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learning objective
• determine the dominant intermolecular forces (IMFs) of organic compounds
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.
The properties of liquids are intermediate between those of gases and solids but are more similar to solids.
Intermolecular forces determine bulk properties such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid.
Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures. (For more information on the behavior of real gases and deviations from the ideal gas law,.)
In this section, we explicitly consider three kinds of intermolecular interactions: There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. The first two are often described collectively as van der Waals forces.
Dipole–Dipole Interactions
Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure \(\PageIndex{1a}\).
These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\PageIndex{1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\PageIndex{1b}\), are attractive intermolecular interactions, whereas those in Figure \(\PageIndex{1d}\) are repulsive intermolecular interactions. Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(2\). On average, however, the attractive interactions dominate.
Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/r, where r is the distance between the ions. Doubling the distance (r → 2r) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/r3, so doubling the distance between the dipoles decreases the strength of the interaction by 23, or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure, whereas \(\ce{NaCl}\), which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(1\).
Table \(1\): Relationships between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Molar Mass
Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K)
C3H6 (cyclopropane) 42 0 240
CH3OCH3 (dimethyl ether) 46 1.30 248
CH3CN (acetonitrile) 41 3.9 355
The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.
Example
Arrange ethyl methyl ether (CH3OCH2CH3), 2-methylpropane [isobutane, (CH3)2CHCH3], and acetone (CH3COCH3) in order of increasing boiling points. Their structures are as follows:
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.
Solution:
The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds.
The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point.
Ethyl methyl ether has a structure similar to H2O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point.
Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point.
Thus we predict the following order of boiling points:
2-methylpropane < ethyl methyl ether < acetone.
This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.
Exercise
Arrange carbon tetrafluoride (CF4), ethyl methyl sulfide (CH3SC2H5), dimethyl sulfoxide [(CH3)2S=O], and 2-methylbutane [isopentane, (CH3)2CHCH2CH3] in order of decreasing boiling points.
Answer
dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)
London Dispersion Forces
Thus far we have considered only interactions between polar molecules, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(2\)).
What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances.
Table \(2\): Normal Melting and Boiling Points of Some Elements and Nonpolar Compounds
Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C)
Ar 40 −189.4 −185.9
Xe 131 −111.8 −108.1
N2 28 −210 −195.8
O2 32 −218.8 −183.0
F2 38 −219.7 −188.1
I2 254 113.7 184.4
CH4 16 −182.5 −161.5
Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(3\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/r6. Doubling the distance therefore decreases the attractive energy by 26, or 64-fold.
Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H2 molecules in part (b) in Figure \(3\), tends to become more pronounced as atomic and molecular masses increase (Table \(2\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones.
For similar substances, London dispersion forces get stronger with increasing molecular size.
The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(4\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(4\) shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C5H12. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C).
All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.
Example
Arrange n-butane, propane, 2-methylpropane [isobutene, (CH3)2CHCH3], and n-pentane in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Determine the intermolecular forces in the compounds and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n-butane (−0.5°C) < n-pentane (36.1°C).
Exercise
Arrange GeH4, SiCl4, SiH4, CH4, and GeCl4 in order of decreasing boiling points.
Answer
GeCl4 (87°C) > SiCl4 (57.6°C) > GeH4 (−88.5°C) > SiH4 (−111.8°C) > CH4 (−161°C)
Hydrogen Bonds
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(5\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H2Te and H2Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.
Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure \(6\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water rather than sinks.
Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cagelike structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.
Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.
Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze.
Example
Considering CH3OH, C2H6, Xe, and (CH3)3N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Given: compounds
Asked for: formation of hydrogen bonds and structure
Strategy:
1. Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.
2. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.
Solution:
A Of the species listed, xenon (Xe), ethane (C2H6), and trimethylamine [(CH3)3N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.
B The one compound that can act as a hydrogen bond donor, methanol (CH3OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows:
Exercise
Considering CH3CO2H, (CH3)3N, NH3, and CH3F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Answer
CH3CO2H and NH3;
Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH3. Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.
Example: Buckyballs
Arrange C60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N2O in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N2O have very similar molar masses (40 and 44 g/mol, respectively), but N2O is polar while Ar is not. Consequently, N2O should have a higher boiling point. A C60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N2O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses:
He (−269°C) < Ar (−185.7°C) < N2O (−88.5°C) < C60 (>280°C) < NaCl (1465°C).
Exercise
Arrange 2,4-dimethylheptane, Ne, CS2, Cl2, and KBr in order of decreasing boiling points.
Answer
KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS2 (46.6°C) > Cl2 (−34.6°C) > Ne (−246°C)
Example
Identify the most significant intermolecular force in each substance.
1. C3H8
2. CH3OH
3. H2S
Solution
a. Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular force for this substance would be dispersion forces.
b. This molecule has an H atom bonded to an O atom, so it will experience hydrogen bonding.
c. Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is bent, so it has a permanent dipole. The most significant force in this substance is dipole-dipole interaction.
Exercise
Identify the most significant intermolecular force in each substance.
1. HF
2. HCl
Answer a
hydrogen bonding
Answer b
dipole-dipole interactions
More complex examples of hydrogen bonding
The hydration of negative ions
When an ionic substance dissolves in water, water molecules cluster around the separated ions. This process is called hydration. Water frequently attaches to positive ions by co-ordinate (dative covalent) bonds. It bonds to negative ions using hydrogen bonds.
If you are interested in the bonding in hydrated positive ions, you could follow this link to co-ordinate (dative covalent) bonding.
The diagram shows the potential hydrogen bonds formed to a chloride ion, Cl-. Although the lone pairs in the chloride ion are at the 3-level and would not normally be active enough to form hydrogen bonds, in this case they are made more attractive by the full negative charge on the chlorine.
However complicated the negative ion, there will always be lone pairs that the hydrogen atoms from the water molecules can hydrogen bond to.
Hydrogen bonding in alcohols
An alcohol is an organic molecule containing an -OH group. Any molecule which has a hydrogen atom attached directly to an oxygen or a nitrogen is capable of hydrogen bonding. Such molecules will always have higher boiling points than similarly sized molecules which don't have an -O-H or an -N-H group. The hydrogen bonding makes the molecules "stickier", and more heat is necessary to separate them.
Ethanol, CH3CH2OH, and methoxymethane, CH3OCH3, are structural isomers with the same molecular formula, C2H6O.
They have the same number of electrons, and a similar length to the molecule. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be much the same. However, ethanol has a hydrogen atom attached directly to an oxygen - and that oxygen still has exactly the same two lone pairs as in a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge.
In methoxymethane, lone pairs on the oxygen are still there, but the hydrogens are not sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur. The boiling points of ethanol and methoxymethane show the dramatic effect that the hydrogen bonding has on the stickiness of the ethanol molecules:
ethanol (with hydrogen bonding) 78.5°C
methoxymethane (without hydrogen bonding) -24.8°C
The hydrogen bonding in the ethanol has lifted its boiling point about 100°C. It is important to realize that hydrogen bonding exists in addition to van der Waals attractions. For example, all the following molecules contain the same number of electrons, and the first two are much the same length. The higher boiling point of the butan-1-ol is due to the additional hydrogen bonding.
Comparing the two alcohols (containing -OH groups), both boiling points are high because of the additional hydrogen bonding due to the hydrogen attached directly to the oxygen - but they are not the same. The boiling point of the 2-methylpropan-1-ol isn't as high as the butan-1-ol because the branching in the molecule makes the van der Waals attractions less effective than in the longer butan-1-ol.
Hydrogen bonding in organic molecules containing nitrogen
Hydrogen bonding also occurs in organic molecules containing N-H groups - in the same sort of way that it occurs in ammonia. Examples range from simple molecules like CH3NH2 (methylamine) to large molecules like proteins and DNA. The two strands of the famous double helix in DNA are held together by hydrogen bonds between hydrogen atoms attached to nitrogen on one strand, and lone pairs on another nitrogen or an oxygen on the other one.
Donors and Acceptors
In order for a hydrogen bond to occur there must be both a hydrogen donor and an acceptor present. The donor in a hydrogen bond is the atom to which the hydrogen atom participating in the hydrogen bond is covalently bonded, and is usually a strongly electronegative atom such as N,O, or F. The hydrogen acceptor is the neighboring electronegative ion or molecule, and must posses a lone electron pair in order to form a hydrogen bond.
Why does a hydrogen bond occur?
Since the hydrogen donor is strongly electronegative, it pulls the covalently bonded electron pair closer to its nucleus, and away from the hydrogen atom. The hydrogen atom is then left with a partial positive charge, creating a dipole-dipole attraction between the hydrogen atom bonded to the donor, and the lone electron pair on the accepton. This results in a hydrogen bond.(see Interactions Between Molecules With Permanent Dipoles)
Types of hydrogen bonds
Hydrogen bonds can occur within one single molecule, between two like molecules, or between two unlike molecules.
Intramolecular hydrogen bonds
Intramolecular hydrogen bonds are those which occur within one single molecule. This occurs when two functional groups of a molecule can form hydrogen bonds with each other. In order for this to happen, both a hydrogen donor an acceptor must be present within one molecule, and they must be within close proximity of each other in the molecule. For example, intramolecular hydrogen bonding occurs in ethylene glycol (C2H4(OH)2) between its two hydroxyl groups due to the molecular geometry.
Intermolecular hydrogen bonds
Intermolecular hydrogen bonds occur between separate molecules in a substance. They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present an in positions in which they can interact.For example, intermolecular hydrogen bonds can occur between NH3 molecules alone, between H2O molecules alone, or between NH3 and H2O molecules.
Properties and effects of hydrogen bonds
On Boiling Point
When we consider the boiling points of molecules, we usually expect molecules with larger molar masses to have higher normal boiling points than molecules with smaller molar masses. This, without taking hydrogen bonds into account, is due to greater dispersion forces (see Interactions Between Nonpolar Molecules). Larger molecules have more space for electron distribution and thus more possibilities for an instantaneous dipole moment. However, when we consider the table below, we see that this is not always the case.
Compound Molar Mass Normal Boiling Point
\(H_2O\) 18 g/mol 373 K
\(HF\) 20 g/mol 292.5 K
\(NH_3\) 17 g/mol 239.8 K
\(H_2S\) 34 g/mol 212.9 K
\(HCl\) 36.4 g/mol 197.9 K
\(PH_3\) 34 g/mol 185.2 K
We see that H2O, HF, and NH3 each have higher boiling points than the same compound formed between hydrogen and the next element moving down its respective group, indicating that the former have greater intermolecular forces. This is because H2O, HF, and NH3 all exhibit hydrogen bonding, whereas the others do not. Furthermore, \(H_2O\) has a smaller molar mass than HF but partakes in more hydrogen bonds per molecule, so its boiling point is consequently higher.
On Viscosity
The same effect that is seen on boiling point as a result of hydrogen bonding can also be observed in the viscosity of certain substances. Those substances which are capable of forming hydrogen bonds tend to have a higher viscosity than those that do not. Substances which have the possibility for multiple hydrogen bonds exhibit even higher viscosities.
Factors preventing Hydrogen bonding
Electronegativity
Hydrogen bonding cannot occur without significant electronegativity differences between hydrogen and the atom it is bonded to. Thus, we see molecules such as PH3, which no not partake in hydrogen bonding. PH3 exhibits a trigonal pyramidal molecular geometry like that of ammmonia, but unlike NH3 it cannot hydrogen bond. This is due to the similarity in the electronegativities of phosphorous and hydrogen. Both atoms have an electronegativity of 2.1, and thus, no dipole moment occurs. This prevents the hydrogen bonding from acquiring the partial positive charge needed to hydrogen bond with the lone electron pair in another molecule. (see Polarizability)
Atom Size
The size of donors and acceptors can also effect the ability to hydrogen bond. This can account for the relatively low ability of Cl to form hydrogen bonds. When the radii of two atoms differ greatly or are large, their nuclei cannot achieve close proximity when they interact, resulting in a weak interaction.
Hydrogen Bonding in Nature
Hydrogen bonding plays a crucial role in many biological processes and can account for many natural phenomena such as the Unusual properties of Water. In addition to being present in water, hydrogen bonding is also important in the water transport system of plants, secondary and tertiary protein structure, and DNA base pairing.
Plants
The cohesion-adhesion theory of transport in vascular plants uses hydrogen bonding to explain many key components of water movement through the plant's xylem and other vessels. Within a vessel, water molecules hydrogen bond not only to each other, but also to the cellulose chain which comprises the wall of plant cells. This creates a sort of capillary tube which allows for capillary action to occur since the vessel is relatively small. This mechanism allows plants to pull water up into their roots. Furthermore,hydrogen bonding can create a long chain of water molecules which can overcome the force of gravity and travel up to the high altitudes of leaves.
Proteins
Hydrogen bonding is present abundantly in the secondary structure of proteins, and also sparingly in tertiary conformation. The secondary structure of a protein involves interactions (mainly hydrogen bonds) between neighboring polypeptide backbones which contain Nitrogen-Hydrogen bonded pairs and oxygen atoms. Since both N and O are strongly electronegative, the hydrogen atoms bonded to nitrogen in one polypeptide backbone can hydrogen bond to the oxygen atoms in another chain and visa-versa. Though they are relatively weak,these bonds offer great stability to secondary protein structure because they repeat a great number of times.
In tertiary protein structure,interactions are primarily between functional R groups of a polypeptide chain; one such interaction is called a hydrophobic interaction. These interactions occur because of hydrogen bonding between water molecules around the hydrophobe and further reinforce conformation.
• Jose Pietri | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.10%3A_Intermolecular_Forces_%28IMFs%29_-_Review.txt |
Learning Objective
• predict the relative boil points of organic compounds
Intermolecular forces (IMFs) can be used to predict relative boiling points. The stronger the IMFs, the lower the vapor pressure of the substance and the higher the boiling point. Therefore, we can compare the relative strengths of the IMFs of the compounds to predict their relative boiling points.
H-bonding > dipole-dipole > London dispersion (van der Waals)
When comparing compounds with the same IMFs, we use size and shape as tie breakers since the London dispersion forces increase as the surface area increases. Since all compounds exhibit some level of London dispersion forces and compounds capable of H-bonding also exhibit dipole-dipole, we will use the phrase "dominant IMF" to communicate the IMF most responsible for the physical properties of the compound.
In the table below, we see examples of these relationships. When comparing the structural isomers of pentane (pentane, isopentane, and neopentane), they all have the same molecular formula C5H12. However, as the carbon chain is shortened to create the carbon branches found in isopentane and neopentane the overall surface area of the molecules decreases. The visual image of MO theory can be helpful in seeing each compound as a cloud of electrons in an all encompassing MO system. Branching creates more spherical shapes noting that the sphere allows the maximum volume with the least surface area. The structural isomers with the chemical formula C2H6O have different dominant IMFs. The H-bonding of ethanol results in a liquid for cocktails at room temperature, while the weaker dipole-dipole of the dimethylether results in a gas a room temperature. In the last example, we see the three IMFs compared directly to illustrate the relative strength IMFs to boiling points.
Boiling points and melting points
The observable melting and boiling points of different organic molecules provides an additional illustration of the effects of noncovalent interactions. The overarching principle involved is simple: the stronger the noncovalent interactions between molecules, the more energy that is required, in the form of heat, to break them apart. Higher melting and boiling points signify stronger noncovalent intermolecular forces.
Consider the boiling points of increasingly larger hydrocarbons. More carbons means a greater surface area possible for hydrophobic interaction, and thus higher boiling points.
As you would expect, the strength of intermolecular hydrogen bonding and dipole-dipole interactions is reflected in higher boiling points. Just look at the trend for hexane (nonpolar London dispersion interactions only ), 3-hexanone (dipole-dipole interactions), and 3-hexanol (hydrogen bonding).
Of particular interest to biologists (and pretty much anything else that is alive in the universe) is the effect of hydrogen bonding in water. Because it is able to form tight networks of intermolecular hydrogen bonds, water remains in the liquid phase at temperatures up to 100 OC, (slightly lower at high altitude). The world would obviously be a very different place if water boiled at 30 OC.
Exercise
1. Based on their structures, rank phenol, benzene, benzaldehyde, and benzoic acid in terms of lowest to highest boiling point.
Solution
By thinking about noncovalent intermolecular interactions, we can also predict relative melting points. All of the same principles apply: stronger intermolecular interactions result in a higher melting point. Ionic compounds, as expected, usually have very high melting points due to the strength of ion-ion interactions (there are some ionic compounds, however, that are liquids at room temperature). The presence of polar and especially hydrogen-bonding groups on organic compounds generally leads to higher melting points. Molecular shape, and the ability of a molecule to pack tightly into a crystal lattice, has a very large effect on melting points. The flat shape of aromatic compounds such as napthalene and biphenyl allows them to stack together efficiently, and thus aromatics tend to have higher melting points compared to alkanes or alkenes with similar molecular weights.
Comparing the melting points of benzene and toluene, you can see that the extra methyl group on toluene disrupts the molecule's ability to stack, thus decreasing the cumulative strength of intermolecular London dispersion forces.
Note also that the boiling point for toluene is 111 oC, well above the boiling point of benzene (80 oC). The key factor for the boiling point trend in this case is size (toluene has one more carbon), whereas for the melting point trend, shape plays a much more important role. This makes sense when you consider that melting involves ‘unpacking’ the molecules from their ordered array, whereas boiling involves simply separating them from their already loose (liquid) association with each other.
If you are taking an organic lab course, you may have already learned that impurities in a crystalline substance will cause the observed melting point to be lower compared to a pure sample of the same substance. This is because impurities disrupt the ordered packing arrangement of the crystal, and make the cumulative intermolecular interactions weaker.
The melting behavior of lipid structures
An interesting biological example of the relationship between molecular structure and melting point is provided by the observable physical difference between animal fats like butter or lard, which are solid at room temperature, and vegetable oils, which are liquid. Both solid fats and liquid oils are based on a ‘triacylglycerol’ structure, where three hydrophobic hydrocarbon chains of varying length are attached to a glycerol backbone through an ester functional group (compare this structure to that of the membrane lipids discussed in section 2.4B).
Interactive 3D image of a saturated triacylglycerol (BioTopics)
Saturated vs mono-unsaturated fatty acid (BioTopics)
In vegetable oils, the hydrophobic chains are unsaturated, meaning that they contain one or more double bonds. Solid animal fat, in contrast, contains saturated hydrocarbon chains, with no double bonds. The double bonds in vegetable oils cause those hydrocarbon chains to be more rigid, and ‘bent’ at an angle (remember that rotation is restricted around double bonds), with the result that they don’t pack together as closely, and thus can be broken apart (ie. melted) more readily. Shown in the figure above is a polyunsaturated fatty acid chain (two double bonds), and you can click on the link to see interactive images of a saturated fatty acid compared to a monounsaturated fatty acid (one double bond).
Exercise
2. Arrange the following compounds in order of decreasing boiling point.
Answer
2.
Contributors
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)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.11%3A_Intermolecular_Forces_and_Relative_Boiling_Point.txt |
Learning Objective
• predict whether a mixture of compounds will a form homogeneous or heterogeneous solution
The type of intermolecular forces (IMFs) exhibited by compounds can be used to predict whether two different compounds can be mixed to form a homogeneous solution (soluble or miscible). Because organic chemistry can perform reactions in non-aqueous solutions using organic solvents. It is important to consider the solvent as a reaction parameter and the solubility of each reagent. With this said, solvent effects are secondary to the sterics and electrostatics of the reactants. Make sure that you do not drown in the solvent.
Solubility
Virtually all of the organic chemistry that you will see in this course takes place in the solution phase. In the organic laboratory, reactions are often run in nonpolar or slightly polar solvents such as toluene (methylbenzene), hexane, dichloromethane, or diethylether. In recent years, much effort has been made to adapt reaction conditions to allow for the use of ‘greener’ (in other words, more environmentally friendly) solvents such as water or ethanol, which are polar and capable of hydrogen bonding. In organic reactions that occur in the cytosolic region of a cell, the solvent is of course water. It is critical for any organic chemist to understand the factors which are involved in the solubility of different molecules in different solvents.
You probably remember the rule you learned in general chemistry regarding solubility: ‘like dissolves like’ (and even before you took any chemistry at all, you probably observed at some point in your life that oil does not mix with water). Let’s revisit this old rule, and put our knowledge of covalent and noncovalent bonding to work.
Imagine that you have a flask filled with water, and a selection of substances that you will test to see how well they dissolve in the water. The first substance is table salt, or sodium chloride. As you would almost certainly predict, especially if you’ve ever inadvertently taken a mouthful of water while swimming in the ocean, this ionic compound dissolves readily in water. Why? Because water, as a very polar molecule, is able to form many ion-dipole interactions with both the sodium cation and the chloride anion, the energy from which is more than enough to make up for energy required to break up the ion-ion interactions in the salt crystal and some water-water hydrogen bonds.
The end result, then, is that in place of sodium chloride crystals, we have individual sodium cations and chloride anions surrounded by water molecules – the salt is now in solution. Charged species as a rule dissolve readily in water: in other words, they are very hydrophilic (water-loving).
Now, we’ll try a compound called biphenyl, which, like sodium chloride, is a colorless crystalline substance (the two compounds are readily distinguishable by sight, however – the crystals look quite different).
Biphenyl does not dissolve at all in water. Why is this? Because it is a very non-polar molecule, with only carbon-carbon and carbon-hydrogen bonds. It is able to bond to itself very well through nonpolar (London dispersion) interactions, but it is not able to form significant attractive interactions with the very polar solvent molecules. Thus, the energetic cost of breaking up the biphenyl-to-biphenyl interactions in the solid is high, and very little is gained in terms of new biphenyl-water interactions. Water is a terrible solvent for nonpolar hydrocarbon molecules: they are very hydrophobic ('water-fearing').
Next, you try a series of increasingly large alcohol compounds, starting with methanol (1 carbon) and ending with octanol (8 carbons).
You find that the smaller alcohols - methanol, ethanol, and propanol - dissolve easily in water. This is because the water is able to form hydrogen bonds with the hydroxyl group in these molecules, and the combined energy of formation of these water-alcohol hydrogen bonds is more than enough to make up for the energy that is lost when the alcohol-alcohol hydrogen bonds are broken up. When you try butanol, however, you begin to notice that, as you add more and more to the water, it starts to form its own layer on top of the water.
The longer-chain alcohols - pentanol, hexanol, heptanol, and octanol - are increasingly non-soluble. What is happening here? Clearly, the same favorable water-alcohol hydrogen bonds are still possible with these larger alcohols. The difference, of course, is that the larger alcohols have larger nonpolar, hydrophobic regions in addition to their hydrophilic hydroxyl group. At about four or five carbons, the hydrophobic effect begins to overcome the hydrophilic effect, and water solubility is lost.
Now, try dissolving glucose in the water – even though it has six carbons just like hexanol, it also has five hydrogen-bonding, hydrophilic hydroxyl groups in addition to a sixth oxygen that is capable of being a hydrogen bond acceptor.
We have tipped the scales to the hydrophilic side, and we find that glucose is quite soluble in water.
We saw that ethanol was very water-soluble (if it were not, drinking beer or vodka would be rather inconvenient!) How about dimethyl ether, which is a constitutional isomer of ethanol but with an ether rather than an alcohol functional group? We find that diethyl ether is much less soluble in water. Is it capable of forming hydrogen bonds with water? Yes, in fact, it is –the ether oxygen can act as a hydrogen-bond acceptor. The difference between the ether group and the alcohol group, however, is that the alcohol group is both a hydrogen bond donor and acceptor.
The result is that the alcohol is able to form more energetically favorable interactions with the solvent compared to the ether, and the alcohol is therefore more soluble.
Here is another easy experiment that can be done (with proper supervision) in an organic laboratory. Try dissolving benzoic acid crystals in room temperature water – you'll find that it is not soluble. As we will learn when we study acid-base chemistry in a later chapter, carboxylic acids such as benzoic acid are relatively weak acids, and thus exist mostly in the acidic (protonated) form when added to pure water.
Acetic acid, however, is quite soluble. This is easy to explain using the small alcohol vs large alcohol argument: the hydrogen-bonding, hydrophilic effect of the carboxylic acid group is powerful enough to overcome the hydrophobic effect of a single methyl group on acetic acid, but not the larger hydrophobic effect of the 6-carbon benzene group on benzoic acid.
Now, try slowly adding some aqueous sodium hydroxide to the flask containing undissolved benzoic acid. As the solvent becomes more and more basic, the benzoic acid begins to dissolve, until it is completely in solution.
What is happening here is that the benzoic acid is being converted to its conjugate base, benzoate. The neutral carboxylic acid group was not hydrophilic enough to make up for the hydrophobic benzene ring, but the carboxylate group, with its full negative charge, is much more hydrophilic. Now, the balance is tipped in favor of water solubility, as the powerfully hydrophilic anion part of the molecule drags the hydrophobic part, kicking and screaming, (if a benzene ring can kick and scream) into solution. If you want to precipitate the benzoic acid back out of solution, you can simply add enough hydrochloric acid to neutralize the solution and reprotonate the carboxylate.
If you are taking a lab component of your organic chemistry course, you will probably do at least one experiment in which you will use this phenomenon to separate an organic acid like benzoic acid from a hydrocarbon compound like biphenyl.
Similar arguments can be made to rationalize the solubility of different organic compounds in nonpolar or slightly polar solvents. In general, the greater the content of charged and polar groups in a molecule, the less soluble it tends to be in solvents such as hexane. The ionic and very hydrophilic sodium chloride, for example, is not at all soluble in hexane solvent, while the hydrophobic biphenyl is very soluble in hexane.
Exercise
1. Vitamins can be classified as water-soluble or fat-soluble (consider fat to be a very non-polar, hydrophobic 'solvent'. Decide on a classification for each of the vitamins shown below.
Solutions
Exercise
2. Both aniline and phenol are insoluble in pure water. Predict the solubility of these two compounds in 10% aqueous hydrochloric acid, and explain your reasoning. Hint – in this context, aniline is basic, phenol is not!
Solutions
Illustrations of solubility concepts: metabolic intermediates, lipid bilayer membranes, soaps and detergents
Because water is the biological solvent, most biological organic molecules, in order to maintain water-solubility, contain one or more charged functional groups. These are most often phosphate, ammonium or carboxylate, all of which are charged when dissolved in an aqueous solution buffered to pH 7.
Sugars often lack charged groups, but as we discussed in our ‘thought experiment’ with glucose, they are quite water-soluble due to the presence of multiple hydroxyl groups.
Some biomolecules, in contrast, contain distinctly nonpolar, hydrophobic components. The ‘lipid bilayer’ membranes of cells and subcellular organelles serve to enclose volumes of water and myriad biomolecules in solution. The lipid (fat) molecules that make up membranes are amphipathic: they have a charged, hydrophilic ‘head’ and a hydrophobic hydrocarbon ‘tail’.
interactive 3D image of a membrane phospholipid (BioTopics)
Notice that the entire molecule is built on a ‘backbone’ of glycerol, a simple 3-carbon molecule with three alcohol groups. In a biological membrane structure, lipid molecules are arranged in a spherical bilayer: hydrophobic tails point inward and bind together by London dispersion forces, while the hydrophilic head groups form the inner and outer surfaces in contact with water.
Interactive 3D Image of a lipid bilayer (BioTopics)
Because the interior of the bilayer is extremely hydrophobic, biomolecules (which as we know are generally charged species) are not able to diffuse through the membrane– they are simply not soluble in the hydrophobic interior. The transport of molecules across the membrane of a cell or organelle can therefore be accomplished in a controlled and specific manner by special transmembrane transport proteins, a fascinating topic that you will learn more about if you take a class in biochemistry.
A similar principle is the basis for the action of soaps and detergents. Soaps are composed of fatty acids, which are long (typically 18-carbon), hydrophobic hydrocarbon chains with a (charged) carboxylate group on one end,
Fatty acids are derived from animal and vegetable fats and oils. In aqueous solution, the fatty acid molecules in soaps will spontaneously form micelles, a spherical structure that allows the hydrophobic tails to avoid contact with water and simultaneously form favorable London dispersion contacts.
Interactive 3D images of a fatty acid soap molecule and a soap micelle (Edutopics)
Because the outside of the micelle is charged and hydrophilic, the structure as a whole is soluble in water. Micelles will form spontaneously around small particles of oil that normally would not dissolve in water (like that greasy spot on your shirt from the pepperoni slice that fell off your pizza), and will carry the particle away with it into solution. We will learn more about the chemistry of soap-making in a later chapter (section 12.4B).
Synthetic detergents are non-natural amphipathic molecules that work by the same principle as that described for soaps. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.12%3A_Intermolecular_Forces_and_Solubilities.txt |
Learning Objective
• distinguish between organic compounds that are H-bond donors versus H-bond acceptors
H-bond donors vs H-bond acceptors
Compounds with H-bonding as their dominant intermolecular force (IMF) are BOTH H-bond donors and H-bond acceptors. They are H-bond donors because they have a highly polar hydrogen atom bonded to a strongly electronegative atom, primarily nitrogen, oxygen, or fluorine (NOF). Because there is an equivalent partial negative charge on the atom bonded to hydrogen (mostly NOF), this atom can accept H-bonds from another atoms. Since H-bond donors are ALWAYS H-bond acceptors, we simplify communication to "H-bond donor". There are two H-bonding interactions for H-bond donors. The strongly electronegative elements (primarily nitrogen, oxygen, and fluorine) will always form a relatively large partial negative charge when bonded with carbon. These elements can accept H-bonds when they are part of the organic molecule. In this situation, there is only one H-bonding interaction. The diagram below illustrates the similarities and differences between H-bond donors and H-bond acceptors. Water and alcohols may serve as both donors and acceptors, whereas ethers, aldehydes, ketones and esters can function only as acceptors. Similarly, primary and secondary amines are both donors and acceptors, but tertiary amines function only as acceptors.
Exercise
1. Classify the compounds below as H-bond donors, H-bond acceptors, or neither.
Answer
1.
Comparing Physical Properties of H-bond Donors vs H-bond Acceptors
Once we are able to recognize compounds that can exhibit intermolecular hydrogen bonding, the relatively high boiling points they exhibit become understandable. The data in the following table serve to illustrate this point.
Compound Formula Mol. Wt. Boiling Point Melting Point
dimethyl ether CH3OCH3 46 –24ºC –138ºC
ethanol CH3CH2OH 46 78ºC –130ºC
propanol CH3(CH2)2OH 60 98ºC –127ºC
diethyl ether (CH3CH2)2O 74 34ºC –116ºC
propyl amine CH3(CH2)2NH2 59 48ºC –83ºC
methylaminoethane CH3CH2NHCH3 59 37ºC
trimethylamine (CH3)3N 59 3ºC –117ºC
ethylene glycol HOCH2CH2OH 62 197ºC –13ºC
acetic acid CH3CO2H 60 118ºC 17ºC
ethylene diamine H2NCH2CH2NH2 60 118ºC 8.5ºC
Alcohols boil cosiderably higher than comparably sized ethers (first two entries), and isomeric 1º, 2º & 3º-amines, respectively, show decreasing boiling points, with the two hydrogen bonding isomers being substantially higher boiling than the 3º-amine (entries 5 to 7). Also, O–H---O hydrogen bonds are clearly stronger than N–H---N hydrogen bonds, as we see by comparing propanol with the amines.
As expected, the presence of two hydrogen bonding functions in a compound raises the boiling point even further. Acetic acid (the ninth entry) is an interesting case. A dimeric species, shown on the right, held together by two hydrogen bonds is a major component of the liquid state. If this is an accurate representation of the composition of this compound then we would expect its boiling point to be equivalent to that of a C4H8O4 compound (formula weight = 120). A suitable approximation of such a compound is found in tetramethoxymethane, (CH3O)4C, which is actually a bit larger (formula weight = 136) and has a boiling point of 114ºC. Thus, the dimeric hydrogen bonded structure appears to be a good representation of acetic acid in the condensed state.
A related principle is worth noting at this point. Although the hydrogen bond is relatively weak (ca. 4 to 5 kcal per mole), when several such bonds exist the resulting structure can be quite robust. The hydrogen bonds between cellulose fibers confer great strength to wood and related materials. For additional information on this subject Click Here. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.14%3A_Organic_Functional_Groups-_H-bond_donors_and_H-b.txt |
Hybridization
2-1
2-2
Longest to shortest bond length: b > a > c
Strongest to weakest bond: c > a > b
2-3
Sigma bonds: 7
Pi bonds: 0
2-4
Sigma bonds: 5
Pi bonds: 1
2-5
Sigma bonds: 3
Pi bonds: 2
Hybridization, Electron Geometry, and Molecular Shape
2-6 Correct answer is (b) sp3, tetrahedral.
2-7
a) Tetrahedral
b) Trigonal bipyramidal
c) Tetrahedral
d) Trigonal planar
2-8
2-9
2-10 Boron has trigonal planar geometry. The hydrogen atoms are at a 120° angle from each other to be as far apart as possible.
Bond Rotation
2-11 This molecule can rotate freely around the middle bond as there are no major steric hindrance interactions.
2-12 This molecule cannot rotate freely around the middle bond as the large bromine substituents attached at the ortho positions of the benzene rings experience significant steric hindrance with each other.
2-13 No; the pi-bond prevents free rotation about the C=C bond.
Polarity of Bonds and Molecules
2-14
2-15
a) 2
b) No dipole moment
c) 1
2-16 True
Intermolecular Forces (IMFs)
2-17
a) Cannot H-bond
b) Can H-bond
c) Can H-bond
d) Can H-bond
e) Cannot H-bond
f) Cannot H-bond
2-18
2-19
a) London Dispersion Forces
b) Dipole-Dipole Interactions
c) Ionic Forces
d) Hydrogen bonding
IMFs and Solubility
2-20
a) Not miscible
b) Miscible
c) Not miscible
d) Soluble
e) Not soluble
f) Soluble
2-21 Caffeine will dissolve in dichloromethane (DCM) significantly more than in hexanes as DCM is a more polar solvent and caffeine is a polar molecule (like dissolves like).
Hydrocarbons and an Introduction to Isomerism
2-22
a) Alkene
b) Alkane
c) Alkyne
d) Alkane
e) Alkene
f) Alkene
2-23
2-24
2-25 It does not have cis/trans configuration, as the triple bond in the compound (CH3)2CHC☰CCH3 holds the four carbons in a straight line due to the sp hybridization of the middle two carbons (which have a linear geometric configuration).
Organic Compounds with Oxygen
2-26
a) Ether
b) Ketone
c) Carboxylic acid
d) Alcohol and Amine
e) Amide
f) Ether and Alkene
2-27
a) Alcohol and Amine (We will learn that the most correct classification for hydroxyl groups bonded to benzene rings is phenol)
b) Alcohol, Ether, Ketone, Amine and Alkene
c) Ester, Ether, Amine and Alkene
2-28
a) Aldehyde and carboxylic acid
b) Alcohol, Ketone, Amine
c) Alcohol, Ketone, Carboxylic acid
2-29
Organic Compounds with Nitrogen
2-30 Compound B has a slight dipole moment due to the cis configuration of the amine groups. Since it has a dipole moment, it experiences dipole-dipole interactions in addition to hydrogen bonding, thus increasing its boiling point.
2-31
2.16: Additional Exercises
Hybridization
2-1 For each of the following compounds, identify the hybridization of each carbon or nitrogen atom with an arrow pointed at it.
2-2 Rank the following bonds in order of decreasing bond length. Then rank the bonds in order from strongest to weakest.
2-3 How many sigma and pi bonds are in a molecule of ethane (C2H6)?
2-4 How many sigma and pi bonds are in a molecule of ethylene (C2H4)?
2-5 How many sigma and pi bonds are in a molecule of acetylene (C2H2)?
Hybridization, Electron Geometry, and Molecular Shape
2-6 What is the hybridization state and geometry of the carbon atom in methane (CH4)?
a) sp, linear
b) sp3, tetrahedral
c) sp2, trigonal planar
d) None of the above
2-7 Identify the electron geometry of the following compounds.
a) H2O
b) PF5
c) NH4+
d) The carbonyl carbon of acetone (CH3)2CO. (Note that double bonds between carbon and oxygen must be recognized.)
2-8 For the following compounds, identify which atoms have sp2 hybridization.
2-9 Draw the orbitals showing the geometric shape of ammonia (NH3). Identify its geometric shape.
2-10 What is the geometric shape of the boron atom in BH3? What is the bond angle of the hydrogen atoms?
Bond Rotation
2-11 Will the following compound experience free rotation around the middle bond? Explain why or why not.
2-12 Will the following compound experience free rotation around the middle bond? Explain why or why not.
2-13 Can a molecule of ethylene experience free rotation around the C=C bond?
Polarity of Bonds and Molecules
2-14 For the following compounds, draw an arrow to show the direction of the dipole moment (if any).
2-15 In the following pairs of compounds, identify the compound with the larger dipole moment (if any).
2-16 True or False: Generally, the larger the difference in electronegativity of connected atoms, the greater the dipole moment.
Intermolecular Forces (IMFs)
2-17 Identify which of the following compounds can form hydrogen bonds.
2-18 For the compounds in the previous problem ( 2-17 above) that can hydrogen bond, draw how they can form those bonds.
2-19 Identify what type of intermolecular force the following compounds experience.
IMFs and Solubility
2-20 Identify whether the following compounds are miscible or soluble in water.
2-21 Identify which solvent, hexanes or dichloromethane (DCM), would be the better solvent to dissolve 3.0 grams of caffeine. Explain your answer.
Hydrocarbons and an Introduction to Isomerism
2-22 Identify whether the following hydrocarbons are alkanes, alkenes, or alkynes.
2-23 For the following compounds, identify the hybridization state of each labeled carbon.
2-24 Draw all possible isomers for the following compounds.
a) C4H10
b) C6H14
c) C3H6
2-25 Does (CH3)2CHCCCH3 show cis/trans isomerism? Explain why or why not.
Organic Compounds with Oxygen
2-26 Identify the functional group(s) of each compound.
2-27 What functional groups are found in the following compounds?
2-28 What oxygen-containing functional groups are present in the following compounds. The nitrogen-containing group is a challenge question.
2-29 Identify whether the functional groups on the following compounds are classified correctly. If not, give the correct classification.
Organic Compounds with Nitrogen
2-30 First, identify which of the following compounds has a dipole moment. Then, predict which of the following compounds will have the higher boiling point and explain why.
2-31 Identify whether the functional groups of the following compounds are classified correctly. If not, give the correct classification. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.15%3A_Solutions_to_Additional_Exercises.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• use R groups to draw generic functional groups - refer to section 3.1
• name alkanes, cycloalkanes, alkenes, alkynes, alkyl halides, ethers, alcohols, amines, benzene and its derivatives, aldehydes, ketones, amines, carboxylic acids, and carboxylic acid derivatives using IUPAC (systematic) and selected common name nomenclature - refer to sections 3.2 - 3.14
• draw the structure of alkanes, cycloalkanes, alkenes, alkynes, alkyl halides, ethers, alcohols, amines, benzene and its derivatives, aldehydes, ketones, amines, carboxylic acids, and carboxylic acid derivatives from the IUPAC (systematic) and selected common names - refer to sections 3.2 - 3.14
• classify alkyl halides, alcohols and amines - refer to sections 3.5, 3.8, and 3.12 respectively.
A Pearl of Wisdom: Most common names were derived from older systems of nomenclature that some may argue were "not systematic at all". However, it is helpful to note that the older systems of nomenclature were often based on shared structural features and/or chemical reactivity. Learning carefully selected common names can offer insights into chemical reactivity and structural patterns. Additionally, there are some common names that are so prevalent, they need to be memorized.
Please note: The nomenclature for organic compounds with sulfur and phosphorus are introduced so that students can interpret a given name and draw the correct structure. Derivation of names can be required by the professor and requires additional instruction.
• 3.1: Generic (Abbreviated) Structures (aka R Groups)
It is not always necessary to draw the entire structure of a compound. The correct use of the "R group" is explained.
• 3.2: Overview of the IUPAC Naming Strategy
The International Union of Pure and Applied Chemistry (IUPAC) names for organic compounds all follow the same set of rules and can have up to four parts. Recognizing the overall pattern can simplify the learning process.
• 3.3: Alkanes
Alkanes form the carbon backbone of all organic compounds.
• 3.4: Cycloalkanes
The rotational limits of cycloalkanes introduce stereochemistry to some compounds. Some disubstituted cycloalkanes can exist as geometric isomers (cis/trans).
• 3.5: Haloalkane - Classification and Nomenclature
The reactivity of the alkyl halides (haloalkanes) can be predicted using their structural classifications of primary, secondary, or tertiary.
• 3.6: Alkenes
The rigid, carbon-carbon double bond (C=C) can also introduce stereochemical considerations in naming.
• 3.7: Alkynes
The linear geometry of carbon-carbon triple bonds simplifies the names of this functional group.
• 3.8: 3.8 Alcohols - Classification and Nomenclature
Alcohols are organic compounds with hydroxyl groups as a unique structural feature. Alcohol classification will be useful as we learn patterns of chemical reactivity. The terms "vicinal" and "germinal" are also explained.
• 3.9: Ethers, Epoxides and Sulfides
Ethers, epoxides, and sulfides all have the heteroatom disrupting the continuous carbon chain. There is no IUPAC suffix for ethers. The alkoxy group is always a substituent.
• 3.10: Benzene and its Derivatives
This section focuses on naming benzene derivatives and the important distinction between a phenyl group and a benzyl group. Phenol nomenclature and with other important benzene derivatives are discussed.
• 3.11: Aldehydes and Ketones
While there are many functional groups that include the carbonyl structural feature, aldehydes and ketones are collectively referred to as "the carbonyls." With their highly similar chemical reactivity, their nomenclature is taught together.
• 3.12: Amines - Classification and Nomenclature
Amines are important weak bases, organic reactants. and biologically active compounds, such as alkaloids. Amine classification is helpful in recognizing patterns of chemical reactivity.
• 3.13: Carboxylic Acids
Carboxylic acids are important natural products and synthetic precursors. The common names are provided for carboxylic acids (1 to 10 carbons) and selected dicarboxylic acids.
• 3.14: The Carboxylic Acid Derivatives
All of these functional groups can be hydrolyzed to form carboxylic acids, so they are collective called the carboxylic acid derivatives.
• 3.15: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 3.16: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
• 3.17: Appendix - IUPAC Nomenclature Rules
This appendix provides a link to the full text of the IUPAC rules of nomenclature for organic compounds.
03: Functional Groups and Nomenclature
learning objective
• use R groups to draw generic functional groups - refer to section 3.1
Drawing Generic (abbreviated) Organic Structures
In chapter 2, we learned to recognize and distinguish between organic functional groups. Often when drawing organic structures, chemists find it convenient to use the letter 'R' to designate part of a molecule outside of the region of interest. "R" represents the "Rest of the Molecule". If we just want to refer in general to a functional group without drawing a specific molecule, for example, we can use 'R groups' to focus attention on the group of interest:
The 'R' group is a convenient way to abbreviate the structures of large biological molecules, especially when we are interested in something that is occurring specifically at one location on the molecule. For example, in chapter 15 when we look at biochemical oxidation-reduction reactions involving the flavin molecule, we will abbreviate a large part of the flavin structure which does not change at all in the reactions of interest:
As an alternative, we can use a 'break' symbol to indicate that we are looking at a small piece or section of a larger molecule. This is used commonly in the context of drawing groups on large polymers such as proteins or DNA.
Finally, 'R' groups can be used to concisely illustrate a series of related compounds, such as the family of penicillin-based antibiotics.
Using abbreviations appropriately is a very important skill to develop when studying organic chemistry in a biological context, because although many biomolecules are very large and complex (and take forever to draw!), usually we are focusing on just one small part of the molecule where a change is taking place.
As a rule, you should never abbreviate any atom involved in a bond-breaking or bond-forming event that is being illustrated: only abbreviate that part of the molecule which is not involved in the reaction of interest.
For example, carbon #2 in the reactant/product below most definitely is involved in bonding changes, and therefore should not be included in the 'R' group.
If you are unsure whether to draw out part of a structure or abbreviate it, the safest thing to do is to draw it out.
