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Introduction
Photon energies associated with the infrared (from 1 to 15 kcal/mole) are not large enough to excite electrons, but may induce vibrational excitation of covalently bonded atoms and groups.
The covalent bonds in molecules are not rigid sticks or rods, such as found in molecular model kits, but are more like stiff springs that can be stretched and bent. The mobile nature of organic molecules was noted in the chapter concerning conformational isomers. We must now recognize that, in addition to the facile rotation of groups about single bonds, molecules experience a wide variety of vibrational motions, characteristic of their component atoms. Consequently, virtually all organic compounds will absorb infrared radiation that corresponds in energy to these vibrations. Infrared spectrometers, similar in principle to the UV-Visible spectrometer described elsewhere, permit chemists to obtain absorption spectra of compounds that are a unique reflection of their molecular structure.
Vibrational Spectroscopy
A molecule composed of n-atoms has 3n degrees of freedom, six of which are translations and rotations of the molecule itself. This leaves 3n-6 degrees of vibrational freedom (3n-5 if the molecule is linear). Vibrational modes are often given descriptive names, such as stretching, bending, scissoring, rocking and twisting. The four-atom molecule of formaldehyde, the gas phase spectrum of which is shown below, provides an example of these terms. If a ball & stick model of formaldehyde is not displayed to the right of the spectrum, press the view ball&stick model button on the right. We expect six fundamental vibrations (12 minus 6), and these have been assigned to the spectrum absorptions. To see the formaldehyde molecule display a vibration, click one of the buttons under the spectrum, or click on one of the absorption peaks in the spectrum.
Covalent bonds in organic molecules are not rigid sticks – rather, they behave more like springs. At room temperature, organic molecules are always in motion, as their bonds stretch, bend, and twist. These complex vibrations can be broken down mathematically into individual vibrational modes, a few of which are illustrated below.
The energy of molecular vibration is quantized rather than continuous, meaning that a molecule can only stretch and bend at certain 'allowed' frequencies. If a molecule is exposed to electromagnetic radiation that matches the frequency of one of its vibrational modes, it will in most cases absorb energy from the radiation and jump to a higher vibrational energy state - what this means is that the amplitude of the vibration will increase, but the vibrational frequency will remain the same. The difference in energy between the two vibrational states is equal to the energy associated with the wavelength of radiation that was absorbed. It turns out that it is the infrared region of the electromagnetic spectrum which contains frequencies corresponding to the vibrational frequencies of organic bonds.
An IR Spectrum
We will use a ketone sample to illustrate this process. The sample is irradiated with infrared light and the carbonyl bond will specifically absorb light with this same frequency, which by equations 4.1 and 4.2 corresponds to a wavelength of 5.83 x 10-6 m and an energy of 4.91 kcal/mol. When the carbonyl bond absorbs this energy, it jumps up to an excited vibrational state.
The value of ΔE - the energy difference between the low energy (ground) and high energy (excited) vibrational states - is equal to 4.91 kcal/mol, the same as the energy associated with the absorbed light frequency. The molecule does not remain in its excited vibrational state for very long, but quickly releases energy to the surrounding environment in form of heat, and returns to the ground state.
With an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. In the spectrophotometer, infrared light with frequencies ranging from about 1013 to 1014 Hz is passed though our sample of cyclohexane. Most frequencies pass right through the sample and are recorded by a detector on the other side.
Our 5.15 x 1013 Hz carbonyl stretching frequency, however, is absorbed by the 2-hexanone sample, and so the detector records that the intensity of this frequency, after having passed through the sample, is something less than 100% of its initial intensity.
The vibrations of a 2-hexanone molecule are not, of course, limited to the simple stretching of the carbonyl bond. The various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these vibrational modes also absorb different frequencies of infrared light.
The power of infrared spectroscopy arises from the observation that different functional groups have different characteristic absorption frequencies. The carbonyl bond in a ketone, as we saw with our 2-hexanone example, typically absorbs in the range of 5.11 - 5.18 x 1013 Hz, depending on the molecule. The carbon-carbon triple bond of an alkyne, on the other hand, absorbs in the range 6.30 - 6.80 x 1013 Hz. The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest. If we pass infrared light through an unknown sample and find that it absorbs in the carbonyl frequency range but not in the alkyne range, we can infer that the molecule contains a carbonyl group but not an alkyne.
Now, let's look at some actual output from IR spectroscopy experiments. Below is the IR spectrum for 2-hexanone.
There are a number of things that need to be explained in order for you to understand what it is that we are looking at. On the horizontal axis we see IR wavelengths expressed in terms of a unit called wavenumber (cm-1), which tells us how many waves fit into one centimeter. On the vertical axis we see ‘% transmittance’, which tells us how strongly light was absorbed at each frequency (100% transmittance means no absorption occurred at that frequency). The solid line traces the values of % transmittance for every wavelength – the ‘peaks’ (which are actually pointing down) show regions of strong absorption. For some reason, it is typical in IR spectroscopy to report wavenumber values rather than wavelength (in meters) or frequency (in Hz). The ‘upside down’ vertical axis, with absorbance peaks pointing down rather than up, is also a curious convention in IR spectroscopy. We wouldn’t want to make things too easy for you!
A calculator for interconverting these frequency and wavelength values is provided on the right. Simply enter the value to be converted in the appropriate box, press "Calculate" and the equivalent number will appear in the empty box.
Infrared spectra may be obtained from samples in all phases (liquid, solid and gaseous). Liquids are usually examined as a thin film sandwiched between two polished salt plates (note that glass absorbs infrared radiation, whereas NaCl is transparent). If solvents are used to dissolve solids, care must be taken to avoid obscuring important spectral regions by solvent absorption. Perchlorinated solvents such as carbon tetrachloride, chloroform and tetrachloroethene are commonly used. Alternatively, solids may either be incorporated in a thin KBr disk, prepared under high pressure, or mixed with a little non-volatile liquid and ground to a paste (or mull) that is smeared between salt plates.
Frequency - Wavelength Converter
Frequency in cm-1
Wavelength in μ
Gas Phase Infrared Spectrum of Formaldehyde, H2C=O
1. View CH2 Asymmetric Stretch
View CH2 Symmetric Stretch
View C=O Stretch
View CH2 Scissoring
View CH2 Rocking
View CH2 Wagging
Ball&Stick Model
Spacefill Model
Stick Model
Motion Off
The exact frequency at which a given vibration occurs is determined by the strengths of the bonds involved and the mass of the component atoms. For a more detailed discussion of these factors Click Here. In practice, infrared spectra do not normally display separate absorption signals for each of the 3n-6 fundamental vibrational modes of a molecule. The number of observed absorptions may be increased by additive and subtractive interactions leading to combination tones and overtones of the fundamental vibrations, in much the same way that sound vibrations from a musical instrument interact. Furthermore, the number of observed absorptions may be decreased by molecular symmetry, spectrometer limitations, and spectroscopic selection rules. One selection rule that influences the intensity of infrared absorptions, is that a change in dipole moment should occur for a vibration to absorb infrared energy. Absorption bands associated with C=O bond stretching are usually very strong because a large change in the dipole takes place in that mode.
Some General Trends:
1. Stretching frequencies are higher than corresponding bending frequencies. (It is easier to bend a bond than to stretch or compress it.)
2. Bonds to hydrogen have higher stretching frequencies than those to heavier atoms.
3. Triple bonds have higher stretching frequencies than corresponding double bonds, which in turn have higher frequencies than single bonds.(Except for bonds to hydrogen).
The general regions of the infrared spectrum in which various kinds of vibrational bands are observed are outlined in the following chart. Note that the blue colored sections above the dashed line refer to stretching vibrations, and the green colored band below the line encompasses bending vibrations. The complexity of infrared spectra in the 1450 to 600 cm-1 region makes it difficult to assign all the absorption bands, and because of the unique patterns found there, it is often called the fingerprint region. Absorption bands in the 4000 to 1450 cm-1 region are usually due to stretching vibrations of diatomic units, and this is sometimes called the group frequency region.
To illustrate the usefulness of infrared absorption spectra, examples for five C4H8O isomers are presented below their corresponding structural formulas. Try to associate each spectrum (A - E) with one of the isomers in the row above it.
Answers | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.02%3A_Infrared_%28IR%29_Spectroscopy.txt |
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).
11.04: Interpretting IR Spectra
Guided IR Spectrum Interpretation
Now, let’s take a look at the IR spectrum for 1-hexanol. There is a very broad ‘mountain’ centered at about 3400 cm-1. This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules.
In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm-1. We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance.
The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm-1 is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens.
Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm-1 due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen.
It is possible to identify other functional groups such as 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. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in this table.
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.
More examples of IR spectra
To illustrate the usefulness of infrared absorption spectra, examples for five C4H8O isomers are presented below their corresponding structural formulas. Try to associate each spectrum with one of the isomers in the row above it. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.03%3A_IR-Active_and_IR-Inactive_Vibrations.txt |
Common Group Frequencies Summary
When analyzing an IR spectrum, it is helpful to overlay the diagram below onto the spectrum with our mind to help recognize functional groups.
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.
Group Frequencies - a closer look
Detailed information about the infrared absorptions observed for various bonded atoms and groups is usually presented in tabular form. The following table provides a collection of such data for the most common functional groups. Following the color scheme of the chart, stretching absorptions are listed in the blue-shaded section and bending absorptions in the green shaded part. More detailed descriptions for certain groups (e.g. alkenes, arenes, alcohols, amines & carbonyl compounds) may be viewed by clicking on the functional class name. Since most organic compounds have C-H bonds, a useful rule is that absorption in the 2850 to 3000 cm-1 is due to sp3 C-H stretching; whereas, absorption above 3000 cm-1 is from sp2 C-H stretching or sp C-H stretching if it is near 3300 cm-1.
Stretching Vibrations
Bending Vibrations
Functional Class
Range (cm-1)
Intensity
Assignment
Range (cm-1)
Intensity
Assignment
Alkanes
2850-3000 str CH3, CH2 & CH
2 or 3 bands
1350-1470
1370-1390
720-725
med
med
wk
CH2 & CH3 deformation
CH3 deformation
CH2 rocking
Alkenes
3020-3100
1630-1680
1900-2000
med
var
str
=C-H & =CH2 (usually sharp)
C=C (symmetry reduces intensity)
C=C asymmetric stretch
880-995
780-850
675-730
str
med
med
=C-H & =CH2
(out-of-plane bending)
cis-RCH=CHR
Alkynes
3300
2100-2250
str
var
C-H (usually sharp)
C≡C (symmetry reduces intensity)
600-700 str C-H deformation
Arenes
3030
1600 & 1500
var
med-wk
C-H (may be several bands)
C=C (in ring) (2 bands)
(3 if conjugated)
690-900 str-med C-H bending &
ring puckering
Alcohols & Phenols
3580-3650
3200-3550
970-1250
var
str
str
O-H (free), usually sharp
O-H (H-bonded), usually broad
C-O
1330-1430
650-770
med
var-wk
O-H bending (in-plane)
O-H bend (out-of-plane)
Amines
3400-3500 (dil. soln.)
3300-3400 (dil. soln.)
1000-1250
wk
wk
med
N-H (1°-amines), 2 bands
N-H (2°-amines)
C-N
1550-1650
660-900
med-str
var
NH2 scissoring (1°-amines)
NH2 & N-H wagging
(shifts on H-bonding)
Aldehydes & Ketones
2690-2840(2 bands)
1720-1740
1710-1720
med
str
str
str
str
str
str
C-H (aldehyde C-H)
C=O (saturated aldehyde)
C=O (saturated ketone)
aryl ketone
α, β-unsaturation
cyclopentanone
cyclobutanone
1350-1360
1400-1450
1100
str
str
med
α-CH3 bending
α-CH2 bending
C-C-C bending
Carboxylic Acids & Derivatives
2500-3300 (acids) overlap C-H
1705-1720 (acids)
1210-1320 (acids)
str
str
med-str
str
str
str
str
str
str
O-H (very broad)
C=O (H-bonded)
O-C (sometimes 2-peaks)
C=O
C=O (2-bands)
O-C
C=O
O-C (2-bands)
C=O (amide I band)
1395-1440
1590-1650
1500-1560
med
med
med
C-O-H bending
N-H (1°-amide) II band
N-H (2°-amide) II band
Nitriles
Isocyanates,Isothiocyanates,
Diimides, Azides & Ketenes
2240-2260
2100-2270
med
med
C≡N (sharp)
-N=C=O, -N=C=S
-N=C=N-, -N3, C=C=O
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• C–H stretch from 3000–2850 cm-1
• C–H bend or scissoring from 1470-1450 cm-1
• C–H rock, methyl from 1370-1350 cm-1
• C–H rock, methyl, seen only in long chain alkanes, from 725-720 cm-1
Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR 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.
In alkenes compounds, each band in the spectrum can be assigned:
• C=C stretch from 1680-1640 cm-1
• =C–H stretch from 3100-3000 cm-1
• =C–H bend from 1000-650 cm-1
Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• –C?C– stretch from 2260-2100 cm-1
• –C?C–H: C–H stretch from 3330-3270 cm-1
• –C?C–H: C–H bend from 700-610 cm-1
The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• C–H stretch from 3100-3000 cm-1
• overtones, weak, from 2000-1665 cm-1
• C–C stretch (in-ring) from 1600-1585 cm-1
• C–C stretch (in-ring) from 1500-1400 cm-1
• C–H "oop" from 900-675 cm-1
Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• O–H stretch, hydrogen bonded 3500-3200 cm-1
• C–O stretch 1260-1050 cm-1 (s)
Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• C=O stretch - aliphatic ketones 1715 cm-1
- ?, ?-unsaturated ketones 1685-1666 cm-1
Figure 8. shows the spectrum of 2-butanone. This is a saturated ketone, and the C=O band appears at 1715.
If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• H–C=O stretch 2830-2695 cm-1
• C=O stretch:
• aliphatic aldehydes 1740-1720 cm-1
• alpha, beta-unsaturated aldehydes 1710-1685 cm-1
Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• C=O stretch
• aliphatic from 1750-1735 cm-1
• ?, ?-unsaturated from 1730-1715 cm-1
• C–O stretch from 1300-1000 cm-1
Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 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.
• O–H stretch from 3300-2500 cm-1
• C=O stretch from 1760-1690 cm-1
• C–O stretch from 1320-1210 cm-1
• O–H bend from 1440-1395 and 950-910 cm-1
Figure 11. shows the spectrum of hexanoic acid.
Organic Nitrogen Compounds
• N–O asymmetric stretch from 1550-1475 cm-1
• N–O symmetric stretch from 1360-1290 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.
• C–H wag (-CH2X) from 1300-1150 cm-1
• C–X stretches (general) from 850-515 cm-1
• C–Cl stretch 850-550 cm-1
• C–Br stretch 690-515 cm-1
The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation table is linked on bottom of page to find other assigned IR peaks.
Exercise
1. What functional groups give the following signals in an IR spectrum?
A) 1700 cm-1
B) 1550 cm-1
C) 1700 cm-1 and 2510-3000 cm-1
2. How can you distinguish the following pairs of compounds through IR analysis?
A) CH3OH (Methanol) and CH3CH2OCH2CH3 (Diethylether)
B) Cyclopentane and 1-pentene.
C)
3. The following spectra is for the accompanying compound. What are the peaks that you can I identify in the spectrum?
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)
4. What absorptions would the following compounds have in an IR spectra?
Answer
1.
2.
A) A OH peak will be present around 3300 cm-1 for methanol and will be absent in the ether.
B) 1-pentene 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
C) Cannot distinguish these two isomers. They both have the same functional groups and therefore would have the same peaks on an IR spectra.
3.
Frequency (cm-1) Functional Group
3200 C≡C-H
2900-3000 C-C-H, C=C-H
2100 C≡C
1610 C=C
(There is also an aromatic undertone region between 2000-1600 which describes the substitution on the phenyl ring.)
4.
A)
Frequency (cm-1) Functional Group
2900-3000 C-C-H, C=C-H
1710 C=O
1610 C=C
1100 C-O
B)
Frequency (cm-1) Functional Group
3200 C≡C-H
2900-3000 C-C-H, C=C-H
2100 C≡C
1710 C=O
C)
Frequency (cm-1) Functional Group
3300 (broad) O-H
2900-3000 C-C-H, C=C-H
2000-1800 Aromatic Overtones
1710 C=O
1610 C=C | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.05%3A_Infrared_Spectra_of_Some_Common_Functional_Groups.txt |
Learning Objectives
After completing this section, you should be able to
1. sketch a simple diagram to show the essential features of a mass spectrometer.
2. identify peaks in a simple mass spectrum, and explain how they arise.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• parent peak (molecular ion peak)
• relative abundance
• mass spectroscopy
• molecular ion (M+·)
• mass-to-charge ratio (m/z)
Study Notes
You may remember from general first-year chemistry how mass spectroscopy has been used to establish the atomic mass and abundance of isotopes.
Mass spectrometers are large and expensive, and usually operated only by fully trained personnel, so you will not have the opportunity to use such an instrument as part of this course. Research chemists often rely quite heavily on mass spectra to assist them in the identification of compounds, and you will be required to interpret simple mass spectra both in assignments and on examinations. Note that in most attempts to identify an unknown compound, chemists do not rely exclusively on the results obtained from a single spectroscopic technique. A combination of chemical and physical properties and spectral evidence is usually employed.
The Mass Spectrometer
In order to measure the characteristics of individual molecules, a mass spectrometer converts them to ions so that they can be moved about and manipulated by external electric and magnetic fields. The three essential functions of a mass spectrometer, and the associated components, are:
1. The ions are sorted and separated according to their mass and charge. The Mass Analyzer
2. The separated ions are then measured, and the results displayed on a chart. The Detector
Because ions are very reactive and short-lived, their formation and manipulation must be conducted in a vacuum. Atmospheric pressure is around 760 torr (mm of mercury). The pressure under which ions may be handled is roughly 10-5 to 10-8 torr (less than a billionth of an atmosphere). Each of the three tasks listed above may be accomplished in different ways. In one common procedure, ionization is effected by a high energy beam of electrons, and ion separation is achieved by accelerating and focusing the ions in a beam, which is then bent by an external magnetic field. The ions are then detected electronically and the resulting information is stored and analyzed in a computer. A mass spectrometer operating in this fashion is outlined in the following diagram. The heart of the spectrometer is the ion source. Here molecules of the sample (black dots) are bombarded by electrons (light blue lines) issuing from a heated filament. This is called an EI (electron-impact) source. Gases and volatile liquid samples are allowed to leak into the ion source from a reservoir (as shown). Non-volatile solids and liquids may be introduced directly. Cations formed by the electron bombardment (red dots) are pushed away by a charged repellor plate (anions are attracted to it), and accelerated toward other electrodes, having slits through which the ions pass as a beam. Some of these ions fragment into smaller cations and neutral fragments. A perpendicular magnetic field deflects the ion beam in an arc whose radius is inversely proportional to the mass of each ion. Lighter ions are deflected more than heavier ions. By varying the strength of the magnetic field, ions of different mass can be focused progressively on a detector fixed at the end of a curved tube (also under a high vacuum).
When a high energy electron collides with a molecule it often ionizes it by knocking away one of the molecular electrons (either bonding or non-bonding). This leaves behind a molecular ion (colored red in the following diagram). Residual energy from the collision may cause the molecular ion to fragment into neutral pieces (colored green) and smaller fragment ions (colored pink and orange). The molecular ion is a radical cation, but the fragment ions may either be radical cations (pink) or carbocations (orange), depending on the nature of the neutral fragment. An animated display of this ionization process will appear if you click on the ion source of the mass spectrometer diagram.
Below is typical output for an electron-ionization MS experiment (MS data below is derived from the Spectral Database for Organic Compounds, a free, web-based service provided by AIST in Japan.
The sample is acetone. On the horizontal axis is the value for m/z (as we stated above, the charge z is almost always +1, so in practice this is the same as mass). On the vertical axis is the relative abundance of each ion detected. On this scale, the most abundant ion, called the base peak, is set to 100%, and all other peaks are recorded relative to this value. For acetone, the base peak corresponds to a fragment with m/z = 43 - . The molecular weight of acetone is 58, so we can identify the peak at m/z = 58 as that corresponding to the molecular ion peak, or parent peak. Notice that there is a small peak at m/z = 59: this is referred to as the M+1 peak. How can there be an ion that has a greater mass than the molecular ion? Simple: a small fraction - about 1.1% - of all carbon atoms in nature are actually the 13C rather than the 12C isotope. The 13C isotope is, of course, heavier than 12C by 1 mass unit. In addition, about 0.015% of all hydrogen atoms are actually deuterium, the 2H isotope. So the M+1 peak represents those few acetone molecules in the sample which contained either a 13C or 2H. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.07%3A_Mass_Spectrometry_-_an_introduction.txt |
Objectives
After completing this section, you should be able to
1. suggest possible molecular formulas for a compound, given the m/z value for the molecular ion, or a mass spectrum from which this value can be obtained.
2. predict the relative heights of the M+·, (M + 1)+·, etc., peaks in the mass spectrum of a compound, given the natural abundance of the isotopes of carbon and the other elements present in the compound.
3. interpret the fragmentation pattern of the mass spectrum of a relatively simple, known compound (e.g., hexane).
4. use the fragmentation pattern in a given mass spectrum to assist in the identification of a relatively simple, unknown compound (e.g., an unknown alkane).
Study Notes
When interpreting fragmentation patterns, you may find it helpful to know that, as you might expect, the weakest carbon-carbon bonds are the ones most likely to break. You might wish to refer to the table of bond dissociation energies when attempting problems involving the interpretation of mass spectra.
This page looks at how fragmentation patterns are formed when organic molecules are fed into a mass spectrometer, and how you can get information from the mass spectrum.
The Origin of Fragmentation Patterns
When the vaporized organic sample passes into the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion - or sometimes the parent ion and is often given the symbol M+ or . The dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionization process.
The molecular ions are energetically unstable, and some of them will break up into smaller pieces. The simplest case is that a molecular ion breaks into two parts - one of which is another positive ion, and the other is an uncharged free radical.
The uncharged free radical will not produce a line on the mass spectrum. Only charged particles will be accelerated, deflected and detected by the mass spectrometer. These uncharged particles will simply get lost in the machine - eventually, they get removed by the vacuum pump.
The ion, X+, will travel through the mass spectrometer just like any other positive ion - and will produce a line on the stick diagram. All sorts of fragmentations of the original molecular ion are possible - and that means that you will get a whole host of lines in the mass spectrum. For example, the mass spectrum of pentane looks like this:
Note
The pattern of lines in the mass spectrum of an organic compound tells you something quite different from the pattern of lines in the mass spectrum of an element. With an element, each line represents a different isotope of that element. With a compound, each line represents a different fragment produced when the molecular ion breaks up.
In the stick diagram showing the mass spectrum of pentane, the line produced by the heaviest ion passing through the machine (at m/z = 72) is due to the molecular ion. The tallest line in the stick diagram (in this case at m/z = 43) is called the base peak. This is usually given an arbitrary height of 100, and the height of everything else is measured relative to this. The base peak is the tallest peak because it represents the commonest fragment ion to be formed - either because there are several ways in which it could be produced during fragmentation of the parent ion, or because it is a particularly stable ion.
Using Fragmentation Patterns
This section will ignore the information you can get from the molecular ion (or ions). That is covered in three other pages which you can get at via the mass spectrometry menu. You will find a link at the bottom of the page.
Example: Pentane
Let's have another look at the mass spectrum for pentane:
What causes the line at m/z = 57?
How many carbon atoms are there in this ion? There cannot 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.
Example: Pentan-3-one
This time the base peak (the tallest peak - and so the commonest fragment ion) is at m/z = 57. But this is not produced by the same ion as the same m/z value peak in pentane.
If you remember, the m/z = 57 peak in pentane was produced by [CH3CH2CH2CH2]+. If you look at the structure of pentan-3-one, it's impossible to get that particular fragment from it.
Work along the molecule mentally chopping bits off until you come up with something that adds up to 57. With a small amount of patience, you'll eventually find [CH3CH2CO]+ - which is produced by this fragmentation:
You would get exactly the same products whichever side of the CO group you split the molecular ion. The m/z = 29 peak is produced by the ethyl ion - which once again could be formed by splitting the molecular ion either side of the CO group.
Peak Heights and Stability
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. We'll look at two common examples of this.
Carbocations (carbonium ions)
Summarizing the most important conclusion from the page on carbocations:
Order of stability of carbocations
primary < secondary < tertiary
Applying the logic of this to fragmentation patterns, it means that a split which produces a secondary carbocation is going to be more successful than one producing a primary one. A split producing a tertiary carbocation will be more successful still. Let's look at the mass spectrum of 2-methylbutane. 2-methylbutane is an isomer of pentane - isomers are molecules with the same molecular formula, but a different spatial arrangement of the atoms.
Look first at the very strong peak at m/z = 43. This is caused by a different ion than the corresponding peak in the pentane mass spectrum. This peak in 2-methylbutane is caused by:
The ion formed is a secondary carbocation - it has two alkyl groups attached to the carbon with the positive charge. As such, it is relatively stable. The peak at m/z = 57 is much taller than the corresponding line in pentane. Again a secondary carbocation is formed - this time, by:
You would get the same ion, of course, if the left-hand CH3 group broke off instead of the bottom one as we've drawn it. In these two spectra, this is probably the most dramatic example of the extra stability of a secondary carbocation.
Acylium ions, [RCO]+
Ions with the positive charge on the carbon of a carbonyl group, C=O, are also relatively stable. This is fairly clearly seen in the mass spectra of ketones like pentan-3-one.
The base peak, at m/z=57, is due to the [CH3CH2CO]+ ion. We've already discussed the fragmentation that produces this.
Note
The more stable an ion is, the more likely it is to form. The more of a particular ion that is formed, the higher will be its peak height.
Using mass spectra to distinguish between compounds
Suppose you had to suggest a way of distinguishing between pentan-2-one and pentan-3-one using their mass spectra.
pentan-2-one CH3COCH2CH2CH3
pentan-3-one CH3CH2COCH2CH3
Each of these is likely to split to produce ions with a positive charge on the CO group. In the pentan-2-one case, there are two different ions like this:
• [CH3CO]+
• [COCH2CH2CH3]+
That would give you strong lines at m/z = 43 and 71. With pentan-3-one, you would only get one ion of this kind:
• [CH3CH2CO]+
In that case, you would get a strong line at 57. You don't need to worry about the other lines in the spectra - the 43, 57 and 71 lines give you plenty of difference between the two. The 43 and 71 lines are missing from the pentan-3-one spectrum, and the 57 line is missing from the pentan-2-one one.
The two mass spectra look like this:
As you've seen, the mass spectrum of even very similar organic compounds will be quite different because of the different fragmentation patterns that can occur. Provided you have a computer data base of mass spectra, any unknown spectrum can be computer analyzed and simply matched against the data base.
Exercise
5. 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?
6. 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
5. C8H10N4O2
C = 12 × 8 = 96
N = 14 × 4 = 56
H = 1 × 10 = 10
O = 2 × 16 = 32
96+56+10+32 = 194 g/mol
6. 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. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.08%3A_Fragmentation_Patterns_in_Mass_Spectrometry.txt |
The Nature of Mass Spectra
A mass spectrum will usually be presented as a vertical bar graph, in which each bar represents an ion having a specific mass-to-charge ratio (m/z) and the length of the bar indicates the relative abundance of the ion. The most intense ion is assigned an abundance of 100, and it is referred to as the base peak. Most of the ions formed in a mass spectrometer have a single charge, so the m/z value is equivalent to mass itself. Modern mass spectrometers easily distinguish (resolve) ions differing by only a single atomic mass unit, and thus provide completely accurate values for the molecular mass of a compound. The highest-mass ion in a spectrum is normally considered to be the molecular ion, and lower-mass ions are fragments from the molecular ion, assuming the sample is a single pure compound.
Atomic mass is given in terms of the unified atomic mass unit (symbol: μ) or dalton (symbol: Da). In recent years there has been a gradual change towards using the dalton in preference to the unified atomic mass unit. The dalton is classified as a "non-SI unit whose values in SI units must be obtained experimentally". It is defined as one twelfth of the rest mass of an unbound atom of carbon-12 in its nuclear and electronic ground state, and has a value of 1.660538782(83)x10-27 kg.
The following diagram displays the mass spectra of three simple gaseous compounds, carbon dioxide, propane and cyclopropane. The molecules of these compounds are similar in size, CO2 and C3H8 both have a nominal mass of 44 Da, and C3H6 has a mass of 42 Da. The molecular ion is the strongest ion in the spectra of CO2 and C3H6, and it is moderately strong in propane. The unit mass resolution is readily apparent in these spectra (note the separation of ions having m/z=39, 40, 41 and 42 in the cyclopropane spectrum). Even though these compounds are very similar in size, it is a simple matter to identify them from their individual mass spectra. By clicking on each spectrum in turn, a partial fragmentation analysis and peak assignment will be displayed. Even with simple compounds like these, it should be noted that it is rarely possible to explain the origin of all the fragment ions in a spectrum. Also, the structure of most fragment ions is seldom known with certainty.
Since a molecule of carbon dioxide is composed of only three atoms, its mass spectrum is very simple. The molecular ion is also the base peak, and the only fragment ions are CO (m/z=28) and O (m/z=16). The molecular ion of propane also has m/z=44, but it is not the most abundant ion in the spectrum. Cleavage of a carbon-carbon bond gives methyl and ethyl fragments, one of which is a carbocation and the other a radical. Both distributions are observed, but the larger ethyl cation (m/z=29) is the most abundant, possibly because its size affords greater charge dispersal. A similar bond cleavage in cyclopropane does not give two fragments, so the molecular ion is stronger than in propane, and is in fact responsible for the the base peak. Loss of a hydrogen atom, either before or after ring opening, produces the stable allyl cation (m/z=41). The third strongest ion in the spectrum has m/z=39 (C3H3). Its structure is uncertain, but two possibilities are shown in the diagram. The small m/z=39 ion in propane and the absence of a m/z=29 ion in cyclopropane are particularly significant in distinguishing these hydrocarbons.
Most stable organic compounds have an even number of total electrons, reflecting the fact that electrons occupy atomic and molecular orbitals in pairs. When a single electron is removed from a molecule to give an ion, the total electron count becomes an odd number, and we refer to such ions as radical cations. The molecular ion in a mass spectrum is always a radical cation, but the fragment ions may either be even-electron cations or odd-electron radical cations, depending on the neutral fragment lost. The simplest and most common fragmentations are bond cleavages producing a neutral radical (odd number of electrons) and a cation having an even number of electrons. A less common fragmentation, in which an even-electron neutral fragment is lost, produces an odd-electron radical cation fragment ion. Fragment ions themselves may fragment further. As a rule, odd-electron ions may fragment either to odd or even-electron ions, but even-electron ions fragment only to other even-electron ions. The masses of molecular and fragment ions also reflect the electron count, depending on the number of nitrogen atoms in the species.
Ions with no nitrogen
or an even # N atoms
odd-electron ions
even-number mass
even-electron ions
odd-number mass
Ions having an
odd # N atoms
odd-electron ions
odd-number mass
even-electron ions
even-number mass
This distinction is illustrated nicely by the follwing two examples. The unsaturated ketone, 4-methyl-3-pentene-2-one, on the left has no nitrogen so the mass of the molecular ion (m/z = 98) is an even number. Most of the fragment ions have odd-numbered masses, and therefore are even-electron cations. Diethylmethylamine, on the other hand, has one nitrogen and its molecular mass (m/z = 87) is an odd number. A majority of the fragment ions have even-numbered masses (ions at m/z = 30, 42, 56 & 58 are not labeled), and are even-electron nitrogen cations. The weak even -electron ions at m/z=15 and 29 are due to methyl and ethyl cations (no nitrogen atoms). The fragmentations leading to the chief fragment ions will be displayed by clicking on the appropriate spectrum. Repeated clicks will cycle the display.
4-methyl-3-pentene-2-one
N,N-diethylmethylamine
When non-bonded electron pairs are present in a molecule (e.g. on N or O), fragmentation pathways may sometimes be explained by assuming the missing electron is partially localized on that atom. A few such mechanisms are shown above. Bond cleavage generates a radical and a cation, and both fragments often share these roles, albeit unequally.
Isotopes
Since a mass spectrometer separates and detects ions of slightly different masses, it easily distinguishes different isotopes of a given element. This is manifested most dramatically for compounds containing bromine and chlorine, as illustrated by the following examples. Since molecules of bromine have only two atoms, the spectrum on the left will come as a surprise if a single atomic mass of 80 Da is assumed for Br. The five peaks in this spectrum demonstrate clearly that natural bromine consists of a nearly 50:50 mixture of isotopes having atomic masses of 79 and 81 Da respectively. Thus, the bromine molecule may be composed of two 79Br atoms (mass 158 Da), two 81Br atoms (mass 162 Da) or the more probable combination of 79Br-81Br (mass 160 Da). Fragmentation of Br2 to a bromine cation then gives rise to equal sized ion peaks at 79 and 81 Da.
bromine
vinyl chloride
methylene chloride
The center and right hand spectra show that chlorine is also composed of two isotopes, the more abundant having a mass of 35 Da, and the minor isotope a mass 37 Da. The precise isotopic composition of chlorine and bromine is:
• Chlorine: 75.77% 35Cl and 24.23% 37Cl
• Bromine: 50.50% 79Br and 49.50% 81Br
The presence of chlorine or bromine in a molecule or ion is easily detected by noticing the intensity ratios of ions differing by 2 Da. In the case of methylene chloride, the molecular ion consists of three peaks at m/z=84, 86 & 88 Da, and their diminishing intensities may be calculated from the natural abundances given above. Loss of a chlorine atom gives two isotopic fragment ions at m/z=49 & 51 Da, clearly incorporating a single chlorine atom. Fluorine and iodine, by contrast, are monoisotopic, having masses of 19 Da and 127 Da respectively. It should be noted that the presence of halogen atoms in a molecule or fragment ion does not change the odd-even mass rules given above.
To make use of a calculator that predicts the isotope clusters for different combinations of chlorine, bromine and other elements Click Here. This application was developed at Colby College.
Isotopic Abundance Calculator
C H N O SSi
Molecular Ion
100%
M + 1
M + 2
Two other common elements having useful isotope signatures are carbon, 13C is 1.1% natural abundance, and sulfur, 33S and 34S are 0.76% and 4.22% natural abundance respectively. For example, the small m/z=99 Da peak in the spectrum of 4-methyl-3-pentene-2-one (above) is due to the presence of a single 13C atom in the molecular ion. Although less important in this respect, 15N and 18O also make small contributions to higher mass satellites of molecular ions incorporating these elements.
The calculator on the right may be used to calculate the isotope contributions to ion abundances 1 and 2 Da greater than the molecular ion (M). Simply enter an appropriate subscript number to the right of each symbol, leaving those elements not present blank, and press the "Calculate" button. The numbers displayed in the M+1 and M+2 boxes are relative to M being set at 100%.
Fragmentation Patterns
The fragmentation of molecular ions into an assortment of fragment ions is a mixed blessing. The nature of the fragments often provides a clue to the molecular structure, but if the molecular ion has a lifetime of less than a few microseconds it will not survive long enough to be observed. Without a molecular ion peak as a reference, the difficulty of interpreting a mass spectrum increases markedly. Fortunately, most organic compounds give mass spectra that include a molecular ion, and those that do not often respond successfully to the use of milder ionization conditions. Among simple organic compounds, the most stable molecular ions are those from aromatic rings, other conjugated pi-electron systems and cycloalkanes. Alcohols, ethers and highly branched alkanes generally show the greatest tendency toward fragmentation.
The mass spectrum of dodecane on the right illustrates the behavior of an unbranched alkane. Since there are no heteroatoms in this molecule, there are no non-bonding valence shell electrons. Consequently, the radical cation character of the molecular ion (m/z = 170) is delocalized over all the covalent bonds. Fragmentation of C-C bonds occurs because they are usually weaker than C-H bonds, and this produces a mixture of alkyl radicals and alkyl carbocations. The positive charge commonly resides on the smaller fragment, so we see a homologous series of hexyl (m/z = 85), pentyl (m/z = 71), butyl (m/z = 57), propyl (m/z = 43), ethyl (m/z = 29) and methyl (m/z = 15) cations. These are accompanied by a set of corresponding alkenyl carbocations (e.g. m/z = 55, 41 &27) formed by loss of 2 H. All of the significant fragment ions in this spectrum are even-electron ions. In most alkane spectra the propyl and butyl ions are the most abundant.
The presence of a functional group, particularly one having a heteroatom Y with non-bonding valence electrons (Y = N, O, S, X etc.), can dramatically alter the fragmentation pattern of a compound. This influence is thought to occur because of a "localization" of the radical cation component of the molecular ion on the heteroatom. After all, it is easier to remove (ionize) a non-bonding electron than one that is part of a covalent bond. By localizing the reactive moiety, certain fragmentation processes will be favored. These are summarized in the following diagram, where the green shaded box at the top displays examples of such "localized" molecular ions. The first two fragmentation paths lead to even-electron ions, and the elimination (path #3) gives an odd-electron ion. Note the use of different curved arrows to show single electron shifts compared with electron pair shifts.
The charge distributions shown above are common, but for each cleavage process the charge may sometimes be carried by the other (neutral) species, and both fragment ions are observed. Of the three cleavage reactions described here, the alpha-cleavage is generally favored for nitrogen, oxygen and sulfur compounds. Indeed, in the previously displayed spectra of 4-methyl-3-pentene-2-one and N,N-diethylmethylamine the major fragment ions come from alpha-cleavages. Further examples of functional group influence on fragmentation are provided by a selection of compounds that may be examined by clicking the left button below. Useful tables of common fragment ions and neutral species may be viewed by clicking the right button.
Assorted Mass Spectra
View Fragment Tables
The complexity of fragmentation patterns has led to mass spectra being used as "fingerprints" for identifying compounds. Environmental pollutants, pesticide residues on food, and controlled substance identification are but a few examples of this application. Extremely small samples of an unknown substance (a microgram or less) are sufficient for such analysis. The following mass spectrum of cocaine demonstrates how a forensic laboratory might determine the nature of an unknown street drug. Even though extensive fragmentation has occurred, many of the more abundant ions (identified by magenta numbers) can be rationalized by the three mechanisms shown above. Plausible assignments may be seen by clicking on the spectrum, and it should be noted that all are even-electron ions. The m/z = 42 ion might be any or all of the following: C3H6, C2H2O or C2H4N. A precise assignment could be made from a high-resolution m/z value (next section).
Odd-electron fragment ions are often formed by characteristic rearrangements in which stable neutral fragments are lost. Mechanisms for some of these rearrangements have been identified by following the course of isotopically labeled molecular ions. A few examples of these rearrangement mechanisms may be seen by clicking the following button.
Assorted Rearrangement Fragmentations
Exercise
7. What are the masses of all the components in the following fragmentations?
Answer
7.
11.10: Determination of the Molecular Formula by High Re
High Resolution Mass Spectrometry
In assigning mass values to atoms and molecules, we have assumed integral values for isotopic masses. However, accurate measurements show that this is not strictly true. Because the strong nuclear forces that bind the components of an atomic nucleus together vary, the actual mass of a given isotope deviates from its nominal integer by a small but characteristic amount (remember E = mc2). Thus, relative to 12C at 12.0000, the isotopic mass of 16O is 15.9949 Da (not 16) and 14N is 14.0031 Da (not 14).
Formula
C6H12 C5H8O C4H8N2
Mass
84.0939 84.0575 84.0688
By designing mass spectrometers that can determine m/z values accurately to four decimal places, it is possible to distinguish different formulas having the same nominal mass. The table on the right illustrates this important feature, and a double-focusing high-resolution mass spectrometer easily distinguishes ions having these compositions. Mass spectrometry therefore not only provides a specific molecular mass value, but it may also establish the molecular formula of an unknown compound.
Tables of precise mass values for any molecule or ion are available in libraries; however, the mass calculator provided below serves the same purpose. Since a given nominal mass may correspond to several molecular formulas, lists of such possibilities are especially useful when evaluating the spectrum of an unknown compound. Composition tables are available for this purpose, and a particularly useful program for calculating all possible combinations of H, C, N & O that give a specific nominal mass has been written by Jef Rozenski. To use this calculator Click Here.
University of ArizonaIowa State University University of Leeds | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/11%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.09%3A__Useful_Patterns_for_Structure_Elucidation.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• explain how 1H NMR spectrometers work - refer to section 12.1
• interpret chemical shifts of 1H NMR spectra as they relate to shielding and deshielding - refer to section 12.2 and 12.14
• explain the delta scale of 1H NMR spectra - refer to section 12.3
• recognize equivalent protons within an organic compound - refer to section 12.4
• correlate functional group structural features with chemical shifts - refer to section 12.5
• determine the proton ratio from 1H NMR spectra peak integration data - refer to section 12.6
• explain and interpret spin-spin splitting patterns in 1H NMR spectra - refer to section 12.7
• explain and interpret spin-spin splitting patterns in 1H NMR spectra - refer to section 12.8
• describes examples of some uses of 1H NMR spectroscopy - refer to section 12.9
• explain how 13C NMR spectrometers work - refer to section 12.10
• interpret the chemical shifts of 13C NMR spectra to determine the structural features of organic compounds - refer to section 12.11 and 12.14
• explain how DEPT (distortionless enhancement by polarization transfer) is used to determine the number of hydrogens bonded to each carbon - refer to section 12.12
• describes some uses of 13C NMR spectroscopy - refer to section 12.13
12: Nuclear Magnetic Resonance Spectroscopy
Nuclear precession, spin states, and the resonance condition
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. So, 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.
NMR Spectrometer
In the absence of an external magnetic field the direction of the spin of the nuclei will be randomly oriented (see figure below left). However, when a sample of these nuclei is place in an external magnetic field, the nuclear spins will adopt specific orientations much as a compass needle responses 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). See figure below right.
When the same sample is placed within the field of a very strong magnet in an NMR instrument (this field is referred to by NMR spectroscopists as the applied field, abbreviated B0 ) each hydrogen will assume one of two possible spin states. In what is referred to as the +½ spin state, the hydrogen's magnetic moment is aligned with the direction of B0, while in the -½ spin state it is aligned opposed to the direction of B0.
Because the +½ spin state is slightly lower in energy, in a large population of organic molecules slightly more than half of the hydrogen atoms will occupy this state, while slightly less than half will occupy the –½ state. The difference in energy between the two spin states increases with increasing strength of B0.This last statement is in italics because it is one of the key ideas in NMR spectroscopy, as we shall soon see.
At this point, we need to look a little more closely at how a proton spins in an applied magnetic field. You may recall playing with spinning tops as a child. When a top slows down a little and the spin axis is no longer completely vertical, it begins to exhibit precessional motion, as the spin axis rotates slowly around the vertical. In the same way, hydrogen atoms spinning in an applied magnetic field also exhibit precessional motion about a vertical axis. It is this axis (which is either parallel or antiparallel to B0) that defines the proton’s magnetic moment. In the figure below, the proton is in the +1/2 spin state.
The frequency of precession (also called the Larmour frequency, abbreviated ωL) is simply the number of times per second that the proton precesses in a complete circle. A proton`s precessional frequency increases with the strength of B0.
If a proton that is precessing in an applied magnetic field is exposed to electromagnetic radiation of a frequency ν that matches its precessional frequency ωL, we have a condition called resonance. In the resonance condition, a proton in the lower-energy +½ spin state (aligned with B0) will transition (flip) to the higher energy –½ spin state (opposed to B0). In doing so, it will absorb radiation at this resonance frequency ν = ωL. This frequency, as you might have already guessed, corresponds to the energy difference between the proton’s two spin states. With the strong magnetic fields generated by the superconducting magnets used in modern NMR instruments, the resonance frequency for protons falls within the radio-wave range, anywhere from 100 MHz to 800 MHz depending on the strength of the magnet.
If the ordered nuclei are now subjected to EM radiation 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 or NMR.
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.
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.
The basic arrangement of an NMR spectrometer is displayed below. A sample (in a small glass tube) is placed between the poles of a strong magnetic. A radio frequency generator pulses the sample and excites the nuclei causing a spin-flip. The spin flip is detected by the detector and the signal sent to a computer where it is processed.
Exercise
1. If in a field strength of 4.7 T, H1 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?
2. 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
1.
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).
2.
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. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.01%3A_Theory_of_Nuclear_Magnetic_Resonance_%28NMR%29.txt |
Objectives
After completing this section, you should be able to
1. explain, in general terms, the origin of shielding effects in NMR spectroscopy.
2. explain the number of peaks occurring in the 1H or 13C NMR spectrum of a simple compound, such as methyl acetate.
3. describe, and sketch a diagram of, a simple NMR spectrometer.
4. explain the difference in time scales of NMR and infrared spectroscopy.
5. predict the number of peaks expected in the 1H or 13C NMR spectrum of a given compound.
Study Notes
Before you go on, make sure that you understand that each signal in the 1H NMR spectrum shown for methyl acetate is due to a different proton environment. The three protons on the same methyl group are equivalent and appear in the spectrum as one signal. However, the two methyl groups are in two different environments (one is more deshielded) and so we see two signals in the whole spectrum (aside from the TMS reference peak).
Methyl acetate has a very simple 1H NMR spectrum, because there is no proton-proton coupling, and therefore no splitting of the signals. In later sections, we discuss splitting patterns in 1H NMR spectra and how they help a chemist determine the structure of organic compounds.
The basics of an NMR experiment
Given that chemically nonequivalent protons have different resonance frequencies in the same applied magnetic field, we can see how NMR spectroscopy can provide us with useful information about the structure of an organic molecule. A full explanation of how a modern NMR instrument functions is beyond the scope of this text, but in very simple terms, here is what happens. First, a sample compound (we'll use methyl acetate) is placed inside a very strong applied magnetic field (B0).
All of the protons begin to precess: the Ha protons at precessional frequency ωa, the Hb protons at ωb.At first, the magnetic moments of (slightly more than) half of the protons are aligned with B0, and half are aligned against B0. Then, the sample is hit with electromagnetic radiation in the radio frequency range. The two specific frequencies which match ωaandωb(i.e. the resonance frequencies) cause those Ha and Hb protons which are aligned with B0 to 'flip' so that they are now aligned against B0. In doing so, the protons absorb radiation at the two resonance frequencies. The NMR instrument records which frequencies were absorbed, as well as the intensity of each absorbance.
In most cases, a sample being analyzed by NMR is in solution. If we use a common laboratory solvent (diethyl ether, acetone, dichloromethane, ethanol, water, etc.) to dissolve our NMR sample, however, we run into a problem – there many more solvent protons in solution than there are sample protons, so the signals from the sample protons will be overwhelmed. To get around this problem, we use special NMR solvents in which all protons have been replaced by deuterium. Recall that deuterium is NMR-active, but its resonance frequency is very different from that of protons, and thus it is `invisible` in 1H-NMR. Some common NMR solvents are shown below.
The Chemical Shift
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, the horizontal axis to frequency (typically the vertical axis is not shown in an NMR spectrum).
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 tetramethylsilane (TMS), a standard reference compound that was added to our sample.
You may be wondering about a few things at this point - why is TMS necessary, and what is the meaning of the `ppm (δ)` label on the horizontal axis? Shouldn't the frequency units be in Hz? Keep in mind that NMR instruments of many different applied field strengths are used in organic chemistry laboratories, and that the proton's resonance frequency range depends on the strength of the applied field. The spectrum above was generated on an instrument with an applied field of approximately 7.1 Tesla, at which strength protons resonate in the neighborhood of 300 million Hz (chemists refer to this as a 300 MHz instrument). If our colleague in another lab takes the NMR spectrum of the same molecule using an instrument with a 2.4 Tesla magnet, the protons will resonate at around 100 million Hz (so we’d call this a 100 MHz instrument). It would be inconvenient and confusing to always have to convert NMR data according to the field strength of the instrument used. Therefore, 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 TMS. This is where the ppm – parts per million – term comes in. 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 (almost all protons in organic molecules have a higher resonance frequency than those in TMS, for reasons we shall explore quite soon).
The two proton groups in our methyl acetate sample are recorded as resonating at frequencies 2.05 and 3.67 ppm higher than TMS. One-millionth (1.0 ppm) of 300 MHz is 300 Hz. Thus 2.05 ppm, on this instrument, corresponds to 615 Hz, and 3.67 ppm corresponds to 1101 Hz. If the TMS protons observed by our 7.1 Tesla instrument resonate at exactly 300,000,000 Hz, this means that the protons in our ethyl acetate samples are resonating at 300,000,615 and 300,001,101 Hz, respectively. Likewise, if the TMS protons in our colleague's 2.4 Tesla instrument resonate at exactly 100 MHz, the methyl acetate protons in her sample resonate at 100,000,205 and 100,000,367 Hz (on the 100 MHz instrument, 1.0 ppm corresponds to 100 Hz). The absolute frequency values in each case are not very useful – they will vary according to the instrument used – but the difference in resonance frequency from the TMS standard, expressed in parts per million, should be the same regardless of the instrument.
Expressed this way, the resonance frequency for a given proton in a molecule is called its chemical shift. A frequently used symbolic designation for chemical shift in ppm is the lower-case Greek letter delta (δ). Most protons in organic compounds have chemical shift values between 0 and 12 ppm from TMS, although values below zero and above 12 are occasionally observed. By convention, the left-hand side of an NMR spectrum (higher chemical shift) is called downfield, and the right-hand direction is called upfield.
In our methyl acetate example we included for illustrative purposes a small amount of TMS standard directly in the sample, as was the common procedure for determining the zero point with older NMR instruments．That practice is generally no longer necessary, as modern NMR instruments are designed to use the deuterium signal from the solvent as a standard reference point, then to extrapolate the 0 ppm baseline that corresponds to the TMS proton signal (in an applied field of 7.1 Tesla, the deuterium atom in CDCl3 resonates at 32 MHz, compared to 300 MHz for the protons in TMS). In the remaining NMR spectra that we will see in this text we will not see an actual TMS signal, but we can always assume that the 0 ppm point corresponds to where the TMS protons would resonate if they were present.
Example
A proton has a chemical shift (relative to TMS) of 4.56 ppm.
1. a) What is its chemical shift, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
2. b) What is its resonance frequency, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
(Assume that in these instruments, the TMS protons resonate at exactly 300 or 200 MHz, respectively)
Solution
Diamagnetic shielding and deshielding
We come now to the question of why nonequivalent protons have different chemical shifts. The chemical shift of a given proton is determined primarily by its immediate electronic environment. Consider the methane molecule (CH4), in which the protons have a chemical shift of 0.23 ppm. The valence electrons around the methyl carbon, when subjected to B0, are induced to circulate and thus generate their own very small magnetic field that opposes B0. This induced field, to a small but significant degree, shields the nearby protons from experiencing the full force of B0, an effect known as local diamagnetic shielding. The methane protons therefore do not experience the full force of B0 - what they experience is called Beff, or the effective field, which is slightly weaker than B0.
Therefore, their resonance frequency is slightly lower than what it would be if they did not have electrons nearby to shield them.
Now consider methyl fluoride, CH3F, in which the protons have a chemical shift of 4.26 ppm, significantly higher than that of methane. This is caused by something called the deshielding effect. Because fluorine is more electronegative than carbon, it pulls valence electrons away from the carbon, effectively decreasing the electron density around each of the protons. For the protons, lower electron density means less diamagnetic shielding, which in turn means a greater overall exposure to B0, a stronger Beff, and a higher resonance frequency. Put another way, the fluorine, by pulling electron density away from the protons, is deshielding them, leaving them more exposed to B0. As the electronegativity of the substituent increases, so does the extent of deshielding, and so does the chemical shift. This is evident when we look at the chemical shifts of methane and three halomethane compounds (remember that electronegativity increases as we move up a column in the periodic table).
To a large extent, then, we can predict trends in chemical shift by considering how much deshielding is taking place near a proton. The chemical shift of trichloromethane is, as expected, higher than that of dichloromethane, which is in turn higher than that of chloromethane.
The deshielding effect of an electronegative substituent diminishes sharply with increasing distance:
The presence of an electronegative oxygen, nitrogen, sulfur, or sp2-hybridized carbon also tends to shift the NMR signals of nearby protons slightly downfield:
Table 2 lists typical chemical shift values for protons in different chemical environments.
Armed with this information, we can finally assign the two peaks in the the 1H-NMR spectrum of methyl acetate that we saw a few pages back. The signal at 3.65 ppm corresponds to the methyl ester protons (Hb), which are deshielded by the adjacent oxygen atom. The upfield signal at 2.05 ppm corresponds to the acetate protons (Ha), which is deshielded - but to a lesser extent - by the adjacent carbonyl group.
Finally, a note on the use of TMS as a standard in NMR spectroscopy: one of the main reasons why the TMS proton signal was chosen as a zero-point is that the TMS protons are highly shielded: silicon is slightly less electronegative than carbon, and therefore donates some additional shielding electron density. Very few organic molecules contain protons with chemical shifts that are negative relative to TMS.
Exercise
3. 2-cholorobutene shows 4 different hydrogen signals. Explain why this is.
Answer
3. The same colors represent the same signal. 4 different colors for 4 different signals. The hydrogen on the alkene would give two different signals. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.02%3A_NMR_Spectra_-_an_introduction_and_overview.txt |
Objectives
After completing this section, you should be able to
1. describe the delta scale used in NMR spectroscopy.
2. perform calculations based on the relationship between the delta value (in ppm), the observed chemical shift (in Hz), and the operating frequency of an NMR spectrometer (in Hz).
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.
In Section 13.9 we discuss 1H NMR chemical shifts in more detail. Although you will eventually be expected to associate the approximate region of a 1H NMR spectrum with a particular type of proton, you are expected to use a general table of 1H NMR chemical shifts such as the one shown in Section 13.9.
Chemical Shifts
The NMR spectra is displayed as a plot of the applied radio frequency versus the absorption. The applied frequency increases from left to right, thus the left side of the plot is the low field, downfield or deshielded side and the right side of the plot is the high field, upfield or shielded side (see the figure below). The concept of shielding will be explained shortly.
The position on the plot at which the nuclei absorbs is called the chemical shift. Since this has an arbitrary value a standard reference point must be used. The two most common standards are TMS (tetramethylsilane, (Si(CH3)4) which has been assigned a chemical shift of zero, and CDCl3 (deuterochloroform) which has a chemical shift of 7.26 for 1H NMR and 77 for 13C NMR.
The scale is commonly expressed as parts per million (ppm) which is independent of the spectrometer frequency. The scale is the delta (δ) scale.
The range at which most NMR absorptions occur is quite narrow. Almost all 1H absorptions occur downfield within 10 ppm of TMS. For 13C NMR almost all absorptions occurs within 220 ppm downfield of the C atom in TMS.
Shielding in NMR
Structural features of the molecule will have an effect on the exact magnitude of the magnetic field experienced by a particular nucleus. This means that H atoms which have different chemical environments will have different chemical shifts. This is what makes NMR so useful for structure determination in organic chemistry. There are three main features that will affect the shielding of the nucleus, electronegativity, magnetic anisotropy of π systems and hydrogen bonding.
Electronegativity
The electrons that surround the nucleus are in motion so they created their own electromagnetic field. This field opposes the the applied magnetic field and so reduces the 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 and result in a larger chemical shift. Compare the 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 increases. This is an effect of the halide atom pulling the electron density away from the methyl group. This exposes the nuclei of both the C and H atoms, "deshielding" the nuclei and shifting the peak downfield.
The effects are cumulative so the presence of more electron withdrawing groups will produce a greater deshielding and therefore a larger chemical shift, i.e.
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 the deshielding can occur further down the chain, i.e.
NMR signal -CH2-CH2-CH2Br
δ (ppm) 1.25 1.69 3.30
Magnetic Anisotropy: Pi 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).
This effect is common for any atoms near a π bond, i.e.
Proton Type Effect Chemical shift (ppm)
C6H5-H highly deshielded 6.5 - 8
C=C-H deshielded 4.5 - 6
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
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 as well as solvation effects, acidity, concentration and temperature make it very difficult to predict the chemical shifts for these atoms.
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
4. The following peaks were from a H1 NMR spectra from a 400 MHz spectrometer. Convert to δ units
A. CHCl3 1451 Hz
B. CH3Cl 610 Hz
C. CH3OH 693 Hz
D. CH2Cl2 1060 Hz
5. 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
4.
A. 3.627 ppm
B. 1.525 ppm
C. 1.732 ppm
D. 2.65 ppm
5.
A. Since TMS is at 0 δ = 0 Hz for reference, the difference between the two would be 630 Hz
B. No not a different chemical shift, but a different frequency would be seen, 840 Hz
C. 840 Hz | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.03%3A_Chemical_Shifts_and_Shielding.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
Study Notes
It is important at this stage to be able to identify equivalent protons in any organic compound given the structure of that compound. Once you know the number of different groups of equivalent protons in a compound, you can predict the number (before coupling) and relative strength of signals. Look at the following examples and make sure you understand how the number and intensity ratio of signals are derived from the structure shown.
Structure Number of Signals Ratio of Signals
CH3OCH2CH2Br 3 A : B : C 3 : 2 : 2
1
3 A : B : C 2 : 2 : 6 (or 1 : 1 : 3)
3 A : B : C 2 : 4 : 2 (or 1 : 2 : 1)
4 A : B : C : D 3 : 2 : 2 : 3
5 A : B : C : D : E 3 : 1 : 1 : 1 : 1
If all protons in all organic molecules had the same resonance frequency in an external magnetic field of a given strength, the information in the previous paragraph would be interesting from a theoretical standpoint, but would not be terribly useful to organic chemists. Fortunately for us, however, 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, for example, there are two ‘sets’ 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.
The ability to recognize chemical equivalancy and nonequivalency among atoms in a molecule will be central to understanding NMR. In each of the molecules below, all protons are chemically equivalent, and therefore will have the same resonance frequency in an NMR experiment.
You might expect that the equitorial 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 equitorial 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 equitorial 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.
Each the molecules in the next figure contains two sets of protons, just like our previous example of methyl acetate, and again in each case the resonance frequency of the Ha protons will be different from that of the Hb protons.
Notice how the symmetry of para-xylene results in there being only two different sets of protons.
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.
When stereochemistry is taken into account, the issue of equivalence vs 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 non-equivalent and have different resonance frequencies. In the alkene and cyclohexene structures below, for example, Ha is trans to the chlorine substituent, while Hb is cis to chlorine.
What is not so intuitive is that diastereotopic hydrogens (section 3.10) on chiral molecules are also non-equivalent:
However, enantiotopic and homotopic hydrogens are chemically equivalent.
Example
How many different sets of protons do the following molecules contain? (count diastereotopic protons as non-equivalent).
Solution
Exercise
6. How many non-equivalent hydrogen are in the following molecules; how many different signals will you see in a H1 NMR spectrum.
A. CH3CH2CH2Br
B. CH3OCH2C(CH3)3
C. Ethyl Benzene
D. 2-methyl-1-hexene
Answer
6. A. 3; B. 3; C. 5; D. 7 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.04%3A_H_NMR_Spectroscopy_and_Proton_Equivalence.txt |
Objectives
After completing this section, you should be able to
1. state the approximate chemical shift (δ) for the following types of protons:
1. aromatic.
2. vinylic.
3. those bonded to carbon atoms which are in turn bonded to a highly electronegative element.
4. those bonded to carbons which are next to unsaturated centres.
5. those bonded to carbons which are part of a saturated system.
2. predict the approximate chemical shifts of each of the protons in an organic compound, given its structure and a table of chemical shift correlations.
Study Notes
You should not attempt to memorize the chemical shifts listed in the table of this section, although it is probable that you will need to refer to it quite frequently throughout the remainder of this course. To fulfil Objective 1, above, you should be familiar with the information presented in the figure of chemical shift ranges for organic compounds. 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. Notice that we shall not try to understand why aromatic protons are deshielded or why alkynyl protons are not deshielded as much as vinylic protons. These phenonomena can be explained, but the focus is on the interpretation of 1H NMR spectra, not on the underlying theory.
1H NMR Chemical Shifts
Chemical shift is associated with the Larmor frequency of a nuclear spin to its chemical environment. Tetramethylsilan[TMS;(CH3)4Si] is generally used for standard to determine chemical shift of compounds: δTMS=0ppm. In other words, frequencies for chemicals are measured for a 1H or 13C nucleus of a sample from the 1H or 13C resonance of TMS. It is important to understand trend of chemical shift in terms of NMR interpretation. The proton NMR chemical shift is affect by nearness to electronegative atoms (O, N, halogen.) and unsaturated groups (C=C,C=O, aromatic). Electronegative groups move to the down field (left; increase in ppm). Unsaturated groups shift to downfield (left) when affecting nucleus is in the plane of the unsaturation, but reverse shift takes place in the regions above and below this plane. 1H chemical shift play a role in identifying many functional groups. Figure 1. indicates important example to figure out the functional groups.
Chemical shift values are in parts per million (ppm) relative to tetramethylsilane.
Hydrogen type
Chemical shift (ppm)
RCH3
0.9 - 1.0
RCH2R
1.2 - 1.7
R3CH
1.5 – 2.0
2.0 – 2.3
1.5 – 1.8
RNH2
1 - 3
ArCH3
2.2 – 2.4
2.3 – 3.0
ROCH3
3.7 – 3.9
3.7 – 3.9
ROH
1 - 5
3.7 – 6.5
5 - 9
ArH
6.0 – 8.7
9.5 – 10.0
10 - 13
Exercise
7. The following have one H1 NMR peak. In each case predict approximately where this peak would be in a spectra.
8. Identify the different equivalent protons in the following molecule and predict their expected chemical shift.
Answer
7. A. 5.20 δ; B. 1.50 δ; C. 6.40 δ; D. 1.00 δ
8. There are 6 different protons in this molecule
The shifts are (close) to the following: (a) 2 δ; (b) 6 δ; (c) 6.5 δ; (d) 7 δ; (e) 7.5 δ; (f) 7 δ | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.05%3A_Functional_Groups_and_Chemical_Shifts_in_H_NMR__Spect.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.
Study Notes
The concept of peak integration is that the area of a given peak in a 1H NMR spectrum is proportional to the number of (equivalent) protons giving rise to the peak. Thus, a peak which is caused by a single, unique proton has an area which measures one third of the area of a peak resulting from a methyl (CH3) group in the same spectrum.
In practice, we do not have to measure these areas ourselves: it is all done electronically by the spectrometer, and an integration curve is superimposed on the rest of the spectrum. The integration curve appears as a series of steps, with the height of each step being proportional to the area of the corresponding absorption peak, and consequently, to the number of protons responsible for the absorption.
As it can be difficult to decide precisely where to start and stop when measuring integrations, you should not expect your ratios to be exact whole numbers.
Signal integration
The computer in an NMR instrument can be instructed to automatically integrate the area under a signal or group of signals. This is very useful, because in 1H-NMR spectroscopy the area under a signal is proportional to the number of hydrogens to which the peak corresponds. The two signals in the methyl acetate spectrum, for example, integrate to approximately the same area, because they both correspond to a set of three equivalent protons.
Take a look next at the spectrum of para-xylene (IUPAC name 1,4-dimethylbenzene):
This molecule has two sets of protons: the six methyl (Ha) protons and the four aromatic (Hb) protons. 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 (along with the actual chemical shift values, which we'll discuss soon) tells us which set of protons corresponds to which NMR signal.
The integration function 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.
Example \(1\)
You take a 1H-NMR spectrum of a mixed sample of acetone (CH3(CO)CH3) and dichloromethane (CH2Cl2). The integral ratio of the two signals (acetone : dichloromethane) is 2.3 to 1. What is the molar ratio of the two compounds in the sample?
Example \(2\)
You take the 1H-NMR spectrum of a mixed sample of 36% para-xylene and 64% acetone in CDCl3 solvent (structures are shown earlier in this chapter). 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)
Exercise
9. Predict how many signals the following molecule would have? Sketch the spectra and estimate the integration of the peaks.
Answer
9. There will be two peaks. Ideal general spectrum shown with integration. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.06%3A_Integration_of_H_NMR_Absorptions-_Proton_Counting.txt |
Objectives
After completing this section, you should be able to
1. explain the spin-spin splitting pattern observed in the 1H NMR spectrum of a simple organic compound, such as chloroethane or 2-bromopropane.
2. interpret the splitting pattern of a given 1H NMR spectrum.
3. determine the structure of a relatively simple organic compound, given its 1H NMR spectrum and other relevant information.
4. use coupling constants to determine which groups of protons are coupling with one another in a 1H NMR spectrum.
5. 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
Study Notes
From what we have learned about 1H NMR spectra so far, we might predict that the spectrum of 1,1,2-trichloroethane, CHCl2CH2Cl, would consist of two peaks—one, at about 2.5-4.0 δ, expected for CH2-halogen compounds 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 it appears to be much more complex. True, we see absorptions in the regions we predicted, but these absorptions appear as a group of two peaks (a doublet) and a group of three peaks (a triplet). This complication, which may be disturbing to a student who longs for the simple life, 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.
