have to put a bit more thought into it - and, in particular, worry about the values which do not always exactly match those in the table!
• The carbon-oxygen double bond in the peak for the ketone group has a slightly lower value than the table suggests for a ketone. There is an interaction (resonance) between the carbon-oxygen and carbon-carbon double bonds in the molecule which affects the value slightly. This will be observed in many conjugated systems. Discrepencies can also happen in more complicated systems.
• The two peaks for the carbons in the carbon-carbon double bond are exactly where they would be expected to be. Notice that they aren't in exactly the same environment, and so do not have the same shift values. The one closer to the carbon-oxygen double bond has the larger value.
• And the methyl group on the end has exactly the sort of value you would expect for one attached to C=O. The table gives a range of 20 - 30, and that's where it is.
One final important thing to notice. There are four carbons in the molecule, but they aren't all the same height. In 13C NMR, you cannot draw any simple conclusions from the heights of the various peaks.
Working out Structures from 13C NMR Spectra
So far, the structures of them molecules have been known and we have just been trying to see the relationship between carbons in particular environments in a molecule and the spectrum produced. Now let's make it a little more difficult - by looking at isomers! How could you tell from just a quick look at a 13C NMR spectrum whether you had propanone or propanal (assuming those were the only options)?
Because these are isomers, each has the same number of carbon atoms, but there is a difference between the environments of the carbons which will make a big impact on the spectra. In propanone, the two carbons in the methyl groups are in exactly the same environment, and so will produce only a single peak. That means that the propanone spectrum will have only 2 peaks - one for the methyl groups and one for the carbon in the C=O group. However, in propanal, all the carbons are in completely different environments, and the spectrum will have three peaks.
Example \(1\)
There are four alcohols with the molecular formula C4H10O.
A.
B.
C.
D.
Which one produced the 13C NMR spectrum below?
Solution
You can do this perfectly well without referring to chemical shift tables at all.
In the spectrum there are a total of three peaks - that means that there are only three different environments for the carbons, despite there being four carbon atoms.
In A and B, there are four totally different environments. Both of these would produce four peaks.
In D, there are only two different environments - all the methyl groups are exactly equivalent. D would only produce two peaks.
That leaves C. Two of the methyl groups are in exactly the same environment - attached to the rest of the molecule in exactly the same way. They would only produce one peak. With the other two carbon atoms, that would make a total of three. The alcohol is C.
Example \(2\)
This follows on from previous example, and also involves an isomer of C4H10O but which isn't an alcohol. Its 13C NMR spectrum is below. Work out what its structure is.
Solution
Because we do not know what sort of structure we are looking at, this time it would be a good idea to look at the shift values. The approximations are perfectly good, and we will work from this table:
carbon environment chemical shift (ppm)
C-C 0 - 50
C-O 50 - 100
C=C 100 - 150
C=O 150 - 200
The peak at 66.75 ppm indicates there is a peak for carbon(s) in a carbon-oxygen single bond. The peak at 15.55 ppm indicates that there is a peak for carbon(s) in a carbon-carbon single bond. That would be consistent with C-C-O in the structure.
It is not an alcohol (you are told that in the question), and the molecular formula is C4H10O. With only two peaks, but four total carbons there must be symmetry in the carbons within the molecule. The only solution to that is to have two identical ethyl groups either side of the oxygen. The compound is ethoxyethane (diethyl ether), CH3CH2OCH2CH3.
Example \(3\)
Using the simplified table of chemical shifts above, work out the structure of the compound with the following 13C NMR spectrum. Its molecular formula is C4H6O2.
13C NMR Spectrum:
Solution
Let's sort out what we've got.
• There are four peaks and four carbons. No two carbons are in exactly the same environment, so all of our carbons are accounted for since there are only four in the molecular formula. This means no symmetry within the molecule.
• The peak at just over 50 must be a carbon attached to an oxygen by a single bond.
• The two peaks around 130 must be the two carbons at either end of a carbon-carbon double bond.
• The peak at just less than 170 is the carbon in a carbon-oxygen double bond.
Putting this together is a matter of playing around with the structures until you have come up with something reasonable. But you can't be sure that you have got the right structure using this simplified table. In this particular case, the spectrum was for the compound:
If you refer back to the more accurate table of chemical shifts towards the top of the page, you will get some better confirmation of this. The relatively low value of the carbon-oxygen double bond peak suggests an ester or acid rather than an aldehyde or ketone.
It can't be an acid because there has to be a carbon attached to an oxygen by a single bond somewhere - apart from the one in the -COOH group. We've already accounted for that carbon atom from the peak at about 170. If it was an acid, you would already have used up both oxygen atoms in the structure in the -COOH group. Without this information, though, you could probably come up with reasonable alternative structures. If you were working from the simplified table in an exam, your examiners would have to allow any valid alternatives. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/06%3A_Carbon-13_NMR_Spectroscopy/6.05%3A_Interpreting_C-13_NMR_Spectra.txt |
Learning Objectives
• understand how 13C NMR data can be used to distinguish between two (or more) possible structures for an unknown organic compound
• determine if you made the correct product
13C NMR spectroscopy derives information that is helpful for structure determination, especially when paired with 1H NMR spectroscopy. 13C allows the organic chemist a way to determine how many non-equivalent carbons are in a molecule of interest. This allows one to understand if there is symmetry in the molecule or not. The chemical shift of each of the resonances in 13C NMR spectra gives information about the electronic environment, which can indicate what type of functional group is present or what type of bond is present. If you add in DEPT, then the organic chemist can determine how many hdyrogens are attached to each of the carbons.
Example \(1\)
The E2 reaction follows Zaitsev's rule, but how do we know that? Let's consider the reaction below:
1) How can you tell the difference between 1-butene and 2-butene in a 13C NMR spectrum?
2) Which product was made based on the 13C NMR spectrum of the product?
13C NMR spectrum of the product:
Solution
1) IR spectroscopy and mass spectrometry would not be helpful in elucidation of this problem. It is very difficult to address the structural differences in either of those spectroscopic methods. 1H NMR could be used, but there would be overlapping peaks, so it would again be difficult. In a 13C NMR spectrum, 1-butene would have 4 different signals in the spectrum where as 2-butene would have 2 distinct signals in the spectrum.
2) With just 2 signals in the spectrum, the product is 2-butene.
Example \(2\)
The 1H NMR spectrum had overlapping peaks, which lead to inconclusive results on if the product, ipsenol, was isolated. The structure of ipsenol is below. A 13C NMR spectrum along with a DEPT-135 and DEPT-90 to identify the product.
13C NMR spectrum:
DEPT-135 spectrum:
DEPT-90 spectrum:
Solution
Carbon NMR along with DEPT can be a great tool in structure determination. The range for carbon NMR is wider than for 1H NMR, so it can help spread the spectrum out and remove overlapping peaks. In the 13C NMR spectrum, there are 10 distinct peaks, so all of the carbons have been accounted for. If we consider the structure for a moment, then it can be seen that there is 1 quaternary carbon, 3 methines (CH), 4 methylenes (CH2), and 2 methyl groups (CH3). In the DEPT-135, the peak at 142.79 ppm has gone away. This is the quaternary carbon, since it is not attached to any hydrogens directly, it is not observed. The negative peaks are methylene groups and there are four - just as expected. There are 5 peaks that are either a methine or methyl group. Finally, the DEPT-90 informs that there are 3 methine groups, which means the peaks at 23.08 ppm and 21.50 ppm would be methyl groups. These 1D-carbon spectra do correlate with the structure.
6.07: Structure Determination Problems with C-13 NMR and 1-H NMR
Learning Objectives
• Solve unknown problems using13C and 1H NMR spectra and molecular formula.
Exercise \(1\)
Which isomer of ortho, meta, or para xylene do you have based on the 13C NMR spectrum?
Answer
There are 5 different carbons in the spectrum with four different aromatic carbons. p-Xylene would have 2 types of aromatic carbons and o-xylene would have 3 types. m-Xylene is the only one with 4 different types of aromatic carbons, which fits this spectrum. The methyl groups would all be similary, so not a point of difference.
Exercise \(2\)
Propose a structure using the spectral data below for C9H10O.
13C broadband decoupled spectrum:
1H NMR spectrum: Integration: 1 (doublet; J = 1 Hz):5 (multiplet):1 (quartet of doublets;J = 7 Hz and 1 Hz):3 (doublet; J = 7 Hz)
Answer
Exercise \(3\)
Propose a structure using the spectral data below for C6H12O2
13C broadband decoupled spectrum:
DEPT-135:
DEPT-90:
1H NMR spectrum: Integration: 3 (singlet):1 (sextet; J = 8 Hz):1 (multiplet;J = 25 Hz and 8 Hz):1(multiplet; J = 25 Hz and 8 Hz):3 (doublet; J = 8 Hz):3 (triplet; J = 8 Hz)
Answer
Structure:
6.S: Summary
Concepts & Vocabulary
6.2 C-13 NMR Spectroscopy- Signal Averaging and FT-NMR
• The magnetic moment of a 13C nucleus is much weaker than that of a proton. This means that NMR signals from 13C nuclei are inherently much weaker than proton signals, which makes 13C NMR harder to acquire good data.
• Chemical shift is similar to 1H, where the environment around the carbon changes for each carbon in the molecule.
• Integration is not done in 13C NMR spectroscopy because the signals for some types of carbons are inherently weaker than for other types.
• Because of the low natural abundance of 13C nuclei, it is very unlikely to find two 13C atoms near each other in the same molecule, which means that spin-spin coupling is not observed between neighboring carbons in a 13C NMR spectrum.
• There is heteronuclear coupling between 13C carbons and the hydrogens to which they are bound, proton-coupled 13C spectra show complex overlapping multiplets, which makes for a very difficult interpretation. For clarity, broadband decoupling is used, which essentially 'turns off' C-H coupling, resulting in a spectrum where all carbon signals are singlets.
6.3 Characteristics of C-13 NMR Spectroscopy
• Carbons resonate from 0-220 ppm relative to the TMS standard, as opposed to only 0-12 ppm for protons. Because of this, 13C signals rarely overlap, meaning we can almost always distinguish separate peaks for each carbon.
• The 13C NMR is used for determining functional groups based on characteristic shift values.
• 13C chemical shifts are greatly affected by electronegative effects and magnetic anisotropy.
6.4 DEPT C-13 NMR Spectroscopy
• Distortionless enhancement by polarization transfer, DEPT, is one of these techniques and making it possible to distinguish between methyl (CH3), methylene (CH2), methine (CH), and quaternary carbons.
• In DEPT, it takes advantage of the 13C to 1H coupling that is removed in broadband-decoupled 13C spectra.
6.5 Interpreting C-13 NMR Spectra
• Chemical shift is a big indicator into what type of carbon is at that resonance.
• Different carbons are carbons in distinct chemical environments and each different carbon will appear at a different resonance.
• Tables of chemical shift data can be used to distinguish different types of carbons.
6.6 Uses of 13C NMR Spectroscopy
• 13C NMR spectroscopy derives information that is helpful for structure determination.
• Scientists use 13C as a way to determine how many non-equivalent carbons are in a molecule of interest.
Skills to Master
• Skill 6.1 Distinguish between different types of carbons in a molecule.
• Skill 6.2 Estimate the chemical shift of carbons.
• Skill 6.3 Know which carbons will be more downfield.
• Skill 6.4 Determine which carbons are attached to hydrogens using DEPT
• Skill 6.5 Solve unknown structure determination problems with 1H and 13C NMR spectroscopy. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/06%3A_Carbon-13_NMR_Spectroscopy/6.06%3A_Uses_of_C-13_NMR_Spectroscopy.txt |
Chapter Objectives and Preview Correlation NMR Spectroscopy; 2-D NMR
Learning Objectives
After completing this chapter, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• solve road-map problems which may require the interpretation of 2D NMR spectra in addition to other spectral data.
• define, and use in context, the key terms introduced in this chapter.
In Section 5 and Section 6, the focus was on one-dimensional (1D) techniques to elucidate structures of organic molecules. Complex molecules like polymers or biomolecules can be hard to elucidate solely with proton and carbon NMR, which while powerful techniques don't quite solve the entire picture. This is where correlation NMR spectroscopy can be used and many of these experiments are two-dimensional (2D) techniques. While there are many correlation NMR spectroscopy experiments that can be run, this chapter will focus on the more common ones to be used.
7.02: Theory
Learning Objectives
• Learn the difference between 1-D and 2-D NMR spectroscopy
In 1971, Jean Jeener introduced the two dimensional variation of NMR spectroscopy. Since then, scientists have applied the concept to develop the many techniques of two-dimensional (2-D) NMR. Correlation itself is not a new concept in NMR. In 1H NMR spectroscopy, the splitting of resonances indicates that groups were "correlated" to each other due to the spins within each group. Spin-spin splitting gives information about the neighboring nuclei and in the case of 1H NMR, it was how many neighboring hydrogens there were in the molecule. This coupling of hydrogens will be one type of coupling 2-D NMR will consider. The use of 2-D NMR allows for better resolution in signals that normally overlap in 1-D NMR.
All the basic NMR theory for 1-D NMR still applies. In 1-D Fourier transform NMR, a magnetic field is applied to a sample, which is then hit with a series of pulsed radiofrequency (rf), as seen in the pulse sequence below. The Fourier transform of the outgoing signal results in a 1-D spectra as a function chemical shift.
Two-dimensional NMR adds additional experimental variables and thus introducing a second dimension to the resulting spectrum. A simple 2-D experiment pulse sequence consists of a relaxation delay, a pulse, a variable time interval (t1), a second pulse, and acquisition (t2). The pulse sequence is repeated many times while varying the length of time (t1) the system is allowed to evolve during the first pulse. In 2-D experiments, the signal detected during acquisition is a function of acquisition time (t2), which has been modulated as a function of the time interval (t1). This means that magnetization evolves around one frequency during t1 and a different frequency during t2. The output once Fourier transformed is a 2-D spectrum with two axes. One axis (v2) represents the nucleus detected during acquisition (t2), while the other axis (v1) can represent the same nucleus or a different nucleus. With two axes, it leads to cross peaks along a diagonal connecting coupled nuclei. Due to the nature of how the experiment is run, magnetization is redistributed equally in both directions (just like spin-spin coupling), the cross peaks will be symmetrically disposed about the diagonal. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/07%3A_Two-Dimensional_NMR_Spectroscopy/7.01%3A_Chapter_Objectives_and_Preview_of_Correlation_NMR_Spectroscopy.txt |
Learning Objectives
• Understand what correlation spectroscopy is and why it is used
• Learn about different types of homonuclear 2-D NMR spectroscopy
The previous sections have discussed one-dimensional NMR techniques, but for particularly complicated molecules it is hard to get the full picture of what is happening. Two-dimensional NMR spectra provide more information about a molecule than one-dimensional NMR spectra in these situations. In this section, homonuclear 2-D NMR spectroscopy is going to be considered and what is meant by homonuclear is looking at the correlation of the same nuclei in a molecule.
Correlation Spectroscopy (COSY)
The most basic form of 2-D NMR is the COSY (COrrelation SpectroscopY) experiment. This experiment looks at 1H coupling to 1H through bonds typically 3 bonds away. It relies on the J-coupling to provide spin-spin correlation to indicate which protons are close to each other on the cross peak. In a 1H-1H COSY experiment, the pulse sequence consists of proton pulses separated by the required evolution period (t1), and the acquisition period (t2). The evolution period is systematically incremented during the repeated pulse sequences. Specifically, it consists of a 90o RF pulse followed by an evolution time and an additional 90o pulse (shown below). The resulting oscillating magnetization (symbolized by decaying the sinusoidal curve) is then acquired during t2
A COSY takes a 1-D 1H NMR and spreads it across two dimensions, which means the peaks are spread out into an array. The spectra of COSY give rise to cross peaks (off diagonal) for all protons that have spin-spin coupling. This means that the peaks off the diagonal are coupled protons. The purpose of a COSY is to determine which protons are coupled to what other protons in the molecule through bonds. In the example below, you will use a simpler molecule, propyl acetate, to understand how to read a COSY and what information you can glean from it.
Example \(1\)
The COSY spectrum for propyl acetate is below:
Assign all of the correlations for propyl acetate and indicate the coupling as geminal, vicinal, or long range.
Solution
The structure of propyl acetate
From the chapter on 1-D 1H NMR spectroscopy, the methyl group attached to the carbonyl will be a singlet. It does not have any neighboring carbons to couple with. The other hydrogens will have coupling neighbors, so while you may only need 1-D 1H NMR to correlate the coupling hydrogens, this knowledge will help flesh out how to read a COSY.
First, the COSY is set out in an array and spread across two dimensions. The peaks are displayed on the axis. In the above spectrum, it is the same proton spectrum for propyl acetate on each of the axis because it is a homonuclear experiment where you look at the same nuclei. The peaks are then plotted against each other. There peak at 3.9 ppm on both axis, so there is a data point at (3.9, 3.9). This is repeated with all the peaks, which ends up being a diagonal. This information along the diagonal is the same information from the 1-D 1H NMR.
The points of interest are those not along the diagonal or the cross peaks. These peaks inform about which hydrogens are coupled to what other hydrogens. There is a cross peak at (0.9, 1.6). There is also a cross peak at (1.6, 0.9) because these protons are coupled to each other and if you recall coupling is reciprocal. The peak at 0.9 ppm corresponds to the methyl group with a neighboring -CH2- and the 1.6 ppm peak is the -CH2- with the neighboring methyl group. However, the peak at 1.6 ppm has another cross peak (1.6, 3.9), which indicates that it is also coupled to the -CH2- on the other side. And of course, there is a cross peak at (3.9, 1.6) since the coupling goes both ways.
All the cross peaks have now been identified and the bonding between these hydrogens is all vicinal coupling.
Total Correlation Spectroscopy (TOCSY)
The next type of 1H to 1H correlation spectroscopy is TOtal Correlation SpectroscopY (TOCSY). In this type of coupling, the information obtained creates correlations between all protons within a given spin system. A molecule can have just one spin system or hundreds in more complex systems. The goal in TOCSY is to transfer the magnetization beyond directly coupled spins. This is not just looking at geminal and vicinal correlations as in COSY, but the entire spin system. The pulse sequence is similar to COSY, but in TOCSY, a mixing period is added to the pulse sequence. This allows the magnetization to be relayed from one spin to its neighbor to its neighbor throughout the entire spin system. The longer this mixing period, the further out the transfer of magnetization can travel with the goal being the entire spin system. The spectra of TOCSY give rise to cross peaks for all protons that are part of a coupled spin network. TOCSY spectra display the entire chain of protons, each coupled to the next.
Example \(2\)
Below is the TOCSY of propyl acetate:
Does what you know of the structure of propyl acetate match what the TOCSY is indicating?
Solution
YES! Remember that TOCSY tells information about the entire spin system. As a reminder, a spin system includes nuclei where spin-spin interaction exists between them. In propyl acetate, this is the chain where the hydrogens are coupled. Just like in COSY, the peaks that show up along the diagonal in a TOCSY is the same information as a 1H NMR spectrum and it is the cross peaks that inform about the chain of connection. The peak at 2.0 ppm has no cross peaks since it is a methyl group with no neighboring hydrogens. The other peaks are much more interesting. There is a cross peak at (0.9, 1.6) just like in COSY. This is because those protons are coupled to each other. There is also a cross peak at (1.6, 0.9) again because the reciprocality of how coupling works. However, there is a new cross (0.9, 3.9) because the methylene at 3.9 ppm is coupled to the methylene at 1.6 ppm. For the 1.6 ppm methylene, there are two cross peaks (1.6, 0.9) and (1.6, 3.9) like in COSY because it is coupled to those protons. Then in the 3.9 ppm methylene, there is again two cross peaks (3.9, 0.9), which is new from COSY and (3.9, 1.6), which was the same as COSY. The cross peak (3.9, 0.9) appears for the same reason the cross peak (0.9, 3.9) appeared when looking at the 0.9 ppm peak because the methylene at 1.6 ppm is coupled to the 0.9 ppm methyl group. The TOCSY does corroborate what we know about the connectivity of propyl acetate.
Nuclear Overhauser Effect (NOE) Correlation Spectroscopy
Thus far, only the coupling of nuclei through bonds has been considered. In this type of coupling, the magnetization of nuclei affect those closely bound to them through the electrons that make up those bonds. This is not the only type of coupling that occurs. Coupling directly between nuclei that are in close spatial proximity to each other also occurs. This is called the Nuclear Overhauser Effect (NOE), and it arises when the spin relaxation of nuclei A is felt by nearby nuclei B, stimulating a corresponding change in magnetization in B. In a typical NMR spectrum, the interference of electrons makes this coupling undetectable. However, a sample can be decoupled to “neutralize” the bond coupling through electrons, allowing the space coupling of the NOE to be detected. This is called NOESY (Nuclear Overhouser Effect SpectroscopY) and is another type of homonuclear NMR. The purpose of NOESY is to determine which signals arise from protons athar are close to each other in space, even if they are not bonded. NOESY can be very useful for looking at stereochemistry and 3-D structure.
Like COSY, the first step is a 90o pulse followed by a variable evolution time. Unlike COSY, however, pulse two actually consists of two 90 degree pulses separated by a short delay. The first pulse converts the bulk magnetization from the transverse plane to the z-plane, eliminating the effect of electron-aided bond coupling. Then, during the ττm, there is cross relaxation between spatially adjacent nuclei. Finally, the last 90 degree pulse converts the space coupling of nuclei into an observable transverse magnetization, which can be detected during t2.The pulse sequence for a NOESY NMR experiment is depicted below.
Just like in COSY, the peaks that show up along the diagonal in a NOESY is the same information as a 1H NMR spectrum and it is the cross peaks that inform about the through space relationships. The cross peaks correlate the spin of one nuclei to that illuminated by the source spin (other nuclei) if nearby. In the example below, you will use a simpler molecule, propyl acetate, to understand how to read a NOESY and what information you can glean from it.
Example \(3\)
Lactide is a product of the fermentation of corn and soybeans; it can be polymerized to make a sort of brittle plastic, PLA. PLA is used for food packaging because it can be composted in industrial and municipal waste management sites. However, there are three isomers of lactide (D-lactide, L-lactide, and Rac-lactide). L-lactide and Rac-lactide are depicted below:
Using the spectral data, do you have L-lactide, or rac-lactide?
1H NMR:
NOESY:
Solution
rac-Lactide is the diastereomer L-lactide (or D-lactide). It has different physical properties, including different NMR spectra. We could carefully compare the spectrum below to reported spectra for rac-LA and L-LA (or D-LA) to see which isomer we have. On the other hand, we could just look at the NOESY spectrum. In rac-LA, the methyl on one end of the molecule is on the same face of the ring as the hydrogen on the other end. NOESY would allow us to see that through-space relationship. We wouldn't see it in L-LA or D-LA, since the two methyl groups are on the same face. Just like in a COSY spectrum, all of the peaks that show up along the diagonal of a NOESY spectrum are simply the ones we would see in a regular 1H spectrum. The peaks that show up off the diagonal tell us about through-space relationships. In this case, the relationship between the methyl hydrogen and the alpha hydrogen suggest we have a sample of rac-LA.
Exercise \(1\)
The magnetic effect of which type of particle must be removed from an NMR experiment in order to observe an NOE?
Answer
Electrons.
Exercise \(2\)
What type of information do you get from:
1. COSY?
2. NOESY?
3. TOCSY?
Answer
1. COSY gives information on which spins are coupled to each other through bonds.
2. NOESY determines which signals arise from protons that are close to each other in space, even if not bonded.
3. TOCSY creates correlations between all the protons within a spin system (not just geminal and vicinal coupling as in COSY).
Exercise \(3\)
The following COSY spectrum is for an isomer of dinitrobenzene. Which isomer is it?
Answer
: ortho isomer | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/07%3A_Two-Dimensional_NMR_Spectroscopy/7.03%3A_Two_Dimensional_Homonuclear_Spectroscopy.txt |
Learning Objectives
• Understand what HSQC is and when to use it
• Understand what HMBC is and when to use it
The previous section discussed the correlation of 1H to 1H, but those are not the only NMR active nuclei in a molecule. Heteronculear 2-D NMR is the correlation between different nuclei, such as a 1H to 13C and heteronuclear 2-D NMR is especially important in biological chemistry, especially in the elucidation of the three-dimensional structure of proteins.
Heteronuclear Single Quantum Coherence
Hetereonuclear Single Quantum Coherence (HSQC) is used to determine the proton to carbon or heteroatom (often nitrogen) single bond correlations. It is also known as heteronuclear multiple quantum coherence (HMQC). While the DEPT may give the same information for 13C to 1H correlations, HSQC is more sensitive, therefore, it may be more adventageous to use this type of experiment in complex situations. In an HSQC experiment, polarization is transferred from a 1H nuclei to a neighboring heteroatom (13C or 15N). This polaraization is then transferred back to the 1H nuclei. The signal from the 1H nuclei is recorded. The pulse sequence for a typical HSQC experiment is detailed below involving 1H and 15N nuclei.
In an HSQC spectrum, a 13C or heteroatom spectrum is displayed on one axis and a 1H spectrum is displayed on the other axis. Cross-peaks show which proton is attached to which carbon or heteroatom. The purpose of a HSQC is to determine which protons are coupled to what other specific carbon or heteroatom in the molecule through bonds.
Example \(1\)
The HSQC of propyl acetate is below:
Assign which hydrogens are attached to which carbons.
Solution
An HSQC experiment spreads things out into two dimensions just like the homonuclear experiments did in the previous section. Looking along the x axis, we the 1H NMR spectrum and along the y axis we see the 13C NMR spectrum displayed. The peak at 171 ppm in the 13C has no cross peak. This means that this carbon does not have any hydrogens attached, therefore, this is our carbonyl carbon. The peak at 77 ppm is residual solvent, in this case CDCl3. The cross peak at (3.9, 66) is a methylene attached to a carbon that is attached to an electron-withdrawing group. If you look at a 13C data table, then you can see where different carbon groups show up. In propyl acetate, oxygen is the most electron withdrawing group, so the piece we have at this cross peak is -CH2-O. The cross peak at (2.0, 21.8) is the methyl group that has no coupling neighboring hydrogens. The next cross peak at (1.6, 20.8) is a methylene group, The final cross peak is (0.9, 10.2) is a methyl group.
Another example HSCQ spectrum from ubiquitin is shown below.
1H15N HSQC spectrum of ubiquitin
Notice the greater clarity of spectra of the HSQC experiment. This is a strong advantage of heteronuclear NMR. In this diagram, each peak corresponds to a cross peak, showing coupling between sets of 1H and 15N nuclei. Each peak represents the 15N—1H of a unique amino acid along the backbone of the amino acid.
Heteronuclear Multiple Bond Correlation (HMBC)
Hetereonuclear Multiple Bond Correlation (HMBC) is used to determine long range 1H to 13C connectivity. This experiment gives the correlation between 1H and 13C when separated by two, three, and even four (if through a conjugated system) bonds away. In an HSQC experiment, direct one bond correlations are suppressed as part of the sequence and like HSQC is proton detected. The time delay in the pulse sequence can be optimized for different coupling constants, J. The signal from the 1H nuclei is recorded. The spectra of HSQC give rise to cross peaks that correlate one nuclei to another. These correlations help determine which protons are coupled to what other specific heteroatom more than one bond away.
Example \(2\)
Below is the HMBC of propyl acetate:
Does the HMBC corroborate with the structure of propyl acetate?
Solution
Yes! While propyl acetate is a simpler structure than typically analyzed by 2D NMR spectroscopy, it helps illuminate how to read the spectra and what information to gather from it. By using the HSQC (above), we already know which hydrogen(s) are directly attached to which carbons. The cross peaks here indicate what other carbons are attached beyond one bond away. Below is propyl acetate with the atoms labeled:
Focus on a column from one hydrogen peak. If we start with the 0.9 ppm methyl group (6), there are two cross peaks in the column (09., 20.8) and (0.9, 65.9). These correspond to a methylene group (5) and the methylene group attached to the oxygen (4). If we move the the methylene (5) at 1.6 ppm, then it has two cross peaks that correspond to C6 and C4. The methyl (1) at 2 ppm has not shown any correlations in previous spectra, but in HMBC there is a cross peak. The methyl (1) is attached to the carbonyl carbon (2). Finally, the methylene at 3.9 ppm has 3 cross peaks. It shows that it is connected through bonds to a methyl (6), a methylene (5), and the carbonyl carbon (2). All of this does match the structure for propyl acetate.
Exercise \(1\)
Analyze the data to determine which of the two isomers (below) we are dealing with.
or
Answer
Exercise \(2\)
What type of information do you get from:
1. HSQC?
2. HMBC?
Answer
1. HSQC determines the proton-carbon single bond correlations. It could also be other heteroatoms to hydrogen single bonds.
2. HMBC gives the correlations between carbon and hydrogen separated by 2 to 4 bonds away. This gives more information about connectivity within the molecule.
7.05: Uses for 2-D NMR Spectroscopy
Learning Objectives
• Learn in what instances scientists use 2-D NMR
• Understand practical applications of 2-D NMR
It has been previously mentioned in this chapter that the major advantage for using 2-D NMR over 1-D NMR is the ability to distinguish between overlapping signals that exist in larger molecules. 2-D NMR is incredibly important in biological and polymer chemistry to elucidate te three-dimensional structure of these large macromolecules. In these cases, HSQC can be used to determine the profile of metabolites in low concentrations (microMolar) accurately. TOCSY has been utilized to show changes in tumor cells and identify biomarkers associated with these cells. Molecular dynamics can be studied using 2-D NMR spectroscopy to map the molecule's internal mobility patterns. With molecular dynamics the loose ends of proteins can be studied and elucidated to learn more about the flexible surface areas often lost in other methods. 2D NMR has many more applications beyond protein NMR, including characterization of pharmaceuticals, temperature dependence of carbohydrate conformations, and metabolomics, to just name a few.
2-D NMR is often used in metabolic profiling like in the example below. This is due to the fact that NMR is not a destructive analysis, quantitative, reproducible and gives a lot of information about the sample.
Using TOCSY, this particular study compared the global metabolic profiles of urine obtained from two types of mice, specifically of wild-type and a knockout. Both 1-D and 2-D NMR experiments were run to determine if statistical differences between the techniques, especially when looking at low abundance metabolites. Both 1-D and 2-D NMR data could differentiate between the two types of mice, but only the 2-D data could be used to show statistically relevant changes in the low abundance metabolites. The con of 2-D NMR data is that it takes longer to obtain the data compared to 1-D NMR data collection, however, the data obtained resulted in a more meaningful and comprehensive metabolic profile, aided in metabolite identifications, and minimized ambiguities in peak assignments.2 This is just one example of how 2-D NMR has been applied. | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/07%3A_Two-Dimensional_NMR_Spectroscopy/7.04%3A_Two_Dimensional_Heteronuclear_NMR_Spectroscopy.txt |
Learning Objectives
• understand how to analyze the different 2-D spectra and what information you can gather from each.
• determine the structure of an unknown molecule using different types of NMR spectroscopy.
This page is devoted to explaining solved structure elucidation problems using different types of 1-D and 2-D NMR spectra to help guide you through different scenarios. The first scenario involves analyzing different types of NMR spectra to determine if you made the desired product.
Example \(1\)
Did you synthesize menthyl anthranilate? The spectra were taken in DMSO-d6. Use the spectra to determine whether the product desired was made.
1H:
COSY:
13C:
DEPT-135:
HMBC:
Solution
With this example, we have the advantage to knowing what our structure is. It would be helpful to look up what menthyl anthranilate's structure is to start before moving on.
Menthyl Anthranilate:
Atoms Numbered in Menthyl Anthranilate:
Step 1: Characterization of the peaks in 1H NMR spectrum by considering chemical shift, integration, spin-spin splitting, and coupling constants. Remember if protons are coupled to each other, they will have the same coupling constant, J.
The deuterated solvent is DMSO and the peak for residual solvent appears at 2.50 ppm, which is why this peak is missing integration. The peak at 0.7 ppm is the methyl group at position 7 on the numbered structure. The integration is 3 and there is no other peak that will have that integration. The next peak at 0.9 ppm are the other methyl groups labeled 9 and 10. Both of these peaks are doublets, which correlates to them having one neighboring hydrogen (positions 6 and 8) and this is the case for both. The next three peaks to come all integrate to 2 protons, which will correspond to positions 1, 2, and 5. The hydrogens at position 5 will be the furthest downfield since it is closest to an electron-withdrawing group. Remember, the closer the hydrogen is to an electron-withdrawing group, the more the magnetic field surrounding the hydrogen is influenced. The peak at 4.8 ppm is H4. It is directly attached to an electron-withdrawing group and it integrates to 1H. The broad peak at 3.4 ppm corresponds to the NH2. Broad peaks in NMR often correspond to an averaging and do not show spin-spin splitting. In this case, the hydrogens attached to nitrogen are exchangeable leading to an averaging of "seeing" a different H spin state depending on if an H is attached or not. The peaks in the 6.5 to 8.0 ppm range are the aromatic protons. COSY will help identify the peaks that were not labeled specifically in the 1H spectrum.
Step 2: Even with the COSY, it is difficult to completely elucidate the structure. The cross peaks in the 6.5-8.0 ppm range are all protons that are part of an aromatic ring. The correlations are indicating that the two groups coming off the ring will be ortho to each other. The correlations indicate that the peak at 7.7 ppm has two other neighbors, which leads to the doublet of doublets. This is the same for the peak at 6.6 ppm, which is also a doublet of doublets. The 7.2 ppm peak is a triplet of doublets, so a total of 3 neighbors and there are 3 cross peaks to confirm this. The peak at 6.5 ppm is also a triplet of doublets and has 3 cross peaks. The only way this pattern can occur is to have ortho groups coming off the aromatic ring. The next protons that can be used to help elucidate the structure come from the peak at 4.8 ppm. This is the methine group attached to the oxygen at C4. The cross peak at (4.8, 1.1) is a methylene group (C5) attached to C4. The cross peak at (4.8, 1.5) indicates a correlation to another methylene group (possibly C2). The cross peak at (4.8, 1.9) is a correlation to another methine group (C3). The peak at 0.9 ppm is a methyl group. The cross peaks of this methyl group indicate that it is attached to a methine group (C6). The remaining correlations are difficult to decipher since it is squished into a small region with overlapping cross peaks.
Step 3: 13C NMR analysis is a necessary step in full structural characterization. However, ¹³C NMR alone does not provide enough information to assign the carbons in the molecule. It can be used to confirm the number of carbons in the molecule and this is where DEPT or 2-D NMR will assist. The 13C spectrum above does indeed show all seventeen carbons. The group of peaks around 40 ppm is DMSO-d6, so will not be counted towards the seventeen. However, the peak at 41 ppm is one of the seventeen total carbons.
Step 4: DEPT will tell what kinds of carbons do we have in the molecule. There are three quaternary carbons (absent from the DEPT spectrum compared to the 13C spectrum), three methylene groups (negative peaks), and eleven methine and methyl groups.
Step 5: In the HMBC, there are cross peaks that extend to our carbonyl at 167 ppm. These cross peaks indicate that the carbonyl is attached to the aromatic ring just as we can see in the structure. It also shows that there is a correlation between the carbonyl carbon and the methine attached to the oxygen, this confirms the ester functionality. The two other quaternary carbons (152 ppm and 110 ppm) also have cross peaks indicating connection as part of the aromatic ring. Because the carbon spectrum is a wider range, it spreads out the cross peaks and there is less overlap and it is easier to analyze each cross peak. As this is done, each one confirms that the structure is indeed menthyl anthranilate.
This process can be used for known and unknown structure elucidation. You can also see that in more complicated molecules, multiple spectra help shine light on all parts of the molecule.
This next example show how you attempt to determine an unknown structure from molecular formula and different types of 1-D and 2-D NMR spectra.
Example \(2\)
Identify the compound C8H10 from the spectra below. The sample was dissolved in CDCl3
1H:
Note: The peak at 0 ppm is TMS and the peak at 1.5 ppm is water. These can be disregarded in analysis.
COSY:
13C:
HSQC:
HMBC:
Solution
Degrees of unsaturation = 4.
1H NMR spectrum: At 1.2 ppm there is a triplet that integrates to 3H. This means the fragment must be -CH2-CH3. The peak at 2.67 ppm is a quartet that integrates to 2H, so this is also a -CH2-CH3. The multiplet at 7.2 ppm that integrates to 4.6 is in the aromatic region and with a degree of unsaturation of 4, these are most likely 5 aromatic protons.
COSY: The COSY corroborates with the fragment of a methylene coupling to a methyl group. This information was determined from the two cross peaks: (1.2, 2.67) and (2.67, 1.2). Coupling is reciprocal, so it makes sense that there is symmetry with the coupling. The aromatic region protons do not couple beyond that system itself.
13C: The 13C indicates that there are 6 different types of carbon nuclei. This means there must be some symmetry within the carbons since there should be a total of 8. The peak at 76 ppm is CDCl3
HSQC: The first cross peak (1.2, 15) corresponds to the methyl group. The next cross peak (2.4, 29) corresponds to the methylene group. The final cross peaks (7.27, 125), (7.27, 128), and (7.27, 128.4) all correspond to the aromatic protons. These carbons are also in the double bond region. There is no cross peak for the carbon at 144 ppm, so this carbon must not have any hydrogens attached to it.
HMBC: Cross peak (1.2, 29) indicates that the methyl group is attached to the methylene group carbon. The cross peak at (2.4, 15) is the reciprocity of this correlation. The methyl hydrogens also have a cross peak at (1.2, 144), so the methylene must be attached to that carbon. When looking down the column for the methylene, there is a cross peak to corroborate this as well as a cross peak at (2.4, 128). This indicates that beyond its attachment to the quarternary carbon is the aromatic system. The cross peaks under the 7.2 ppm peak confirm this.
Final structure: | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/07%3A_Two-Dimensional_NMR_Spectroscopy/7.06%3A_Interpreting_2-D_NMR_Spectra.txt |
Learning Objectives
• Solve unknown problems using a variety of spectra and the molecular formula.
Exercise \(1\)
Propose a structure using the spectral data below for C10H14O. Note: You may need to check for solvent peaks. This sample was dissolved in CDCl3
IR:
1H:
COSY:
NOESY:
13C:
HSQC:
HMBC:
Answer
Exercise \(2\)
Present an analysis of the following data and propose a structure.
MW: 131 amu
The full 1H NMR spectrum in D2O:
An expansion:
13C NMR:
COSY:
HMBC:
Answer
Example \(3\)
Propose a structure using the spectral data below for C10H14O. Note: You may need to check for solvent peaks. This sample was dissolved in CDCl3.
Mass Spectrum:
1H:
COSY:
13C:
HSQC:
HMBC:
7.S: Summary
7.1: Chapter Objectives and Preview of Correlation NMR Spectroscopy
• Complex molecules can be hard to elucidate solely with 1-D NMR because it doesn't quite solve the entire picture. This is where correlation NMR spectroscopy can be used.
7.2 Theory
• The splitting of resonances indicates that groups were "correlated" to each other due to the spins within each group.
• A simple 2-D experiment pulse sequence consists of a relaxation delay, a pulse, a variable time interval (t1), a second pulse, and acquisition (t2).
• In 2-D experiments, the signal detected during acquisition is a function of acquisition time (t2), which has been modulated as a function of the time interval (t1). This means that magnetization evolves around one frequency during t1 and a different frequency during t2.
• The output once Fourier transformed is a 2-D spectrum with two axes.
• One axis (v2) represents the nucleus detected during acquisition (t2), while the other axis (v1) can represent the same nucleus or a different nucleus.
• With two axes, it leads to cross peaks along a diagonal connecting coupled nuclei.
7.3 Two Dimensional Homonuclear NMR Spectroscopy
• Homonuclear 2-D NMR spectroscopy is looking at the correlation of the same nuclei in a molecule.
• COSY looks at 1H coupling to 1H through bonds typically 3 bonds away and relies on the J-coupling to provide spin-spin correlation to indicate which protons are close to each other.
• TOCSY obtains correlations between all protons within a given spin system and begin to chain together fragments of a molecule.
• A molecule can have just one spin system or hundreds in more complex systems.
• The goal in TOCSY is to transfer the magnetization beyond directly coupled spins.
• NOESY determines which signals arise from protons athar are close to each other in space, even if they are not bonded.
7.4 Two Dimensional Heteronuclear NMR Spectroscopy
• Heteronculear 2-D NMR is the correlation between different nuclei, such as a 1H to 13C.
• HSQC is used to determine the proton to carbon or heteroatom (often nitrogen) single bond correlations.
• The purpose of a HSQC is to determine which protons are coupled to what other specific carbon or heteroatom in the molecule through bonds.
• HMBC is used to determine long range 1H to 13C connectivity.
• HMBC gives the correlation between 1H and 13C when separated by two, three, and even four (if through a conjugated system) bonds away.
7.5 Uses of 2-DNMR Spectroscopy
• Complex structure elucidation often requires 2-D NMR spectroscopy.
• HSQC can be used to determine the profile of metabolites in low concentrations (microMolar) accurately.
• TOCSY has been utilized to show changes in tumor cells and identify biomarkers associated with these cells.
• Molecular dynamics can be studied using 2-D NMR spectroscopy to map the molecule's internal mobility patterns.
Skills to Master
• Skill 7.1 Distinguish between 1-D and 2-D techniques.
• Skill 7.2 Learn to read COSY, TOCSY, NOESY, HSQC, HMBC spectra.
• Skill 7.3 Know what type of experiment to use to gain the information needed.
• Skill 7.4 Understand applications of 2-D NMR spectroscopy.
• Skill 7.5 Solve unknown structure determination problems with 2-D spectroscopy.
8.01 Chapter Objectives and Preview
Learning Objectives
After completing this chapter, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• solve road-map problems which may require the interpretation of spectral data.
This chapter is devoted to using your skills obtained in the previous sections to practice structure determination. There are five structure elucidation problems in this chapter using a variety of different spectral data to help solve the structures.
8.02 Problem 1
Learning Objectives
• Determine the structure for the following problem.
Exercise \(1\)
Solve the structure using the following spectral data.
Molecular Formula: C6H12O3
IR Spectrum:
1H NMR Spectrum:
Integration: 2:1:2:2:2:3
Answer
8.03 Problem 2
Learning Objectives
• Determine the structure for the following problem.
Exercise \(1\)
Solve the structure using the following spectral data.
Molecular Formula: C8H7BrO
13C Spectrum:
1H NMR Spectrum:
Integration: 1:1:1:1:1:2 (left to right)
Chemical Shift Splitting Coupling Constant (Hz)
3.8 doublet 2
7 triplet of doublets 7 and 2
7.2 doublet of doublets 7 and 2
7.3 triplet of doublets 7 and 2
7.5 doublet of doublets 7 and 2
9.8 triplet 2
Answer
8.04 Problem 3
Learning Objectives
• Determine the structure for the following problem.
Exercise \(1\)
Solve the structure using the following spectral data.
Molecular Formula: C6H12O
IR spectrum:
13C Spectrum:
1H Spectrum:
Integration: 1:1:1:1:1:1:3:3
COSY:
Answer
8.05 Problem 4
Learning Objectives
• Determine the structure for the following problem.
Exercise \(1\)
Solve the structure using the following spectral data.
Mass Spectrum:
13C Spectrum:
DEPT-135 Spectrum:
1H NMR Spectrum:
Integration: 1:1:1:1:3:9
1H NMR Spectrum Zoomed in from 7 ppm to 8.3 ppm:
COSY:
HSQC:
Answer
8.06 Problem 5
Learning Objectives
• Determine the structure for the following problem.
Exercise \(1\)
Solve the structure using the following spectral data.
Mass Spectrum:
IR Spectrum:
1H NMR Spectrum:
Integration: 1:1:1:1:2:2:3
Zoomed in on Aromatic Region:
COSY:
HMBC:
Answer | textbooks/chem/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/07%3A_Two-Dimensional_NMR_Spectroscopy/7.07%3A_2-D_NMR_Problems.txt |
Wöhler synthesis of Urea in 1828 heralded the birth of modern chemistry. The Art of synthesis is as old as Organic chemistry itself. Natural product chemistry is firmly rooted in the science of degrading a molecule to known smaller molecules using known chemical reactions and conforming the assigned structure by chemical synthesis from small, well known molecules using well established synthetic chemistry techniques. Once this art of synthesizing a molecule was mastered, chemists attempted to modify bioactive molecules in an attempt to develop new drugs and also to unravel the mystery of biomolecular interactions. Until the middle of the 20th Century, organic chemists approached the task of synthesis of molecules as independent tailor made projects, guided mainly by chemical intuition and a sound knowledge of chemical reactions. During this period, a strong foundation was laid for the development of mechanistic principles of organic reactions, new reactions and reagents. More than a century of such intensive studies on the chemistry of carbohydrates, alkaloids, terpenes and steroids laid the foundation for the development of logical approaches for the synthesis of molecules.
The job of a synthetic chemist is akin to that of an architect (or civil engineer). While the architect could actually see the building he is constructing, a molecular architect called Chemist is handicapped by the fact that the molecule he is synthesizing is too small to be seen even through the most powerful microscope developed to date. With such a limitation, how does he ‘see’ the developing structure? For this purpose, a chemist makes use of spectroscopic tools. How does he cut, tailor and glue the components on a molecule that he cannot see? For this purpose chemists have developed molecular level tools called Reagents and Reactions. How does he clean the debris and produce pure molecules? This feat is achieved by crystallization, distillation and extensive use of Chromatography techniques. A mastery over several such techniques enables the molecular architect (popularly known as organic chemist) to achieve the challenging task of synthesizing the mirade molecular structures encountered in Natural Products Chemistry, Drug Chemistry and modern Molecular Materials. In this task, he is further guided by several ‘thumb rules’ that chemists have evolved over the past two centuries. The discussions on the topics Name Reactions, Reagents for synthesis, Spectroscopy and Chromatography are beyond the scope of this write-up. Let us begin with a brief look at some of the important ‘Rules’ in organic chemistry that guide us in planning organic synthesis. We would then discuss Protection and Deprotection of some important functional groups. We could then move on to the Logic of planning Organic Synthesis.
2.01: Baldwin’s Rule for Ring Closure Reactions
There are a few rules that provide guidelines for planning strategies in organic synthesis. These rules and guidelines have come from the keen observations of chemists after looking into several examples from their own research and other published work in the literature. These observations are to be treated as thumb-rules to be applied with caution. They may not be applicable for all situations. Nonetheless, they serve as guidelines to avoid pit-falls in planning. Since all these rules are governed by the underlying principles of mechanistic organic chemistry and stereochemistry, these basic mechanistic principles are the touchstones against which the conclusions reached are to be tested. Most of the rules that are useful in planning synthesis are collected here for convenience.
• 2.1: Baldwin’s Rule for Ring Closure Reactions
J.E. Baldwin proposed a set of rules for ring closure reactions. He suggested that the rules are applicable to reactive intermediates as well and supported his views with several examples from literature and special experiment designed to test the validity of the rules.
• 2.2: Bredt's Rule
Bredt's Rule states that bridged ring systems cannot have a double bond at the bridgehead position. Bretd’s Rule cautions us on the type of rings that could bear a double bond.
• 2.3: Cram's Rule and Prelog's Rule
Cram defined a Reactive conformation, as the least energy conformation in which the chemical reaction takes place. An extension of Cram's idea of reactive conformation to chiral esters of α -ketoesters(pyruvates) is the Prelog's Rule reported in 19533. It generally relates to Grignard addition to chiral pyruvates made using chiral alcohols .
• 2.4: Hofmann’s Rule and Zaitsev’s Rule
In reactions like Hofmann’s Exhaustive Methylation – Elimination reactions, the least substituted olefin is generally formed as a major product. This is called the Hofmann’s Rule. All such reactions bear charged leaving groups like –NR3+ or –SR2+ and involve strong bases. The Zaitsev’s Rule draws our attention to the alternate possibility. On elimination of HX, the more stable olefin is obtained.
• 2.5: Markovnikov Rule
Polar addition of H+X¯ to olefins proceed in such a way that the negative component adds to the more stable carbonium ion intermediate .
02: Rules and Guidelines Governing Organic Synthesis
J.E. Baldwin proposed a set of rules for ring closure reactions. He suggested that the rules are applicable to reactive intermediates as well and supported his views with several examples from literature and special experiment designed to test the validity of the rules.
The ring closure reaction of a reactive intermediate could be one of the three types. An attack could be on a triple bond center (called Digonal center, Dig-), a double bond center (called Trigonal center, Trig-) or at a single bond center (called Tetrahedral center, Tet-). The attacking species could be a carbanion, a carbonium ion or a free radical. The attack could be endo- or exo-. With respect to the newly developing ring in the transition state, when the pair of electrons in the displaced bond is exo- to the developing ring, the transition state is described as exo- attack. When they form part of the newly developing ring (or transition state) the system is described as endo- attack (Fig 2.2.1).
For each pair of reactive center, there could be two modes of attack – endo- or exo- as shown for the attack of an anion (Fig 2.2.2).
Reactions take place only when the orbitals concerned overlap efficiently. For this purpose, Baldwin suggested the following geometry (Fig 2.2.3). In a ring formation reaction, the optimal geometry could be achieved only when the length of the chain (a tether of atoms) connecting the reactive centers have a minimum optimal length.
Based on this criterion, Baldwin suggested the following rules. The number (3 to 7) in the nomenclature refers to the number of atoms in the chain that leads to the proposed cyclic transition state.
Baldwins Rules
1. All Exo-Tet reactions are favored reactions
2. All Endo-Tet are disfavored reactions
3. All Exo-Trig reactions are favored reactions
4. 3 to 5-Endo-Trig reactions are disfavored reactions
5. 6 and 7-Endo-Trig reactions are favored reactions
6. 3 to 7-Endo-Dig reactions are favored reactions
7. 3 to 4-Exo-Dig reactions are disfavored reaction
8. 5 to 7-Exo-Dig are favored reactions
These complex rules are simple to apply, but difficult to remember without a suitable ‘memory aid’. E. Juaristi 2 (in his web page http://www.relaq.mx/RLQ/EusebioJuaristi_vitae.htm) suggests the following mnemonics for all the Disfavored Reactions (the Stop sign). A modified version is presented here (Fig 2.2.4). Note that the numbers are progressively increasing. The STOP sign and the wagon would probably be easier to remember.
An alternate summary of the Baldwin’s Rules is provided by Clayden et. al., in their inimitable text book. This table is now easy to remember (Fig 2.2.5). You have to remember 5-Endo-trig and 4-Exo-Dig as key points for Disfavored reactions.
Baldwin cited several examples in support of these rules. Scientists soon attempted to validate the proposed rules. Steric and electronic factors appear to modify these conclusions. In most of the studies, the rules were generally applicable. Let us look at the following interesting study (Fig 2.2.6). A 6-endo-Tet reaction should be disfavored. Such reactions do not form a ring, but appear to pass through a cyclic transition state leading to the cleavage of a sigma bond, while a new sigma bond is formed.
Using an ingenious double labeling experiment shown below, Eschenmoser clearly established that this reaction is intermolecular (Fig 2.2.7). He took an equimolar mixture of hexadeutero- and normal starting materials and conducted the reaction. If the reaction were only intramolecular, the product would be both hexadeutero- and no deutero- products only. The actual product distribution was as shown in Figure 2.2.7, which is in keeping with intermolecular mechanism only.
The constraint in such cases lies obviously in the length of the chain bearing the reactive groups. The angle of attack of 1800 is not attainable in a six-membered ring. King et al.,showed that as the length of the chain (tether) increased, intramolecular reaction became more feasible as shown in Figure 2.2.8. An intramolecular reaction became feasible only when the teather allowed a ten-membered ring transition state.
Another Disfavored cyclization is 5-Endo-Trig. The following Michael-type cyclization is of considerable interest (Fig 2.2.9). The base catalyzed intramolecular addition does not proceed as expected. However, the acid catalyzed reaction proceeds. A close look at an alternate mechanism suggests that a 5-Exo-Trig path becomes available under acid catalyzed conditions .
Baldwin sites three closely related reactions (Fig 2.2.10) to support his rules. The oxygen analogue failed to cyclize as anticipated for 5-Endo-Trig reactions. However, the thiol analogue cyclized readily to give the thiophene ring. This is attributed to the fact that the larger atom requires entirely different bond angles not envisaged in Baldwin’s Rules. The nitrogen analogue investigated preferred an alternate path. Why does the nitrogen prefer amidation and not Michael-type cyclization? Could it be due to an unreactive olefin? This question was answered through an intermolecular ‘control reaction’ shown in Fig 2.2.11. The amine moiety preferred a Michael-type addition to amidation.
This lack of cyclization was therefore attributed to the fact that the lone pair cannot attack the pi-orbitals due to wrong orientation in one conformation and the distances involved in the other conformation (see Figure 2.2.12).
Applicability of Baldwin’s 5-Endo-Trig restriction was verified through a carefully planned retro-reaction (Fig 2.2.13). The Deuterium labeling experiment proved the fact that anion formation was effective. Lack of cleavage reaction proved that the above conclusions (5-Endo-Trig restriction) are valid in reverse reactions as well.
The fact that 5-Exo-Trig is favored is seen in several reactions. The fact that 6-Endo-Trig reactions are favored is also well documented (Fig 2.2.14).
The 5-Endo-Dig cyclization looks awkward on paper (Fig 2.2.15). But the reaction proceeds to completion. A close look at the orbitals concerned explains the reaction. The orientation of one of the pi-orbitals is just right for the reaction to occur. Compare this with the orientation shown in Figure 2.2.12.
How about the acetal formation reaction of ketones with ethylene glycol (Fig 2.2.16)? Is this a case of failure of Baldwin’s Rules as in several carbonium ion reactions?
These rules are generally applicable. However, there are several exceptions as well. A proper approach would be to use these ‘thumb rules’ as a guideline while planning synthetic schemes and not use them as inviolable rules. The following points may be kept in mind while applying these rules.
1. The rules suggest only the ‘favored paths’. This expression ‘favored / disfavored’ should not be read as ‘allowed / disallowed’. Under suitable conditions, the alternate high-energy path may be opened, either partially or exclusively.
2. When a large atom from the Second Group of the periodic table is involved, the angle requirements of these atoms may vary. In such cases, the rules may not be applicable. See Fig 2.2.10 for one such example.
3. Special modified Baldwin’s Rules have been evolved for enolate anions. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/01%3A_Synthesis_of_Organic_Molecules.txt |
In its original form, Bredt's Rule stated that bridged ring systems (like camphane (Figure 1.2.1.1A) and pinane (Figure 1.2.1.1B)) (Fig 1.2.1.1) cannot have a double bond at the bridgehead position (the points marked by bold dots in structures ‘A’ and ‘B’). This rule came from observations on dehydration of alcohols in these ring systems. When you look at these molecules carefully, you could see that the bridged rings are made out of a larger ring (shown by thick lines in the Figure s) bearing a bridge at specified points. Most of the rings studied by Bredt had six-membered ring as the largest ring. The constraint dictated by Bredt's Rule has now been attributed to the fact that the small and common rings can accommodate only a cis- double bond. A bridgehead olefin demands a trans- geometry at the olefin. Hence the rule was applicable to almost all naturally occurring bridged ring systems known at that time.
The scope of this rule has been investigated in detail. Medium sized rings are large enough to accommodate a trans- double bond. Hence the bicyclic rings 1.2.1.1F and 1.2.1.1G, bearing a cyclooctane ring as the outer ring, were synthesized and were indeed found to be stable. On the other hand, the isomeric cycloheptene ring system 1.2.1.1H was unstable. Faweet (1950) suggested that the S value, which is a summation of the numbers found in the nomenclature (m + n + o = S), would determine the stability of the ring system. Bicyclic ring systems with a bridgehead double bond having S value less than 9 would be highly strained. As the S value increases, the strain decreases.
Bretd’s Rule cautions us on the type of rings that could bear a double bond. When a synthetic intermediate or a transition state in a mechanism demands such an intermediate, one should exercise caution on the position of the olefin. For example, Prelog (1948, 1949) attempted an aldol condensation on the ring systems shown (Fig 1.2.1.2). When n = 5, both products bicyclo[6,4,0]alkene (S = 10) (C), and
Fig 1.2.1.2
bicyclo[5,3,1]alkene (S=9) (B) were formed. On the other hand, when n=>6, the main product was a bicyclo[6,3,1]alkene system (B) (S=10). When n = <5 product B was not observed. Note that these models were based on reversible aldol condensation reactions (equilibrium reactions) and therefore correspond to the thermodynamic stability of the product. Under forced conditions however, the rule may not hold, as seen in the following example (Fig 1.2.1.3).
Fig 1.2.1.3
2.03: Cram's Rule and Prelog's Rule
A carbonyl carbon is sp2 hybridized. This means that the carbonyl carbon and the three other atoms attached to it would be in one plane. A nucleophile attacking the carbonyl carbon in molecules like 1A in Figure 1.2.1 could attack from either side of this plane with equal ease. Have a close look at the aldehyde shown in Figure 1.2.1.The phenyl group is also flat. Nucleophilic attack on benzaldehyde could take place from either side with equal ease. Since a new asymmetric center is now created by this reaction, both enantiomers could be formed with equal ease, resulting in a racemic mixture.
In complex organic molecules, this seldom happens. When the $\alpha$ - carbon is asymmetric (see 1B), the nucleophile would experience more steric hindrance from one side, leading to unequal synthesis of the two enantiomers. This fact was first recognized by D. Cram in 1952. After analyzing several published reactions, he put forth an explanation based on one proposed conformation. He set forth to first define a Reactive conformation, as the least energy conformation in which the chemical reaction takes place. Thus, for 1B there could be several conformations, out of which chemical reactions prefer to proceed via the conformation having the least steric strain (Fig 1.2.2).
Studies in Stereochemistry. X. The Rule of "Steric Control of Asymmetric Induction" in the Syntheses of Acyclic Systems Donald J. Cram, Fathy Ahmed Abd Elhafez, J.Am. Chem. Soc. 74, 5828(1952).
Cram's Rule
• The existing asymmetric center would have a Small, Medium and Large group, denoted S,M and L respectively.
• In the reactive conformation, the carbonyl group would orient itself in such a way that it will rest between the Small group and the Medium group.
• The attacking nucleophile would prefer to attack from the side of the small group, resulting in the predominant formation of one diastereomer in the product.
This is now known as Cram's Rule. When the starting material is a pure enantiomer, the product mixture would show predominance of one enantiomer. He supported his argument with a set of his own experiments shown in Figure 1.2.3
The predominance of one diastereomer in such reactions could be explained on the basis of Cram's Rules. This rule proved to be effective in predicting the major product in most of the asymmetric syntheses. However, several exceptions were soon observed. They were explained on the basis of further conformational arguments.
Felkin pointed out that the reactive conformations of Cram had severe eclipsing strain between R group on carbonyl center and L group on the $\alpha\!$ - carbon (see 2B in Figure 1.2.2). He progressively increased the bulk of the R group and observed that the diastereomeric excess could swing from erythro 74% to threo 96% as the bulk increased (Fig 1.2.4).
Torsional strain involving partial bonds. The stereochemistry of the lithium aluminium hydride reduction of some simple open-chain ketones. Marc Cherest, Hugh Felkin and Nicole Prudent Tetrahedron Letters, 9, 2199 (1968).
He proposed a staggered conformation, wherein the nucleophile approached the carbonyl carbon from the least hindered side. The major difference in these two approaches is depicted in Figure 1.2.5.
Figure 1.2.5
The flaw in the Felkin's model is that it is not true for aldehydes, where R=H. An improvement over Felkin's model was the Felkin-Nguyen (Felkin-Anh) model, which suggested that the nucleophile would attack the sp2 center of the carbonyl at 950 to 1050 angle relative to the carbonyl bond axis, favoring an attack from the least hindered direction (Figure 1.2.6). This differed from the earlier models that suggested a perpendicular attack.
Figure 1.2.6
What happens when the $\alpha\!$- carbon bears atoms like 'O','N' or 'S'? The first thought would be that the nucleophile would prefer the side opposite to the electronegative group. This would lead to a Felkin-Anh transition state. This would be true only when the reagent does not have additional complexing site. For example, in sodium borohydride the sodium ion cannot from a bivalent complex. On the other hand, zinc borohydride could form a divalent chelation complex between the carbonyl oxygen and the electronegative atom. This is called Cram-chelation complex, which leads to a Cram-chelate product. The three arguments discussed so far are shown in Figure 1.2.7.
Figure 1.2.7
Two examples are shown in Figure 1.2.8 to depict that Cram's chelation control could lead to a reversal in selectivity. Notwithstanding these refinements to the original Crams's Rule, the fact remains that this line of extending conformational arguments to reactive conformations suggested by Cram has resulted in greater understanding of nucleophilic additions to aldehydes and ketones.
Figure 1.2.8
Prelog's Rule
An extension of Cram's idea of reactive conformation to chiral esters of $\alpha$-ketoesters(pyruvates) is the Prelog's Rule reported in 19533. It generally relates to Grignard addition to chiral pyruvates made using chiral alcohols (Fig 1.2.9).
Figure 1.2.9
The rule has been applied for asymmetric synthesis of $\alpha$-hydroxyacids and for assigning the configuration of secondary and tertiary alcohols. The anti configurational arrangement of the two $\alpha$-carbonyl moieties could be rationalized. The negative end of these dipoles would prefer to be as far removed as possible. The two lone pairs would sit on ether oxygen like the 'Rabit Ears'. The keto-carbonyl would orient between the two ears. This will place the bonds shown in red in the same plane as thr keto-carbonyl group. The attack from the side of the small (S) group is an extension of Cram's Rules. The asymmetric induction could be at times poor due to the large distance between the reaction center and the asymmetric center inducing asymmetry at the developing chiral center.