Exercise
1. a) If you intend to draw out the chemical details of a reaction in which the methyl ester functional group of cocaine (see earlier figure) was converted to a carboxylate plus methanol, what would be an appropriate abbreviation to use for the cocaine structure (assuming that you only wanted to discuss the chemistry specifically occurring at the ester group)?
b) Below is the (somewhat complicated) reaction catalyzed by an enzyme known as 'Rubisco', by which plants 'fix' carbon dioxide. Carbon dioxide and the oxygen of water are colored red and blue respectively to help you see where those atoms are incorporated into the products. Propose an appropriate abbreviation for the starting compound (ribulose 1,5-bisphosphate), using two different 'R' groups, R1 and R2.
Solutions to exercises
3.02: Overview of the IUPAC Naming Strategy
learning objectives
• name alkanes, cycloalkanes, alkenes, alkynes, alkyl halides, ethers, alcohols, amines, benzene and its derivatives, aldehydes, ketones, amines, carboxylic acids, and carboxylic acid derivatives using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkanes, cycloalkanes, alkenes, alkynes, alkyl halides, ethers, alcohols, amines, benzene and its derivatives, aldehydes, ketones, amines, carboxylic acids, and carboxylic acid derivatives from the IUPAC (systematic) and selected common names
Overview of the IUPAC System for Naming Organic Compounds
The International Union of Pure and Applied Chemistry (IUPAC) has established the rules of nomenclature of all chemical compounds. IUPAC nomenclature can also be called "systematic" nomenclature because there is an overall system and structure to the names. This section provides an overview of the general naming strategy and structure for organic compounds.
Naming organic compounds according to the IU{AC system requires up to four pieces of information
1. recognize & prioritize the functional group(s) present
2. identify & number the longest continuous carbon chain to give the highest ranking group the lowest possible number
3. cite the substituents (branches) alphabetically using the numbering determined above
4. recognize & classify any stereochemistry (E/Z, R/S, cis/trans, etc)
With these four pieces of information, the IUPAC name is written using the format below. This same format applies to ALL the organic compounds.
Recognize & Prioritize the Functional Group(s) Present
The IUPAC Rules of Organic Nomenclature assume that the following table is understood and memorized.
Identify & Number the Longest Continuous Carbon Chain with the Highest Priority Group
The longest continuous carbon chain (parent) is named using the Homologous Series, as well as any carbon branches. The suffixes and location within the name distinguish between the parent and the branches.
When Alkenes and Alkynes have Lower Priority
The hydrocarbon suffixes differ in the letter preceeding the "n". When alkenes and alkynes occur in compounds with higher priority functional groups, then the distinction between hydrocarbons is communicated with a single letter: "e", or "y" for alkenes and alkynes, respectively, An example is shown below to illustrate the application of this rule.
substituents (branches) alphabetically
The major substituents are listed above and need to be memorized. There are a few additional substituents that will introduced later in the text.
Stereochemistry
Distinguishing spatial orientations of atoms (stereochemistry) are communicated at the beginning of the name using the appropriate symbols, such as E/Z, R/S, cis/trans, etc. This nomenclature will be discussed when it is possible for a functional group. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.01%3A_Generic_%28Abbreviated%29_Structures_%28aka_R_Groups%29.txt |
learning objectives
• name alkanes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkanes from IUPAC (systematic) and selected common names
Alkanes are hydrocarbons that can be described by the general formula CnH2n+2. They consist only of carbon and hydrogen and contain only single bonds. Alkanes are also known as "saturated hydrocarbons."
The following table contains the systematic names for the first twenty straight chain alkanes. It will be important to familiarize yourself with these names because they will be the basis for naming many other organic molecules throughout your course of study.
Name Molecular Formula Condensed Structural Formula
Methane CH4 CH4
Ethane C2H6 CH3CH3
Propane C3H8 CH3CH2CH3
Butane C4H10 CH3(CH2)2CH3
Pentane C5H12 CH3(CH2)3CH3
Hexane C6H14 CH3(CH2)4CH3
Heptane C7H16 CH3(CH2)5CH3
Octane C8H18 CH3(CH2)6CH3
Nonane C9H20 CH3(CH2)7CH3
Decane C10H22 CH3(CH2)8CH3
Undecane C11H24 CH3(CH2)9CH3
Dodecane C12H26 CH3(CH2)10CH3
Tridecane C13H28 CH3(CH2)11CH3
Tetradecane C14H30 CH3(CH2)12CH3
Pentadecane C15H32 CH3(CH2)13CH3
Hexadecane C16H34 CH3(CH2)14CH3
Heptadecane C17H36 CH3(CH2)15CH3
Octadecane C18H38 CH3(CH2)16CH3
Nonadecane C19H40 CH3(CH2)17CH3
Eicosane C20H42 CH3(CH2)18CH3
Carbon Atom Classifications
To assign the prefixes sec-, which stands for secondary, and tert-, for tertiary, it is important that we first learn how to classify carbon atoms. If a carbon is attached to only one other carbon, it is called a primary carbon. If a carbon is attached to two other carbons, it is called a seconday carbon. A tertiary carbon is attached to three other carbons and last, a quaternary carbon is attached to four other carbons. These terms are summarixed with an example in the table below.
Classification Example
methyl CH4
primary
secondary
tertiary
Using Common Names with Branched Alkanes
Certain branched alkanes have common names that are still widely used today. These common names make use of prefixes, such as iso-, sec-, tert-, and neo-.
Isoalkanes
The prefix iso-, which stands for isomer, is commonly given to 2-methyl alkanes. In other words, if there is methyl group located on the second carbon of a carbon chain, we can use the prefix iso-. The prefix will be placed in front of the alkane name that indicates the total number of carbons as in isopentane which is the same as 2-methylbutane and isobutane which is the same as 2-methylpropane. The pattern is illustrated below
Sec- and Tert-alkanes
Secondary and tertiary alkanes can be further distinguished from their "iso-counter parts" by applying comparing the carbon classifications. The common names for three and four carbon branches are summarized below. Notice that the "iso" prefix is joined directly to the alkyl name. When alphabetizing branches, the "i" is considered. For "sec" and "tert", the prefix is separated from the alkyl name and is NOT considered when alphabetizing branches.
For some compounds, the common names bring a simple a simple elegance to the experience.
Neo-alkanes
A five carbon alkane and the corresponding five carbon branch can form a structural pattern commonly known as neopentane and neopentyl respectively. The prefix neo- can also be applied to larger alkanes as shown below.
Alkyl Groups
An alkyl group is formed by removing one hydrogen from the alkane chain and is described by the formula CnH2n+1. The removal of this hydrogen results in a stem change from -ane to -yl. Take a look at the following examples.
The same approach can be used with any of the alkanes in the table above and with common names.
Alkyl Group Common Names that sound like Alkenes
In long hydrocarbon chains there can be many -CH2- groups in a series. These internal -(CH2)n- groups are names using the homologous series stem with the suffice "ene" even though there are NO carbon-carbon double bonds present. For example, the common name for dichloromethane, CH2Cl2, is methylene chloride; and the common names for the major ingredients in antifreeze are ethylene glycol (CH2(OH)CH2OH) and propylene glycol (CH2(OH)CH2(OH)CH3), in spite of the fact that there are no carbon-carbon double bonds in any of these three compounds.
Three Basic Principles of Naming
1. Choose the longest, most substituted carbon chain containing a functional group.
2. A carbon bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number.
3. Take the alphabetical order into consideration; that is, after applying the first two rules given above, make sure that your substitutes and/or functional groups are written in alphabetical order.
Example
Solution
Rule #1 Choose the longest, most substituted carbon chain containing a functional group. This example does not contain any functional groups, so we only need to be concerned with choosing the longest, most substituted carbon chain. The longest carbon chain has been highlighted in red and consists of eight carbons.
Rule #2 Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. Because this example does not contain any functional groups, we only need to be concerned with the two substitutes present, that is, the two methyl groups. If we begin numbering the chain from the left, the methyls would be assigned the numbers 4 and 7, respectively. If we begin numbering the chain from the right, the methyls would be assigned the numbers 2 and 5. Therefore, to satisfy the second rule, numbering begins on the right side of the carbon chain as shown below. This gives the methyl groups the lowest possible numbering.
In this example, there is no need to utilize the third rule. Because the two substitutes are identical, neither takes alphabetical precedence with respect to numbering the carbons. This concept will become clearer in the next example.
Example
Solution
Rule #1 Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2 Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. In this example, numbering the chain from the left or the right would satisfy this rule. If we number the chain from the left, bromine and chlorine would be assigned the second and sixth carbon positions, respectively. If we number the chain from the right, chlorine would be assigned the second position and bromine would be assigned the sixth position. In other words, whether we choose to number from the left or right, the functional groups occupy the second and sixth positions in the chain. To select the correct numbering scheme, we need to utilize the third rule.
Rule #3 After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Example
Solution
Rule #1 Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine, and one substitute, the methyl group. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2 Carbons bonded to a functional group must have the lowest possible carbon number. After taking functional groups into consideration, any substitutes present must have the lowest possible carbon number. This particular example illustrates the point of difference principle. If we number the chain from the left, bromine, the methyl group and chlorine would occupy the second, fifth and sixth positions, respectively. This concept is illustrated in the second drawing below. If we number the chain from the right, chlorine, the methyl group and bromine would occupy the second, third and sixth positions, respectively, which is illustrated in the first drawing below. The position of the methyl, therefore, becomes a point of difference. In the first drawing, the methyl occupies the third position. In the second drawing, the methyl occupies the fifth position. To satisfy the second rule, we want to choose the numbering scheme that provides the lowest possible numbering of this substitute. Therefore, the first of the two carbon chains shown below is correct.
Therefore, the first numbering scheme is the appropriate one to use.
Once you have determined the correct numbering of the carbons, it is often useful to make a list, including the functional groups, substitutes, and the name of the parent chain.
Parent chain: heptane 2-Chloro 3-Methyl 6-Bromo
6-bromo-2-chloro-3-methylheptane
Exercises
Write the IUPAC (systematic) name for each of the compounds below. The parent chains have numbered for the first two compounds to help you begin.
a)
b)
c)
Solutions
a) 9-chloro-7-ethyl-2,2,4-tromethyldecane
b) 3-chloro-5-ethyl-4,4-dimethylheptane
c) 2-bromo-6-ethyloctane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.03%3A_Alkanes.txt |
learning objectives
• name cycloalkanes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of cycloalkanes from IUPAC (systematic) and selected common names
Cycloalkanes are cyclic hydrocarbons, meaning that the carbons of the molecule are arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds). There are also polycyclic alkanes, which are molecules that contain two or more cycloalkanes that are joined, forming multiple rings.
Introduction
Many organic compounds found in nature or created in a laboratory contain rings of carbon atoms with distinguishing chemical properties; these compounds are known as cycloalkanes. Cycloalkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds, but in cycloalkanes, the carbon atoms are joined in a ring. The smallest cycloalkane is cyclopropane.
If you count the carbons and hydrogens, you will see that they no longer fit the general formula \(C_nH_{2n+2}\). By joining the carbon atoms in a ring,two hydrogen atoms have been lost. The general formula for a cycloalkane is \(C_nH_{2n}\). Cyclic compounds are not all flat molecules. All of the cycloalkanes, from cyclopentane upwards, exist as "puckered rings". Cyclohexane, for example, has a ring structure that looks like this:
In addition to being saturated cyclic hydrocarbons, cycloalkanes may have multiple substituents or functional groups that further determine their unique chemical properties. The most common and useful cycloalkanes in organic chemistry are cyclopentane and cyclohexane, although other cycloalkanes varying in the number of carbons can be synthesized. Understanding cycloalkanes and their properties are crucial in that many of the biological processes that occur in most living things have cycloalkane-like structures.
Glucose (6 carbon sugar) Ribose (5 carbon sugar)
Cholesterol (polycyclic)
Although polycyclic compounds are important, they are highly complex and typically have common names accepted by IUPAC. However, the common names do not generally follow the basic IUPAC nomenclature rules. The general formula of the cycloalkanes is \(C_nH_{2n}\) where \(n\) is the number of carbons. The naming of cycloalkanes follows a simple set of rules that are built upon the same basic steps in naming alkanes. Cyclic hydrocarbons have the prefix "cyclo-".
Contents
For simplicity, cycloalkane molecules can be drawn in the form of skeletal structures in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens.
same as same as
Cycloalkane Molecular Formula Basic Structure
Cyclopropane C3H6
Cyclobutane C4H8
Cyclopentane C5H10
Cyclohexane C6H12
Cycloheptane C7H14
Cyclooctane C8H16
Cyclononane C9H18
Cyclodecane C10H20
IUPAC Rules for Nomenclature
1. Determine the cycloalkane to use as the parent chain. The parent chain is the one with the highest number of carbon atoms. If there are two cycloalkanes, use the cycloalkane with the higher number of carbons as the parent chain.
2. If there is an alkyl straight chain that has a greater number of carbons than the cycloalkane, then the alkyl chain must be used as the primary parent chain. Cycloalkane acting as a substituent to an alkyl chain has an ending "-yl" and, therefore, must be named as a cycloalkyl.
Cycloalkane Cycloalkyl
cyclopropane cyclopropyl
cyclobutane cyclobutyl
cyclopentane cyclopentyl
cyclohexane cyclohexyl
cycloheptane cycloheptyl
cyclooctane cyclooctyl
cyclononane cyclononanyl
cyclodecane cyclodecanyl
Example \(1\):
The longest straight chain contains 10 carbons, compared with cyclopropane, which only contains 3 carbons. Because cyclopropane is a substituent, it would be named a cyclopropyl-substituted alkane.
3) Determine any functional groups or other alkyl groups.
4) Number the carbons of the cycloalkane so that the carbons with functional groups or alkyl groups have the lowest possible number. A carbon with multiple substituents should have a lower number than a carbon with only one substituent or functional group. One way to make sure that the lowest number possible is assigned is to number the carbons so that when the numbers corresponding to the substituents are added, their sum is the lowest possible.
(1+3=4) NOT (1+5=6)
5) When naming the cycloalkane, the substituents and functional groups must be placed in alphabetical order.
(ex: 2-bromo-1-chloro-3-methylcyclopentane)
6) Indicate the carbon number with the functional group with the highest priority according to alphabetical order. A dash"-" must be placed between the numbers and the name of the substituent. After the carbon number and the dash, the name of the substituent can follow. When there is only one substituent on the parent chain, indicating the number of the carbon atoms with the substituent is not necessary.
(ex: 1-chlorocyclohexane or cholorocyclohexane is acceptable)
7) If there is more than one of the same functional group on one carbon, write the number of the carbon two, three, or four times, depending on how many of the same functional group is present on that carbon. The numbers must be separated by commas, and the name of the functional group that follows must be separated by a dash. When there are two of the same functional group, the name must have the prefix "di". When there are three of the same functional group, the name must have the prefix "tri". When there are four of the same functional group, the name must have the prefix "tetra". However, these prefixes cannot be used when determining the alphabetical priorities.
There must always be commas between the numbers and the dashes that are between the numbers and the names.
Example \(2\)
(2-bromo-1,1-dimethylcyclohexane)
Notice that "f" of fluoro alphabetically precedes the "m" of methyl. Although "di" alphabetically precedes "f", it is not used in determining the alphabetical order.
Example 3
(2-fluoro-1,1,-dimethylcyclohexane NOT 1,1-dimethyl-2-fluorocyclohexane)
8) If the substituents of the cycloalkane are related by the cis or trans configuration, then indicate the configuration by placing "cis-" or "trans-" in front of the name of the structure.
Blue=Carbon Yellow=Hydrogen Green=Chlorine
Notice that chlorine and the methyl group are both pointed in the same direction on the axis of the molecule; therefore, they are cis.
cis-1-chloro-2-methylcyclopentane
9) After all the functional groups and substituents have been mentioned with their corresponding numbers, the name of the cycloalkane can follow.
Summary
1. Determine the parent chain: the parent chain contains the most carbon atoms.
2. Number the substituents of the chain so that the sum of the numbers is the lowest possible.
3. Name the substituents and place them in alphabetical order.
4. If stereochemistry of the compound is shown, indicate the orientation as part of the nomenclature.
5. Cyclic hydrocarbons have the prefix "cyclo-" and have an "-alkane" ending unless there is an alcohol substituent present. When an alcohol substituent is present, the molecule has an "-ol" ending.
Glossary
• alkyl: A structure that is formed when a hydrogen atom is removed from an alkane.
• cyclic: Chemical compounds arranged in the form of a ring or a closed chain form.
• cycloalkanes: Cyclic saturated hydrocarbons with a general formula of CnH(2n). Cycloalkanes are alkanes with carbon atoms attached in the form of a closed ring.
• functional groups: An atom or groups of atoms that substitute for a hydrogen atom in an organic compound, giving the compound unique chemical properties and determining its reactivity.
• hydrocarbon: A chemical compound containing only carbon and hydrogen atoms.
• saturated: All of the atoms that make up a compound are single bonded to the other atoms, with no double or triple bonds.
• skeletal structure: A simplified structure in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens.
Exercises
1. Name the following structures. (Note: The structures are complex for practice purposes and may not be found in nature.)
a) b) c) d) e) f)
g)
2. Draw the following structures.
a) 1,1-dibromo-5-fluoro-3-butyl-7-methylcyclooctane
b) trans-1-bromo-2-chlorocyclopentane
c) 1,1-dibromo-2,3-dichloro-4-propylcyclobutane
d) 2-methyl-1-ethyl-1,3-dipropylcyclopentane
e) cycloheptane-1,3,5-triol
3. Name the following structures.
Blue=Carbon Yellow=Hydrogen Red=Oxygen Green=Chlorine
a) b) c) d) e)
f) g)
Solutions
1. a) cyclodecane
b) chlorocyclopentane or 1-chlorocyclopentane
c) trans-1-chloro-2-methylcycloheptane
d) 3-cyclopropyl-6-methyldecane
e) cyclopentylcyclodecane or 1-cyclopentylcyclodecane
f) 1,3-dibromo-1-chloro-2-fluorocycloheptane
g) 1-cyclobutyl-4-isopropylcyclohexane
2.
a) b) c) d) e)
3. a) cyclohexane
b) cyclohexanol
c) chlorocyclohexane
d) cyclopentylcyclohexane
e) 1-chloro-3-methylcyclobutane
f) 2,3-dimethylcyclohexanol
g) cis-1-propyl-2-methylcyclopentane
Contributors and Attributions
• Pwint Zin
• Jim Clark (ChemGuide) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.04%3A_Cycloalkanes.txt |
learning objectives
• classify alkyl halides as primary, secondary, or tertiary
• name alkyl halides using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkyl halides from IUPAC (systematic) and selected common names
The haloalkanes, also known as alkyl halides, are a group of chemical compounds comprised of an alkane with one or more hydrogens replaced by a halogen atom (fluorine, chlorine, bromine, or iodine). There is a fairly large distinction between the structural and physical properties of haloalkanes and the structural and physical properties of alkanes. As mentioned above, the structural differences are due to the replacement of one or more hydrogens with a halogen atom. The differences in physical properties are a result of factors such as electronegativity, bond length, bond strength, and molecular size. A few representative alklyl halides are shown below.
Alkyl halides are a versatile and useful functional group for multi-step organic synthesis. The reactivity of the alkyl halides can be predicted using their structural classifications. To communicate the three different structures, the terms primary, secondary, and tertiary are used. The classification is determined by the number of carbons bonded to the carbon bearing the halide. This classification strategy is analogous to the one used for alcohols and is explained in further detail below.
Classification of Alkyl Halides
Functional group classifications are based on the bonding patterns of the atoms involved. There is only one neutral bonding pattern for halogens (three lone pairs and a single bond) so the halogens cannot be used to determine their classification. To determine the classification of alkyl halides, the bonding pattern of the carbon bonded to the halogen is used as shown in the diagram below.
Primary alkyl halides
In a primary (1°) halogenoalkane, the carbon which carries the halogen atom is only attached to one other alkyl group.Some examples of primary alkyl halides 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 halogen. There is an exception to this: CH3Br and the other methyl halides are often counted as primary alkyl halides even though there are no alkyl groups attached to the carbon with the halogen on it.
Secondary alkyl halides
In a secondary (2°) halogenoalkane, the carbon with the halogen attached is joined directly to two other alkyl groups, which may be the same or different. Examples:
Tertiary alkyl halides
In a tertiary (3°) halogenoalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
IUPAC and Common Nomenclature
The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane) so the nomenclature system is closely related to the system for alkanes. The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide.
Common Name Format: alkyl name + halide name
The IUPAC system uses the name of the parent alkane with a prefix indicating the halogen substituents, preceded by number indicating the substituent’s location. The prefixes are fluoro-, chloro-, bromo-, and iodo-. Thus CH3CH2Cl has the common name ethyl chloride and the IUPAC name chloroethane. For simple halo alkanes, the IUPAC name includes the three parts shown below.
IUPAC Name Format: locator # + halo prefix + parent alkane
Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names.
Example
Give the common and IUPAC names for each compound.
1. CH3CH2CH2Br
2. (CH3)2CHCl
Solution
1. The alkyl group (CH3CH2CH2–) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane.
2. The alkyl group [(CH3)2CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro-) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane.
Exercise
Give common and IUPAC names for each compound.
1. CH3CH2I
• CH3CH2CH2CH2F
• Exercise
Give the IUPAC name for each compound.
3.
4.
Solution
3. The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
4. The parent alkane is hexane. Methyl (CH3) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.
Exercise
Give the IUPAC name for each compound.
5.
6.
Solutions
5. 2-choro-3-methylbutane
6. 1-bromo-2-chloro-4-methylpentane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.05%3A_Haloalkane_-_Classification_and_Nomenclature.txt |
learning objectives
• name alkenes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkenes from IUPAC (systematic) and selected common names
Alkenes contain carbon-carbon double bonds and are unsaturated hydrocarbons with the molecular formula is CnH2n. Be aware - this is also the same molecular formula ratio as cycloalkanes as shown in the example below.
Introduction
The parent structure is the longest chain containing both carbon atoms of the double bond. The two carbon atoms of a double bond and the four atoms attached to them lie in a plane, with bond angles of approximately 120° A double bond consists of one sigma bond formed by overlap of sp2 hybrid orbitals and one pi bond formed by overlap of parallel 2 p orbitals.
The carbon atoms sharing the double bond can be referred to as the "vinyl carbons". This common name arose because alkenes the source for vinyl polymers.
Pi Bond Rigidity & Geometric Isomers
The rigidity of the pi bond in alkenes creates the possibility of stereoisomers called geometric isomers. To name alkenes, it may be neccessary to communicate the stereochemistry of the structure using the cis/trans or E/Z systems.
cis Isomers
The two largest groups are on the same side of the double bond.
trans Isomers
The two largest groups are on opposite sides of the double bond.
E/Z nomenclature
When the cis/trans system is ambiguous, the E/Z system can be used where E = entgegan ("trans") and Z = zusamen ("cis"). The E/Z system is used to prioritize when there are 3 or 4 different non-hydrogen atoms are attached to the vinyl carbons (carbons sharing the double bond). This system bases priority on the atomic number (Z) and/or atomic mass (A) of the atoms bonded to the vinyl carbons. An atom attached by a multiple bond is counted once for each bond. If there is a tie in priority, then move to the next atom along each chain until a difference occurs. Atomic number has higher priority than atomic mass. Atomic mass is used to establish priority for isotopes, therefore deuterium (D) has higher priority than hydrogen (H).
For example, when comparing atoms bonded to the vinylic carbons in the the compound below,
we would rank the priority as
fluorine atom > propyl group > ethyl group > methyl
Z = 9 > 3 x C chain > 2 x C chain > 1 x C chain
and name the compound ((2E)-3-ethyl-2-fluorohex-2-ene.
The double bond of the allylic group creates higher priority over a simple propyl grup such that -CH2-CH=CH2 > -CH2CH2CH3.
For straight chain alkenes, it is the same basic rules as nomenclature of alkanes except change the suffix to "-ene."
1. Find the Longest Carbon Chain that Contains the Carbon Carbon double bond. If you have two ties for longest Carbon chain, and both chains contain a Carbon Carbon double bond, then identify the most substituted chain.
2. Give the lowest possible number to the Carbon Carbon double bond.
a) Do not need to number cycloalkenes because it is understood that the double bond is in the one position.
b) Alkenes that have the same molecular formula but the location of the double bonds are different means they are constitutional isomers.
c) Functional Groups with higher priority determine the suffix
3. Add substituents and their position to the alkene as prefixes. Of course remember to give the lowest numbers possible. And remember to name them in alphabetical order when writting them.
4. Next is identifying stereoisomers - cis/trans. when there are only two non hydrogen attachments to the alkene then use cis and trans to name the molecule.
For example, the compound below is a cis isomer. It has both the substituents going upward. This molecule would be called (cis) 5-chloro-3-heptene.
5. Next is identifying stereoisomers - E/Z if cis/trans is ambiguous or skip 4 above and jump straight to E/Z.
For example, if we look at the "trans" alternative lo the previous compound, the cis/trans system cannot be applied.
cis or trans?
This molecule would be called (3E)-4-chlorohept-3-ene. It is E because the ethyl group (-CH2CH3) has the higher priority for the vicinal carbon on the left and the chlorine atom has the higher priority for the vicinal carbon on the right and these two groups are on opposite sides of the double bond.
6. An example of functional group priorities in nomenclature is that the hydroxyl group gets precedence (has higher priority) over the double bond.
Therefore, alkenes containing alcohol groups are called alkenols with the sufffix --enol.
7. Lastly remember that alkene substituents are called alkenyl. Suffix --enyl.
Here is a chart containing the systemic name for the first twenty straight chain alkenes.
Name Molecular formula
Ethene C2H4
Propene C3H6
Butene C4H8
Pentene C5H10
Hexene C6H12
Heptene C7H14
Octene C8H16
Nonene C9H18
Decene C10H20
Undecene C11H22
Dodecene C12H24
Tridecene C13H26
Tetradecene C14H28
Pentadecene C15H30
Hexadecene C16H32
Heptadecene C17H34
Octadecene C18H36
Nonadecene C19H38
Eicosene C20H40
Did you notice how there is no methene? Because it is impossible for a carbon to have a double bond with nothing.
Common names
The carbon atoms sharing the double bond can be referred to as the "vinyl carbons". The carbon atoms adjacent to the vinyl carbon atoms are called "allylic carbons". These carbon atoms have unique reactivity because of the potential for interaction with the pi bond.
Overall, remove the -ane suffix and add -ylene.
There are a couple of unique ones like ethenyl's common name is vinyl and 2-propenyl's common name is allyl, that you should know are...
• vinyl substituent H2C=CH-
• allyl substituent H2C=CH-CH2-
• allene molecule H2C=C=CH2
• isoprene is shown below
Endocyclic & Exocyclic Alkenes
Endocyclic double bonds have both carbons in the ring and exocyclic double bonds have only one carbon as part of the ring.
Cyclopentene is an example of an endocyclic double bond.
Methylenecylopentane is an example of an exocyclic double bond.
For example, when naming the compound below, the methyl group is considered when numbering the double bond.
The compound can be named 1-methylcyclobutene or 1-methylcyclobut-1-ene.
When naming this next compound, the ethenyl group is considered when numbering the double bond.
The IUPAC name for the compound can be 1-ethenylcyclohexene or 1-ethenylcyclohex-1-ene. The common name would be 1-vinylcyclohexene. For the compound below, the name is 2-vinyl-1,3-cyclohexadiene.
Exercise
1. Give the IUPAC name for the following compounds. When stereochemistry is included, write the name using both the cis/trans and E/Z names if possible.
2. Draw the bond-line structures for the following compounds.
a) trans-2-pentene
b) (Z)-3-heptene
c) 4-methyl-2-pentene
d) (Z)-5-Chloro-3-ethly-4-hexen-2-ol.
Answer
1. a) 1-pentene or pent-1-ene
b) 2-ethyl-1-hexene or 2-ethylhex-1-ene (parent chain must include the double bond)
c) cis-2-hexene or (Z)-2-hexene or (2Z)-hex-2-ene
d) (2E)-3-methylpent-2-ene or (E)-3-methyl-2-pentene (cis/trans cannnt be applied)
e) trans-2-methyl-4-octene or (4E)-2-methyloct-4-ene or (E)-2-methyl-4-octene (branch breaks the tie in numbering the parent chain since both directions begin the double bond at carbon 4.
2.
Contributors and Attributions
• S. Devarajan (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.06%3A_Alkenes.txt |
learning objectives
• name alkynes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkynes from IUPAC (systematic) and selected common names
Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of \(C_nH_{2n-2}\). They are unsaturated hydrocarbons. Like alkenes have the suffix –ene, alkynes use the ending –yne; this suffix is used when there is only one alkyne in the molecule.
Straight Chain Alkynes
Here are the molecular formulas and names of the first ten carbon straight chain alkynes.
Name
Molecular Formula
Ethyne
C2H2
Propyne
C3H4
1-Butyne
C4H6
1-Pentyne
C5H8
1-Hexyne
C6H10
1-Heptyne
C7H12
1-Octyne
C8H14
1-Nonyne
C9H16
1-Decyne
C10H18
Naming Alkynes
Like previously mentioned, the IUPAC rules are used for the naming of alkynes.
Rule 1
Find the longest carbon chain that includes both carbons of the triple bond.
Rule 2
Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes.
For example:
4-chloro-6-diiodo-7-methyl-2-nonyne
Rule 3
After numbering the longest chain with the lowest number assigned to the alkyne, label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order.
For example:
1-triiodo-4-dimethyl-2-nonyne
If there is an alcohol present in the molecule, number the longest chain starting at the end closest to it, and follow the same rules. However, the suffix would be –ynol, because the alcohol group takes priority over the triple bond.
5- methyl-7-octyn-3-ol
When there are two triple bonds in the molecule, find the longest carbon chain including both the triple bonds. Number the longest chain starting at the end closest to the triple bond that appears first. The suffix that would be used to name this molecule would be –diyne.
For example:
4-methyl-1,5-octadiyne
Rule 4
Substituents containing a triple bond are called alkynyl.
For example:
1-chloro-1-ethynyl-4-bromocyclohexane
Here is a table with a few of the alkynyl substituents:
Name
Molecule
Ethynyl
-C?CH
2- Propynyl
-CH2C?CH
2-Butynyl
-CH3C?CH2CH3
Rule 5
A molecule that contains both double and triple bonds is called an alkenyne. The chain can be numbered starting with the end closest to the functional group that appears first. For example:
6-ethyl-3-methyl-1,4-nonenyne
Common Names
The more commonly used name for ethyne is acetylene, which used industrially.
Similar to the allylic carbon position of alkenes, the carbons bonded to the alkyne carbons are called "propargyl" carbons and also have differences in chemical reactivity because of the interaction of the two pi bonds with the propargyl carbons.
HC≡CH–CH2 Propargyl group
Exercise
1. Briefly identify the important differences between an alkene and an alkyne. How are they similar?
2. The alkene (CH3)2CHCH2CH=CH2 is named 4-methyl-1-pentene. What is the name of (CH3)2CHCH2C≡CH?
3. Do alkynes show cis-trans isomerism? Explain.
4. Draw the bond-line structure & write the condensed structural formula for each compound except (a). For part (a) write the condensed formula and full Lewis (Kekule) structure.
a) acetylene
b) 3-octyne
c) 3-methyl-1-hexyne
d) 4,4-dimethyl-2-pentyne
e) trans-3-hepten-1-yne
5. Give the IUPAC (Systematic) name for each compound.
a) CH3CH2CH2C≡CH
b) CH3CH2CH2C≡CCH3
Answer
1. Alkenes have double bonds; alkynes have triple bonds. Both bonds are rigid and do not undergo rotation, however, the pi bonds allow both alkenes and alkynes to undergo addition reactions.
2. 4-methyl-1-pentyne
3. No; a triply bonded carbon atom can form only one other bond and has linear electron geometry so there are no "sides". Allkenes have two groups attached to each inyl carbon with a trigonal planar electron geometry that creates the possibility of cis-trans isomerism.
4.
5. a) 1-pentyne or pent-1-yne
b) 2-hexyne or hex-2-yne
c) 2-methylnon-4-yne or 2-methyl-4-nonyne
d) (4E)-hex-4-en-1-yne or (E)-hex-4-en-1-yne (alkynes have higher priority over alkenes if they occur sooner in the parent chain)
e) pent-1-en-4-yne (alkenes have higher priority over alkynes when they have the same position in the parent chain)
f) pent-4yn-2-ol (alcohols have higher priority than alkynes)
Reference
1. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. 5th Edition. New York: W. H. Freeman & Company, 2007.
Contributors and Attributions
• A. Sheth and S. Sujit (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.07%3A_Alkynes.txt |
learning objectives
• classify alcohols as primary, secondary, or tertiary
• name alkanes using IUPAC (systematic) and selected common name nomenclature
• draw the structure of alkanes from IUPAC (systematic) and selected common namesAlcohol Classification
Alcohol Classification
Alcohols may can be classified as primary, , secondary, & tertiary, , in the same manner as alkyl halides. This terminology refers to alkyl substitution of the carbon atom bearing the hydroxyl group (colored blue in the illustration). This classification system is based on the neutral bonding pattern for oxygen. The structure of alcohols requires one of the oxygen bonds to form with hydrogen and the other oxygen bond to form with carbon. All oxygen atoms of alcohols look the same. To distinguish between alcohol classifications, we must look at the carbon atom bonded to the hydroxyl group. The bonding pattern of this carbon atom determines the classification of the alcohol.
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:
Nomenclature - IUPAC Introduction & Common Names
Alcohols are designated by an ol suffix if they are the highest priority functinal group. For example, ethanol contains a hydroxyl group to form an alcohol as shown in the follwowing condensed formula: CH3CH2OH. Note that a locator number is not needed on a two-carbon chain, but on longer chains the location of the hydroxyl group determines chain numbering and must be specified in the name. For example: CH3CH(OH)CH2CH2CH3 is 2-pentanol or pentan-2-ol.
Other examples of IUPAC nomenclature are shown below, together with the common names often used for some of the simpler compounds. For the mono-functional alcohols, this common system consists of naming the alkyl group followed by the word alcohol.
Many functional groups have a characteristic suffix designator, and only one such suffix (other than "ene" and "yne") may be used in a name. When the hydroxyl functional group is present together with a function of higher nomenclature priority, it must be cited and located by the prefix hydroxy and an appropriate number. For example, lactic acid has the IUPAC name 2-hydroxypropanoic acid.
The terms "vicinal" and "geminal" can be applied to any two functional groups that are part of the same compound. Typically, these terms are first encountered with alcohols. Vicinal is used to describe the structure of a compound in which the two groups are bonded to neighboring carbons. Geminal is used when both functional groups are bonded to the same carbon. In Latin, "gemini" means twins. In the same way that twins are connected to the same mother, geminal groups are bonded to the same carbon. Similarly, the vicinal groups are in vicinity of each other.
Naming Alcohols
1. Find the longest chain containing the hydroxy group (OH). If there is a chain with more carbons than the one containing the OH group it will be named as a subsitutent.
2. Place the OH on the lowest possible number for the chain. With the exception of carbonyl groups such as ketones and aldehydes, the alcohol or hydroxy groups have first priority for naming.
3. When naming a cyclic structure, the -OH is assumed to be on the first carbon unless the carbonyl group is present, in which case the later will get priority at the first carbon.
4. When multiple -OH groups are on the cyclic structure, number the carbons on which the -OH groups reside.
5. Remove the final e from the parent alkane chain and add -ol. When multiple alcohols are present use di, tri, et.c before the ol, after the parent name. ex. 2,3-hexandiol. If a carbonyl group is present, the -OH group is named with the prefix "hydroxy," with the carbonyl group attached to the parent chain name so that it ends with -al or -one.
Examples
Ethane: CH3CH3 ----->Ethanol: (the alcohol found in beer, wine and other consumed sprits)
Secondary alcohol: 2-propanol
Other functional groups on an alcohol: 3-bromo-2-pentanol
Cyclic alcohol (two -OH groups): cyclohexan-1,4-diol
Other functional group on the cyclic structure: 3-hexeneol (the alkene is in bold and indicated by numbering the carbon closest to the alcohol)
A complex alcohol:4-ethyl-3hexanol (the parent chain is in red and the substituent is in blue)
Exercise
1. Give the IUPAC (Systematic) name for each compound. For parts (a)-(c), classify the alcohols as primary, secondary, or tertiary.
Part (d) is a challenge question and sneak preview of coming attractions.
2. Draw the bond line structures, condensed strucutre, and name all the alcohols with the molecular formula C3H9O.
3. Oleic acid, a commonly occurring fatty acid in vegetable oils has the structure below.
a) Describe the stereochemistry of oleic acid - cis or trans?
b) Write the condensed formula for oleic acid.
4. Give the IUPAC name for each compound. For parts (a), (b), (d), and (e) classify the alcohol as primary, secondary, tertiary, or allylic. Part (c) is also a challenge question.
Answer
1. a) 2-butanol or butan-2-ol; secondary
b) 2-methylcyclohexan-1-ol or 2-methyl-1-cyclohexanol; secondary
c) cyclopentylmethanol (alcohol is higher priority over carbon chain or ring size); primary
d) 2-(butan2-yl)-6-tert-butyl-4-(propan-2-yl)phenol or 2-sec-butyl-6-t-butyl-4-isopropylphenol (some common names are recognized by IUPAC)
2.
3. a) cis alkene
b) CH3(CH2)6CHCH(CH2)6CO2H
4. a) 2-methyl-1-cyclohexanol or 2-methyl-cyclohexan-1-ol (no stereochemistry communicated even though it is possible); secondary
b) 1-methyl-1-cyclohexanol or 1-methyl-cyclohexan-1-ol (no steereochemicstry because both groups are bonded to the same carbon); tertiary
c) 4-nitrophenol
d) 2-propen-1-ol (alcohols have higher priority than alkenes so determine numbering and suffix); allylic
e) (1S, 2S)-2-methylcyclopentan-1-ol (For students who have learned about chirality); secondary
There are two equivalent answers for students who have not yet learned about chirlatiy:
trans-1-methyl-2-cyclpentanol or trans-1-methylcyclopentan-1-ol
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.08%3A_3.8_Alcohols_-__Classification_and_Nomenclature.txt |
learning objectives
• name ethers, epoxides, and sulfides using IUPAC (systematic) and selected common name nomenclature
• draw the structure of ethers, epoxides, and sulfides from IUPAC (systematic) and selected common names
Note: Heterocyclic oxygen compounds are included for the sake of completion. Their nomenclature may or may not be required by the professor are requires additional instruction. Make sure to ask.
Ethers
Ethers are compounds having two alkyl or aryl groups bonded to an oxygen atom, as in the formula R1–O–R2. The ether functional group does not have a characteristic IUPAC nomenclature suffix, so it is necessary to designate it as a substituent. To do so the common alkoxy substituents are given names derived from their alkyl component (below):
Alkyl Group Name Alkoxy Group Name
CH3 Methyl CH3O– Methoxy
CH3CH2 Ethyl CH3CH2O– Ethoxy
(CH3)2CH– Isopropyl (CH3)2CHO– Isopropoxy
(CH3)3C– tert-Butyl (CH3)3CO– tert-Butoxy
C6H5 Phenyl C6H5O– Phenoxy
The smaller, shorter alkyl group becomes the alkoxy substituent. The larger, longer alkyl group side becomes the alkane base name. Each alkyl group on each side of the oxygen is numbered separately. The numbering priority is given to the carbon closest to the oxgen. The alkoxy side (shorter side) has an "-oxy" ending with its corresponding alkyl group. For example, CH3CH2CH2CH2CH2-O-CH2CH2CH3 is 1-propoxypentane. If there is cis or trans stereochemistry, the same rule still applies.
Example
Examples are: CH3CH2OCH2CH3, diethyl ether (sometimes referred to as ether), and CH3OCH2CH2OCH3, ethylene glycol dimethyl ether (glyme).
Common names
Simple ethers are given common names in which the alkyl groups bonded to the oxygen are named in alphabetical order followed by the word "ether". The top left example shows the common name in blue under the IUPAC name. Many simple ethers are symmetrical, in that the two alkyl substituents are the same. These are named as "dialkyl ethers". If we read the word "ether", the author is most likely communicating the compound CH3CH2OCH2CH3, ethoxyethane (diethyl ether), but we do not know with certainty - another example of the importance of accurate nomenclature.
Epoxides
An epoxide is a cyclic ether with three ring atoms. These rings approximately define an equilateral triangle, which makes it highly strained. The strained ring makes epoxides more reactive than other ethers. Simple epoxides are named from the parent compound ethylene oxide or oxirane, such as in chloromethyloxirane. As a functional group, epoxides feature the epoxy prefix, such as in the compound 1,2-epoxycycloheptane, which can also be called cycloheptene epoxide, or simply cycloheptene oxide.
A generic epoxide.
The chemical structure of the epoxide glycidol, a common chemical intermediate
A polymer formed by reacting epoxide units is called a polyepoxide or an epoxy. Epoxy resins are used as adhesives and structural materials. Polymerization of an epoxide gives a polyether, for example ethylene oxide polymerizes to give polyethylene glycol, also known as polyethylene oxide.
Sulfides (Thioethers)
A thioether is a functional group in organosulfur chemistry with the connectivity C-S-C as shown below. Like many other sulfur-containing compounds, volatile thioethers have foul odors.[1] A thioether is similar to an ether except that it contains a sulfur atom in place of the oxygen. The grouping of oxygen and sulfur in the periodic table suggests that the chemical properties of ethers and thioethers are somewhat similar.
General structure of a thioether with the blue marked functional group.
Nomenclature
Thioethers are sometimes called sulfides, especially in the older literature and this term remains in use for the names of specific thioethers. The two organic substituents are indicated by the prefixes. (CH3)2S is called dimethylsulfide. Some thioethers are named by modifying the common name for the corresponding ether. For example, C6H5SCH3 is methyl phenyl sulfide, but is more commonly called thioanisole, since its structure is related to that for anisole, C6H5OCH3.
Structure and properties
Thioether is an angular functional group, the C-S-C angle approaching 90°. The C-S bonds are about 180 pm.
Thioethers are characterized by their strong odors, which are similar to thiol odor. This odor limits the applications of volatile thioethers. In terms of their physical properties they resemble ethers but are less volatile, higher melting, and less hydrophilic. These properties follow from the polarizability of the divalent sulfur center, which is greater than that for oxygen in ethers.
Heterocycles with Oxygen
In cyclic ethers (heterocycles), one or more carbons are replaced with oxygen. Often, it's called heteroatoms, when carbon is replaced by an oxygen or any atom other than carbon or hydrogen. In this case, the stem is called the oxacycloalkane, where the prefix "oxa-" is an indicator of the replacement of the carbon by an oxygen in the ring. These compounds are numbered starting at the oxygen and continues around the ring. For example,
If a substituent is an alcohol, the alcohol has higher priority. However, if a substituent is a halide, ether has higher priority. If there is both an alcohol group and a halide, alcohol has higher priority. The numbering begins with the end that is closest to the higher priority substituent. There are ethers that are contain multiple ether groups that are called cyclic polyethers or crown ethers. These are also named using the IUPAC system.
Thiophenes
Thiophenes are a special class of thioether-containing heterocyclic compounds. Because of their aromatic character, they are non-nucleophilic. The nonbonding electrons on sulfur are delocalized into the π-system. As a consequence, thiophene exhibits few properties expected for a thioether - thiophene is non-nucleophilic at sulfur and, in fact, is sweet-smelling. Upon hydrogenation, thiophene gives tetrahydrothiophene, C4H8S, which indeed does behave as a typical thioether.
Example
Examples of ethers include CH3CH2OCH2CH3, diethyl ether (sometimes referred to as ether), and CH3OCH2CH2OCH3, ethylene glycol dimethyl ether (glyme).
Common names
Simple ethers are given common names in which the alkyl groups bonded to the oxygen are named in alphabetical order followed by the word "ether". The top left example shows the common name in blue under the IUPAC name. Many simple ethers are symmetrical, in that the two alkyl substituents are the same. These are named as "dialkyl ethers".
• anisole (try naming anisole by the other two conventions.