Spin-spin coupling is often one of the more challenging topics for organic chemistry students to master. Remember the n + 1 rule and the associated coupling patterns.
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. Rather than being a complication, however, 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, we show enlargements of individual signals 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 Hb signal 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 trichloromethane example, the Ha and Hb protons are spin-coupled to each other. Here's how it works, looking first at the Ha signal: in addition to being shielded by nearby valence electrons, each of the Ha protons is also influenced by the small magnetic field generated by Hb next door (remember, each spinning proton is like a tiny magnet). The magnetic moment of Hb will be aligned with B0 in (slightly more than) half of the molecules in the sample, while in the remaining half of the molecules it will be opposed to B0. The Beff ‘felt’ by Ha is a slightly weaker if Hb is aligned against B0, or slightly stronger if Hb is aligned with B0. In other words, in half of the molecules Ha is shielded by Hb (thus the NMR signal is shifted slightly upfield) and in the other half Ha is deshielded by Hb(and the NMR signal shifted slightly downfield). What would otherwise be a single Ha peak has been split into two sub-peaks (a doublet), one upfield and one downfield of the original signal. These ideas an be illustrated by a splitting diagram, as shown below.
Now, let's think about the Hbsignal. 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. Third and fourth, Ha1 could be with B0 and Ha2 opposed, or Ha1opposed 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.
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.
Now, consider the spectrum for ethyl acetate:
We see an unsplit 'singlet' peak at 1.833 ppm that corresponds to the acetyl (Ha) hydrogens – this is similar to the signal for the acetate hydrogens in methyl acetate that we considered earlier. This signal is unsplit because there are no adjacent hydrogens on the molecule. 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 Hbhydrogens give rise to a quartet signal at 3.915 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.
Example
1. Explain, using left and right arrows to illustrate the possible combinations of nuclear spin states for the Hc hydrogens, why the Hb signal in ethyl acetate is split into a quartet.
2. The integration ratio of doublets is 1:1, and of triplets is 1:2:1. What is the integration ratio of the Hb quartet in ethyl acetate? (Hint – use the illustration that you drew in part a to answer this question.)
Solution
By now, you probably have recognized the 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. Thus the two Hb hydrogens in ethyl acetate split the Hc signal into a triplet, and the three Hc hydrogens split the Hb signal into a quartet. 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 `neighbors`. 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, 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 details about how we quantify coupling interactions is provided in section 5.5B). 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.
Below are a few more examples of chemical shift and splitting pattern information for some relatively simple organic molecules.
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
Example
How many proton signals would you expect to see in the 1H-NMR spectrum of triclosan (a common antimicrobial agent found in detergents)? For each of the proton signals, predict the splitting pattern. Assume that you see only 3-bond coupling.
Solutions
Example
Predict the splitting pattern for the 1H-NMR signals corresponding to the protons at the locations indicated by arrows (the structure is that of the neurotransmitter serotonin).
Solutions
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, between two adjacent sub-peaks in a split signal. For our doublet in the 1,1,2-trichloroethane spectrum, for example, the two 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.
When we look closely at the triplet signal in 1,1,2-trichloroethane, we see that the coupling constant - the `gap` between subpeaks - is 6.1 Hz, the same as for the doublet. This is an important concept! 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, this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets.
Coupling constants between proton sets on neighboring sp3-hybridized carbons is typically in the region of 6-8 Hz. With protons bound to sp2-hybridized carbons, coupling constants can range from 0 Hz (no coupling at all) to 18 Hz, depending on the bonding arrangement.
For vinylic hydrogens in a trans configuration, we see coupling constants in the range of 3J = 11-18 Hz, while cis hydrogens couple in the 3J = 6-15 Hz range. The 2-bond coupling between hydrogens bound to the same alkene carbon (referred to as geminal hydrogens) is very fine, generally 5 Hz or lower. Ortho hydrogens on a benzene ring couple at 6-10 Hz, while 4-bond coupling of up to 4 Hz is sometimes seen between meta hydrogens.
Fine (2-3 Hz) coupling is often seen between an aldehyde proton and a three-bond neighbor. Table 4 lists typical constant values.
Exercise
10. Predict the splitting patterns of the following molecules:
11. Draw the following according to the criteria given.
A. C3H5O; two triplet, 1 doublet
B. C4H8O2; three singlets
C. C5H12; one singlet
12. The following spectrum is for C3H8O. Determine the structure.
A triplet; B singlet; C sextet; D triplet
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 3 December 2016)
Answer
10.
A. H: Doublet. H: Septet
B. H: Doublet, H: Triplet
C. H: Singlet, H: Quartet, H: Triplet
11.
These are just some drawings, more may be possible.
12.
Note: Remember, chemically equivalent protons do not couple with one another to give spin-spin splitting. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.07%3A_Spin-Spin_Splitting_in_H_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.
Key Terms
Make certain that you can define, and use in context, the key term below.
• tree diagram
Study Notes
We saw the effects of spin-spin coupling on the appearance of a 1H NMR signal. These effects can be further complicated when that signal is coupled to several different protons. For example, BrCH2CH2CH2Cl would produce three signals. The hydrogens at C1 and C3 would each be triplets because of coupling to the two hydrogens on C2. However, the hydrogen on C2 “sees” two different sets of neighbouring hydrogens, and would therefore produce a triplet of triplets.
Another effect that can complicate a spectrum is the “closeness” of signals. If signals accidently overlap they can be difficult to identify. In the example above, we expected a triplet of triplets. However, if the coupling is identical (or almost identical) between the hydrogens on C2 and the hydrogens on both C1 and C3, one would observe a quintet in the 1H NMR spectrum. [You can try this yourself by drawing a tree diagram of a triplet of triplets assuming, first, different coupling constants, and then, identical coupling constants.] 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:
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.
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.
In examinations, you will be given a range of information (IR, MS, UV data and empirical formulae) to aid you with your structural determination using 1H NMR spectroscopy. For example, you may be asked to determine the structure of C6H12O given the following spectra:
Infrared spectrum: 3000 cm−1 and 1720 cm−1 absorptions are both strong
1H NMR δ (ppm) Protons Multiplicity
0.87 6 doublet
1.72 1 broad multiplet
2.00 3 singlet
2.18 2 doublet
To answer this question, you note that the infrared spectrum of C6H12O shows \$\ce{\sf{C-H}}\$ stretching (3000 cm−1) and \$\ce{\sf{C-O}}\$ stretching (1720 cm−1). Now you have to piece together the information from the 1H NMR spectrum. Notice the singlet with three protons at 2.00 ppm. This signal indicates a methyl group that is not coupled to other protons. It could possibly mean the presence of a methyl ketone functional group.
The signal at 1.72 ppm is a broad multiplet, suggesting that a carbon with a single proton is beside carbons with several different protons.
The doublet signal at 2.18 ppm implies that a \$\ce{\sf{-CH2-}}\$ group is attached to a carbon having only one proton.
The six protons showing a doublet at 0.87 ppm indicate two equivalent methyl groups attached to a carbon with one proton.
Whenever you see a signal in the 0.7-1.3 ppm range that is a multiplet of three protons (3, 6, 9) it is most likely caused by equivalent methyl groups.
Using trial and error, and with the above observations, you should come up with the correct structure.
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. 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.
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 Hbsignal:
Example
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
When constructing a splitting diagram to analyze complex coupling patterns, it is usually easier to show the larger splitting first, followed by the finer splitting (although the reverse would give the same end result).
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. In the spectrum of 1,1,3-trichloropropane, for example, we would expect the signal for Hb to be split into a triplet by Ha, and again into doublets by Hc, resulting in a 'triplet of doublets'.
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.
For similar reasons, the Hc peak in the spectrum of 2-pentanone appears as a sextet, split by the five combined Hb and Hd protons. Technically, this 'sextet' could be considered to be a 'triplet of quartets' with overlapping sub-peaks.
Example
What splitting pattern would you expect for the signal coresponding to Hb in the molecule below? Assume that Jab ~ Jbc. Draw a splitting diagram for this signal, and determine the relative integration values of each subpeak.
Solution
In many cases, it is difficult to fully analyze a complex splitting pattern. In the spectrum of toluene, for example, if we consider only 3-bond coupling we would expect the signal for Hb to be a doublet, Hd 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 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!)
Practice Unknowns
1. Given the information below, draw the structures of compounds A through D.
1. An unknown compound A was prepared as follows:
Mass spectrum:
base peak m/e = 39
parent peak m/e = 54
1H NMR spectrum:
δ (ppm) Relative Area Multiplicity
1.0 2 triplet
5.4 1 quintet
2. Unknown compound B has the molecular formula C7H6O2.
Infrared spectrum:
3200 cm−1 (broad) and 1747 cm−1 (strong) absorptions
1H NMR spectrum:
δ (ppm) Protons
6.9 2
7.4 2
9.8 1
10.9 1
Hint: Aromatic ring currents deshield all proton signals just outside the ring.
3. Unknown compound C shows no evidence of unsaturation and contains only carbon and hydrogen.
Mass spectrum:
parent peak m/e = 68
1H NMR spectrum:
δ (ppm) Relative Area Multiplicity
1.84 3 triplet
2.45 1 septet
Hint: Think three dimensionally!
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
Answers
Exercise
13. In the following molecule, the C2 is coupled with both the vinyl, C1, and the alkyl C3. Draw the splitting tree diagram.
Answer
13. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.08%3A_More_Complex_Spin-Spin_Splitting_Patterns.txt |
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.
One of the most complicated problems to deal with is the analysis of a mixture. This situation is not uncommon when students run reactions in lab and analyse the data.
• Sometimes the spectra show 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.
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, 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 structure. There must be six times as many acetone molecules as benzene molecules in the sample.
There are six protons in the benzene, and they should all show up near 7 ppm. There are six protons in acetone, and 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 benzenes.
Similarly, if you have decided that you can identify two sets of peaks in the 1H spectrum, analysing 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.
However, 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.
We will look at two examples of sample mixtures that could arise in lab. Results like these are pretty common events in the labIn the first example, a student tried to carry out the following reaction, a borohydride reduction of an aldehyde. The borohydride should give a hydride anion to the C=O carbon; washing with water should then supply a proton to the oxygen, giving an alcohol.
Her reaction produced the following spectrum.
(simulated data)
From this data, she produced the table below.
Notice how she calculated that ratio. She found a peak in molecule 1, the aldehyde, that she was pretty sure corresponded to the aldehydic hydrogen, the H attached to the C=O; in other words, the CH=O. She found another peak from molecule 2, the alcohol, that she was pretty sure represented the two hydrogens on the carbon attached to oxygen, the CH2-O.
The integrals for those two peaks are equal. They are both 2H in her table. However, she notes that within each molecule, the first integral really represents 1H and the second represents 2H. That means there must be twice as many of molecule 1 as there are molecule 2. That way, there would be 2 x CH=O, and its integral would be the same as the 1 x CH2-O in the other molecule.
One way to approach this kind of problem is to:
• choose one peak from each of the two compounds you want to compare.
• decide how many hydrogens each peak is supposed to represent in a molecule. Is it supposed to be a CH2, a CH, a CH3?
• divide the integral value for that peak by that number of hydrogens it is supposed to represent in a molecule.
• compare the two answers (integral A / ideal # H) vs (integral B / ideal # H).
• the ratio of those two answers is the ratio of the two molecules in the sample.
So there is twice as much aldehyde as alcohol in the mixture. In terms of these two compounds alone, she has 33% alcohol and 66% aldehyde. That's ( 1/(1+2) ) x100% for the alcohol, and ( 2/(1+2) ) x100% for the aldehyde. That calculation just represents the amount of individual component divided by the total of the components she wants to compare.
There are a number of things to take note of here.
• Her reaction really didn't work very well. She still has majority starting material, not product.
• She will get a good grade on this lab. Although the experiment didn't work well, she has good data, and she has analyzed it very clearly.
• She has separated her data table into different sections for different compounds. Sometimes that makes it easier to analyze things.
• She has noted the actual integral data (she may have measured the integral with a ruler) and also converted it into a more convenient ratio, based on the integral for a peak that she felt certain about.
• She went one step further, and indicated the internal integration ratio within each individual compound.
• She calculated the % completion of the reaction using the integral data for the reactant and product, and she made clear what part of the data she used for that calculation. A similar procedure could be done if a student were just trying to separate two components in a mixture rather than carry out a reaction.
• She also calculated the overall purity of the mixture, including a solvent impurity that she failed to remove.
• However, CHCl3 is not included in her analysis of purity. CHCl3 really isn't part of her sample; it was just present in the NMR solvent, so it doesn't represent anything in the material she ended up with at the end of lab.
Another student carried out a similar reaction, shown below. He also finished the reaction by washing with water, but because methanol is soluble in water, he had to extract his product out of the water. He chose to use dichloromethane for that purpose.
He obtained the following data.
From this data, he constructed the following table.
There are some things to learn about this table, too.
• Does the integration ratio really match the integral data? Or is this just wishful thinking?
• This table might reflect what he wants to see in the data. But what else could be in the data?
• CHCl3 is often seen in NMR spectra if CDCl3 is used for the NMR sample. It's there, at 7.2 ppm.
• "Leftover" or residual solvent is very common in real lab data. There it is, CH2Cl2 from the extraction, at 5.4 ppm.
• What about water? Sometimes people don't dry their solutions properly before evaporating the solvent. There is probably water around 1.5 to 1.6 ppm here.
This student might not get a very good grade; the sample does not even show up in the spectrum, so he lost it somewhere. But his analysis is also poor, so he will really get a terrible grade.
Example
Three students performed a synthesis of a fragrant ester, ethyl propanoate, CH3CH2CO2CH2CH3. During their reactions, they each used a different solvent. The students were able to see peaks in the NMR spectrum for ethyl propanoate, as well as peaks for chloroform (CHCl3, in the CDCl3 they used to make their NMR samples).
• See the first student's spectrum.
• See the second student's spectrum.
• See the third student's spectrum.
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).
1. What is the ratio of leftover solvent to ethyl propanoate in each sample?
2. What is the percent of each sample that is leftover solvent
Exercise
14. How can H1 NMR determine products? For example, how can you tell the difference between the products of this reaction?
Answer
14. Yes, you are able to determine the difference in the spectra. For the 2-chloro compound will have multiple quartets while the 1-chloro compound will only have a quintet and a triplet for the signals in the ring. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.09%3A_Uses_of_H_NMR_Spectroscopy.txt |
The basics of 13C-NMR spectroscopy
The magnetic moment of a 13C nucleus is much weaker than that of a proton, meaning that NMR signals from 13C nuclei are inherently much weaker than proton signals. This, combined 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. 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. Peak integration is generally not useful in 13C-NMR spectroscopy, except when investigating molecules that have been enriched with 13C isotope (see section 5.6B).
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.
Example \(1\)
How many sets of non-equivalent carbons are there in each of the molecules shown in exercise 5.1?
Example \(2\)
How many sets of non-equivalent carbons are there in:
1. toluene
2. 2-pentanone
3. para-xylene
4. triclosan
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, and thus we do not see spin-spin coupling between neighboring carbons in a 13C-NMR spectrum. There is, however, 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. For clarity, chemists generally use a technique called broadband decoupling, which essentially 'turns off' C-H coupling, resulting in a spectrum in which all carbon signals are singlets. Below is the proton-decoupled13C-NMR spectrum of ethyl acetate, showing the expected four signals, one for each of the carbons.
While broadband decoupling results in a much simpler spectrum, useful information about the presence of neighboring protons is lost. However, another modern NMR technique called DEPT (Distortionless Enhancement by Polarization Transfer) allows us to determine how many hydrogens are bound to each carbon. For example, a DEPT experiment tells us that the signal at 171 ppm in the ethyl acetate spectrum is a quaternary carbon (no hydrogens bound, in this case a carbonyl carbon), that the 61 ppm signal is from a methylene (CH2) carbon, and that the 21 ppm and 14 ppm signals are both methyl (CH3) carbons. The details of the DEPT experiment are beyond the scope of this text, but DEPT information will often be provided along with 13C spectral data in examples and problems.
12.11: Chemical Shifts and Interpreting C NMR Spectra
13C NMR Chemical Shifts
The Carbon NMR is used for determining functional groups using characteristic shift values. 13C chemical shift is affect by electronegative effect and steric effect. If an H atoms in an alkane is replace by substituent X, electronegative atoms (O, N, halogen), ?-carbon and ?-carbon shift to downfield (left; increase in ppm) while ?-carbon shifts to upfield. The steric effect is observed in acyclic and clyclic system, which leads to downshifted chemical shifts. Figure 9 shows typical 13C chemical shift regions of the major chemical class.
Spin-Spin splitting
Comparing the 1H NMR, there is a big difference thing in the 13C NMR. The 13C-13Cspin-spin splitting rarely exit between adjacent carbons because 13C is naturally lower abundant (1.1%)
• 13C-1H Spin coupling: 13C-1H Spin coupling provides useful information about the number of protons attached a carbon atom. In case of one bond coupling (1JCH), -CH, -CH2, and CH3 have respectively doublet, triplet, quartets for the 13C resonances in the spectrum. However, 13C-1H Spin coupling has an disadvantage for 13C spectrum interpretation. 13C-1H Spin coupling is hard to analyze and reveal structure due to a forest of overlapping peaks that result from 100% abundance of 1H.
• Decoupling: Decoupling is the process of removing 13C-1H coupling interaction to simplify a spectrum and identify which pair of nuclei is involved in the J coupling. The decoupling 13C spectra shows only one peak(singlet) for each unique carbon in the molecule(Fig 10.). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.
Fig 10. Decoupling in the 13C NMR | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.10%3A_C_NMR_Spectroscopy.txt |
Distortions Enhancement by Polarization Transfer (DEPT)
DEPT is used for distinguishing between a CH3 group, a CH2 group, and a CH group. The proton pulse is set at 45°, 90°, or 135° in the three separate experiments. The different pulses depend on the number of protons attached to a carbon atom. Fig 11. is an example about DEPT spectrum.
Fig 11. DEPT spectrum of n-isobutlybutrate
While broadband decoupling results in a much simpler spectrum, useful information about the presence of neighboring protons is lost. However, another modern NMR technique called DEPT (Distortionless Enhancement by Polarization Transfer) allows us to determine how many hydrogens are bound to each carbon. For example, a DEPT experiment tells us that the signal at 171 ppm in the ethyl acetate spectrum is a quaternary carbon (no hydrogens bound, in this case a carbonyl carbon), that the 61 ppm signal is from a methylene (CH2) carbon, and that the 21 ppm and 14 ppm signals are both methyl (CH3) carbons. The details of the DEPT experiment are beyond the scope of this text, but DEPT information will often be provided along with 13C spectral data in examples and problems.
Below are two more examples of 13C NMR spectra of simple organic molecules, along with DEPT information.
Example 13.5.2
Give peak assignments for the 13C-NMR spectrum of methyl methacrylate, shown above.
Solution
One of the greatest advantages of 13C-NMR compared to 1H-NMR is the breadth of the spectrum - recall that 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, and we can almost always distinguish separate peaks for each carbon, even in a relatively large compound containing carbons in very similar environments. In the proton spectrum of 1-heptanol, for example, only the signals for the alcohol proton (Ha) and the two protons on the adjacent carbon (Hb) are easily analyzed. The other proton signals overlap, making analysis difficult.
In the 13C spectrum of the same molecule, however, 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 13.5.3
13C-NMR (and DEPT) data for some common biomolecules are shown below (data is from the Aldrich Library of 1H and 13C NMR). Match the NMR data to the correct structure, and make complete peak assignments.
• spectrum a: 168.10 ppm (C), 159.91 ppm (C), 144.05 ppm (CH), 95.79 ppm (CH)
• spectrum b: 207.85 ppm (C), 172.69 ppm (C), 29.29 ppm (CH3)
• spectrum c: 178.54 ppm (C), 53.25 ppm (CH), 18.95 ppm (CH3)
• spectrum d: 183.81 ppm (C), 182. 63 ppm (C), 73.06 ppm (CH), 45.35 ppm (CH2)
Solution
13C NMR Chemical Shifts
The Carbon NMR is used for determining functional groups using characteristic shift values. 13C chemical shift is affect by electronegative effect and steric effect. If an H atoms in an alkane is replace by substituent X, electronegative atoms (O, N, halogen), ?-carbon and ?-carbon shift to downfield (left; increase in ppm) while ?-carbon shifts to upfield. The steric effect is observed in acyclic and clyclic system, which leads to downshifted chemical shifts. Figure 9 shows typical 13C chemical shift regions of the major chemical class.
Spin-Spin splitting
Comparing the 1H NMR, there is a big difference thing in the 13C NMR. The 13C-13Cspin-spin splitting rarely exit between adjacent carbons because 13C is naturally lower abundant (1.1%)
• 13C-1H Spin coupling: 13C-1H Spin coupling provides useful information about the number of protons attached a carbon atom. In case of one bond coupling (1JCH), -CH, -CH2, and CH3 have respectively doublet, triplet, quartets for the 13C resonances in the spectrum. However, 13C-1H Spin coupling has an disadvantage for 13C spectrum interpretation. 13C-1H Spin coupling is hard to analyze and reveal structure due to a forest of overlapping peaks that result from 100% abundance of 1H.
• Decoupling: Decoupling is the process of removing 13C-1H coupling interaction to simplify a spectrum and identify which pair of nuclei is involved in the J coupling. The decoupling 13C spectra shows only one peak(singlet) for each unique carbon in the molecule(Fig 10.). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.
Fig 10. Decoupling in the 13C NMR | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.12%3A_C_NMR_Spectroscopy_and_DEPT.txt |
The interpretation of 13C NMR spectra does not form a part of Chemistry 350; hence, you may omit Section 13.7. Interested students may wish to read this section for enrichment purposes.
Features of a C-13 NMR spectrum
Butane shows two different peaks in the 13C NMR spectrum, below. Note that: the chemical shifts of these peaks are not very different from methane. The carbons in butane are in a similar environment to the one in methane.
• there are two distinct carbons in butane: the methyl, or CH3, carbon, and the methylene, or CH2, carbon.
• the methyl carbon absorbs slightly upfield, or at lower shift, around 10 ppm.
• the methylene carbon absorbs at slightly downfield, or at higher shift, around 20 ppm.
• other factors being equal, methylene carbons show up at slightly higher shift than methyl carbons.
In the 13C NMR spectrum of pentane (below), you can see three different peaks, even though pentane just contains methyl carbons and methylene carbons like butane. As far as the NMR spectrometer is concerned, pentane contains three different kinds of carbon, in three different environments. That result comes from symmetry.
Symmetry is an important factor in spectroscopy. Nature says:
• atoms that are symmetry-inequivalent can absorb at different shifts.
• atoms that are symmetry-equivalent must absorb at the same shift.
To learn about symmetry, take a model of pentane and do the following:
• make sure the model is twisted into the most symmetric shape possible: a nice "W".
• choose one of the methyl carbons to focus on.
• rotate the model 180 degrees so that you are looking at the same "W" but from the other side.
• note that the methyl you were focusing on has simply switched places with the other methyl group. These two carbons are symmetry-equivalent via two-fold rotation.
Animation NMR1. A three-dimensional model of pentane. Grab the model with the mouse and rotate it so that you are convinced that the second and fourth carbons are symmetry-equivalent, but the third carbon is not.
By the same process, you can see that the second and fourth carbons along the chain are also symmetry-equivalent. However, the middle carbon is not; it never switches places with the other carbons if you rotate the model. There are three different sets of inequivalent carbons; these three groups are not the same as each other according to symmetry.
Example \(1\)
Determine how many inequivalent carbons there are in each of the following compounds. How many peaks do you expect in each 13C NMR spectrum?
Practically speaking, there is only so much room in the spectrum from one end to the other. At some point, peaks can get so crowded together that you can't distinguish one from another. You might expect to see ten different peaks in eicosane, a twenty-carbon alkane chain, but when you look at the spectrum you can only see seven different peaks. That may be frustrating, because the experiment does not seem to agree with your expectation. However, you will be using a number of methods together to minimize the problem of misleading data.
The C-13 NMR spectrum for ethanol
This is a simple example of a C-13 NMR spectrum. Don't worry about the scale for now - we'll look at that in a minute.
Note
Note: The NMR spectra on this page have been produced from graphs taken from the Spectral Data Base System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan.
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. The two lines are in different places in the NMR spectrum because they need different external magnetic fields to bring them in to resonance at a particular radio frequency.
The C-13 NMR spectrum for a more complicated compound
This is the C-13 NMR spectrum for 1-methylethyl propanoate (also known as isopropyl propanoate or isopropyl propionate).
This time there are 5 lines in the spectrum. That means that there must be 5 different environments for the carbon atoms in the compound. Is that reasonable from the structure?
Well - if you count the carbon atoms, there are 6 of them. So why only 5 lines? In this case, two of the carbons are in exactly the same environment. They are attached to exactly the same things. Look at the two CH3 groups on the right-hand side of the molecule.
You might reasonably ask why the carbon in the CH3 on the left is not also in the same environment. Just like the ones on the right, the carbon is attached to 3 hydrogens and another carbon. But the similarity is not exact - you have to chase the similarity along the rest of the molecule as well to be sure.
The carbon in the left-hand CH3 group is attached to a carbon atom which in turn is attached to a carbon with two oxygens on it - and so on down the molecule. That's not exactly the same environment as the carbons in the right-hand CH3 groups. They are attached to a carbon which is attached to a single oxygen - and so on down the molecule. We'll look at this spectrum again in detail on the next page - and look at some more similar examples as well. This all gets easier the more examples you look at.
For now, all you need to realize is that each line in a C-13 NMR spectrum recognizes a carbon atom in one particular environment in the compound. If two (or more) carbon atoms in a compound have exactly the same environment, they will be represented by a single line.
Note
You might wonder why all this works, since only about 1% of carbon atoms are C-13. These are the only ones picked up by this form of NMR. If you had a single molecule of ethanol, then the chances are only about 1 in 50 of there being one C-13 atom in it, and only about 1 in 10,000 of both being C-13.
But you have got to remember that you will be working with a sample containing huge numbers of molecules. The instrument can pick up the magnetic effect of the C-13 nuclei in the carbon of the CH3 group and the carbon of the CH2 group even if they are in separate molecules. There's no need for them to be in the same one. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.13%3A_Uses_of_C_NMR_Spectroscopy.txt |
Additional NMR Examples
For each molecule, predict the number of signals in the 1H-NMR and the 13C-NMR spectra (do not count split peaks - eg. a quartet counts as only one signal). Assume that diastereotopic groups are non-equivalent.
P5.2: For each of the 20 common amino acids, predict the number of signals in the proton-decoupled 13C-NMR spectrum.
P5.3: Calculate the chemical shift value (expressed in Hz, to one decimal place) of each sub-peak on the 1H-NMR doublet signal below. Do this for:
a) a spectrum obtained on a 300 MHz instrument
b) a spectrum obtained on a 100 MHz instrument
P5.4: Consider a quartet signal in an 1H-NMR spectrum obtained on a 300 MHz instrument. The chemical shift is recorded as 1.7562 ppm, and the coupling constant is J = 7.6 Hz. What is the chemical shift, expressed to the nearest 0.1 Hz, of the furthest downfield sub-peak in the quartet? What is the resonance frequency (again expressed in Hz) of this sub-peak?)
P5.5: One easily recognizable splitting pattern for the aromatic proton signals from disubstituted benzene structures is a pair of doublets. Does this pattern indicate ortho, meta, or para substitution?
P5.6 :Match spectra below to their corresponding structures A-F.
Structures:
Spectrum 1
δ
splitting
integration
4.13
q
2
2.45
t
2
1.94
quintet
1
1.27
t
3
Spectrum 2
δ
splitting
integration
3.68
s
3
2.99
t
2
1.95
quintet
1
Spectrum 3
δ
splitting
integration
4.14
q
1
2.62
s
1
1.26
t
1.5
Spectrum 4
δ
splitting
integration
4.14
q
4
3.22
s
1
1.27
t
6
1.13
s
9
Spectrum 5
δ
splitting
integration
4.18
q
1
1.92
q
1
1.23
t
1.5
0.81
t
1.5
Spectrum 6
δ
splitting
integration
3.69
s
1.5
2.63
s
1
P5.7: Match spectra 7-12 below to their corresponding structures G-L .
Structures:
Spectrum 7:
δ
splitting
integration
9.96
d
1
5.88
d
1
2.17
s
3
1.98
s
3
Spectrum 8:
δ
splitting
integration
9.36
s
1
6.55
q
1
2.26
q
2
1.99
d
3
0.96
t
3
Spectrum 9:
δ
splitting
integration
9.57
s
1
6.30
s
1
6.00
s
1
1.84
s
3
Spectrum 10:
δ
splitting
integration
9.83
t
1
2.27
d
2
1.07
s
9
Spectrum 11:
δ
splitting
integration
9.75
t
1
2.30
dd
2
2.21
m
1
0.98
d
6
Spectrum 12:
δ
splitting
integration
8.08
s
1
4.13
t
2
1.70
m
2
0.96
t
3
P5.8: Match the 1H-NMR spectra 13-18 below to their corresponding structures M-R .
Structures:
Spectrum 13:
δ
splitting
integration
8.15
d
1
6.33
d
1
Spectrum 14: 1-723C (structure O)
δ
splitting
integration
6.05
s
1
2.24
s
3
Spectrum 15:
δ
splitting
integration
8.57
s (b)
1
7.89
d
1
6.30
d
1
2.28
s
3
Spectrum 16:
δ
splitting
integration
9.05
s (b)
1
8.03
s
1
6.34
s
1
5.68
s (b)
1
4.31
s
2
Spectrum 17:
δ
splitting
integration
7.76
d
1
7.57
s (b)
1
6.44
d
1
2.78
q
2
1.25
t
3
Spectrum 18:
δ
splitting
integration
4.03
s
1
2.51
t
1
2.02
t
1
P5.9: Match the 1H-NMR spectra 19-24 below to their corresponding structures S-X.
Structures:
Spectrum 19:
δ
splitting
integration
9.94
s
1
7.77
d
2
7.31
d
2
2.43
s
3
Spectrum 20:
δ
splitting
integration
10.14
s
2
8.38
s
1
8.17
d
2
7.75
t
1
Spectrum 21:
δ
splitting
integration
9.98
s
1
7.81
d
2
7.50
d
2
Spectrum 22:
δ
splitting
integration
7.15-7.29
m
2.5
2.86
t
1
2.73
t
1
2.12
s
1.5
Spectrum 23:
δ
splitting
integration
7.10
d
1
6.86
d
1
3.78
s
1.5
3.61
s
1
2.12
s
1.5
Spectrum 24:
δ
splitting
integration
7.23-7.30
m
1
3.53
s
1
P5.10: Match the 1H-NMR spectra 25-30 below to their corresponding structures AA-FF.
Structures:
Spectrum 25:
δ
splitting
integration
9.96
s
1
7.79
d
2
7.33
d
2
2.72
q
2
1.24
t
3
Spectrum 26:
δ
splitting
integration
9.73
s
1
7.71
d
2
6.68
d
2
3.06
s
6
Spectrum 27:
δ
splitting
integration
7.20-7.35
m
10
5.12
s
1
2.22
s
3
Spectrum 28:
δ
splitting
integration
8.08
s
1
7.29
d
2
6.87
d
2
5.11
s
2
3.78
s
3
Spectrum 29:
δ
splitting
integration
7.18
d
1
6.65
m
1.5
3.2
q
2
1.13
t
3
Spectrum 30:
δ
splitting
integration
8.32
s
1
4.19
t
2
2.83
t
2
2.40
s
3
P5.11: Match the 1H-NMR spectra 31-36 below to their corresponding structures GG-LL
Structures:
Spectrum 31:
δ
splitting
integration
6.98
d
1
6.64
d
1
6.54
s
1
4.95
s
1
2.23
s
3
2.17
s
3
Spectrum 32:
δ
splitting
integration
7.08
d
1
6.72
d
1
6.53
s
1
4.81
s
1
3.15
7-tet
1
2.24
s
3
1.22
d
6
Spectrum 33:
δ
splitting
integration
7.08
d
2
6.71
d
2
6.54
s
1
3.69
s
3
3.54
s
2
Spectrum 34:
δ
splitting
integration
9.63
s
1
7.45
d
2
6.77
d
2
3.95
q
2
2.05
s
3
1.33
t
3
Spectrum 35:
δ
splitting
integration
9.49
s
1
7.20
d
2
6.49
d
2
4.82
s
2
1.963
s
3
Spectrum 36:
δ
splitting
integration
9.58
s(b)
1
9.31
s
1
7.36
d
1
6.67
s
1
6.55
d
1
2.21
s
3
2.11
s
3
P5.12: Use the NMR data given to deduce structures.
a ) Molecular formula: C5H8O
1H-NMR:
δ
splitting
integration
9.56
s
1
6.25
d (J~1 Hz)
1
5.99
d (J~1 Hz)
1
2.27
q
2
1.18
t
3
13C-NMR
δ
DEPT
194.60
CH
151.77
C
132.99
CH2
20.91
CH2
11.92
CH3
b) Molecular formula: C7H14O2
1H-NMR:
δ
splitting
integration
3.85
d
2
2.32
q
2
1.93
m
1
1.14
t
3
0.94
d
6
13C-NMR
δ
DEPT
174.47
C
70.41
CH2
27.77
CH
27.64
CH2
19.09
CH3
9.21
CH3
c) Molecular formula: C5H12O
1H-NMR:
δ
splitting
integration
3.38
s
2H
2.17
s
1H
0.91
s
9H
13C-NMR
δ
DEPT
73.35
CH2
32.61
C
26.04
CH3
d) Molecular formula: C10H12O
1H-NMR:
δ
splitting
integration
7.18-7.35
m
2.5
3.66
s
1
2.44
q
1
1.01
t
1.5
13C-NMR
δ
DEPT
208.79
C
134.43
C
129.31
CH
128.61
CH
126.86
CH
49.77
CH2
35.16
CH2
7.75
CH3
P5.13:
13C-NMR data is given for the molecules shown below. Complete the peak assignment column of each NMR data table.
a)
δ
DEPT
carbon #
161.12
CH
65.54
CH2
21.98
CH2
10.31
CH3
b)
δ
DEPT
carbon #
194.72
C
149.10
C
146.33
CH
16.93
CH2
14.47
CH3
12.93
CH3
c)
δ
DEPT
carbon #
171.76
C
60.87
CH2
58.36
C
24.66
CH2
14.14
CH3
8.35
CH3
d)
δ
DEPT
carbon #
173.45
C
155.01
C
130.34
CH
125.34
C
115.56
CH
52.27
CH3
40.27
CH2
e)
δ
DEPT
carbon #
147.79
C
129.18
CH
115.36
CH
111.89
CH
44.29
CH2
12.57
CH3
P5.14: You obtain the following data for an unknown sample. Deduce its structure.
1H-NMR:
13C-NMR:
Mass Spectrometry:
P5.15:You take a 1H-NMR spectrum of a sample that comes from a bottle of 1-bromopropane. However, you suspect that the bottle might be contaminated with 2-bromopropane. The NMR spectrum shows the following peaks:
δ
splitting
integration
4.3
septet
0.0735
3.4
triplet
0.661
1.9
sextet
0.665
1.7
doublet
0.441
1.0
triplet
1.00
How badly is the bottle contaminated? Specifically, what percent of the molecules in the bottle are 2-bromopropane?
Challenge problems
C5.1: All of the 13C-NMR spectra shown in this chapter include a signal due to CDCl3, the solvent used in each case. Explain the splitting pattern for this signal.
C5.2: Researchers wanted to investigate a reaction which can be catalyzed by the enzyme alcohol dehydrogenase in yeast. They treated 4'-acylpyridine (1) with living yeast, and isolated the alcohol product(s) (some combination of 2A and 2B).
a) Will the products 2A and 2B have identical or different 1H-NMR spectra? Explain.
b) Suggest a 1H-NMR experiment that could be used to determine what percent of starting material (1) got turned into product (2A and 2B).
c) With purified 2A/2B, the researchers carried out the subsequent reaction shown below to make 3A and 3B, known as 'Mosher's esters'. Do 3A and 3B have identical or different 1H-NMR spectra? Explain.
d) Explain, very specifically, how the researchers could use 1H-NMR to determine the relative amounts of 2A and 2B formed in the reaction catalyzed by yeast enzyme.
12.15: Sample NMR Spectra
Sample 1H-NMR Spectra
List of Animated 1H-NMR Spectra
Bromoethane 1-bromopropane 2-propanol 3-bromopropene propanal
Phenol acetone propanoic acid ethyl acetate 2-propenamide
For all spectra click on a peak to highlight the protons responsible for the peak.
More spectra can be found at Animated Spectra
To see the integratals, right click on the spectra to open the menu, go to "view" and check the integrate" box. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/12%3A_Nuclear_Magnetic_Resonance_Spectroscopy/12.14%3A__More_NMR_Examples.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• distinguish between alcohols, phenols, enols, and carboxylic acids - refer to section 13.1
• classify alcohols as primary, secondary, or tertiary - refer to section 13.2
• predict relative physical properties of alcohols, such as relative boiling points and solubility in a specified solvent - refer to section 13.3
• determine the structure of alcohols and phenols from spectroscopic data - refer to section 13.4
• predict the relative acidity of alcohols - refer to section 13.5
• use resonance to explain why phenols are more acidic than alcohols - refer to section 13.5
• specify the base needed to ionize an alcohol or phenol - refer to section 13.5
• predict the products and specify the reagents for alcohol and diol synthesis from alkyl halides, alkenes, and alkynes from the previous chapters - refer to section 13.6
• predict the products and specify the reagents to synthesize alcohols from the reduction of carbonyls - refer to section 13.7
• predict the products and specify the reagents to prepare Grignard and organolithium reagents - refer to section 13.8
• predict the products and specify the reagents for alcohol synthesis from organometallic reagents with aldehydes, ketones, esters, acyl halides, & epoxides - section 13.9
• distinguish between the structure and reactivity of thiols and sulfides - refer to section 13.10
• explain the commercial synthesis of alcohols - refer to section 13.11
Please note: IUPAC nomenclature and important common names of alcohols were explained in Chapter 3.
• 13.1: Introduction to Structure and Synthesis of Alcohols
Alcohols, phenols, enols, and carboxylic acids all contain hydroxyl groups. It is essential to distinguish between the three functional groups. Because of the structural similarities between alcohols and phenols, their similarities and differences are emphasized.
• 13.2: Classification of Alcohols
Alcohols are classified by the bonding pattern of the carbon bonded to the hydroxyl group. Alcohol classification is helpful in discerning patterns of reactivity.
• 13.3: Physical Properties of Alcohols
Alcohols are the first functional group we are studying in detail that is capable of H-bonding. The effects of increased polarity and stronger intermolecular forces on the physical properties of alcohols relative to alkanes are discussed.
• 13.4: Spectroscopy of Alcohols
The hydroxyl group plays an important role in the spectroscopy of alcohols and phenols.
• 13.5: Acidity of Alcohols and Phenols
Phenols are weakly acidic (pKa = 10) because of their resonance stabilized conjugate base, phenoxide. Alcohols are considered neutral with pKa values similar to water (pKa = 14). The concepts used to predict relative acidity are explained in Chapter 1.
• 13.6: Synthesis of Alcohols - Review
Through the first ten chapters, we have learned to synthesize alcohols from alkyl halides via nucleophilic substitution (SN2 & SN1) and from alkenes using a variety of pathways determined by regiochemistry and stereochemistry. Gentle oxidation of the alkenes can also be used to synthesize diols.
• 13.7: Reduction of the Carbonyl Group - Synthesis of 1º and 2º Alcohols
Aldehydes, ketones, carboxylic acids, and esters can all be reduced to form alcohols. An important pattern of chemical reactivity is introduced when we notice that aldehydes and ketones often use the same reagents, where as carboxylic acids and esters require different reagents to create similar reactivity.
• 13.8: Organometallic Reagents
Grignard (RMgX) and organolithium (RLi) reagents are made from alkyl halides in aprotic solvents.
• 13.9: Organometallic Reagents in Alcohol Synthesis
Organometallic reagents can react with aldehydes, ketones, acyl halides, esters, and epoxides to synthesize alcohols with an increased number of carbon atoms in the product. Building larger organic molecules is a useful skill for multi-step synthesis.
• 13.10: Thiols (Mercaptans)
Thiols are the sulfur-analogs to alcohols. The major difference is the larger atomic size of sulfur relative to oxygen which creates differences in relative acidity and effectiveness as nucleophiles.
• 13.11: Commercially Important Alcohols
Both industrial and fermentation processes for alcohol synthesis are discussed.
• 13.12: 13.12 Additional Exercises
This section has additional exercises for the key learning objectives of the chapter.
• 13.13: Solutions to Additional Exercises
This section has the solutions to the Additional Practice Problems in the previous section.
13: Structure and Synthesis of Alcohols
Introduction
Alcohols, phenols, enols, and carboxylic acids all contain a hydroxyl group. However, the chemistry of these four different functional groups are different.
Alcohols
Molecules of alcohols contain one or more hydroxyl groups (OH groups) substituted for hydrogen atoms along the carbon chain.
The structure of the simplest alcohol, methanol (methyl alcohol), can be derived from that of methane by putting an OH in place of one of the H’s:
Methanol is also called wood alcohol because it can be obtained by heating wood in the absence of air, a process called destructive distillation. Methanol vapor given off when the wood is heated can be condensed to a liquid by cooling below its boiling point of 65°C. Methanol is highly toxic. In 1986, there were six deaths of residents of Peerless Lake, Alberta, brought about by drinking photocopier fluid which contained methanol (or methyl hydrate as it is often called in press reports). In 2000, more than 100 people died in El Salvador after black marketeers sold discarded liquor bottles that had been refilled with a methanol mixture. Indeed the problem has repeated itself globally and so often that in 2014 the World Health Organization released an information note warning of methanol poisoning outbreaks which “occur when methanol is added to illicitly‑ or informally‑produced alcoholic drinks.”
The second member of the alcohol family is ethanol (ethyl alcohol)― the substance we commonly call alcohol. Ethanol is also known as grain alcohol because it is obtained when grain or sugar ferments.
Fermentation refers to a chemical reaction which is speeded up by enzymes and occurs in the absence of air. (Enzymes, catalysts which occur naturally in yeasts and other living organisms, are discussed in more detail elsewhere.) Almost everyone is aware that the alcohol present in alcoholic beverages is ethanol (also called ethyl alcohol or grain alcohol). However, many people do not realize that in its pure state, or in solutions of high concentration, this substance is poisonous. In the laboratory one may find containers labeled “absolute ethanol,” “95% ethanol” and “denatured ethanol.” The acquisition of ethanol by laboratories, and its subsequent disposal, is carefully monitored by provincial authorities. On no account should one consider drinking laboratory ethanol, even after it has been diluted to a concentration equivalent to that found in beer. Denatured alcohol is ethanol to which appropriate quantities of poisonous or nauseating substances (such as methanol) have been added. Ethanol is used as a solvent, in some special fuels, in antifreeze, and to manufacture a number of other chemicals. You are probably most familiar with it as a component of alcoholic beverages. Ethanol makes up 3 to 6 percent of beer, 12 to 15 percent of most wines, and 49 to 59 percent of distilled liquor. (The “proof” of an alcoholic beverage is just twice the percentage of ethanol.) Alcohol’s intoxicating effects are well known, and it is a mild depressant. Prolonged overuse can lead to liver damage.
A third commonly encountered alcohol, isopropyl alcohol (“rubbing alcohol” or 2‑propanol), is also toxic. It has the ability to kill germs and has a temporary lubricating effect during the rubbing process. Unlike methanol, 2‑propanol is not absorbed through the skin; therefore it poses less of a health hazard.
Phenols
A phenol is an organic compound in which a hydroxyl group is directly bonded to one of the carbon atoms of an aromatic ring. The chemical behavior of phenols is different in some respects from that of the alcohols, so it is sensible to treat them as a similar but characteristically distinct group.
Until the late nineteenth century, a person undergoing surgery had to face the fact that he or she might suffer the consequences of what we now know to be bacterial infection, contracted during the course of the operation. The physicians of the time did not know that bacteria existed, and had no way to counter the problems that bacteria caused. In 1867, Joseph Lister, who had learned of the existence of bacteria as a result of research done by Louis Pasteur, began using solutions of phenol to clean wounds and surgical instruments. The phenol solution was an effective antiseptic, killing bacteria, and as a result, a patient’s chances of surviving surgery improved greatly. Phenol itself was rather strong for these purposes—it burns healthy tissue—and substitutes were eventually found. One such substitute, used today in throat lozenges and mouthwashes, is 4‑n‑hexylresorcinol.
Enols
When a hydroxyl group is bonded to a vinyl carbon, the functional group is called an enol. As discussed in Chapter 10, alkyne hydration can result in enol formation. Enols undergo tautomerization to form ketones, In most cases, the keto-form is more stable and predominates the equilibrium. Phenol is an important exception to this trend.
Carboxylic Acids
In carboxylic acids, the hydroxyl group is bonded to a carbonyl carbon. As the name implies, this functional group is acidic and can readily donate the proton on the hydroxyl group.
Exercise
1. Classify the following compounds as alcohols, phenols, enols, or carboxylic acids. One compound can be classified by two of the options.
Answer
1.
a) carboxylic acid
b) phenol and enol
c) alcohol | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.01%3A_Introduction_to_Structure_and_Synthesis_of_Alcohols.txt |
Alcohol classification is an application of the neutral bonding patterns for organic compounds. Oxygen can only form two bonds. The alcohol functional group requires that one of these bonds form with hydrogen to create the hydroxyl group and the other bond needs to be with carbon to create an alcohol. All of the oxygen atoms of all the alcohols look the same, so a different distinction is needed. To classify alcohols, we look at the carbon atom bonded to the hydroxyl group.
methyl alcohol = no (0) other carbon atoms (CH3OH)
primary alcohols (1°) = 1 other carbon atoms
secondary alcohols (2°) = 2 other carbon atom
tertiary alcohols (3°) = 3 other carbon atoms
Primary alcohols
In a primary (1°) alcohol, the carbon which carries the -OH group is only attached to one alkyl group. Some examples of primary alcohols include:
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH2 group holding the -OH group. There is an exception to this. Methanol, CH3OH, is counted as a primary alcohol even though there are no alkyl groups attached to the carbon with the -OH group on it.
Secondary alcohols
In a secondary (2°) alcohol, the carbon with the -OH group attached is joined directly to two alkyl groups, which may be the same or different. Examples:
Tertiary alcohols
In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Exercise
2. Classify the following alcohols as primary, secondary, or tertiary.
Answer
2.
a) secondary
b) primary
c) tertiary
13.03: Physical Properties of Alcohols
Boiling Points
The chart below shows the boiling points of the following simple primary alcohols with up to 4 carbon atoms:
These boiling points are compared with those of the equivalent alkanes (methane to butane) with the same number of carbon atoms.
Notice that:
• The boiling points of the alcohols increase as the number of carbon atoms increases.
The patterns in boiling point reflect the patterns in intermolecular attractions.
Hydrogen bonding
Hydrogen bonding occurs between molecules in which a hydrogen atom is attached to a strongly electronegative element: fluorine, oxygen or nitrogen. In the case of alcohols, hydrogen bonds occur between the partially-positive hydrogen atoms and lone pairs on oxygen atoms of other molecules.
The hydrogen atoms are slightly positive because the bonding electrons are pulled toward the very electronegative oxygen atoms. In alkanes, the only intermolecular forces are van der Waals dispersion forces. Hydrogen bonds are much stronger than these, and therefore it takes more energy to separate alcohol molecules than it does to separate alkane molecules. This the main reason for higher boiling points in alcohols.
The effect of van der Waals forces
• Comparison between alkanes and alcohols: Even without any hydrogen bonding or dipole-dipole interactions, the boiling point of the alcohol would be higher than the corresponding alkane with the same number of carbon atoms.
Compare ethane and ethanol:
Ethanol is a longer molecule, and the oxygen atom brings with it an extra 8 electrons. Both of these increase the size of the van der Waals dispersion forces, and subsequently the boiling point. A more accurate measurement of the effect of the hydrogen bonding on boiling point would be a comparison of ethanol with propane rather than ethane. The lengths of the two molecules are more similar, and the number of electrons is exactly the same.
Solubility of alcohols in water
Small alcohols are completely soluble in water; mixing the two in any proportion generates a single solution. However, solubility decreases as the length of the hydrocarbon chain in the alcohol increases. At four carbon atoms and beyond, the decrease in solubility is noticeable; a two-layered substance may appear in a test tube when the two are mixed.
Consider ethanol as a typical small alcohol. In both pure water and pure ethanol the main intermolecular attractions are hydrogen bonds.
In order to mix the two, the hydrogen bonds between water molecules and the hydrogen bonds between ethanol molecules must be broken. Energy is required for both of these processes. However, when the molecules are mixed, new hydrogen bonds are formed between water molecules and ethanol molecules.
The energy released when these new hydrogen bonds form approximately compensates for the energy needed to break the original interactions. In addition, there is an increase in the disorder of the system, an increase in entropy. This is another factor in deciding whether chemical processes occur. Consider a hypothetical situation involving 5-carbon alcohol molecules.
The hydrocarbon chains are forced between water molecules, breaking hydrogen bonds between those water molecules. The -OH ends of the alcohol molecules can form new hydrogen bonds with water molecules, but the hydrocarbon "tail" does not form hydrogen bonds. This means that many of the original hydrogen bonds being broken are never replaced by new ones.
In place of those original hydrogen bonds are merely van der Waals dispersion forces between the water and the hydrocarbon "tails." These attractions are much weaker, and unable to furnish enough energy to compensate for the broken hydrogen bonds. Even allowing for the increase in disorder, the process becomes less feasible. As the length of the alcohol increases, this situation becomes more pronounced, and thus the solubility decreases.
Exercise
Use 1-octanol and ethanol, shown below, to answer the following questions.
3. Which compound is more water soluble?
4. Which compound has the highest boiling point?
5. Explain why the answers above are not in conflict using your understanding of intermolecular forces, relative boiling points, and solubility.
Answer
3. ethanol
4. octanol
5. While both compounds exhibit H-bonding, the smaller, hydrophobic carbon chain of ethanol results in higher water solubility of ethanol while the longer carbon chain of octanol increases the surface area resulting in a higher boiling point. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.02%3A_Classification_of_Alcohols.txt |
Infrared Spectroscopy
If you look at an IR spectrum of 1-butanol, you will see:
• there are sp3 C-H stretching and CH2 bending modes at 2900 and 1500 cm-1.
• there is a strong C-O stretching mode near 1000 cm-1.
• there is a very large peak around 3400 cm-1. O-H peaks are usually very broad like this one.
Peak shapes are sometimes very useful in recognizing what kind of bond is present. The rounded shape of most O-H stretching modes occurs because of hydrogen bonding between different hydroxy groups. Because protons are shared to varying extent with neighboring oxygens, the covalent O-H bonds in a sample of alcohol all vibrate at slightly different frequencies and show up at slightly different positions in the IR spectrum. Instead of seeing one sharp peak, you see a whole lot of them all smeared out into one broad blob. Since C-H bonds don't hydrogen bond very well, you don't see that phenomenon in an ether, and an O-H peak is very easy to distinguish in the IR spectrum.
Alcohols
Where is the -O-H peak? This is very confusing! Different sources quote totally different chemical shifts for the hydrogen atom in the -OH group in alcohols - often inconsistently. For example:
• The Nuffield Data Book quotes 2.0 - 4.0, but the Nuffield text book shows a peak at about 5.4.
• The OCR Data Sheet for use in their exams quotes 3.5 - 5.5.
• A reliable degree level organic chemistry text book quotes1.0 - 5.0, but then shows an NMR spectrum for ethanol with a peak at about 6.1.
• The SDBS database (used throughout this site) gives the -OH peak in ethanol at about 2.6.
The problem seems to be that the position of the -OH peak varies dramatically depending on the conditions - for example, what solvent is used, the concentration, and the purity of the alcohol - especially on whether or not it is totally dry.
A clever way of picking out the -OH peak
If you measure an NMR spectrum for an alcohol like ethanol, and then add a few drops of deuterium oxide, D2O, to the solution, allow it to settle and then re-measure the spectrum, the -OH peak disappears! By comparing the two spectra, you can tell immediately which peak was due to the -OH group.
The reason for the loss of the peak lies in the interaction between the deuterium oxide and the alcohol. All alcohols, such as ethanol, are very, very slightly acidic. The hydrogen on the -OH group transfers to one of the lone pairs on the oxygen of the water molecule. The fact that here we've got "heavy water" makes no difference to that.
The negative ion formed is most likely to bump into a simple deuterium oxide molecule to regenerate the alcohol - except that now the -OH group has turned into an -OD group.
Deuterium atoms don't produce peaks in the same region of an NMR spectrum as ordinary hydrogen atoms, and so the peak disappears. You might wonder what happens to the positive ion in the first equation and the OD- in the second one. These get lost into the normal equilibrium which exists wherever you have water molecules - heavy or otherwise.
\[2D_2O \rightleftharpoons D_3O^+ + OH^-\]
The lack of splitting with -OH groups
Unless the alcohol is absolutely free of any water, the hydrogen on the -OH group and any hydrogens on the next door carbon don't interact to produce any splitting. The -OH peak is a singlet and you don't have to worry about its effect on the next door hydrogens.
The left-hand cluster of peaks is due to the CH2 group. It is a quartet because of the 3 hydrogens on the next door CH3 group. You can ignore the effect of the -OH hydrogen. Similarly, the -OH peak in the middle of the spectrum is a singlet. It hasn't turned into a triplet because of the influence of the CH2 group.
Mass Spectra Fragmentation Patterns
The fragmentation of molecular ions into an assortment of fragment ions is a mixed blessing. The nature of the fragments often provides a clue to the molecular structure, but if the molecular ion has a lifetime of less than a few microseconds it will not survive long enough to be observed. Without a molecular ion peak as a reference, the difficulty of interpreting a mass spectrum increases markedly. Fortunately, most organic compounds give mass spectra that include a molecular ion, and those that do not often respond successfully to the use of milder ionization conditions. Among simple organic compounds, the most stable molecular ions are those from aromatic rings, other conjugated pi-electron systems and cycloalkanes. Alcohols, ethers and highly branched alkanes generally show the greatest tendency toward fragmentation.
The presence of a functional group, particularly one having a heteroatom Y with non-bonding valence electrons (Y = N, O, S, X etc.), can dramatically alter the fragmentation pattern of a compound. This influence is thought to occur because of a "localization" of the radical cation component of the molecular ion on the heteroatom. After all, it is easier to remove (ionize) a non-bonding electron than one that is part of a covalent bond. By localizing the reactive moiety, certain fragmentation processes will be favored. These are summarized in the following diagram, where the green shaded box at the top displays examples of such "localized" molecular ions. The first two fragmentation paths lead to even-electron ions, and the elimination (path #3) gives an odd-electron ion. Note the use of different curved arrows to show single electron shifts compared with electron pair shifts.
3-Pentanol shows three significant fragment ions. Alpha-fragmentation (loss of an ethyl radical) forms the m/z=59 base peak. Loss of water from this gives a m/z=41 fragment, and loss of ethene from m/z=59 gives a m/z=31 fragment.
Exercise
6. From mass spectroscopy analysis it was determined that a compound has the general formula C5H12O. Given the following 1H NMR spectrum, draw the structure. The integration values of each group of signals is given on the spectrum.
7. Given that alcohols are relatively acidic and the protons transfer in solution, what would you expect to happen to the NMR spectrum if D2O was used as a solvent.
8. From mass spectroscopy analysis it was determined that a compound has the general formula C3H8O. Given the following 1H NMR spectrum, draw the structure. The integration values of each group of signals is given on the spectrum.
Answer
6.
7.
The alcohol proton signal's intensity in the 1H NMR would be expected to diminish and likely disappear. This is due to the fact that NMR can only probe the spin changes of nuclei with an odd number of protons.
8.
1-propanol | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.04%3A_Spectroscopy_of_Alcohols.txt |
Overview - Aqueous vs Organic Solvents
In aqueous solutions, phenols are weakly acidic and lower the pH of a solution. Sodium hydroxide can be used to fully deprotonate a phenol. Water soluble alcohols do not change the pH of the solution and are considered neutral. Aqueous solutions of sodium hydroxide can NOT deprotonate alcohols to a high enough concentration to be synthetically useful.
In solutions of organic solvents, more extreme reaction conditions can be created. Sodium metal can be added to an alcohol in an organic solvent system to fully deprotonate the alcohol to form alkoxide ions.
Acidity of Alcohols
Several important chemical reactions of alcohols involving the O-H bond or oxygen-hydrogen bond only and leave the carbon-oxygen bond intact. An important example is salt formation with acids and bases. Alcohols, like water, are both weak bases and weak acids. The acid ionization constant (Ka) of ethanol is about 10~18, slightly less than that of water. Ethanol can be converted to its conjugate base by the conjugate base of a weaker acid such as ammonia {Ka — 10~35), or hydrogen (Ka ~ 10-38). It is convenient to employ sodium metal or sodium hydride, which react vigorously but controllably with alcohols:
The order of acidity of various liquid alcohols generally is water > primary > secondary > tertiary ROH. By this we mean that the equilibrium position for the proton-transfer reaction lies more on the side of ROH as R is changed from primary to secondary to tertiary; therefore, tert-butyl alcohol is considered less acidic than ethanol:
\[ ROH + OH^- \rightleftharpoons RO^- + HOH\]
However, in the gas phase the order of acidity is reversed, and the equilibrium position for lies increasingly on the side of the alkoxide as R is changed from primary to secondary to tertiary, tert-butyl alcohol is therefore more acidic than ethanol in the gas phase. This seeming contradiction appears more reasonable when one considers what effect solvation (or the lack of it) has on equilibria. In solution, the larger alkoxide ions, probably are less well solvated than the smaller ions, because fewer solvent molecules can be accommodated around the negatively charged oxygen in the larger ions:
Acidity of alcohols therefore decreases as the size of the conjugate base increases. However, “naked” gaseous ions are more stable the larger the associated R groups, probably because the larger R groups can stabilize the charge on the oxygen atom better than the smaller R groups. They do this by polarization of their bonding electrons, and the bigger the group, the more polarizable it is.
Basicity of Alcohols
Alcohols are bases similar in strength to water and accept protons from strong acids. An example is the reaction of methanol with hydrogen bromide to give methyloxonium bromide, which is analogous to the formation of hydroxonium bromide with hydrogen bromide and water:
Acidity of Phenol
Compounds like alcohols and phenol which contain an -OH group attached to a hydrocarbon are very weak acids. Alcohols are so weakly acidic that, for normal lab purposes, their acidity can be virtually ignored. However, phenol is sufficiently acidic for it to have recognizably acidic properties - even if it is still a very weak acid. A hydrogen ion can break away from the -OH group and transfer to a base. For example, in solution in water:
Phenol is a very weak acid and the position of equilibrium lies well to the left. Phenol can lose a hydrogen ion because the phenoxide ion formed is stabilised to some extent. The negative charge on the oxygen atom is delocalised around the ring. The more stable the ion is, the more likely it is to form. One of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring.
This overlap leads to a delocalization which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localized on the oxygen, but is spread out around the whole ion.
Spreading the charge around makes the ion more stable than it would be if all the charge remained on the oxygen. However, oxygen is the most electronegative element in the ion and the delocalized electrons will be drawn towards it. That means that there will still be a lot of charge around the oxygen which will tend to attract the hydrogen ion back again. That is why phenol is only a very weak acid.
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
In this reaction, the hydrogen ion has been removed by the strongly basic hydroxide ion in the sodium hydroxide solution.
Acids react with the more reactive metals to give hydrogen gas. Phenol is no exception - the only difference is the slow reaction because phenol is such a weak acid. Phenol is warmed in a dry tube until it is molten, and a small piece of sodium added. There is some fizzing as hydrogen gas is given off. The mixture left in the tube will contain sodium phenoxide.
Acidity of Substituted Phenols
Substitution of the hydroxyl hydrogen atom is even more facile with phenols, which are roughly a million times more acidic than equivalent alcohols. This phenolic acidity is further enhanced by electron-withdrawing substituents ortho and para to the hydroxyl group, as displayed in the following diagram. The alcohol cyclohexanol is shown for reference at the top left. It is noteworthy that the influence of a nitro substituent is over ten times stronger in the para-location than it is meta, despite the fact that the latter position is closer to the hydroxyl group. Furthermore additional nitro groups have an additive influence if they are positioned in ortho or para locations. The trinitro compound shown at the lower right is a very strong acid called picric acid.
Comparing the Acidity of Alcohols with Phenols
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. Formulas illustrating this electron delocalization will be displayed when the "Resonance Structures" button beneath the previous diagram is clicked. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
Exercise
9. Arrange the following compounds in order of decreasing acidity when they are in solution.
10. Specify the base needed to deprotonate each reactant.
Answer
9. B > C > A
10. a) Na or NaH or NNH2
b) NaOH or KOH or LiOH
Contributors
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.05%3A_Acidity_of_Alcohols_and_Phenols.txt |
Alcohols are prepared by SN2 & SN1 (solvolysis) reactions
Alkyl halides can be converted to alcohols by using SN2 reactions with OH- as a nucleophile. Substrates that undergo substitution by SN1 reaction can be converted to alcohols using water as the nucleophile (and it can even be the solvent). Recall that SN1 reactions are promoted in polar, protic solvents.
Alcohols from alkenes
Oxymercuration is a special electrophilic addition. It is anti-stereospecific and regioselective. Regioselectivity is a process in which the substituents choses one direction it prefers to be attached to over all the other possible directions. The good thing about this reaction is that there are no carbocation rearrangement due to stabilization of the reactive intermediate. Similar stabilization is also seen in bromination addition to alkenes.
Carbocation rearrangement is a process in which the carbocation intermediate can form a more stable ion. With carbocation rearrangement, the reaction would not be able to hydrate quickly under mild conditions and be produced in high yields. This reaction is very fast and proceeds with 90% yield.
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from BH3 or BHR2) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene bouble bond. Furthermore, the borane acts as a lewis acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The Hydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific (syn addition), meaning that the hydroboration takes place on the same face of the double bond, this leads cis stereochemistry.
Diols from alkenes
Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the syn-hydroxylation reaction described above. Proton transfer from the acid catalyst generates the conjugate acid of the epoxide, which is attacked by nucleophiles such as water in the same way that the cyclic bromonium ion described above undergoes reaction. The result is anti-hydroxylation of the double bond, in contrast to the syn-stereoselectivity of the earlier method. In the following equation this procedure is illustrated for a cis-disubstituted epoxide, which, of course, could be prepared from the corresponding cis-alkene. This hydration of an epoxide does not change the oxidation state of any atoms or groups.
Osmium tetroxide oxidizes alkenes to give glycols through syn addition. A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons.
The reaction with \(OsO_4\) is a concerted process that has a cyclic intermediate and no rearrangements. Vicinal syn dihydroxylation complements the epoxide-hydrolysis sequence which constitutes an anti dihydroxylation of an alkene. When an alkene reacts with osmium tetroxide, stereocenters can form in the glycol product. Cis alkenes give meso products and trans alkenes give racemic mixtures.
13.07: Reduction of the Carbonyl Group - Synthesis of 1 and 2 Al
Reduction of Aldehydes and Ketones
The most common sources of the hydride nucleophile are lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). Note! The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminum is less electronegative than boron, the Al-H bond in LiAlH4 is more polar, thereby, making LiAlH4 a stronger reducing agent.
Addition of a hydride anion (H:-) to an aldehyde or ketone gives an alkoxide anion, which upon protonation yields the corresponding alcohol. Aldehydes produce 1º-alcohols and ketones produce 2º-alcohols.
In metal hydrides reductions the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the sodium borohydride reduction the methanol solvent system achieves this hydrolysis automatically. In the lithium aluminum hydride reduction water is usually added in a second step. The lithium, sodium, boron and aluminum end up as soluble inorganic salts at the end of either reaction. Note! LiAlH4 and NaBH4 are both capable of reducing aldehydes and ketones to the corresponding alcohol.
Formally, that process is referred to as a reduction. Reduction generally means a reaction in which electrons are added to a compound; the compound that gains electrons is said to be reduced. Because hydride can be thought of as a proton plus two electrons, we can think of conversion of a ketone or an aldehyde to an alcohol as a two-electron reduction. A carbonyl (aldehyde or ketone) plus two electrons and two protons becomes an alcohol.
Example 1
Mechanism
This mechanism is for a LiAlH4 reduction. The mechanism for a NaBH4 reduction is the same except methanol is the proton source used in the second step.
1) Nucleopilic hydride anion reacts with the electrophilic carbonyl carbon forcing the pi electrons onto the electronegative oxygen atom.
2) The alkoxide is protonated.
Biological Reduction
Addition to a carbonyl by a semi-anionic hydride, such as NaBH4, results in conversion of the carbonyl compound to an alcohol. The hydride from the BH4- anion acts as a nucleophile, adding H- to the carbonyl carbon. A proton source can then protonate the oxygen of the resulting alkoxide ion, forming an alcohol.