2.04: Hofmann’s Rule and Zaitsev’s Rule
In reactions like Hofmann’s Exhaustive Methylation – Elimination reactions, the least substituted olefin is generally formed as a major product. This is called the Hofmann’s Rule. All such reactions bear charged leaving groups like –NR3+ or –SR2+ and involve strong bases. The Zaitsev’s Rule (or Saytzeff rule) draws our attention to the alternate possibility. On elimination of HX, the more stable olefin is obtained (Fig 2.3.1). The apparent contradiction in this set of rules is easily resolved through a critical look at the mechanisms involved in these two sets of reaction conditions.
There could be two reasons for such preferences. Ingold (1960) and Bunnett (1969) suggested that a positively charged leaving group increases the acidity of the β-protons. A substituent at the β-position could hyperconjugatively decrease the acidity of the β-proton. Consequently, a terminal methyl group (this has no alkyl substituent) is more acidic than the internal methine proton (bearing at least one alkyl substituent). When the leaving group is a halogen, the mechanism shifts towards E1. Under these conditions, the stability of the developing double bond becomes important and this leads to the thermodynamically more stable product. The school of H.C Brown had suggested (1956) that steric factors govern such elimination reactions. The charged leaving groups are large compared to neutral leaving groups.
The larger leaving groups like –NR3+ and –SR2+ give more Hoffmann product than smaller groups like halogens. The bulkiness of the base also increases the Hoffmann product at the cost of the Zaitsav product. The situation appears to be more complex. When the base strength was increased without increasing the bulk at the reaction site (X-C6H4-O¯), the Hoffmann product increased at the cost of the Zaitsav product (Froemsdorf (1966,67)). This suggests an E1cB mechanism, where the acidity of the β proton is important. Thus the mechanism (and therefore the products composition) could be altered by factors such as the size of the leaving group, size of the base, nature of the leaving group and the strength of the base.
2.05: Markovnikov Rule
Polar addition of HBr to olefins proceeds in such a way, the carbon that is rich in hydrogen becomes richer. In mechanistic terms, we could restate the rules as follows: Polar addition of H+X¯ to olefins proceed in such a way that the negative component adds to the more stable carbonium ion intermediate (Fig 2.4.1).
HBr and HI easily undergo free radical addition, promoted by light or heat. Free radical additions give the anti-Markovnikov product. Clean Markovnikov products are obtained when such reactions are carried out in polar solvents and care is taken to avoid light. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/02%3A_Rules_and_Guidelines_Governing_Organic_Synthesis/2.02%3A_Bredt%27s_Rule.txt |
Once a Target Molecule is chosen for synthesis, one could sit down and device several routes for its synthesis. On what criteria do you select a Target and how do you arrive at a synthetic route? The answer depends on the overall goal of your project.
For a natural product chemist, he might have isolated and determined the structure of a new molecule. He may need to synthesize the molecule to prove the structure. While working on structure elucidation, the route chosen for synthesis should unambiguously establish the part of the structure you are working on. Each step is chosen on this structure-criterion alone. Here the length of the route and the cost of the chemicals are not important.
An elegant (enantiopure) synthesis of some complex structure is the dream of an academician working in the university laboratory. He is often more concerned with developing new routes, new reactions and new mechanistic principles. His concern is to develop new horizons and give good training to the young chemists. He is seldom worried about the cost of his research. He has time at his command and hopefully enough money to pursue his passion. He is judged by the quality (and quantity) of his research publications and the quality of training he has imparted to the students. Having a patent is an added feather to his cap.
A pharmaceutical chemist and a material chemist are more interested in developing versatile and fast synthetic routes for a chosen molecule. Their efforts are directed towards the synthesis of a large number of closely related molecules, within a short time. Such a chemist is looking into Structure-Activity Studies, aimed at developing new drug molecules or molecules with special properties. He is often judged by the number of such active molecules that he has discovered and patents held in his name and not just by the number of new molecules synthesized by him or the elegance of the synthetic route. A publication to his credit is an added feather to his cap. In his endeavours, the cost and the efficiency of the synthetic routes are not the criteria for research. He believes that once the ‘right molecule’ is discovered, more efficient routes could always be generated at a suitable date. His art is directed towards fast discovery of molecules with the right properties.
An industrial chemist is most concerned with the ‘cost’ of synthesis of the molecule. His efforts are directed towards development of economical synthetic procedures, which includes not only the cost of the chemicals but also the cost of waste treatment, recycling and environmental cleaning. He selects the molecules on the basis of their economic value (net profit for his company). He should be concerned about eco-friendly reactions and procedures. In general, a development chemist in an industrial R&D laboratory looks at very large-scale (typically several kilogram batch) reactions, their reproducibility, safety and cost parameters. His focus is on the commercial value of his product, the profitability and the patents.
Therefore, the target molecule and the route chosen depend of the hat the chemist is wearing. We do have examples of amazing chemists who efficiently juggle with more than one hat at the same time. We take our proverbial hats off for those versatile and multifaceted chemists. This is because, for an efficient operation, different tasks demand different skill-sets and use of different sets of databases. Nonetheless, the underlying chemistry is same in all these activities. Of course, occasionally one could be creative, versatile and cost-effective at the same time. In the following pages we would look into the broad principles governing the art of organic syntheses. We have already discussed some ‘rules’ that govern synthesis. We would now explore guidelines that concern the logic of organic syntheses. We would then discuss several interesting syntheses to illustrate these principles. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/03%3A_Criteria_for_Selection_of_the_Synthetic_Route.txt |
‘There could be ART in Organic Synthesis’ declared the inimitable monarch of organic synthesis, Professor R.B. Woodward. His school unveiled several elegant approaches covering a variety of complex structures and broke new grounds to define the art of organic synthesis. ‘If organic synthesis is a branch of science, what is the LOGIC of organic synthesis?’ marveled several others. The development of the concept of logical approaches towards synthesis has been evolving over the past several decades. A few stalwarts focused their attention on this theme and attempted to evolve a pattern to define this logic. There is no doubt that all of us who dabble with synthesis contribute our small bit in the magnificent direction. A few names stand out in our minds for their outstanding contributions. Notable contributions came from the schools of J.A. Marshal, E.J. Wenkert, G. Stock, S Hanessian, E.E. van Tamalen, S. Masamune, R.B. Woodward, E.J. Corey and several others. More focused on this theme were the contributions from the school of E.J. Corey.
The period 1960 – 1990 witnessed the evolution of this thought and the concept bloomed into a full-fledged topic that now merits a separate space in college curriculum. Earlier developments focused on the idea of ANTITHETIC APPROACHES and perfected the art of DISCONNECTION via RETROSYNTHESIS. This led to logical approaches for the construction of SYNTHETIC TREES that summarized various possible approaches for the proposed Target structure. All disconnections may not lead to good routes for synthesis. Once the synthetic tree was constructed, the individual branches were analyzed critically. The reactions involved were looked into, to study their feasibility in the laboratory, their mechanistic pathways were analyzed to understand the conformational and stereochemical implications on the outcome of each step involved and the time / cost factors of the proposed routes were also estimated. The possible areas of pitfall were identified and the literature was critically scanned to make sure that the steps contemplated were already known or feasible on the basis of known chemistry. In some cases, model compounds were first constructed to study the feasibility of the particular reaction, before embarking on the synthesis of the complex molecular architecture. Thus a long process of logical planning is now put in place before the start of the actual synthetic project. In spite of all these careful and lengthy preparations, an experienced chemist is still weary of the Damocles Sword of synthesis viz., the likely failure of a critical step in the proposed route(s), resulting in total failure of the entire project. All achievements are 10% inspiration and 90% perspiration. For these brave molecular engineers, sometimes also called chemists, these long-drawn programs and possible perils of failures are still worth, for the perspiration is enough reward.
A sound knowledge of mechanistic organic chemistry, detailed information on the art and science of functional group transformations, bond formation and cleavage reactions, mastery over separation and purification techniques and a sound knowledge of spectroscopic analysis are all essential basics for the synthesis of molecules. A synthetic chemist should also be aware of developments in synthetic strategies generated over the years for different groups of compounds, which include Rules and guidelines governing synthesis. Since organic chemistry has a strong impact on the development of other sister disciplines like pharmacy, biochemistry and material science, an ability to understand one or more of these areas and interact with them using their terminologies is also an added virtue for a synthetic chemist. With achievements from synthesis of strained molecules (once considered difficult (if not impossible) to synthesize, to the synthesis of complex, highly functionalized and unstable molecules, an organic chemist could now confidently say that he could synthesize any molecule that is theoretically feasible. This is the current status of the power of organic synthesis. Based on the task assigned to the chemist, he would select a Target molecule for investigation and devise suitable routes for synthesis.
Protection and Deprotection Strategies in Organic syntheses
For the manipulation of functional groups and formation of new covalent bonds we make use of a large number of Reagents and Name Reactions. In complex organic syntheses, the starting materials and intermediates in the synthetic scheme often have more than one reactive functional group. A few such multifunctional building blocks are shown below to illustrate this point (Fig 4.1.1 ). While working on such complex
molecules, it is often necessary to protect some groups to enable selective working at the desired locations only. Organic chemists have heavily relied on such protection / deprotection strategies and have diligently developed protecting (masking) and deprotection (unmasking) protocols. We would discuss some of the important protecting groups in this chapter.
Before proceed further, it must be emphasized here that this protocol should be applied only after alternate options have been critically analyzed. This is because protection / deprotection strategy involves an increase of at least two more critical steps, adding to the length of the synthesis and consequent drop in overall yields of the desired compound. In large-scale reactions, this leads to a huge impact to the Atom Economy and pollution cost of the synthetic process. All this translates into an increase in the overall cost of the final drug molecule.
Protection Strategies
Group / Site Selective Reagent: Protection / deprotection is not always required whenever you see a multiplicity of functional groups. You could solve the selectivity issue by using site selective reactions / reagents. By choosing an appropriate selective reagent to suit the scheme on hand, you could selectively attack only one of the reactive sites. Consider an olefinic ketone (Fig 4.1.2). Sodium borohydride reduction in methanol as solvent could selectively reduce the keto- group to a secondary alcohol
leaving the olefin undisturbed. On the other hand, diborane reagent in THF as solvent would be a reagent of choice when the selective reduction at the olefin moiety is desired. Diborane reduction of an olefin is several times faster than reduction of ketones. The oxidative cleavage of borane product is also selective. Thus, you can avoid the protection / deprotection strategy by employing a selective reagent. In C – C bond formation reactions we come across several such site-selective reagents. One such reagent widely used in research is the Wittig reagent. They attack the aldehyde or ketone selectively in the presence of ester, nitrile. olefin etc..
Selective Protection
In the case of a molecule like 4.1.3A (Fig 4.1.3) bearing an olefin and a carboxylic acid, the –COOH group is several times more reactive than the olefin towards diborane reduction.
Hydroboration / oxidation reduces the acid to a primary alcohol, leaving the olefin unaffected. On the other hand, if you need a selective reduction of olefin, the acid group has to be processed through a selective protection / deprotection sequence as shown in (Fig 4.1.3)
Compound 4.1.4A illustrates several important points in Protection / Deprotection protocol. Both the functional groups could react with a Grignard Reagent. Carboxylic acid group would first react with one mole of the Grignard Reagent to give a carboxylate anion salt. This anion does not react any further with the reagent. When two moles of Grignard Reagent are added to the reaction mixture, the second mole attacks the ketone to give a tertiary alcohol. On aqueous work-up, the acid group is regenerated. Thus, the first mole of the reagent provides a selective transient protection for the –COOH group. Once the acid group is esterified, such selectivity towards this reagent is lost. The reagent attacks at both sites. If reaction is desired only at the ester site, the keto- group should be selectively protected as an acetal. In the next step, the grignard reaction is carried out. Now the reagent has only one group available for reaction. On treatment with acid, the ketal protection in the intermediate compound is also hydrolyzed to regenerated the keto- group.
Orthogonal Protection or Differential Protection
Orthogonal protection is a strategy that allows deprotection of multiple protective groups one at a time, each with a dedicated set of reagents and / reaction conditions without affecting the other. This technique is best illustrated with peptide bond formation and associated deprotection reactions. An amino acid has two functional groups –NH2 and –COOH. When two amino acids (A and B) react under conditions for the peptide bond condensation reaction, a mixture of 4 dipeptides (at least) could be formed as shown below.
$\ce{ A + B \rightarrow A-A + A-B + B-A + B-B}$
If we are interested in only one product A – B, we have to do selective protections and selective deprotections in a proper sequence. Consider the following peptide bond formation reaction.
In order to get only one product A – B, we should protect the N – terminal of ‘A’ and C – terminal of ‘B’. Let us look closely at two different dipeptide formation schemes. In the following sequence, the C – terminal is protected in two different ways for one amino acid. For the second amino acid, the N – terminal is protected with an acid labile Boc- protection.
In the next step, the two monoprotected amino acids are coupled as shown below.
Take a close look at both the products. In the first product, both protections are acid sensitive. If the final product desired is the protection-free dipeptide, this is indeed a short route.
If the desired product is a mono-protected dipeptide, then selective deprotection is the preferred reaction. This is feasible only when we use starting compounds that are differentially protected. This is called Orthogonal Protection.
Similar techniques are available for other functional groups as well. Let us now learn more on Protection / Deprotection for some important functional groups.
Protection of R – COOH Group
In the introduction, we have seen that carboxylate ion lends protection to an attack of Grignard reagents at this carbonyl carbon. However, this is not sufficient for a vast variety of reagents. Meyer’s 2-oxazolines mask an acid function while activating the α- position for lithiation reaction. The use of this group as protection for –COOH group is rare.
Protection of Aldehydes and Ketones
Since alcohols, aldehydes and ketones are the most frequently manipulated functional groups in organic synthesis, a great deal of work has appeared in their protection / deprotection strategies. In this discussion let us focus on the classes of protecting groups rather than an exhaustive treatment of all the protections.
Acetals
There are two general methods for the introduction of this protection. Transketalation is the method of choice when acetals (ketals) with methanol are desired. Acetone is the by-product, which has to be removed to shift the equilibrium to the right hand side. This is achieved by refluxing with a large excess of the acetonide reagent. Acetone formed is constantly distilled. In the case of cyclic diols, the water formed is continuously removed using a Dean-Stork condenser (Fig 4.1.6).
The rate of formation of ketals from ketones and 1,2-ethanediol (ethylene glycol), 1,3-propanediol and 2,2-dimethyl-1,3-propanediol are different. So is the deketalation reaction. This has enabled chemists to selectively work at one center. The following examples from steroid chemistry illustrate these points (Fig 4.1.7).
The demand for Green Chemistry processes has prompted search for new green procedures. Some examples from recent literature are given here (Fig 4.1.8).
Thioketals
Compared with their oxygen analogues, thioketals markedly differ in their chemistry. The formation as well as deprotection is promoted by suitable Lewis acids. The thioacetals are markedly stable under deketalation conditions, thus paving way for selective operations at two different centers. When conjugated ketones are involved, the ketal formation (as well as deprotection) proceeds with double bond migration. On the other hand, thioketals are formed and deketalated without double bond migration (Fig 4.1.9).
Protection of Amino groups (-NH2 &–NH)
N-Acetyl (N – COCH3), N – Benzoyl (N – COPh) Protections
These are the classical protecting groups for primary and secondary amines. The reagents are cheap and the protocol is simple. Such amides generally need drastic conditions for deprotection, though the yields are generally good (Fig 4.1.10). A standard procedure is refluxing in aqueous alkali or aqueous mineral acid. Due to the drastic conditions, care should be exercised in this procedure to ensure racemi zation is avoided. Amides are generally crystalline solids that are easily purified by crystallization. When the protection is introduced at the early stages of a long synthetic scheme and a very stable protection is desired (as in nucleotide synthesis) an amide is the most preferred protection.
Several more labile amide bonds have been investigated. The amides of trifluoroacetic acid are of special interest. The introduction as well as cleavage is simple and mild {Fig 4.1.11).
A recent report in amide hydrolysis is given below.
N – Phthaloyl Protection (N – Pht)
Mechanism for NaBH4 Reduction of N – Pht
N – Carboxylic acid Esters as protective groups
As described above, the amide bonds are very strong. On the other hand, the ester bonds are easily cleaved by mild base conditions. A carboethoxy protection on amine has an amide bond as well as an ester bond. Since N – COOH groups obtained on hydrolysis are very unstable, this protection provides a large family of protective groups for primary and secondary amines.
N – Carboethoxycarbonly (N – COOEt) and Carbobenzyloxycarbonyl (N – COOCH2Ph) (N – Cbz or N – Z) Protections:
These groups are easily introduced using the corresponding chloroformate esters. Anhydrides or mixed anhydrides under mild basic conditions. Both these protections could be removed under prolonged stirring with base at room temperature. Though mild, some racemisation is sometimes observed. The N – Cbz protection has an added advantage in that it could be easily cleaved under hydrogenolysis conditions (Fig 4.1.14). N – Cbz Protection is however stable to acidic conditions. Compare this with –Boc protection discussed below.
Tert-Butyloxycarbonyl Protection ( N – COOBut, N – Boc)
The Tert-Butyloxycarbonyl Protection could be introduced and removed under very mild acid conditions. This protection is stable to alkali and hydrogenolysis (Fig 4.1.15). Thus, N – Z and N – Boc are complimentary as protective groups.
N – Fluoromethyleneoxycarbonyl Protection (Fmoc)
This UV active protecting group is very popular in Solid Phase Peptide Synthesis (SPPS) protocols. Protection as well as deprotection steps proceed under mild conditions in good yields (Fig 4.1.16).
The mechanism for Fmoc deprotection is shown in (Fig 4.1.17)
N – Silylation
Silylation is a common protection for active hydrogen on heteroatoms. In the case of N – Si bond, quaternary ammonium fluorides cleave this bond (Fig 4.1.18).
N – Tosylation (N – Tos)
This protection is very stable. N – Tosylation is easily carried out through acid chloride procedure. It is cleaved by solvated electron cleavage reaction. When this group is attached to a primary amine, the –NH group becomes very acidic (Fig 4.1.19).
Protection of – OH Groups
Acetates ( – Ac) and benzoates (– Obz)):
The – OH group protection chemistry has been extensively investigated. The classical protection is the formation of esters of aliphatic and aromatic carboxylic acids. Aromatic esters are comparatively difficult to hydrolyze under mild base condition. This provides an opportunity for selective deprotection protocols (Fig 4.1.20). Note that this protection is sensitive to acid as well as base conditions.
Methyl (– OMe) and Benzyl (R – OBn) Ethers
An ether group is one amongst the most stable functional groups. Hence, this group has been the most favored protecting group. Deprotection was a problem. In the early part of the twentieth century, the only procedure was refluxing with aqueous HI or HBr. In recent years several new procedures have appeared for effective removal under mild conditions. The special feature of the benzyl ethers is that this protection is readily removed under neutral hydrogenolysis conditions (Fig 4.1.21). Substituents like – OMe or – NO2 could be introduced on the benzene ring to modify the reactivity at the protection site.
When an olefin could compete at the hydrogenolysis procedure, the following sequence appears to be an alternate procedure (Fig 4.1.22).
Allyl ether is a recent introduction in – OH protection. The versatility of this protection could be seen in the following examples (Fig 4.1.23).
Silyl Ethers (R – OSiR3)
The oxygen – silicon sigma bond is stable to lithium and Grignard reagents, nucleophiles and hydride reagents but very unstable to water and mild aqueous acid and base conditions. A silyl ether of secondary alcohol is less reactive than that of a primary alcohol. The O – trimethylsilyl (O – SiMe3) was first protection of this class. (Fig 4.1.24).
Replacement of methyl group with other alkyl and aryl groups gives a large variety of silyl ether with varying degrees of stability towards hydrolysis (Fig 4.1.25).
The following examples illustrate the selectivity in formation and hydrolysis of this group (Fig 4.1.26).
Tetrahydropyranyl ether (– OTHP) and Tetrahydrofuranyl ether (– OTHF)
These protective groups for alcohols are in fact acetals. They are synthesized using the dihydropyran (DHP) and dihydrofuran (DHF) respectively. They behave like acetals in their stability and cleavage (Fig 4.1.27). The rate of formation and cleavage for these two groups differ, which finds application for differential protection of alcohols.
These protective groups found extensive use in synthesis. However, two major drawbacks were soon observed.
1. A new stereopoint is generated while introducing this protection. Though it is not relevant from the point of view of the target molecule, in chiral molecules this created diastereomer problems in spectroscopy (NMR and MS) and chromatography.
2. These ethers occasionally caused explosions in hydroboration procedures due to peroxide formation. The diastereomer problem was solved by the introduction of O-methyleneoxymethyl ether ( – O – MOM) and O – methyleneoxybenzyl ether (R – O – MOB) (Fig 4.1.28). Several other modifications are now available.
Protection of vic – Diols
On reaction with benzaldehyde or acetone with suitable acid catalyst, vic-diols form cyclic acetals. This in fact is a proof for the existence of vic-diols in the molecule. They are acetal protections and therefore behave as acetals in their chemistry (Fig 4.1.29)
Conclusion
The above discussions are just a glimpse of the vast literature on this topic. When more than one competing functional groups are present in a molecule, it may be necessary to introduce at least one protection and one deprotection step in the synthetic scheme. This adds not only to the length of the synthetic scheme, but also to the cost of the final compound. With growing awareness in Green Chemistry, chemists have been trying to reduce this protocol to a minimum or preferably avoid this altogether. Several protection free syntheses of natural products are known in the literature. We would discuss this topic at the end of this chapter.
Further Reading
1. Greene T. W., Protective Groups in Organic Synthesis. Wiley. N. Y., (1980), (1991).
2. Smith M. B., Organic Synthesis, McGraw-Hill Inc, N. Y., (1994).
3. Djerassi C., Steroid Reactions –An Outline for Organic Chemists, Holden-Day nc. San Francisco )1963).
4. Advanced Organic Chemistry: Principles Tools and Logic of Synthesis, R. Balaji Rao, Vishal Publishing Co., Jalandhar, India (2012).
5. Amino Acids, Peptides and Proteins in Organic Chemistry, Vol 4, Ed by Andrew B. Hughes (2011) Wiley-VCH; Protection Reactions, V.V. Sureshbabu and N. Narendra page 1 – 97.
4.2 Disconnection of bonds
Having chosen the TARGET molecule for synthesis, the next exercise is to draw out synthetic plans that would summarize all reasonable routes for its synthesis. During the past few decades, chemists have been working on a process called RETROSYNTHESIS. Retrosynthesis could be described as a logical Disconnection at strategic bonds in such a way that the process would progressively lead to easily available starting material(s) through several synthetic plans. Each plan thus evolved, describes a ‘ROUTE’ based on a retrosynthesis. Each disconnection leads to a simplified structure. The logic of such disconnections forms the basis for the retroanalysis of a given target molecule. Natural products have provided chemists with a large variety of structures, having complex functionalities and stereochemistry. This area has provided several challenging targets for development of these concepts. The underlining principle in devising logical approaches for synthetic routes is very much akin to the following simple problem. Let us have a look of the following big block, which is made by assembling several small blocks (Fig 4.2.1). You could easily see that the large block could be broken down in different ways and then reassembled to give the same original block.
Now let us try and extend the same approach for the synthesis of a simple molecule. Let us look into three possible ‘disconnections’ for a cyclohexane ring as shown in Figure 4.2.2.
In the above analysis we have attempted to develop three ways of disconnecting the six membered ring. Have we thus created three pathways for the synthesis of cyclohexane ring? Do such disconnections make chemical sense? The background of an organic chemist should enable him to read the process as a chemical reaction in the reverse (or ‘retro-‘) direction. The dots in the above structures could represent a carbonium ion, a carbanion, a free radical or a more complex reaction (such as a pericyclic reaction or a rearrangement). Applying such chemical thinking could open up several plausible reactions. Let us look into path b, which resulted from cleavage of one sigma bond. An anionic cyclisation route alone exposes several candidates as suitable intermediates for the formation of this linkage. The above analysis describes only three paths out of the large number of alternate cleavage routes that are available. An extended analysis shown below indicates more such possibilities (Fig 4.2.3). Each such intermediate could be subjected to further disconnection process and the process continued until we reach a reasonably small, easily available starting materials. Thus, a complete ‘SYNTHETIC TREE’ could be constructed that would summarize all possible routes for the given target molecule.
4.3 Efficiency of a route
A route is said to be efficient when the ‘overall yield’ of the total process is the best amongst all routes investigated. This would depend not only on the number of steps involved in the synthesis, but also on the type of strategy followed. The strategy could involve a ‘linear syntheses’ involving only consequential steps or a ‘convergent syntheses’ involving fewer consequential steps. Figure 4.3.1 shown below depicts a few patterns that could be recognized in such synthetic trees. When each disconnection process leads to only one feasible intermediate and the process proceeds in this fashion
all the way to one set of starting materials (SM), the process is called a Linear Synthesis. On the other hand, when an intermediate could be disconnected in two or more ways leading to different intermediates, branching occurs in the plan. The processes could be continued all the way to SMs. In such routes different branches of the synthetic pathways converge towards an intermediate. Such schemes are called Convergent Syntheses.
The flow charts shown below (Fig 4.3.2) depicts a hypothetical 5-step synthesis by the above two strategies. Assuming a very good yield (90%) at each step (this is rarely seen in real projects), a linier synthesis gives 59% overall yield, whereas a convergent synthesis gives 73% overall yield for the same number of steps..
4.4 Problem of substituents and stereoisomers
The situation becomes more complex when you consider the possibility of unwanted isomers generated at different steps of the synthesis. The overall yield drops down considerably for the synthesis of the right isomer. Reactions that yield single isomers (Diastereospecific reactions) in good yields are therefore preferred. Some reactions like the Diels Alder Reaction generate several stereopoints (points at which stereoisomers are generated) simultaneously in one step in a highly predictable manner. Such reactions are highly valued in planning synthetic strategies because several desirable structural features are introduced in one step. Where one pure enantiomer is the target, the situation is again complex. A pure compound in the final step could still have 50% unwanted enantiomer, thus leading to a drastic drop in the efficiency of the route. In such cases, it is desirable to separate the optical isomers as early in the route as possible, along the synthetic route. This is the main merit of the Chiron Approach, in which the right starting material is chosen from an easily available, cheap ‘chiral pool’. We would discuss this aspect after we have understood the logic of planning syntheses. Given these parameters, you could now decide on the most efficient route for any given target.
Molecules of interest are often more complex than the plain cyclohexane ring discussed above. They may have substituents and functional groups at specified points and even specific stereochemical points. Construction of a synthetic tree should ideally accommodate all these parameters to give efficient routes. Let us look into a slightly more complex example shown in Figure 4.4.1 . The ketone 4.4.1A is required as an intermediate in a synthesis. Unlike the plain cyclohexane discussed above, the substitution pattern and the keto- group in this molecule impose some restrictions on disconnection processes.
Cleavage a: This route implies attack of an anion of methylisopropylketone on a bromo-component. Cleavage b: This route implies simple regiospecific methylation of a larger ketone that bears all remaining structural elements. Cleavage c: This route implies three different possibilities. Route C-1 envisages an acylonium unit, which could come from an acid halide or an ester. Route C-2 implies an umpolung reaction at the acyl unit. Route C-3 suggests an oxidation of a secondary alcohol, which could be obtained through a Grignard-type reaction. Cleavage d: This implies a Micheal addition.
Each of these routes could be further developed backwards to complete the synthetic tree. These are just a few plausible routes to illustrate an important point that the details on the structure would restrict the possible cleavages to some strategic points. Notable contributions towards planning organic syntheses came from E.J. Corey’s school. These developments have been compiles by Corey in a book by the title LOgIC OF CHEMICAL SYNTHESIS. These and several related presentations on this topic should be taken as guidelines. They are devised after analyzing most of the known approaches published in the literature and identifying a pattern in the logic. They need not restrict the scope for new possibilities. Some of the important strategies are outlined below.
4.5 Preliminary scan
When a synthetic chemist looks at the given Target, he should first ponder on some preliminary steps to simplify the problem on hand. Is the molecule polymeric? See whether the whole molecule could be split into monomeric units, which could be coupled by a known reaction. This is easily seen in the case of peptides, nucleotides and organic polymers. This could also be true to other natural products. In molecules like C-Toxiferin 1 (4.5.1A) (Fig 4.5.1), the point of dimerisation is obvious. In several other cases, a deeper insight is required to identify the monomeric units, as is the case with Usnic acid (4.5.1B). In the case of the macrolide antibiotic Nonactin (4.5.1C), this strategy reduces the possibilities to the synthesis of a monomeric unit (4.5.1D). The overall structure has S4 symmetry and is achiral even though assembled from chiral precursors. Both (+)-nonactic acid and (−)-nonactic acid (4.5.1D) are needed to construct the macrocycle and they are joined head-to-tail in an alternating (+)-(−)-(+)-(−) pattern. (see J. Am. Chem. Soc., 131, 17155 (2009) and references cited therein).
Is a part of the structure already solved? Critical study of the literature may often reveal that the same molecule or a closely related one has been solved. R.B. Woodward synthesized (4.5.2C) as a key intermediate in an elegant synthesis of Reserpine (4.5.2A). The same intermediate compound (4.5.2C) became the key starting compound for Velluz et.al., in the synthesis of Deserpidine (4.5.2B) (Fig 4.5.2).
Such strategies reduce the time taken for the synthesis of new drug candidates. These strategies are often used in natural product chemistry and drug chemistry. Once the preliminary scan is complete, the target molecule could be disconnected at Strategic Bonds.
4.6 Strategic Bonds, Retrons and Transforms
STRATEGIC BONDS are the bonds that are cleaved to arrive at suitable Starting Materials (SM) or SYNTHONS. For the purpose of bond disconnection, Corey has suggested that the structure could be classified according to the sub-structures generated by known chemical reactions. He called the sub-structures RETRONS and the chemical transformations that generate these Retrons were called TRANSFORMS. A short list of Transforms and Retrons are given below (TABLE 4.6.1). Note that when Transforms generate Retrons, the product may have new STEREOPOINTS (stereochemical details) generated that may need critical appraisal.
The structure of the target could be such that the Retron and the corresponding Transforms could be easily visualized and directly applied. In some cases, the Transforms or the Retrons may not be obvious. In several syntheses, transformations do not simplify the molecule, but they facilitate the process of synthesis. For example, a keto- group could be generated through modification of a -CH-NO2 unit through a Nef reaction. This generates a new set of Retron / transforms pair. A few such transforms are listed below, along with the nomenclature suggested by Corey (Fig 4.6.2).
A Rearrangement Reaction could be a powerful method for generating suitable new sub-structures. In the following example, a suitable Pinacol Retron, needed for the rearrangement is obtained through an acyloin transform (Fig 4.6.3). Such rearrangement Retrons are often not obvious to inexperienced eyes.
Some transforms may be necessary to protect (acetals for ketones), modify (reduction of a ketone to alcohol to avoid an Aldol condensation during a Claisen condensation) or transpose a structural element such as a stereopoint (e.g. SN2 inversion, epimerization etc.,) or shifting a functional group. Such transforms do not simplify the given structural unit. At times, activation at specific points on the structure may be introduced to bring about a C-C bond formation and later the extra group may be removed. For example, consider the following retrosynthesis in which an extra ester group has been introduced to facilitate a Dieckmann Retron. In complex targets, combinations of such strategies could prove to be a very productive strategy in planning retrosynthesis. Witness the chemical modification strategy shown below for an efficient stereospecific synthesis of a trisubstituted olefin (Fig 4.6.4)
Figure 4.6.4 Examples for FGA / FGR strategies for complex targets
Amongst the molecular architectures, the bridged-rings pose a complex challenge in Structure-Based disconnection procedures. Corey has suggested guidelines for efficient disconnections of strategic bonds.
A bond cleavage for retrosynthesis should lead to simplified structures, preferably bearing five- or six-membered rings. The medium and large rings are difficult to synthesize stereospecifically. Amongst the common rings, a six-membered ring is easily approached and manipulated to large and small rings. Simultaneous cleavage of two bonds, suggesting cycloaddition – retrons are often more efficient. Some cleavages of strategic bonds are shown in Figure 4.6.5, suggesting good and poor cleavage strategies based on this approach. However, these guidelines are not restrictive.
Identifying Retron – Transform sets in a given target molecule is therefore a critical component in retrosynthesis. Such an approach could often generate several synthetic routes. The merit of this approach is that starting materials do not prejudice this logic. Retrosyntheses thus developed could throws open several routes that need further critical scrutiny on the basis of known facts.
Identification of Retrons / Transforms sets provided the prerequisite for computer assisted programs designed for generating retrosynthetic routes. A list of Retrons and the corresponding transforms were interlinked and the data was stored in the computer. All known reactions were thus analyzed for their Retron / Transform characteristics and documented. The appropriate literature citations were also documented and linked. Based on these inputs, computer programs were designed to generate retrosynthetic routes for any given structure. Several such programs are now available in the market to help chemists generate synthetic strategies. Given any structure, these programs generate several routes. Once the scientist identifies the specific routes of interest for further analysis, the program generates detailed synthetic steps, reagents required and the appropriate citations. In spite of such powerful artificial intelligence, the intelligence and intuitive genius of a chemist is still capable of generating a new strategy, not yet programmed. Again, human intelligence is still a critical input for the analysis the routes generated using a computer. Based on the experience of the chemists’ team, their projected aim of the project and facilities available, the routes are further screened.
4.7 Elaboration of the concepts
Short lists of syntheses that exemplify retroanalysis strategies devised through powerful transforms are given below. Several syntheses from natural product chemistry are later discussed in this chapter, which further illustrates these points.
Retrosynthesis based on Diels-Alder Transform; (E.J. Corey et.al., J.A.C.S. (1972), 94, 2549). Fumagillol (4.7.1A) presents 4 stereocentres and sensitive functionalities.
Simplification of the functional groups first exposed a vic-diol. This site could come from an olefin D. Further retroanlysis led to a structurally simplified target sequence B to F. A cyclohexene ring system is suitable for a powerful DA Transform. This step generated two stereocentres in one reaction and also an olefin in the correct position for hydroxylation. The key intermediate C could also be generated through a functional group transform leading to G. This provided scope for a new set of starting materials using another DA Transform. The Retrosynthetic analysis and the actual synthesis are shown in Figure 4.7.1.
Synthetic protocol reported by Corey is outlined in Figure 4.7.2
For the synthesis of Estrone, an interesting DA Transform strategy was devised by Kametani et.al.. The retrosynthetic strategy is depicted in Fig 4.7.3. The required diene precursor was generated via cyclo-reversion reaction of a cyclobutene unit (T. Kametani et.al., Tetrahedron, (1981), 37, 3).
The crucial stereospecific trisubstituted olefins on Squalene (4.6.4B) were synthesized using a Claisen Retron 4.7.4A (Fig 4.7.4). Note the double Claisen approach in this strategy.
The biogenetic-type cyclisation of olefins provides scope for application of Mechanistic Transform or transforms based on mechanistic considerations. A cleaver introduction of a chiral centre provided an efficient route for generating several enantiopure chiral centres in one step using this strategy (Fig 4.7.5).
4.8 The problem of enantiomers
In these lengthy discussions above, we learnt about disconnection approaches. We said that stereocentres could introduce special challenges in planning efficient synthetic routes. Let us look at the molecule Biotin to understand disconnection strategies and problem of stereocentres.
Baker established the structure of Biotin in 1947 through an unambiguous synthesis of the molecule. A retroanalysis of the synthetic scheme is as shown below (Fig 4.8.1).
The SM chosen and the synthetic approach clearly established the atom connectivities and the overall structure of the compound. However, the route made not attempt to synthesize one pure isomer because the actual stereochemistry was not established at that time. The route yielded all the eight stereoisomers (3 asymmetric centers). These isomers were carefully separated. In 1952 the biologically active isomer was identified as the all cis- enantiomer (+)-Biotoin. At this stage, several groups reported the stereospecific synthesis of the all cis- isomer exclusively (Fig 4.8.2). The following retroanalysis depicts three such attempts. Note that these efforts were directed towards the synthesis of the racemate and not the pure (+)- isomer of Biotin.
These approaches solved the problem of diastereomeric purity. But they still left a mixture of two unresolved enentiomer viz., (±)-Biotin. To obtain a pure enantiomer in excellent yields, you have to resolve the racemic mixtures at appropriate stages. Alternately one could resort to asymmetric synthesis at all crucial stages. A still better approach would be to start from a chiral SM, which has most of the stereocentres in the correct fashion. This elegant approach is called the Chiron approach. When carefully executed, such procedures yield very pure enantiomer as the final product. Two such approaches for (+)-Biotin is shown below. In the first approach a chiral amino acid cysteine is chosen because it has one key asymmetric center, the sulphur moiety and a carboxylic acid in the correct positions (Fig 4.8.3). In this example the choice of the SM is quite obvious. Note the cleaver introduction of the second nitrogen and the cyclisation step leading to the formation of the tetrahydrothiophene ring. Also note that the yield of Cram vs anti-Cram (chelation) products could be influenced by a choice of the reagent. These kinds of insights come only through a thorough knowledge of this particular reaction.
Fig. 4.8.3
In the second example chosen here, the choice of the SM as the appropriate chiron is not obvious but hidden. Such an analysis demands a more critical insight into the concerned stereocentres.
Conclusion
An emerging concept in the Logic In Synthesis is deliberate planning of Green Synthetic Pathways. The logic of retroanalysis is same as discussed above. The only differentiating point is that the criteria for selection of synthetic route discussed earlier would now analyses the same synthetic tree through a Green Chemistry window to select only those routes that have maximum Green aspects. The green chemistry goal is enforced through inclusion the Twelve Principles of Green Chemistry. This could be done by embracing one or more of the following techniques - Use of Green Energy Sources like Microwave, Sonochemistry, Photochemistry etc., solvent free syntheses, using easily recoverable new solvent and eco-friendly solvents, reusable catalysts in syntheses and schemes that avoid protecting group chemistry. Most of the chemistry used in Green Chemistry is not really new to chemists. Chemistry is now revisited due to the environment consciousness that has now crept into industrial chemistry and society at large. Most of the chemistry is buried in two centuries of chemical literature. Several new discoveries in reagents have appeared in recent years. Now Chemists have to become more alert to this awakening to environmental damages caused by chemical activities on this globe.
The above discussions are meant only to illustrate the major steps involved in retrosynthetic analysis of a molecule. Thorough knowledge of synthetic tools, mechanisms and stereochemistry are essential prerequisites for a chemist to venture into the synthesis of complex molecules. Needless to add that all these efforts have to be suitably backed up by a team of chemists, having a rigorous training in laboratory techniques, a first-hand experience on several organic reactions / reagents and thorough knowledge of purification techniques, spectroscopic techniques and not the least, a good knowledge of search techniques to scan and retrieve requisite information from the vast chemical literature accumulated since the dawn of modern chemistry.
Retroanalysis of Some Interesting Molecules
Let us now dwell deep into a few select structures chosen from natural product chemistry and see how these structures have been tackled through different synthetic strategies. We would start with a simple molecule – Disparlure – with only two asymmetric centers. The course would end with a flovour of some Green Chemistry based syntheses to draw the attention of students to this newly emerging concepts and concerns. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/04%3A_The_Logic_of_Synthesis.txt |
The gypsy moth (Porthytria dispar) is a serious pest of the forests. In 1976 B.A. Bierl et.al., (Science, 170,88 (1970)) isolated the sex pheromone from extracts of 78,000 tips of the last two abdominal segments of female moths. The structure was assigned as 5.1. Later, the precursor molecule – the cis-olefin was also isolated from the same source.
A laboratory bioassay from synthetic materials showed that just 2 pg of 5.1 was enough to elicit bioactivity. Since the availability of the molecule from natural sources was very minute even for structure elucidation problems and study its anticipated role as pest control molecule, there was intense interest in an efficient synthesis of this molecule. Some disconnections for this simple molecule are depicted in Figure 5.2.
Epoxides could be made from corresponding olefins. In this case, the olefin should be Z-olefin. When synthesis of such olefins are not stereospecific, direct epoxidation using peroxides would yield a mixture of α- and β-epoxide from both isomeric olefins. To avoid such mixtures at the last stage, one should introduce selectivity at an early stage of the synthesis.
The first attempt was directed towards synthesis of the appropriate olefin and epoxidation (B.A. Bierl et.al., (Science, 170, 88 (1970)). The stereoselectivity was unsatisfactory (Fig 5.3). This necessitated extensive purification.
The ratio of cis- / trans- isomers in Wittig Olefination reaction could be altered by modification of reagents and reaction parameters. H.T. Bestmann et.al., (Chem. Ber., 109, 3375 (1976)) were able to improve the synthesis by modifying the Wittig reaction conditions (Fig 5.4).
Pure cis- olefins could be obtained by catalytic reduction of acetylenes (Angew. Chem., Int. Ed., 11, 60 (1972). Klunenberg et.al., (Angew. Chem., Int. Ed., 17, 47 (1978) took advantage of the cis- olefin moieties in 1,5-cyclooctadiene by selective oxidation of one double bond. The chains were introduced by sequential Kolbe electrolysis (Fig 5.5).
Synthesis of Optically pure Disparlures
Epoxidation of olefins yield only racemates unless the epoxidation step involves an asymmetric synthesis. Synthesis of pure (+)- and (-)- isomers could be achieved in three ways.
1. Resolution of a racemate: This could be a method of choice when both enantiomers are needed for SAR studies. All antipodes of the compound would be available through identical synthetic pathways.
2. Asymmetric synthesis of appropriate intermediates: When only one of the antipodes is desired, this process provides a wide range of synthetic possibilities for investigation. When a large number of closely related compounds are the targets, this method is a better choice.
3. Sourcing the chiral intermediate from a suitable chiral pool: Once the chiral target is clear, this could be a method of choice. In the case of Disparlure, the amount that could be isolated from gypsy moth was so small that even the optical rotation could not be determined. Several workers have reported synthesis of optically pure (+)- and (-)- Disparlure and its structural isomers. The first report came from S. Iwaki et.al., (J. Am. Chem. Soc., 96, 7842 (1974)). They started with L-(+)-Glutamic acid and resolved the intermediate diastereomeric alcohol-lactones by repeated crystallization technique (Fig 5.6). Their synthesis was not stereospecific. SAR studies revealed that the cis-(+)- isomer was most effective.
Mori et.al.,( Tet. Lett., 3953 (1976); Tetrahedron, 53, 833 (1979)) soon followed with a synthesis of (+)- and (-)- Disparlures starting from L-(+)-tartaric acid (Fig 5.7).
This synthesis had the merit that some of the chiral intermediated were crystallisable and therefore amenable for easy purification to very pure intermediates and pure final products. Their intermediates were subjected to critical spectral analysis to assess their purity. Thus, their bio-assays gave more reliable data. A synthesis of (+)- and (-)- Disparlure from another chiral synthon - isopropylidene D- and L- erythroses - was reported by Alexandros E. Koumbis et.al., (Tetrahedron Letters, 46, 4353 (2005)) (Fig 5.8).
A successful synthesis of (+)- Disparlure by the application of Sharpless epoxidation was reported by Kossier B. E.. et.al., (J. Am. Chem. Soc., 103, 464 (1981)) (Fig 5.9).
Synthesis of all four isomers in a very pure form came from the school of Sharpless E. B. ( Tet. Lett., 6411 (1992)). Using both the chiral hydroxylation agents, they reported an efficient synthesis of all enantiomers (Fig 5.10). The efficiency of the asymmetric synthesis was as high as 95% and gave 100% pure intermediated by crystallization. The overall process was very efficient. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/05%3A_Strategies_in_Disparlure_Synthesis.txt |
(-)-Menthol is amongst the most important perfume / flavor chemical, extensively used in pharmaceuticals, cosmetics, toothpastes, chewing gums and toiletries. Out a the estimated total production of about 20,000 m.tons, natural menthol accounts to about 13 m.tons, the rest coming from synthetic sources. The natural source - oil of Mentha Arvensis - being erratic due to dependence on monsoon, the demand for synthetic menthol is on the increase. The manufacturing processes chosen for discussions here demonstrate three important methodologies used in industry for the synthesis of chiral compounds. A summary of some of the known processes is provided in Figure 6.1.
Symrise Process (formerly know as Haarmann & Reimer process) (US Patent 3,943,181 (Mar 9 1976)) – In this process (Fig 6.2), thymol is synthesized from m-cresol. Catalytic hydrogenation gave a mixture of Menthols from which menthols were first obtained as a racemic mixture by careful fractional distillation. The residual mixture was epimerised to increase the content of racemic menthol using a patented catalytic process. The breakthrough in the process is the resolution of the benzoate ester of the racemate by recrystallization by a process of seeding the concentrate with one pure epimer. The mother liquor that was now rich in the (+) isomer was recycled by taking it back to the distillation cycle. In this process, overall yield of (-)-menthol is about 90%.
Takasago Process: In this process a (S)-DINAP catalysed isomerization is the key step Fig 6.3. Addition of lithium amide to Myrcene gave an addition compound that was isomerised using a chiral ruthenium catalyst. Hydrolysis of the resulting enamine gave an aldehyde citronellal in high enantiomeric purity. This was cyclized by Lewis catalyst. Catalytic reduction of the olefin gave (-)-Menthol1.
BASF Process: BASF has already set up processes for the synthesis of a series of terpenes starting from butene. In view of the high demand for (-)-Menthol, they extended the product chain to (-)-Menthol as well. The scheme for the synthesis of the product chain is shown in Figure 6.4.
Fig 6.4
Extention of the Citral value chain to (+)-Citronellal, (-)-Isopulegol and finally to (-)-Menthol gave a range of value added products. Note that these processes have taken advantage of the developments in catalytic processes in recent years. Details of the catalysts are not made public.
07: Strategies in Longfolene Synthesis
The synthesis of Longifolene has held the fascination of synthetic organic chemists for several decades. Since the compound was available in a pure form from natural sources in sufficient quantities, the fascination was purely academic. During structure elucidation studies, it was observed that this bridged structure underwent a host of migration reactions. These rearrangements were of interest both from theoretical and practical points of view. The concept of ‘disconnection’ and ‘retrosynthesis’ that was evolving around 1960’ led to the development of Logic in Organic Synthesis during this period. The structure of Longifolene was a happy exploration ground for those grand masters who were involved in these developments. For this reason synthesis of Longifolene has been closely associated with these developments. While analyzing such structures, Corey had suggested a few ‘strategic bond disconnection’
for logical approaches towards synthesis (Fig 7.1). Please note that these suggestions were meant to provide guidelines and therefore need not restrict any further innovations. It was further pointed out that any one of these disconnections could lead to several paths for the construction of synthetic trees. For example, let us consider one such disconnections at bond ‘a’ shown in Fig 7.2. This could throw open four branches on the synthetic tree. Out of these, Corey first selected the Michael strategy because one such cyclisation was already known in the chemistry of Santonin.
Execution of this concept by Corey ( J. Am. Chem. Soc., 83, 2151 (1961); ibid.,86, 478 (1964)) is shown in Fig 7.3.
L.W. Oppolzer et.al., set up the longifolene ring system through a rearrangement transform. He made use of De Mayo reaction, which is a photocyclisation – retroaldol sequence shown below (Fig 7.4).
As shown in the synthetic scheme Figure 7.5, a [2+2] cycloaddition reaction on C gave D, which exposed the retroaldol components after an hydrogenolysis. Another useful feature of this synthesis is the utilization hydrogenolysis reaction on cyclopropane to expose a gem-dimethyl group. A chiral starting material A yielded (+)-Longifolene in 25% overall yield.
A carbene insertion strategy was reported by A.G. Schutz et.al., (J. Org. Chem., 50, 915(1985)). The retroanalysis is shown in Figure 7.6.
The synthetic scheme is shown in Figure 7.7.
The Diels-Alder strategy indicated in the introduction to this section on Longifolene was demonstrated by Fallis et.al., (Fig 7.8)
Another Diels-Alder strategy came from Ho and Liu (Fig 7.9). Note the utilization of the exo- cyclic olefin by converting this unit to the required seven membered ring moiety.
A cationic cyclisation strategy for Longifolene skeleton reported by Johnson’s school has some interesting features. Figure 7.10 depicts the cyclisation reaction that forms the key step in this scheme.
The detailed synthetic plan is shown in Figure 7.11.
Note that this cyclization still leaves the methyl group on the wrong carbon. The unwanted –OH group was removed via a Lewis acid complexed hydride transfer reaction. The acid catalyzed isomerisation of the double bond is followed by a hydroxylation- oxidation sequence to expose a carbonyl group. The quaternary methyl group was then introduced via., an enolate. The exocyclic methylene was reintroduced to complete the synthesis. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/06%3A_Strategies_in_(-)-Menthol_Synthesis.txt |
Cedrene represents a very complex tricyclic sesquiterpene. Such complexity called for ingenious approaches for the total synthesis of the ring system. There are several syntheses reported for this ring system. We would discuss a few approaches. Some strategic bond cleavages are shown as broken lines in Figure 8.1.
Disconnection at ‘a’: G. Stork et.al., (J. Am. Chem. Soc., 77, 1078 (1955); ibid, 83, 3114 (1961)) took advantage of the fact that fused five membered rings would be cis- fused. Such a system would have a crowded ‘concave phase’. These two features formed the basis for Stork’s synthesis of Cedrene. Their synthetic scheme is shown in Figure 8.2. The first alkylation set in motion the steric out come of the remaining steps.
Disconnection at ‘b’: For his 1969 Cedrene synthesis (J. Am. Chem. Soc., 91, 1557 (1969)), Corey set up the spiro-ring system (cleavage of bond ‘b’). A p- alkylation of phenolate gave the spiro ring (J. Am. Chem. Soc., 84, 788 (1962)). Lewis catalyzed cylisation of the enolate completed the skeleton (Figure 8.3).
Disconnection at ‘b’ and ‘d’: Based on biogenetic cyclization concept, Corey achieved a one-step cyclization of the A and B rings (Figure 8.4) onto a preformed C ring (Tet. Lett., 2455 (1972)). The key for success in this scheme is the prior formation of bond ‘d’ followed by the formation of the ‘b’ bond. Note the role of a cyclopropyl ketone that orchestrated the development of a carbonium ion followed by an incipient carbanion to complete the A and B rings in this order.
Biogenetic type cyclization: A remarkable biogenetic type cyclization of Nerolidol to Cedrene was reported by Anderson et.al.,(Tet. Lett., 2455 (1972). The cyclizations could be accomplished in two steps. With formic acid a six membered ring is first formed. Further treatments with triflouroacetic acid completed the synthesis. The overall yield was very moderate (Figure 8.5).
Disconnection at ‘e’ and ‘d’ bonds: In Anderson’s biogenetic-type cyclization, formation of the six membered C ring was first realized. A six membered ring system also served as a key step for free radical cyclization reported by Hee-Yoon Lee. Tandem radical cyclization was their central theme. Such radical cyclizations have been reported for the synthesis of natural products. Hee-Yoon Lee reported application of this strategy to the synthesis of α-Cedrene (Ter. Lett., 7713 (1998)). The success of this scheme hinged on two factors.
1. Selective formation of a radical for formation of ‘d’ bond to accomplish the B ring
2. The acceptor carbon should generate a new free radical in the process to complete the A ring. The first task was achieved through incorporation of a xanthane unit and the second crucial task was accomplished via., N-aziridinylimine group. The synthetic scheme is shown in Figure 8.6.
Disconnection at ‘b’ and ‘d’: An approach by Chen et.al., (Tet. Lett., 2961 (1993)) relied on the free radical cyclization reaction on a suitably fuctionalized C ring (Figure 8.7). The synthesis of C ring is achieved via., a DA reaction. Knoevenagal reaction placed the required chain for completing the A and B rings. Tin hydride reduction generated a free radical at the site of the nitro group, which underwent a tandem cyclization to complete the A and B rings in that order.
A Diels-Alder approach to to tricyclic cedrene skeleton was reported by Breitholle et.al., (Can. J. Chem., 54, 1991 (1976) (Figure 8.8). Alkylation of cyclobutadiene to the requisite chain gave 8.8A after equilibration. The DA reaction proceeded in 36% yield to give a mixture of isomers. The ketone 8.8B did not undergo ring expantion with diazomethane. The ring expansion was finally achieved via the methylene amine 8.8C via diazotisation. The fact that two ring expansion products 8.8D and 8.8E were formed with 8.8D as the major product suggests that the amine 8.8F was the major isomer. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/08%3A_Strategies_in_Cedrene_Synthesis.txt |
The structure of Reserpine was solved by 1953. R.B. Woodward’s group reported the first synthesis of Reserpine in 1956 (J. Am. Chem. Soc., 78, 2023, 2657 (1956); Tetrahedron, 2, 1 (1958)). His scholarly analysis clearly displayed aspects of retroanalysis, which was just evolving at that time. This synthesis commands admiration for the way he used conformational analysis and stereoelectronic effects to precisely develop the stereopoints in this exceedingly complex problem for that time. He recognized that the E-ring has a dense array of 5 asymmetric centers in a six membered E ring. His disconnection of reserpine led him to the key intermediate C. We could formalize his retroanalysis as shown in Figure 9.1.
Another brilliant piece of conformational analysis could be seen in the way he converted Isoreserpine to Reserpine by introducing conformational strain in an otherwise comfortable molecule.
Synthesis of Woodward’s Aldehyde (9.1C): In a cleaver execution by Woodward’s group (Fig 9.2), all the required carbons for the D/E rings and three of the five asymmetric centers were created by one Diels-Alder reaction (9.2B). Note the simple dissymmetry in one component – methyl acrylate – could precisely place three asymmetric centers in a row in a correct fashion. This cycloaddition reaction developed a concave phase and a convex phase1 in the product that guided further modifications on this intermediate. The hydride reagent in the next step delivered the hydride from the less hindered convex phase, placing the –OH group in the concave phase. This facilitated the formation of a five-membered lactone ring (9.2C). The alternate six-membered lactone ring was presumably more strained. Bromination on 9.2C with molecular bromine formed the bromonium ion complex from the convex phase, while the –OH group could enter from the concave phase to form an ether (9.2D). Treatment with sodium methoxide displaced the bromine from the convex phase, possibly via elimination – addition route (9.2E). The next bromonium ion complex again proceeded from the convex phase, with the water molecule entering from the concave phase to enable a trans-diaxial opening of the bromonium complex (9.2F). After one oxidation step, the molecule was now set for a complex double elimination reaction, with the zinc attacking two centers. An attack at the bromide center opened the ether ring while another attack at the carbonyl group opened the lactone ring. This complex Tandem Reaction placed all the five asymmetric centers. Opening the unsaturated ring gave the key intermediate 9.2I.
Completion of the A, B, C, D and E rings: Condensation of 9.2I with o-methoxytryptamine followed by Pictet-Spengler condensation led to the formation of Isoreserpine and not Reserpine. Woodward reasoned that this was due to the fact that the C, D and E rings in all-trans geometry, had all substituents in the stable equatorial orientation in Isoreserpine. Woodward executed the isomerisation at C3 in an ingenious way. Aqueous alkali hydrolyzed the acetate – ester functions, which was than lactonised under DCC conditions. Thus, an unstable all-axial E ring was locked in as a lactone (9.2M). This forced the molecule in a crowded unstable state. On acid catalyzed isomerisation with the high boiling carboxylic acid, the C3 position isomerized to the reserpine configuration. On transesterification, the correct isomer was formed with a free –OH ready for final acylation.
G. Metha’s Synthesis of E ring: G. Metha et.al., (J. Chem. Soc., Perkin Trans. 1, 1319 (2000)) controlled the stereochemistry on the E ring through a bicyclo[2.2.1]heptane system as shown in Figure 9.3.
Stork’s synthesis: Gilbert Stork (J. Am. Chem. Soc., 127, 16255 (2005) reasoned that the stereochemistry at C3 in the (AB ABDE ABCDE) cyclisation route of Woodward could be controlled if an iminium intermediate 9.4 could be formed first. This would then orient a chair-like folding of the tether chain (potential C ring) for an axial attack on the iminium ion to give a stereoselective C/D ring closure to form reserpine. They reasoned that the intermediate 9.5 would serve as the key intermediate.
9.4 (left) and 9.5 (right)
After some unsuccessful attempts, Stork’s school completed the synthesis of 9.5 using a route shown in Figure 9.6. Condensation of the hexynal 9.6A with the lithium enolate of methyl methoxy acetate followed by benzenesulphonyl chloride and further heating with DBU gave the conjugated methoxy ester with the (Z) isomer 9.6B as the dominant product. Trimethylsilyl chloride trapped the diene ketene acetal 9.6C without purification of the intermediate product. D.A reaction with maleic anhydride followed with aqueous THF gave the decarboxylated acid that was esterified to give 9.6E. A free-radical cyclisation with tributylstannane in refluxing t-butanol and subsequent workup led to an epimeric mixture that was equilibrated with base to the desired isomer 9.6G. Reduction of the keto- group with L-Selectride gave the axial alcohol. This alcohol was the inverted via mesylation and treatment with cesium acetate to give 9.6I. Reduction with LAH, selective tosylation of the primary alcohol and silylation of the secondary alcohol and ozonolysis gave 9.6L via 9.6K. Opening of the 5-membered ring was achieved by trapping the kinetic enolate as TMS derivative 9.6M and ozonolysis and final esterification gave 9.6N. This key intermediate was also synthesized via an alternate more efficient route.
Unfortunately, condensation of 9.5 with the indole precursor gave isoreserpine as the major product (Fig 9.7). The researchers reasoned that this unexpected result was most probably due to formation of the
C ring prior to cyclisation. In order to drive the reaction towards a DE ring formation, they decided to trap the intermediate imine as the cyanoamine 9.8A. This was achieved by adding an excess of potassium cyanide to the reaction mixture
The expected product 9.8A with an axial cyano- group was the sole product. Refluxing the aminonitrile in acetonitril as solvent again gave (±) methyl isoreserpine precursor as the major product (Fig 9.9).Since the expected chair-like folding and an axial attack of the indole moiety on the iminium ion moiety appeared to be on sound srereoelectronic grounds, the authors reasoned that the cyanide anion on the α- phase of the molecule formed a tight ion pair with the immonium ion
under these conditions.This anion prevented axial attack on a chair-like folding of the chain, forcing a boat like conformation prior to cyclisation followed by an axial attack to give isoreserpine as a major product.
In such an event, allowing the cyanide counter ion to escape from the influence of the iminium ion by the use of a polar solvent should clean the reaction site for a chair-axial attack. When the nitrilamine 9.8A was treated 10%
solution of 1 N HCl in THF solvent, the precursor for (±) methyl reserpine was the product in 90% yield. This was eventually converted to reserpine by known procedures. Based on these results, another very short chiral route has been reported by the authors. Thus, persistence and critical mechanistic reasoning at every point of failure led to a successful direct synthesis of the correct isomer.
S. Hanessean et.al., (J.Org. Chem.,62, 465 (1997) applied the Chiron approach to the key intermediate 9.10B suitable for the CD ring formation in a E ABE ABCDE approach discussed so far.
Quinic acid 9.10F had most of the chiral centers in the proper orientation. However it needed modification at the potential C20, C15 and C16 centres. Retroanalysis of Hanessian is shown in Figure 9.10. The actual synthesis is shown in Figure 9.11. Lactonisation, followed by selective benzyl protection and methylation of the remaining OH groups gave the compound B. Oxidation was followed by methanolysis of lactone ring led to elimination to give the conjugaten ketone D. Protection of the C18 – OH as TBDMS followed by vinylation using Grignard reagent placed the potential carboxylic acid function. The t-alcohol thus generated provided an anchor the stereospecific introduction of the needed two carbon chain. A reductive free radical cyclisation placed the carbon chain in the correct stereochemistry. The correct isomer at C20 (I) was then moved to the next step. This key intermediate L led to the reserpine and isoreserpine precursors 9.11M and 9.11M’ in the ratio 1.4 : 1 respectively. These steps are shown in Figure 9.11. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/09%3A_Strategies_in_Reserpine_Synthesis.txt |
Chemical Synthesis of Prostaglandins witnessed phenomenal activity during the 1960’s and 70’s. During this period, organic chemistry saw intensive development in ‘disconnection’ and ‘Logic’ as primary tools for synthesis. This period also saw development of several new reagents for stereoselective synthesis. The complexity of the structure of PG skeleton posed a great challenge for synthesis. The fact that molecules belonging to this family held great potential as drug candidates but were available only in minute quantities from natural sources was the main reason for the intense activity in the chemical synthesis and skeletal modifications for SAR studies. The numbering system on the skeleton and the main structural features of this family of molecules could be seen in Figure 10.1
Placing a keto- group at C9 is a delicate operation because such β- hydroxy ketones would readily undergo dehydration to give a PGA skeleton. On a five membered ring, PGA system could again undergo ready isomerisation to PGB, a stable ring system, probably via a PGC skeleton. On reduction with sodium borohydride, the keto- group at C9 is reduced to a mixture 9α- and the unnatural 9β- epimers. The natural PGF skeleton has an 9α- configuration for the –OH group. In the natural PGF skeleton we have four asymmetric centers on a five membered ring. As you know well, a five membered ring is conformationally very flexible. Hence setting up precise stereochemistry on this ring posed a great challenge during the 60’s. There could be three main strategies for the construction of five membered ring systems (Fig 10.2).
1. An open chain could be cyclized to a ring.
2. A suitable cycloalkane ring could be ring expanded or contracted to a five membered ring.
3. A suitable bicyclo[l,m,n]ring system could be opened to give a five membered ring.
Let us look at a few outstanding synthetic efforts that successfully met these challenges in Prostaglandin chemistry.