• oxirane or 1,2-epoxyethane, ethylene oxide, dimethylene oxide, oxacyclopropane,
• furan (this compound is aromatic)
tetrahydrofuran
oxacyclopentane, 1,4-epoxybutane, tetramethylene oxide,
• dioxane or 1,4-dioxacyclohexane
Exercise
Give the IUPAC and common name (if possible) for each compound respectively.
Answer
1. ethoxyethane; diethyl ether
2. 2-ethoxy-2-methyl-propane; etheyl t-butyl ether (ethyl tert-butyl ether)
3. cis-1-ethoxy-2-methoxycyclopentane; no common name possible
4. 1-ethoxy-1-methylcyclohexane; no common name possible
5. 1,2-epoxyethane; ethylene oxide or dimethylene oxide or oxacyclopropane or oxirane
6. 2,2-dimethyloxirane
Contributors and Attributions
• Wikipedia (used with permission) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.09%3A_Ethers_Epoxides_and_Sulfides.txt |
learning objectives
• name benzene and its derivatives using IUPAC (systematic) and selected common name nomenclature
• draw the structures of benzene and its derivatives from IUPAC (systematic) and selected common names
Benzene, as shown on the left, is an organic aromatic compound with many interesting properties. Unlike aliphatic (straight chain carbons) or other cyclic organic compounds, the structure of benzene (3 conjugated π bonds) allows benzene and its derived products to be useful in fields such as health, laboratory synthesis, and other applications such as rubber synthesis. Benzene is unique because we can write a condensed formula for its ringed structure, C6H6.
Introduction
Benzene derived products are well known to be pleasantly fragrant. For this reason, organic compounds containing benzene rings were classified as being "aromatic" (sweet smelling) amongst scientists in the early 19th century when a relation was established between benzene derived compounds and sweet/spicy fragrances. There is a misconception amongst the scientific community, however, that all aromatics are sweet smelling and that all sweet smelling compounds would have a benzene ring in its structure. This is false, since non-aromatic compounds, such as camphor, extracted from the camphor laurel tree, release a strong, minty aroma, yet it lacks the benzene ring in its structure (See figure 1). On the other hand, benzene itself gives off a rather strong and unpleasant smell that would otherwise invalidate the definition of an aromatic (sweet-smelling) compound. Despite this inconsistency, however, the term aromatic continues to be used today in order to designate molecules with benzene-like rings in their structures. For a modern, chemical definition of aromaticity, refer to sections Aromaticity and Hückel's Rule.
Figure 1. Top-view of camphor, along with its monoterpene unit. Notice how camphor lacks the benzene ring to be "aromatic".
Many aromatic compounds are however, sweet/pleasant smelling. Eugenol, for example, is extracted from essential oils of cloves and it releases a spicy, clove-like aroma used in perfumes. In addition, it is also used in dentistry as an analgesic.
Figure 2. Eugenol, an aromatic compound extracted from clove essential oils. Used in perfumes and as an analgesic.
The benzene ring is labeled in red in the eugenol molecule.
Is it cyclohexane or is it benzene?
Due to the similarity between benzene and cyclohexane, the two is often confused with each other in beginning organic chemistry students.
If you were to count the number of carbons and hydrogens in cyclohexane, you will notice that its molecular formula is C6H12. Since the carbons in the cyclohexane ring is fully saturated with hydrogens (carbon is bound to 2 hydrogens and 2 adjacent carbons), no double bonds are formed in the cyclic ring. In contrast, benzene is only saturated with one hydrogen per carbon, leading to its molecular formula of C6H6. In order to stabilize this structure, 3 conjugated π (double) bonds are formed in the benzene ring in order for carbon to have four adjacent bonds.
In other words, cyclohexane is not the same as benzene! These two compounds have different molecular formulas and their chemical and physical properties are not the same. The hydrogenation technique can be used by chemists to convert from benzene to cyclohexane by saturating the benzene ring with missing hydrogens.
IMPORTANT NOTE: A special catalyst is required to hydrogenate benzene rings due to its unusual stability and configuration. Normal catalytic hydrogenation techniques will not hydrogenate benzene and yield any meaningful products.
What about Resonance?
Benzene can be drawn a number of different ways. This is because benzene's conjugated pi electrons freely resonate within the cyclic ring, thus resulting in its two resonance forms.
Figure 4. The figure to the left shows the two resonance forms of benzene. The delocalized electrons are moved from one carbon to the next, thus providing stabilization energy. Ring structures stabilized by the movement of delocalized electrons are sometimes referred to as arenes.
As the electrons in the benzene ring can resonate within the ring at a fairly high rate, a simplified notation is often used to designate the two different resonance forms. This notation is shown above, with the initial three pi bonds (#1, #2) replaced with an inner ring circle (#3). Alternatively, the circle within the benzene ring can also be dashed to show the same resonance forms (#4).
The Phenyl Group (The Foundation of Benzene Derivatives)
The phenyl group is formed by removing one hydrogen from benzene to create the fragment is C6H5. NOTE: Although the molecular formula of the phenyl group is C6H5, the phenyl group would always have something attached to where the hydrogen was removed. Thus, the formula is often written as Ph-R, where Ph refers to the phenyl group and R refers to the R group attached where the hydrogen was removed.
Figure 5. Figure demonstrating the removal of hydrogen to form the phenyl group.
Different R groups on the phenyl group allows different benzene derivatives to be formed. Phenol, Ph-OH, or C6H5OH, for example, is formed when an alcohol (-OH) group displaces a hydrogen atom on the benzene ring. Benzene, for this very same reason, can be formed from the phenyl group by reattaching the hydrogen back its place of removal. Thus benzene, similar to phenol, can be abbreviated Ph-H, or C6H6.
As you can see above, these are only some of the many possibilities of the benzene derived products that have special uses in human health and other industrial fields.
Nomenclature of Benzene Derived Compounds
Unlike aliphatic organics, nomenclature of benzene-derived compounds can be confusing because a single aromatic compound can have multiple possible names (such as common and systematic names) be associated with its structure. In these sections, we will analyze some of the ways these compounds can be named.
Simple Benzene Naming
Some common substituents, like NO2, Br, and Cl, can be named this way when it is attached to a phenyl group. Long chain carbons attached can also be named this way. The general format for this kind of naming is:
(positions of substituents (if >1)- + # (di, tri, ...) + substituent)n + benzene.
For example, chlorine (Cl) attached to a phenyl group would be named chlorobenzene (chloro + benzene). Since there is only one substituent on the benzene ring, we do not have to indicate its position on the benzene ring (as it can freely rotate around and you would end up getting the same compound.)
Figure 7. Example of simple benzene naming with chlorine and NO2 as substituents.
Figure 8. More complicated simple benzene naming examples - Note that standard nomenclature priority rules are applied here, causing the numbering of carbons to switch.
Ortho-, Meta-, Para- (OMP) Nomenclature for Disubstituted Benzenes
Instead of using numbers to indicate substituents on a benzene ring, ortho- (o-), meta- (m-), or para (p-) can be used in place of positional markers when there are two substituents on the benzene ring (disubstituted benzenes). They are defined as the following:
• ortho- (o-): 1,2- (next to each other in a benzene ring)
• meta- (m): 1,3- (separated by one carbon in a benzene ring)
• para- (p): 1,4- (across from each other in a benzene ring)
Continuing with the example above in figure 8 (1,3-dichlorobenzene), we can use the ortho-, meta-, para- nomenclature to transform the chemical name into m-dichlorobenzene, as shown in the figure below.
Figure 10 . Transformation of 1,3-dichlorobenzene into m-dichlorobenzene.
Here are some other examples of ortho-, meta-, para- nomenclature used in context:
However, the substituents used in ortho-, meta-, para- nomenclature do not have to be the same. For example, we can use chlorine and a nitro group as substituents in the benzene ring.
In conclusion, these can be pieced together into a summary diagram, as shown below:
Derivatives as Parent Names
In addition to simple benzene naming and OMP nomenclature, benzene derivtives are also sometimes used as the "parent" in the name of a larger compound. name.
For example, phenol (C6H5OH) is the parent name of the compound below because hydroxyl groups have higher nomenclature priority than halides. The chlorine atom is considered a branch at the ortho- position to the hydroxyl group. Accordingly, the compound is named 2-chlorophenol or o-chlorophenol.
Figure 14. An example showing phenol as a base in its chemical name. Note how benzene no longer serves as a base when an OH group is added to the benzene ring.
Alternatively, we can use the numbering system to indicate this compound. When the numbering system is used, the carbon where the substituent is attached on the base will be given the first priority and named as carbon #1 (C1). The normal priority rules then apply in the nomenclature process (give the rest of the substituents the lowest numbering as you could).
Figure 15. The naming process for 2-chlorophenol (o-chlorophenol). Note that 2-chlorophenol = o-chlorophenol.
Below is a list of commonly seen benzene-derived compounds. Some of these mono-substituted compounds (labeled in red and green), such as phenol or toluene, can be used in place of benzene for the chemical's base name.
Figure 16. Common benzene derived compounds with various substituents.
Common vs. Systematic (IUPAC) Nomenclature
According to the indexing preferences of the Chemical Abstracts, phenol, benzaldehyde, and benzoic acid (labeled in red in Figure 16) are some of the common names that are retained in the IUPAC (systematic) nomenclature. Other names such as toluene, styrene, naphthalene, or phenanthrene can also be seen in the IUPAC system in the same way. While the use of other common names are usually acceptable in IUPAC, their use are discouraged in the nomenclature of compounds.
Nomenclature for compounds which has such discouraged names will be named by the simple benzene naming system. An example of this would include toluene derivatives like TNT. (Note that toluene by itself is retained by the IUPAC nomenclature, but its derivatives, which contains additional substituents on the benzene ring, might be excluded from the convention). For this reason, the common chemical name 2,4,6-trinitrotoluene, or TNT, as shown in figure 17, would not be advisable under the IUPAC (systematic) nomenclature.
In order to correctly name TNT under the IUPAC system, the simple benzene naming system should be used:
Figure 18. Systematic (IUPAC) name of 2,4,6-trinitrotoluene (common name), or TNT.
Note that the methyl group is individually named due to the exclusion of toluene from the IUPAC nomenclature.
Figure 19. The common name 2,4-dibromophenol, is shared by the IUPAC systematic nomenclature.
Only substituents phenol, benzoic acid, and benzaldehyde share this commonality.
Since the IUPAC nomenclature primarily rely on the simple benzene naming system for the nomenclature of different benzene derived compounds, the OMP (ortho-, meta-, para-) system is not accepted in the IUPAC nomenclature. For this reason, the OMP system will yield common names that can be converted to systematic names by using the same method as above. For example, o-Xylene from the OMP system can be named 1,2-dimethylbenzene by using simple benzene naming (IUPAC standard).
The Phenyl and Benzyl Groups
The Phenyl Group
As mentioned previously, the phenyl group (Ph-R, C6H5-R) can be formed by removing a hydrogen from benzene and attaching a substituent to where the hydrogen was removed. To this phenomenon, we can name compounds formed this way by applying this rule: (phenyl + substituent). For example, a chlorine attached in this manner would be named phenyl chloride, and a bromine attached in this manner would be named phenyl bromide. (See below diagram)
Figure 20. Naming of Phenyl Chloride and Phenyl Bromide
While compounds like these are usually named by simple benzene type naming (chlorobenzene and bromobenzene), the phenyl group naming is usually applied to benzene rings where a substituent with six or more carbons is attached, such as in the diagram below.
Figure 21. Diagram of 2-phenyloctane.
Although the diagram above might be a little daunting to understand at first, it is not as difficult as it seems after careful analysis of the structure is made. By looking for the longest chain in the compound, it should be clear that the longest chain is eight (8) carbons long (octane, as shown in green) and that a benzene ring is attached to the second position of this longest chain (labeled in red). As this rule suggests that the benzene ring will act as a function group (a substituent) whenever a substituent of more than six (6) carbons is attached to it, the name "benzene" is changed to phenyl and is used the same way as any other substituents, such as methyl, ethyl, or bromo. Putting it all together, the name can be derived as: 2-phenyloctane (phenyl is attached at the second position of the longest carbon chain, octane).
The Benzyl Group
The benzyl group (abbv. Bn), similar to the phenyl group, is formed by manipulating the benzene ring. In the case of the benzyl group, it is formed by taking the phenyl group and adding a CH2 group to where the hydrogen was removed. Its molecular fragment can be written as C6H5CH2-R, PhCH2-R, or Bn-R. Nomenclature of benzyl group based compounds are very similar to the phenyl group compounds. For example, a chlorine attached to a benzyl group would simply be called benzyl chloride, whereas an OH group attached to a benzyl group would simply be called benzyl alcohol.
Figure 22. Benzyl Group Nomenclature
Additionally, other substituents can attach on the benzene ring in the presence of the benzyl group. An example of this can be seen in the figure below:
Figure 23. Nomenclature of 2,4-difluorobenzyl chloride.
Similar to the base name nomenclature system, the carbon in which the base substituent is attached on the benzene ring is given the first priority and the rest of the substituents are given the lowest number order possible. Under this consideration, the above compound can be named: 2,4-difluorobenzyl chloride.
Commonly Named Benzene Compounds Nomenclature Summary Flowchart
Summary Flowchart (Figure 24). Summary of nomenclature rules used in commonly benzene derived compounds.
As benzene derived compounds can be extremely complex, only compounds covered in this article and other commonly named compounds can be named using this flowchart.
Determination of Common and Systematic Names using Flowchart
To demonstrate how this flowchart can be used to name TNT in its common and systematic (IUPAC) name, a replica of the flowchart with the appropriate flow paths are shown below:
Exercise
1. True or False? The compound below contains a benzene ring and thus is aromatic.
2. Benzene unusual stability is caused by a combination of the _________________ conjugated pi bonds in its cyclic ring.
specify the number
3. Menthol, a topical analgesic used in many ointments for the relief of pain, releases a peppermint aroma upon exposure to the air. Based on this conclusion, can you imply that a benzene ring is present in its chemical structure? Why or why not?
4. Predict the product of the reaction below or explain why no reaction occurs.
5. At normal conditions, benzene has ___ resonance structures.
6. Which of the following name(s) is/are correct for the following compound?
a) nitrohydride benzene
b) phenylamine
c) phenylamide
d) aniline
e) nitrogenhydrogen benzene
f) All of the above is correct
7. Convert 1,4-dimethylbenzene into its common name.
8. TNT's common name is: ______________________________
9. Name the following compound using OMP nomenclature:
10. Draw the structure of 2,4-dinitrotoluene.
11. Give the IUPAC name the following compound:
12. Which of the following is the correct name for the following compound?
a) 3,4-difluorobenzyl bromide
b) 1,2-difluorobenzyl bromide
c) 4,5-difluorobenzyl bromide
d) 1,2-difluoroethyl bromide
e) 5,6-difluoroethyl bromide
f) 4,5-difluoroethyl bromide
13. a) True or False? Benzyl chloride can be abbreviated Bz-Cl.
b) What is the condensed formula for benzyl chloride?
c) What is the condensed formula for phenyl chloride?
14. Benzoic Acid has what R group attached to its phenyl functional group?
15. True or False? A single aromatic compound can have multiple names indicating its structure.
16. List the corresponding positions for the OMP system (o-, m-, p-).
17. A scientist has conducted an experiment on an unknown compound. He was able to determine that the unknown compound contains a cyclic ring in its structure as well as an alcohol (-OH) group attached to the ring. What is the unknown compound?
a) Cyclohexanol
b) Cyclicheptanol
c) Phenol
d) Methanol
e) Bleach
f) Cannot determine from the above information
18. Which of the following statements is false for the compound, phenol?
a) Phenol is a benzene derived compound.
b) Phenol can be made by attaching an -OH group to a phenyl group.
c) Phenol is highly toxic to the body even in small doses.
d) Phenol can be used as a catalyst in the hydrogenation of benzene into cyclohexane.
e) Phenol is used as an antiseptic in minute doses.
f) Phenol is amongst one of the three common names retained in the IUPAC nomenclature.
Answer
1. False, this compound does not contain a benzene ring in its structure.
2. 3
3. No, a substance that is fragrant does not imply a benzene ring is in its structure. See camphor example (figure 1)
4. No reaction, benzene requires a special catalyst to be hydrogenated due to its unusual stability given by its three conjugated pi bonds.
5. 2
6. b, d
7. p-xylene
8. 2,4,6-trinitrotoluene
9. p-chloronitrobenzene
10.
11. 4-phenylheptane
12. a) 3,4-difluorobenzyl bromide
13. a) False, the correct abbreviation for the benzyl group is Bn, not Bz. The correct abbreviation for Benzyl chloride is Bn-Cl.
b) C6H5CH2Cl.
c) C6H5Cl.
14. COOH or CO2H
15. True. TNT, for example, has the common name 2,4,6-trinitrotoluene and its systematic name is 2-methyl-1,3,5-trinitrobenzene.
16. Ortho - 1,2 ; Meta - 1,3 ; Para - 1,4
17. The correct answer is f). We cannot determine what structure this is since the question does not tell us what kind of cyclic ring the -OH group is attached on. Just as cyclohexane can be cyclic, benzene and cycloheptane can also be cyclic. The chemical formula would allow a determination.
18. d
• David Lam | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.10%3A_Benzene_and_its_Derivatives.txt |
learning objectives
• name aldehydes and ketones using IUPAC (systematic) and selected common name nomenclature
• draw the structure of aldehydes and ketones from IUPAC (systematic) and selected common names
Aldehydes and ketones contain the carbonyl group. Aldehydes derive their name from the dehydration of alcohols. Aldehydes contain the carbonyl group bonded to at least one hydrogen atom. Ketones contain the carbonyl group bonded to two carbon atoms.
Aldehydes and ketones are organic compounds which incorporate a carbonyl functional group, C=O. The carbon atom of this group has two remaining bonds that may be occupied by hydrogen, alkyl or aryl substituents. If at least one of these substituents is hydrogen, the compound is an aldehyde. If neither is hydrogen, the compound is a ketone. When writing the condensed formulas for aldehydes and ketones, it is important to note that the carbonyl bond is not drawn. It must be recognized. The generic condensed formula for aldehydes is RCHO (CHO is our aldehyde CHUM) and RCOR' for ketones (no cute memorization aid - if you have one please share it.)
Naming Aldehydes
The IUPAC system of nomenclature assigns a characteristic suffix -al to aldehydes. For example, H2C=O is methanal, more commonly called formaldehyde. Since an aldehyde carbonyl group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. There are several simple carbonyl containing compounds which have common names which are retained by IUPAC.
Also, there is a common method for naming aldehydes and ketones. For aldehydes common parent chain names, similar to those used for carboxylic acids, are used and the suffix –aldehyde is added to the end. In common names of aldehydes, carbon atoms near the carbonyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
If the aldehyde moiety (-CHO) is attached to a ring the suffix –carbaldehyde is added to the name of the ring. The carbon attached to this moiety will get the #1 location number in naming the ring.
Summary of Aldehyde Nomenclature rules
1. Aldehydes take their name from their parent alkane chains. The -e is removed from the end and is replaced with -al.
2. The aldehyde funtional group is given the #1 numbering location and this number is not included in the name.
3. For the common name of aldehydes start with the common parent chain name and add the suffix -aldehyde. Substituent positions are shown with Greek letters.
4. When the -CHO functional group is attached to a ring the suffix -carbaldehyde is added, and the carbon attached to that group is C1.
Example 1
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Aldehyde Common Names to Memorize
Aldehydes often called the formyl groups. There are some common names that are still used and need to be memorized. Recognizing the patterns can be helpful.
Naming Ketones
The IUPAC system of nomenclature assigns a characteristic suffix of -one to ketones. A ketone carbonyl function may be located anywhere within a chain or ring, and its position is usually given by a location number. Chain numbering normally starts from the end nearest the carbonyl group. Very simple ketones, such as propanone and phenylethanone do not require a locator number, since there is only one possible site for a ketone carbonyl function. The common names for ketones are formed by naming both alkyl groups attached to the carbonyl then adding the suffix -ketone. The attached alkyl groups are arranged in the name alphabetically.
Summary of Ketone Nomenclature rules
1. Ketones take their name from their parent alkane chains. The ending -e is removed and replaced with -one.
2. The common name for ketones are simply the substituent groups listed alphabetically + ketone.
3. Some common ketones are known by their generic names. Such as the fact that propanone is commonly referred to as acetone.
Example 2
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Ketone Common Names to Memorize
There are some common names that are still used and need to be memorized. Recognizing the patterns can be helpful.
Naming Aldehydes and Ketones in the Same Molecule
As with many molecules with two or more functional groups, one is given priority while the other is named as a substituent. Because aldehydes have a higher priority than ketones, molecules which contain both functional groups are named as aldehydes and the ketone is named as an "oxo" substituent. It is not necessary to give the aldehyde functional group a location number, however, it is usually necessary to give a location number to the ketone.
Example 3
Naming Dialdehydes and Diketones
For dialdehydes the location numbers for both carbonyls are omitted because the aldehyde functional groups are expected to occupy the ends of the parent chain. The ending –dial is added to the end of the parent chain name.
Example 4
For diketones both carbonyls require a location number. The ending -dione or -dial is added to the end of the parent chain.
Example 5
Naming Cyclic Ketones and Diketones
In cyclic ketones the carbonyl group is assigned location position #1, and this number is not included in the name, unless more than one carbonyl group is present. The rest of the ring is numbered to give substituents the lowest possible location numbers. Remember the prefix cyclo is included before the parent chain name to indicate that it is in a ring. As with other ketones the –e ending is replaced with the –one to indicate the presence of a ketone.
With cycloalkanes which contain two ketones both carbonyls need to be given a location numbers. Also, an –e is not removed from the end, but the suffix –dione is added.
Example 6
Naming Carbonyls and Hydroxyls in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains an alcohol functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of alcohols the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
Example 7
Naming Carbonyls and Alkenes in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains analkene functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. When carbonyls are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an -en ending to indicate the presence of an alkene)-(the location number of the carbonyl if a ketone is present)-(either an –one or and -anal ending).
Remember that the carbonyl has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Example 8
Aldehydes and Ketones as Fragments
• Alkanoyl is the common name of the fragment, though the older naming, acyl, is still widely used.
• Formyl is the common name of the fragment.
• Acety is the common name of the CH3-C=O- fragment.
Example 9
Exercise
1. Give the IUPAC name for each compound and write the condensed formulas for parts (a), (b), (c), (d), (e), (f), (h), (i), and (l).
2) Draw the bond-line structure and write the condensed formula {except for (b), (d) and (h)} corresponding to each name:
A) butanal
B) 2-hydroxycyclopentanone
C) 2,3-pentanedione
D) 1,3-cyclohexanedione
E) 4-hydoxy-3-methyl-2-butanone
F) (E) 3-methyl-2-hepten-4-one
G) 3-oxobutanal
H) cis-3-bromocyclohexanecarboaldehyde
I) butanedial
J) trans-2-methyl-3-hexenal
Answer
Solutions
1.
a) 3,4-dimethylhexanal; CH3CH2CH(CH3)CH(CH3)CH2CHO
b) 5-bromo-2-pentanone; CH2BrCH2CH2COCH3
c) 2,4-hexanedione; CH3COCH2COCH2CH3
d) cis-3-Penenal; cis-CH3CHHCH2CHO
e) 6-methyl-5-hepten-3-one; CH3C(CH3)CHCH2COCH2CH3 or (CH3)2CCHCH2COCH2CH3
f) 3-hydroxy-2,4-pentanedione; CH3OCH(OH)COCH3
g) 1,2-cyclobutanedione
h) 2-methyl-propanedial; CHOCH(CH3)CHO
i) 3-methyl-5-oxo-hexanal; CH3OCH2CH(CH3)CH2CHO
j) cis-2,3-dihydroxycyclohexanone
k) 3-Bromo-2-methylcyclopentanecarboaldehyde
l) 3-bromo-2-methylpropanal; CHOCH(CH3)CH2Br
2. condensed formulas below and bond-line structures to the right
a) CH3CH2CH2CHO
c) CH3COCOCH2CH3
e) CH2(OH)CH(CH3)COCH3
f) CH3CHC(CH3)COCH2CH2CH3
g) CH3COCH2CHO
i) CHOCH2CH2CHO
j) CH3CH2CHCHCH(CH3)CHO | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.11%3A_Aldehydes_and_Ketones.txt |
learning objectives
• classify amines as primary, secondary, tertiary, quaternary, or heterocyclic
• name amines using IUPAC (systematic) and selected common name nomenclature
• draw the structure of amines from IUPAC (systematic) and selected common names
Amine bases 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.
Nomenclature
Amines are derivatives of ammonia in which one or more of the hydrogens has been replaced by an alkyl or aryl group. Amino compounds can be named either as derivatives of ammonia or as amino-substituted compounds:
The nomenclature of amines is further complicated by the fact that several different nomenclature systems exist, and there is no clear preference for one over the others. The four compounds shown in the top row of the following diagram are all C4H11N isomers. The first two are classified as 1º-amines, since only one alkyl group is bonded to the nitrogen; however, the alkyl group is primary in the first example and tertiary in the second. The third and fourth compounds in the row are 2º and 3º-amines respectively. The bottom row shows the structures for some common amines that need to be memorized.
• The Chemical Abstract Service has adopted a nomenclature system in which the suffix -amine is attached to the root alkyl name. For 1º-amines such as butanamine (first example) this is analogous to IUPAC alcohol nomenclature (-ol suffix). The additional nitrogen substituents in 2º and 3º-amines are designated by the prefix N- before the group name. These CA names are colored magenta in the diagram.
• Finally, a common system for simple amines names each alkyl substituent on nitrogen in alphabetical order, followed by the suffix -amine. These are the names given in the last row (colored black).
To be consistent and logical in naming amines as substituted ammonias, they strictly should be called alkanamines and arenamines, according to the nature of the hydrocarbon grouping. Unfortunately, the term alkylamine is used very commonly in place of alkanamine, while a host of trivial names are used for arenamines. We shall try to indicate both the trivial and the systematic names where possible. Some typical amines, their names, and their physical properties are listed in he Table below. The completely systematic names give in the Table illustrate the difficulty one gets into by using completely systematic names, and why simpler but less systematic names continue to be used for common compounds. A good example is \(\ce{N}\),\(\ce{N}\)-dibutylbutamine versus tributylamine. The special ways of naming heterocyclic amines can be referenced in the appendix of this chapter.
Common Amines and Their Properties
Alkaloids
Many biologically important compounds are amines. Alkaloids are amines synthesized by plants to protect them from being eaten. Humans primarily use alkaloids medicinally as pain killers. All alkaloids are toxic and addictive. The Greeks killed Socrates with (S)-coniine. Mild cases of alkaloid poisoning produce psychological effects resembling peacefulness, euphoria or hallucinations.
Ammonium Salts
A nitrogen bonded to four alkyl groups will necessarily be positively charged, and is called a 4º-ammonium cation. For example, (CH3)4N(+) Br(–) is tetramethylammonium bromide. Salts of amines with inorganic or organic acids are named as substituted ammonium salts, except when the nitrogen is part of a ring system. Examples are
\(^1\)Note the use of azonia to denote the cationic nitrogen in the ring, whereas aza is used for neutral nitrogen.
Perhaps the most noteworthy aspect of ammonium salts is that they have low odor and are water soluble. These qualities are explored more fully in the amine chapter
Heterocyclic Amine Nomenclature
Heterocyclic amines are amines in which the nitrogen is part of a ring that contains at least one double bond. Many aromatic and heterocyclic amines are known by unique common names, the origins of which are often unknown to the chemists that use them frequently. Since these names are not based on a rational system, it is necessary to memorize them. There is a systematic nomenclature of heterocyclic compounds, but it will not be discussed here. For further details, refer to the appendix of this chapter for the full IUPAC rules of organic compound nomenclature.
Exercise
1. Draw the bond-line structure for each compound and write the condensed structural formula for parts (a) & (d) - (g).
a) 3-bromo-pentan-2-amine
b) cyclopentanamine
c) trans-3-ethylcyclohexanamine
d) sec-butyl tert-butyl amine
e) N,N-dimethyl-3-pentanamine
f) 4-methyl-2-hexanamine
g) 6-bromo-4-amino-2-heptanol
2. Give the IUPAC name and condensed structural formula for each compound.
Answer
1.
2.
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)/03%3A_Functional_Groups_and_Nomenclature/3.12%3A_Amines_-_Classification_and_Nomenclature.txt |
learning objectives
• name carboxylic acids using IUPAC (systematic) and selected common name nomenclature
• draw the structure of carboxylic acids from IUPAC (systematic) and selected common names
The IUPAC system of nomenclature assigns a characteristic suffix to these classes. The –e ending is removed from the name of the parent chain and is replaced -anoic acid. Since a carboxylic acid group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. Many carboxylic acids are called by the common names that were chosen by chemists to usually describe the origin of the compound. In common names of aldehydes, carbon atoms near the carboxyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
Formula
Common Name
Source
IUPAC Name
Melting Point
Boiling Point
HCO2H formic acid ants (L. formica) methanoic acid 8.4 ºC 101 ºC
CH3CO2H acetic acid vinegar (L. acetum) ethanoic acid 16.6 ºC 118 ºC
CH3CH2CO2H propionic acid milk (Gk. protus prion) propanoic acid -20.8 ºC 141 ºC
CH3(CH2)2CO2H butyric acid butter (L. butyrum) butanoic acid -5.5 ºC 164 ºC
CH3(CH2)3CO2H valeric acid valerian root pentanoic acid -34.5 ºC 186 ºC
CH3(CH2)4CO2H caproic acid goats (L. caper) hexanoic acid -4.0 ºC 205 ºC
CH3(CH2)5CO2H enanthic acid vines (Gk. oenanthe) heptanoic acid -7.5 ºC 223 ºC
CH3(CH2)6CO2H caprylic acid goats (L. caper) octanoic acid 16.3 ºC 239 ºC
CH3(CH2)7CO2H pelargonic acid pelargonium (an herb) nonanoic acid 12.0 ºC 253 ºC
CH3(CH2)8CO2H capric acid goats (L. caper) decanoic acid 31.0 ºC 219 ºC
Naming carboxyl groups added to a ring
When a carboxyl group is added to a ring the suffix -carboxylic acid is added to the name of the cyclic compound. The ring carbon attached to the carboxyl group is given the #1 location number.
Naming carboxylates
Salts of carboxylic acids are named by writing the name of the cation followed by the name of the acid with the –ic acid ending replaced by an –ate ending. This is true for both the IUPAC and Common nomenclature systems.
Naming carboxylic acids which contain other functional groups
Carboxylic acids are given the highest nomenclature priority by the IUPAC system. This means that the carboxyl group is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of molecules containing carboxylic acid and alcohol functional groups the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
In the case of molecules containing a carboxylic acid and aldehydes and/or ketones functional groups the carbonyl is named as a "Oxo" substituent.
In the case of molecules containing a carboxylic acid an amine functional group the amine is named as an "amino" substituent.
When carboxylic acids are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an –enoic acid ending to indicate the presence of an alkene and carboxylic acid)
Remember that the carboxylic acid has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Naming dicarboxylic acids
For dicarboxylic acids the location numbers for both carboxyl groups are omitted because both functional groups are expected to occupy the ends of the parent chain. The ending –dioic acid is added to the end of the parent chain.
Common Names for selected dicarboxylic acids
The following common names for these selected dicarboxylic acids are important to memorize; they are prevalent in biochemistry or industrial applications.
The saying, "Oh my, such good apple pie!", can help us remember these common names by correlating the first letters of each word with the common names:
oxalic acid, malonic acid, succinic acid, glutaric acid, adipic acid, and phthalic.
Exercise
1. Draw the bond-line structure and write the condensed structural formula for each compound.
a) octanoic acid
b) 4-hydroxypentanoic acid
c) cis-4-hexenoioc acid or cis-hex-4-enoic acid
d) (E)-5-bromo-3-heptenoic acid or (E)-5-bromohept-3-enoic acid.
e) 2-aminopropanoic acid
2. Give the IUPAC name and condensed structural formula for each compound.
Answer
1.
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.13%3A_Carboxylic_Acids.txt |
learning objectives
• name acid halides, anhydrides, esters, amides, nitriles, and dicarboxylic acids using IUPAC (systematic) and selected common name nomenclature
• draw the structure of acid halides, anhydrides, esters, amides, and nitriles from IUPAC (systematic) and selected common names
Note: Nomenclature of thioesters and phosphoesters is also discussed. Ask the professor if this information is required for your course.
Introduction
The important classes of organic compounds known as alcohols, phenols, ethers, amines and halides consist of alkyl and/or aryl groups bonded to hydroxyl, alkoxyl, amino and halo substituents respectively. If these same functional groups are attached to an acyl group (RCO–) their properties are substantially changed, and they are designated as carboxylic acid derivatives. Carboxylic acids have a hydroxyl group bonded to an acyl group, and their functional derivatives are prepared by replacement of the hydroxyl group with substituents, such as halo, alkoxyl, amino and acyloxy. The carboxylic acid derivatives can all be hydrolyzed to carboxylic acids. The specific reaction conditions are discussed in the corresponding chapter later in this text, however, the shared pattern of chemical reactivity is summarized in the diagram below.
IUPAC Nomenclature - One Pattern, so Many Variations
Apply the IUPAC nomenclature format to carboxylic acid derivatives as summarized below using the suffix or substituent names listed in the table. Some students find esters challenging to name. Carboxylates can be described as independent ions, but require a contain to form compounds. It can be helpful to think of esters as "alkylated carboxylates": identify and name the carboxylate, this name is preceded by the alkyl group branch on the carboxyl oxygen.
Functional Group Structure Suffix Name Substituent Name
carboxylic acid -oic acid carboxy-
carboxylate -oate see above
ester -oate alkoxycarbonyl-
dicarboxylic acid -dioic acid not applicable
acyl halide -oyl halide not applicable
anhydride
-anhydride
not applicable
amide -amide amido-
nitrile -nitrile cyano-
Common names
Most common names were derived from older systems of nomenclature that some may argue were "not systematic at all". However, it is helpful to note that the older systems of nomenclature were often based on shared structural features and/or chemical reactivity. Understanding the older nomenclature systems can offer insights into chemical reactivity and structural patterns. There are some common names that are so prevalent, they need to be memorized. Common names frequently exist when the group bonded to the carbonyl carbon is a methyl group (indicated with "acet" or "acetyl" in the common name) or a hydrogen atom (indicated with "formyl" or "form"). For example CH3C≡N is ethanenitrile (or acetonitrile) and HCONH2 is methanamide (or formamide).
The common names for compounds with carbonyl groups often use Greek letters to specify the carbon position relative to the carbonyl carbon.
Example
In this text, we will learn about the patterns of reactivity for compounds with for beta-hydroxy carbonyl structures like the one shown below.
Nomenclature of acid halides
The nomenclature of acid halides starts with the name of the corresponding carboxylic acid. The –ic acid ending is removed and replaced with the ending -yl followed by the name of the halogen with an –ide ending. This is true for both common and IUPAC nomenclature. The carbonyl carbon is given the #1 location number. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
Example
Nomenclature of Anhydrides
The acid anhydride functional group results when two carboxylic acids combine and lose water (anhydride = without water). Symmetrical acid anhydrides are named like carboxylic acids except the ending -acid is replaced with -anhydride. This is true for both the IUPAC and Common nomenclature.
Symmetrical anhydrides
A symmetrical anhydride is a carboxylic acid anhydride that has the following general structural formula.
where R1=R2= hydrogen atoms, alkyl groups, aryl groups
Unsymmetrical Anhydrides
A mixed or unsymmetrical anhydride is a carboxylic acid anhydride that has the following general structural formula.
where R1≠R2, but are hydrogen atoms, alkyl groups, aryl groups. When naming unsymmetrical acid anhydrides, name both using alkanoic general method and then put the two names alphabetically. Hence, first name each component and alphabetically arranged them followed by spaces and then the word anhydride.
propanoic anhydride
ethanoic propanoic anhydride
Exercises
1. Draw the bond-line structure for benzoic anhydride.
Solution:
2. What is the common name for the compound below?
Solution: acetic benzoic anhydride
Common anhydride names to know
acetic anhydride (Try to name this anhydride by the proper name. J )
succinic anhydride (Try to name this anhydride by the proper name. J )
Nomenclature of Esters
Esters are made from a carboxylic acid and an alcohol.
Esters are named as if the alkyl chain from the alcohol is a substituent. No number is assigned to this alkyl chain. This is followed by the name of the parent chain form the carboxylic acid part of the ester with an –e remove and replaced with the ending –oate.
Example
Exercises
3. Draw the bond-line structure for phenyl hexanoate.
Solution
4. What is the IUPAC name for the compound below?
Solution
methyl benzoate
Nomenclature of Lactones (Cyclic Esters)
Cyclic esters are called lactones. A Greek letter identifies the location of the alkyl oxygen relative to the carboxyl carbonyl group.
Nomenclature of Amides
Primary amides
Primary amides are named by changing the name of the acid by dropping the -oic acid or -ic acid endings and adding -amide. The carbonyl carbon is given the #1 location number. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
methanamide or formamide (left), ethanamide or acetamide (center) , benzamide (right)
Exercises
5. Draw the bond-line structure for 3-chloro-N-ethylbenzamide.
Solution
6. What is the IUPAC name the compound below?
Solution
pentanamide
Secondary amides
Secondary amides are named by using an upper case N to designate that the alkyl group is on the nitrogen atom. Alkyl groups attached to the nitrogen are named as substituents. The letter N is used to indicate they are attached to the nitrogen. Tertiary amides are named in the same way.
N-methylpropanamide
Exercises
7. Draw the bond-line structure for N,N-dimethylformamide.
Solution
8. What are the IUPAC and common names the compound below?
Solution
N-phenylethanamide and N-phenylacetamide, respectively
Cyclic amides are called lactams. A Greek letter identifies the location of the nitrogen on the alkyl chain relative to the carboxyl carbonyl group.
Nomenclature of Nitriles
Name the parent alkane (include the carbon atom of the nitrile as part of the parent) followed with the word -nitrile. The carbon in the nitrile is given the #1 location position. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain. A nitrile substituent, e.g. on a ring, is named carbonitrile.
include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
Example
(3-methylbutanenitrile (or isovaleronitrile) cyclopentanecarbonitrile
Nomenclature of Thioesters
Thiosters are made from a carboxylic acid and an thiol.
Thioesters are named as if the alkyl chain from the alcohol is a substituent. No number is assigned to this alkyl chain. This is followed by the name of the parent chain from the carboxylic acid part of the thioester named as an alkane with the ending –thiooate.
Example
Nomenclature of Phosphates
Phosphoryl groups are derivatives of phosphoric acid, a strong acid that is commonly used in the laboratory. The fully deprotonated conjugate base of phosphoric acid is called a phosphate ion, or inorganic phosphate (often abbreviated 'Pi'). When two phosphate groups are linked to each other, the linkage is referred to as a 'phosphate anhydride', and the ion is called 'inorganic pyrophosphate' (abbreviation PPi).
When a phosphate ion is attached to a carbon atom on an organic molecule, the chemical linkage is referred to as a phosphate ester, and the whole species is called an organic monophosphate. Glucose-6-phosphate is an example.
If an organic molecule is linked to two or three phosphate groups, the resulting species are called organic diphosphates and organic triphosphates.
Isopententyl diphosphate and adenosine triphosphate (ATP) are good examples:
Oxygen atoms in phosphate groups are referred to either 'bridging' and 'non-bridging', depending on their position. An organic diphosphate has two bridging and five non-bridging oxygens.
When a single phosphate is linked to two organic groups, the term 'phosphate diester' is used. The backbone of DNA is composed of phosphate diesters.
The term 'phosphoryl group' is a general way to refer to all of the phosphate-based groups mentioned in the paragraphs above.
Recall that phosphate groups on organic structures are sometimes abbreviated simply as 'P', a convention that we will use throughout this text. For example, glucose-6-phosphate and isopentenyl diphosphate are often depicted as shown below. Notice that the 'P' abbreviation includes the oxygen atoms and negative charges associated with the phosphate groups.
Exercise
1. Name the following compounds using IUPAC conventions
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Answer
1.
a) 3-methylpentanoyl chloride
b) 2-cyclopentylacetamide
c) propyl2-methylpropanoate
d) cyclohexylbutanoate
e) tert-butylcyclopentanecarboxylate
f) 1-methylbutylcyclopentane carboxylate
g) N-methyl-3-butenamide
h) (S)-2-hydroxypropanoyl phosphate
i) propyl 2,3-dimethyl-2-butenethioate | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.14%3A__The_Carboxylic_Acid_Derivatives.txt |
Alkane Nomenclature
3-1 Draw the structure for each of the following compounds
(a) sec-butylcyclohexane
(b) 2,2,7-trimethyloctane
(c) 3-ethyl-2,4,5-trimethylhexane
(d) 1-cyclopropylbutane
(e) 3-ethyl-2,5-dimethyl-4-propylheptane
(f) 1-tert-butyl-4-iodocyclohexane
(g) trans-1,4-diisopropylcyclohexane
(h) 3-bromo-1-cyclopropyl-2-methylbutane
(i) Z-1,2-dichloro-1-methylcyclopentane
(j) 1-chloro-3-ethyl-3-methylhexane
(k) Norbornane (Bicyclo[2.2.1]heptane)
(l) 1-[2-(bromomethyl)cyclopropyl]-4-tert-butylcyclohexane
3-2 Give the IUPAC names of the following alkanes.
(a) (CH3)2CHCH2CH2CH(CH2CH3)CH(CH3)CH2CH3
(b)
(c)
(d)
(e)
(f)
(g)
(h)
3-3 The following names are incorrect or incomplete, but they represent the real structures. Draw each structures and name it correctly.
(a) 2-ethyl-3-methylhexane
(b) 2-isopropylpentane
(c) 2-bromo-5-isopropylheptane
(d) 4-ethyl-3-isopropylheptane
(e) 2,4-trimethylpentane
(f) 1-chloro-2-ethyl-5-methylcyclohexane
Alkyl Halide Nomenclature
3-4 Draw the structures of these following compounds
a) 2-chlorohexane
b) 1-bromo-2-chlorocyclopentane
c) Isobutyl chloride
d) 5-chloro-2,3-dimethylhexane
e) (2S) 2-bromohexane
3-5 Give IUPAC names to the following compounds
a)
b)
c)
d)
3-6 Classify each of the following compounds as primary, secondary, or tertiary alkyl halides
a)
b)
c)
d)
Alkene Nomenclature
3-7 Give the IUPAC name for the following alkene structures.
3-8 Give the structures of the following compounds.
a) [1,1'-bi(cyclohexan)]-1-ene
b) 2,3-dimethylbut-2-ene
c) (4E)-4-ethylidene-2-methylhept-1-ene
d) [(1E,3E)-hexa-1,3-dien-1-yl]cyclopentane
3-9 State whether the following compounds are Z, E, or neither.
3-10 Give the IUPAC name for the following structures.
Alkyne Nomenclature
3-11 Draw the bond-line for these following compounds:
a) 3-hexyne
b) 2-bromo-5-octyne
c) 3,3-dimethyl-6-decyne
d) Cyclopentylacetylene
e) 2-chloronon-4-en-6-yne
f) 2,4-heptadiyne
3-12 Give IUPAC names for the following compounds
a) b) c) d)
Alcohol and Phenol Nomenclature
3-13 Classify each alcohol as primary, secondary, or tertiary.
3-14 Give a systematic (IUPAC) name for each alcohol.
3-15 Draw the structures of the following compounds.
a) 2-methylhexan-1-ol
b) 2-methylpentan-2-ol
c) 2-chlorophenol
d) 4-(chloromethyl)hexan-1-ol
e) cyclopent-2-en-1-ol
f) 3-bromo-2-(2-bromoethyl)pentan-1-ol
3-16 Give a systematic (IUPAC) name for each phenol.
Ethers
3-17 Give the IUPAC name for the following chemical structures.
(g) (h) (i)
3-18 Draw structures of the following.
(a) 3-isopropoxypentane (b) 1-(4-chlorophenoxy)-3-methylbenzene (c) 2-(tert-butoxy)-2-methylpropane
3-19 Name the following ethers and sulfides.
(a) (b) (c) (d) (e)
Benzene and its Derivatives
3-20 State wither the following is para, meta, or ortho substituted.