Aldehydes, ketones and alcohols are very common features in biological molecules. Converting between these compounds is a frequent event in many biological pathways. However, semi-anionic compounds like sodium borohydride don't exist in the cell. Instead, a number of biological hydride donors play a similar role.
NADH is a common biological reducing agent. NADH is an acronym for nicotinamide adenine dinucleotide hydride. Insetad of an anionic donor that provides a hydride to a carbonyl, NADH is actually a neutral donor. It supplies a hydride to the carbonyl under very specific circumstances. In doing so, it forms a cation, NAD+. However, NAD+ is stabilized by the fact that its nicotinamide ring is aromatic; it was not aromatic in NADH.
Reduction of Carboxylic Acids and Esters
Carboxylic acids can be converted to 1o alcohols using Lithium aluminum hydride (LiAlH4). Note that NaBH4 is not strong enough to convert carboxylic acids or esters to alcohols. An aldehyde is produced as an intermediate during this reaction, but it cannot be isolated because it is more reactive than the original carboxylic acid.
Esters can be converted to 1o alcohols using LiAlH4, while sodium borohydride (\(NaBH_4\)) is not a strong enough reducing agent to perform this reaction.
Reduction Reaction Summary
The table below summarize the reduction reactions covered so far in our text. It is important to distinguish between functional group reactivity as we add more multiple-step synthetic pathways.
Exercise
11.
Give the aldehyde, ketone, or carboxyllic acid (there can be multiple answers) that could be reduced to form the following alcohols.
(a) (b) (c) (d)
12.
Given the following alcohol, draw the structure from which it could be derived using only NaBH4
(a) (b) (c) (d)
Answer
11.
(a) (b) (c) (d)
12.
Note, NaBH4 is only a strong enough reducing agent to reduce ketones and aldehydes.
(a) (b) (c) (d) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.06%3A_Synthesis_of_Alcohols_-_Review.txt |
Introduction
A Grignard reagent has a formula $\ce{RMgX}$ where $\ce{X}$ is a halogen, and $\ce{R}$ is an alkyl or aryl (based on a benzene ring) group. For the purposes of this page, we shall take R to be an alkyl group. A typical Grignard reagent might be $\ce{CH3CH2MgBr}$. Organolithium reagents have the chemical formula RLi. A typical reagent might be $\ce{CH3CH2Li}$.
Formation of Organometallic Reagents
Many organometallic reagents are commercially available, however, it is often necessary to make then. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination).
• An Alkyl Lithium Reagent
$\ce{R3C-X} + \ce{2Li} \rightarrow \ce{R3C-Li} + \ce{LiX}$
• A Grignard Reagent
$\ce{R3C-X} + \ce{Mg} \rightarrow \ce{R3C-MgX}$
• An Organocuprate Reagent
Halide reactivity in these reactions increases in the order: Cl < Br < I and Fluorides are usually not used. The alkyl magnesium halides described in the second reaction are called Grignard Reagents after the French chemist, Victor Grignard, who discovered them and received the Nobel prize in 1912 for this work. The other metals mentioned above react in a similar manner, but Grignard and Alky Lithium Reagents most widely used. Although the formulas drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation. Organocuprate reagents have limited reactivity and will be used for ketone synthesis.
Organometallic Reagents are Strong Nucleophiles
The bond between the carbon atom and the metal atom is polar. Therefore, organometallic reagents are strong nucleophiles. Using the Grignard reagent as an example, the carbon is more electronegative than magnesium, so the bonding pair of electrons is pulled towards the carbon creating a partial negative charge. Grignard reagents are strong nucleophiles. Nucleophilic carbon atoms are very useful in building carbon chains in multiple step systhesis.
Grignard reactions create the possibility for substitution reactions at vinylic carbons. This reaction pathway is very useful since vinyl halides cannot react by the SN1 and SN2 mechanisms.
Organometallic Reagents and Protic Solvents (like water)
Everything must be perfectly dry because organometallic reagents react with water (see below) or any protic solvent. Reactions using the Grignard reagent must use an ether as the solvent. Organolithium reactions also require aprotic solvents, but ethers are not required and alkanes can be used as solvents. The resulting reaction mixture is used directly for the next reaction. There are no separation and isolation procedures between reaction steps. Organometallic reagents react with water or any protic solvent to produce alkanes. For this reason, everything has to be very dry during the preparation above. The term dry means that no water or other protonated solvents are present. There is still a liquid ether solvent.
A suitable solvent must be used. For alkyl lithium formation pentane or hexane are usually used. Diethyl ether can also be used but the subsequent alkyl lithium reagent must be used immediately after preparation due to an interaction with the solvent. Ethyl ether or THF are essential for Grignard reagent formation. Lone pair electrons from two ether molecules form a complex with the magnesium in the Grignard reagent (As pictured below). This complex helps stabilize the organometallic and increases its ability to react.
These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). This change, shown below, makes alkyl lithium and Grignard reagents excellent nucleophiles and useful reactants in synthesis.
Example:
Common Organometallic Reagents
Exercises
13.
Predict the product or specify the missing regent(s) in the reactions below.
Answer
13. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.08%3A_Organometallic_Reagents.txt |
Introduction
The nucleophilic carbon atoms of organometallic reagents react with the electrophilic carbon atoms of aldehydes, ketones, acyl halides, esters, and epoxides to build larger carbon chains. In the process, an alcohol is formed. The ability to build larger organic molecules is an important and useful skill for multi-step synthesis. These various reaction pathways are summarized below showing the Grignard reagent.
Organometallic Reactions with Aldehydes and Ketones
Because organometallic reagents react as their corresponding carbanion, they are excellent nucleophiles. The basic reaction involves the nucleophilic reaction of the carbanionic carbon in the organometallic reagent with the electrophilic carbon in the carbonyl to form alcohols.
Both Grignard and Organolithium Reagents will perform these reactions.
Addition to formaldehyde gives 1o alcohols.
Addition to aldehydes gives 2o alcohols.
Addition to ketones gives 3o alcohols
Examples
Mechanism for the Addition to Carbonyls
The mechanism for a Grignard agent is shown. The mechanism for an organolithium reagent is the same.
1) Nucleophilic reaction
2) Protonation
Grignard reagents convert esters to 3o alcohols
After the first Grignard reaction, the carbonyl reforms creating a ketones which can then react with the Grignard. In effect, the Grignard reagent adds twice.
Mechanism
1) Nucleophilic reaction
2) Carbonyl reforms with leaving group removal
3) Nucleophilic reaction
4) Protonation
Example
Grignard reagents convert Acyl Halides to 3o alcohols
Grignard reagents react with acyl halides similar to the reaction with esters. The first reaction produces a ketone which them undergoes a second reaction to form a tertiary alcohol following the analogous mechanism shown above for esters.
The reaction of benzoyl chloride with a Grignard reagent is shown below as an example.
Grignard Reactions with Epoxides
Another important route for producing an alcohol from a Grignard reagent involves the reaction of the Grignard reagent with ethylene oxide to produce a primary alcohol containing two more carbon atoms than the original Grignard reagent.
The first step of the mechanism is shown below. With the second step following the protonation step common to the other reaction pathways studied in this section.
Limitation of Organometallic Reagents
As discussed above, Grignard and organolithium reagents are powerful bases. Because of this they cannot be used as nucleophiles on compounds which contain acidic hydrogens. If they are used they will act as a base and deprotonate the acidic hydrogen rather than act as a nucleophile and attack the carbonyl. A partial list of functional groups which cannot be used are: alcohols, amides, 1o amines, 2o amines, carboxylic acids, and terminal alkynes.
Exercises
14. Predict the products of the reactions below.
15. If allylmagnesium chloride were added to a solution of the following compound and then worked-up with acid, the product would contain a chiral center. Would the product be a racemic mixture or an enatiomerically pure product? Draw both enantiomers.
16. What combination of carbonyl compound and grignard (use MgBr) reagent would yield the following alcohols (after workup)?
(a) (b) (c) (d)
17. The following epoxide can be transformed into an alcohol using a grignard reagent, take for example allylmagnesium chloride. Draw the product of the treatment of this epoxide with this grignard after being worked up with H2O. Note the stereochemistry and also remember that benzyllic carbons are good Sn2 electrophiles.
18. How might you prepare the following molecules from esters and Grignards?
(a)
(b)
(c)
Answer
14.
15.
The result would be a racemic mixture of the following.
16.
(a)
(b)
(c)
(d)
17.
18.
(a)
(b)
(c) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.09%3A_Organometallic_Reagents_in_Alcohol_Synthesis.txt |
Objectives
After completing this section, you should be able to
1. Nomenclature and Reactivity
1. write the IUPAC name of a thiol, given its Kekulé, condensed or shorthand structure.
2. draw the structure of a thiol, given its IUPAC name.
3. write an equation to represent the formation of a thiol by the reaction of hydrosulfide anion with an alkyl halide.
4. write an equation to illustrate the preparation of a thiol by the reaction of thiourea with an alkyl halide.
2. write an equation to show the interconversion between thiols and disulfides.
1. write the name of a sulfide, given its structure.
2. draw the structure of a sulfide, given its name.
3. write an equation showing how a sulfide may be prepared by the reaction of a thiolate anion on an alkyl halide.
4. identify the product from the reaction of a given alkyl halide with a given thiolate anion.
5. identify the reagents necessary to prepare a given sulfide.
6. write an equation to illustrate the formation of a trialkylsulfonium salt from a sulfide and an alkyl halide.
Key Terms
• disulfide
• mercapto group
• (organic) sulfide
• sulfone
• sulfoxide
• thiol
• thiolate anion
• trialkylsulfonium ion (trialkylsulfonium salt)
Study Notes
The chemistry of sulfur-containing organic compounds is often omitted from introductory organic chemistry courses. However, we have included a short section on these compounds, not for the sake of increasing the amount of material to be digested, but because much of the chemistry of these substances can be predicted from a knowledge of their oxygen-containing analogues. A thiol is a compound which contains an SH functional group. The -SH group itself is called a mercapto group. A disulfide is a compound containing an -S-S- linkage. (Organic) sulfides have the structure R-S-R′, and are therefore the sulfur analogues of ethers. The nomenclature of sulfides can be easily understood if one understands the nomenclature of the corresponding ethers. Notice that the term “thio” is also used in inorganic chemistry. For example, SO42 is the sulfate ion; while S2O32, in which one of the oxygen atoms of a sulfate ion has been replaced by a sulfur atom, is called thiosulfate. Thiolate anions, RS- , are analogous to alkoxy anions, RO- . Thiolate anions are better nucleophiles than are alkoxy anions (see Section 11.5, pages 389-394 of the textbook). If you have trouble understanding why trialkylsulfonium ions are formed, think of them as being somewhat similar to the hydronium ions that are formed by protonating water:
Later we shall see examples of tetraalkylammonium ions, R4N+, which again may be regarded as being similar to hydronium ions. sulfoxides and sulfones are obtained by oxidizing organic sulfides. You need not memorize the methods used to carry out these oxidations.
Table 18.1, below, provides a quick comparison of oxygen-containing and sulfur-containing organic compounds.
Oxidation States of Sulfur Compounds
Oxygen assumes only two oxidation states in its organic compounds (–1 in peroxides and –2 in other compounds). Sulfur, on the other hand, is found in oxidation states ranging from –2 to +6, as shown in the following table (some simple inorganic compounds are displayed in orange).
Thiols
Thiols, which are also called mercaptans, are analogous to alcohols. They are named in a similar fashion as alcohols except the suffix -thiol is used in place of -ol. By itself the -SH group is called a mercapto group.
Thiols are usually prepared by using the hydrosulfide anion (-SH) as a neucleophile in an SN2 reaction with alkyl halides.
On problem with this reaction is that the thiol product can undergo a second SN2 reaction with an additional alkyl halide to produce a sulfide side product. This problem can be solved by using thiourea, (NH2)2C=S, as the nucleophile. The reaction first produces an alkyl isothiourea salt and an intermediate. This salt is then hydrolyzed by a reaction with aqueous base.
Disulfides
Oxidation of thiols and other sulfur compounds changes the oxidation state of sulfur rather than carbon. We see some representative sulfur oxidations in the following examples. In the first case, mild oxidation converts thiols to disufides. An equivalent oxidation of alcohols to peroxides is not normally observed. The reasons for this different behavior are not hard to identify. The S–S single bond is nearly twice as strong as the O–O bond in peroxides, and the O–H bond is more than 25 kcal/mole stronger than an S–H bond. Thus, thermodynamics favors disulfide formation over peroxide.
Disulfide bridges in proteins
Disulfide (sulfur-sulfur) linkages between two cysteine residues are an integral component of the three-dimensional structure of many proteins. The interconversion between thiols and disulfide groups is a redox reaction: the thiol is the reduced state, and the disulfide is the oxidized state.
Notice that in the oxidized (disulfide) state, each sulfur atom has lost a bond to hydrogen and gained a bond to a sulfur - this is why the disulfide state is considered to be oxidized relative to the thiol state.
The redox agent that mediates the formation and degradation of disulfide bridges in most proteins is glutathione, a versatile coenzyme that we have met before in a different context (section 14.2A). Recall that the important functional group in glutathione is the thiol, highlighted in blue in the figure below. In its reduced (free thiol) form, glutathione is abbreviated 'GSH'.
In its oxidized form, glutathione exists as a dimer of two molecules linked by a disulfide group, and is abbreviated 'GSSG'.
A new disulfide in a protein forms via a 'disulfide exchange' reaction with GSSH, a process that can be described as a combination of two SN2-like attacks. The end result is that a new cysteine-cysteine disulfide forms at the expense of the disulfide in GSSG.
In its reduced (thiol) state, glutathione can reduce disulfides bridges in proteins through the reverse of the above reaction.
Disulfide bridges exist for the most part only in proteins that are located outside the cell. Inside the cell, cysteines are kept in their reduced (free thiol) state by a high intracellular concentration of GSH, which in turn is kept in a reduced state (ie. GSH rather than GSSG) by a flavin-dependent enzyme called glutathione reductase.
Disulfide bridges in proteins can also be directly reduced by another flavin-dependent enzyme called 'thioredoxin'. In both cases, NADPH is the ultimate electron donor, reducing FAD back to FADH2 in each catalytic cycle.
In the biochemistry lab, proteins are often maintained in their reduced (free thiol) state by incubation in buffer containing an excess concentration of b-mercaptoethanol (BME) or dithiothreitol (DTT). These reducing agents function in a manner similar to that of GSH, except that DTT, because it has two thiol groups, forms an intramolecular disulfide in its oxidized form.
Sulfides
Sulfur analogs of ethers are called sulfides. The chemical behavior of sulfides contrasts with that of ethers in some important ways. Since hydrogen sulfide (H2S) is a much stronger acid than water (by more than ten million fold), we expect, and find, thiols to be stronger acids than equivalent alcohols and phenols. Thiolate conjugate bases are easily formed, and have proven to be excellent nucleophiles in SN2 reactions of alkyl halides and tosylates.
R–S(–) Na(+) + (CH3)2CH–Br (CH3)2CH–S–R + Na(+) Br(–)
Although the basicity of ethers is roughly a hundred times greater than that of equivalent sulfides, the nucleophilicity of sulfur is much greater than that of oxygen, leading to a number of interesting and useful electrophilic substitutions of sulfur that are not normally observed for oxygen. Sulfides, for example, react with alkyl halides to give ternary sulfonium salts (equation # 1) in the same manner that 3º-amines are alkylated to quaternary ammonium salts. Although equivalent oxonium salts of ethers are known, they are only prepared under extreme conditions, and are exceptionally reactive.
sulfides are named using the same rules as ethers except sulfide is used in the place of ether. For more complex substance alkylthio is used instead of alkoxy.
SAM methyltransferases
The most common example of sulfonium ions in a living organism is the reaction of S-Adenosylmethionine. Some of the most important examples of SN2 reactions in biochemistry are those catalyzed by S-adenosyl methionine (SAM) – dependent methyltransferase enzymes. We have already seen, in chapter 6 and again in chapter 8, how a methyl group is transferred in an SN2 reaction from SAM to the amine group on the nucleotide base adenosine:
Another SAM-dependent methylation reaction is catalyzed by an enzyme called catechol-O-methyltransferase. The substrate here is epinephrine, also known as adrenaline.
Notice that in this example, the attacking nucleophile is an alcohol rather than an amine (that’s why the enzyme is called an O-methyltransferase). In both cases, though, a basic amino acid side chain is positioned in the active site in just the right place to deprotonate the nucleophilic group as it attacks, increasing its nucleophilicity. The electrophile in both reactions is a methyl carbon, so there is little steric hindrance to slow down the nucleophilic attack. The methyl carbon is electrophilic because it is bonded to a positively-charged sulfur, which is a powerful electron withdrawing group. The positive charge on the sulfur also makes it an excellent leaving group, as the resulting product will be a neutral and very stable sulfide. All in all, in both reactions we have a reasonably good nucleophile, an electron-poor, unhindered electrophile, and an excellent leaving group.
Because the electrophilic carbon in these reactions is a methyl carbon, a stepwise SN1-like mechanism is extremely unlikely: a methyl carbocation is very high in energy and thus is not a reasonable intermediate to propose. We can confidently predict that this reaction is SN2. Does this SN2 reaction occur, as expected, with inversion of stereochemistry? Of course, the electrophilic methyl carbon in these reactions is achiral, so inversion is not apparent. To demonstrate inversion, the following experiment has been carried out with catechol-O-methyltransferase:
Here, the methyl group of SAM was made to be chiral by incorporating hydrogen isotopes tritium (3H, T) and deuterium (2H, D). The researchers determined that the reaction occurred with inversion of configuration, as expected for an SN2 displacement (J. Biol. Chem. 1980, 255, 9124).
Sulfides can be easily oxidized. Reacting a sulfide with hydrogen peroxide, H2O2, as room termpeature produces a sulfoxide (R2SO). The oxidation can be continued by reaction with a peroxyacid to produce the sulfone (R2SO2)
A common example of a sulfoxide is the solvent dimethyl sulfoxide (DMSO). DMSO is polar aprotic solvent.
Figure AB16.3. DMSO is a very polar, aprotic solvent. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.10%3A_Thiols_%28Mercaptans%29.txt |
This page looks at the manufacture of alcohols by the direct hydration of alkenes, concentrating mainly on the hydration of ethene to make ethanol. It then compares that method with making ethanol by fermentation.
Manufacturing alcohols from alkenes
Ethanol is manufactured by reacting ethene with steam. The catalyst used is solid silicon dioxide coated with phosphoric(V) acid. The reaction is reversible.
Only 5% of the ethene is converted into ethanol at each pass through the reactor. By removing the ethanol from the equilibrium mixture and recycling the ethene, it is possible to achieve an overall 95% conversion. A flow scheme for the reaction looks like this:
The manufacture of other alcohols from alkenes
Some - but not all - other alcohols can be made by similar reactions. The catalyst used and the reaction conditions will vary from alcohol to alcohol. The reason that there is a problem with some alcohols is well illustrated with trying to make an alcohol from propene, CH3CH=CH2. In principle, there are two different alcohols which might be formed:
You might expect to get either propan-1-ol or propan-2-ol depending on which way around the water adds to the double bond. In practice what you get is propan-2-ol. If you add a molecule H-X across a carbon-carbon double bond, the hydrogen nearly always gets attached to the carbon with the most hydrogens on it already - in this case the CH2 rather than the CH. The effect of this is that there are bound to be some alcohols which it is impossible to make by reacting alkenes with steam because the addition would be the wrong way around.
Making ethanol by fermentation
This method only applies to ethanol and you cannot make any other alcohol this way. The starting material for the process varies widely, but will normally be some form of starchy plant material such as maize (US: corn), wheat, barley or potatoes. Starch is a complex carbohydrate, and other carbohydrates can also be used - for example, in the lab sucrose (sugar) is normally used to produce ethanol. Industrially, this wouldn't make sense. It would be silly to refine sugar if all you were going to use it for was fermentation. There is no reason why you should not start from the original sugar cane, though.
The first step is to break complex carbohydrates into simpler ones. For example, if you were starting from starch in grains like wheat or barley, the grain is heated with hot water to extract the starch and then warmed with malt. Malt is germinated barley which contains enzymes which break the starch into a simpler carbohydrate called maltose, \(C_{12}H_{22}O_{11}\). Maltose has the same molecular formula as sucrose but contains two glucose units joined together, whereas sucrose contains one glucose and one fructose unit.
Yeast is then added and the mixture is kept warm (say 35°C) for perhaps several days until fermentation is complete. Air is kept out of the mixture to prevent oxidation of the ethanol produced to ethanoic acid (vinegar). Enzymes in the yeast first convert carbohydrates like maltose or sucrose into even simpler ones like glucose and fructose, both \(C_6H_{12}O_6\), and then convert these in turn into ethanol and carbon dioxide. You can show these changes as simple chemical equations, but the biochemistry of the reactions is much, much more complicated than this suggests.
\[ C_{12}H_{22}O_{11} + H_2O \longrightarrow 2C_6H_{12}O_6 \]
\[ C_6H_{12}O_6 \longrightarrow 2CH_3CH_2OH + 2CO2\]
Yeast is killed by ethanol concentrations in excess of about 15%, and that limits the purity of the ethanol that can be produced. The ethanol is separated from the mixture by fractional distillation to give 96% pure ethanol. For theoretical reasons (minimum boiling point azeotrope), it is impossible to remove the last 4% of water by fractional distillation.
Fermentation Hydration of ethene
Table 1.1.1: A comparison of fermentation with the direct hydration of ethene
Type of process A batch process. Everything is put into a container and then left until fermentation is complete. That batch is then cleared out and a new reaction set up. This is inefficient. A continuous flow process. A stream of reactants is passed continuously over a catalyst. This is a more efficient way of doing things.
Rate of reaction Very slow. Very rapid.
Quality of product Produces very impure ethanol which needs further processing Produces much purer ethanol.
Reaction conditions Uses gentle temperatures and atmospheric pressure. Uses high temperatures and pressures, needing lots of energy input.
Use of resources Uses renewable resources based on plant material. Uses finite resources based on crude oil. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.11%3A_Commercially_Important_Alcohols.txt |
Physical Properties of Alcohols
13-1 Identify which compound is more acidic. Explain your reasoning for each choice.
13-2 Identify which is the most acidic proton in the following compounds. Explain your reasoning for each choice.
13-3 Draw all possible resonance forms of the conjugate base of phenol.
13-4 List the following compounds in order from most to least acidic.
13-5 Predict which compound of each pair is more soluble in water and explain your reasoning.
1. butan-1-ol or pentan-1-ol
2. phenol or cyclohexanol
3. octan-1,3-diol or octan-1-ol
4. 1-chlorohexane or hexan-1-ol
13-6 Predict which compound has the higher boiling point and explain your reasoning.
1. water or ethanol
2. butan-1-ol or octan-1-ol
3. hexan-2-ol or hexan-2-one
Synthesis of Alcohols
13-7 Show a possible way to synthesize the following alcohols.
13-8 Give the product of each reaction.
13-9 Give the product of each reaction.
13-10 Give the product of each reaction.
13-11 Give the product of each reaction.
Organometallic Reagents for Alcohol Synthesis
13-12 Draw the products of the following reactions.
13-13 Draw the products of the following reactions.
13-14 What is the final product of the following reaction.
13-15 Draw the mechanism for question 13-14.
13-16 Identify the product of the following reaction and explain why that is the correct answer.
Addition of Organometallic Reagents to Carbonyl Compounds
13-17 Give the product(s) of the following reactions. Include stereochemistry when necessary.
13-18 Show a possible carbonyl compound that was used to make the following alcohols through a Grignard reaction.
13-19 For the following compounds, identify the Grignard reagent used and the initial methyl ester compound.
13-20 Give the products of the following reactions.
Reduction of the Carbonyl Group
13-21 Identify whether the initial compound is undergoing oxidation or reduction.
13-22 Give the product of each reaction.
13-23 Give the product of each reaction (same starting molecule), making sure to specify where each proton ends up in the final product.
13-24 Give the mechanism for the following hydride reduction reaction.
13-25 Draw the structures for A and B.
Thiols (Mercaptans)
13-26 Name the following compounds following IUPAC nomenclature.
13-27 Identify the product of the following reaction. Include stereochemistry if appropriate.
13-28 Identify the product of the following reaction.
13.13: Solutions to Additional Exercises
Physical Properties of Alcohols
13-1
1. Compound 1 is more acidic than compound 2, since its conjugate base is stabilized by resonance.
2. Compound 1 is more acidic than compound 2 as the proton in question is bonded to a very electronegative atom (oxygen). When comparing the conjugate bases of both compounds, oxygen can stabilize a negative charge far better than the carbon atom of compound 2, allowing it to be a more stable conjugate base and stronger acid.
3. Compound 1 is more acidic than compound 2, since its conjugate base is stabilized by resonance.
4. Compound 2 is more acidic than compound 1. The halogen on compound 2 helps stabilize the negative charge of the conjugate base by withdrawing electron density through induction.
13-2
1. Proton 2 is more acidic than proton 1. The conjugate base formed by removing proton 2 is more stable than the conjugate base formed by removing proton 1, due to the strong induction effects by the halogens which withdraw electron density and stabilize the negative charge.
2. Proton 1 is more acidic than proton 2. The conjugate base formed by removing proton 1 is stabilized by resonance.
13-3 Resonance forms of the phenol conjugate base:
13-4
13-5
1. butan-1-ol is more soluble in water because it has a smaller hydrophobic region compared to pentan-1-ol, allowing butan-1-ol to interact with water better.
2. phenol is more soluble in water than cyclohexanol because of the more polar character of its ring. phenol is able to interact with water better than cyclohexanol due to the conjugated pi-system of electrons in its ring, which which gives it a more ionic character.
3. octan-1,3-diol is more soluble in water as it has two hydroxy groups, allowing it to form more hydrogen bonds and interact with water better than octan-1-ol.
4. hexan-1-ol is more soluble in water as it can hydrogen bond compared to alkyl halides, such as 1-chlorohexane, which are insoluble in water.
13-6
1. Water has a higher boiling point compared to ethanol as it participates in more hydrogen bonding with other water molecules, thus requiring more energy to break the intermolecular attractions between water molecules.
2. octan-1-ol has the higher boiling point compared to butan-1-ol. Both alcohols can H-bond, however the longer hydrophobic carbon chain tail of octan-1-ol experiences more van der Waal interactions compared to the shorter hydrophobic region of butan-1-ol leading to a higher boiling point.
3. Since hexan-1-ol can H-bond, it has a higher boiling point than hexan-2-one, which cannot H-bond.
13-7
13-8
13-9
13-10
13-11
Organometallic Reagents for Alcohol Synthesis
13-12
13-13
13-14
13-15
13-16
When you react an acid with an organometallic reagent, you will get a salt since organometallics are strong bases and will deprotonate the acid, before even getting a chance to attack the carbonyl carbon.
13-17
13-18
13-19
13-20
13-21
1. Reduction
2. Oxidation
3. Oxidation
4. Reduction
13-22
13-23
13-24
13-25
13-26
13-27
13-28 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/13%3A_Structure_and_Synthesis_of_Alcohols/13.12%3A_13.12_Additional_Exercises.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• predict the products, specify the reagents, and determine the mechanism of the reactions of alcohols with
1. hydrohalic acids (refer to section 14.1)
2. phosphorous halides (refer to section 14.2)
3. thionyl chloride (refer to section 14.2)
4. carboxylic acids, acid chlorides, and tosyl choride (refer to section 14.3)
5. dehydrating reagents – H2SO4/heat or POCl3/pyridine (refer to section 14.4)
6. oxidizing agents (refer to section 14.6)
7. sodium and potassium (refer to section 14.11)
• apply the most efficient and effective oxidizing agents (refer to section 14.6)
• determine the alcohol classification using laboratory experiments (refer to section 14.7)
• predict the products and specify the reagents of alcohol protecting group reactions (refer to section 14.9)
• predict the products and specify the reagents for diol cleavage reactions (refer to section 14.10)
• predict the products, specify the reagents for alkoxide ion reactions (refer to section 14.11)
• describe selected alcohol oxidation reactions in biology (refer to section 14.12)
• determine multiple-step synthetic pathways using alcohols (chapters 1-10 and 13-14)
14: Reactions of Alcohols
Conversion of Alcohols into Alkyl Halides
When alcohols react with a hydrogen halide, a substitution takes place producing an alkyl halide and water:
• The order of reactivity of alcohols is 3° > 2° > 1° methyl.
• The order of reactivity of the hydrogen halides is HI > HBr > HCl (HF is generally unreactive).
The reaction is acid catalyzed. Alcohols react with the strongly acidic hydrogen halides HCl, HBr, and HI, but they do not react with nonacidic NaCl, NaBr, or NaI. Primary and secondary alcohols can be converted to alkyl chlorides and bromides by allowing them to react with a mixture of a sodium halide and sulfuric acid:
Because Cl- is a weaker nucleophile than Br-, the reaction with HCl requires a catalyst such as ZnCl2 as shown below.
Mechanisms of the Reactions of Alcohols with HX
Secondary, tertiary, allylic, and benzylic alcohols appear to react by a mechanism that involves the formation of a carbocation, in an \(S_N1\) reaction with the protonated alcohol acting as the substrate.
The \(S_N1\) mechanism is illustrated by the reaction tert-butyl alcohol and aqueous hydrochloric acid (\(H_3O^+\), \(Cl^-\) ). The first two steps in this \(S_n1\) substitution mechanism are protonation of the alcohol to form an oxonium ion. Although the oxonium ion is formed by protonation of the alcohol, it can also be viewed as a Lewis acid-base complex between the cation (\(R^+\)) and \(H_2O\). Protonation of the alcohol converts a poor leaving group (OH-) to a good leaving group water, H2O, which makes the dissociation step of the \(S_N1\) mechanism more favorable.
In step 3, the carbocation reacts with a nucleophile (a halide ion) to complete the substitution.
When we convert an alcohol to an alkyl halide, we carry out the reaction in the presence of acid and in the presence of halide ions, and not at elevated temperature. Halide ions are good nucleophiles (they are much stronger nucleophiles than water), and since halide ions are present in high concentration, most of the carbocations react with an electron pair of a halide ion to form a more stable species, the alkyl halide product. The overall result is an SN1 reaction.
Not all acid-catalyzed conversions of alcohols to alkyl halides proceed through the formation of carbocations. Primary alcohols and methanol react to form alkyl halides under acidic conditions by an SN2 mechanism.
In these reactions the function of the acid is to produce a protonated alcohol. The halide ion then displaces a molecule of water (a good leaving group) from carbon; this produces an alkyl halide:
Again, acid is required. Although halide ions (particularly iodide and bromide ions) are strong nucleophiles, they are not strong enough to carry out substitution reactions with alcohols themselves. Direct displacement of the hydroxyl group does not occur because the leaving group would have to be a strongly basic hydroxide ion:
We can see now why the reactions of alcohols with hydrogen halides are acid-promoted.
Carbocation rearrangements are extremely common in organic chemistry reactions are are defined as the movement of a carbocation from an unstable state to a more stable state through the use of various structural reorganizational "shifts" within the molecule. Once the carbocation has shifted over to a different carbon, we can say that there is a structural isomer of the initial molecule. However, this phenomenon is not as simple as it sounds.
The most common methods for converting 1º- and 2º-alcohols to the corresponding chloro and bromo alkanes (i.e. replacement of the hydroxyl group) are treatments with thionyl chloride and phosphorus tribromide, respectively. These reagents are generally preferred over the use of concentrated HX due to the harsh acidity of these hydrohalic acids and the carbocation rearrangements associated with their use. The alcohol reactions with thionyl chloride or phosphorus tribromide are discussed in the next section.
Exercises
1. Predict the product of each reaction below.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/14%3A_Reactions_of_Alcohols/14.01%3A_Reactions_of_Alcohols_with_Hydrohalic_Acids.txt |
Carbocation containment
Synthetic organic chemists use phosphorus tribromide and thionyl chloride to convert an alcohol into a better leaving group without carbocation rearrangement.
Despite their general usefulness, phosphorous tribromide and thionyl chloride have shortcomings. Hindered 1º- and 2º-alcohols react sluggishly with the former, and may form rearrangement products, as noted in the following equation.
Below, an abbreviated mechanism for the reaction is displayed. The initially formed trialkylphosphite ester may be isolated if the HBr byproduct is scavenged by base. In the presence of HBr a series of acid-base and SN2 reactions take place, along with the transient formation of carbocation intermediates. Rearrangement (pink arrows) of the carbocations leads to isomeric products.
Reaction of thionyl chloride with chiral 2º-alcohols has been observed to proceed with either inversion or retention. In the presence of a base such as pyridine, the intermediate chlorosulfite ester reacts to form an "pyridinium" salt, which undergoes a relatively clean SN2 reaction to the inverted chloride. In ether and similar solvents the chlorosulfite reacts with retention of configuration, presumably by way of a tight or intimate ion pair. This is classified as an SNi reaction (nucleophilic substitution internal). The carbocation partner in the ion pair may also rearrange. These reactions are illustrated by the following equations. An alternative explanation for the retention of configuration, involving an initial solvent molecule displacement of the chlorosulfite group (as SO2 and chloride anion), followed by chloride ion displacement of the solvent moiety, has been suggested. In this case, two inversions lead to retention.
Example: Conversion of Alcohols to Alkyl Chlorides
There’s one important thing to note here: see the stereochemistry? It’s been inverted.*(white lie alert – see below) That’s an important difference between SOCl2 and tosyl chloirde, TsCl, which leaves the stereochemistry alone. The TsCl reaction is studied in section 12.3.
Mechanisms
Since the reaction proceeds through a backside SN2 reaction, there is inversion of configuration at the carbon
The PBr3 reaction is thought to involve two successive SN2-like steps:
Notice that these reactions result in inversion of stereochemistry in the resulting alkyl halide.
Exercises
2. Draw the mechanism of the reaction of thinoylchloride with cyclohexanol, given below.
3. Draw the expected product of the reaction of cylohexanol with the following reagents.
(a) SOCl2 (b) PBr3
Answer
2.
3.
(a) (b)
14.03: Alcohol conversion to Esters - Tosylate and Carboxylate
The poor leaving group of alcohols can be overcome by converting the hydroxyl group to a tosylate ester, an excellent leaving group. The tosylate ester undergoes subsequent reactions (typically SN1 or SN2) as part of a multiple step synthesis.
The synthesis of carboxylate esters (the other ester) is commonly the final step of a synthetic pathway.
Tosylate Ester Formation
We can transform an alcohol group into a sulfonic ester using para-toluene sulfonyl chloride (Ts-Cl) or methanesulfonyl chloride (Ms-Cl), creating what is termed an organic tosylate or mesylate:
Notice that unlike the halogenation reactions of alcohols with thionyl chloride or phosphorous tribromide, conversion of an alcohol to a tosylate or mesylate proceeds with retention of configuration at the electrophilic carbon.
Tosylate/mesylate groups are excellent leaving groups in nucleophilic substitution reactions, due to resonance delocalization of the developing negative charge on the leaving oxygen.
The laboratory synthesis of isopentenyl diphosphate - the 'building block' molecule used by nature for the construction of isoprenoid molecules such as cholesterol and b-carotene - was accomplished by first converting the alcohol into an organic tosylate (step 1), then displacing the tosylate group with an inorganic pyrophosphate nucleophile (step 2) (J. Org. Chem 1986, 51, 4768).
The major reactive species of tosylate chemistry are summarized below.
Exercise
4. Predict the structures of A and B in the following reaction.
Answer
4.
Conversion of Alcohols into Esters
Acid chlorides react with alcohols to form esters
Example \(1\)
Exercise
5. Predict the products or specify the reagents for the following reactions.
Answer
5.
Contributors and Attributions
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/14%3A_Reactions_of_Alcohols/14.02%3A_Reactions_with_Phosphorus_Halides_and_Thionyl_Chloride.txt |
Dehydration of Alcohols to Yield Alkenes
One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo E1 or E2 mechanisms to lose water and form a double bond. The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
The required range of reaction temperature decreases with increasing substitution of the hydroxy-containing carbon:
• 1° alcohols: 170° - 180°C
• 2° alcohols: 100°– 140 °C
• 3° alcohols: 25°– 80°C
If the reaction is not sufficiently heated, the alcohols do not dehydrate to form alkenes, but react with one another to form ethers (e.g., the Williamson Ether Synthesis).
Alcohols are amphoteric; they can act as both acid or base. The lone pair of electrons on oxygen atom makes the –OH group weakly basic. Oxygen can donate two electrons to an electron-deficient proton. Thus, in the presence of a strong acid, R—OH acts as a base and protonates into the very acidic alkyloxonium ion +OH2 (The pKa value of a tertiary protonated alcohol can go as low as -3.8). This basic characteristic of alcohol is essential for its dehydration reaction with an acid to form alkenes.
Mechanism for the Dehydration of Alcohol into Alkene
Different types of alcohols may dehydrate through a slightly different mechanism pathway. However, the general idea behind each dehydration reaction is that the –OH group in the alcohol donates two electrons to H+ from the acid reagent, forming an alkyloxonium ion. This ion acts as a very good leaving group which leaves to form a carbocation. The deprotonated acid (the base) then reacts with the hydrogen adjacent to the carbocation and form a double bond.
Primary alcohols undergo bimolecular elimination (E2 mechanism) while secondary and tertiary alcohols undergo unimolecular elimination (E1 mechanism). The relative reactivity of alcohols in dehydration reactions is ranked as follows:
Methanol < primary < secondary < tertiary
Primary alcohols dehydrate through the E2 mechanism. The hydroxyl oxygen donates two electrons to a proton from sulfuric acid (H2SO4), forming an alkyloxonium ion. Then the conjugate base, HSO4, reacts with one of the adjacent (beta) hydrogen atoms while the alkyloxonium ion leaves in a concerted process, forming a double bond.
Secondary and tertiary alcohols dehydrate through the E1 mechanism. Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, in this case the ion leaves first and forms a carbocation as the reaction intermediate. The water molecule (which is a stronger base than the HSO4- ion) then abstracts a proton from an adjacent carbon to form a double bond. Notice in the mechanism below that the alkene formed depends on which proton is abstracted: the red arrows show formation of the more substituted 2-butene, while the blue arrows show formation of the less substituted 1-butene. Recall that according to Zaitsev's Rule, the more substituted alkenes are formed preferentially because they are more stable than less substituted alkenes. Additinally, trans alkenes are more stable than cis alkenes and are also the major product formed. For the example below, the trans diastereomer of the 2-butene product is most abundant.
Dehydration reaction of secondary alcohol
The dehydration mechanism for a tertiary alcohol is analogous to that shown above for a secondary alcohol.
The E2 elimination of 3º-alcohols under relatively non-acidic conditions may be accomplished by treatment with phosphorous oxychloride (POCl3) in pyridine. This procedure is also effective with hindered 2º-alcohols, but for unhindered and 1º-alcohols an SN2 chloride ion substitution of the chlorophosphate intermediate competes with elimination. Examples of these and related reactions are given in the following figure. The first equation shows the dehydration of a 3º-alcohol. The predominance of the non-Zaitsev product (less substituted double bond) is presumed due to steric hindrance of the methylene group hydrogen atoms, which interferes with the approach of base at that site. The second example shows two elimination procedures applied to the same 2º-alcohol. The first uses the single step POCl3 method, which works well in this case because SN2 substitution is retarded by steric hindrance. The second method is another example in which an intermediate sulfonate ester confers halogen-like reactivity on an alcohol. In every case the anionic leaving group is the conjugate base of a strong acid.
Practice Problems (aka Exercises)
Exercises
6. Starting with cyclohexanol, describe how you would prepare cyclohexene.
7. In the dehydration of 1-methylcyclohexanol, which product is favored?
8.
In the dehydration of this diol the resulting product is a ketone. Draw the mechanism of its formation. (Hint a rearrangement occurs)
9.
Draw an arrow pushing mechanism for the acid catalyzed dehydration of the following alcohol, make sure to draw both potential mechanisms. Assume no rearrangement for the first two product mechanisms. Which of these two would likely be the major product? If there was a rearrangement, draw the expected major product.
Answer
6. H2SO4 with heat since there are no concerns about C+ rearrangement
7. The more substituted alkene is favored, as more substituted alkenes are relatively lower in energy.
8.
This reaction is known as the Pinacol rearrangement.
Note how the carbocation after the rearrangement is resonance stabilized by the oxygen
9. Note: While the mechanism is instructive for the first part of the this answer. The carbocation rearrangement would occur and determine the major and minor products as explained in the second part of this answer.
The major product of this mechanism would be the more highly substituted alkene, or the product formed from the red arrows.
Note: With the secondary carbocation adjacent a tertiary carbon center, a 1,2 hydride shift (rearrangement) would occur to form a tertiary carbocation and vcompound below would be the major product. The minor product being the same product as the one formed from the red arrows.
• Jeffrey Ma | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/14%3A_Reactions_of_Alcohols/14.04%3A_Dehydration_Reactions_of_Alcohols.txt |
Introduction
You are undoubtedly already familiar with the general idea of oxidation and reduction: you learned in general chemistry that when a compound or atom is oxidized it loses electrons, and when it is reduced it gains electrons. You also know that oxidation and reduction reactions occur in pairs: if one species is oxidized, another must be reduced at the same time - thus the term 'redox reaction'.
Most of the redox reactions you have seen previously in general chemistry probably involved the flow of electrons from one metal to another, such as the reaction between copper ion in solution and metallic zinc:
$Cu^{+2}_{(aq)} + Zn_{(s)} \rightarrow Cu_{(s)} + Zn^{+2}_{(aq)} \tag{16.1.1}$
In organic chemistry, redox reactions look a little different. Electrons in an organic redox reaction often are transferred in the form of a hydride ion - a proton and two electrons. Because they occur in conjunction with the transfer of a proton, these are commonly referred to as hydrogenation and dehydrogenation reactions: a hydride plus a proton adds up to a hydrogen (H2) molecule. Be careful - do not confuse the terms hydrogenation and dehydrogenation with hydration and dehydration - the latter refer to the gain and loss of a water molecule (and are not redox reactions), while the former refer to the gain and loss of a hydrogen molecule.
Oxidation and Reduction - The Organic Chemistry View
When a carbon atom in an organic compound loses a bond to hydrogen and gains a new bond to a heteroatom (or to another carbon), we say the compound has been dehydrogenated, or oxidized. A very common biochemical example is the oxidation of an alcohol to a ketone or aldehyde:
When a carbon atom loses a bond to hydrogen and gains a bond to a heteroatom (or to another carbon atom), it is considered to be an oxidative process because hydrogen, of all the elements, is the least electronegative. Thus, in the process of dehydrogenation the carbon atom undergoes an overall loss of electron density - and loss of electrons is oxidation.
Conversely, when a carbon atom in an organic compound gains a bond to hydrogen and loses a bond to a heteroatom (or to another carbon atom), we say that the compound has been hydrogenated, or reduced. The hydrogenation of a ketone to an alcohol, for example, is overall the reverse of the alcohol dehydrogenation shown above. Illustrated below is another common possibility, the hydrogenation (reduction) of an alkene to an alkane.
Hydrogenation results in higher electron density on a carbon atom(s), and thus we consider process to be one of reduction of the organic molecule.
Notice that neither hydrogenation nor dehydrogenation involves the gain or loss of an oxygen atom. Reactions which do involve gain or loss of one or more oxygen atoms are usually referred to as 'oxygenase' and 'reductase' reactions.
For the most part, when talking about redox reactions in organic chemistry we are dealing with a small set of very recognizable functional group transformations. It is therefore very worthwhile to become familiar with the idea of 'oxidation states' as applied to organic functional groups. By comparing the relative number of bonds to hydrogen atoms, we can order the familiar functional groups according to oxidation state. We'll take a series of single carbon compounds as an example. Methane, with four carbon-hydrogen bonds, is highly reduced. Next in the series is methanol (one less carbon-hydrogen bond, one more carbon-oxygen bond), followed by formaldehyde, formate, and finally carbon dioxide at the highly oxidized end of the group.
This pattern holds true for the relevant functional groups on organic molecules with two or more carbon atoms:
Alkanes are highly reduced, while alcohols - as well as alkenes, ethers, amines, sulfides, and phosphate esters - are one step up on the oxidation scale, followed by aldehydes/ketones/imines and epoxides, and finally by carboxylic acid derivatives (carbon dioxide, at the top of the oxidation list, is specific to the single carbon series).
Notice that in the series of two-carbon compounds above, ethanol and ethene are considered to be in the same oxidation state. You know already that alcohols and alkenes are interconverted by way of addition or elimination of water. When an alcohol is dehydrated to form an alkene, one of the two carbons loses a C-H bond and gains a C-C bond, and thus is oxidized. However, the other carbon loses a C-O bond and gains a C-C bond, and thus is considered to be reduced. Overall, therefore, there is no change to the oxidation state of the molecule.
You should learn to recognize when a reaction involves a change in oxidation state in an organic reactant. Looking at the following transformation, for example, you should be able to quickly recognize that it is an oxidation: an alcohol functional group is converted to a ketone, which is one step up on the oxidation ladder.
Likewise, this next reaction involves the transformation of a carboxylic acid derivative (a thioester) first to an aldehyde, then to an alcohol: this is a double reduction, as the substrate loses two bonds to heteroatoms and gains two bonds to hydrogens.
An acyl transfer reaction (for example the conversion of an acyl phosphate to an amide) is not considered to be a redox reaction - the oxidation state of the organic molecule is does not change as substrate is converted to product, because a bond to one heteroatom (oxygen) has simply been traded for a bond to another heteroatom (nitrogen).
It is important to be able to recognize when an organic molecule is being oxidized or reduced, because this information tells you to look for the participation of a corresponding redox agent that is being reduced or oxidized- remember, oxidation and reduction always occur in tandem! We will soon learn in detail about the most important biochemical and laboratory redox agents.
Exercises
10. Indicate whether the following reactions are oxidations [O], reductions [H], hydrations, or dehydrations.
Answer
10.
a) reduction
b) hydration
c) oxidation
d) dehydration
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/14%3A_Reactions_of_Alcohols/14.05%3A_Oxidation_States_of_Alcohols_and_Related_Functional_Groups.txt |
Oxidizing agents
The oxidizing agent commonly shown is a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \tag{17.7.1}$
• K2Cr2O7 potassium dichromate
• CrO3 Chromium Trioxide
• Pyridinium chlorochromate (PCC) is a milder version of chromic acid that is suitable for converting a primary alcohol into an aldehyde without oxidizing it all the way to a carboxylic acid. This reagent is being replaced in laboratories by Dess‑Martin periodinane (DMP), which has several practical advantages over PCC, such as producing higher yields and requiring less rigorous reaction conditions. DMP is named after Daniel Dess and James Martin, who developed it in 1983. Both reagents are used along with H2SO4, H2O
Primary alcohols
Primary alcohols can be oxidized to either aldehydes or carboxylic acids depending on the reaction conditions. In the case of the formation of carboxylic acids, the alcohol is first oxidized to an aldehyde which is then oxidized further to the acid.
Full oxidation to carboxylic acids
You need to use an excess of the oxidizing agent and make sure that the aldehyde formed as the half-way product stays in the mixture. The alcohol is heated under reflux with an excess of the oxidizing agent. When the reaction is complete, the carboxylic acid is distilled off. The full equation for the oxidation of ethanol to ethanoic acid is:
$3CH_3CH_2OH + 2Cr_2O_7^{2-} + 16H+ \rightarrow 3CH_3COOH + 4Cr^{3+} + 11H_2O \tag{17.7.1}$
The more usual simplified version looks like this:
$CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O \tag{17.7.2}$
Alternatively, you could write separate equations for the two stages of the reaction - the formation of ethanal and then its subsequent oxidation.
$CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O \tag{17.7.3}$
$CH_3CHO + [O] \rightarrow CH_3COOH \tag{17.7.4}$
This is what is happening in the second stage:
Secondary alcohols
Secondary alcohols are oxidized to ketones - and that's it. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute sulfuric acid, you get propanone formed. Playing around with the reaction conditions makes no difference whatsoever to the product. Using the simple version of the equation and showing the relationship between the structures:
If you look back at the second stage of the primary alcohol reaction, you will see that an oxygen "slotted in" between the carbon and the hydrogen in the aldehyde group to produce the carboxylic acid. In this case, there is no such hydrogen - and the reaction has nowhere further to go.
Tertiary alcohols
Tertiary alcohols are not oxidized by acidified sodium or potassium dichromate(VI) solution - there is no reaction whatsoever. If you look at what is happening with primary and secondary alcohols, you will see that the oxidizing agent is removing the hydrogen from the -OH group, and a hydrogen from the carbon atom attached to the -OH. Tertiary alcohols don't have a hydrogen atom attached to that carbon.
You need to be able to remove those two particular hydrogen atoms in order to set up the carbon-oxygen double bond.
Oxidation of Primary Alcohols by PCC - a closer look
Pyridinium chlorochromate (PCC) is a milder version of chromic acid.
PCC oxidizes alcohols one rung up the oxidation ladder, from primary alcohols to aldehydes and from secondary alcohols to ketones. Unlike chromic acid, PCC will not oxidize aldehydes to carboxylic acids. Similar to or the same as: $CrO_3$ and pyridine (the Collins reagent) will also oxidize primary alcohols to aldehydes. Here are two examples of PCC in action.
• If you add one equivalent of PCC to either of these alcohols, you obtain the oxidized version. The byproducts (featured in grey) are Cr(IV) as well as pyridinium hydrochloride.
• One has to be careful with the amount of water present in the reaction. If water were present, it can ad to the aldehyde to make the hydrate, which could be further oxidized by a second equivalent of PCC were it present. This is not a concern with ketones, since there is no H directly bonded to C.
How does it work? Oxidation reactions of this sort are actually a kind of elimination reaction. We’re going from a carbon-oxygen single bond to a carbon-oxygen double bond. The elimination reaction can occur because we’re putting a good leaving group on the oxygen, namely the chromium, which will be displaced when the neighboring C-H bond is broken with a base.
In the first step, the oxygen on the chromium reacts with the alcohol hydroxy group to form the Cr-O bond. Secondly, a proton on the (now positive) OH is transferred to one of the oxygens of the chromium, possibly through the intermediacy of the pyridinium salt. A chloride ion is then displaced, in a reaction reminiscent of a 1,2 elimination reaction, to form what is known as a chromate ester.
The C-O double bond is formed when a base removes the proton on the carbon adjacent to the oxygen. [aside: I've drawn the base as Cl- although there are certainly other species which could also act as bases here (such as an alcohol). It is also possible for pyridine to be used as the base here, although only very low concentrations of the deprotonated form will be present under these acidic conditions.] The electrons from the C-H bond move to form the C-O bond, and in the process break the O-Cr bond, and Cr(VI) becomes Cr(IV) in the process (drawn here as O=Cr(OH)2 ).
Real life notes: If you end up using PCC in the lab, don’t forget to add molecular sieves or Celite or some other solid to the bottom of the flask, because otherwise you get a nasty brown tar that is a real major pain to clean up. The toxicity and mess associated with chromium has spurred the development of other alternatives like TPAP, IBX, DMP, and a host of other neat reagents you generally don’t learn about until grad school.
Examples
Exercises
11. Draw the alcohol that the following ketones/aldehydes would have resulted from if oxidized. What oxidant could be used?
(a) (b)
12. Show the products of the oxidation of 1-propanol and 2-propanol with chromic acid in aqueous solution.
Answer
11. Any oxidant capable of oxidizing an alcohol to a ketone would work, such as the Jones reagent (CrO3, H2SO4, H2O), PCC, or Dess-Martin periodinane.
(a)
(b) Since this is a primary alcohol, there are some precautions necessary to avoid formation of the carboxyllic acid. Milder oxidants such as the Dess-Martin periodinane, and also PCC (there is no water to form the carboxyllic acid) would work.
12. The answers are correlated with the question.
Contributors and Attributions
James Ashenhurst (MasterOrganicChemistry.com) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/14%3A_Reactions_of_Alcohols/14.06%3A_Oxidation_Reactions_of_Alcohols.txt |
Using alcohol reactivity to distinguish between classifications
The presence of an alcohol can be determined with test reagents that react with the -OH group. The initial test to identify alcohols is to take the neutral liquid, free of water and add solid phosphorus(V) chloride. A a burst of acidic steamy hydrogen chloride fumes indicate the presence of an alcohol. Subsequent tests are needed to distinguish between alcohol classifications.
Determining the tertiary alcohol
A few drops of the alcohol are added to a test tube containing potassium dichromate(VI) solution acidified with dilute sulfuric acid. The tube is warmed in a hot water bath.
After heating, the following colors are observed:
In the case of a primary or secondary alcohols, the orange solution turns green. The Schiff's test will need to be performed to distinguish between the primary and secondary alcohols. With a tertiary alcohol, there is no color change.
Schiff's Reagent - Distinguishing between the primary and secondary alcohols
Schiff's reagent is a fuchsian dye decolorized by passing sulfur dioxide through it. In the presence of even small amounts of an aldehyde, it turns bright magenta. It must, however, be used absolutely cold, because ketones react with it very slowly to give the same color. Heat obviously causes a faster color change, but is potentially confusing because of the competing ketone reaction. While warming the reaction mixture in the hot water bath, pass any vapors produced through some Schiff's reagent.
• If the Schiff's reagent quickly becomes magenta, then an aldehyde was produced from a primary alcohol.
• If there is no color change in the Schiff's reagent, or only a trace of pink color within a minute or so, then no aldehyde was produced and a primary alcohol is not present.
A secondary alcohol is identified by the color change with the acidified potassium dichromate(VI) solution and the absence of a color change with the Schiff's reagent might.
Exercise
13. The chromic acid oxidation test and Schiff's test are performed on the three alcohols shown below. Describe the expected test results.
Answer
13.
a) The chromic acid solution turns green, but the Schiff's reagent remains colorless.
b) The chromic acid solution turns green, and the Schiff's reagent turns magenta.
c) The chromic acid solution remains orance, and the Schiffs reagent remains colorless.
14.08: Protection of Alcohols
Introduction
Often during the synthesis of complex molecules, one functional group in a molecule interferes with an intended reaction on a second functional group on the same molecule. An excellent example is the fact that a Grignard reagent can't be prepared from halo alcohol because the C-Mg bond is not compatible with the acidic -OH group.
When situations like this occurs, chemists circumvent the problem by protecting the interfering functional group.
Functional group protection involves three steps:
1. Blocking the interfering functionality by introducing a protecting group.
2. Performing the intended reaction.
3. Removing the protecting group and reforming the original functional group.
There are several methods for protecting an alcohol, however, the most common is the reaction with a chlorotrialkylsilane, Cl-SiR3 This reactions forms a trialkylsilyl ether, R'-O-SiR3. Chlorotrimethylsilane is often used in conjuction with a base, such as triethylamine, The base helps to form the alkoxide anion and remove the HCl produced by the reaction.
Example
The silyl ether protecting group can be removed by reaction with an aqueous acid or the fluoride ion.
By utilizing a protecting group a Grignad reagent can be formed and reacted on a halo alcohol.
1) Protect the Alcohol
2) Form the Grignard Reagent
3) Perform the Grignard Reaction
4) Deprotection
Exercise
14. Propose a multiple-step synthesis to transform 4-bromo-1-butanol into 5-methylhexane-1,5-diol.
Answer
14.
14.09: Cleavage of Diols
Glycol Cleavage
The vicinal glycols prepared by alkene hydroxylation (reaction with osmium tetroxide or permanganate) are cleaved to aldehydes and ketones in high yield by the action of lead tetraacetate (Pb(OAc)4) or periodic acid (HIO4). This oxidative cleavage of a carbon-carbon single bond provides a two-step, high-yield alternative to ozonolysis, that is often preferred for small scale work involving precious compounds. A general equation for these oxidations is shown below. As a rule, cis-glycols react more rapidly than trans-glycols, and there is evidence for the intermediacy of heterocyclic intermediates (as shown), although their formation is not necessary for reaction to occur.
Exercise
15. Predict the product of the reaction below.
Answer
15.
14.10: Reactions of Alkoxides
Introduction
The hydrogen atom of a hydroxyl group is ionizable and can be replaced by other substituents as illustrated in the reactions below. The first reaction shows simple alcohols with sodium (and sodium hydride). The second reaction shows the isotopic exchange that occurs when mixing an alcohol with deuterium oxide (heavy water). This exchange, which is catalyzed by acid or base, is rapid under normal conditions because it is difficult to avoid traces of these catalysts in most experimental systems.
2 R–O–H + 2 Na 2 R–O(–)Na(+) + H2
R–O–H + D2O R–O–D + D–O–H
The mechanism by which these substitution reactions proceed is straightforward. The oxygen atom of an alcohol is nucleophilic; therefore, it is prone to react with electrophiles. The resulting "onium" intermediate then loses a proton to a base, forming the substitution product. If a strong electrophile is not present, then the nucleophilicity of the oxygen may be enhanced by conversion to its conjugate base (an alkoxide). This powerful nucleophile then reacts with weak electrophiles. These two variations of the substitution mechanism are illustrated in the following diagram.
Williamson Ether Synthesis
Alkyl substitution of the hydroxyl group creates ethers. This reaction provides examples of both strong electrophilic substitution (first equation below) and weak electrophilic substitution (second equation). The latter SN2 reaction is known as the Williamson ether synthesis and is generally only used with 1º alkyl halide reactants because the strong alkoxide base leads to E2 elimination with 2º and 3º alkyl halides.
Exercise
16. Show how you would use the Williamson Ether synthesis to make 1-ethoxy-2-methylpropane from isobutyl alcohol and ethanol.
Answer
16. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/14%3A_Reactions_of_Alcohols/14.07%3A_Determining_Alcohol_Classifications_in_the_Lab_-_alternate_reactions.txt |
Foundation
All reactions which involve electron flow are considered oxidation-reduction reactions. The basic definition can be defined as: One reactant is oxidized (loses electrons), while another is reduced (gains electrons). A couple of basic oxidation-reduction or "redox" example's are given here.
Example 1
The reaction of magnesium metal with oxygen, involves the oxidation of magnesium
$2Mg(s) + O_2(g)→ 2MgO(s) \label{1}$
Since the magnesium solid is oxidized, we expect to see a loss of electrons. Similarly, since oxygen must therefore be reduced, we should see a gain of electrons.
As the magnesium is oxidized there is a loss of 2 electrons while simultaneously, oxygen gains those two electrons. Another example of a redox reaction is with the two gasses CO2 and H2. This redox reaction also demonstrates the importance of implementing "oxidation numbers" in the methodology of redox reactions, allowing for the determination of which reactant is being reduced and which reactant is being oxidized.
Example 2
The reaction of carbon dioxide gas with hydrogen gas, involving the oxidation of hydrogen
$CO_2 (g) + H_2 (g) → 2CO (g) + H_2O (g) \label{3}$
Since the hydrogen gas is being oxidized (reductant), we expect to see an overall loss of electrons for the resulting molecule. Similarly, we expect to see a gain in the overall number of electrons for the resulting molecule of the oxidant (CO2).
Here it is possible to infer that the carbon of CO2 is being reduced by review of its unique oxidation number. Such that, C (of CO2) goes from an oxidation number of +4 to C (of CO) having an oxidation number of +2, representing a loss of two electrons. Similarly, H2 is noted as going from an oxidation number of 0 to +1, or gaining one electron in a reduction process. For more information on oxidation numbers, review the following link: Oxidation-Reduction Reactions
A Basic Biological Model
The flow of electrons is a vital process that provides the necessary energy for the survival of all organisms. The primary source of energy that drives the electron flow in nearly all of these organisms is the radiant energy of the sun, in the form of electromagnetic radiation or Light. Through a series of nuclear reactions, the sun is able to generate thermal energy (which we can feel as warmth) from electromagnetic radiation (which we perceive as light). However, the particular wavelength of the electromagnetic spectrum we are able to detect with the human eye is only between 400 and 700 nm in wavelength. It should therefore be noted that the visible part of the electromagnetic spectrum is actually a small percentage of the whole; where a much greater percentage remains undetectable for the human eye.
In physics, the use of the term "light" refers to electromagnetic radiation of any wavelength, independent of its detectability for the human eye. For plants, the upper and lower ends of the visible spectrum are the wavelengths that help drive the process of splitting water (H2O) during photosynthesis, to release its electrons for the biological reduction of carbon dioxide (CO2) and the release of diatomic oxygen (O2) to the atmosphere. It is through the process of photosynthesis that plants are able to use the energy from light to convert carbon dioxide and water into the chemical energy storage form called glucose.
Plants represent one of the most basic examples of biological oxidation and reduction. The chemical conversion of carbon dioxide and water into sugar (glucose) and oxygen is a light-driven reduction process:
$6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2 \label{5}$
The process by which non photosynthetic organisms and cells obtain energy, is through the consumption of the energy rich products of photosynthesis. By oxidizing these products, electrons are passed along to make the products carbon dioxide, and water, in an environmental recycling process. The process of oxidizing glucose and atmospheric oxygen allowed energy to be captured for use by the organism that consumes these products of the plant. The following reaction represents this process:
$C_6H_{12}O_6 + O_2 \rightarrow 6CO_2 + 6H_2O +Energy \label{6}$
It is therefore through this process that heterotrophs (most generally "animals" which consume other organisms obtain energy) and autotrophs (plants which are able to produce their own energy) participate in an environmental cycle of exchanging carbon dioxide and water to produce energy containing glucose for organismal oxidation and energy production, and subsequently allowing the regeneration of the byproducts carbon dioxide and water, to begin the cycle again. Therefore, these two groups of organisms have been allowed to diverge interdependently through this natural life cycle.
Physical Chemistry's Understanding
Biological oxidation-reduction reactions, or simply biological oxidations utilize multiple stages or processes of oxidation to produce large amounts of Gibbs energy, which is used to synthesize the energy unit called adenosine triphosphate or ATP. To efficiently produce ATP, the process of glycolysis must be near an abundance of oxygen. Since glycolysis by nature is not an efficient process, if it lacks sufficient amounts of oxygen the end product pyruvate, is reduced to lactate with NADH as the reducing agent. However, in a more favorable aerobic process, the degradation of glucose through glycolysis proceeds with two additional processes known as the citric acid cycle and the terminal respiratory chain; yielding the end products carbon dioxide and water, which we exhale with each breath.
The products NADH and FADH2 formed during glycolysis and the citric acid cycle are able to reduce molecular oxygen (O2) thereby releasing large amounts of Gibbs energy used to make ATP. The process by which electrons are transferred from NADH or FADH2 to O2 by a series of electron transfer carriers, is known as oxidative phosphorylation. It is through this process that ATP is able to form as a result of the transfer of electrons.
Thee specific examples of redox reactions that are used in biological processes, involving the transfer of electrons and hydrogen ions as follows. During some biological oxidation reactions, there is a simultaneous transfer of hydrogen ions with electrons (1). In other instances, hydrogen ions may be lost by the substance being oxidized while transferring only its electrons to the substance being reduced (2). A third type of biological oxidation might involve only a transfer of electrons (3). It should be noted that biological oxidation rarely proceeds in a direct manner, and generally involves complex mechanisms of several enzymes. The outline below recaps the three processes of biological oxidation stated above, in descending order.
Table 1: Transfer of hydrogen ions and electrons for the general reaction scheme of A + B with intermediate stage shown
Reactants Intermediate Stage Products
AH2 + B [A + 2H+ + 2e- + B] A + BH2
AH2 + B [A + 2H+ + 2e- + B] A + B2- + 2H+
A2- + B [A + 2e- + B] A + B2-
In the last stage of the metabolic process (the terminal respiratory chain), the sequence by which electrons are carried is determined by relative redox potentials. The carrier molecules used to transfer electrons in this stage are called cytochromes, which are an electron-carrying protein containing a heme group. The iron atom of each cytochrome molecule can exist either in the oxidized (Fe3+) or reduced (Fe2+) form. Within the terminal respiratory chain, each carrier molecule alternates between the reduced state and the oxidized state, with molecular oxygen as the final electron acceptor at the end.
It is through the knowledge of redox potentials, that the knowledge of biological processes can be further expanded. The standard reduction potential is denoted as Eo' and is often based on the hydrogen electrode scale of pH 7, rather than pH 0, a common reference point for listed values. Moreover, the superscript symbol ( o ) denotes standard-state conditions, while the adjacent superscript symbol ( ' ) denotes the pH scale of 7 for biochemical processes.
It therefore becomes possible to trace the energy transfer in cells back to the fundamental flow of electrons from one particular molecule to another. Where this electron flow occurs via the physics principle of higher potential to lower potential; similar to a ball rolling down a hill, as opposed to the opposite direction. All of these reactions involving electron flow can be attributed to the basic definition of the oxidation-reduction pathway stated above.