Cyclization of open chain precursors
Setting up a series of asymmetric centers on an open chain is as great a challenge as setting them up on a five membered ring. Corey’s early attempt in 1968 was to design a suitable six membered ring, open up the ring to a chain and then cyclize the chain regiospecifically to the required five membered ring ( J. Am. Chem. Soc., 90, 3245 (1968)). Their retroanalysis and synthesis are shown in Figure 10.3. The first DA reaction set two crucial stereocentres that guide the remaining stereopoints on the five membered ring formed by aldol reaction. This synthesis is a good example for convergent synthesis. The starting materials come by two different routes shown in Figure 10.4.
The Kojima’s disconnection of the cyclopentane ring at C8 – C12 bond provided an open chain. Careful planning of the functional groups on the proposed chain enabled his group to plan the stereocentres in a more direct way. Their retroanalysis and synthetic route are shown in Figure 10.5.
Cyclopentane ring precursor
The difficult problem of setting up stereocentres on a five membered ring was solved elegantly by Corey et.al.,(Tet. Lett., 311 (1970)) (Fig 10.6). Observe the artistic precision with which the substituents are woven into a five membered ring with the aid of iodolactonisation and epoxide ring formation reactions. An example for mastery in the Art of Synthesis.
F.S. Alvarez et.al., took advantage of the steric constraints due to eclipsing strains in five membered rings and wove three asymmetric centers in a row on a five membered ring (J. Am. Chem. Soc., 94, 7823 (1972)) (Fig 10.6).
Cyclohexane ring Precursor
Starting from all cis-cyclohexan1,3,5-triol, Woodward’s school demonstrated their excellence in the Art of Organic Synthesis. The first step is the differential protection with glyoxalic acid (Fig 10.8). Note an interesting architectural design aspect. This chain of two carbon protecting group is eventually incorporated into the main structure. Note that this also accomplishes another very difficult task viz., placing a reactive 2 carbon chain into the crowded concave phase. The solvolysis of the mesylate is assisted by the neighbouring olefin. Note the elegant planning of the ring contraction step.
In 1973, Corey revealed a synthesis (Tet. Lett., 309 (1973)) based on six membered ring involving a ring expansion and a ring contraction to achieve the goal (Fig 10.9).
Bicycloalkane Approaches
There are several syntheses based on such strategy to derive the stereochemical advantage of a bicycloalkane ring system. Here we shall discuss the Bicycloheptane strategy of Corey (J. Am. Chem. Soc., 91, 5675 (1969): Ann. N.Y. Acad. Sci., 180, 24 (1971)). This route provides entry to all natural and unnatural PGs. It provides facility for separation of enantiomers at a very early stage using Amphetamine salt procedure on the first chiral acid-alcohol. This scheme has been scaled up to multigram-scales. Corey’s retroanalysis is shown in Figure 10.10. This retroanalysis led to DA reaction.
The first problem was the synthesis of a cyclopentadiene with the alkyl group at the methylene carbon. Anion routes for alkylation of cyclopentadiene mostly led to a mixture of cyclopentadienes, due to easy isomerisation of the olefin. The problem was solved by alkylation with thallium anion (Fig 10.11). The desired cyclopentadiene was obtained in about 97% yield. The next challenge was in the DA reaction.
Ketenes do not undergo Diels-Alder cycloaddition. They give cyclobutane products even with dienes. Using a ‘masked ketene precursor’ this problem was solved as shown in Figure 10.12. The bicyclo[2.2.1]heptane thus formed underwent Baeyer-Villiger oxidation as expected. Saponification of the lactone gave a five membered ring with three asymmetric centers in place. The fourth center was generated through an iodolactonisation reaction.
Note the stereochemistry of this iodolactonisation step. After removal of the halogen, the –OH at C11 was protected as p-phenylbenzoyl ester. This ester not only gave a crystalline derivative for purification (a useful feature during large scale reactions) but also served as a shielding agent to induce C15-(S)– OH at a later stage. Hydrogenolysis of the C13 – OR to – OH followed by Collins oxidation gave the C13 aldehyde suitable for a Wittig reaction. The α,β-unsaturated ketone was reduced selectively to C15-(S)– OH using Zinc borohydride. This selectivity has been attributed the shielding effect of the bulky aroyl group from the front side, allowing the hydride to enter stereoselectively. Reduction of the ester group with trialkylaluminium hydride gave the hemiacetal for the final Wittig. Note the differential protections used in this synthesis at different stages. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/10%3A_Strategies_in_Prostaglandins_Synthesis.txt |
The main skeletal features in steroid rings are depicted by the two figures shown in Figure 11.1. The same numbering system is maintained even while describing part structures obtained as synthetic intermediates.
This convention helps us to follow the development of structural features through long synthetic schemes. Students should also become familiar with another convention followed by chemists to categorize synthetic schemes, originally evolved for steroids. Similar descriptions are also found in alkaloid chemistry. ‘An $AB \rightarrow ABC \rightarrow ABCD$ Approach’ would mean that a naphthalene skeleton (either aromatic or suitable perhydro- skeleton) is chosen as SM. The C ring is then constructed on the AB rings. The D ring is then formed by ring closure. An example to this strategy is Bechmann’s synthesis (1940). Such descriptions do not indicate any details like the substitution patterns on the SM or synthetic intermediates nor do they spill light on stereochemical details. These details are discussed in the synthetic schemes.
Bechmann’s Synthesis (1940) of Equilenin
This classical synthesis exploits the chemistry of the naphthalene ring. 1-naphthylamine-2-sulphonic acid was chosen as the AB ring (Fig 11.2). The sulphonic
acid moiety was converted to a phenol and protected as the methyl ether. The amine moiety was converted to iodide through diazotisation. The C and D rings were then built on the AB rings
Woodward’s synthesis of Cholesterol:
Woodward’s synthesis of Cholesterol (J. Am. Chem. Soc., 73, 2403, 3547, 3548 (1951); ibid, 74, 4223 (1952) (Fig 11.3) could be described as $C \rightarrow CD \rightarrow BCD \rightarrow ABCD$ Approach. Since the D ring remains D-homo until the last step of ring construction and the required 5-membered ring was obtained only after a ring contraction, it could also be termed as $C \rightarrow BC \rightarrow ABC \rightarrow ABCD$ Approach. Remember that this molecule has 8 asymmetric centers and could therefore have 28 (256) optical isomers. This synthesis envisaged the synthesis of just one set of diastereomers. This stereospecific synthesis incorporated all the stereopoints in a stereo- and regiospecific manner. The D ring provides an anchor for variations in the chain. The double bond on the C ring allowed an opening for the synthesis of cortisone. The legends on the arrows show the reactions.
Synthesis of Estrone
Estrone has attracted several chemists as a target for executing new methodologies on this complex yet useful steroid. A very popular method for the introduction of the D ring, followed by cyclization of the C ring in steroid synthesis was introduced by Torgov (1950, 63). His synthesis of Estrone is shown in Figure 11.4. His procedure for $AB \rightarrow ABD \rightarrow ABCD$ Approach became very popular. Several modifications were later developed to further improve this methodology.
Bartlet used a mechanistic transform in his synthesis of Estrone. He applied a biogenetic-type cyclisation for his $\ce{A -> AD -> ABCD}$ approach shown in Figure 11.5. A furan ring served as a masked 1,4-diketone needed for the D ring. Also note the rearrangement transform used for the stereospecific introduction of the angular methyl group at C13 position in the last step of the synthesis.
Hughes (1960) relied on aldol transforms for his A AD ABCD Approach. Here he made use of the reliable reduction protocols developed by several workers for controlling the stereochemistry at ring junctions (Fig 11.6)
P. Hermann et.al., (J. Org. Chem., 73, 6202 (2008)) had attempted a sequential Media:Ring Closure Reactions (RCR) strategy for the stereospecific synthesis of Estrone. These workers relied on the CP2ZrBU2 catalyst studied extensively in their laboratory and developed a short 9-step synthesis from known diene (A AB ABC ABCD Approach)(Fig 11.7). Cyclisation of the B ring proceeded satisfactorily. Cylisation of the C ring was sensitive to the halogen. After several attempt, cyclisation proceeded well with the vinyl fluoride. The formation of the D ring using Zr catalyst was unsuccessful. The ring closure of the D ring was finally accomplished using the second-generation Media:Grubbs catalyst. The final modification of the D-ring has been already reported by Bartlett P.A et.al., ( J. Am. Chem. Soc., 95, 7501 (1973) thereby completing a formal synthesis of Estrone.
Marko Weimar et.al., (J. Org. Chem., 75, 2718 (2010)) have reported a very successful DA Transform for the formation of CD ring from Dane’s diene as AB ring and a D ring as dienoplile (Fig 11.8) (AB ABCD Approach; see references cited for similar approaches).
Key for the success of this strategy was the metal-free chiral catalyst, which they successfully tailored by the introduction of three H-bonding sites. The catalyst that worked efficiently was the Media:axially chiral amidine salt 11.9.
They have suggested that this ligand formed a Host-Guest complex with the dienoplile as depicted in 11.10.
Under these conditions, the Diels-Alder reaction proceeded in excellent yield. This compound was converted to Estrone as shown in Figure 11.11
Synthesis of Cortisone
Cortisone attracted the attention of several synthetic chemists, because this wonder drug was available only in minute quantities from animal sources. Two challenging features in the structure of cortisone were the keto- group at C11 and the 1,2,3- oxygenation pattern at the two-carbon side chain at C17.
Sarrett (J. Am. Chem. Soc., 74, 4974 (1952); J. Am. Chem. Soc., 76, 5031 (1954)) used an ABC ABCD Approach. His starting molecule had all the features needed for the A, B and C rings of Cortisone. It had a C11 –OH group and a conveniently placed ketone for building the D ring. They exploited the known stereoelectronic constraints of anion reactions on a rigid 6-membered ring to construct the C/D trans ring junction (Fig 11.12). This cyclisation has two noteworthy features. The olefin served as a masked ketone. Unlike Woodward’s Cholesterol synthesis (Fig 11.3), which had a competing site for anion formation, in this scheme only one anion is feasible leading to predictable product.
An interesting new approach in steroid synthesis came from Y. Horiguchi (J. Org. Chem., 51, 4325 (1986)). This could be described as CD BCD ABCD Approach. Note that the C11 oxygenation and the carbons needed for the A ring came through a single oxidation step, as envisaged in their retroanalysis shown in Figure 11.13.
A detailed synthetic scheme of Horiguchi is shown in Figure 11.14. The strategy for formation of A as well as the D ring is of interest in this synthesis.
Nemoto’s retroanalysis (J. Org. Chem., 55, 5625 (1990)) provided an interesting electrocyclic ring opening – cycloaddition strategy for a B BCD ABCD Approach (Fig 11.15).
The synthetic scheme is shown in Figure 11.16. Note the versatile chiral auxiliary in the first step that serves several useful bond-forming purposes in addition to guiding three asymmetric center on C/D rings.
The fascination for mastering the steroid skeleton still continues unabated. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/11%3A_Strategies_in_Steroids_Synthesis.txt |
Chlorophyll, the most conspicuous of natural products, has held the fancy of organic chemists for more than two century, not only for its complex structure but also for its photochemical roll in the production of food that sustains all living things on this planet. Isolation and structure elucidation work began during the end of nineteenth century. This monumental activity that started with Willstätter, culminated in elucidation of the complete structure for Chlorophyll ‘a’ only by the middle of twentieth century. Most of these outstanding works, stretched over half a century, took place when modern organic chemistry was at its infancy. The synthesis of Chlorophyll ‘a’ by R. B. Woodward1 is acclaimed was an outstanding achievement in organic synthesis and ranks amongst the shinning gems in synthesis. The preliminary analysis for the synthesis, the persistent planned attach on this complex, delicate chemistry by his school and the logic of the famous Woodwardian approach are all good lessons for any discerning student of Organic Synthesis.
Analysis before initiating the synthesis
In the cited lecture, Woodward starts his discussions on synthesis with a series of questions that delineates not only the critical features of the structure under attach, but also arrives at a plausible target and approach for synthesis of this complex molecule. The structure of Chlorophyll ‘α’ is shown in (Fig 12.1).
Earlier work had established that the magnesium atom and the phytyl group were easy to remove and put in place through well established chemistry. Chlorophyll ‘α’ belongs to the green coloured pigment family chlorins (12.2). Simple chlorins are readily oxidized (lost their ‘extra’ hydrogens) to give the more conjugated porphyrins (12.3). But some chlorins like chlorophyll do so only under drastic conditions. The second feature that drew their attention is the carbocyclic five membered fused ring attached to ring III. A close study of the molecular models suggested that a porphyrin, which had a string of substituents on C5, C6, Cγ, C7 and C8 positions, would be highly crowded as shown in (12.3) and (12.4). They noted that chlorins and porphyrins that had a substitution at Cγ and a carboxyl group at C6, loose CO2 with ease, while absence of Cγ subsitution
endeared the C6 carbonyl increased stability. All these features led them to conclude that in porphyrins, the space at C7 / Cγ and Cγ / C6 are most crowded. Decarboxylation reduces this strain partially(12.5). Presence of substituent at Cγ increases the strain for C7 / Cγ space as well. This crowding is perhaps partly relieved by formation of the carbocyclic ring at Cγ / C6. Furthermore, the strain at C7 / Cγ could be relieved by introduction of the ‘extra’ hydrogen at C7 and C8. The same strain factors would influence the trans- arrangement for substituents at C7 and C8. Such detailed analysis of the given structure not only helped them to understand the given structure but also suggested that the ‘Target’ for synthesis could be a porphylin like (12.6). Once it is synthesized, such a structure could be coerced to move on to take up the ‘extra’ hydrogens and move to (12.7). Some questions remained. The vinyl group on the first ring was considered as very sensitive to withstand the rigour of the projected synthesis.
Hence, it was replaced with an equivalent aminopropane chain. As shown in structure (12.7), the residue needed at Cγ is an acetic acid moiety. However, mechanistic analysis of such a porphyrin (12.8)suggested that such a unit would readily eliminate due to an expected ‘electronic factor’ (actually, we would now say ‘enamine-like activity’). They decided to make this unit a propionic acid residue to avoid this instability. We have now arrived at a target structure (12.9). Based on the known pyrrole chemistry, they decided to make the Right Hand Side and the Left Hand Side of the molecule independently and condense these units into a tetramer. Thus, they arrived at the following four monomer units (12.10).
RHS Unit: The pyrrole units II and III were combined to give the expected dimer unit in good yields. Acylation with β-carbomethoxypropionyl chloride gave the RHS unit (12.17).
With this RHS unit (12.17) on hand, they tried the crucial cyclisation with a readily available model LHS unit (12.18). On condensation under acid conditions, followed by oxidation with iodine, the required porphyrin (12.19) could be obtained in acceptable 25% yield.
Encouraged with this result, they went ahead to synthesize the actual LHS unit (12.26).
With the LHS unit ready at hand, they looked at the cyclisation of these two units. They realized that this condensation could give two products due to two different orientations of the reactants. Though the yield of this condensation product was comparable to previous condensation, they saw some drawbacks.
The yield was not acceptable for their projected purpose. Furthermore, this route resulted in the formation of isomeric products, whose structure assignment posed problems. Such ‘inelegance’ (an expression used by RBW in his lecture) was not acceptable for their group. Hence, they decided to freeze the mobility of the two units by linking the amine and the aldehyde units into a Schiff base (12.29).
This masterly defined craftsmanship in molecular engineering was, however, not easy to achieve in practice. Pyrrole aldehydes were generally unreactive and could be converted to Schiff bases only under acid catalysis or buffered conditions. The problem arose from the LHS unit that proved to be very sensitive to such reaction conditions. No trace of the expected condensation product was observed. After several experiments, activation at aldehyde unit via Schiff bases followed by condensation with LHS unit via base exchange became a viable approach. But the discovery that such Schiff bases could be
smoothly converted to thioaldehyde (12.30) provided a breakthrough. This thioaldehyde smoothly condensed with the LHS unit under neutral conditions in good yield. This ‘extraordinarily sensitive’ Schiff base, on acid treatment gave the cation (12.31), which was isolated as the dibromide. This dication was immediately oxidized with iodine to (12.32) and isolated as the acetamide (12.33) on acylation with acetic anhydride. In spite of the sensitivity of the intermediates and quick subsequent steps, the porphyrin (12.33) could be obtained in good yield and could be scaled up to several gram scale.
When the porphyrin (12.33) was heated in acetic acid in air, migration of the first hydrogen occurred to give (12.34). On heating in acetic acid at 110 0C, the chain at Cγ cyclised at C7 to give the reduced ring unit IV . Note the orientation of the two chains on ring IV is trans as expected. At this stage, attention was shifted to the vinyl group on ring I . This was readily achieved by Hoffmann exhaustive methylation-elimination sequence shown above (12.36).
On exposure to light and air, the extra ring readily cleaved to give a pyruvic ester and a formyl group (12.37). The extra pyruvyl moiety was readily cleaved in methanolic KOH, which led to a methoxy lactone (12.380. Dry HCl gave the hamiacetal that could be resolved using quinine salt method to give (+)-Chlorin-5 (12.39). Diazomethane esterification exposed the aldehyde unit (12.40). Treatment of the formyl compound (12.40) with HCN in triethyl amine gave the cyanolactone (12.41), which on reduction, esterification and hydrolysis with methanolic HCl gave (12.44), the precursor for a Dieckmann cyclisation product.
From here to (-)-Clorophyll α was already known. A Diekmann followed by introduction of the phytyl group by enzymztic methods completed the synthesis (Fortchr. Chem. Forsch., 2, 538 (1952)).
To this day, this masterpiece in organic synthesis has several feature that hold the interest for lovers of Art and Logic in organic synthesis. | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/12%3A_Woodwards_Synthesis_of_Chlorophyll.txt |
The total synthesis of Vitamin B12 was accomplished in 1973 by a grand collaboration between R. B. Woodward’s group at Harvard University (USA) and A. Eschenmoser’s group, Swiss Federal Institute of Technology (ETH), Zürich, Switzerland. It took about twelve years and more than two dozen senior scientists to complete this gigantic task. The achievement is variously eulogized by organic chemists – monumental achievement in the annals of organic synthetic chemistry; a breakthrough; a mile stone in organic synthesis; unrivaled even after 40 years; a task no less adventurous than concurring Mount Everest. At the time the adventure started it was the most formidable challenge in synthetic chemistry, which few would have dared. The announcement of its synthesis marked the coming of age of synthetic organic chemistry. Woodward and Eschenmoser worked in close collaboration and in competition during this historic pace. In the process, they achieved not only an astounding synthesis, but also opened several new fields for future investigations. Woodward Hoffmann Rule is the most famous of the offshoots. While studying this synthesis, a student should also ponder over the thoroughness in their planning of all aspects of the scheme and initiation of suitable basic studies well in advance to facilitate the main scheme at appropriate junctures. This long introduction is just in tune with the length of the scheme, time taken and magnificent achievements.
Retroanalysis of Vitamin \(B_{12}\) synthesis
Figure 13.1 shows the structure of Vit B12 and the main structural features / challenges of the complex molecule.
In his inimitable style, R. B. Woodward draws attention to these challenges in the beginning of his lectures and to the fact that it took close to 50 years to establish the structure of Vit B12 by chemical degradation and finally by X-Ray diffraction studies by Dorothy Hodgkin in 1956. One should note that most of the chemical degradation studies, remarkable as they might be, happened during the early years of the twentieth century, when modern organic chemistry was in its making. Spectroscopic data, if any, were conspicuous by their absence and most of the degradation and synthetic chemistry were too harsh by modern (meaning 1960s) standards for this delicate molecule. Hence, this was a daunting scenario. This meant that a whole lot of synthetic chemistry had to be reinvented to suit these challenges. Such aspects were thoroughly planned in and contingency plans were put in place well in advance. Discerning students could see a glimpse of this planning in this short presentation. The ideas on retroanalysis, as we understand today, were in the developing stages during the sixties. The retroanalyses we use here are later-day additions by other scientists, based on the facts (lectures, papers etc.,) published by the two groups. All such citations are included at the end of the write-up.
The synthetic target was identified as Cobyric Acid, because this compound was a natural product and had been converted to Vit B12 by Bernhauer K., et.al., (1960)(Figure 13.2). Hence total synthesis of Cobyric Acid would amount to a formal synthesis of Vit B12.
Cobyric acid had seven carboxylic acid side chains, out of which four were propionic acid moieties, one on each heterocyclic ring. The main challenge was to differentiate the propionic acid chain on the D ring from the other acetic acid and propionic acid moieties. It was therefore decided that this odd acid moiety would be masked as nitrile (Figure 13.3). That still leaves a daunting task of differentiation, which we could address later. It was first decided to view the molecule as made up of two halves – the Eastern Side and the Western Side. The first disconnection was at the A/B ring junction at the methylene bridge (1.13.3A). Cleavage of the second bridge at C/D ring junctions gave the Eastern Half as Thiodextrolin in charge of Eschenmoser’s group at Zurich and the Western Half as Cyanobromide in charge of Woodward’s group at Harvard (US).
Retroanalysis of cyanobromide
This western half has a formidable array of six contiguous stereocentres on an eight-carbon frame. Note that the stereocentres were planned on the basis of known stereoselectivities and the six membered rings were built to provide the propionic acid chains (Figure 13.4). The nitrogen for the A ring came from an indole, whose benzene ring gave the side chains for the A ring. The D ring nitrogen came via a Beckmann rearrangement. Corrnorsterone (1.13.4A) was the key intermediate (corner stone) that held all the stereocentres and chains on the Western Half.
Synthesis of Western Half
The required enantiopure 1,2,3-trimethylyclopentene unit came from camphorquinone as shown in Figure 13.5.
Through another convergent synthesis, a five membered ring was fused to indole at C2 – C3 bond and resolved as shown in Figure 13.6.
The (+)- enantiomer was actually needed for target synthesis. The useless (-)- enentiomer was used as a model compound (for this was "just about the only kind of model study which we regard as wholly reliable” – RBW).
Fragments A and B were combined and then processed to Corrnorsterone as shown in Figure 13.7.
Note that the Beckmann ring expansion process set in motion a cascade of reaction, leading to a Claisen condensation and D ring formation, all in one step. The six membered imide carbonyl was also cleaved and placed an acetate chain on the D ring. In spite of this elaborate planning and execution, the process yielded a mixture of epimers at propionic acid side chain on the A ring, the required isomer being a minor component of the mixture. The major undesired product does not cleave at the amide bond due to unfavourable steric compression at the developing side chains (Figure 13.8).
This unfavorable steric problem was however soon solved. On hydrolysis under strong base conditions, the amide ring opened and the propionic acid side chain isomerised to the less strained isomer. This could be then be acidified and esterified to β-corrnorsterone, with a recovery of 90% of the desired isomer (Figure 13.9).
This correct isomer was treated with a mixture of methanol and thiophenol under acid conditions (Figure 13.10). This set two processes in motion. The thiophenol attacked the ketone and activated this center, while the methanol oxygen attacked the amide bond leading to an ester and a thioenol ether.
Ozonolysis of the thioenol ether at – 90°C cleaved the olefin unit to the aldehyde-thioester compound (Figure 13.11). An interesting new chemistry evolved here. While thioesters are less reactive to acid hydrolysis and showed comparable reactivity with oxygen nucleophiles, nitrogen nucleophiles were unique. The thioester reacted much faster then normal oxyesters to give amides. Thus, the thioester was exclusively cleaved to amide with ammonia, leaving three methyl esters untouched. The aldehyde moiety was then selectively converted to alcohol and then mesylated under mixed anhydride / pyridine conditions. This sequence also converted the amide moiety to nitrile. The mesylate was then converted to bromide to give the key intermediate cyanobromide.
The Zurich group was simultaneously working on the synthesis of the Eastern Half named Thiodextrolin. The fragments for B and C rings were planned via., a single intermediate (1.13.12).
The synthesis started with a Diels-Alder reaction to secure two asymmetric centers properly and the racemate was resolved. The pure enantiomer was followed through the scheme to obtain the B ring segment. The same intermediate yielded the C ring fragment as well (Figure 13.13).
To connect two such fragments Eschenmoser had developed two sulphide contraction procedures. The mechanisms of the processes are shown below.
Using the oxidative coupling / sulfur extrusion procedure, they coupled the B and C rings as shown in Figure 13.15.
Woodward’s group also developed a new synthesis for the C ring starting from (+)-Camphorquinone. The scheme is shown in Figure 13.16.
Though great care had been spared for the stereocentres at all points, the problem of stereoisomers could not be avoided. The crystalline thiodextrollin that was synthesized was actually a mixture of two stereoisomers at the propionic acid moiety of the B ring. Though they had purified the mixture at this point, it was of no value because this stereocentre was due for further disturbances at later stages. The mixture was taken forward for the first coupling at C / D bridge. After considerable effort that lasted over a year, they were first coupled at the Southern end using the alkylation / extrusion procedure (Figure 13.17). Note that the first product of alkylation was a thioether Type I, which readily isomerised to thioether Type II. The product was named Cyanocorrigenolide. This isomerisation disturbed the stereocentre at the C ring.
The next phase was the formation of A / B bridge. The C / D coupled compound was first treated with phosphorous pentasulphide followed by trimethyloxonium fluoroborate (Me3OBF4) (Figure 13.18). This procedure replaced the oxygens on the A and B rings by sulfur and finally to the S-methyl derivative. Dimethylamine in methnol cleaved the thiolactone ring selectively to a dimethylacetamide chain and a terminal olefin. The olefin was rather unstable. This was to be converted immediately to the cobalt complex. This procedure was not easy. Under several conditions, the cobalt metal ion catalyzed further reactions leading to extensive “destruction” of the compound. After several experiments it was observed that cobalt chloride or iodide in THF was unique for smooth cobaltation. The complexation process brought the A and B rings in close proximity. A base catalysed reaction then enabled formation of the bridge and removal of the sulfur moiety the best condition being DBN catalysed cyclization.
Note that the overall transformation interfered with the asymmetric center at C ring. Nonetheless, the A / B bridge was finally in place.
The Zurich group also came up with an alternate Zn-complex procedure along the same lines. All these manipulations were indeed harsh to the (three) epimerisable propionic acid chains on the A, B and C rings. The final product was purified by TLC (“plate chromatography”) and critically analyzed by HPLC (a new chromatographic tool at that time). The UV chromophores in all the products were of great help in this chromatography.
The synthesis has now two major milestones to pass. There are three active methine bridges in the molecule. The bridge at B / C rings had to be ‘protected’ from methylation. This was achieved by an oxidative lactone formation reaction at the B ring (I2, AcOH) (Figure 13.19). This new quaternary center and the existing quaternary carbon at C12 together exerted a steric congestion around C10. The chloromethyl ether entered the C5 and C15 centres exclusively. Ra-Ni hydrogenolysis cleaved the thioethers and the lactone ring in one step. This step was followed by esterification.
Conc. sulfuric acid converted the nitrile to amide (Figure 13.20).At this stage, they faced the difficult task of selective cleavage of amide, in the presence of six esters in the molecule. After extensive parallel experimentations, the Harvard group rediscovered an efficient selective cleavage of the amide moiety in N2O2 reagent.
The Zurich group also came out with a “diabolically cleaver scheme” for selective hydrolysis of the amide group in the presence of ester groups. This scheme is shown in Figure 13.21. However the former was preferred due to its simplicity and better yields. Nonetheless, Eschenmoser’s solution stands testimony to human ingenuity.
The very last step of this long synthesis posed a major problem that needs special mention. Amidation of esters with ammonia was the only remaining step for the final assault on the synthesis of Cobyric acid. A closer look reveals that the task may not be that easy. The ester moieties (in particular the acetate units) were in crowed environments. Parallel to these developments, model studies on very similar molecules were in progress. Based on these studies, when the hexamethylester 1 acid was treated with ammonia in ethylene glycol at 75°C for 30 hours, the product obtained was not cobyric acid but a pseudocobyric acid, whose structure was established as dehydrocobyric acid. This product could be purified only by HPLC. This was one among the products obtained by earlier workers. But they had no idea about such a complication. This mystery took several critical studies to solve. Throughout all these studies, great care was exercised to see that the solvents were well deoxygenated before use. Oxygen was strictly avoided from the reaction atmosphere. The source for this oxidative cyclization was attributed to cobalt, which was suspected to oxidatively bond to the C9 position, facilitating the amide anion to cyclise. This unwanted reaction took a long time to solve. Several studies were aimed at ammonialysis under reducing conditions and also to open the lactam ring under reduction conditions.
Finally a few milligrams of ammonium acetate were added to the reaction mixture to prevent the formation of amide anion, which was suspected to be the culprit for cyclization (Figure 13.22). This trick helped. The reaction was complete within 10 hours, with cobyric acid as the only product in good yields. This cobyric acid was identical in all respects, particularly in HPLC, with the natural product, thus ending this long journey for the formal synthesis of Vitamin B12.
This extraordinary adventure in organic synthesis is noted for several most significant achievements.
• Developments in Corrin chemistry
• Synthetic strategy
• Development of several new methodologies
• Woodward-Hoffmann Rules
Even after half a century this Vit B12 synthesis remains unrivaled and continues to inspire generations of chemists.
Further Reading
1. Recent advances in the chemistry of natural products, Woodward, R.B. Pure & Appl. Chem., 17, 519 (1968).
2. Eschenmoser, A. Pure Appl. Chem., 297–316 (1963).
3. Eschenmoser, A. Angew. Chem. Int. Ed., 5 (1988)
4. Recent advances in the chemistry of natural products, Woodward, R.B. Pure & Appl. Chem., 25, 283 (1971).
5. The total synthesis of vitamin B12, R. B. Woodward, Pure & Appl. Chem., 33, 145 (1973).
6. Natural Product Synthesis and Vitamin B12, Eschenmoser, A.; Wintner, C.E. Science, 196, 1410 (1977).
7. Crowfoot-Hodgkin, D. et al. Nature, 178, 64 (1956).
8. Friedrich, W., Eds., Walter de Gruyter: Berlin, 37, (1979).
9. Targets, Strategies, Methods. Weinheim, Germany: VCH, (1996).
10. The Art and Science of Total Synthesis at the Dawn of the Twenty-First Century, K. C. Nicolaou, Dionisios Vourloumis, Nicolas Winssinger, and Phil S. Baran, Angew. Chem. Int. Ed., 39, 44 (2000).
11. Vitamin B12 : An Epic Adventure in Total Synthesis, by Neil Garg, January 29, 2002.
12. The Asymmetric Total Synthesis of Vitamin B12, Nathan S. Werner, Denmark Group Meeting, September 28th, 2010.
13. Nicolaou, K.C.; Sorensen, E. J. Vitamin B12. Classics in Total Synthesis, VCH: New York, 100, (2003). | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/13%3A_Synthesis_of_Vitamin_B.txt |
During the past two centuries, the art organic synthesis has seen enormous progress in several disciplines of organic chemistry. Soon after the strides made by stalwarts like Kekule, Fischer and Robison during late 19th century and early 20th century, chemistry witnessed the dawn of mechanistic organic chemistry. The era of spectroscopy and reagents soon followed to compliment and advance mechanistic organic chemistry and synthesis to new heights. During the middle of twentieth century, advances in Separation Science took over the classical art of purification. The last three decades of the century saw the birth of Logic in Synthesis. All these developments augured well for new developments in drug discovery leading to large-scale synthesis of several complex organic molecules. However, this hectic pace of developments in industrial chemistry has taken place at the cost of environment that is so vital for survival of life on earth. In chemical industry, this realization came about only after disasters like the one witnessed at Bhopal, India jolted the chemical world. The onus of development of new molecules rests on us (chemists) and therefore the onus of responsible development is once again on our shoulders. Clean Chemistry (meaning Green Chemistry) depends on several factors – clean starting materials, clean reagents, clean solvents, clean product, clean energy and (not the least) clean processes. While all these subtitles are discussed under Green Chemistry, we would restrict our present discussions to one aspect of the Green Process viz., protection free syntheses because this falls under our present discussions ‘Logic of Synthesis’.
• Advanced Organic Chemistry: Principles Tools and Logic of Synthesis, Vishal Publishing Co., Jalandhar, India (2012).
The Logic of Synthesis opens up a systematic methodology for the development of Synthetic Trees. When properly devised, a synthetic tree contains all plausible routes for the synthesis of the Target compound. The route chosen depends on the need and constraints of the developer scientist. Hence, the choice of clean chemistry starts here. The routes should be analyzed from the windows for Green Chemistry. One such green window is the lookout for a Protection Free synthetic route. Why Protection Free Route? An industrially feasible route has to be very efficient to be economical. By imposing such a condition - Protection Free - to the choice of routes, it should possible to arrive at such routes. Each protection / deprotection sequence adds two steps to the scheme. Thus, more the protecting groups, less is the overall yield. In addition, each protection / deprotection step adds to the byproduct load released into the environment. Thus, protection based planning has a double-edged sword hidden at every such step. To minimize loss (of precious final product) protection / deprotection sequence should therefore be treated as an avoidable technique to be used only sparingly. Always remember that protection methodology provides such a sense of security, one tends to abuse and over use such a facility. Once a chemist settles in such a comfort zone, one tends to spend less time / effort to look for alternate approaches.
of routes, it should possible to arrive at such routes. Each protection / deprotection sequence adds two steps to the scheme. Thus, more the protecting groups, less is the overall yield. In addition, each protection / deprotection step adds to the byproduct load released into the environment. Thus, protection based planning has a double-edged sword hidden at every such step. To minimize loss (of precious final product) protection / deprotection sequence should therefore be treated as an avoidable technique to be used only sparingly. Always remember that protection methodology provides such a sense of security, one tends to abuse and over use such a facility. Once a chemist settles in such a comfort zone, one tends to spend less time / effort to look for alternate approaches.
Is it possible to avoid it? It should be possible to deliberately select such routes from the Synthetic Tree or device such a tree based on this additional criteria. One could select only those routes that do not call for protection or has a minimum of such techniques.
• One could add filtration criteria while devising synthetic trees to avoid such conflicting groups. This would indeed greatly shrink the size of the tree and thus limit the number of choices.
• The scheme could be devised with a minimum of conflicting groups. This would enhance the number of synthetic schemes and thus provide more choice to the scientist.
• Wherever such conflicts arise, one should first look for suitable reagents that are chemoselective and / or regioselective, instead of resorting to the comfort of protection. Advances in reagents / reactions are such that we now have several selective reagents and their number is increasing by each passing day.
Once a conscious decision is taken that protection free syntheses are ‘elegant’ in addition to being economical, one is more likely to come up with suitable schemes. Since industrial scale syntheses are mainly guided by cost factors, it is not surprising to find several industrial processes that are protection free syntheses. Protection / deprotection should be treated as a case of mind set rather than limitations of chemistry. Having said this in such strong terms, let us also concede that it may not be feasible to completely avoid protection / deprotection strategies. In support of these statements, we would soon look at some successful protection free synthesis in this chapter. In 2007, Phil S. Baran et.al., reported the total synthesis of some marine natural products without using protecting groups. While discussing the synthesis, the authors suggest that the following guidelines would be helpful while planning retro-analysis under this philosophy.
1. Redox reactions that do not lead the C – C bond formation should be minimized.
2. The percentage of C – C bond formation events in the synthetic steps should be maximized.
3. Disconnections should be made to maximize convergence.
4. The overall oxidation level of intermediates should linearly escalate during assembly of the molecular framework.
5. Wherever possible cascade or tandem reactions should be designed.
6. The innate reactivity of the functional groups should be exploited to avoid protection strategies.
7. If necessary one should invent new methodologies to discover new aspects of chemical reactivities.
8. Biomimectic chemistry could provide useful guidelines when dealing with natural products.
In an effort to prove their point, they have reported an efficient route for the synthesis of (+)-Ambiguine H. This synthesis is shown below (Fig 14.1). The strategic bond disconnections envisaged by these workers were guided on the above guidelines.
Since a route chosen is a deliberate choice of the researcher, it may not be fair to compare research publications based on different motives. I would be like comparing apples and oranges.Nature 446, 404 (2007) and references cited therein.
The proposed strategic disconnections for ambiguine H are shown below (Fig 14.1).
The first target was (-)-Hapalindole U for which the t-prenyl unit is not required. The top six membered ring unit came from the well-known terpene through four steps shown in (Fig 14.2)
The α- position to the ketone in A was linked the C3 of indole without protection, using a reaction developed in their laboratory (J. Am. Chem. Soc., 127, 7459 (2004)) (Fig 14.3). The second coupling was attempted with several catalysts. The free radical Heck conditions failed. They led mainly to cyclisation at C2 of indole unit. Probing further, the reaction succeeded by using the technique of slow addition of Herrman’s catalyst. In the absence of NH group on indole, such reactions failed to give this coupling. Now conversion to the natural product (-)-Hapalindole U (A) needed stereoselective transformation of the carbonyl unit to isonitrile unit. This was achieved by stereocontroled, microwave assisted reductive amination followed by formylation and dehydration reactions to give (-)-Hapalindole U (A). All these reactions could be scaled up to the level of several grams. (-)-Hapalindole U was made in sufficient quantities and stored.
The next target (+)-Ambiguine H was a very unstable compound, which was made as and when needed, using the following reactions.
(-)-Hapalindole U had several very reactive units in proximity. Introducing a prenyl unit into the crowded C2 region of the indole unit was very difficult (Fig 14.4). Treatment of t-BuOCl followed by prenyl-9-BBN led to a complex series of reaction, resulting in a crystalline product B. The formation of this product was rationalized by a cascade reaction mechanism shown in figure. Photolysis of B led to a Norrish-type cleavage, followed by rearrangement that got rid of all the extra groups on the skeleton to yield Ambiguine H. The reaction was not allowed to reach completion because the product was also sensitive to photolysis. The reported yield is based on recovered B.
The insistence that no protection protocols would be permitted in the scheme gave not only a short route to the target compound, but also led to the use of Rearrangement Transforms, use of Cascade Reactions and led to new discoveries in chemical reactivities.
In old chemistry, you would come across several such complex targets (for that time), whose synthesis were achieved without (or with very few) protection / deprotection sequences. There are other syntheses reported in recent years that have followed a ‘no protection’ barrier in planning the syntheses . We could visit two more such synthesis in this chapter.
Kessane, a constituent of Japanese Valarian root is of interest due to its sedative and anxiolitic effect. In 2003, Booker-Milburn K. I. and co-workers reported an industrial synthesis of this compound in 8 steps following the theme ‘no protection / deprotection’. The outline of the synthesis is presented in (Fig 14.5)
• Chem. Int. Ed. Engl., 34, 1370 (1995); Hoffman R. W, Synthesis, 21, 3531 (2006)
• Booker-Milburn K.I.; Jenkins H.; Charmant J. P. H.; Mohr Peter Org. Lett., 5, 3309 (2003).
The use of transition metal reagents could aid in the construction of C – C bonds without the need for protective groups, as demonstrated by Baldwin’s group in 2006 (Fig 14.6). This synthesis of Bureothin was achieved in 7 steps with no protection protocol.
• Jacobson, M. F., Moses, J. E., Adlington, R. M. and Baldwin, J. E., Tetrahedron, 62, 1675 (2006).
Some rare molecular engineering marvels like R.B. Woodward’s synthesis of PGF (Fig 1.10.7) open up new concepts on ‘Protection’ in molecular synthesis. Note the way he incorporated a protecting group scaffold into the molecular mainframe. Unlike the usual way of thinking of protection / deprotection as an isolated operation, these workers subsequently incorporated this protecting unit into the molecular frame, thereby achieving ‘carbon economy’ well ahead to his time. These concepts of green chemistry came decades later. Also revisit some of the biogenetic type cyclisation reaction discussed earlier in the light of Green Chemistry.
Conclusion
Any doubt on feasibility of protection free synthesis could be laid to rest. However, at the present state of developments in common synthetic methodologies, the protection free methodology has its limitations too. At present, it appears inconceivable to bring about a medium sized peptide or nucleotide with such a protection free protocols. This could also be true to several other syntheses of complex molecules. The main thesis of this essay is that it is practical to conceive such protection free synthetic protocols. This falls within the call of Green Chemistry. Though this has been the watchword in industrial synthesis, Baran’s 2007 paper coined the idiom “Protection-free synthesis’ and drew attention to such a possibility. Since then several papers have appeared with this idiom. One can be assured that such a basic green concept cannot be a passing fade. It is more likely to mark a corner stone in the logic of organic synthesis.
Further Readings
1. Organic Synthesis; Michael B. Smith, McGraw-Hill, Internationl Edition(1994).
2. The Logic of Chemical Synthesis, E.J. Corey & Xue-Min Cheng, John Wiley & Sons(1989).
3. The Total Synthesis of Natural Products, Volume 4; Edited by John ApSimon, John Wiley & Sons(1981).
4. Advanced Organic Chemistry: Reactions and Synthesis; Part A & B; Francis A. Corey, Richard J. Sundberg, Springer(2007).
5. Total Synthesis of Natural Products: Total Synthesis 1980-1994, Volume 11, Part2; Michael C. Pirrung, Andrew T. Morehead & Bruce G. Young, John Wiley & Sons(2000).
6. Advanced Organic Chemistry: Principles Tools and Logic of Organic Synthesis, R. Balaji Rao, Vishal Publications. Jalandhar 144008 (PB), India (2012). | textbooks/chem/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/14%3A_Green_Chemistry_-_Protection-Free_Organic_Synthesis.txt |
• Basic Electronic Structure of Atoms
This page explores how you write electronic structures for atoms using s, p, and d notation. It assumes that you know about simple atomic orbitals - at least as far as the way they are named, and their relative energies.
• Case Study: Electron Configuration of Mn vs. Cu
• d-orbital Occupation and Electronic Configurations
• Electronic Configurations Intro
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to electro
• Electronic Structure and Orbitals
• Hund's Rules
Hund's rule states that: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin). When assigning electrons to orbitals, an electron first seeks to fill all the orbitals with similar energy (also referred to as degenerate orbitals) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible.
• Pauli Exclusion Principle
The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins. This means if one is assigned an up-spin ( +1/2), the other must be down-spin (-1/2).
• Spin Pairing Energy
Spin pairing energy refers to the energy associated with paired electrons sharing one orbital and its effect on the molecules surrounding it. Electron pairing determining the direction of spin depends on several laws founded by chemists over the years such as Hund's law, the Aufbau principle, and Pauli's exclusion principle. An overview of the different types laws associated with the electron pairing rules.
• Spin Quantum Number
The Spin Quantum Number (s) is a value (of 1/2) that describes the angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum. Because angular momentum is a vector, the Spin Quantum Number (s) has both a magnitude (1/2) and direction (+ or -). This vector is called the magnetic spin quantum number (ms).
• The Aufbau Process
The Aufbau model lets us take an atom and make predictions about its properties. All we need to know is how many protons it has (and how many electrons, which is the same as the number of protons for a neutral atom). We can predict the properties of the atom based on our vague idea of where its electrons are and, more importantly, the energy of those electrons.
• The Octet Rule
The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds.
• The Order of Filling 3d and 4s Orbitals
This page looks at some of the problems with the usual way of explaining the electronic structures of the d-block elements based on the order of filling of the d and s orbitals. The way that the order of filling of orbitals is normally taught gives an easy way of working out the electronic structures of elements. However, it does throw up problems when you come to explain various properties of the transition elements. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.01%3A_The_Structure_of_an_Atom.txt |
• Basic Electronic Structure of Atoms
This page explores how you write electronic structures for atoms using s, p, and d notation. It assumes that you know about simple atomic orbitals - at least as far as the way they are named, and their relative energies.
• Case Study: Electron Configuration of Mn vs. Cu
• d-orbital Occupation and Electronic Configurations
• Electronic Configurations Intro
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to electro
• Electronic Structure and Orbitals
• Hund's Rules
Hund's rule states that: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin). When assigning electrons to orbitals, an electron first seeks to fill all the orbitals with similar energy (also referred to as degenerate orbitals) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible.
• Pauli Exclusion Principle
The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins. This means if one is assigned an up-spin ( +1/2), the other must be down-spin (-1/2).
• Spin Pairing Energy
Spin pairing energy refers to the energy associated with paired electrons sharing one orbital and its effect on the molecules surrounding it. Electron pairing determining the direction of spin depends on several laws founded by chemists over the years such as Hund's law, the Aufbau principle, and Pauli's exclusion principle. An overview of the different types laws associated with the electron pairing rules.
• Spin Quantum Number
The Spin Quantum Number (s) is a value (of 1/2) that describes the angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum. Because angular momentum is a vector, the Spin Quantum Number (s) has both a magnitude (1/2) and direction (+ or -). This vector is called the magnetic spin quantum number (ms).
• The Aufbau Process
The Aufbau model lets us take an atom and make predictions about its properties. All we need to know is how many protons it has (and how many electrons, which is the same as the number of protons for a neutral atom). We can predict the properties of the atom based on our vague idea of where its electrons are and, more importantly, the energy of those electrons.
• The Octet Rule
The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds.
• The Order of Filling 3d and 4s Orbitals
This page looks at some of the problems with the usual way of explaining the electronic structures of the d-block elements based on the order of filling of the d and s orbitals. The way that the order of filling of orbitals is normally taught gives an easy way of working out the electronic structures of elements. However, it does throw up problems when you come to explain various properties of the transition elements. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.02%3A_How_Electrons_in_an_Atom_are_Distributed.txt |
There are many types of chemical bonds and forces that bind molecules together. The two most basic types of bonds are characterized as either ionic or covalent. In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same electronegativity share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons.
Introduction
Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion. Ionic bonds require an electron donor, often a metal, and an electron acceptor, a nonmetal.
Ionic bonding is observed because metals have few electrons in their outer-most orbitals. By losing those electrons, these metals can achieve noble gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in their valence shells tend to readily accept electrons to achieve noble gas configuration. In ionic bonding, more than 1 electron can be donated or received to satisfy the octet rule. The charges on the anion and cation correspond to the number of electrons donated or received. In ionic bonds, the net charge of the compound must be zero.
This sodium molecule donates the lone electron in its valence orbital in order to achieve octet configuration. This creates a positively charged cation due to the loss of electron.
This chlorine atom receives one electron to achieve its octet configuration, which creates a negatively charged anion.
The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of the electrostatic attraction between the particles. At the ideal interatomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ion.
Example \(1\): Chloride Salts
In this example, the sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0.
In this example, the magnesium atom is donating both of its valence electrons to chlorine atoms. Each chlorine atom can only accept 1 electron before it can achieve its noble gas configuration; therefore, 2 atoms of chlorine are required to accept the 2 electrons donated by the magnesium. Notice that the net charge of the compound is 0.
Covalent Bonding
Covalent bonding is the sharing of electrons between atoms. This type of bonding occurs between two atoms of the same element or of elements close to each other in the periodic table. This bonding occurs primarily between nonmetals; however, it can also be observed between nonmetals and metals.
If atoms have similar electronegativities (the same affinity for electrons), covalent bonds are most likely to occur. Because both atoms have the same affinity for electrons and neither has a tendency to donate them, they share electrons in order to achieve octet configuration and become more stable. In addition, the ionization energy of the atom is too large and the electron affinity of the atom is too small for ionic bonding to occur. For example: carbon does not form ionic bonds because it has 4 valence electrons, half of an octet. To form ionic bonds, Carbon molecules must either gain or lose 4 electrons. This is highly unfavorable; therefore, carbon molecules share their 4 valence electrons through single, double, and triple bonds so that each atom can achieve noble gas configurations. Covalent bonds include interactions of the sigma and pi orbitals; therefore, covalent bonds lead to formation of single, double, triple, and quadruple bonds.
Example \(2\): \(PCl_3\)
In this example, a phosphorous atom is sharing its three unpaired electrons with three chlorine atoms. In the end product, all four of these molecules have 8 valence electrons and satisfy the octet rule.
Bonding in Organic Chemistry
Ionic and covalent bonds are the two extremes of bonding. Polar covalent is the intermediate type of bonding between the two extremes. Some ionic bonds contain covalent characteristics and some covalent bonds are partially ionic. For example, most carbon-based compounds are covalently bonded but can also be partially ionic. Polarity is a measure of the separation of charge in a compound. A compound's polarity is dependent on the symmetry of the compound and on differences in electronegativity between atoms. Polarity occurs when the electron pushing elements, found on the left side of the periodic table, exchanges electrons with the electron pulling elements, on the right side of the table. This creates a spectrum of polarity, with ionic (polar) at one extreme, covalent (nonpolar) at another, and polar covalent in the middle.
Both of these bonds are important in organic chemistry. Ionic bonds are important because they allow the synthesis of specific organic compounds. Scientists can manipulate ionic properties and these interactions in order to form desired products. Covalent bonds are especially important since most carbon molecules interact primarily through covalent bonding. Covalent bonding allows molecules to share electrons with other molecules, creating long chains of compounds and allowing more complexity in life.
Problems
1. Are these compounds ionic or covalent?
2. In the following reactions, indicate whether the reactants and products are ionic or covalently bonded.
a)
b) Clarification: What is the nature of the bond between sodium and amide? What kind of bond forms between the anion carbon chain and sodium?
c)
Solutions
• 1) From left to right: Covalent, Ionic, Ionic, Covalent, Covalent, Covalent, Ionic.
• 2a) All products and reactants are ionic.
• 2b) From left to right: Covalent, Ionic, Ionic, Covalent, Ionic, Covalent, Covalent, Ionic.
• 2c) All products and reactants are covalent. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.03%3A_Ionic_and_Covalent_Bonds.txt |
Objectives
After completing this section, you should be able to
1. describe the physical significance of an orbital.
2. list the atomic orbitals from 1s to 3d in order of increasing energy.
3. sketch the shapes of s and p orbitals.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• nodal plane
• node
• orbital
• quantum mechanics
• wave function
For a refresher on quantum numbers view this text
Atomic Orbitals
An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of having an electron.
One way of representing electron probability distributions was illustrated in Figure 6.5.2 for the 1s orbital of hydrogen. Because Ψ2 gives the probability of finding an electron in a given volume of space (such as a cubic picometer), a plot of Ψ2 versus distance from the nucleus (r) is a plot of the probability density. The 1s orbital is spherically symmetrical, so the probability of finding a 1s electron at any given point depends only on its distance from the nucleus. The probability density is greatest at r = 0 (at the nucleus) and decreases steadily with increasing distance. At very large values of r, the electron probability density is very small but not zero.
In contrast, we can calculate the radial probability (the probability of finding a 1s electron at a distance r from the nucleus) by adding together the probabilities of an electron being at all points on a series of x spherical shells of radius r1, r2, r3,…, rx − 1, rx. In effect, we are dividing the atom into very thin concentric shells, much like the layers of an onion (part (a) in Figure \(1\)), and calculating the probability of finding an electron on each spherical shell. Recall that the electron probability density is greatest at r = 0 (part (b) in Figure \(1\)), so the density of dots is greatest for the smallest spherical shells in part (a) in Figure \(1\). In contrast, the surface area of each spherical shell is equal to 4πr2, which increases very rapidly with increasing r (part (c) in Figure \(1\)). Because the surface area of the spherical shells increases more rapidly with increasing r than the electron probability density decreases, the plot of radial probability has a maximum at a particular distance (part (d) in Figure \(1\)). Most important, when r is very small, the surface area of a spherical shell is so small that the total probability of finding an electron close to the nucleus is very low; at the nucleus, the electron probability vanishes (part (d) in Figure \(1\)).
For the hydrogen atom, the peak in the radial probability plot occurs at r = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the n = 1 orbit. Thus the most probable radius obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the quantum mechanical Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle.
Figure \(2\) compares the electron probability densities for the hydrogen 1s, 2s, and 3s orbitals. Note that all three are spherically symmetrical. For the 2s and 3s orbitals, however (and for all other s orbitals as well), the electron probability density does not fall off smoothly with increasing r. Instead, a series of minima and maxima are observed in the radial probability plots (part (c) in Figure \(2\)). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability.
s Orbitals
Three things happen to s orbitals as n increases (Figure \(2\)):
1. They become larger, extending farther from the nucleus.
2. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude.
3. For a given atom, the s orbitals also become higher in energy as n increases because of their increased distance from the nucleus.
Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron density, as was shown for the hydrogen 1s, 2s, and 3s orbitals in part (b) in Figure \(2\). Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2s and 3s orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding.
p Orbitals
Only s orbitals are spherically symmetrical. As the value of l increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2p subshell has l = 1, with three values of ml (−1, 0, and +1), there are three 2p orbitals.
The electron probability distribution for one of the hydrogen 2p orbitals is shown in Figure \(3\). Because this orbital has two lobes of electron density arranged along the z axis, with an electron density of zero in the xy plane (i.e., the xy plane is a nodal plane), it is a 2pz orbital. As shown in Figure \(4\), the other two 2p orbitals have identical shapes, but they lie along the x axis (2px) and y axis (2py), respectively. Note that each p orbital has just one nodal plane. In each case, the phase of the wave function for each of the 2p orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the phase of the wave that describes the electron motion, not to positive or negative charges.
The surfaces shown enclose 90% of the total electron probability for the 2px, 2py, and 2pz orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2p orbital. The phase of the wave function is positive (orange) in the region of space where x, y, or z is positive and negative (blue) where x, y, or z is negative.
Just as with the s orbitals, the size and complexity of the p orbitals for any atom increase as the principal quantum number n increases. The shapes of the 90% probability surfaces of the 3p, 4p, and higher-energy p orbitals are, however, essentially the same as those shown in Figure \(4\).
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.05%3A_Atomic_Orbitals.txt |
Hybridization was introduced to explain molecular structure when the valence bond theory failed to correctly predict them. It is experimentally observed that bond angles in organic compounds are close to 109o, 120o, or 180o. According to Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs repel each other and the bonds and lone pairs around a central atom are generally separated by the largest possible angles.
Introduction
Carbon is a perfect example showing the value of hybrid orbitals. Carbon's ground state configuration is:
According to Valence Bond Theory, carbon should form two covalent bonds, resulting in a CH2, because it has two unpaired electrons in its electronic configuration.However, experiments have shown that \(CH_2\) is highly reactive and cannot exist outside of a reaction. Therefore, this does not explain how CH4 can exist. To form four bonds the configuration of carbon must have four unpaired electrons.
One way CH4 can be explained is, the 2s and the 3 2p orbitals combine to make four, equal energy sp3 hybrid orbitals. That would give us the following configuration:
Now that carbon has four unpaired electrons it can have four equal energy bonds. The hybridization of orbitals is favored because hybridized orbitals are more directional which leads to greater overlap when forming bonds, therefore the bonds formed are stronger. This results in more stable compounds when hybridization occurs.
The next section will explain the various types of hybridization and how each type helps explain the structure of certain molecules.
sp3 hybridization
sp3 hybridization can explain the tetrahedral structure of molecules. In it, the 2s orbitals and all three of the 2p orbitals hybridize to form four sp3 orbitals, each consisting of 75% p character and 25% s character. The frontal lobes align themselves in the manner shown below. In this structure, electron repulsion is minimized.
Energy changes occurring in hybridization
Hybridization of an s orbital with all three p orbitals (px , py, and pz) results in four sp3 hybrid orbitals. sp3 hybrid orbitals are oriented at bond angle of 109.5o from each other. This 109.5o arrangement gives tetrahedral geometry (Figure 4).
Example: sp3 Hybridization in Methane
Because carbon plays such a significant role in organic chemistry, we will be using it as an example here. Carbon's 2s and all three of its 2p orbitals hybridize to form four sp3 orbitals. These orbitals then bond with four hydrogen atoms through sp3-s orbital overlap, creating methane. The resulting shape is tetrahedral, since that minimizes electron repulsion.
Hybridization
Lone Pairs: Remember to take into account lone pairs of electrons. These lone pairs cannot double bond so they are placed in their own hybrid orbital. This is why H2O is tetrahedral. We can also build sp3d and sp3d2 hybrid orbitals if we go beyond s and p subshells.
sp2 hybridization
sp2 hybridization can explain the trigonal planar structure of molecules. In it, the 2s orbitals and two of the 2p orbitals hybridize to form three sp orbitals, each consisting of 67% p and 33% s character. The frontal lobes align themselves in the trigonal planar structure, pointing to the corners of a triangle in order to minimize electron repulsion and to improve overlap. The remaining p orbital remains unchanged and is perpendicular to the plane of the three sp2 orbitals.
Energy changes occurring in hybridization
Hybridization of an s orbital with two p orbitals (px and py) results in three sp2 hybrid orbitals that are oriented at 120o angle to each other (Figure 3). Sp2 hybridization results in trigonal geometry.
Example: sp2 Hybridization in Aluminum Trihydride
In aluminum trihydride, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals that align themselves in the trigonal planar structure. The three Al sp2 orbitals bond with with 1s orbitals from the three hydrogens through sp2-s orbital overlap.
Example: sp2 Hybridization in Ethene
Similar hybridization occurs in each carbon of ethene. For each carbon, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals. These hybridized orbitals align themselves in the trigonal planar structure. For each carbon, two of these sp orbitals bond with two 1s hydrogen orbitals through s-sp orbital overlap. The remaining sp2 orbitals on each carbon are bonded with each other, forming a bond between each carbon through sp2-sp2 orbital overlap. This leaves us with the two p orbitals on each carbon that have a single carbon in them. These orbitals form a ? bonds through p-p orbital overlap, creating a double bond between the two carbons. Because a double bond was created, the overall structure of the ethene compound is linear. However, the structure of each molecule in ethene, the two carbons, is still trigonal planar.
sp Hybridization
sp Hybridization can explain the linear structure in molecules. In it, the 2s orbital and one of the 2p orbitals hybridize to form two sp orbitals, each consisting of 50% s and 50% p character. The front lobes face away from each other and form a straight line leaving a 180° angle between the two orbitals. This formation minimizes electron repulsion. Because only one p orbital was used, we are left with two unaltered 2p orbitals that the atom can use. These p orbitals are at right angles to one another and to the line formed by the two sp orbitals.
Energy changes occurring in hybridization
Figure 1: Notice how the energy of the electrons lowers when hybridized.
These p orbitals come into play in compounds such as ethyne where they form two addition? bonds, resulting in in a triple bond. This only happens when two atoms, such as two carbons, both have two p orbitals that each contain an electron. An sp hybrid orbital results when an s orbital is combined with p orbital (Figure 2). We will get two sp hybrid orbitals since we started with two orbitals (s and p). sp hybridization results in a pair of directional sp hybrid orbitals pointed in opposite directions. These hybridized orbitals result in higher electron density in the bonding region for a sigma bond toward the left of the atom and for another sigma bond toward the right. In addition, sp hybridization provides linear geometry with a bond angle of 180o.
Example: sp Hybridization in Magnesium Hydride
In magnesium hydride, the 3s orbital and one of the 3p orbitals from magnesium hybridize to form two sp orbitals. The two frontal lobes of the sp orbitals face away from each other forming a straight line leading to a linear structure. These two sp orbitals bond with the two 1s orbitals of the two hydrogen atoms through sp-s orbital overlap.
Hybridization
Example: sp Hybridization in Ethyne
The hybridization in ethyne is similar to the hybridization in magnesium hydride. For each carbon, the 2s orbital hybridizes with one of the 2p orbitals to form two sp hybridized orbitals. The frontal lobes of these orbitals face away from each other forming a straight line. The first bond consists of sp-sp orbital overlap between the two carbons. Another two bonds consist of s-sp orbital overlap between the sp hybridized orbitals of the carbons and the 1s orbitals of the hydrogens. This leaves us with two p orbitals on each carbon that have a single carbon in them. This allows for the formation of two ? bonds through p-p orbital overlap. The linear shape, or 180° angle, is formed because electron repulsion is minimized the greatest in this position.
Hybridization
Problems
Using the Lewis Structures, try to figure out the hybridization (sp, sp2, sp3) of the indicated atom and indicate the atom's shape.
1. The carbon.
2. The oxygen.
3. The carbon on the right.
Answers
1. sp2- Trigonal Planar
The carbon has no lone pairs and is bonded to three hydrogens so we just need three hybrid orbitals, aka sp2.
2. sp3 - Tetrahedral
Don't forget to take into account all the lone pairs. Every lone pair needs it own hybrid orbital. That makes three hybrid orbitals for lone pairs and the oxygen is bonded to one hydrogen which requires another sp3 orbital. That makes 4 orbitals, aka sp3.
3. sp - Linear
The carbon is bonded to two other atoms, that means it needs two hybrid orbitals, aka sp.
An easy way to figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom. Use this method to go over the above problems again and make sure you understand it. It's a lot easier to figure out the hybridization this way.
Contributors
• Harpreet Chima (UCD), Farah Yasmeen | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.07%3A_How_Single_Bonds_Are_Formed_in_Organic_Compounds.txt |
Hybridization was introduced to explain molecular structure when the valence bond theory failed to correctly predict them. It is experimentally observed that bond angles in organic compounds are close to 109o, 120o, or 180o. According to Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs repel each other and the bonds and lone pairs around a central atom are generally separated by the largest possible angles.
Introduction
Carbon is a perfect example showing the value of hybrid orbitals. Carbon's ground state configuration is:
According to Valence Bond Theory, carbon should form two covalent bonds, resulting in a CH2, because it has two unpaired electrons in its electronic configuration.However, experiments have shown that \(CH_2\) is highly reactive and cannot exist outside of a reaction. Therefore, this does not explain how CH4 can exist. To form four bonds the configuration of carbon must have four unpaired electrons.
One way CH4 can be explained is, the 2s and the 3 2p orbitals combine to make four, equal energy sp3 hybrid orbitals. That would give us the following configuration:
Now that carbon has four unpaired electrons it can have four equal energy bonds. The hybridization of orbitals is favored because hybridized orbitals are more directional which leads to greater overlap when forming bonds, therefore the bonds formed are stronger. This results in more stable compounds when hybridization occurs.