3-21 Name the following compounds.
3-22 Draw the following structures
1. p-chloroiodobenzene
2. m-bromotoluene
3. p-chloroaniline
4. 1,3,5-trimethylbenzene
3-23 Give the IUPAC name for the following benzene derivatives.
3-24 Draw the following molecules given by their IUPAC nomenclature.
1. 5-formyl-2-hydroxybenzoic acid
2. 1,2-diethyl-3-fluorobenzene
3. 4-amino-3-ethyl-5-methylphenol
3-25 Rank the functional groups on the following molecule in order of priority (highest to lowest).
Aldehydes and Ketones
3-26 For the following molecules, give the IUPAC nomenclature.
Amines
3-27 Draw the structures given by the following IUPAC names.
1. 4-methylpentan-2-amine
2. N-ethyl-N-methylpropan-1-amine
3. 6-aminocyclohex-1-ene-1-carboxylic acid
4. (dimethylamino)acetonitrile
Carboxylic Acids
3-28 Give the correct IUPAC nomenclature for the following compounds.
Carboxylic Acid Derivatives
3-29 Name the following compounds using IUPAC conventions
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
3.16: Solutions to Additional Exercises
Alkane Nomenclature
3-1 The structure that corresponds with each name is drawn below.
(a) sec-butylcyclohexane
(b) 2,2,7-trimethyloctane
(c) 3-ethyl-2,4,5-trimethylhexane
(d) 1-cyclopropylbutane
(e) 3-ethyl-2,5-dimethyl-4-propylheptane
(f) 1-tert-butyl-4-iodocyclohexane
(g) trans-1,4-diisopropylcyclohexane
(h) 3-bromo-1-cyclopropyl-2-methylbutane
(i) Z-1,2-dichloro-1-methylcyclopentane
(j) 1-chloro-3-ethyl-3-methylhexane
(k) Norbornane (Bicyclo[2.2.1]heptane)
(l) 1-[2-(bromomethyl)cyclopropyl]-4-tert-butylcyclohexane
3-2 Give the IUPAC names of the following alkanes.
(a) 5-ethyl-2,6-dimethyloctane
(b) 3,4,7-trimethyl-6-(1-methylethyl)nonane
(c) 4-(1-methylpropyl)-2,3-dimethyloctane
(d) 1,3-dimethyl-2-(1-methylethyl)cyclobutane
(e) 1-tert-butyl-4-ethyl-2-methylcyclopentane
(f) trans-1,3-diethylcyclohexane
(g) cis-1,4-di(1-methylethyl)cyclononane
(h) 3-methylhexane
3-3 The following names are incorrect or incomplete, but they represent the real structures. Draw each structures and name it correctly.
(a)
(b)
(c)
(d)
(e)
(f)
Alkyl Halide Nomenclature
3-4
a)
b)
c)
d)
e)
3-5
a) 3-chloropentane
b) 1-chloro-3-methylcyclohexane
c) 5-chloro-2,2-dimethylhexane
d) (1R,2R)-1-bromo-2-methylcyclohexane
3-6
a) Secondary
b) Primary
c) Tertiary
d) Tertiary
3-7
3-8
3-9
a) Z
b) Neither
c) E
3-10
Alkyne Nomenclature
3-11
a)
b)
c)
d)
e)
f)
3-12
a) 1-chloropent-3-yne
b) 2,2-dimethyl-6-methyl-hex-4-yne
c) cyclohexylhept-1-yne
d) 4-bromo-2-chloro-3-methylnon-7-yne
Alcohol and Phenol Nomenclature
3-13
a) Primary
b) Secondary
c) Tertiary
d) Secondary
3-14
3-15
3-16
Ethers
3-17
(g) oxydicyclopentane
(h) 2-phenyloxirane
(i) 1-cyclohexylethane-1-thiol
3-18
(a) (b) (c)
3-19
(a) diisopropylsulfide (b) 1,3-dimethoxybenzene (c) 2-Methyltetrahydro-2H-pyran (d) methyl 3-sulfanylbenzoate (e) methyl(phenyl)sulfide
Benzene and its Derivatives
3-20
A – meta; B – para; C – ortho
3-21
1. 1,3-Dibromobenzene
2. 1-phenyl-4-methylhexane
3. 1,4-Dichloro-2,5-dimethylbenzene
4. 2-methyl-1,3,5-trinitrobenzene. (Also known as trinitrotoluene, or TNT)
13-22
3-23
3-24
3-25 In order of priority: Benzoyl bromide > Amino > Nitro
3-26
3-27
3-28
Carboxylic Acid Derivatives
3-29
1. 3-methylpentanoyl chloride
2. 2-cyclopentylacetamide
3. propyl 2-methylpropanoate
4. cyclohexylbutanoate
5. tert-butyl cyclopentanecarboxylate
6. 1-methylbutylcyclopentane carboxylate
7. N-methyl-3-butenamide
8. (S)-2-hydroxypropanoyl phosphate
9. propyl 2,3-dimethyl-2-butenethioate
3.17: Appendix - IUPAC Nomenclature Rules
Wikipedia Summary
Full Text of IUPAC Rules
Nomenclature 101 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/03%3A_Functional_Groups_and_Nomenclature/3.15%3A__Additional_Exercises.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• distinguish between the different hydrocarbon functional groups - refer to section 4.1
• explain & predict the physical properties of alkanes including relative bp and solubility in a mixture - refer to section 4.2
• interpret and draw the rotation about a carbon-carbon single bond using Newman projections and sawhorse structures - refer to section 4.3 - 4.5
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for butane, higher alkanes, cyclohexane, mono-substituted cyclohexanes, and disubstituted cyclohexanes - refer to sections 4.3, 4.3, 4.4, 4.5, and 4.7, 4.8, and 4.10 respectively
• explain the partial rotation of carbon-carbon single bonds in rings - refer to section 4.6
• explain ring strain and its relationship to cycloalkane stability - refer to section 4.6
• draw cyclohexane conformations (chair & boat) - refer to section 4.7
• draw mono-substituted cyclohexane conformers (chair only) - refer to section 4.8
• identify & draw the geometric (cis/trans) isomers of cycloalkanes - refer to section 4.9
• draw di-substituted cyclohexane conformers (chair only) - refer to section 4.10
• recognize and draw the three ways to join two rings - refer to section 4.11
• describe the uses and sources of alkanes - refer to section 4.12
• recognize and distinguish between the two major reactions of alkanes (combustion and halogenation) - refer to section 4.13
• 4.1: Hydrocarbon Functional Groups
Hydrocarbons are organic compounds that consist entirely of carbon and hydrogen atoms. Hydrocarbons can form different functional groups based upon their carbon bonding patterns as alkanes, alkenes, alkynes, or arenes.
• 4.2: Physical Properties of Alkanes
Alkanes are not very reactive and have little biological activity; alkanes are colorless, odorless non-polar compounds.
• 4.3: Structure and Conformations of Alkanes
The carbon-carbon single bonds of alkanes rotate freely. Conformers are the same molecule shown with different sigma bond rotations. Newman projections are one way to communicate bond rotation.
• 4.4: Conformations of Butane
The conformations of butane are studied to introduce the language and energetic considerations of single bond rotation when alkyl group interactions can occur.
• 4.5: Conformations of Higher Alkanes
Pentane and higher alkanes have conformational preferences similar to ethane and butane. Each dihedral angle tries to adopt a staggered conformation and each internal C-C bond attempts to take on an anti conformation to minimize the potential energy of the molecule.
• 4.6: Cycloalkanes and Ring Strain
For cyclic alkanes, only partial rotation of carbon-carbon single bonds can occur. The actual shape of the carbon ring distorts from the traditional geometric shapes to reduce steric hindrance and ring strain to lower the overall potential energy of the molecule.
• 4.7: Cyclohexane Conformations
Cyclohexane rings are notably stable. Understanding the conformations of cyclohexane and their relative energies is helpful when studying the chemistry of simple carbohydrates (monosaccharides).
• 4.8: Conformations of Monosubstituted Cyclohexanes
For monosubstituted cyclohexanes, the axial or equatorial orientation of the substituent influences the overall potential energy of the conformation.
• 4.9: Cis-trans Isomerism in Cycloalkanes
Stereoisomerism is possible for cycloalkanes with two different substituent groups (not counting other ring atoms). Cis and trans isomers are unique compounds.
• 4.10: Conformations of Disubstituted Cyclohexanes
Because six-membered rings are so common among natural and synthetic compounds and its conformational features are rather well understood, we shall focus on the six-membered cyclohexane ring to study the energetic relationship of conformation and overall potential energy.
• 4.11: Joined Rings
Two or more rings can be fused together into a bicyclic or spirocyclic system. The steroid fused ring system is a notable example.
• 4.12: Uses and Sources of Alkanes
The primary sources for alkanes are oil and natural gas. Alkanes are important raw materials for the chemical industry and are used as fuels for motors.
• 4.13: Reactions of Alkanes - a Brief Overview
Alkanes (the most basic of all organic compounds) undergo very few reactions. The two reactions of more importaces is combustion and halogenation, (i.e., substitution of a single hydrogen on the alkane for a single halogen) to form a haloalkane. The halogen reaction is very important in organic chemistry because it opens a gateway to further chemical reactions.
• 4.14: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 4.15: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
04: Structure and Stereochemistry of Alkanes
Learning Objective
• distinguish between the different hydrocarbon functional groups
Saturated vs. Unsaturated Molecules
Hydrocarbons are organic compounds that contain only carbon and hydrogen. The broadest distinction between hydrocarbons is whether they are saturated and unsaturated. Saturated hydrocarbons only contain carbon-carbon single bonds with the maximum number of hydrogens relative to the number of carbon atoms. It can be said that the carbon atoms are "saturated" with hydrogen atoms in the same way a saturated solution has dissolved the maximum amount of solute. Hydrocarbons that contain pi bonds as carbon-carbon double or triple bonds are classified as unsaturated hydrocarbons. Unsaturation indicates that some of the carbon-hydrogen bonds were lost to from pi bonds between carbon atoms. There are less than the maximum number os hydrogens relative to the number of carbon atoms.
1. Saturated hydrocarbons (alkanes) are the simplest of the hydrocarbon species. They are composed entirely of single bonds and are saturated with hydrogen. Saturated hydrocarbons are the basis of petroleum fuels and are found as either linear or branched species.The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane:
2. Unsaturated hydrocarbons have one or more double or triple bonds between carbon atoms. Those with double bond are called alkenes and those with one double bond have the formula $C_nH_{2n}$ (assuming non-cyclic structures). Those containing triple bonds are called alkynes, with general formula CnH2n-2.
The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene:
The smallest alkyne is ethyne, which is also known as acetylene:
3. For now, we will focus on benzene as the representative aromatic hydrocarbon. Aromatic compounds were first noted for their strong aromas and low chemical reactivity compared to other saturated hydrocarbons. Aromatic compounds will be discussed in greater detail in the second semester to organic chemistry.
Hydrocarbon Functional Groups
The four distinct hydrocarbon functional groups are: alkanes, alkenes, alkynes and arenes. Aromatic compounds derive their names from the fact that many of these compounds in the early days of discovery were grouped because they were oils with fragrant odors.
Alkanes are organic compounds that consist entirely of single-bonded carbon and hydrogen atoms and lack any other functional groups. Alkanes have the general formula CnH2n+2. Alkanes can be subdivided into the following three groups: the linear straight-chain alkanes, branched alkanes, and cycloalkanes. Alkanes are also saturated hydrocarbons. Alkanes are the simplest and least reactive hydrocarbon species containing only carbons and hydrogens. The distinguishing feature of an alkane, making it distinct from other compounds that also exclusively contain carbon and hydrogen, is its lack of unsaturation. That is to say, it contains no double or triple bonds, which are highly reactive in organic chemistry. Though not totally devoid of reactivity, their lack of reactivity under most laboratory conditions makes them a relatively uninteresting, though very important component of organic chemistry. As you will learn about later, the energy confined within the carbon-carbon bond and the carbon-hydrogen bond is quite high and their rapid oxidation produces a large amount of heat, typically in the form of fire.
The general formula for saturated hydrocarbons is CnH2n+2(assuming non-cyclic structures) as shown in hexane (C6H14) below.
Cycloalkanes are hydrocarbons containing one or more carbon rings to which hydrogen atoms are attached. The general formula for a cyclic hydrocarbon containing one ring is CnH2n as shown in cyclohexane (C6H12) below.
Akenes contain at least one carbon-carbon double bond and alkynes contain at least one carbon-carbon triple bond. Alkenes have the general formula CnH2n. Alkynes have the general formula CnH2n-2. The ratio of carbon to hydrogen increased because hydrogen atoms are replaced with pi bonds as shown in trans-2-butene (C4H8) and 2-butyne (C4H6) below. Since both double and triple bonds include pi bonds, Alkenes and alkynes share similar chemical reactivity.
Aromatic hydrocarbons, also known as arenes, are hydrocarbons that have at least one aromatic ring. Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds:
Calculating Degrees of Unsaturation (DU)
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation (DU) is useful information. Knowing the degrees of unsaturation tells us the combined number of pi bonds and rings within a compound which makes it easier to figure out the molecular structure.
Degree of Unsaturation (DU) can be calculated with the equation below and the molecular formula
DU= (2C+2+N-X-H)/2
where: C is the number of carbons; N is the number of nitrogens; X is the number of halogens (F, Cl, Br, I); and H is the number of hydrogens from the molecular formula.
As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6.
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
DU
Possible combinations of rings/ bonds
# of rings
# of double bonds
# of triple bonds
1
1
0
0
0
1
0
2
2
0
0
0
2
0
0
0
1
1
1
0
3 3 0 0
2 1 0
1 2 0
0 1 1
0 3 0
1 0 1
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example
What is the Degree of Unsaturation for benzene?
Solution - "Thinking it through"
The molecular formula for benzene is C6H6. Thus,
DU= 4, where C=6, N=0,X=0, and H=6. 1 DU can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
Even though there are other possible structures with a DU = 4, like the ones shown below. We will learn the benzene rings have unusual stability and occur frequently in the world of organic chemistry. When the DU for a compound is > 4, we can assume the presence of at least one benzene ring.
Exercise
1. Are the following molecules saturated or unsaturated:
1. (b.) (c.) (d.) C10H6N4
2. Using the molecules from 1., give the degrees of unsaturation for each.
3. Calculate the degrees of unsaturation for the following molecular formulas:
1. (a.) C9H20 (b.) C7H8 (c.) C5H7Cl (d.) C9H9NO4
4. Using the molecular formulas from 3, are the molecules unsaturated or saturated.
5. Using the molecular formulas from 3, if the molecules are unsaturated, how many rings/double bonds/triple bonds are predicted?
Answer
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3. Use the formula to solve
(a.) 0
(b.) 4
(c.) 2
(d.) 6
4.
(a.) saturated
(b.) unsaturated
(c.) unsaturated
(d.) unsaturated
5.
(a.) 0 (Remember-a saturated molecule only contains single bonds)
(b.) The molecule can contain any of these combinations (i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
• Kim | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.01%3A_Hydrocarbon_Functional_Groups.txt |
Learning Objective
• explain & predict the physical properties of alkanes including relative bp and solubility in a mixture
Overview
Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless non-polar compounds. The relative weak London dispersion forces of alkanes result in gaseous substances for short carbon chains, volatile liquids with densities around 0.7 g/mL for moderate carbon chains, and solids for long carbon chains. The differences in the physical states occurs because there is a direct relationship between the size and shape of molecules and the strength of the intermolecular forces (IMFs).
Because alkanes have relatively predictable physical properties and undergo relatively few chemical reactions other than combustion, they serve as a basis of comparison for the properties of many other organic compound families. Let’s consider their physical properties first.
Boiling Points
Table \(1\) describes some of the properties of some of the first 10 straight-chain alkanes. Because alkane molecules are nonpolar, they are insoluble in water, which is a polar solvent, but are soluble in nonpolar and slightly polar solvents. Consequently, alkanes themselves are commonly used as solvents for organic substances of low polarity, such as fats, oils, and waxes. Nearly all alkanes have densities less than 1.0 g/mL and are therefore less dense than water (the density of H2O is 1.00 g/mL at 20°C). These properties explain why oil and grease do not mix with water but rather float on its surface.
Table \(1\): Physical Properties of Some Alkanes
Molecular Name Formula Melting Point (°C) Boiling Point (°C) Density (20°C)* Physical State (at 20°C)
methane CH4 –182 –164 0.668 g/L gas
ethane C2H6 –183 –89 1.265 g/L gas
propane C3H8 –190 –42 1.867 g/L gas
butane C4H10 –138 –1 2.493 g/L gas
pentane C5H12 –130 36 0.626 g/mL liquid
hexane C6H14 –95 69 0.659 g/mL liquid
octane C8H18 –57 125 0.703 g/mL liquid
decane C10H22 –30 174 0.730 g mL liquid
*Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure.
The boiling points for the "straight chain" isomers and isoalkanes isomers are shown to demonstrate that branching decreases the surfaces area, weakens the IMFs, and lowers the boiling point.
This next diagrams summarizes the physical states of the first six alkanes. The first four alkanes are gases at room temperature, and solids do not begin to appear until about \(C_{17}H_{36}\), but this is imprecise because different isomers typically have different melting and boiling points. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers!
Cycloalkanes have boiling points that are approximately 20 K higher than the corresponding straight chain alkane.
There is not a significant electronegativity difference between carbon and hydrogen, thus, there is not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be Van der Waals dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size.
Where you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules.
Example: Structure dependent Boiling Points
For example, the boiling points of the three isomers of \(C_5H_{12}\) are:
• pentane: 309.2 K
• 2-methylbutane: 301.0 K
• 2,2-dimethylpropane: 282.6 K
The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"!
Solubility
Alkanes (both alkanes and cycloalkanes) are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds.
Solubility in Water
When a molecular substance dissolves in water, the following must occur:
• break the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces.
• break the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds.
Breaking either of these attractions requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is relatively negligible; this is not true of the hydrogen bonds in water.
As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water.; the alkane does not dissolve.
The energy only description of solvation is an oversimplification because entropic effects are also important when things dissolve.
The lack of water solubility can lead to environmental concerns when oils are spilled into natural bodies of water as shown below.
Oil Spills. Crude oil coats the water’s surface in the Gulf of Mexico after the Deepwater Horizon oil rig sank following an explosion. The leak was a mile below the surface, making it difficult to estimate the size of the spill. One liter of oil can create a slick 2.5 hectares (6.3 acres) in size. This and similar spills provide a reminder that hydrocarbons and water don’t mix. Source: Photo courtesy of NASA Goddard / MODIS Rapid Response Team, http://www.nasa.gov/topics/earth/features/oilspill/oil-20100519a.html.
Solubility in organic solvents
In most organic solvents, the primary forces of attraction between the solvent molecules are Van der Waals - either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility.
Looking Closer: Gas Densities and Fire Hazards
Table \(1\) indicates that the first four members of the alkane series are gases at ordinary temperatures. Natural gas is composed chiefly of methane, which has a density of about 0.67 g/L. The density of air is about 1.29 g/L. Because natural gas is less dense than air, it rises. When a natural-gas leak is detected and shut off in a room, the gas can be removed by opening an upper window. On the other hand, bottled gas can be either propane (density 1.88 g/L) or butanes (a mixture of butane and isobutane; density about 2.5 g/L). Both are much heavier than air (density 1.2 g/L). If bottled gas escapes into a building, it collects near the floor. This presents a much more serious fire hazard than a natural-gas leak because it is more difficult to rid the room of the heavier gas.
Also shown in Table \(1\) are the boiling points of the straight-chain alkanes increase with increasing molar mass. This general rule holds true for the straight-chain homologs of all organic compound families. Larger molecules have greater surface areas and consequently interact more strongly; more energy is therefore required to separate them. For a given molar mass, the boiling points of alkanes are relatively low because these nonpolar molecules have only weak dispersion forces to hold them together in the liquid state.
Looking Closer: An Alkane Basis for Properties of Other Compounds
An understanding of the physical properties of the alkanes is important in that petroleum and natural gas and the many products derived from them—gasoline, bottled gas, solvents, plastics, and more—are composed primarily of alkanes. This understanding is also vital because it is the basis for describing the properties of other organic and biological compound families. For example, large portions of the structures of lipids consist of nonpolar alkyl groups. Lipids include the dietary fats and fatlike compounds called phospholipids and sphingolipids that serve as structural components of living tissues. These compounds have both polar and nonpolar groups, enabling them to bridge the gap between water-soluble and water-insoluble phases. This characteristic is essential for the selective permeability of cell membranes.
Tripalmitin (a), a typical fat molecule, has long hydrocarbon chains typical of most lipids. Compare these chains to hexadecane (b), an alkane with 16 carbon atoms.
Exercise
1. Without referring to a table, predict which has a higher boiling point—hexane or octane. Explain.
2. If 25 mL of hexane were added to 100 mL of water in a beaker, which of the following would you expect to happen? Explain.
1. Hexane would dissolve in water.
2. Hexane would not dissolve in water and would float on top.
3. Hexane would not dissolve in water and would sink to the bottom of the container.
3. Without referring to a table or other reference, predict which member of each pair has the higher boiling point.
1. pentane or butane
2. heptane or nonane
4. For which member of each pair is hexane a good solvent?
1. pentane or water
2. sodium chloride or soybean oil
Answer
1. octane because of its greater molar mass
2. b; Hexane is insoluble in water and is less dense than water so it floats on top.
3. a) pentane
b) nonane
4. a) pentane
b) soybean oil | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.02%3A_Physical_Properties_of_Alkanes.txt |
Learning Objective
• interpret and draw the rotation about a carbon-carbon single bond using Newman projections and sawhorse structures
• correlate energies of conformations with rotational energy diagrams
Single Bond Rotation and Conformational Isomerism
Conformational isomerism involves rotation about sigma bonds, and does not involve any differences in the connectivity or geometry of bonding. Two or more structures that are categorized as conformational isomers, or conformers, are really just two of the exact same molecule that differ only in terms of the angle about one or more sigma bonds. The carbon-carbon single bonds of alkanes rotate freely. Conformers are the same molecule shown with different sigma bond rotations. Newman projections are one way to communicate bond rotation.
Ethane Conformations
Although there are seven sigma bonds in the ethane molecule, rotation about the six carbon-hydrogen bonds does not result in any change in the shape of the molecule because the hydrogen atoms are essentially spherical. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations.
In order to better visualize these different conformations, it is convenient to use a drawing convention called the Newman projection. In a Newman projection, we look lengthwise down a specific bond of interest – in this case, the carbon-carbon bond in ethane. When there are multiple carbons,then we specify the bond of interest using the carbon numbers from the IUPAC name. We depict the ‘front’ atom as a dot, and the ‘back’ atom as a larger circle.
The six carbon-hydrogen bonds are shown as solid lines protruding from the two carbons at 120°angles, which is what the actual tetrahedral geometry looks like when viewed from this perspective and flattened into two dimensions.
The lowest energy conformation of ethane, shown in the figure above, is called the ‘staggered’ or 'anti' conformation, in which all of the C-H bonds on the front carbon are positioned at dihedral angles of 60°relative to the C-H bonds on the back carbon. In this conformation, the distance between the bonds (and the electrons in them) is maximized.
If we now rotate the front CH3 group 60°clockwise, the molecule is in the highest energy ‘eclipsed' conformation, where the hydrogens on the front carbon are as close as possible to the hydrogens on the back carbon.
This is the highest energy conformation because of unfavorable interactions between the electrons in the front and back C-H bonds. The energy of the eclipsed conformation is approximately 3 kcal/mol higher than that of the staggered conformation.
Another 60°rotation returns the molecule to a second eclipsed conformation. This process can be continued all around the 360°circle, with three possible eclipsed conformations and three staggered conformations, in addition to an infinite number of variations in between.
The carbon-carbon bond is not completely free to rotate – there is indeed a small, 3 kcal/mol barrier to rotation that must be overcome for the bond to rotate from one staggered conformation to another. This rotational barrier is not high enough to prevent constant rotation except at extremely cold temperatures. However, at any given moment the molecule is more likely to be in a staggered conformation - one of the rotational ‘energy valleys’ - than in any other state.
Free Rotations Do Not Exist in Ethane
The carbon-carbon bond is not completely free to rotate – there is indeed a small, 3 kcal/mol barrier to rotation that must be overcome for the bond to rotate from one staggered conformation to another. This rotational barrier is not high enough to prevent constant rotation except at extremely cold temperatures. However, at any given moment the molecule is more likely to be in a staggered conformation - one of the rotational ‘energy valleys’ - than in any other state. The potential energy associated with the various conformations of ethane varies with the dihedral angle of the bonds, as shown below.
The potential energy associated with the various conformations of ethane varies with the dihedral angle of the bonds.
Although the conformers of ethane are in rapid equilibrium with each other, the 3 kcal/mol energy difference leads to a substantial preponderance of staggered conformers (> 99.9%) at any given time. The animation below illustrates the relationship between ethane's potential energy and its dihedral angle
Exercise
1. Draw the Newman projections for the staggered and eclipsed conformers of propane along the C1-C2 axis.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.03%3A_Structure_and_Conformations_of_Alkanes.txt |
Learning Objective
• interpret and draw the rotation about a carbon-carbon single bond using Newman projections and sawhorse structures
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for butane
Butane Conformations
Now let's consider butane, with its four-carbon chain. There are now three rotating carbon-carbon bonds to consider, but we will focus on the middle bond between C2 and C3. Below are two representations of butane in a conformation which puts the two CH3 groups (C1 and C4) in the eclipsed position, with the two C-C bonds at a 0o dihedral angle.
If we rotate the front, (blue) carbon by 60° clockwise, the butane molecule is now in a staggered conformation.
This is more specifically referred to as the gauche conformation of butane. Notice that although they are staggered, the two methyl groups are not as far apart as they could possibly be.
A further rotation of 60° gives us a second eclipsed conformation (B) in which both methyl groups are lined up with hydrogen atoms.
One more 60 rotation produces another staggered conformation called the anti conformation, where the two methyl groups are positioned opposite each other (a dihedral angle of 180o).
As with ethane, the staggered conformations of butane are energy 'valleys', and the eclipsed conformations are energy 'peaks'. However, in the case of butane there are two different valleys, and two different peaks. The gauche conformation is a higher energy valley than the anti conformation due to steric strain, which is the repulsive interaction caused by the two bulky methyl groups being forced too close together. Clearly, steric strain is lower in the anti conformation. In the same way, steric strain causes the eclipsed A conformation - where the two methyl groups are as close together as they can possibly be - to be higher in energy than the two eclipsed B conformations.
The diagram below summarizes the relative energies for the various eclipsed, staggered, and gauche conformations.
The following diagram illustrates the change in potential energy that occurs with rotation about the C2–C3 bond at smaller rotational increments.
Potential curve vs dihedral angle of the C2-C3 bond of butane.
Because the anti conformation (staggered) is lowest in energy (and also simply for ease of drawing), it is conventional to draw open-chain alkanes in a 'zigzag' form, which implies anti conformation at all carbon-carbon bonds. The figure below shows, as an example, a Newman projection looking down the C2-C3 bond of octane.
Exercise
1: Using free rotation around C-C single bonds, show that (R,S) and (S,R)-tartaric acid are identical molecules.
2: Draw a Newman projection, looking down the C2-C3 bond, of 1-butene in the conformation shown below (C2 should be your front carbon).
Solutions to exercises
Online lectures from Kahn Academy
Newman projections part I
Newman projections part II
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
4.05: Conformations of Higher Alkanes
Learning Objective
• interpret and draw the rotation about a carbon-carbon single bond using Newman projections and sawhorse structures
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for higher alkanes
Pentane and Higher Alkanes
Pentane and higher alkanes have conformational preferences similar to ethane and butane. Each dihedral angle tries to adopt a staggered conformation and each internal C-C bond attempts to take on an anti conformation to minimize the potential energy of the molecule. The most stable conformation of any unbranched alkane follows these rules to take on zigzag shapes:
Let's analyze the staggered conformations of pentane in more detail, considering conformations about the \$C_2 –C_3 \$ and \$C_3 –C_4 \$ bonds. shows a few possible permutations. The most stable conformation is anti at both bonds, whereas less stable conformations contain gauche interactions. One gauche-gauche conformer is particularly unfavorable because methyl groups are aligned with parallel bonds in close proximity. This conformation is called syn. This type of steric hindrance across five atoms is called a syn-pentane interaction. Syn-pentane interactions have an energetic cost of about 3.6 kcal/mol relative to the anti-anti conformation and are therefore disfavored.
Exercises
1. Draw Newman projections of the eclipsed and staggered conformations of propane.
2. Draw a Newman projection, looking down the C2-C3 bond, of 1-butene in the conformation shown below.
3. Draw the energy diagram for the rotation of the bond highlighted in pentane.
Answer
1.
.
2.
3. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.04%3A_Conformations_of_Butane.txt |
Learning Objective
• explain the partial rotation of carbon-carbon single bonds in rings
• explain ring strain and its relationship to cycloalkane stability
Cycloalkanes (aka Rings)
Cycloalkanes have one or more rings of carbon atoms. The simplest examples of this class consist of a single, unsubstituted carbon ring, and these form a homologous series similar to the unbranched alkanes. The IUPAC names of the first five members of this series are given in the following table. The last column gives the general formula for a cycloalkane of any size. If a simple unbranched alkane is converted to a cycloalkane two hydrogen atoms, one from each end of the chain, must be lost. Hence the general formula for a cycloalkane composed of n carbons is CnH2n. Although a cycloalkane has two fewer hydrogens than the equivalent alkane, each carbon is bonded to four other atoms so such compounds are still considered to be saturated with hydrogen.
Table: Examples of Simple Cycloalkanes
Name Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cycloalkane
Molecular
Formula
C3H6 C4H8 C5H10 C6H12 C7H14 CnH2n
Structural
Formula
(CH2)n
Line
Formula
Although the customary line drawings of simple cycloalkanes are geometrical polygons, the actual shape of these compounds in most cases is very different. Cyclic systems are a little different from open-chain systems. In an open chain, any bond can be rotated 360 degrees, going through many different conformations. Complete rotation isn't possible in a cyclic system, because the parts that you would be trying to twist away from each other would still be connected together. Cyclic systems have fewer "degrees of freedom" than aliphatic systems; they have "restricted rotation". Because of the restricted rotation of cyclic systems, most of them have much more well-defined shapes than their aliphatic counterparts. Let's take a look at the basic shapes of some common rings. Many biologically important compounds are built around structures containing rings, so it's important that we become familiar with them. In nature, three- to six-membered rings are frequently encountered, so we'll focus on those.
The Baeyer Theory and the Experimental Evidence of Ring Strain
Many of the properties of cyclopropane and its derivatives are similar to the properties of alkenes. In 1890, the famous German organic chemist, A. Baeyer, suggested that cyclopropane and cyclobutane derivatives are different from cyclopentane and cyclohexane, because their C—C—C angles cannot have the tetrahedral value of 109.5°. At the same time, Baeyer hypothesized that the difficulties encountered in synthesizing cycloalkane rings from C7 upward was the result of the angle strain that would be expected if the large rings were regular planar polygons (see Table 12-3). Baeyer also believed that cyclohexane had a planar structure like that shown in Figure 12-2, which would mean that the bond angles would have to deviate 10.5° from the tetrahedral value. However, in 1895, the then unknown chemist H. Sachse suggested that cyclohexane exists in the strain-free chair and boat forms discussed in Section 12-3. This suggestion was not accepted at the time because it led to the prediction of several possible isomers for compounds such as chlorocyclohexane (cf. Exercise 12-4). The idea that such isomers might act as a single substance, as the result of rapid equilibration, seemed like a needless complication, and it was not until 1918 that E. Mohr proposed a definitive way to distinguish between the Baeyer and Sachse cyclohexanes. As will be discussed in Section 12-9, the result, now known as the Sachse-Mohr theory, was complete confirmation of the idea of nonplanar large rings.
Table: Strain in Cycloalkane Rings and Heats of Combustion of Cycloalkanes
Compound n Angle Strain at each CH2 Heat of Combustion ΔHo (kcal/mol) Heat of Combustion ΔHo per CH2/N (kcal/mol) Total Strain (kcal/mol)
ethene 2 109.5 337.2 168.6 22.4
cyclopropane 3 49.5 499.9 166.6 27.7
cyclobutane 4 19.5 655.9 164.0 26.3
cyclopentane 5 1.5 793.4 158.7 6.5
cyclohexane 6 10.5 944.8 157.5 0.4
cycloheptane 7 19.1 1108.1 158.4 6.3
cyclooctane 8 25.5 1268.9 158.6 9.7
cyclononane 9 30.5 1429.5 158.8 12.9
cyclodecane 10 34.5 1586.1 158.6 12.1
cyclopentadecane 15 46.5 2362.5 157.5 1.5
open chain alkane 157.4 -
Ring Strain in Cycloalkanes
Ring Strain occurs because the carbons in cycloalkanes are sp3 hybridized, which means that they do not have the expected ideal bond angle of 109.5o ; this causes an increase in the potential energy because of the desire for the carbons to be at an ideal 109.5o. An example of ring strain can be seen in the diagram of cyclopropane below in which the bond angle is 60o between the carbons.
The reason for ring strain can be seen through the tetrahedral carbon model. The C-C-C bond angles in cyclopropane (diagram above) (60o) and cyclobutane (90o) are much different than the ideal bond angle of 109.5o. This bond angle causes cyclopropane and cyclobutane to have a high ring strain. However, molecules, such as cyclohexane and cyclopentane, would have a much lower ring strain because the bond angle between the carbons is much closer to 109.5o.
Below are some examples of cycloalkanes. Ring strain can be seen more prevalently in the cyclopropane and cyclobutane models
Below is a chart of cycloalkanes and their respective heats of combustion ( ΔHcomb). The ΔHcomb value increases as the number of carbons in the cycloalkane increases (higher membered ring), and the ΔHcomb/CH2 ratio decreases. The increase in ΔHcomb can be attributed to the greater amount of London Dispersion forces. However, the decrease in ΔHcomb/CH2can be attributed to a decrease in the ring strain.
Certain cycloalkanes, such as cyclohexane, deal with ring strain by forming conformers. A conformer is a stereoisomer in which molecules of the same connectivity and formula exist as different isomers, in this case, to reduce ring strain. The ring strain is reduced in conformers due to the rotations around the sigma bonds.
Other Types of Strain
There are many different types of strain that occur with cycloalkanes. In addition to ring strain, there is also transannular strain, eclipsing, or torsional strain and bond angle strain.Transannular strain exists when there is steric repulsion between atoms. Eclipsing (torsional) strain exists when a cycloalkane is unable to adopt a staggered conformation around a C-C bond, and bond angle strain is the energy needed to distort the tetrahedral carbons enough to close the ring. The presence of angle strain in a molecule indicates that there are bond angles in that particular molecule that deviate from the ideal bond angles required (i.e., that molecule has conformers).
Cyclopropane
A three membered ring has no rotational freedom whatsoever, so the three carbon atoms in cyclopropane are all constrained to lie in the same plane at the corners of an equilateral triangle. The 60º bond angles are much smaller than the optimum 109.5º angles of a normal tetrahedral carbon atom, and the resulting angle strain dramatically influences the chemical behavior of this cycloalkane. Cyclopropane also suffers substantial eclipsing strain, since all the carbon-carbon bonds are fully eclipsed.
Furthermore, if you look at a model you will find that the neighboring C-H bonds (C-C bonds, too) are all held in eclipsed conformations.
Cyclopropane is always at maximum torsional strain. This strain can be illustrated in a line drawing of cyclopropane as shown from the side. In this oblique view, the dark lines mean that those sides of the ring are closer to you.
However, the ring isn't big enough to introduce any steric strain, which does not become a factor until we reach six membered rings. Until that point, rings are not flexible enough for two atoms to reach around and bump into each other.
The really big problem with cyclopropane is that the C-C-C bond angles are all too small.
• All the carbon atoms in cyclopropane appear to be tetrahedral.
• These bond angles ought to be 109 degrees.
• The angles in an equilateral triangle are actually 60 degrees, about half as large as the optimum angle.
• This factor introduces a huge amount of strain in the molecule, called ring strain.
Cyclobutane
Cyclobutane is a four membered ring. In two dimensions, it is a square, with 90 degree angles at each corner. Cyclobutane reduces some bond-eclipsing strain by folding (the out-of-plane dihedral angle is about 25º), but the total eclipsing and angle strain remains high. Cyclopentane has very little angle strain (the angles of a pentagon are 108º), but its eclipsing strain would be large (about 10 kcal/mol) if it remained planar. Consequently, the five-membered ring adopts non-planar puckered conformations whenever possible.
However, in three dimensions, cyclobutane is flexible enough to buckle into a "butterfly" shape, relieving torsional strain a little bit. When it does that, the bond angles get a little worse, going from 90 degrees to 88 degrees.
In a line drawing, this butterfly shape is usually shown from the side, with the near edges drawn using darker lines.
• With bond angles of 88 rather than 109 degrees, cyclobutane has a lot of ring strain, but less than in cyclopropane.
• Torsional strain is still present, but the neighbouring bonds are not exactly eclipsed in the butterfly.
• Cyclobutane is still not large enough that the molecule can reach around to cause crowding. Steric strain is very low.
• Cyclobutanes are a little more stable than cyclopropanes and are also a little more common in nature.
Cyclopentane
Cyclopentanes are even more stable than cyclobutanes, and they are the second-most common paraffinic ring in nature, after cyclohexanes. In two dimensions, a cyclopentane appears to be a regular pentagon.
In three dimensions, there is enough freedom of rotation to allow a slight twist out of this planar shape. In a line drawing, this three-dimensional shape is drawn from an oblique view, just like cyclobutane.
• The ideal angle in a regular pentagon is about 107 degrees, very close to a tetrahedral bond angle.
• Cyclopentane distorts only very slightly into an "envelope" shape in which one corner of the pentagon is lifted up above the plane of the other four, and as a result, ring strain is entirely removed.
• The envelope removes torsional strain along the sides and flap of the envelope. However, the neighbouring carbons are eclipsed along the "bottom" of the envelope, away from the flap. There is still some torsional strain in cyclopentane.
• Again, there is no steric strain in this system.
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring).
Exercise \(1\)
1. If cyclobutane were to be planar how many H-H eclipsing interactions would there be, and assuming 4 kJ/mol per H-H eclipsing interaction what is the strain on this “planar” molecule?
2. In the two conformations of cis-cyclopentane one is more stable than the other. Explain why this is.
Answer
1. There are 8 eclipsing interactions (two per C-C bond). The extra strain on this molecule would be 32 kJ/mol (4 kJ/mol x 8).
2. The first conformation is more stable. Even though the methyl groups are cis in the model on the left, they are eclipsing due the conformation, therefore increasing the strain within the molecule. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.06%3A_Cycloalkanes_and_Ring_Strain.txt |
Learning Objective
• draw cyclohexane conformations (chair & boat)
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for cyclohexane
Introduction
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring).
Cyclohexane Conformations (aka Chair Flips)
Cyclohexane is rapidly rotating between the two most stable conformations known as the chair conformations in what is called the "Chair Flip" shown below.
Several other notable cyclohexane conformations occur during the transition from one chair conformer to the other - the boat, the twist, and the half-chair. The relative energies of the conformations is a direct reflection of their relative stabilities. These structural and energetic relationships are summarized in the conformational energy diagram for cyclohexane below.
The Chair Conformation - a closer look
Since the chair conformation has the lowest potential energy, it is the most relevant to the conformation of cyclohexane. On careful examination of a chair conformation of cyclohexane, we find that the twelve hydrogens are not structurally equivalent. Six of them are located about the periphery of the carbon ring, and are termed equatorial. The other six are oriented above and below the approximate plane of the ring (three in each location), and are termed axial because they are aligned parallel to the symmetry axis of the ring.
In the figure above, the equatorial hydrogens are colored blue, and the axial hydrogens are in bold. Since there are two equivalent chair conformations of cyclohexane in rapid equilibrium, all twelve hydrogens have 50% equatorial and 50% axial character. The figure below illustrates how to convert a molecular model of cyclohexane between two different chair conformations - this is something that you should practice with models. Notice that a 'ring flip' causes equatorial hydrogens to become axial, and vice-versa.
How to draw stereo bonds ("up" and "down" bonds)
There are various ways to show these orientations. The solid (dark) "up wedge" I used is certainly common. Some people use an analogous "down wedge", which is light, to indicate a down bond; unfortunately, there is no agreement as to which way the wedge should point, and you are left relying on the lightness of the wedge to know it is "down". The "down bond" avoids this wedge ambiguity, and just uses some kind of light line. The down bond I used (e.g., in Figure 5B) is a dashed line; IUPAC encourages a series of parallel lines, something like What I did is a variation of what is recommended by IUPAC:
• In ISIS/Draw, the "up wedge" and "down bond" that I used, along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the tool button for ordinary C-C bonds.
• In Symyx Draw, the "up wedge" and "down bond", along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the "Chain" tool button.
• ChemSketch provides up and down wedges, but not the simple up and down bonds discussed above. The wedges are available from the second toolbar across the top. For an expanded discussion of using these wedges, see the section of my ChemSketch Guide on Stereochemistry: Wedge bonds.
As always, the information provided on these pages in intended to help you get started. Each program has more options for drawing bonds than discussed here. When you feel the need, look around!
How to Draw chairs
Most of the structures shown on this page were drawn with the free program ISIS/Draw. I have posted a guide to help you get started with ISIS/Draw. ISIS/Draw provides a simple cyclohexane (6-ring) hexagon template on the toolbar across the top. It provides templates for various 6-ring chair structures from the Templates menu; choose Rings. There are templates for simple chairs, without substituents (e.g., Fig 1B), and for chairs showing all the substituents (e.g., Fig 2B). In either case, you can add, delete, or change things as you wish. Various kinds of stereo bonds (wedges and bars) are available by clicking the left-side tool button that is just below the regular C-C single bond button. It may have a wedge shown on it, but this will vary depending on how it has been used. To choose a type of stereo bond, click on the button and hold the mouse click; a new menu will appear to the right of the button.
The free drawing program Symyx Draw, the successor to ISIS/Draw, provides similar templates and tools. A basic chair structure is provided on the default template bar that is shown. More options are available by choosing the Rings template. See my page Symyx Draw for a general guide for getting started with this program.
The free drawing program ChemSketch provides similar templates and tools. To find the special templates for chairs, go to the Templates menu, choose Template Window, and then choose "Rings" from the drop-down menu near upper left. See my page ChemSketch for a general guide for getting started with this program.
If you want to draw chair structures by hand (and if you are going on in organic chemistry, you should)... Be careful. The precise zigs and zags, and the angles of substituents are all important. Your textbook may offer you some hints for how to draw chairs. A short item in the Journal of Chemical Education offers a nice trick, showing how the chair can be thought of as consisting of an M and a W. The article is V Dragojlovic, A method for drawing the cyclohexane ring and its substituents. J Chem Educ 78:923, 7/01. (I thank M Farooq Wahab, Chemistry, Univ Karachi, for suggesting that this article be noted here.)
Aside from drawing the basic chair, the key points in adding substituents are:
• Axial groups alternate up and down, and are shown "vertical".
• Equatorial groups are approximately horizontal, but actually somewhat distorted from that, so that the angle from the axial group is a bit more than a right angle -- reflecting the common 109 degree bond angle.
• As cautioned before, it is usually easier to draw and see what is happening at the four corners of the chair than at the two middle positions. Try to use the corners as much as possible.
Because axial bonds are parallel to each other, substituents larger than hydrogen generally suffer greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which the larger substituents assume equatorial orientation.
When the methyl group in the structure above occupies an axial position it suffers steric crowding by the two axial hydrogens located on the same side of the ring.
The conformation in which the methyl group is equatorial is more stable, and thus the equilibrium lies in this direction
Exercise
Questions
1. Consider the conformations of cyclohexane, chair, boat, twist boat. Order them in increasing strain in the molecule.
2. Draw two conformations of cyclohexyl amine (C6H11NH2). Indicate axial and equatorial positions.
3. Draw the two isomers of 1,4-dihydroxylcyclohexane, identify which are equatorial and axial.
4. In the following molecule, label which are equatorial and which are axial, then draw the chair flip (showing labels 1,2,3).