Contributors and Attributions
• Brent Younglove (Hope)
No description | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/14%3A_Reactions_of_Alcohols/14.11%3A_Biological_Oxidation_-_An_Introduction.txt |
14-1 What is the IUPAC name for the product of the following reaction?
a) 1-(3-methylcyclopentyl)ethan-1-ol
b) (3-methylcyclopentyl)acetaldehyde
c) 1-(3-methylcyclopentyl)ethan-1-one
d) 1-(3-methylcyclopentyl)ethane-1,2-diol
14-2 Convert 3-chlorocyclohexanol to the following products. Any of these products can be used as the reactant in any subsequent part.
(a) 3-chlorocyclohexane
(b) 3-chlorocyclohexyl tosylate
(c) 3-chlorohexanone
d) sodium 3-chlorocyclohexan-1-olate
(e) 3-chloro-1-methylcyclohexanol
(f) 1-bromo-3-chlorocyclohexane
(g) 3-chlorocyclohexyl acetate
(h) 1-chloro-3-ethoxycyclohexane
14-3 Show how you would synthesis the chloride, bromide, and iodide from the corresponding alcohols
(a) 1-halopentane (halo=chloro, bromo, iodo)
(b) halocyclobutane
(c) 1-halo-1-ethylcyclopentane
(d) 1-halo-2-propylcyclopentane
14-4 Predict the major products of the following reactions. Clearly indicate stereochemistry where appropriate.
(a) (R)-pentan-2-ol + TsCl in pyridine
(b) (R)-2-pentyl tosylate + NaBr
(c) cyclopentanol + CrO3/H2SO4
(d) 2-cyclopenttylethanol +CrO3.pyridine.HCl
(e) 2-cyclopenttylethanol+ CrO3/H2SO4
(f) 1-propanol + HCl/ZnCl2
(g) 2-methylpropan-2-ol +HBr
(h) ethanol + CH3MgCl
(i) potassium tert-butoxide + ethyl iodide
(j) tert-butyl tosylate + sodium ethoxide
(k) 1-methylcyclohexanol + H2SO4/heat
(l) product from (k) + OsO4/H2O2, then HIO4
(m) sodium cyclohexoxide + 1-iodopropane
(n) sodium ethoxide + isopropyl tosylate
(o) cyclopentylmethanol + DMSO+ oxalyl chloride
(p) cyclopropanol + DMP reagent
14-5 Propose an efficient synthesis for each of the following transformation
(a) (b)
(c) (d)
14-6 Predict the major products of sulfuric acid catalyzed dehydration
(a) butan-1-ol
(b) 2-methyl-3-pentanol
(c) cyclohexanol
(d) 1-cyclopentylethanol
(d) cyclohexylmethanol
(e) 2-methylcyclohexanol
14-7 Predict the products of the following ester synthesis reactions
(a) CH3CH2COOH + CH3CH2CH2OH
(b) CH3CH2OH + HNO3
(c) CH3OH +H3PO4
(d)
(e)
14-8 Show how you would convert (R)-2-pentanol to:
(a) (S)-2-chloropentane
(b) (R)-2-bromopentane
(c) (S)-2-pentanol
14-9 When 3-methyl-2-butanol reacts with concentrated aqueous HBr, the major product is 2-bromo-2-methylbutane
a) Propose a plausible mechanism for the above reaction
b) Show how you would convert 3-methyl-2-butanol into 2-bromo-3-methylbutane:
14-10 Predict the major products when trans-2-ethylcyclopentanol reacts with the following reagents. Include stereochemistry if necessary.
(a) PBr3
(b) SOCl2
(c) Lucas reagent
(d) concentrated HBr
(e) TsCl/py then NaCN
(f)TsCl/py then NaOEt
14-11 Using an alcohol of your choice, show how you would synthesis each following compound.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
14-12 Describe chemical tests that can be used to distinguish the following pairs of compounds. Include the reagents, reaction conditions, observations, and chemical equations in your answers.
(a) 2-propanol and 2-methyl-2-propanol
(b) 1-propanol and 2-propanol
(c) cyclopentanol and cyclopentene
(d) cyclopentanol and 1-cyclopentylethanone
(e) cyclopentanone and 1-methylcyclopentanol
14-13 Draw important resonance structures for the following compounds
(a) (b) (c)
14-14 The following sequence of reaction transforms alcohol A to 3,4-diethyl-2,5-dimethylhexan-3-ol. Propose structure for compounds A, B, C, D, and E.
14-15 Consider the following transformation. Identify the structures of compound X, Y, Z, W, and V.
14-16 Show how each of the following compounds can be synthesized. You might use any alcohol containing five or fewer carbon atoms as your starting materials.
(a) (b) (c)
(d) (e) (f)
(g) (h)
14-17 Show how you would synthesize the following compound. Only use alcohols containing four or fewer carbons as your organic materials. You might use any necessary solvents and inorganic reagents.
14-18
a) The above transformation does not work because of a common conceptual error. What is the conceptual error implicit in this transformation?
b) Show how you could accomplish the transformation in good yield?
14-19 X and Y are constitutional isomers of molecular formula C3H6O. Given the following results with four chemical test, propose structures and assign IUPAC names for X and Y.
SOCl2
K2Cr2O7
Br2 (liquid)
Tollens’ reagent
Compound X
No Rxn
Orange -> Green
No Rxn
No Rxn
Compound Y
Bubbles
Orange -> Green
Decolorize
Grey precipitate of Silver
14-20 The Williamson ether synthesis converts an alkyl halide or tosylate to an ether. Would the following synthesis be possible? If not, explain why not and show an alternative synthesis that would be more likely to work.
14.13: Solutions to Additional Exercises
14-1
C.
14-2
(a) (b)
(c) (d)
(e)
(f)
(g)
(h)
14-3
(a)
(b)
(c)
(d)
14-4
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
14-5
(a)
(b)
(c)
(d)
14-6
(a) (b) (c)
(d) (e) (f)
14-7
(a)
(b)
(c)
(d)
(e)
14-8
(a)
(b)
(c)
14-9
(a)
(b)
14-10
(a) (b) (c)
(d) (e) (f)
14-11
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
14-12
(a) We can use potassium permanganate solution to distinguish between 2-propanol and 2-methyl-2-propanol. In acidic condition, KMnO4 oxidizes 2-propanol into acetone which forms the MnO2 brown precipitate and vanishes KMnO4 purple. As tertiary alcohol cannot be oxidized, 2-methyl-2-propanol remains purple.
(b) 1-propanol and 2-propanol first need to be oxidized into propanal and acetone respectively.
Note: we use pyridinium chlorochromate (PCC) in methylene chloride CH2Cl2 to produce aldehyde without further oxidation.
Fehling’s test then can be used to determine the presence of an aldehyde. Propanal reacts with Fehling’s reagent (Cu2+ in basic solution), forming a brick-red precipitate Cu2O, while acetone cannot react to Fehling’s solution, remaining a deep transparent blue color.
(c) We can use Bromine test to distinguished between cyclopentanol and cyclopentene. Bromine reacts rapidly with cyclopentene, in which the reddish brown color disappears quickly without forming HBr gas bubble. Cyclopentanol does not react with bromine.
(d) Besides KMnO4, K2Cr2O7 in acidic condition is another oxidizing agent that can be used to distinguish between cyclopentanol and cyclopentanone. Acidified K2Cr2O7 oxidizes cyclopentanol into cyclopentanone. Evidence for the reaction is the orange solution (Cr2O72-) turns green solution (Cr3+).
1-cyclopentylethanone cannot be oxidized, remaining the orange solution.
(e) Sodium metal can be used to distinguish between cyclopentanone and 1-methylcyclopentanol. 1-methylcyclopentanol reacts with Na, forming sodium 1-methylcyclopentanolate and releasing H2 bubbles. Cyclopentanone does not react with sodium metal.
14-13
(a)
(b)
(c)
14-14
14-15
14-16
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
14-17
14-18
a) Alcohol functional group typically has pKa of 16 while the pKa of a terminal alkyne is usually about 25. The strong base NaNH2 would deprotonate the stronger acid, which in this case is the terminal alkyne. The resulting alkoxide then react with the alkyl halide CH3CH2Cl
b)
14-19
14-20
Williamson ether synthesis is an SN2 reaction, which favors strong nucleophile and a primary substrate for back-side attack. Since a tertiary alcohol is given, the resulting alkyl halide is also tertiary, which is sterically hindered for SN2 reaction to occur.
The alkoxide then would function as a base, and an elimination reaction would happen instead of SN2 reaction.
An alternative synthesis that is more likely to occur involving the reaction between a tertiary alkoxide and a primary alkyl halide: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/14%3A_Reactions_of_Alcohols/14.12%3A__Additional_Exercises.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• predict relative boiling points and solubilities of ethers (refer to section 15.1)
• explain how ether solvents stabilize electrophilic reagents (refer to section 15.1)
• determine the structures of ethers from their spectra, and explain their characteristic absorptions and fragmentations (refer to section 15.2)
• devise efficient laboratory synthesis of ethers and epoxides, including:
a) Williamson ether synthesis (refer to section 15.3)
b) alkoximercuration-demercuration (refer to section 15.4)
c) peroxyacid epoxidation (refer to Chapter 9 section 12)
d) base-promoted cyclization of halohydrins (refer to section 15.7)
• predict the products or reactions of ethers and epoxides, including:
a) acidic cleavage of ethers (refer to section 15.5)
b) opening of epoxides (refer to section 15.8)
c) reactions of epoxides with organometallic reagents (refer to section 15.10)
• explain how Crown ethers solvate metal cations (refer to section 15.10)
• explain the reaction of epoxy monomers to form the adhesive resin (refer to section 15.11)
• describe the structure and reactive of sulfides (refer to section 15.12)
• use your knowledge of chemical reactivity to propose mechanisms and products for similar reactions you have never seen before (chapters to date)
• propose multiple-step syntheses using all of the reactions studied through this chapter (chapters to date)
Please note: IUPAC nomenclature and important common names of alcohols were explained in Chapter 3.
15: Ethers Epoxides and Thioethers
Comparisons of Physical Properties of Alcohols and Ethers
Ether molecules have no hydrogen atom on the oxygen atom (that is, no OH group). Therefore there is no intermolecular hydrogen bonding between ether molecules, and ethers therefore have quite low boiling points for a given molar mass. Ether molecules do have an oxygen atom, however, and engage in hydrogen bonding with water molecules. Consequently, an ether has about the same solubility in water as the alcohol that is isomeric with it. For example, dimethyl ether and ethanol (both having the molecular formula C2H6O) are completely soluble in water, whereas diethyl ether and 1-butanol (both C4H10O) are barely soluble in water (8 g/100 mL of water). Indeed, ethers have boiling points about the same as those of alkanes of comparable molar mass and much lower than those of the corresponding alcohols as shown in the table below.
Table. Comparison of Boiling Points of Alkanes, Alcohols, and Ethers
Condensed Structural Formula Name Molar Mass Boiling Point (°C) Intermolecular Hydrogen Bonding in Pure Liquid?
CH3CH2CH3 propane 44 –42 no
CH3OCH3 dimethyl ether 46 –25 no
CH3CH2OH ethyl alcohol 46 78 yes
CH3CH2CH2CH2CH3 pentane 72 36 no
CH3CH2OCH2CH3 diethyl ether 74 35 no
CH3CH2CH2CH2OH butyl alcohol 74 117 yes
Ethers are Good Solvents for Many Organic Reactions
Ethers can only accept H-bonds, while alcohols are both H-bond donors and acceptors. The ability of ethers to accept H-bonds combined with the London forces of the alkyl groups bonded to the oxygen allows ethers to be excellent solvents for a wide range of organic compounds. The low chemical reactivity of ethers also makes ethers a preferred solvent for many organic reactions. Additionally, the high volatility of ethers allows for their evaporation when isolating reaction products.
Exercise
1. Draw the bond-line structures and arrange the following ethers in order of increasing boiling point: 1-propoxybutane, diethyl ether, 1-ethoxybutane, dibutyl ether.
2. Arrange the following ethers in order of increasing water solubility: 1-propoxybutane, diethyl ether, 1-ethoxybutane, dibutyl ether.
Answer
1.
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/15%3A_Ethers_Epoxides_and_Thioethers/15.01%3A_Physical_Properties_of_Ethers.txt |
Infrared Spectroscopy
Oxygen forms two bonds. An oxygen atom could be found in between two carbons, as in dibutyl ether.
If you look at an IR spectrum of dibutyl ether, you will see:
• there are the usual sp3 C-H stretching and CH2 bending modes at 2900 and 1500 cm-1.
• there is a strong peak near 1000 cm-1. This peak is due to the C-O stretching vibration.
NMR Spectroscopy
• Hydrogens on carbon adjacent to the ether show up in the region of 3.4-4.5 ppm.
• Similar peaks in epoxides are shifted to a slightly higher field than other ethers. Hydrogens on carbons in and epoxide show up at 2.5 to 3.5 ppm.
The 1H NMR spectrum of dipropyl ether shows three signals with the triplet at 3.37 ppm assigned to the -CH2- beside the ether and the other two signals upfield (1.59 and 0.93 ppm). Notice the protons closer to the electron withdrawing oxygen atom are further downfield indicating some deshielding. Protons at (A) and (C) are each coupled to two equivalent (B) protons. So, each of these signals appears as a triplet. The (B) protons in turn are coupled to a set of two and three equivalent protons and you would therefore formally expect a quartet of triplets. However, because the coupling constants are very similar, the signal appears as a sextet. Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 28 June 2017)
Exercise
2. A mixture of ethers was separated into two fractions: A and B. Elemental analysis reveals that the fractions are structural isomers: 72% C, 12% H, and 16% O. The IR spectra for both fractions show a couple weak bands near 3050 cm-1, several stronger bands around 2950 cm-1, and a strong, sharp band near 1204 cm-1. The proton and 13C NMR spectra for each fraction are shown below. Give the common name and draw the bond-line structure for each fraction and correlate the NMR signals with their respective atoms.
Answer
2.
15.03: The Williamson Ether Synthesis
One important procedure, known as the Williamson Ether Synthesis, proceeds by an SN2 reaction of an alkoxide nucleophile with an alkyl halide. There are four reaction pathways possible between an alkoxide and alkyl halide: SN2, SN1, E2, and E1. To maximize the amount of ether produced by the SN2 mechanism, use a 1º alkyl halide as the electrophile because the strong alkoxide base leads to E2 elimination with 2º and 3º alkyl halides.
The reactions below show how alkoxide and alkyl halide structure can influence the products when applied to an unsymmetrical ether. Two different combinations of reactants are possible. Of these one is usually better than the other for ether synthesis. The first reaction gives a better and cleaner yield of benzyl isopropyl ether, while the second reaction generates considerable elimination product.
Exercises
3. When preparing ethers using the Williamson ether synthesis, what factors are important when considering the nucleophile and the electrophile?
4. How would you synthesize the following ethers? Keep in mind there are multiple ways. The Williamson ether synthesis, alkoxymercuration of alkenes, and also the acid catalyzed substitution.
(a) (b) (c) (d) (e)
5. Draw the electron arrow pushing mechanism for the formation of diethyl ether in the previous problem.
Answer
3. The nucleophile ideally should be very basic, yet not sterically hindered. This will minimize any elimination reactions. The electrophile should have the characteristics of a good SN2 electrophile, preferably primary to minimize any elimination reactions.
4. The Williamson ether syntheses require added catalytic base. Also, most of the halides can be interchanged, say for example for a -Br or a -Cl. Although, typically -I is the best leaving group.
(a) (b) (c) (d) Note, there is only one ether (also called a silyl ether, and often used as an alcohol protecting group.) The other group is an ester.
(e)
5. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/15%3A_Ethers_Epoxides_and_Thioethers/15.02%3A_Spectroscopy_of_Ethers.txt |
Introduction
Acid-catalyzed ether synthesis from alkenes is limited by carbocation stability. Carbocation rearrangement can occur to form a more stable ion as shown in the example below.
Acid-catalyzed ether synthesis using the alkoxymercuration-demercuration reaction pathway reliably produces the Markovnikov product without carbocation rearrangment as shown in the example below.
Alkoxymercuration-demercuration is a two step pathway used to produce ethers that proceeds in a Markovnikov manner and is stereospecific (anti addition). The two steps of alkoxymerecuration-demercuraton take place on opposite faces of the double bond creating trans stereochemistry.
Alkoxymercuration-Demercuration Mechanism
This reaction follows electrophilic addition mechanism we have learned. The major difference is that a mercurium ion bridge stabilizes the carbocation intermediate so that it cannot rearrange. Metals are electropositive. Mercury carries a partial positive charge in the acetate complex and is the electrophile. During the first step of this mechanism, the pi electrons form a bond to mercury while the lone pair on the mercury simultaneously bonds to the other vinyl carbon creating a mercurium ion bridge. The mercurium ion forms in conjunction with the loss of an acetate ion. The mercurium ion stabilizes the carbocation so that it does not rearrange. In the second step of this mechanism, an alcohol molecule reacts with the most substituted carbon to open the mercurium ion bridge. The third step of this mechanism is a proton transfer to a solvent alcohol molecule to neutralize the addition product. The fourth step of the reaction pathway is the reduction of the organomercury intermediate with sodium borohydride under basic conditions. The mechanism of the fourth step is beyond the scope of first year organic chemistry.
Notice that overall, the alkoxymercuration - demercuration mechanism follows Markovnikov's regioselectivity with the OR group attached to the most substituted carbon and the H attached to the least substituted carbon. The reaction is useful, because strong acids are not required and carbocation rearrangements are avoided because no discreet carbocation intermediate forms.
Exercise
6. Show how 3-methyl-2-isopropoxypentane may be synthesized by
a) Williamson ether synthesis
b) Alkoxymercuration-demercuration
c) Which synthesis is better? Why?
Answer
6.
a)
b)
c) The alkoxymercuration-demercuration is more effective because the Williamson Synthesis would use a bulky base and secondary alkyl halide. These reactants would favor the elimination mechanism. Additionally, the alkene can under go carbocation rearrangement, so the mercurium ion stabilization of the reactive intermediate is needed.
15.05: Acidic Cleavage of Ethers
The most common reaction of ethers is cleavage of the C–O bond by strong acids. This may occur by SN1 or E1 mechanisms for 3º-alkyl groups or by an SN2 mechanism for 1º-alkyl groups.
Some examples are shown in the following diagram. The conjugate acid of the ether is an intermediate in all these reactions, just as conjugate acids were intermediates in certain alcohol reactions.
The first two reactions proceed by a sequence of SN2 steps in which the iodide or bromide anion displaces an alcohol in the first step, and then converts the conjugate acid of that alcohol to an alkyl halide in the second. Since SN2 reactions are favored at least hindered sites, the methyl group in example #1 is cleaved first. The 2º-alkyl group in example #3 is probably cleaved by an SN2 mechanism, but the SN1 alternative cannot be ruled out. The phenol formed in this reaction does not react further, since SN2, SN1 and E1 reactions do not take place on aromatic rings. The last example shows the cleavage of a 3º-alkyl group by a strong acid. Acids having poorly nucleophilic conjugate bases are often chosen for this purpose so that E1 products are favored.
Exercise
7. Draw the bond-line structures of the product(s) for each reaction below.
Answer
7.
15.06: Autoxidation of Ethers
Over time ethers that are exposed to air will autoxidize to peroxides which are potentially explosive. The general reaction is shown below.
For example, diisopropyl ether will autoxidize to the products shown in the reaction below.
How to avoid explosions.
1. buy ether in small quantities
2. keep containers tightly sealed
3. use opened containers promptly
4. discard suspect containers
15.07: Synthesis of Epoxides
Epoxides (also known as oxiranes) are three-membered ring structures in which one of the vertices is an oxygen and the other two are carbons.
The most important and simplest epoxide is ethylene oxide which is prepared on an industrial scale by catalytic oxidation of ethylene by air.
Ethylene oxide is used as an important chemical feedstock in the manufacturing of ethylene glycol, which is used as antifreeze, liquid coolant and solvent. In turn, ethylene glycol is used in the production of polyester and polyethylene terephthalate (PET) the raw material for plastic bottles.
Peroxyacid reactions with Alkenes
Oxacyclopropane synthesis by peroxycarboxylic acid requires an alkene and a peroxycarboxylic acid as well as an appropriate solvent. The peroxycarboxylic acid has the unique property of having an electropositive oxygen atom on the COOH group. The reaction is initiated by the electrophilic oxygen atom reacting with the nucleophilic carbon-carbon double bond. The mechanism involves a concerted reaction with a four-part, circular transition state. The result is that the originally electropositive oxygen atom ends up in the oxacyclopropane ring and the COOH group becomes COH.
Intramolecular Williamson Ether Synthesis via Halohydrins
Epoxides can also be synthesized by the treatment of a halohydrin with a base. This causes an intramolecular Williamson ether synthesis.
Exercise
8. What reagents would you use to perform the following transformations?
(a)
(b)
(c)
(d)
Answer
8.
(a)
(b) Note the cis addition
(c)
An oxidation to an alcohol through hydroboration, and subsequent substitution with 2-bromopropane could also work, but this route provides the least likelihood of an elimination reaction occurring.
(d)
Lindlar's catalyst reduces alkynes to cis/Z alkenes. This stereochemistry is retained after epoxidation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/15%3A_Ethers_Epoxides_and_Thioethers/15.04%3A_Alkoxymercuration-Demercuration_Synthesis_of_Ethers.txt |
Epoxide ring-opening reactions - SN1 vs. SN2, regioselectivity, and stereoselectivity
The ring-opening reactions of epoxides provide an excellent review of the differences between SN1 and SN2 reactions. Both mechanisms are good examples of regioselective reactions. In a regioselective reaction, two (or more) different constitutional isomers are possible as products, but one is formed preferentially (or sometimes exclusively). Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions. For the SN1 mechanism, the stability of the charged intermediate determines the regioselectivity. For the concerted SN2 mechanism, sterics are the dominating consideration. If the epoxide is asymmetric, the structure of the product will vary according to which mechanism dominates. When an asymmetric epoxide undergoes solvolysis in basic methanol, ring-opening occurs by an SN2 mechanism, and the less substituted carbon is the site of nucleophilic reaction, leading to what we will refer to as product B:
Conversely, when solvolysis occurs in acidic methanol, the reaction occurs by a mechanism with substantial SN1 character, and the more substituted carbon is the site of reaction. As a result, product A predominates.
Let us examine the basic, SN2 case first. The leaving group is an alkoxide anion, because there is no acid available to protonate the oxygen prior to ring opening. An alkoxide is a poor leaving group, and thus the ring is unlikely to open without a 'push' from the nucleophile.
The nucleophile itself is a potent, deprotonated, negatively charged methoxide ion. When a nucleophilic substitution reaction involves a poor leaving group and a powerful nucleophile, it is very likely to proceed by an SN2 mechanism.
What about the electrophile? There are two electrophilic carbons in the epoxide, but the best target for the nucleophile in an SN2 reaction is the carbon that is least hindered. This accounts for the observed regiochemical outcome. Like in other SN2 reactions, bimolecular, nucleophilic substitution reactions take place from the backside, resulting in inversion at the electrophilic carbon.
The acid-catalyzed epoxide ring-opening reaction mechanism is analogous to the formation of the bromonium ion in halogenation of alkenes and mercurium ion formation in oxymercuration/demercuratioin or alkoxymercuration/demercuration. First, the oxygen is protonated, creating a good leaving group (step 1 below) . Then the carbon-oxygen bond begins to break (step 2) and positive charge begins to build up on the more substituted carbon (recall carbocation stability).
Unlike in an SN2 reaction, the nucleophile reacts with the electrophilic carbon (step 3) before a complete carbocation intermediate has a chance to form.
The reaction takes place preferentially from the backside (like in an SN2 reaction) because the carbon-oxygen bond is still to some degree in place, and the oxygen blocks reaction from the front side. Notice, however, how the regiochemical outcome is different from the base-catalyzed reaction: in the acid-catalyzed process, the nucleophile reacts with the more substituted carbon because this carbon that holds a greater degree of positive charge.
Example
Predict the major product(s) of the ring opening reaction that occurs when the epoxide shown below is treated with:
1. ethanol and a small amount of sodium hydroxide
2. ethanol and a small amount of sulfuric acid
Hint: be sure to consider both regiochemistry and stereochemistry!
Answer
Anti Dihydroxylation
Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the syn-hydroxylation reaction. Proton transfer from the acid catalyst generates the conjugate acid of the epoxide, which is attacked by nucleophiles such as water in the same way that the cyclic bromonium ion described above undergoes reaction. The result is anti-hydroxylation of the double bond. In the following equation this procedure is illustrated for a cis-disubstituted epoxide, which can be prepared from the corresponding cis-alkene. This hydration of an epoxide does not change the oxidation state of any atoms or groups.
Addition of HX
Epoxides can also be opened by other anhydrous acids (HX) to form a trans halohydrin. When both the epoxide carbons are either primary or secondary the halogen anion will react with the less substituted carbon and an SN2 like reaction. However, if one of the epoxide carbons is tertiary, the halogen anion will primarily react with the tertiary carbon in a SN1 like reaction.
Example
In the first example, the epoxide is formed by a secondary and primary carbon. The reaction with the Cl- nucleophile proceeds via the SN2 mechanism and reacts with the least substituted carbon.
In the second example, the epoxide is formed by a tertiary and primary carbon. The reaction with Cl- nucleophile proceeds via the SN1 like mechanism and reacts with the most substituted carbon because it carries the greater partial positive charge.
Exercise
9. Given the following, predict the product assuming only the epoxide is affected. (Remember stereochemistry)
10. Predict the product of the following, similar to above but a different nucleophile is used and not in acidic conditions. (Remember stereochemistry)
11. Epoxides are often very useful reagents to use in synthesis when the desired product is a single stereoisomer. If the following alkene were reacted with an oxyacid to form an epoxide, would the result be a enantiomerically pure? If not, what would it be?
Answer
9.
Note that the stereochemistry has been inverted
10.
11.
First, look at the symmetry of the alkene. There is a mirror plane, shown here.
Then, think about the mechanism of epoxidation with an oxyacid, take for example mCPBA. The mechanism is concerted, so the original cis stereochemistry is not changed. This leads to "two" epoxides.
However, these two mirror images are actually identical due to the mirror plane of the cis geometry. It is a meso compound, so the final result is a single stereoisomer, but not a single enantiomer. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/15%3A_Ethers_Epoxides_and_Thioethers/15.08%3A_Opening_of_Epoxides.txt |
Grignard Reactions with Epoxides
Grignard reactions with ethylene oxide produce a primary alcohol containing two more carbon atoms than the original Grignard reagent.
The first step of the mechanism is shown below. With the second step following the protonation step common to the other reaction pathways studied in this section. This reaction follows the same SN2 mechanism as the opening of epoxide rings under basic conditions since Grignard reagents are both strong nucleophiles and strong bases.
Exercise
12. Given the following, predict the product assuming only the epoxide is affected. (Remember stereochemistry)
13. Predict the product of the following, similar to above but a different nucleophile is used and not in acidic conditions. (Remember stereochemistry)
14. Epoxides are often very useful reagents to use in synthesis when the desired product is a single stereoisomer. If the following alkene were reacted with an oxyacid to form an epoxide, would the result be a enantiomerically pure? If not, what would it be?
Answer
12.
Note that the stereochemistry has been inverted
13.
14.
First, look at the symmetry of the alkene. There is a mirror plane, shown here.
Then, think about the mechanism of epoxidation with an oxyacid, take for example mCPBA. The mechanism is concerted, so the original cis stereochemistry is not changed. This leads to "two" epoxides.
However, these two mirror images are actually identical due to the mirror plane of the cis geometry. It is a meso compound, so the final result is a single stereoisomer, but not a single enantiomer.
15.10: Crown Ethers
A “crown ether ” is a cyclic ether containing several (i.e., 4, 5, 6 or more) oxygen atoms. It is possible to dissolve ionic compounds in organic solvents using crown ethers. Cyclic polyether with four or more oxygen atoms separated by two or three carbon atoms. All crown ethers have a central cavity that can accommodate a metal ion coordinated to the ring of oxygen atoms., cyclic compounds with the general formula (OCH2CH2)n. Crown ethers are named using both the total number of atoms in the ring and the number of oxygen atoms. Thus 18-crown-6 is an 18-membered ring with six oxygen atoms (part (a) in Figure 18.7.1 ). The cavity in the center of the crown ether molecule is lined with oxygen atoms and is large enough to be occupied by a cation, such as K+. The cation is stabilized by interacting with lone pairs of electrons on the surrounding oxygen atoms. Thus crown ethers solvate cations inside a hydrophilic cavity, whereas the outer shell, consisting of C–H bonds, is hydrophobic. Crown ethers are useful for dissolving ionic substances such as KMnO4 in organic solvents such as isopropanol [(CH3)2CHOH] (Figure 18.7.1). The availability of crown ethers with cavities of different sizes allows specific cations to be solvated with a high degree of selectivity.
Figure 2: Effect of a Crown Ether on the Solubility of KMnO4 in Benzene. Normally which is intensely purple, is completely insoluble in benzene which has a relatively low dielectric constant. In the presence of a small amount of crown ether, KMnO4 dissolves in benzene as shown by the reddish purple color caused by the permanganate ions in solution.
Cryptands (from the Greek kryptós, meaning “hidden”) are compounds that can completely surround a cation with lone pairs of electrons on oxygen and nitrogen atoms (Figure 18.7.1b). The number in the name of the cryptand is the number of oxygen atoms in each strand of the molecule. Like crown ethers, cryptands can be used to prepare solutions of ionic compounds in solvents that are otherwise too nonpolar to dissolve them.
15.11: Epoxy Resins - The Advent of Modern Glues
A very useful group of adhesives and plastics is based on condensation polymers of bisphenol A and chloromethyloxacyclopropane (epichlorohydrin). The first step in the formation of epoxy resins is to form a prepolymer by condensation polymerization of the sodium salt of bisphenol A with the epoxide:
The formation of a prepolymer involves two different kinds of reactions. One is an $S_\text{N}2$-type displacement, and the other is oxide-ring opening of the product by attack of more bisphenol A. Usually, for practical purposes the degree of polymerization $n$ of the prepolymer is small (5 to 12 units).
The epoxy prepolymer can be cured, that is, converted to a three-dimensional network, in several different ways. A trifunctional amine, such as $\ce{NH_2CH_2CH_2NHCH_2CH_2NH_2}$, can be mixed in and will extend the chain of the polymer and form cross-links by reacting with the oxide rings:
Alternatively, a polybasic acid anhydride can be used to link the chains through combination with secondary alcohol functions and then the oxide rings.
Contributors and Attributions
• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/15%3A_Ethers_Epoxides_and_Thioethers/15.09%3A_Reactions_of_Epoxides_with_Grignard_and_Organolithium_Reagents.txt |
Thiols and sulfides are the "sulfur equivalent" of alcohols and ethers. You can replace the oxygen atom of an alcohol with a sulfur atom to make a thiol; similarly, you can replace the oxygen atom in an ether with S to make the corresponding alkyl sulfide. This is because thiols contain the C-S-H functional group, while sulfides contain the C-S-C group.
Oxidation States of Sulfur Compounds
Oxygen assumes only two oxidation states in its organic compounds (–1 in peroxides and –2 in other compounds). Sulfur, on the other hand, is found in oxidation states ranging from –2 to +6, as shown in the following table (some simple inorganic compounds are displayed in orange).
Thiols
Thiols are often called “mercaptans,” a reference to the Latin term mercurium captans(capturing mercury), since the -SH group forms strong bonds with mercury and its ions. Thiols are analogous to alcohols. They are named in a similar fashion as alcohols except the suffix -thiol is used in place of -ol. By itself the -SH group is called a mercapto group.
Thiols are usually prepared by using the hydrosulfide anion (-SH) as a nucleophile in an SN2 reaction with alkyl halides.
On problem with this reaction is that the thiol product can undergo a second SN2 reaction with an additional alkyl halide to produce a sulfide side product. This problem can be solved by using thiourea, (NH2)2C=S, as the nucleophile. The reaction first produces an alkyl isothiourea salt and an intermediate. This salt is then hydrolyzed by a reaction with aqueous base.
Disulfides
Oxidation of thiols and other sulfur compounds changes the oxidation state of sulfur rather than carbon. We see some representative sulfur oxidations in the following examples. In the first case, mild oxidation converts thiols to disufides. An equivalent oxidation of alcohols to peroxides is not normally observed. The reasons for this different behavior are not hard to identify. The S–S single bond is nearly twice as strong as the O–O bond in peroxides, and the O–H bond is more than 25 kcal/mole stronger than an S–H bond. Thus, thermodynamics favors disulfide formation over peroxide.
Disulfide bridges in proteins
Disulfide (sulfur-sulfur) linkages between two cysteine residues are an integral component of the three-dimensional structure of many proteins. The interconversion between thiols and disulfide groups is a redox reaction: the thiol is the reduced state, and the disulfide is the oxidized state.
Notice that in the oxidized (disulfide) state, each sulfur atom has lost a bond to hydrogen and gained a bond to a sulfur - this is why the disulfide state is considered to be oxidized relative to the thiol state. The redox agent that mediates the formation and degradation of disulfide bridges in most proteins is glutathione, a versatile coenzyme that we have met before in a different context (section 14.2A). Recall that the important functional group in glutathione is the thiol, highlighted in blue in the figure below. In its reduced (free thiol) form, glutathione is abbreviated 'GSH'.
In its oxidized form, glutathione exists as a dimer of two molecules linked by a disulfide group, and is abbreviated 'GSSG'.
A new disulfide in a protein forms via a 'disulfide exchange' reaction with GSSH, a process that can be described as a combination of two SN2-like attacks. The end result is that a new cysteine-cysteine disulfide forms at the expense of the disulfide in GSSG.
In its reduced (thiol) state, glutathione can reduce disulfides bridges in proteins through the reverse of the above reaction.
Disulfide bridges exist for the most part only in proteins that are located outside the cell. Inside the cell, cysteines are kept in their reduced (free thiol) state by a high intracellular concentration of GSH, which in turn is kept in a reduced state (ie. GSH rather than GSSG) by a flavin-dependent enzyme called glutathione reductase.
Disulfide bridges in proteins can also be directly reduced by another flavin-dependent enzyme called 'thioredoxin'. In both cases, NADPH is the ultimate electron donor, reducing FAD back to FADH2 in each catalytic cycle.
In the biochemistry lab, proteins are often maintained in their reduced (free thiol) state by incubation in buffer containing an excess concentration of b-mercaptoethanol (BME) or dithiothreitol (DTT). These reducing agents function in a manner similar to that of GSH, except that DTT, because it has two thiol groups, forms an intramolecular disulfide in its oxidized form.
Sulfides
Sulfur analogs of ethers are called sulfides. Sulfides are less common than thiols as naturally occurring compounds. However, sulfides—especially disulfides (C-S-S-C)—have important biological functions, mainly in reducing agents (antioxidants). The chemical behavior of sulfides contrasts with that of ethers in some important ways. Since hydrogen sulfide (H2S) is a much stronger acid than water (by more than ten million fold), we expect, and find, thiols to be stronger acids than equivalent alcohols and phenols. Thiolate conjugate bases are easily formed, and have proven to be excellent nucleophiles in SN2 reactions of alkyl halides and tosylates.
R–S(–) Na(+) + (CH3)2CH–Br (CH3)2CH–S–R + Na(+) Br(–)
Although the basicity of ethers is roughly a hundred times greater than that of equivalent sulfides, the nucleophilicity of sulfur is much greater than that of oxygen, leading to a number of interesting and useful electrophilic substitutions of sulfur that are not normally observed for oxygen. Sulfides, for example, react with alkyl halides to give ternary sulfonium salts (equation # 1) in the same manner that 3º-amines are alkylated to quaternary ammonium salts. Although equivalent oxonium salts of ethers are known, they are only prepared under extreme conditions, and are exceptionally reactive.
sulfides are named using the same rules as ethers except sulfide is used in the place of ether. For more complex substance alkylthio is used instead of alkoxy.
SAM methyltransferases
The most common example of sulfonium ions in a living organism is the reaction of S-Adenosylmethionine. Some of the most important examples of SN2 reactions in biochemistry are those catalyzed by S-adenosyl methionine (SAM) – dependent methyltransferase enzymes. We have already seen, in chapter 6 and again in chapter 8, how a methyl group is transferred in an SN2 reaction from SAM to the amine group on the nucleotide base adenosine:
Another SAM-dependent methylation reaction is catalyzed by an enzyme called catechol-O-methyltransferase. The substrate here is epinephrine, also known as adrenaline.
Notice that in this example, the attacking nucleophile is an alcohol rather than an amine (that’s why the enzyme is called an O-methyltransferase). In both cases, though, a basic amino acid side chain is positioned in the active site in just the right place to deprotonate the nucleophilic group as it attacks, increasing its nucleophilicity. The electrophile in both reactions is a methyl carbon, so there is little steric hindrance to slow down the nucleophilic attack. The methyl carbon is electrophilic because it is bonded to a positively-charged sulfur, which is a powerful electron withdrawing group. The positive charge on the sulfur also makes it an excellent leaving group, as the resulting product will be a neutral and very stable sulfide. All in all, in both reactions we have a reasonably good nucleophile, an electron-poor, unhindered electrophile, and an excellent leaving group.
Because the electrophilic carbon in these reactions is a methyl carbon, a stepwise SN1-like mechanism is extremely unlikely: a methyl carbocation is very high in energy and thus is not a reasonable intermediate to propose. We can confidently predict that this reaction is SN2. Does this SN2 reaction occur, as expected, with inversion of stereochemistry? Of course, the electrophilic methyl carbon in these reactions is achiral, so inversion is not apparent. To demonstrate inversion, the following experiment has been carried out with catechol-O-methyltransferase:
Here, the methyl group of SAM was made to be chiral by incorporating hydrogen isotopes tritium (3H, T) and deuterium (2H, D). The researchers determined that the reaction occurred with inversion of configuration, as expected for an SN2 displacement (J. Biol. Chem. 1980, 255, 9124).
Sulfides can be easily oxidized. Reacting a sulfide with hydrogen peroxide, H2O2, as room termpeature produces a sulfoxide (R2SO). The oxidation can be continued by reaction with a peroxyacid to produce the sulfone (R2SO2)
A common example of a sulfoxide is the solvent dimethyl sulfoxide (DMSO). DMSO is polar aprotic solvent.
15.13: Additional Exercises
Industrial Synthesis of Ethers using Bimolecular Condensation
15-1 Give the final product of the following reaction.
15-2 Draw the mechanism for the reaction in the previous problem, 15-1.
Synthesis of Ethers by Alkoxymercuration-Demercuration
15-3 Predict the product of the following reaction and give its correct IUPAC nomenclature.
15-4 Propose a route of synthesis (that includes an alkoxymercuration-demercuration reaction) for the following ether starting with a carbon molecule containing no more than 3 carbons.
15-5 Predict the products of the following reactions.
15-6 Give the final product for the following reaction.
Cleavage of Ethers
15-7 Choose the correct IUPAC name for the product of the following reaction.
a) Butan-1-ol
b) 1-iodobutane
c) 1,4-diiodobutane
d) 2-iodooxolane
Synthesis of Epoxides
15-8 Choose the correct IUPAC nomenclature of the product of the following reaction and provide its structure.
a) 2,2-dimethyl-3-(propan-2-yl)oxirane
b) 2,4-dimethylpentan-2-ol
c) 2,4-dimethylpent-2-en-3-ol
d) 2-methyl-2-(2-methylpropyl)oxirane
15-9 Predict the products of the following reactions.
15-10 Predict the product of the following reaction. Be sure to include proper stereochemistry.
Reactions of Epoxides
15-11 Provide the structure of the product of the following reaction. Be sure to include proper stereochemistry.
15-12 Predict the product of the following reaction.
15-13 Suggest the structure of the starting brominated molecule that was used in the following reaction to make the final product.
15-14 Provide the final products of the following reactions.
15-15 Provide the final product of the following reaction.
15-16 Choose the correct product of the following reaction.
15-17 Suggest a route of synthesis (that includes an epoxide intermediate) for the following compound, starting with ethane and using any other necessary reagents.
15-18 Predict the products of the following reactions. Be sure to include stereochemistry where applicable.
Reactions of Epoxides with Grignard and Organolithium Reagents
15-19 Predict the product of the following reaction and give the proper IUPAC nomenclature.
15-20 Predict the product of the following reaction.
15-21 Choose the correct IUPAC nomenclature of a product of the following reaction.
a) heptan-4-one
b) heptan-3-one
c) 3-aminoheptan-3-ol
d) (3Z)-hept-3-en-3-ol
15.14: Solutions to Additional Exercises
Industrial Synthesis of Ethers using Bimolecular Condensation
15-1:
15-2:
Synthesis of Ethers by Alkoxymercuration-Demercuration
15-3:
15-4:
Possible route of synthesis:
15-5:
15-6:
Cleavage of Ethers
15-7:
Answer: C
Synthesis of Epoxides
15-8:
Answer: A
15-9:
15-10:
Reactions of Epoxides
15-11:
15-12:
15-13:
15-14:
15-15:
15-16:
Answer: D
15-17:
Possible route of synthesis:
15-18:
Reactions of Epoxides with Grignard and Organolithium Reagents
15-19:
15-20:
15-21:
Answer: B | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/15%3A_Ethers_Epoxides_and_Thioethers/15.12%3A_Thioethers_%28Sulfides%29_and_Silyl_Ethers.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• construct & interpret MO diagrams of ethene, butadiene and allylic systems (refer to section 16.1)
• recognize reactions that are enhanced by resonance stabilization of the allylic intermediate (refer to section 16.2)
• predict the products and specify the reagents for electrophilic addition reactions (EAR) of conjugated dienes (refer to section 16.3)
• specify reaction conditions to promote thermodynamic or kinetic control of the reaction mechanism; correlate these conditions to reaction energy diagrams (section 16.4)
• predict the products and specify the reagents for bimolecular substitution reactions (SN2) of allylic halides (refer to section 16.5)
• predict the products of Diels-Alder reactions with stereochemistry, including the orientation of cycloaddition with asymmetrical reagents (refer to sections 16.6 and 16.7)
• develop mechanisms to explain the observed products of 1,2- & 1,4- addition reactions, including the resonance forms of the stabilized intermediates (refer to section 16.6)
• use MO theory to predict whether cycloaddition reactions will be thermally or photochemically allowed (refer to section 16.6 and 16.7)
• recognize the effect of conjugation on UV absorption (refer to section 16.9 and 16.10)
• use Beer’s Law in UV absorption calculations (refer to section 16.9 and 16.10)
• explain how light, the conjugation of double bonds, and the stereochemistry of double bonds contribute to visualizing color
16: Conjugated Systems Orbital Symmetry and Ultraviolet Spectroscopy
Introduction
Conjugated dienes are characterized by alternating carbon-carbon double bonds separated by carbon-carbon single bonds. Cumulated dienes are characterized by adjacent carbon-carbon double bonds. While conjugated dienes are energetically more stable than isolated double bonds. Cumulated double bonds are unstable. The chemistry of cumulated double bonds can be explored in advance organic chemistry courses. The chemistry of isolated alkenes is covered in Chapters 8 and 9 of this LibreText. The chemistry of conjugated double bonds is the focus of this chapter.
Conjugated Diene Stability
Conjugated dienes are more stable than non conjugated dienes (both isolated and cumulated) due to factors such as delocalization of charge through resonance and hybridization energy. This stability can be seen in the differences in the energies of hydrogenation between isolated and conjugated alkenes. Since the higher the heat of hydrogenation the less stable the compound, it is shown below that conjugated dienes (~54 kcal) have a lower heat of hydrogenation than their isolated (~60 kcal) and cumulated diene (~70 kcal) counterparts.
Here is an energy diagram comparing different types of bonds with their heats of hydrogenation to show relative stability of each molecule:
Stability of Conjugated Dienes - The Resonance Explanation
Conjugated double bonds are separated by a single bond. 1,3-dienes are an excellent example of a conjugated system. Each carbon in 1,3 dienes is sp2 hybridized and therefore has one p orbital. The four p orbitals in 1,3-butadiene overlap to form a conjugated system.
The resonance structure shown below gives a good understanding of how the pi electrons are delocalized across the four carbons in this conjugated diene. This delocalization of electrons stablizes the conjugated diene:
Stability of Conjugated Dienes - The Molecular Orbitals Explanation
A molecular orbital model for 1,3-butadiene is shown below. Note that the lobes of the four p-orbital components in each pi-orbital are colored differently and carry a plus or minus sign. This distinction refers to different phases, defined by the mathematical wave equations for such orbitals. Regions in which adjacent orbital lobes undergo a phase change are called nodes. Orbital electron density is zero in such regions. Thus a single p-orbital has a node at the nucleus, and all the pi-orbitals shown here have a nodal plane that is defined by the atoms of the diene. This is the only nodal surface in the lowest energy pi-orbital, π1. Higher energy pi-orbitals have an increasing number of nodes. Since 1,3-butadiene has four pi electrons. The two bonding molecular orbitals are filled to explain the measurable stability of conjugated double bonds.
Exercise
1. The heat of hydrogenation for allene is about 300 kJ/mol. Order a conjugated diene, a non-conjugated diene, and allene in increasing stability.
Answer
1. allene < non-conjugated diene < conjugated diene (most stable) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/16%3A_Conjugated_Systems_Orbital_Symmetry_and_Ultraviolet_Spectroscopy/16.01%3A_Stability_of_Conjugated_Dien.txt |
Resonance and Allylic Carbocation Stability
Conjugation occurs when p orbital on three or more adjacent atoms can overlap. Conjugation tends to stabilize molecules. Allylic carbocations are a common conjugated system. The resonance structures below help explain the stability of allylic carbocations. The true structure of the conjugated allyl carbocation is a hybrid of of the two resonance structure so the positive charge is delocalized over the two terminal carbons. This delocalization stablizes the allyl carbocation making it more stable than a normal primary carbocation.
The positive charge of a carbocation is contained in a p orbital of a sp2 hybridized carbon. This allows for overlap with double bonds. The positive charge is more stable because it is spread over 2 carbons.
Molecular Orbitals and Allylic Carbocation Stability
The stability of the carbocation of propene is due to a conjugated π electron system. A "double bond" doesn't really exist. Instead, it is a group of 3 adjacent, overlapping, non-hybridized p orbitals we call a conjugated π electron system. You can clearly see the interactions between all three of the p orbitals from the three carbons resulting in a really stable cation. It all comes down to where the location of the electron-deficient carbon is.
Molecular orbital descriptions can explain allylic stability in yet another way using 2-propenyl. Fig.6
If we just take the π molecular orbital and not any of the s, we get three of them. π1 is bonding with no nodes, π2 is nonbonding (In other words, the same energy as a regular p-orbital) with a node, and π3 is antibonding with 2 nodes (none of the orbitals are interacting). The first two electrons will go into the π1 molecular orbital, regardless of whether it is a cation, radical, or anion. If it is a radical or anion, the next electron goes into the π2 molecular orbital. The last anion electron goes into the nonbonding orbital also. So no matter what kind of carbon center exists, no electron will ever go into the antibonding orbital.
The Bonding orbitals are the lowest energy orbitals and are favorable, which is why they are filled first. Even though the nonbonding orbitals can be filled, the overall energy of the system is still lower and more stable due to the filled bonding molecular orbitals.
This figure also shows that π2 is the only molecular orbital where the electrion differs, and it is also where a single node passes through the middle. Because of this, the charges of the molecule are mainly on the two terminal carbons and not the middle carbon.
This molecular orbital description can also illustrate the stability of allylic carbon centers in figure 7.
The π bonding orbital is lower in energy than the nonbonding p orbital. Since every carbon center shown has two electrons in the lower energy, bonding π orbitals, the energy of each system is lowered overall (and thus more stable), regardless of cation, radical, or anion.
Allylic 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.
This can also explain why allylic radicals are much more stable than secondary or even tertiary carbocations. This is all due to the positioning of the pi orbitals and ability for overlap to occur to strengthen the single bond between the two double bonds.
Exercise
2. Draw the bond-line structure for the most stable carbocation that can be formed from each hydrocarbon below. Arrange the carbocations in order of decreasing stability.
Answer
2. Carbocations in order of decreasing stability. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/16%3A_Conjugated_Systems_Orbital_Symmetry_and_Ultraviolet_Spectroscopy/16.02%3A_Allylic_Cations.txt |
OBJECTIVES
• After completing this section, you should be able to
• write an equation for the addition of one or two mole equivalents of a halogen or a hydrogen halide to a nonconjugated diene.
• write an equation for the addition of one or two mole equivalents of a halogen or a hydrogen halide to a conjugated diene.
• write the mechanism for the addition of one mole equivalent of hydrogen halide to a conjugated diene, and hence account for the formation of 1,2- and 1,4-addition products.
• explain the stability of allylic carbocations in terms of resonance.
• draw the resonance contributors for a given allylic carbocation.
• predict the products formed from the reaction of a given conjugated diene with one mole equivalent of halogen or hydrogen halide.
• predict which of the possible 1,2- and 1,4-addition products is likely to predominate when one mole equivalent of a hydrogen halide is reacted with a given conjugated diene.
• use the concept of carbocation stability to explain the ratio of the products obtained when a given conjugated diene is reacted with one mole equivalent of hydrogen halide.
KEY TERMS
• Make certain that you can define, and use in context, the key terms below.
• 1,2-addition
• 1,4-addition
STUDY NOTES
Notice that the numbers used in the expressions 1,2-addition and 1,4-addition do not refer to the positions of the carbon atoms in the diene molecule. Here, 1,2 indicates two neighbouring carbon atoms, while 1,4 indicates two carbon atoms which are separated in the carbon chain by two additional carbon atoms. Thus in 1,2- and 1,4-additions to 2,4-hexadiene, the additions actually occur at carbons 2 and 3, and 2 and 5, respectively.
The term “monoadduct” should be interpreted as meaning the product or products formed when one mole of reagent adds to one mole of substrate. In the objectives above, this process is referred to as the addition of one mole equivalent (or one mol equiv).
In Section 7.9 we saw that electrophilic addition to a simple alkene would follow Markovnikov’s rule, where the stability of the carbocation intermediate would increase: primary < secondary < tertiary. With conjugated dienes the allylic carbocation intermediately generated has different resonance forms. The following scheme represents the mechanism for the addition of HBr to 1,3-butadiene (at 0°C). Note the resonance contributors for the allylic carbocation intermediate and that the product resulting from the secondary cation is generated in higher yield than from the primary cation as you might expect from our discussions until now. However, in the next section you will see that the resulting product ratio can be drastically affected by a number of reaction conditions, including temperature.
The reactions of 1,3-butadiene are reasonably typical of conjugated dienes. The compound undergoes the usual reactions of alkenes, such as catalytic hydrogenation or radical and polar additions, but it does so more readily than most alkenes or dienes that have isolated double bonds. Furthermore, the products frequently are those of 1,2 and 1,4 addition:
Formation of both 1,2- and 1,4-addition products occurs not only with halogens, but also with other electrophiles such as the hydrogen halides. The mechanistic course of the reaction of 1,3-butadiene with hydrogen chloride is shown in Equation 13-1. The first step, as with alkenes, is formation of a carbocation. However, with 1,3-butadiene, if the proton is added to C1C1 (but not C2C2), the resulting cation has a substantial delocalization energy, with the charge distributed over two carbons (review Sections 6-5 and 6-5C if this is not clear to you). Attack of Cl⊖Cl⊖ as a nucleophile at one or the other of the positive carbons yields the 1,2- or the 1,4- addition product:
An important feature of reactions in which 1,2 and 1,4 additions occur in competition with one another is that the ratio of the products can depend on the temperature, the solvent, and also on the total time of reaction.
Direct vs Conjugate Addition
The description of direct versus conjugate addition uses numbers localized within the conjugate system and have nothing to do with the numbering system used to determine the IUPAC name for a conjugated diene. The reactions of 1,3-butadiene are reasonably typical of conjugated dienes and illustrate the difference in the numbering system to describe the reaction versus the IUPAC nomenclature numbers.
Mechanism for the Electrophilic Addition to Conjugate Dienes
The mechanism below explains the formation and distribution of addition products to conjugated dienes using 1,3-butadiene as an example. The first step, as with isolated alkenes, is the formation of a carbocation. For 1,3-butadiene, the proton is added to form the allylic, resonance stabilized carbocation intermediate. The resulting cation has a substantial delocalization energy, with the charge distributed over two carbons. The nucleophile reacts with both carbons, but favors the carbon bearing the larger partial positive charge. The reaction yields both the 1,2- or the 1,4- addition products. The more stable the intermediate produces the greater the percentage of the final products as shown in the mechanism below.
Formation of both 1,2- and 1,4-addition products occurs not only with hydrohalic acids, but with halogens, catalytic hydrogenation or radical, and other polar additions associated with the electrophilic addition reactions of isolated alkenes.
In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Exercise
3. Give the 1,2 and the 1,4 products of the addition of one equivalent of HBr to 1,3-hexa-diene.
4. Look at the previous addition reaction of HBr with a diene. Consider the transition states, predict which of them would be the major products and which will be the minor.
Answer
3.
4. The products i-iii all show a secondary cation intermediate which is more stable than primary. Therefore those would be major products and the iv product would be the minor product.
Contributors and Attributions
• Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University)
• Prof. Steven Farmer (Sonoma State University)
• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/16%3A_Conjugated_Systems_Orbital_Symmetry_and_Ultraviolet_Spectroscopy/16.03%3A_Electrophilic_Additions_to_C.txt |
Thermodynamic vs Kinetic Control
Upon electrophilic addition, the conjugated diene forms a mixture of two products—the kinetic product and the thermodynamic product—whose ratio is determined by the conditions of the reaction. A reaction yielding more thermodynamic product is under thermodynamic control, and likewise, a reaction that yields more kinetic product is under kinetic control. The reaction of one equivalent of hydrogen bromide with 1,3-butadiene gives different ratios of products under different reaction conditions to illustrate the difference between thermodynamic and kinetic control.
The green mechanism arrows show the formation of the kinetically favored 1,2-addition product. As shown in the reaction energy diagram below the reaction, the 1,2-addition reaction has a smaller activation energy and faster reaction rate. This faster reaction rate is what led to the term "kinetic control". This reaction is favored by low temperatures where the activation energy becomes the primary barrier to chemical reactivity.
The blue mechanism arrows show the formation of the 1,4-addition product, the thermodynamically favored product. As shown in the reaction energy diagram below the reaction, the product of the 1,4-addition reaction is lower in potential energy. Its formation is favored by reactions at high temperatures where there is adequate thermodynamic energy to overcome all of the activation energy barriers. This reaction is favored by elevated temperatures which led to the term "thermodynamic control".
Reaction Energy Diagram for 1,3-butadiene + HBr
The table below summarizes the empirically derived reactivity patterns for conjugated dienes at four different reaction conditions. Becoming familiar with the reactivity data and patterns in this table helps us build wisdom for determining the optimum reaction conditions when competing mechanisms are possible.
Table : Conjugated Dienes: Kinetic vs. Thermodynamic Conditions
Temperature Kinetic or Thermodynamically Controlled Speed of Reaction 1,2-adduct : 1,4-adduct Ratio
-15 °C Kinetic Fast 70:30
0 °C Kinetic Fast 60:40
40 °C Thermodynamic Slow 15:85
60 °C Thermodynamic Slow 10:90
A Warning: Not every reaction has different thermodynamic and kinetic products!
Note that not every reaction has an energy profile diagram like Figure \(1\), and not every reaction has different thermodynamic and kinetic products! If the transition states leading to the formation of \(\ce{C}\) (e.g., TC1, and TC2) were to be higher in energy than that leading to \(\ce{B}\) (e.g., TB1, and TB2), then \(\ce{B}\) would simultaneously be both the thermodynamic and kinetic product. There are plenty of reactions in which the more stable product (thermodynamic) is also formed faster (kinetic).
Exercise
5. Consider the reaction with 1,3-buta-diene reacting with HCl. Propose a mechanism for the reaction.
Answer
5.
16.05: SN2 Reactions of Allylic Hal
SN2 Reactions of Allylic Halides and Tosylates
Allylic halides and tosylates are excellent electrophiles for bimolecular nucleophilic substitution reactions (SN2).
They exhibit faster SN2 reactivity than secondary alkyl halides because the bimolecular transition state is stabilized by hyperconjugation between the orbital of the nucleophile and the conjugated pi bond of the allylic group as shown in the diagram below.
Exercise
6. Arrange the compounds 3-bromopentane, bromobenzene, and 3-bromo-1-propene in order of decreasing SN2 reactivity using their bond-line structures.
Answer
6.
16.06: The Diels-Alder (4 2) C
The Diels-Alder (4+2) Cycloaddition Reaction
A cycloaddition reaction is the concerted bonding together of two independent pi-electron systems to form a new ring of atoms. When this occurs, two pi-bonds are converted to two sigma-bonds, the simplest example being the hypothetical combination of two ethene molecules to give cyclobutane. This does not occur under normal conditions, but the cycloaddition of 1,3-butadiene to cyanoethene (acrylonitrile) does, and this is an example of the Diels-Alder reaction. The following diagram illustrates two cycloadditions, and introduces several terms that are useful in discussing reactions of this kind.
In the hypothetical ethylene dimerization on the left, each reactant molecule has a pi-bond (colored orange) occupied by two electrons. The cycloaddition converts these pi-bonds into new sigma-bonds (colored green), and this transformation is then designated a [2+2] cycloaddition, to enumerate the reactant pi-electrons that change their bonding location.
The Diels-Alder reaction is an important and widely used method for making six-membered rings, as shown on the right. The reactants used in such reactions are a conjugated diene, simply referred to as the diene, and a double or triple bond co-reactant called the dienophile, because it combines with (has an affinity for) the diene. The Diels-Alder cycloaddition is classified as a [4+2] process because the diene has four pi-electrons that shift position in the reaction and the dienophile has two.
Diels-Alder Mechanism
The Diels-Alder reaction is a single step process, so the diene component must adopt an s-cisconformation in order for the end carbon atoms (#1 & #4) to bond simultaneously to the dienophile. For many acyclic dienes the s-trans conformer is more stable than the s-cis conformer (due to steric crowding of the end groups), but the two are generally in rapid equilibrium, permitting the use of all but the most hindered dienes as reactants in Diels-Alder reactions. In order for a Diels-Alder reaction to occur, the diene molecule must adopt what is called the s-cis conformation:
In its usual form, the diene component is electron rich, and the best dienophiles are electron poor due to electron withdrawing substituents such as CN, C=O & NO2. The initial bonding interaction reflects this electron imbalance, with the two new sigma-bonds being formed simultaneously, but not necessarily at equal rates. Essentially, this process involves overlap of the 2p orbitals on carbons 1 and 4 of the diene with the two 2p orbitals on the sp2-hybridized carbons of the dienophile. Both of these new overlaps form 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.
Since the diene takes the role of the nucleophile, electron donating group increase the reactivity of the diene. While the dienophile takes the role of the electrophile, electronwithdrawing groups increase the reactivity of the dienophile. The reactions below are examples of the Diels-Alder reaction.
Exercise
7. Draw the bond-line structures for the reactions below.
Answer
7. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/16%3A_Conjugated_Systems_Orbital_Symmetry_and_Ultraviolet_Spectroscopy/16.04%3A_Kinetic_versus_Thermodynamic.txt |
Diels-Alder Reactions are Stereospecific
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:
We noted earlier that addition reactions of alkenes often exhibited stereoselectivity, in that the reagent elements in some cases added syn and in other cases anti to the the plane of the double bond. Both reactants in the Diels-Alder reaction may demonstrate stereoisomerism, and when they do it is found that the relative configurations of the reactants are preserved in the product (the adduct). The following drawing illustrates this fact for the reaction of 1,3-butadiene with (E)-dicyanoethene. The trans relationship of the cyano groups in the dienophile is preserved in the six-membered ring of the adduct. Likewise, if the terminal carbons of the diene bear substituents, their relative configuration will be retained in the adduct. Using the earlier terminology, we could say that bonding to both the diene and the dienophile is syn. An alternative description, however, refers to the planar nature of both reactants and terms the bonding in each case to be suprafacial (i.e. to or from the same face of each plane). This stereospecificity also confirms the synchronous nature of the 1,4-bonding that takes place.
Diene Substituents and Diels-Alder Reactivity
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.
Bicyclic Ring Formation and the Exo- and Endo- Positions
Cyclic dienes that are ‘locked’ in the s-cis conformation 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 reaction with 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 three examples of Diels-Alder reactions that have been reported in recent years:
Other Pericyclic Reactions
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.
Diels-Alder Reaction Summary
The essential characteristics of the Diels-Alder cycloaddition reaction may be summarized as follows:
1. The reaction always creates a new six-membered ring. When intramolecular, another ring may also be formed.
2. The diene component must be able to assume a s-cis conformation.
3. Electron withdrawing groups on the dienophile facilitate reaction.
4. Electron donating groups on the diene facilitate reaction.
5. Steric hindrance at the bonding sites may inhibit or prevent reaction.
6. The reaction is stereospecific with respect to substituent configuration in both the dienophile and the diene.
These features are illustrated by the following eight examples, one of which does not give a Diels-Alder cycloaddition.
There is no reaction in example D because this diene cannot adopt an s-cis orientation. In examples B, C, F, G & H at least one of the reactants is cyclic so that the product has more than one ring, but the newly formed ring is always six-membered. In example B the the same cyclic compound acts as both the diene colored blue) and the dienophile (colored red). The adduct has three rings, two of which are the five-membered rings present in the reactant, and the third is the new six-membered ring (shaded light yellow). Example C has an alkyne as a dienophile (colored red), so the adduct retains a double bond at that location. This double bond could still serve as a dienophile, but in the present case the diene is sufficiently hindered to retard a second cycloaddition. The quinone dienophile in reaction F has two dienophilic double bonds. However, the double bond with two methyl substituents is less reactive than the unsubstituted dienophile due in part to the electron donating properties of the methyl groups and in part to steric hindrance. The stereospecificity of the Diels-Alder reaction is demonstrated by examples A, E & H. In A & H the stereogenic centers lie on the dienophile, whereas in E these centers are on the diene. In all cases the configuration of the reactant is preserved in the adduct.
Exercise
8. Of the following dienes, which are S-trans and which are s-cis? Of those that are s-trans, are they able to rotate to become s-cis?
9. Predict the product of the following reaction.
Answer
8.
A) s-trans, unable to rotate to become s-cis
B) s-cis
C) s-trans, can rotate to become s-cis.
9.
16.08: Diene Polymers - Natural and
Conjugated dienes (alkenes with two double bonds and a single bond in between) can be polymerized to form important compounds like rubber. This takes place, in different forms, both in nature and in the laboratory. Interactions between double bonds on multiple chains leads to cross-linkage which creates elasticity within the compound.
Polymerization of 1,3-Butadiene
For rubber compounds to be synthesized, 1,3-butadiene must be polymerized. Below is a simple illustration of how this compound is formed into a chain. The 1,4 polymerization is much more useful to polymerization reactions.
Above, the green structures represent the base units of the polymers that are synthesized and the red represents the bonds between these units which form these polymers. Whether the 1,3 product or the 1,4 product is formed depends on whether the reaction is thermally or kinetically controlled.
Natural Rubber
Natural rubber is an addition polymer that is obtained as a milky white fluid known as latex from a tropical rubber tree. Natural rubber is from the monomer isoprene (2-methyl-1,3-butadiene), which is a conjugated diene hydrocarbon as mentioned above. In natural rubber, most of the double fonds formed in the polymer chain have the Z configuration, resulting in natural rubber's elastomer qualities.
Charles Goodyear accidentally discovered that by mixing sulfur and rubber, the properties of the rubber improved in being tougher, resistant to heat and cold, and increased in elasticity. This process was later called vulcanization after the Roman god of fire. Vulcanization causes shorter chains to cross link through the sulfur to longer chains. The development of vulcanized rubber for automobile tires greatly aided this industry.
Synthetic Rubber
The most important synthetic rubber is Neoprene which is produced by the polymerization of 2-chloro-1,3-butadiene.
In this illustration, the dashed lines represent repetition of the same base units, so both the products and reactants are polymers. The reaction proceeds with a mechanism similar to the Friedel-Crafts mechanism. Cross-linkage between the chlorine atom of one chain and the double bond of another contributes to the overall elasticity of neoprene. This cross-linkage occurs as the chains lie next to each other at random angles, and the attractions between double bonds prevent them from sliding back and forth.
References
1. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. New York: W. H. Freeman & Company, 2007.
2. Buehr, Walter. Rubber: Natural and Synthetic. Morrow, 1964.
Exercise
10. Draw out the mechanism for the natural synthesis of rubber from 3-methyl-3-butenyl pyrophosphate and 2-methyl-1,3-butadiene. Show the movement of electrons with arrows.
11. Draw a segment for the polymer that may be made from 2-tert-butyl-1,3-butadiene.
12. Propose the mechanism for the acid catalyzed polymerization of 2-methyl-1,3-butadiene.
Answer
10.
11.