The next section will explain the various types of hybridization and how each type helps explain the structure of certain molecules.
sp3 hybridization
sp3 hybridization can explain the tetrahedral structure of molecules. In it, the 2s orbitals and all three of the 2p orbitals hybridize to form four sp3 orbitals, each consisting of 75% p character and 25% s character. The frontal lobes align themselves in the manner shown below. In this structure, electron repulsion is minimized.
Energy changes occurring in hybridization
Hybridization of an s orbital with all three p orbitals (px , py, and pz) results in four sp3 hybrid orbitals. sp3 hybrid orbitals are oriented at bond angle of 109.5o from each other. This 109.5o arrangement gives tetrahedral geometry (Figure 4).
Example: sp3 Hybridization in Methane
Because carbon plays such a significant role in organic chemistry, we will be using it as an example here. Carbon's 2s and all three of its 2p orbitals hybridize to form four sp3 orbitals. These orbitals then bond with four hydrogen atoms through sp3-s orbital overlap, creating methane. The resulting shape is tetrahedral, since that minimizes electron repulsion.
Hybridization
Lone Pairs: Remember to take into account lone pairs of electrons. These lone pairs cannot double bond so they are placed in their own hybrid orbital. This is why H2O is tetrahedral. We can also build sp3d and sp3d2 hybrid orbitals if we go beyond s and p subshells.
sp2 hybridization
sp2 hybridization can explain the trigonal planar structure of molecules. In it, the 2s orbitals and two of the 2p orbitals hybridize to form three sp orbitals, each consisting of 67% p and 33% s character. The frontal lobes align themselves in the trigonal planar structure, pointing to the corners of a triangle in order to minimize electron repulsion and to improve overlap. The remaining p orbital remains unchanged and is perpendicular to the plane of the three sp2 orbitals.
Energy changes occurring in hybridization
Hybridization of an s orbital with two p orbitals (px and py) results in three sp2 hybrid orbitals that are oriented at 120o angle to each other (Figure 3). Sp2 hybridization results in trigonal geometry.
Example: sp2 Hybridization in Aluminum Trihydride
In aluminum trihydride, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals that align themselves in the trigonal planar structure. The three Al sp2 orbitals bond with with 1s orbitals from the three hydrogens through sp2-s orbital overlap.
Example: sp2 Hybridization in Ethene
Similar hybridization occurs in each carbon of ethene. For each carbon, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals. These hybridized orbitals align themselves in the trigonal planar structure. For each carbon, two of these sp orbitals bond with two 1s hydrogen orbitals through s-sp orbital overlap. The remaining sp2 orbitals on each carbon are bonded with each other, forming a bond between each carbon through sp2-sp2 orbital overlap. This leaves us with the two p orbitals on each carbon that have a single carbon in them. These orbitals form a ? bonds through p-p orbital overlap, creating a double bond between the two carbons. Because a double bond was created, the overall structure of the ethene compound is linear. However, the structure of each molecule in ethene, the two carbons, is still trigonal planar.
sp Hybridization
sp Hybridization can explain the linear structure in molecules. In it, the 2s orbital and one of the 2p orbitals hybridize to form two sp orbitals, each consisting of 50% s and 50% p character. The front lobes face away from each other and form a straight line leaving a 180° angle between the two orbitals. This formation minimizes electron repulsion. Because only one p orbital was used, we are left with two unaltered 2p orbitals that the atom can use. These p orbitals are at right angles to one another and to the line formed by the two sp orbitals.
Energy changes occurring in hybridization
Figure 1: Notice how the energy of the electrons lowers when hybridized.
These p orbitals come into play in compounds such as ethyne where they form two addition? bonds, resulting in in a triple bond. This only happens when two atoms, such as two carbons, both have two p orbitals that each contain an electron. An sp hybrid orbital results when an s orbital is combined with p orbital (Figure 2). We will get two sp hybrid orbitals since we started with two orbitals (s and p). sp hybridization results in a pair of directional sp hybrid orbitals pointed in opposite directions. These hybridized orbitals result in higher electron density in the bonding region for a sigma bond toward the left of the atom and for another sigma bond toward the right. In addition, sp hybridization provides linear geometry with a bond angle of 180o.
Example: sp Hybridization in Magnesium Hydride
In magnesium hydride, the 3s orbital and one of the 3p orbitals from magnesium hybridize to form two sp orbitals. The two frontal lobes of the sp orbitals face away from each other forming a straight line leading to a linear structure. These two sp orbitals bond with the two 1s orbitals of the two hydrogen atoms through sp-s orbital overlap.
Hybridization
Example: sp Hybridization in Ethyne
The hybridization in ethyne is similar to the hybridization in magnesium hydride. For each carbon, the 2s orbital hybridizes with one of the 2p orbitals to form two sp hybridized orbitals. The frontal lobes of these orbitals face away from each other forming a straight line. The first bond consists of sp-sp orbital overlap between the two carbons. Another two bonds consist of s-sp orbital overlap between the sp hybridized orbitals of the carbons and the 1s orbitals of the hydrogens. This leaves us with two p orbitals on each carbon that have a single carbon in them. This allows for the formation of two ? bonds through p-p orbital overlap. The linear shape, or 180° angle, is formed because electron repulsion is minimized the greatest in this position.
Hybridization
Problems
Using the Lewis Structures, try to figure out the hybridization (sp, sp2, sp3) of the indicated atom and indicate the atom's shape.
1. The carbon.
2. The oxygen.
3. The carbon on the right.
Answers
1. sp2- Trigonal Planar
The carbon has no lone pairs and is bonded to three hydrogens so we just need three hybrid orbitals, aka sp2.
2. sp3 - Tetrahedral
Don't forget to take into account all the lone pairs. Every lone pair needs it own hybrid orbital. That makes three hybrid orbitals for lone pairs and the oxygen is bonded to one hydrogen which requires another sp3 orbital. That makes 4 orbitals, aka sp3.
3. sp - Linear
The carbon is bonded to two other atoms, that means it needs two hybrid orbitals, aka sp.
An easy way to figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom. Use this method to go over the above problems again and make sure you understand it. It's a lot easier to figure out the hybridization this way.
Contributors
• Harpreet Chima (UCD), Farah Yasmeen | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.09%3A_How_a_Triple_Bond_is_Formed%3A_The_Bonds_in_Ethyne.txt |
Solutions to exercises
17.1A: The geometry and relative stability of carbon radicals
As organic chemists, we are particularly interested in radical intermediates in which the unpaired electron resides on a carbon atom. Experimental evidence indicates that the three bonds in a carbon radical have trigonal planar geometry, and therefore the carbon is considered to be sp2-hybridized with the unpaired electron occupying the perpendicular, unhybridized 2pzorbital. Contrast this picture with carbocation and carbanion intermediates, which are both also trigonal planar but whose 2pz orbitals contain zero or two electrons, respectively.
The trend in the stability of carbon radicals parallels that of carbocations (section 8.4B): tertiary radicals, for example, are more stable than secondary radicals, followed by primary and methyl radicals. This should make intuitive sense, because radicals, like carbocations, can be considered to be electron deficient, and thus are stabilized by the electron-donating effects of nearby alkyl groups. Benzylic and allylic radicals are more stable than alkyl radicals due to resonance effects - an unpaired electron can be delocalized over a system of conjugated pi bonds. An allylic radical, for example, can be pictured as a system of three parallel 2pz orbitals sharing three electrons.
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Exercise 17.1: Draw the structure of a benzylic radical compound, and then draw a resonance form showing how the radical is stabilized.
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With enough resonance stabilization, radicals can be made that are quite unreactive. One example of an inert organic radical structure is shown below.
In this molecule, the already extensive resonance stabilization is further enhanced by the ability of the chlorine atoms to shield the radical center from external reagents. The radical is, in some sense, inside a protective 'cage'.
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Exercise 17.2: Draw a resonance contributor of the structure above in which the unpaired electron is located on a chlorine atom.
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17.1B: The diradical character of triplet oxygen
You may be surprised to learn that molecular oxygen (O2) often reacts like a radical species - or more accurately, like a diradical, with two separate unpaired valence electrons. This puzzling phenomenon is best explained by molecular orbital theory (you may want to go back to chapter 2 at this point to review basic MO theory).
In molecular orbital energy diagram form, the configuration of O2 looks like this:
When the molecular orbitals of O2 are filled up with electrons according to Hund's rule (section 1.1C) , the HOMOs (Highest Occupied Molecular Orbitals) are the two antibonding π*2p orbitals, each holding a single, unpaired electron. This electron configuration, which describes oxygen in its lowest energy (ground) state, is referred to as the triplet state - the oxygen in the air around you istriplet oxygen. A higher energy state, in which the two highest energy electrons are paired in the same π*2p orbital, is called the singlet state of oxygen.
It is the last two unpaired electrons in triplet oxygen that react in radical-like fashion. For this reason, triplet O2 is often depicted by a Lewis structure as a diradical, with a single covalent bond, four lone pairs, and two unpaired electrons. The excited singlet state is often depicted as a doubly-bonded molecule.
These Lewis-dot depictions, while only approximations and unsatisfactory in many respects, can nonetheless be helpful in illustrating how O2 reacts. In any event, the molecular orbital picture of O2 is not simply an academic exercise: the diradical model for triplet oxygen is key to understanding its reactivity in many enzymatic and non-enzymatic contexts. We will see one example of triplet oxygen reactivity in section 17.2D.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.10%3A_Bonding_in_the_Methyl_Cation_the_Methyl_Radical_and_the_Methyl_Anion.txt |
Learning Objectives
• Identify three special properties of water that make it unusual for a molecule of its size, and explain how these result from hydrogen bonding.
• Explain what is meant by hydrogen bonding and the molecular structural features that bring it about.
• Describe the "structure", such as it is, of liquid water.
• Sketch out structural examples of hydrogen bonding in three small molecules other than H2O.
• Describe the roles of hydrogen bonding in proteins and in DNA.
Most students of chemistry quickly learn to relate the structure of a molecule to its general properties. Thus we generally expect small molecules to form gases or liquids, and large ones to exist as solids under ordinary conditions. And then we come to H2O, and are shocked to find that many of the predictions are way off, and that water (and by implication, life itself) should not even exist on our planet! In this section we will learn why this tiny combination of three nuclei and ten electrons possesses special properties that make it unique among the more than 15 million chemical species we presently know.
In water, each hydrogen nucleus is covalently bound to the central oxygen atom by a pair of electrons that are shared between them. In H2O, only two of the six outer-shell electrons of oxygen are used for this purpose, leaving four electrons which are organized into two non-bonding pairs. The four electron pairs surrounding the oxygen tend to arrange themselves as far from each other as possible in order to minimize repulsions between these clouds of negative charge. This would ordinarily result in a tetrahedral geometry in which the angle between electron pairs (and therefore the H-O-H bond angle) is 109.5°. However, because the two non-bonding pairs remain closer to the oxygen atom, these exert a stronger repulsion against the two covalent bonding pairs, effectively pushing the two hydrogen atoms closer together. The result is a distorted tetrahedral arrangement in which the H—O—H angle is 104.5°.
Water's large dipole moment leads to hydrogen bonding
The H2O molecule is electrically neutral, but the positive and negative charges are not distributed uniformly. This is illustrated by the gradation in color in the schematic diagram here. The electronic (negative) charge is concentrated at the oxygen end of the molecule, owing partly to the nonbonding electrons (solid blue circles), and to oxygen's high nuclear charge which exerts stronger attractions on the electrons. This charge displacement constitutes an electric dipole, represented by the arrow at the bottom; you can think of this dipole as the electrical "image" of a water molecule.
Opposite charges attract, so it is not surprising that the negative end of one water molecule will tend to orient itself so as to be close to the positive end of another molecule that happens to be nearby. The strength of this dipole-dipole attraction is less than that of a normal chemical bond, and so it is completely overwhelmed by ordinary thermal motions in the gas phase. However, when the H2O molecules are crowded together in the liquid, these attractive forces exert a very noticeable effect, which we call (somewhat misleadingly) hydrogen bonding. And at temperatures low enough to turn off the disruptive effects of thermal motions, water freezes into ice in which the hydrogen bonds form a rigid and stable network.
Notice that the hydrogen bond (shown by the dashed green line) is somewhat longer than the covalent O—H bond. It is also much weaker, about 23 kJ mol–1 compared to the O–H covalent bond strength of 492 kJ mol–1.
Forty-one anomalies of water" — some of them rather esoteric.
Water has long been known to exhibit many physical properties that distinguish it from other small molecules of comparable mass. Although chemists refer to these as the "anomalous" properties of water, they are by no means mysterious; all are entirely predictable consequences of the way the size and nuclear charge of the oxygen atom conspire to distort the electronic charge clouds of the atoms of other elements when these are chemically bonded to the oxygen.
Boiling point
The most apparent peculiarity of water is its very high boiling point for such a light molecule. Liquid methane CH4 (molecular weight 16) boils at –161°C. As you can see from this diagram, extrapolation of the boiling points of the various Group 16 hydrogen compounds to H2O suggests that this substance should be a gas under normal conditions.
Surface Tension
Compared to most other liquids, water also has a high surface tension. Have you ever watched an insect walk across the surface of a pond? The water strider takes advantage of the fact that the water surface acts like an elastic film that resists deformation when a small weight is placed on it. (If you are careful, you can also "float" a small paper clip or steel staple on the surface of water in a cup.) This is all due to the surface tension of the water. A molecule within the bulk of a liquid experiences attractions to neighboring molecules in all directions, but since these average out to zero, there is no net force on the molecule. For a molecule that finds itself at the surface, the situation is quite different; it experiences forces only sideways and downward, and this is what creates the stretched-membrane effect.
The distinction between molecules located at the surface and those deep inside is especially prominent in H2O, owing to the strong hydrogen-bonding forces. The difference between the forces experienced by a molecule at the surface and one in the bulk liquid gives rise to the liquid's surface tension. This drawing highlights two H2O molecules, one at the surface, and the other in the bulk of the liquid. The surface molecule is attracted to its neighbors below and to either side, but there are no attractions pointing in the 180° solid angle angle above the surface. As a consequence, a molecule at the surface will tend to be drawn into the bulk of the liquid. But since there must always be some surface, the overall effect is to minimize the surface area of a liquid.
The geometric shape that has the smallest ratio of surface area to volume is the sphere, so very small quantities of liquids tend to form spherical drops. As the drops get bigger, their weight deforms them into the typical tear shape.
Ice floats on water
The most energetically favorable configuration of H2O molecules is one in which each molecule is hydrogen-bonded to four neighboring molecules. Owing to the thermal motions described above, this ideal is never achieved in the liquid, but when water freezes to ice, the molecules settle into exactly this kind of an arrangement in the ice crystal. This arrangement requires that the molecules be somewhat farther apart then would otherwise be the case; as a consequence, ice, in which hydrogen bonding is at its maximum, has a more open structure, and thus a lower density than water.
Here are three-dimensional views of a typical local structure of water (left) and ice (right.) Notice the greater openness of the ice structure which is necessary to ensure the strongest degree of hydrogen bonding in a uniform, extended crystal lattice. The more crowded and jumbled arrangement in liquid water can be sustained only by the greater amount of thermal energy available above the freezing point.
When ice melts, the more vigorous thermal motion disrupts much of the hydrogen-bonded structure, allowing the molecules to pack more closely. Water is thus one of the very few substances whose solid form has a lower density than the liquid at the freezing point. Localized clusters of hydrogen bonds still remain, however; these are continually breaking and reforming as the thermal motions jiggle and shove the individual molecules. As the temperature of the water is raised above freezing, the extent and lifetimes of these clusters diminish, so the density of the water increases.
At higher temperatures, another effect, common to all substances, begins to dominate: as the temperature increases, so does the amplitude of thermal motions. This more vigorous jostling causes the average distance between the molecules to increase, reducing the density of the liquid; this is ordinary thermal expansion.
Because the two competing effects (hydrogen bonding at low temperatures and thermal expansion at higher temperatures) both lead to a decrease in density, it follows that there must be some temperature at which the density of water passes through a maximum. This temperature is 4° C; this is the temperature of the water you will find at the bottom of an ice-covered lake in which this most dense of all water has displaced the colder water and pushed it nearer to the surface.
Structure of Liquid Water
The nature of liquid water and how the H2O molecules within it are organized and interact are questions that have attracted the interest of chemists for many years. There is probably no liquid that has received more intensive study, and there is now a huge literature on this subject. The following facts are well established:
• H2O molecules attract each other through the special type of dipole-dipole interaction known as hydrogen bonding
• a hydrogen-bonded cluster in which four H2Os are located at the corners of an imaginary tetrahedron is an especially favorable (low-potential energy) configuration, but...
• the molecules undergo rapid thermal motions on a time scale of picoseconds (10–12 second), so the lifetime of any specific clustered configuration will be fleetingly brief.
A variety of techniques including infrared absorption, neutron scattering, and nuclear magnetic resonance have been used to probe the microscopic structure of water. The information garnered from these experiments and from theoretical calculations has led to the development of around twenty "models" that attempt to explain the structure and behavior of water. More recently, computer simulations of various kinds have been employed to explore how well these models are able to predict the observed physical properties of water.
This work has led to a gradual refinement of our views about the structure of liquid water, but it has not produced any definitive answer. There are several reasons for this, but the principal one is that the very concept of "structure" (and of water "clusters") depends on both the time frame and volume under consideration. Thus, questions of the following kinds are still open:
• How do you distinguish the members of a "cluster" from adjacent molecules that are not in that cluster?
• Since individual hydrogen bonds are continually breaking and re-forming on a picosecond time scale, do water clusters have any meaningful existence over longer periods of time? In other words, clusters are transient, whereas "structure" implies a molecular arrangement that is more enduring. Can we then legitimately use the term "clusters" in describing the structure of water?
• The possible locations of neighboring molecules around a given H2O are limited by energetic and geometric considerations, thus giving rise to a certain amount of "structure" within any small volume element. It is not clear, however, to what extent these structures interact as the size of the volume element is enlarged. And as mentioned above, to what extent are these structures maintained for periods longer than a few picoseconds?
In the 1950's it was assumed that liquid water consists of a mixture of hydrogen-bonded clusters (H2O)n in which n can have a variety of values, but little evidence for the existence of such aggregates was ever found. The present view, supported by computer-modeling and spectroscopy, is that on a very short time scale, water is more like a "gel" consisting of a single, huge hydrogen-bonded cluster. On a 10–12-10–9 sec time scale, rotations and other thermal motions cause individual hydrogen bonds to break and re-form in new configurations, inducing ever-changing local discontinuities whose extent and influence depends on the temperature and pressure.
Ice
Ice, like all solids, has a well-defined structure; each water molecule is surrounded by four neighboring H2Os. two of these are hydrogen-bonded to the oxygen atom on the central H2O molecule, and each of the two hydrogen atoms is similarly bonded to another neighboring H2O.
Ice forms crystals having a hexagonal lattice structure, which in their full development would tend to form hexagonal prisms very similar to those sometimes seen in quartz. This does occasionally happen, and anyone who has done much winter mountaineering has likely seen needle-shaped prisms of ice crystals floating in the air. Under most conditions, however, the snowflake crystals we see are flattened into the beautiful fractal-like hexagonal structures that are commonly observed.
Snowflakes
The H2O molecules that make up the top and bottom plane faces of the prism are packed very closely and linked (through hydrogen bonding) to the molecules inside. In contrast to this, the molecules that make up the sides of the prism, and especially those at the hexagonal corners, are much more exposed, so that atmospheric H2O molecules that come into contact with most places on the crystal surface attach very loosely and migrate along it until they are able to form hydrogen-bonded attachments to these corners, thus becoming part of the solid and extending the structure along these six directions. This process perpetuates itself as the new extensions themselves acquire a hexagonal structure.
Why is ice slippery?
At temperatures as low as 200 K, the surface of ice is highly disordered and water-like. As the temperature approaches the freezing point, this region of disorder extends farther down from the surface and acts as a lubricant.
The illustration is taken from from an article in the April 7, 2008 issue of C&EN honoring the physical chemist Gabor Somorjai who pioneered modern methods of studying surfaces.
"Pure" water
To a chemist, the term "pure" has meaning only in the context of a particular application or process. The distilled or de-ionized water we use in the laboratory contains dissolved atmospheric gases and occasionally some silica, but their small amounts and relative inertness make these impurities insignificant for most purposes. When water of the highest obtainable purity is required for certain types of exacting measurements, it is commonly filtered, de-ionized, and triple-vacuum distilled. But even this "chemically pure" water is a mixture of isotopic species: there are two stable isotopes of both hydrogen (H1 and H2, the latter often denoted by D) and oxygen (O16 and O18) which give rise to combinations such as H2O18, HDO16, etc., all of which are readily identifiable in the infrared spectra of water vapor. And to top this off, the two hydrogen atoms in water contain protons whose magnetic moments can be parallel or antiparallel, giving rise to ortho- and para-water, respectively. The two forms are normally present in a o/p ratio of 3:1.
The amount of the rare isotopes of oxygen and hydrogen in water varies enough from place to place that it is now possible to determine the age and source of a particular water sample with some precision. These differences are reflected in the H and O isotopic profiles of organisms. Thus the isotopic analysis of human hair can be a useful tool for crime investigations and anthropology research.
More about hydrogen bonding
Hydrogen bonds form when the electron cloud of a hydrogen atom that is attached to one of the more electronegative atoms is distorted by that atom, leaving a partial positive charge on the hydrogen. Owing to the very small size of the hydrogen atom, the density of this partial charge is large enough to allow it to interact with the lone-pair electrons on a nearby electronegative atom. Although hydrogen bonding is commonly described as a form of dipole-dipole attraction, it is now clear that it involves a certain measure of electron-sharing (between the external non-bonding electrons and the hydrogen) as well, so these bonds possess some covalent character.
Hydrogen bonds are longer than ordinary covalent bonds, and they are also weaker. The experimental evidence for hydrogen bonding usually comes from X-ray diffraction studies on solids that reveal shorter-than-normal distances between hydrogen and other atoms.
Hydrogen bonding in small molecules
The following examples show something of the wide scope of hydrogen bonding in molecules.
Ammonia (mp –78, bp –33°C) is hydrogen-bonded in the liquid and solid states.
Hydrogen bonding is responsible for ammonia's remarkably high solubility in water.
Many organic (carboxylic) acids form hydrogen-bonded dimers in the solid state.
Here the hydrogen bond acceptor is the π electron cloud of a benzene ring. This type of interaction is important in maintaining the shape of proteins.
Hydrogen fluoride (mp –92, bp 33°C) is another common substance that is strongly hydrogen-bonded in its condensed phases.
The bifluoride ion (for which no proper Lewis structure can be written) can be regarded as a complex ion held together by the strongest hydrogen bond known: about 155 kJ mol–1.
"As slow as molasses in the winter!" Multiple hydroxyl groups provide lots of opportunities for hydrogen bonding and lead to the high viscosities of substances such as glycerine and sugar syrups.
Hydrogen bonding in biopolymers
Hydrogen bonding plays an essential role in natural polymers of biological origin in two ways:
• Hydrogen bonding between adjacent polymer chains (intermolecular bonding);
• Hydrogen bonding between different parts of the same chain (intramolecular bonding;
• Hydrogen bonding of water molecules to –OH groups on the polymer chain ("bound water") that helps maintain the shape of the polymer.
The examples that follow are representative of several types of biopolymers.
Cellulose
Cellulose is a linear polymer of glucose (see above), containing 300 to over 10,000 units, depending on the source. As the principal structural component of plants (along with lignin in trees), cellulose is the most abundant organic substance on the earth. The role of hydrogen bonding is to cross-link individual molecules to build up sheets as shown here. These sheets than stack up in a staggered array held together by van der Waals forces. Further hydrogen-bonding of adjacent stacks bundles them together into a stronger and more rigid structure.
Proteins
These polymers made from amino acids R—CH(NH2)COOH depend on intramolecular hydrogen bonding to maintain their shape (secondary and tertiary structure) which is essential for their important function as biological catalysts (enzymes). Hydrogen-bonded water molecules embedded in the protein are also important for their structural integrity.
The principal hydrogen bonding in proteins is between the -N—H groups of the "amino" parts with the -C=O groups of the "acid" parts. These interactions give rise to the two major types of the secondary structure which refers to the arrangement of the amino acid polymer chain:
[images]
beta-sheet
Although carbon is not usually considered particularly electronegative, C—H----X hydrogen bonds are also now known to be significant in proteins.
DNA (Deoxyribonucleic acid)
Who you are is totally dependent on hydrogen bonds! DNA, as you probably know, is the most famous of the biopolymers owing to its central role in defining the structure and function of all living organisms. Each strand of DNA is built from a sequence of four different nucleotide monomers consisting of a deoxyribose sugar, phosphate groups, and a nitrogenous base conventionally identified by the letters A,T, C and G. DNA itself consists of two of these polynucleotide chains that are coiled around a common axis in a configuration something like the protein alpha helix depicted above. The sugar-and-phosphate backbones are on the outside so that the nucleotide bases are on the inside and facing each other. The two strands are held together by hydrogen bonds that link a nitrogen atom of a nucleotide in one chain with a nitrogen or oxygen on the nucleotide that is across from it on the other chain.
Efficient hydrogen bonding within this configuration can only occur between the pairs A-T and C-G, so these two complementary pairs constitute the "alphabet" that encodes the genetic information that gets transcribed whenever new protein molecules are built. Water molecules, hydrogen-bonded to the outer parts of the DNA helix, help stabilize it. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.11%3A_The_Bonds_in_Water.txt |
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
What is the first step that you must do in order to draw the electron dot formula for ammonia, NH3?
The first step in drawing the electron dot formula for ammonia is to determine the number of bonding electrons for each of the atoms. The arabic numeral above the element's column in the periodic table gives you that number.
What are the numbers of bonding electrons for the nitrogen atom and the hydrogen atoms?
The nitrogen atom has 5 electrons and each of the three hydrogen atoms has 1 electron. The total number of electrons for ammonia will therefore be 8 electrons.
Draw a hydrogen atom next to a nitrogen atom and place a pair of electrons between these two atoms.
Now draw a second hydrogen atom next to the nitrogen atom and place a second pair of electrons between these atoms.
Now draw a third hydrogen atom next to the nitrogen atom and place a third pair of electrons between these atoms.
You have now used 6 of the 8 electrons. Place the remaining two electrons on the nitrogen atom. Remember that each hydrogen atom should not have more than 2 electrons and the nitrogen atom should have a total of 8 electrons.
The electron dot formula for ammonia is now complete. Note that the nitrogen atom now has three bonds. This is the normal number of bonds for a nitrogen atom when it does not have a charge.
Click here to see an animated gif of this process...
1.13: The Bond in a Hydrogen Halide
This page discusses the acidity of the hydrogen halides: hydrogen fluoride, hydrogen chloride, hydrogen bromide and hydrogen iodide. It begins by describing their physical properties and synthesis and then explains what happens when they react with water to make acids such as hydrofluoric acid and hydrochloric acid.
Physical properties
The hydrogen halides are colorless gases at room temperature, producing steamy fumes in moist air. The boiling points of these compounds are shown in the figure below:
Hydrogen fluoride has an abnormally high boiling point for a molecule of its size (293 K or 20°C), and can condense under cool conditions. This is due to the fact that hydrogen fluoride can form hydrogen bonds. Because fluorine is the most electronegative of all the elements, the fluorine-hydrogen bond is highly polarized. The hydrogen atom carries a high partial positive charge (δ+); the fluorine is fairly negatively charged (δ-).
In addition, each fluorine atom has 3 lone pairs of electrons. Fluorine's outer electrons are at the n=2 level, and the lone pairs represent small, highly charged regions of space. Hydrogen bonds form between the δ+ hydrogen on one HF molecule and a lone pair on the fluorine of another one.The figure below illustrates this association:
The other hydrogen halides do not form hydrogen bonds because the larger halogens are not as electronegative as fluorine; therefore, the bonds are less polar. In addition, their lone pairs are at higher energy levels, so the halogen does not carry such an intensely concentrated negative charge; therefore, other hydrogen atoms are not attracted as strongly.
Making hydrogen halides
There are several ways of synthesizing hydrogen halides; the method considered here is the reaction between an ionic halide, like sodium chloride, and an acid like concentrated phosphoric(V) acid, H3PO4, or concentrated sulfuric acid.
Making hydrogen chloride
When concentrated sulfuric acid is added to sodium chloride under cold conditions, the acid donates a proton to a chloride ion, forming hydrogen chloride. In the gas phase, it immediately escapes from the system.
$Cl^- + H_2SO_4 \rightarrow HCl + HSO_4^- \nonumber$
The full equation for the reaction is:
$NaCl + H_2SO_4 \rightarrow HCl + NaHSO_4 \nonumber$
Sodium bisulfate is also formed in the reaction. Concentrated phosphoric(V) acid reacts similarly, according to the following equation:
$Cl^- + H_3PO_4 \rightarrow HCl + H_2PO_4^- \nonumber$
The full ionic equation showing the formation of the salt, sodium biphosphate(V), is given below:
$NaCl + H_3PO_4 \rightarrow HCl + NaH_2PO_4 \nonumber$
Making other hydrogen halides
All hydrogen halides can be formed by the same method, using concentrated phosphoric(V) acid.
Concentrated sulfuric acid, however, behaves differently. Hydrogen fluoride can be made with sulfuric acid, but hydrogen bromide and hydrogen iodide cannot.
The problem is that concentrated sulfuric acid is an oxidizing agent, and as well as producing hydrogen bromide or hydrogen iodide, some of the halide ions are oxidized to bromine or iodine. Phosphoric acid does not have this ability because it is not an oxidant.
The acidity of the hydrogen halides
Hydrogen chloride as an acid
By the Bronsted-Lowry definition of an acid as a proton donor, hydrogen chloride is an acid because it transfers protons to other species. Consider its reaction with water.
Hydrogen chloride gas is soluble in water; its solvated form is hydrochloric acid. Hydrogen chloride fumes in moist air are caused by hydrogen chloride reacting with water vapor in the air to produce a cloud of concentrated hydrochloric acid.
A proton is donated from the hydrogen chloride to one of the lone pairs on a water molecule.
A coordinate (dative covalent) bond is formed between the oxygen and the transferred proton.
The equation for the reaction is the following:
$H_2O + HCl \rightarrow H_3O^+ + Cl^- \nonumber$
The H3O+ ion is the hydroxonium ion (also known as the hydronium ion or the oxonium ion). This is the normal form of protons in water; sometimes it is shortened to the proton form, H+(aq), for brevity.
When hydrogen chloride dissolves in water (to produce hydrochloric acid), almost all the hydrogen chloride molecules react in this way. Hydrochloric acid is therefore a strong acid. An acid is strong if it is fully ionized in solution.
Hydrobromic acid and hydriodic acid as strong acids
Hydrogen bromide and hydrogen iodide dissolve in (and react with) water in exactly the same way as hydrogen chloride does. Hydrogen bromide forms hydrobromic acid; hydrogen iodide gives hydriodic acid. Both of these are also strong acids.
Hydrofluoric acid as an exception
By contrast, although hydrogen fluoride dissolves freely in water, hydrofluoric acid is only a weak acid; it is similar in strength to organic acids like methanoic acid. The complicated reason for this is discussed below.
The bond enthalpy of the H-F bond
Because the fluorine atom is so small, the bond enthalpy (bond energy) of the hydrogen-fluorine bond is very high. The chart below gives values for all the hydrogen-halogen bond enthalpies:
bond enthalpy
(kJ mol-1)
H-F +562
H-Cl +431
H-Br +366
H-I +299
In order for ions to form when the hydrogen fluoride reacts with water, the H-F bond must be broken. It would seem reasonable to say that the relative reluctance of hydrogen fluoride to react with water is due to the large amount of energy needed to break that bond, but this explanation does not hold.
The energetics of the process from HX(g) to X-(aq)
The energetics of this sequence are of interest:
All of these terms are involved in the overall enthalpy change as you convert HX(g) into its ions in water.
However, the terms involving the hydrogen are the same for every hydrogen halide. Only the values for the red terms in the diagram need be considered. The values are tabulated below:
bond enthalpy of HX
(kJ mol-1)
electron affinity of X
(kJ mol-1)
hydration enthalpy of X-
(kJ mol-1)
sum of these
(kJ mol-1)
HF +562 -328 -506 -272
HCl +431 -349 -364 -282
HBr +366 -324 -335 -293
HI +299 -295 -293 -289
There is virtually no difference in the total HF and HCl values.
The large bond enthalpy of the H-F bond is offset by the large hydration enthalpy of the fluoride ion. There is a very strong attraction between the very small fluoride ion and the water molecules. This releases a lot of heat (the hydration enthalpy) when the fluoride ion becomes wrapped in water molecules.
Other attractions in the system
The energy terms considered previously have concerned HX molecules in the gas phase. To reach a more correct explanation, the molecules must first be considered as unreacted aqueous HX molecules. The equation for this is given below:
The equation is incorporated into an improved energy cycle as follows:
Unfortunately, values for the first step in the reaction are not readily available. However, in each case, the initial separation of the HX from water molecules is endothermic. Energy is required to break the intermolecular attractions between the HX molecules and water.
That energy is much greater for hydrogen fluoride because it forms hydrogen bonds with water. The other hydrogen halides experience only the weaker van der Waals dispersion forces or dipole-dipole attractions.
The overall enthalpy changes (including all the stages in the energy cycle) for the reactions are given in the table below:
$HX(aq) + H_2O (l) \rightarrow H_3O^+ (aq) + X^- (aq) \nonumber$
enthalpy change
(kJ mol-1)
HF -13
HCl -59
HBr -63
H-I -57
The enthalpy change for HF is much smaller in magnitude than that for the other three hydrogen halides, but it is still negative exothermic change. Therefore, more information is needed to explain why HF is a weak acid.
Entropy and free energy considerations
The free energy change, not the enthalpy change, determines the extent and direction of a reaction.
Free energy change is calculated from the enthalpy change, the temperature of the reaction and the entropy change during the reaction.
For simplicity, entropy can be thought of as a measure of the amount of disorder in a system. Entropy is given the symbol S. If a system becomes more disordered, then its entropy increases. If it becomes more ordered, its entropy decreases.
The key equation is given below:
In simple terms, for a reaction to happen, the free energy change must be negative. But more accurately, the free energy change can be used to calculate a value for the equilibrium constant for a reaction using the following expression:
The term Ka is the equilibrium constant for the reaction below:
$HX(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + X^-(aq) \nonumber$
The values for TΔS (needed to calculate ΔG) for the four reactions at a temperature of 298 K are tabulated below:
TS
(kJ mol-1)
HF -29
HCl -13
HBr -4
H-I +4
Notice that at the top of the group, the systems become more ordered when the HX reacts with the water. The entropy of the system (the amount of disorder) decreases, particularly for the hydrogen fluoride.
The reason for this is that the very strong attraction between H3O+ and F-(aq) imposes a lot of order on the system, as does the attraction between the water molecules and the various ions present. These attractions are each greatest for the small fluoride ions.
The total effect on the free energy change, and therefore the value of the equilibrium constant, can now be considered. These values are calculated in the following table:
H
(kJ mol-1)
TS
(kJ mol-1)
G
(kJ mol-1)
Ka
(mol dm-3)
HF -13 -29 +16 1.6 x 10-3
HCl -59 -13 -46 1.2 x 108
HBr -63 -4 -59 2.2 x 1010
HI -57 +4 -61 5.0 x 1010
The values for these estimated equilibrium constants for HCl, HBr and HI are so high that the reaction can be considered "one-way". The ionization is virtually 100% complete. These are all strong acids, increasing in strength down the group.
By contrast, the estimated Ka for hydrofluoric acid is small. Hydrofluoric acid only ionizes to a limited extent in water. Therefore, it is a weak acid.
The estimated value for HF in the table can be compared to the experimental value:
• Experimental value: 5.6 x 10-4 mol dm-3
• Estimated value: 1.6 x 10-3 mol dm-3
These values differ by an order of magnitude, but because of the logarithmic relationship between the free energy and the equilibrium constant, a very small change in ΔG has a very large effect on Ka.
To have the values in close agreement, ΔG would have to increase from +16 to +18.5 kJ mol-1. Given the uncertainty in the values used to calculate ΔG, the difference between the calculated value and the experimental value could easily fall within this range.
Summary: Why is hydrofluoric acid a weak acid?
The two main factors are:
• Entropy decreases dramatically when the hydrogen fluoride reacts with water. This is particularly noticeable with hydrogen fluoride because the attraction of the small fluoride ions produced imposes significant order on the surrounding water molecules and nearby hydronium ions. The effect decreases with larger halide ions.
• Very strong hydrogen bonding exists between the hydrogen fluoride molecules and water molecules. This costs a large amount of energy to break. This effect does not occur in the other hydrogen halides.
Contributors and Attributions
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.12%3A_The_Bonds_in_Ammonia_and_in_the_Ammonium_Ion.txt |
Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. For example, in diatomic nitrogen, N≡N, the bond order is 3; in acetylene, H−C≡C−H, the carbon-carbon bond order is also 3, and the C−H bond order is 1. Bond order and bond length indicate the type and strength of covalent bonds between atoms. Bond order and length are inversely proportional to each other: when bond order is increased, bond length is decreased.
Introduction
Chemistry deals with the way in which subatomic particles bond together to form atoms. Chemistry also focuses on the way in which atoms bond together to form molecules. In the atomic structure, electrons surround the atomic nucleus in regions called orbitals. Each orbital shell can hold a certain number of electrons. When the nearest orbital shell is full, new electrons start to gather in the next orbital shell out from the nucleus, and continue until that shell is also full. The collection of electrons continues in ever widening orbital shells as larger atoms have more electrons than smaller atoms. When two atoms bond to form a molecule, their electrons bond them together by mixing into openings in each others' orbital shells. As with the collection of electrons by the atom, the formation of bonds by the molecule starts at the nearest available orbital shell opening and expand outward.
Bond Order
Bond order is the number of bonding pairs of electrons between two atoms. In a covalent bond between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. To determine the bond order between two covalently bonded atoms, follow these steps:
1. Draw the Lewis structure.
2. Determine the type of bonds between the two atoms.
• 0: No bond
• 1: Single bond
• 2: double bond
• 3: triple bond
If the bond order is zero, the molecule cannot form. The higher bond orders indicate greater stability for the new molecule. In molecules that have resonance bonding, the bond order does not need to be an integer.
Example $1$: $CN^-$
Determine the bond order for cyanide, CN-.
Solution
1) Draw the Lewis structure.
2) Determine the type of bond between the two atoms.
Because there are 3 dashes, the bond is a triple bond. A triple bond corresponds to a bond order of 3.
Example $2$: $H_2$
Determine the bond order for hydrogen gas, H2.
Solution
1) Draw the Lewis structure.
2) Determine the type of bond between the two atoms.
There is only one pair of shared electrons (or dash), indicating is a single bond, with a bond order of 1.
Polyatomic molecules
If there are more than two atoms in the molecule, follow these steps to determine the bond order:
1. Draw the Lewis structure.
2. Count the total number of bonds.
3. Count the number of bond groups between individual atoms.
4. Divide the number of bonds between atoms by the total number of bond groups in the molecule.
Example $3$: $NO_3^-$
Determine the bond order for nitrate, $NO_3^-$.
Solution
1) Draw the Lewis structure.
2) Count the total number of bonds.
4
The total number of bonds is 4.
3) Count the number of bond groups between individual atoms.
3
The number of bond groups between individual atoms is 3.
4) Divide the number of bonds between individual atoms by the total number of bonds.
$\dfrac{4}{3}= 1.33$
The bond order is 1.33
Example $4$: $NO^+_2$
Determine the bond order for nitronium ion: $NO_2^+$.
Solution
1) Draw the Lewis Structure.
2) Count the total number of bonds.
4
The total number of bonds is 4.
3) Count the number of bond groups between individual atoms.
2
The number of bond groups between atoms is 2.
4) Divide the bond groups between individual atoms by the total number of bonds.
$\frac{4}{2} = 2$
The bond order is 2.
A high bond order indicates more attraction between electrons. A higher bond order also means that the atoms are held together more tightly. With a lower bond order, there is less attraction between electrons and this causes the atoms to be held together more loosely. Bond order also indicates the stability of the bond. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability.
Bond order increases across a period and decreases down a group.
Bond Length
Bond length is defined as the distance between the centers of two covalently bonded atoms. The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms. Bond length is reported in picometers. Therefore, bond length increases in the following order: triple bond < double bond < single bond.
To find the bond length, follow these steps:
1. Draw the Lewis structure.
2. Look up the chart below for the radii for the corresponding bond.
3. Find the sum of the two radii.
4
Determine the carbon-to-chlorine bond length in CCl4.
Solution
Using Table A3, a C single bond has a length of 75 picometers and that a Cl single bond has a length of 99 picometers. When added together, the bond length of a C-Cl bond is approximately 174 picometers.
2
Determine the carbon-oxygen bond length in CO2.
Solution
Using Table A3, we see that a C double bond has a length of 67 picometers and that an O double bond has a length of 57 picometers. When added together, the bond length of a C=O bond is approximately 124 picometers.
Because the bond length is proportional to the atomic radius, the bond length trends in the periodic table follow the same trends as atomic radii: bond length decreases across a period and increases down a group.
Problems
1. What is the bond order of $O_2$?
2. What is the bond order of $NO_3^-$?
3. What is the carbon-nitrogen bond length in $HCN$?
4. Is the carbon-to-oxygen bond length greater in $CO_2$ or $CO$?
5. What is the nitrogen-fluoride bond length in $NF_3$?
Solutions
1. First, write the Lewis structure for $O_2$.
There is a double bond between the two oxygen atoms; therefore, the bond order of the molecule is 2.
2. The Lewis structure for NO3- is given below:
To find the bond order of this molecule, take the average of the bond orders. N=O has a bond order of two, and both N-O bonds have a bond order of one. Adding these together and dividing by the number of bonds (3) reveals that the bond order of nitrate is 1.33.
3. To find the carbon-nitrogen bond length in HCN, draw the Lewis structure of HCN.
The bond between carbon and nitrogen is a triple bond, and a triple bond between carbon and nitrogen has a bond length of approximately 60 + 54 =114 pm.
4. From the Lewis structures for CO2 and CO, there is a double bond between the carbon and oxygen in CO2 and a triple bond between the carbon and oxygen in CO.
Referring to the table above, a double bond between carbon and oxygen has a bond length of approximately 67 + 57 = 124 pm and a triple bond between carbon and oxygen has a bond length of approximately 60 + 53 =113 pm. Therefore, the bond length is greater in CO2.
Another method makes use of the fact that the more electron bonds between the atoms, the tighter the electrons are pulling the atoms together. Therefore, the bond length is greater in CO2.
5. To find the nitrogen-to-fluorine bond length in NF3, draw the Lewis structure.
The bond between fluorine and nitrogen is a single bond. From the table above, a single bond between fluorine and nitrogen has a bond length of approximately 64 + 71 =135 pm.
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1.16: An Introduction to Acids and Bases
An acid–base reaction is a chemical reaction that occurs between an acid and a base. Several theoretical frameworks provide alternative conceptions of the reaction mechanisms and their application in solving related problems; these are called acid–base theories, for example, Brønsted–Lowry acid–base theory. Their importance becomes apparent in analyzing acid–base reactions for gaseous or liquid species, or when acid or base character may be somewhat less apparent.
Thumbnail: Warning picture used with dangerous acids and dangerous bases. Bases are the opposites of acids. (Public Domain). | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.14%3A_Summary%3A_Hybridization_Bond_Lengths_Bond_Strengths_and_Bond_Angles.txt |
Learning Objectives
• To define the pH scale as a measure of acidity of a solution
• Tell the origin and the logic of using the pH scale.
• Apply the same strategy for representing other types of quantities such as pKa, pKb, pKw.
Auto-Ionization of Water
Because of its amphoteric nature (i.e., acts as both an acid or a base), water does not always remain as $H_2O$ molecules. In fact, two water molecules react to form hydronium and hydroxide ions:
$\ce{ 2 H_2O (l) \rightleftharpoons H_3O^+ (aq) + OH^{−} (aq)} \label{1}$
This is also called the self-ionization of water. The concentration of $H_3O^+$ and $OH^-$ are equal in pure water because of the 1:1 stoichiometric ratio of Equation $\ref{1}$. The molarity of H3O+ and OH- in water are also both $1.0 \times 10^{-7} \,M$ at 25° C. Therefore, a constant of water ($K_w$) is created to show the equilibrium condition for the self-ionization of water. The product of the molarity of hydronium and hydroxide ion is always $1.0 \times 10^{-14}$ (at room temperature).
$K_w= [H_3O^+][OH^-] = 1.0 \times 10^{-14} \label{2}$
Equation $\ref{2}$ also applies to all aqueous solutions. However, $K_w$ does change at different temperatures, which affects the pH range discussed below.
$H^+$ and $H_3O^+$ is often used interchangeably to represent the hydrated proton, commonly call the hydronium ion.
Equation \ref{1} can also be written as
$H_2O \rightleftharpoons H^+ + OH^- \label{3}$
As expected for any equilibrium, the reaction can be shifted to the reactants or products:
• If an acid ($H^+$) is added to the water, the equilibrium shifts to the left and the $OH^-$ ion concentration decreases
• If base ( $OH^-$) is added to water, the equilibrium shifts to left and the $H^+$ concentration decreases.
pH and pOH
Because the constant of water, Kw is $1.0 \times 10^{-14}$ (at 25° C), the $pK_w$ is 14, the constant of water determines the range of the pH scale. To understand what the pKw is, it is important to understand first what the "p" means in pOH and pH. The addition of the "p" reflects the negative of the logarithm, $-\log$. Therefore, the pH is the negative logarithm of the molarity of H, the pOH is the negative logarithm of the molarity of $\ce{OH^-}$, and the $pK_w$ is the negative logarithm of the constant of water:
\begin{align} pH &= -\log [H^+] \label{4a} \[4pt] pOH &= -\log [OH^-] \label{4b} \[4pt] pK_w &= -\log [K_w] \label{4c} \end{align}
At room temperature,
$K_w =1.0 \times 10^{-14} \label{4d}$
So
\begin{align} pK_w &=-\log [1.0 \times 10^{-14}] \label{4e} \[4pt] &=14 \end{align}
Using the properties of logarithms, Equation $\ref{4e}$ can be rewritten as
$10^{-pK_w}=10^{-14}. \label{4f}$
The equation also shows that each increasing unit on the scale decreases by the factor of ten on the concentration of $\ce{H^{+}}$. Combining Equations \ref{4a} - \ref{4c} and \ref{4e} results in this important relationship:
$pK_w= pH + pOH = 14 \label{5b}$
Equation \ref{5b} is correct only at room temperature since changing the temperature will change $K_w$.
The pH scale is logarithmic, meaning that an increase or decrease of an integer value changes the concentration by a tenfold. For example, a pH of 3 is ten times more acidic than a pH of 4. Likewise, a pH of 3 is one hundred times more acidic than a pH of 5. Similarly a pH of 11 is ten times more basic than a pH of 10.
Properties of the pH Scale
From the simple definition of pH in Equation \ref{4a}, the following properties can be identified:
• This scale is convenient to use, because it converts some odd expressions such as $1.23 \times 10^{-4}$ into a single number of 3.91.
• This scale covers a very large range of $\ce{[H+]}$, from 0.1 to 10-14. When $\ce{[H+]}$ is high, we usually do not use the pH value, but simply the $\ce{[H+]}$. For example, when $\mathrm{[H^+] = 1.0}$, pH = 0. We seldom say the pH is 0, and that is why you consider pH = 0 such an odd expression. A pH = -0.30 is equivalent to a $\ce{[H+]}$ of 2.0 M. Negative pH values are only for academic exercises. Using the concentrations directly conveys a better sense than the pH scales.
• The pH scale expands the division between zero and 1 in a linear scale or a compact scale into a large scale for comparison purposes. In mathematics, you learned that there are infinite values between 0 and 1, or between 0 and 0.1, or between 0 and 0.01 or between 0 and any small value. Using a log scale certainly converts infinite small quantities into infinite large quantities.
• The non-linearity of the pH scale in terms of $\ce{[H+]}$ is easily illustrated by looking at the corresponding values for pH between 0.1 and 0.9 as follows:
pH 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
[H+] 1 0.79 0.63 0.50 0.40 0.32 0.25 0.20 0.16 0.13
• Because the negative log of $\ce{[H+]}$ is used in the pH scale, the pH scale usually has positive values. Furthermore, the larger the pH, the smaller the $\ce{[H+]}$.
The Effective Range of the pH Scale
It is common that the pH scale is argued to range from 0-14 or perhaps 1-14, but neither is correct. The pH range does not have an upper nor lower bound, since as defined above, the pH is an indication of concentration of H+. For example, at a pH of zero the hydronium ion concentration is one molar, while at pH 14 the hydroxide ion concentration is one molar. Typically the concentrations of H+ in water in most solutions fall between a range of 1 M (pH=0) and 10-14 M (pH=14). Hence a range of 0 to 14 provides sensible (but not absolute) "bookends" for the scale. One can go somewhat below zero and somewhat above 14 in water, because the concentrations of hydronium ions or hydroxide ions can exceed one molar. Figure $1$ depicts the pH scale with common solutions and where they are on the scale.
Quick Interpretation
• If pH >7, the solution is basic. The pOH should be looked in the perspective of OH- instead of H+. Whenever the value of pOH is less than 7, then it is considered basic. And therefore there are more OH- than H+ in the solution.
• At pH 7, the substance or solution is at neutral and means that the concentration of H+ and OH- ion is the same.
• If pH < 7, the solution is acidic. There are more H+ than OH- in an acidic solution.
• The pH scale does not have an upper nor lower bound.
Example $1$
If the concentration of $NaOH$ in a solution is $2.5 \times 10^{-4}\; M$, what is the concentration of $H_3O^+$?
Solution
We can assume room temperature, so
$1.0 \times 10^{-14} = [H_3O^+][OH^-] \nonumber$
to find the concentration of H3O+, solve for the [H3O+].
$\dfrac{1.0 \times 10^{-14}}{[OH^-]} = [H_3O^+]$
$\dfrac{1.0 \times 10^{-14}}{2.5 \times 10^{-4}} = [H_3O^+] = 4.0 \times 10^{-11}\; M$
Example $2$
1. Find the pH of a solution of 0.002 M of HCl.
2. Find the pH of a solution of 0.00005 M NaOH.
Solution
1. The equation for pH is -log [H+]
$[H^+]= 2.0 \times 10^{-3}\; M \nonumber$
$pH = -\log [2.0 \times 10^{-3}] = 2.70 \nonumber$
1. The equation for pOH is -log [OH-]
$[OH^-]= 5.0 \times 10^{-5}\; M \nonumber$
$pOH = -\log [5.0 \times 10^{-5}] = 4.30 \nonumber$
$pK_w = pH + pOH \nonumber$
and
$pH = pK_w - pOH \nonumber$
then
$pH = 14 - 4.30 = 9.70 \nonumber$
Example $3$: Soil
If moist soil has a pH of 7.84, what is the H+ concentration of the soil solution?
Solution
$pH = -\log [H^+] \nonumber$
$7.84 = -\log [H^+] \nonumber$
$[H^+] = 1.45 \times 10^{-8} M \nonumber$
Hint
Place -7.84 in your calculator and take the antilog (often inverse log or 10x) = 1.45 x 10-8M
Proper Definition of pH
The pH scale was originally introduced by the Danish biochemist S.P.L. Sørenson in 1909 using the symbol pH. The letter p is derived from the German word potenz meaning power or exponent of, in this case, 10. In 1909, S.P.L. Sørenson published a paper in Biochem Z in which he discussed the effect of H+ ions on the activity of enzymes. In the paper, he invented the term pH (purported to mean pondus hydrogenii in Latin) to describe this effect and defined it as the $-\log[H^+]$. In 1924, Sørenson realized that the pH of a solution is a function of the "activity" of the H+ ion and not the concentration. Thus, he published a second paper on the subject. A better definition would be
$pH = -\log\,a\{\ce{H^{+}}\}$
where $a\{H^+\}$ denotes the activity (an effective concentration) of the H+ ions. The activity of an ion is a function of many variables of which concentration is one.
• Concentration is abbreviated by using square brackets, e.g., $[H_3O^+]$ is the concentration of hydronium ion in solution.
• Activity is abbreviated by using "a" with curly brackets, e.g., $a\{H_3O^+\}$ is the activity of hydronium ions in solution
Because of the difficulty in accurately measuring the activity of the $\ce{H^{+}}$ ion for most solutions the International Union of Pure and Applied Chemistry (IUPAC) and the National Bureau of Standards (NBS) has defined pH as the reading on a pH meter that has been standardized against standard buffers. The following equation is used to calculate the pH of all solutions:
\begin{align} pH &= \dfrac{F(E-E_{standard})}{RT\;\ln 10} + pH_{standard} \label{6a} \[4pt] &= \dfrac{5039.879 (E-E_{standard})}{T} + pH_{standard} \label{6b} \end{align}
with
• $R$ is the ideal gas constant,
• $F$ is the Faraday's constant, and
• $T$ is absolute temperature (in K)
The activity of the H+ ion is determined as accurately as possible for the standard solutions used. The identity of these solutions vary from one authority to another, but all give the same values of pH to ± 0.005 pH unit. The historical definition of pH is correct for those solutions that are so dilute and so pure the H+ ions are not influenced by anything but the solvent molecules (usually water).
When measuring pH, [H+] is in units of moles of H+ per liter of solution. This is a reasonably accurate definition at low concentrations (the dilute limit) of H+. At very high concentrations (10 M hydrochloric acid or sodium hydroxide, for example,) a significant fraction of the ions will be associated into neutral pairs such as H+Cl, thus reducing the concentration of “available” ions to a smaller value which we will call the effective concentration. It is the effective concentration of H+ and OH that determines the pH and pOH. The pH scale as shown above is called sometimes "concentration pH scale" as opposed to the "thermodynamic pH scale". The main difference between both scales is that in thermodynamic pH scale one is interested not in H+concentration, but in H+activity. What a person measures in the solution is just activity, not the concentration. Thus it is thermodynamic pH scale that describes real solutions, not the concentration one.
For solutions in which ion concentrations don't exceed 0.1 M, the formulas pH = –log [H+] and pOH = –log[OH] are generally reliable, but don't expect a 10.0 M solution of a strong acid to have a pH of exactly –1.00! However, this definition is only an approximation (albeit very good under most situations) of the proper definition of pH, which depends on the activity of the hydrogen ion:
$pH= -\log a\{H^+\} \approx -\log [H^+] \label{7}$
The activity is a measure of the "effective concentration" of a substance, is often related to the true concentration via an activity coefficient, $\gamma$:
$a{H^+}=\gamma [H^+] \label{8}$
Calculating the activity coefficient requires detailed theories of how charged species interact in solution at high concentrations (e.g., the Debye-Hückel Theory). In most solutions the pH differs from the -log[H+ ] in the first decimal point. The following table gives experimentally determined pH values for a series of HCl solutions of increasing concentration at 25 °C.
Table $1$: HCl Solutions with corresponding pH values. Data taken from Christopher G. McCarty and Ed Vitz, Journal of Chemical Education, 83(5), 752 (2006) and G.N. Lewis, M. Randall, K. Pitzer, D.F. Brewer, Thermodynamics (McGraw-Hill: New York, 1961; pp. 233-34).
Molar Concentration of $HCl$ pH defined as Concentration Experimentally Determined pH Relative Deviation
0.00050 3.30 3.31 0.3%
0.0100 2 2.04 1.9%
0.100 1 1.10 9%
0.40 0.39 0.52 25%
7.6 -0.88 -1.85 52%
While the pH scale formally measures the activity of hydrogen ions in a substance or solution, it is typically approximated as the concentration of hydrogen ions; this approximation is applicable only under low concentrations.
Living Systems
Molecules that make up or are produced by living organisms usually function within a narrow pH range (near neutral) and a narrow temperature range (body temperature). Many biological solutions, such as blood, have a pH near neutral. pH influences the structure and the function of many enzymes (protein catalysts) in living systems. Many of these enzymes have narrow ranges of pH activity. Cellular pH is so important that death may occur within hours if a person becomes acidotic (having increased acidity in the blood). As one can see pH is critical to life, biochemistry, and important chemical reactions. Common examples of how pH plays a very important role in our daily lives are given below:
• Water in swimming pool is maintained by checking its pH. Acidic or basic chemicals can be added if the water becomes too acidic or too basic.
• Whenever we get a heartburn, more acid build up in the stomach and causes pain. We needs to take antacid tablets (a base) to neutralize excess acid in the stomach.
• The pH of blood is slightly basic. A fluctuation in the pH of the blood can cause in serious harm to vital organs in the body.
• Certain diseases are diagnosed only by checking the pH of blood and urine.
• Certain crops thrive better at certain pH range.
• Enzymes activate at a certain pH in our body.
Table $2$: pH in Living Systems
Compartment pH
Gastric Acid 1
Lysosomes 4.5
Granules of Chromaffin Cells 5.5
Human Skin 5.5
Urine 6
Neutral H2O at 37 °C 6.81
Cytosol 7.2
Cerebrospinal Fluid 7.3
Blood 7.43-7.45
Mitochondrial Matrix 7.5
Pancreas Secretions 8.1
Problems
1. In a solution of $2.4 \times 10^{-3} M$ of HI, find the concentration of $OH^-$.
2. Determine the pH of a solution that is 0.0035 M HCl.
3. Determine the [H3O+] of a solution with a pH = 5.65
4. If the pOH of NH3, ammonia, in water is 4.74. What is the pH?
5. Pepsin, a digestive enzyme in our stomach, has a pH of 1.5. Find the concentration of OH- in the stomach.
Solutions
1. We use the dissociation of water equation to find [OH-].
Kw = [H3O+][OH-] = 1.0 X 10-14
Solve for [OH-]
[OH-] = (1.0 X 10-14)/ [H3O+]
Plug in the molarity of HI and solve for OH-.
[OH-] = (1.0 X 10-14)/ [2.4 X 10-3] = 4.17 X 10-12 M.
2. pH = -log[H3O+]
Plug the molarity of the HCl in and solve for pH.
pH = -log[0.0035] = 2.46
3. pH = -log[H3O+]
Plug in the pH and solve for [H3O+]
5.65 = -log[H3O+]
Move the negative sign to the pH. -5.65 = log[H3O+]
10-5.65=10log[H3O+] = 2.24 X 10-6 M
4. pH + pOH = 14
Solve for pH.
14 - pOH = pH
14 - 4.74 = pH = 9.26
5. There are several ways to do this problem.
Answer 1.
pH + pOH = 14
Solve for pOH.
pOH = 14 - pH
pOH = 14 - 1.5 = 12.5
When the pOH is solved, solve for the concentration by using log.
pOH = -log[OH-]
12.5 = -log[OH-]
-12.5 = log[OH-]
10-12.5 = 10log[OH-] = 3.16 X 10-13 M.
Answer 2.
pH = -log[H+]
Plug in the pH and solve for the molarity of H+ of pepsin.
1.5 = -log[H+]
-1.5 = log[H+]
10-1.5 = 10log[H+] = [H+]= 0.032
Use the concentration of H+ to solve for the concentration of OH-.
[H+][OH-] = 1.0 X 10-14
Plug in the [H+] and solve for [OH-].
[OH-] = (1.0 X 10-14)/[H3O+]
[OH-] = (1.0 X 10-14)/(0.032) = 3.125 X 10-14 M | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.17%3A_pka_and_pH.txt |
This page explains the acidity of simple organic acids and looks at the factors which affect their relative strengths.
Organic acids as weak acids
For the purposes of this topic, we are going to take the definition of an acid as "a substance which donates hydrogen ions (protons) to other things". We are going to get a measure of this by looking at how easily the acids release hydrogen ions to water molecules when they are in solution in water.
An acid in solution sets up this equilibrium:
A hydroxonium ion is formed together with the anion (negative ion) from the acid.
This equilibrium is sometimes simplified by leaving out the water to emphasise the ionisation of the acid.
If you write it like this, you must include the state symbols - "(aq)". Writing H+(aq) implies that the hydrogen ion is attached to a water molecule as H3O+. Hydrogen ions are always attached to something during chemical reactions.
The organic acids are weak in the sense that this ionisation is very incomplete. At any one time, most of the acid will be present in the solution as un-ionised molecules. For example, in the case of dilute ethanoic acid, the solution contains about 99% of ethanoic acid molecules - at any instant, only about 1% have actually ionised. The position of equilibrium therefore lies well to the left.
Comparing the strengths of weak acids
The strengths of weak acids are measured on the pKa scale. The smaller the number on this scale, the stronger the acid is.
Three of the compounds we shall be looking at, together with their pKa values are:
Remember - the smaller the number the stronger the acid. Comparing the other two to ethanoic acid, you will see that phenol is very much weaker with a pKa of 10.00, and ethanol is so weak with a pKa of about 16 that it hardly counts as acidic at all!
Why are these acids acidic?
In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X":
So . . . if the same bond is being broken in each case, why do these three compounds have such widely different acid strengths?
Differences in acid strengths between carboxylic acids, phenols and alcohols
Two of the factors which influence the ionization of an acid are:
• the strength of the bond being broken,
• the stability of the ions being formed.
In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar. The most important factor in determining the relative acid strengths of these molecules is the nature of the ions formed. You always get a hydroxonium ion - so that's constant - but the nature of the anion (the negative ion) varies markedly from case to case.
Example \(1\): Ethanoic Acid
Ethanoic acid has the structure:
The acidic hydrogen is the one attached to the oxygen. When ethanoic acid ionises it forms the ethanoate ion, CH3COO-.