Answer
1. Chair < Twist Boat < Boat (most strain)
2.
3.
4. Original conformation: 1 = axial, 2 = equatorial, 3 = axial
Flipped chair now looks like this.
Contributors and Attributions
Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University)
Prof. Steven Farmer (Sonoma State University)
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
>Robert Bruner (http://bbruner.org) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.07%3A_Cyclohexane_Conformations.txt |
Learning Objective
• draw mono-substituted cyclohexane conformers (chair only)
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for butane, higher alkanes, cyclohexane, mono-substituted cyclohexanes, and disubstituted cyclohexanes
Introduction
Because axial bonds are parallel to each other, substituents larger than hydrogen generally suffer greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which the larger substituents assume equatorial orientation.
When the methyl group in the structure above occupies an axial position it suffers steric crowding by the two axial hydrogens located on the same side of the ring. The conformation in which the methyl group is equatorial is more stable, and thus the equilibrium lies in this direction.
In examining possible structures for monosubstituted cyclohexanes, it is useful to follow two principles:
1. Chair conformations are generally more stable than other possibilities.
2. Substituents on chair conformers prefer to occupy equatorial positions due to the increased steric hindrance of axial locations.
Experimental Measurements of Steric Hindrance
The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane may be determined by the conformational equilibrium of the compound. The corresponding equilibrium constant is related to the energy difference between the conformers and collecting such data allows us to evaluate the relative tendency of substituents to exist in an equatorial or axial location.
Looking at the energy values the table, it is clear that the apparent "size" of a substituent (in terms of its preference for equatorial over axial orientation) is influenced by its width and bond length to cyclohexane, as evidenced by the fact that an axial vinyl group is less hindered than ethyl, and iodine slightly less than chlorine.
A Selection of AG° Values for the Change from Axial to Equatorial Orientation of Substituents for Monosubstituted Cyclohexanes
Substituent \(-\Delta{G}^o\) kcal/mol Substituent \(-\Delta{G}^o\) kcal/mol
\(\ce{CH_3\bond{-}}\) 1.7 \(\ce{O_2N\bond{-}}\) 1.1
\(\ce{CH_2H_5\bond{-}}\) 1.8 \(\ce{N#C\bond{-}}\) 0.2
\(\ce{(CH_3)_2CH\bond{-}}\) 2.2 \(\ce{CH_3O\bond{-}}\) 0.5
\(\ce{(CH_3)_3C\bond{-}}\) \(\geq 5.0\) (CH3)3C- 0.7
\(\ce{F\bond{-}}\) 0.3 F- 1.3
\(\ce{Cl\bond{-}}\) 0.5 \(\ce{C_6H_5\bond{-}}\) 3.0
\(\ce{Br\bond{-}}\) 0.5
\(\ce{I\bond{-}}\) 0.5
Exercise
1. In the molecule, cyclohexyl ethyne there is little steric strain, why?
Answer
1.
The ethyne group is linear and therefore does not affect the hydrogens in the 1,3 positions to say to the extent as a bulkier or a bent group (e.g. ethene group) would. This leads to less of a strain on the molecule.
Contributors and Attributions
>Robert Bruner (http://bbruner.org)
4.09: Cis-trans Isomerism in Cycloalkanes
Learning Objective
• identify & draw the geometric (cis/trans) isomers of cycloalkanes
Geometric Isomerism of Cycloalkanes
The carbon ring of cycloalkanes forms a pseudo-plane that can be used to assign the relative orientation of atoms or substituents bonded to the ring (stereochemistry). One side of the ring is called "up" while the other side is called "down". By agreement, chemists use heavy, wedge-shaped bonds to indicate a substituent located above the average plane of the ring (up), and a hatched line for bonds to atoms or groups located below the ring (down).
Disubstituted cycloalkane stereoisomers may be designated by nomenclature prefixes such as cis and trans. Cis and trans isomers are also called "geometric isomers".
For the cis isomer, both substituents are aobe or below the carbon ring. For the trans isomer, one substituent is above the ring while the other substituent is below the ring.
The cis and trans isomers for 1,2-dibromocyclopentane are shown as an example below.
While the carbon-carbon single bonds of the rings can rotate partially, there is NO way to inter-convert between the cis and trans isomers. Cis and trans isomers are unique compounds with their own unique melting points, boiling points, densities, etc.
Further explanation:
In general, if any two sp3 carbons in a ring have two different substituent groups (not counting other ring atoms) stereoisomerism is possible. This is similar to the substitution pattern that gives rise to stereoisomers in alkenes; indeed, one might view a double bond as a two-membered ring. Four other examples of this kind of stereoisomerism in cyclic compounds are shown below.
If more than two ring carbons have different substituents (not counting other ring atoms) the stereochemical notation distinguishing the various isomers becomes more complex. However, we can always state the relationship of any two substituents using cis or trans. For example, in the trisubstitutued cyclohexane below, we can say that the methyl group is cis to the ethyl group, and trans to the chlorine. We can also say that the ethyl group is trans to the chlorine. We cannot, however, designate the entire molecule as a cis or trans isomer.
Exercise
1. Draw the following molecules:
trans-1,3-dimethylcyclohexane
trans-1,2-dibromocyclopentane
cis-1,3-dichlorocyclobutane
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.08%3A_Conformations_of_Monosubstituted_Cyclohexanes.txt |
Learning Objective
• draw di-substituted cyclohexane conformers (chair only)
• correlate energies of conformations with rotational energy diagrams and predict the most stable conformations for disubstituted cyclohexanes
Conformational Structures of Disubstituted Cyclohexanes
In a sample of cyclohexane, the two identical chair conformers are present in equal concentration, and the hydrogens are all equivalent (50% equatorial & 50% axial) due to rapid interconversion of the conformers. When the cyclohexane ring bears a substituent, the two chair conformers are not the same. In one conformer the substituent is axial, in the other it is equatorial. Due to steric hindrance in the axial location, substituent groups prefer to be equatorial and that chair conformer predominates in the equilibrium.
When cycloalkanes have two substituents on different ring carbon atoms, then a pair of configurational stereoisomers exist. Now we must examine the way in which favorable ring conformations influence the properties of the configurational isomers. Remember, configurational stereoisomers are stable, unique chemical compounds, whereas, conformational isomers are different rotations of the same compound. In examining possible structures for disubstituted cyclohexane, it is useful to follow two principles:
1. Substituents on chair conformers prefer to occupy equatorial positions due to the increased steric hindrance of axial locations.
The following equations and formulas illustrate how the presence of two or more substituent on a cyclohexane ring perturbs the interconversion of the two chair conformers in ways that can be predicted. When there is a potential energy difference between the conformers, then the lower energy conformation is favored as indicated by the equilibrium reaction arrows.
1,1-dimethylcyclohexane
1-t-butyl-1-methylcyclohexane
cis-1,2-dimethylcyclohexane
trans-1,2-dimethylcyclohexane
cis-1,3-dimethylcyclohexane
trans-1,3-dimethylcyclohexane
cis-1,4-dimethylcyclohexane
trans-1,4-dimethylcyclohexane
In the case of 1,1-disubstituted cyclohexanes, one of the substituents must necessarily be axial and the other equatorial, regardless of which chair conformer is considered. Since the substituents are the same in 1,1-dimethylcyclohexane, the two conformers are identical and present in equal concentration. In 1-t-butyl-1-methylcyclohexane the t-butyl group is much larger than the methyl, and that chair conformer in which the larger group is equatorial will be favored in the equilibrium( > 99%). Consequently, the methyl group in this compound is almost exclusively axial in its orientation.
In the cases of 1,2-, 1,3- and 1,4-disubstituted compounds the analysis is a bit more complex. It is always possible to have both groups equatorial, but whether this requires a cis-relationship or a trans-relationship depends on the relative location of the substituents. As we count around the ring from carbon #1 to #6, the uppermost bond on each carbon changes its orientation from equatorial (or axial) to axial (or equatorial) and back. It is important to remember that the bonds on a given side of a chair ring-conformation always alternate in this fashion. Therefore, it should be clear that for cis-1,2-disubstitution, one of the substituents must be equatorial and the other axial; in the trans-isomer both may be equatorial. Because of the alternating nature of equatorial and axial bonds, the opposite relationship is true for 1,3-disubstitution (cis is all equatorial, trans is equatorial/axial). Finally, 1,4-disubstitution reverts to the 1,2-pattern.
The conformations of some substituted cyclohexanes may be examined as interactive models by .
It can be helpful to add the hydrogen atoms at the axial positions to help recognize the equatorial position.
Exercise
1. Draw the two chair conformations for cis-1-ethyl-2-methylcyclohexane using bond-line structures and indicate the more energetically favored conformation.
2. Draw the most stable conformation for trans-1-ethyl-3-methylcyclohexane using bond-line structures.
3. Draw the most stable conformation for trans-1-t-butyl-4-methylcyclohexane using bond-line structures.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.10%3A_Conformations_of_Disubstituted_Cyclohexanes.txt |
Learning Objective
• recognize, classify, and draw the three ways to join two rings
Bicyclic Ring Systems
There are three ways to join two rings. If two rings share two or more atoms, then the structure is called a bicyclic compound. If the two rings share a single atom, then the structure is called a spirocyclic compound. Examples of each way to join rings is shown below.
Bicyclic Compounds
Fused rings share two adjacent carbon atoms. Decalin is a fused bicyclic compound. Its IUPAC name is bibyclo[4.4.0]decane to communicate the bonding arrangement.
Bridged rings share two non-adjacent carbon atoms and one or more carbon atoms between them. Bicyclo[2.2.1]heptane shows the difference between the bridgehead carbons and the bridge carbons.
Spirocyclic Compounds
A spirobicycloalkane is a molecule in which only one carbon atom is shared by the two rings in the molecule. The carbon atom shared by the two rings is called the spirocarbon. A chain of bonds originating and ending at the spirocarbon is called a bridge. The compound below is named spiro[4.5]decane to communicate the number of carbons in each bridge with the spirocarbon.
Steroids
Steroids include such well known compounds as cholesterol, sex hormones, birth control pills, cortisone, and anabolic steroids.
The best known and most abundant steroid in the body is cholesterol. Cholesterol is formed in brain tissue, nerve tissue, and the blood stream. It is the major compound found in gallstones and bile salts. Cholesterol also contributes to the formation of deposits on the inner walls of blood vessels. These deposits harden and obstruct the flow of blood. This condition, known as atherosclerosis, results in various heart diseases, strokes, and high blood pressure.
Much research is currently underway to determine if a correlation exists between cholesterol levels in the blood and diet. Not only does cholesterol come from the diet, but cholesterol is synthesized in the body from carbohydrates and proteins as well as fat. Therefore, the elimination of cholesterol rich foods from the diet does not necessarily lower blood cholesterol levels. Some studies have found that if certain unsaturated fats and oils are substituted for saturated fats, the blood cholesterol level decreases. The research is incomplete on this problem.
Structures of Sex Hormones
Sex hormones are also steroids. The primary male hormone, testosterone, is responsible for the development of secondary sex characteristics. Two female sex hormones, progesterone and estrogen or estradiol control the ovulation cycle. Notice that the male and female hormones have only slight differences in structures, but yet have very different physiological effects.
Testosterone promotes the normal development of male genital organs ans is synthesized from cholesterol in the testes. It also promotes secondary male sexual characteristics such as deep voice, facial and body hair.
Estrogen, along with progesterone regulates changes occurring in the uterus and ovaries known as the menstrual cycle. For more details see Birth Control. Estrogen is synthesized from testosterone by making the first ring aromatic which results in mole double bonds, the loss of a methyl group and formation of an alcohol group.
Exercise
1. Someone stated that trans-decalin is more stable than cis-decalin. Explain why this is incorrect.
Answer
1. Cis-decalin has fewer steric interactions than trans-decalinbecause each ring can assume the chair form in both conformations. Working with models can be helpful.
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)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.11%3A_Joined_Rings.txt |
Learning Objective
• describe the uses and sources of alkanes
Occurrence
The most important sources for alkanes are oil and natural gas. Oil is a mixture of liquid alkanes and other hydrocarbons. Higher alkanes (which are solid) occur as residues from oil distillation ("tar"). One of the largest natural deposits of solid alkanes is in an asphalt lake known as the Pitch Lake in Trinidad and Tobago. Natural gas contains primarily methane (70-90%) with some ethane, propane and butane; some gas sources deliver up to 8% CO2. Traces of methane (about 0.00017% or 1.7 ppm) occur in the Earth's atmosphere, the content in the oceans is negligible due to the low solubility of methane in water.(1)
Use of Alkanes
Alkanes are important raw materials of the chemical industry and the principal constituent of gasoline and lubricating oils. Natural gas mainly contains methane and ethane and is used for heating and cooking purposes and for power utilities (gas turbines). For transportation purposes, natural gas may be liquefied by applying pressure and cooling it (LNG = liqid natural gas). The Sultanate of Oman, for example, exports most of its natural gas as LNG - see the LNG plant at Qalhat which has been designed to liquefy 6.6 million tons natural gas per year. Crude oil is separated into its components by fractional distillation at oil refineries. The different "fractions" of crude oil have different boiling points and consist mostly of alkanes of similar chain lengths (the higher the boiling point the more carbon atoms the components of a particular fraction contain - see the list of alkanes for details about the boiling points).
The following table provides a short survey of the different fractions of crude oil:
C3..C4 Propane and butane can be liquefied at fairly low pressures, and are used, for example, in the propane gas burner, or as propellants in aerosol sprays. Butane in used in cigarette lighters (where the pressure at room temperature is about 2 bar).
C5..C8 The alkanes from pentane to octane are highly volatile liquids and good solvents for nonpolar substances. They are used as fuels in internal combustion engines.
C9..C16 Alkanes from nonane to hexadecane are liquids of higher viscosity, being used in diesel and aviation fuel (kerosene). The higher melting points of these alkanes can cause problems at low temperatures and in polar regions, where the fuel becomes too viscous.
C17..C35 Alkanes with 17 to 35 carbon atoms form the major components of lubricating oil. They also act as anti-corrosive agents, as their hydrophobic nature protects the metal surface from contact with water. Solid alkanes also find use as paraffin wax in candles(2).
>C35 Alkanes with a chain length above 35 carbon atoms are found in bitumen (as it is used in road surfacing). These higher alkanes have little chemical and commercial value and are usually split into lower alkanes by cracking.
Notes:
(1) Methane can co-crystallize with water at high pressures and low temperatures, forming a solid methane hydrate. The energy content of the known submarine methane hydrate fields exceeds that of all known natural gas and oil deposits put together.
(2) Paraffin wax should not be mixed up with true animal or plant wax, which consist of esters of various carboxylic acids and alcohols. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.12%3A_Uses_and_Sources_of_Alkanes.txt |
Learning Objective
• recognize and distinguish between the two major reactions of alkanes - combustion and halogenation
Combustion
Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide and water. It is quite important that you can write properly balanced equations for these reactions, because they often come up as a part of thermochemistry calculations. Some are easier than others. For example, with alkanes, the ones with an even number of carbon atoms are marginally harder than those with an odd number!
Example: Propane Combustion
For example, with propane (C3H8), you can balance the carbons and hydrogens as you write the equation down. Your first draft would be:
$C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$
Counting the oxygens leads directly to the final version:
$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
Example: Butane Combustion
With butane (C4H10), you can again balance the carbons and hydrogens as you write the equation down.
$C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O$
Counting the oxygens leads to a slight problem - with 13 on the right-hand side. The simple trick is to allow yourself to have "six-and-a-half" O2 molecules on the left.
$C_4H_{10} + 6\dfrac{1}{2}\, O_2 \rightarrow 4CO_2 + 5H_2O$
If that offends you, double everything:
$2C_4H_{10} + 13 O_2 \rightarrow 8CO_2 + 10 H_2O$
The hydrocarbons become harder to ignite as the molecules get bigger. This is because the bigger molecules don't vaporize so easily - the reaction is much better if the oxygen and the hydrocarbon are well mixed as gases. If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.
Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame. Incomplete combustion (where there is not enough oxygen present) can lead to the formation of carbon or carbon monoxide. As a simple way of thinking about it, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets whatever is left over! The presence of glowing carbon particles in a flame turns it yellow, and black carbon is often visible in the smoke. Carbon monoxide is produced as a colorless poisonous gas.
Note: Why carbon monoxide is poisonous
Oxygen is carried around the blood by hemoglobin, which unfortunately binds to exactly the same site on the hemoglobin that oxygen does. The difference is that carbon monoxide binds irreversibly (or very strongly) - making that particular molecule of hemoglobin useless for carrying oxygen. If you breath in enough carbon monoxide you will die from a sort of internal suffocation.
Halogenation of Alkanes
Halogenation is the replacement of one or more hydrogen atoms in an organic compound by a halogen (fluorine, chlorine, bromine or iodine). Unlike the complex transformations of combustion, the halogenation of an alkane appears to be a simple substitution reaction in which a C-H bond is broken and a new C-X bond is formed. The chlorination of methane, shown below, provides a simple example of this reaction.
CH4 + Cl2 + energy → CH3Cl + HCl
Since only two covalent bonds are broken (C-H & Cl-Cl) and two covalent bonds are formed (C-Cl & H-Cl), this reaction seems to be an ideal case for mechanistic investigation and speculation. However, one complication is that all the hydrogen atoms of an alkane may undergo substitution, resulting in a mixture of products, as shown in the following unbalanced equation. The relative amounts of the various products depend on the proportion of the two reactants used. In the case of methane, a large excess of the hydrocarbon favors formation of methyl chloride as the chief product; whereas, an excess of chlorine favors formation of chloroform and carbon tetrachloride.
CH4 + Cl2 + energy → CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl
In the presence of a flame, the reactions are rather like the fluorine one - producing a mixture of carbon and the hydrogen halide. The violence of the reaction drops considerably as you go from fluorine to chlorine to bromine. The interesting reactions happen in the presence of ultra-violet light (sunlight will do). These are photochemical reactions that happen at room temperature. We'll look at the reactions with chlorine, although the reactions with bromine are similar, but evolve more slowly.
Substitution reactions happen in which hydrogen atoms in the methane are replaced one at a time by chlorine atoms. You end up with a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane.
The original mixture of a colorless and a green gas would produce steamy fumes of hydrogen chloride and a mist of organic liquids. All of the organic products are liquid at room temperature with the exception of the chloromethane which is a gas.
If you were using bromine, you could either mix methane with bromine vapor, or bubble the methane through liquid bromine - in either case, exposed to UV light. The original mixture of gases would, of course, be red-brown rather than green. One would not choose to use these reactions as a means of preparing these organic compounds in the lab because the mixture of products would be too tedious to separate. The mechanisms for the reactions are explained on separate pages.
Larger alkanes and chlorine
You would again get a mixture of substitution products, but it is worth just looking briefly at what happens if only one of the hydrogen atoms gets substituted (monosubstitution) - just to show that things aren't always as straightforward as they seem! For example, with propane, you could get one of two isomers:
If chance was the only factor, you would expect to get three times as much of the isomer with the chlorine on the end. There are 6 hydrogens that could get replaced on the end carbon atoms compared with only 2 in the middle. In fact, you get about the same amount of each of the two isomers. If you use bromine instead of chlorine, the great majority of the product is where the bromine is attached to the center carbon atom.
Cycloalkanes
The reactions of the cycloalkanes are generally just the same as the alkanes, with the exception of the very small ones - particularly cyclopropane. In the presence of UV light, cyclopropane will undergo substitution reactions with chlorine or bromine just like a non-cyclic alkane. However, it also has the ability to react in the dark. In the absence of UV light, cyclopropane can undergo addition reactions in which the ring is broken. For example, with bromine, cyclopropane gives 1,3-dibromopropane.
This can still happen in the presence of light - but you will get substitution reactions as well. The ring is broken because cyclopropane suffers badly from ring strain. The bond angles in the ring are 60° rather than the normal value of about 109.5° when the carbon makes four single bonds. The overlap between the atomic orbitals in forming the carbon-carbon bonds is less good than it is normally, and there is considerable repulsion between the bonding pairs. The system becomes more stable if the ring is broken.
Exercise
1. Classify the following reactions as combustion or halogenation.
Answer
1. a) halogentation
b) combustion | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.13%3A_Reactions_of_Alkanes_-_a_Brief_Overview.txt |
Structural and Geometric Isomerism
4-1
a) There are five alkane isomers of hexane C6H14. Draw and name all of them.
b) The heat of combustion of hexane is 4163.2 kJ/mol. Heat of combustion of neohexane is 4159.5 kJ/mol. Predict the relative stability of these two compounds?
c) Draw and name all cycloalkane isomers of C5H10, including all possible geometric (cis-trans) stereoisomers.
4-2 Which of the following structures represent the same compound? Name the structures given in part (a), (d), (e), (f), (g)
a)
b)
c)
d)
e)
f)
g)
4-3 Each of the following descriptions applies to more than one alkane. In each case, draw and name two structures that match the description.
(a) a sec-butylheptane
(b) a trans-dimethylcyclobutane
(c) a cis-di-tert-butylcyclohexane
(d) an isopropyloctane
(e) a (1,2-dimethylpropyl)cycloalkane
(f) a bicycloheptane
4-4 Write structures for a homologous series of alcohols (R-OH) having from one to five carbons.
4-5 In each pair of compound, which compound has the higher boiling point? Explain your reasoning.
(a) Nonane or 3-ethylhexane
(b) Pentane or 2-methylbutane
(c) Octane or 2,2,4-trimethylpentane
4-6 There are four isomeric four-carbon alkyl groups. Draw them, give their systematic names and label the degree of substitution (primary, secondary, or tertiary) of the head carbon atom which is bonded to the main chain.
4-7 Draw Newman projection of the most stable conformation of the following compounds as viewed from the indicated bond.
(a) 3-methylhexane viewed at C3-C4 bond
(b) 2,2-dimethylbutane viewed at C2-C3 bond
4-8
(a) Draw two chair conformations of trans-1,2-dimethylcyclohexane and label all position as (a) for axial or (e) for equatorial.
(b) Determine the higher-energy and the lower-energy conformations
(c) Calculate the energy difference in these two conformations
4-9 Draw the two chair conformations of each compound and label the substituents as axial or equatorial. In each case, determine which conformation is more stable.
(a) cis-1-ethyl-4-methylcyclohexane
(b) trans-1-ethyl-4-methylcyclohexane
(c) cis-1-bromo-3-methylcyclohexane
(d) trans-1-bromo-3-methylcyclohexane
(e) cis-1-methyl-2-isopropylcychlohexane
(f) trans-1-methyl-2-isopropylchyclohexane
4-10 Glucose with molecular formula C6H12O6 is by far the most abundant sugar in nature. Glucose can take form as an open chain or as can be closed into a ring form. Below are chair conformations of α and β D-glucose. Using what you know about the conformational energy of substituted cyclohexane, predict which of the two isomers predominates in equilibrium. Explain your reasoning.
4-11 Provide a line drawing corresponding to each of the following Newman projections and name them using IUPAC rules.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
4-12 Draw Newman projections along the C3-C4 bond to show the most stable and the least stable conformation of 2,3,5-trimethylhexane.
4-13 In β-D-glucose, the hydroxyl group in C1 position is cis to the CH2OH group in C5 position, as shown in the figure below.
There are two chair conformations of β-D-glucose. Draw them and identify which conformation is more stable.
4.15: Solutions to Additional Exercises
Structural and Geometric Isomerism
4-1 a)
b) The isomer that releases the least energy is the most stable, so neohexane is more stable than hexane.
c)
4-2
a) The following structures all represent hexane.
The following structure represents 3-methylpentane:
b) Both structures represent cis-2-pentene:
and
Both structures represent trans-2-pentene:
and
Both structures represent 1-pentene:
and
c) Both structures represent 3-ethyl-3-heptene:
and
These three structures all represent 3-ethyl-2-heptene:
d) Both structures represent (3R)-3-methylhenxane:
All four structures represent (3S)-3-methylhenxane:
e) Both structures represent trans-1,2-dichlorocyclobutane:
All four structures represent cis-1,2-dichlorocyclobutane:
f) All three structures represent trans-1,2-cyclohexanediol:
All three structures represent cis-1,2-cyclohexanediol:
Both structures represent trans-1,4-cyclohexanediol Cis-1,4-cyclohexanediol:
g) Both structures represent 3,4-dimethylpentan-2-ol:
All three structures represent 2,4-dimethylhexan-3-ol:
4-3
(a) a dimethylnonane
(b) a trans-dimethylcyclobutane
(c) a cis-di-tert-butylcyclohexane
(d) an isopropyloctane
(e) a (1,2-dimethylpropyl)cycloalkane
(f) a bicycloheptane
Bicyclo[2.2.1]heptane Bicyclo[3.1.1]heptane
4-4
4-5
(a) Nonane has the higher boiling point because it has the higher molecular weight. Recall that higher molecular weight compounds have more surface area, and therefore they have stronger London dispersion forces. As a result, higher molecular weight compounds have the higher boiling temperatures.
(b) Pentane has the higher boiling point. Pentane has a straight chain while 2-methylbutane is branched. Compared to a straight-chain isomer, a branched hydrocarbon has a lower boiling temperature because of its smaller surface area.
(c) Octane has the higher boiling point because 2,2,4-trimethylpentane is highly branched while octane is a straight-chain hydrocarbon.
4-6
4-7
(a)
(b)
4-8
(c) The first conformer has one gauche interaction between the –CH3 groups, so the strain energy of this conformer is 0.9 kcal/mol. The second conformer has four 1,3-diaxial interaction between H and –CH3 groups, so its strain energy is 4 x 0.9 = 3.6 kcal/mol. Therefore, the energy difference in these two conformations is 3.6 – 0.9 =2.7 kcal/mol.
4-9
(a)
(b)
(c)
(d)
(e)
(f)
4-10
In α-D-glucopyranose, the hydroxyl group at C1 occupies an axial position. In β-D-glucopyranose, the hydroxyl group at C1 of occupies an equatorial position, which is the more stable structure. So the β forms predominates in equilibrium.
4.11
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
4.12
4.13 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.14%3A__Additional_Exercises.txt |
Learning Objectives
After reading the chapter and completing ALL the exercises and homework, a student can be able to:
• recognize and distinguish between the four major types of organic reactions (additions, eliminations, substitutions, and rearrangements) - refer to section 5.1
• accurately and precisely use reaction mechanism notation and symbols including curved arrows to show the flow of electrons - refer to section 5.2
• identify nucleophiles and electrophiles in polar reactions - refer to section 5.3
• perform calculations using the equation $ΔGº = –RT \ln K = –2.303 RT \log_{10} K$ and explain the relationship between equilibrium and free energy - refer to section 5.4
• calculate reaction enthalpies from bond dissociation energies - refer to section 5.5
• draw Reaction Energy Diagrams from the thermodynamic and kinetic data/information - refer to se5.6
• use a Reaction Energy Diagram to discuss transition states, Ea, intermediates & rate determining step - refer to section 5.6
• draw the transition states & intermediates of a reaction - refer to section 5.6
• describe the structure & relative stabilities of carbocations, free radicals and carbanions - refer to sections 5.7 - 5.9 respectively
• Explain the mechanism & energetics of the free-radical halogenation of alkanes - refer to section 5.10
• Predict the products of chlorination & bromination reactions of alkanes based on relative reactivity and selectivity - refer to section 5.11
• describe the similarities and differences between reactions performed in the lab with biochemical reactions - refer to section 5.12
• 5.1: Types of Organic Reactions
The four main classes of organic reactions are additions, eliminations, substitutions, and rearrangements.
• 5.2: Reaction Mechanism Notation and Symbols
Arrows are used by chemists to communicate electron flow in mechanisms, reaction completion/equilibrium, and resonance relationships. It is important to use accuracy when selecting the type of arrow for reactions and precision in drawing the location of the arrow head and tail for the curved arrows of electron flow.
• 5.3: Polar Reactions- the Dance of the Nucleophile and Electrophile
Sterics and electronics are the underlying driving forces for polar organic reactions. The electron rich nucleophile (Nu:) reacts with the electron poor electrophile through a variety of pathways that can be limited and/or influenced by steric hindrance. We explore and learn the polar reaction pathways in subsequent chapters.
• 5.4: Describing a Reaction - Equilibrium and Free Energy Changes
The relationship between equilibrium and free energy is reviewed quantitatively and applied to organic reactions conceptually.
• 5.5: Homolytic Cleavage and Bond Dissociation Energies
The products of homolytic cleavage are radicals and the energy that is required to break the bond homolytically is called the Bond Dissociation Energy (BDE) and is a measure of the strength of the bond.
• 5.6: Reaction Energy Diagrams and Transition States
Reaction energy diagrams efficiently and effectively communicate the thermodynamics and kinetics of chemical reactions in a single diagram. They are a useful tool in learning organic chemistry.
• 5.7: Reactive Intermediates - Carbocations
A carbocation is a cation in which carbon has an empty p orbital and bears a positive charge creating a highly reactive intermediate. Comparing the relative stability of reaction intermediates helps elucidate reaction mechanisms and predict major and minor products.
• 5.8: Reactive Intermediates - Radicals
A radical (more precisely, a free radical) is an atom, molecule, or ion that has unpaired valence electron (half filled orbital) creating a highly reactive intermediate.
• 5.9: Reactive Intermediates- Carbanions and Carbon Acids
A carbanion is an anion in which carbon has an unshared pair of electrons and bears a negative charge creating a highly reactive intermediate.
• 5.10: The Free-Radical Halogenation of Alkanes
Free radical halogenation of alkanes is the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. This reaction is very important in organic chemistry because it opens a gateway to further chemical reactions. We will apply the reaction concepts discussed in this chapter to this reaction to show how empirical data supports these theories.
• 5.11: Reactivity and Selectivity
In general, high reactivity correlates with low selectivity and vice versa. Depending on the structure of the substrate, reaction conditions can be optimized for high reactivity or high selectivity and occasionally for both.
• 5.12: A Comparison between Biological Reactions and Laboratory Reactions
Biochemical reactions occur within our body fluids at a typical pH of 7.4 and temperature of 98.6C. Our biochemistry relies on enzymes to catalyze physiological reactions within this narrow range of environmental conditions. Synthetic organic chemists can create extreme conditions within reactions flasks to catalyze and promote chemical reactions.
• 5.13: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 5.14: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
05: An Introduction to Organic Reactions using Free Radical Halogenation of Alkanes
Learning Objective
• recognize and distinguish between the four major types of organic reactions (additions, eliminations, substitutions, and rearrangements)
Introduction
If you scan any organic textbook you will encounter what appears to be a very large, often intimidating, number of reactions. These are the "tools" of a chemist, and to use these tools effectively, we must organize them in a sensible manner and look for patterns of reactivity that permit us make plausible predictions. Most of these reactions occur at special sites of reactivity known as functional groups, and these constitute one organizational scheme that helps us catalog and remember reactions.
Ultimately, the best way to achieve proficiency in organic chemistry is to understand how reactions take place, and to recognize the various factors that influence their course.
First, we identify four broad classes of reactions based solely on the structural change occurring in the reactant molecules. This classification does not require knowledge or speculation concerning reaction paths or mechanisms. The four main reaction classes are additions, eliminations, substitutions, and rearrangements.
Addition Reaction
Elimination Reaction
Substitution Reaction
Rearrangement Reaction
In an addition reaction the number of σ-bonds in the substrate molecule increases, usually at the expense of one or more π-bonds. The reverse is true of elimination reactions, i.e.the number of σ-bonds in the substrate decreases, and new π-bonds are often formed. Substitution reactions, as the name implies, are characterized by replacement of an atom or group (Y) by another atom or group (Z). Aside from these groups, the number of bonds does not change. A rearrangement reaction generates an isomer, and again the number of bonds normally does not change.
The examples illustrated above involve simple alkyl and alkene systems, but these reaction types are general for most functional groups, including those incorporating carbon-oxygen double bonds and carbon-nitrogen double and triple bonds. Some common reactions may actually be a combination of reaction types.
Example: substitution Reaction
The reaction of an ester with ammonia to give an amide, as shown below, appears to be a substitution reaction ( Y = CH3O & Z = NH2 ); however, it is actually two reactions, an addition followed by an elimination.
Example: Addition reaction
The addition of water to a nitrile does not seem to fit any of the above reaction types, but it is simply a slow addition reaction followed by a rapid rearrangement, as shown in the following equation. Rapid rearrangements of this kind are called tautomerizations.
Exercise
1. Classify each reaction as addition, elimination, substitution, or rearrangement.
Answer
1. A = Substitution; B = Elimination; C = Addition | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.01%3A_Types_of_Organ.txt |
Learning Objective
• accurately and precisely use reaction mechanism notation and symbols including curved arrows to show the flow of electrons
The Arrow Notation in Mechanisms
Since chemical reactions involve the breaking and making of bonds, a consideration of the movement of bonding (and non-bonding) valence shell electrons is essential to this understanding. It is now common practice to show the movement of electrons with curved arrows, and a sequence of equations depicting the consequences of such electron shifts is termed a mechanism. In general, two kinds of curved arrows are used in drawing mechanisms:
A full head on the arrow indicates the movement or shift of an electron pair:
A partial head (fishhook) on the arrow indicates the shift of a single electron:
The use of these symbols in bond-breaking and bond-making reactions is illustrated below. If a covalent single bond is broken so that one electron of the shared pair remains with each fragment, as in the first example, this bond-breaking is called homolysis. If the bond breaks with both electrons of the shared pair remaining with one fragment, as in the second and third examples, this is called heterolysis.
Bond-Breaking Bond-Making
Other Arrow Symbols
Chemists also use arrow symbols for other purposes, and it is essential to use them correctly.
The Reaction Arrow
The Equilibrium Arrow
The Resonance Arrow
The following equations illustrate the proper use of these symbols:
Reactive Intermediates
The products of bond breaking, shown above, are not stable in the usual sense, and cannot be isolated for prolonged study. Such species are referred to as reactive intermediates, and are believed to be transient intermediates in many reactions. The general structures and names of four such intermediates are given below.
A pair of widely used terms, related to the Lewis acid-base notation, should also be introduced here.
• Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile.
• Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile (or Lewis acid).
Using these definitions, it is clear that carbocations ( called carbonium ions in the older literature ) are electrophiles and carbanions are nucleophiles. Carbenes have only a valence shell sextet of electrons and are therefore electron deficient. In this sense they are electrophiles, but the non-bonding electron pair also gives carbenes nucleophilic character. As a rule, the electrophilic character dominates carbene reactivity. Carbon radicals have only seven valence electrons, and may be considered electron deficient; however, they do not in general bond to nucleophilic electron pairs, so their chemistry exhibits unique differences from that of conventional electrophiles. Radical intermediates are often called free radicals.
The importance of electrophile / nucleophile terminology comes from the fact that many organic reactions involve at some stage the bonding of a nucleophile to an electrophile, a process that generally leads to a stable intermediate or product. Reactions of this kind are sometimes called ionic reactions, since ionic reactants or products are often involved. Some common examples of ionic reactions and their mechanisms may be examined below.
The shapes ideally assumed by these intermediates becomes important when considering the stereochemistry of reactions in which they play a role. A simple tetravalent compound like methane, CH4, has a tetrahedral configuration. Carbocations have only three bonds to the charge bearing carbon, so it adopts a planar trigonal configuration. Carbanions are pyramidal in shape ( tetrahedral if the electron pair is viewed as a substituent), but these species invert rapidly at room temperature, passing through a higher energy planar form in which the electron pair occupies a p-orbital. Radicals are intermediate in configuration, the energy difference between pyramidal and planar forms being very small. Since three points determine a plane, the shape of carbenes must be planar; however, the valence electron distribution varies.
Ionic Reactions
The principles and terms introduced in the previous sections can now be summarized and illustrated by the following three examples. Reactions such as these are called ionic or polar reactions, because they often involve charged species and the bonding together of electrophiles and nucleophiles. Ionic reactions normally take place in liquid solutions, where solvent molecules assist the formation of charged intermediates.
The substitution reaction shown on the left can be viewed as taking place in three steps. The first is an acid-base equilibrium, in which HCl protonates the oxygen atom of the alcohol. The resulting conjugate acid then loses water in a second step to give a carbocation intermediate. Finally, this electrophile combines with the chloride anion nucleophile to give the final product.
The addition reaction shown on the left can be viewed as taking place in two steps. The first step can again be considered an acid-base equilibrium, with the pi-electrons of the carbon-carbon double bond functioning as a base. The resulting conjugate acid is a carbocation, and this electrophile combines with the nucleophilic bromide anion.
The elimination reaction shown on the left takes place in one step. The bond breaking and making operations that take place in this step are described by the curved arrows. The initial stage may also be viewed as an acid-base interaction, with hydroxide ion serving as the base and a hydrogen atom component of the alkyl chloride as an acid.
There are many kinds of molecular rearrangements called isomerizations. The examples shown on the left are from an important class called tautomerization or, more specifically, keto-enol tautomerization. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) 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.
Exercise
1. Add curved arrows to explain the indicated reactivity and classify the reaction as "homolytic cleavage" or "heterolytic cleavage".
2. Add the correct arrow to each expression below using your knowledge of chemistry.
3. Classify the following reactions as substituion, addition, elimination, or tautomerization (an example of isomerization).
Answer
1.
2.
3. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.02%3A_Reaction_Mecha.txt |
Learning Objective
• identify nucleophiles and electrophiles in polar reactions
• relate bond polarity to chemical reactivity
Nucleophiles and Electrophiles
The reactants of polar reactions are often called the "nucleophile" and "electrophile". These terms are related to Lewis acid-base notation, so it can be helpful to apply and transfer the knowledge and wisdom gained from this definiation of acid-base chemistry.
• Electrophile (Lewis Acid): An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile
• Nucleophile (Lewis Base): An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile
Nucleophile
Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates. More specifically in laboratory reactions, halide and azide (N3-) anions are commonly seen acting as nucleophiles.
Enolate ions are the most common carbon nucleophiles in biochemical reactions, while the cyanide ion (CN-) is just one example of a carbon nucleophile commonly used in the laboratory. Hydrocarbons carbons with pi bonds can also be nucleophiles. Reactions with carbon nucleophiles will be dealt with in later chapters In this chapter, we will concentrate on non-carbon nucleophiles.
When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Electrophiles
In the vast majority of polar reactions, the electrophilic atom is a carbon atom bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom with partial positive charge is an attractive target an electron-rich nucleophile. Electrophiles can be challenging to recognize because their partial positive charge is hidden in polar bonds and/or resonance. Allkyl halides and carbonyl groups are useful electrophiles for synthetic organic chemistry.
Electrophilicity of Alkyl Halides
With respect to electronegativity, halogens are more electronegative than carbons. This results in a carbon-halogen bond that is polarized. As shown in the image below, carbon atom has a partial positive charge, while the halogen has a partial negative charge.
Allkyl halides are useful electrophiles for synthetic organic chemistry. Of the four halogens, fluorine is the most electronegative and iodine the least. That means that the electron pair in the carbon-fluorine bond will be dragged most towards the halogen end. Looking at the methyl halides as simple examples:
The following image shows the relative electronegativity of the halogens. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
You might have thought that either of these would be more effective in the case of the carbon-fluorine bond with the quite large amounts of positive and negative charge already present. But that's not so - quite the opposite is true! The thing that governs the reactivity is the strength of the bonds which have to be broken. If is difficult to break a carbon-fluorine bond, but easy to break a carbon-iodine one. The relative electrophilicity of alkyl halides is summarized below.
Electrophilicity of the Carbonyl Group
The carbon atom of the carbonyl group (C=O) is electrophilic because the carbon-oxygen double bond is polar and one of the resonance contributors is ionized with a full positiv echarge on the carbonyl carbon. Oxygen is more electronegative than carbon, so electron density is higher on the oxygen side of the bond and lower on the carbon side. Both of these factors combine to increase the electrophilicity of carbonyl groups. Carbonyl chemistry is studied in greater detail in the second semester of organic chemistry.
Exercise
1. Recognizing organic compounds as nucleophiles or electrophiles is an important first step in recognizing and learning patterns of chemical reactivity. Classify the following compounds as nucleophiles or electrophiles.
a) methoxide (CH3O-)
b) formaldehye (CH2O)
c) bromocyclopentane
d) water
e) sodium cyanide
f) methanamine (CH3NH2)
Answer
a) charged nucleophile
b) electrophile (Carbonyl carbon has partial positive charge.)
c) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
d) neutral nucleophile
e) charge nucleophile (Don't let the cation distract us from the CN-)
f) neutral nucleophile (The lone pair electrons on the nitrogen are nucleophilic in the same way they are Lewis bases (electron donators). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.03%3A_Polar_Reaction.txt |
Learning Objective
• perform calculations using the equation $ΔGº = –RT \ln K = –2.303 RT \log_{10} K \label{eq2}$ and explain the relationship between equilibrium and free energy
Equilibrium Constant
For the hypothetical chemical reaction:
$aA + bB \rightleftharpoons cC + dD$
the equilibrium constant is defined as:
$K_C = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$
where the notation [A] signifies the molar concentration of species A. Note that the expression for the equilibrium constant includes only solutes and gases; pure solids and liquids do not appear in the expression. For example, the equilibrium expression for the reaction
$\ce{CaH_2 (s) + 2H_2O (g) \rightleftharpoons Ca(OH)_2 (s) + 2H_2 (g)}$
is the following:
$K_C = \dfrac{[H_2]^2}{[H_2O]^2}$
Observe that the gas-phase species $\ce{H2O}$ and $\ce{H2}$ appear in the expression but the solids $\ce{CaH2}$ and $\ce{Ca(OH)2}$ do not appear.
The equilibrium constant is most readily determined by allowing a reaction to reach equilibrium, measuring the concentrations of the various solution-phase or gas-phase reactants and products, and substituting these values into the Law of Mass Action.
Free Energy
The interaction between enthalpy and entropy changes in chemical reactions is best observed by studying their influence on the equilibrium constants of reversible reactions. To this end a new thermodynamic function called Free Energy (or Gibbs Free Energy), symbol $ΔG$, is defined as shown in the first equation below. Two things should be apparent from this equation. First, in cases where the entropy change is small, $ΔG ≅ ΔH$. Second, the importance of $ΔS$ in determining $ΔG$ increases with increasing temperature.
$ΔGº = ΔHº – TΔSº$
where the temperature is measured in absolute temperature (K).
The free energy function provides improved insight into the thermodynamic driving forces that influence reactions. A negative $ΔGº$ is characteristic of an exergonic reaction, one which is thermodynamically favorable and often spontaneous, as is the melting of ice at 1 ºC. Likewise a positive $ΔGº$ is characteristic of an endergonic reaction, one which requires an input of energy from the surroundings.
For an example of the relationship of free energy to enthalpy consider the decomposition of cyclobutane to ethene, shown in the following equation. The standard state for all the compounds is gaseous.
This reaction is endothermic, but the increase in number of molecules from one (reactants) to two (products) results in a large positive ΔSº.
At 25 ºC (298 K):
ΔGº = 19 kcal/mol – 298(43.6) cal/mole = 19 – 13 kcal/mole = +6 kcal/mole.
Thus, the entropy change opposes the enthalpy change, but is not sufficient to change the sign of the resulting free energy change, which is endergonic. Indeed, cyclobutane is perfectly stable when kept at room temperature.
Because the entropy contribution increases with temperature, this energetically unfavorable transformation can be made favorable by raising the temperature. At 200 ºC (473 K),
\begin{align} ΔGº &= 19\, kcal/mol – 473(43.6)\, cal/mole \[4pt] &= 19 – 20.6\, kcal/mole \[4pt] &= –1.6 kcal/mole.\end{align}
This is now an exergonic reaction, and the thermal cracking of cyclobutane to ethene is known to occur at higher temperatures.