12. The initial step is an addition of a hydrogen from the acid, followed by the polymerization. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/16%3A_Conjugated_Systems_Orbital_Symmetry_and_Ultraviolet_Spectroscopy/16.07%3A_Diels-Alder_Stereochemistry.txt |
The Electromagnetic Spectrum
The visible spectrum constitutes but a small part of the total radiation spectrum. Most of the radiation that surrounds us cannot be seen, but can be detected by dedicated sensing instruments. This electromagnetic spectrum ranges from very short wavelengths (including gamma and x-rays) to very long wavelengths (including microwaves and broadcast radio waves). The following chart displays many of the important regions of this spectrum, and demonstrates the inverse relationship between wavelength and frequency (shown in the top equation below the chart).
The energy associated with a given segment of the spectrum is proportional to its frequency. The bottom equation describes this relationship, which provides the energy carried by a photon of a given wavelength of radiation.
To obtain specific frequency, wavelength and energy values use this calculator.
• Indigo: 420 - 440 nm
• Green: 490 - 570 nm
• Orange: 585 - 620 nm
• Red: 620 - 780 nm
When white light passes through or is reflected by a colored substance, a characteristic portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary color to the wavelength(s) absorbed. This relationship is demonstrated by the color wheel shown below. Here, complementary colors are diametrically opposite each other. Thus, absorption of 420-430 nm light renders a substance yellow, and absorption of 500-520 nm light makes it red. Green is unique in that it can be created by absoption close to 400 nm as well as absorption near 800 nm.
Early humans valued colored pigments, and used them for decorative purposes. Many of these were inorganic minerals, but several important organic dyes were also known. These included the crimson pigment, kermesic acid, the blue dye, indigo, and the yellow saffron pigment, crocetin. A rare dibromo-indigo derivative, punicin, was used to color the robes of the royal and wealthy. The deep orange hydrocarbon carotene is widely distributed in plants, but is not sufficiently stable to be used as permanent pigment, other than for food coloring. A common feature of all these colored compounds, displayed below, is a system of extensively conjugated $\pi$-electrons.
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/mole, and the near ultraviolet region, out to 200 nm, extends this energy range to 143 kcal/mole. 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". A diagram showing the various kinds of electronic excitation that may occur in organic molecules is shown on the left. Of the six transitions outlined, only the two lowest energy ones (left-most, colored blue) 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 isoprene spectrum shown below. Since isoprene is colorless, it does not absorb in the visible part of the spectrum and this region is not displayed on the graph. Absorbance usually ranges from 0 (no absorption) to 2 (99% absorption), and is precisely defined in context with spectrometer operation.
Electronic transitions
Let’s take as our first example the simple case of molecular hydrogen, H2. As you may recall from section 2.1A, 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 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. 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, 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.
space-filling view
1,3-butadiene is the simplest example of a system of conjugated pi bonds. 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, while the C6-C7 double bond 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. (Hydrogenation is a reaction type that we will learn much more about in chapter 15: essentially, it is the process of adding a hydrogen molecule - two protons and two electrons - to a p bond). 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 oxygen and nitrogen atoms as well as carbon. In the metabolism of fat molecules, some of the key reactions involve alkenes that are conjugated to carbonyl groups.
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.
Example 1
Identify all conjugated and isolated double bonds in the structures below. For each conjugated pi system, specify the number of overlapping p orbitals, and how many pi electrons are shared among them.
Solution: Look for sp3 hybridized carbons to find disruptions in conjugation.
ExAMPLE 2
Identify all isolated and conjugated pi bonds in lycopene, the red-colored compound in tomatoes. How many pi electrons are contained in the conjugated pi system?
Solution: There are 11 conjugated pi bonds for a total of 22 pi electrons and 2 isolated pi bonds.
Exercise
13. What is the energy range for 300 nm to 500 nm in the ultraviolet spectrum? How does this compare to energy values from NMR and IR spectroscopy?
Answer
13.
E = hc/λ
E = (6.62 × 10−34 Js)(3.00 × 108 m/s)/(3.00 × 10−7 m)
E = 6.62 × 10−19 J
The range of 3.972 × 10-19 to 6.62 × 10-19 joules. This energy range is greater in energy than the in NMR and IR. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/16%3A_Conjugated_Systems_Orbital_Symmetry_and_Ultraviolet_Spectroscopy/16.09%3A_Structure_Determination_in_C.txt |
UV Spectroscopy and Pi Electron Transitions between the HOMO and LUMO
The ultraviolet absorption maximum of a conjugated molecule is dependent upon the extent of conjugation. As the conjugation increases, the Molecular Orbital energy decreases so that the pi electron transitions occur in the UV and visible regions of the electromagnetic spectrum. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores.
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 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.
Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are bonding, while the upper two are anti-bonding.
Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.
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.
Looking at UV-vis spectra
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 the same as for IR spectroscopy: 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+ (we'll learn what it does in section 16.4) This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems.
You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one 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) rather than in cm-1 as is the convention in IR spectroscopy.
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 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'. This contains the same information as the 'percent transmittance' number used in IR spectroscopy, just expressed in slightly different terms. 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.
Exercise
14. Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well).
Solution
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
15. How large is the π - π* transition in 4-methyl-3-penten-2-one?
Solution
Exercise
16. Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer.
Solution
Exercise
17. Which of the following would show UV absorptions in the 200-300 nm range?
Answer
17. B would be the only one to show in that range. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/16%3A_Conjugated_Systems_Orbital_Symmetry_and_Ultraviolet_Spectroscopy/16.10%3A_Interpreting_Ultraviolet_Spe.txt |
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 the eye and stimulates its photoreceptors.
Physical Characteristics of Light
[1]
The energy of light can be determined from its wavelength. The energy of light increases from long wavelength to short wavelength. The visible spectrum ranges from 400 nm to 700 nm.
Energy converting chemicals
Light energy can convert chemical 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 convert 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.
Colored molecules
The conjugated double bonds in beta-carotene produce the orange color in carrots. The conjugated double bonds in lycopene produce the red color in tomatoes.
ß carotene
lycopene
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 describe the ways cells react and response 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.
16.12: Additional Exercises
General Review
16-1 Identify which of the following dienes are isolated, conjugated, or cumulated.
16-2 Identify which pathway gives the thermodynamic or the kinetic product and provide a reason for your answer.
16-3 Draw the resonance structures of the following molecule.
16-4 Provide the final product of the following reaction.
16-5 Predict the final product for the following reactions.
16-6 For the following compound, predict which bonds could be broken to give the most probable Diels-Alder dienes that reacted to make the compound. Provide the resulting dienes from the prediction.
16-7 Give the final product of the following reaction chain..
1,2- and 1,4-Addition to Conjugated Dienes
16-8 Predict all possible products of the following reaction.
16-9 Identify the kinetic and thermodynamic products of the previous problem, 16-8.
16-10 Predict the kinetic product(s) of the following reaction and provide proper IUPAC nomenclature.
Allylic Radicals
16-11 Draw arrows to show the movement of electrons in the following allylic radical.
16-12 Show the products of the following reactions.
16-13 Choose the correct answer of the following reaction.
SN2 Displacement Reactions of Allylic Halides and Tosylates
16-14 Predict the products of the following reactions.
16-15 Choose the correct IUPAC nomenclature of the product of the following reaction and provide its structure.
a) (3R)-3-ethoxycyclohex-1-ene
b) (3S)-3-ethoxycyclohex-1-ene
c) ethoxycyclohexane
d) cyclohexa-1,3-diene
16-16 Predict the product of the following reaction.
Diels-Alder Reactions
16-17 Predict the products of the following Diels-Alder reactions.
16-18 Predict the product of the following reaction.
16-19 Choose the correct answer.
16.13: Solutions to Additional Exer
General Review
16-1
1. Conjugated
2. Cumulated
3. Isolated
16-2 Pathway A leads to the kinetic product. Since the intermediate formed during this pathway is the most stable intermediate, it forms the fastest and will always give the kinetic product. Pathway B forms the thermodynamic product, which is the most stable final product.
16-3
16-4
16-5
16-6
16-7
1,2- and 1,4-Addition to Conjugated Dienes
16-8:
16-9:
16-10:
Allylic Radicals
16-11:
16-12:
16-13:
Answer: C
SN2 Displacement Reactions of Allylic Halides and Tosylates
16-14:
16-15:
Answer:
16-16:
Diels-Alder Reactions
16-17:
16-18:
16-19:
Answer: B | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/16%3A_Conjugated_Systems_Orbital_Symmetry_and_Ultraviolet_Spectroscopy/16.11%3A_Conjugation_Color_and_the_Ch.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• summarize the discovery of the structure of benzene (refer to section 17.1)
• predict the physical properties of aromatic compounds using IMFs (refer to section 17.2)
• apply resonance and MO Theory to the structure of benzene (refer to section 17.3)
• apply MO Theory to cyclobutadiene (refer to section 17.4)
• apply resonance to aromatic compounds and ions (refer to sections 17.5 and 17.6)
• use Hückel’s rule to predict whether a given cyclic compound or ion is aromatic, antiaromatic or nonaromatic (refer to sections 17.5 and 17.6)
• for heterocycles, determine whether the lone pairs of the heteroatoms occupy p orbitals or sp2 orbitals (refer to sections 17.5 and 17.7)
• for heterocyclic amines, and predict whether the nitrogen atom is weakly or strongly basic (refer to sections 17.5 and 17.7)
• use Huckel's Rule to prdict whether polycyclic aromatic hydrocarbons are aromatic (refer to setions 17.5 and 17.8)
• use IR, NMR, UV and mass spectra to determine the structures of aromatic compounds (refer to section 17.9)
• given an aromatic compound, predict the important features of its spectra (refer to section 17.9)
Please note: IUPAC nomenclature and important common names of alcohols were explained in Chapter 3.
17: Aromatic Compounds
The 100 Year Mystery of Benzene
It took humans over 100 years to determine and confirm the structure of benzene. Why did it take so long? Why was there such a curiosity? The 1:1 ratio of carbon to hydrogen in the empirical formula and low chemical reactivity of benzene were a paradox to chemists in the early 1800's.
In 1825, Michael Faraday isolated an oily residue of gas lamps. Faraday called this liquid "bicarburet of hydrogen" and measured the boiling point to be 80°C. Additionally, Faraday determined the empirical formula to be CH. About nine years later, Eilhard Mitscherlich synthesized the same compound from benzoic acid and lime (CaO).
During the mid to late 1800's, several possible structures (shown below) were proposed for benzene.
It was not until the 1930's that Kekule's structure was confirmed by X-ray and electron diffraction. During the end of Kekule's career he revealed that the structure came to him in a vision after enjoying a glass or two of wine by the fire in his favorite chair. His inspiration for the structure of benzene was derived from an ouroboros in the flames.
17.02: The Structure and Properties of Benzene and its Derivatives
Benzene
Benzene, C6H6, is the simplest member of a large family of hydrocarbons, called aromatic hydrocarbons. These compounds contain ring structures and exhibit bonding that must be described using the resonance hybrid concept of valence bond theory or the delocalization concept of molecular orbital theory. (To review these concepts, refer to the earlier chapters on chemical bonding). The resonance structures for benzene, C6H6, are:
There are many derivatives of benzene. The hydrogen atoms can be replaced by many different substituents. Aromatic compounds more readily undergo substitution reactions than addition reactions; replacement of one of the hydrogen atoms with another substituent will leave the delocalized double bonds intact. The following are typical examples of substituted benzene derivatives:
Toluene and xylene are important solvents and raw materials in the chemical industry. Styrene is used to produce the polymer polystyrene.
Example: Structure of Aromatic Hydrocarbons
One possible isomer created by a substitution reaction that replaces a hydrogen atom attached to the aromatic ring of toluene with a chlorine atom is shown here. Draw two other possible isomers in which the chlorine atom replaces a different hydrogen atom attached to the aromatic ring:
Solution
Since the six-carbon ring with alternating double bonds is necessary for the molecule to be classified as aromatic, appropriate isomers can be produced only by changing the positions of the chloro-substituent relative to the methyl-substituent:
Exercise
1. Draw three isomers of a six-membered aromatic ring compound substituted with two bromine atoms.
Answer
Larger Arenes
Most arenes that contain a single six-membered ring are volatile liquids, such as benzene and the xylenes, although some arenes with substituents on the ring are solids at room temperature. In the gas phase, the dipole moment of benzene is zero, but the presence of electronegative or electropositive substituents can result in a net dipole moment that increases intermolecular attractive forces and raises the melting and boiling points. For example, 1,4-dichlorobenzene, a compound used as an alternative to naphthalene in the production of mothballs, has a melting point of 52.7°C, which is considerably greater than the melting point of benzene (5.5°C).
Certain aromatic hydrocarbons, such as benzene and benz[a]pyrene, are potent liver toxins and carcinogens. In 1775, a British physician, Percival Pott, described the high incidence of cancer of the scrotum among small boys used as chimney sweeps and attributed it to their exposure to soot. His conclusions were correct: benz[a]pyrene, a component of chimney soot, charcoal-grilled meats, and cigarette smoke, was the first chemical carcinogen to be identified.
Although arenes are usually drawn with three C=C bonds, benzene is about 150 kJ/mol more stable than would be expected if it contained three double bonds. This increased stability is due to the delocalization of the π electron density over all the atoms of the ring. Compared with alkenes, arenes are poor nucleophiles. Consequently, they do not undergo addition reactions like alkenes; instead, they undergo a variety of electrophilic aromatic substitution reactions that involve the replacement of –H on the arene by a group –E, such as –NO2, –SO3H, a halogen, or an alkyl group, in a two-step process. The first step involves addition of the electrophile (E) to the π system of benzene, forming a carbocation. In the second step, a proton is lost from the adjacent carbon on the ring:
The carbocation formed in the first step is stabilized by resonance.
Arenes undergo substitution reactions rather than elimination because of increased stability arising from delocalization of their π electron density.
Many substituted arenes have potent biological activity. Some examples include common drugs and antibiotics such as aspirin and ibuprofen, illicit drugs such as amphetamines and peyote, the amino acid phenylalanine, and hormones such as adrenaline as shown below.
Aspirin (antifever activity), ibuprofen (antifever and anti-inflammatory activity), and amphetamine (stimulant) have pharmacological effects. Phenylalanine is an amino acid. Adrenaline is a hormone that elicits the “fight or flight” response to stress.
Physical Properties
The physical properties of aromatic compounds are similar to other hydrocarbons. As hydrocarbons, the dominant IMF is the London Dispersion Force. This relatively weak IMF results in more volatile compounds which led to the term "aromatic". Chemists can frequently recognize the presence of an aromatic compound by simply smelling its aroma. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/17%3A_Aromatic_Compounds/17.01%3A_Introduction-_The_Discovery_of_Benzene.txt |
Benzene Structure
Among the many distinctive features of benzene, its aromaticity is the major contributor to why it is so unreactive. This section will try to clarify the theory of aromaticity and why aromaticity gives unique qualities that make these conjugated alkenes inert to compounds such as Br2 and even hydrochloric acid. It will also go into detail about the unusually large resonance energy due to the six conjugated carbons of benzene.
The delocalization of the p-orbital carbons on the sp2 hybridized carbons is what gives the aromatic qualities of benzene.
This diagram shows one of the molecular orbitals containing two of the delocalized electrons, which may be found anywhere within the two "doughnuts". The other molecular orbitals are almost never drawn.
• Benzene (\(C_6H_6\)) is a planar molecule containing a ring of six carbon atoms, each with a hydrogen atom attached.
• The six carbon atoms form a perfectly regular hexagon. All of the carbon-carbon bonds have exactly the same lengths - somewhere between single and double bonds.
• There are delocalized electrons above and below the plane of the ring, which makes benzene particularly stable.
• Benzene resists addition reactions because those reactions would involve breaking the delocalization and losing that stability.
Because of the aromaticity of benzene, the resulting molecule is planar in shape with each C-C bond being 1.39 Å in length and each bond angle being 120°. You might ask yourselves how it's possible to have all of the bonds to be the same length if the ring is conjugated with both single (1.47 Å) and double (1.34 Å), but it is important to note that there are no distinct single or double bonds within the benzene. Rather, the delocalization of the ring makes each count as one and a half bonds between the carbons which makes sense because experimentally we find that the actual bond length is somewhere in between a single and double bond. Finally, there are a total of six p-orbital electrons that form the stabilizing electron clouds above and below the aromatic ring.
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.
The High Stability of 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
Since benzene has six pi electrons, all of the bonding MOs are filled which indicates a highly stable molecule.
Exercise
2. The molecule shown, p-methylpyridine, has similar properties to benzene (flat, 120° bond angles). Draw the pi-orbitals for this compound.
Answer
2. The nitrogen has a lone pair of electrons perpendicular to the ring.
17.04: The Molecular Orbital Picture of Cyclobutadiene
Molecular Orbital Diagram for Cyclobutadiene
Cyclobutadiene is so unstable that its physical properties have not been reliably measured. The diagram below helps explain why cyclobutadiene is very unstable. With four pi electrons, both non-bonding Molecular Orbitals are singly occupied. Cyclobutadiene is so unstable relative to cyclobutane, that it is described as "antiaromatic". This term will be further explored in the next section. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/17%3A_Aromatic_Compounds/17.03%3A_Resonance_and_the_Molecular_Orbitals_of_Benzene.txt |
In 1931, German chemist and physicist Erich Hückel proposed a theory to help determine if a planar ring molecule would have aromatic properties. His rule states that if a cyclic, planar molecule has 4n+2 $π$ electrons, it is considered aromatic. This rule would come to be known as Hückel's Rule.
Huckel's Rule: Aromatic, Antiaromatic, and Nonaromatic
Huckel's Rule is a set of algorithms that combine the number of $\pi$ electrons ($N$) and the physical structure of the ring system to determine whether the molecule is aromatic, antiaromatic, or nonaromatic.
The number of $\pi$ electrons in an aromatic system can be determined by the following algorithm:
$N = 4n +2$
where $n$ is an integer.
The number of $\pi$ electrons in an antiaromatic system can be determined by the following algorithm:
$N = 4n$
where $n$ is an integer.
If a compound does not have a continuous ring of conjugated p orbitals in a planar conformation, then it is nonaromatic.
Huckel's Rule is a useful first step in evaluating the potential for a ringed molecule to be aromatic. The planar requirement of the ring may require further investigation.
Four Criteria for Aromaticity
When deciding if a compound is aromatic, go through the following checklist. If the compound does not meet all the following criteria, it is likely not aromatic.
1. The molecule is cyclic (a ring of atoms)
2. The molecule is planar (all atoms in the molecule lie in the same plane)
3. The molecule is fully conjugated (p orbitals at every atom in the ring)
4. The molecule has 4n+2 $π$ electrons (n=0 or any positive integer)
Why 4n+2 $\pi$ Electrons?
According to Hückel's Molecular Orbital Theory, a compound is particularly stable if all of its bonding molecular orbitals are filled with paired electrons. This is true of aromatic compounds, meaning they are quite stable. With aromatic compounds, 2 electrons fill the lowest energy molecular orbital, and 4 electrons fill each subsequent energy level (the number of subsequent energy levels is denoted by n), leaving all bonding orbitals filled and no anti-bonding orbitals occupied. This gives a total of 4n+2 $\pi$ electrons. You can see how this works with the molecular orbital diagram for the aromatic compound, benzene, below. Benzene has 6 $\pi$ electrons. Its first 2 $\pi$ electrons fill the lowest energy orbital, and it has 4 $\pi$ electrons remaining. These 4 fill in the orbitals of the succeeding energy level. Notice how all of its bonding orbitals are filled, but none of the anti-bonding orbitals have any electrons.
To apply the 4n+2 rule, first count the number of $\pi$ electrons in the molecule. Then, set this number equal to $4n+2$ and solve for $n$. If is 0 or any positive integer (1, 2, 3,...), the rule has been met. For example, benzene has six $\pi$ electrons:
\begin{align*} 4n + 2 &= 6 \ 4n &= 4 \ n &= 1 \end{align*}
For benzene, we find that $n=1$, which is a positive integer, so the rule is met.
How Can You Tell Which Electrons are $\pi$ Electrons?
Perhaps the toughest part of Hückel's Rule is figuring out which electrons in the compound are actually $\pi$ electrons. Once this is figured out, the rule is quite straightforward. $\pi$ electrons lie in p orbitals and $sp^2$ hybridized atoms have 1 p orbital each. So if every carbon atom in the cyclic compound is $sp^2$ hybridized, this means the molecule is fully conjugated (has 1 p orbital at each atom), and the electrons in these p orbitals are the $\pi$ electrons. A simple way to know if an atom is $sp^2$ hybridized is to see if it has 3 attached atoms and no lone pairs of electrons. This video provides a very nice tutorial on how to determine an atom's hybridization. In a cyclic hydrocarbon compound with alternating single and double bonds, each carbon is attached to 1 hydrogen and 2 other carbons. Therefore, each carbon is $sp^2$ hybridized and has a p orbital. Let's look at our previous example, benzene:
Each double bond (π bond) always contributes 2 $\pi$ electrons. Benzene has 3 double bonds, so it has 6 $\pi$ electrons.
Aromatic Ions
Hückel's Rule also applies to ions. As long as a compound has 4n+2 $\pi$ electrons, it does not matter if the molecule is neutral or has a charge. For example, cyclopentadienyl anion is an aromatic ion. How do we know that it is fully conjugated? That is, how do we know that each atom in this molecule has 1 p orbital? Let's look at the following figure. Carbons 2-5 are sp2 hybridized because they have 3 attached atoms and have no lone electron pairs. What about carbon 1? Another simple rule to determine if an atom is sp2 hybridized is if an atom has 1 or more lone pairs and is attached to an sp2 hybridized atom, then that atom is sp2 hybridized also. This video explains the rule very clearly. Therefore, carbon 1 has a p orbital. Cyclopentadienyl anion has 6 $\pi$ electrons and fulfills the 4n+2 rule.
Heterocyclic Aromatic Compounds
So far, you have encountered many carbon homocyclic rings, but compounds with elements other than carbon in the ring can also be aromatic, as long as they fulfill the criteria for aromaticity. These molecules are called heterocyclic compounds because they contain 1 or more different atoms other than carbon in the ring. A common example is furan, which contains an oxygen atom. We know that all carbons in furan are sp2 hybridized. But is the oxygen atom sp2 hybridized? The oxygen has at least 1 lone electron pair and is attached to an sp2 hybridized atom, so it is sp2 hybridized as well. Notice how oxygen has 2 lone pairs of electrons. How many of those electrons are $\pi$ electrons? An sp2 hybridized atom only has 1 p orbital, which can only hold 2 electrons, so we know that 1 electron pair is in the p orbital, while the other pair is in an sp2 orbital. So, only 1 of oxygen's 2 lone electron pairs are $\pi$ electrons. Furan has 6 $\pi$ electrons and fulfills the 4n+2 rule.
A Common Misconception
A very common misconception is that hybridization can be used to predict the geometry, or that hybridization somehow involves an energy cost associated with 'promoting' electrons into the hybrid orbitals. This is entirely wrong. Hybridization is always determined by geometry. You can only assign hybridization states to an atom if you already know its geometry, based on some experimental or theoretical evidence. The geometry of the oxygen in furan is trigonal planar and therefore the hybridization must be $sp^2$.
The specific rule is that if you have an $sp^2$ conjugated system, the lone pair will be involved if it makes the system more stable. In this case, conferring Hückel $4n+2$ aromaticity. For furan with two lone pairs on the oxygen atom, if we count electrons from the carbon atoms, we have 4 (one per carbon). So adding two electrons from one of the lone pairs will give 6 = 4(1)+2, so Hückel rule is applicable and furan is aromatic.
Exercise
3. Using the criteria for aromaticity, determine if the following molecules are aromatic:
Answer
3.
Cpd 1: Aromatic - only 1 of S's lone pairs counts as $\pi$ electrons, so there are 6 $\pi$ electrons, n=1 Not aromatic - not fully conjugated, top C is sp3 hybridized
Cpd 2: Not aromatic - top C is sp2 hybridized, but there are 4 $\pi$ electrons, n=1/2
Cpd 3: Aromatic - N is using its 1 p orbital for the electrons in the double bond, so its lone pair of electrons are not $\pi$ electrons, there are 6 $\pi$ electrons, n=1
Cpd 4: Aromatic - there are 6 $\pi$ electrons, n=1
Cpd 5: Aromatic - there are 6 $\pi$ electrons, n=1 because the N assumes sp2 hybridization.
Cpd 6: Not aromatic - all atoms are sp2 hybridized, but only 1 of S's lone pairs counts as $\pi$ electrons, so there 8 $\pi$ electrons, n=1.5
Cpd 7: Not aromatic - there are 4 $\pi$ electrons, n=1/2
Cpd 8: Aromatic - only 1 of N's lone pairs counts as $\pi$ electrons, so there are 6 $\pi$ electrons, n=1
Cpd 9: Not aromatic - not fully conjugated, top C is sp3 hybridized
Cpd 10: Aromatic - O is using its 1 p orbital for the elections in the double bond, so its lone pair of electrons are not $\pi$ electrons, there are 6 $\pi$ electrons, n=1 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/17%3A_Aromatic_Compounds/17.05%3A_Aromaticity_and_Huckel%27s_Rule.txt |
Cyclic anions and cations can be aromatic if they follow Huckel's Rule. Since aromatic ions have increased stability, it is important to recognize their formation when predicting reaction products.
Charged Aromatic Compounds
Carbanions and carbocations may also show aromatic stabilization. Some examples are:
The three-membered ring cation has 2 $\pi$-electrons and is surprisingly stable, considering its ring strain. Cyclopentadiene is as acidic as ethanol, reflecting the stability of its 6 π-electron conjugate base. Salts of cycloheptatrienyl cation (tropylium ion) are stable in water solution, again reflecting the stability of this 6 $\pi$-electron cation.
Exercise
4. Draw the resonance structures for cycloheptatriene anion. Are all bonds equivalent? How many lines (signals) would you see in a H1 C13 NMR?
5. The following reaction occurs readily. Propose a reason why this occurs?
Answer
4. All protons and carbons are the same, so therefore each spectrum will only have one signal each.
5. The ring becomes aromatic with the addition of two electrons. Thereby obeying the 4n+2 rule.
17.07: Heterocyclic Aromatic Compounds - a closer look
Aromatic Heterocycles
While benzene is the archetypical aromatic compound, many unsaturated cyclic compounds have exceptional properties that we now consider characteristic of "aromatic" systems. The aromatic heterocycle pyridine is similar to benzene, and is often used as a weak base for scavenging protons. In the bonding picture for pyridine, the nitrogen is sp2-hybridized, with two of the three sp2 orbitals forming sigma overlaps with the sp2 orbitals of neighboring carbon atoms, and the third nitrogen sp2 orbital containing the lone pair. The unhybridized p orbital contains a single electron, which is part of the 6 pi-electron system delocalized around the ring.
Pyrrole, furan, and thiophene have heterocyclic five-membered rings, in which the heteroatom has at least one pair of non-bonding valence shell electrons. By hybridizing this heteroatom to a sp2 state, a p-orbital occupied by a pair of electrons and oriented parallel to the carbon p-orbitals is created. The resulting planar ring meets the first requirement for aromaticity, and the π-system is occupied by 6 electrons, 4 from the two double bonds and 2 from the heteroatom, thus satisfying the Hückel Rule.
Four additional examples of heterocyclic aromatic compounds are shown below.
The first example is azulene, a blue-colored 10 π-electron aromatic hydrocarbon isomeric with naphthalene. The second and third compounds are heterocycles having aromatic properties. Pyridine has a benzene-like six-membered ring incorporating one nitrogen atom. The non-bonding electron pair on the nitrogen is not part of the aromatic π-electron sextet, and may bond to a proton or other electrophile without disrupting the aromatic system. In the case of thiophene, a sulfur analog of furan, one of the sulfur electron pairs (colored blue) participates in the aromatic ring π-electron conjugation. The last compound is imidazole, a heterocycle having two nitrogen atoms. Note that only one of the nitrogen non-bonding electron pairs is used for the aromatic π-electron sextet. The other electron pair is weakly basic and behaves similarly to the electron pair in pyridine.
The Two Nitrogens of Imidazole
Imidazole is another important example of an aromatic heterocycle found in biomolecules - the side chain of the amino acid histidine contains an imidazole ring. In imidazole, one nitrogen is 'pyrrole-like' (the lone pair contributes to the aromatic sextet) and one nitrogen is 'pyridine-like' (the lone pair is located in an sp2 orbital, and is not part of the aromatic sextet). The diagram below shows how the lone pair electrons on the nitrogen atoms differ.
Basicity versus Aromaticity
When the nitrogen atom of an aromatic heterocycle contains a pi bond, then the lone pair occupies an sp2 orbital and is available to react as a weak base. We can view these nitrogens as being "pyridine like". When the nitrogen atom of an aromatic heterocycle has single bonds only, then the nitrogen is still sp2 hybridized, but the lone pair occupies the unhybridized p orbital to create aromaticity. This lone pair is part of the conjugated pi electron system and is not available to react as a weak base.
Exercise
6. Draw the orbitals of thiophene to show that is aromatic.
7. The following ring is called a thiazolium ring. Describe how it is aromatic.
8. The nitrogenous bases for the nucleotides of DNA and RNA are shown below. Determine which nitrogen atoms are weak bases and which nitrogen atoms have lone pairs contributing to the aromaticity of the compound.
Answer
6. This drawing shows it has 6 electrons in the pi-orbital.
7. Similar to the last question, the drawing shows that there is only 6 electrons in the pi-system.
8. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/17%3A_Aromatic_Compounds/17.06%3A_Aromatic_Ions_-_a_closer_look.txt |
Aromatic Compounds with more than one ring
Benzene rings may be joined together (fused) to give larger polycyclic aromatic compounds. A few examples are drawn below, together with the approved numbering scheme for substituted derivatives. The peripheral carbon atoms (numbered in all but the last three examples) are all bonded to hydrogen atoms. Unlike benzene, all the C-C bond lengths in these fused ring aromatics are not the same, and there is some localization of the pi-electrons.
The six benzene rings in coronene are fused in a planar ring; whereas the six rings in hexahelicene are not joined in a larger ring, but assume a helical turn, due to the crowding together of the terminal ring atoms. This helical configuration renders the hexahelicene molecule chiral, and it has been resolved into stable enantiomers.
Exercise
9. This is an isomer of naphthalene. Is it aromatic? Draw a resonance structure for it.
Answer
9. Yes, it is aromatic. 4n+2 pi-electrons.
17.09: Spectroscopy of Aromatic Compounds
The chemical shifts of aromatic protons
Some protons resonate much further downfield than can be accounted for simply by the deshielding effect of nearby electronegative atoms. Vinylic protons (those directly bonded to an alkene carbon) and aromatic (benzylic) protons are dramatic examples.
We'll consider the aromatic proton first. Recall that in benzene and many other aromatic structures, a sextet of pelectrons is delocalized around the ring. When the molecule is exposed to B0, these pelectrons begin to circulate in a ring current, generating their own induced magnetic field that opposes B0. In this case, however, the induced field of the pelectrons does not shield the benzylic protons from B0 as you might expect– rather, it causes the protons to experience a stronger magnetic field in the direction of B0 – in other words, it adds to B0 rather than subtracting from it.
To understand how this happens, we need to understand the concept of diamagnetic anisotropy (anisotropy means `non-uniformity`). So far, we have been picturing magnetic fields as being oriented in a uniform direction. This is only true over a small area. If we step back and take a wider view, however, we see that the lines of force in a magnetic field are actually anisotropic. They start in the 'north' direction, then loop around like a snake biting its own tail.
If we are at point A in the figure above, we feel a magnetic field pointing in a northerly direction. If we are at point B, however, we feel a field pointing to the south.
In the induced field generated by the aromatic ring current, the benzylic protons are at the equivalent of ‘point B’ – this means that the induced current in this region of space is oriented in the same direction as B0.
In total, the benzylic protons are subjected to three magnetic fields: the applied field (B0) and the induced field from the pelectrons pointing in one direction, and the induced field of the non-aromatic electrons pointing in the opposite (shielding) direction. The end result is that benzylic protons, due to the anisotropy of the induced field generated by the ring current, appear to be highly deshielded. Their chemical shift is far downfield, in the 6.5-8 ppm region.
Characteristic NMR Absorption of Benzene Derivatives
Hydrogens directly attached to an arene ring show up about 7-9 PPM in the NMR. This is called the aromatic region. Hydrogen environments directly bonded to an arene ring show up about 2.5 PPM.
Charateristic IR Absorption of Benzene Derivatives
Arenes have absorption bands in the 650-900 cm−1 region due to bending of the C–H bond out of the plane of the ring. The exact placement of these absorptions can indicate the pattern of substitution on a benzene ring. However, this is beyond the scope of introductory organic chemistry. Arenes also possess a characteristic absorption at about 3030-3100 cm−1 as a result of the aromatic C–H stretch. It is somewhat higher than the alkyl C–H stretch (2850–2960 cm−1), but falls in the same region as olefinic compounds. Two bands (1500 and 1660 cm−1) caused by C=C in plane vibrations are the most useful for characterization as they are intense and are likely observed.
In aromatic compounds, each band in the spectrum can be assigned:
• C–H stretch from 3100-3000 cm-1
• overtones, weak, from 2000-1665 cm-1
• C–C stretch (in-ring) from 1600-1585 cm-1
• C–C stretch (in-ring) from 1500-1400 cm-1
• C–H "oop" from 900-675 cm-1
Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Exercise
10. A straw-colored oily liquid with a bitter almond odor and burning nut-like taste was isolated from the exudate of the castor sacs of mature North American beavers. The oily is slightly soluble in water with a boiling point of 197oC. Elemental analysis results are as follows: 68.84% C, 4.96% H, and 26.20% O. Name, draw the bond-line structure, and correlate the structure with the IR, 1H NMR, and 13C NMR spectral data below.
Answer
10.
17.10: Additional Exercises
General Review
17-1 State whether the following molecules are aromatic, antiaromatic or nonaromatic.
17-2 Predict which of the following nitrogen-containing heterocycles is more acidic than the other. Provide a reason for your choices.
17-3 For the following bases of DNA, identify the heterocyclic compound that makes up the core structure of each one - purine or pyrimidine?
17-4 Provide the proper IUPAC names for the following compounds.
17.11: Solutions to Additional Exercises
General Review
17-1
1. Aromatic
2. Antiaromatic
3. Nonaromatic
4. Aromatic
5. Aromatic
6. Nonaromatic
17-2
Compound A will be more acidic than compound B due to the aromaticity of its conjugate base. When compound A loses its proton, the new lone pair of electrons can delocalize with the double bonds in the ring, whereas when compound B is deprotonated, its new lone pair of electrons is localized on the nitrogen. Compound B is the stronger base.
17-3
Thymine and Cytosine are derivatives of purine. Adenine and Guanine are derivatives of pyrimidine.
17-4 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/17%3A_Aromatic_Compounds/17.08%3A_Polycyclic_Aromatic_Hydrocarbons.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• propose mechanisms for Electrophilic Aromatic Substitution Reactions (EAS): halogenation, nitration, sulfonation, and Friedel-Crafts Alkylation & Acylation (sections 18.1 to 18.5)
• predict products and specify reagents for Electrophilic Aromatic Substitution Reactions (EAS): halogenation, nitration, sulfonation, and Friedel-Crafts Alkylation & Acylation (sections 18.1 to 18.5)
• draw resonance structures of the sigma complexes resulting from EAS rxns of substituted aromatic rings (sections 18.1 to 18.5)
• draw reaction energy diagrams for EAS reactions (sections 18.1 to 18.5)
• explain why substituents are activating or deactivating and o,p-directors or m-directors (section 18.6)
• list the major substituents in their EAS activation “pecking order” (section 18.6)
• predict the products of side chain reactions: oxidation of catechols and alkyl substituents, bromination of benzylic carbons, SN1 and SN2 rxns at the benzylic carbon, reduction of carbonyls, and reduction of nitro groups (sections 18.7 and 18.12)
• design multiple step syntheses that use substituent effects to create the desired isomers of multi-substituted aromatic compounds (sections 18.8 and 18.9)
• predict the products of Nucleophilic Aromatic Substitution Reactions (NAS): addition-elimination and elimination-addition (benzyne) (sections 18.10 and 18.11)
• propose mechanisms for Nucleophilic Aromatic Substitution Reactions (NAS): addition-elimination and elimination-addition (benzyne) (sections 18.10 and 18.11)
18: Reactions of Aromatic Compounds
Benzene and Electrophilic Aromatic Substitution Reactions
While it took chemists many years to determine the structure of benzene and its derivatives, chemists recognized this class of compounds by their distinct aromas and low reactivity compared to isolated alkenes.
With the stability created by the conjugated pi electron system, the lack of chemical reactivity is not surprising. However, over time chemists found ways to catalyze reactions of benzene and its derivatives. With its strong electronic character, benzene is inherently electrophilic. The majority of the reactions for benzene are Electrophilic Aromatic Substitution reactions. One of the benzene hydrogen atoms can be substituted for a different group with electrophilic properties followed by restoration of the stable aromatic ring.
Examples of Electophilic Aromatic Substitution (EAS)
Many substitution reactions of benzene have been observed, the five most useful are listed below (chlorination and bromination are the most common halogenation reactions). Since the reagents and conditions employed in these reactions are electrophilic, these reactions are commonly referred to as Electrophilic Aromatic Substitution. The catalysts and co-reagents serve to generate the strong electrophilic species needed to effect the initial step of the substitution. The specific electrophile believed to function in each type of reaction is listed in the right hand column.
Reaction Type Typical Equation Electrophile E(+)
Halogenation: C6H6 + Cl2 & heat
FeCl3 catalyst
——> C6H5Cl + HCl
Chlorobenzene
Cl(+) or Br(+)
Nitration: C6H6 + HNO3 & heat
H2SO4 catalyst
——> C6H5NO2 + H2O
Nitrobenzene
NO2(+)
Sulfonation: C6H6 + H2SO4 + SO3
& heat
——> C6H5SO3H + H2O
Benzenesulfonic acid
SO3H(+)
Alkylation:
Friedel-Crafts
C6H6 + R-Cl & heat
AlCl3 catalyst
——> C6H5-R + HCl
An Arene
R(+)
Acylation:
Friedel-Crafts
C6H6 + RCOCl & heat
AlCl3 catalyst
——> C6H5COR + HCl
An Aryl Ketone
RCO(+)
General Mechanism
Electrophilic Aromatic Substitution (EAS) reactions include the same three mechanistic steps. Step 1 is needed to create a strong enough Electrophile to create reactivity with the pi electrons of benzene. Because the sigma complex is resonance stabilized, carbocation rearrangement is not a consideration for this intermediate.
Step 1: Formation of a Strong Electrophile
Step 2: Benzene pi electrons form a sigma bond with the Strong Electrophile to create the "sigma complex", a resonance stabilized, charged intermediate
Step 3: Deprotonation of the sigma complex to reform the aromatic ring
The generic mechanism shared by all EAS reactions is shown below.
Activating and Deactivation Groups
As we study the EAS reactions, we will learn that some substituents increase the reactivity of the benzene ring for EAS reactions and are called "activating groups". Other substituents decrease the reactivity of the benzene ring for EAS reactions and are called "deactivating groups". These groups can be recognized by their effects on the stability of the sigma complex. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/18%3A_Reactions_of_Aromatic_Compounds/18.01%3A_Electrophilic_Aromatic_Substitution_%28EAS%29.txt |
A Mechanism for Halogenation of Benzene
A three-step mechanism is common for many electrophilic aromatic substitution reactions. In the first step, a strong electrophile is created to entice the pi electrons of the aromatic ring to react. In the second, slow or rate-determining step a pair of pi electrons from the benzene form a sigma-bond with the electrophile generating a positively charged sigma complex (the benzenonium intermediate for halogenation). In the third, fast step, a proton is removed from the sigma complex producing a halogenated benzene ring. The steps are illustrated below.
Step 1: Formation of a strong electrophile, in this case an electrophilic bromine cation.
Step 2: Pi electrons of benzene react with the bromine cation to form the sigma comoplex, resonance stabilized benzenonium intermediate. This step is the rate determining step.
Step 3: Deprotonation of the benzenonium intermediate (sigma complex) to restore aromaticity.
The reaction energy diagram below shows Steps 2 and 3 of the mechanism since these steps involve benzene.
Exercises
1. What reagents would you need to get the given product?
What reagents would you need to gete given product
2. What product would result from the given reagents?
3. What is the major product given the reagents below?
4. Draw the formatin of Cl+ from AlCl3 and Cl2
5. Draw the mechanism of the reaction between Cl+ and a benzene.
Answer
1. Cl2 and AlCl3 or Cl2 and FeCl3
2. No Reaction
3.
4.
5.
18.03: Nitration of Benzene (an EAS Reaction)
Nitration of Benzene
Sulfuric acid catalyzes the nitration of benzene. It is important to note the chemical formula for the nitrate group bonded to benzene is -NO2. The chemical formula and name are assigned from an organic chemistry perspective which does not align with the inorganic perspective.
Step 1: Nitric acid (HNO3) is protonated by sulfuric acid which causes the loss of a water molecule and formation of a nitronium ion, a strong electrophile.
Steps 2 and 3: Two pi electrons from benzene form a sigma bond with the nitronium ion to create the sigma complex. Bisulfate deprotonates the sigma complex to restore the aromatic ring as shown below.
Exercise
6. Draw an energy diagram for the nitration of benzene. Draw the intermediates, starting materials, and products. Label the transition states.
Answer
6.
18.04: Sulfonation of Benzene (an EAS Reaction)
Sulfonation of Benzene
Sulfonation is a reversible reaction that produces benzenesulfonic acid by adding sulfur trioxide and fuming sulfuric acid. It is important to note that the chemical formula of the sulfonic group is -SO3H. The reaction is reversed by adding hot aqueous acid to benzenesulfonic acid to produce benzene.
Mechanism
To produce benzenesulfonic acid from benzene, fuming sulfuric acid and sulfur trioxide are added. Fuming sulfuric acid, also refered to as oleum, is a concentrated solution of dissolved sulfur trioxide in sulfuric acid. The sulfur in sulfur trioxide is electrophilic because the oxygens pull electrons away from it because oxygen is very electronegative. The benzene reacts with the sulfur of sulfur trixoxide to form the sigma complex. A subsequent proton transfer occurs to produce benzenesulfonic acid. All three steps are shown together in the mechanism below.
Reverse Sulfonation
Sulfonation of benzene is a reversible reaction. Sulfur trioxide readily reacts with water to produce sulfuric acid and heat. Therefore, by adding heat to benzenesulfonic acid in diluted aqueous sulfuric acid the reaction is reversed.
The reversibility of the sulfonation reaction creates an opportunity to prepare deuterated benzene. Isotopically labeled reagents can be useful in determining reaction mechanisms since the C-D bond is stronger than the C-H bond.
Exercise
7. What is/are the required reagent(s)for the following reaction:
8. What is the product of the following reaction:
9. Why is it important that the nitration of benzene by nitric acid occurs in sulfuric acid?
10. Write a detailed mechanism for the sulfonation of benzene, including all resonance forms.
Answer
7. SO3 and H2SO4 (fuming)
8.
9. Sulfuric acid is needed in order for a good electrophile to form. Sulfuric acid protonates nitric acid to form the nitronium ion (water molecule is lost). The nitronium ion is a very good electrophile and is open to attack by benzene. Without sulfuric acid the reaction would not occur.
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Friedel-Crafts Alkylation
The Fridedel-Crafts Alkylation reaction froms alkyl benzenes from alkyl halides. The usefulness of this reaction is limited, because it can be difficult to stop the reaction at a single alkylation. Additionally, a carbocation intermediate is produced in Step 1 which brings the potential for carbocation rearrangements (ominous theme music).
The mechanism takes place as follows:
Step 1: A carbocation is created to form the electrophile. This steps activates the haloalkane. Secondary and tertiary halides only form the free carbocation in this step.
Step 2: The pi electrons from benzene react with the electrophile to form the resonance stabilized alkylbenzenium ion.
Step 3: Any Lewis Base reacts picks up the hydrogen from the allkylbenzenium ion to reform the aromatic ring.
The finish step shown above is the two products.
The reactivity of haloalkanes increases as you move up the periodic table and increase polarity. This means that an RF haloalkane is most reactive followed by RCl then RBr and finally RI. This means that the Lewis acids used as catalysts in Friedel-Crafts Alkylation reactions tend have similar halogen combinations such as BF3, SbCl5, AlCl3, SbCl5, and AlBr3, all of which are commonly used in these reactions.
Some limitations of Friedel-Crafts Alkylation
There are possibilities of carbocation rearrangements when you are trying to add a carbon chain greater than two carbons. The rearrangements occur due to hydride shifts and methyl shifts. For example, the product of a Friedel-Crafts Alkylation will show an iso rearrangement when adding a three carbon chain as a substituent. One way to resolve these problems is through Friedel-Crafts Acylation.
Also, the reaction will only work if the ring you are adding a substituent to is not deactivated. Friedel-Crafts fails when used with compounds such as nitrobenzene and other strong deactivating systems.
Friedel-Crafts reactions cannot be preformed then the aromatic ring contains a NH2, NHR, or NR2 substituent. The lone pair electrons on the amines react with the Lewis acid AlCl3. This places a positive charge next to the benzene ring, which is so strongly activating that the Friedel-Crafts reaction cannot occur.
Lastly, Friedel-Crafts alkylation can undergo polyalkylation. The reaction adds an electron donating alkyl group, which activates the benzene ring to further alkylation.
This problem does not occur during Friedel-Crafts Acylation because an acyl group is deactivating. The prevents further acylations.
Friedel-Crafts Acylation
There is an additional reaction step for Fridel-Crafts Acylation. The acyl group of the product complexes with the aluminum chloride. Water is added to isolate the acyl benzene final product.
1st Reaction
Mechanism Step 1: Acylium ion formation
Mechanism Step 2: Pi electrons of benzene react with the acylium ion to form the sigma complex, resonance stabilized acylbenzenium intermediate:
Mechanism Step 3: Deprotonation of the sigma comlex to restore aromaticity.
During the third step, AlCl4 returns to remove a proton from the benzene ring, which enables the ring to return to aromaticity. In doing so, the original AlCl3 is regenerated for use again, along with HCl. Most importantly, we have the first part of the final product of the reaction, which is a ketone. The product forms a complex with aluminum chloride as shown below.
2nd Reaction: Water is added to liberate the final product as the acylbenzene:
Friedel-Crafts Acylations offer several synthetic advantages over Friedel-Crafts Alkylation. These advantages provide greater control over the production of reaction products. The acylium ion is stabilized by resonance, so no carbocation rearrangement occurs. Additionally, acyl groups are deactivating with for EAS reactions, so the product does not undergo further reactions. However, Friedel-Crafts Acylations do not work with nitrobenzenes or other deactivated benzene rings. The concept of deactivated will be more fully explored in the next two sections of this chapter.
Exercise
11. Which of the following will NOT undergo a rearrangement in a Friedel-Crafts reaction?
12. Suggest an acyl chloride that was used to make the following compounds:
Answer
11. A, B, and E will not undergo a rearrangement.
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Important Note:
Recognizing substituents as Electron Donating or Withdrawing is a useful skill for evaluating reaction mechanisms. For Electrophilic Aromatic Substitution (EAS) reactions, the rate determining step is the formation of a positively charged sigma complex. In future reactions, the intermediate may have a negative charge. While the electron donating and withdrawing properties of a substituent are inherent within the substituent, their effect on the stability of an intermediate and the reaction rate depends on the charge of the intermediate.
Substituents and their Directing Effects in EAS Reactions
Electron donating groups (D) direct the reaction to the ortho- or para-position, which means the electrophile substitutes for the hydrogen on carbon 2 or carbon 4 relative to the donating group. The withdrawing group directs the reaction to the meta position, which means the electrophile substitutes for the hydrogen on carbon 3 relative to the withdrawing group. The halogens are an exception to this pattern. The halogens are a deactivating group that direct ortho or para substitution.
Examples of electron donating groups in the relative order from the most activating group to the least activating:
-NH2, -NR2 > -OH, -OR> -NHCOR> -CH3 and other alkyl groups with R as alkyl groups (CnH2n+1)
Examples of electron withdraing groups in the relative order from the most deactivating to the least deactivating:
-NO2, -CF3> -COR, -CN, -CO2R, -SO3H > Halogens with R as alkyl groups (CnH2n+1)
ortho-, para-Directors via Resonance
Groups that donate electrons through resonance are ortho-, para-directors for EAS reactions. Methoxybenzene (anisole) will be used to demonstrate the ortho-, para-direction of substituents that stabilize the sigma complex through resonance. The nitronium ion (O=N+=O) will be used to represent the Electrophile (E+).
The ortho- and para-directed mechanisms for the nitration of anisole are shown below. When the nitro group adds at the ortho or para position, the stability of the sigma complex is increased by the presence of a fourth resonance form . The greater the stability of the sigma complex causes the ortho and para products for form faster than meta. Generally, the para-product is favored over the ortho-product because of steric effects even though there are two ortho- positions.
Mechanism for ortho-directed product formation
Mechanism for para-directed product formation
ortho-, para-Directors via Induction
Alkyl groups are ortho-, para-directors for EAS reactions. Toluene will be used to demonstrate the ortho-, para-direction of substituents that stabilize the sigma complex through induction. The nitronium ion (O=N+=O) will be used to represent the Electrophile (E+). Since the inductive effect is weaker than resonance, we can see that a small percentage of the meta product is also isolated.
Looking at the stability of the resonance structures of the sigma complex in the reaction mechanism for nitration of toluene explains why the ortho- and para- substitutions are the major products. When the nitro group adds at the ortho or para position, the methyl group stabilizes the transition state through induction electron donation which favors the formation of the ortho- and para- products. As seen with the resonance directed products, the para product is favored because of steric effects even though there are two ortho- positions.
Mechanism for ortho-directed product formation
Mechanism for para-directed product formation
meta Directors - the Electron Withdrawing Groups
Electron withdrawing groups destabilize the sigma complex and deactivate benzene rings to EAS reactions. For electron withdrawing groups, all of the sigma complexes are destabilized. The meta-position is the least destabilized and produces the largest percentage of the reaction products.
Acetophenone will be used to demonstrate the reactivity of meta-directors using the sigma complexes below. Acyl groups are resonance deactivators.
Ortho and para reactions produce a resonance structure that places the arenium cation next to an additional cation at the carbonyl carbon. This close proximity of partial positive charges destabilizes the sigma complex and slows down ortho and para reaction.
By default the meta product forms faster because the destabilizing effects are reduced through greater physical separation of the partial positive charges.
Substituents and Electrophilic Aromatic Substitution (EAS) Reaction Rates
Since sigma complex formation is the rate determining step of EAS reactions, benzene derivatives are divided into two groups based on how the substituent stabilizes or destabilizes the positively charged sigma complex. The EAS reaction of a substituted ring with an activating group is faster than benzene. On the other hand, a substituted ring with a deactivated group is slower than benzene. Activating groups speed up the EAS reaction by either resonance or inductive electron donation (typically R groups). For resonance, unpaired electrons can be donated to stabilize the positive charge of the sigma complex in the transition state. Stabilizing the intermediate, speeds up the reaction by lowering the activating energy. Inductive electron donation by R groups is an analogous, yet weaker effect than resonance. Inductive electron donation helps to stabilize the sigma complex and speed up (activate) the reaction. Deactivating groups withdraw the electrons away from the carbocation of the sigma complex causing destabilization and increasing the activation energy which slows down (deactivates) the reaction.
• Activated rings: the substituents on the ring donate electrons and increase EAS reaction rates
• Examples of electron donating groups in the relative order from the most activating group to the least activating:
-NH2, -NR2 > -OH, -OR> -NHCOR> -CH3 and other alkyl groups with R as alkyl groups (CnH2n+1)
The reaction energy diagram illustrating the substituent effect of electron donating groups (D:) on EAS reaction rates is shown below.
• Deactivated rings: the substituents on the ring withdraw electrons and decrease EAS reaction rates
• Examples of electron withdraing groups in the relative order from the most deactivating to the least deactivating:
-NO2, -CF3> -COR, -CN, -CO2R, -SO3H > Halogens with R as alkyl groups (CnH2n+1)
The reaction energy diagram illustrating the substituent effect of electron withdrawing groups (W) on EAS reaction rate is shown below.
• The Halogen Paradox: Deactivators that are ortho, para-directors
Halogens deactivate rings to subsequent EAS reactions. The order of reactivity of the benzene rings toward the electrophilic substitution when it is substituted with a halogen groups, follows the order of electronegativity.
F> Cl > Br > I
The ring that is substituted with the most electronegative halogen is the most reactive ring ( less deactivating substituent) and the ring that is substituted with the least electronegatvie halogen is the least reactive ring ( more deactivating substituent ). The size of the halogen also effects the reactivity of the benzene ring - as the size of the halogen increases, the reactivity of the ring decreases.
However, the lone pair electrons on the halogen atoms are still available for resonance delocalization in the sigma complex causing ortho-, para-direction of the electrophile. The reaction energy diagram below resolves these contradictory aspects of EAS reactions of halogenated benzene derivatives.
Outside Links
In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Exercises
13. Predict the direction of the electrophile substition on these rings:
14. Which nitration product is going to form faster?
nitration of aniline or nitration of nitrobenzene?
15. Predict the product of the following two sulfonation reactions:
A.
16. Classify these two groups as activating or deactivating groups:
A. alcohol
B. ester
17. By which effect does trichloride effect a monosubstituted ring?
18. Trichloromethylbenzene has a strong concentration of electrons at the methyl substituent. Comparing this compound with toluene, which is more reactive toward electrophilic substitution?
19. The following compound is less reactive towards electrophilic substitution than aniline? Explain.
20. Consider the intermediates of the following molecule during an electrophilic substitution. Draw resonance structures for ortho, meta, and para reactions.
Answer
13. The first substitution is going to be ortho and/or para substitution since we have a halogen subtituent. The second substition is going to be ortho and/or para substitution also since we have an alkyl substituent.
14. The nitration of aniline is going to be faster than the nitration of nitrobenzene, since the aniline is a ring with NH2 substituent and nitrobenzene is a ring with NO2 substiuent. As described above NH2is an activating group which speeds up the reaction and NO2 is deactivating group that slows down the reaction.
15.
A. the product is
B. the product is
16.
A. alcohol is an activating group.
B. ester is a deactivating group.
17. Trichloride deactivate a monosubstitued ring by inductive effect.
18. The trichloromethyl group is an electron donor into the benzene ring, therefore making it more stable and therefore more reactive compared to electrophilic substitution.
19. As seen in resonance the electron density is also localized off of the ring, thereby deactivating it compared to aniline.
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Oxidation of Alkyl Side Chains
There is another reaction which occurs at the atom directly attached to an aromatic ring. This is known as "side chain oxidation." When a compound which has an alkyl group directly attached to an aryl group is treated with a strong oxidizing agent like potassium permanganate (KMnO4) or Jones Reagent (CrO3/H2SO4), the benzylic carbon is oxidized to a carboxylic acid group which remains attached to the aryl group. Any other carbon-carbon bonds in the alkyl group are broken. For the oxidation reaction, the number of carbon atoms in the alkyl side chain does not matter, However, the benzylic carbon must have at least one benzylic hydrogen attached. Thus, tertiary carbons attached to an aromatic ring are not affected by these reactions. It should be noted that during this reaction an ortho/para directing alkyl group is converted to a meta directing carboxylic.
Add tertiary no reaction
Two other examples of this reaction are given below, and illustrate its usefulness in preparing substituted benzoic acids.
Add straing chain and tertirary.
The mechanism of this reaction is obscure, but the fact that it specifically requires that there be a benzylic C-H bond suggests that breaking this bond is essential. Any intermediate that might be formed by breaking this bond will be stabilized by resonance with the aryl group, which provides an explanation for the specificity of attack at the benzylic position. Such reactions also occur in a biological context. Enzymes oxidize alkyl side chains on aromatic rings as part of making such compounds soluble enough to be eliminated.
Bromination of the Benzylic Carbon
The bromination reaction is the N-bromosuccinamide (NBS) radical, substitution reaction previously studied. As with the oxidation reaction, one benzylic hydrogen is needed so that it can be substituted with bromine. Examples of both reactions are shown below.
The brominating reagent, N-bromosuccinimide (NBS), has proven useful for achieving allylic or benzylic substitution in CCl4 solution at temperatures below its boiling point (77 ºC). One such application is shown in the second equation. the allylic bromination with NBS is analagous to the alkane halogenation reaction (Section 10.3) since it also occurs as a radical chain reaction. The NBS serves as the source for the bromine, which is used in the initiation step to create a bromine radical that then abstracts a proton from the allylic position in the propagation step. The radical created then reacts with the NBS to to become bromiated and the cycle continues until it is terminated.
The predominance of allylic substitution over other positions comes down to bond dissociation energies. The relative bond dissociation energies are shown in the table at the top of this section. The C-H bond that we are focusing on as the point of difference for each of the energies shows that the allylic C-H bond has a strength of about 88 kcal/mol. This means that the allylic radical created is more stable than a typical alkyl radical with the same substitution by about 9 kcal/mol. Therefore, this radical is the most likely one to form and thus react.
The benzylic C-H bonds weaker than most sp3 hybridized C-H. This is because the radical formed from homolysis is resonance stabilized.
Resonance stabilization of the benzylic radical
Because of the weak C-H bonds, benzylic hydrogens can form benzylic halides under radical conditions.
NBS as a Bromine Source
NBS (N-bromosuccinimide) is the most commonly used reagent to produce low concentrations of bromine. When suspended in tetrachloride (CCl4), NBS reacts with trace amounts of HBr to produce a low enough concentration of bromine to facilitate the allylic bromination reaction.
Allylic Bromination Mechanism
Step 1: Initiation
Once the pre-initiation step involving NBS produces small quantities of Br2, the bromine molecules are homolytically cleaved by light to produce bromine radicals.
Step 2 and 3: Propagation
Step 4: Termination
Acyl Side Chain Reductions
Since there are several limitations to the Friedel-Crafts alkylation that are not observed with the Friedel-Crafts acylation, reduction of the acyl side chain to an alkyl side chain is a useful reaction for multiple step synthesis. The Wolff-Kishner reaction reduces the carbon groups (aldehydes and ketones) to alkanes and is not limited to acyl groups bonded to benzene rings. This acyl reduction reaction is also useful because acyl groups are deactivating, meta-directors, and alkyl groups are activating, ortho-, para-directors which adds flexibility to multiple step synthesis strategies.
Exercise
21. Draw the bond-line structures for the product(s) of the following reactions.
Answer
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Synthetic Considerations
To develop multiple step syntheses for di-substituted benzene derivatives, the regiochemistry of the substituents will determine the order of the reactions. The directing effects of the benzene substituents are summarized below.
The limitations of the Friedel-Crafts reactions must also be considered. Friedel-Crafts alkylation and acylation reactions can only occur on benzene rings or benzene rings with ortho-, para-directors (activated rings or rings with halogens). Even though both acyl and nitro groups are meta-directors, benzene would need to be acylated before it is nitrated. To synthesize 3-nitroanisole [1-(3-nitrophenyl)ethan-1-one], the top reaction sequence is needed. The bottom reaction sequence will not produce the desired product as shown in the two synthetic pathways below.
Exercise
22. Starting with benzene and using any synthetic reagents, propose a multiple step synthesis for each of the following compounds.
Answer
22.
18.09: Trisubstituted Benzenes - Effects of Multiple Substituents
Orientational Interaction of Substituents
When a benzene ring has two substituent groups, each exerts an influence on subsequent substitution reactions. The activation or deactivation of the ring can be predicted more or less by the sum of the individual effects of these substituents. The site at which a new substituent is introduced depends on the orientation of the existing groups and their individual directing effects. We can identify two general behavior categories, as shown in the following table. Thus, the groups may be oriented in such a manner that their directing influences act in concert, reinforcing the outcome; or are opposed (antagonistic) to each other. Note that the orientations in each category change depending on whether the groups have similar or opposite individual directing effects.
Antagonistic or Non-Cooperative
Reinforcing or Cooperative
D = Electron Donating Group (ortho/para-directing)
W = Electron Withdrawing Group (meta-directing)
Reinforcing or Cooperative Substitutions
The products from substitution reactions of compounds having a reinforcing orientation of substituents are easier to predict than those having antagonistic substituents. For example, the six equations shown below are all examples of reinforcing or cooperative directing effects operating in the expected manner. Symmetry, as in the first two cases, makes it easy to predict the site at which substitution is likely to occur. Note that if two different sites are favored, substitution will usually occur at the one that is least hindered by ortho groups.
The first three examples have two similar directing groups in a meta-relationship to each other. In examples 4 through 6, oppositely directing groups have an ortho or para-relationship. The major products of electrophilic substitution, as shown, are the sum of the individual group effects. The strongly activating hydroxyl (–OH) and amino (–NH2) substituents favor dihalogenation in examples 5 and six.
Antagonistic or Non-Cooperative Substitutions
Substitution reactions of compounds having an antagonistic orientation of substituents require a more careful analysis. If the substituents are identical, as in example 1 below, the symmetry of the molecule will again simplify the decision. When one substituent has a pair of non-bonding electrons available for adjacent charge stabilization, it will normally exert the product determining influence, examples 2, 4 & 5, even though it may be overall deactivating (case 2). Case 3 reflects a combination of steric hindrance and the superior innate stabilizing ability of methyl groups relative to other alkyl substituents. Example 6 is interesting in that it demonstrates the conversion of an activating ortho/para-directing group into a deactivating meta-directing "onium" cation [–NH(CH3)2(+) ] in a strong acid environment.
In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Exercise
23. Predict the products of the following reactions:
Answer
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A Nucleophilic Aromatic Displacement Reactions of Aryl Halides
The carbon-halogen bonds of aryl halides are like those of alkenyl halides in being much stronger than those of alkyl halides. The simple aryl halides generally are resistant to reaction with nucleophiles in either SN1 or SN2 reactions. However, this low reactivity can be changed dramatically by changes in the reaction conditions and the structure of the aryl halide. In fact, nucleophilic displacement becomes quite rapid
1. when the aryl halide is activated by substitution with strongly electron-attracting groups such as N02, and
2. when very strongly basic nucleophilic reagents are used.
Addition-Elimination Mechanism of Nucleophilic Substitution of Aryl Halides
Although the simple aryl halides are inert to the usual nucleophilic reagents, considerable activation is produced by strongly electron-attracting substituents provided these are located in either the ortho or para positions, or both. For example, the displacement of chloride ion from 1-chloro-2,4-dinitrobenzene by dimethylamine occurs readily in ethanol solution at room temperature. Under the same conditions chlorobenzene completely fails to react; thus the activating influence of the two nitro groups amounts to a factor of at least 108:
In general, the reactions of activated aryl halides closely resemble the SN2-displacement reactions of aliphatic halides. The same nucleophilic reagents are effective (e.g., CH3O-, HO-, and RNH2); the reactions are second order overall (first order in halide and first order in nucleophile); and for a given halide the stronger the nucleophile, the faster the reaction. However, there must be more than a subtle difference in mechanism because an aryl halide is unable, to pass through the same type of transition state as an alkyl halide in SN2 displacements.
The generally accepted mechanism of nucleophilic aromatic substitution of aryl halides carrying activating groups involves two steps that are closely analogous to those described for alkenyl and alkynyl halides. The first step involves the nucleophile Y:- reacting with the carbon bearing the halogen substituent to form an intermediate carbanion (Equation 14-3). The aromatic system is destroyed on forming the anion, and the carbon at the reaction site changes from planar (sp2 bonds) to tetrahedral (sp3 bonds).