You might reasonably suppose that the structure of the ethanoate ion was as below, but measurements of bond lengths show that the two carbon-oxygen bonds are identical and somewhere in length between a single and a double bond.
To understand why this is, you have to look in some detail at the bonding in the ethanoate ion. Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond. Pi bonds are made by sideways overlap between p orbitals on the carbon and the oxygen.
In an ethanoate ion, one of the lone pairs on the negative oxygen ends up almost parallel to these p orbitals, and overlaps with them
This leads to a delocalised pi system over the whole of the -COO- group, rather like that in benzene.
All the oxygen lone pairs have been left out of this diagram to avoid confusion. Because the oxygens are more electronegative than the carbon, the delocalised system is heavily distorted so that the electrons spend much more time in the region of the oxygen atoms.
So where is the negative charge in all this? It has been spread around over the whole of the -COO- group, but with the greatest chance of finding it in the region of the two oxygen atoms. Ethanoate ions can be drawn simply as:
The dotted line represents the delocalisation. The negative charge is written centrally on that end of the molecule to show that it isn't localised on one of the oxygen atoms. The more you can spread charge around, the more stable an ion becomes. In this case, if you delocalise the negative charge over several atoms, it is going to be much less attractive to hydrogen ions - and so you are less likely to re-form the ethanoic acid.
Example \(2\): Phenol
Phenols have an -OH group attached directly to a benzene ring. Phenol itself is the simplest of these with nothing else attached to the ring apart from the -OH group.
When the hydrogen-oxygen bond in phenol breaks, you get a phenoxide ion, C6H5O-.
Delocalisation also occurs in this ion. This time, one of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring.
This overlap leads to a delocalisation which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localised on the oxygen, but is spread out around the whole ion.
Why then is phenol a much weaker acid than ethanoic acid?
Think about the ethanoate ion again. If there wasn't any delocalisation, the charge would all be on one of the oxygen atoms, like this:
But the delocalisation spreads this charge over the whole of the COO group. Because oxygen is more electronegative than carbon, you can think of most of the charge being shared between the two oxygens (shown by the heavy red shading in this diagram).
If there wasn't any delocalisation, one of the oxygens would have a full charge which would be very attractive towards hydrogen ions. With delocalisation, that charge is spread over two oxygen atoms, and neither will be as attractive to a hydrogen ion as if one of the oxygens carried the whole charge.
That means that the ethanoate ion won't take up a hydrogen ion as easily as it would if there wasn't any delocalisation. Because some of it stays ionised, the formation of the hydrogen ions means that it is acidic.
In the phenoxide ion, the single oxygen atom is still the most electronegative thing present, and the delocalised system will be heavily distorted towards it. That still leaves the oxygen atom with most of its negative charge.
What delocalisation there is makes the phenoxide ion more stable than it would otherwise be, and so phenol is acidic to an extent.
However, the delocalisation hasn't shared the charge around very effectively. There is still lots of negative charge around the oxygen to which hydrogen ions will be attracted - and so the phenol will readily re-form. Phenol is therefore only very weakly acidic.
Example \(3\): Ethanol
Ethanol, CH3CH2OH, is so weakly acidic that you would hardly count it as acidic at all. If the hydrogen-oxygen bond breaks to release a hydrogen ion, an ethoxide ion is formed:
This has nothing at all going for it. There is no way of delocalising the negative charge, which remains firmly on the oxygen atom. That intense negative charge will be highly attractive towards hydrogen ions, and so the ethanol will instantly re-form.
Since ethanol is very poor at losing hydrogen ions, it is hardly acidic at all.
Variations in acid strengths between different carboxylic acids
You might think that all carboxylic acids would have the same strength because each depends on the delocalization of the negative charge around the -COO- group to make the anion more stable, and so more reluctant to re-combine with a hydrogen ion.
In fact, the carboxylic acids have widely different acidities. One obvious difference is between methanoic acid, HCOOH, and the other simple carboxylic acids:
pKa
HCOOH 3.75
CH3COOH 4.76
CH3CH2COOH 4.87
CH3CH2CH2COOH 4.82
Remember that the higher the value for pKa, the weaker the acid is.
Why is ethanoic acid weaker than methanoic acid? It again depends on the stability of the anions formed - on how much it is possible to delocalise the negative charge. The less the charge is delocalised, the less stable the ion, and the weaker the acid.
The methanoate ion (from methanoic acid) is:
The only difference between this and the ethanoate ion is the presence of the CH3 group in the ethanoate.
But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the -COO- group. Any build-up of charge will make the ion less stable, and more attractive to hydrogen ions.
Ethanoic acid is therefore weaker than methanoic acid, because it will re-form more easily from its ions.
The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of propanoic acid and butanoic acid are very similar to ethanoic acid. The acids can be strengthened by pulling charge away from the -COO- end. You can do this by attaching electronegative atoms like chlorine to the chain.
As the next table shows, the more chlorines you can attach the better:
pKa
CH3COOH 4.76
CH2ClCOOH 2.86
CHCl2COOH 1.29
CCl3COOH 0.65
Trichloroethanoic acid is quite a strong acid.
Attaching different halogens also makes a difference. Fluorine is the most electronegative and so you would expect it to be most successful at pulling charge away from the -COO- end and so strengthening the acid.
pKa
CH2FCOOH 2.66
CH2ClCOOH 2.86
CH2BrCOOH 2.90
CH2ICOOH 3.17
The effect is there, but isn't as great as you might expect.
Finally, notice that the effect falls off quite quickly as the attached halogen gets further away from the -COO- end. Here is what happens if you move a chlorine atom along the chain in butanoic acid.
pKa
CH3CH2CH2COOH 4.82
CH3CH2CHClCOOH 2.84
CH3CHClCH2COOH 4.06
CH2ClCH2CH2COOH 4.52
The chlorine is effective at withdrawing charge when it is next-door to the -COO- group, and much less so as it gets even one carbon further away.
This page explains why simple organic bases are basic and looks at the factors which affect their relative strengths. For A'level purposes, all the bases we are concerned with are primary amines - compounds in which one of the hydrogens in an ammonia molecule, NH3, is replaced either by an alkyl group or a benzene ring.
Ammonia as a weak base
All of the compounds we are concerned with are derived from ammonia and so we'll start by looking at the reason for its basic properties. For the purposes of this topic, we are going to take the definition of a base as "a substance which combines with hydrogen ions (protons)". We are going to get a measure of this by looking at how easily the bases take hydrogen ions from water molecules when they are in solution in water.
Ammonia in solution sets up this equilibrium:
An ammonium ion is formed together with hydroxide ions. Because the ammonia is only a weak base, it doesn't hang on to the extra hydrogen ion very effectively and so the reaction is reversible. At any one time, about 99% of the ammonia is present as unreacted molecules. The position of equilibrium lies well to the left.
The ammonia reacts as a base because of the active lone pair on the nitrogen. Nitrogen is more electronegative than hydrogen and so attracts the bonding electrons in the ammonia molecule towards itself. That means that in addition to the lone pair, there is a build-up of negative charge around the nitrogen atom. That combination of extra negativity and active lone pair attracts the new hydrogen from the water.
Comparing the strengths of weak bases
The strengths of weak bases are measured on the pKb scale. The smaller the number on this scale, the stronger the base is. Three of the compounds we shall be looking at, together with their pKb values are:
Remember - the smaller the number the stronger the base. Comparing the other two to ammonia, you will see that methylamine is a stronger base, whereas phenylamine is very much weaker.
• Methylamine is typical of aliphatic primary amines - where the -NH2 group is attached to a carbon chain. All aliphatic primary amines are stronger bases than ammonia.
• Phenylamine is typical of aromatic primary amines - where the -NH2 group is attached directly to a benzene ring. These are very much weaker bases than ammonia.
Explaining the differences in base strengths
Two of the factors which influence the strength of a base are:
• the ease with which the lone pair picks up a hydrogen ion,
• the stability of the ions being formed.
Why are aliphatic primary amines stronger bases than ammonia?
Methylamine
Methylamine has the structure:
The only difference between this and ammonia is the presence of the CH3 group in the methylamine. But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the nitrogen atom. That extra negativity around the nitrogen makes the lone pair even more attractive towards hydrogen ions.
Making the nitrogen more negative helps the lone pair to pick up a hydrogen ion. What about the effect on the positive methylammonium ion formed? Is this more stable than a simple ammonium ion? Compare the methylammonium ion with an ammonium ion:
In the methylammonium ion, the positive charge is spread around the ion by the "electron-pushing" effect of the methyl group. The more you can spread charge around, the more stable an ion becomes. In the ammonium ion there isn't any way of spreading the charge.
To summarize:
• The nitrogen is more negative in methylamine than in ammonia, and so it picks up a hydrogen ion more readily.
• The ion formed from methylamine is more stable than the one formed from ammonia, and so is less likely to shed the hydrogen ion again.
Taken together, these mean that methylamine is a stronger base than ammonia.
The other aliphatic primary amines
The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of the other aliphatic primary amines are very similar to methylamine. For example:
pKb
CH3NH2 3.36
CH3CH2NH2 3.27
CH3CH2CH2NH2 3.16
CH3CH2CH2CH2NH2 3.39
Why are aromatic primary amines much weaker bases than ammonia?
An aromatic primary amine is one in which the -NH2 group is attached directly to a benzene ring. The only one you are likely to come across is phenylamine. Phenylamine has the structure:
The lone pair on the nitrogen touches the delocalized ring electrons . . .
. . . and becomes delocalized with them:
That means that the lone pair is no longer fully available to combine with hydrogen ions. The nitrogen is still the most electronegative atom in the molecule, and so the delocalized electrons will be attracted towards it, but the intensity of charge around the nitrogen is nothing like what it is in, say, an ammonia molecule.
The other problem is that if the lone pair is used to join to a hydrogen ion, it is no longer available to contribute to the delocalisation. That means that the delocalization would have to be disrupted if the phenylamine acts as a base. Delocalization makes molecules more stable, and so disrupting the delocalization costs energy and will not happen easily.
Taken together - the lack of intense charge around the nitrogen, and the need to break some delocalization - this means that phenylamine is a very weak base indeed. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.18____Organic_Acids_and_Bases.txt |
/*<![CDATA[*/ MathJax.Hub.Config({ TeX: { equationNumbers: { autoNumber: "all", formatNumber: function (n) { return 3.6 + '.' + n } }, Macros: { PageIndex: ["{3.6. #1}",1] } } }); /*]]>*/A: Defining the acidity constant
You are no doubt aware that some acids are stronger than others. The relative acidity of different compounds or functional groups – in other words, their relative capacity to donate a proton to a common base under identical conditions – is quantified by a number called the acidity constant, abbreviated Ka. The common base chosen for comparison is water.
We will consider acetic acid as our first example. If we make a dilute solution of acetic acid in water, an acid-base reaction occurs between the acid (proton donor) and water (proton acceptor).
Acetic acid is a weak acid, so the equilibrium favors reactants over products - it is thermodynamically 'uphill', as indicated in the figure above by the relative length of the forward and reverse reaction arrows, and in the reaction coordinate diagram below in which products are higher energy than reactants.
As you know, the equilibrium constant Keq is defined as:
Every equilibrium constant expressions is actually a ratio of the activities of all species involved in the reaction. To avoid the use of activities, and to simplify experimental measurements, the equilibrium constant of concentrations approximates the activities of solutes and gases in dilute solutions with their respective molarities. However, the activities of solids, pure liquids, and solvents are not approximated with their molarities. Instead these activities are defined to have a value equal to 1 (one).
Thus, if we acknowledge that the activity of water in a dilute solution is approximated with the value of unity (1), we can divide by 1 to get the common form of the expression for Ka, the acid constant for acetic acid:
$K_a = K_{eq} = \dfrac{a_{H_3O^+}· a_{CH_3COO^-}}{a_{CH_3COOH} · a_{H_2O}} ≈ \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH](1)} = \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$
In fact, for an dilute aqueous solution, the activity of water is approximated with the value of 1, so the generic dissociation constant for a given acid HA or HB+ is expressed as:
The value of Ka for acetic acid is 1.75 x 10-5 - much less than 1, indicating there is much more acetic acid in solution at equilibrium than acetate and hydronium ions.
Conversely, sulfuric acid, with a Ka of approximately 109, or hydrochloric acid, with a Ka of approximately 107, both undergo essentially complete dissociation in water: they are very strong acids.
A number like 1.75 x 10- 5 is not very easy either to say, remember, or visualize, so chemists usually use a more convenient term to express relative acidity. The pKa value of an acid is simply the log (base 10) of its Ka value.
pKa = -log Ka Ka = 10-pKa
Doing the math, we find that the pKa of acetic acid is 4.8. The pKa of sulfuric acid is -10, and of hydrochloric acid is -7. The use of pKa values allows us to express the relative acidity of common compounds and functional groups on a numerical scale of about –10 (for a very strong acid) to 50 (for a compound that is not acidic at all). The lower the pKa value, the stronger the acid.
The ionizable (proton donating or accepting) functional groups relevant to biological organic chemistry generally have pKa values ranging from about 5 to about 20. The most important of these are summarized below, with very rough pKa values for the conjugate acid forms. More acidic groups with pKa values near zero are also included for reference.
Approximate pKa values to know
hydronium ion (H3O+) : 0
protonated alcohol: 0
protonated carbonyl: 0
carboxylic acids: 5
protonated imines: 7
protonated amines: 10
phenols: 10
thiols: 10
water: 14
alcohols: 15-18
alpha-carbon acids*: 20
*alpha-carbon acids will be discussed later in this chapter
You are strongly recommended to commit these rough values to memory now - then if you need a more precise value, you can always look it up in a pKa table.
Caution! pKa is not the same as pH!
It is important to realize that pKa is not the same thing as pH: the former is an inherent property of a compound or functional group, while the latter is a measure of hydronium ion concentration in a given aqueous solution:
pH = -log [H3O+]
Knowing pKa values not only allows us to compare acid strength, it also allows us to compare base strength. The key idea to remember is this: the stronger the conjugate acid, the weaker the conjugate base. We can determine that hydroxide ion is a stronger base than ammonia (NH3), because ammonium ion (NH4+, pKa = 9.2) is a stronger acid than water (pKa = 14.0).
Exercise 7.2.1
Which is the stronger base, CH3O- or CH3S-? Acetate ion or ammonia? Hydroxide ion or acetate ion?
Solution
Let's put our understanding of the pKa concept to use in the context of a more complex molecule. For example, what is the pKa of the compound below?
We need to evaluate the potential acidity of four different types of protons on the molecule, and find the most acidic one. The aromatic protons are not all acidic - their pKa is about 45. The amine group is also not acidic, its pKa is about 35. (Remember, uncharged amines are basic: it is positively-charged protonated amines, with pKa values around 10, that are weakly acidic.) The alcohol proton has a pKa of about 15, and the phenol proton has a pKa of about 10: thus, the most acidic group on the molecule above is the phenol. (Be sure that you can recognize the difference between a phenol and an alcohol - remember, in a phenol the OH group is bound directly to the aromatic ring). If this molecule were to react with one molar equivalent of a strong base such as sodium hydroxide, it is the phenol proton which would be donated to form a phenolate anion.
Exercise 7.2.2
Identify the most acidic functional group on each of the molecules below, and give its approximate pKa.
Solution
B: Using pKa values to predict reaction equilibria
By definition, the pKa value tells us the extent to which an acid will react with water as the base, but by extension we can also calculate the equilibrium constant for a reaction between any acid-base pair. Mathematically, it can be shown that:
Keq = 10ΔpKa
. . . where ΔpKa = (pKa of product acid minus pKa of reactant acid).
Consider a reaction between methylamine and acetic acid:
The first step is to identify the acid species on either side of the equation, and look up or estimate their pKa values. On the left side, the acid is of course acetic acid while on the right side the acid is methyl ammonium ion (in other words, methyl ammonium ion is the acid in the reaction going from right to left). We can look up the precise pKa values in table 7 (at the back of the book), but we already know (because we have this information memorized, right?!) that the pKa of acetic acids is about 5, and methyl ammonium is about 10. More precise values are 4.8 and 10.6, respectively.
Without performing any calculations at all, you should be able to see that this equilibrium lies far to the right-hand side: acetic acid has a lower pKa, meaning it is a stronger acid than methyl ammonium, and thus it wants to give up its proton more than methyl ammonium does. Doing the math, we see that
Keq = 10ΔpKa = 10(10.6 – 4.8) = 105.8 = 6.3 x 105
So Keq is a very large number (much greater than 1) and the equilibrium for the reaction between acetic acid and methylamine lies far to the right-hand side of the equation, just as we had predicted. This also tells us that the reaction has a negative Gibbs free energy change, and is thermodynamically favorable.
If you had just wanted to quickly approximate the value of Keq without benefit of precise pKa information or a calculator, you could have approximated pKa ~ 5 (for the carboxylic acid) and pKa ~10 (for the ammonium ion) and calculated in your head that the equilibrium constant should be somewhere in the order of 105.
Exercise 7.2.3
Show the products of the following acid-base reactions, and roughly estimate the value of Keq.
Solution
C: Organic molecules in buffered solution: the Henderson-Hasselbalch equation
The environment inside a living cell, where most biochemical reactions take place, is an aqueous buffer with pH ~ 7. Recall from your General Chemistry course that a buffer is a solution of a weak acid and its conjugate base. The key equation for working with buffers is the Henderson-Hasselbalch equation:
The Henderson-Hasselbalch equation:
The equation tells us that if our buffer is an equimolar solution of a weak acid and its conjugate base, the pH of the buffer will equal the pKa of the acid (because the log of 1 is equal to zero). If there is more of the acid form than the base, then of course the pH of the buffer is lower than the pKa of the acid.
Exercise 7.2.4
What is the pH of an aqueous buffer solution that is 30 mM in acetic acid and 40 mM in sodium acetate? The pKa of acetic acid is 4.8.
Solution
The Henderson-Hasselbalch equation is particularly useful when we want to think about the protonation state of different biomolecule functional groups in a pH 7 buffer. When we do this, we are always assuming that the concentration of the biomolecule is small compared to the concentration of the buffer components. (The actual composition of physiological buffer is complex, but it is primarily based on phosphoric and carbonic acids).
Imagine an aspartic acid residue located on the surface of a protein in a human cell. Being on the surface, the side chain is in full contact with the pH 7 buffer surrounding the protein. In what state is the side chain functional group: the protonated state (a carboxylic acid) or the deprotonated state (a carboxylate ion)? Using the Henderson-Hasselbalch equation, we fill in our values for the pH of the buffer and a rough pKa approximation of pKa = 5 for the carboxylic acid functional group. Doing the math, we find that the ratio of carboxylate to carboxylic acid is about 100 to 1: the carboxylic acid is almost completely ionized (in the deprotonated state) inside the cell. This result extends to all other carboxylic acid groups you might find on natural biomolecules or drug molecules: in the physiological environment, carboxylic acids are almost completely deprotonated.
Now, let's use the equation again, this time for an amine functional group, such as the side chain of a lysine residue: inside a cell, are we likely to see a neutral amine (R-NH2) or an ammonium cation (R-NH3+?) Using the equation with pH = 7 (for the biological buffer) and pKa = 10 (for the ammonium group), we find that the ratio of neutral amine to ammonium cation is about 1 to 100: the group is close to completely protonated inside the cell, so we will see R-NH3+, not R-NH2.
We can do the same rough calculation for other common functional groups found in biomolecules. It will be very helpful going forward to commit the following to memory:
At physiological pH:
Carboxylic acids are deprotonated (in the carboxylate anion form)
Amines are protonated (in the ammonium cation form)
Thiols, phenols, alcohols, and amides are uncharged
Imines are a mixture of the protonated (cationic) and deprotonated (neutral) states
We will talk about the physiological protonation state of phosphate groups in chapter 9.
Exercise 7.2.5
The molecule below is not drawn in a reasonable protonation state for pH 7. Redraw it in with the functional groups in the correct protonation state.
Solution
While we are most interested in the state of molecules at pH 7, the Henderson-Hasselbalch equation can of course be used to determine the protonation state of functional groups in solutions buffered to other pH levels. The exercises below provide some practice in this type of calculation.
Exercise 7.2.6
What is the ratio of acetate ion to neutral acetic acid when a small amount of acetic acid (pKa = 4.8) is dissolved in a buffer of pH 2.8? pH 3.8? pH 4.8? pH 5.8? pH 6.8?
Solution
Exercise 7.2.7
Would you expect phenol to be soluble in an aqueous solution buffered to pH 2? pH 7? pH 12? Explain your answer.
Solution
Exercise 7.2.8
Exercise 7.2.8: What is the approximate net charge on a tetrapeptide Cys-Asp-Lys-Glu in pH 7 buffer?
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.19%3A_How_to_Predict_the_Outcome_of_an_Acid-Base_Reaction.txt |
Now that we know how to quantify the strength of an acid or base, our next job is to gain an understanding of the fundamental reasons behind why one compound is more acidic or more basic than another. This is a big step: we are, for the first time, taking our knowledge of organic structure and applying it to a question of organic reactivity. Many of the ideas that we’ll see for the first here will continue to apply throughout the book as we tackle many other organic reaction types.
First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table. We’ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules with the same functionalities, for example the side chains of the amino acids alanine (alkane), lysine (amine), and serine (alcohol).
Horizontal periodic trend in acidity and basicity
We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable (weaker) the conjugate base, the stronger the acid. Look at where the negative charge ends up in each conjugate base. In the conjugate base of ethane, the negative charge is borne by a carbon atom, while on the conjugate base of methylamine and ethanol the negative charge is located on a nitrogen and an oxygen, respectively. Remember that electronegativity also increases as we move from left to right along a row of the periodic table, meaning that oxygen is the most electronegative of the three atoms, and carbon the least.
The more electronegative an atom, the better able it is to bear a negative charge. Weaker bases have negative charges on more electronegative atoms; stronger bases have negative charges on less electronegative atoms.
Thus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl carbanion anion is the least stable (highest energy, most basic). Conversely, ethanol is the strongest acid, and ethane the weakest acid.
When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity. This is best illustrated with the haloacids and halides: basicity, like electronegativity, increases as we move up the column.
Vertical periodic trend in acidity and basicity
Conversely, acidity in the haloacids increases as we move down the column.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly larger volume:
This illustrates a fundamental concept in organic chemistry:
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ over a larger area.
We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts. For now, we are applying the concept only to the influence of atomic radius on base strength. Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than a carboxylic acid. HI, with a pKa of about -9, is almost as strong as sulfuric acid.
More importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols. The pKa of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pKa for the alcohol group on the serine side chain is on the order of 17.
Remember the concept of 'driving force' that we learned about in chapter 6? Recall that the driving force for a reaction is usually based on two factors: relative charge stability, and relative total bond energy. Let's see how this applies to a simple acid-base reaction between hydrochloric acid and fluoride ion:
HCl + F- HF + Cl-
We know that HCl (pKa -7) is a stronger acid than HF (pKa 3.2), so the equilibrium for the reaction lies on the product side: the reaction is exergonic, and a 'driving force' pushes reactant to product.
What explains this driving force? Consider first the charge factor: as we just learned, chloride ion (on the product side) is more stable than fluoride ion (on the reactant side). This partially accounts for the driving force going from reactant to product in this reaction: we are going from less stable ion to a more stable ion.
What about total bond energy, the other factor in driving force? If you consult a table of bond energies, you will see that the H-F bond on the product side is more energetic (stronger) than the H-Cl bond on the reactant side: 565 kJ/mol vs 427 kJ/mol, respectively). This also contributes to the driving force: we are moving from a weaker (less stable) bond to a stronger (more stable) bond.
B: Resonance effects
In the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements. Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about the same element acting as the proton donor/acceptor. The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups.
Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are strikingly different. What makes a carboxylic acid so much more acidic than an alcohol? As before, we begin by considering the stability of the conjugate bases, remembering that a more stable (weaker) conjugate base corresponds to a stronger acid.
In both species, the negative charge on the conjugate base is located on oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: two resonance contributors can be drawn for the conjugate base, and the negative charge can be delocalized (shared) over two oxygen atoms. In the ethoxide ion, by contrast, the negative charge is localized, or ‘locked’ on the single oxygen – it has nowhere else to go. This makes the ethoxide ion much less stable.
Recall the important general statement that we made a little earlier: 'Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.' Now, we are seeing this concept in another context, where a charge is being ‘spread out’ (in other words, delocalized) by resonance, rather than simply by the size of the atom involved.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pKa units between ethanol and acetic acid (and remember, pKa is a log expression, so we are talking about a fator of 1012 between the Ka values for the two molecules!)
The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but not basic when it is part of an amide group. Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a minor but still important resonance contributor in which the nitrogen lone pair is part of a pi bond.
Whereas the lone pair of an amine nitrogen is ‘stuck’ in one place, the lone pair on an amide nitrogen is delocalized by resonance. Notice that in this case, we are extending our central statement to say that electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too ‘comfortable’ being part of the delocalized pi bonding system. The lone pair on an amine nitrogen, by contrast, is not so comfortable - it is not part of a delocalized pi system, and is available to form a bond with any acidic proton that might be nearby.
If an amide group is protonated, it will be at the oxygen rather than the nitrogen.
Exercise 7.3.1
a) Draw the Lewis structure of nitric acid, HNO3.
b) Nitric acid is a strong acid - it has a pKa of -1.4. Make a structural argument to account for its strength. Your answer should involve the structure of nitrate, the conjugate base of nitric acid.
Solution
Exercise 7.3.2
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Exercise 7.3.3
(challenging!) Often it requires some careful thought to predict the most acidic proton on a molecule. Ascorbic acid, also known as Vitamin C, has a pKa of 4.1 - the fact that this is in the range of carboxylic acids suggest to us that the negative charge on the conjugate base can be delocalized by resonance to two oxygen atoms. Which if the four OH protons on the molecule is most acidic? Draw the structure of ascorbate, the conjugate base of ascorbic acid, then draw a second resonance contributor showing how the negative charge is delocalized to a second oxygen atom. Hint - try removing each OH group in turn, then use your resonance drawing skills to figure out whether or not delocalization of charge can occur.
Solution
C: Inductive effects
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
The presence of the chlorine atoms clearly increases the acidity of the carboxylic acid group, but the argument here does not have to do with resonance delocalization, because no additional resonance contributors can be drawn for the chlorinated molecules. Rather, the explanation for this phenomenon involves something called the inductive effect. A chlorine atom is more electronegative than a hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. In effect, the chlorine atoms are helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect. In this context, the chlorine substituent can be referred to as an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects.
Because the inductive effect depends on electronegativity, fluorine substituents have a more pronounced pKa-lowered effect than chlorine substituents.
In addition, the inductive takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.
Exercise 7.3.4
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.22%3A_How_Substituents_Affect_the_Strength_of_an_Acid.txt |
Now that we know how to quantify the strength of an acid or base, our next job is to gain an understanding of the fundamental reasons behind why one compound is more acidic or more basic than another. This is a big step: we are, for the first time, taking our knowledge of organic structure and applying it to a question of organic reactivity. Many of the ideas that we’ll see for the first here will continue to apply throughout the book as we tackle many other organic reaction types.
First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table. We’ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules with the same functionalities, for example the side chains of the amino acids alanine (alkane), lysine (amine), and serine (alcohol).
Horizontal periodic trend in acidity and basicity
We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable (weaker) the conjugate base, the stronger the acid. Look at where the negative charge ends up in each conjugate base. In the conjugate base of ethane, the negative charge is borne by a carbon atom, while on the conjugate base of methylamine and ethanol the negative charge is located on a nitrogen and an oxygen, respectively. Remember that electronegativity also increases as we move from left to right along a row of the periodic table, meaning that oxygen is the most electronegative of the three atoms, and carbon the least.
The more electronegative an atom, the better able it is to bear a negative charge. Weaker bases have negative charges on more electronegative atoms; stronger bases have negative charges on less electronegative atoms.
Thus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl carbanion anion is the least stable (highest energy, most basic). Conversely, ethanol is the strongest acid, and ethane the weakest acid.
When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity. This is best illustrated with the haloacids and halides: basicity, like electronegativity, increases as we move up the column.
Vertical periodic trend in acidity and basicity
Conversely, acidity in the haloacids increases as we move down the column.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly larger volume:
This illustrates a fundamental concept in organic chemistry:
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ over a larger area.
We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts. For now, we are applying the concept only to the influence of atomic radius on base strength. Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than a carboxylic acid. HI, with a pKa of about -9, is almost as strong as sulfuric acid.
More importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols. The pKa of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pKa for the alcohol group on the serine side chain is on the order of 17.
Remember the concept of 'driving force' that we learned about in chapter 6? Recall that the driving force for a reaction is usually based on two factors: relative charge stability, and relative total bond energy. Let's see how this applies to a simple acid-base reaction between hydrochloric acid and fluoride ion:
HCl + F- HF + Cl-
We know that HCl (pKa -7) is a stronger acid than HF (pKa 3.2), so the equilibrium for the reaction lies on the product side: the reaction is exergonic, and a 'driving force' pushes reactant to product.
What explains this driving force? Consider first the charge factor: as we just learned, chloride ion (on the product side) is more stable than fluoride ion (on the reactant side). This partially accounts for the driving force going from reactant to product in this reaction: we are going from less stable ion to a more stable ion.
What about total bond energy, the other factor in driving force? If you consult a table of bond energies, you will see that the H-F bond on the product side is more energetic (stronger) than the H-Cl bond on the reactant side: 565 kJ/mol vs 427 kJ/mol, respectively). This also contributes to the driving force: we are moving from a weaker (less stable) bond to a stronger (more stable) bond.
B: Resonance effects
In the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements. Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about the same element acting as the proton donor/acceptor. The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups.
Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are strikingly different. What makes a carboxylic acid so much more acidic than an alcohol? As before, we begin by considering the stability of the conjugate bases, remembering that a more stable (weaker) conjugate base corresponds to a stronger acid.
In both species, the negative charge on the conjugate base is located on oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: two resonance contributors can be drawn for the conjugate base, and the negative charge can be delocalized (shared) over two oxygen atoms. In the ethoxide ion, by contrast, the negative charge is localized, or ‘locked’ on the single oxygen – it has nowhere else to go. This makes the ethoxide ion much less stable.
Recall the important general statement that we made a little earlier: 'Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.' Now, we are seeing this concept in another context, where a charge is being ‘spread out’ (in other words, delocalized) by resonance, rather than simply by the size of the atom involved.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pKa units between ethanol and acetic acid (and remember, pKa is a log expression, so we are talking about a fator of 1012 between the Ka values for the two molecules!)
The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but not basic when it is part of an amide group. Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a minor but still important resonance contributor in which the nitrogen lone pair is part of a pi bond.
Whereas the lone pair of an amine nitrogen is ‘stuck’ in one place, the lone pair on an amide nitrogen is delocalized by resonance. Notice that in this case, we are extending our central statement to say that electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too ‘comfortable’ being part of the delocalized pi bonding system. The lone pair on an amine nitrogen, by contrast, is not so comfortable - it is not part of a delocalized pi system, and is available to form a bond with any acidic proton that might be nearby.
If an amide group is protonated, it will be at the oxygen rather than the nitrogen.
Exercise 7.3.1
a) Draw the Lewis structure of nitric acid, HNO3.
b) Nitric acid is a strong acid - it has a pKa of -1.4. Make a structural argument to account for its strength. Your answer should involve the structure of nitrate, the conjugate base of nitric acid.
Solution
Exercise 7.3.2
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Exercise 7.3.3
(challenging!) Often it requires some careful thought to predict the most acidic proton on a molecule. Ascorbic acid, also known as Vitamin C, has a pKa of 4.1 - the fact that this is in the range of carboxylic acids suggest to us that the negative charge on the conjugate base can be delocalized by resonance to two oxygen atoms. Which if the four OH protons on the molecule is most acidic? Draw the structure of ascorbate, the conjugate base of ascorbic acid, then draw a second resonance contributor showing how the negative charge is delocalized to a second oxygen atom. Hint - try removing each OH group in turn, then use your resonance drawing skills to figure out whether or not delocalization of charge can occur.
Solution
C: Inductive effects
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
The presence of the chlorine atoms clearly increases the acidity of the carboxylic acid group, but the argument here does not have to do with resonance delocalization, because no additional resonance contributors can be drawn for the chlorinated molecules. Rather, the explanation for this phenomenon involves something called the inductive effect. A chlorine atom is more electronegative than a hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. In effect, the chlorine atoms are helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect. In this context, the chlorine substituent can be referred to as an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects.
Because the inductive effect depends on electronegativity, fluorine substituents have a more pronounced pKa-lowered effect than chlorine substituents.
In addition, the inductive takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.
Exercise 7.3.4
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.23%3A_An_Introduction_to_Delocalized_Electrons.txt |
Now that we know how to quantify the strength of an acid or base, our next job is to gain an understanding of the fundamental reasons behind why one compound is more acidic or more basic than another. This is a big step: we are, for the first time, taking our knowledge of organic structure and applying it to a question of organic reactivity. Many of the ideas that we’ll see for the first here will continue to apply throughout the book as we tackle many other organic reaction types.
First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table. We’ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules with the same functionalities, for example the side chains of the amino acids alanine (alkane), lysine (amine), and serine (alcohol).
Horizontal periodic trend in acidity and basicity
We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable (weaker) the conjugate base, the stronger the acid. Look at where the negative charge ends up in each conjugate base. In the conjugate base of ethane, the negative charge is borne by a carbon atom, while on the conjugate base of methylamine and ethanol the negative charge is located on a nitrogen and an oxygen, respectively. Remember that electronegativity also increases as we move from left to right along a row of the periodic table, meaning that oxygen is the most electronegative of the three atoms, and carbon the least.
The more electronegative an atom, the better able it is to bear a negative charge. Weaker bases have negative charges on more electronegative atoms; stronger bases have negative charges on less electronegative atoms.
Thus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl carbanion anion is the least stable (highest energy, most basic). Conversely, ethanol is the strongest acid, and ethane the weakest acid.
When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity. This is best illustrated with the haloacids and halides: basicity, like electronegativity, increases as we move up the column.
Vertical periodic trend in acidity and basicity
Conversely, acidity in the haloacids increases as we move down the column.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly larger volume:
This illustrates a fundamental concept in organic chemistry:
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ over a larger area.
We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts. For now, we are applying the concept only to the influence of atomic radius on base strength. Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than a carboxylic acid. HI, with a pKa of about -9, is almost as strong as sulfuric acid.
More importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols. The pKa of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pKa for the alcohol group on the serine side chain is on the order of 17.
Remember the concept of 'driving force' that we learned about in chapter 6? Recall that the driving force for a reaction is usually based on two factors: relative charge stability, and relative total bond energy. Let's see how this applies to a simple acid-base reaction between hydrochloric acid and fluoride ion:
HCl + F- HF + Cl-
We know that HCl (pKa -7) is a stronger acid than HF (pKa 3.2), so the equilibrium for the reaction lies on the product side: the reaction is exergonic, and a 'driving force' pushes reactant to product.
What explains this driving force? Consider first the charge factor: as we just learned, chloride ion (on the product side) is more stable than fluoride ion (on the reactant side). This partially accounts for the driving force going from reactant to product in this reaction: we are going from less stable ion to a more stable ion.
What about total bond energy, the other factor in driving force? If you consult a table of bond energies, you will see that the H-F bond on the product side is more energetic (stronger) than the H-Cl bond on the reactant side: 565 kJ/mol vs 427 kJ/mol, respectively). This also contributes to the driving force: we are moving from a weaker (less stable) bond to a stronger (more stable) bond.
B: Resonance effects
In the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements. Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about the same element acting as the proton donor/acceptor. The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups.
Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are strikingly different. What makes a carboxylic acid so much more acidic than an alcohol? As before, we begin by considering the stability of the conjugate bases, remembering that a more stable (weaker) conjugate base corresponds to a stronger acid.
In both species, the negative charge on the conjugate base is located on oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: two resonance contributors can be drawn for the conjugate base, and the negative charge can be delocalized (shared) over two oxygen atoms. In the ethoxide ion, by contrast, the negative charge is localized, or ‘locked’ on the single oxygen – it has nowhere else to go. This makes the ethoxide ion much less stable.
Recall the important general statement that we made a little earlier: 'Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.' Now, we are seeing this concept in another context, where a charge is being ‘spread out’ (in other words, delocalized) by resonance, rather than simply by the size of the atom involved.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pKa units between ethanol and acetic acid (and remember, pKa is a log expression, so we are talking about a fator of 1012 between the Ka values for the two molecules!)
The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but not basic when it is part of an amide group. Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a minor but still important resonance contributor in which the nitrogen lone pair is part of a pi bond.
Whereas the lone pair of an amine nitrogen is ‘stuck’ in one place, the lone pair on an amide nitrogen is delocalized by resonance. Notice that in this case, we are extending our central statement to say that electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too ‘comfortable’ being part of the delocalized pi bonding system. The lone pair on an amine nitrogen, by contrast, is not so comfortable - it is not part of a delocalized pi system, and is available to form a bond with any acidic proton that might be nearby.
If an amide group is protonated, it will be at the oxygen rather than the nitrogen.
Exercise 7.3.1
a) Draw the Lewis structure of nitric acid, HNO3.
b) Nitric acid is a strong acid - it has a pKa of -1.4. Make a structural argument to account for its strength. Your answer should involve the structure of nitrate, the conjugate base of nitric acid.
Solution
Exercise 7.3.2
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Exercise 7.3.3
(challenging!) Often it requires some careful thought to predict the most acidic proton on a molecule. Ascorbic acid, also known as Vitamin C, has a pKa of 4.1 - the fact that this is in the range of carboxylic acids suggest to us that the negative charge on the conjugate base can be delocalized by resonance to two oxygen atoms. Which if the four OH protons on the molecule is most acidic? Draw the structure of ascorbate, the conjugate base of ascorbic acid, then draw a second resonance contributor showing how the negative charge is delocalized to a second oxygen atom. Hint - try removing each OH group in turn, then use your resonance drawing skills to figure out whether or not delocalization of charge can occur.
Solution
C: Inductive effects
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
The presence of the chlorine atoms clearly increases the acidity of the carboxylic acid group, but the argument here does not have to do with resonance delocalization, because no additional resonance contributors can be drawn for the chlorinated molecules. Rather, the explanation for this phenomenon involves something called the inductive effect. A chlorine atom is more electronegative than a hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. In effect, the chlorine atoms are helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect. In this context, the chlorine substituent can be referred to as an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects.
Because the inductive effect depends on electronegativity, fluorine substituents have a more pronounced pKa-lowered effect than chlorine substituents.
In addition, the inductive takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.
Exercise 7.3.4
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.24%3A_A_Summary_of_the_Factors_that_Determine_Acid_Strength.txt |
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Blood as a Buffer
Buffer solutions are extremely important in biology and medicine because most biological reactions and enzymes need very specific pH ranges in order to work properly.
• Henderson-Hasselbalch Approximation
The Henderson-Hasselbalch approximation allows us one method to approximate the pH of a buffer solution.
• How Does A Buffer Maintain pH?
A buffer is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit.
• Introduction to Buffers
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Preparing Buffer Solutions
When it comes to buffer solution one of the most common equation is the Henderson-Hasselbalch approximation. An important point that must be made about this equation is it's useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations.
Thumbnail: Simulated titration of an acidified solution of a weak acid (pKa = 4.7) with alkali. (Public Domain; Lasse Havelund).
1.27: Lewis Acids and Bases
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Write the equation for the proton transfer reaction involving a Brønsted-Lowry acid or base, and show how it can be interpreted as an electron-pair transfer reaction, clearly identifying the donor and acceptor.
• Give an example of a Lewis acid-base reaction that does not involve protons.
• Write equations illustrating the behavior of a given non-aqueous acid-base system.
The Brønsted-Lowry proton donor-acceptor concept has been one of the most successful theories of Chemistry. But as with any such theory, it is fair to ask if this is not just a special case of a more general theory that could encompass an even broader range of chemical science. In 1916, G.N. Lewis of the University of California proposed that the electron pair is the dominant actor in acid-base chemistry. The Lewis theory did not become very well known until about 1923 (the same year that Brønsted and Lowry published their work), but since then it has been recognized as a very powerful tool for describing chemical reactions of widely different kinds and is widely used in organic and inorganic chemistry. According to Lewis,
• An acid is a substance that accepts a pair of electrons, and in doing so, forms a covalent bond with the entity that supplies the electrons.
• A base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared.
In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles.
Proton-Transfer Reactions Involve Electron-Pair Transfer
Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place.
This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points:
• The arrow shows the movement of a proton from the hydronium ion to the hydroxide ion.
• Note carefully that the electron-pairs themselves do not move; they remain attached to their central atoms. The electron pair on the base is "donated" to the acceptor (the proton) only in the sense that it ends up being shared with the acceptor, rather than being the exclusive property of the oxygen atom in the hydroxide ion.
• Although the hydronium ion is the nominal Lewis acid here, it does not itself accept an electron pair, but acts merely as the source of the proton that coordinates with the Lewis base.
The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a Lewis acid-base reaction from an oxidation-reduction reaction, in which a physical transfer of one or more electrons from donor to acceptor does occur. The product of a Lewis acid-base reaction is known formally as an "adduct" or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" H2O. The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with.
Another example, showing the autoprotolysis of water. Note that the conjugate base is also the adduct.
Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the hydrolysis of the ammonium ion.
Because $\ce{HF}$ is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, F accepts a proton from water, which is transformed into a hydroxide ion.
The bisulfite ion is amphiprotic and can act as an electron donor or acceptor.
Acid-base Reactions without Transferring Protons
The major utility of the Lewis definition is that it extends the concept of acids and bases beyond the realm of proton transfer reactions. The classic example is the reaction of boron trifluoride with ammonia to form an adduct:
$\ce{BF_3 + NH_3 \rightarrow F_3B-NH_3}$
One of the most commonly-encountered kinds of Lewis acid-base reactions occurs when electron-donating ligands form coordination complexes with transition-metal ions.
Exercise $1$
Here are several more examples of Lewis acid-base reactions that cannot be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction.
1. $\ce{Al(OH)_3 + OH^{–} \rightarrow Al(OH)_4^–}$
2. $\ce{SnS_2 + S^{2–} \rightarrow SnS_3^{2–}}$
3. $\ce{Cd(CN)_2 + 2 CN^– \rightarrow Cd(CN)_4^{2+}}$
4. $\ce{AgCl + 2 NH_3 \rightarrow Ag(NH_3)_2^+ + Cl^–}$
5. $\ce{Fe^{2+} + NO \rightarrow Fe(NO)^{2+}}$
6. $\ce{Ni^{2+} + 6 NH_3 \rightarrow Ni(NH_3)_5^{2+}}$
Applications to organic reaction mechanisms
Although organic chemistry is beyond the scope of these lessons, it is instructive to see how electron donors and acceptors play a role in chemical reactions. The following two diagrams show the mechanisms of two common types of reactions initiated by simple inorganic Lewis acids:
In each case, the species labeled "Complex" is an intermediate that decomposes into the products, which are conjugates of the original acid and base pairs. The electric charges indicated in the complexes are formal charges, but those in the products are "real".
In reaction 1, the incomplete octet of the aluminum atom in $\ce{AlCl3}$ serves as a better electron acceptor to the chlorine atom than does the isobutyl part of the base. In reaction 2, the pair of non-bonding electrons on the dimethyl ether coordinates with the electron-deficient boron atom, leading to a complex that breaks down by releasing a bromide ion.
Non-aqueous Protonic Acid-Base Systems
We ordinarily think of Brønsted-Lowry acid-base reactions as taking place in aqueous solutions, but this need not always be the case. A more general view encompasses a variety of acid-base solvent systems, of which the water system is only one (Table $1$). Each of these has as its basis an amphiprotic solvent (one capable of undergoing autoprotolysis), in parallel with the familiar case of water.
The ammonia system is one of the most common non-aqueous system in Chemistry. Liquid ammonia boils at –33° C, and can conveniently be maintained as a liquid by cooling with dry ice (–77° C). It is a good solvent for substances that also dissolve in water, such as ionic salts and organic compounds since it is capable of forming hydrogen bonds. However, many other familiar substances can also serve as the basis of protonic solvent systems as Table $1$ indicates:
Table $1$: Popular Solvent systems
solvent
autoprotolysis reaction
pKap
water 2 H2O → H3O+ + OH 14
ammonia 2 NH3 → NH4+ + NH2 33
acetic acid 2 CH3COOH → CH3COOH2+ + CH3COO 13
ethanol 2 C2H5OH → C2H5OH2+ + C2H5O 19
hydrogen peroxide 2 HO-OH → HO-OH2+ + HO-O 13
hydrofluoric acid 2 HF → H2F+ + F 10
sulfuric acid 2 H2SO4 → H3SO4+ + HSO4 3.5
One use of nonaqueous acid-base systems is to examine the relative strengths of the strong acids and bases, whose strengths are "leveled" by the fact that they are all totally converted into H3O+ or OH ions in water. By studying them in appropriate non-aqueous solvents which are poorer acceptors or donors of protons, their relative strengths can be determined. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Covalent_Bonding/1.26%3A_Buffer_Solutions.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
02: Acids and Bases
An acid–base reaction is a chemical reaction that occurs between an acid and a base. Several theoretical frameworks provide alternative conceptions of the reaction mechanisms and their application in solving related problems; these are called acid–base theories, for example, Brønsted–Lowry acid–base theory. Their importance becomes apparent in analyzing acid–base reactions for gaseous or liquid species, or when acid or base character may be somewhat less apparent.
Thumbnail: Warning picture used with dangerous acids and dangerous bases. Bases are the opposites of acids. (Public Domain).
2.2: pka and pH
Learning Objectives
• To define the pH scale as a measure of acidity of a solution
• Tell the origin and the logic of using the pH scale.
• Apply the same strategy for representing other types of quantities such as pKa, pKb, pKw.
Auto-Ionization of Water
Because of its amphoteric nature (i.e., acts as both an acid or a base), water does not always remain as $H_2O$ molecules. In fact, two water molecules react to form hydronium and hydroxide ions:
$\ce{ 2 H_2O (l) \rightleftharpoons H_3O^+ (aq) + OH^{−} (aq)} \label{1}$
This is also called the self-ionization of water. The concentration of $H_3O^+$ and $OH^-$ are equal in pure water because of the 1:1 stoichiometric ratio of Equation $\ref{1}$. The molarity of H3O+ and OH- in water are also both $1.0 \times 10^{-7} \,M$ at 25° C. Therefore, a constant of water ($K_w$) is created to show the equilibrium condition for the self-ionization of water. The product of the molarity of hydronium and hydroxide ion is always $1.0 \times 10^{-14}$ (at room temperature).
$K_w= [H_3O^+][OH^-] = 1.0 \times 10^{-14} \label{2}$
Equation $\ref{2}$ also applies to all aqueous solutions. However, $K_w$ does change at different temperatures, which affects the pH range discussed below.
$H^+$ and $H_3O^+$ is often used interchangeably to represent the hydrated proton, commonly call the hydronium ion.
Equation \ref{1} can also be written as
$H_2O \rightleftharpoons H^+ + OH^- \label{3}$
As expected for any equilibrium, the reaction can be shifted to the reactants or products:
• If an acid ($H^+$) is added to the water, the equilibrium shifts to the left and the $OH^-$ ion concentration decreases
• If base ( $OH^-$) is added to water, the equilibrium shifts to left and the $H^+$ concentration decreases.
pH and pOH
Because the constant of water, Kw is $1.0 \times 10^{-14}$ (at 25° C), the $pK_w$ is 14, the constant of water determines the range of the pH scale. To understand what the pKw is, it is important to understand first what the "p" means in pOH and pH. The addition of the "p" reflects the negative of the logarithm, $-\log$. Therefore, the pH is the negative logarithm of the molarity of H, the pOH is the negative logarithm of the molarity of $\ce{OH^-}$, and the $pK_w$ is the negative logarithm of the constant of water:
\begin{align} pH &= -\log [H^+] \label{4a} \[4pt] pOH &= -\log [OH^-] \label{4b} \[4pt] pK_w &= -\log [K_w] \label{4c} \end{align}
At room temperature,
$K_w =1.0 \times 10^{-14} \label{4d}$
So
\begin{align} pK_w &=-\log [1.0 \times 10^{-14}] \label{4e} \[4pt] &=14 \end{align}
Using the properties of logarithms, Equation $\ref{4e}$ can be rewritten as
$10^{-pK_w}=10^{-14}. \label{4f}$
The equation also shows that each increasing unit on the scale decreases by the factor of ten on the concentration of $\ce{H^{+}}$. Combining Equations \ref{4a} - \ref{4c} and \ref{4e} results in this important relationship:
$pK_w= pH + pOH = 14 \label{5b}$
Equation \ref{5b} is correct only at room temperature since changing the temperature will change $K_w$.
The pH scale is logarithmic, meaning that an increase or decrease of an integer value changes the concentration by a tenfold. For example, a pH of 3 is ten times more acidic than a pH of 4. Likewise, a pH of 3 is one hundred times more acidic than a pH of 5. Similarly a pH of 11 is ten times more basic than a pH of 10.
Properties of the pH Scale
From the simple definition of pH in Equation \ref{4a}, the following properties can be identified:
• This scale is convenient to use, because it converts some odd expressions such as $1.23 \times 10^{-4}$ into a single number of 3.91.
• This scale covers a very large range of $\ce{[H+]}$, from 0.1 to 10-14. When $\ce{[H+]}$ is high, we usually do not use the pH value, but simply the $\ce{[H+]}$. For example, when $\mathrm{[H^+] = 1.0}$, pH = 0. We seldom say the pH is 0, and that is why you consider pH = 0 such an odd expression. A pH = -0.30 is equivalent to a $\ce{[H+]}$ of 2.0 M. Negative pH values are only for academic exercises. Using the concentrations directly conveys a better sense than the pH scales.
• The pH scale expands the division between zero and 1 in a linear scale or a compact scale into a large scale for comparison purposes. In mathematics, you learned that there are infinite values between 0 and 1, or between 0 and 0.1, or between 0 and 0.01 or between 0 and any small value. Using a log scale certainly converts infinite small quantities into infinite large quantities.
• The non-linearity of the pH scale in terms of $\ce{[H+]}$ is easily illustrated by looking at the corresponding values for pH between 0.1 and 0.9 as follows:
pH 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
[H+] 1 0.79 0.63 0.50 0.40 0.32 0.25 0.20 0.16 0.13
• Because the negative log of $\ce{[H+]}$ is used in the pH scale, the pH scale usually has positive values. Furthermore, the larger the pH, the smaller the $\ce{[H+]}$.
The Effective Range of the pH Scale
It is common that the pH scale is argued to range from 0-14 or perhaps 1-14, but neither is correct. The pH range does not have an upper nor lower bound, since as defined above, the pH is an indication of concentration of H+. For example, at a pH of zero the hydronium ion concentration is one molar, while at pH 14 the hydroxide ion concentration is one molar. Typically the concentrations of H+ in water in most solutions fall between a range of 1 M (pH=0) and 10-14 M (pH=14). Hence a range of 0 to 14 provides sensible (but not absolute) "bookends" for the scale. One can go somewhat below zero and somewhat above 14 in water, because the concentrations of hydronium ions or hydroxide ions can exceed one molar. Figure $1$ depicts the pH scale with common solutions and where they are on the scale.
Quick Interpretation
• If pH >7, the solution is basic. The pOH should be looked in the perspective of OH- instead of H+. Whenever the value of pOH is less than 7, then it is considered basic. And therefore there are more OH- than H+ in the solution.
• At pH 7, the substance or solution is at neutral and means that the concentration of H+ and OH- ion is the same.
• If pH < 7, the solution is acidic. There are more H+ than OH- in an acidic solution.
• The pH scale does not have an upper nor lower bound.
Example $1$
If the concentration of $NaOH$ in a solution is $2.5 \times 10^{-4}\; M$, what is the concentration of $H_3O^+$?
Solution
We can assume room temperature, so
$1.0 \times 10^{-14} = [H_3O^+][OH^-] \nonumber$
to find the concentration of H3O+, solve for the [H3O+].
$\dfrac{1.0 \times 10^{-14}}{[OH^-]} = [H_3O^+]$
$\dfrac{1.0 \times 10^{-14}}{2.5 \times 10^{-4}} = [H_3O^+] = 4.0 \times 10^{-11}\; M$
Example $2$
1. Find the pH of a solution of 0.002 M of HCl.
2. Find the pH of a solution of 0.00005 M NaOH.
Solution
1. The equation for pH is -log [H+]
$[H^+]= 2.0 \times 10^{-3}\; M \nonumber$
$pH = -\log [2.0 \times 10^{-3}] = 2.70 \nonumber$
1. The equation for pOH is -log [OH-]
$[OH^-]= 5.0 \times 10^{-5}\; M \nonumber$
$pOH = -\log [5.0 \times 10^{-5}] = 4.30 \nonumber$
$pK_w = pH + pOH \nonumber$
and
$pH = pK_w - pOH \nonumber$
then
$pH = 14 - 4.30 = 9.70 \nonumber$
Example $3$: Soil
If moist soil has a pH of 7.84, what is the H+ concentration of the soil solution?
Solution
$pH = -\log [H^+] \nonumber$
$7.84 = -\log [H^+] \nonumber$
$[H^+] = 1.45 \times 10^{-8} M \nonumber$
Hint
Place -7.84 in your calculator and take the antilog (often inverse log or 10x) = 1.45 x 10-8M
Proper Definition of pH
The pH scale was originally introduced by the Danish biochemist S.P.L. Sørenson in 1909 using the symbol pH. The letter p is derived from the German word potenz meaning power or exponent of, in this case, 10. In 1909, S.P.L. Sørenson published a paper in Biochem Z in which he discussed the effect of H+ ions on the activity of enzymes. In the paper, he invented the term pH (purported to mean pondus hydrogenii in Latin) to describe this effect and defined it as the $-\log[H^+]$. In 1924, Sørenson realized that the pH of a solution is a function of the "activity" of the H+ ion and not the concentration. Thus, he published a second paper on the subject. A better definition would be
$pH = -\log\,a\{\ce{H^{+}}\}$
where $a\{H^+\}$ denotes the activity (an effective concentration) of the H+ ions. The activity of an ion is a function of many variables of which concentration is one.
• Concentration is abbreviated by using square brackets, e.g., $[H_3O^+]$ is the concentration of hydronium ion in solution.
• Activity is abbreviated by using "a" with curly brackets, e.g., $a\{H_3O^+\}$ is the activity of hydronium ions in solution
Because of the difficulty in accurately measuring the activity of the $\ce{H^{+}}$ ion for most solutions the International Union of Pure and Applied Chemistry (IUPAC) and the National Bureau of Standards (NBS) has defined pH as the reading on a pH meter that has been standardized against standard buffers. The following equation is used to calculate the pH of all solutions:
\begin{align} pH &= \dfrac{F(E-E_{standard})}{RT\;\ln 10} + pH_{standard} \label{6a} \[4pt] &= \dfrac{5039.879 (E-E_{standard})}{T} + pH_{standard} \label{6b} \end{align}
with
• $R$ is the ideal gas constant,
• $F$ is the Faraday's constant, and
• $T$ is absolute temperature (in K)
The activity of the H+ ion is determined as accurately as possible for the standard solutions used. The identity of these solutions vary from one authority to another, but all give the same values of pH to ± 0.005 pH unit. The historical definition of pH is correct for those solutions that are so dilute and so pure the H+ ions are not influenced by anything but the solvent molecules (usually water).
When measuring pH, [H+] is in units of moles of H+ per liter of solution. This is a reasonably accurate definition at low concentrations (the dilute limit) of H+. At very high concentrations (10 M hydrochloric acid or sodium hydroxide, for example,) a significant fraction of the ions will be associated into neutral pairs such as H+Cl, thus reducing the concentration of “available” ions to a smaller value which we will call the effective concentration. It is the effective concentration of H+ and OH that determines the pH and pOH. The pH scale as shown above is called sometimes "concentration pH scale" as opposed to the "thermodynamic pH scale". The main difference between both scales is that in thermodynamic pH scale one is interested not in H+concentration, but in H+activity. What a person measures in the solution is just activity, not the concentration. Thus it is thermodynamic pH scale that describes real solutions, not the concentration one.
For solutions in which ion concentrations don't exceed 0.1 M, the formulas pH = –log [H+] and pOH = –log[OH] are generally reliable, but don't expect a 10.0 M solution of a strong acid to have a pH of exactly –1.00! However, this definition is only an approximation (albeit very good under most situations) of the proper definition of pH, which depends on the activity of the hydrogen ion:
$pH= -\log a\{H^+\} \approx -\log [H^+] \label{7}$
The activity is a measure of the "effective concentration" of a substance, is often related to the true concentration via an activity coefficient, $\gamma$:
$a{H^+}=\gamma [H^+] \label{8}$
Calculating the activity coefficient requires detailed theories of how charged species interact in solution at high concentrations (e.g., the Debye-Hückel Theory). In most solutions the pH differs from the -log[H+ ] in the first decimal point. The following table gives experimentally determined pH values for a series of HCl solutions of increasing concentration at 25 °C.
Table $1$: HCl Solutions with corresponding pH values. Data taken from Christopher G. McCarty and Ed Vitz, Journal of Chemical Education, 83(5), 752 (2006) and G.N. Lewis, M. Randall, K. Pitzer, D.F. Brewer, Thermodynamics (McGraw-Hill: New York, 1961; pp. 233-34).
Molar Concentration of $HCl$ pH defined as Concentration Experimentally Determined pH Relative Deviation
0.00050 3.30 3.31 0.3%
0.0100 2 2.04 1.9%
0.100 1 1.10 9%
0.40 0.39 0.52 25%
7.6 -0.88 -1.85 52%
While the pH scale formally measures the activity of hydrogen ions in a substance or solution, it is typically approximated as the concentration of hydrogen ions; this approximation is applicable only under low concentrations.
Living Systems
Molecules that make up or are produced by living organisms usually function within a narrow pH range (near neutral) and a narrow temperature range (body temperature). Many biological solutions, such as blood, have a pH near neutral. pH influences the structure and the function of many enzymes (protein catalysts) in living systems. Many of these enzymes have narrow ranges of pH activity. Cellular pH is so important that death may occur within hours if a person becomes acidotic (having increased acidity in the blood). As one can see pH is critical to life, biochemistry, and important chemical reactions. Common examples of how pH plays a very important role in our daily lives are given below:
• Water in swimming pool is maintained by checking its pH. Acidic or basic chemicals can be added if the water becomes too acidic or too basic.
• Whenever we get a heartburn, more acid build up in the stomach and causes pain. We needs to take antacid tablets (a base) to neutralize excess acid in the stomach.
• The pH of blood is slightly basic. A fluctuation in the pH of the blood can cause in serious harm to vital organs in the body.
• Certain diseases are diagnosed only by checking the pH of blood and urine.
• Certain crops thrive better at certain pH range.
• Enzymes activate at a certain pH in our body.
Table $2$: pH in Living Systems
Compartment pH
Gastric Acid 1
Lysosomes 4.5
Granules of Chromaffin Cells 5.5
Human Skin 5.5
Urine 6
Neutral H2O at 37 °C 6.81
Cytosol 7.2
Cerebrospinal Fluid 7.3
Blood 7.43-7.45
Mitochondrial Matrix 7.5
Pancreas Secretions 8.1
Problems
1. In a solution of $2.4 \times 10^{-3} M$ of HI, find the concentration of $OH^-$.
2. Determine the pH of a solution that is 0.0035 M HCl.
3. Determine the [H3O+] of a solution with a pH = 5.65
4. If the pOH of NH3, ammonia, in water is 4.74. What is the pH?
5. Pepsin, a digestive enzyme in our stomach, has a pH of 1.5. Find the concentration of OH- in the stomach.
Solutions
1. We use the dissociation of water equation to find [OH-].
Kw = [H3O+][OH-] = 1.0 X 10-14
Solve for [OH-]
[OH-] = (1.0 X 10-14)/ [H3O+]
Plug in the molarity of HI and solve for OH-.
[OH-] = (1.0 X 10-14)/ [2.4 X 10-3] = 4.17 X 10-12 M.
2. pH = -log[H3O+]
Plug the molarity of the HCl in and solve for pH.
pH = -log[0.0035] = 2.46
3. pH = -log[H3O+]
Plug in the pH and solve for [H3O+]
5.65 = -log[H3O+]
Move the negative sign to the pH. -5.65 = log[H3O+]
10-5.65=10log[H3O+] = 2.24 X 10-6 M
4. pH + pOH = 14
Solve for pH.
14 - pOH = pH
14 - 4.74 = pH = 9.26
5. There are several ways to do this problem.
Answer 1.
pH + pOH = 14
Solve for pOH.
pOH = 14 - pH
pOH = 14 - 1.5 = 12.5
When the pOH is solved, solve for the concentration by using log.
pOH = -log[OH-]
12.5 = -log[OH-]
-12.5 = log[OH-]
10-12.5 = 10log[OH-] = 3.16 X 10-13 M.
Answer 2.
pH = -log[H+]
Plug in the pH and solve for the molarity of H+ of pepsin.
1.5 = -log[H+]
-1.5 = log[H+]
10-1.5 = 10log[H+] = [H+]= 0.032
Use the concentration of H+ to solve for the concentration of OH-.
[H+][OH-] = 1.0 X 10-14
Plug in the [H+] and solve for [OH-].
[OH-] = (1.0 X 10-14)/[H3O+]
[OH-] = (1.0 X 10-14)/(0.032) = 3.125 X 10-14 M | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/02%3A_Acids_and_Bases/2.1%3A_An_Introduction_to_Acids_and_Bases.txt |
This page explains the acidity of simple organic acids and looks at the factors which affect their relative strengths.