$ΔGº = –RT \ln K = –2.303 RT \log_{10} K \label{eq2}$
where R = 1.987 cal/ K mole T = temperature in K and K = equilibrium constant
Note
Equation \ref{eq2} is important because it demonstrates the fundamental relationship of $ΔGº$ to the equilibrium constant, $K$. Because of the negative logarithmic relationship between these variables, a negative ΔGº generates a K>1, whereas a positive ΔGº generates a K<1. When ΔGº = 0, K = 1. Furthermore, small changes in ΔGº produce large changes in K. A change of 1.4 kcal/mole in ΔGº changes K by approximately a factor of 10. This interrelationship may be explored with the calculator on the right. Entering free energies outside the range -8 to 8 kcal/mole or equilibrium constants outside the range 10-6 to 900,000 will trigger an alert, indicating the large imbalance such numbers imply.
Applications to Organic Reactions
The equation below can also be useful without performing any calculations.
$ΔGº = –RT \ln K = –2.303 RT \log_{10} K \label{eq2}$
Conceptually, this equation helps us compare the energetics of reaction mechanisms to predict the major products. For example, if ΔG° < 0, then K > 1, and products are favored over reactants. If ΔG° > 0, then K < 1, and reactants are favored over products. If ΔG° = 0, then K = 1, and the system is at equilibrium. Recognizing the underlying energetics of equilibrium, the stability of charged reactants and products can be used to predict reaction equilibrium. Reaction conditions can also be adjusted or controlled to shift the equilibrium in the desired direction by a range of experimental methods. A theoretical understanding of the reaction free energy and equilibrium helps us predict and design the optimum reaction conditions for a desired product.
Exercises
1. At 155°C, the equilibrium constant, Keq, for the reaction
$\ce{\sf{CH3CO2H + CH3CH2OH <=> CH3CO2CH2CH3 + H2O}}$
has a value of 4.0. Calculate ΔG° for this reaction at 155°C.
2. Acetylene (C2H2) can be converted into benzene (C6H6) according to the equation:
$\ce{\sf{3H-C#C-H (g) <=> C6H6 (l)}}$
At 25°C, ΔG° for this reaction is −503 kJ and ΔH° is −631 kJ. Determine ΔS° and indicate whether the size of ΔS° agrees with what you would have predicted simply by looking at the chemical equation.
Answer
Δ G ° = − R T ln K eq = − ( 8.314 J ⋅ K − 1 ⋅ mol − 1 ) ( 428 K ) ln ( 4.0 ) = − ( 8.314 J ⋅ K − 1 ⋅ mol − 1 ) ( 428 K ) ( 1.386 ) = − 4.9 × 10 3 J ⋅ mol − 1 = − 4.9 KJ ⋅ mol − 1 Δ G ° = Δ H ° − T Δ S ° Δ S ° = ( Δ H ° − Δ G ° ) T = − 631 kJ − ( − 503 kJ ) 298 K = − 128 kJ 298 K = − 0.430 KJ ⋅ mol − 1 = − 430 J ⋅ mol − 1
The entropy change is negative, as one would expect from looking at the chemical equation, since three moles of reactants yield one mole of product; that is, the system becomes much more “ordered” as it goes from reactants to products. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.04%3A_Describing_a_R.txt |
Learning Objective
• calculate reaction enthalpies from bond dissociation energies
Introduction
The homolytic bond dissociation energy is the amount of energy needed to break apart one mole of covalently bonded gases into a pair of radicals. The SI units used to describe bond energy are kiloJoules per mole of bonds (kJ/Mol). It indicates how strongly the atoms are bonded to each other. Solvation is the interaction between solvent molecules and the ions or molecules dissolved in that solvent.
Breaking a covalent bond between two partners, A-B, can occur either heterolytically, where the shared pair of electron goes with one partner or another
$A-B \rightarrow A^+ + B:^-$
or
$A-B \rightarrow A:^- + B^+ $
or homolytically, where one electron stays with each partner.
$A-B \rightarrow A^• + B^•$
The products of homolytic cleavage are radicals and the energy that is required to break the bond homolytically is called the Bond Dissociation Energy (BDE) and is a measure of the strength of the bond.
Calculation of the BDE
The BDE for a molecule A-B is calculated as the difference in the enthalpies of formation of the products and reactants for homolysis
$BDE = \Delta_fH(A^•) + \Delta_fH(B^•) - \Delta_fH(A-B)$
Officially, the IUPAC definition of bond dissociation energy refers to the energy change that occurs at 0 K, and the symbol is $D_o$. However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation enthalpy, which generally refers to the enthalpy change at room temperature (298K). Although there are technically differences between BDEs at 0 K and 298 K, those difference are not large and generally do not affect interpretations of chemical processes.
Bond Breakage/Formation
Bond dissociation energy (or enthalpy) is a state function and consequently does not depend on the path by which it occurs. Therefore, the specific mechanism in how a bond breaks or is formed does not affect the BDE. Bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using Hess's Law can be used to estimate reaction enthalpies.
Example: Chlorination of Methane
Consider the chlorination of methane
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$
the overall reaction thermochemistry can be calculated exactly by combining the BDEs for the bonds broken and bonds formed
CH4 → CH3• + H• BDE(CH3-H)
Cl2 → 2Cl• BDE(Cl2)\]
H• + Cl• → HCl -BDE(HCl)
CH3• + Cl• → CH3Cl -BDE(CH3-Cl)
---------------------------------------------------
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$
$\Delta H = BDE(R-H) + BDE(Cl_2) - BDE(HCl) - BDE(CH_3-Cl)$
Because reaction enthalpy is a state function, it does not matter what reactions are combined to make up the overall process using Hess's Law. However, BDEs are convenient to use because they are readily available.
Alternatively, BDEs can be used to assess individual steps of a mechanism. For example, an important step in free radical chlorination of alkanes is the abstraction of hydrogen from the alkane to form a free radical.
RH + Cl• → R• + HCl
The energy change for this step is equal to the difference in the BDEs in RH and HCl
$\Delta H = BDE(R-H) - BDE(HCl)$
This relationship shows that the hydrogen abstraction step is more favorable when BDE(R-H) is smaller. The difference in energies accounts for the selectivity in the halogenation of hydrocarbons with different types of C-H bonds.
Table 6.8.1: Representative C-H BDEs in Organic Molecules
R-H Do, kJ/mol D298, kJ/mol R-H Do, kJ/mol D298, kJ/mol
CH3-H 432.7±0.1 439.3±0.4 H2C=CH-H 456.7±2.7 463.2±2.9
CH3CH2-H 423.0±1.7 C6H5-H 465.8±1.9 472.4±2.5
(CH3)2CH-H 412.5±1.7 HCCH 551.2±0.1 557.8±0.3
(CH3)3C-H 403.8±1.7
H2C=CHCH2-H 371.5±1.7
HC(O)-H 368.6±0.8 C6H5CH2-H 375.3±2.5
CH3C(O)-H 374.0±1.2
Trends in C-H BDEs
It is important to remember that C-H BDEs refer to the energy it takes to break the bond, and is the difference in energy between the reactants and the products. Therefore, it is not appropriate to interpret BDEs solely in terms of the "stability of the radical products" as is often done.
Analysis of the BDEs shown in the table above shows that there are some systematic trends:
1. BDEs vary with hybridization: Bonds with sp3 hybridized carbons are weakest and bonds with sp hybridized carbons are much stronger. The vinyl and phenyl C-H bonds are similar, reflecting their sp2 hybridization. The correlation with hybridization can be viewed as a reflection of the C-H bond lengths. Longer bonds formed with sp3 orbitals are consequently weaker. Shorter bonds formed with orbitals that have more s-character are similarly stronger.
2. C-H BDEs vary with substitution: Among sp3 hybridized systems, methane has the strongest C-H bond. C-H bonds on primary carbons are stronger than those on secondary carbons, which are stronger than those on tertiary carbons.
Interpretation of C-H BDEs for sp3 Hybridized Carbons
The interpretation of the BDEs in saturated molecules has been subject of recent controversy. As indicated above, the variation in BDEs with substitution has traditionally been interpreted as reflecting the stabilities of the alkyl radicals, with the assessment that more highly substituted radicals are more stable, as with carbocations. Although this is a popular explanation, it fails to account fo the fact the bonds to groups other than H do not show the same types of variation.
R BDE(R-CH3) BDE(R-Cl) BDE(R-Br) BDE(R-OH)
CH3- 377.0±0.4 350.2±0.4 301.7±1.3 385.3±0.4
CH3CH2-
372.4±1.7
354.8±2.1 302.9±2.5 393.3±1.7
(CH3)2CH- 370.7±1.7 356.5±2.1 309.2±2.9 399.6±1.7
(CH3)3C- 366.1±1.7 355.2±2.9 303.8±2.5 400.8±1.7
Therefore, although C-CH3 bonds get weaker with more substitution, the effect is not nearly as large as that observed with C-H bonds. The strengths of C-Cl and C-Br bonds are not affected by substitution, despite the fact that the same radicals are formed as when breaking C-H bonds, and the C-OH bonds in alcohols actually increase with more substitution.
Gronert has proposed that the variation in BDEs is alternately explained as resulting from destabilization of the reactants due to steric repulsion of the substituents, which is released in the nearly planar radicals.1 Considering that BDEs reflect the relative energies of reactants and products, either explanation can account for the trend in BDEs.
Another factor that needs to be considered is the electronegativity. The Pauling definition of electronegativity says that the bond dissociation energy between unequal partners is going to be dependent on the difference in electrongativities, according to the expression
$D_o(A-B) = \dfrac{D_o(A-A) + D_o(B-B)}{2} + (X_A - X_B)^2$
where $X_A$ and $X_B$ are the electronegativities and the bond energies are in eV. Therefore, the variation in BDEs can be interpreted as reflecting variation in the electronegativities of the different types of alkyl fragments.
There is likely some merit in all three interpretations. Since Gronert's original publication of his alternate explanation, there have been many desperate attempts to defend the radical stability explanation.
Exercise
1. Given that ΔH° for the reaction
CH4(g) + 4F2(g) -> CF4(g) + 4HF(g)
is −1936 kJ, use the following data to calculate the average bond energy of the $\ce{{\sf{C-F}} bonds in CF4. Bond Average Bond Energy \(\ce{\sf{C-H}}$ 413 kJ · mol−1
$\ce{\sf{F-F}}$ 155 kJ · mol−1
$\ce{\sf{H-F}}$ 567 kJ · mol−1
2. Calculate ΔH° for the reactions given below.
1. CH3CH2OCH3 + HI -> CH3CH2OH + CH3I
2. CH3Cl + NH3 -> CH3NH2 + HCl
Answer
$1.$CH4(g) + 4 F2(g) CF4(g) + 4HF(g) ΔH°=416kcal
Bonds broken:
4 mol C-H bonds $×\text{}\frac{\left(413\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=1652\text{\hspace{0.17em}}\text{kJ}$
4 mol F-F bonds $×\text{}\frac{\left(155\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=620\text{\hspace{0.17em}}\text{kJ}$
Bonds formed:
4 mol C-F bonds $×\text{}\frac{\left(x\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=4x\text{\hspace{0.17em}}\text{kJ}$
(where x = the average energy of one mole of C-F bonds in CF4, expressed in kJ)
4 mol H-F bonds $×\text{}\frac{\left(567\text{\hspace{0.17em}}\text{kJ}\right)}{\left(1\text{\hspace{0.17em}}\text{mol}\right)}=2268\text{\hspace{0.17em}}\text{kJ}$
$\begin{array}{rl}\Delta H°& =\Delta H°\left(\text{bonds broken}\right)-\Delta H°\left(\text{bonds formed}\right)\ & =\left(1652\text{\hspace{0.17em}}\text{kJ}+620\text{\hspace{0.17em}}\text{kJ}\right)-\left(4x+2268\text{\hspace{0.17em}}\text{kJ}\right)\ & =1652\text{\hspace{0.17em}}\text{kJ}+620\text{\hspace{0.17em}}\text{kJ}-4x-2268\text{\hspace{0.17em}}\text{kJ}\ & =-1936\text{\hspace{0.17em}}\text{kJ}\end{array}$
Thus,
$\begin{array}{rl}4x& =1936\text{\hspace{0.17em}}\text{kJ}-2268\text{\hspace{0.17em}}\text{kJ}+620\text{\hspace{0.17em}}\text{kJ}+1652\text{\hspace{0.17em}}\text{kJ}\ & =1940\text{\hspace{0.17em}}\text{kJ}\end{array}$
and
$\begin{array}{rl}x& =\frac{1940\text{\hspace{0.17em}}\text{kJ}}{4\text{\hspace{0.17em}}\text{mol}}\ & =385\text{\hspace{0.17em}}\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$
The average energy of a C-F bond in CF4 is 385 kJ · mol-1
2. CH3Cl + NH3 -> CH3NH2 + HCl
$\dfrac { \begin{array}{cc} \textrm{Reactant bonds broken} & D \[6pt] \ce{CH3-Cl} & {351 \,\, \textrm{kJ/mol}} \ \ce{NH2-H} & {449 \,\, \textrm{kJ/mol}} \ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Reactant bonds broken}}} & {800 \,\, \textrm{kJ/mol}} \end{array} } \quad \dfrac { \begin{array}{cc} \textrm{Product bonds formed} & D \[6pt] \ce{CH3-NH2} & {335 \,\, \textrm{kJ/mol}} \ \ce{H-Cl} & {432 \,\, \textrm{kJ/mol}} \ \end{array} } { \begin{array}{cc} {\phantom{\textrm{Product bonds formed}}} & {767 \,\, \textrm{kJ/mol}} \end{array} }$
\begin{align*} \Delta H^{\circ} & = D_{\textrm{bonds broken}} + D_{\textrm{bonds formed}}\ & = {800 \,\, \textrm{kJ/mol} - 767 \,\, \textrm{kJ/mol}} \ & = {+33 \,\, \textrm{kJ/mol}} \end{align*}
Further Reading
MasterOrganicChemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.05%3A_Homolytic_Clea.txt |
Learning Objective
• draw Reaction Energy Diagrams from the thermodynamic and kinetic data/information
• use a Reaction Energy Diagram to discuss transition states, Ea, intermediates & rate determining step
• draw the transition state of a reaction
You may recall from general chemistry that it is often convenient to describe chemical reactions with energy diagrams. In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products. The energy diagram for a typical one-step reaction might look like this:
Despite its apparent simplicity, this energy diagram conveys some very important ideas about the thermodynamics and kinetics of the reaction. Recall that when we talk about the thermodynamics of a reaction, we are concerned with the difference in energy between reactants and products, and whether a reaction is ‘downhill’ (exergonic, energy releasing) or ‘uphill (endergonic, energy absorbing). When we talk about kinetics, on the other hand, we are concerned with the rate of the reaction, regardless of whether it is uphill or downhill thermodynamically.
First, let’s review what this energy diagram tells us about the thermodynamics of the reaction illustrated by the energy diagram above. The energy level of the products is lower than that of the reactants. This tells us that the change in standard Gibbs Free Energy for the reaction (Δrnx) is negative. In other words, the reaction is exergonic, or ‘downhill’. Recall that the Δrnx term encapsulates both Δrnx, the change in enthalpy (heat) and Δrnx , the change in entropy (disorder):
$ΔG˚ = ΔH˚- TΔS˚$
where T is the absolute temperature in Kelvin. For chemical processes where the entropy change is small (~0), the enthalpy change is essentially the same as the change in Gibbs Free Energy. Energy diagrams for these processes will often plot the enthalpy (H) instead of Free Energy for simplicity.
The standard Gibbs Free Energy change for a reaction can be related to the reaction's equilibrium constant ($K_{eq}\_) by a simple equation: $ΔG˚ = -RT \ln K_{eq}$ where: • Keq = [product] / [reactant] at equilibrium • R = 8.314 J×K-1×mol-1 or 1.987 cal× K-1×mol-1 • T = temperature in Kelvin (K) If you do the math, you see that a negative value for Δrnx (an exergonic reaction) corresponds - as it should by intuition - to Keq being greater than 1, an equilibrium constant which favors product formation. In a hypothetical endergonic (energy-absorbing) reaction the products would have a higher energy than reactants and thus Δrnx would be positive and Keq would be less than 1, favoring reactants. Now, let's move to kinetics. Look again at the energy diagram for exergonic reaction: although it is ‘downhill’ overall, it isn’t a straight downhill run. First, an ‘energy barrier’ must be overcome to get to the product side. The height of this energy barrier, you may recall, is called the ‘activation energy’ (ΔG). You may have been taught to use the term “activated complex” rather than “transition state,” as the two are often used interchangeably. Similarly, the activation energy of a reaction is often represented by the symbol Eact or Ea. The activation energy is what determines the kinetics of a reaction: the higher the energy hill, the slower the reaction. At the very top of the energy barrier, the reaction is at its transition state (TS), which is the point at which the bonds are in the process of breaking and forming. The transition state is an ‘activated complex’: a transient and dynamic state that, unlike more stable species, does not have any definable lifetime. It may help to imagine a transition state as being analogous to the exact moment that a baseball is struck by a bat. Transition states are drawn with dotted lines representing bonds that are in the process of breaking or forming, and the drawing is often enclosed by brackets. Here is a picture of a likely transition state for a substitution reaction between hydroxide and chloromethane: $CH_3Cl + HO^- \rightarrow CH_3OH + Cl^-$ This reaction involves a collision between two molecules: for this reason, we say that it has second order kinetics. The rate expression for this type of reaction is: rate = k[reactant 1][reactant 2] . . . which tells us that the rate of the reaction depends on the rate constant k as well as on the concentration of both reactants. The rate constant can be determined experimentally by measuring the rate of the reaction with different starting reactant concentrations. The rate constant depends on the activation energy, of course, but also on temperature: a higher temperature means a higher k and a faster reaction, all else being equal. This should make intuitive sense: when there is more heat energy in the system, more of the reactant molecules are able to get over the energy barrier. Here is one more interesting and useful expression. Consider a simple reaction where the reactants are A and B, and the product is AB (this is referred to as a condensation reaction, because two molecules are coming together, or condensing). If we know the rate constant k for the forward reaction and the rate constant kreverse for the reverse reaction (where AB splits apart into A and B), we can simply take the quotient to find our equilibrium constant \(K_{eq}$:
This too should make some intuitive sense; if the forward rate constant is higher than the reverse rate constant, equilibrium should lie towards products.
Exercise
1. Which reaction is faster, ΔG = + 55 kJ/mol or ΔG = + 75 kJ/mol?
Answer
1. The + 55 kJ/mol reaction is the faster reaction. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.06%3A_Reaction_Energ.txt |
Learning Objective
• describe the structure & relative stabilities of carbocations
Carbocations and their Stability
A carbocation is an ion with a positively-charged carbon atom. A carbocation is very electron-poor, and thus anything which donates electron density to the center of electron poverty will help to stabilize it. Conversely, a carbocation will be destabilized by an electron withdrawing group.
Alkyl groups are electron donating and carbocation-stabilizing because the electrons around the neighboring carbons are drawn towards the nearby positive charge, thus slightly reducing the electron poverty of the positively-charged carbon. What this means is that, in general, more substituted carbocations are more stable: a tert-butyl carbocation, for example, is more stable than an isopropyl carbocation. Primary carbocations are highly unstable and not often observed as reaction intermediates; methyl carbocations are even less stable.
It is not accurate to say, however, that carbocations with higher substitution are always more stable than those with less substitution. Just as electron-donating groups can stabilize a carbocation, electron-withdrawing groups act to destabilize carbocations. Carbonyl groups are electron-withdrawing by inductive effects, due to the polarity of the C=O double bond. It is possible to demonstrate in the laboratory (see section 16.1D) that carbocation A below is more stable than carbocation B, even though A is a primary carbocation and B is secondary.
The difference in stability can be explained by considering the electron-withdrawing inductive effect of the ester carbonyl. Recall that inductive effects - whether electron-withdrawing or donating - are relayed through covalent bonds and that the strength of the effect decreases rapidly as the number of intermediary bonds increases. In other words, the effect decreases with distance. In species B the positive charge is closer to the carbonyl group, thus the destabilizing electron-withdrawing effect is stronger than it is in species A.
Stabilization of a carbocation can also occur through resonance effects, and as we have already discussed in the acid-base chapter, resonance effects as a rule are more powerful than inductive effects. Consider the simple case of a benzylic carbocation:
This carbocation is comparatively stable. In this case, electron donation is a resonance effect. Three additional resonance structures can be drawn for this carbocation in which the positive charge is located on one of three aromatic carbons. The positive charge is not isolated on the benzylic carbon, rather it is delocalized around the aromatic structure: this delocalization of charge results in significant stabilization. As a result, benzylic and allylic carbocations (where the positively charged carbon is conjugated to one or more non-aromatic double bonds) are significantly more stable than even tertiary alkyl carbocations.
Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in section 7.4). Consider the two pairs of carbocation species below:
In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually destabilize the positive charge, because they are electron withdrawing by induction.
Finally, vinylic carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any reaction.
Carbocation Rearrangements
Carbocations typically undergo rearrangement reactions from less stable structures to equally stable or more stable ones with rate constants in excess of 109/sec. This fact complicates synthetic pathways to many compounds, so it is important to look for carbocation rearrangements anytime they are formed. It is possible for either a neighboring hydrogen atom or methyl group to shift to the carbocation to create a more stable intermediate. In the 1,2-hydride shift shown below, the secondary carbocatin rearranges to a more stable tertiary carbocation. The numbers, 1,2- refer to the vicinal location of the rearrangement, not the nomenclature numbers.
In this next example, the methyl group shifts to stabilize the carbocation.
Exercise
1. In which of the structures below is the carbocation expected to be more stable? Explain.
2. Draw a resonance structure of the crystal violet cation in which the positive charge is delocalized to one of the nitrogen atoms.
State which carbocation in each pair below is more stable, or if they are expected to be approximately equal. Explain your reasoning.
Answer
1. In the carbocation on the left, the positive charge is located in a position relative to the nitrogen such that the lone pair of electrons on the nitrogen can be donated to fill the empty orbital. This is not possible for the carbocation species on the right.
2.
3.
a) 1 (tertiary vs. secondary carbocation)
b) equal
c) 1 (tertiary vs. secondary carbocation)
d) 2 (positive charge is further from electron-withdrawing fluorine)
e) 1 (lone pair on nitrogen can donate electrons by resonance)
f) 1 (allylic carbocation – positive charge can be delocalized to a second carbon)
History
The history of carbocations dates back to 1891 when G. Merling[8] reported that he added bromine to tropylidene (cycloheptatriene) and then heated the product to obtain a crystalline, water-soluble material, \(C_7H_7Br\). He did not suggest a structure for it; however, Doering and Knox[9] convincingly showed that it was tropylium (cycloheptatrienylium) bromide. This ion is predicted to be aromatic by Hückel's rule.
In 1902, Norris and Kehrman independently discovered that colorless triphenylmethanol gives deep-yellow solutions in concentrated sulfuric acid. Triphenylmethyl chloride similarly formed orange complexes with aluminium and tin chlorides. In 1902, Adolf von Baeyer recognized the salt-like character of the compounds formed.
He dubbed the relationship between color and salt formation halochromy, of which malachite green is a prime example.
Carbocations are reactive intermediates in many organic reactions. This idea, first proposed by Julius Stieglitz in 1899,[10] was further developed by Hans Meerwein in his 1922 study[11][12] of the Wagner-Meerwein rearrangement. Carbocations were also found to be involved in the \(S_N1\) reaction, the \(E1\0 reaction, and in rearrangement reactions such as the Whitmore 1,2 shift. The chemical establishment was reluctant to accept the notion of a carbocation and for a long time the Journal of the American Chemical Society refused articles that mentioned them.
The first NMR spectrum of a stable carbocation in solution was published by Doering et al.[13] in 1958. It was the heptamethylbenzenium ion, made by treating hexamethylbenzene with methyl chloride and aluminium chloride. The stable 7-norbornadienyl cation was prepared by Story et al. in 1960[14] by reacting norbornadienyl chloride with silver tetrafluoroborate in sulfur dioxide at −80 °C. The NMR spectrum established that it was non-classically bridged (the first stable non-classical ion observed).
In 1962, Olah directly observed the tert-butyl carbocation by nuclear magnetic resonance as a stable species on dissolving tert-butyl fluoride in magic acid. The NMR of the norbornyl cation was first reported by Schleyer et al.[15] and it was shown to undergo proton-scrambling over a barrier by Saunders et al.[16]
References
1. Hansjörg Grützmacher, Christina M. Marchand (1997), "Heteroatom stabilized carbenium ions", Coordination Chemistry Reviews, volume 163, pages 287-344. doi:10.1016/S0010-8545(97)00043-X
2. Gold Book definition carbonium ion HTML
3. George A. Olah (1972), "Stable carbocations. CXVIII. General concept and structure of carbocations based on differentiation of trivalent (classical) carbenium ions from three-center bound penta- of tetracoordinated (nonclassical) carbonium ions. Role of carbocations in electrophilic reactions." Journal of the American Chemical Society, volume 94, issue 3, pages 808–820 doi:10.1021/ja00758a020
4. Organic chemistry 5th Ed. John McMurry ISBN 0-534-37617-7
5. Organic Chemistry, Fourth Edition Paula Yurkanis Bruice ISBN 0-13-140748-1
6. Clayden, Jonathan; Greeves, Nick; Warren, Stuart; Wothers, Peter (2001). Organic Chemistry (1st ed.). Oxford University Press. ISBN 978-0-19-850346-0.
7. Organic Chemistry by Marye Anne Fox and James K. Whitesell ISBN 0-7637-0413-X
8. Chem. Ber. 24, 3108 1891
9. The Cycloheptatrienylium (Tropylium) Ion W. Von E. Doering and L. H. Knox J. Am. Chem. Soc.; 1954; 76(12) pp 3203 - 3206; doi:10.1021/ja01641a027
10. On the Constitution of the Salts of Imido-Ethers and other Carbimide Derivatives; Am. Chem. J. 21, 101; ISSN: 0096-4085
11. H. Meerwein and K. van Emster, Berichte, 1922, 55, 2500.
12. Rzepa, H. S.; Allan, C. S. M. (2010). "Racemization of Isobornyl Chloride via Carbocations: A Nonclassical Look at a Classic Mechanism". Journal of Chemical Education 87 (2): 221. Bibcode:2010JChEd..87..221R. doi:10.1021/ed800058c. edit
13. The 1,1,2,3,4,5,6-heptamethylbenzenonium ion W. von E. Doering and M. Saunders H. G. Boyton, H. W. Earhart, E. F. Wadley and W. R. Edwards G. Laber Tetrahedron Volume 4, Issues 1-2 , 1958, Pages 178-185 doi:10.1016/0040-4020(58)88016-3
14. The 7-norbornadienyl carbonium ion Paul R. Story and Martin Saunders J. Am. Chem. Soc.; 1960; 82(23) pp 6199 - 6199; doi:10.1021/ja01508a058
15. Stable Carbonium Ions. X.1 Direct Nuclear Magnetic Resonance Observation of the 2-Norbornyl Cation Paul von R. Schleyer, William E. Watts, Raymond C. Fort, Melvin B. Comisarow, and George A. Olah J. Am. Chem. Soc.; 1964; 86(24) pp 5679 - 5680; doi:10.1021/ja01078a056
16. Stable Carbonium Ions. XI.1 The Rate of Hydride Shifts in the 2-Norbornyl Cation Martin Saunders, Paul von R. Schleyer, and George A. Olah J. Am. Chem. Soc.; 1964; 86(24) pp 5680 - 5681; doi:10.1021/ja01078a057
17. George A. Olah - Nobel Lecture
18. Nuclear magnetic double resonance studies of the dimethylcyclopropylcarbinyl cation. Measurement of the rotation barrier David S. Kabakoff, , Eli. Namanworth J. Am. Chem. Soc. 1970, 92 (10), pp 3234–3235 doi:10.1021/ja00713a080
19. Stable Carbonium Ions. XVII.1a Cyclopropyl Carbonium Ions and Protonated Cyclopropyl Ketones Charles U. Pittman Jr., George A. Olah J. Am. Chem. Soc., 1965, 87 (22), pp 5123–5132 doi:10.1021/ja00950a026
20. F.A. Carey, R.J. Sundberg Advanced Organic Chemistry Part A 2nd Ed. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.07%3A_Reactive_Inter.txt |
Learning Objective
• describe the structure & relative stabilities of free radicals
Radicals
In chemistry, a radical (more precisely, a free radical) is an atom, molecule, or ion that has unpaired valence electrons or an open electron shell, and therefore may be seen as having one or more "dangling" covalent bonds.
With some exceptions, these "dangling" bonds make free radicals highly chemically reactive towards other substances, or even towards themselves: their molecules will often spontaneously dimerize or polymerize if they come in contact with each other. Most radicals are reasonably stable only at very low concentrations in inert media or in a vacuum.
A notable example of a free radical is the hydroxyl radical (HO•), a molecule that is one hydrogen atom short of a water molecule and thus has one bond "dangling" from the oxygen. Two other examples are the carbene molecule (:CH2), which has two dangling bonds; and the superoxide anion (•O−2), the oxygen molecule O2 with one extra electron, which has one dangling bond. In contrast, the hydroxyl anion (HO−), the oxide anion (O2−) and thecarbenium cation (CH+3) are not radicals, since the bonds that may appear to be dangling are in fact resolved by the addition or removal of electrons.
Free radicals may be created in a number of ways, including synthesis with very dilute or rarefied reagents, reactions at very low temperatures, or breakup of larger molecules. The latter can be affected by any process that puts enough energy into the parent molecule, such as ionizing radiation, heat, electrical discharges, electrolysis, and chemical reactions. Indeed, radicals are intermediate stages in many chemical reactions.
Free radicals play an important role in combustion, atmospheric chemistry, polymerization, plasma chemistry, biochemistry, and many other chemical processes. In living organisms, the free radicals superoxide and nitric oxideand their reaction products regulate many processes, such as control of vascular tone and thus blood pressure. They also play a key role in the intermediary metabolism of various biological compounds. Such radicals can even be messengers in a process dubbed redox signaling. A radical may be trapped within a solvent cage or be otherwise bound.
Until late in the 20th century the word "radical" was used in chemistry to indicate any connected group of atoms, such as a methyl group or a carboxyl, whether it was part of a larger molecule or a molecule on its own. The qualifier "free" was then needed to specify the unbound case. Following recent nomenclature revisions, a part of a larger molecule is now called a functional group or substituent, and "radical" now implies "free". However, the old nomenclature may still occur in the literature.
Formation
The formation of radicals may involve breaking of covalent bonds homolytically, a process that requires significant amounts of energy. For example, splitting H2 into 2H· has a ΔH° of +435 kJ/mol, and Cl2 into 2Cl· has a ΔH° of +243 kJ/mol. This is known as the homolytic bond dissociation energy, and is usually abbreviated as the symbol ΔH°. The bond energy between two covalently bonded atoms is affected by the structure of the molecule as a whole, not just the identity of the two atoms. Likewise, radicals requiring more energy to form are less stable than those requiring less energy. Homolytic bond cleavage most often happens between two atoms of similar electronegativity. In organic chemistry this is often the O-O bond in peroxide species or O-N bonds. Sometimes radical formation is spin-forbidden, presenting an additional barrier. However, propagation is a very exothermic reaction. Likewise, although radical ions do exist, most species are electrically neutral. Radicals may also be formed by single electron oxidation or reduction of an atom or molecule. An example is the production of superoxide by the electron transport chain. Early studies of organometallic chemistry, especially tetra-alkyl lead species by F.A. Paneth and K. Hahnfeld in the 1930s supported heterolytic fission of bonds and a radical based mechanism.
Depiction in chemical reactions
In chemical equations, free radicals are frequently denoted by a dot placed immediately to the right of the atomic symbol or molecular formula as follows:
$\mathrm{Cl}_2 \; \xrightarrow{UV} \; {\mathrm{Cl} \cdot} + {\mathrm{Cl} \cdot}$
Chlorine gas can be broken down by ultraviolet light to form atomic chlorine radicals.
Radical reaction mechanisms use single-headed arrows to depict the movement of single electrons:
The homolytic cleavage of the breaking bond is drawn with a 'fish-hook' arrow to distinguish from the usual movement of two electrons depicted by a standard curly arrow. It should be noted that the second electron of the breaking bond also moves to pair up with the attacking radical electron; this is not explicitly indicated in this case.
Relative Stability
Radical alkyl intermediates are stabilized by similar physical processes to carbocations: as a general rule, the more substituted the radical center is, the more stable it is. This directs their reactions. Thus, formation of a tertiary radical (R3C·) is favored over secondary (R2HC·), which is favored over primary (RH2C·). Likewise, radicals next to functional groups such as carbonyl, nitrile, and ether are more stable than tertiary alkyl radicals.
Exercise
1. State which carbon radical (free radical) in each pair below is more stable or if they are expected to have comparable stability. Explain your reasoning.
Answer
1.
a) Cpd 1: Tertiary radicals are more stable than secondary radicals.
b) Cpds 1 and 2 are both secondary so they have comparable stability.
c) Cpd 1: Tertiary radicals are more stable than secondary radicals with similar effects from the Cl atom.
d) Cpd 2: Both compounds are secondary, but positive charge is further from electron-withdrawing chlorine on Cpd 2.
e) Cpd 1: Lone pair on nitrogen can donate electrons by resonance.
f) Cpd 1: Secondary allylic radicals are more stable than tertiary radicals. (Primary allylic radicals are comparable in stability to tertiary radicals.) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.08%3A_Reactive_Inter.txt |
Carbanions
A carbanion is an anion in which carbon has an unshared pair of electrons and bears a negative charge usually with three substituents for a total of eight valence electrons.[1] The carbanion exists in a trigonal pyramidal geometry. Formally, a carbanion is the conjugate base of a carbon acid.
$\ce{R_3C-H + B^- \rightarrow R_3C^- + H-B}$
where B stands for the base. A carbanion is one of several reactive intermediates in organic chemistry.
A carbanion is a nucleophile, which stability and reactivity determined by several factors:
1. The inductive effect. Electronegative atoms adjacent to the charge will stabilize the charge;
2. Hybridization of the charge-bearing atom. The greater the s-character of the charge-bearing atom, the more stable the anion;
3. The extent of conjugation of the anion. Resonance effects can stabilize the anion. This is especially true when the anion is stabilized as a result of aromaticity.
A carbanion is a reactive intermediate and is encountered in organic chemistry for instance in the E1cB elimination reaction and in organometallic chemistry in for instance a Grignard reaction or in alkyl lithium chemistry. Stable carbanions do however exist. In 1984 Olmstead presented the lithium crown ether salt of the triphenylmethyl carbanion from triphenylmethane, n-butyllithium and 12-crown-4 at low temperatures:[2]
Adding n-butyllithium to triphenylmethane in THF at low temperatures followed by 12-crown-4 results in a red solution and the salt complex precipitates at −20 °C. The central C-C bond lengths are 145 pm with the phenyl ring propelled at an average angle of 31.2°. This propeller shape is less pronounced with a tetramethylammonium counterion.[3] One tool for the detection of carbanions in solution is proton NMR.[4] A spectrum of cyclopentadiene in DMSO shows four vinylic protons at 6.5 ppm and two methylene bridge protons at 3 ppm whereas the cyclopentadienyl anion has a single resonance at 5.50 ppm.
Carbon acids
Any molecule containing a C-H can lose a proton forming the carbanion. Hence any hydrocarbon containing C-H bonds can be considered an acid with a corresponding pKa value. Methane is certainly not an acid in its classical meaning yet its estimated pKa is 56. Compare this to acetic acid with pKa 4.76. The same factors that determine the stability of the carbanion also determine the order in pKa in carbon acids. These values are determined for the compounds either in water in order to compare them to ordinary acids, indimethyl sulfoxide in which the majority of carbon acids and their anions are soluble or in the gas phase. With DMSO the acidity window for solutes is limited to its own pKa of 35.5.
Table 1. Carbon acid acidities in pKa in DMSO [5]. Reference acids in bold.
name formula structural formula pKa
Methane CH4 ~ 56
Ethane C2H6 ~ 50
Anisole C7H8O ~ 49
Cyclopentane C5H10 ~ 45
Propene C3H6 ~ 44
Benzene C6H6 ~ 43
Toluene C6H5CH3 ~ 43
Dimethyl sulfoxide (CH3)2SO 35.5
Diphenylmethane C13H12 32.3
Aniline C6H5NH2 30.6
Triphenylmethane C19H16 30.6
Xanthene C13H10O 30
Ethanol C2H5OH 29.8
Phenylacetylene C8H6 28.8
Thioxanthene C13H10S 28.6
Acetone C3H6O 26.5
Acetylene C2H2 25
Benzoxazole C7H5NO 24.4
Fluorene C13H10 22.6
Indene C9H8 20.1
Cyclopentadiene C5H6 18
Malononitrile C3H2N2 11.2
Hydrogen cyanide HCN 9.2
Acetylacetone C5H8O2 8.95
Dimedone C8H12O2 5.23
Meldrum's acid C6H8O4 4.97
Acetic acid CH3COOH 4.76
Barbituric acid C4H2O3(NH)2 4.01
Trinitromethane HC(NO2)3 0.17
Fulminic acid HCNO -1.07
Carborane superacid HCHB11Cl11 -9
Note that the anions formed by ionization of acetic acid, ethanol or aniline are not carbanions.
Starting from methane in Table 1, the acidity increases:
• when the anion is aromatic, either because the added electron causes the anion to become aromatic (as in indene and cyclopentadiene), or because the negative charge on carbon can be delocalized over several already-aromatic rings (as in triphenylmethane or the carborane superacid).
• when the carbanion is surrounded by strongly electronegative groups, through the partial neutralisation of the negative charge (as in malononitrile).
• when the carbanion is immediately next to a carbonyl group. The α-protons of carbonyl groups are acidic because the negative charge in the enolate can be partially distributed in the oxygen atom. Meldrum's acid and barbituric acid, historically named acids, are in fact a lactone and a lactam respectively, but their acidic carbon protons make them acidic. The acidity of carbonyl compounds is an important driving force in many organic reactions such as the aldol reaction.
Chiral carbanions
With the molecular geometry for a carbanion described as a trigonal pyramid the question is whether or not carbanions can display chirality, because if the activation barrier for inversion of this geometry is too low any attempt at introducing chirality will end inracemization, similar to the nitrogen inversion. However, solid evidence exists that carbanions can indeed be chiral for example in research carried out with certain organolithium compounds.
The first ever evidence for the existence of chiral organolithium compounds was obtained in 1950. Reaction of chiral 2-iodooctane with sec-butyllithium in petroleum ether at −70 °C followed by reaction with dry ice yielded mostly racemic 2-methylbutyric acid but also an amount of optically active 2-methyloctanoic acid which could only have formed from likewise optical active 2-methylheptyllithium with the carbon atom linked to lithium the carbanion:[6]
On heating the reaction to 0 °C the optical activity is lost. More evidence followed in the 1960s. A reaction of the cis isomer of 2-methylcyclopropyl bromide with sec-butyllithium again followed by carboxylation with dry ice yielded cis-2-methylcyclopropylcarboxylic acid. The formation of the trans isomer would have indicated that the intermediate carbanion was unstable.[7]
In the same manner the reaction of (+)-(S)-l-bromo-l-methyl-2,2-diphenylcyclopropane with n-butyllithium followed by quench with methanol resulted in product with retention of configuration:[8]
Of recent date are chiral methyllithium compounds:[9]
The phosphate 1 contains a chiral group with a hydrogen and a deuterium substituent. The stannyl group is replaced by lithium to intermediate 2 which undergoes a phosphate-phosphorane rearrangement to phosphorane 3 which on reaction with acetic acid givesalcohol 4. Once again in the range of −78 °C to 0 °C the chirality is preserved in this reaction sequence.[10]
History
A carbanionic structure first made an appearance in the reaction mechanism for the benzoin condensation as correctly proposed by Clarke and Lapworth in 1907.[11] In 1904 Schlenk prepared Ph3C-NMe4+ in a quest for pentavalent nitrogen (fromTetramethylammonium chloride and Ph3CNa) [12] and in 1914 he demonstrated how triarylmethyl radicals could be reduced to carbonions by alkali metals.[13] The phrase carbanion was introduced by Wallis and Adams in 1933 as the negatively charged counterpart of the carbonium ion. [14][15]
External links
• Large database of Bordwell pKa values at www.chem.wisc.edu Link
• Large database of Bordwell pKa values at daecr1.harvard.edu Link | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.09%3A_Reactive_Inter.txt |
Alkanes (the most basic of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. This reaction is very important in organic chemistry because it opens a gateway to further chemical reactions.
Introduction
While the reactions possible with alkanes are few, there are many reactions that involve haloalkanes. In order to better understand the mechanism (a detailed look at the step by step process through which a reaction occurs), we will closely examine the chlorination of methane. When methane (CH4) and chlorine (Cl2) are mixed together in the absence of light at room temperature nothing happens. However, if the conditions are changed, so that either the reaction is taking place at high temperatures (denoted by Δ) or there is ultra violet irradiation, a product is formed, chloromethane (CH3Cl).
Energetics
Why does this reaction occur? Is the reaction favorable? A way to answer these questions is to look at the change in enthalpy ($\Delta{H}$) that occurs when the reaction takes place.
ΔH = (Energy put into reaction) – (Energy given off from reaction)
If more energy is put into a reaction than is given off, the ΔH is positive, the reaction is endothermic and not energetically favorable. If more energy is given off in the reaction than was put in, the ΔH is negative, the reaction is said to be exothermic and is considered favorable. The figure below illustrates the difference between endothermic and exothermic reactions.
ΔH can also be calculated using bond dissociation energies (ΔH°):
$\Delta{H} = \sum \Delta{H^°} \text{ of bonds broken} - \sum \Delta{H^°} \text{ of bonds formed}$
Let’s look at our specific example of the chlorination of methane to determine if it is endothermic or exothermic:
Since, the ΔH for the chlorination of methane is negative, the reaction is exothermic. Energetically this reaction is favorable. In order to better understand this reaction we need to look at the mechanism ( a detailed step by step look at the reaction showing how it occurs) by which the reaction occurs.
Radical Chain Mechanism
The reaction proceeds through the radical chain mechanism. The radical chain mechanism is characterized by three steps: initiation, propagation and termination. Initiation requires an input of energy but after that the reaction is self-sustaining. The first propagation step uses up one of the products from initiation, and the second propagation step makes another one, thus the cycle can continue until indefinitely.
Step 1: Initiation
Initiation breaks the bond between the chlorine molecule (Cl2). For this step to occur energy must be put in, this step is not energetically favorable. After this step, the reaction can occur continuously (as long as reactants provide) without input of more energy. It is important to note that this part of the mechanism cannot occur without some external energy input, through light or heat.
Step 2: Propagation
The next two steps in the mechanism are called propagation steps. In the first propagation step, a chlorine radical combines with a hydrogen on the methane. This gives hydrochloric acid (HCl, the inorganic product of this reaction) and the methyl radical. In the second propagation step more of the chlorine starting material (Cl2) is used, one of the chlorine atoms becomes a radical and the other combines with the methyl radical.
The first propagation step is endothermic, meaning it takes in heat (requires 2 kcal/mol) and is not energetically favorable. In contrast the second propagation step is exothermic, releasing 27 kcal/mol. Once the reaction is initiated, the exothermic energy released from the second propagation step provides the activation energy for the first propagation step creating a cyclic chain reaction following Le Chatelier's principle until termination.
Step 3: Termination
In the termination steps, all the remaining radicals combine (in all possible manners) to form more product (CH3Cl), more reactant (Cl2) and even combinations of the two methyl radicals to form a side product of ethane (CH3CH3).
Limitations of the Chlorination
The chlorination of methane or any other alkane does not necessarily stop after one chlorination. It may actually be very hard to get a monosubstituted chloromethane. Instead di-, tri- and even tetra-chloromethanes are formed. One way to avoid this problem is to use a much higher concentration of methane or other alkane in comparison to chloride. This reduces the chance of a chlorine radical running into a chloromethane and starting the mechanism over again to form a dichloromethane. Through this method of controlling product ratios one is able to have a relative amount of control over the product.
Exercises
1. Compounds other than chlorine and methane can react via free-radical halogenation. Write out the complete mechanism for the monobromination of ethane.
2. Explain how the energetically unfavorable first propagation step can continue to occur without the input of energy from an external source.
3. Which step of the radical chain mechanism requires outside energy? What can be used as this energy?
4. Use the table provided below to calculate the change in enthalpy for the monobromination of ethane.
Compound Bond Dissociation Energy (kcal/mol)
CH3CH2-H 101
CH3CH2-Br 70
H-Br 87
Br2 46
Solutions
1.