In the second step, loss of an anion, X- or Y-, regenerates an aromatic system. If X- is lost, the overall reaction is nucleophilic displacement of X by Y. In the case of a neutral nucleophilic reagent, Y or HY, the reaction sequence would be the same except for the necessary adjustments in the charge of the intermediate:
Why is this reaction pathway generally unfavorable for the simple aryl halides? The answer is that the anionic intermediate is too high in energy to be formed at any practical rate. Not only has the anion lost the aromatic stabilization of the benzene ring, but its formation results in transfer of negative charge to the ring carbons, which themselves are not very electronegative:
However, when strongly electron-attracting groups are located on the ring at the ortho-para positions, the intermediate anion is stabilized by delocalization of electrons from the ring carbons to more favorable locations on the substituent groups. As an example, consider the displacement of bromine by -OCH3 in the reaction of 4-bromonitrobenzene and methoxide ion:
The anionic intermediate formed by addition of methoxide ion to the aryl halide can be described by the valence-bond structures 5a-5d. Of these structures 5d is especially important because the charge is transferred from the ring carbons to the electronegative oxygen of the nitro substituent:
Substituents in the meta positions have much less effect on the reactivity of an aryl halide because delocalization of electrons to the substituent is not possible. No formulas can be written analogous to 5c and 5d in which the negative charges are both on atoms next to positive nitrogen, and
In a few instances, stable compounds resembling the postulated reaction intermediate have been isolated. One classic example is the complex 7 (isolated by J. Meisenheimer), which is the product of the reaction of either the methyl aryl ether 6 with potassium ethoxide, or the ethyl aryl ether 8 and potassium methoxide:
Exercise
24. Propose a mechanism for the following reaction:
Answer
24.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/18%3A_Reactions_of_Aromatic_Compounds/18.10%3A_Nucleophilic_Aromatic_Substitution_-_The_Addition-Elimantion_.txt |
Elimination-Addition Mechanism of Nucleophilic Aromatic Substitution via Benzyne (Arynes)
The reactivities of aryl halides, such as the halobenzenes, are exceedingly low toward nucleophilic reagents that normally effect displacements with alkyl halides and activated aryl halides. Substitutions do occur under forcing conditions of either high temperatures or very strong bases. For example, chlorobenzene reacts with sodium hydroxide solution at temperatures around $340^\text{o}$ and this reaction was once an important commercial process for the production of benzenol (phenol):
In addition, aryl chlorides, bromides, and iodides can be converted to areneamines $\ce{ArNH_2}$ by the conjugate bases of amines. In fact, the reaction of potassium amide with bromobenzene is extremely rapid, even at temperatures as low as $-33^\text{o}$ with liquid ammonia as solvent:
However, displacement reactions of this type differ from the previously discussed displacements of activated aryl halides in that rearrangement often occurs. That is, the entering group does not always occupy the same position on the ring as that vacated by the halogen substituent. For example, the hydrolysis of 4-chloromethylbenzene at $340^\text{o}$ gives an equimolar mixture of 3- and 4-methylbenzenols:
Even more striking is the exclusive formation of 3-methoxybenzenamine in the amination of 2-chloromethoxybenzene. Notice that this result is a violation of the principle of least structural change (Section 1-1H):
The mechanism of this type of reaction has been studied extensively, and much evidence has accumulated in support of a stepwise process, which proceeds first by base-catalyzed elimination of hydrogen halide $\left( \ce{HX} \right)$ from the aryl halide - as illustrated below for the amination of bromobenzene:
Elimination
The product of the elimination reaction is a highly reactive intermediate $9$ called benzyne, or dehydrobenzene, which differs from benzene in having two less hydrogen and an extra bond between two ortho carbons. Benzyne reacts rapidly with any available nucleophile, in this case the solvent, ammonia, to give an addition product:
Addition
The rearrangements in these reactions result from the reaction of the nucleophile at one or the other of the carbons of the extra bond in the intermediate. With benzyne the symmetry is such that no rearrangement would be detected. With substituted benzynes isomeric products may result. Thus 4-methylbenzyne, $10$, from the reaction of hydroxide ion with 4-chloro-1-methylbenzene gives both 3- and 4-methylbenzenols:
In the foregoing benzyne reactions the base that produces the benzyne in the elimination step is derived from the nucleophile that adds in the addition step. This need not always be so, depending on the reaction conditions. In fact, the synthetic utility of aryne reactions depends in large part of the success with which the aryne can be generated by one reagent but captured by another. One such method will be discussed in Section 14-10C and involves organometallic compounds derived from aryl halides. Another method is to generate the aryne by thermal decomposition of a 1,2-disubstituted arene compound such as $11$, in which both substituents are leaving groups - one leaving with an electron pair, the other leaving without:
When $11$ decomposes in the presence of an added nucleophile, the benzyne intermediate is trapped by the nucleophile as it is formed. Or, if a conjugated diene is present, benzyne will react with it by a [4 + 2] cycloaddition. In the absence of other compounds with which it can react, benzyne will undergo [2 + 2] cycloaddition to itself:
Exercise
25. When p-chlorotoluene is reacted with NaOH, two products are seen. While when m-chlorotoluene is reacted with NaOH, three products are seen. Explain this.
Answer
25. You need to look at the benzyne intermediates. The para substituted only allows for two products, while the para produces two different alkynes which give three different products.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
18.12: Reduction of Aromatic Compounds
Reduction of Aromatic Compounds
Although it does so less readily than simple alkenes or dienes, benzene adds hydrogen at high pressure in the presence of Pt, Pd or Ni catalysts. The product is cyclohexane and the heat of reaction provides evidence of benzene's thermodynamic stability. Substituted benzene rings may also be reduced in this fashion, and hydroxy-substituted compounds, such as phenol, catechol and resorcinol, give carbonyl products resulting from the fast ketonization of intermediate enols. Nickel catalysts are often used for this purpose, as noted in the following equations.
Reduction of Nitro Groups and Aryl Ketones Substituents on Benzene Rings
Electrophilic nitration and Friedel-Crafts acylation reactions introduce deactivating, meta-directing substituents on an aromatic ring. The attached atoms are in a high oxidation state, and their reduction converts these electron withdrawing functions into electron donating amino and alkyl groups. Reduction is easily achieved either by catalytic hydrogenation (H2 + catalyst), or with reducing metals in acid. Examples of these reductions are shown here, equation 6 demonstrating the simultaneous reduction of both functions. Note that the butylbenzene product in equation 4 cannot be generated by direct Friedel-Crafts alkylation due to carbocation rearrangement. The zinc used in ketone reductions, such as 5, is usually activated by alloying with mercury (a process known as amalgamation).
Several alternative methods for reducing nitro groups to amines are known. These include zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most cases.
The Birch Reduction
Another way of adding hydrogen to the benzene ring is by treatment with the electron rich solution of alkali metals, usually lithium or sodium, in liquid ammonia. See examples of this reaction, which is called the Birch Reduction. The Birch reduction is the dissolving-metal reduction of aromatic rings in the presence of an alcohol.
Mechanism:
Exercise
26. How would you make the following from benzene and an acid chloride?
Answer
26. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/18%3A_Reactions_of_Aromatic_Compounds/18.11%3A__NAS_Reactions_-_the_Elimination-Addition_%28Benzyne%29_Mecha.txt |
18-1 Draw the resonance structures for benzaldehyde to show the electron-withdrawing group.
18-2 Draw the resonance structures for methoxybenzene to show the electron-donating group.
18-3 How would make the following compounds from benzene?
A) m-bromonitrobenzene
B) m-bromoethylbenzene
18-4 There is something wrong with the following reaction, what is it?
18-5 Choose the correct answer.
a) 1-methyl-3-(2-methylprop-1-en-1-yl)benzene
b) 2-chloro-4-methyl-1-(2-methylprop-1-en-1-yl)benzene
c) 1-chloro-3-(2-methylprop-1-en-1-yl)benzene
d) 1-(1-chloroethenyl)-3-methylbenzene
18-6 Predict the final product of the following reaction chain.
18-7 Provide the final product for the following reactions.
18-8 For the following reaction chain, provide the intermediate and final product(s).
18-9 Give the IUPAC name for the final product(s) of the previous problem, 18-8.
18-10 For the following reaction chain, provide the intermediate and final product(s).
18-11 Give the final product of the following reactions.
Halogenation, Nitration, and Sulfonation of Benzene
18-12 Predict the products of the following reactions.
18-13 Give the IUPAC nomenclature and structure of the product of the following reaction.
18-14 Choose the correct answer that describes the best route of synthesis of the following molecule.
a) Chlorination, sulfonation, nitration
b) Sulfonation, nitration, chlorination
c) Nitration, sulfonation, chlorination
Activating, Ortho-, Para-Directing Substituents
18-15 For the following compounds, point to the position(s) on the ring that are most likely to have a substituent added.
18-16 Predict the major product of the following reactions.
18-17 Provide the correct IUPAC nomenclature and structure of the product of the following reaction.
Deactivating, Meta-Directing Substituents
18-18 Predict the products of the following reactions.
18-19 Predict the product of the following reaction.
18-20 Choose the pathway that will lead to the product formed on the right.
Halogen Substitutes: Deactivating, but Ortho, Para-Directing
18-21 Choose the correct IUPAC nomenclature of one of the products of the following reaction.
a) 3-bromobenzene-1-sulfonic acid
b) 4-bromobenzene-1-sulfonic acid
c) 5-bromobenzene-1,3-disulfonic acid
d) 4-bromophenyl hydrogen sulfate
18-22 Predict the products of the following reactions.
18-23 Propose a route of synthesis for the following compound, starting with chlorobenzene (assume any desired intermediate compounds can be isolated for use in subsequent steps).
Effects of Multiple Substituents on Electrophilic Aromatic Substitution
18-24 For the following compounds, identify which substituent is the stronger activating group and predict the position(s) of a subsequent electrophilic aromatic substitution.
18-25 Predict all possible singly chlorinated products of the following reaction.
18-26 Rank the following compounds in order from slowest to fastest to go through an electrophilic aromatic substituent reaction.
Friedel-Crafts Alkylation/Acylation
18-27 Predict the products of the following reactions.
18-28 Explain whether or not the following reaction is the best way to synthesize propylbenzene and if not, propose a better route of synthesis.
18-29 Choose the correct answer and if a product is formed, provide the structure of the product.
a) No reaction
b) 4-amino-2-methylbenzonitrile
c) 4-amino-2-ethylbenzonitrile
d) N-(4-cyano-2-ethylphenyl)acetamide
Nucleophilic Aromatic Substitution
18-30 Predict the product of the following reaction and provide the correct IUPAC nomenclature.
18-31 Predict the product of the following reaction.
18-32 Suggest a route of synthesis to make N-hydroxy-3,5-dimethylaniline from 1-bromo-4-nitrobenzene.
Aromatic Substitutions Using Organometallic Reagents
18-33 Provide the structure and IUPAC nomenclature of the product of the following reaction.
18-34 Suggest a route of synthesis to make 2-bromo-4-chlorobenzoic acid from 4-nitrobenzoic acid.
18-35 Choose the correct IUPAC nomenclature for the product of the following reaction.
a) 2-bromo-5-nitrobenzoic acid
b) 2-bromo-5-cyanobenzoic acid
c) 3-cyano-5-nitrobenzoyl bromide
d) 3-bromo-5-cyanobenzoic acid
Side-Chain Reactions of Benzene Derivatives
18-36 Predict the products of the following reactions.
18-37 Choose the correct structure of the product of the following reaction.
18-38 Provide the intermediate and final products of the following reactions.
18.14: Solutions to Additional Exercises
18-1
18-2
18-3 This is just one possible way to synthesize it.
18-4 The bromine should be in the meta position. Right now it is in the ortho position, from perhaps having the ethyl group present first and then the having it substituted there. BUT the ethyl group is last to form, and the aldehyde and nitro groups would both encourage a meta substitution.
18-5 Answer: A
18-6
18-7
18-8
18-9 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene
18-10
18-11
Halogenation, Nitration, and Sulfonation of Benzene
18-12:
18-13:
18-14:
Answer: A
Activating, Ortho-, Para-Directing Substituents
18-15:
18-16:
18-17:
Deactivating, Meta-Directing Substituents
18-18:
18-19:
18-20:
Answer: B
Halogen Substitutes: Deactivating, but Ortho, Para-Directing
18-21:
Answer: B
18-22:
18-23:
Effects of Multiple Substituents on Electrophilic Aromatic Substitution
18-24:
18-25:
18-26:
Friedel-Crafts Alkylation/Acylation
18-27:
18-28:
Friedel-Crafts alkylation using 1-chloropropane is not the best way to synthesize propylbenzene. You will end up with (propan-2-yl)benzene as your main product due to a hydride shift occurring during an intermediate step. A better route of synthesis may be Friedel-Crafts acylation using propanoyl chloride to make 1-phenylpropan-1-one, followed by a Clemmensen reduction to obtain the final product.
18-29:
Answer: D
Nucleophilic Aromatic Substitution
18-30:
18-31:
18-32:
Possible route of synthesis:
Aromatic Substitutions Using Organometallic Reagents
18-33:
18-34:
Possible route of synthesis:
18-35:
Answer: D
Side-Chain Reactions of Benzene Derivatives
18-36:
18-37:
Answer: A
18-38: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/18%3A_Reactions_of_Aromatic_Compounds/18.13%3A_Additional_Exercises.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• describe the structure and physical properties of aldehydes and ketones (section 19.1)
• determine the structure of aldehydes and ketones from their elemental analysis and spectral data (MS, IR 1H NMR & 13C NMR) (section 19.2)
• predict the products and specify the reagents to synthesize aldehydes and ketones for reactions studied to date (section 19.3)
• predict the products and specify the reagents to synthesize aldehydes and ketones for new reactions (section 19.4)
• write the general mechanism for nucleophilic addition reactions with aldehydes and ketones (sections 19.5 to 19.11, 19.13, & 19.15)
• predict the relative reactivity of carbonyl compounds to nucleophilic addition reactions (sections 19.5 to 19.13, & 19.15)
• predict the relative equilibrium constant & rates of hydration for aldehydes and ketones (section 19.6)
• show the general mechanism for the Wittig reaction (section 19.13)
• predict the products and specify the reagents for oxidation and reduction reactions of aldehydes and ketones (section 19.14 and 19.15)
• combine the reactions studied to date to develop efficient and effective multiple-step synthesis including the use of acetals/ketals as protecting groups (sect 19.12)
Please note: IUPAC nomenclature and important common names of aldehydes and ketones were explained in Chapter 3.
19: Ketones and Aldehydes
The Carbonyls: the Aldehyes and Ketones
While there are several functional groups that include a carbonyl structural feature, the term "carbonyls" is used to describe aldehydes and ketones.
Aldehydes and ketones share a great deal of chemical reactivity so it makes sense to talk about these two functional groups within the same chapter. Aldehydes and ketones are both polar molecules that are H-bond acceptors. Following the "4 to 6 Rule", aldehydes and ketones with short carbon chains are soluble in water. Aldehydes and ketones are typically liquids with densities of approximately 0.8 g/mL. Aldehydes and ketones are prevalent in common household substances as shown in the table below.
Aldehydes and Ketones are Electrophiles
The carbon of the carbonyl group is electrophilic because of resonance as shown below.
19.02: Spectroscopy of Ketones and Aldehydes
IR Spectra
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
C=O stretch
• aliphatic ketones 1715 cm-1
• alpha, beta-unsaturated ketones 1685-1666 cm-1
Figure 8. shows the spectrum of 2-butanone. This is a saturated ketone, and the C=O band appears at 1715.
If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
H–C=O stretch 2830-2695 cm-1
C=O stretch
• aliphatic aldehydes 1740-1720 cm-1
• alpha, beta-unsaturated aldehydes 1710-1685 cm-1
Figure 9. shows the spectrum of butyraldehyde.
NMR Spectra
Hydrogens attached to carbon adjacent to the sp2 hybridized carbon in aldehydes and ketones usually show up 2.0-2.5 ppm.
.
Aldehyde hydrogens are highly deshielded and appear far downfield as 9-10 ppm.
Chemical shift of each protons is predicted by 1H chemical shift ranges (Ha): chemical shift of methyl groups (1.1 ppm). (Hb) The chemical shift of the -CH- group move downfield due to effect an adjacent aldehyde group: (2.4 ppm). The chemical shift of aldehyde hydrogen is highly deshielded (9.6 ppm).
4) Splitting pattern is determined by (N+1) rule: Ha is split into two peaks by Hb(#of proton=1). Hb has the septet pattern by Ha (#of proton=6). Hc has one peak.(Note that Hc has doublet pattern by Hb due to vicinal proton-proton coupling.)
Mass Spectra
Much of the utility in electron-ionization MS comes from the fact that the radical cations generated in the electron-bombardment process tend to fragment in predictable ways. Detailed analysis of the typical fragmentation patterns of different functional groups is beyond the scope of this text, but it is worthwhile to see a few representative examples, even if we don’t attempt to understand the exact process by which the fragmentation occurs. We saw, for example, that the base peak in the mass spectrum of acetone is m/z = 43. This is the result of cleavage at the ‘alpha’ position - in other words, at the carbon-carbon bond adjacent to the carbonyl. Alpha cleavage results in the formation of an acylium ion (which accounts for the base peak at m/z = 43) and a methyl radical, which is neutral and therefore not detected.
After the parent peak and the base peak, the next largest peak, at a relative abundance of 23%, is at m/z = 15. This, as you might expect, is the result of formation of a methyl cation, in addition to an acyl radical (which is neutral and not detected).
A common fragmentation pattern for larger carbonyl compounds is called the McLafferty rearrangement:
The mass spectrum of 2-hexanone shows a 'McLafferty fragment' at m/z = 58, while the propene fragment is not observed because it is a neutral species (remember, only cationic fragments are observed in MS). The base peak in this spectrum is again an acylium ion.
When alcohols are subjected to electron ionization MS, the molecular ion is highly unstable and thus a parent peak is often not detected. Often the base peak is from an ‘oxonium’ ion.
Other functional groups have predictable fragmentation patterns as well. By carefully analyzing the fragmentation information that a mass spectrum provides, a knowledgeable spectrometrist can often ‘put the puzzle together’ and make some very confident predictions about the structure of the starting sample.
Click here for examples of compounds listed by functional group, which demonstrate patterns which can be seen in mass spectra of compounds ionized by electron impact ionization.
Exercise
1. a) What are the masses of all the components in the following fragmentation reactions?
b) A mixture was separated into three fractions: A, B, and C. Elemental analysis reveals that the fractions are structural isomers with the following composition: 69.72% C, 11.70% H, and 18.58% O. The IR spectra for all fractions show several moderate bands around 2950 cm-1, and a strong band near 1700 cm-1. The proton and 13C NMR spectra for each fraction are shown below. Give the common name and draw the bond-line structure for each fraction and correlate the NMR signals with their respective atoms.
Answer
1. a)
b) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/19%3A_Ketones_and_Aldehydes/19.01%3A_Carbonyl_Compound_Structure_and_Properties.txt |
So far this text has discussed aldehyde and ketone synthesis from the ozolysis of alkenes, hydration of alkynes, oxidation of alcohols, and Friedel-Crafts acylation of benzene rings.
Hydration of an alkyne to form aldehydes
Anti-Markovnikov addition of a hydroxyl group to an alkyne forms an aldehyde. The addition of a hydroxyl group to an alkyne causes tautomerization which subsequently forms a carbonyl.
Hydration of an alkyne to form ketones
The addition of a hydroxyl group to an alkyne causes tautomerization which subsequently forms a carbonyl. Markovnikov addition of a hydroxyl group to an alkyne forms a ketone.
Oxidation of 2o alcohols to form ketones
Typically uses Jones reagent (CrO3 in H2SO4) but many other reagents can be used
Friedel-Crafts acylation to form a ketone
Exercise
2. Specify the reagents to complete the reaction map below.
Answer
2.
19.04: 19.4 New Synthesis of Aldehydes and Ketones
Ester, Acid Chloride, and Nitrile Reduction to form Aldehydes
The reduction of esters, acid chlorides, and nitriles require reducing agents that are derivatives of lithium aluminum hydride (LiAlH4). For esters and nitriles, LiAlH4 is modified into the organometallic reagent diisobutyl aluminum hydride which can be represented as DIBAL or DIBAL-H or DIBAH or DIBALH. To reduce acid chlorides, t-butoxide groups are combined with LiAlH4 to form lithium tritert-butoxy aluminum hydride.
Carboxylic Acids can be converted to Aldehydes
Carboxylic acids cannot be reduced directly to aldehydes. Carboxylic acids can be converted to acid chlorides using thionyl chloride which can then be reduced to aldehydes using LiAlH(O-t-Bu)3.
Grignard reagents react with Nitriles to form Ketones
Nitriles can also be used to synthesize ketones when they react with Grignards as shown below.
Organocuprate reagents react with Acid Chlorides to form Ketones
Organocuprate reagents are the least reactive of the organometallic reagents studied so far. While we learned to synthesize alcohols by reacting Grignard reagents with aldehydes and ketones, organocuprates will not react with aldehydes and ketones.
Grignard reagents will keep reacting with the product of the acid chloride reaction.
Organocuprate reactions with acid chlorides stop at the ketone as shown below.
Exercise
3. Complete the reaction map below.
Answer
3.
19.05: Nucleophilic Addition Reactions of Ketones and Aldehydes
Carbonyls are Electrophiles
Before we consider in detail the reactivity of aldehydes and ketones, we need to look back and remind ourselves of what the bonding picture looks like in a carbonyl. Carbonyl carbons are sp2 hybridized, with the three sp2 orbitals forming soverlaps with orbitals on the oxygen and on the two carbon or hydrogen atoms. These three bonds adopt trigonal planar geometry. The remaining unhybridized 2p orbital on the central carbonyl carbon is perpendicular to this plane, and forms a ‘side-by-side’ pbond with a 2p orbital on the oxygen.
The carbon-oxygen double bond is polar: oxygen is more electronegative than carbon, so electron density is higher on the oxygen side of the bond and lower on the carbon side. Recall that bond polarity can be depicted with a dipole arrow, or by showing the oxygen as holding a partial negative charge and the carbonyl carbon a partial positive charge.
A third way to illustrate the carbon-oxygen dipole is to consider the two main resonance contributors of a carbonyl group: the major form, which is what you typically see drawn in Lewis structures, and a minor but very important contributor in which both electrons in the pbond are localized on the oxygen, giving it a full negative charge. The latter depiction shows the carbon with an empty 2p orbital and a full positive charge.
The result of carbonyl bond polarization, however it is depicted, is straightforward to predict. The carbon, because it is electron-poor, is an electrophile: it is a great target for attack by an electron-rich nucleophilic group. Because the oxygen end of the carbonyl double bond bears a partial negative charge, anything that can help to stabilize this charge by accepting some of the electron density will increase the bond’s polarity and make the carbon more electrophilic. Very often a general acid group serves this purpose, donating a proton to the carbonyl oxygen.
The same effect can also be achieved if a Lewis acid, such as a magnesium ion, is located near the carbonyl oxygen.
Nucelophilic Addition to a Carbonyl
When a nucleophile reacts with the carbonyl carbon of an aldehyde or ketone, there is no leaving group – the incoming nucleophile simply ‘pushes’ the electrons in the pi bond up to the oxygen. After the carbonyl has reacted with the nucleophile, the negatively charged oxygen has the capacity to act as a nucleophile. The nucleophile can be charged or neutral. However, most commonly the oxygen acts as a base, abstracting a proton from a nearby acid group in the solvent or enzyme active site. The nucelophiiles studied in this chapter are water (H2O), cyanide (CN-), Grignard reagent (RMgX), amines (and ammonia), hydrazine (N2H4), alcohols (ROH), and phosphorus ylides (R3P=CRH).
The generic mechanism for a strong nucleophile generally occurs under basic conditions as shown below.
A closer look at the tetrahedral intermediate shows us that if the carbonyl reforms, then the original aldehyde or ketone is reformed.
It is possible for the nucleophile to repeatedly add and leave the carbonyl group. Protonation of the tetrahedral intermediate to form the nucleophilic additon product is favored by the low activation energy of proton transfer reactions.
For reaction with weak nucleophiles generally occurs under acidic conditions to increase the electrophilicity of the carbonyl group as resonance form below illustrates.
While the net result of the reaction is similar, the mechanism is slightly different due to the order of the proton transfer reactions.
Relative Reactivity of Carbonyl Compounds to Nucleophilic Addition
In general aldehydes are more reactive than ketones because of the lack of stabilizing alkly groups. The primary carbocation formed in the in the polarizing resonance structure of an aldehyde (discussed above) is less stable and therefore more reactive than the secondary carbocation formed by a ketone.
Exercise
4. Compare the mechanisms of
an SN2 reaction between 2-bromobutane and cyanide
tetrahedral complex formation between 2-butanone and cyanide.
Answer
4. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/19%3A_Ketones_and_Aldehydes/19.03%3A_Review_of_Ketone_and_Aldehyde_Synthesis.txt |
It has been demonstrated that water, in the presence of an acid or a base, adds rapidly to the carbonyl function of aldehydes and ketones establishing a reversible equilibrium with a hydrate (geminal-diol or gem-diol). The word germinal or gem comes from the Latin word for twin, geminus.
Example
Reversibility of the Reaction
Isolation of gem-diols is difficult because the reaction is reversibly. Removal of the water during a reaction can cause the conversion of a gem-diol back to the corresponding carbonyl.
Factors Affecting the Gem-diol Equilibrium
In most cases the resulting gem-diol is unstable relative to the reactants and cannot be isolated. Exceptions to this rule exist, one being formaldehyde where the weaker pi-component of the carbonyl double bond, relative to other aldehydes or ketones, and the small size of the hydrogen substituents favor addition. Thus, a solution of formaldehyde in water (formalin) is almost exclusively the hydrate, or polymers of the hydrate. The addition of electron donating alkyl groups stabilized the partial positive charge on the carbonyl carbon and decreases the amount of gem-diol product at equilibrium. Because of this ketones tend to form less than 1% of the hydrate at equilibrium. Likewise, the addition of strong electron-withdrawing groups destabilizes the carbonyl and tends to form stable gem-diols.
The relative equilbrium between the carbonyl and gem-diol builds understanding about steric effects in reaction mechanisms. This relationship is summarized below.
Two examples of this are chloral, and 1,2,3-indantrione. It should be noted that chloral hydrate is a sedative and has been added to alcoholic beverages to make a “Knock-out” drink also called a Mickey Finn. Also, ninhydrin is commonly used by forensic investigators to resolve finger prints.
Mechanism of Gem-diol Formation
The mechanism is catalyzed by the addition of an acid or base. Note! This may speed up the reaction but is has not effect on the equilibriums discussed above. Basic conditions speed up the reaction because hydroxide is a better nucleophilic than water. Acidic conditions speed up the reaction because the protonated carbonyl is more electrophilic.
Basic conditions
1) Nucleophilic hydroxide reacts with carbonyl carbon
2) Protonation of the alkoxide
Acidic conditions
1) Protonation of the carbonyl
2) Nucleophilic water reacts with carbonyl carbon
3) Deprotonation
Exercise
5. Draw the expected products of the following reactions.
6. Of the following pairs of molecules which would you expect to form a larger percentage of gem-diol at equilibrium? Please explain your answer.
7. Would you expect the following molecule to form appreciable amount of gem-diol in water? Please explain your answer.
Answer
5.
6. The compound on the left would. Fluorine is more electronegative than bromine and would remove more electron density from the carbonyl carbon. This would destabilize the carbonyl allowing for more gem-diol to form.
7. Although ketones tend to not form gem-diols this compound exists almost entirely in the gem-diol form when placed in water. Ketones tend to not form gem-diols because of the stabilizing effect of the electron donating alkyl group. However, in this case the electron donating effects of alkyl group is dominated by the presence of six highly electronegative fluorines. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/19%3A_Ketones_and_Aldehydes/19.06%3A_Nucleophilic_Addition_of_Water_%28Hydration%29.txt |
Cyanohydrins have the structural formula of R2C(OH)CN. The “R” on the formula represents an alkyl, aryl, or hydrogen. To form a cyanohydrin, a hydrogen cyanide adds reversibly to the carbonyl group of an organic compound thus forming a hydroxyalkanenitrile adducts (commonly known and called as cyanohydrins).
The reaction of aldehydes and ketones with hydrogen cyanide
Hydrogen cyanide adds across the carbon-oxygen double bond in aldehydes and ketones to produce compounds known as hydroxynitriles. For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile:
With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile:
The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism.
Mechanism of Cyanohydrin Formation
Acid-catalyzed hydrolysis of silylated cyanohydrins has recently been shown to give cyanohydrins instead of ketones; thus an efficient synthesis of cyanohydrins has been found which works with even highly hindered ketones.
Acetone Cyanohydrins
Acetone cyanohydrins (ACH) have the structural formula of (CH3)2C(OH)CN. It is an organic compound serves in the production of methyl methacrylate (also known as acrylic).
It is classified as an extremely hazardous substance, since it rapidly decomposes when it's in contact with water. In ACH, sulfuric acid is treated to give the sulfate ester of the methacrylamid. Preparations of other cyanohydrins are also used from ACH: for HACN to Michael acceptors and for the formylation of arenas. The treatment with lithium hydride affords anhydrous lithium cyanide.
Acetylide Ions (RCC-)
The reactivity of acetylide ions with aldehydes and ketones follows the mechanism for cyanide. For example, 2-butanone reacts with acetylide as shown in the reaction below.
Exercise
8. Mandelonitrile occurs in pits of some fruits. Glycolonitrile is the simplest cyanohydrin. Their structures are shown below.
Madelonitrile glycolonitrile
Propose the reactants for mandelonitrile and glycolonitrile.
Answer
8.
19.08: Nucleophilic Addition of Grignards
Common Organometallic Reagents
The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, the former being stronger than the latter. These same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to a halide anion, and the carbon bonds to the metal which has characteristics similar to a carbanion (R:-). Some common organometallic reagents are shown below
Reaction of Organometallic Reagents with Various Carbonyls
Because organometallic reagents react as their corresponding carbanion, they are excellent nucleophiles. The basic reaction involves the nucleophilic attack of the carbanionic carbon in the organometallic reagent with the electrophilic carbon in the carbonyl to form alcohols.
Both Grignard and Organolithium Reagents will perform these reactions going from Reactants to Products by this simplified pattern:
Addition to formaldehyde gives 1o alcohols
Addition to aldehydes gives 2o alcohols
Addition to ketones gives 3o alcohols
Addition to carbon dioxide (CO2) forms a carboxylic acid
Example
Mechanism for the Addition to Carbonyls
The mechanism for a Grignard agent is shown. The mechanism for an organometallic reagent is the same.
1) Nucleophilic reaction with carbonyl carbon
2) Protonation of tetrahedral intermediate
Organometallic Reagents as Bases
These reagents are very strong bases (pKa's of saturated hydrocarbons range from 42 to 50). Although not usually done with Grignard reagents, organolithium reagents can be used as strong bases. Both Grignard reagents and organolithium reagents react with water to form the corresponding hydrocarbon. This is why so much care is needed to insure dry glassware and solvents when working with organometallic reagents.
In fact, the reactivity of Grignard reagents and organolithium reagents can be exploited to create a new method for the conversion of halogens to the corresponding hydrocarbon (illustrated below). The halogen is converted to an organometallic reagent and then subsequently reacted with water to from an alkane.
Conjugate base anions of terminal alkynes (acetylide anions) are nucleophiles, and can do both nucleophilic substitution and nucleophilic addition reactions.
Limitation of Organometallic Reagents
As discussed above, Grignard and organolithium reagents are powerful bases. Because of this they cannot be used as nucleophiles on compounds which contain acidic hydrogens. If they are used they will act as a base and deprotonate the acidic hydrogen rather than act as a nucleophile and react with the carbonyl. A partial list of functional groups which cannot be used are: alcohols, amides, 1o amines, 2o amines, carboxylic acids, and terminal alkynes.
Exercises
9. Please write the product of the following reactions.
10. Please indicate the starting material required to produce the product.
11. Please give a detailed mechanism and the final product of this reaction
12. Please show two sets of reactants which could be used to synthesize the following molecule using a Grignard reaction.
Answers
9.
10.
11.
Nucleophilic addition reaction
Protonation
12. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/19%3A_Ketones_and_Aldehydes/19.07%3A_Nucleophilic_Addition_of_Cyanide_and_Acetylide.txt |
Reaction with Primary Amines to form Imines
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled. The rate at which these imine compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. At high pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H2O. At low pH most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Converting reactants to products simply
Mechanism of imine formation
1) Nucleophilic addition reaction
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation to form neutral final product
Reversibility of imine forming reactions
Imines can be hydrolyzed back to the corresponding primary amine under acidic aqueous conditons.
Reactions involving other reagents of the type Y-NH2
Imines are sometimes difficult to isolate and purify due to their sensitivity to hydrolysis. Consequently, other reagents of the type Y–NH2 have been studied, and found to give stable products (R2C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed in the following table, together with the structures and names of their carbonyl reaction products. Hydrazones are used as part of the Wolff-Kishner reduction and will be discussed in more detail in another module.
With the exception of unsubstituted hydrazones, these derivatives are easily prepared and are often crystalline solids - even when the parent aldehyde or ketone is a liquid. Since melting points can be determined more quickly and precisely than boiling points, derivatives such as these are useful for comparison and identification of carbonyl compounds. It should be noted that although semicarbazide has two amino groups (–NH2) only one of them is a reactive amine. The other is amide-like and is deactivated by the adjacent carbonyl group.
Exercise
13. Draw the products of the following reactions.
14. Draw the structure of the reactant needed to produce the indicated product.
Answer
13.
14.
Contributors and Attributions
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Reaction with Secondary Amines to form Enamines
Most aldehydes and ketones react with 2º-amines to give products known as enamines. It should be noted that, like acetal formation, these are acid-catalyzed reversible reactions in which water is lost. Consequently, enamines are easily converted back to their carbonyl precursors by acid-catalyzed hydrolysis.
Mechanism
1) Nuleophilic addition reaction
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation to neutralize final product
Reversibility of Enamines
Exercise
15. Draw the products for the following reactions.
16. Draw the missing reactant to complete each reaction below.
Answer
15.
16.
19.10: Nucleophilic Addition of Hydrazine (Wolff-Kishner Reaction)
Aldehydes and ketones can be converted to a hydrazine derivative by reaction with hydrazine. These "hydrazones" can be further converted to the corresponding alkane by reaction with base and heat. These two steps can be combined into one reaction called the Wolff-Kishner Reduction which represents a general method for converting aldehydes and ketones into alkanes. Typically a high boiling point solvent, such as ethylene glycol, is used to provide the high temperatures needed for this reaction to occur. Note! Nitrogen gas is produced as part of this reaction.
Reaction of Aldehydes or Ketones with Hydrazine Produces a Hydrazone
Reaction with a Base and Heat Converts a Hydrazone to an Alkane
Both Reactions Together Produces the Wolff-Kishner Reduction
The Wolff-Kishner reaction for cyclopentanone is shown below.
Mechanism of the Wolff-Kishner Reduction
1) Deprotonation of Nitrogen
2) Protonation of the Carbon
3) Deprotonation of Nitrogen
4) Protonation of Carbon
Exercise
17. Draw the products for the following reactions.
Answer
17.
19.11: Nucelophilic Addition of Alcohols (Acetal Formation)
Introduction
It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones to form geminal-diol. In a similar reaction alcohols add reversibly to aldehydes and ketones to form hemiacetals (hemi, Greek, half). This reaction can continue by adding another alcohol to form an acetal. Hemiacetals and acetals are important functional groups because they appear in sugars.
hemiacetal acetal
To achieve effective hemiacetal or acetal formation, two additional features must be implemented. First, an acid catalyst must be used because alcohol is a weak nucleophile; and second, the water produced with the acetal must be removed from the reaction by a process such as a molecular sieves or a Dean-Stark trap. The latter is important, since acetal formation is reversible. Indeed, once pure hemiacetal or acetals are obtained they may be hydrolyzed back to their starting components by treatment with aqueous acid and an excess of water.
Formation of Hemiacetals
Example: Formation of Hemiacetals
Example: Hemiacetal Reversibility
Formation of Acetals
Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents (or an excess amount) of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term. It is important to note that a hemiacetal is formed as an intermediate during the formation of an acetal.
Example: Formation of Acetals
Example: Acetal Reversibility
Mechanism for Hemiacetal and Acetal Formation
The mechanism shown here applies to both acetal and hemiacetal formation
1) Protonation of the carbonyl
2) Nucleophilic additional reaction by the alcohol
3) Deprotonation to form a hemiacetal
4) Protonation of the alcohol
5) Removal of water
6) Nucleophilic addition reaction by the alcohol
7) Deprotonation by water
Exercise
18. Draw the products for the following reactions.
Answer
18.
19.12: Acetals as Protecting Groups
Acetals as Protecting Groups
The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments. As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with ethers in general. Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) metallo-hydride, alkyl and aryl reagents. If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented.
In the following example we would like a Grignard reagent to react with the ester and not the ketone. This cannot be done without a protecting group because Grignard reagents react with esters and ketones.
Exercise
19. Specify the reagents to perform the following chemical transformations.
Answer
19. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/19%3A_Ketones_and_Aldehydes/19.09%3A_Nucelophilic_Addition_of_Amines_%28Imine_and_Enamine_Formation%29.txt |
Organophosphorus ylides react with aldehydes or ketones to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Preparation of Phosphorus Ylides
It has been noted that dipolar phosphorus compounds are stabilized by p-d bonding. This bonding stabilization extends to carbanions adjacent to phosphonium centers, and the zwitterionic conjugate bases derived from such cations are known as ylides. An ylide is defined as a compound with opposite charges on adjacent atoms both of which have complete octets. For the Wittig reaction discussed below an organophosphorus ylide, also called Wittig reagents, will be used. The ability of phosphorus to hold more than eight valence electrons allows for a resonance structure to be drawn forming a double bonded structure.
The stabilization of the carbanion provided by the phosphorus causes an increase in acidity (pKa ~35). Very strong bases, such as butyl lithium, are required for complete formation of ylides.
The ylides shown here are all strong bases. Like other strongly basic organic reagents, they are protonated by water and alcohols, and are sensitive to oxygen. Water decomposes phosphorous ylides to hydrocarbons and phosphine oxides, as shown.
Although many ylides are commercially available it is often necessary to create them synthetically. Ylides can be synthesized from an alkyl halide and a trialkyl phosphine. Typically triphenyl phosphine is used to synthesize ylides. Because a SN2 reaction is used in the ylide synthesis methyl and primary halides perform the best. Secondary halides can also be used but the yields are generally lower. This should be considered when planning out a synthesis which involves a synthesized Wittig reagent.
1) SN2 reaction
2) Deprotonation
The Wittig Reaction
The most important use of ylides in synthesis comes from their reactions with aldehydes and ketones, which are initiated in every case by a covalent bonding of the nucleophilic alpha-carbon to the electrophilic carbonyl carbon. Ylides react to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Going from reactants to products simplified
Mechanism of the Wittig reaction
Following the initial carbon-carbon bond formation, two intermediates have been identified for the Wittig reaction, a dipolar charge-separated species called a betaine and a four-membered heterocyclic structure referred to as an oxaphosphatane. Cleavage of the oxaphosphatane to alkene and phosphine oxide products is exothermic and irreversible.
1) Nucleophillic reaction with the carbonyl
2) Formation of a 4 membered ring
3) Formation of the alkene
Limitation of the Wittig reaction
If possible both E and Z isomer of the double bond will be formed. This should be considered when planning a synthesis involving a Wittig Reaction.
Exercise
20. Draw the products of the following reactions.
21. Please indicate the starting material required to produce the product.
22. Draw the structure of the oxaphosphetane which is made during the mechanism of the reaction given that produces product C.
23. Draw the structure of the betaine which is made during the mechanism of the reaction given that produces product D.
24. Propose a detailed mechanism and the final product of this reaction
25. It has been shown that reacting and epoxide with triphenylphosphine forms an alkene. Propose a mechanism for this reaction. Review the section on epoxide reactions if you need help.
Answer
20.
21.
22.
23.
24.
Nucleophillic reaction with the carbonyl
Formation of a 4 membered ring
Formation of the alkene
25.
Nucleophillic reaction with the epoxide
Formation of a 4 membered ring
Formation of the alkene | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/19%3A_Ketones_and_Aldehydes/19.13%3A_Nucelophilic_Addition_of_Phosphorus_Ylides_%28The_Wittig_Reaction%29.txt |
Oxidation and Carbonyls
Aldehydes are so easily oxidized that oxidation reactions can be an unwanted side reaction. Isolated ketones cannot be oxidized. The poly-hydroxy ketones of monosaccharides can be oxidized as seen in carbohydrate chemistry.
Tollen's Test
The Tollen's Test differentiates between aldehydes and ketones based on one of their few differences in chemical reactivity. Aldehydes can be oxidized and ketones cannot. In the Tollen's Test, silver ions (Ag+) oxidize aldehydes to carboxylic acids and are reduces to silver metals which can form a beautiful coating on the inside of the reaction flask.
Exercise
26. Draw and name the oxidation products for the following compounds.
a) benzaldehyde
b) acetaldehyde
c) 3-hydroxypentanal
Answer
26.
19.15: Reductions of Ketones and Aldehydes
Carbonyl Reduction Reactions
Aldehydes and ketones can be partially reduced to alcohols with sodium borohydride or Raney nickel. Of course, LiAlH4 could be used in stead of NaBH4. However, LiAlH4 is a stronger reducing agent so some benefits of selectivity are lost. Depending on the presence of more than one functional group within a single molecule, reaction selectivity can be useful. Aldehydes and ketones can be fully reduced to alkanes the Wolff-Kishner or Clemmensen Reduction.
Hydride Reduction Mechanism
The hydride reduction mechanism follows the pattern we have learned for other nucleophilic addition reactions.
Exercise
27. Complete the reaction map below.
Answer
27.
19.16: Additional Exercises
General Review
19-1 Give the product of the following reaction.
19-2 For each of the following reactions, give the final product.
19-3 For the following reaction, give the final product.
19-4 For the following reactions, identify the side favored at equilibrium.
19-5 Identify the correct product of the following reaction.
19-6 Give the final product of the following chain of reactions.
Synthesis of Ketones from Carboxylic Acids
19-7 Provide the structures of the products of the following reactions.
19-8 Provide the structure and IUPAC nomenclature of the product of the following reaction.
19-9 Choose the correct answer.
Synthesis of Ketones and Aldehydes from Acid Chlorides, Esters, and Nitriles
19-10 Provide the structures of the products of the following reactions.
19-11 Choose the correct product of the following reaction.
19-12 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) butanal
b) 2-methylhexan-3-one
c) pentan-2-one
d) 2-methylpentan-2-ol
Reactions of Ketones and Aldehydes
19-13 Provide the products of the following reactions.
19-14 Suggest a way to make the following compound from butanol. Use any necessary reagents.
19-15 Choose the correct product of the following reaction.
a) 2-chloropentane
b) 1-chlorobutan-1-ol
c) 2-chlorobutanoic acid
d) butanoyl chloride
The Wittig Reaction
19-16 Predict the structure of the product of the following reaction.
19-17 Provide the product of the following reaction.
19-18 Choose the correct product of the following reaction.
Formation of Cyanohydrins
19-19 Predict the structure of the product of the following reaction.
19-20 Provide the structure of the product of the following reaction.
19-21 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) 3-ethylhex-1-yn-3-ol
b) 2-ethyl-2-hydroxypentanenitrile
c) 3-(aminomethyl)hexan-3-ol
d) butanoyl cyanide
Formation of Imines
19-22 Predict the product of the following reaction.
19-23 Provide the structure of the product of the following reaction.
19-24 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) (4Z)-N-butyloctan-4-imine
b) (4Z)-N-propyloctan-4-imine
c) (4Z)-5-propylnon-4-en-1-amine
d) 4-(butylamino)octan-4-ol
Condensations with Hydroxylamine and Hydrazines
19-25 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) 1-ethyl-2-propylhydrazine
b) N-propoxyethanamine
c) (1E)-N-ethylbut-1-en-1-amine
d) (2E)-1-ethyl-2-propylidenehydrazine
19-26 Provide the structure of the product of the following reaction.
19-27 Provide the structure of the products of the following reactions.
Formation and Use of Acetals
19-28 Provide the structure of the resulting acetal.
19-29 Choose the correct structure of the product of the following reaction.
19-30 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) (dimethoxymethyl)benzene
b) benzoic acid
c) phenylmethanediol
d) (methoxymethyl)benzene
Oxidation of Aldehydes and Reductions of Ketones and Aldehydes
19-31 Draw the structure of the product of the following oxidation reaction.
19-32 Draw the structures of the products of the following reduction reactions.
19-33 Predict the product of the following reaction and provide its IUPAC nomenclature.
19.17: Solutions to Additional Exercises
General Review
19-1
19-2
19-3
19-4
1. B
2. A
19-5 Answer: C
19-6
Synthesis of Ketones from Carboxylic Acids
19-7:
19-8:
19-9:
Answer: A
Synthesis of Ketones and Aldehydes from Acid Chlorides, Esters, and Nitriles
19-10:
19-11:
Answer: A
19-12:
Answer: C
Reactions of Ketones and Aldehydes
19-13:
19-14:
19-15:
Answer: D
The Wittig Reaction
19-16:
19-17:
19-18:
Answer: B
Formation of Cyanohydrins
19-19:
19-20:
19-21:
Answer: B
Formation of Imines
19-22:
19-23:
19-24:
Answer: A
Condensations with Hydroxylamine and Hydrazines
19-25:
Answer: D
18-26:
19-27:
Formation and Use of Acetals
19-28:
19-29:
Answer: C
19-30:
Answer: A
Oxidation of Aldehydes and Reductions of Ketones and Aldehydes
19-31:
19-32:
19-33: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/19%3A_Ketones_and_Aldehydes/19.14%3A_Oxidation_of_Aldehydes.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• describe the structure and physical properties of amines and ammonium salts (section 20.1)
• explain and predict the relative basicity of amines using resonance, hybridization, substituent effects, and aromaticity (section 20.2)
• determine the structure of amines from their elemental analysis and spectral data (MS, IR 1H NMR & 13C NMR) (section 20.3)
• predict the products and specify the reagents to synthesize amines (section 20.4)
• predict the products and specify the reagents to synthesize primary amines (section 20.5)
• predict the products and specify the reagents for reactions of amines with
• aldehydes & ketones (section 20.6)
• alkyl halides and tosylates (section 20.6)
• acyl chlorides (section 20.6)
• sulfonyl chlorides (section 20.6)
• nitrous acid (section 20.7)
• oxidizing agents via Cope Elimination (section 20.9)
• explain the activating effects of aryl amines during electrophilic aromatic substitution reactions (section 20.7)
• use amides as protecting groups in multiple step synthesis (section 20.7)
• use diazonium salts to design multiple step syntheses using the Sandmeyer reactions (section 20.7)
• specify the reagents and predict the products for Hofmann Elimination reactions (section 20.8)
• Specify reagents for chemical transformations using all of the reactions studied to date
• combine the reactions studied to date to develop efficient and effective multiple-step synthesis including the use of amides as protecting groups
Please note: IUPAC nomenclature and important common names of amines were explained in Chapter 3.
20: Amines
Structure
Amines typically have three bonds and one pair of lone pair electrons. This makes the nitrogen sp3 hybridized, trigonal pyramidal, with a bond angle of roughly 109.5o.
Stereochemistry
Single-bonded nitrogen is pyramidal in shape, with the non-bonding electron pair pointing to the unoccupied corner of a tetrahedral region. Since the nitrogen in these compounds is bonded to three different groups, it is stereogenic and its configuration is chiral. The non-identical mirror-image configurations are illustrated in the following diagram (the remainder of the molecule is represented by R, and the electron pair is colored yellow). If these configurations were stable, there would be four additional stereoisomers of ephedrine and pseudoephedrine. However, pyramidal nitrogen is normally not configurationally stable. It rapidly inverts its configuration (equilibrium arrows) by passing through a planar, sp2-hybridized transition state, leading to a mixture of interconverting R and S configurations. If the nitrogen atom were the only chiral center in the molecule, a 50:50 (racemic) mixture of R and S configurations would exist at equilibrium. If other chiral centers are present, as in the ephedrin isomers, a mixture of diastereomers will result. The take-home message is that nitrogen does not contribute to isolable stereoisomers.
Amines are classified according to the number of alkyl or aryl groups attached to nitrogen. Amines are classified differently from alkyl halides and alcohols because nitrogen has a neutral bonding pattern of three bonds with a single lone pair. To classify amines, we look at the nitrogen atom of the amine and count the number of alkyl groups bonded to it. This number is the classification of the amine.
There are two additional classifications of amines. When the nitrogen is double bonded to carbon, then it is call am imine. When nitrogen is part of a ring that includes double bonds, then it is classified as heterocyclic, as seen in the aromatic nitrogen bases shown below.
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.
Odor
The free base form of amines can be quite odiferous. The foul smell of dying flesh is primarily from the amines released during decomposition of the proteins in an organism. The common names for the amines below emphasize this aspect of amines. It is also useful to note that common names frequently indicate the presence of an amine with the "ine" suffix.
The smell of amines can be reduced by reacting them with strong acids to form the ammonium salts. For example, the acid from lemons can be used to disguise the smell of fish that is past optimum freshness. While the free base forms of amines can be thermally unstable and smelly, the ammonium salt formed from the conjugate acid of the amine have increased thermal stability and reduced odor. If the amine is not soluble as a free base, its ammonium salt will be water soluble.
Exercise
1. Draw the structures for the following amines in order of decreasing water solubility: methanamine, 2-octanamine, 2-butanamine.
2. Draw the for the following amines in order of decreasing boiling point: cyclohexanamine, 2-octanamine, 2-butanamine.
3. Classify the following amines.
Answer
1.
2.
3. a) primary b) secondary c) secondary d) tertiary | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/20%3A_Amines/20.01%3A_Structure_and_Physical_Properties_of_Amines.txt |
Basicity of nitrogen groups
When evaluating the relative basicity of several nitrogen-containing functional groups: amines, amides, anilines, imines, and nitriles, the central question is: how reactive (and thus how basic) is the lone pair on the nitrogen? In other words, how much does that lone pair want to break away from the nitrogen nucleus and form a new bond with a hydrogen?
Comparing the basicity of alkyl amines to ammonia
Alkyl groups donate electrons to the more electronegative nitrogen. This inductive effect makes the electron density on the alkylamine's nitrogen greater than the nitrogen of ammonium. Correspondingly, primary, secondary, and tertiary alkyl amines are more basic than ammonia. Inductive effects are also moderated by the increased steric hindrance of alkyl groups (R grps).
Comparing the basicity of alkylamines to amides
With an alkyl amine the lone pair electron is localized on the nitrogen. However, the lone pair electron on an amide are delocalized between the nitrogen and the oxygen through resonance. This makes amides much less basic compared to alkylamines.
When an amide reacts with an acid, the protonation occurs at the carbonyl oxygen and not the nitrogen. The cation resulting from oxygen protonation is resonance stabilized., while the cation resulting for the protonation of nitrogen is not resonance stabilized.
Basicity of heterocyclic amines
When a nitrogen atom is incorporated directly into an aromatic ring, its basicity depends on the bonding context. In a pyridine ring, for example, the nitrogen lone pair occupies an sp2-hybrid orbital, and is not part of the aromatic sextet - it is essentially an imine nitrogen. Its electron pair is available for forming a bond to a proton, and thus the pyridine nitrogen atom is somewhat basic.
In a pyrrole ring, in contrast, the nitrogen lone pair is part of the aromatic sextet. This means that these electrons are very stable right where they are (in the aromatic system), and are much less available for bonding to a proton (and if they do pick up a proton, the aromic system is destroyed). For these reasons, pyrrole nitrogens are not strongly basic.
The aniline, pyridine, and pyrrole examples are good models for predicting the reactivity of nitrogen atoms in more complex ring systems (a huge diversity of which are found in nature). The tryptophan side chain, for example, contains a non-basic 'pyrrole-like' nitrogen, while adenine (a DNA/RNA base) contains all three types.
The lone pair electrons on the nitrogen of a nitrile are contained in a sp hybrid orbital. The 50% s character of an sp hybrid orbital means that the electrons are close to the nucleus and therefore not significantly basic.
Base Strength and pKa Values
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).
Compound
NH3
pyridine
aniline
4-nitroaniline
pyrrole
CH3C≡N
acetonitrile
pKa 11.0 10.7 10.7 9.3 5.2 4.6 1.0 0.0 -1.0 -10.
The first four compounds in the table above are all weak bases. The last five compounds are significantly less basic to neutral and even acidic as a consequence of two possible factors:
• orbital hybridization
• electron delocalization through resonance.
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, 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. For aniline and 4-nitroaniline, the nitrogen lone pair is stabilized through hyperconjugation with the aromatic ring. Pyrrole exhibits exceptional delocalization of the nitrogen electron pair because of its incorporation into the aromatic ring.
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.
Strong bases have weak conjugate acids, and weak bases have strong conjugate acids.
Amine Extraction in the Laboratory
Extraction is often employed in organic chemistry to purify compounds. Liquid-liquid extractions take advantage of the difference in solubility of a substance in two immiscible liquids (e.g. ether and water). The two immiscible liquids used in an extraction process are (1) the solvent in which the solids are dissolved, and (2) the extracting solvent. The two immiscible liquids are then easily separated using a separatory funnel. For amines one can take advantage of their basicity by forming the protonated salt (RNH2+Cl), which is soluble in water. The salt will extract into the aqueous phase leaving behind neutral compounds in the non-aqueous phase. The aqueous layer is then treated with a base (NaOH) to regenerate the amine and NaCl. A second extraction-separation is then done to isolate the amine in the non-aqueous layer and leave behind NaCl in the aqueous layer.
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.
Ammonium Salt Formation and Water Solubility
Many pharmaceuticals include amine functional groups. Sometimes the basicity of these amine groups is used to increase the water solubility of a drug so that it can be administered orally or intravenously. Ammonium salts are also thermally more stable and have less odor than their "free base" conjugates. The antihistamine, pseudoephedrine, reacts with hydrochloric acid to form the water soluble ammonium salt as shown below.
Exercise
4. Select the more basic compound from each of the following pairs of compounds.
(a)
(b)
(c)
5. The 4-methylbenzylammonium ion has a pKa of 9.51, and the butylammonium ion has a pKa of 10.59. Which is more basic? What's the pKb for each compound?
6. The structures for indole and imidazole are shown below. One compound has a pKa of -2 while the other compound has a pKa of 7. Assign the pKa values to the corresponding compound and explain your reasoning.
Answer
4.
(a)
(b)
(c)
5. The butylammonium is more basic. The pKb for butylammonium is 3.41, the pKb for 4-methylbenzylammonium is 4.49.
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NMR
The hydrogens attached to an amine show up ~ 0.5-5.0 ppm. The location is dependent on the amount of hydrogen bonding and the sample's concentration.
The hydrogens on carbons directly bonded to an amine typically appear ~2.3-3.0 ppm.
Addition of D2O will normally cause all hydrogens on non-carbon atoms to exchange with deuteriums, thus making these resonances "disappear." Addition of a few drops of D2O causing a signal to vanish can help confirm the presence of -NH.
IR
The infrared spectrum of aniline is shown beneath the following table. Some of the characteristic absorptions for C-H stretching and aromatic ring substitution are also marked, but not colored.
Amine Class
Stretching Vibrations
Bending Vibrations
Primary (1°)
The N-H stretching absorption is less sensitive to hydrogen bonding than are O-H absorptions. In the gas phase and in dilute CCl4 solution free N-H absorption is observed in the 3400 to 3500 cm-1 region. Primary aliphatic amines display two well-defined peaks due to asymmetric (higher frequency) and symmetric N-H stretching, separated by 80 to 100 cm-1. In aromatic amines these absorptions are usually 40 to 70 cm-1 higher in frequency. A smaller absorption near 3200 cm-1 (shaded orange in the spectra) is considered to be the result of interaction between an overtone of the 1600 cm-1 band with the symmetric N-H stretching band.
C-N stretching absorptions are found at 1200 to 1350 cm-1 for aromatic amines, and at 1000 to 1250 cm-1 for aliphatic amines.
Strong in-plane NH2 scissoring absorptions at 1550 to 1650 cm-1, and out-of-plane wagging at 650 to 900 cm-1 (usually broad) are characteristic of 1°-amines.
Secondary (2°)
Secondary amines exhibit only one absorption near 3420 cm-1. Hydrogen bonding in concentrated liquids shifts these absorptions to lower frequencies by about 100 cm-1. Again, this absorption appears at slightly higher frequency when the nitrogen atom is bonded to an aromatic ring.
The C-N absorptions are found in the same range, 1200 to 1350 cm-1(aromatic) and 1000 to 1250 cm-1 (aliphatic) as for 1°-amines.
A weak N-H bending absorption is sometimes visible at 1500 to 1600 cm-1. A broad wagging absorption at 650 to 900 cm-1 may be discerned in liquid film samples.
Tertiary (3°)
No N-H absorptions. The C-N absorptions are found in the same range, 1200 to 1350 cm-1 (aromatic) and 1000 to 1250 cm-1 (aliphatic) as for 1°-amines.
Aside from the C-N stretch noted on the left, these compounds have spectra characteristic of their alkyl and aryl substituents.
Mass Spectrometry and the Nitrogen Rule
The nitrogen rule states that a molecule that has no or even number of nitrogen atoms has an even nominal mass, whereas a molecule that has an odd number of nitrogen atoms has an odd nominal mass.
eg. 1:
eg. 2:
eg. 3:
Exercise
7. Oh no! The labels have fallen off two samples: Q and R. The elemental analysis for the samples indicated the following composition: compound Q is 81.15% C, 8.34% H, and 10.52% O and compound R is 71.08% C, 6.72% H, 10.36% N, and 11.84% O. Fortunately, we can analyze the samples using IR and 1H NMR spectroscopy. Name and draw the bond-line structures for compounds Q and R using the information provided. Support your answer by correlating the spectral data to the compound structures.
Answer
7. Vial 1 contains compound R which is acetanilide. Vial 2 contains compound Q which is N-ethyl-3-methylaniline.
20.04: Synthesis of Amines
Reductive Amination of Aldehydes and Ketones (Carbonyls)
Aldehydes and ketones can be converted into 1o, 2o and 3o amines using reductive amination. The reaction takes place in two parts. The first step is the nucleophiic addition of the carbonyl group to form an imine. The second step is the reduction of the imine to an amine using an reducing agent. A reducing agent commonly used for this reaction is sodium cyanoborohydride (NaBH3CN).
The nitrogen gains a bond to carbon during this reaction sequence. When carbonyls react with ammonia, a primary amine is produced. The reaction pattern continues for each amine classification. For example, pyrrolidine reacts with 2-butanone to produce the imine, which can be reduced by LiAlH4, sodium cyanoborohydride (NaBH3CN), or H2 with an active metal catalyst to produce a tertiary amine.
Amide Reduction to 1°, 2° or 3° Amines using LiAlH4
There is a direct correlation between the structure of the amide and the structure of the amine produced. Primary amides are reduced to primary amines. Secondary amides are reduced to secondary amines. Tertiary amides are reduced to tertiary amines. Lithium aluminum hydride is a stronger reducing agent than sodium borohydride, which is not strong enough for this reaction.
Exercise
8. Add the missing reactants/products to the following reactions.
Answer
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Preparation of Primary Amines
Although direct alkylation of ammonia by alkyl halides leads to 1º-amines, alternative procedures are preferred in many cases. These methods require two steps, but they provide pure product, usually in good yield. The general strategy is to first form a carbon-nitrogen bond by reacting a nitrogen nucleophile with a carbon electrophile. The following table lists several general examples of this strategy in the rough order of decreasing nucleophilicity of the nitrogen reagent. In the second step, extraneous nitrogen substituents that may have facilitated this bonding are removed to give the amine product.
Nitrogen
Reactant
Carbon
Reactant
1st Reaction
Type
Initial Product
2nd Reaction
Conditions
2nd Reaction
Type
Final Product
N3(–) RCH2-X or
R2CH-X
SN2 RCH2-N3 or
R2CH-N3
LiAlH4 or
4 H2 & Pd
Hydrogenolysis RCH2-NH2 or
R2CH-NH2
C6H5C2O2NH / OH- RCH2-X or
R2CH-X
SN2 RCH2-NC2O2C6H5 or
R2CH-NC2O2C6H5
NaOH / heat Hydrogenolysis RCH2-NH2 or
R2CH-NH2
CN(–) RCH2-X or
R2CH-X
SN2 RCH2-CN or
R2CH-CN
LiAlH4 Reduction RCH2-CH2NH2 or
R2CH-CH2NH2
NH3 RCH=O or
R2C=O
Addition /
Elimination
RCH=NH or
R2C=NH
H2 & Ni
or NaBH3CN
Reduction RCH2-NH2 or
R2CH-NH2
NH3 RCOX Addition /
Elimination
RCO-NH2 LiAlH4 Reduction RCH2-NH2
NH2CONH2
(urea)
R3C(+) SN1 R3C-NHCONH2 NaOH soln. Hydrolysis R3C-NH2
A specific example of each general class is provided in the diagram below. In the first two, an anionic nitrogen species undergoes an SN2 reaction with a modestly electrophilic alkyl halide reactant. For example #2, the Gabriel synthesis is shown. The alkaline conditions deprotonate phthalmide to create a strong nucleophile for SN2 reactions with alkyl halides. Example #3 also starts with an SN2 reaction of cyanide with an alkyl halide following by reduction of the cyano group to form a primary amine that extends the carbon system of the alkyl halide by a methylene group (CH2). In all three of these methods 3º-alkyl halides cannot be used because the major reaction path is an E2 elimination.
The methods illustrated by examples #4 and #5 proceed by the reaction of ammonia, or equivalent nitrogen nucleophiles, with the electrophilic carbon of a carbonyl group. A full discussion of carbonyl chemistry is presented in an independent chapter, but for present purposes it is sufficient to recognize that the C=O double bond is polarized so that the carbon atom is electrophilic. Nucleophilic addition to aldehydes and ketones is often catalyzed by acids. Acid halides and anhydrides are even more electrophilic, and do not normally require catalysts to react with nucleophiles. The reaction of ammonia with aldehydes or ketones occurs by a reversible addition-elimination pathway to give imines (compounds having a C=N function). These intermediates are not usually isolated, but are reduced as they are formed (i.e. in situ). Acid chlorides react with ammonia to give amides, also by an addition-elimination path, and these are reduced to amines by LiAlH4.
The 6th example is a specialized procedure for bonding an amino group to a 3º-alkyl group (none of the previous methods accomplishes this). Since a carbocation is the electrophilic species, rather poorly nucleophilic nitrogen reactants can be used. Urea, the diamide of carbonic acid, fits this requirement nicely. The resulting 3º-alkyl-substituted urea is then hydrolyzed to give the amine. One important method of preparing 1º-amines, especially aryl amines, uses a reverse strategy. Here a strongly electrophilic nitrogen species (NO2(+)) bonds to a nucleophilic carbon compound. This nitration reaction gives a nitro group that can be reduced to a 1º-amine by any of several reduction procedures.
Hofmann rearrangement
Hofmann rearrangement, also known as Hofmann degradation and not to be confused with Hofmann elimination, is the reaction of a primary amide with a halogen (chlorine or bromine) in strongly basic (sodium or potassium hydroxide) aqueous medium, which converts the amide to a primary amine. For example:
Mechanism:
Curtius Rearrangement
The Curtius rearrangement involves an acyl azide.
The mechanism of the Curtius rearrangement involves the migration of an -R group form the carbonyl carbon to the the neighboring nitrogen.
Reduction of Nitro Groups
Several methods for reducing nitro groups to amines are known. These include catalytic hydrogenation (H2 + catalyst), zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most case
Reduction of Nitriles
Nitriles can be converted to primary amines by reaction with lithium aluminum hydride. During this reaction the hydride nucleophile reacts with the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water.
Exercise
9. Complete the reaction map below to build reaction patterns for multiple step synthesis.
Answer
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Amines as Nucleophiles
Amines seldom serve as leaving groups in nucleophilic substitution or base-catalyzed elimination reactions. Indeed, they are even less effective in this role than are hydroxyl and alkoxyl groups. While we will see another section that it is possible to coax the amine to serve as a leaving group. As weak bases, amines are good nucleophiles.
Amines and Carbonyls
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled. The rate at which these imine compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. At high pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H2O. At low pH most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Imine formation is reversible. Imines can be hydrolyzed back to the corresponding primary amine under acidic aqueous conditions.
Most aldehydes and ketones react with 2º-amines to give products known as enamines.
It should be noted that, like acetal and imine formation, these are acid-catalyzed reversible reactions in which water is lost. Consequently, enamines are easily converted back to their carbonyl precursors by acid-catalyzed hydrolysis.
Amines and Acid Chlorides
Acid chlorides react with ammonia, 1o amines and 2o amines to form amides.
Examples:
Amines and Sulfonyl Chloride
The sulfonyl group is the sulfur-analog to the carbonyl group. Both groups contain an electrophilic carbonyl carbon with chloride as an excellent leaving group. Because sulfur is a third shell element it can form "expanded octets".
Amines react with sulfonyl groups to form sulfonamides. Sulfonamides are used as antimicrobial agents therapeutically and called sulfa drugs. The reaction to form sulfonamides occurs under alkaline conditions to keep the amine nucleophilic. Any time amines are present in an aqueous solution, measurable hydroxide is present. The mechanism for the sulfonation reaction is analogous to the acylation mechanism as is shown below.
The end of this chapter includes some additional information on sulfonamides.
It is instructive to examine nitrogen substitution reactions using common alkyl halides as the electrophiles. Thus, reaction of a primary alkyl bromide with a large excess of ammonia yields the corresponding 1º-amine, presumably by an SN2 mechanism. The hydrogen bromide produced in the reaction combines with some of the excess ammonia, giving ammonium bromide as a by-product. Water does not normally react with 1º-alkyl halides to give alcohols, so the enhanced nucleophilicity of nitrogen relative to oxygen is clearly demonstrated.