Organic acids as weak acids
For the purposes of this topic, we are going to take the definition of an acid as "a substance which donates hydrogen ions (protons) to other things". We are going to get a measure of this by looking at how easily the acids release hydrogen ions to water molecules when they are in solution in water.
An acid in solution sets up this equilibrium:
A hydroxonium ion is formed together with the anion (negative ion) from the acid.
This equilibrium is sometimes simplified by leaving out the water to emphasise the ionisation of the acid.
If you write it like this, you must include the state symbols - "(aq)". Writing H+(aq) implies that the hydrogen ion is attached to a water molecule as H3O+. Hydrogen ions are always attached to something during chemical reactions.
The organic acids are weak in the sense that this ionisation is very incomplete. At any one time, most of the acid will be present in the solution as un-ionised molecules. For example, in the case of dilute ethanoic acid, the solution contains about 99% of ethanoic acid molecules - at any instant, only about 1% have actually ionised. The position of equilibrium therefore lies well to the left.
Comparing the strengths of weak acids
The strengths of weak acids are measured on the pKa scale. The smaller the number on this scale, the stronger the acid is.
Three of the compounds we shall be looking at, together with their pKa values are:
Remember - the smaller the number the stronger the acid. Comparing the other two to ethanoic acid, you will see that phenol is very much weaker with a pKa of 10.00, and ethanol is so weak with a pKa of about 16 that it hardly counts as acidic at all!
Why are these acids acidic?
In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X":
So . . . if the same bond is being broken in each case, why do these three compounds have such widely different acid strengths?
Differences in acid strengths between carboxylic acids, phenols and alcohols
Two of the factors which influence the ionization of an acid are:
• the strength of the bond being broken,
• the stability of the ions being formed.
In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar. The most important factor in determining the relative acid strengths of these molecules is the nature of the ions formed. You always get a hydroxonium ion - so that's constant - but the nature of the anion (the negative ion) varies markedly from case to case.
Example \(1\): Ethanoic Acid
Ethanoic acid has the structure:
The acidic hydrogen is the one attached to the oxygen. When ethanoic acid ionises it forms the ethanoate ion, CH3COO-.
You might reasonably suppose that the structure of the ethanoate ion was as below, but measurements of bond lengths show that the two carbon-oxygen bonds are identical and somewhere in length between a single and a double bond.
To understand why this is, you have to look in some detail at the bonding in the ethanoate ion. Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond. Pi bonds are made by sideways overlap between p orbitals on the carbon and the oxygen.
In an ethanoate ion, one of the lone pairs on the negative oxygen ends up almost parallel to these p orbitals, and overlaps with them
This leads to a delocalised pi system over the whole of the -COO- group, rather like that in benzene.
All the oxygen lone pairs have been left out of this diagram to avoid confusion. Because the oxygens are more electronegative than the carbon, the delocalised system is heavily distorted so that the electrons spend much more time in the region of the oxygen atoms.
So where is the negative charge in all this? It has been spread around over the whole of the -COO- group, but with the greatest chance of finding it in the region of the two oxygen atoms. Ethanoate ions can be drawn simply as:
The dotted line represents the delocalisation. The negative charge is written centrally on that end of the molecule to show that it isn't localised on one of the oxygen atoms. The more you can spread charge around, the more stable an ion becomes. In this case, if you delocalise the negative charge over several atoms, it is going to be much less attractive to hydrogen ions - and so you are less likely to re-form the ethanoic acid.
Example \(2\): Phenol
Phenols have an -OH group attached directly to a benzene ring. Phenol itself is the simplest of these with nothing else attached to the ring apart from the -OH group.
When the hydrogen-oxygen bond in phenol breaks, you get a phenoxide ion, C6H5O-.
Delocalisation also occurs in this ion. This time, one of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring.
This overlap leads to a delocalisation which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localised on the oxygen, but is spread out around the whole ion.
Why then is phenol a much weaker acid than ethanoic acid?
Think about the ethanoate ion again. If there wasn't any delocalisation, the charge would all be on one of the oxygen atoms, like this:
But the delocalisation spreads this charge over the whole of the COO group. Because oxygen is more electronegative than carbon, you can think of most of the charge being shared between the two oxygens (shown by the heavy red shading in this diagram).
If there wasn't any delocalisation, one of the oxygens would have a full charge which would be very attractive towards hydrogen ions. With delocalisation, that charge is spread over two oxygen atoms, and neither will be as attractive to a hydrogen ion as if one of the oxygens carried the whole charge.
That means that the ethanoate ion won't take up a hydrogen ion as easily as it would if there wasn't any delocalisation. Because some of it stays ionised, the formation of the hydrogen ions means that it is acidic.
In the phenoxide ion, the single oxygen atom is still the most electronegative thing present, and the delocalised system will be heavily distorted towards it. That still leaves the oxygen atom with most of its negative charge.
What delocalisation there is makes the phenoxide ion more stable than it would otherwise be, and so phenol is acidic to an extent.
However, the delocalisation hasn't shared the charge around very effectively. There is still lots of negative charge around the oxygen to which hydrogen ions will be attracted - and so the phenol will readily re-form. Phenol is therefore only very weakly acidic.
Example \(3\): Ethanol
Ethanol, CH3CH2OH, is so weakly acidic that you would hardly count it as acidic at all. If the hydrogen-oxygen bond breaks to release a hydrogen ion, an ethoxide ion is formed:
This has nothing at all going for it. There is no way of delocalising the negative charge, which remains firmly on the oxygen atom. That intense negative charge will be highly attractive towards hydrogen ions, and so the ethanol will instantly re-form.
Since ethanol is very poor at losing hydrogen ions, it is hardly acidic at all.
Variations in acid strengths between different carboxylic acids
You might think that all carboxylic acids would have the same strength because each depends on the delocalization of the negative charge around the -COO- group to make the anion more stable, and so more reluctant to re-combine with a hydrogen ion.
In fact, the carboxylic acids have widely different acidities. One obvious difference is between methanoic acid, HCOOH, and the other simple carboxylic acids:
pKa
HCOOH 3.75
CH3COOH 4.76
CH3CH2COOH 4.87
CH3CH2CH2COOH 4.82
Remember that the higher the value for pKa, the weaker the acid is.
Why is ethanoic acid weaker than methanoic acid? It again depends on the stability of the anions formed - on how much it is possible to delocalise the negative charge. The less the charge is delocalised, the less stable the ion, and the weaker the acid.
The methanoate ion (from methanoic acid) is:
The only difference between this and the ethanoate ion is the presence of the CH3 group in the ethanoate.
But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the -COO- group. Any build-up of charge will make the ion less stable, and more attractive to hydrogen ions.
Ethanoic acid is therefore weaker than methanoic acid, because it will re-form more easily from its ions.
The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of propanoic acid and butanoic acid are very similar to ethanoic acid. The acids can be strengthened by pulling charge away from the -COO- end. You can do this by attaching electronegative atoms like chlorine to the chain.
As the next table shows, the more chlorines you can attach the better:
pKa
CH3COOH 4.76
CH2ClCOOH 2.86
CHCl2COOH 1.29
CCl3COOH 0.65
Trichloroethanoic acid is quite a strong acid.
Attaching different halogens also makes a difference. Fluorine is the most electronegative and so you would expect it to be most successful at pulling charge away from the -COO- end and so strengthening the acid.
pKa
CH2FCOOH 2.66
CH2ClCOOH 2.86
CH2BrCOOH 2.90
CH2ICOOH 3.17
The effect is there, but isn't as great as you might expect.
Finally, notice that the effect falls off quite quickly as the attached halogen gets further away from the -COO- end. Here is what happens if you move a chlorine atom along the chain in butanoic acid.
pKa
CH3CH2CH2COOH 4.82
CH3CH2CHClCOOH 2.84
CH3CHClCH2COOH 4.06
CH2ClCH2CH2COOH 4.52
The chlorine is effective at withdrawing charge when it is next-door to the -COO- group, and much less so as it gets even one carbon further away.
This page explains why simple organic bases are basic and looks at the factors which affect their relative strengths. For A'level purposes, all the bases we are concerned with are primary amines - compounds in which one of the hydrogens in an ammonia molecule, NH3, is replaced either by an alkyl group or a benzene ring.
Ammonia as a weak base
All of the compounds we are concerned with are derived from ammonia and so we'll start by looking at the reason for its basic properties. For the purposes of this topic, we are going to take the definition of a base as "a substance which combines with hydrogen ions (protons)". We are going to get a measure of this by looking at how easily the bases take hydrogen ions from water molecules when they are in solution in water.
Ammonia in solution sets up this equilibrium:
An ammonium ion is formed together with hydroxide ions. Because the ammonia is only a weak base, it doesn't hang on to the extra hydrogen ion very effectively and so the reaction is reversible. At any one time, about 99% of the ammonia is present as unreacted molecules. The position of equilibrium lies well to the left.
The ammonia reacts as a base because of the active lone pair on the nitrogen. Nitrogen is more electronegative than hydrogen and so attracts the bonding electrons in the ammonia molecule towards itself. That means that in addition to the lone pair, there is a build-up of negative charge around the nitrogen atom. That combination of extra negativity and active lone pair attracts the new hydrogen from the water.
Comparing the strengths of weak bases
The strengths of weak bases are measured on the pKb scale. The smaller the number on this scale, the stronger the base is. Three of the compounds we shall be looking at, together with their pKb values are:
Remember - the smaller the number the stronger the base. Comparing the other two to ammonia, you will see that methylamine is a stronger base, whereas phenylamine is very much weaker.
• Methylamine is typical of aliphatic primary amines - where the -NH2 group is attached to a carbon chain. All aliphatic primary amines are stronger bases than ammonia.
• Phenylamine is typical of aromatic primary amines - where the -NH2 group is attached directly to a benzene ring. These are very much weaker bases than ammonia.
Explaining the differences in base strengths
Two of the factors which influence the strength of a base are:
• the ease with which the lone pair picks up a hydrogen ion,
• the stability of the ions being formed.
Why are aliphatic primary amines stronger bases than ammonia?
Methylamine
Methylamine has the structure:
The only difference between this and ammonia is the presence of the CH3 group in the methylamine. But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the nitrogen atom. That extra negativity around the nitrogen makes the lone pair even more attractive towards hydrogen ions.
Making the nitrogen more negative helps the lone pair to pick up a hydrogen ion. What about the effect on the positive methylammonium ion formed? Is this more stable than a simple ammonium ion? Compare the methylammonium ion with an ammonium ion:
In the methylammonium ion, the positive charge is spread around the ion by the "electron-pushing" effect of the methyl group. The more you can spread charge around, the more stable an ion becomes. In the ammonium ion there isn't any way of spreading the charge.
To summarize:
• The nitrogen is more negative in methylamine than in ammonia, and so it picks up a hydrogen ion more readily.
• The ion formed from methylamine is more stable than the one formed from ammonia, and so is less likely to shed the hydrogen ion again.
Taken together, these mean that methylamine is a stronger base than ammonia.
The other aliphatic primary amines
The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of the other aliphatic primary amines are very similar to methylamine. For example:
pKb
CH3NH2 3.36
CH3CH2NH2 3.27
CH3CH2CH2NH2 3.16
CH3CH2CH2CH2NH2 3.39
Why are aromatic primary amines much weaker bases than ammonia?
An aromatic primary amine is one in which the -NH2 group is attached directly to a benzene ring. The only one you are likely to come across is phenylamine. Phenylamine has the structure:
The lone pair on the nitrogen touches the delocalized ring electrons . . .
. . . and becomes delocalized with them:
That means that the lone pair is no longer fully available to combine with hydrogen ions. The nitrogen is still the most electronegative atom in the molecule, and so the delocalized electrons will be attracted towards it, but the intensity of charge around the nitrogen is nothing like what it is in, say, an ammonia molecule.
The other problem is that if the lone pair is used to join to a hydrogen ion, it is no longer available to contribute to the delocalisation. That means that the delocalization would have to be disrupted if the phenylamine acts as a base. Delocalization makes molecules more stable, and so disrupting the delocalization costs energy and will not happen easily.
Taken together - the lack of intense charge around the nitrogen, and the need to break some delocalization - this means that phenylamine is a very weak base indeed. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/02%3A_Acids_and_Bases/2.3%3A_Organic_Acids_and_Bases.txt |
/*<![CDATA[*/ MathJax.Hub.Config({ TeX: { equationNumbers: { autoNumber: "all", formatNumber: function (n) { return 3.6 + '.' + n } }, Macros: { PageIndex: ["{3.6. #1}",1] } } }); /*]]>*/A: Defining the acidity constant
You are no doubt aware that some acids are stronger than others. The relative acidity of different compounds or functional groups – in other words, their relative capacity to donate a proton to a common base under identical conditions – is quantified by a number called the acidity constant, abbreviated Ka. The common base chosen for comparison is water.
We will consider acetic acid as our first example. If we make a dilute solution of acetic acid in water, an acid-base reaction occurs between the acid (proton donor) and water (proton acceptor).
Acetic acid is a weak acid, so the equilibrium favors reactants over products - it is thermodynamically 'uphill', as indicated in the figure above by the relative length of the forward and reverse reaction arrows, and in the reaction coordinate diagram below in which products are higher energy than reactants.
As you know, the equilibrium constant Keq is defined as:
Every equilibrium constant expressions is actually a ratio of the activities of all species involved in the reaction. To avoid the use of activities, and to simplify experimental measurements, the equilibrium constant of concentrations approximates the activities of solutes and gases in dilute solutions with their respective molarities. However, the activities of solids, pure liquids, and solvents are not approximated with their molarities. Instead these activities are defined to have a value equal to 1 (one).
Thus, if we acknowledge that the activity of water in a dilute solution is approximated with the value of unity (1), we can divide by 1 to get the common form of the expression for Ka, the acid constant for acetic acid:
$K_a = K_{eq} = \dfrac{a_{H_3O^+}· a_{CH_3COO^-}}{a_{CH_3COOH} · a_{H_2O}} ≈ \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH](1)} = \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$
In fact, for an dilute aqueous solution, the activity of water is approximated with the value of 1, so the generic dissociation constant for a given acid HA or HB+ is expressed as:
The value of Ka for acetic acid is 1.75 x 10-5 - much less than 1, indicating there is much more acetic acid in solution at equilibrium than acetate and hydronium ions.
Conversely, sulfuric acid, with a Ka of approximately 109, or hydrochloric acid, with a Ka of approximately 107, both undergo essentially complete dissociation in water: they are very strong acids.
A number like 1.75 x 10- 5 is not very easy either to say, remember, or visualize, so chemists usually use a more convenient term to express relative acidity. The pKa value of an acid is simply the log (base 10) of its Ka value.
pKa = -log Ka Ka = 10-pKa
Doing the math, we find that the pKa of acetic acid is 4.8. The pKa of sulfuric acid is -10, and of hydrochloric acid is -7. The use of pKa values allows us to express the relative acidity of common compounds and functional groups on a numerical scale of about –10 (for a very strong acid) to 50 (for a compound that is not acidic at all). The lower the pKa value, the stronger the acid.
The ionizable (proton donating or accepting) functional groups relevant to biological organic chemistry generally have pKa values ranging from about 5 to about 20. The most important of these are summarized below, with very rough pKa values for the conjugate acid forms. More acidic groups with pKa values near zero are also included for reference.
Approximate pKa values to know
hydronium ion (H3O+) : 0
protonated alcohol: 0
protonated carbonyl: 0
carboxylic acids: 5
protonated imines: 7
protonated amines: 10
phenols: 10
thiols: 10
water: 14
alcohols: 15-18
alpha-carbon acids*: 20
*alpha-carbon acids will be discussed later in this chapter
You are strongly recommended to commit these rough values to memory now - then if you need a more precise value, you can always look it up in a pKa table.
Caution! pKa is not the same as pH!
It is important to realize that pKa is not the same thing as pH: the former is an inherent property of a compound or functional group, while the latter is a measure of hydronium ion concentration in a given aqueous solution:
pH = -log [H3O+]
Knowing pKa values not only allows us to compare acid strength, it also allows us to compare base strength. The key idea to remember is this: the stronger the conjugate acid, the weaker the conjugate base. We can determine that hydroxide ion is a stronger base than ammonia (NH3), because ammonium ion (NH4+, pKa = 9.2) is a stronger acid than water (pKa = 14.0).
Exercise 7.2.1
Which is the stronger base, CH3O- or CH3S-? Acetate ion or ammonia? Hydroxide ion or acetate ion?
Solution
Let's put our understanding of the pKa concept to use in the context of a more complex molecule. For example, what is the pKa of the compound below?
We need to evaluate the potential acidity of four different types of protons on the molecule, and find the most acidic one. The aromatic protons are not all acidic - their pKa is about 45. The amine group is also not acidic, its pKa is about 35. (Remember, uncharged amines are basic: it is positively-charged protonated amines, with pKa values around 10, that are weakly acidic.) The alcohol proton has a pKa of about 15, and the phenol proton has a pKa of about 10: thus, the most acidic group on the molecule above is the phenol. (Be sure that you can recognize the difference between a phenol and an alcohol - remember, in a phenol the OH group is bound directly to the aromatic ring). If this molecule were to react with one molar equivalent of a strong base such as sodium hydroxide, it is the phenol proton which would be donated to form a phenolate anion.
Exercise 7.2.2
Identify the most acidic functional group on each of the molecules below, and give its approximate pKa.
Solution
B: Using pKa values to predict reaction equilibria
By definition, the pKa value tells us the extent to which an acid will react with water as the base, but by extension we can also calculate the equilibrium constant for a reaction between any acid-base pair. Mathematically, it can be shown that:
Keq = 10ΔpKa
. . . where ΔpKa = (pKa of product acid minus pKa of reactant acid).
Consider a reaction between methylamine and acetic acid:
The first step is to identify the acid species on either side of the equation, and look up or estimate their pKa values. On the left side, the acid is of course acetic acid while on the right side the acid is methyl ammonium ion (in other words, methyl ammonium ion is the acid in the reaction going from right to left). We can look up the precise pKa values in table 7 (at the back of the book), but we already know (because we have this information memorized, right?!) that the pKa of acetic acids is about 5, and methyl ammonium is about 10. More precise values are 4.8 and 10.6, respectively.
Without performing any calculations at all, you should be able to see that this equilibrium lies far to the right-hand side: acetic acid has a lower pKa, meaning it is a stronger acid than methyl ammonium, and thus it wants to give up its proton more than methyl ammonium does. Doing the math, we see that
Keq = 10ΔpKa = 10(10.6 – 4.8) = 105.8 = 6.3 x 105
So Keq is a very large number (much greater than 1) and the equilibrium for the reaction between acetic acid and methylamine lies far to the right-hand side of the equation, just as we had predicted. This also tells us that the reaction has a negative Gibbs free energy change, and is thermodynamically favorable.
If you had just wanted to quickly approximate the value of Keq without benefit of precise pKa information or a calculator, you could have approximated pKa ~ 5 (for the carboxylic acid) and pKa ~10 (for the ammonium ion) and calculated in your head that the equilibrium constant should be somewhere in the order of 105.
Exercise 7.2.3
Show the products of the following acid-base reactions, and roughly estimate the value of Keq.
Solution
C: Organic molecules in buffered solution: the Henderson-Hasselbalch equation
The environment inside a living cell, where most biochemical reactions take place, is an aqueous buffer with pH ~ 7. Recall from your General Chemistry course that a buffer is a solution of a weak acid and its conjugate base. The key equation for working with buffers is the Henderson-Hasselbalch equation:
The Henderson-Hasselbalch equation:
The equation tells us that if our buffer is an equimolar solution of a weak acid and its conjugate base, the pH of the buffer will equal the pKa of the acid (because the log of 1 is equal to zero). If there is more of the acid form than the base, then of course the pH of the buffer is lower than the pKa of the acid.
Exercise 7.2.4
What is the pH of an aqueous buffer solution that is 30 mM in acetic acid and 40 mM in sodium acetate? The pKa of acetic acid is 4.8.
Solution
The Henderson-Hasselbalch equation is particularly useful when we want to think about the protonation state of different biomolecule functional groups in a pH 7 buffer. When we do this, we are always assuming that the concentration of the biomolecule is small compared to the concentration of the buffer components. (The actual composition of physiological buffer is complex, but it is primarily based on phosphoric and carbonic acids).
Imagine an aspartic acid residue located on the surface of a protein in a human cell. Being on the surface, the side chain is in full contact with the pH 7 buffer surrounding the protein. In what state is the side chain functional group: the protonated state (a carboxylic acid) or the deprotonated state (a carboxylate ion)? Using the Henderson-Hasselbalch equation, we fill in our values for the pH of the buffer and a rough pKa approximation of pKa = 5 for the carboxylic acid functional group. Doing the math, we find that the ratio of carboxylate to carboxylic acid is about 100 to 1: the carboxylic acid is almost completely ionized (in the deprotonated state) inside the cell. This result extends to all other carboxylic acid groups you might find on natural biomolecules or drug molecules: in the physiological environment, carboxylic acids are almost completely deprotonated.
Now, let's use the equation again, this time for an amine functional group, such as the side chain of a lysine residue: inside a cell, are we likely to see a neutral amine (R-NH2) or an ammonium cation (R-NH3+?) Using the equation with pH = 7 (for the biological buffer) and pKa = 10 (for the ammonium group), we find that the ratio of neutral amine to ammonium cation is about 1 to 100: the group is close to completely protonated inside the cell, so we will see R-NH3+, not R-NH2.
We can do the same rough calculation for other common functional groups found in biomolecules. It will be very helpful going forward to commit the following to memory:
At physiological pH:
Carboxylic acids are deprotonated (in the carboxylate anion form)
Amines are protonated (in the ammonium cation form)
Thiols, phenols, alcohols, and amides are uncharged
Imines are a mixture of the protonated (cationic) and deprotonated (neutral) states
We will talk about the physiological protonation state of phosphate groups in chapter 9.
Exercise 7.2.5
The molecule below is not drawn in a reasonable protonation state for pH 7. Redraw it in with the functional groups in the correct protonation state.
Solution
While we are most interested in the state of molecules at pH 7, the Henderson-Hasselbalch equation can of course be used to determine the protonation state of functional groups in solutions buffered to other pH levels. The exercises below provide some practice in this type of calculation.
Exercise 7.2.6
What is the ratio of acetate ion to neutral acetic acid when a small amount of acetic acid (pKa = 4.8) is dissolved in a buffer of pH 2.8? pH 3.8? pH 4.8? pH 5.8? pH 6.8?
Solution
Exercise 7.2.7
Would you expect phenol to be soluble in an aqueous solution buffered to pH 2? pH 7? pH 12? Explain your answer.
Solution
Exercise 7.2.8
Exercise 7.2.8: What is the approximate net charge on a tetrapeptide Cys-Asp-Lys-Glu in pH 7 buffer?
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/02%3A_Acids_and_Bases/2.4%3A_How_to_Predict_the_Outcome_of_an_Acid-Base_Reaction.txt |
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Blood as a Buffer
Buffer solutions are extremely important in biology and medicine because most biological reactions and enzymes need very specific pH ranges in order to work properly.
• Henderson-Hasselbalch Approximation
The Henderson-Hasselbalch approximation allows us one method to approximate the pH of a buffer solution.
• How Does A Buffer Maintain pH?
A buffer is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit.
• Introduction to Buffers
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Preparing Buffer Solutions
When it comes to buffer solution one of the most common equation is the Henderson-Hasselbalch approximation. An important point that must be made about this equation is it's useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations.
Thumbnail: Simulated titration of an acidified solution of a weak acid (pKa = 4.7) with alkali. (Public Domain; Lasse Havelund).
2.9: Lewis Acids and Bases
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Write the equation for the proton transfer reaction involving a Brønsted-Lowry acid or base, and show how it can be interpreted as an electron-pair transfer reaction, clearly identifying the donor and acceptor.
• Give an example of a Lewis acid-base reaction that does not involve protons.
• Write equations illustrating the behavior of a given non-aqueous acid-base system.
The Brønsted-Lowry proton donor-acceptor concept has been one of the most successful theories of Chemistry. But as with any such theory, it is fair to ask if this is not just a special case of a more general theory that could encompass an even broader range of chemical science. In 1916, G.N. Lewis of the University of California proposed that the electron pair is the dominant actor in acid-base chemistry. The Lewis theory did not become very well known until about 1923 (the same year that Brønsted and Lowry published their work), but since then it has been recognized as a very powerful tool for describing chemical reactions of widely different kinds and is widely used in organic and inorganic chemistry. According to Lewis,
• An acid is a substance that accepts a pair of electrons, and in doing so, forms a covalent bond with the entity that supplies the electrons.
• A base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared.
In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles.
Proton-Transfer Reactions Involve Electron-Pair Transfer
Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place.
This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points:
• The arrow shows the movement of a proton from the hydronium ion to the hydroxide ion.
• Note carefully that the electron-pairs themselves do not move; they remain attached to their central atoms. The electron pair on the base is "donated" to the acceptor (the proton) only in the sense that it ends up being shared with the acceptor, rather than being the exclusive property of the oxygen atom in the hydroxide ion.
• Although the hydronium ion is the nominal Lewis acid here, it does not itself accept an electron pair, but acts merely as the source of the proton that coordinates with the Lewis base.
The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a Lewis acid-base reaction from an oxidation-reduction reaction, in which a physical transfer of one or more electrons from donor to acceptor does occur. The product of a Lewis acid-base reaction is known formally as an "adduct" or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" H2O. The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with.
Another example, showing the autoprotolysis of water. Note that the conjugate base is also the adduct.
Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the hydrolysis of the ammonium ion.
Because $\ce{HF}$ is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, F accepts a proton from water, which is transformed into a hydroxide ion.
The bisulfite ion is amphiprotic and can act as an electron donor or acceptor.
Acid-base Reactions without Transferring Protons
The major utility of the Lewis definition is that it extends the concept of acids and bases beyond the realm of proton transfer reactions. The classic example is the reaction of boron trifluoride with ammonia to form an adduct:
$\ce{BF_3 + NH_3 \rightarrow F_3B-NH_3}$
One of the most commonly-encountered kinds of Lewis acid-base reactions occurs when electron-donating ligands form coordination complexes with transition-metal ions.
Exercise $1$
Here are several more examples of Lewis acid-base reactions that cannot be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction.
1. $\ce{Al(OH)_3 + OH^{–} \rightarrow Al(OH)_4^–}$
2. $\ce{SnS_2 + S^{2–} \rightarrow SnS_3^{2–}}$
3. $\ce{Cd(CN)_2 + 2 CN^– \rightarrow Cd(CN)_4^{2+}}$
4. $\ce{AgCl + 2 NH_3 \rightarrow Ag(NH_3)_2^+ + Cl^–}$
5. $\ce{Fe^{2+} + NO \rightarrow Fe(NO)^{2+}}$
6. $\ce{Ni^{2+} + 6 NH_3 \rightarrow Ni(NH_3)_5^{2+}}$
Applications to organic reaction mechanisms
Although organic chemistry is beyond the scope of these lessons, it is instructive to see how electron donors and acceptors play a role in chemical reactions. The following two diagrams show the mechanisms of two common types of reactions initiated by simple inorganic Lewis acids:
In each case, the species labeled "Complex" is an intermediate that decomposes into the products, which are conjugates of the original acid and base pairs. The electric charges indicated in the complexes are formal charges, but those in the products are "real".
In reaction 1, the incomplete octet of the aluminum atom in $\ce{AlCl3}$ serves as a better electron acceptor to the chlorine atom than does the isobutyl part of the base. In reaction 2, the pair of non-bonding electrons on the dimethyl ether coordinates with the electron-deficient boron atom, leading to a complex that breaks down by releasing a bromide ion.
Non-aqueous Protonic Acid-Base Systems
We ordinarily think of Brønsted-Lowry acid-base reactions as taking place in aqueous solutions, but this need not always be the case. A more general view encompasses a variety of acid-base solvent systems, of which the water system is only one (Table $1$). Each of these has as its basis an amphiprotic solvent (one capable of undergoing autoprotolysis), in parallel with the familiar case of water.
The ammonia system is one of the most common non-aqueous system in Chemistry. Liquid ammonia boils at –33° C, and can conveniently be maintained as a liquid by cooling with dry ice (–77° C). It is a good solvent for substances that also dissolve in water, such as ionic salts and organic compounds since it is capable of forming hydrogen bonds. However, many other familiar substances can also serve as the basis of protonic solvent systems as Table $1$ indicates:
Table $1$: Popular Solvent systems
solvent
autoprotolysis reaction
pKap
water 2 H2O → H3O+ + OH 14
ammonia 2 NH3 → NH4+ + NH2 33
acetic acid 2 CH3COOH → CH3COOH2+ + CH3COO 13
ethanol 2 C2H5OH → C2H5OH2+ + C2H5O 19
hydrogen peroxide 2 HO-OH → HO-OH2+ + HO-O 13
hydrofluoric acid 2 HF → H2F+ + F 10
sulfuric acid 2 H2SO4 → H3SO4+ + HSO4 3.5
One use of nonaqueous acid-base systems is to examine the relative strengths of the strong acids and bases, whose strengths are "leveled" by the fact that they are all totally converted into H3O+ or OH ions in water. By studying them in appropriate non-aqueous solvents which are poorer acceptors or donors of protons, their relative strengths can be determined. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/02%3A_Acids_and_Bases/2.8%3A_Buffer_Solutions.txt |
Homework Problems
Homework Solutions
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03: An Introduction to Organic Compounds: Nomenclature Physical Properties and Representation of Structure
The names of all alkanes end with -ane. Whether or not the carbons are linked together end-to-end in a ring (called cyclic alkanes or cycloalkanes) or whether they contain side chains and branches, the name of every carbon-hydrogen chain that lacks any double bonds or functional groups will end with the suffix -ane.
Alkanes with unbranched carbon chains are simply named by the number of carbons in the chain. The first four members of the series (in terms of number of carbon atoms) are named as follows:
1. CH4 = methane = one hydrogen-saturated carbon
2. C2H6 = ethane = two hydrogen-saturated carbons
3. C3H8 = propane = three hydrogen-saturated carbons
4. C4H10 = butane = four hydrogen-saturated carbons
Alkanes with five or more carbon atoms are named by adding the suffix -ane to the appropriate numerical multiplier, except the terminal -a is removed from the basic numerical term. Hence, C5H12 is called pentane, C6H14 is called hexane, C7H16 is called heptane and so forth.
Straight-chain alkanes are sometimes indicated by the prefix n- (for normal) to distinguish them from branched-chain alkanes having the same number of carbon atoms. Although this is not strictly necessary, the usage is still common in cases where there is an important difference in properties between the straight-chain and branched-chain isomers: e.g. n-hexane is a neurotoxin while its branched-chain isomers are not.
IUPAC nomenclature
The IUPAC nomenclature is a system on which most organic chemists have agreed to provide guidelines to allow them to learn from each others' works. Nomenclature, in other words, provides a foundation of language for organic chemistry.
Number of Hydrogen to Carbons
This equation describes the relationship between the number of hydrogen and carbon atoms in alkanes:
H = 2C + 2
where "C" and "H" are used to represent the number of carbon and hydrogen atoms present in one molecule. If C = 2, then H = 6.
Many textbooks put this in the following format:
CnH2n+2
where "Cn" and "H2n+2" represent the number of carbon and hydrogen atoms present in one molecule. If Cn = 3, then H2n+2 = 2(3) + 2 = 8. (For this formula look to the "n" for the number, the "C" and the "H" letters themselves do not change.)
Progressively longer hydrocarbon chains can be made and are named systematically, depending on the number of carbons in the longest chain.
The following table contains the systematic names for the first twenty straight chain alkanes. It will be important to familiarize yourself with these names because they will be the basis for naming many other organic molecules throughout your course of study.
Drawing Hydrocarbons
Recall that when carbon makes four bonds, it adopts the tetrahedral geometry. In the tetrahedral geometry, only two bonds can occupy a plane simultaneously. The other two bonds point in back or in front of this plane. In order to represent the tetrahedral geometry in two dimensions, solid wedges are used to represent bonds pointing out of the plane of the drawing toward the viewer, and dashed wedges are used to represent bonds pointing out of the plane of the drawing away from the viewer. Consider the following representation of the molecule methane:
Figure 1: Two dimensional representation of methane
In the above drawing, the two hydrogens connected by solid lines, as well as the carbon in the center of the molecule, exist in a plane (specifically, the plane of the computer monitor / piece of paper, etc.). The hydrogen connected by a solid wedge points out of this plane toward the viewer, and the hydrogen connected by the dashed wedge points behind this plane and away from the viewer.
In drawing hydrocarbons, it can be time-consuming to write out each atom and bond individually. In organic chemistry, hydrocarbons can be represented in a shorthand notation called a skeletal structure. In a skeletal structure, only the bonds between carbon atoms are represented. Individual carbon and hydrogen atoms are not drawn, and bonds to hydrogen are not drawn. In the case that the molecule contains just single bonds (sp3 bonds), these bonds are drawn in a "zig-zag" fashion. This is because in the tetrahedral geometry all bonds point as far away from each other as possible, and the structure is not linear. Consider the following representations of the molecule propane:
Figure 2: Full structure of propane and skeletal structure of propane
Only the bonds between carbons have been drawn, and these have been drawn in a "zig-zag" manner. Note that there is no representation of hydrogens in a skeletal structure. Since, in the absence of double or triple bonds, carbon makes four bonds total, the presence of hydrogens is implicit. Whenever an insufficient number of bonds to a carbon atom are specified in the structure, it is assumed that the rest of the bonds are made to hydrogens. For example, if the carbon atom makes only one explicit bond, there are three hydrogens implicitly attached to it. If it makes two explicit bonds, there are two hydrogens implicitly attached, etc. Note also that two lines are sufficient to represent three carbon atoms. It is the bonds only that are being drawn out, and it is understood that there are carbon atoms (with three hydrogens attached!) at the terminal ends of the structure.
Alkyl Groups
Alkanes can be described by the general formula CnH2n+2. An alkyl group is formed by removing one hydrogen from the alkane chain and is described by the formula CnH2n+1. The removal of this hydrogen results in a stem change from -ane to -yl. Take a look at the following examples.
The same concept can be applied to any of the straight chain alkane names provided in the table above.
Name Molecular Formula Condensed Structural Formula
Methane CH4 CH4
Ethane C2H6 CH3CH3
Propane C3H8 CH3CH2CH3
Butane C4H10 CH3(CH2)2CH3
Pentane C5H12 CH3(CH2)3CH3
Hexane C6H14 CH3(CH2)4CH3
Heptane C7H16 CH3(CH2)5CH3
Octane C8H18 CH3(CH2)6CH3
Nonane C9H20 CH3(CH2)7CH3
Decane C10H22 CH3(CH2)8CH3
Undecane C11H24 CH3(CH2)9CH3
Dodecane C12H26 CH3(CH2)10CH3
Tridecane C13H28 CH3(CH2)11CH3
Tetradecane C14H30 CH3(CH2)12CH3
Pentadecane C15H32 CH3(CH2)13CH3
Hexadecane C16H34 CH3(CH2)14CH3
Heptadecane C17H36 CH3(CH2)15CH3
Octadecane C18H38 CH3(CH2)16CH3
Nonadecane C19H40 CH3(CH2)17CH3
Eicosane C20H42 CH3(CH2)18CH3
Using Common Names with Branched Alkanes
Certain branched alkanes have common names that are still widely used today. These common names make use of prefixes, such as iso-, sec-, tert-, and neo-. The prefix iso-, which stands for isomer, is commonly given to 2-methyl alkanes. In other words, if there is methyl group located on the second carbon of a carbon chain, we can use the prefix iso-. The prefix will be placed in front of the alkane name that indicates the total number of carbons. Examples:
• isopentane which is the same as 2-methylbutane
• isobutane which is the same as 2-methylpropane
To assign the prefixes sec-, which stands for secondary, and tert-, for tertiary, it is important that we first learn how to classify carbon molecules. If a carbon is attached to only one other carbon, it is called a primary carbon. If a carbon is attached to two other carbons, it is called a seconday carbon. A tertiary carbon is attached to three other carbons and last, a quaternary carbon is attached to four other carbons. Examples:
• 4-sec-butylheptane (30g)
• 4-tert-butyl-5-isopropylhexane (30d); if using this example, may want to move sec/tert after iso disc
The prefix neo- refers to a substituent whose second-to-last carbon of the chain is trisubstituted (has three methyl groups attached to it). A neo-pentyl has five carbons total. Examples:
• neopentane
• neoheptane
Alkoxy Groups
Alkoxides consist of an organic group bonded to a negatively charged oxygen atom. In the general form, alkoxides are written as RO-, where R represents the organic substituent. Similar to the alkyl groups above, the concept of naming alkoxides can be applied to any of the straight chain alkanes provided in the table above.
Three Principles of Naming
1. Choose the longest, most substituted carbon chain containing a functional group.
2. A carbon bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number.
3. Take the alphabetical order into consideration; that is, after applying the first two rules given above, make sure that your substitutes and/or functional groups are written in alphabetical order.
Example 1
What is the name of the following molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example does not contain any functional groups, so we only need to be concerned with choosing the longest, most substituted carbon chain. The longest carbon chain has been highlighted in red and consists of eight carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. Because this example does not contain any functional groups, we only need to be concerned with the two substitutes present, that is, the two methyl groups. If we begin numbering the chain from the left, the methyls would be assigned the numbers 4 and 7, respectively. If we begin numbering the chain from the right, the methyls would be assigned the numbers 2 and 5. Therefore, to satisfy the second rule, numbering begins on the right side of the carbon chain as shown below. This gives the methyl groups the lowest possible numbering.
Rule 3: In this example, there is no need to utilize the third rule. Because the two substitutes are identical, neither takes alphabetical precedence with respect to numbering the carbons. This concept will become clearer in the following examples.
Example 2
What is the name of the following molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. In this example, numbering the chain from the left or the right would satisfy this rule. If we number the chain from the left, bromine and chlorine would be assigned the second and sixth carbon positions, respectively. If we number the chain from the right, chlorine would be assigned the second position and bromine would be assigned the sixth position. In other words, whether we choose to number from the left or right, the functional groups occupy the second and sixth positions in the chain. To select the correct numbering scheme, we need to utilize the third rule.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Example 3
What is the name of the follow molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine, and one substitute, the methyl group. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. After taking functional groups into consideration, any substitutes present must have the lowest possible carbon number. This particular example illustrates the point of difference principle. If we number the chain from the left, bromine, the methyl group and chlorine would occupy the second, fifth and sixth positions, respectively. This concept is illustrated in the second drawing below. If we number the chain from the right, chlorine, the methyl group and bromine would occupy the second, third and sixth positions, respectively, which is illustrated in the first drawing below. The position of the methyl, therefore, becomes a point of difference. In the first drawing, the methyl occupies the third position. In the second drawing, the methyl occupies the fifth position. To satisfy the second rule, we want to choose the numbering scheme that provides the lowest possible numbering of this substitute. Therefore, the first of the two carbon chains shown below is correct.
Therefore, the first numbering scheme is the appropriate one to use.
Once you have determined the correct numbering of the carbons, it is often useful to make a list, including the functional groups, substitutes, and the name of the parent chain.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Parent chain: heptane 2-Chloro 3-Methyl 6-Bromo
6-bromo-2-chloro-3-methylheptane
Problems
What is the name of the follow molecules? | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/03%3A_An_Introduction_to_Organic_Compounds%3A_Nomenclature_Physical_Properties_and_Representation_of_Structure/3.01%3A_How_Alkyl_Substituents_Are_Named.txt |
Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). The halogen atoms significantly alters the physical properties of the molecules including electronegativity, bond length, bond strength, and molecular size.
3.07: The Physical Properties of Alkanes Alkyl
Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). The halogen atoms significantly alters the physical properties of the molecules including electronegativity, bond length, bond strength, and molecular size.
Boiling Point and Water Solubility
It is instructive to compare the boiling points and water solubility of amines with those of corresponding alcohols and ethers. The dominant factor here is hydrogen bonding, and the first table below documents the powerful intermolecular attraction that results from -O-H---O- hydrogen bonding in alcohols (light blue columns). Corresponding -N-H---N- hydrogen bonding is weaker, as the lower boiling points of similarly sized amines (light green columns) demonstrate. Alkanes provide reference compounds in which hydrogen bonding is not possible, and the increase in boiling point for equivalent 1º-amines is roughly half the increase observed for equivalent alcohols.
Compound CH3CH3 CH3OH CH3NH2 CH3CH2CH3 CH3CH2OH CH3CH2NH2
Mol.Wt. 30 32 31 44 46 45
Boiling
Point ºC
-88.6º 65º -6.0º -42º 78.5º 16.6º
The second table illustrates differences associated with isomeric 1º, 2º & 3º-amines, as well as the influence of chain branching. Since 1º-amines have two hydrogens available for hydrogen bonding, we expect them to have higher boiling points than isomeric 2º-amines, which in turn should boil higher than isomeric 3º-amines (no hydrogen bonding). Indeed, 3º-amines have boiling points similar to equivalent sized ethers; and in all but the smallest compounds, corresponding ethers, 3º-amines and alkanes have similar boiling points. In the examples shown here, it is further demonstrated that chain branching reduces boiling points by 10 to 15 ºC.
Compound CH3(CH2)2CH3 CH3(CH2)2OH CH3(CH2)2NH2 CH3CH2NHCH3 (CH3)3CH (CH3)2CHOH (CH3)2CHNH2 (CH3)3N
Mol.Wt. 58 60 59 59 58 60 59 59
Boiling
Point ºC
-0.5º 97º 48º 37º -12º 82º 34º
The water solubility of 1º and 2º-amines is similar to that of comparable alcohols. As expected, the water solubility of 3º-amines and ethers is also similar. These comparisons, however, are valid only for pure compounds in neutral water. The basicity of amines (next section) allows them to be dissolved in dilute mineral acid solutions, and this property facilitates their separation from neutral compounds such as alcohols and hydrocarbons by partitioning between the phases of non-miscible solvents.
Basicity of Amines
A review of basic acid-base concepts should be helpful to the following discussion. Like ammonia, most amines are Brønsted and Lewis bases, but their base strength can be changed enormously by substituents. It is common to compare basicity's quantitatively by using the pKa's of their conjugate acids rather than their pKb's. Since pKa + pKb = 14, the higher the pKa the stronger the base, in contrast to the usual inverse relationship of pKa with acidity. Most simple alkyl amines have pKa's in the range 9.5 to 11.0, and their water solutions are basic (have a pH of 11 to 12, depending on concentration). The first four compounds in the following table, including ammonia, fall into that category.
The last five compounds (colored cells) are significantly weaker bases as a consequence of three factors. The first of these is the hybridization of the nitrogen. In pyridine the nitrogen is sp2 hybridized, and in nitriles (last entry) an sp hybrid nitrogen is part of the triple bond. In each of these compounds (shaded red) the non-bonding electron pair is localized on the nitrogen atom, but increasing s-character brings it closer to the nitrogen nucleus, reducing its tendency to bond to a proton.
Compound
NH3 CH3C≡N
pKa 11.0 10.7 10.7 9.3 5.2 4.6 1.0 0.0 -1.0 -10.0
Secondly, aniline and p-nitroaniline (first two green shaded structures) are weaker bases due to delocalization of the nitrogen non-bonding electron pair into the aromatic ring (and the nitro substituent). This is the same delocalization that results in activation of a benzene ring toward electrophilic substitution. The following resonance equations, which are similar to those used to explain the enhanced acidity of ortho and para-nitrophenols illustrate electron pair delocalization in p-nitroaniline. Indeed, aniline is a weaker base than cyclohexyl amine by roughly a million fold, the same factor by which phenol is a stronger acid than cyclohexanol. This electron pair delocalization is accompanied by a degree of rehybridization of the amino nitrogen atom, but the electron pair delocalization is probably the major factor in the reduced basicity of these compounds. A similar electron pair delocalization is responsible for the very low basicity (and nucleophilic reactivity) of amide nitrogen atoms (last green shaded structure). This feature was instrumental in moderating the influence of amine substituents on aromatic ring substitution, and will be discussed further in the section devoted to carboxylic acid derivatives.
Conjugated amine groups influence the basicity of an existing amine. Although 4-dimethylaminopyridine (DMAP) might appear to be a base similar in strength to pyridine or N,N-dimethylaniline, it is actually more than ten thousand times stronger, thanks to charge delocalization in its conjugate acid. The structure in the gray box shows the locations over which positive charge (colored red) is delocalized in the conjugate acid. This compound is often used as a catalyst for acyl transfer reactions.
Finally, the very low basicity of pyrrole (shaded blue) reflects the exceptional delocalization of the nitrogen electron pair associated with its incorporation in an aromatic ring. Indole (pKa = -2) and imidazole (pKa = 7.0), see above, also have similar heterocyclic aromatic rings. Imidazole is over a million times more basic than pyrrole because the sp2 nitrogen that is part of one double bond is structurally similar to pyridine, and has a comparable basicity.
Although resonance delocalization generally reduces the basicity of amines, a dramatic example of the reverse effect is found in the compound guanidine (pKa = 13.6). Here, as shown below, resonance stabilization of the base is small, due to charge separation, while the conjugate acid is stabilized strongly by charge delocalization. Consequently, aqueous solutions of guanidine are nearly as basic as are solutions of sodium hydroxide.
The relationship of amine basicity to the acidity of the corresponding conjugate acids may be summarized in a fashion analogous to that noted earlier for acids:
Strong bases have weak conjugate acids, and weak bases have strong conjugate acids.
Acidity of Amines
We normally think of amines as bases, but it must be remembered that 1º and 2º-amines are also very weak acids (ammonia has a pKa = 34). In this respect it should be noted that pKa is being used as a measure of the acidity of the amine itself rather than its conjugate acid, as in the previous section. For ammonia this is expressed by the following hypothetical equation:
NH3 + H2O ____> NH2(–) + H2O-H(+)
The same factors that decreased the basicity of amines increase their acidity. This is illustrated by the following examples, which are shown in order of increasing acidity. It should be noted that the first four examples have the same order and degree of increased acidity as they exhibited decreased basicity in the previous table. The first compound is a typical 2º-amine, and the three next to it are characterized by varying degrees of nitrogen electron pair delocalization. The last two compounds (shaded blue) show the influence of adjacent sulfonyl and carbonyl groups on N-H acidity. From previous discussion it should be clear that the basicity of these nitrogens is correspondingly reduced.
Compound C6H5SO2NH2
pKa 33 27 19 15 10 9.6
The acids shown here may be converted to their conjugate bases by reaction with bases derived from weaker acids (stronger bases). Three examples of such reactions are shown below, with the acidic hydrogen colored red in each case. For complete conversion to the conjugate base, as shown, a reagent base roughly a million times stronger is required.
C6H5SO2NH2 + KOH C6H5SO2NH(–) K(+) + H2O a sulfonamide base
(CH3)3COH + NaH (CH3)3CO(–) Na(+) + H2 an alkoxide base
(C2H5)2NH + C4H9Li (C2H5)2N(–) Li(+) + C4H10 an amide base
Important Reagent Bases
The significance of all these acid-base relationships to practical organic chemistry lies in the need for organic bases of varying strength, as reagents tailored to the requirements of specific reactions. The common base sodium hydroxide is not soluble in many organic solvents, and is therefore not widely used as a reagent in organic reactions. Most base reagents are alkoxide salts, amines or amide salts. Since alcohols are much stronger acids than amines, their conjugate bases are weaker than amide bases, and fill the gap in base strength between amines and amide salts. In the following table, pKa again refers to the conjugate acid of the base drawn above it.
Base Name Pyridine Triethyl
Amine
Hünig's Base Barton's
Base
Potassium
t-Butoxide
Sodium HMDS LDA
Formula (C2H5)3N (CH3)3CO(–) K(+) [(CH3)3Si]2N(–) Na(+) [(CH3)2CH]2N(–) Li(+)
pKa 5.3 10.7 11.4 14 19 26 35.7
Pyridine is commonly used as an acid scavenger in reactions that produce mineral acid co-products. Its basicity and nucleophilicity may be modified by steric hindrance, as in the case of 2,6-dimethylpyridine (pKa=6.7), or resonance stabilization, as in the case of 4-dimethylaminopyridine (pKa=9.7). Hünig's base is relatively non-nucleophilic (due to steric hindrance), and like DBU is often used as the base in E2 elimination reactions conducted in non-polar solvents. Barton's base is a strong, poorly-nucleophilic, neutral base that serves in cases where electrophilic substitution of DBU or other amine bases is a problem. The alkoxides are stronger bases that are often used in the corresponding alcohol as solvent, or for greater reactivity in DMSO. Finally, the two amide bases see widespread use in generating enolate bases from carbonyl compounds and other weak carbon acids.
Nonionic Superbases
An interesting group of neutral, highly basic compounds of nitrogen and phosphorus have been prepared, and are referred to as superbases. To see examples of these compounds Click Here. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/03%3A_An_Introduction_to_Organic_Compounds%3A_Nomenclature_Physical_Properties_and_Representation_of_Structure/3.06%3A_The_Structures_of_Alkyl_Halides_Alcohols_.txt |
Before we begin our exploration of stereochemistry and chirality, we first need to consider the subject of conformational isomerism, which has to do with rotation about single bonds.
We learned in section 2.1 that single bonds in organic molecules are free to rotate, due to the 'end-to-end' (sigma) nature of their orbital overlap. Consider the carbon-oxygen bond in ethanol, for example: with a 180o rotation about this bond, the shape of the molecule would look quite different:
Or ethane: rotation about the carbon-carbon sigma bond results in many different possible three-dimensional arrangements of the atoms.
These different arrangements, resulting from sigma bond rotation, are referred to in organic chemistry as conformations. Any one specific conformation is called a conformational isomer, or conformer.
In order to better visualize different conformations of a molecule, it is convenient to use a drawing convention called the Newman projection. In a Newman projection, we look lengthwise down a specific bond of interest – in this case, the carbon-carbon bond in ethane. We depict the ‘front’ atom as a dot, and the ‘back’ atom as a larger circle.
The six carbon-hydrogen bonds are shown as solid lines protruding from the two carbons. Note that we do not draw bonds as solid or dashed wedges in a Newman projection.
Looking down the C-C bond in this way, the angle formed between a C-H bond on the front carbon and a C-H bond on the back carbon is referred to as a dihedral angle. (The dihedral angle between the hour hand and the minute hand on a clock is 0o at noon, 90o at 3:00, and so forth).
The lowest energy conformation of ethane, shown in the figure above, is called the ‘staggered’ conformation: all of the dihedral angles are 60o, and the distance between the front and back C-H bonds is maximized.
If we now rotate the front CH3 group 60° clockwise, the molecule is in the highest energy ‘eclipsed' conformation, where the dihedral angles are all 0o (we stagger the bonds slightly in our Newman projection drawing so that we can see them all).
The energy of the eclipsed conformation, where the electrons in the front and back C-H bonds are closer together, is approximately 12 kJ/mol higher than that of the staggered conformation.
Another 60° rotation returns the molecule to a second staggered conformation. This process can be continued all around the 360° circle, with three possible eclipsed conformations and three staggered conformations, in addition to an infinite number of conformations in between these two extremes.
Now let's consider butane, with its four-carbon chain. There are now three rotating carbon-carbon bonds to consider, but we will focus on the middle bond between C2 and C3. Below are two representations of butane in a conformation which puts the two CH3 groups (C1 and C4) in the eclipsed position, with the two C-C bonds at a 0o dihedral angle.
If we rotate the front, (blue) carbon by 60° clockwise, the butane molecule is now in a staggered conformation.
This is more specifically referred to as the gauche conformation of butane. Notice that although they are staggered, the two methyl groups are not as far apart as they could possibly be.
A further rotation of 60° gives us a second eclipsed conformation (B) in which both methyl groups are lined up with hydrogen atoms.
One more 60 rotation produces another staggered conformation called the anti conformation, where the two methyl groups are positioned opposite each other (a dihedral angle of 180o).
As with ethane, the staggered conformations of butane are energy 'valleys', and the eclipsed conformations are energy 'peaks'. However, in the case of butane there are two different valleys, and two different peaks. The gauche conformation is a higher energy valley than the anti conformation due to steric strain, which is the repulsive interaction caused by the two bulky methyl groups being forced too close together. Clearly, steric strain is lower in the anti conformation. In the same way, steric strain causes the eclipsed A conformation - where the two methyl groups are as close together as they can possibly be - to be higher in energy than the two eclipsed B conformations.
The diagram below summarizes the relative energies for the various eclipsed, staggered, and gauche conformations.
Because the anti conformation is lowest in energy (and also simply for ease of drawing), it is conventional to draw open-chain alkanes in a 'zigzag' form, which implies anti conformation at all carbon-carbon bonds. The figure below shows, as an example, a Newman projection looking down the C2-C3 bond of octane.
Exercise 3.1: Draw Newman projections of the lowest and highest energy conformations of propane.
Exercise 3.2: Draw a Newman projection, looking down the C2-C3 bond, of 1-butene in the conformation shown below (C2 should be your front carbon).
Solutions to exercises
Online lectures from Kahn Academy
Newman projections part I
Newman projections part II
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/03%3A_An_Introduction_to_Organic_Compounds%3A_Nomenclature_Physical_Properties_and_Representation_of_Structure/3.08%3A_Rotation_Occurs_About_Carbon-Carbon_Singl.txt |
Cycloalkanes are very important in components of food, pharmaceutical drugs, and much more. However, to use cycloalkanes in such applications, we must know the effects, functions, properties, and structures of cycloalkanes. Cycloalkanes are alkanes that are in the form of a ring; hence, the prefix cyclo-. Stable cycloalkanes cannot be formed with carbon chains of just any length. Recall that in alkanes, carbon adopts the sp3 tetrahedral geometry in which the angles between bonds are 109.5°. For some cycloalkanes to form, the angle between bonds must deviate from this ideal angle, an effect known as angle strain. Additionally, some hydrogen atoms may come into closer proximity with each other than is desirable (become eclipsed), an effect called torsional strain. These destabilizing effects, angle strain and torsional strain are known together as ring strain. The smaller cycloalkanes, cyclopropane and cyclobutane, have particularly high ring strains because their bond angles deviate substantially from 109.5° and their hydrogens eclipse each other. Cyclopentane is a more stable molecule with a small amount of ring strain, while cyclohexane is able to adopt the perfect geometry of a cycloalkane in which all angles are the ideal 109.5° and no hydrogens are eclipsed; it has no ring strain at all. Cycloalkanes larger than cyclohexane have ring strain and are not commonly encountered in organic chemistry.
Ring Strain and the Structures of Cycloalkanes
There are many forms of cycloalkanes, such as cyclopropane, cyclobutane, cyclopentane, cyclohexane, among others. The process of naming cycloalkanes is the same as naming alkanes but the addition of the prefix cyclo- is required. Cyclobutane is in a form of a square, which is highly unfavorable and unstable (this will be explained soon). There are different drawings for cyclobutane, but they are equivalent to each other. Cyclobutane can reduce the ring string by puckering the square cyclobutane. Cyclopentane takes the shape of a pentagon and cyclohexane is in the shape of a hexagon.
Chair Conformation of Cyclohexane - Equitorial and Axial
There are two ways to draw cyclohexane because it can be in a hexagon shape or in a different conformational form called the chair conformation and the boat conformation.
• The chair conformation drawing is more favored than the boat because of the energy, the steric hindrance, and a new strain called the transannular strain.
• The boat conformation is not the favored conformation because it is less stable and has a steric repulsion between the two H's, shown with the pink curve. This is known as the transannular strain, which means that the strain results from steric crowding of two groups across a ring. The boat is less stable than the chair by 6.9 kcal/mol. The boat conformation, however, is flexible, and when we twist one of the C-C bonds, it reduces the transannular strain.
• When we twist the C-C bond in a boat, it becomes a twisted boat.
Some Conformations of Cyclohexane Rings. (William Reusch, MSU)
Although there are multiple ways to draw cyclohexane, the most stable and major conformer is the chair because is has a lower activation barrier from the energy diagram.
Conformational Energy Profile of Cyclohexane. (William Reusch, MSU).
The transition state structure is called a half chair. This energy diagram shows that the chair conformation is lower in energy; therefore, it is more stable. The chair conformation is more stable because it does not have any steric hindrance or steric repulsion between the hydrogen bonds. By drawing cyclohexane in a chair conformation, we can see how the H's are positioned. There are two positions for the H's in the chair conformation, which are in an axial or an equitorial formation.
This is how a chair conformation looks, but you're probably wondering which H's are in the equitorial and axial form. Here are more pictures to help.
These are hydrogens in the axial form.
These hydrogens are in an equitorial form. Of these two positions of the H's, the equitorial form will be the most stable because the hydrogen atoms, or perhaps the other substituents, will not be touching each other. This is the best time to build a chair conformation in an equitorial and an axial form to demonstrate the stability of the equitorial form.
Ring Strain
Cycloalkanes tend to give off a very high and non-favorable energy, and the spatial orientation of the atoms is called the ring strain. When atoms are close together, their proximity is highly unfavorable and causes steric hindrance. The reason we do not want ring strain and steric hindrance is because heat will be released due to an increase in energy; therefore, a lot of that energy is stored in the bonds and molecules, causing the ring to be unstable and reactive. Another reason we try to avoid ring strain is because it will affect the structures and the conformational function of the smaller cycloalkanes. One way to determine the presence of ring strain is by its heat of combustion. By comparing the heat of combustion with the value measured for the straight chain molecule, we can determine the stability of the ring. There are two types of strain, which are eclipsing/torsional strain and bond angle strain. Bond angle strain causes a ring to have a poor overlap between the atoms, resulting in weak and reactive C-C bonds. An eclipsed spatial arrangement of the atoms on the cycloalkanes results in high energy.
With so many cycloalkanes, which ones have the highest ring strain and are very unlikely to stay in its current form? The figures below show cyclopropane, cyclobutane, and cyclopentane, respectively. Cyclopropane is one of the cycloalkanes that has an incredibly high and unfavorable energy, followed by cyclobutane as the next strained cycloalkane. Any ring that is small (with three to four carbons) has a significant amount of ring strain; cyclopropane and cyclobutane are in the category of small rings. A ring with five to seven carbons is considered to have minimal to zero strain, and typical examples are cyclopentane, cyclohexane, and cycloheptane. However, a ring with eight to twelve carbons is considered to have a moderate strain, and if a ring has beyond twelve carbons, it has minimal strain.
There are different types of ring strain:
• Transannular strain isdefined as the crowding of the two groups in a ring.
• Eclipsing strain, also known as torsional strain, is intramolecular strain due to the bonding interaction between two eclipsed atoms or groups.
• Bond angle strain is present when there is a poor overlap between the atoms. There must be an ideal bond angle to achieve the maximum bond strength and that will allow the overlapping of the atomic/hybrid orbitals.
Cyclohexane
Most of the time, cyclohexane adopts the fully staggered, ideal angle chair conformation. In the chair conformation, if any carbon-carbon bond were examined, it would be found to exist with its substituents in the staggered conformation and all bonds would be found to possess an angle of 109.5°.
Cyclohexane in the chair conformation. (William Reusch, MSU).
In the chair conformation, hydrogen atoms are labeled according to their location. Those hydrogens which exist above or below the plane of the molecule (shown with red bonds above) are called axial. Those hydrogens which exist in the plane of the molecule (shown with blue bonds above) are called equatorial.
Although the chair conformation is the most stable conformation that cyclohexane can adopt, there is enough thermal energy for it to also pass through less favorable conformations before returning to a different chair conformation. When it does so, the axial and equatorial substituents change places. The passage of cyclohexane from one chair conformation to another, during which the axial substituents switch places with the equatorial substituents, is called a ring flip.
Methylcyclohexane
Methylcyclohexane is cyclohexane in which one hydrogen atom is replaced with a methyl group substituent. Methylcyclohexane can adopt two basic chair conformations: one in which the methyl group is axial, and one in which it is equatorial. Methylcyclohexane strongly prefers the equatorial conformation. In the axial conformation, the methyl group comes in close proximity to the axial hydrogens, an energetically unfavorable effect known as a 1,3-diaxial interaction (Figure 3). Thus, the equatorial conformation is preferred for the methyl group. In most cases, if the cyclohexane ring contains a substituent, the substituent will prefer the equatorial conformation.
Methylcyclohexane in the chair conformation. (William Reusch, MSU).
Problems
1. Trans- 1,2-dimethylcyclopropane is more stable than cis-1,2-dimethylcyclopropane. Why? Drawing a picture of the two will help your explanation.
2. Out of all the cycloalkanes, which one has the most ring strain and which one is strain free? Explain.
3. Which of these chair conformations are the most stable and why?
3.
1. What does it mean when people say "increase in heat leads to increase in energy" and how does that statement relate to ring strains?
2. Why is that the bigger rings have lesser strains compared to smaller rings?
Answers
1. The cis isomer suffers from steric hindrance and has a larger heat of combustion.
2. Cyclopropane- ring strain. Cyclohexane chair conformation- ring strain free.
3. Top one is more stable because it is in an equitorial conformation. When assembling it with the OChem Molecular Structure Tool Kit, equitorial formation is more spread out.
4. When there is an increase in heat there will be an increase of energy released therefore there will be a lot of energy stored in the bond and molecule making it unstable.