2. The exothermic energy released from the second propagation step provides the activation energy for the first propagation step creating a cyclic chain reaction following Le Chatelier's principle until termination.
3. The initiation step requires energy from heat o lit. For maximum photoefficiency, the wavelength of light is correlated with bond being homolytically cleaved.
4.
References
1. Matyjaszewski, Krzysztof, Wojciech Jakubowski, Ke Min, Wei Tang, Jinyu Huang, Wade A. Braunecker, and Nicolay V. Tsarevsky. "Diminishing Catalyst Concentration in Atom Transfer Radical Polymerization with Reducing Agents." Science 72 (1930): 379-90.
2. Phillips, Francis C. "# Researches upon the Chemical Properties of Gases." Researches upon the Chemical Properties of Gases 17 (1893): 149-236.
Contributors and Attributions
• Kristen Kelley and Britt Farquharson | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.10%3A_The_Free-Radic.txt |
Comparing Reactivity
Given the knowledge that a particular reaction will proceed at a suitable rate, a host of practical considerations are necessary for satisfactory operation. These considerations include interference by possible side reactions that give products other than those desired, the ease of separation of the desired product from the reaction mixture, and costs of materials, apparatus, and labor. We shall consider these problems in connection with the important synthetic reactions discussed in this book.
The chlorination of saturated hydrocarbons can be induced by light, but also can be carried out at temperatures of about $300^\text{o}$ in the dark. Under such circumstances the mechanism is similar to that of light-induced chlorination, except that the chlorine atoms are formed by thermal dissociation of chlorine molecules. Solid carbon surfaces catalyze thermal chlorination, possibly by aiding in the cleavage of the chlorine molecules.
Direct monohalogenation of saturated hydrocarbons works satisfactorily only with chlorine and bromine. For the general reaction
the calculated $\Delta H^\text{0}$ value is negative and very large for fluorine, negative and moderate for chlorine and bromine, and positive for iodine (see Table 4-7). With fluorine, the reaction evolves so much heat that it may be difficult to control, and products from cleavage of carbon-carbon as well as of carbon-hydrogen bonds may be obtained. The only successful, direct fluorination procedure for hydrocarbons involves diffusion of minute amounts of fluorine mixed with helium into liquid or solid hydrocarbons at low temperatures, typically $-78^\text{o}$ (Dry Ice temperature). As fluorination proceeds, the concentration of fluorine can be increased. The process is best suited for preparation of completely fluorinated compounds, and it has been possible to obtain in this way amounts of $\left( CF_3 \right)_4C$ and $\left( CF_3 \right)_3 C-C \left( CF_3 \right)_3$ from 2,2-dimethylpropane and 2,2,3,3-tetramethylbutane corresponding to $10$-$15\%$ yields based on the fluorine used.
Bromine generally is much less reactive toward hydrocarbons than chlorine is, both at high temperatures and with activation by light. Nonetheless, it usually is possible to brominate saturated hydrocarbons successfully. Iodine is unreactive.
Table: Calculated Heat of Reaction for Halogenation fo Hydrocarbons
Halogen (X) $\Delta H^o$ (kcal/mole)a
F -116
Cl -27
Br -10
I 13
aCalculated from the bond energies of Table 4-3.
The chlorination of methane does not have to stop with the formation of chloromethane (methyl chloride). It is usual when chlorinating methane to obtain some of the higher chlorination products: dichloromethane (methylene chloride), trichloromethane (chloroform), and tetrachloromethane (carbon tetrachloride):
In practice, one can control the degree of substitution to a considerable extent by controlling the methane-chlorine ratio. For example, for monochlorination to predominate, a high methane-chlorine ratio is necessary such that the chlorine atoms react with $CH_4$ and not with $CH_3Cl$.
Selectivity in Alkane Halogenation
For propane and higher hydrocarbons for which more than one monosubstitution product is generally possible, difficult separation problems bay arise when a particular product is desired. For example, the chlorination of 2-methylbutane $3$ at $300^\text{o}$ gives all four possible monosubstitution products. On a purely statistical basis, we may expect the ratio of products from 2-methylbutane to correlate with the number of available hydrogens at the various positions of substitution in the ratio 6:1:2:3 ($50\%$:$8\%$:$17\%$:$25\%$:). However, as can be seen from the strengths of bonds between hydrogen and primary, secondary, and tertiary carbons are not the same and we would expect the weaker $C-H$ bonds to preferentially react with $Cl \cdot$. As such, the proportion of the tertiary halide is about three times that expected on a statistical basis which is in accord with our expectation that the tertiary $C-H$ bond of 2-methylbutane should be the weakest of the $C-H$ bonds.
Bromine atoms are far more selective than chlorine atoms. This is not unexpected because is endothermic, whereas corresponding reactions with a chlorine atoms usually are exothermic (data from Table 4-6). Bromine removes only those hydrogens that are relatively weakly bonded to a carbon atom. As predicted, attack of $Br \cdot$ on 2-methylbutane leads mostly to 2-bromo-2-methylbutane, some secondary bromide, and essentially no primary bromides:
When the structure of the alkane is symmetrical, then the fast reactivity of chlorination can be used for efficiency. When the structure of the alkane can produce a range of monohalogenated products, then the selectivity of bromination can be used to produce the most stable product in the greatest percentage.
Exercises
1. Specify the optimum halogenation conditions (Cl2/heat or Br2/heat) to produce the indicated major product.
Solution
1.
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)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.11%3A_Reactivity_and.txt |
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Key Terms
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Study Notes
This section is a brief (but perhaps interesting) overview of some of the key differences between reactions performed in the lab and those in living systems. At this point, do not concern yourself with memorizing large biological molecules and reactions.
The active site
A critical element in the three-dimensional structure of any enzyme is the presence of an ‘active site’, which is a pocket, usually located in the interior of the protein, that serves as a docking point for the enzyme’s substrate(s) (‘substrate’ is the term that biochemists use for a reactant molecule in an enzyme-catalyzed reaction). It is inside the active site pocket that enzymatic catalysis occurs. Shown below is an image of the glycolytic enzyme fructose-1,6-bisphosphate aldolase, with its substrate bound inside the active site pocket.
When the substrate binds to the active site, a large number of noncovalent interactions form with the amino acid residues that line the active site. The shape of the active site, and the enzyme-substrate interactions that form as a result of substrate binding, are specific to the substrate-enzyme pair: the active site has evolved to 'fit' one particular substrate and to catalyze one particular reaction. Other molecules do not fit in this active site nearly so well as fructose 1,6-bisphosphate.
Here are two close-up views of the same active site pocket, showing some of the specific hydrogen-bonding interactions between the substrate and active site amino acids. The first image below is a three-dimensional rendering directly from the crystal structure data. The substrate is shown in 'space-filling' style, while the active site amino acids are shown in the 'ball and stick' style. Hydrogens are not shown. The color scheme is grey for carbon, red for oxygen, blue for nitrogen, and orange for phosphorus.
Below is a two-dimensional picture of the substrate (colored red) surrounded by hydrogen-bonding active site amino acids. Notice that both main chain and side chain groups contribute to hydrogen bonding: in this figure, main chain H-bonding groups are colored blue, and side chain H-bonding groups are colored green.
Looking at the last three images should give you some appreciation for the specific manner in which a substrate fits inside its active site.
Transition state stabilization
One of the most important ways that an enzyme catalyzes any given reaction is through entropy reduction: by bringing order to a disordered situation (remember that entropy is a component of Gibbs Free Energy, and thus a component of the activation energy). Let’s turn again to our previous example (from section 6.1C) of a biochemical nucleophilic substitution reaction, the methylation of adenosine in DNA. The reaction is shown below with non-reactive sections of the molecules depicted by variously shaped 'bubbles' for the sake of simplicity.
In order for this reaction to occur, the two substrates (reactants) must come into contact in precisely the right way. If they are both floating around free in solution, the likelihood of this occurring is very small – the entropy of the system is simply too high. In other words, this reaction takes place very slowly without the help of a catalyst.
Here’s where the enzyme’s active site pocket comes into play. It is lined with various functional groups from the amino acid main and side chains, and has a very specific three-dimensional architecture that has evolved to bind to both of the substrates. If the SAM molecule, for example, diffuses into the active site, it can replace its (favorable) interactions with the surrounding water molecules with (even more favorable) new interactions with the functional groups lining the active site.
In a sense, SAM is moving from one solvent (water) to another 'solvent' (the active site), where many new energetically favorable interactions are possible. Remember: these new contacts between SAM and the active site groups are highly specific to SAM and SAM alone – no other molecule can ‘fit’ so well in this precise active site environment, and thus no other molecule will be likely to give up its contacts to water and bind to the active site.
The second substrate also has a specific spot reserved in the active site. (Because in this case the second substrate is a small segment of a long DNA molecule, the DNA-binding region of the active site is more of a 'groove' than a 'pocket').
So now we have both substrates bound in the active site. But they are not just bound in any random orientation – they are specifically positioned relative to one another so that the nucleophilic nitrogen is held very close to the electrophilic carbon, with a free path of attack. What used to be a very disordered situation – two reactants diffusing freely in solution – is now a very highly ordered situation, with everything set up for the reaction to proceed. This is what is meant by entropy reduction: the entropic component of the energy barrier has been lowered.
Looking a bit deeper, though, it is not really the noncovalent interaction between enzyme and substrate that are responsible for catalysis. Remember: all catalysts, enzymes included, accelerate reactions by lowering the energy of the transition state. With this in mind, it should make sense that the primary job of an enzyme is to maximize favorable interactions with the transition state, not with the starting substrates. This does not imply that enzyme-substrate interactions are not strong, rather that enzyme-TS interactions are far stronger, often by several orders of magnitude. Think about it this way: if an enzyme were to bind to (and stabilize) its substrate(s) more tightly than it bound to (and stabilized) the transition state, it would actually slow down the reaction, because it would be increasing the energy difference between starting state and transition state. The enzyme has evolved to maximize favorable noncovalent interactions to the transition state: in our example, this is the state in which the nucleophilic nitrogen is already beginning to attack the electrophilic carbon, and the carbon-sulfur bond has already begun to break.
In many enzymatic reactions, certain active site amino acid residues contribute to catalysis by increasing the reactivity of the substrates. Often, the catalytic role is that of acid and/or base. In our DNA methylation example, the nucleophilic nitrogen is deprotonated by a nearby aspartate side chain as it begins its nucleophilic attack on the methyl group of SAM. We will study nucleophilicity in greater detail in chapter 8, but it should make intuitive sense that deprotonating the amine increases the electron density of the nitrogen, making it more nucleophilic. Notice also in the figure below that the main chain carbonyl of an active site proline forms a hydrogen bond with the amine, which also has the effect of increasing the nitrogen's electron density and thus its nucleophilicity (Nucleic Acids Res. 2000, 28, 3950).
How does our picture of enzyme catalysis apply to multi-step reaction mechanisms? Although the two-step nucleophilic substitution reaction between tert-butyl chloride and hydroxide (section 6.1C) is not a biologically relevant process, let’s pretend just for the sake of illustration that there is a hypothetical enzyme that catalyzes this reaction.
The same basic principles apply here: the enzyme binds best to the transition state. But therein lies the problem: there are two transition states! To which TS does the enzyme maximize its contacts?
Recall that the first step – the loss of the chloride leaving group to form the carbocation intermediate – is the slower, rate-limiting step. It is this step that our hypothetical enzyme needs to accelerate if it wants to accelerate the overall reaction, and it is thus the energy of TS1 that needs to be lowered.
By Hammond’s postulate, we also know that the intermediate I is a close approximation of TS1. So the enzyme, by stabilizing the intermediate, will also stabilize TS1 (as well as TS2) and thereby accelerate the reaction.
If you read scientific papers about enzyme mechanisms, you will often see researchers discussing how an enzyme stabilizes a reaction intermediate. By virtue of Hammond's postulate, they are, at the same time, talking about how the enzyme lowers the energy of the transition state.
An additional note: although we have in this section been referring to SAM as a 'substrate' of the DNA methylation reaction, it is also often referred to as a coenzyme, or cofactor. These terms are used to describe small (relative to protein and DNA) biological organic molecules that bind specifically in the active site of an enzyme and help the enzyme to do its job. In the case of SAM, the job is methyl group donation. In addition to SAM, we will see many other examples of coenzymes in the coming chapters, a number of which - like ATP (adenosine triphosphate), coenzyme A, thiamine, and flavin - you have probably heard of before. The full structures of some common coenzymes are shown in table 6 in the tables section.
5.13: Additional Ex
Kinetics and the Rate Equation
5-1 For a chemical reaction, what is the rate equation used to correlate?
5-2 Write the rate equation that describes the rate of the following reaction.
aA + bB → cC + dD
5-3 What is the overall order of the following reaction with multiple reactants?
Rate = k [A]1[B]1/2
Halogenation of Alkanes
5-4 For the following compounds, give all possible monochlorinated derivatives.
5-5 For the following compounds, identify the major product of free-radical bromination.
5-6 Explain why radical bromination is significantly more selective than radical chlorination.
5.14: Solutions to
Kinetics and the Rate Equation
5-1 The rate equation is an experimentally derived equation that explains the relationship between the concentration of reactants and the rate of the reaction.
5-2 Rate = k [A]m[B]n
5-3 Overall order = 1.5
Halogenation of Alkanes
5-4
5-5
5-6 Radical bromination is more selective because of its slightly higher activation energy required to break a C-H bond during the propagation steps (when the bromine radical abstracts a proton from the substrate). Though the difference in activation energy is not huge (Cl = ~1 kcal/mol and Br = ~3 kcal/mol), it leads to a significant difference in selectivity. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes/5.12%3A_A_Comparison_b.txt |
Learning Objectives
After reading this chapter and completing the exercises and homework, a student can be able to:
• recognize and classify molecules as chiral or achiral and identify planes of symmetry - refer to section 6.1
• draw, interpret, and convert between perspective formulae and Fischer projections for chiral compounds - refer to section 6.2
• name chiral compounds using (R) & (S) nomenclature - refer to section 6.3
• recognize and classify diastereomers and meso compounds - refer to section 6.4 and 6.5 respectively
• explain how physical properties differ for different types of stereoisomers - refer to section ?????
• distinguish and discern the structural and chemical relationships between isomeric compounds - refer to section 6.6
• define and explain the lack of optical activity of racemic mixtures - refer to section 6.7
• determine the percent composition of an enantiomeric mixture from polarimetry data and the for specific rotation formula - refer to section 6.7
• explain how to resolve (separate) a pair of enantiomers - refer to section 6.8
• interpret the stereoisomerism of compounds with three or more chiral centers - refer to section 6.9
• compare and contrast absolute configuration with relative configuration - refer to section 6.10
• interpret the stereoisomerism of compounds with nitrogen, phosphorus, or sulfur as chiral centers - refer to section 6.11
• recognize and explain biochemical applications of chirality - refer to section 6.12
• describe Jean Baptiste Biot and Louis Pasteur's contributions to the understanding of optical isomers - refer to section 6.13
• 6.1: Chirality
Chiral carbons are tetrahedral carbons bonded to four unique groups. At first glance, many carbons may look alike, but upon closer inspection, we can discern their differences.
• 6.2: Fischer Projections to communicate Chirality
Converting between perspective formula structures (wedges and dashes) and Fischer projections can be useful when evaluating stereochemistry, especially for carbohydrate chemistry.
• 6.3: Absolute Configuration and the (R) and (S) System
The absolute configuration of chiral centers as R or S is determined by applying the Cahn-Ingold-Prelog rules.
• 6.4: Diastereomers - more than one chiral center
Diastereomers are stereoisomers with two or more chiral centers that are not enantiomers. Diastereomers have different physical properties (melting points, boiling points, and densities). Depending on the reaction mechanism, diastereomers can produce different stereochemical products.
• 6.5: Meso Compounds
A meso compound is an achiral compound that has two or more chiral centers. Molecular symmetry allows the mirror images to super-impose so that they are not enantiomers.
• 6.6: Isomerism Summary Diagram
A simple diagram is helpful in distinguishing between the different types of isomers that are possible.
• 6.7: Optical Activity and Racemic Mixtures
Optical activity is one of the few ways to distinguish between enantiomers. A racemic mixture is a 50:50 mixture of two enantiomers. Racemic mixtures were an interesting experimental discovery because two optically active samples were combined to create an optically INACTIVE sample.
• 6.8: Resolution (Separation) of Enantiomers
The most commonly used procedure for separating enantiomers is to convert them to a mixture of diastereomeric salts that can be separated based on their differences in their physical properties. After separation, the isolated D or the L enantiomer can be recovered.
• 6.9: Stereochemistry of Molecules with Three or More Asymmetric Carbons
As the number of chiral carbons increases, the number of stereoisomers also increases. This sections shows a short cut for compounds with three or more stereocenters.
• 6.10: Absolute and Relative Configuration - the distinction
The absolute configuration at a chiral center in a molecule is a time-independent and unambiguous symbolic description of the spatial arrangement of ligands (atoms) around it. The relative configuration is the experimentally determined relationship between two enantiomers even though we may not know the absolute configuration.
• 6.11: Chirality at Nitrogen, Phosphorus, and Sulfur
Chirality can also occur with central atoms other than carbon.
• 6.12: Biochemistry of Enantiomers
Biological activity and chirality are strongly correlated. The section explores a few examples.
• 6.13: The Discovery of Enantiomers
The initial work carried out by Biot and Pasteur contributed to the concepts of chirality.
• 6.14: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 6.15: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
06: Stereochemistry at Tetrahedral Centers
Learning Objective
• recognize and classify molecules as chiral or achiral and identify planes of symmetry
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existence of these molecules are determined by concept known as chirality.
Introduction
Organic compounds, molecules created around a chain of carbon atom (more commonly known as carbon backbone), play an essential role in the chemistry of life. These molecules derive their importance from the energy they carry, mainly in a form of potential energy between atomic molecules. Since such potential force can be widely affected due to changes in atomic placement, it is important to understand the concept of an isomer, a molecule sharing same atomic make up as another but differing in structural arrangements. This article will be devoted to a specific isomers called stereoisomers and its property of chirality (Figure 1).
The concepts of steroisomerism and chirality command great deal of importance in modern organic chemistry, as these ideas helps to understand the physical and theoretical reasons behind the formation and structures of numerous organic molecules, the main reason behind the energy embedded in these essential chemicals. In contrast to more well-known constitutional isomerism, which develops isotopic compounds simply by different atomic connectivity, stereoisomerism generally maintains equal atomic connections and orders of building blocks as well as having same numbers of atoms and types of elements.
What, then, makes stereoisomers so unique? To answer this question, the learner must be able to think and imagine in not just two-dimensional images, but also three-dimensional space. This is due to the fact that stereoisomers are isomers because their atoms are different from others in terms of spatial arrangement.
Spatial Arrangement
First and foremost, one must understand the concept of spatial arrangement in order to understand stereoisomerism and chirality. Spatial arrangement of atoms concern how different atomic particles and molecules are situated about in the space around the organic compound, namely its carbon chain. In this sense, spatial arrangement of an organic molecule are different another if an atom is shifted in any three-dimensional direction by even one degree. This opens up a very broad possibility of different molecules, each with their unique placement of atoms in three-dimensional space .
Stereoisomers
Stereoisomers are, as mentioned above, contain different types of isomers within itself, each with distinct characteristics that further separate each other as different chemical entities having different properties. Type called entaniomer are the previously-mentioned mirror-image stereoisomers, and will be explained in detail in this article. Another type, diastereomer, has different properties and will be introduced afterwards.
The Many Synonyms of the Chiral Carbon
Be aware - all of the following terms can be used to describe a chiral carbon.
chiral carbon = asymmetric carbon = optically active carbon = stereo carbon
Enantiomers
This type of stereoisomer is the essential mirror-image, non-superimposable type of stereoisomer introduced in the beginning of the article. Figure 3 provides a perfect example; note that the gray plane in the middle demotes the mirror plane.
Note that even if one were to flip over the left molecule over to the right, the atomic spatial arrangement will not be equal. This is equivalent to the left hand - right hand relationship, and is aptly referred to as 'handedness' in molecules. This can be somewhat counter-intuitive, so this article recommends the reader try the 'hand' example. Place both palm facing up, and hands next to each other. Now flip either side over to the other. One hand should be showing the back of the hand, while the other one is showing the palm. They are not same and non-superimposable.
This is where the concept of chirality comes in as one of the most essential and defining idea of stereoisomerism.
Chirality
Chirality essentially means 'mirror-image, non-superimposable molecules', and to say that a molecule is chiral is to say that its mirror image (it must have one) is not the same as it self. Whether a molecule is chiral or achiral depends upon a certain set of overlapping conditions. Figure 4 shows an example of two molecules, chiral and achiral, respectively. Notice the distinct characteristic of the achiral molecule: it possesses two atoms of same element. In theory and reality, if one were to create a plane that runs through the other two atoms, they will be able to create what is known as bisecting plane: The images on either side of the plan is the same as the other (Figure 4).
In this case, the molecule is considered 'achiral'. In other words, to distinguish chiral molecule from an achiral molecule, one must search for the existence of the bisecting plane in a molecule. All chiral molecules are deprive of bisecting plane, whether simple or complex.
As a universal rule, no molecule with different surrounding atoms are achiral. Chirality is a simple but essential idea to support the concept of stereoisomerism, being used to explain one type of its kind. The chemical properties of the chiral molecule differs from its mirror image, and in this lies the significance of chilarity in relation to modern organic chemistry.
Exercise 1
Identify the following as either a constitutional isomer or stereoisomer. If stereoisomer, determine if it is an enantiomer or diastereomer. Explain the reason behind the answer. Also mark chirality for each molecule.
a) b) c)
Solutions
a) achiral
b) chiral
c) chiral
Exercise 2
Identify the chiral centers in each of the following:
Solutions
Contributors and Attributions
• Dan Chong
• Jonathan Mooney (McGill University)
6.02: Fischer Projections to communicate Chirality
Learning Objective
• draw, interpret, and convert between perspective formulae and Fischer projections for chiral compounds
Perspective Formulas and Fischer Projections
So far, we have communicated the stereochemical orentation of compounds using the wedges and dashes of perspective formulas. For example, the perspective formula for (R)-Lactic acid is shown below.
A Fischer projection is a convention used to depict stereochemistry in two dimensions. The horizontal bonds are seen as wedges and the vertical bonds are seen as dashed lines as shown below in the example below for glyceraldehyde.
Converting between Perspective and Fischer Formulas
It can be useful to convert between perspective formulas and Fischer projections. Below is one approach using (R)-lactic acid as an example..
Step 1: Hold the molecule so that
a) the chiral center is on the plane of the paper,
b) two bonds are coming out of the plane of the paper and are on a horizontal plane,
c) the two remaining bonds are going into the plane of the paper and are on a vertical plane
Step 2: Push the two bonds coming out of the plane of the paper onto the plane of the paper.
Step 3: Pull the two bonds going into the plane of the paper onto the plane of the paper.
Step 4: Omit the chiral atom symbol for convenience. This is the Fischer Projection of (R)-Lactic acid.
The stereochemical formula for (R)-lactic acid can be drawn using either method. To build this skill, we begin by drawing the structures and converting them step wise. Models can also be helpful. Eventually, we will be able to mentally conversion between these two structures.
Exercise 1
1. Convert each compound into the alternate sterochemical structure (Perspective Fischer).
Answer
See also D,L-convention. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.01%3A_Chirality.txt |
Learning Objective
• name chiral compounds using (R) & (S) nomenclature
USE YOUR MODELING KIT: Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S).
IF YOU DO NOT HAVE A MODELING KIT: remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models.
If you have a modeling kit use it as you read through this section and work the practice problems.
Introduction and the Cahn-Ingold-Prelog rules of Priority
To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The letters "R" and "S" are determined by applying the Cahn-Ingold-Prelog (CIP) rules. The optical activity (+/-) can also be communicated in the name, but must be empirically derived. There are also biochemical conventions for carbohydrates (sugars) and amino acids (the building blocks of proteins).
The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and, as such, is also often called the Cahn-Ingold-Prelog rules. In addition to the CIP system, there are two ways of experimentally determining the absolute configuration of an enantiomer:
1. X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer.
2. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction.
However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of optical rotation, although different for the two enantiomers of a chiral molecule,at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes.
The Cahn-Ingold-Prelog rules of priority are based on the atomic numbers of the atoms of interest. For chirality, the atoms of interest are the atoms bonded to the chiral carbon.
1. The atom with higher atomic number has higher priority (I > Br > Cl > S > P > F > O > N > C > H).
2. When comparing isotopes, the atom with the higher mass number has higher priority [18O > 16O or 15N > 14N or 13C > 12C or T (3H) > D (2H) > H].
3. When there is a tie in (2) above, establish relative priority by proceeding to the next atom(s) along the chain until the first difference is observed.
Multiple bonds are treated as if each bond of the multiple bond is bonded to a unique atom. For example, the ethenyl group (CH2=CH) has higher priority than the ethyl group (CH3CH2). The ethenyl carbon priority is "two" bonds to carbon atoms and one bond to a hydrogen atom compared with the ethyl carbon that has only one bond to a carbon atom and two bonds to two hydrogen atoms. Similarly, the carbon-carbon triple bond of acetylene would give it higher CIP priority than the ethenyl group as summarized below.
Stereocenters are labeled R or S
The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S.
Consider the diagram above on the left: a curved arrow is drawn counter-clockwise (c-cw) from the highest priority substituent (1) to the lowest priority substituent (4) in the S-configuration ("Sinister" → Latin= "left"). The counterclockwise direction can be recognized by the movement left when leaving the 12 o' clock position. Now consider the diagram above on the right where a curved arrow is drawn clockwise (cw) from the highest priority substituent (1) to the lowest priority substituent (4) in the R configuration ("Rectus" → Latin= "right"). The R or S is then added as a prefix, in parenthesis, to the name of the enantiomer of interest. A locator number is required if there is more than one chiral center. Otherwise, the person reading the name is expected to recognize the chiral center.
Example 1
The two chiral compounds below are drawn to emphasize the chiral carbon with the full chemical name below each structure.
Absolute Configurations of Perspective Formulas
Chemists need a convenient way to distinguish one stereoisomer from another. The Cahn-Ingold-Prelog system is a set of rules that allows us to unambiguously define the stereochemical configuration of any stereocenter, using the designations 'R ’ (from the Latin rectus, meaning right-handed) or ' S ’ (from the Latin sinister, meaning left-handed).
The rules for this system of stereochemical nomenclature are, on the surface, fairly simple.
Rules for assigning an R/S designation to a chiral center
1: Assign priorities to the four substituents, with #1 being the highest priority and #4 the lowest. Priorities are based on the atomic number.
2: Trace a circle from #1 to #2 to #3.
3: Determine the orientation of the #4 priority group. If it is oriented into the plane of the page (away from you), go to step 4a. If it is oriented out of the plane of the page (toward you) go to step 4b.
4a: (#4 group pointing away from you): a clockwise circle in part 2 corresponds to the R configuration, while a counterclockwise circle corresponds to the S configuration.
4b: (#4 group pointing toward you): a clockwise circle in part 2 corresponds to the S configuration, while a counterclockwise circle corresponds to the R configuration.
We’ll use the 3-carbon sugar glyceraldehyde as our first example. The first thing that we must do is to assign a priority to each of the four substituents bound to the chiral center. We first look at the atoms that are directly bonded to the chiral center: these are H, O (in the hydroxyl), C (in the aldehyde), and C (in the CH2OH group).
Assigning R/S configuration to glyceraldehyde:
Two priorities are easy: hydrogen, with an atomic number of 1, is the lowest (#4) priority, and the hydroxyl oxygen, with atomic number 8, is priority #1. Carbon has an atomic number of 6. Which of the two ‘C’ groups is priority #2, the aldehyde or the CH2OH? To determine this, we move one more bond away from the chiral center: for the aldehyde we have a double bond to an oxygen, while on the CH2OH group we have a single bond to an oxygen. If the atom is the same, double bonds have a higher priority than single bonds. Therefore, the aldehyde group is assigned #2 priority and the CH2OH group the #3 priority.
With our priorities assigned, we look next at the #4 priority group (the hydrogen) and see that it is pointed back away from us, into the plane of the page - thus step 4a from the procedure above applies. Then, we trace a circle defined by the #1, #2, and #3 priority groups, in increasing order. The circle is clockwise, which by step 4a tells us that this carbon has the ‘R’ configuration, and that this molecule is (R)-glyceraldehyde. Its enantiomer, by definition, must be (S)-glyceraldehyde.
Next, let's look at one of the enantiomers of lactic acid and determine the configuration of the chiral center. Clearly, H is the #4 substituent and OH is #1. Owing to its three bonds to oxygen, the carbon on the acid group takes priority #2, and the methyl group takes #3. The #4 group, hydrogen, happens to be drawn pointing toward us (out of the plane of the page) in this figure, so we use step 4b: The circle traced from #1 to #2 to #3 is clockwise, which means that the chiral center has the S configuration.
The drug thalidomide is an interesting - but tragic - case study in the importance of stereochemistry in drug design. First manufactured by a German drug company and prescribed widely in Europe and Australia in the late 1950's as a sedative and remedy for morning sickness in pregnant women, thalidomide was soon implicated as the cause of devastating birth defects in babies born to women who had taken it. Thalidomide contains a chiral center, and thus exists in two enantiomeric forms. It was marketed as a racemic mixture: in other words, a 50:50 mixture of both enantiomers.
Let’s try to determine the stereochemical configuration of the enantiomer on the left. Of the four bonds to the chiral center, the #4 priority is hydrogen. The nitrogen group is #1, the carbonyl side of the ring is #2, and the –CH2 side of the ring is #3.
The hydrogen is shown pointing away from us, and the prioritized substituents trace a clockwise circle: this is the R enantiomer of thalidomide. The other enantiomer, of course, must have the S configuration.
Although scientists are still unsure today how thalidomide works, experimental evidence suggests that it was actually the R enantiomer that had the desired medical effects, while the S enantiomer caused the birth defects. Even with this knowledge, however, pure (R)-thalidomide is not safe, because enzymes in the body rapidly convert between the two enantiomers - we will see how that happens in chapter 12.
As a historical note, thalidomide was never approved for use in the United States. This was thanks in large part to the efforts of Dr. Frances Kelsey, a Food and Drug officer who, at peril to her career, blocked its approval due to her concerns about the lack of adequate safety studies, particularly with regard to the drug's ability to enter the bloodstream of a developing fetus. Unfortunately, though, at that time clinical trials for new drugs involved widespread and unregulated distribution to doctors and their patients across the country, so families in the U.S. were not spared from the damage caused.
Very recently a close derivative of thalidomide has become legal to prescribe again in the United States, with strict safety measures enforced, for the treatment of a form of blood cancer called multiple myeloma. In Brazil, thalidomide is used in the treatment of leprosy - but despite safety measures, children are still being born with thalidomide-related defects.
Exercise 1.: Determine the stereochemical configurations of the chiral centers in the biomolecules shown below.
Exercise 2.: Should the (R) enantiomer of malate have a solid or dashed wedge for the C-O bond in the figure below?
Exercise 3.: Using solid or dashed wedges to show stereochemistry, draw the (R) enantiomer of ibuprofen and the (S) enantiomer of 2-methylerythritol-4-phosphate (structures are shown earlier in this chapter without stereochemistry).
Solutions to exercises
Absolute Configurations of Fischer Projections
To determine the absolute configuration of a chiral center in a Fisher projection, use the following two-step procedure.
Step 1
Assign priority numbers to the four ligands (groups) bonded to the chiral center using the CIP priority system.
Step 2 - vertical option
If the lowest priority ligand is on a Vertical bond, then it is pointing away from the viewer.
Trace the three highest-priority ligands starting at the highest-priority ligand (① → ② → ) in the direction that will give a Very correct answer.
In the compound below, the movement is clockwise indicating an R-configuration. The complete IUPAC name for this compound is (R)-butan-2-ol.
Step 2 - horizontal option
If the lowest-priority ligand is on a Horizontal bond, then it is pointing toward the viewer.
Trace the three highest-priority ligands starting at the highest-priority ligand (① → ② → ) in the direction that will give a Horribly wrong answer. Note in the table below that the configurations are reversed from the first example.
In the compound below, the movement is clockwise (R) which is Horribly wrong, so the actual configuration is S. The complete IUPAC name for this compound is (S)-butan-2-ol.
Manipulating Fischer Projections with NO Change to Configuration
A Fischer projection restricts a three-dimensional molecule into two dimensions. Consequently, there are limitations as to the operations that can be performed on a Fischer projection without changing the absolute configuration at chiral centers. The operations that do not change the absolute configuration at a chiral center in a Fischer projections can be summarized as two rules.
Rule 1: Rotation of the Fischer projection by 180º in either direction without lifting it off the plane of the paper does not change the absolute configuration at the chiral center.
Rule 2: Rotation of three ligands on the chiral center in either direction, keeping the remaining ligand in place, does not change the absolute configuration at the chiral center.
Manipulating Fischer Projections with Change to Configuration
The operations that do change the absolute configuration at a chiral center in a Fischer projection can be summarized as two rules.
Rule 1: Rotation of the Fischer projection by 90º in either direction changes the absolute configuration at the chiral center.
Rule 2: Interchanging any two ligands on the chiral center changes the absolute configuration at the chiral center.
The above rules assume that the Fischer projection under consideration contains only one chiral center. However, with care, they can be applied to Fischer projections containing any number of chiral centers.
Exercise 1
Classify the following compounds as R or S?
Solution
1. S: I > Br > F > H. The lowest priority substituent, H, is already going towards the back. It turns left going from I to Br to F, so it's a S.
2. R: Br > Cl > CH3 > H. You have to switch the H and Br in order to place the H, the lowest priority, in the back. Then, going from Br to Cl, CH3 is turning to the right, giving you a R.
3. Neither R or S: This molecule is achiral. Only chiral molecules can be named R or S.
4. R: OH > CN > CH2NH2 > H. The H, the lowest priority, has to be switched to the back. Then, going from OH to CN to CH2NH2, you are turning right, giving you a R.
5. (5) S: \(\ce{-COOH}\) > \(\ce{-CH_2OH}\) > \(\ce{C#CH}\) > \(\ce{H}\). Then, going from \(\ce{-COOH}\) to \(\ce{-CH_2OH}\) to \(\ce{-C#CH}\) you are turning left, giving you a S configuration.
Exercises
6. Orient the following so that the least priority (4) atom is paced behind, then assign stereochemistry (R or S).
7. Draw (R)-2-bromobutan-2-ol.
8. Assign R/S to the following molecule.
Solutions
6.
A = S; B = R
7.
8. The stereo center is R. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.03%3A_Absolute_Configuration_and_the_%28R%29_and_%28S%29_Syst.txt |
Learning Objective
• recognize and classify diastereomers
Diastereomers are stereoisomers with two or more chiral centers that are not enantiomers. Diastereomers have different physical properties (melting points, boiling points, and densities). Depending on the reaction mechanism, diastereomers can produce different stereochemical products.
Introduction
So far, we have been analyzing compounds with a single chiral center. Next, we turn our attention to those which have multiple chiral centers. We'll start with some stereoisomeric four-carbon sugars with two chiral centers.
To avoid confusion, we will simply refer to the different stereoisomers by capital letters.
Look first at compound A below. Both chiral centers in have the R configuration (you should confirm this for yourself!). The mirror image of Compound A is compound B, which has the S configuration at both chiral centers. If we were to pick up compound A, flip it over and put it next to compound B, we would see that they are not superimposable (again, confirm this for yourself with your models!). A and B are nonsuperimposable mirror images: in other words, enantiomers.
Now, look at compound C, in which the configuration is S at chiral center 1 and R at chiral center 2. Compounds A and C are stereoisomers: they have the same molecular formula and the same bond connectivity, but a different arrangement of atoms in space (recall that this is the definition of the term 'stereoisomer). However, they are not mirror images of each other (confirm this with your models!), and so they are not enantiomers. By definition, they are diastereomers of each other.
Notice that compounds C and B also have a diastereomeric relationship, by the same definition.
So, compounds A and B are a pair of enantiomers, and compound C is a diastereomer of both of them. Does compound C have its own enantiomer? Compound D is the mirror image of compound C, and the two are not superimposable. Therefore, C and D are a pair of enantiomers. Compound D is also a diastereomer of compounds A and B.
This can also seem very confusing at first, but there some simple shortcuts to analyzing stereoisomers:
Stereoisomer shortcuts
If all of the chiral centers are of opposite R/S configuration between two stereoisomers, they are enantiomers.
If at least one, but not all of the chiral centers are opposite between two stereoisomers, they are diastereomers.
(Note: these shortcuts to not take into account the possibility of additional stereoisomers due to alkene groups: we will come to that later)
Here's another way of looking at the four stereoisomers, where one chiral center is associated with red and the other blue. Pairs of enantiomers are stacked together.
We know, using the shortcut above, that the enantiomer of RR must be SS - both chiral centers are different. We also know that RS and SR are diastereomers of RR, because in each case one - but not both - chiral centers are different.
Diastereomers vs. Enantiomers in Wine Chemistry
Tartaric acid, C4H6O6, is an organic compound that can be found in grape, bananas, and in wine. The structures of tartaric acid itself is really interesting. Naturally, it is in the form of (R,R) stereocenters. Artificially, it can be in the meso form (R,S), which is achiral. R,R tartaric acid is enantiomer to is mirror image which is S,S tartaric acid and diasteromers to meso-tartaric acid (Figure 5.6.2).
(R,R) and (S,S) tartaric acid have similar physical properties and reactivity. However, meso-tartaric acid have different physical properties and reactivity. For example, melting point of (R,R) & (S,S) tartaric is about 170 degree Celsius, and melting point of meso-tartaric acid is about 145 degree Celsius.
Diastereomers vs. Enantiomers in Sugar Chemistry
D-erythrose is a common four-carbon sugar.
A note on sugar nomenclature: biochemists use a special system to refer to the stereochemistry of sugar molecules, employing names of historical origin in addition to the designators 'D' and 'L'. You will learn about this system if you take a biochemistry class. We will use the D/L designations here to refer to different sugars, but we won't worry about learning the system.
As you can see, D-erythrose is a chiral molecule: C2 and C3 are stereocenters, both of which have the R configuration. In addition, you should make a model to convince yourself that it is impossible to find a plane of symmetry through the molecule, regardless of the conformation. Does D-erythrose have an enantiomer? Of course it does – if it is a chiral molecule, it must. The enantiomer of erythrose is its mirror image, and is named L-erythrose (once again, you should use models to convince yourself that these mirror images of erythrose are not superimposable).
Notice that both chiral centers in L-erythrose both have the S configuration.
Note
In a pair of enantiomers, all of the chiral centers are of the opposite configuration.
What happens if we draw a stereoisomer of erythrose in which the configuration is S at C2 and R at C3? This stereoisomer, which is a sugar called D-threose, is not a mirror image of erythrose. D-threose is a diastereomer of both D-erythrose and L-erythrose.
The definition of diastereomers is simple: if two molecules are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers, then they are diastereomers by default. In practical terms, this means that at least one - but not all - of the chiral centers are opposite in a pair of diastereomers. By definition, two molecules that are diastereomers are not mirror images of each other.
L-threose, the enantiomer of D-threose, has the R configuration at C2 and the S configuration at C3. L-threose is a diastereomer of both erythrose enantiomers.
Erythronolide B, a precursor to the 'macrocyclic' antibiotic erythromycin, has 10 stereocenters. It’s enantiomer is that molecule in which all 10 stereocenters are inverted.
In total, there are 210 = 1024 stereoisomers in the erythronolide B family: 1022 of these are diastereomers of the structure above, one is the enantiomer of the structure above, and the last is the structure above.
We know that enantiomers have identical physical properties and equal but opposite degrees of specific rotation. Diastereomers, in theory at least, have different physical properties – we stipulate ‘in theory’ because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to separate them. In addition, the specific rotations of diastereomers are unrelated – they could be the same sign or opposite signs, and similar in magnitude or very dissimilar.
Exercises
1. Draw the structures of L-galactose (the enantiomer of D-galactose) and two more diastereomers of D-glucose (one should be an epimer).
2. Determine the stereochemistry of the following molecule:
Answer
1.
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.04%3A_Diastereomers_-_more_than_one_chiral_center.txt |
Learning Objective
• recognize and classify meso compounds
A meso compound is an achiral compound that has chiral centers. It is superimposed on its mirror image and is optically inactive although it contains two or more stereocenters.
Introduction
In general, a meso compound should contain two or more identical substituted stereocenters. Also, it has an internal symmetry plane that divides the compound in half. These two halves reflect each other by the internal mirror. The stereochemistry of stereocenters should "cancel out". What it means here is that when we have an internal plane that splits the compound into two symmetrical sides, the stereochemistry of both left and right side should be opposite to each other, and therefore, result in optically inactive. Cyclic compounds may also be meso.
Identification
If A is a meso compound, it should have two or more stereocenters, an internal plane, and the stereochemistry should be R and S.
1. Look for an internal plane, or internal mirror, that lies in between the compound.
2. The stereochemistry (e.g. R or S) is very crucial in determining whether it is a meso compound or not. As mentioned above, a meso compound is optically inactive, so their stereochemistry should cancel out. For instance, R cancels S out in a meso compound with two stereocenters.
trans-1,2-dichloro-1,2-ethanediol
(meso)-2,3-dibromobutane
Tips: An interesting thing about single bonds or sp3-orbitals is that we can rotate the substituted groups that attached to a stereocenter around to recognize the internal plane. As the molecule is rotated, its stereochemistry does not change. For example:
Another case is when we rotate the whole molecule by 180 degree. Both molecules below are still meso.
Remember the internal plane here is depicted on two dimensions. However, in reality, it is three dimensions, so be aware of it when we identify the internal mirror.
Example \(1\):
1 has a plane of symmetry (the horizontal plane going through the red broken line) and, therefore, is achiral; 1 has chiral centers. Thus, 1 is a meso compound.
Example \(2\):
This molecules has a plane of symmetry (the vertical plane going through the red broken line perpendicular to the plane of the ring) and, therefore, is achiral, but has has two chiral centers. Thus, its is a meso compound.
Exercise \(1\)
Which of the following are meso-compounds:
1. C – 2,3-dibromobutane
2. D – 2,3-dibromopentane
Answer
Compounds A and C are meso.
Other Examples of meso compounds
Meso compounds can exist in many different forms such as pentane, butane, heptane, and even cyclobutane. They do not necessarily have to be two stereocenters, but can have more.
The chiral centers in the preceding examples have all been different. In the case of 2,3-dihydroxybutanedioic acid, known as tartaric acid, the two chiral centers have the same four substituents and are equivalent. As a result, two of the four possible stereoisomers of this compound are identical due to a plane of symmetry, so there are only three stereoisomeric tartaric acids. Two of these stereoisomers are enantiomers and the third is an achiral diastereomer, called a meso compound. Meso compounds are achiral (optically inactive) diastereomers of chiral stereoisomers. Investigations of isomeric tartaric acid salts, carried out by Louis Pasteur in the mid 19th century, were instrumental in elucidating some of the subtleties of stereochemistry. Some physical properties of the isomers of tartaric acid are given in the following table.
(+)-tartaric acid: [α]D = +13º m.p. 172 ºC
(–)-tartaric acid: [α]D = –13º m.p. 172 ºC
meso-tartaric acid: [α]D = 0º m.p. 140 ºC
Fischer projection formulas provide a helpful view of the configurational relationships within the structures of these isomers. In the following illustration a mirror line is drawn between formulas that have a mirror-image relationship. In demonstrating the identity of the two meso-compound formulas, remember that a Fischer projection formula may be rotated 180º in the plane.
Optical Activity Analysis
When the optical activity of a meso compound is attempted to be determined with a polarimeter, the indicator will not show (+) or (-). It simply means there is no certain direction of rotation of the polarized light, neither levorotatory (-) and dexorotatory (+).
Achiral Diastereomers (meso-Compounds)
Exercise \(1\)
Beside meso, there are also other types of molecules: enantiomer, diastereomer, and identical. Determine if the following molecules are meso.