2 RCH2Br + NH3 (large excess) RCH2NH2 + NH4(+) Br(–)
It follows that simple amines should also be more nucleophilic than their alcohol or ether equivalents. If, for example, we wish to carry out an SN2 reaction of an alcohol with an alkyl halide to produce an ether (the Williamson synthesis), it is necessary to convert the weakly nucleophilic alcohol to its more nucleophilic conjugate base for the reaction to occur. In contrast, amines react with alkyl halides directly to give N-alkylated products. Since this reaction produces HBr as a co-product, hydrobromide salts of the alkylated amine or unreacted starting amine (in equilibrium) will also be formed.
2 RNH2 + C2H5Br RNHC2H5 + RNH3(+) Br(–) RNH2C2H5(+) Br(–) + RNH2
Unfortunately, the direct alkylation of 1º or 2º-amines to give a more substituted product does not proceed cleanly. If a 1:1 ratio of amine to alkyl halide is used, only 50% of the amine will react because the remaining amine will be tied up as an ammonium halide salt (remember that one equivalent of the strong acid HX is produced). If a 2:1 ratio of amine to alkylating agent is used, as in the above equation, the HX issue is solved, but another problem arises. Both the starting amine and the product amine are nucleophiles. Consequently, once the reaction has started, the product amine competes with the starting material in the later stages of alkylation, and some higher alkylated products are also formed. Even 3º-amines may be alkylated to form quaternary (4º) ammonium salts. When tetraalkyl ammonium salts are desired, as shown in the following example, Hünig's base may be used to scavenge the HI produced in the three SN2 reactions. Steric hindrance prevents this 3º-amine (Hünig's base) from being methylated.
C6H5NH2 + 3 CH3I + Hünig's base C6H5N(CH3)3(+) I(–) + HI salt of Hünig's base
You get a complicated series of reactions on heating primary amines with halogenoalkanes to give a mixture of products - probably one of the most confusing sets of reactions you will meet at this level. The products of the reactions include secondary and tertiary amines and their salts, and quaternary ammonium salts.
Making secondary amines and their salts
In the first stage of the reaction, you get the salt of a secondary amine formed. For example if you started with ethylamine and bromoethane, you would get diethylammonium bromide
In the presence of excess ethylamine in the mixture, there is the possibility of a reversible reaction. The ethylamine removes a hydrogen from the diethylammonium ion to give free diethylamine - a secondary amine.
Making tertiary amines and their salts
But it doesn't stop here! The diethylamine also reacts with bromoethane - in the same two stages as before. This is where the reaction would start if you reacted a secondary amine with a halogenoalkane.
In the first stage, you get triethylammonium bromide.
There is again the possibility of a reversible reaction between this salt and excess ethylamine in the mixture.
The ethylamine removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine - triethylamine.
Making a quaternary ammonium salt
The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).
This time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here.
Exercise
10. Draw the products of the following reactions.
11. Draw the structure of the reactant needed to produce the indicated product.
12. Draw the products for the following reactions.
13. Draw the missing reactant to complete each reaction below.
Answer
10.
11.
12.
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Overreaction of Aniline
Arylamines are very reactive towards electrophilic aromomatic substitution. The strongest activating and ortho/para-directing substituents are the amino (-NH2) and hydroxyl (-OH) groups. Direct nitration of phenol (hydroxybenzene) by dilute nitric acid gives modest yields of nitrated phenols and considerable oxidative decomposition to tarry materials; aniline (aminobenzene) is largely destroyed. Monobromination of both phenol and aniline is difficult to control, with di- and tri-bromo products forming readily.
Because of their high nucleophilic reactivity, aniline and phenol undergo substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. The mixed halogen iodine chloride (ICl) provides a more electrophilic iodine moiety, and is effective in iodinating aromatic rings having less powerful activating substituents.
C6H5–NH2 + I2 + NaHCO3 p-I–C6H4–NH2 + NaI + CO2 + H2O
In addition to overreactivity, we have previously seen that Friedel-Crafts reactions employing AlCl3 catalyst do not work with aniline. A salt complex forms and prevents electrophilic aromatic substitution.
Both this problem and the aniline overreactivity can be circumvented by first going through the corresponding amide.
Modifying the Influence of Strong Activating Groups
By acetylating the heteroatom substituent on aniline, its activating influence can be substantially attenuated. For example, acetylation of aniline gives acetanilide (first step in the following equation), which undergoes nitration at low temperature, yielding the para-nitro product in high yield. The modifying acetyl group can then be removed by acid-catalyzed hydrolysis (last step), to yield para-nitroaniline. Although the activating influence of the amino group has been reduced by this procedure, the acetyl derivative remains an ortho/para-directing and activating substituent.
C6H5–NH2 + (CH3CO)2O
pyridine (a base)
C6H5–NHCOCH3
HNO3 , 5 ºC
p-O2N–C6H4–NHCOCH3
H3O(+) & heat
p-O2N–C6H4–NH
The following diagram illustrates how the acetyl group acts to attenuate the overall electron donating character of oxygen and nitrogen. The non-bonding valence electron pairs that are responsible for the high reactivity of these compounds (blue arrows) are diverted to the adjacent carbonyl group (green arrows). However, the overall influence of the modified substituent is still activating and ortho/para-directing.
.
Diazonium Ions and their Salts
A resonance description of diazonium ions shows that the positive charge is delocalized over the two nitrogen atoms. It is not possible for nucleophiles to bond to the inner nitrogen, but bonding (or coupling) of negative nucleophiles to the terminal nitrogen gives neutral azo compounds. As shown in the following equation, this coupling to the terminal nitrogen should be relatively fast and reversible. The azo products may exist as E / Z stereoisomers. In practice it is found that the E-isomer predominates at equilibrium.
Unless these azo products are trapped or stabilized in some manner, reversal to the diazonium ion and slow nucleophilic substitution at carbon (with irreversible nitrogen loss) will be the ultimate course of reaction, as described in the previous section. For example, if phenyldiazonium bisufate is added rapidly to a cold solution of sodium hydroxide a relatively stable solution of sodium phenyldiazoate (the conjugate base of the initially formed diazoic acid) is obtained. Lowering the pH of this solution regenerates phenyldiazoic acid (pKa ca. 7), which disassociates back to the diazonium ion and eventually undergoes substitution, generating phenol.
C6H5N2(+) HSO4(–) + NaOH (cold solution) C6H5N2–OH + NaOH (cold) C6H5N2–O(–) Na(+)
phenyldiazonium bisulfate phenyldiazoic acid sodium phenyldiazoate
Aryl diazonium salts may be reduced to the corresponding hydrazines by mild reducing agents such as sodium bisulfite, stannous chloride or zinc dust. The bisulfite reduction may proceed by an initial sulfur-nitrogen coupling, as shown in the following equation.
Ar-N2(+) X(–)
NaHSO3
Ar-N=N-SO3H
NaHSO3
Ar-NH-NH-SO3H
H2O
Ar-NH-NH2 + H2SO4
Diazonium Salts: The Sandmeyer Reactions
Aryl diazonium salts are important intermediates. They are prepared in cold (0 º to 10 ºC) aqueous solution, and generally react with nucleophiles with loss of nitrogen. Some of the more commonly used substitution reactions are shown in the following diagram. Since the leaving group (N2) is thermodynamically very stable, these reactions are energetically favored. Those substitution reactions that are catalyzed by cuprous salts are known as Sandmeyer reactions. Fluoride substitution occurs on treatment with BF4(–), a reaction known as the Schiemann reaction. Stable diazonium tetrafluoroborate salts may be isolated, and on heating these lose nitrogen to give an arylfluoride product. The top reaction with hypophosphorus acid, H3PO2, is noteworthy because it achieves the reductive removal of an amino (or nitro) group. Unlike the nucleophilic substitution reactions, this reduction probably proceeds by a radical mechanism.
These aryl diazonium substitution reactions significantly expand the tactics available for the synthesis of polysubstituted benzene derivatives. Consider the following options:
1. The usual precursor to an aryl amine is the corresponding nitro compound. A nitro substituent deactivates an aromatic ring and directs electrophilic substitution to meta locations.
2. Reduction of a nitro group to an amine may be achieved in several ways. The resulting amine substituent strongly activates an aromatic ring and directs electrophilic substitution to ortho & para locations.
3. The activating character of an amine substituent may be attenuated by formation of an amide derivative (reversible), or even changed to deactivating and meta-directing by formation of a quaternary-ammonium salt (irreversible).
4. Conversion of an aryl amine to a diazonium ion intermediate allows it to be replaced by a variety of different groups (including hydrogen), which may in turn be used in subsequent reactions.
The following examples illustrate some combined applications of these options to specific cases. You should try to conceive a plausible reaction sequence for each. Once you have done so, you may check suggested answers below.
Answer 1:
It should be clear that the methyl substituent will eventually be oxidized to a carboxylic acid function. The timing is important, since a methyl substituent is ortho/para-directing and the carboxyl substituent is meta-directing. The cyano group will be introduced by a diazonium intermediate, so a nitration followed by reduction to an amine must precede this step.
Answer 2:
The hydroxyl group is a strong activating substituent and would direct aromatic ring chlorination to locations ortho & para to itself, leading to the wrong product. As an alternative, the nitro group is not only meta-directing, it can be converted to a hydroxyl group by way of a diazonium intermediate. The resulting strategy is self evident.
Answer 3:
Selective introduction of a fluorine is best achieved by treating a diazonium intermediate with boron tetrafluoride anion. To get the necessary intermediate we need to make p-nitroaniline. Since the nitro substituent on the starting material would direct a new substituent to a meta-location, we must first reduce it to an ortho/para-directing amino group. Amino groups are powerful activating substituents, so we deactivate it by acetylation before nitration. The acetyl substituent also protects the initial amine function from reaction with nitrous acid later on. It is removed in the last step.
Answer 4:
Polybromination of benzene would lead to ortho/para substitution. In order to achieve the mutual meta-relationship of three bromines, it is necessary to introduce a powerful ortho/para-directing prior to bromination, and then remove it following the tribromination. An amino group is ideal for this purpose. Reductive removal of the diazonium group may be accomplished in several ways (three are shown).
Answer 5:
The propyl substituent is best introduced by Friedel-Crafts acylation followed by reduction, and this cannot be carried out in the presence of a nitro substituent. Since an acyl substituent is a meta-director, it is logical to use this property to locate the nitro and chloro groups before reducing the carbonyl moiety. The same reduction method can be used to reduce both the nitro group (to an amine) and the carbonyl group to propyl. We have already seen the use of diazonium intermediates as precursors to phenols.
Answer 6:
Aromatic iodination can only be accomplished directly on highly activated benzene compounds, such as aniline, or indirectly by way of a diazonium intermediate. Once again, a deactivated amino group is the precursor of p-nitroaniline (prb.#3). This aniline derivative requires the more electrophilic iodine chloride (ICl) for ortho-iodination because of the presence of a deactivating nitro substituent. Finally, the third iodine is introduced by the diazonium ion procedure.
Substitution reactions of diazonium ions
Diazonium ions are present in solutions such as benzenediazonium chloride solution. They contain an -N2+ group. In the case of benzenediazonium chloride, this is attached to a benzene ring. Benzenediazonium chloride looks like this:
In this set of reactions of the diazonium ion, the -N2+ group is replaced by something else. The nitrogen is released as nitrogen gas.
Substitution by an -OH group
To get this reaction, all you need to do is warm the benzenediazonium chloride solution. The diazonium ion reacts with the water in the solution and phenol is formed - either in solution or as a black oily liquid (depending on how much is formed). Nitrogen gas is evolved.
This is the same reaction that you get if you react phenylamine with nitrous acid in the warm. The diazonium ion is formed first and then immediately reacts with the water in the solution to give phenol.
Substitution by an iodine atom
This is a good example of the use of diazonium salts to substitute things into a benzene ring which are otherwise quite difficult to attach. (That's equally true of the previous reaction, by the way.) If you add potassium iodide solution to the benzenediazonium chloride solution in the cold, nitrogen gas is given off, and you get oily droplets of iodobenzene formed.
There is a simple reaction between the diazonium ions and the iodide ions from the potassium iodide solution.
Coupling reactions of diazonium ions
In the substitution reactions above, the nitrogen in the diazonium ion is lost. In the rest of the reactions on this page, the nitrogen is retained and used to make a bridge between two benzene rings.
The most important application of diazo coupling reactions is electrophilic aromatic substitution of activated benzene derivatives by diazonium electrophiles. The products of such reactions are highly colored aromatic azo compounds that find use as synthetic dyestuffs, commonly referred to as azo dyes. Azobenzene (Y=Z=H) is light orange; however, the color of other azo compounds may range from red to deep blue depending on the nature of the aromatic rings and the substituents they carry. Azo compounds may exist as cis/trans isomer pairs, but most of the well-characterized and stable compounds are trans.
The reaction with phenol
Phenol is dissolved in sodium hydroxide solution to give a solution of sodium phenoxide.
The solution is cooled in ice, and cold benzenediazonium chloride solution is added. There is a reaction between the diazonium ion and the phenoxide ion and a yellow-orange solution or precipitate is formed. The product is one of the simplest of what are known as azo compounds, in which two benzene rings are linked by a nitrogen bridge.
The reaction with naphthalen-2-ol
Naphthalen-2-ol is also known as 2-naphthol or beta-naphthol. It contains an -OH group attached to a naphthalene molecule rather than to a simple benzene ring. Naphthalene has two benzene rings fused together. The reaction is done under exactly the same conditions as with phenol. The naphthalen-2-ol is dissolved in sodium hydroxide solution to produce an ion just like the phenol one. This solution is cooled and mixed with the benzenediazonium chloride solution.
An intense orange-red precipitate is formed - another azo compound.
The reaction with phenylamine (aniline)
Some liquid phenylamine is added to a cold solution of benzenediazonium chloride, and the mixture is shaken vigorously. A yellow solid is produced.
These strongly coloured azo compounds are frequently used as dyes known as azo dyes. The one made from phenylamine (aniline) is known as "aniline yellow" (amongst many other things - see note above). Azo compounds account for more than half of modern dyes.
The use of an azo dye as an indicator - methyl orange
Azo compounds contain a highly delocalized system of electrons which takes in both benzene rings and the two nitrogen atoms bridging the rings. The delocalization can also extend to things attached to the benzene rings as well. If white light falls on one of these molecules, some wavelengths are absorbed by these delocalized electrons. The colour you see is the result of the non-absorbed wavelengths. The groups which contribute to the delocalization (and so to the absorption of light) are known as a chromophore.
Modifying the groups present in the molecule can have an effect on the light absorbed, and so on the color you see. You can take advantage of this in indicators. Methyl orange is an azo dye which exists in two forms depending on the pH:
As the hydrogen ion is lost or gained there is a shift in the exact nature of the delocalisation in the molecule, and that causes a shift in the wavelength of light absorbed. Obviously that means that you see a different colour. When methyl orange is added, a hydrogen ion attaches to give the red form. Methyl orange is red in acidic solutions (in fact solutions of pH less than 3.1). If an alkali is added or hydrogen ions are removed, then the yellow form if generated. Methyl orange is yellow at pH's greater than 4.4.
In between, at some point there will be equal amounts of the red and yellow forms and so methyl orange looks orange.
Some examples of azo coupling reactions are shown below. A few simple rules are helpful in predicting the course of such reactions:
1. At acid pH (< 6) an amino group is a stronger activating substituent than a hydroxyl group (i.e. a phenol). At alkaline pH (> 7.5) phenolic functions are stronger activators, due to increased phenoxide base concentration.
2. Coupling to an activated benzene ring occurs preferentially para to the activating group if that location is free. Otherwise ortho-coupling will occur.
3. Naphthalene normally undergoes electrophilic substitution at an alpha-location more rapidly than at beta-sites; however, ortho-coupling is preferred. See the diagram for examples of α / β notation in naphthalenes.
You should try to conceive a plausible product structure for each of the following couplings.
Exercise
14. Propose a synthesis for the following compound via benzene and any amine you may require.
15. Proposes synthesis for each of the following compounds via benzene.
(a) N,N-Diethylaniline
(b) p-Bromoaniline
(c) m-Bromoaniline
(d) 2,4-Diethylaniline
16. Propose a synthesis for each of the following molecules from benzene via the diazonium ion.
(a) p-Chlorobenzoic acid
(b) m-Chlorobenzoic acid
(c) m-Dichlorobenzene
(d) p-Ethylbenzoic acid
(e) 1,2,4-Trichlorobenzene
Classify the following alcohols as primary, secondary, or tertiary.
Answer
14.
15.
(a) 1. HNO3, H2SO4; 2. Zn(Hg), HCl; 3. EtBr
(b) 1. HNO3, H2SO4; 2. Zn(Hg), HCl; 3. (CH3CO)O2; 4. Br2, FeBr3; 5. H2O, NaOH
(c) 1. HNO3, H2SO4; 2. Br2, FeBr3; 3. Zn(Hg), HCl
(d) 1. HNO3, H2SO4; 2. Zn(Hg), HCl; 3. (CH3CO)O2; 4. EtCl, AlCl3; 5. H2O, NaOH
16.
(a) 1. CH3CH2Cl, AlCl3; 2. HNO3, H2SO4; 3. SnCl2; 4. NaNO2, H2SO4; 5. CuBr; 6. KMnO4, H2O
(b) 1. HNO3, H2SO4; 2. Cl2, FeCl3; 3. SnCl2, H3O+; 4. NaNO2, H2SO4; 5. CuCN; 6. H3O+
(c) 1. HNO3, H2SO4; 2. Cl2, FeCl3; 3. SnCl2; 4. NaNO2, H2SO4; 5. CuCl
(d) 1. CH3CH2Cl, AlCl3: 2. HNO3, H2SO4; 3. SnCl2; 4. NaNO2, H2SO4; 5. CuCN; 6. H3O+
(e) 1. HNO3, H2SO4; 2. H2/PtO2; 3. (CH3CO)2O; 4. 2 Cl2; 5. H2O, NaOH; 6. NaNO2, H2SO4; 7. CuCl | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/20%3A_Amines/20.07%3A_Reactions_of_Arylamines.txt |
Hofmann Elimination
Elimination reactions of 4º-ammonium salts are termed Hofmann eliminations. Since the counter anion in most 4º-ammonium salts is halide, this is often replaced by the more basic hydroxide ion through reaction with silver hydroxide (or silver oxide). The resulting hydroxide salt must then be heated (100 - 200 ºC) to effect the E2-like elimination of a 3º-amine. Example #1 below shows a typical Hofmann elimination. Obviously, for an elimination to occur one of the alkyl substituents on nitrogen must have one or more beta-hydrogens, as noted earlier in examining elimination reactions of alkyl halides.
In example #2 above, two of the alkyl substituents on nitrogen have beta-hydrogens, all of which are on methyl groups (colored orange & magenta). The chief product from the elimination is the alkene having the more highly substituted double bond, reflecting not only the 3:1 numerical advantage of those beta-hydrogens, but also the greater stability of the double bond.
Example #3 illustrates two important features of the Hofmann elimination:
1. Simple amines are easily converted to the necessary 4º-ammonium salts by exhaustive alkylation, usually with methyl iodide (methyl has no beta-hydrogens and cannot compete in the elimination reaction). Exhaustive methylation is shown again in example #4.
2. When a given alkyl group has two different sets of beta-hydrogens available to the elimination process (colored orange & magenta here), the major product is often the alkene isomer having the less substituted double bond.
The tendency of Hofmann eliminations to give the less-substituted double bond isomer is commonly referred to as the Hofmann Rule, and contrasts strikingly with the Zaitsev Rule formulated for dehydrohalogenations and dehydrations. In cases where other activating groups, such as phenyl or carbonyl, are present, the Hofmann Rule may not apply. Thus, if 2-amino-1-phenylpropane is treated in the manner of example #3, the product consists largely of 1-phenylpropene (E & Z-isomers).
To understand why the base-induced elimination of 4º-ammonium salts behaves differently from that of alkyl halides it is necessary to reexamine the nature of the E2 transition state, first described for dehydrohalogenation. The energy diagram shown earlier for a single-step bimolecular E2 mechanism is repeated below.
The E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the bond to the leaving group (X) is substantially broken relative to the other bond changes, the transition state approaches that for an E1 reaction (initial ionization followed by a fast second step). At the other extreme, if the acidity of the beta-hydrogens is enhanced, then substantial breaking of C–H may occur before the other bonds begin to be affected. For most simple alkyl halides it was proper to envision a balanced transition state, in which there was a synchronous change in all the bonds. Such a model was consistent with the Zaitsev Rule.
When the leaving group X carries a positive charge, as do the 4º-ammonium compounds discussed here, the inductive influence of this charge will increase the acidity of both the alpha and the beta-hydrogens. Furthermore, the 4º-ammonium substituent is much larger than a halide or hydroxyl group and may perturb the conformations available to substituted beta-carbons. It seems that a combination of these factors acts to favor base attack at the least substituted (least hindered and most acidic) set of beta-hydrogens. The favored anti orientation of the leaving group and beta-hydrogen, noted for dehydrohalogenation, is found for many Hofmann eliminations; but syn-elimination is also common, possibly because the attraction of opposite charges orients the hydroxide base near the 4º-ammonium leaving group.
Three additional examples of the Hofmann elimination are shown in the following diagram. Example #1 is interesting in two respects. First, it generates a 4º-ammonium halide salt in a manner different from exhaustive methylation. Second, this salt is not converted to its hydroxide analog prior to elimination. A concentrated aqueous solution of the halide salt is simply dropped into a refluxing sodium hydroxide solution, and the volatile hydrocarbon product is isolated by distillation.
Example #2 illustrates an important aspect of the Hofmann elimination. If the nitrogen atom is part of a ring, then a single application of this elimination procedure does not remove the nitrogen as a separate 3º-amine product. In order to sever the nitrogen function from the molecule, a second Hofmann elimination must be carried out. Indeed, if the nitrogen atom was a member of two rings (fused or spiro), then three repetitions of the Hofmann elimination would be required to sever the nitrogen from the remaining molecular framework.
Example #3 is noteworthy because the less stable trans-cyclooctene is the chief product, accompanied by the cis-isomer. An anti-E2-transition state would necessarily give the cis-cycloalkene, so the trans-isomer must be generated by a syn-elimination. The cis-cyclooctene produced in this reaction could also be formed by a syn-elimination. Cyclooctane is a conformationally complex structure. Several puckered conformations that avoid angle strain are possible, and one of the most stable of these is shown on the right. Some eclipsed bonds occur in all these conformers, and transannular hydrogen crowding is unavoidable. Since the trimethylammonium substituent is large (about the size of tert-butyl) it will probably assume an equatorial-like orientation to avoid steric crowding. An anti-E2 transition state is likely to require an axial-like orientation of this bulky group, making this an unfavorable path.
Exercise
17. Draw the product for a Hoffman elimination for each of the following molecules.
(a)
(b)
(c)
(d)
18. Draw the product of a Hoffman elimination for the following molecule.
Answer
17.
(a)
(b)
(c)
(d)
18. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/20%3A_Amines/20.08%3A_The_Hofmann_Elimination-_Amines_as_Leaving_Groups.txt |
When a tertiary amine oxide bearing one or more beta hydrogens is heated, it is converted to an alkene. The reaction is known as Cope elimination or Cope reaction, not to be confused with Cope Rearrangement. For example:
The net reaction is 1,2-elimination, hence the name Cope elimination.
mechanism: Cope elimination is an intramolecular E2 reaction. It is also a pericyclic reaction.
Intermolecular E2 reactions occur preferentially from the conformation of the substrate in which the leaving group and the beta hydrogen abstracted by the base are antiperiplanar, which is not possible in intramolecular E2 reactions in which the base is built into the leaving group because the basic atom is too far away from the beta hydrogen anti to the leaving group. Intramolecular E2 reactions occur preferentially from the conformation of the substrate in which the leaving group and the beta hydrogen abstracted by the base are synperiplanar. The basic atom and the beta hydrogen abstracted by it are closest to each other in this conformation. For example:
mechanism:
Cope elimination is regioselective. Unlike intermolecular E2 reactions, it does not follow Zaitsev’s rule; the major product is always the least stable alkene, i.e., the alkene with the least highly substituted double bond. For example:
This trend is most likely due to the fact that the less highly substituted β-carbon bears more hydrogen atoms than the more highly substituted one; at a given moment, in a sample of the substrate, there are more molecules in which a hydrogen atom on the less highly substituted beta carbon is synperiplanar to the leaving group than there are in which a hydrogen atom on the more highly substituted beta carbon is.
20.10: Sulfa Drugs - a closer look
Sulfa Drug Synthesis
Sulfa drugs are an important group of synthetic antimicrobial agents (pharmaceuticals) that contain the sulfonamide group. The synthesis of sulfanilamide (a sulfa drug) illustrates how the reactivity of aniline can be modified to make possible an electrophilic aromatic substitution. The corresponding acetanilide undergoes chlorosulfonation. The resulting 4-acetamidobenzenesulfanyl chloride is treated with ammonia to replace the chlorine with an amino group and affords 4-acetamidobenzenesulfonamide. The subsequent hydrolysis of the sulfonamide produces the sulfanilamide.
Their use introduced and substantiated the concept of metabolic antagonism. Sulfonamides, as antimetabolites, compete with para-aminobenzoic acid (PABA) for incorporation into folic acid. The action of sulfonamides illustrates the principle of selective toxicity where some difference between mammal cells and bacterial cells is exploited. All cells require folic acid for growth. Folic acid (as a vitamin is in food) diffuses or is transported into human cells. However, folic acid cannot cross bacterial cell walls by diffusion or active transport. For this reason bacteria must synthesize folic acid from p-aminobenzoic acid. Sulfonamides or sulfa drugs have the following general structures as shown below.
Sulfanilamide which was the first compound used of this type has H's at R1 and R4. To date about 15,000 sulfonamide derivatives, analogues, and related compounds have been synthesized. This has lead to the discovery of many useful drugs which are effective for diuretics, antimalerial and leprosy agents, and antithyroid agents. The basic structure of sulfonamide cannot be modified if it is to be an effective competitive "mimic" for p-aminobenzoic acid. Essential structural features are the benzene ring with two substituents para to each other; an amino group in the fourth position; and the singly substituted 1-sulfonamido group.
Mechanism for Action
Normally folic acid is synthesized in two steps in bacteria by the top reaction on the left. If A sulfa drug is used, the first enzyme is not to specific and can use the sulfonamide in the first reaction. This reaction produces the product containing pteridine and the sulfa drug. The next and final step is the reaction PABA + with glutamic acid to make folic acid. If the sulfa drug has been substituted for the PABA, then the final enzyme is inhibited and no folic acid is produced.
Recent studies indicate that substituents on the N(1) nitrogen may play the role of competing for a site on the enzyme surface reserved for the glutamate residue in p-aminobenzoic acid-glutamate through one of the following two ways:
1. Direct competition in the linking of PABA-glutamate with the pteridine derivative.
2. Indirect interference with the coupling of glutamate to dihydropteroic acid.
Questions
1. In your own words explain how the sulfa drug works including enzyme inhibition, folic acid, and antimetabolite.
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/20%3A_Amines/20.09%3A_Oxidation_of_Amines_-_The_Cope_Elimination.txt |
General Review
20-1 Predict which amine is more basic and provide a reason for your answer.
20-2 Give the product of the following reaction.
20-3 Predict the final product of the following reaction chain and give its IUPAC name.
20-4 Propose a route of synthesis from nitrobenzene to the given product. Assume the given molecule is the major product, and for the purposes of this problem, ignore its isomers.
20-5 Choose the correct answer that describes the product of the following Cope elimination reaction.
a) N,N,2-trimethylpropan-1-iminium
b) N,N,2-trimethylpropan-1-amine
c) 2-methylprop-1-ene and N-hydroxy-N-methylmethanamine
d) N-hydroxy-N,2-dimethylpropan-1-amine
20-6 Explain why the following reaction might not be the best way to synthesize ethanamine.
Basicity and Effects of Amines
20-7 Draw all possible resonance structures for aniline and cyclohexanamine.
20-8 Identify which of the following nitroaniline isomers is the most basic and give a reason for your answer.
20-9 For the following compounds, identify which substituents are pi-acceptors of the electrons from the amine group (if applicable) and if they are, draw their resonance structure to show the movement of electrons.
Aromatic Substitution of Arylamines and Pyridin
20-10 Explain why the following arylamine needs to be turned into an amide before a Friedel-Crafts acylation and then predict the final product of the reaction.
20-11 Predict the products of the following reactions.
20-12 Predict the product of the following reaction and provide the correct IUPAC nomenclature.
Alkylation and Acylation of Amines
20-13 Predict the product of the following acylation reaction.
20-14 Suggest a route of synthesis for the following compound, starting with benzoyl chloride.
20-15 Choose the correct product of the following reaction.
Formation of Sulfonamides
20-16 Choose the correct structure of the product of the following reaction.
20-17 Provide the structure of the intermediate compound and final product of the following reaction.
20-18 Predict the product of the following reaction.
Amines as Leaving Groups: The Hofmann Elimination
20-19 Predict the major alkene product of the following Hofmann elimination reaction and give the proper IUPAC nomenclature.
20-20 Choose the correct product of the following reaction.
20-21 Propose a route of synthesis from pentan-1-amine to pentanal (include a Hofmann elimination reaction).
Oxidation of Amines: The Cope Elimination
20-22 Predict the structure and give the proper IUPAC nomenclature of the product of the following reaction.
20-23 Propose a route of synthesis for the following compound, starting with cyclohexanecarboxylic acid and include a Cope elimination reaction.
20-24 Predict the structure of the product of the following reaction.
Reactions of Amines with Nitrous Acid
20-25 Predict the product of the following reaction and provide the correct IUPAC nomenclature.
20-26 Predict the products of the following reactions.
20-27 Suggest a route of synthesis for the following product, starting with aniline.
Synthesis of Amines by Reductive Amination and Acylation-Reduction
20-28 Predict the product of the following reaction and provide its IUPAC nomenclature.
20-29 Identify which route of synthesis is the better way to make N-(cyclopentylmethyl)- -N-methylethanamine and then show the intermediate molecules for the correct path.
20-30 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) N-propylbenzamide
b) phenyl(propylamino)methanol
c) N-benzylpropan-1-amine
d) benzamide
20.12: Solutions to Additional Exercises
20-1 Amine A is the more basic of the two amines. Since its lone pair of electrons cannot resonance into the ring like that of amine B, it is more basic. Amine B can delocalize its electrons, making it a weaker base but a stronger acid.
20-2
20-3
20-4
20-5 Answer: C
20-6 This reaction is not the best way to synthesize ethanamine because of the high ratio of mixed products obtained. Because the reaction occurs so fast, we are unable to stop the reaction at only the primary alkylation; the amine will continue on to make many secondary and tertiary amines in addition to our desired product.
Basicity and Effects of Amines
20-7:
Cyclohexanamine has no resonance structure that can contribute to delocalizing its lone pair of electrons.
20-8:
Unlike on the para- and ortho-nitroaniline isomers, m-nitroaniline’s nitro and amine groups cannot form any resonance structures to delocalize their pi-electrons with each other, making m-nitroaniline the most basic.
20-9:
a) and d) have no pi-acceptors directly attached to the amine.
Aromatic Substitution of Arylamines and Pyridine
20-10:
If a protecting group is not placed on the amino group of aniline, it will form a complex with AlCl3 during the acylation step of the reaction and prevent the reaction from occurring. By placing a protecting group on the amine, we still maintain an activated ring that can give us ortho- or para- substituted products, but won’t interfere with the reaction itself.
20-11:
20-12:
Alkylation and Acylation of Amines
20-13:
20-14:
20-15:
Answer: A
Formation of Sulfonamides
20-16:
Answer: C
20-17:
20-18:
Amines as Leaving Groups: The Hofmann Elimination
20-19:
20-20:
Answer: D
20-21:
Possible route of synthesis:
Oxidation of Amines: The Cope Elimination
20-22:
20-23:
Possible route of synthesis:
20-24:
Reactions of Amines with Nitrous Acid
20-25:
20-26:
20-27:
Synthesis of Amines by Reductive Amination and Acylation-Reduction
20-28:
20-29:
Answer: B
20-30:
Answer: C | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/20%3A_Amines/20.11%3A_Additional_Exercises.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• describe the structure and physical properties of carboxylic acids and carboxylate salts (section 21.1)
• explain and predict the relative acidity of carboxylic acids using resonance, hybridization, and substituent effects (section 21.2)
• determine the structure of carboxylic acids from their elemental analysis and spectral data (MS, IR 1H NMR & 13C NMR) (section 21.3)
• predict the products and specify the reagents to synthesize carboxylic acids (section 21.4)
• recognize and classify the major reactions of carboxylic acids (section 21.5)
• show the general mechanism for Nucleophilic Acyl Substitution Reactions (section 21.5)
• predict the products and specify the reagents for reactions of carboxylic acids with
• sulfonyl chlorides (section 21.5)
• alcohols (section 21.6)
• diazomethane (section 21.7)
• amines (section 21.8)
• reducing agents (section 21.9)
• combine the reactions studied to date to develop efficient and effective multiple-step synthesis
Please note: IUPAC nomenclature and important common names of carboxylic acids were explained in Chapter 3.
It can useful is often required to memorize the structures for the following common names: formic acid, acetic acid, acetyl chloride, acetic anhydride, acetic formic anhydride, ethyl acetate, sodium and potassium salts of formate, acetate, and benzoate, acetamide, benzamide, acetonitrile, benzonitrile, carbonic acid, oxalic acid, malonic acid, succinic acid, glutaric acid, adipic acid, and phthalic acid
21: Carboxylic Acids
Structure of the carboxyl acid group
Carboxylic acids are organic compounds which incorporate a carboxyl functional group, CO2H. The name carboxyl comes from the fact that a carbonyl and a hydroxyl group are attached to the same carbon.
The carbon and oxygen in the carbonyl are both sp2 hybridized which give a carbonyl group a basic trigonal shape. The hydroxyl oxygen is also sp2 hybridized which allows one of its lone pair electrons to conjugate with the pi system of the carbonyl group. This make the carboxyl group planar an can represented with the following resonance structure.
Carboxylic acids are named such because they can donate a hydrogen to produce a carboxylate ion. The factors which affect the acidity of carboxylic acids are discussed in the next section of this chapter.
Physical Properties of Some Carboxylic Acids
The table at the beginning of this page gave the melting and boiling points for a homologous group of carboxylic acids having from one to ten carbon atoms. The boiling points increased with size in a regular manner, but the melting points did not. Unbranched acids made up of an even number of carbon atoms have melting points higher than the odd numbered homologs having one more or one less carbon. This reflects differences in intermolecular attractive forces in the crystalline state. In the table of fatty acids we see that the presence of a cis-double bond significantly lowers the melting point of a compound. Thus, palmitoleic acid melts over 60º lower than palmitic acid, and similar decreases occur for the C18 and C20 compounds. Again, changes in crystal packing and intermolecular forces are responsible.
The factors that influence the relative boiling points and water solubilities of various types of compounds were discussed earlier. In general, dipolar attractive forces between molecules act to increase the boiling point of a given compound, with hydrogen bonds being an extreme example. Hydrogen bonding is also a major factor in the water solubility of covalent compounds To refresh your understanding of these principles Click Here. The following table lists a few examples of these properties for some similar sized polar compounds (the non-polar hydrocarbon hexane is provided for comparison).
The first five entries all have oxygen functional groups, and the relatively high boiling points of the first two is clearly due to hydrogen bonding. Carboxylic acids have exceptionally high boiling points, due in large part to dimeric associations involving two hydrogen bonds. A structural formula for the dimer of acetic acid is shown here. When the mouse pointer passes over the drawing, an electron cloud diagram will appear. The high boiling points of the amides and nitriles are due in large part to strong dipole attractions, supplemented in some cases by hydrogen bonding.
Formula
IUPAC Name
Molecular Weight
Boiling Point
Water Solubility
Physical Properties of Some Organic Compounds
CH3(CH2)2CO2H butanoic acid 88 164 ºC very soluble
CH3(CH2)4OH 1-pentanol 88 138 ºC slightly soluble
CH3(CH2)3CHO pentanal 86 103 ºC slightly soluble
CH3CO2C2H5 ethyl ethanoate 88 77 ºC moderately soluble
CH3CH2CO2CH3 methyl propanoate 88 80 ºC slightly soluble
CH3(CH2)2CONH2 butanamide 87 216 ºC soluble
CH3CON(CH3)2 N,N-dimethylethanamide 87 165 ºC very soluble
CH3(CH2)4NH2 1-aminobutane 87 103 ºC very soluble
CH3(CH2)3CN pentanenitrile 83 140 ºC slightly soluble
CH3(CH2)4CH3 hexane 86 69 ºC insoluble
Carboxylate Salts
The water solubility of carboxylic acids is determined by the ratio of carboxyl groups to the the number of carbon atoms in the molecule following the "4 to 6 Rule". As seen with amines, water solubility of carboxylic acids can be increased when they are ionized. Typically a strong base is used to deprotonate the carboxylic acid and drive the reaction to completion as shown below.
Exercise
1. Use acid-base chemistry and differences in water solubility to separate 1-octanol from octanoic acid using the following solutions: 1 M NaOH, ether, and 6 M HCl and any lab equipment.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/21%3A_Carboxylic_Acids/21.01%3A__Structure_and_Properties_of_Carboxylic_Acids_and_their_Salts.txt |
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 pKa, the stronger the acid. Comparing the other two to ethanoic acid, we 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! The pKa of ethanol is about 17, while the pKa of acetic acid is about 5: this is a 1012-fold difference in the two acidity constants. In both compounds, the acidic proton is bonded to an oxygen atom. How can they be so different in terms of acidity? We begin by considering the conjugate bases.
In both species, the negative charge on the conjugate base is held by an oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: a resonance contributor can be drawn in which the negative charge is localized on the second oxygen of the group. The two resonance forms for the conjugate base are equal in energy. What this means is that the negative charge on the acetate ion is not located on one oxygen or the other: rather it is shared between the two. Chemists use the term ‘delocalization of charge’ to describe this situation. In the ethoxide ion, by contrast, the negative charge is ‘locked’ on the single oxygen – it has nowhere else to go.
Recall the fundamental idea that electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one atom. Here, 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. The acetate ion is that much more stable than the ethoxide ion, all due to the effects of resonance delocalization.
Resonance Effects on the Acidity of Carboxylic Acids
Why should the presence of a carbonyl group adjacent to a hydroxyl group have such a profound effect on the acidity of the hydroxyl proton? To answer this question we must return to the nature of acid-base equilibria and the definition of pKa , illustrated by the general equations given below.
We know that an equilibrium favors the thermodynamically more stable side, and that the magnitude of the equilibrium constant reflects the energy difference between the components of each side. In an acid base equilibrium the equilibrium always favors the weaker acid and base (these are the more stable components). Water is the standard base used for pKa measurements; consequently, anything that stabilizes the conjugate base (A:(–)) of an acid will necessarily make that acid (H–A) stronger and shift the equilibrium to the right. Both the carboxyl group and the carboxylate anion are stabilized by resonance, but the stabilization of the anion is much greater than that of the neutral function, as shown in the following diagram. In the carboxylate anion the two contributing structures have equal weight in the hybrid, and the C–O bonds are of equal length (between a double and a single bond). This stabilization leads to a markedly increased acidity, as illustrated by the energy diagram displayed by clicking the "Toggle Display" button.
Inductive Effects on Relative Acidity
The resonance effect described here is undoubtedly the major contributor to the exceptional acidity of carboxylic acids. However, inductive effects also play a role. For example, alcohols have pKa's of 16 or greater but their acidity is increased by electron withdrawing substituents on the alkyl group. The following diagram illustrates this factor for several simple inorganic and organic compounds (row #1), and shows how inductive electron withdrawal may also increase the acidity of carboxylic acids (rows #2 & 3). The acidic hydrogen is colored red in all examples. Water is less acidic than hydrogen peroxide because hydrogen is less electronegative than oxygen, and the covalent bond joining these atoms is polarized in the manner shown. Alcohols are slightly less acidic than water, due to the poor electronegativity of carbon, but chloral hydrate, Cl3CCH(OH)2, and 2,2,2,-trifluoroethanol are significantly more acidic than water, due to inductive electron withdrawal by the electronegative halogens (and the second oxygen in chloral hydrate). In the case of carboxylic acids, if the electrophilic character of the carbonyl carbon is decreased the acidity of the carboxylic acid will also decrease. Similarly, an increase in its electrophilicity will increase the acidity of the acid. Acetic acid is ten times weaker an acid than formic acid (first two entries in the second row), confirming the electron donating character of an alkyl group relative to hydrogen, as noted earlier in a discussion of carbocation stability. Electronegative substituents increase acidity by inductive electron withdrawal. As expected, the higher the electronegativity of the substituent the greater the increase in acidity (F > Cl > Br > I), and the closer the substituent is to the carboxyl group the greater is its effect (isomers in the 3rd row). Substituents also influence the acidity of benzoic acid derivatives, but resonance effects compete with inductive effects. The methoxy group is electron donating and the nitro group is electron withdrawing (last three entries in the table of pKa values).
A closer look at the effects of electron-withdrawing and electron-donating groups on the stability of the conjugate bases can be seen in the pKa values of benzoic acid as shown in the table below. The conjugate base of benzoic acid is stabilized by electron-withdrawing groups. This makes the acid more acidic
Electron-withdrawing groups deactivate the benzene ring to electrophilic reactions and make benzoic acids more acidic.
The conjugate base of benzoic acid is destabilized by electron-donating groups. This makes the acid less acidic
Electron-donating groups activate the benzene ring to electrophilic reactions and make benzoic acids less acidic.
Exercise
2. Draw the bond-line structures and arrange the following compounds in order of increasing acidity: 4-nitrobenzoic acid; 4-(methylamino)benzoic acid; p-chlorobenzoic acid; 4-(dimethylamino)benzoic acid.
Answer
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/21%3A_Carboxylic_Acids/21.02%3A_Acidity_of_Carboxylic_Acids.txt |
IR
The carboxyl group is associated with two characteristic infrared stretching absorptions which change markedly with hydrogen bonding. The spectrum of a CCl4 solution of propionic acid (propanoic acid), shown below, is illustrative. Carboxylic acids exist predominantly as hydrogen bonded dimers in condensed phases. The O-H stretching absorption for such dimers is very strong and broad, extending from 2500 to 3300 cm-1.
This absorption overlaps the sharper C-H stretching peaks, which may be seen extending beyond the O-H envelope at 2990, 2950 and 2870 cm-1. The smaller peaks protruding near 2655 and 2560 are characteristic of the dimer. In ether solvents a sharper hydrogen bonded monomer absorption near 3500 cm-1 is observed, due to competition of the ether oxygen as a hydrogen bond acceptor. The carbonyl stretching frequency of the dimer is found near 1710 cm-1, but is increased by 25 cm-1 or more in the monomeric state. Other characteristic stretching and bending absorptions are marked in the spectrum.
NMR
The combination of anisotropy and electronegativity causes the O-H hydrogen in a carboxylic acid to be very deshielded.
Hydrogen environments adjacent to a carboxylic acid are shifted to the region of 2.5-3.0 ppm.Deshielding occurs due to the fact that the sp2 hybridized carbon the the carboxylic acid is more electronegative than a sp3 hybridized carbon.
Exercise
2. Sample W is a reactant for a wide range of biochemical processes. Sample X was isolated from vanilla beans. Elemental analysis indicated the compounds are structural isomers with the composition: 54.52% C, 9.16% H and 36.32% O. The IR spectrum for each compound showed a broad absorption from 3500 - 2500 cm-1 and a strong band near 1710 cm-1. The 1H NMR is being serviced, so only the 13C NMR spectra shown below were available.
Name and draw the bond-line structures for Samples W and X and correlate the 13C NMR spectral signals to their respective compounds.
Answer
2.
21.04: Synthesis of Carboxylic Acids
Carboxylic Acid Synthesis - Review
The carbon atom of a carboxyl group has a high oxidation state. It is not surprising, therefore, that many of the chemical reactions used for their preparation are oxidations. Such reactions have been discussed in previous sections of this text, and the following diagram summarizes most of these. To review the previous discussion of any of these reaction classes simply click on the number (1 to 4) or descriptive heading for the group.
Carboxylic Acid Synthesis - New
Two other useful procedures for preparing carboxylic acids involve hydrolysis of nitriles and carboxylation of organometallic intermediates. As shown in the following diagram, both methods begin with an organic halogen compound and the carboxyl group eventually replaces the halogen. Both methods require two steps, but are complementary in that the nitrile intermediate in the first procedure is generated by a SN2 reaction, in which cyanide anion is a nucleophilic precursor of the carboxyl group. The hydrolysis may be either acid or base-catalyzed, but the latter give a carboxylate salt as the initial product.
In the second procedure the electrophilic halide is first transformed into a strongly nucleophilic metal derivative, and this adds to carbon dioxide (an electrophile). The initial product is a salt of the carboxylic acid, which must then be released by treatment with strong aqueous acid.
An existing carboxylic acid may be elongated by one methylene group, using a homologation procedure called the Arndt-Eistert reaction. To learn about this useful method Click Here.
Hydrolysis of Carboxylic Acid Derivatives and Nitriles
In this chapter we learn that all of the carboxylic acid derivatives can be synthesized from carboxylic acids. These reactions tend to be more useful for multiple-step syntheses to build large and complex molecules. The carboxylic acid derivatives along can be hydrolyzed to produce carboxylic acids. These hydrolysis reactions have limited use in multiple-step synthesis because the acidic proton can be problematic for many organic reactions. Biochemically, hydrolysis reactions are very important in the metabolism of food, drugs, and other nutrients. Hydrolysis can occur under acidic or basic conditions that determine the ionization of the carboxylic acid. Reactions under basic conditions will require a final neutralization step with dilute H+ to recover the carboxylic acid. To reinforce our awareness of the pH sensitivity of carboxylic acids, both reaction maps are shown below.
• Acid Hydrolysis of the Carboxylic Acid Derivatives
Helpful Hint: Different professors tend to use different proton sources. It is helpful to recognize reagents for their role in a reaction. For example, H2SO4, HCl, H3PO4, CH3CO2H, and p-TSA are all sources of H+. It does not matter which one is used for catalysis or neutralization. Nitric acid is oxidizing, so it is typically only used in oxidation reactions. As we continue learning more reactions/reagents, it can help to group them by their reactivity: acids, bases, neutral, oxidizing, reducing, protic, aprotic, etc. For example, peroxides are oxidizing - whether is it H2O2 or MCPBA (m-chloroperoxybenzoic acid).
• Basic Hydrolysis of the Carboxylic Acid Derivatives
Exercise
4. Complete the reactions below.
Answer
4. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/21%3A_Carboxylic_Acids/21.03%3A_Spectroscopy_of_Carboxylic_Acids.txt |
Carboxylic Acid Reactions Overview
Four general reaction categories are represented here: (1) As carboxylic acid deprotonates quite readily, it is quite easy to form a carboxylate salt or to substitute the hydroxyl hydrogen. (2) The category of nucleophilic acyl substitution represents the substitution of the whole hydroxyl group, which we will see later in more detail leads to several carboxylic acid derivatives (e.g. acid halides, esters, amides, thioesters, acid anhydrides etc.). (3) Like other carbonyl compounds, carboxylic acids can be reduced by reagents like LiAlH4. (4) While the proton on the carbon alpha to the carbonyl group is not as acidic as the hydroxyl hydrogen, it can be removed leading to substitution at the alpha site.
The scheme summarizes some of the general reactions that carboxylic acids undergo. The reaction details can be found in other sections of this text.
Nucleophilic Acyl Substitution Reactions and the Carboxylic Acid Derivatives
In subsequent sections, the reaction details to synthesize acyl chlorides, anhydrides, esters, and amides from carboxylic acids will be discussed. Because acyl chlorides, anhydrides, esters, and amides can all be synthesized from carboxylic acids, these functional groups are called "carboxylic acid derivatives".
These reactions all share the same general mechanism. The nucleophile reacts with the electrophilic carbonyl carbon to produce the "Tetrahedral Intermediate or TI". The carbonyl reforms forcing the loss of the leaving group (LG) to produce the "Nucleophilic Acyl Substitution Product". The reaction mechanism depends on the pH of the reaction conditions. The reaction mechanisms for both acidic and basic conditions are shown below.
Exercise
5. Draw the bond-line structures for the products and classify each reaction using the terms from the top diagram:
acid-base, NAS (Nucleophilic Acyl Substitutioin), gentle reduction, strong reduction, or alpha-substitution.
Answer
5.
21.06: Condensation of Acids with Alcohols- The Fischer Esterification
Carboxylic acids can react with alcohols to form esters in a process called Fischer esterification. An acid catalyst is required and the alcohol is also used as the reaction solvent. The oxygen atoms are color-coded in the reaction below to help understand the reaction mechanism.
For example, butanoic acid reacts with methanol to synthsize methylbutanoate. It is important to note that any proton source can be used as the catalyst. Sulfuric acid is shown in the example below.
Mechanism
1) Protonation of the carbonyl by the acid. The carbonyl is now activated toward nucleophilic reactions.
2) Nucleophilic reaction at the carbonyl
3) Proton transfer
4) Water leaves
5) Deprotonation
Isotopic Labeling
Evidence to support the Fischer esterfication mechanism comes from isotopic labeling experiments with oxygen-18. If the reaction is carried out with oxygen-18 labeled alcohol, the isotope is found exclusively in the ester and not the water generated.
Exercise
6. Draw the bond-line structures for the products of the following reactions.
Answer
6.
21.07: Methyl Ester Synthesis Using Diazomethane
Diazomethane, CH2N2, is a yellow, poisonous, potentially explosive compound, which is a gas at room temperature. The structure of diazomethane is explained using three resonance forms.
Conversion of carboxylic acids to methyl esters
Carboxylic acids react with diazomethane to produce methyl esters. Because of the high reactivity of diazomethane, it is produced in-situ and then immediately reacted with the carboxylic acid to produce the methyl ester.
The first step of the mechanism is a simple acid-base reaction to deprotonate the carboxylic acid. The carboxylate is then the nucleophile of an SN2 reaction with protonated diazomethane to produce the methyl ester with nitrogen gas as a leaving group. It is important to keep reaction vessels vented when gases are produced to avoid explosions.
21.08: Condensation of Acids with Amines
Conversion of Carboxylic Acids to Amides
The direct reaction of a carboxylic acid with an amine would be expected to be difficult because the basic amine would deprotonate the carboxylic acid to form a highly unreactive carboxylate. However when the ammonium carboxylate salt is heated to a temperature above 100 oC water is driven off and an amide is formed.
Conversion of Carboxylic acids to amide using DCC as an activating agent
The direct conversion of a carboxylic acid to an amide is difficult because amines are basic and tend to convert carboxylic acids to their highly unreactive carboxylates. In this reaction the carboxylic acid adds to the DCC molecule to form a good leaving group which can then be displaced by an amine during nucleophilic substitution.
DCC induced coupling to form an amide linkage is an important reaction in the synthesis of peptides.
Mechanism
1) Deprotonation
2) Nucleophilic reaction with carboxylate acting as the nucleophile
3) Nucleophilic reaction with the amine acting as the nucleophile
4) Proton transfer
5) Leaving group removal
Exercise
7. Complete the reaction map below proposing two different ways to synthesize benzoic acid from benzene.
Answer
7.
21.09: Reduction of Carboxylic Acids
Carboxylic acids can be converted to 1o alcohols using Lithium aluminum hydride (LiAlH4)
Note that NaBH4 is not strong enough to convert carboxylic acids or esters to alcohols. An aldehyde is produced as an intermediate during this reaction, but it cannot be isolated because it is more reactive than the original carboxylic acid.
For example, butanoic acid can be reduced to butanol when reacted with lithium aluminum hydride as shown below.
Possible Mechanism
There is not complete agreement on the mechanism for this reaction. However, the mechanism below is considered probable by many chemists.
1) Deprotonation
2) Nucleopilic reaction by the hydride anion
3) Leaving group removal
4) Nucleopilic reaction by the hydride anion
5) The alkoxide is protonated
Exercise
8. Using benzyl bromide and sodium cyanide as the only source of carbons, propose a synthetic strategy to produce 2-phenyl-ethan-1-ol.
Answer
8. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/21%3A_Carboxylic_Acids/21.05%3A_Reactions_of_Carboxylic_Acids_Overview.txt |
Carboxylic acids are widespread in nature, often combined with other functional groups. Simple alkyl carboxylic acids, composed of four to ten carbon atoms, are liquids or low melting solids having very unpleasant odors. The fatty acids are important components of the biomolecules known as lipids, especially fats and oils. As shown in the following table, these long-chain carboxylic acids are usually referred to by their common names, which in most cases reflect their sources. A mnemonic phrase for the C10 to C20 natural fatty acids capric, lauric, myristic, palmitic, stearic and arachidic is: "Curly, Larry & Moe Perform Silly Antics" (note that the names of the three stooges are in alphabetical order).
Interestingly, the molecules of most natural fatty acids have an even number of carbon atoms. Analogous compounds composed of odd numbers of carbon atoms are perfectly stable and have been made synthetically. Since nature makes these long-chain acids by linking together acetate units, it is not surprising that the carbon atoms composing the natural products are multiples of two. The double bonds in the unsaturated compounds listed on the right are all cis (or Z).
FATTY ACIDS
Saturated
Formula
Common Name
Melting Point
CH3(CH2)10CO2H lauric acid 45 ºC
CH3(CH2)12CO2H myristic acid 55 ºC
CH3(CH2)14CO2H palmitic acid 63 ºC
CH3(CH2)16CO2H stearic acid 69 ºC
CH3(CH2)18CO2H arachidic acid 76 ºC
Unsaturated
Formula
Common Name
Melting Point
CH3(CH2)5CH=CH(CH2)7CO2H palmitoleic acid 0 ºC
CH3(CH2)7CH=CH(CH2)7CO2H oleic acid 13 ºC
CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2H linoleic acid -5 ºC
CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7CO2H linolenic acid -11 ºC
CH3(CH2)4(CH=CHCH2)4(CH2)2CO2H arachidonic acid -49 ºC
The following formulas are examples of other naturally occurring carboxylic acids. The molecular structures range from simple to complex, often incorporate a variety of other functional groups, and many are chiral.
Aspirin, Arachidonic Acid, and Prostaglandins
Prostaglandins were first discovered and isolated from human semen in the 1930s by Ulf von Euler of Sweden. Thinking they had come from the prostate gland, he named them prostaglandins. It has since been determined that they exist and are synthesized in virtually every cell of the body. Prostaglandins, are like hormones in that they act as chemical messengers, but do not move to other sites, but work right within the cells where they are synthesized. Prostaglandins are unsaturated carboxylic acids, consisting of of a 20 carbon skeleton that also contains a five member ring. They are biochemically synthesized from the fatty acid, arachidonic acid. See the graphic on the left. The unique shape of the arachidonic acid caused by a series of cis double bonds helps to put it into position to make the five member ring.
Structure of prostaglandin E2 (PGE2)
Prostaglandins are unsaturated carboxylic acids, consisting of of a 20 carbon skeleton that also contains a five member ring and are based upon the fatty acid, arachidonic acid. There are a variety of structures one, two, or three double bonds. On the five member ring there may also be double bonds, a ketone, or alcohol groups.
There are a variety of physiological effects associated with prostaglandins including:
1. Activation of the inflammatory response, production of pain, and fever. When tissues are damaged, white blood cells flood to the site to try to minimize tissue destruction. Prostaglandins are produced as a result.
2. Blood clots form when a blood vessel is damaged. A type of prostaglandin called thromboxane stimulates constriction and clotting of platelets. Conversely, PGI2, is produced to have the opposite effect on the walls of blood vessels where clots should not be forming.
3. Certain prostaglandins are involved with the induction of labor and other reproductive processes. PGE2 causes uterine contractions and has been used to induce labor.
4. Prostaglandins are involved in several other organs such as the gastrointestinal tract (inhibit acid synthesis and increase secretion of protective mucus), increase blood flow in kidneys, and leukotriens promote constriction of bronchi associated with asthma.
Ball-and-stick model of the aspirin molecule, as found in the solid state. Single-crystal X-ray diffraction data from Kim, Y.; Machida, K.; Taga, T.; Osaki, K. (1985). "Structure Redetermination and Packing Analysis of Aspirin Crystal". Chem. Pharm. Bull. 33 (7): 2641-2647. ISSN 1347-5223.
Effects of Aspirin and other Pain Killers on Prostaglandin Production
When you see that prostaglandins induce inflammation, pain, and fever, what comes to mind but aspirin. Aspirin blocks an enzyme called cyclooxygenase, COX-1 and COX-2, which is involved with the ring closure and addition of oxygen to arachidonic acid converting to prostaglandins. The acetyl group on aspirin is hydrolzed and then bonded to the alcohol group of serine as an ester. This has the effect of blocking the channel in the enzyme and arachidonic can not enter the active site of the enzyme. By inhibiting or blocking this enzyme, the synthesis of prostaglandins is blocked, which in turn relives some of the effects of pain and fever. Aspirin is also thought to inhibit the prostaglandin synthesis involved with unwanted blood clotting in coronary heart disease. At the same time an injury while taking aspirin may cause more extensive bleeding.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook
21.11: Additional Exercises
General Review
21-1 Provide the products for the following reactions.
21-2 Provide the proper IUPAC name for the product of the following reaction.
Synthesis of Carboxylic Acids
21-3 For the following reactions, predict the final product and provide their proper IUPAC nomenclature.
21-4 Propose another method of synthesis (starting with cyclopentane) to make the product in the previous question, 21-3.a.
21-5 Choose the correct alkene that was oxidatively cleaved by hot, concentrated potassium permanganate to form 3-methylbutanoic acid.
a) (4E/Z)-oct-4-ene
b) (4E/Z)-2-methyloct-4-ene
c) (4E/Z)-2,7-dimethyloct-4-ene
d) (3E/Z)-2,5-dimethylhex-3-ene
Reactions of Carboxylic Acids and Derivatives: Nucleophilic Acyl Substitution
21-6 Provide the structures of the products of the following reactions.
21-7 Provide the structure of the product that forms when propanoic acid reacts with thionyl chloride, then an excess of CH3CH2MgBr and finally followed by an acid workup.
21-8 Choose the correct product of the following reaction.
Condensation of Acids with Alcohols: the Fischer Esterification
21-9 Predict the product of the following Fischer Esterification reaction.
21-10 Choose the correct product of the following Fischer Esterification reaction.
21-11 Predict the product of the following reaction (use a benzene ring with a total of two para-oriented substituents).
Condensation of Acids with Amines: Direct Synthesis of Amides
21-12 Give the structure of the product of the following reaction.
21-13 Choose the correct IUPAC nomenclature of the product and provide its structure.
a) benzyl 4-bromohexanoate
b) (4E)-N-benzylhex-4-enamide
c) N-benzyl-4-bromohexanamide
d) 4-bromo-N-phenylhexanamide
21-14 Provide the structure of the final product.
Alkylation of Carboxylic Acids to Form Ketones
21-15 Predict the product of the following reaction.
21-16 Explain why two equivalents of organolithium reagent is necessary to alkylate carboxylic acids (see problem 20-1).
21-17 Give the products of the following reactions.
21.12: Solutions to Additional Exercises
General Review
21-1
21-2 1-[4-(hydroxymethyl)phenyl]ethan-1-one
Synthesis of Carboxylic Acids
21-3:
21-4:
Possible route of synthesis:
21-5:
Answer: C
(E or Z orientation allowed)
Reactions of Carboxylic Acids and Derivatives: Nucleophilic Acyl Substitution
21-6:
21-7:
21-8:
Answer: B
Condensation of Acids with Alcohols: the Fischer Esterification
21-9:
21-10:
Answer: D
21-11:
Condensation of Acids with Amines: Direct Synthesis of Amides
21-12:
21-13:
Answer: C
21-14:
Alkylation of Carboxylic Acids to Form Ketones
21-15:
21-16:
Two equivalents of the organolithium reagent is necessary for the alkylation of carboxylic acids because one equivalent is used to form a salt with the carboxylic acid and the other equivalent is the nucleophile that adds to the carbonyl carbon.
An example of the intermediate from problem 20-1.
21-17: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/21%3A_Carboxylic_Acids/21.10%3A_Biochemically_Interesting_Carboxylic_Acids.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• describe the structure and physical properties of carboxylic acid derivatives and nitriles (section 22.1)
• determine the structure of carboxylic acid derivatives and nitriles from their elemental analysis and spectral data (MS, IR 1H NMR & 13C NMR) (section 22.2)
• predict the products and specify the reagents to interconvert between a carboxylic acid and its derivatives (section 22.3)
• predict the products and specify the reagents to hydrolyze carboxylic acid derivatives (22.4)
• predict the products and specify the reagents for transesterification reactions (section 22.5)
• predict the products and specify the reagents for reduction reactions of carboxylic acid derivatives (section 22.6)
• predict the products and specify the reagents for organometallic reactions with carboxylic acid derivatives (section 22.7)
• predict the products and specify the reagents for the synthesis and reactions of
• acyl chlorides (section 22.4)
• anhydrides (section 22.5)
• esters (section 22.6)
• amides (section 22.7)
• nitriles (section 22.8)
• thioesters (section 22.9)
• step-growth (condensation) polymers via ester and amide bonds (section 22.10)
• discuss the chemistry of beta-lactams and biological acylation (section 22.11 and 22.12 respectively)
• combine the reactions studied to date to develop efficient and effective multiple-step synthesis
Please note: IUPAC nomenclature and important common names of carboxylic acid derivatives and nitriles were explained in Chapter 3.
22: Carboxylic Acid Derivatives and Nitriles
The Structure of Carboxylic Acid Derivatives
Carboxylic acid and its derivatives are functional groups with closely related chemistry.
The carboxylic acid derivatives all include an acyl group, an R-group bonded to a carbonyl carbon. The "other group" bonded to the carbonyl carbon distinguishes the derivatives from each other and has a strong influence on the relative reactivity between the derivatives. This "other group" takes the role of the "Leaving Group or LG" in Nucleophilic Acyl Substitution (NAS) reactions.
Although there are many types of carboxylic acid derivatives known, this chapter focuses on four: acid halides (acyl halides), acid anhydrides, esters, and amides. Another common feature of carboxylic acid derivatives is that they can be hydrolyzed back to the original carboxylic acid. For this reason, nitriles are sometimes grouped with the carboxylic acid derivatives and will also be discussed in this chapter. The hydrolysis reaction for acetonitrile is shown below as an example.
Physical Properties of Carboxylic Acid Derivatives
The intermolecular forces of the respective acyl group combine with the size and structure of the R-group to determine the physical properties of the the carboxylic acid derivatives as shown in the summary below.
Structure and Properties of Nitriles
The electronic structure of nitriles is very similar to that of an alkyne with the main difference being the presence of a set of lone pair electrons on the nitrogen. Both the carbon and the nitrogen are sp hydridized which leaves them both with two p orbitals which overlap to form the two $\pi$ bond in the triple bond. The R-C-N bond angle in and nitrile is 180o which give a nitrile functional group a linear shape.
The lone pair electrons on the nitrogen are contained in a sp hybrid orbital which makes them much less basic and an amine. The 50% character of an sp hybrid orbital close to the nucleus and therefore less basic compared to other nitrogen containing compounds such as amines.
The presence of an electronegative nitrogen causes nitriles to be very polar molecules. Consequently, nitriles tend to have higher boiling points than molecules with a similar size.
Exercise
1. Design an extraction separation strategy to separate acetaminophen from cocaine using ether, 1 M HCl, and 3 M NaHCO3. The structures for acetaminophen and cocaine are shown below.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.01%3A_Structure_and_Physical_Properties_of_Acid_Derivative.txt |
Infrared Spectroscopy (IR)
While all of the carboxylic acid derivatives include a carbonyl group, the heteroatoms that characterize the derivative can be used to distinguish between the derivatives. Additionally, there is a useful correlation between the reactivity of the carboxylic acid derivatives and their carbonyl stretching frequencies. Thus, the very reactive acyl halides and anhydrides absorb at frequencies significantly higher, while the relatively unreactive amides absorb at lower frequencies. The IR spectral characteristics that can be used to determine the identity of carboxylic acid derivatives are listed below. Infrared spectra of many carboxylic acid derivatives will be displayed in the figure below the table by clicking the appropriate buttons presented there.
Carbonyl Derivative
Carbonyl Absorption
Comments
Acyl Halides (RCOX)
X = F
X = Cl
X = Br
C=O stretch
1860 ± 20 cm-1
1800 ± 15
1800 ± 15
Conjugation lowers the C=O frequencies reported here, as with aldehydes & ketones.
In acyl chlorides a lower intensity shoulder or peak near 1740 cm-1 is due to an overtone interaction.
Acid Anhydride, (RCO)2O
acyclic
6-membered ring
5-membered ring
C=O stretch (2 bands)
1750 & 1820 cm-1
1750 &1820
1785 & 1865
Conjugation lowers the C=O frequencies reported here, as with aldehydes & ketones.
The two stretching bands are separated by 60 ± 30 cm-1, and for acyclic anhydrides the higher frequency (asymmetric stretching) band is stronger than the lower frequency (symmetric) absorption.