5. Smaller rings are more compacts, which leads to steric hindrance and the angles for these smaller rings are harder to get ends to meet. Bigger rings tend to have more space and that the atoms attached to the ring won't be touching each other as much as atoms attached to the smaller ring. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/03%3A_An_Introduction_to_Organic_Compounds%3A_Nomenclature_Physical_Properties_and_Representation_of_Structure/3.09%3A_Some_Cycloalkanes_Have_Angle_Strain.txt |
Cycloalkanes are very important in components of food, pharmaceutical drugs, and much more. However, to use cycloalkanes in such applications, we must know the effects, functions, properties, and structures of cycloalkanes. Cycloalkanes are alkanes that are in the form of a ring; hence, the prefix cyclo-. Stable cycloalkanes cannot be formed with carbon chains of just any length. Recall that in alkanes, carbon adopts the sp3 tetrahedral geometry in which the angles between bonds are 109.5°. For some cycloalkanes to form, the angle between bonds must deviate from this ideal angle, an effect known as angle strain. Additionally, some hydrogen atoms may come into closer proximity with each other than is desirable (become eclipsed), an effect called torsional strain. These destabilizing effects, angle strain and torsional strain are known together as ring strain. The smaller cycloalkanes, cyclopropane and cyclobutane, have particularly high ring strains because their bond angles deviate substantially from 109.5° and their hydrogens eclipse each other. Cyclopentane is a more stable molecule with a small amount of ring strain, while cyclohexane is able to adopt the perfect geometry of a cycloalkane in which all angles are the ideal 109.5° and no hydrogens are eclipsed; it has no ring strain at all. Cycloalkanes larger than cyclohexane have ring strain and are not commonly encountered in organic chemistry.
Ring Strain and the Structures of Cycloalkanes
There are many forms of cycloalkanes, such as cyclopropane, cyclobutane, cyclopentane, cyclohexane, among others. The process of naming cycloalkanes is the same as naming alkanes but the addition of the prefix cyclo- is required. Cyclobutane is in a form of a square, which is highly unfavorable and unstable (this will be explained soon). There are different drawings for cyclobutane, but they are equivalent to each other. Cyclobutane can reduce the ring string by puckering the square cyclobutane. Cyclopentane takes the shape of a pentagon and cyclohexane is in the shape of a hexagon.
Chair Conformation of Cyclohexane - Equitorial and Axial
There are two ways to draw cyclohexane because it can be in a hexagon shape or in a different conformational form called the chair conformation and the boat conformation.
• The chair conformation drawing is more favored than the boat because of the energy, the steric hindrance, and a new strain called the transannular strain.
• The boat conformation is not the favored conformation because it is less stable and has a steric repulsion between the two H's, shown with the pink curve. This is known as the transannular strain, which means that the strain results from steric crowding of two groups across a ring. The boat is less stable than the chair by 6.9 kcal/mol. The boat conformation, however, is flexible, and when we twist one of the C-C bonds, it reduces the transannular strain.
• When we twist the C-C bond in a boat, it becomes a twisted boat.
Some Conformations of Cyclohexane Rings. (William Reusch, MSU)
Although there are multiple ways to draw cyclohexane, the most stable and major conformer is the chair because is has a lower activation barrier from the energy diagram.
Conformational Energy Profile of Cyclohexane. (William Reusch, MSU).
The transition state structure is called a half chair. This energy diagram shows that the chair conformation is lower in energy; therefore, it is more stable. The chair conformation is more stable because it does not have any steric hindrance or steric repulsion between the hydrogen bonds. By drawing cyclohexane in a chair conformation, we can see how the H's are positioned. There are two positions for the H's in the chair conformation, which are in an axial or an equitorial formation.
This is how a chair conformation looks, but you're probably wondering which H's are in the equitorial and axial form. Here are more pictures to help.
These are hydrogens in the axial form.
These hydrogens are in an equitorial form. Of these two positions of the H's, the equitorial form will be the most stable because the hydrogen atoms, or perhaps the other substituents, will not be touching each other. This is the best time to build a chair conformation in an equitorial and an axial form to demonstrate the stability of the equitorial form.
Ring Strain
Cycloalkanes tend to give off a very high and non-favorable energy, and the spatial orientation of the atoms is called the ring strain. When atoms are close together, their proximity is highly unfavorable and causes steric hindrance. The reason we do not want ring strain and steric hindrance is because heat will be released due to an increase in energy; therefore, a lot of that energy is stored in the bonds and molecules, causing the ring to be unstable and reactive. Another reason we try to avoid ring strain is because it will affect the structures and the conformational function of the smaller cycloalkanes. One way to determine the presence of ring strain is by its heat of combustion. By comparing the heat of combustion with the value measured for the straight chain molecule, we can determine the stability of the ring. There are two types of strain, which are eclipsing/torsional strain and bond angle strain. Bond angle strain causes a ring to have a poor overlap between the atoms, resulting in weak and reactive C-C bonds. An eclipsed spatial arrangement of the atoms on the cycloalkanes results in high energy.
With so many cycloalkanes, which ones have the highest ring strain and are very unlikely to stay in its current form? The figures below show cyclopropane, cyclobutane, and cyclopentane, respectively. Cyclopropane is one of the cycloalkanes that has an incredibly high and unfavorable energy, followed by cyclobutane as the next strained cycloalkane. Any ring that is small (with three to four carbons) has a significant amount of ring strain; cyclopropane and cyclobutane are in the category of small rings. A ring with five to seven carbons is considered to have minimal to zero strain, and typical examples are cyclopentane, cyclohexane, and cycloheptane. However, a ring with eight to twelve carbons is considered to have a moderate strain, and if a ring has beyond twelve carbons, it has minimal strain.
There are different types of ring strain:
• Transannular strain isdefined as the crowding of the two groups in a ring.
• Eclipsing strain, also known as torsional strain, is intramolecular strain due to the bonding interaction between two eclipsed atoms or groups.
• Bond angle strain is present when there is a poor overlap between the atoms. There must be an ideal bond angle to achieve the maximum bond strength and that will allow the overlapping of the atomic/hybrid orbitals.
Cyclohexane
Most of the time, cyclohexane adopts the fully staggered, ideal angle chair conformation. In the chair conformation, if any carbon-carbon bond were examined, it would be found to exist with its substituents in the staggered conformation and all bonds would be found to possess an angle of 109.5°.
Cyclohexane in the chair conformation. (William Reusch, MSU).
In the chair conformation, hydrogen atoms are labeled according to their location. Those hydrogens which exist above or below the plane of the molecule (shown with red bonds above) are called axial. Those hydrogens which exist in the plane of the molecule (shown with blue bonds above) are called equatorial.
Although the chair conformation is the most stable conformation that cyclohexane can adopt, there is enough thermal energy for it to also pass through less favorable conformations before returning to a different chair conformation. When it does so, the axial and equatorial substituents change places. The passage of cyclohexane from one chair conformation to another, during which the axial substituents switch places with the equatorial substituents, is called a ring flip.
Methylcyclohexane
Methylcyclohexane is cyclohexane in which one hydrogen atom is replaced with a methyl group substituent. Methylcyclohexane can adopt two basic chair conformations: one in which the methyl group is axial, and one in which it is equatorial. Methylcyclohexane strongly prefers the equatorial conformation. In the axial conformation, the methyl group comes in close proximity to the axial hydrogens, an energetically unfavorable effect known as a 1,3-diaxial interaction (Figure 3). Thus, the equatorial conformation is preferred for the methyl group. In most cases, if the cyclohexane ring contains a substituent, the substituent will prefer the equatorial conformation.
Methylcyclohexane in the chair conformation. (William Reusch, MSU).
Problems
1. Trans- 1,2-dimethylcyclopropane is more stable than cis-1,2-dimethylcyclopropane. Why? Drawing a picture of the two will help your explanation.
2. Out of all the cycloalkanes, which one has the most ring strain and which one is strain free? Explain.
3. Which of these chair conformations are the most stable and why?
3.
1. What does it mean when people say "increase in heat leads to increase in energy" and how does that statement relate to ring strains?
2. Why is that the bigger rings have lesser strains compared to smaller rings?
Answers
1. The cis isomer suffers from steric hindrance and has a larger heat of combustion.
2. Cyclopropane- ring strain. Cyclohexane chair conformation- ring strain free.
3. Top one is more stable because it is in an equitorial conformation. When assembling it with the OChem Molecular Structure Tool Kit, equitorial formation is more spread out.
4. When there is an increase in heat there will be an increase of energy released therefore there will be a lot of energy stored in the bond and molecule making it unstable.
5. Smaller rings are more compacts, which leads to steric hindrance and the angles for these smaller rings are harder to get ends to meet. Bigger rings tend to have more space and that the atoms attached to the ring won't be touching each other as much as atoms attached to the smaller ring. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/03%3A_An_Introduction_to_Organic_Compounds%3A_Nomenclature_Physical_Properties_and_Representation_of_Structure/3.10%3A_Conformers_of_Cyclohexane.txt |
Generally, there are two types of inorganic compounds that can be formed: ionic compounds and molecular compounds. Nomenclature is the process of naming chemical compounds with different names so that they can be easily identified as separate chemicals. Inorganic compounds are compounds that do not deal with the formation of carbohydrates, or simply all other compounds that do not fit into the description of an organic compound. For example, organic compounds include molecules with carbon rings and/or chains with hydrogen atoms (see picture below). Inorganic compounds, the topic of this section, are every other molecule that does not include these distinctive carbon and hydrogen structures.
Compounds between Metals and Nonmetals (Cation and Anion)
Compounds made of a metal and nonmetal are commonly known as Ionic Compounds, where the compound name has an ending of –ide. Cations have positive charges while anions have negative charges. The net charge of any ionic compound must be zero which also means it must be electrically neutral. For example, one Na+ is paired with one Cl-; one Ca2+ is paired with two Br-. There are two rules that must be followed through:
• The cation (metal) is always named first with its name unchanged
• The anion (nonmetal) is written after the cation, modified to end in –ide
Table 1: Cations and Anions:
+1 Charge +2 Charge -1 Charge -2 Charge -3 Charge -4 Charge
Group 1A elements Group 2A elements Group 7A elements Group 6A elements Group 5A elements Group 4A elements
Hydrogen: H+ Beryllium: Be2+ Hydride: H- Oxide: O2- Nitride: N3- Carbide: C4-
Lithium: Li+ Magnesium: Mg2+ Fluoride: F- Sulfide: S2- Phosphide: P3-
Soduim: Na+ Calcium: Ca2+ Chloride: Cl-
Potassium: K+ Strontium: Sr2+ Bromide: Br-
Rubidium: Rb+ Barium: Ba2+ Iodide: I-
Cesium: Cs+
Example 1
Na+ + Cl- = NaCl; Ca2+ + 2Br- = CaBr2
Sodium + Chlorine = Sodium Chloride; Calcium + Bromine = Calcium Bromide
The transition metals may form more than one ion, thus it is needed to be specified which particular ion we are talking about. This is indicated by assigning a Roman numeral after the metal. The Roman numeral denotes the charge and the oxidation state of the transition metal ion. For example, iron can form two common ions, Fe2+ and Fe3+. To distinguish the difference, Fe2+ would be named iron (II) and Fe3+ would be named iron (III).
Table of Transition Metal and Metal Cations:
+1 Charge +2 Charge +3 Charge +4 Charge
Copper(I): Cu+ Copper(II): Cu2+ Aluminum: Al3+ Lead(IV): Pb4+
Silver: Ag+ Iron(II): Fe2+ Iron(III): Fe3+ Tin(IV): Sn4+
Cobalt(II): Co2+ Cobalt(III): Co3+
Tin(II): Sn2+
Lead(II): Pb2+
Nickel: Ni2+
Zinc: Zn2+
Example 2
Ions: Fe2++ 2Cl- Fe3++ 3Cl-
Compound: FeCl2 FeCl3
Nomenclature Iron (II) Chloride Iron (III) Chloride
However, some of the transition metals' charges have specific Latin names. Just like the other nomenclature rules, the ion of the transition metal that has the lower charge has the Latin name ending with -ous and the one with the the higher charge has a Latin name ending with -ic. The most common ones are shown in the table below:
Transition Metal Ion with Roman Numeral Latin name
Copper (I): Cu+ Cuprous
Copper (II): Cu2+ Cupric
Iron (II): Fe2+ Ferrous
Iron (III): Fe3+ Ferric
Lead (II): Pb2+ Plumbous
Lead (IV): Pb4+ Plumbic
Mercury (I): Hg22+ Mercurous
Mercury (II): Hg2+ Mercuric
Tin (II): Sn2+ Stannous
Tin (IV): Sn4+ Stannic
Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion. Instead of using Roman numerals, the different ions can also be presented in plain words. The metal is changed to end in –ous or –ic.
• -ous ending is used for the lower oxidation state
• -ic ending is used for the higher oxidation state
Example 3
Compound Cu2O CuO FeCl2 FeCl3
Charge Charge of copper is +1 Charge of copper is +2 Charge of iron is +2 Charge of iron is +3
Nomenclature Cuprous Oxide Cupric Oxide Ferrous Chloride Ferric Chloride
However, this -ous/-ic system is inadequate in some cases, so the Roman numeral system is preferred. This system is used commonly in naming acids, where H2SO4 is commonly known as Sulfuric Acid, and H2SO3 is known as Sulfurous Acid.
Compounds between Nonmetals and Nonmetals
Compounds that consist of a nonmetal bonded to a nonmetal are commonly known as Molecular Compounds, where the element with the positive oxidation state is written first. In many cases, nonmetals form more than one binary compound, so prefixes are used to distinguish them.
# of Atoms 1 2 3 4 5 6 7 8 9 10
Prefixes Mono- Di- Tri- Tetra- Penta- Hexa- Hepta- Octa- Nona- Deca-
Example 4
CO = carbon monoxide BCl3 = borontrichloride
CO2 = carbon dioxide N2O5 =dinitrogen pentoxide
The prefix mono- is not used for the first element. If there is not a prefix before the first element, it is assumed that there is only one atom of that element.
Binary Acids
Although HF can be named hydrogen fluoride, it is given a different name for emphasis that it is an acid. An acid is a substance that dissociates into hydrogen ions (H+) and anions in water. A quick way to identify acids is to see if there is an H (denoting hydrogen) in front of the molecular formula of the compound. To name acids, the prefix hydro- is placed in front of the nonmetal modified to end with –ic. The state of acids is aqueous (aq) because acids are found in water.
Some common binary acids include:
HF (g) = hydrogen fluoride -> HF (aq) = hydrofluoric acid
HBr (g) = hydrogen bromide -> HBr (aq) = hydrobromic acid
HCl (g) = hydrogen chloride -> HCl (aq) = hydrochloric acid
H2S (g) = hydrogen sulfide -> H2S (aq) = hydrosulfuricacid
It is important to include (aq) after the acids because the same compounds can be written in gas phase with hydrogen named first followed by the anion ending with –ide.
Example 5
hypo____ite ____ite ____ate per____ate
ClO- ClO2- ClO3- ClO4-
hypochlorite chlorite chlorate perchlorate
---------------->
As indicated by the arrow, moving to the right, the following trends occur:
Increasing number of oxygen atoms
Increasing oxidation state of the nonmetal
(Usage of this example can be seen from the set of compounds containing Cl and O)
This occurs because the number of oxygen atoms are increasing from hypochlorite to perchlorate, yet the overall charge of the polyatomic ion is still -1. To correctly specify how many oxygen atoms are in the ion, prefixes and suffixes are again used.
Polyatomic Ions
In polyatomic ions, polyatomic (meaning two or more atoms) are joined together by covalent bonds. Although there may be a element with positive charge like H+, it is not joined with another element with an ionic bond. This occurs because if the atoms formed an ionic bond, then it would have already become a compound, thus not needing to gain or loose any electrons. Polyatomic anions are more common than polyatomic cations as shown in the chart below. Polyatomic anions have negative charges while polyatomic cations have positive charges. To indicate different polyatomic ions made up of the same elements, the name of the ion is modified according to the example below:
Table: Common Polyatomic ions
Name: Cation/Anion Formula
Ammonium ion NH4+
Hydronium ion
H3O+
Acetate ion
C2H3O2-
Arsenate ion
AsO43-
Carbonate ion
CO32-
Hypochlorite ion
ClO-
Chlorite ion
ClO2-
Chlorate ion
ClO3-
Perchlorate ion
ClO4-
Chromate ion
CrO42-
Dichromate ion
Cr2O72-
Cyanide ion
CN-
Hydroxide ion
OH-
Nitrite ion
NO2-
Nitrate ion
NO3-
Oxalate ion
C2O42-
Permanganate ion
MnO4-
Phosphate ion
PO43-
Sulfite ion
SO32-
Sulfate ion
SO42-
Thiocyanate ion
SCN-
Thiosulfate ion
S2O32-
To combine the topic of acids and polyatomic ions, there is nomenclature of aqueous acids. Such acids include sulfuric acid (H2SO4) or carbonic acid (H2CO3). To name them, follow these quick, simple rules:
1. If the ion ends in -ate and is added with an acid, the acid name will have an -ic ending. Examples: nitrate ion (NO3-) + H+ (denoting formation of acid) = nitric acid (HNO3)
2. If the ion ends in -ite and is added with an acid, then the acid name will have an -ous ending. Example: nitite ion (NO2-) + H+ (denoting formation of acid) = nitrous acid (HNO2)
Problems
1. What is the correct formula for Calcium Carbonate?
a. Ca+ + CO2-
b. CaCO2-
c. CaCO3
d. 2CaCO3
2. What is the correct name for FeO?
a. Iron oxide
b. Iron dioxide
c. Iron(III) oxide
d. Iron(II) oxide
3. What is the correct name for Al(NO3)3?
a. Aluminum nitrate
b. Aluminum(III) nitrate
c. Aluminum nitrite
d. Aluminum nitrogen trioxide
4. What is the correct formula of phosphorus trichloride?
a. P2Cl2
b. PCl3
c. PCl4
d. P4Cl2
5. What is the correct formula of lithium perchlorate?
a. Li2ClO4
b. LiClO2
c. LiClO
d. None of these
6. Write the correct name for these compounds.
a. BeC2O4:
b. NH4MnO4:
c. CoS2O3:
7. What is W(HSO4)5?
8. How do you write diphosphorus trioxide?
9. What is H3P?
10. By adding oxygens to the molecule in number 9, we now have H3PO4? What is the name of this molecule?
Answer
1.C; Calcium + Carbonate --> Ca2+ + CO32- --> CaCO3
2.D; FeO --> Fe + O2- --> Iron must have a charge of +2 to make a neutral compound --> Fe2+ + O2- --> Iron(II) Oxide
3.A; Al(NO3)3 --> Al3+ + (NO3-)3 --> Aluminum nitrate
4.B; Phosphorus trichloride --> P + 3Cl --> PCl3
5.D, LiClO4; Lithium perchlorate --> Li+ + ClO4- --> LiClO4
6. a. Beryllium Oxalate; BeC2O4 --> Be2+ + C2O42- --> Beryllium Oxalate
b. Ammonium Permanganate; NH4MnO4 --> NH4+ + MnO4- --> Ammonium Permanganate
c. Cobalt (II) Thiosulfate; CoS2O3 --> Co + S2O32- --> Cobalt must have +2 charge to make a neutral compund --> Co2+ + S2O32- --> Cobalt(II) Thiosulfate
7. Tungsten (V) hydrogen sulfate
8. P2O3
9. Hydrophosphoric Acid
10. Phosphoric Acid
Contributors and Attributions
• Pui Yan Ho (UCD), Alex Moskaluk (UCD), Emily Nguyen (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/03%3A_An_Introduction_to_Organic_Compounds%3A_Nomenclature_Physical_Properties_and_Representation_of_Structure/3.13%3A_Fused_Cyclohexane_Rings.txt |
Homework Problems
Homework Solutions
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI
04: Alkenes: Structure Nomenclature and an Introduction to Reactivity
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of $\pi$ bonds and/or cyclic rings.
Saturated vs. Unsaturated Molecules
In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated.
CH3CH2CH3 1-methyoxypentane
Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s).
CH3CH=CHCH3 3-chloro-5-octyne
Calculating Degrees of Unsaturation (DoU)
Degree of Unsaturation (DoU) is also known as Double Bond Equivalent. If the molecular formula is given, plug in the numbers into this formula:
$DoU= \dfrac{2C+2+N-X-H}{2}$
• $C$ is the number of carbons
• $N$ is the number of nitrogens
• $X$ is the number of halogens (F, Cl, Br, I)
• $H$ is the number of hydrogens
As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6.
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
DoU
Possible combinations of rings/ bonds
# of rings
# of double bonds
# of triple bonds
1
1
0
0
0
1
0
2
2
0
0
0
2
0
0
0
1
1
1
0
3 3 0 0
2 1 0
1 2 0
0 1 1
0 3 0
1 0 1
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example: Benzene
What is the Degree of Unsaturation for Benzene?
Solution
The molecular formula for benzene is C6H6. Thus,
DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DoB of 4 and have the same molecular formula as benzene.
Problems
1. Are the following molecules saturated or unsaturated:
1. (b.) (c.) (d.) C10H6N4
2. Using the molecules from 1., give the degrees of unsaturation for each.
3. Calculate the degrees of unsaturation for the following molecular formulas:
1. (a.) C9H20 (b.) C7H8 (c.) C5H7Cl (d.) C9H9NO4
4. Using the molecular formulas from 3, are the molecules unsaturated or saturated.
5. Using the molecular formulas from 3, if the molecules are unsaturated, how many rings/double bonds/triple bonds are predicted?
Answers
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3. Use the formula to solve
(a.) 0
(b.) 4
(c.) 2
(d.) 6
4.
(a.) saturated
(b.) unsaturated
(c.) unsaturated
(d.) unsaturated
5.
(a.) 0 (Remember-a saturated molecule only contains single bonds)
(b.) The molecule can contain any of these combinations (i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
Contributors
• Kim Quach (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/04%3A_Alkenes%3A_Structure_Nomenclature_and_an_Introduction_to_Reactivity/4.1%3A_Molecular_Formulas_and_the_Degree_of_Unsaturation.txt |
Homework Problems
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05: The Reactions of Alkenes and Alkynes: An Introduction to Multistep Synthesis
In the vast majority of the nucleophilic substitution reactions you will see in this and other organic chemistry texts, the electrophilic atom is a carbon which is bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom is an attractive target for something that is electron-rich, i.e. a nucleophile. However, we must also consider the effect of steric hindrance on electrophilicity. In addition, we must discuss how the nature of the electrophilic carbon, and more specifically the stability of a potential carbocationic intermediate, influences the SN1 vs. SN2 character of a nucleophilic substitution reaction.
8.4A: Steric effects on electrophilicity
Consider two hypothetical SN2 reactions: one in which the electrophile is a methyl carbon and another in which it is tertiary carbon.
Because the three substituents on the methyl carbon electrophile are tiny hydrogens, the nucleophile has a relatively clear path for backside attack. However, backside attack on the tertiary carbon is blocked by the bulkier methyl groups. Once again, steric hindrance - this time caused by bulky groups attached to the electrophile rather than to the nucleophile - hinders the progress of an associative nucleophilic (SN2) displacement.
The factors discussed in the above paragraph, however, do not prevent a sterically-hindered carbon from being a good electrophile - they only make it less likely to be attacked in a concerted SN2 reaction. Nucleophilic substitution reactions in which the electrophilic carbon is sterically hindered are more likely to occur by a two-step, dissociative (SN1) mechanism. This makes perfect sense from a geometric point of view: the limitations imposed by sterics are significant mainly in an SN2 displacement, when the electrophile being attacked is a sp3-hybridized tetrahedral carbon with its relatively ‘tight’ angles of 109.4o. Remember that in an SN1 mechanism, the nucleophile attacks an sp2-hybridized carbocation intermediate, which has trigonal planar geometry with ‘open’ 120 angles.
With this open geometry, the empty p orbital of the electrophilic carbocation is no longer significantly shielded from the approaching nucleophile by the bulky alkyl groups. A carbocation is a very potent electrophile, and the nucleophilic step occurs very rapidly compared to the first (ionization) step.
8.4B: Stability of carbocation intermediates
We know that the rate-limiting step of an SN1 reaction is the first step - formation of the this carbocation intermediate. The rate of this step – and therefore, the rate of the overall substitution reaction – depends on the activation energy for the process in which the bond between the carbon and the leaving group breaks and a carbocation forms. According to Hammond’s postulate (section 6.2B), the more stable the carbocation intermediate is, the faster this first bond-breaking step will occur. In other words, the likelihood of a nucleophilic substitution reaction proceeding by a dissociative (SN1) mechanism depends to a large degree on the stability of the carbocation intermediate that forms.
The critical question now becomes, what stabilizes a carbocation?
Think back to Chapter 7, when we were learning how to evaluate the strength of an acid. The critical question was “how stable is the conjugate base that results when this acid donates its proton"? In many cases, this conjugate base was an anion – a center of excess electron density. Anything that can draw some of this electron density away– in other words, any electron withdrawing group – will stabilize the anion.
So if it takes an electron withdrawing group to stabilize a negative charge, what will stabilize a positive charge? An electron donating group!
A positively charged species such as a carbocation is very electron-poor, and thus anything which donates electron density to the center of electron poverty will help to stabilize it. Conversely, a carbocation will be destabilized by an electron withdrawing group.
Alkyl groups – methyl, ethyl, and the like – are weak electron donating groups, and thus stabilize nearby carbocations. What this means is that, in general, more substituted carbocations are more stable: a tert-butyl carbocation, for example, is more stable than an isopropyl carbocation. Primary carbocations are highly unstable and not often observed as reaction intermediates; methyl carbocations are even less stable.
Alkyl groups are electron donating and carbocation-stabilizing because the electrons around the neighboring carbons are drawn towards the nearby positive charge, thus slightly reducing the electron poverty of the positively-charged carbon.
It is not accurate to say, however, that carbocations with higher substitution are always more stable than those with less substitution. Just as electron-donating groups can stabilize a carbocation, electron-withdrawing groups act to destabilize carbocations. Carbonyl groups are electron-withdrawing by inductive effects, due to the polarity of the C=O double bond. It is possible to demonstrate in the laboratory (see section 16.1D) that carbocation A below is more stable than carbocation B, even though A is a primary carbocation and B is secondary.
The difference in stability can be explained by considering the electron-withdrawing inductive effect of the ester carbonyl. Recall that inductive effects - whether electron-withdrawing or donating - are relayed through covalent bonds and that the strength of the effect decreases rapidly as the number of intermediary bonds increases. In other words, the effect decreases with distance. In species B the positive charge is closer to the carbonyl group, thus the destabilizing electron-withdrawing effect is stronger than it is in species A.
In the next chapter we will see how the carbocation-destabilizing effect of electron-withdrawing fluorine substituents can be used in experiments designed to address the question of whether a biochemical nucleophilic substitution reaction is SN1 or SN2.
Stabilization of a carbocation can also occur through resonance effects, and as we have already discussed in the acid-base chapter, resonance effects as a rule are more powerful than inductive effects. Consider the simple case of a benzylic carbocation:
This carbocation is comparatively stable. In this case, electron donation is a resonance effect. Three additional resonance structures can be drawn for this carbocation in which the positive charge is located on one of three aromatic carbons. The positive charge is not isolated on the benzylic carbon, rather it is delocalized around the aromatic structure: this delocalization of charge results in significant stabilization. As a result, benzylic and allylic carbocations (where the positively charged carbon is conjugated to one or more non-aromatic double bonds) are significantly more stable than even tertiary alkyl carbocations.
Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in section 7.4). Consider the two pairs of carbocation species below:
In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually destabilize the positive charge, because they are electron withdrawing by induction.
Finally, vinylic carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any reaction.
Example 8.10
Exercise 8.10: In which of the structures below is the carbocation expected to be more stable? Explain.
For the most part, carbocations are very high-energy, transient intermediate species in organic reactions. However, there are some unusual examples of very stable carbocations that take the form of organic salts. Crystal violet is the common name for the chloride salt of the carbocation whose structure is shown below. Notice the structural possibilities for extensive resonance delocalization of the positive charge, and the presence of three electron-donating amine groups.
Example 8.11
Draw a resonance structure of the crystal violet cation in which the positive charge is delocalized to one of the nitrogen atoms.
Solution
When considering the possibility that a nucleophilic substitution reaction proceeds via an SN1 pathway, it is critical to evaluate the stability of the hypothetical carbocation intermediate. If this intermediate is not sufficiently stable, an SN1 mechanism must be considered unlikely, and the reaction probably proceeds by an SN2 mechanism. In the next chapter we will see several examples of biologically important SN1 reactions in which the positively charged intermediate is stabilized by inductive and resonance effects inherent in its own molecular structure.
Example 8.12
State which carbocation in each pair below is more stable, or if they are expected to be approximately equal. Explain your reasoning.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
5.15: An Introduction to Multistep Synthesis
under construction | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/05%3A_The_Reactions_of_Alkenes_and_Alkynes%3A_An_Introduction_to_Multistep_Synthesis/5.02%3A_Carbocation_Stability_Depends_on_the_Number_of_Alkyl_Groups_Attache.txt |
Homework Problems
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06: Isomers and Stereochemistry
Several organic compounds may have identical compositions but will have widely different physical and chemical properties because the arrangement of the atoms is different. Isomers and identical compounds both have the same number of each kind of element in a formula. A simple count will establish this fact.
Introduction
As a consequence of the double bond, some alkene compounds exhibit a unique type of isomerism. Rotation around a single bond occurs readily, while rotation around a double bond is restricted. The pi bond prevents rotation because of the electron overlap both above and below the plane of the atoms.
A single bond is analogous to two boards nailed together with one nail. A double bond is analogous to two boards nailed together with two nails. In the first case you can twist the boards, while in the second case you cannot twist them.
Geometric Isomers are compounds with different spatial arrangements of groups attached to the carbons of a double bond. In alkenes, the carbon-carbon double bond is rigidly fixed. Even though the attachment of atoms is the same, the geometry (the way the atoms "see" each other) is different.
When looking for geometric isomers, a guiding principle is that there MUST BE TWO DIFFERENT "GROUPS" ON EACH CARBON OF THE DOUBLE BOND. A "group" can be hydrogen, alkyls, halogens, etc.
Identical compounds may appear to have different arrangements as written, but closer examination by rotation or turning will result in the molecules being superimposed. If they are super impossible or if they have identical names, then the two compounds are in fact identical.
Isomers of compounds have a different arrangement of the atoms. Isomer compounds will differ from identical compounds by the arrangement of the atoms. See example below.
Both compounds have the same number of atoms, C5H12. They are isomers because in the left molecule the root is 4 carbons with one branch. In the right molecule, the root is 3 carbons with 2 branches. They are isomers because they have the same number of atoms but different arrangements of those atoms.
Completely different compounds: If the number of each element is different, the two compounds are merely completely different. A simple count of the atoms will reveal them as different.
1,2-dichlorethene
In the example on the left, the chlorine atoms can be opposite or across from each other in which case it is called the "trans" isomer. If the the chlorine atoms are next to or adjacent each other, the isomer is called " cis".
If one carbon of the double bond has two identical groups such as 2 H's or 2 Cl's or 2 CH3 etc. there cannot be any geometric isomers.
2-butene
Consider the longest chain containing the double bond: If two groups (attached to the carbons of the double bond) are on the same side of the double bond, the isomer is a cis alkene. If the two groups lie on opposite sides of the double bond, the isomer is a trans alkene. One or more of the "groups" may or may not be part of the longest chain. In the case on the left, the "group" is a methyl - but is actually part of the longest chain.
A common mistake is to name this compound as 1,2-dimethylethene. Look at all carbons for the longest continuous chain - the root is 4 carbons - butene.
Problems
For the structures below:
a. Draw both cis/trans isomers, if any, of the structure based upon the name.
b. Look at the graphic and state whether the compound is cis, trans, or not cis/trans isomers.
• 2-methyl-2-butene
• 3-methyl-2-pentene
• 1-pentene
Contributors
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.01%3A_Cis-Trans_Isomers_Result_from_Restricted_Rotation.txt |
To better understand the molecular origins of the ideal gas law,
$PV=nRT$
the basics of the Kinetic Molecular Theory of Gases (KMT) should be understood. This model is used to describe the behavior of gases. More specifically, it is used to explain macroscopic properties of a gas, such as pressure and temperature, in terms of its microscopic components, such as atoms. Like the ideal gas law, this theory was developed in reference to ideal gases, although it can be applied reasonably well to real gases.
In order to apply the kinetic model of gases, five assumptions are made:
1. Gases are made up of particles with no defined volume but with a defined mass. In other words their volume is miniscule compared to the distance between themselves and other molecules.
2. Gas particles undergo no intermolecular attractions or repulsions. This assumption implies that the particles possess no potential energy and thus their total energy is simply equal to their kinetic energies.
3. Gas particles are in continuous, random motion.
4. Collisions between gas particles are completely elastic. In other words, there is no net loss or gain of kinetic energy when particles collide.
5. The average kinetic energy is the same for all gases at a given temperature, regardless of the identity of the gas. Furthermore, this kinetic energy is proportional to the absolute temperature of the gas.
Temperature and KMT
The last assumption can be written in equation form as:
$KE = \dfrac{1}{2}mv^2 = \dfrac{3}{2}k_BT$
where
• $k_B$ is Boltzmann's constant (kB = 1.381×10-23 m2 kg s-2 K-1) and
• $T$ is the absolution temperature (in Kelvin)
This equation says that the speed of gas particles is related to their absolute temperature. In other words, as their temperature increases, their speed increases, and finally their total energy increases as well. However, it is impossible to define the speed of any one gas particle. As such, the speeds of gases are defined in terms of their root-mean-square speed.
Pressure and KMT
The macroscopic phenomena of pressure can be explained in terms of the kinetic molecular theory of gases. Assume the case in which a gas molecule (represented by a sphere) is in a box, length L (Figure 1). Through using the assumptions laid out above, and considering the sphere is only moving in the x-direction, we can examine the instance of the sphere colliding elastically with one of the walls of the box.
The momentum of this collision is given by p=mv, in this case p=mvx, since we are only considering the x dimension. The total momentum change for this collision is then given by
$mv_x - m(-v_x) = 2mv_x$
Given that the amount of time it takes between collisions of the molecule with the wall is L/vx we can give the frequency of collisions of the molecule against a given wall of the box per unit time as vx/2L. One can now solve for the change in momentum per unit of time:
$(2mv_x)(v_x/2L) = mv_x^2/L$
Solving for momentum per unit of time gives the force exerted by an object (F=ma=p/time). With the expression that F=mvx2/L one can now solve for the pressure exerted by the molecular collision, where area is given as the area of one wall of the box, A=L2:
$P=\dfrac{F}{A}$
$P=\dfrac{mv_x^2}{[L(L^2)}$
The expression can now be written in terms of the pressure associated with collisions from N number of molecules:
$P=\dfrac{Nmv_x^2}{V}$
This expression can now be adjusted to account for movement in the x, y and z directions by using mean-square velocity for three dimensions and a large value of N. The expression now is written as:
$P={\dfrac{Nm\overline{v}^2}{3V}}$
This expression now gives pressure, a macroscopic quality, in terms of atomic motion. The significance of the above relationship is that pressure is proportional to the mean-square velocity of molecules in a given container. Therefore, as molecular velocity increases so does the pressure exerted on the container.
Contributors and Attributions
• Sevini Shahbaz, Andrew Cooley | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.05%3A_Asymmetric_Centers_and_Stereocenters.txt |
When examining the ideal gas laws in conjunction with the kinetic theory of gases, we gain insights into the behavior of ideal gas. We can then predict how gas particles behaviors such as gas molecular speed, effusion rates, distances traveled by gas molecules. Graham's Law, which was formulated by the Scottish physical chemist Thomas Graham, is an important law that connects gas properties to the kinetic theory of gases.
Introduction
The Kinetic Molecular Theory states that the average energy of molecules is proportional to absolute temperature as illustrated by the following equation:
$e_K=\dfrac{3}{2}\dfrac{R}{N_A}T$
where
• ek is the average translation kinetic energy,
• R is the gas constant,
• NA is Avogadro's number, and
• T is temperature in Kelvins.
Since R and NA are constants, this means that the Kelvin temperature (T) of a gas is directly proportional to the average kinetic energy of its molecules. This means that at a given temperature, different gases (for example He or O2) will the same average kinetic energy.
Graham's Law
Gas molecules move constantly and randomly throughout the volume of the container they occupy. When examining the gas molecules individually, we see that not all of the molecules of a particular gas at a given temperature move at exactly the same speed. This means that each molecule of a gas have slightly different kinetic energy. To calculate the average kinetic energy (eK) of a sample of a gas, we use an average speed of the gas, called the root mean square speed (urms).
$e_K=\dfrac{1}{2}m{u_{rms}^2}$
with
• eK is the kinetic energy measures in Joules
• m is mass of a molecule of gas (kg)
• urms is the root mean square speed (m/s)
The root mean square speed, urms, can be determined from the temperature and molar mass of a gas.
$u_{rms}=\sqrt{\dfrac{3RT}{M}}$
with
• R the ideal gas constant (8.314 kg*m2/s2*mol*K)
• T temperature (Kelvin)
• M molar mass (kg/mol)
When examining the root mean square speed equation, we can see that the changes in temperature (T) and molar mass (M) affect the speed of the gas molecules. The speed of the molecules in a gas is proportional to the temperature and is inversely proportional to molar mass of the gas. In other words, as the temperature of a sample of gas is increased, the molecules speed up and the root mean square molecular speed increases as a result.
Graham's Law states that the rate of effusion of two different gases at the same conditions are inversely proportional to the square roots of their molar masses as given by the following equation:
$\dfrac {\it{Rate\;of\;effusion\; of \;A} }{ \it{Rate \;of \;effusion\; of \;B}}=\dfrac{(u_rms)_A}{(u_rms)_B}=\dfrac{\sqrt {3RT/M_A}}{\sqrt{3RT/M_B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}$
In according with the Kinetic Molecular Theory, each gas molecule moves independently. However, the net rate at which gas molecules move depend on their average speed. By examining the equation above, we can conclude that the heavier the molar mass of the gas molecules slower the gas molecules move. And conversely, lighter the molar mass of the gas molecules the faster the gas molecules move.
Limitations of Graham's Law
Graham's Law can only be applied to gases at low pressures so that gas molecules escape through the tiny pinhole slowly. In addition, the pinhole must be tiny so that no collisions occur as the gas molecules pass through. Since Graham's Law is an extension of the Ideal Gas Law, gases that follows Graham's Law also follows the Ideal Gas Law.
Molecular Effusion
The random and rapid motion of tiny gas molecules results in effusion. Effusion is the escape of gas molecules through a tiny hole or pinhole.
The behavior of helium gas in balloons is an example of effusion. The balloons are made of latex which is porous material that the small helium atom can effuse through. The helium inside a newly inflated balloon will eventually effuse out. This is the reason why balloons will deflate after a period of time. Molecular speeds are also used to explain why small molecules (such as He) diffuse more rapidly than larger molecules (O2). That is the reason why a balloon filled with helium gas will deflate faster than a balloon filled with oxygen gas.
The effusion rate, r, is inversely proportional to the square root of its molar mass, M.
$r\propto\sqrt{\dfrac{1}{M}}$
When there are two different gases the equation for effusion becomes
$\dfrac {\it{Rate\;of\;effusion\; of \;A} }{ \it{Rate \;of \;effusion\; of \;B}}=\dfrac{\sqrt {3RT/M_A}}{\sqrt{3RT/M_B}}$
$M_A$ is the molar mass of gas A, $M_B$ is the molar mass of gas B, $T$ Temperature in Kelvin, $R$ is the ideal gas constant.
From the equation above,
Rates of effusion of two gases
The relative rates of effusion of two gases at the same temperature is given as:
$\dfrac{\it{Rate\;of\;effusion\;of\;gas\;}\mathrm 1}{\it{Rate\;of \;effusion\;of\;gas\;}\mathrm 2}=\dfrac{\sqrt{M_2}}{\sqrt{M_1}}$
Units used to express rate of effusion includes: moles/seconds, moles/minutes, grams/seconds, grams/minutes.
Relative distances of two gases
The relative distances traveled by the two gases is given as:
$\dfrac{Distance\;traveled\;by\;gas\;1}{Distance\;traveled\;by\;gas\;2}=\dfrac{\sqrt{M_2}}{\sqrt{M_1}}$
By examine the Graham's law as stated above, we can conclude that a lighter gas will effuse or travel more rapidly than a heavier gas. Mathematically speaking, a gas with smaller molar mass will effuse faster than a gas with larger molar mass under the same condition.
Molecular Diffusion
Similar to effusion, the process of diffusion is the spread of gas molecules through space or through a second substance such as the atmosphere.
Diffusion has many useful applications. Here is an example of diffusion that is use in everyday households. Natural gas is odorless and used commercially daily. An undetected leakage can be very dangerous as it is highly flammable and can cause an explosion when it comes in contact with an ignition source. In addition, the long term breathing of natural gas can lead to asphyxiation. Fortunately, chemists have discovered a way to easily detect natural gas occur leak by adding a small quantity of a gaseous organic sulfur compound named methyl mercaptan, CH3SH, to the natural gas. When a leak happens, the diffusion of the odorous methyl mercaptan in the natural gas will serve as a sign of warning.
General guideline in solving effusion problems
The following flowchart outlines the steps in solving quantitative problems involving Graham's Law. It can be used as a general guideline.
Example $1$
Calculate the root mean square speed, $u_{rms}$, in m/s of helium at $30 ^o C$.
Solution
Start by converting the molar mass for helium from g/mol to kg/mol.
$M=(4.00\;g/mol)\times\dfrac{1\;kg}{1000\;g}$
$M=4.00\times10^{-3}\;kg/mol$
Now, using the equation for $u_{rms}$ substitute in the proper values for each variable and perform the calculation.
$u_{rms}=\sqrt{\dfrac{3RT}{M}}$
$u_{rms}=\sqrt{\dfrac{3 \times (8.314\; kg\;m^2/s^2*mol*K)(303K)}{4.00 \times 10^{-3}kg/mol}}$
$u_{rms}=1.37 \times 10^3 \;m/s$
Example $2$
What is the ratio of $u_{rms}$ values for helium vs. xenon at $30^oC$. Which is higher and why?
Solution
There are two approaches to solve this problem: the hard way and the easy way
Hard way:
The $u_{rms}$ speed of helium is calculated from the above example.
First convert the molar mass of xenon from g/mol to kg/mol as we did for helium in example 1
$M_{Xe}=(131.3\;g/mol)\times\dfrac{1\;kg}{1000\;g}$
$M=0.1313\;kg/mol$
Now, using the equation for the urms, insert the given and known values and solve for the variable of interest.
$u_{rms}=\sqrt{\dfrac{3(8.314\; kg\;m^2/s^2*mol*K)(303\;K)}{0.1313\;kg/mol}}$
$u_{rms}=2.40 \times 10^2\;m/s$
Compare the two values for xenon and helium and decide which is greater.
• $u_{Xe}=2.4 \times 10^2 \;m/s$
• $u_{He}=1.37 \times10^3 \;m/s$
So the ratio of RMS speeds is
$\dfrac{u_{Xe}}{u_{He}} \approx 0.18$
Helium has the higher $u_{rms}$ speed. This is in according with Graham's Law, because helium atoms are much lighter than xenon atoms.
Easy Way:
Since the temperature is the same for both gases, only the square root of the ratio of molar mass is needed to be calculated.
$\sqrt{\dfrac{M_{He}}{M_{Xe}}} = \sqrt{\dfrac{4.00 \; g/mol}{131.3\;g/mol }} \approx 0.18$
In either approach, helium has a faster RMS speed than xenon and this is due exclusively to its smaller mass.
Example $3$
If oxygen effuses from a container in 5.00 minutes, what is the molecular weight of a gas with the same given quantity of molecules effusing from the same container in 4.00 min?
Solution
Let oxygen be gas A and its rate is 1/5.00 minutes because it takes that much time for a certain quantity of oxygen to effuse and its molecular weight is 32 grams/mole (O2 (g)). Let the unknown gas be B and its rate is 1/4.00 minutes and Mb be the molecular weight of the unknown gas.
First, choose the appropriate equation,
$\dfrac{\it{Rate\;effusion\;of\;gas\;A}}{\it{Rate\;effusion\;of\;gas\;B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}$
Second, using algebra, solve for the variable of interest on one side of the equation and plug in known and given values from the problem.
$\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A} )(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}$
$\sqrt{M_B}=\dfrac{(1/5.00\;minutes)(\sqrt{32\;grams/mole})}{1/4.00\;minutes}$
$\sqrt{M_B}=4.525$
${M_B}=(4.525)^2=20.5\;grams/mole$
Example $4$
Oxygen, O2 (g), effuses from a container at the rate of 3.64 mL/sec, what is the molecular weight of a gas effusing from the same container under identical conditions at 4.48 mL/sec?
Solution
Let label oxygen as gas A. Therefore, rate of effusion of gas A is equaled to 3.64 mL/sec and the molecular weight is 32 grams/mole. The unknown gas is B and its rate of effusion is 4.48 mL/sec. We are solving for the molecular weight of gas B which is labeled as MB.
First, choose the appropriate equation,
$\dfrac{\it{Rate\;effusion\;of\;gas\;A}}{\it{Rate\;effusion\;of\;gas\;B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}$
$\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A} )(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}$
Second, using algebra, solve for the variable of interest on one side of the equation and plug in known and given values from the problem.
$\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A})(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}$
$\sqrt{M_B}=\dfrac{(3.64\;mL/sec)(\sqrt{32\;grams/mole})}{4.48\;mL/sec}$
$\sqrt{M_B}=4.50$
${M_B}=(4.50)^2=20.2\;grams/mole$
Example $5$
Which answer(s) are true when comparing 1.0 mol O2 (g) at STP (Standard Temperature and Pressure) and 0.50 mole of S2 (g) at STP?
The two gases have equal through the same orifice, a tiny opening:
1. average molecular kinetic energies
2. root-mean-square speeds
3. masses
4. volumes
5. densities
6. effusion rates
Solution
The average kinetic energy of gas molecules depends on only the Temperature as exemplifies in this equation:
$e_K=\dfrac{3}{2}\dfrac{R}{N_A}T$
In addition, the mass of 0.50 mole of S2 is the same as that of 1.0 mole of O2. The rest of the choices are false.
Therefore, the correct answers are (a) and (c).
• Tram Anh Dao | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.06%3A_How_to_Draw_Enantiomers.txt |
To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The method for this is formally known as R/S nomenclature.
Introduction
The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and is also often called the Cahn-Ingold-Prelog rules. In addition to the Cahn-Ingold system, there are two ways of experimentally determining the absolute configuration of an enantiomer:
1. X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer.
2. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction.
However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of optical rotation, although different for the two enantiomers of a chiral molecule, at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes.
Stereocenters are labeled R or S
The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S.
Consider the first picture: a curved arrow is drawn from the highest priority (1) substituent to the lowest priority (4) substituent. If the arrow points in a counterclockwise direction (left when leaving the 12 o' clock position), the configuration at stereocenter is considered S ("Sinister" → Latin= "left"). If, however, the arrow points clockwise,(Right when leaving the 12 o' clock position) then the stereocenter is labeled R ("Rectus" → Latin= "right"). The R or S is then added as a prefix, in parenthesis, to the name of the enantiomer of interest. For example: (R)-2-Bromobutane and (S)-2,3- Dihydroxypropanal.
Sequence rules to assign priorities to substituents
Before applying the R and S nomenclature to a stereocenter, the substituents must be prioritized according to the following rules:
Rule 1
First, examine at the atoms directly attached to the stereocenter of the compound. A substituent with a higher atomic number takes precedence over a substituent with a lower atomic number. Hydrogen is the lowest possible priority substituent, because it has the lowest atomic number.
1. When dealing with isotopes, the atom with the higher atomic mass receives higher priority.
2. When visualizing the molecule, the lowest priority substituent should always point away from the viewer (a dashed line indicates this). To understand how this works or looks, imagine that a clock and a pole. Attach the pole to the back of the clock, so that when when looking at the face of the clock the pole points away from the viewer in the same way the lowest priority substituent should point away.
3. Then, draw an arrow from the highest priority atom to the 2nd highest priority atom to the 3rd highest priority atom. Because the 4th highest priority atom is placed in the back, the arrow should appear like it is going across the face of a clock. If it is going clockwise, then it is an R-enantiomer; If it is going counterclockwise, it is an S-enantiomer.
When looking at a problem with wedges and dashes, if the lowest priority atom is not on the dashed line pointing away, the molecule must be rotated.
Remember that
• Wedges indicate coming towards the viewer.
• Dashes indicate pointing away from the viewer.
Rule 2
If there are two substituents with equal rank, proceed along the two substituent chains until there is a point of difference. First, determine which of the chains has the first connection to an atom with the highest priority (the highest atomic number). That chain has the higher priority.
If the chains are similar, proceed down the chain, until a point of difference.
For example: an ethyl substituent takes priority over a methyl substituent. At the connectivity of the stereocenter, both have a carbon atom, which are equal in rank. Going down the chains, a methyl has only has hydrogen atoms attached to it, whereas the ethyl has another carbon atom. The carbon atom on the ethyl is the first point of difference and has a higher atomic number than hydrogen; therefore the ethyl takes priority over the methyl.
Rule 3
If a chain is connected to the same kind of atom twice or three times, check to see if the atom it is connected to has a greater atomic number than any of the atoms that the competing chain is connected to.
• If none of the atoms connected to the competing chain(s) at the same point has a greater atomic number: the chain bonded to the same atom multiple times has the greater priority
• If however, one of the atoms connected to the competing chain has a higher atomic number: that chain has the higher priority.
Example 2
A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below:
However:
Remember that being double or triple bonded to an atom means that the atom is connected to the same atom twice. In such a case, follow the same method as above.
Caution!!
Keep in mind that priority is determined by the first point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant.
When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.
After all your substituents have been prioritized in the correct manner, you can now name/label the molecule R or S.
1. Put the lowest priority substituent in the back (dashed line).
2. Proceed from 1 to 2 to 3. (it is helpful to draw or imagine an arcing arrow that goes from 1--> 2-->3)
3. Determine if the direction from 1 to 2 to 3 clockwise or counterclockwise.
i) If it is clockwise it is R.
ii) if it is counterclockwise it is S.
USE YOUR MODELING KIT: Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S).
IF YOU DO NOT HAVE A MODELING KIT: remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models.
If you have a modeling kit use it to help you solve the following practice problems.
Exercise \(1\)
Are the following R or S?
Answer
1. S: I > Br > F > H. The lowest priority substituent, H, is already going towards the back. It turns left going from I to Br to F, so it's a S.
2. R: Br > Cl > CH3 > H. You have to switch the H and Br in order to place the H, the lowest priority, in the back. Then, going from Br to Cl, CH3 is turning to the right, giving you a R.
3. Neither R or S: This molecule is achiral. Only chiral molecules can be named R or S.
4. R: OH > CN > CH2NH2 > H. The H, the lowest priority, has to be switched to the back. Then, going from OH to CN to CH2NH2, you are turning right, giving you a R. (5)
5. S: \(\ce{-COOH}\) > \(\ce{-CH_2OH}\) > \(\ce{C#CH}\) > \(\ce{H}\). Then, going from \(\ce{-COOH}\) to \(\ce{-CH_2OH}\) to \(\ce{-C#CH}\) you are turning left, giving you a S configuration. | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.07%3A_Naming_Enantiomers_by_the_RS_System.txt |
What do you notice about these three pictures? Count the number of left gloves and right gloves.
6 left and 6 right gloves, correct?
What about this one:
I count 8 right gloves, 4 left gloves. So there’s a slight excess of right gloves here.
Finally, this figure:
ONLY right hand gloves here. 12 right gloves, zero left gloves.
Application to organic chemistry?
Gloves are chiral objects. That is, they lack an internal plane of symmetry. Left gloves and right gloves are mirror images of each other, but they can’t be superimposed. In chemistry, there’s a word we have to describe a pair of non-superimposable mirror images – they’re called enantiomers.
Tying it back to the drawings, we can have three types of situations.
1. Racemic Mixture: In the first drawing, we have an equal number of left and right gloves (i.e. enantiomers). This is called a racemic mixture of enantiomers.
2. Enantiomeric excess: In the second drawing, we have an excess of right gloves compared to left gloves. In a situtation like this we can say we have an “enantiomeric excess” of gloves, or alternatively, the mixture is “enantioenriched” in the right-hand glove. [We can also calculate the "excess" here: the mixture is 66% right and 33% left - so we have a 33% "excess" of the right-hand enantiomer].
3. Enantiomeric pure: In the third drawing, we have only right-hand gloves. This is said to be an “enantiomerically pure” mixture of gloves, since we have only one enantiomer present.
To tie it back to chemistry, let’s say we have a solution of a chiral molecule, like 2-butanol, which can exist as either the (R)-enantiomer or the (S)-enantiomer.
• A solution containing equal amounts of (R)-2-butanol and (S)-2-butanol is a racemic mixture.
• A solution containing an excess of either the (R)-enantiomer or the (S)-enantiomer would be enantioenriched.
• A solution containing only the (R)-enantiomer or the (S)-enantiomer will be enantiomerically pure.
Contributors
James Ashenhurst (MasterOrganicChemistry.com)
• A big thanks to Agnieszka at IlluScientia for the glove drawings.
4.5.1 Enantiomeric Excess and Optical Purity
Enantiomeric Excess
For non-racemic mixtures of enantiomers, one enantiomer is more abundant than the other. The composition of these mixtures is described by the enantiomeric excess, which is the difference between the relative abundance of the two enantiomers. Therefore, if a mixture contains 75% of the R enantiomer and 25% S, the enantiomeric excess if 50%. Similarly, a mixture that is 95% of one enantiomer, the enantiomeric excess is 90%, etc.
Enantiomeric excess is useful because it reflects the optical activity of the mixture. The standard optical rotation by the mixture ($[\alpha]_{mix}$) is equal to the product of the standard optical rotation of the major isomer ($[\alpha]_{major}$) and the enantiomeric excess ($EE$):
$[\alpha]_{mix} = EE \times [\alpha]_{major}$
In the same way, the enantiomeric excess in a mixture can be measured if the optical rotation of the pure enantiomer is known.
Diastereomeric Excess
A similar approach can be used to describe mixtures of diastereomers, resulting in the diastereomeric excess.
4.5.2 Resolution of Enantiomers
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent.
Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. The following diagram illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible.
To learn more about chemical procedures for achieving resolution Click Here.
Cliffs Notes
Web Pages
Further Reading on Enantiomeric Excess and Optical Purity
Carey 4th Edition On-Line Activity
5.13: How to Name Isomers with More than One Asymmetric Center
Diastereomers are stereoisomers that are not related as object and mirror image and are not enantiomers. Unlike enatiomers which are mirror images of each other and non-sumperimposable, diastereomers are not mirror images of each other and non-superimposable. Diastereomers can have different physical properties and reactivity. They have different melting points and boiling points and different densities. They have two or more stereocenters.
Introduction
It is easy to mistake between diasteromers and enantiomers. For example, we have four steroisomers of 3-bromo-2-butanol. The four possible combination are SS, RR, SR and RS (Figure 1). One of the molecule is the enantiomer of its mirror image molecule and diasteromer of each of the other two molecule (SS is enantiomer of RR and diasteromer of RS and SR). SS's mirror image is RR and they are not superimposable, so they are enantiomers. RS and SR are not mirror image of SS and are not superimposable to each other, so they are diasteromers.
Figure 1
Diastereomers vs. Enantiomers vs. Meso Compounds
Tartaric acid, C4H6O6, is an organic compound that can be found in grape, bananas, and in wine. The structures of tartaric acid itself is really interesting. Naturally, it is in the form of (R,R) stereocenters. Artificially, it can be in the meso form (R,S), which is achiral. R,R tartaric acid is enantiomer to is mirror image which is S,S tartaric acid and diasteromers to meso-tartaric acid (figure 2).
(R,R) and (S,S) tartaric acid have similar physical properties and reactivity. However, meso-tartaric acid have different physical properties and reactivity. For example, melting point of (R,R) & (S,S) tartaric is about 170 degree Celsius, and melting point of meso-tartaric acid is about 145 degree Celsius.
Figure 2
To identify meso, meso compound is superimposed on its mirror image, and has an internal plane that is symmetry (figure 3). Meso-tartaric acid is achiral and optically unactive.
Problems
Identify which of the following pair is enantiomers, diastereomers or meso compounds.
Answer
1. Diasteromers
2. Identical
3. Meso
4. Enantiomers | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.10%3A_Enantiomeric_Excess.txt |
The earth's stratospheric ozone layer plays a critical role in absorbing ultraviolet radiation emitted by the sun. In the last thirty years, it has been discovered that stratospheric ozone is depleting as a result of anthropogenic pollutants. There are a number of chemical reactions that can deplete stratospheric ozone; however, some of the most significant of these involves the catalytic destruction of ozone by halogen radicals such as chlorine and bromine.
Introduction
The atmosphere of the Earth is divided into five layers. In order of closest and thickest to farthest and thinnest the layers are listed as follows: troposphere, stratosphere, mesosphere, thermosphere and exosphere. The majority of the ozone in the atmosphere resides in the stratosphere, which extends from six miles above the Earth’s surface to 31 miles. Humans rely heavily on the absorption of ultraviolet B rays by the ozone layer because UV-B radiation causes skin cancer and can lead to genetic damage. The ozone layer has historically protected the Earth from the harmful UV rays, although in recent decades this protection has diminished due to stratospheric ozone depletion.
Ozone depletion is largely a result of man-made substances. Humans have introduced gases and chemicals into the atmosphere that have rapidly depleted the ozone layer in the last century. This depletion makes humans more vulnerable to the UV-B rays which are known to cause skin cancer as well as other genetic deformities. The possibility of ozone depletion was first introduced by scientists in the late 1960's as dreams of super sonic transport began to become a reality. Scientists had long been aware that nitric oxide (NO) can catalytically react with ozone ($O_3$) to produce $O_2$ molecules; however, $NO$ molecules produced at ground level have a half life far too short to make it into the stratosphere. It was not until the advent of commercial super sonic jets (which fly in the stratosphere and at an altitude much higher then conventional jets) that the potential for $NO$ to react with stratospheric ozone became a possibility. The threat of ozone depletion from commercial super sonic transport was so great that it is often cited as the main reason why the US federal government pulled support for its development in 1971. Fear of ozone depletion was abated until 1974 when Sherwood Rowland and Mario Molina discovered that chlorofluorocarbons could be photolyzed by high energy photons in the stratosphere. They discovered that this process could releasing chlorine radicals that would catalytically react with $O_3$ and destroy the molecule. This process is called the Rowland-Molina theory of $O_3$ depletion.
The Chapman Cycle
The stratosphere is in a constant cycle with oxygen molecules and their interaction with ultraviolet rays. This process is considered a cycle because of its constant conversion between different molecules of oxygen. The ozone layer is created when ultraviolet rays react with oxygen molecules (O2) to create ozone (O3) and atomic oxygen (O). This process is called the Chapman cycle.
Step 1: An oxygen molecules is photolyzed by solar radiation, creating two oxygen radicals:
$h\nu + O_2 \rightarrow 2O^. \nonumber$
Step 2: Oxygen radicals then react with molecular oxygen to produce ozone:
$O_2 + O^. \rightarrow O_3 \nonumber$
Step 3: Ozone then reacts with an additional oxygen radical to form molecular oxygen:
$O_3 + O^. \rightarrow 2O_2 \nonumber$
Step 4: Ozone can also be recycled into molecular oxygen by reacting with a photon:
$O_3 + h\nu \rightarrow O_2 + O^. \nonumber$
It is important to keep in mind that ozone is constantly being created and destroyed by the Chapman cycle and that these reactions are natural processes, which have been taking place for millions of years. Because of this, the thickness the ozone layer at any particular time can vary greatly. It is also important to know that O2 is constantly being introduced into the atmosphere through photosynthesis, so the ozone layer has the capability of regenerating itself.
Chemistry of Ozone Depletion
CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle:
$Cl + O_3 \rightarrow ClO + O_2 \tag{step 1}$
$ClO + O^. \rightarrow Cl + O_2 \tag{step 2}$
$O_3 + O^. \rightarrow 2O_2 \tag{Overall reaction}$
Chlorine is able to destroy so much of the ozone because it acts as a catalyst. Chlorine initiates the breakdown of ozone and combines with a freed oxygen to create two oxygen molecules. After each reaction, chlorine begins the destructive cycle again with another ozone molecule. One chlorine atom can thereby destroy thousands of ozone molecules. Because ozone molecules are being broken down they are unable to absorb any ultraviolet light so we experience more intense UV radiation at the earths surface.
From 1985 to 1988, researchers studying atmospheric properties over the south pole continually noticed significantly reduced concentrations of ozone directly over the continent of Antarctica. For three years it was assumed that the ozone data was incorrect and was due to some type of instrument malfunction. In 1988, researchers finally realized their error and concluded that an enormous hole in the ozone layer had indeed developed over Antarctica. Examination of NASA satellite data later showed that the hole had begun to develop in the mid 1970's.
The ozone hole over Antarctica is formed by a slew of unique atmospheric conditions over the continent that combine to create an ideal environment for ozone destruction.
• Because Antarctica is surrounded by water, winds over the continent blow in a unique clockwise direction creating a so called "polar vortex" that effectively contains a single static air mass over the continent. As a result, air over Antarctica does not mix with air in the rest of the earth's atmosphere.
• Antarctica has the coldest winter temperatures on earth, often reaching -110 F. These chilling temperatures result in the formation of polar stratospheric clouds (PSC's) which are a conglomeration of frozen H2O and HNO3. Due to their extremely cold temperatures, PSC's form an electrostatic attraction with CFC molecules as well as other halogenated compounds
As spring comes to Antarctica, the PSC's melt in the stratosphere and release all of the halogenated compounds that were previously absorbed to the cloud. In the antarctic summer, high energy photons are able to photolyze the halogenated compounds, freeing halogen radicals that then catalytically destroy O3. Because Antarctica is constantly surrounded by a polar vortex, radical halogens are not able to be diluted over the entire globe. The ozone hole develops as result of this process.
Resent research suggests that the strength of the polar vortex from any given year is directly correlated to the size of the ozone hole. In years with a strong polar vortex, the ozone hole is seen to expand in diameter, whereas in years with a weaker polar vortex, the ozone hole is noted to shrink
Ozone Depleting Substances
The following substances are listed as ozone depleting substances under Title VI of the United State Clean Air Act:
Table $1$: Ozone Depleting Substances And Their Ozone-Depletion Potential. Taken directly from the Clean Air Act, as of June 2010.
Substance Ozone- depletion potential
chlorofluorocarbon-11 (CFC–11) 1.0
chlorofluorocarbon-12 (CFC–12) 1.0
chlorofluorocarbon-13 (CFC–13) 1.0
chlorofluorocarbon-111 (CFC–111) 1.0
chlorofluorocarbon-112 (CFC–112) 1.0
chlorofluorocarbon-113 (CFC–113) 0.8
chlorofluorocarbon-114 (CFC–114) 1.0
chlorofluorocarbon-115 (CFC–115) 0.6
chlorofluorocarbon-211 (CFC–211) 1.0
chlorofluorocarbon-212 (CFC–212) 1.0
chlorofluorocarbon-213 (CFC–213) 1.0
chlorofluorocarbon-214 (CFC–214) 1.0
chlorofluorocarbon-215 (CFC–215) 1.0
chlorofluorocarbon-216 (CFC–216) 1.0
chlorofluorocarbon-217 (CFC–217) 1.0
halon-1211 3.0
halon-1301 10.0
halon-2402 6.0
carbon tetrachloride 1.1
methyl chloroform 0.1
hydrochlorofluorocarbon-22 (HCFC–22) 0.05
hydrochlorofluorocarbon-123 (HCFC–123) 0.02
hydrochlorofluorocarbon-124 (HCFC–124) 0.02
hydrochlorofluorocarbon-141(b) (HCFC–141(b)) 0.1
hydrochlorofluorocarbon-142(b) (HCFC–142(b)) 0.06
General Questions
• What are the causes of the depletion of our ozone layer?
• the release of free radicals, the use of CFC's, the excessive burning of fossil fuels
• What is the chemical reaction that displays how ozone is created?
• UV + O2 -> 2O + heat, O2 + O -> O3, O3 + O -> 2O2
• Which reactions demonstrate the destruction of the ozone layer?
• Cl + O3 ------> ClO + O2 and ClO + O ------> Cl + O
• How do CFC's destroy the ozone layer?
• the atomic chlorine freed from CFC reacts in a catalytic manner with ozone and atomic oxygen to make more oxygen molecules
• Why should regulations be enforced now in regards to pollution and harmful chemicals?
• without regulation, the production and use of chemicals will run out of hand and do irreversible damage to the stratosphere
• What type of atom in the CFC molecule is most destructive to the ozone?
• chlorine
• In which layer of the atmosphere does the ozone layer?
• the stratosphere, the second closest to the Earth's surface
• What cycle is responsible for ozone in the stratosphere?
• the Chapman cycle
• What factor is responsible for breaking up stable molecules?
• ultraviolet rays from the sun | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.14%3A__Reactions_of_Compounds_that_Contain_an_Asymmetric_Center.txt |
The earth's stratospheric ozone layer plays a critical role in absorbing ultraviolet radiation emitted by the sun. In the last thirty years, it has been discovered that stratospheric ozone is depleting as a result of anthropogenic pollutants. There are a number of chemical reactions that can deplete stratospheric ozone; however, some of the most significant of these involves the catalytic destruction of ozone by halogen radicals such as chlorine and bromine.
Introduction
The atmosphere of the Earth is divided into five layers. In order of closest and thickest to farthest and thinnest the layers are listed as follows: troposphere, stratosphere, mesosphere, thermosphere and exosphere. The majority of the ozone in the atmosphere resides in the stratosphere, which extends from six miles above the Earth’s surface to 31 miles. Humans rely heavily on the absorption of ultraviolet B rays by the ozone layer because UV-B radiation causes skin cancer and can lead to genetic damage. The ozone layer has historically protected the Earth from the harmful UV rays, although in recent decades this protection has diminished due to stratospheric ozone depletion.
Ozone depletion is largely a result of man-made substances. Humans have introduced gases and chemicals into the atmosphere that have rapidly depleted the ozone layer in the last century. This depletion makes humans more vulnerable to the UV-B rays which are known to cause skin cancer as well as other genetic deformities. The possibility of ozone depletion was first introduced by scientists in the late 1960's as dreams of super sonic transport began to become a reality. Scientists had long been aware that nitric oxide (NO) can catalytically react with ozone ($O_3$) to produce $O_2$ molecules; however, $NO$ molecules produced at ground level have a half life far too short to make it into the stratosphere. It was not until the advent of commercial super sonic jets (which fly in the stratosphere and at an altitude much higher then conventional jets) that the potential for $NO$ to react with stratospheric ozone became a possibility. The threat of ozone depletion from commercial super sonic transport was so great that it is often cited as the main reason why the US federal government pulled support for its development in 1971. Fear of ozone depletion was abated until 1974 when Sherwood Rowland and Mario Molina discovered that chlorofluorocarbons could be photolyzed by high energy photons in the stratosphere. They discovered that this process could releasing chlorine radicals that would catalytically react with $O_3$ and destroy the molecule. This process is called the Rowland-Molina theory of $O_3$ depletion.
The Chapman Cycle
The stratosphere is in a constant cycle with oxygen molecules and their interaction with ultraviolet rays. This process is considered a cycle because of its constant conversion between different molecules of oxygen. The ozone layer is created when ultraviolet rays react with oxygen molecules (O2) to create ozone (O3) and atomic oxygen (O). This process is called the Chapman cycle.
Step 1: An oxygen molecules is photolyzed by solar radiation, creating two oxygen radicals:
$h\nu + O_2 \rightarrow 2O^. \nonumber$
Step 2: Oxygen radicals then react with molecular oxygen to produce ozone:
$O_2 + O^. \rightarrow O_3 \nonumber$
Step 3: Ozone then reacts with an additional oxygen radical to form molecular oxygen:
$O_3 + O^. \rightarrow 2O_2 \nonumber$
Step 4: Ozone can also be recycled into molecular oxygen by reacting with a photon:
$O_3 + h\nu \rightarrow O_2 + O^. \nonumber$
It is important to keep in mind that ozone is constantly being created and destroyed by the Chapman cycle and that these reactions are natural processes, which have been taking place for millions of years. Because of this, the thickness the ozone layer at any particular time can vary greatly. It is also important to know that O2 is constantly being introduced into the atmosphere through photosynthesis, so the ozone layer has the capability of regenerating itself.
Chemistry of Ozone Depletion
CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle:
$Cl + O_3 \rightarrow ClO + O_2 \tag{step 1}$
$ClO + O^. \rightarrow Cl + O_2 \tag{step 2}$
$O_3 + O^. \rightarrow 2O_2 \tag{Overall reaction}$
Chlorine is able to destroy so much of the ozone because it acts as a catalyst. Chlorine initiates the breakdown of ozone and combines with a freed oxygen to create two oxygen molecules. After each reaction, chlorine begins the destructive cycle again with another ozone molecule. One chlorine atom can thereby destroy thousands of ozone molecules. Because ozone molecules are being broken down they are unable to absorb any ultraviolet light so we experience more intense UV radiation at the earths surface.
From 1985 to 1988, researchers studying atmospheric properties over the south pole continually noticed significantly reduced concentrations of ozone directly over the continent of Antarctica. For three years it was assumed that the ozone data was incorrect and was due to some type of instrument malfunction. In 1988, researchers finally realized their error and concluded that an enormous hole in the ozone layer had indeed developed over Antarctica. Examination of NASA satellite data later showed that the hole had begun to develop in the mid 1970's.
The ozone hole over Antarctica is formed by a slew of unique atmospheric conditions over the continent that combine to create an ideal environment for ozone destruction.
• Because Antarctica is surrounded by water, winds over the continent blow in a unique clockwise direction creating a so called "polar vortex" that effectively contains a single static air mass over the continent. As a result, air over Antarctica does not mix with air in the rest of the earth's atmosphere.
• Antarctica has the coldest winter temperatures on earth, often reaching -110 F. These chilling temperatures result in the formation of polar stratospheric clouds (PSC's) which are a conglomeration of frozen H2O and HNO3. Due to their extremely cold temperatures, PSC's form an electrostatic attraction with CFC molecules as well as other halogenated compounds
As spring comes to Antarctica, the PSC's melt in the stratosphere and release all of the halogenated compounds that were previously absorbed to the cloud. In the antarctic summer, high energy photons are able to photolyze the halogenated compounds, freeing halogen radicals that then catalytically destroy O3. Because Antarctica is constantly surrounded by a polar vortex, radical halogens are not able to be diluted over the entire globe. The ozone hole develops as result of this process.
Resent research suggests that the strength of the polar vortex from any given year is directly correlated to the size of the ozone hole. In years with a strong polar vortex, the ozone hole is seen to expand in diameter, whereas in years with a weaker polar vortex, the ozone hole is noted to shrink
Ozone Depleting Substances
The following substances are listed as ozone depleting substances under Title VI of the United State Clean Air Act:
Table $1$: Ozone Depleting Substances And Their Ozone-Depletion Potential. Taken directly from the Clean Air Act, as of June 2010.
Substance Ozone- depletion potential
chlorofluorocarbon-11 (CFC–11) 1.0
chlorofluorocarbon-12 (CFC–12) 1.0
chlorofluorocarbon-13 (CFC–13) 1.0
chlorofluorocarbon-111 (CFC–111) 1.0
chlorofluorocarbon-112 (CFC–112) 1.0
chlorofluorocarbon-113 (CFC–113) 0.8
chlorofluorocarbon-114 (CFC–114) 1.0
chlorofluorocarbon-115 (CFC–115) 0.6
chlorofluorocarbon-211 (CFC–211) 1.0
chlorofluorocarbon-212 (CFC–212) 1.0
chlorofluorocarbon-213 (CFC–213) 1.0
chlorofluorocarbon-214 (CFC–214) 1.0
chlorofluorocarbon-215 (CFC–215) 1.0
chlorofluorocarbon-216 (CFC–216) 1.0
chlorofluorocarbon-217 (CFC–217) 1.0
halon-1211 3.0
halon-1301 10.0
halon-2402 6.0
carbon tetrachloride 1.1
methyl chloroform 0.1
hydrochlorofluorocarbon-22 (HCFC–22) 0.05
hydrochlorofluorocarbon-123 (HCFC–123) 0.02
hydrochlorofluorocarbon-124 (HCFC–124) 0.02
hydrochlorofluorocarbon-141(b) (HCFC–141(b)) 0.1
hydrochlorofluorocarbon-142(b) (HCFC–142(b)) 0.06
General Questions
• What are the causes of the depletion of our ozone layer?
• the release of free radicals, the use of CFC's, the excessive burning of fossil fuels
• What is the chemical reaction that displays how ozone is created?
• UV + O2 -> 2O + heat, O2 + O -> O3, O3 + O -> 2O2
• Which reactions demonstrate the destruction of the ozone layer?
• Cl + O3 ------> ClO + O2 and ClO + O ------> Cl + O
• How do CFC's destroy the ozone layer?
• the atomic chlorine freed from CFC reacts in a catalytic manner with ozone and atomic oxygen to make more oxygen molecules
• Why should regulations be enforced now in regards to pollution and harmful chemicals?
• without regulation, the production and use of chemicals will run out of hand and do irreversible damage to the stratosphere
• What type of atom in the CFC molecule is most destructive to the ozone?
• chlorine
• In which layer of the atmosphere does the ozone layer?
• the stratosphere, the second closest to the Earth's surface
• What cycle is responsible for ozone in the stratosphere?
• the Chapman cycle
• What factor is responsible for breaking up stable molecules?
• ultraviolet rays from the sun | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.15%3A_Using_Reactions_that_Do_Not_Break_Bonds_to_an_Asymmetric_Center_to_Determine_Relative_Configurations.txt |
We turn now to concept of chirality that formed the basis of the story about Louis Pasteur in the beginning of this chapter. Recall that the term chiral, from the Greek work for 'hand', refers to anything which cannot be superimposed on its own mirror image. Your hands, of course, are chiral - you cannot superimpose your left hand on your right, and you cannot fit your left hand into a right-handed glove (which is also a chiral object). Another way of saying this is that your hands do not have a mirror plane of symmetry: you cannot find any plane which bisects your hand in such a way that one side of the plane is a mirror image of the other side. Chiral objects do not have a plane of symmetry.
Your face, on the other hand is achiral - lacking chirality - because, some small deviations notwithstanding, you could superimpose your face onto its mirror image. If someone were to show you a mirror image photograph of your face, you could line the image up, point-for-point, with your actual face. Your face has a plane of symmetry, because the left side is the mirror image of the right side.
What Pasteur, Biot, and their contemporaries did not yet fully understand when Pasteur made his discovery of molecular chirality was the source of chirality at the molecular level. It stood to reason that a chiral molecule is one that does not contain a plane of symmetry, and thus cannot be superimposed on its mirror image. We now know that chiral molecules contain one or more chiral centers, which are almost always tetrahedral (sp3-hybridized) carbons with four different substituents. Consider the cartoon molecule A below: a tetrahedral carbon, with four different substituents denoted by balls of four different colors (for the time being, don't worry about exactly what these substituents could be - we will see real examples very soon).
The mirror image of A, which we will call B, is drawn on the right side of the figure, and an imaginary mirror is in the middle. Notice that every point on A lines up through the mirror with the same point on B: in other words, if A looked in the mirror, it would see B looking back.
Now, if we flip compound A over and try to superimpose it point for point on compound B, we find that we cannot do it: if we superimpose any two colored balls, then the other two are misaligned.
A is not superimposable on its mirror image (B), thus by definition A is a chiral molecule. It follows that B also is not superimposable on its mirror image (A), and thus it is also a chiral molecule. Also notice in the figure below (and convince yourself with models) that neither A nor B has an internal plane of symmetry.
A and B are stereoisomers: molecules with the same molecular formula and the same bonding arrangement, but a different arrangement of atoms in space. There are two types of stereoisomers: enantiomers and diastereomers. Enantiomers are pairs of stereoisomers which are mirror images of each other: thus, A and B are enantiomers. It should be self-evident that a chiral molecule will always have one (and only one) enantiomer: enantiomers come in pairs. Enantiomers have identical physical properties (melting point, boiling point, density, and so on). However, enantiomers do differ in how they interact with polarized light (we will learn more about this soon) and they may also interact in very different ways with other chiral molecules - proteins, for example. We will begin to explore this last idea in later in this chapter, and see many examples throughout the remainder of our study of biological organic chemistry.
Diastereomers are stereoisomers which are not mirror images of each other. For now, we will concentrate on understanding enantiomers, and come back to diastereomers later.
We defined a chiral center as a tetrahedral carbon with four different substituents. If, instead, a tetrahedral carbon has two identical substituents (two black atoms in the cartoon figure below), then of course it still has a mirror image (everything has a mirror image, unless we are talking about a vampire!) However, it is superimposable on its mirror image, and has a plane of symmetry.
This molecule is achiral (lacking chirality). Using the same reasoning, we can see that a trigonal planar (sp2-hybridized) carbon is also not a chiral center.
Notice that structure E can be superimposed on F, its mirror image - all you have to do is pick E up, flip it over, and it is the same as F. This molecule has a plane of symmetry, and is achiral.
Let's apply our general discussion to real molecules. For now, we will limit our discussion to molecules with a single chiral center. It turns out that tartaric acid, the subject of our chapter introduction, has two chiral centers, so we will come back to it later.
Consider 2-butanol, drawn in two dimensions below.
Carbon #2 is a chiral center: it is sp3-hybridized and tetrahedral (even though it is not drawn that way above), and the four things attached to is are different: a hydrogen, a methyl (-CH3) group, an ethyl (-CH2CH3) group, and a hydroxyl (OH) group. Let's draw the bonding at C2 in three dimensions, and call this structure A. We will also draw the mirror image of A, and call this structure B.
When we try to superimpose A onto B, we find that we cannot do it. A and B are both chiral molecules, and they are enantiomers of each other.
2-propanol, unlike 2-butanol, is not a chiral molecule. Carbon #2 is bonded to two identical substituents (methyl groups), and so it is not a chiral center.
Notice that 2-propanol is superimposable on its own mirror image.
When we look at very simple molecules like 2-butanol, it is not difficult to draw out the mirror image and recognize that it is not superimposable. However, with larger, more complex molecules, this can be a daunting challenge in terms of drawing and three-dimensional visualization. The easy way to determine if a molecule is chiral is simply to look for the presence of one or more chiral centers: molecules with chiral centers will (almost always) be chiral. We insert the 'almost always' caveat here because it is possible to come up with the exception to this rule - we will have more to say on this later, but don't worry about it for now.
Here's another trick to make your stereochemical life easier: if you want to draw the enantiomer of a chiral molecule, it is not necessary to go to the trouble of drawing the point-for-point mirror image, as we have done up to now for purposes of illustration. Instead, keep the carbon skeleton the same, and simply reverse the solid and dashed wedge bonds on the chiral carbon: that accomplishes the same thing. You should use models to convince yourself that this is true, and also to convince yourself that swapping any two substituents about the chiral carbon will result in the formation of the enantiomer.
Here are four more examples of chiral biomolecules, each one shown as a pair of enantiomers, with chiral centers marked by red dots.
Here are some examples of achiral biomolecules - convince yourself that none of them contain a chiral center:
When looking for chiral centers, it is important to recognize that the question of whether or not the dashed/solid wedge drawing convention is used is irrelevant. Chiral molecules are sometimes drawn without using wedges (although obviously this means that stereochemical information is being omitted). Conversely, wedges may be used on carbons that are not chiral centers – look, for example, at the drawings of glycine and citrate in the figure above.
Can a chiral center be something other than a tetrahedral carbon with four different substituents? The answer to this question is 'yes' - however, these alternative chiral centers are very rare in the context of biological organic chemistry, and outside the scope of our discussion here.
You may also have wondered about amines: shouldn't we consider a secondary or tertiary amine to be a chiral center, as they are tetrahedral and attached to four different substituents, if the lone-pair electrons are counted as a 'substituent'? Put another way, isn't an amine non-superimposable on its mirror image?
The answer: yes it is, in the static picture, but in reality, the nitrogen of an amine is rapidly and reversibly inverting, or turning inside out, at room temperature.
If you have trouble picturing this, take an old tennis ball and cut it in half. Then, take one of the concave halves and flip it inside out, then back again: this is what the amine is doing. The end result is that the two 'enantiomers' if the amine are actually two rapidly interconverting forms of the same molecule, and thus the amine itself is not a chiral center. This inversion process does not take place on a tetrahedral carbon, which of course has no lone-pair electrons.
Exercise 3.8: Locate all of the chiral centers (there may be more than one in a molecule). Remember, hydrogen atoms bonded to carbon usually are not drawn in the line structure convention - but they are still there!
Exercise 3.9:
a) Draw two enantiomers of i) mevalonate and ii) serine.
b) Are the two 2-butanol structures below enantiomers?
Exercise 3.10: Label the molecules below as chiral or achiral, and locate all chiral centers.
Solutions to exercises
Kahn Academy video tutorials
Chirality
Enantiomers
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.17%3A_Nitrogen_and_Phosphorus_Atoms_Can_Be_Asymmetric_Centers.txt |
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5.20: The Stereochemistry of Enzyme-Catalyzed Reactions
While challenging to understand and visualize, the stereochemistry concepts we have explored in this chapter are integral to the study of living things. The vast majority of biological molecules contain chiral centers and/or stereogenic alkene groups. Most importantly, proteins are chiral, which of course includes all of the the enzymes which catalyze the chemical reactions of a cell, the receptors which transmit information within or between cells, and the antibodies which bind specifically to potentially harmful invaders. You know from your biology classes that proteins, because they fold up into a specific three dimensional shape, are able to very specifically recognize and bind to other organic molecules. The ligand or substrate bound by a particular protein could be a small organic molecule such as pyruvate all the way up to a large biopolymer such as a specific region of DNA, RNA, or another protein. Because they are chiral molecules, proteins are very sensitive to the stereochemistry of their ligands: a protein may bind specifically to (R)-glyceraldehyde, for example, but not bind to (S)-glyceraldehyde, just as your right hand will not fit into a left-handed baseball glove (see end of chapter for a link to an animation illustrating this concept).
The over-the-counter painkiller ibuprofen is currently sold as a racemic mixture, but only the S enantiomer is effective, due to the specific way it is able to bind to and inhibit the action of prostaglandin H2 synthase, an enzyme in the body's inflammation response process.
The R enantiomer of ibuprofen does not bind to prostaglandin H2 synthase in the same way as the S enantiomer, and as a consequence does not exert the same inhibitory effect on the enzyme's action (see Nature Chemical Biology 2011, 7, 803 for more details). Fortunately, (R)-ibuprofen apparently does not cause any harmful side effects, and is in fact isomerized gradually by an enzyme in the body to (S)-ibuprofen.
Earlier in this chapter we discussed the tragic case of thalidomide, and mentioned that it appears that it is specifically the S enantiomer which caused birth defects. Many different proposals have been made over the past decades to try to explain the teratogenic (birth defect-causing) effect of the drug, but a clear understanding still evades the scientific community. In 2010, however, a team in Japan reported evidence that thalidomide binds specifically to a protein called 'thereblon'. Furthermore, when production of thereblon is blocked in female zebra fish, developmental defects occur in her offspring which are very similar to the defects caused by the administration of thalidomide, pointing to the likelihood that thalidomide binding somehow inactivates the protein, thus initiating the teratogenic effect. (http://news.sciencemag.org/2010/03/t...-partner-crime)
You can, with a quick trip to the grocery store, directly experience the biological importance of stereoisomerism. Carvone is a chiral, plant-derived molecule that contributes to the smell of spearmint in the R form and caraway (a spice) in the S form.
Although details are not known, the two enantiomers presumably interact differently with one or more smell receptor proteins in your nose, generating the transmission of different chemical signals to the olfactory center of your brain.
Exercise 3.28: Ephedrine, found in the Chinese traditional medicine ma huang, is a stimulant and appetite suppressant. Both pseudoephedrine and levomethamphetamine are active ingredients in over-the-counter nasal decongestants. Methamphetamine is a highly addictive and illegal stimulant, and is usually prepared in illicit 'meth labs' using pseudoephedrine as a starting point.
What is the relationship between ephedrine and pseudoephedrine? Between methamphetamine and levomethamphetamine? Between pseudoephedrine and methamphetamine? Your choices are: not isomers, constitutional isomers, diastereomers, enantiomers, or same molecule.
Solutions to exercises
Enzymes are very specific with regard to the stereochemistry of the reactions they catalyze. When the product of a biochemical reaction contains a chiral center or a stereogenic alkene, with very few exceptions only one stereoisomer of the product is formed. In the glycolysis pathway, for example, the enzyme triose-phosphate isomerase catalyzes the reversible interconversion between dihydroxyacetone (which is achiral) and (R)-glyceraldehyde phosphate. The (S)-glyceraldehyde enantiomer is not formed by this enzyme in the left-to-right reaction, and is not used as a starting compound in the right-to-left reaction - it does not 'fit' in the active site of the enzyme.
In the isoprenoid biosynthesis pathway, two five-carbon building-block molecules combine to form a ten-carbon chain containing an E-alkene group. The enzyme does not catalyze formation of the Z diastereomer.
In chapters 9-17 of this book, and continuing on into your study of biological and organic chemistry, you will be learning about how enzymes are able to achieve these feats of stereochemical specificity. If you take a more advanced class in organic synthesis, you will also learn how laboratory chemists are figuring out ingenious ways to exert control over the stereochemical outcomes of nonenzymatic reactions, an area of chemistry that is particularly important in the pharmaceutical industry.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.19%3A_The_Stereochemistry_of_Electrophilic_addition_Reactions_of_Alkenes.txt |
Chiral molecules, as we learned in the introduction to this chapter, have an interesting optical property. You may know from studying physics that light waves are oscillating electric and magnetic fields. In ordinary light, the oscillation is randomly oriented in an infinite number of planes. When ordinary light is passed through a polarizer, all planes of oscillation are filtered out except one, resulting in plane-polarized light.
A beam of plane-polarized light, when passed through a sample of a chiral compound, interacts with the compound in such a way that the angle of oscillation will rotate. This property is called optical activity.
If a compound rotates plane polarized light in the clockwise (+) direction, it is said to be dextrorotatory, while if it rotates light in the counterclockwise (-) direction it is levorotatory. (We mentioned L- and D-amino acids in the previous section: the L-amino acids are levorotatory). The magnitude of the observed optical activity is dependent on temperature, the wavelength of light used, solvent, concentration of the chiral sample, and the path length of the sample tube (path length is the length that the plane-polarized light travels through the chiral sample). Typically, optical activity measurements are made in a 1 decimeter (10 cm) path-length sample tube at 25 °C, using as a light source the so-called “D-line” from a sodium lamp, which has a wavelength of 589 nm. The specific rotation of a pure chiral compound at 25° is expressed by the expression:
. . . where alpha(obs) is the observed rotation, l is path length in decimeters, and c is the concentration of the sample in grams per 100 mL. In other words, the specific rotation of a chiral compound is the optical rotation that is observed when 1 g of the compound is dissolved in enough of a given solvent to make 100 mL solution, and the rotation is measured in a 1 dm cuvette at 25 oC using light from a sodium lamp.
Every chiral molecule has a characteristic specific rotation, which is recorded in the chemical literature as a physical property just like melting point or density. Different enantiomers of a compound will always rotate plane-polarized light with an equal but opposite magnitude. (S)-ibuprofen, for example, has a specific rotation of +54.5o (dextrorotatory) in methanol, while (R)-ibuprofen has a specific rotation of -54.5o. There is no relationship between chiral compound's R/S designation and the direction of its specific rotation. For example, the S enantiomer of ibuprofen is dextrorotatory, but the S enantiomer of glyceraldehyde is levorotatory.
A 50:50 mixture of two enantiomers (a racemic mixture) will have no observable optical activity, because the two optical activities cancel each other out. In a structural drawing, a 'squigly' bond from a chiral center indicates a mixture of both R and S configurations.
Chiral molecules are often labeled according to whether they are dextrorotatory or levorotatory as well as by their R/S designation. For example, the pure enantiomers of ibuprofen are labeled (S)-(+)-ibuprofen and (R)-(-)-ibuprofen, while (±)-ibuprofen refers to the racemic mixture, which is the form in which the drug is sold to consumers.
Exercise 3.14: The specific rotation of (R)-limonene is +11.5o in ethanol. What is the expected observed rotation of a sample of 6.00 g (S)-limonene dissolved in ethanol to a total volume of 80.0 mL in a 1.00 dm (10.0 cm) pathlength cuvette?
Exercise 3.15: The specific rotation of (S)-carvone is +61°, measured 'neat' (pure liquid sample, no solvent). The optical rotation of a mixture of R and S carvone is measured at -23°. Which enantiomer is in excess in the mixture?
Solutions to exercises
All of the twenty natural amino acids except glycine have a chiral center at their alpha-carbon (recall that basic amino acid structure and terminology was introduced in section 1.3). Virtually all of the amino acids found in nature, both in the form of free amino acids or incorporated into peptides and proteins, have what is referred to in the biochemical literature as the 'L' configuration:
The 'L' indicates that these amino acid stereoisomers are levorotatory. All but one of the 19 L-amino acids have S stereochemistry at the a-carbon, using the rules of the R/S naming system.
D-amino acids (the D stands for dextrorotatory) are very rare in nature, but we will learn about an interesting example of a peptide containing one D-amino acid residue later in chapter 12.
Exercise 3.16: Which of the 20 common L-amino acids found in nature has the R configuration? Refer to the amino acid table for structures.
Solutions to exercises
Khan Academy video tutorial on optical acitivity
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/06%3A_Isomers_and_Stereochemistry/5.21%3A__Enantiomers_Can_Be_Distinguished_by_Biological_Molecules.txt |
Homework Problems
Homework Solutions
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07: Delocalized Electrons and Their Effect on Stability Reactivity and pKa (Ultraviolet and Visible Spectroscopy)
Kekulé was the first to suggest a sensible structure for benzene. The carbons are arranged in a hexagon, and he suggested alternating double and single bonds between them. Each carbon atom has a hydrogen attached to it.
This diagram is often simplified by leaving out all the carbon and hydrogen atoms! In diagrams of this sort, there is a carbon atom at each corner. You have to count the bonds leaving each carbon to work out how many hydrogens there are attached to it. In this case, each carbon has three bonds leaving it. Because carbon atoms form four bonds, that means you are a bond missing - and that must be attached to a hydrogen atom.
Problems with the Kekulé structure
Although the Kekulé structure was a good attempt in its time, there are serious problems with it regarding chemistry, structure and stability.
The Kekulé structure has problems with the chemistry. Because of the three double bonds, you might expect benzene to have reactions like ethene - only more so! Ethene undergoes addition reactions in which one of the two bonds joining the carbon atoms breaks, and the electrons are used to bond with additional atoms. Benzene rarely does this. Instead, it usually undergoes substitution reactions in which one of the hydrogen atoms is replaced by something new.
The Kekulé structure has problems with the shape. Benzene is a planar molecule (all the atoms lie in one plane), and that would also be true of the Kekulé structure. The problem is that C-C single and double bonds are different lengths.
• C-C (0.154 nm)
• C=C (0.134 nm)
That would mean that the hexagon would be irregular if it had the Kekulé structure, with alternating shorter and longer sides. In real benzene all the bonds are exactly the same - intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon.
The Kekulé structure has problems with the stability of benzene. Real benzene is a lot more stable than the Kekulé structure would give it credit for. Every time you do a thermochemistry calculation based on the Kekulé structure, you get an answer which is wrong by about 150 kJ mol-1. This is most easily shown using enthalpy changes of hydrogenation. Hydrogenation is the addition of hydrogen to something. If, for example, you hydrogenate ethene you get ethane:
$\ce{CH_2=CH_2 + H_2 \rightarrow CH_3CH_3} \tag{1}$
In order to do a fair comparison with benzene (a ring structure) we're going to compare it with cyclohexene. Cyclohexene, C6H10, is a ring of six carbon atoms containing just one C=C.
When hydrogen is added to this, cyclohexane, C6H12, is formed. The "CH" groups become CH2 and the double bond is replaced by a single one. The structures of cyclohexene and cyclohexane are usually simplified in the same way that the Kekulé structure for benzene is simplified - by leaving out all the carbons and hydrogens.
In the cyclohexane case, for example, there is a carbon atom at each corner, and enough hydrogens to make the total bonds on each carbon atom up to four. In this case, then, each corner represents CH2. The hydrogenation equation could be written:
The enthalpy change during this reaction is -120 kJ mol-1. In other words, when 1 mole of cyclohexene reacts, 120 kJ of heat energy is evolved. Where does this heat energy come from? When the reaction happens, bonds are broken (C=C and H-H) and this costs energy. Other bonds have to be made, and this releases energy. Because the bonds made are stronger than those broken, more energy is released than was used to break the original bonds and so there is a net evolution of heat energy.
If the ring had two double bonds in it initially (cyclohexa-1,3-diene), exactly twice as many bonds would have to be broken and exactly twice as many made. In other words, you would expect the enthalpy change of hydrogenation of cyclohexa-1,3-diene to be exactly twice that of cyclohexene - that is, -240 kJ mol-1.
In fact, the enthalpy change is -232 kJ mol-1 - which isn't far off what we are predicting. Applying the same argument to the Kekulé structure for benzene (what might be called cyclohexa-1,3,5-triene), you would expect an enthalpy change of -360 kJ mol-1, because there are exactly three times as many bonds being broken and made as in the cyclohexene case.
In fact what you get is -208 kJ mol-1 - not even within distance of the predicted value! This is very much easier to see on an enthalpy diagram. Notice that in each case heat energy is released, and in each case the product is the same (cyclohexane). That means that all the reactions "fall down" to the same end point.
Heavy lines, solid arrows and bold numbers represent real changes. Predicted changes are shown by dotted lines and italics. The most important point to notice is that real benzene is much lower down the diagram than the Kekulé form predicts. The lower down a substance is, the more energetically stable it is. This means that real benzene is about 150 kJ mol-1 more stable than the Kekulé structure gives it credit for. This increase in stability of benzene is known as the delocalization energy or resonance energy of benzene.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(Ultraviolet_and_Visible_Spectroscopy)/7.01%3A_Delocalized_Electrons_Explain_Benzen.txt |
Because of the low hydrogen to carbon ratio in aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C7H6O2 compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C6H6, which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Benzene:
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
The Molecular Orbitals of Benzene
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(Ultraviolet_and_Visible_Spectroscopy)/7.02%3A_The_Bonding_in_Benzene.txt |
Because of the low hydrogen to carbon ratio in aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C7H6O2 compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C6H6, which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Benzene:
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
The Molecular Orbitals of Benzene
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(Ultraviolet_and_Visible_Spectroscopy)/7.03%3A_Resonance_Contributors_and_the_Reson.txt |
Resonance is a mental exercise within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It involves constructing multiple Lewis structures that, when combined, represent the full electronic structure of the molecule. Resonance structures are used when a single Lewis structure cannot fully describe the bonding; the combination of possible resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. In general, molecules with multiple resonance structures will be more stable than one with fewer and some resonance structures contribute more to the stability of the molecule than others - formal charges aid in determining this.
Introduction
Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several resonance structures. The nuclear skeleton of the Lewis Structure of these resonance structures remains the same, only the electron locations differ. Such is the case for ozone ($\ce{O3}$), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Let's motivate the discussion by building the Lewis structure for ozone.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The electrons appear to "shift" between different resonance structures and while not strictly correct as each resonance structure is just a limitation of using the Lewis structure perspective to describe these molecules. A more accurate description of the electron structure of the molecule requires considering multiple resonance structures simultaneously.
Delocalization and Resonance Structures Rules
1. Resonance structures should have the same number of electrons, do not add or subtract any electrons. (check the number of electrons by simply counting them).
2. Each resonance structures follows the rules of writing Lewis Structures.
3. The hybridization of the structure must stay the same.
4. The skeleton of the structure can not be changed (only the electrons move).
5. Resonance structures must also have the same number of lone pairs.
"Pick the Correct Arrow for the Job"
Most arrows in chemistry cannot be used interchangeably and care must be given to selecting the correct arrow for the job.
• $\leftrightarrow$: A double headed arrow on both ends of the arrow between Lewis structures is used to show resonance
• $\rightleftharpoons$: Double harpoons are used to designate equilibria
• $\rightharpoonup$: A single harpoon on one end indicates the movement of one electron
• $\rightarrow$: A double headed arrow on one end is used to indicate the movement of two electrons
Example $2$: Carbonate Ion
Identify the resonance structures for the carbonate ion: $\ce{CO3^{2-}}$.
Solution
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32− is an average of three resonance structures.
Using Formal Charges to Identify viable Resonance Structures
While each resonance structure contributes to the total electronic structure of the molecule, they may not contribute equally. Assigning Formal charges to atoms in the molecules is one mechanism to identify the viability of a resonance structure and determine its relative magnitude among other structures. The formal charge on an atom in a covalent species is the net charge the atom would bear if the electrons in all the bonds to the atom were equally shared. Alternatively the formal charge on an atom in a covalent species is the net charge the atom would bear if all bonds to the atom were nonpolar covalent bonds. To determine the formal charge on a given atom in a covalent species, use the following formula:
$\text{Formal Charge} = (\text{number of valence electrons in free orbital}) - (\text{number of lone-pair electrons}) - \frac{1}{2} (\text{ number bond pair electrons}) \label{FC}$
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally (eg. benzene)
Example $3$: Thiocyanate Ion
Consider the thiocyanate ($CNS^-$) ion.
Solution
1. Find the Lewis Structure of the molecule. (Remember the Lewis Structure rules.)
2. Resonance: All elements want an octet, and we can do that in multiple ways by moving the terminal atom's electrons around (bonds too).
3. Assign Formal Charges via Equation \ref{FC}.
Formal Charge = (number of valence electrons in free orbital) - (number of lone-pair electrons) - ( $\frac{1}{2}$ number bond pair electrons)
Remember to determine the number of valence electron each atom has before assigning Formal Charges
C = 4 valence e-, N = 5 valence e-, S = 6 valence e-, also add an extra electron for the (-1) charge. The total of valence electrons is 16.
4. Find the most ideal resonance structure. (Note: It is the one with the least formal charges that adds up to zero or to the molecule's overall charge.)
5. Now we have to look at electronegativity for the "Correct" Lewis structure.
The most electronegative atom usually has the negative formal charge, while the least electronegative atom usually has the positive formal charges.
It is useful to combine the resonance structures into a single structure called the Resonance Hybrid that describes the bonding of the molecule. The general approach is described below:
1. Draw the Lewis Structure & Resonance for the molecule (using solid lines for bonds).
2. Where there can be a double or triple bond, draw a dotted line (-----) for the bond.
3. Draw only the lone pairs found in all resonance structures, do not include the lone pairs that are not on all of the resonance structures.
Example $4$: Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule ($\ce{C6H6}$) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Example $5$: Nitrate Ion
Draw the possible resonance structures for the Nitrate ion $\ce{NO_3^{-}}$.
Solution
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the $\ce{NO_3^{-}}$ ion to be somewhat shorter than a single bond.
Problems
1. True or False, The picture below is a resonance structure?
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
3. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
4. The resonance for HPO32-, and the formal charges (in red).
5. The resonance for CHO21-, and the formal charges (in red).
6. The resonance hybrid for PO43-, hybrid bonds are in red.
7. The resonance hybrid for NO3-, hybrid bonds are in red.
Contributors and Attributions
• Sharon Wei (UCD), Liza Chu (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(Ultraviolet_and_Visible_Spectroscopy)/7.04%3A_How_to_Draw_Resonance_Contributors.txt |
Resonance is a mental exercise within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It involves constructing multiple Lewis structures that, when combined, represent the full electronic structure of the molecule. Resonance structures are used when a single Lewis structure cannot fully describe the bonding; the combination of possible resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. In general, molecules with multiple resonance structures will be more stable than one with fewer and some resonance structures contribute more to the stability of the molecule than others - formal charges aid in determining this.
Introduction
Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several resonance structures. The nuclear skeleton of the Lewis Structure of these resonance structures remains the same, only the electron locations differ. Such is the case for ozone ($\ce{O3}$), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Let's motivate the discussion by building the Lewis structure for ozone.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The electrons appear to "shift" between different resonance structures and while not strictly correct as each resonance structure is just a limitation of using the Lewis structure perspective to describe these molecules. A more accurate description of the electron structure of the molecule requires considering multiple resonance structures simultaneously.
Delocalization and Resonance Structures Rules
1. Resonance structures should have the same number of electrons, do not add or subtract any electrons. (check the number of electrons by simply counting them).
2. Each resonance structures follows the rules of writing Lewis Structures.
3. The hybridization of the structure must stay the same.
4. The skeleton of the structure can not be changed (only the electrons move).
5. Resonance structures must also have the same number of lone pairs.
"Pick the Correct Arrow for the Job"
Most arrows in chemistry cannot be used interchangeably and care must be given to selecting the correct arrow for the job.
• $\leftrightarrow$: A double headed arrow on both ends of the arrow between Lewis structures is used to show resonance
• $\rightleftharpoons$: Double harpoons are used to designate equilibria
• $\rightharpoonup$: A single harpoon on one end indicates the movement of one electron
• $\rightarrow$: A double headed arrow on one end is used to indicate the movement of two electrons
Example $2$: Carbonate Ion
Identify the resonance structures for the carbonate ion: $\ce{CO3^{2-}}$.
Solution
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32− is an average of three resonance structures.
Using Formal Charges to Identify viable Resonance Structures
While each resonance structure contributes to the total electronic structure of the molecule, they may not contribute equally. Assigning Formal charges to atoms in the molecules is one mechanism to identify the viability of a resonance structure and determine its relative magnitude among other structures. The formal charge on an atom in a covalent species is the net charge the atom would bear if the electrons in all the bonds to the atom were equally shared. Alternatively the formal charge on an atom in a covalent species is the net charge the atom would bear if all bonds to the atom were nonpolar covalent bonds. To determine the formal charge on a given atom in a covalent species, use the following formula:
$\text{Formal Charge} = (\text{number of valence electrons in free orbital}) - (\text{number of lone-pair electrons}) - \frac{1}{2} (\text{ number bond pair electrons}) \label{FC}$
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally (eg. benzene)
Example $3$: Thiocyanate Ion
Consider the thiocyanate ($CNS^-$) ion.
Solution
1. Find the Lewis Structure of the molecule. (Remember the Lewis Structure rules.)
2. Resonance: All elements want an octet, and we can do that in multiple ways by moving the terminal atom's electrons around (bonds too).
3. Assign Formal Charges via Equation \ref{FC}.
Formal Charge = (number of valence electrons in free orbital) - (number of lone-pair electrons) - ( $\frac{1}{2}$ number bond pair electrons)
Remember to determine the number of valence electron each atom has before assigning Formal Charges
C = 4 valence e-, N = 5 valence e-, S = 6 valence e-, also add an extra electron for the (-1) charge. The total of valence electrons is 16.
4. Find the most ideal resonance structure. (Note: It is the one with the least formal charges that adds up to zero or to the molecule's overall charge.)
5. Now we have to look at electronegativity for the "Correct" Lewis structure.
The most electronegative atom usually has the negative formal charge, while the least electronegative atom usually has the positive formal charges.
It is useful to combine the resonance structures into a single structure called the Resonance Hybrid that describes the bonding of the molecule. The general approach is described below:
1. Draw the Lewis Structure & Resonance for the molecule (using solid lines for bonds).
2. Where there can be a double or triple bond, draw a dotted line (-----) for the bond.
3. Draw only the lone pairs found in all resonance structures, do not include the lone pairs that are not on all of the resonance structures.
Example $4$: Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule ($\ce{C6H6}$) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Example $5$: Nitrate Ion
Draw the possible resonance structures for the Nitrate ion $\ce{NO_3^{-}}$.
Solution
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the $\ce{NO_3^{-}}$ ion to be somewhat shorter than a single bond.
Problems
1. True or False, The picture below is a resonance structure?
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
3. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
4. The resonance for HPO32-, and the formal charges (in red).
5. The resonance for CHO21-, and the formal charges (in red).
6. The resonance hybrid for PO43-, hybrid bonds are in red.
7. The resonance hybrid for NO3-, hybrid bonds are in red.
Contributors and Attributions
• Sharon Wei (UCD), Liza Chu (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(Ultraviolet_and_Visible_Spectroscopy)/7.05%3A_The_Predicted_Stabilities_of_Resonan.txt |
Resonance is a mental exercise within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It involves constructing multiple Lewis structures that, when combined, represent the full electronic structure of the molecule. Resonance structures are used when a single Lewis structure cannot fully describe the bonding; the combination of possible resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. In general, molecules with multiple resonance structures will be more stable than one with fewer and some resonance structures contribute more to the stability of the molecule than others - formal charges aid in determining this.
Introduction
Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several resonance structures. The nuclear skeleton of the Lewis Structure of these resonance structures remains the same, only the electron locations differ. Such is the case for ozone ($\ce{O3}$), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Let's motivate the discussion by building the Lewis structure for ozone.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The electrons appear to "shift" between different resonance structures and while not strictly correct as each resonance structure is just a limitation of using the Lewis structure perspective to describe these molecules. A more accurate description of the electron structure of the molecule requires considering multiple resonance structures simultaneously.
Delocalization and Resonance Structures Rules
1. Resonance structures should have the same number of electrons, do not add or subtract any electrons. (check the number of electrons by simply counting them).
2. Each resonance structures follows the rules of writing Lewis Structures.
3. The hybridization of the structure must stay the same.
4. The skeleton of the structure can not be changed (only the electrons move).
5. Resonance structures must also have the same number of lone pairs.
"Pick the Correct Arrow for the Job"
Most arrows in chemistry cannot be used interchangeably and care must be given to selecting the correct arrow for the job.
• $\leftrightarrow$: A double headed arrow on both ends of the arrow between Lewis structures is used to show resonance
• $\rightleftharpoons$: Double harpoons are used to designate equilibria
• $\rightharpoonup$: A single harpoon on one end indicates the movement of one electron
• $\rightarrow$: A double headed arrow on one end is used to indicate the movement of two electrons
Example $2$: Carbonate Ion
Identify the resonance structures for the carbonate ion: $\ce{CO3^{2-}}$.
Solution
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32− is an average of three resonance structures.
Using Formal Charges to Identify viable Resonance Structures
While each resonance structure contributes to the total electronic structure of the molecule, they may not contribute equally. Assigning Formal charges to atoms in the molecules is one mechanism to identify the viability of a resonance structure and determine its relative magnitude among other structures. The formal charge on an atom in a covalent species is the net charge the atom would bear if the electrons in all the bonds to the atom were equally shared. Alternatively the formal charge on an atom in a covalent species is the net charge the atom would bear if all bonds to the atom were nonpolar covalent bonds. To determine the formal charge on a given atom in a covalent species, use the following formula:
$\text{Formal Charge} = (\text{number of valence electrons in free orbital}) - (\text{number of lone-pair electrons}) - \frac{1}{2} (\text{ number bond pair electrons}) \label{FC}$
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally (eg. benzene)
Example $3$: Thiocyanate Ion
Consider the thiocyanate ($CNS^-$) ion.
Solution
1. Find the Lewis Structure of the molecule. (Remember the Lewis Structure rules.)
2. Resonance: All elements want an octet, and we can do that in multiple ways by moving the terminal atom's electrons around (bonds too).
3. Assign Formal Charges via Equation \ref{FC}.
Formal Charge = (number of valence electrons in free orbital) - (number of lone-pair electrons) - ( $\frac{1}{2}$ number bond pair electrons)
Remember to determine the number of valence electron each atom has before assigning Formal Charges
C = 4 valence e-, N = 5 valence e-, S = 6 valence e-, also add an extra electron for the (-1) charge. The total of valence electrons is 16.
4. Find the most ideal resonance structure. (Note: It is the one with the least formal charges that adds up to zero or to the molecule's overall charge.)
5. Now we have to look at electronegativity for the "Correct" Lewis structure.
The most electronegative atom usually has the negative formal charge, while the least electronegative atom usually has the positive formal charges.
It is useful to combine the resonance structures into a single structure called the Resonance Hybrid that describes the bonding of the molecule. The general approach is described below:
1. Draw the Lewis Structure & Resonance for the molecule (using solid lines for bonds).
2. Where there can be a double or triple bond, draw a dotted line (-----) for the bond.
3. Draw only the lone pairs found in all resonance structures, do not include the lone pairs that are not on all of the resonance structures.
Example $4$: Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule ($\ce{C6H6}$) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Example $5$: Nitrate Ion
Draw the possible resonance structures for the Nitrate ion $\ce{NO_3^{-}}$.
Solution
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the $\ce{NO_3^{-}}$ ion to be somewhat shorter than a single bond.
Problems
1. True or False, The picture below is a resonance structure?
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
3. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
4. The resonance for HPO32-, and the formal charges (in red).
5. The resonance for CHO21-, and the formal charges (in red).
6. The resonance hybrid for PO43-, hybrid bonds are in red.
7. The resonance hybrid for NO3-, hybrid bonds are in red.
Contributors and Attributions
• Sharon Wei (UCD), Liza Chu (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(Ultraviolet_and_Visible_Spectroscopy)/7.06%3A_Delocalized_Energy_Is_the_Additional.txt |
In the vast majority of the nucleophilic substitution reactions you will see in this and other organic chemistry texts, the electrophilic atom is a carbon which is bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom is an attractive target for something that is electron-rich, i.e. a nucleophile. However, we must also consider the effect of steric hindrance on electrophilicity. In addition, we must discuss how the nature of the electrophilic carbon, and more specifically the stability of a potential carbocationic intermediate, influences the SN1 vs. SN2 character of a nucleophilic substitution reaction.
8.4A: Steric effects on electrophilicity
Consider two hypothetical SN2 reactions: one in which the electrophile is a methyl carbon and another in which it is tertiary carbon.
Because the three substituents on the methyl carbon electrophile are tiny hydrogens, the nucleophile has a relatively clear path for backside attack. However, backside attack on the tertiary carbon is blocked by the bulkier methyl groups. Once again, steric hindrance - this time caused by bulky groups attached to the electrophile rather than to the nucleophile - hinders the progress of an associative nucleophilic (SN2) displacement.
The factors discussed in the above paragraph, however, do not prevent a sterically-hindered carbon from being a good electrophile - they only make it less likely to be attacked in a concerted SN2 reaction. Nucleophilic substitution reactions in which the electrophilic carbon is sterically hindered are more likely to occur by a two-step, dissociative (SN1) mechanism. This makes perfect sense from a geometric point of view: the limitations imposed by sterics are significant mainly in an SN2 displacement, when the electrophile being attacked is a sp3-hybridized tetrahedral carbon with its relatively ‘tight’ angles of 109.4o. Remember that in an SN1 mechanism, the nucleophile attacks an sp2-hybridized carbocation intermediate, which has trigonal planar geometry with ‘open’ 120 angles.
With this open geometry, the empty p orbital of the electrophilic carbocation is no longer significantly shielded from the approaching nucleophile by the bulky alkyl groups. A carbocation is a very potent electrophile, and the nucleophilic step occurs very rapidly compared to the first (ionization) step.
8.4B: Stability of carbocation intermediates
We know that the rate-limiting step of an SN1 reaction is the first step - formation of the this carbocation intermediate. The rate of this step – and therefore, the rate of the overall substitution reaction – depends on the activation energy for the process in which the bond between the carbon and the leaving group breaks and a carbocation forms. According to Hammond’s postulate (section 6.2B), the more stable the carbocation intermediate is, the faster this first bond-breaking step will occur. In other words, the likelihood of a nucleophilic substitution reaction proceeding by a dissociative (SN1) mechanism depends to a large degree on the stability of the carbocation intermediate that forms.
The critical question now becomes, what stabilizes a carbocation?
Think back to Chapter 7, when we were learning how to evaluate the strength of an acid. The critical question was “how stable is the conjugate base that results when this acid donates its proton"? In many cases, this conjugate base was an anion – a center of excess electron density. Anything that can draw some of this electron density away– in other words, any electron withdrawing group – will stabilize the anion.
So if it takes an electron withdrawing group to stabilize a negative charge, what will stabilize a positive charge? An electron donating group!
A positively charged species such as a carbocation is very electron-poor, and thus anything which donates electron density to the center of electron poverty will help to stabilize it. Conversely, a carbocation will be destabilized by an electron withdrawing group.
Alkyl groups – methyl, ethyl, and the like – are weak electron donating groups, and thus stabilize nearby carbocations. What this means is that, in general, more substituted carbocations are more stable: a tert-butyl carbocation, for example, is more stable than an isopropyl carbocation. Primary carbocations are highly unstable and not often observed as reaction intermediates; methyl carbocations are even less stable.
Alkyl groups are electron donating and carbocation-stabilizing because the electrons around the neighboring carbons are drawn towards the nearby positive charge, thus slightly reducing the electron poverty of the positively-charged carbon.
It is not accurate to say, however, that carbocations with higher substitution are always more stable than those with less substitution. Just as electron-donating groups can stabilize a carbocation, electron-withdrawing groups act to destabilize carbocations. Carbonyl groups are electron-withdrawing by inductive effects, due to the polarity of the C=O double bond. It is possible to demonstrate in the laboratory (see section 16.1D) that carbocation A below is more stable than carbocation B, even though A is a primary carbocation and B is secondary.
The difference in stability can be explained by considering the electron-withdrawing inductive effect of the ester carbonyl. Recall that inductive effects - whether electron-withdrawing or donating - are relayed through covalent bonds and that the strength of the effect decreases rapidly as the number of intermediary bonds increases. In other words, the effect decreases with distance. In species B the positive charge is closer to the carbonyl group, thus the destabilizing electron-withdrawing effect is stronger than it is in species A.
In the next chapter we will see how the carbocation-destabilizing effect of electron-withdrawing fluorine substituents can be used in experiments designed to address the question of whether a biochemical nucleophilic substitution reaction is SN1 or SN2.
Stabilization of a carbocation can also occur through resonance effects, and as we have already discussed in the acid-base chapter, resonance effects as a rule are more powerful than inductive effects. Consider the simple case of a benzylic carbocation:
This carbocation is comparatively stable. In this case, electron donation is a resonance effect. Three additional resonance structures can be drawn for this carbocation in which the positive charge is located on one of three aromatic carbons. The positive charge is not isolated on the benzylic carbon, rather it is delocalized around the aromatic structure: this delocalization of charge results in significant stabilization. As a result, benzylic and allylic carbocations (where the positively charged carbon is conjugated to one or more non-aromatic double bonds) are significantly more stable than even tertiary alkyl carbocations.
Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in section 7.4). Consider the two pairs of carbocation species below:
In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually destabilize the positive charge, because they are electron withdrawing by induction.
Finally, vinylic carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any reaction.
Example 8.10
Exercise 8.10: In which of the structures below is the carbocation expected to be more stable? Explain.
For the most part, carbocations are very high-energy, transient intermediate species in organic reactions. However, there are some unusual examples of very stable carbocations that take the form of organic salts. Crystal violet is the common name for the chloride salt of the carbocation whose structure is shown below. Notice the structural possibilities for extensive resonance delocalization of the positive charge, and the presence of three electron-donating amine groups.
Example 8.11
Draw a resonance structure of the crystal violet cation in which the positive charge is delocalized to one of the nitrogen atoms.
Solution
When considering the possibility that a nucleophilic substitution reaction proceeds via an SN1 pathway, it is critical to evaluate the stability of the hypothetical carbocation intermediate. If this intermediate is not sufficiently stable, an SN1 mechanism must be considered unlikely, and the reaction probably proceeds by an SN2 mechanism. In the next chapter we will see several examples of biologically important SN1 reactions in which the positively charged intermediate is stabilized by inductive and resonance effects inherent in its own molecular structure.
Example 8.12
State which carbocation in each pair below is more stable, or if they are expected to be approximately equal. Explain your reasoning.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
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Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(Ultraviolet_and_Visible_Spectroscopy)/7.07%3A_Examples_That_Show_How_Delocalized_E.txt |
Resonance effects involving aromatic structures can have a dramatic influence on acidity and basicity. Notice, for example, the difference in acidity between phenol and cyclohexanol.
Looking at the conjugate base of phenol, we see that the negative charge can be delocalized by resonance to three different carbons on the aromatic ring.
Although these are all minor resonance contributors (negative charge is placed on a carbon rather than the more electronegative oxygen), they nonetheless have a significant effect on the acidity of the phenolic proton. Essentially, the benzene ring is acting as an electron-withdrawing group by resonance.
As we begin to study in detail the mechanisms of biological organic reactions, we’ll see that the phenol side chain of the amino acid tyrosine (see table 5 at the back of the book), with its relatively acidic pKaof 9-10, often acts as a catalytic proton donor/acceptor in enzyme active sites.
Exercise 7.4.1
Draw the conjugate base of 2-napthol (the major resonance contributor), and on your drawing indicate with arrows all of the atoms to which the negative charge can be delocalized by resonance.
Solution
The base-stabilizing effect of an aromatic ring can be accentuated by the presence of an additional electron-withdrawing substituent, such as a carbonyl. For the conjugate base of the phenol derivative below, an additional resonance contributor can be drawn in which the negative formal charge is placed on the carbonyl oxygen.
Now the negative charge on the conjugate base can be spread out over two oxygens (in addition to three aromatic carbons). The phenol acid therefore has a pKa similar to that of a carboxylic acid, where the negative charge on the conjugate base is also delocalized to two oxygen atoms. The ketone group is acting as an electron withdrawing group - it is 'pulling' electron density towards itself, through both inductive and resonance effects.
Exercise 7.4.2
The position of the electron-withdrawing substituent relative to the phenol hydroxyl is very important in terms of its effect on acidity. Which of the two substituted phenols below is more acidic? Use resonance drawings to explain your answer.
Solution
Exercise 7.4.3
Rank the four compounds below from most acidic to least.
Solution
Exercise 7.4.4
Nitro groups are very powerful electron-withdrawing groups. The phenol derivative picric acid (2,4,6 -trinitrophenol) has a pKa of 0.25, lower than that of trifluoroacetic acid. Use a resonance argument to explain why picric acid has such a low pKa.
Solution
Consider the acidity of 4-methoxyphenol, compared to phenol:
Notice that the methoxy group increases the pKa of the phenol group - it makes it less acidic. Why is this? At first inspection, you might assume that the methoxy substituent, with its electronegative oxygen, would be an electron-withdrawing group by induction. That is correct, but only to a point. The oxygen atom does indeed exert an electron-withdrawing inductive effect, but the lone pairs on the oxygen cause the exact opposite effect – the methoxy group is an electron-donating group by resonance. A resonance contributor can be drawn in which a formal negative charge is placed on the carbon adjacent to the negatively-charged phenolate oxygen.
Because of like-charge repulsion, this destabilizes the negative charge on the phenolate oxygen, making it more basic. It may help to visualize the methoxy group ‘pushing’ electrons towards the lone pair electrons of the phenolate oxygen, causing them to be less 'comfortable' and more reactive. The example above is a somewhat confusing but quite common situation in organic chemistry - a functional group, in this case a methoxy group, is exerting both an inductive effect and a resonance effect, but in opposite directions (the inductive effect is electron-withdrawing, the resonance effect is electron-donating). As a general rule a resonance effect is more powerful than an inductive effect - so overall, the methoxy group is acting as an electron donating group. A good rule of thumb to remember:
When resonance and induction compete, resonance usually wins!
Exercise 7.4.5
Rank the three compounds below from lowest pKa to highest, and explain your reasoning. Hint - think about both resonance and inductive effects!
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(Ultraviolet_and_Visible_Spectroscopy)/7.09%3A_How_Delocalized_Electrons_Affect_pKa.txt |
We end this chapter with a discussion of a type of reaction that is different from anything we have seen before. In the Diels-Alder cycloaddition reaction, a conjugated diene reacts with an alkene to form a ring structure.
In a Diels-Alder reaction, the alkene reacting partner is referred to as the dienophile. Essentially, this process involves overlap of the 2p orbitals on carbons 1 and 4 of the diene with 2p orbitals on the two sp2-hybridized carbons of the dienophile. Both of these new overlaps end up forming new sigma bonds, and a new pi bond is formed between carbon 2 and 3 of the diene.
One of the most important things to understand about this process is that it is concerted – all of the electron rearrangement takes place at once, with no carbocation intermediates.
The Diels-Alder reaction is enormously useful for synthetic organic chemists, not only because ring-forming reactions are useful in general but also because in many cases two new stereocenters are formed, and the reaction is inherently stereospecific. A cis dienophile will generate a ring with cis substitution, while a trans dienophile will generate a ring with trans substitution:
In order for a Diels-Alder reaction to occur, the diene molecule must adopt what is called the s-cis conformation:
The s-cis conformation is higher in energy than the s-trans conformation, due to steric hindrance. For some dienes, extreme steric hindrance causes the s-cis conformation to be highly strained, and for this reason such dienes do not readily undergo Diels-Alder reactions.
Cyclic dienes, on the other hand, are ‘locked’ in the s-cis conformation, and are especially reactive. The result of a Diels-Alder reaction involving a cyclic diene is a bicyclic structure:
Here, we see another element of stereopecificity: Diels-Alder reactions with cyclic dienes favor the formation of bicyclic structures in which substituents are in the endo position.
The endo position on a bicyclic structure refers to the position that is inside the concave shape of the larger (six-membered) ring. As you might predict, the exo position refers to the outside position.
The rate at which a Diels-Alder reaction takes place depends on electronic as well as steric factors. A particularly rapid Diels-Alder reaction takes place between cyclopentadiene and maleic anhydride.
We already know that cyclopentadiene is a good diene because of its inherent s-cis conformation. Maleic anhydride is also a very good dienophile, because the electron-withdrawing effect of the carbonyl groups causes the two alkene carbons to be electron-poor, and thus a good target for attack by the pi electrons in the diene.
In general, Diels-Alder reactions proceed fastest with electron-donating groups on the diene (eg. alkyl groups) and electron-withdrawing groups on the dienophile.
Alkynes can also serve as dienophiles in Diels-Alder reactions:
Below are just three examples of Diels-Alder reactions that have been reported in recent years:
The Diels-Alder reaction is just one example of a pericyclic reaction: this is a general term that refers to concerted rearrangements that proceed though cyclic transition states. Two well-studied intramolecular pericyclic reactions are known as the Cope rearrangement . . .
. . .and the Claisen rearrangement (when an oxygen is involved):
Notice that the both of these reactions require compounds in which two double bonds are separated by three single bonds.
Pericyclic reactions are rare in biological chemistry, but here is one example: the Claisen rearrangement catalyzed by chorismate mutase in the aromatic amino acid biosynthetic pathway.
The study of pericyclic reactions is an area of physical organic chemistry that blossomed in the mid-1960s, due mainly to the work of R.B. Woodward, Roald Hoffman, and Kenichi Fukui. The Woodward-Hoffman rules for pericyclic reactions (and a simplified version introduced by Fukui) use molecular orbital theory to explain why some pericyclic processes take place and others do not. A full discussion is beyond the scope of this text, but if you go on to study organic chemistry at the advanced undergraduate or graduate level you are sure to be introduced to this fascinating area of inquiry.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(Ultraviolet_and_Visible_Spectroscopy)/7.12%3A_The_Diels-Adler_Reaction_Is_a_14-Add.txt |
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