Cyclic anhydrides also display two carbonyl stretching absorptions, but the lower frequency band is the strongest.
One or two -CO-O-CO- stretching bands are observed in the 1000 to 1300 cm-1 region.
Esters & Lactones (RCOOR')
esters
6-membered lactone
5-membered lactone
4-membered lactone
C=O stretch
1740 cm ± 10 cm-1
1740 cm ± 10
1765 cm± 5
1840 cm ± 5
Conjugation lowers the C=O frequencies reported here, as with aldehydes & ketones
Strong CO-O stretching absorptions (one ot two) are found from 1150 to 1250 cm-1
Amides & Lactams (RCONR2)
1° & 2°-amides
3°-amides
6-membered lactams
5-membered lactams
4-membered lactams
C=O bands
1510 to 1700 cm-1 (2 bands)
1650± 15 (one band)
1670 ± 10 (one band)
1700 ± 15
1745 ± 15
The effect of conjugation is much less than for aldehydes & ketones.
The higher frequency absorption (1665± 30) is called the Amide I band. The lower frequency Amide II band (1620± 30 in 1° amides & 1530± 30 in 2° amides) is largely due to N-H bending trans to the carbonyl oxygen. In concentrated samples this absorption is often obscured by the stronger amide I absorption. Hydrogen bonded association shifts some of these absorptions, as well as the prominent N-H stretching absorptions.
N-H stretch: 3170 to 3500 cm-1. Two bands for 1°-amides, one for 2°-amides.
NMR Spectra
For NMR, there are a few spectral characteristics that can help identify the carboxylic acid derivative. The protons on carbons adjacent to carbonyls absorb at ~2.0-2.5 ppm. For amides, the N-H protons attached to primary and secondary amines absorb at ~7.5-8.5. For 13C NMR, the carbonyl carbon in carboxylic acid derivatives shows up betweem ~ 160-180 ppm with the carbon in a nitrile appearing ~ 115-120 ppm in their 13C NMR because of its sp hybridization.
Exercise
2. Esters are known for their sweet, fruity aromas. Compound P smells like pineapple and is used as a flavor enhancer for orange juices. Compound Q is responsible for the sweet smell of Red Delicious apples. Compounds P and Q are structural isomers with the following composition: 62.04% C, 0.41% H, and 27.55% O. The IR spectrum for each compound includes several moderate bands around 2940 cm-1, a strong band near 1740 cm-1, and a moderate band near 1200 cm-1. The 1H and 13C NMR spectrum for each compound is shown below.
Name, draw the bond-line structure, and correlate the NMR signals to their respective compound.
Answer
2.
22.03: Interconversion of Acid Derivatives by Nucleophilic
General reaction
Carboxylic acid derivatives are electrophilic and can react with nucleophiles to form nucleophilic acyl substitution products. The driving force of these reactions is the stability of the leaving group shown as :L- below.
General mechanism
The nucleophile reacts with the electrophilic carbonyl carbon to form the tetrahedral intermediate. When the carbonyl reforms, the leaving group is lost to form the substitution product as shown in the mechanism below.
1) Nucleophilic reaction at the carbonyl
2) Carbonyl reforms and leaving group is removed
Relative Reactivity to Nucleophilic Acyl Substitution
The relative reactivity of carbonyl compounds toward nucleophile substitutions is related to the stability of the leaving group - the more stable the leaving group, the more favorable the substitution reaction. Evaluating leaving group stability is analogous to evaluating conjugate base stability as shown in the table below.
Although aldehydes and ketones also contain carbonyls, their chemistry is distinctly different because they do not contain suitable leaving groups. Once a tetrahedral intermediate is formed, aldehydes and ketones cannot reform their carbonyls because the carbide (RC-) and hydride (H-) leaving groups are too unstable. Therefore, aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
Integrating all of this information into the single table below summarizes the relative reactivity of carboxylic acids and their derivatives.
Acid Derivative Interconversion
From this understanding, multiple step synthesis strategies can be developed. The derivatives with the most stable leaving groups can be used to synthesize the derivatives with the least stable leaving groups as illustrated in the diagram below.
While it may appear from diagram above that the acyl chlorides are the "source" of the acid derivatives, acyl chlorides are so highly reactive that it is common to convert the carboxylic acid to the acid chloride and then immediately form the derivative. From a laboratory synthesis perspective, the reaction sequence begins with the carboxylic acid.
Exercise
3. Using ethanol as the only source of carbons in the final products, show how to synthesize ethyl acetate and N-ethylacetamide.
Answer
3. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.02%3A_Spectroscopy_of_Carboxylic_Acid_Derivatives.txt |
Please Note: The terms "acid halide" and "acyl halide" are synonymous and are both used in this text. In biochemistry, the term "acyl" is used more frequently.
Acid Halide Synthesis
Carboxylic acids react with thionyl chloride (SOCl2) or oxalyl chloride (C2O2Cl2) to form acid chlorides. Typically the reactions occur in the presence of a proton scavanger like pyridine to minimize unwanted side reactions. During the reaction the hydroxyl group of the carboxylic acid is converted to a chlorosulfite intermediate making it a better leaving group. The chloride anion produced during the reaction acts a nucleophile.
Analogous to the reactions of primary and secondary alcohols with PBr3 to produce the corresponding alkyl bromide, acid bromides can be formed from the reaction of phosphorous tribromide with carboxylic acids.
Acyl Halide Reactivity
Acyl halides can be hydrolyzed to carboxylic acids and converted to carboxylic acid derivatives. Acid halides can also undergo reduction reactions and reactions with Grignard reagents and organolithium cuprates along with Fridel-Crafts acylation of benzene. The reaction map below summarizes the reactivity of acyl halides.
Acid Halide Hydrolysis
The hydrolysis reaction of acid chlorides is shown below.
The hydrolysis of butonyl chloride is shown below as an example.
Example: Acyl Chloride Hydrolysis
Carboxylic Acid Derivative Synthesis from Acyl Chlorides
Carboxylic acid derivatives can be synthesized from acyl chlorides via the nucleophilic acyl substitution mechanism previously discussed.
Anhydride Synthesis
Acid chlorides react with carboxylic acids to form acid anhydrides as shown in the reaction below.
The synthesis of benzoic anhydride from benzoyl chloride and benzoic acid is shown as an example.
Example: Anhydride Synthesis from Acyl Chlorides
Ester Synthesis
Acid chlorides react with alcohols for form esters are shown in the reaction below.
The synthesis of ethyl benzoate from benzoyl chloride and ethanol is shown as an example.
Example: Ester Synthesis from Acid Chlorides
Amide Synthesis
Acid chlorides react with "ammonia, 1o amines and 2o amines" to form amides as shown in the reaction below.
The reaction requires 1 equivalent of the "ammonia/1o or 2o amine" with a proton scavenger (aka base) like pyridine OR two equivalents of "ammonia/1o or 2o amine". The additional equivalent the nucleophile or base is needed to maintain the nucleophilic character of "ammonia/1o or 2o amine". As shown in the mechanism below, the amide is briefly protonated after the carbonyl reforms from the tetrahedral complex. Since amides are considered neutral with no significant basicity, the "ammonia/1o or 2o amine" quickly accepts their proton and is no longer a nucleophile. The second equivalent of "ammonia/1o or 2o amine" restores the concentration of the nucleophile. For clarity, the mechanism below is shown with ammonia as the nucleophile.
Mechanism
The syntheses of acetamide and N-ethylacetamide from acetyl chloride are shown as examples.
Example: Amide Synthesis from Acyl Chlorides
Acid Chloride Reduction
Acid chlorides can be fully reduced to primary alcohols using either sodium borohydride or lithium aluminum hydride. Acid chlorides can be partially reduced to aldehydes using the lithium tri-tert-butoxyaluminum hydride (LiAlH(O-t-Bu)3. These reactions are summarized in the reaction map below.
The syntheses of 1-hexanol and hexanal from hexanoyl chloride are shown as examples.
Example: Reduction Acyl Chlorides
Acid Chloride Reactions with Organometallic Compounds
Grignard reagents
Acid chlorides react with Grignard reagents to produce tertiary alcohols. Two equivalents of the Grignard reagent are needed because the first equivalent reacts to form a ketone which then reacts with the second equivalent. Because of the high reactivity of the Grigard reagent, the reaction can NOT be stopped at the ketone.
The syntheses of 3-phenylpentan-3-ol from benzoyl chloride is shown as an example.
Example: Acyl Chloride Reactions with Grignard Reagents
Organolithium Cuprates
Organolithium cuprate reagents are less reactive than Grignard reagents and can convert acid chlorides to ketones as shown below.
The synthesis of 1-phenylpropan-1-one from benzoyl chloride is shown as an example.
Example: Acyl Chloride Reactions with Organolithium Cuprates
Friedel-Crafts Acylation of Benzene
Benzene rings can be acylated via the Friedel-Crafts acylation reaction with acid chlorides in the presence of aluminum chloride followed by an aqueous work-up as shown below.
The synthesis of 1-(4-tert-butylphenyl)butan-1-one from t-butylbenzene and butonyl chloride is shown as an example.
Example: Friedel-Crafts Acylation
Exercise
4. Draw the mechanism for the following reaction
5. Propose a synthesis of the following molecules from an acid chloride and an amide.
(a)
(b)
(c)
Answer
4.
5.
a) Acetyl chloride and dimethylamine
b) Benzoyl chloride and ethylamine
c) Acetyl chloride and ammonia
22.05: Acid Anhydride Chemistry
Synthesis of Acid Anhydrides
Acid chlorides react with carboxylic acids to form anhydrides as shown in the reaction below.
Some cyclic anhydrides can be synthesized from the corresponding dicarboxylic acid with gentle heating. The example below shows the reaction of glutaric acid to form a cyclic anhydride.
Example: Acid Anhydride Synthesis
Acid Anhydride Reactivity
Acid anhydrides undergo hydrolysis and nucleophilic acyl substitution reactions.
Acid Anhydride Hydrolysis
Acid anhydrides readily hydrolyze to carboxylic acids. In many cases, this reaction is an unwanted side reaction and steps will be taken in the lab to keep the system "dry" (aka water free). The presence of pyridine facilitates proton transfers during the reaction. The hydrolysis reaction for benzoic anhydride is shown below.
The mechanism is analogous to the mechanism for ester synthesis from acid anhydrides and is shown below is detail.
Nuclephilic Acyl Substitution Reactions from Acid Anhydrides
Carboxylic acid derivatives can be synthesized from acid anhydrides via the nucleophilic acyl substitution mechanism previously discussed.
Ester Synthesis
Acid anhydrides react with alcohols to produce esters as shown in the reaction below. The reactions of anhydrides frequently use pyridine as a solvent.
A carboxylic acid is also produced, but is not considered a synthetic product. The ester is considered the "product of interest".
The synthesis of methyl benzoate from benzoic anhydride and methanol is shown in the example.
Example: Ester Sythesis
The mechanism follows the nucleophililc acyl substitution mechanism as previously discussed and reviewed below.
1) Nucleophilic Alcohol reacts with Electrophilic Carbonyl
2) Deprotonation by pyridine
3) Leaving group removal
4) Protonation of the carboxylate
Amide Synthesis
Acid Anhydrides react with amines to form amides. As seen with acid halide reactions, a second equivalent of the amine must be present for the reaction to proceed.
Example: Amide Synthesis
The mechanism for amide synthesis is analogous to the mechanism for ester formation. The only minor difference is that a second equivalent of the amine or ammonia is used instead of the pyridine.
1) Nucleophilic Amine reacts with Electrophilic Carbonyl
2) Deprotonation by the amine
3) Leaving group removal
Exercise
6. Draw out the mechanism for the following reaction.
7. Draw the product of the reaction between these two molecules.
Answer
6.
7. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.04%3A_Acid_Halide_Chemistry.txt |
Introduction
Esters are readily synthesized and naturally abundant. Esters are frequently the source of flavors and aromas in many fruits and flowers.
Esters also make up the bulk of animal fats and vegetable oils—glycerides (fatty acid esters of glycerol). Soap is produced by a saponification (basic hydrolysis) reaction of a fat or oil.
Esters are also present in a number of important biological molecules and have several commercial and synthetic application. For example, polyester molecules make excellent fibers and are used in many fabrics. A knitted polyester tube, which is biologically inert, can be used in surgery to repair or replace diseased sections of blood vessels. PET is used to make bottles for soda pop and other beverages. It is also formed into films called Mylar. When magnetically coated, Mylar tape is used in audio- and videocassettes. Synthetic arteries can be made from PET, polytetrafluoroethylene, and other polymers.
The most important polyester, polyethylene terephthalate (PET), is made from terephthalic acid and ethylene glycol monomers:
Synthesis of Esters
Carboxylic acids can react with alcohols to form esters in the presence of an acid catalyst as shown in the reaction below.
Acid chlorides react with alcohols to form esters as shown in the reaction below.
Acid anhydrides also react with alcohols to form esters as shown in the reaction below.
As an example, the synthesis of banana oil (isoamyl acetate) is an example of these two reactions.
Example: Ester Synthesis
Esters can be also be synthesized from trans-esterification reactions. Trans-esterification is discussed in the next section on Reactivity of Esters
Reactivity of Esters
Esters can be hydrolyzed to carboxylic acids under acidic or basic conditions. Basic hydrolysis can be used to convert fats and oils into soap and is called a saponification reaction. Esters can be converted to amides via an aminolysis reaction. Esters can undergo trans-esterification reactions to form different esters by applying LeChatlier's principle to this equilibrium reaction. Esters can be reduced to form alcohols or aldehydes depending on the reducing agent. Esters also react with organometallic compounds to form tertiary alcohols. The reaction map for esters is shown below.
Ester Hydrolysis
Ester hydrolysis requires an acid catalyst or base promotion to occur. Esters are less reactive than acyl halides and acid anhydrides because the alkoxide group is a poor leaving group with its negative charge fully localized on a single oxygen atom.
Ester Hydrolysis - Acid Catalyzed
Esters can be cleaved back into a carboxylic acid and an alcohol by reaction with water and a catalytic amount of acid as shown in the reaction below.
The acid catalyzed hydrolysis of ethyl benzoate is shown below as an example.
Example: Acid Catalyzed Ester Hydrolysis
The mechanism for the acid catalyzed hydrolysis reaction begins with protonation of the carbonyl oxygen to increase the reactivity of the ester. The nucleophilic water reacts with the electrophilic carbonyl carbon atom to form the tetrahedral intermediate. Proton transfer reactions occur to create a good leaving group when the carbonyl reforms. The complete mechanism is shown below.
1) Protonation of the Carbonyl
2) Nucleophilic reaction by water
3) Proton transfer
4) Leaving group removal
Ester Hydrolysis - Base promoted
Esters can be cleaved back into a carboxylic acid and an alcohol by reaction with water and a base. The reaction is called a saponification from the Latin sapo which means soap. The name comes from the fact that soap used to me made by the ester hydrolysis of fats. Due to the basic conditions, a carboxylate ion is made rather than a carboxylic acid. The hydroxide ions are consumed in the reaction so it is described as "base promoted".
The base promoted hydrolysis of ethyl benzoate is shown below as an example.
Example: Base Promoted Hydrolysis of Esters
The mechanism for the base promoted hydrolysis reaction begins with the nucleophilic hydroxide reacting with the electrophilic carbonyl carbon atom to form the tetrahedral intermediate. The carbonyl reforms with the loss of the alkoxide leaving group. The alkoxide then deprotonates the resulting carboxylic acid. The complete mechanism is shown below.
1) Nucleophilic reaction by hydroxide
2) Leaving group removal
3) Deprotonation
Nuclephilic Acyl Substitution Reactions from Esters
Carboxylic acid derivatives can be synthesized from esters via the nucleophilic acyl substitution mechanism previously discussed.
Ester Synthesis: Trans-Esterification
Trans-esterification is the conversion of a carboxylic acid ester into a different carboxylic acid ester. When an ester is placed in a large excess of an alcohol along with presence of either an acid or a base there can be an exchange of alkoxy groups. The large excess of alcohol is used to drive the reaction forward. The most common method of trans-esterification is the reaction of the ester with an alcohol in the presence of an acid catalyst as shown below.
The trans-esterification of ethyl acetate to methyl acetate and methyl benzoate to ethyl benzoate are shown below as examples.
Example: Trans-esterification Reactions
Under acidic conditions, the reaction mechanism begins with protonation of the carbonyl oxygen which increases the reactivity of the ester. An alcohol then reactions with protonated ester to form the tetrahedral intermediate. After several proton transfers, the carbonyl reforms to produce a new ester. The complete mechanism is shown below for the trans-esterification of ethyl actetae to methyl acetate.
Since both the reactants and the products are an ester and an alcohol, the reaction is reversible and the equilibrium constant is close to one. Consequently, the Le Chatelier’s principle has to be exploited to drive the reaction to completion. The simplest way to do so is to use the alcohol as the solvent as well.
Under basic conditions, the mechanism begins with the nucleophilic reaction of the alkoxide with the carbonyl carbon to produce the tetrahedral intermediate. The carbonyl reforms with the loss of the leaving group to produce a new ester.
1) Nucleophilic reaction by an alkoxide
2) Leaving group removal
Aminolysis: Conversion of Esters into Amides
Esters react with ammonia and 1o or 2o alkyl amines to yield amides in a reaction called aminolysis.
The aminolysis of ethyl benzoate is shown below as an example. The mechanism for this reaction is analogous to the base promoted hydrolysis reaction of esters shown above.
Example: Aminolysis of Esters
Ester Reduction Reactions
Ester Reduction to a 1o Alcohol
Esters can be converted to 1o alcohols using LiAlH4, while sodium borohydride (NaBH4) is not a strong enough reducing agent to perform this reaction.
The reduction of ethyl benzoate to benzyl alcohol and ethanol is shown as an example.
Example: Ester Reduction to a 1o Alcohol
The mechanism begins with a hydride nucleophile reacting with the ester carbonyl carbon to form the tetrahedral intermediate. The carbonyl reforms to produce an aldehyde with the loss of the alkoxide ion. The resulting aldehyde undergoes a subsequent reaction with a hydride nucleophile to form another tetrahedral intermediate. The carbonyl is not able to reform, because there are no stable leaving groups. Therefore, the alkoxide (tetrahedral intermediate) is protonated to produce a primary alcohol. The complete mechanism is shown below.
1) Nucleophilic reaction by the hydride
2) Leaving group removal
3) Nucleopilic reaction by the hydride anion
4) The alkoxide is protonated
Ester Reduction to an Aldehyde
Esters can be converted to aldehydes using diisobutylaluminum hydride (DIBAH). The reaction is usually carried out at -78 oC to prevent reaction with the aldehyde product.
The reduction of methyl benzoate to form benzaldehyde is shown as an example. The mechanism is analogous to the LiAlH4 mechanism shown above with the important difference that the reaction stops after the aldehyde is produced because the DIBAH reducing agent is not strong enough to reduce the aldehyde at low temperatures.
Example: Ester Reduction to an Aldehyde
Ester Reactions with Organometallic Compounds
Grignard reagents
Esters react with Grignard reagents to form tertiary alcohols. This reaction is analogous to the reaction discussed for acid chlorides with Grignard reagents. The first equivalent of the Grignard reagent produces a ketone which reacts with the second equivalent of the Grignard reagent to produce a tertiary alcohol. In effect, the Grignard reagent adds twice as shown in the reaction below.
The reaction of methyl benzoate with a Grignard reagent to produce 3-phenyl-3-pentanol.
Example: Ester Reaction with a Grignard Reagent
The mechanism begins with a carbide nucleophile from the Grignard reagent reacting with the ester carbonyl carbon to form the tetrahedral intermediate. The carbonyl reforms to produce a ketone with the loss of the alkoxide ion. The resulting ketone undergoes a subsequent reaction with a carbide nucleophile from the Grignard reagent to form another tetrahedral intermediate. The carbonyl is not able to reform, because there are no stable leaving groups. Therefore, the alkoxide (tetrahedral intermediate) is protonated to produce a tertiary alcohol. The complete mechanism is shown below.
1) Nucleophilic reaction
2) Leaving group removal
3) Nucleophilic reaction
4) Protonation
Exercise
8. Why is the alkaline hydrolysis of an ester not a reversible process? Why doesn't the reaction with a hydroxide ion and a carboxylic acid produce an ester?
9. Draw the product of the reaction between the following molecule and LiAlH4, and the product of the reaction between the following molecule and DIBAL.
10. Prepare the following molecules from esters and Grignards?
(a)
(b)
(c)
Answer
8. The reaction between a carboxylic acid and a hydroxide ion is an acid base reaction, which produces water and a carboxylate anion.
9.
10.
(a)
(b)
(c)
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.06%3A_Ester_Chemistry.txt |
Synthesis of Amides
There are five synthetic routes to produce amides: nitrile conversion and the acyl nucleophilic substitution reactions of acid halides, acid anhydrides, and carboxylic acids.
Nitriles can be converted to amides. This reaction can be acid or base catalyzed
Carboxylic acid can be converted to amides by using DCC as an activating agent
Direct conversion of a carboxylic acid to an amide by reaction with an amine.
Acid chlorides react with ammonia, 1o amines and 2o amines to form amides
Acid Anhydrides react with ammonia, 1o amines and 2o amines to form amides
Hydrolysis of Amides
Hydrolysis under acidic conditions
Taking acetamide (ethanamide) as a typical amide. If acetamide is heated with a dilute acid (such as dilute hydrochloric acid), acetic acid is formed together with ammonium ions. So, if you were using hydrochloric acid, the final solution would contain ammonium chloride and acetic acid.
Hydrolysis under alkaline conditions
Also, if acetamide is heated with sodium hydroxide solution, ammonia gas is given off and you are left with a solution containing sodium acetate.
Peptide hydrolysis
Peptide hydrolysis of proteins is amide hydrolysis. What biologists and biochemists call a peptide link (in proteins, for example) is what chemists call an amide link. Apply either hydrolysis reaction above to the dipeptide below to produce two amino acids. The amines in the products are shown in their protonated form because this hydrolysis reaction was performed under acidic conditions.
Reduction of Amides into Amines
Amides can be converted to 1°, 2° or 3° amines using LiAlH4 followed by an aqueous work-up. Alkyl groups attached to the amide nitrogen do not affect the reaction. The amine classification correlates with the amide as shown in the reaction summary below.
The reductions of propanamide and N,N-dimethylpropanamide are shown as examples.
Example: Amide Reductions
The mechanism begins with nucleophilic hydride reacting with the carbonyl carbon to produce the tetrahedral intermediate. An imine forms in concert with the loss on the leaving group. A second hydride nucleophile reacts with the imine carbon to produce the final product.
1) Nucleophilic reaction by the hydride
2) Imine formation with loss of leaving group
3) Nucleophilic reaction by the hydride
Exercise
11. How would you prepare the following compounds from N-Propypl benzamide?
(a)
(b)
(c)
12.
Propose a synthesis for the following.
Answer
11.
a) NaOH, H2O
b) NaOH, H2O, then LiAlH4
c) LiAlH4
12.
22.08: Nitrile Chemistry
Interesting Nitriles
One of the most common occurrences of nitriles is nitrile rubber. Nitrile rubber is a synthetic copolymer of acrylonitrile and butadiene. This form of rubber is highly resistant to chemicals and is used to make protective gloves, hoses and seals.
Synthesis of Nitriles
Nitriles can be synthesized from the reaction of nucleophilic cyanide with electrophilic groups, such as the carbonyls (aldehydes and ketones) and alkyl halides that are suitable for SN2 reactions. Amides can react with thionyl chloride to produce nitriles.
Addition of cyanide (-:C≡N) to an aldehyde or ketone forms a cyanohydrin.
Nitriles are formed by an SN2 reaction between an alkyl bromide and sodium cyanide
Primary (1o) amides can be converted to nitriles by dehydration with thionyl chloride (or other dehydrating agents like P2O5, or POCl3).
For the reaction of primary amides with thionyl chloride, the mechanism begins with the lone pair of the nitrogen atom formimg a protonated imine and pushing the pi electrons of the carbonyl to form a sigma bond with the sulfur of thionyl chloride. The sulfonyl bond reforms in concert with the loss of the leaving group (Cl-). The protonated imine is neutralized by any base. The nitrile is produced by one last deprotonation reaction with a loss of sulfur dioxide and chloride as the leaving groups.
The complete mechanism is shown below.
1) Protonated imine formation with Nucleophilic reaction of carbonyl pi bond
2) Leaving group removal
3) Deprotonation
4) Deprotonation & Leaving group removal
Reactivity of Nitriles
The carbon in a nitrile is electrophilic because a resonance structure can be drawn which places a positive charge on it. Because of this the triple bond of a nitrile accepts a nucleophile in a manner similar to a carbonyl.
Nitriles can be converted to carboxylic acid with heating in sulfuric acid. During the reaction an amide intermediate is formed.
The hydrolysis of cyclopentanecarbonitrile is shown below as an example.
Example: Nitrile Hydrolysis
Note that the presence of water is understood.
The mechanism begins with the protonation of the nitrile to make it more electrophilic to nucleophilic water. Once the water has reacted with the nitrile carbon, proton transfers occur to produce a resonance stabilized intermediate. Water acts as a weak base to deprotonate the carbonyl to form the amide which is hydrolyzed to the carboxylic acid.
1) Protonation
2) Nucleophilic reaction by water
3) Proton Transfer
4) Resonance
5) Deprotonation
6) Further hydrolysis of the amide shown in amide section of this chapter.
Nitrile Reduction
Nitriles can be reduced to primary amines with lithium aluminum hydride followed by an aqueous work-up. During this reaction the hydride nucleophile reacts with the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation, the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water to neutralize the reaction environment. The general reaction is shown below.
The reduction of cyclopentanecarbonitrile is shown below as an example.
Example: Nitrile Reduction
The mechanism begins with the nucleophilic hydride reacting with the electrophilc carbon of the nitrile to form an anionic aluminum complex. A second hydride nucleophile reacts with the same electrophilic carbon to form a tetrahedral complex. Protonation by addition of water produces the primary amine in its neutral form.
1) Nucleophilic reaction by the Hydride
2) Second nucleophilic reaction by the hydride.
3) Protonation by addition of water to give an amine
Organometallic Reaction with Nitriles
Grignard reagents can react with nitriles to form an imine salt that can be hydrolyzed to form a ketone as shown in the reaction below.
The reaction of benzonitrile with the methyl-Grignard reagent to form acetophenone is shown below as an example.
Example: Nitrile reaction with a Grignard Reagent
The mechanism begins with the nucleophilic Grignard Reagent reacting with the electrophilic carbon of the nitrile to form an imine salt. The imine salt is hydrolyzed to produce a ketone through a series of nucleophilic and proton transfer reactions. The complete mechanism is shown below for those who are curious.
1) Nucleophilic reaction by the Grignard Reagent
2) Protonation
3) Protonation
4) Nucleophilic reaction by water
5) Proton Transfer
6) Leaving group removal
7) Deprotonation
Exercise
13. Propose two different synthetic routes to convert benzyl bromide into 2-phenyl acetic acid.
Answer
13. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.07%3A_Amide_Chemistry.txt |
Introduction to thioesters and Coenzyme A
In the metabolism of lipids (fats and oils), thioesters are the principal form of activated carboxylate groups. They are employed as acyl carriers, assisting with the transfer of acyl groups such as fatty acids from one acyl X substrate to another.
The ‘acyl X group’ in a thioester is a thiol. The most important thiol compound used to make thioesters is called coenzyme A, which has the following structure:
Coenzyme A is often abbreviated HSCoA, in order to emphasize that it is the thiol sulfur that provides the critical thioester linkage to acyl groups. When fuel (carbohydrate and fat) is broken down in your body, it is eventually converted to a simple two-carbon unit called acetyl CoA, which is essentially a thioester derivative of acetic acid:
Activation of fatty acids by coenzyme A: a thioesterification reaction
In the biologically active form of fatty acids, the carboxylate groups have been converted to thioesters using coenzyme A. For example, the activated form of the C16 fatty acid palmitate is:
Let’s take a look at how this activation takes place, in a reaction catalyzed by an enzyme called acyl CoA synthetase. You already know that carboxylates are not themselves good substrates for acyl substitution reactions, and must be activated. Thus, you might predict that the first step of this reaction requires ATP to make a high-energy acyl phosphate intermediate. In fact, the activated carboxylate in this case is an acyl-AMP, formed in the same way as the acyl-AMP intermediate in the asparagine synthetase reaction.
The activated acyl-AMP intermediate is then attacked by the thiol sulfur of coenzyme A, and the AMP group is expelled to form the fatty acyl CoA.
Transfer of fatty acyl groups to glycerol: a thioester to ester substitution
The -SCoA thioester form of the fatty acid is a good substrate for a number of metabolic transformations. This is the form of fatty acid, for example, that is oxidized and broken down for energy in the mitochondria of your cells. Fatty acyl CoA also serves as substrate for the construction of triacylglycerol, which is the fat molecule that your body uses to store energy in fat cells. Recall that triacylglycerol is composed of a glycerol ‘backbone’ connected to three fatty acid groups through ester linkages.
The reaction in which a fatty acid acyl group is linked to glycerol represents the conversion of a thioester (fatty acyl CoA) to an ester. First, however, a transthioesterification reaction occurs. A transthioesterification is merely the conversion of one thioester to another. In the case of monoacylglycerolacyltransferase, the fatty acyl group first trades its thioester link to coenzyme A for another thioester link to a cysteine residue in the active site of the enzyme. It is a common strategy for enzymes to first form a covalent link to one substrate before catalyzing the principle chemical reaction.
The fatty acyl group is now ready to be transferred to glycerol, trading its thioester linkage to the cysteine for a new ester linkage to one of the alcohol groups on glycerol. The attacking nucelophile in this reaction is of course the alcohol oxygen of monoacylglycerol.
Because esters are more stable than thioesters, this is an energetically downhill reaction.
Transthioesterification reactions
In the previous section we saw one example of a transthioesterification. Another important transthioesterification reaction involves acetyl CoA, the activated form of acetic acid and the basic two-carbon building block for fats and oils. Before it can be incorporated into a growing fatty acid molecule, acetyl CoA must first be linked to a so-called ‘acyl carrier protein’ (ACP). The acetyl group is linked to the acyl carrier protein via a thiol group on a carrier molecule that is covalently attached to the protein.
Notice that the structure of this carrier group (called phosphopantetheine) is identical to the region of coenzyme A (structure shown earlier in this section) near the thiol group. Once attached to the ACP, the two-carbon acetyl group condenses with another acyl group (which is also attached to its own ACP), and the fatty acid chain begins to grow.
Finally, a transthioesterification is the final step in one of the most important and well-studied reactions in animal metabolism: the conversion of pyruvate to acetyl CoA by a cluster of enzymes called the pyruvate dehydrogenase complex.
The overall reaction looks simple, but is actually quite complex and involves several intermediate species. The final step in the process is a transthioesterification, involving a dithiol molecule called lipoamide:
We will look more closely at the complete biochemical transformation catalyzed by the pyruvate dehydrogenase complex in section 16.12B.
Hydrolysis of thioesters
The acyl group of a thioester can be transferred to a water molecule in a hydrolysis reaction, resulting in a carboxylate. An example of thioester hydrolysis is the conversion of (S)-citryl CoA to citrate in the citric acid cycle (also known as the Krebs cycle).
Reactivity of thioesters and acyl phosphates
Thioesters are reactive among the biologically relevant acyl groups. However, thioesters are not as reactive as an acid chlorides or acid anhydrides.
Relative reactivity of biologically relevant acyl groups | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.09%3A_Thioesters-_Biological_Carboxylic_Acid_Derivatives.txt |
Polyamides
Just as the reaction of a diol and a diacid forms a polyester, the reaction of a diacid and a diamine yields a polyamide. The two difunctional monomers often employed are adipic acid and 1,6-hexanediamine. The monomers condense by splitting out water to form a new product, which is still difunctional and thus can react further to yield a polyamide polymer.
Some polyamides are known as nylons. Nylons are among the most widely used synthetic fibers—for example, they are used in ropes, sails, carpets, clothing, tires, brushes, and parachutes. They also can be molded into blocks for use in electrical equipment, gears, bearings, and valves.
Polyesters
A commercially important esterification reaction is condensation polymerization, in which a reaction occurs between a dicarboxylic acid and a dihydric alcohol (diol), with the elimination of water. Such a reaction yields an ester that contains a free (unreacted) carboxyl group at one end and a free alcohol group at the other end. Further condensation reactions then occur, producing polyester polymers.
The most important polyester, polyethylene terephthalate (PET), is made from terephthalic acid and ethylene glycol monomers:
Polyester molecules make excellent fibers and are used in many fabrics. A knitted polyester tube, which is biologically inert, can be used in surgery to repair or replace diseased sections of blood vessels. PET is used to make bottles for soda pop and other beverages. It is also formed into films called Mylar. When magnetically coated, Mylar tape is used in audio- and videocassettes. Synthetic arteries can be made from PET, polytetrafluoroethylene, and other polymers.
Condensation Polymers
A large number of important and useful polymeric materials are not formed by chain-growth processes involving reactive species such as radicals, but proceed instead by conventional functional group transformations of polyfunctional reactants. These polymerizations often (but not always) occur with loss of a small byproduct, such as water, and generally (but not always) combine two different components in an alternating structure. The polyester Dacron and the polyamide Nylon 66, shown here, are two examples of synthetic condensation polymers, also known as step-growth polymers. In contrast to chain-growth polymers, most of which grow by carbon-carbon bond formation, step-growth polymers generally grow by carbon-heteroatom bond formation (C-O & C-N in Dacron & Nylon respectively). Although polymers of this kind might be considered to be alternating copolymers, the repeating monomeric unit is usually defined as a combined moiety.
Examples of naturally occurring condensation polymers are cellulose, the polypeptide chains of proteins, and poly(β-hydroxybutyric acid), a polyester synthesized in large quantity by certain soil and water bacteria. Formulas for these will be displayed below by clicking on the diagram.
Characteristics of Condensation Polymers
Condensation polymers form more slowly than addition polymers, often requiring heat, and they are generally lower in molecular weight. The terminal functional groups on a chain remain active, so that groups of shorter chains combine into longer chains in the late stages of polymerization. The presence of polar functional groups on the chains often enhances chain-chain attractions, particularly if these involve hydrogen bonding, and thereby crystallinity and tensile strength. The following examples of condensation polymers are illustrative.
Note that for commercial synthesis the carboxylic acid components may actually be employed in the form of derivatives such as simple esters. Also, the polymerization reactions for Nylon 6 and Spandex do not proceed by elimination of water or other small molecules. Nevertheless, the polymer clearly forms by a step-growth process. Some Condensation Polymers
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.10%3A_Polyamides_and_Polyesters-__Step-Growth_Polymers.txt |
The Mechanism of Action of β-Lactam Antibiotics
Antibiotics are specific chemical substances derived from or produced by living organisms that are capable of inhibiting the life processes of other organisms. The first antibiotics were isolated from microorganisms but some are now obtained from higher plants and animals. Over 3,000 antibiotics have been identified but only a few dozen are used in medicine. Antibiotics are the most widely prescribed class of drugs comprising 12% of the prescriptions in the United States. The penicillins were the first antibiotics discovered as natural products from the mold Penicillium.
Introduction
In 1928, Sir Alexander Fleming, professor of bacteriology at St. Mary's Hospital in London, was culturing Staphylococcus aureus. He noticed zones of inhibition where mold spores were growing. He named the mold Penicillium rubrum. It was determined that a secretion of the mold was effective against Gram-positive bacteria.
Penicillins as well as cephalosporins are called beta-lactam antibiotics and are characterized by three fundamental structural requirements: the fused beta-lactam structure (shown in the blue and red rings, a free carboxyl acid group (shown in red bottom right), and one or more substituted amino acid side chains (shown in black). The lactam structure can also be viewed as the covalent bonding of pieces of two amino acids - cysteine (blue) and valine (red).
Penicillin-G where R = an ethyl pheny group, is the most potent of all penicillin derivatives. It has several shortcomings and is effective only against gram-positive bacteria. It may be broken down in the stomach by gastric acids and is poorly and irregularly absorbed into the blood stream. In addition many disease producing staphylococci are able to produce an enzyme capable of inactivating penicillin-G. Various semisynthetic derivatives have been produced which overcome these shortcomings.
Powerful electron-attracting groups attached to the amino acid side chain such as in phenethicillin prevent acid attack. A bulky group attached to the amino acid side chain provides steric hindrance which interferes with the enzyme attachment which would deactivate the pencillins i.e. methicillin. Refer to Table 2 for the structures. Finally if the polar character is increased as in ampicillin or carbenicillin, there is a greater activity against Gram-negative bacteria.
Penicillin Mode of Action
All penicillin derivatives produce their bacteriocidal effects by inhibition of bacterial cell wall synthesis. Specifically, the cross linking of peptides on the mucosaccharide chains is prevented. If cell walls are improperly made cell walls allow water to flow into the cell causing it to burst. Resemblances between a segment of penicillin structure and the backbone of a peptide chain have been used to explain the mechanism of action of beta-lactam antibiotics. The structures of a beta-lactam antibiotic and a peptide are shown on the left for comparison. Follow the trace of the red oxygens and blue nitrogen atoms.
Gram-positive bacteria possess a thick cell wall composed of a cellulose-like structural sugar polymer covalently bound to short peptide units in layers.The polysaccharide portion of the peptidoglycan structure is made of repeating units of N-acetylglucosamine linked b-1,4 to N-acetylmuramic acid (NAG-NAM). The peptide varies, but begins with L-Ala and ends with D-Ala. In the middle is a dibasic amino acid, diaminopimelate (DAP). DAP (orange) provides a linkage to the D-Ala (gray) residue on an adjacent peptide.
The bacterial cell wall synthesis is completed when a cross link between two peptide chains attached to polysaccharide backbones is formed. The cross linking is catalyzed by the enzyme transpeptidase. First the terminal alanine from each peptide is hydrolyzed and secondly one alanine is joined to lysine through an amide bond.
Penicillin binds at the active site of the transpeptidase enzyme that cross-links the peptidoglycan strands. It does this by mimicking the D-alanyl-D-alanine residues that would normally bind to this site. Penicillin irreversibly inhibits the enzyme transpeptidase by reacting with a serine residue in the transpeptidase. This reaction is irreversible and so the growth of the bacterial cell wall is inhibited. Since mammal cells do not have the same type of cell walls, penicillin specifically inhibits only bacterial cell wall synthesis.
Bacterial Resistance
As early as the 1940s, bacteria began to combat the effectiveness of penicillin. Penicillinases (or beta-lactamases) are enzymes produced by structurally susceptable bacteria which renders penicillin useless by hydrolysing the peptide bond in the beta-lactam ring of the nucleus. Penicillinase is a response of bacterial adaptation to its adverse environment, namely the presence of a substance which inhibits its growth. Many other antibiotics are also rendered ineffective because of this same type of resistance.
Severe Allergic Shock
It is estimated that between 300-500 people die each year from penicillin-induced anaphylaxis, a severe allergic shock reaction to penicillin. In afflicted individuals, the beta-lactam ring binds to serum proteins, initiating an IgE-mediated inflammatory response. Penicillin and ala-ala peptide - Chime in new window
Cephalosporins
Cephalosporins are the second major group of beta-lactam antibiotics. They differ from penicillins by having the beta-lactam ring as a 6 member ring. The other difference, which is more significant from a medicinal chemistry stand point, is the existence of a functional group (R) at position 3 of the fused ring system. This now allows for molecular variations to effect changes in properties by diversifying the groups at position 3.
The first member of the newer series of beta-lactams was isolated in 1956 from extracts of Cephalosporium acremonium, a sewer fungus. Like penicillin, cephalosporins are valuable because of their low toxicity and their broad spectrum of action against various diseases. In this way, cephalosporin is very similar to penicillin. Cephalosporins are one of the most widely used antibiotics, and economically speaking, has about 29% of the antibiotic market. The cephalosporins are possibly the single most important group of antibiotics today and are equal in importance to penicillin.
The structure and mode of action of the cephalosporins are similar to that of penicillin. They affect bacterial growth by inhibiting cell wall synthesis, in Gram-positive and negative bacteria. Some brand names include: cefachlor, cefadroxil, cefoxitin, ceftriaxone. Cephalexin - Chime in new window
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.11%3A__Beta-Lactams-_An_Application.txt |
Glutamine synthetase
The carboxylate functional group is a very unreactive substrate for an enzyme-catalyzed acyl substitution reactions. How, then, does a living system accomplish an ‘uphill’ reaction such as the one shown below, where glutamate (a carboxylate) is converted to glutamine (an amide)?
It turns out that this conversion is not carried out directly. Rather, the first conversion is from a carboxylate (the least reactive acyl transfer substrate) to an acyl phosphate (the most reactive acyl transfer substrate). This transformation requires a reaction that we are familiar with from chapter 10: phosphorylation of a carboxylate oxygen with ATP as the phosphate donor.
Note that this is just one of the many ways that ATP is used as a energy storage unit: in order to make a high energy acyl phosphate molecule from a low energy carboxylate, the cell must ‘spend’ the energy of one ATP molecule.
The acyl phosphate version of glutamate is now ready to be converted directly to an amide (glutamine) via a nucleophilic acyl substitution reaction, as an ammonia molecule attacks the carbonyl and the phosphate is expelled.
Overall, this reaction can be written as:
Asparagine synthetase
Another common form of activated carboxylate group is an acyl adenosine phosphate. Consider another amino acid reaction, the conversion of aspartate to asparagine. In the first step, the carboxylate group of aspartate must be activated:
Once again, ATP provides the energy for driving the uphill reaction. This time, however, the activated carboxylate takes the form of an acyl adenosine (mono)phosphate. All that has happened is that the carboxylate oxygen has attacked the a-phosphate of ATP rather than the g-phosphate.
The reactive acyl-AMP version of aspartate is now ready to be converted to an amide (asparagine) via nucleophilic attack by ammonia. In the case of glutamine synthase, the source of ammonia was free ammonium ion in solution. In the case of asparagine synthase, the NH3 is derived from the hydrolysis of glutamine (this is simply another acyl substitution reaction):
The hydrolysis reaction is happening in the same enzyme active site – as the NH3 is expelled in the hydrolysis of glutamine, it immediately turns around and acts as the nucleophile in the conversion of aspartyl-AMP to asparagine:
Keep in mind that the same enzyme is also binding ATP and using it to activate aspartate – this is a busy construction zone!
Overall, this reaction can be written in condensed form as:
The use of glutamine as a ‘carrier’ for ammonia is a fairly common strategy in metabolic pathways. This strategy makes sense, as it allows cells to maintain a constant source of NH3 for reactions that require it, without the need for high solution concentrations of free ammonia.
Glycinamide ribonucleotide synthetase
One of the early steps in the construction of purine bases (the adenine and guanine bases in DNA and RNA) involves an acyl substitution reaction with an acyl phosphate intermediate. In this case, the attacking nucleophile is not ammonia but a primary amine. The strategy, however, is similar to that of glutamine synthase. The carboxylate group on glycine is converted to an acyl phosphate, at the cost of one ATP molecule. The acyl group is then transferred to 5-phosphoribosylamine, resulting in an amide product.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
22.13: Additional Exercises
General Review
22-1 Suggest a carboxylic acid and an acid derivative that could be reacted together to form the following molecule.
22-2 For the following reaction, predict the product if:
• only one equivalent of the Grignard reagent was used (and the product could be isolated)
• if two equivalents were used (followed by an acid workup)
22-3 Provide the final products of the following reactions.
22-4 Predict the final product of the following reaction.
Interconversion of Acid Derivatives by Nucleophilic Acyl Substitution
22-5 Predict the interconverted acid derivatives of the following reactions.
22-6 Predict the structure of the product and give its IUPAC nomenclature.
22-7 Choose the correct answer for the product of the following reaction.
Transesterification
22-8 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) ethyl butanoate
b) propan-2-yl butanoate
c) dipropyl carbonate
d) propyl butanoate
22-9 Explain why transesterification can be done under acidic or basic conditions.
22-10 Give the products of the following transesterification reactions.
Hydrolysis of Carboxylic Acid Derivatives
22-11 Provide the correct structure of the product of the following hydrolysis reaction.
22-12 Provide the structure of all the products resulting from the hydrolysis of the following triglyceride.
22-13 Provide the structures and IUPAC nomenclature of the products.
Reduction of Acid Derivatives
22-14 Give the structure of the product of the following reaction.
22-15 Provide the structure of all the products (including leaving groups) formed in the following reaction.
22-16 Choose the correct answer that gives the products of a fully reduced 1,3-dimethyl-1,3-diazinan-2-one.
Reactions of Acid Derivatives with Organometallic Reagents
22-17 Predict the product of the following reaction.
22-18 Choose the correct IUPAC nomenclature of the product of the following reaction.
a) 6-amino-5-chlorohexan-2-one
b) 6-amino-5-chloro-2-methylhexan-2-ol
c) 6-amino-4-chloro-2-methylhexan-2-ol
d) 4-chloro-2-methylpiperidin-2-ol
22-19 Decide whether or not the following reaction is the best way to obtain the final product. If not, suggest a better route of synthesis.
22.14: Solutions to Additional Exercises
General Review
22-1
Possible set of reactants:
22-2
22-3
22-4
Interconversion of Acid Derivatives by Nucleophilic Acyl Substitution
22-5:
22-6:
22-7:
Answer: B
Transesterification
22-8:
Answer: D
22-9:
Under acidic conditions, the carbonyl oxygen atom is protonated, making it a better electrophile for the reaction to occur. Under basic conditions, the alcohol we are trying to add is deprotonated, making it a better nucleophile.
22-10:
Hydrolysis of Carboxylic Acid Derivatives
22-11:
22-12:
22-13:
Reduction of Acid Derivatives
22-14:
22-15:
22-16:
Answer: C
Reactions of Acid Derivatives with Organometallic Reagents
22-17:
22-18:
Answer: C
22-19:
A possibly better route of synthesis: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/22%3A_Carboxylic_Acid_Derivatives_and_Nitriles/22.12%3A__Biological_Acylation_Reactions.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• predict the relative acidity of the a-hydrogens on various carbonyl compounds (section 23.1)
• explain or predict the equilibrium of enol-keto tautomers (section 23.2)
• predict the products and specify the reagents for the following reactions
• Halogenation of the a-carbon of aldehydes and ketones (section 23.3 and 23.4)
• Halogenation of the a-carbon of carboxylic acids (Hell-Vollhard-Zelinski) (section 23.3 and 23.5)
• Alkylation of the a-carbon of carbonyl compounds via the LDA pathway (section 23.3 and 23.6)
• Alkylation of the a-carbon of aldehydes and ketones via the enamine intermediate (section 23.3 and 3.7)
• Aldol addition and condensation reactions – 2 aldehydes, 2 ketones, 1 aldehyde with 1 ketone (section 23.3 and 23.8)
• Claisen condensation reactions – 2 esters or 1 ester with 1 ketone (section 23.3 and 23.9)
• Diekmann condensation reactions (intramolecular Claisen) - (section 23.9)
• Conjugate Addition a.k.a. Michael reaction (section 23.3 and 23.10)
• Robinson annulation (section 23.10)
• Decarboxylation of 3-oxocarboxylic acids (section 23.3 and 23.12)
• Malonic ester synthesis of carboxylic acids
• Acetoacetic ester synthesis of methyl ketones
Designing synthesis using all of the reactions through this chapter with an emphasis on increasing the size of the carbon backbone by forming new carbon-carbon bonds
23: Alpha Substitutions and Condensations of Carbonyl Compounds
Acidity of Alpha Hydrogens
Alkyl hydrogen atoms bonded to a carbon atom in a a (alpha) position relative to a carbonyl group display unusual acidity. While the pKa values for alkyl C-H bonds is typically on the order of 40-50, pKa values for these alpha hydrogens is more on the order of 19-20. This can most easily be explained by resonance stabilization of the product carbanion, as illustrated in the diagram below.
In the presence of a proton source, the product can either revert back into the starting ketone or aldehyde or can form a new product, the enol. The equilibrium reaction between the ketone or aldehyde and the enol form is commonly referred to as "keto-enol tautomerism". The ketone or aldehyde is generally strongly favored in this reaction.
Because carbonyl groups are sp2 hybridized the carbon and oxygen both have unhybridized p orbitals which can overlap to form the C=O $\pi$ bond.
The presence of these overlapping p orbitals gives $\alpha$ hydrogens (Hydrogens on carbons adjacent to carbonyls) special properties. In particular, $\alpha$ hydrogens are weakly acidic because the conjugate base, called an enolate, is stabilized though conjugation with the $\pi$ orbitals of the carbonyl. The effect of the carbonyl is seen when comparing the pKa for the $\alpha$ hydrogens of aldehydes (~16-18), ketones (~19-21), and esters (~23-25) to the pKa of an alkane (~50).
Of the two resonance structures of the enolate ion the one which places the negative charge on the oxygen is the most stable. This is because the negative change will be better stabilized by the greater electronegativity of the oxygen.
Relative Acidity of Alpha Hydrogens
The acidity of alpha hydrogens varies by carbonyl functional group as shown in the table below. Evaluating the stability of the conjugate bases can explain the differences in the relative acidity of the alpha hydrogens.
Exercise
1. Draw the bond line structure for each compound in the table above including all relevant resonance forms to explain the relative acidity.
Answer
1.
Contributors and Attributions
Prof. Steven Farmer (Sonoma State University)
• Clarke Earley (Department of Chemistry, Kent State University Stark Campus) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.01%3A__Relative_Acidity_of_alpha-Hydrog.txt |
Introduction
Because of the acidity of the alpha-hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
Mechanism for Enol and Enolate Formation
Under acidic conditions, the enol tautomer forms. Under basic conditions, the enolate tautomer forms. Both the enol and enolate are nucleophiles that can undergo subsequent reactions. The mechanism for both acidic and basic reaction conditions are shown below.
Acid conditions
1) Protonation of the Carbonyl
2) Enol formation
Basic conditions
1) Enolate formation
2) Protonation
For reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing SN2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pKa > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases that have been used for enolate anion formation are: NaH (sodium hydride, pKa > 45), NaNH2 (sodium amide, pKa = 34), and LiN[CH(CH3)2]2 (lithium diisopropylamide, LDA, pKa 36). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane.
Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application, one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced.
The presence of these overlapping p orbitals gives $\alpha$ hydrogens (Hydrogens on carbons adjacent to carbonyls) special properties. In particular, $\alpha$ hydrogens are weakly acidic because the conjugate base, called an enolate, is stabilized though conjugation with the $\pi$ orbitals of the carbonyl. The effect of the carbonyl is seen when comparing the pKa for the $\alpha$ hydrogens of aldehydes (~16-18), ketones (~19-21), and esters (~23-25) to the pKa of an alkane (~50).
Of the two resonance structures of the enolate ion the one which places the negative charge on the oxygen is the most stable. This is because the negative change will be better stabilized by the greater electronegativity of the oxygen.
Functional Group Structure pKa
carboxylic acid HO–(C=O)R 5
nitro RCH2–NO2 9
β-diketone * R(O=C)–CH2–(C=O)R 9
β-ketoester * R(O=C)–CH2–(C=O)OR 11
β-diester * RO(O=C)–CH2–(C=O)OR 13
amide RNH–(C=O)R 15
alcohol RCH2–OH 16
aldehyde RCH2–(C=O)H 17
ketone RCH2–(C=O)R 20
thioester RCH2–(C=O)SR 21
ester RCH2–(C=O)OR 25
nitrile RCH2–C≡N 25
sulfone RCH2–SO2R 25
amide RCH2–(C=O)N(CH3)2 30
alkane CH3–R 50
* Note methylene groups bridging between two electron withdrawing groups are more acidic than alpha protons next to only one carbonyl group.
Examples
If the formed enolate is stabilized by more than one carbonyl it is possible to use a weaker base such as sodium ethoxide.
NaOCH2CH3 = Na+ -OCH2CH3 = NaOEt
Because of the acidity of α hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
Exercises
2. Draw the enol forms of the following molecules
1. 4-methylcyclohexanone
2. Ethyl thioactetate
3. Methyl acetate
4. Butanal
5. Propionic Acid
6. 1-phenyl-2-butanone
3. How many acid protons do each of the molecules from the previous question have? Label them.
4. Draw all of the monoenol forms for the following molecule. Which ones are most stable? Why?
Answers
2.
(a)
(b)
(c)
(d)
(e)
(f)
3.
(a)
(b)
(c)
(d)
(e)
(f)
4.
The ability to resonate stabilizes this enol form.
This enol has no resonance forms and is therefore less stable. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.02%3A_Enols_Enolate_Ions_and_Tautomeriz.txt |
Overview
The reactivity of the alpha-carbon can be grouped into three main categories: alpha-substitution, condensation, and decarboxylation. Alpha,beta-unsaturated carbonyls can undergo conjugate addition reactions that are called Michael Additions when the nucleophile is an alpha-carbon. Because the reactants, reagents and products can contain multiple functional groups, it is helpful to initially focus on the overall conversions between functional groups BEFORE digging into the details of each reaction pathway. It is also useful to label the alpha and beta carbons to help follow the reactivity. An overview of the reactions that will be studied in this chapter are shown below.
23.04: Alpha Halogenation of Carbonyls
Ketones with alpha hydrogens can undergo a substitution reaction with halogens. This reaction occurs because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate. For this reaction Cl2, Br2 or I2 can be used as the halogens.
General reaction
Example 1
Aldehydes are oxidized by the halogens so this reaction pathway is not synthetically useful. For example, when benzaldehyde is added to either set of reagents described above for ketones, the majore product is benzoic acid.
Acid Catalyzed Mechanism
Under acidic conditions the reaction occurs thought the formation of an enol which then reacts with the halogen.
1) Protonation of the carbonyl
2) Enol formation
3) SN2 reaction
4) Deprotonation
Kinetic studies provide some evidence for the mechanism shown above. The rate law for the alpha-halogenation of a ketone can be given by:
rate = [ketone][H+]
The implication is that the rate determining step is dependent on the concentrations of the ketone and acid catalyst and therefore associated with the enol formation part of the mechanism. The halogen does not even appear in the rate law. Indeed, the overall rate is completely independent of the concentration of the halogen and suggests the halogenation step occurs rapidly.
Base Promoted Mechanism
Under basic conditions the enolate forms and then reacts with the halogen.
Note! This is base promoted and not base catalyzed because an entire equivalent of base is required.
It is difficult to stop the base promoted reaction after a single substitution, so acidic conditons are used when a monohalo product is required.
1) Enolate formation
2) SN2 reaction
The Haloform Qualitative Reaction to Identify Methyl Ketones
The overreaction during base promotion of alpha halogenation is used as a qualitative test called the haloform reaction to identify methyl ketones. Under basic conditions, subsequent halogenation reactions occur because the halogenated product is more reactive than the starting material due to the electron withdrawing effect of the halogen. The halogen inductively stabilizes the conjugate base and increases the relative acidity of the remaining alpha-carbons. Halogenations occur at the alpha-carbon until the haloform becomes a leaving group and is observed as a precipitate as shown in the example below.
Deuterium Exchange
Due to the acidic nature of α hydrogens they can be exchanged with deuterium by reaction with D2O (heavy water). The process is accelerated by the addition of an acid or base; an excess of D2O is required. The end result is the complete exchange of all α hydrogens with deuteriums.
Example 2
Mechanism in basic conditions
1) Enolate Formation
2) Deuteration
Example Question
Draw the product for the following reactions.
Solutions to example question
Exercises
5. Draw the products of the following reactions
6. Draw out the mechanism for the following reaction.
7. How might you form 2-hepten-4-one from 4-heptanone?
Answer
5.
6.
7. 1) Br2, H3O+; 2) Pyridine, Heat | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.03%3A_Reaction_Overview.txt |
Although the alpha bromination of some carbonyl compounds, such as aldehydes and ketones, can be accomplished with Br2 under acidic conditions, the reaction will generally not occur with acids, esters, and amides. This is because only aldehydes and ketones enolize to a sufficient extent to allow the reaction to occur. However, carboxylic acids, can be brominated in the alpha position with a mixture of Br2 and PBr3 in a reaction called the Hell-Volhard-Zelinskii (HVZ) reaction.
For example, pentanoic acid can be converted to 2-bromopentanoic acid as shown in the example below.
The mechanism of this reaction involves an acid bromide enol instead of the expected carboxylic acid enol. The reaction starts with the reaction of the carboxylic acid with PBr3 to form the acid bromide and HBr. The HBr then catalyzes the formation of the acid bromide enol which subsequently reacts with Br2 to give alpha bromination. Lastly, the acid bromide reacts with water to reform the carboxylic acid.
Exercise
8. Draw the bond-line structure for the product of each reaction below.
Answer
8.
23.06: Alkylation of the alpha-Carbon vi
Alpha Alkylation
A strong base, such as lithium diisopropyl amide (LDA), sodium hydride, or sodium amide, creates the nucleophilic enolate ion which reacts with an alkyl halide suitable for the SN2 reactivity to form an alpha-alkylated product.
Example 1: Alpha Alkylation
Mechanism
The mechanism begins with enolate formation. The resulting enolate is the nucleophile in an SN2 reaction with a suitable alkyl halide.
1) Enolate formation
2) SN2 reaction
Example Question
Write the structure of the product for the following reactions.
Enolate of Unsymmetrical Carbonyl Compounds
Now let’s consider what happens when an unsymmetrical carbonyl is treated with a base. In the case displayed below there are two possible enolates which can form. The removal of the 2o hydrogen forms the kinetic enolate and is formed faster because it is less substituted and thereby less sterically hindered. The removal of the 3o hydrogen forms the thermodynamic enolate which is more stable because it is more substituted.
Kinetic Enolates
Kinetic enolates are formed when a strong bulky base like LDA is used. The bulky base finds the 2o hydrogen less sterically hindered and preferable removes it. Low temperature are typically used when forming the kinetic enolate to prevent equilibration to the more stable thermodynamic enolate. Typically a temperature of -78 oC is used.
Thermodynamic Enolates
The thermodynamic enolate is favored by conditions which allow for equilibration. The thermodynamic enolate is usually formed by using a strong base at room temperature. At equilibrium the lower energy of the thermodynamic enolate is preferred, so that the more stable, more stubstituted enolate is formed.
Exercises
9. How might you prepare the following compounds from an alkylation reaction?
(a)
(b)
(c)
(d)
(e)
(f)
Answer
9.
(a)
(b)
(c)
(d)
(e)
(f)
23.07: Alkylation of the Alpha-Carbon v
Overview of the Stork Enamine Reaction
The reaction conditions for the direct alkylation of the alpha carbon with LDA or other very strong base are quite harsh. Many organic compounds cannot withstand the reaction environment at synthetically useful amounts. Therefore, an alternate synthetic pathway was developed by Gilbert Stork of Columbia University. Some of the advantages of using an enamine over an enolate are that enamines are neutral, easier to prepare, and usually prevent the overreaction problems plagued by enolates. As shown in the example below, the aldehyde or ketone can be recovered from the enamine via a hydrolysis reaction.
Example
Reversible
Enamines act as nucleophiles in a fashion similar to enolates. Because of this, enamines can be used as synthetic equivalents as enolates in many reactions. This process requires a three steps: 1) Formation of the enamine, 2) Reaction with an eletrophile to form an iminium salt, 3) Hydrolysis of the iminium salt to reform the aldehyde or ketone.
Typically we use the following 2o amines for enamine reactions
Alkylation of an Enamine
Enamined undergo an SN2 reaction with reactive alkyl halides to give the iminium salt. The iminium salt can be hydrolyzed back into the carbonyl.
Individual steps
1) Formation of an enamine
2) SN2 Alkylation
3) Reform the carbonyl by hydrolysis
All three steps together:
Exercises
10. Draw the product of the reaction with the enamine prepared from cyclopentanone and pyrrolidine, and the following molecules.
(a)
(b)
(c)
11. Propose a synthesis for the following compounds via an enamine.
(a)
(b)
Answers
10.
(a)
(b)
(c)
11.
(a) cyclopentanone enamine + 2-cyanopropene
(b) cyclohexanone enamine + ethyl acrylate | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.05%3A_Bromination_of_Acids-_The_HVZ_Rea.txt |
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. Due to the carbanion like nature of enolates they can add to carbonyls in a similar manner as Grignard reagents. For this reaction to occur at least one of the reactants must have alpha hydrogens.
The aldol reactions for acetaldehyde and acetone are shown as examples.
Example: Aldol Reactions
Aldol Reaction Mechanism
Step 1: Enolate formation
Step 2: Nucleophilic reaction by the enolate
Step 3: Protonation
Aldol Condensation: the dehydration of aldol products to synthesize α, β unsaturated carbonyls (enones)
The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this $\beta$-elimination reaction is an α,β-unsaturated aldehyde or ketone. Base-catalyzed elimination occurs with heating. The additional stability provided by the conjugated carbonyl system of the product makes some aldol reactions thermodynamically driven and mixtures of stereoisomers (E & Z) are obtained from some reactions. Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water, are termed Condensations. Hence, the following examples are properly referred to as aldol condensations. Overall the general reaction involves a dehydration of an aldol product to form an alkene:
Going from reactants to products simply
Example: Aldol Condensation from an Aldol Reaction Product
1) Form enolate
2) Form enone
Aldol Condensation Acid Catalyzed Mechanism
Under acidic conditions an enol is formed and the hydroxy group is protonated. Water is expelled by either and E1 or E2 reaction.
When performing both reactions together always consider the aldol product first then convert to the enone. Note! The double bond always forms in conjugation with the carbonyl.
Example: Aldol Condensation Directly from the Ketones or Aldehydes
Aldol Reactions in Multiple Step Synthesis
Aldol reactions are excellent methods for the synthesis of many enones or beta hydroxy carbonyls. Because of this, being able to predict when an aldol reaction might be used in a synthesis in an important skill. This accomplished by mentally breaking apart the target molecule and then considering what the starting materials might be.
Fragments which are easily made by an aldol reaction
Steps to 'reverse' the aldol reaction (from the final aldol product towards identifying the starting compounds).
1) From an enone break the double bond and form two single bonds. Place and OH on the bond furthest from the carbonyl and an H on the bond closest to the carbonyl.
2) From the aldol product break the C-C bond between the alpha carbon and the carbon attached to the OH. Then turn the OH into a carbonyl and add an hydrogen to the other carbon.
Example: Determining the Reactant when given the Aldol Condensation Product
What reactant must be used to make the following molecule using an aldol condensation?
Solution
Mixed Aldol Reactions and Condensations
The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called crossed or mixed reactions, and under certain conditions such crossed aldol condensations can be effective.
Example: Mixed Aldol Reaction (One Product)
The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this most mixed aldol reactions are usually not performed unless one reactant has no alpha hydrogens.
The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult.
ACH2CHO + BCH2CHO + NaOH AA + BB + AB + BA
Example: Products of a Mixed Aldol Reaction
The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction.
Example: Claisen-Schmidt Reaction
Intramolecular aldol reaction
Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. In most cases two sets of $\alpha$ hydrogens need to be considered. As with most ring forming reaction five and six membered rings are preferred (less ring strain).
As with other aldol reaction the addition of heat causes an aldol condensation to occur.
Exercise
12. Draw the bond-line structures for the products of the reactions below. Note: One of the reactions is a poorly designed aldol condensation producing four different products.
Answer
12.
23.09: The Claisen Condensation Reaction
Because esters can contain alpha hydrogens, they can undergo a condensation reaction similar to the aldol reaction called a Claisen Condensation. In a fashion similar to the aldol, one ester acts as a nucleophile while a second ester acts as the electrophile. During the reaction a new carbon-carbon bond is formed. The product is a β-keto ester. A major difference with the aldol reaction is the fact that hydroxide cannot be used as a base because it could possibly react with the ester. Instead, an alkoxide version of the alcohol used to synthesize the ester is used to prevent transesterification side products.
Claisen Condensation
The Claisen condensation reactions of methyl acetate and methyl propanoate are shown as examples.
Example: Claisen Condensation
Claisen Condensation Mechanism
1) Enolate formation
2) Nucleophilic reaction
3) Removal of leaving group
Crossed Claisen Condensation
Claisen condensations between different ester reactants are called Crossed Claisen reactions. Crossed Claisen reactions in which both reactants can serve as donors and acceptors generally give complex mixtures. Because of this most Crossed Claisen reactions are usually not performed unless one reactant has no alpha hydrogens.
Example: Crossed Claisen Condensation
Dieckmann Condensation
A diester can undergo an intramolecular reaction called a Dieckmann condensation.
Example: Dieckman Condensation
Mechanism
1. Dieckmann, W. Ber. Dtsch. Chem. Ges. 1894, 27, 102–103.
Exercise
13. Draw the bond-line structures for the products of the following reactions.
Answer
13. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/23%3A_Alpha_Substitutions_and_Condensations_of_Carbonyl_Compounds/23.08%3A_The_Aldol_Reaction_and_Condensati.txt |
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