Answer
A C, D, E are meso compounds.
6.06: Isomerism Summary Diagram
Learning Objective
• distinguish and discern the structural and chemical relationships between isomeric compounds
The various types of isomers have been introduced and explored over several chapters. It can be helpful to review, compare, and contrast all of the forms of isomerism to build our skills of discernment. A brief review of each type of isomerism follows the summary diagram. See the respective chapter for a complete explanation.
Conformational Isomers
The rotation of C–C single bonds both carbon chains creates conformers (the same compound shown in different rotations). Consequently, many different arrangements of the atoms are possible, each corresponding to different degrees of rotation. Differences in three-dimensional structure resulting from rotation about a σ bond are called differences in conformation, and each different arrangement is called a conformational isomer (or conformer). While complete rotation of C-C single bonds is not possible in rings. The freedom of bond movement does allow the rings to assume different conformations, such as the chair and boat for 6-membered rings.
Structural (Constitutional) Isomers
Unlike conformational isomers, structural isomers differ in connectivity, as illustrated below for 1-propanol and 2-propanol. Although these two alcohols have the same molecular formula C3H8O, the position of the –OH group differs creating a unique compounds with differences in their physical and chemical properties.
Consider, for example, the following five structures represented by the formula C5H12. In the conversion of one structural isomer to another, at least one bond must be broken and reformed at a different position in the molecule.
Structures (a) and (d) above represent the same compound, n-pentane. Structures (b) and (c) represent the same compound, 2-methylbutane. No bonds need to be broken and reformed to convert between (a) and (d) or between (b) and (c). The molecules are simply rotated 180° about a vertical axis. Structure (e) is named 2,2-dimethylpropane. There are only three structural isomers possible with the chemical formula C5H12: n-pentane, 2-methylbutane, and 2,2-dimethylpropane. Structural isomers have distinct physical and chemical properties.
Stereoisomers
Enantiomers are pairs of compounds that are non-superimposable images. When there are two or more chiral centers in a compounds, the diatereomers can exist. Diastereomers are stereoisomers that are NOT enantiomers. Enatiomers share all physical properties except for their interaction with plane polarized light. Diastereomers have different physical properties (melting points and boiling points and densities).
Exercise
1. What kind of isomers are the following pairs? Note: It can be difficult to answer this question directly from the names. It can be helpful to draw the structures.
1. (R)-5-chlorohexene and 6-chlorohexene
2. (2R,3R)-dibromohexane and (2R,3S)-dibromohexane
Answer
1.
a. Structural Isomers
b. Diastereomers | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.05%3A_Meso_Compounds.txt |
Learning Objective
• define and explain the lack of optical activity of racemic mixtures
• determine the percent composition of an enantiomeric mixture from polarimetry data and the for specific rotation formula
Racemic Mixtures (Racimates)
A racemic mixture is a 50:50 mixture of two enantiomers. Racemic mixtures were an interesting experimental discovery because two optically active samples can be combined in a 1:1 ratio to create an optically INACTIVE sample. Polarimetry is used to measure optical activity. The history and theoretical foundation are discussed below.
Identifying and distinguishing enantiomers is inherently difficult, since their physical and chemical properties are largely identical. Fortunately, a nearly two hundred year old discovery by the French physicist Jean-Baptiste Biot has made this task much easier. This discovery disclosed that the right- and left-handed enantiomers of a chiral compound perturb plane-polarized light in opposite ways. This perturbation is unique to chiral molecules, and has been termed optical activity.
Polarimetry
Plane-polarized light is created by passing ordinary light through a polarizing device, which may be as simple as a lens taken from polarizing sun-glasses. Such devices transmit selectively only that component of a light beam having electrical and magnetic field vectors oscillating in a single plane. The plane of polarization can be determined by an instrument called a polarimeter, shown in the diagram below.
Monochromatic (single wavelength) light, is polarized by a fixed polarizer next to the light source. A sample cell holder is located in line with the light beam, followed by a movable polarizer (the analyzer) and an eyepiece through which the light intensity can be observed. In modern instruments an electronic light detector takes the place of the human eye. In the absence of a sample, the light intensity at the detector is at a maximum when the second (movable) polarizer is set parallel to the first polarizer (α = 0º). If the analyzer is turned 90º to the plane of initial polarization, all the light will be blocked from reaching the detector.
Chemists use polarimeters to investigate the influence of compounds (in the sample cell) on plane polarized light. Samples composed only of achiral molecules (e.g. water or hexane), have no effect on the polarized light beam. However, if a single enantiomer is examined (all sample molecules being right-handed, or all being left-handed), the plane of polarization is rotated in either a clockwise (positive) or counter-clockwise (negative) direction, and the analyzer must be turned an appropriate matching angle, α, if full light intensity is to reach the detector. In the above illustration, the sample has rotated the polarization plane clockwise by +90º, and the analyzer has been turned this amount to permit maximum light transmission.
The observed rotations ($\alpha$) of enantiomers are opposite in direction. One enantiomer will rotate polarized light in a clockwise direction, termed dextrorotatory or (+), and its mirror-image partner in a counter-clockwise manner, termed levorotatory or (–). The prefixes dextro and levo come from the Latin dexter, meaning right, and laevus, for left, and are abbreviated d and l respectively. If equal quantities of each enantiomer are examined , using the same sample cell, then the magnitude of the rotations will be the same, with one being positive and the other negative. To be absolutely certain whether an observed rotation is positive or negative it is often necessary to make a second measurement using a different amount or concentration of the sample. In the above illustration, for example, α might be –90º or +270º rather than +90º. If the sample concentration is reduced by 10%, then the positive rotation would change to +81º (or +243º) while the negative rotation would change to –81º, and the correct α would be identified unambiguously.
Since it is not always possible to obtain or use samples of exactly the same size, the observed rotation is usually corrected to compensate for variations in sample quantity and cell length. Thus it is common practice to convert the observed rotation, α, to a specific rotation, by the following formula:
$[\alpha]_D = \dfrac{\alpha}{l c} \tag{5.3.1}$
where
• $[\alpha]_D$ is the specific rotation
• $l$ is the cell length in dm
• $c$ is the concentration in g/ml
• $D$ designates that the light used is the 589 line from a sodium lamp
Compounds that rotate the plane of polarized light are termed optically active. Each enantiomer of a stereoisomeric pair is optically active and has an equal but opposite-in-sign specific rotation. Specific rotations are useful in that they are experimentally determined constants that characterize and identify pure enantiomers. For example, the lactic acid and carvone enantiomers discussed earlier have the following specific rotations.
Carvone from caraway: [α]D = +62.5º this isomer may be referred to as (+)-carvone or d-carvone
Carvone from spearmint: [α]D = –62.5º this isomer may be referred to as (–)-carvone or l-carvone
Lactic acid from muscle tissue: [α]D = +2.5º this isomer may be referred to as (+)-lactic acid or d-lactic acid
Lactic acid from sour milk: [α]D = –2.5º this isomer may be referred to as (–)-lactic acid or l-lactic acid
A 50:50 mixture of enantiomers has no observable optical activity. Such mixtures are called racemates or racemic modifications, and are designated (±). When chiral compounds are created from achiral compounds, the products are racemic unless a single enantiomer of a chiral co-reactant or catalyst is involved in the reaction. The addition of HBr to either cis- or trans-2-butene is an example of racemic product formation (the chiral center is colored red in the following equation).
CH3CH=CHCH3 + HBr (±) CH3CH2CHBrCH3
Chiral organic compounds isolated from living organisms are usually optically active, indicating that one of the enantiomers predominates (often it is the only isomer present). This is a result of the action of chiral catalysts we call enzymes, and reflects the inherently chiral nature of life itself. Chiral synthetic compounds, on the other hand, are commonly racemates, unless they have been prepared from enantiomerically pure starting materials.
There are two ways in which the condition of a chiral substance may be changed:
1. A racemate may be separated into its component enantiomers. This process is called resolution.
2. A pure enantiomer may be transformed into its racemate. This process is called racemization.
Enantiomeric Excess
The "optical purity" is a comparison of the optical rotation of a pure sample of unknown stereochemistry versus the optical rotation of a sample of pure enantiomer. It is expressed as a percentage. If the sample only rotates plane-polarized light half as much as expected, the optical purity is 50%.
Because R and S enantiomers have equal but opposite optical activity, it naturally follows that a 50:50 racemic mixture of two enantiomers will have no observable optical activity. If we know the specific rotation for a chiral molecule, however, we can easily calculate the ratio of enantiomers present in a mixture of two enantiomers, based on its measured optical activity. When a mixture contains more of one enantiomer than the other, chemists often use the concept of enantiomeric excess (ee) to quantify the difference. Enantiomeric excess can be expressed as:
For example, a mixture containing 60% R enantiomer (and 40% S enantiomer) has a 20% enantiomeric excess of R: ((60-50) x 100) / 50 = 20 %.
Example
The specific rotation of (S)-carvone is (+)61°, measured 'neat' (pure liquid sample, no solvent). The optical rotation of a neat sample of a mixture of R and S carvone is measured at (-)23°. Which enantiomer is in excess, and what is its ee? What are the percentages of (R)- and (S)-carvone in the sample?
Solution
The observed rotation of the mixture is levorotary (negative, counter-clockwise), and the specific rotation of the pure S enantiomer is given as dextrorotary (positive, clockwise), meaning that the pure R enantiomer must be levorotary, and the mixture must contain more of the R enantiomer than of the S enantiomer.
Rotation (R/S Mix) = [Fraction(S) × Rotation (S)] + [Fraction(R) × Rotation (R)]
Let Fraction (S) = x, therefore Fraction (R) = 1 – x
Rotation (R/S Mix) = x[Rotation (S)] + (1 – x)[Rotation (R)]
–23 = x(+61) + (1 – x)(–61)
Solve for x: x = 0.3114 and (1 – x) = 0.6885
Therefore the percentages of (R)- and (S)-carvone in the sample are 68.9% and 31.1%, respectively.
ee = [(% more abundant enantiomer – 50) × 100]/50
= [68.9 – 50) × 100]/50 = 37.8%
Chiral molecules are often labeled according to the sign of their specific rotation, as in (S)-(+)-carvone and (R)-(-)-carvone, or (±)-carvone for the racemic mixture. However, there is no relationship whatsoever between a molecule's R/S designation and the sign of its specific rotation. Without performing a polarimetry experiment or looking in the literature, we would have no idea that (-)-carvone has the R configuration and (+)-carvone has the S configuration.
Separation of Chiral Compounds
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used in this technique are known as '
Exercise 1
A sample with a concentration of 0.3 g/mL was placed in a cell with a length of 5 cm. The resulting rotation at the sodium D line was +1.52°. What is the [α]D?
Solution
5 cm = 0.5 dm
[α]D = α/(c x l) = +1.52/(0.3 x 0.5) = +10.1° | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.07%3A_Optical_Activity_and_Racemic_Mixtures.txt |
Learning Objective
• explain how to resolve (separate) a pair of enantiomers
Introduction and Overview
A racemic mixture is a 50:50 mixture of two enantiomers. Because they are mirror images, each enantiomer rotates plane-polarized light in an equal but opposite direction and is optically inactive. If the enantiomers are separated, the mixture is said to have been resolved. A common experiment in the laboratory component of introductory organic chemistry involves the resolution of a racemic mixture.
The dramatic biochemical consequences of chirality are illustrated by the use, in the 1950s, of the drug Thalidomide, a sedative given to pregnant women to relieve morning sickness. It was later realized that while the (+)‑form of the molecule, was a safe and effective sedative, the (−)‑form was an active teratogen. The drug caused numerous birth abnormalities when taken in the early stages of pregnancy because it contained a mixture of the two forms.
Chiral Resolution
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used in this technique are known as 'Moscher's esters', after Harry Stone Moscher, a chemist who pioneered the method at Stanford University.
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent.
Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. Figure 5.8.1 illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible.
Because the physical properties of enantiomers are identical, they seldom can be separated by simple physical methods, such as fractional crystallization or distillation. It is only under the influence of another chiral substance that enantiomers behave differently, and almost all methods of resolution of enantiomers are based upon this fact. We include here a discussion of the primary methods of resolution
Chiral Amines as Resolving Agents and Resolution of Racemic Acids
The most commonly used procedure for separating enantiomers is to convert them to a mixture of diastereomers that will have different physical properties: melting point, boiling point, solubility, and so on (Section 5-5). For example, if you have a racemic or D,L mixture of enantiomers of an acid and convert this to a salt with a chiral base having the D configuration, the salt will be a mixture of two diastereomers, (D acid . D base) and (L acid . D base). These diastereomeric salts are not identical and they are not mirror images. Therefore they will differ to some degree in their physical properties, and a separation by physical methods, such as crystallization, may be possible. If the diastereomeric salts can be completely separated, the acid regenerated from each salt will be either exclusively the D or the L enantiomer:
Resolution of chiral acids through the formation of diastereomeric salts requires adequate supplies of suitable chiral bases. Brucine, strychnine, and quinine frequently are used for this purpose because they are readily available, naturally occurring chiral bases. Simpler amines of synthetic origin, such as 2-amino- 1 -butanol, amphetamine, and 1 -phenylethanamine, also can be used, but first they must be resolved themselves.
Resolution of Racemic Bases
Chiral acids, such as (+)-tartaric acid, (-)-malic acid, (-)-mandelic acid, and (+)-camphor- 10-sulfonic acid, are used for the resolution of a racemic base.
The principle is the same as for the resolution of a racemic acid with a chiral base, and the choice of acid will depend both on the ease of separation of the diastereomeric salts and, of course, on the availability of the acid for the scale of the resolution involved. Resolution methods of this kind can be tedious, because numerous recrystallizations in different solvents may be necessary to progressively enrich the crystals in the less-soluble diastereomer. To determine when the resolution is complete, the mixture of diastereomers is recrystallized until there is no further change in the measured optical rotation of the crystals. At this stage it is hoped that the crystalline salt is a pure diastereomer from which one pure enantiomer can be recovered. The optical rotation of this
enantiomer will be a maximum value if it is "optically" pure because any amount of the other enantiomer could only reduce the magnitude of the measured rotation $\alpha$.
Resolution of Racemic Alcohols
To resolve a racemic alcohol, a chiral acid can be used to convert the alcohol to a mixture of diastereomeric esters. This is not as generally useful as might be thought because esters tend to be liquids unless they are very high-molecularweight compounds. If the diastereomeric esters are not crystalline, they must be separated by some other method than fractional crystallization (for instance, by chromatography methods, Section 9-2). Two chiral acids that are useful resolving agents for alcohols are:
The most common method of resolving an alcohol is to convert it to a half-ester of a dicarboxylic acid, such as butanedioic (succinic) or 1,2-benzenedicarboxylic (phthalic) acid, with the corresponding anhydride. The resulting half-ester has a free carboxyl function and may then be resolvable with a chiral base, usually brucine:
Other Methods of Resolution
Qne of the major goals in the field of organic chemistry is the development of reagents with the property of "chiral recognition" such that they can effect a clean separation of enantiomers in one operation without destroying either of the enantiomers. We have not achieved that ideal yet, but it may not be far in the future. Chromatographic methods (Section 9-2), whereby the stationary phase is a chiral reagent that adsorbs one enantiomer more strongly than the other, have been used to resolve racemic compounds, but such resolutions seldom have led to both pure enantiomers on a preparative scale. Other methods, called kinetic resolutions, are excellent when applicable. The procedure takes advantage of differences in reaction rates of enantiomers with chiral reagents. One enantiomer may react more rapidly, thereby leaving an excess of the other enantiomer behind. For example, racemic tartaric acid
can be resolved with the aid of certain penicillin molds that consume the dextrorotatory enantiomer faster than the levorotatory enantiomer. As a result, almost pure (-)-tartaric acid can be recovered from the mixture:
(±)-tartaric acid + mold $\rightarrow$ (-)-tartaric acid + more mold
A disadvantage of resolutions of this type is that the more reactive enantiomer usually is not recoverable from the reaction mixture.
The crystallization procedure employed by Pasteur for his classical resolution of (±)-tartaric acid (Section 5-1C) has been successful only in a very few cases. This procedure depends on the formation of individual crystals of each enantiomer. Thus if the crystallization of sodium ammonium tartrate is carried out below 27", the usual racemate salt does not form; a mixture of crystals of the (+) and (-) salts forms instead. The two different kinds of crystals, which are related as an object to its mirror image, can be separated manually with the aid of a microscope and subsequently may be converted to the tartaric acid enantiomers by strong acid. A variation on this method of resolution is the seeding of a saturated solution of a racemic mixture with crystals of one pure enantiomer in the hope of causing crystallization of just that one enantiomer, thereby leaving the other in solution. Unfortunately, very
few practical resolutions have been achieved in this way.
Even when a successful resolution is achieved, some significant problems remain. For instance, the resolution itself does not provide information on the actual configuration of the (+) or (-) enantiomer. This must be determined by other means (see Section 19-5). Also, it is not possible to tell the enantiomeric purity (optical purity) of the resolved enantiomers without additional information. This point is discussed further in the next section.
Exercise
1. Indicate the reagents you would use to resolve the following compounds. Show the reactions involved and specify the physical method you believe would be the best to separate the diastereomers.
1. 1 -phenyl-2-propanamine
2. 2,3-pentadienedioic acid
3. 1 -phenylethanol
Solutions:
1.
a. React 1-phenyl-2-propanamine racemic mixture with a chiral acid such as (+)-tartaric acid (R, R).
Reaction will produce a mixture of diastereomeric salts (i.e. R, R, R and S, R, R).
Separate diastereomers through crystallization.
Treat salt with strong base (e.g. KOH) to recover the pure enantiomeric amine.
b. React 2,3-pentadienedioic acid mixture with a chiral base such as (R)‑1‑phenylethylamine.
Reaction will produce a mixture of diastereomeric salts.
Separate diastereomers through crystallization.
Treat salt with strong acid (e.g. HCl) to recover the pure enantiomer acid.
c. React 1-phenylethanol mixture with 1,2-benzenedicarboxylic anhydride.
Reaction will produce a mixture of diastereomeric salts.
Separate diastereomers through crystallization.
Then alkaline hydrolysis treatment to recover the pure enantiomeric alcohol.
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)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.08%3A_Resolution_%28Separation%29_of_Enantiomers.txt |
Learning Objective
• interpret the stereoisomerism of compounds with three or more chiral centers
Possible Number of Stereoisomers
In general, a structure with n stereocenters will have 2n different stereoisomers. (We are not considering, for the time being, the stereochemistry of double bonds – that will come later). For example, let's consider the glucose molecule in its open-chain form (recall that many sugar molecules can exist in either an open-chain or a cyclic form). There are two enantiomers of glucose, called D-glucose and L-glucose. The D-enantiomer is the common sugar that our bodies use for energy. It has n = 4 stereocenters, so therefore there are 2n = 24 = 16 possible stereoisomers (including D-glucose itself).
In L-glucose, all of the stereocenters are inverted relative to D-glucose. That leaves 14 diastereomers of D-glucose: these are molecules in which at least one, but not all, of the stereocenters are inverted relative to D-glucose. One of these 14 diastereomers, a sugar called D-galactose, is shown above: in D-galactose, one of four stereocenters is inverted relative to D-glucose. Diastereomers which differ in only one stereocenter (out of two or more) are called epimers. D-glucose and D-galactose can therefore be refered to as epimers as well as diastereomers.
The epimer term is useful because in biochemical pathways, compounds with multiple chiral centers are isomerized at one specific center by enzymes known as epimerases. Two examples of epimerase-catalyzed reactions are below.
Now, let's extend our analysis to a sugar molecule with three chiral centers. Going through all the possible combinations, we come up with eight total stereoisomers - four pairs of enantiomers.
Let's draw the RRR stereoisomer. Being careful to draw the wedge bonds correctly so that they match the RRR configurations, we get:
Now, using the above drawing as our model, drawing any other stereoisomer is easy. If we want to draw the enantiomer of RRR, we don't need to try to visualize the mirror image, we just start with the RRR structure and invert the configuration at every chiral center to get SSS.
Try making models of RRR and SSS and confirm that they are in fact nonsuperimposable mirror images of each other.
There are six diastereomers of RRR. To draw one of them, we just invert the configuration of at least one, but not all three, of the chiral centers. Let's invert the configuration at chiral center 1 and 2, but leave chiral center 3 unchanged. This gives us the SSR configuration.
One more definition at this point: diastereomers which differ at only a single chiral center are called epimers. For example, RRR and SRR are epimers:
The RRR and SSR stereoisomers shown earlier are diastereomers but not epimers because they differ at two of the three chiral centers.
Example \(1\)
1. Draw the structure of the enantiomer of the SRS stereoisomer of the sugar used in the previous example.
2. List (using the XXX format, not drawing the structures) all of the epimers of SRS.
3. List all of the stereoisomers that are diastereomers, but not epimers, of SRS.
Solutions to exercises
Solution
Add text here.
Example \(2\)
The sugar below is one of the stereoisomers that we have been discussing.
The only problem is, it is drawn with the carbon backbone in a different orientation from what we have seen. Determine the configuration at each chiral center to determine which stereoisomer it is.
Exercise \(3\)
Draw the enantiomer of the xylulose-5-phosphate structure in the previous figure.
Exercise \(4\)
The structure of the amino acid D-threonine, drawn without stereochemistry, is shown below. D-threonine has the (S) configuration at both of its chiral centers. Draw D-threonine, it's enantiomer, and its two diastereomers.
Answer
Solutions to exercises
Comparing Stereoisomerism with Structural Isomerism
D-glucose and D-fructose are not stereoisomers, because they have different bonding connectivity: glucose has an aldehyde group, while fructose has a ketone. The two sugars do, however, have the same molecular formula, so by definition they are constitutional isomers.
D-glucose and D-ribose are not isomers of any kind, because they have different molecular formulas.
Exercise 5: Identify the relationship between each pair of structures. Your choices are: not isomers, constitutional isomers, diastereomers but not epimers, epimers, enantiomers, or same molecule
Exercise 6: Identify the relationship between each pair of structures. Hint - figure out the configuration of each chiral center.
Solutions to exercises
Kahn Academy video tutorial on stereoisomeric relationships
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)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.09%3A_Stereochemistry_of_Molecules_with_Three_or_More_Asymmet.txt |
Learning Objective
• compare and contrast absolute configuration with relative configuration
Absolute Configuration
The absolute configuration at a chiral center in a molecule is a time-independent and unambiguous symbolic description of the spatial arrangement of ligands (groups) bonded to the chiral center.
The chiral centers in 1 and 2 bear the same ligands: a,b,d, and e. However, 1 and 2 are not superimposable on each other, meaning that the arrangement of ligands around the chiral center in 1 and in 2 is different. 1 and 2 are mirror images of each other, meaning that the arrangement of ligands around the chiral center in 1 is the exact opposite of that in 2. Chiral centers in 1and 2 are said to have opposite absolute configurations.
According to R,S convention, if the absolute configuration at the chiral center in 1 is R, that at the chiral center in 2 is S or vice versa.
Relative Configuration
The relative configuration is the experimentally determined relationship between two enantiomers even though we may not know the absolute configuration. The sign of rotation of plane-polarized light by an enantiomer is not easily related to its configuration. This is true even for substances with very similar structures. Thus, given lactic acid, $\ce{CH_3CHOHCO_2H}$, with a specific rotation $+3.82^\text{o}$, and methyl lactate, $\ce{CH_3CHOHCO_2CH_3}$, with a specific rotation $-8.25^\text{o}$, we cannot tell from the rotation alone whether the acid and ester have the same or a different arrangement of groups about the chiral center. Their relative configurations have to be obtained by other means.
If we convert $\left( + \right)$-lactic acid into its methyl ester, we can be reasonably certain that the ester will be related in configuration to the acid, because esterification should not affect the configuration about the chiral carbon atom. It happens that the methyl ester so obtained is levorotatory, so we know that $\left( + \right)$-lactic acid and $\left( - \right)$-methyl lactate have the same relative configuration at the asymmetric carbon, even if they possess opposite signs of optical rotation. However, we still do not know the absolute configuration; that is, we are unable to tell which of the two possible configurations of lactic acid, $2a$ or $2b$, corresponds to the dextro or $\left( + \right)$-acid and which to the levo or $\left( - \right)$-acid:
Until 1956, the absolute configuration of no optically active compound was known. Instead, configurations were assigned relative to a standard, glyceraldehyde, which originally was chosen by E. Fischer (around 1885) for the purpose of correlating the configuration of carbohydrates. Fischer arbitrarily assigned the configuration $3a$ to dextrorotatory glyceraldehyde, which was known as $D$-$\left( + \right)$-glyceraldehyde. The levorotatory enantiomer, $3b$, is designated as $L$-$\left( - \right)$-glyceraldehyde. (If you are unsure of the terminology $D$ and $L$, or of the rules for writing Fischer projection formulas, review Sections 5-3C and 5-4.)
The configurations of many compounds besides sugars now have been related to glyceraldehyde, including $\alpha$-amino acids, terpenes, steroids, and other biochemically important substances. Compounds whose configurations are related to $D$-$\left( + \right)$-glyceraldehyde are said to belong to the $D$ series, and those related to $L$-$\left( - \right)$-glyceraldehyde belong to the $L$ series.
At the time the choice of absolute configuration for glyceraldehyde was made, there was no way of knowing whether the configuration of $\left( + \right)$-glyceraldehyde was in reality $3a$ or $3b$. However, the choice had a $50\%$ chance of being correct, and we now know that $3a$, the $D$ configuration, is in fact the correct configuration of $\left( + \right)$-glyceraldehyde. This was established through use of a special x-ray crystallographic technique, which permitted determination of the absolute disposition of the atoms in space of sodium rubidium $\left( + \right)$-tartrate. The configuration of $\left( + \right)$-tartaric acid (Section 5-5) previously had been shown by chemical means to be opposite to that of $\left( + \right)$-glyceraldehyde. Consequently the absolute configuration of any compound now is known once it has been correlated directly or indirectly with glyceraldehyde. For example,
Chemical transformation showing how the configuration of natural $\left( + \right)$-alanine has been related to $L$-$\left( + \right)$-lactic acid and hence to $L$-$\left( - \right)$-glyceraldehyde. The transformations shown involve two $S_\text{N}2$ reactions, each of which is stereospecific and inverts the configuration (Section 8-5). Reduction of the azide group leaves the configuration unchanged.
When there are several chiral carbons in a molecule, the configuration at one center usually is related directly or indirectly to glyceraldehyde, and the configurations at the other centers are determined relative to the first. Thus in the aldehyde form of the important sugar, $\left( + \right)$-glucose, there are four chiral centers, and so there are $2^4 = 16$ possible stereoisorners. The projection formula of the isomer that corresponds to the aldehyde form of natural glucose is $4$. By convention for sugars, the configuration of the highest-numbered chiral carbon is referred to glyceraldehyde to determine the overall configuration of the molecule. For glucose, this atom is $\ce{C5}$, next to the $\ce{CH_2OH}$ group, and has the hydroxyl group on the right. Therefore, naturally occurring glucose, which has a $\left( + \right)$ rotation, belongs to the $D$ series and is properly called $D$-$\left( + \right)$-glucose:
However, the configurations of $\alpha$-amino acids possessing more than one chiral carbon are determined by the lowest-numbered chiral carbon, which is the carbon alpha to the carboxyl group. Thus, even though the natural $\alpha$-amino acid, threonine, has exactly the same kind of arrangement of substituents as the natural sugar, threose, threonine by the amino-acid convention belongs to the $L$-series, whereas threose by the sugar convention belongs to the $D$-series:
A serious ambiguity arises for compounds such as the active tartaric acids. If the amino-acid convention is used, $\left( + \right)$-tartaric acid falls in the $D$ series; by the sugar convention, it has the $L$ configuration. One way out of this dilemma is to use the subscripts $s$ and $g$ to denote the amino-acid or carbohydrate conventions, respectively. Then the absolute configuration of $\left( + \right)$-tartaric acid can be designated as either $D_s$-$\left( + \right)$-tartaric acid of $L_g$-$\left( + \right)$-tartaric acid.
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."
6.11: Chirality at Nitrogen Phosphorus and Sulfur
Learning Objective
• interpret the stereoisomerism of compounds with nitrogen, phosphorus, or sulfur as chiral centers
Stereogenic Nitrogen
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, 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.
Asymmetric quaternary ammonium groups are also chiral. Amines, however, are not chiral, because they rapidly invert, or turn ‘inside out’, at room temperature.
The phosphorus center of phosphate ion and organic phosphate esters, for example, is tetrahedral, and thus is potentially a stereocenter.
We will see in chapter 10 how researchers, in order to investigate the stereochemistry of reactions at the phosphate center, incorporated sulfur and/or 17O and 18O isotopes of oxygen (the ‘normal’ isotope is 16O) to create chiral phosphate groups. Phosphate triesters are chiral if the three substituent groups are different. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.10%3A_Absolute_and_Relative_Configuration_-_the_distinction.txt |
Learning Objective
• recognize and explain biochemical applications of chirality
Some Chiral Organic Biomolecules
There are a number of important biomolecules that could occur as enantiomers, including amino acids and sugars. In most cases, only one enantiomer occurs (although some fungi, for example, are able to produce mirror-image forms of these compounds). We will look later at some of these biomolecules, but first we will look at a compound that occurs naturally in both enantiomeric forms.
Carvone is a secondary metabolite. That means it is a naturally-occurring compound that is not directly connected to the very basic functions of a cell, such as self-replication or the production of energy. The role of secondary metabolites in nature is often difficult to determine. However, these compounds often play roles in self-defense, acting as deterrents against competitor species in a sort of small-scale chemical warfare scenario. They are also frequently used in communications; this role has been studied most extensively among insects, which use lots of compounds to send information to each other.
The two naturally-occurring enantiomers of carvone.
Carvone is produced in two enantiomeric forms. One of these forms, called (-)-carvone, is found in mint leaves, and it is a principal contributor to the distinctive odor of mint. The other form, (+)-carvone, is found in caraway seeds. This form has a very different smell, and is typically used to flavor rye bread and other Eastern European foods.
Note that (+)-carvone is the same thing as (S)-carvone. The (+) designation is based on its positive optical rotation value, which is experimentally measured. The (S) designation is determined by the Cahn-Ingold-Prelog rules for designating stereochemistry, which deal with looking at the groups attached to a chiral center and assigning priority based on atomic number. However, carvone's chiral center actually has three carbons attached to it; they all have the same atomic number. We need a new rule to break the tie.
• If two substituent groups have the same atomic number, go one bond further to the next atom.
• If there is a difference among the second tier of atoms, stop.
• The group in which you have encountered a higher atomic number gets the highest priority.
• If there is not a clear difference, proceed one additional bond to the next set of atoms, and so on, until you find a difference.
In carvone, this decision tree works as follows:
• The chiral center is connected to a H, a C, a C and a C.
• The H is lowest priority.
• One C eventually leads to a C=O. However, at the second bond from the chiral center, this C is connected to a C and two H's.
• A second C is also part of the six-membered ring, but the C=O is farther away in this direction. At the second bond from the chiral center, this C is connected to a C and two H's, just like the first one.
• The third C is part of a little three-carbon group attached to the six-membered ring. At the second bond from the chiral center, it is connected to only one H and has two bonds to another C (this is counted as two bonds to C and one to H).
• Those first two carbon groups are identical so far.
• However, the third group is different; it has an extra bond to C, whereas the others have an extra bond to H. C has a higher atomic number than H, so this group has higher priority.
• The second-highest priority is the branch that reaches the oxygen at the third bond from the chiral center.
Comparing atoms step-by-step to assign configuration.
How different, exactly, are these two compounds, (+)- and (-)-carvone? Are they completely different isomers, with different physical properties? In most ways, the answer is no. These two compounds have the same appearance (colorless oil), the same boiling point (230 °C), the same refractive index (1.499) and specific gravity (0.965). However, they have optical rotations that are almost exactly opposite values.
• Two enantiomers have the same physical properties.
• Enantiomers have opposite optical rotations.
Clearly they have different biological properties; since they have slightly different odors, they must fit into slightly different nasal receptors, signaling to the brain whether the person next to you is chewing a stick of gum or a piece of rye bread. This different shape complimentarity is not surprising, just as it isn't surprising that a left hand only fits into a left handed baseball glove and not into a right handed one.
Thalidomide.
There are other reasons that we might concern ourselves with an understanding of enantiomers, apart from dietary and olfactory preferences. Perhaps the most dramatic example of the importance of enantiomers can be found in the case of thalidomide. Thalidomide was a drug commonly prescribed during the 1950's and 1960's in order to alleviate nausea and other symptoms of morning sickness. In fact, only one enantiomer of thalidomide had any therapeutic effect in this regard. The other enantiomer, apart from being therapeutically useless in this application, was subsequently found to be a teratogen, meaning it produces pronounced birth defects. This was obviously not a good thing to prescribe to pregnant women. Workers in the pharmaceutical industry are now much more aware of these kinds of consequences, although of course not all problems with drugs go undetected even through the extensive clinical trials required in the United States. Since the era of thalidomide, however, a tremendous amount of research in the field of synthetic organic chemistry has been devoted to methods of producing only one enantiomer of a useful compound and not the other. This effort probably represents the single biggest aim of synthetic organic chemistry through the last quarter century.
• Enantiomers may have very different biological properties.
• Obtaining enantiomerically pure compounds is very important in medicine and the pharmaceutical industry.
Exercises
1. Draw the two enantiomeric forms of 2-butanol, CH3CH(OH)CH2CH3. Label their configurations.
2. Sometimes, compounds have many chiral centers in them. For the following compounds, identify four chiral centers in each, mark them with asterisks, and identify each center as R or S configuration.
The following is the structure of dysinosin A, a potent thrombin inhibitor that consequently prevents blood clotting.
Ginkgolide B (below) is a secondary metabolite of the ginkgo tree, extracts of which are used in Chinese medicine.
Sanglifehrin A, shown below, is produced by a bacteria that may be found in the soil of coffee plantations in Malawi. It is also a promising candidate for the treatment of organ transplant patients owing to its potent immuno-suppressant activity.
Solution
1.
2.
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)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.12%3A_Biochemistry_of_Enantiomers.txt |
Learning Objective
• describe Jean Baptiste Biot and Louis Pasteur's contributions to the understanding of optical isomers
Because enantiomers have identical physical and chemical properties in achiral environments, separation of the stereoisomeric components of a racemic mixture or racemate is normally not possible by the conventional techniques of distillation and crystallization. In some cases, however, the crystal habits of solid enantiomers and racemates permit the chemist (acting as a chiral resolving agent) to discriminate enantiomeric components of a mixture. As background for the following example, it is recommended that the section on crystal properties be reviewed.
Tartaric acid, its potassium salt known in antiquity as "tartar", has served as the locus of several landmark events in the history of stereochemistry. In 1832 the French chemist Jean Baptiste Biot observed that tartaric acid obtained from tartar was optically active, rotating the plane of polarized light clockwise (dextrorotatory). An optically inactive, higher melting, form of tartaric acid, called racemic acid was also known. A little more than a decade later, young Louis Pasteur conducted a careful study of the crystalline forms assumed by various salts of these acids. He noticed that under certain conditions, the sodium ammonium mixed salt of the racemic acid formed a mixture of enantiomorphic hemihedral crystals; a drawing of such a pair is shown on the right. Pasteur reasoned that the dissymmetry of the crystals might reflect the optical activity and dissymmetry of its component molecules. After picking the different crystals apart with a tweezer, he found that one group yielded the known dextrorotatory tartaric acid measured by Biot; the second led to a previously unknown levorotatory tartaric acid, having the same melting point as the dextrorotatory acid. Today we recognize that Pasteur had achieved the first resolution of a racemic mixture, and laid the foundation of what we now call stereochemistry.
Optical activity was first observed by the French physicist Jean-Baptiste Biot. He concluded that the change in direction of plane-polarized light when it passed through certain substances was actually a rotation of light, and that it had a molecular basis. His work was supported by the experimentation of Louis Pasteur. Pasteur observed the existence of two crystals that were mirror images in tartaric acid, an acid found in wine. Through meticulous experimentation, he found that one set of molecules rotated polarized light clockwise while the other rotated light counterclockwise to the same extent. He also observed that a mixture of both, a racemic mixture (or racemic modification), did not rotate light because the optical activity of one molecule canceled the effects of the other molecule. Pasteur was the first to show the existence of chiral molecules.
6.14: Additional Exercises
6-1 For the following compounds, star (*) each chiral center, if any.
6-2 For the following compounds, identify the R or S configuration of each chiral carbon atom.
6-3 Draw out the following molecules, including stereocenters.
a) (2R,4S,6R)-2-bromo-6-chloro-4-methylheptane
b) (4R)-4-bromopent-1-ene
c) (1R,2R,3S)-1-fluoro-2,3-dimethylcyclohexane
d) (3S)-3-methylcyclopent-1-ene
(R) and (S) Nomenclature of Asymmetric Carbon Atoms
6-4 For the following compounds, assign R or S configurations for each stereocenter.
6-5 For the following compounds, assign R or S configurations for each stereocenter.
6-6 Identify each molecule as either (R)- or (S)-Limonene.
Chiral Compounds Without Asymmetric Atoms
6-7 Explain why the following compound is optically active.
6-8 Does the following compound contain a chiral center? Is it a chiral molecule?
6-9 Why is this biaryl compound shown below considered chiral, despite having no chiral center?
Fischer Projections and Diastereomers
6-10 For the following Fischer projections, identify the configuration (R or S) of all chiral centers (some atoms may not be chiral centers).
6-11 For the following pairs of compounds, identify whether they are enantiomers, diastereomers, or the same compound.
6-12 For the following pairs of compounds, identify whether they are enantiomers, diastereomers, or the same compound.
Meso Compounds
6-13 For the following compounds, identify whether they are meso or not meso.
6-14 Are meso compounds optically active? Explain your answer.
6-15 Is the following compound meso or not meso?
6.15: Solutions to Additional Exercises
6-1
6-2
6-3
6-4
6-5
6-6
Chiral Compounds Without Asymmetric Atoms
6-7 Though the molecule does not contain a chiral carbon, it is chiral as it is non-superimposable on its mirror image due to its twisted nature (the twist comes from the structure of the double bonds needing to be at 90° angles to each other, preventing the molecule from being planar). This allows it to be optically active.
6-8 The molecule does not contain a chiral center; however, it is a chiral molecule as it is non-superimposable on its mirror image.
6-9 In the case of this biaryl molecule, the large bulky substituents, located at the ortho positions relative to the sigma bond in the middle, experience enough steric interference with each other to create a large energy barrier to free rotation around the C-C sigma bond. Thus, the molecule cannot freely rotate to its other conformations and is non-superimposable on its mirror image.
6-10
6-11
a) Enantiomers
b) Diastereomers
c) Diastereomers
d) Same compound
6-12
a) Enantiomers
b) Diastereomers
c) Diastereomers
Meso Compounds
6-13
a) Meso
b) Not meso
c) Meso
d) Meso
e) Meso
f) Not meso
6-14 Meso compounds are not optically active as they can be superimposed on their mirror images, making them achiral (which are not optically active). The stereocenters on one half of the molecule will rotate light one direction, while the other half of the molecule will rotate light the opposite direction, giving a net rotation of zero and making the molecule optically inactive.
6-15 The compound is meso, since the opposing stereocenters have opposite absolute configurations and if the molecule is rotated about the sigma bond in the middle, you can see that the two halves of the compound are mirror images of each other. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/06%3A_Stereochemistry_at_Tetrahedral_Centers/6.13%3A_The_Discovery_of_Enantiomers.txt |
Learning Objectives
After reading the chapter and completing the exercises and homework, a student can be able to:
• classify alkyl halides - refer to section 7.1
• predict relative boiling points and solubility of alkyl halides - refer to section 7.1
• discuss the common uses of alkyl halides - refer to section 7.2
• specify the reagents for the most efficient synthesis of alkyl halides using free-radical halogenation of alkanes (Chapter 5) or allylic halogenation of alkenes with NBS - refer to section 7.3
• apply the alpha and beta labels to alkyl halides for substitution and elimination reactions - refer to section 7.4
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN1, SN2, E1 & E2 reactions - refer to sections 7.5, 7.6, 7.8, 7.13, and 7.15
• use Zaitsev’s rule to predict major and minor products of elimination reactions including halocyclohexanes - refer to sections 7.14, 7.15, and 7.16
• predict the products and specify the reagents for SN1, SN2, E1 and E2 reactions with stereochemistry - refer to sections 7.6, 7.7, 7.9, 7.14, 7.15, 7.19
• propose mechanisms for SN1, SN2, E1 and E2 reactions - refer to sections 7.5, 7.6, 7.7, 7.8, 7.9, 7.13, 7.14, 7.15, 7.19
• draw, interpret, and apply Reaction Energy Diagrams for SN1, SN2, E1 and E2 reactions - refer to sections 7.5, 7.6, 7.7, 7.8, 7.9, 7.13, 7.14, 7.15, 7.19
• predict carbocation rearrangements in 1st order reactions - refer to section 7.10
• explain and apply Hammond's Postulate to substitution reactions - refer to section 7.11
• explain how the kinetic isotope effect (KIE) can be used to elucidate reaction mechanisms - refer to section 7.17
• distinguish 1st or 2nd order substitution and elimination reactions - refer to sections 7.12 and 7.18
• discuss the importance of leaving groups in biological substitution reactions - refer to section 7.20
• discuss enzymatic elimination reactions of histidine - refer to section 7.21
• 7.1: Alkyl Halides - Structure and Physical Properties
Alkyl halides are classified based upon the structure of the carbon atom bonded to the halogen. Common names and physical properties are discussed.
• 7.2: Common Uses of Alkyl Halides
Halogen containing organic compounds are relatively rare in terrestrial plants and animals, with the thyroid hormones T3 and T4 as notable exceptions. Alkyl halides are excellent electrophiles and quickly become an o-chem student's best friend for synthetic pathways.
• 7.3: Preparation of Alkyl Halides
Alkyl halides can be readily synthesized from alkanes, alkenes, and alcohols. This section expands the ways we can brominate tetrahedral carbons to the allylic position of alkenes.
• 7.4: Reactions of Alkyl Halides- Substitution and Elimination
The two major reaction pathways for alkyl halides (substitution and elimination) are introduced.
• 7.5: The Sₙ2 Reaction
The SN2 mechanism is described mechanistically and kinetically as a one-step (concerted) reaction between two reactants (bimolecular) that inverts the configuration of the carbon at the reactive site. The terms nucleophile, electrophile, and leaving group are explained by application to SN2 reactions.
• 7.6: Characteristics of the Sₙ2 Reaction
In order of decreasing importance, the factors impacting SN2 reaction pathways are the structure of the alkyl halide, the strength of the nucleophile, the stability of the leaving group, and the type of solvent.
• 7.7: Stereochemistry of the SN2 Reaction
The SN2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product.
• 7.8: The Sₙ1 Reaction
In the SN1 reaction, the solvent helps pull apart the halogen and carbon to form a halide and carbocation. A nucleophile can now form a bond with the carbocation to create a new product. The mechanism is explained with stereochemistry and reaction kinetics.
• 7.9: Characteristics of the Sₙ1 Reaction
The formation and stability of the carbocation intermediate strongly influence the SN1 mechanism. The structure of the alkyl halide, the stability of the leaving group, and the type of solvent influence the reaction pathway. Since the nucleophile is not involved in the rate determining step, the strength of the nucleophile has low importance.
• 7.10: Rearrangements of the Carbocation and Sₙ1 Reactions
Carbocation rearrangements are structural reorganizational shifts within an ion to a more stable state (lower energy).
• 7.11: The Hammond Postulate and Transition States
The Hammond postulate states that a transition state resembles the structure of the nearest stable species and helps explain the product distribution differences observed between exergonic and endergonic reactions.
• 7.12: Comparison of SN1 and SN2 Reactions
In comparing the SN1 and SN2 mechanisms, the structure of the alkyl halide (electrophile), the strength of the nucleophile, and the reaction solvent are the primary considerations. The leaving group will have a similar effect for both reactions, so it is not of interest when comparing the mechanistic pathways.
• 7.13: Characteristics of the E2 Reaction
E2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold. In E2 reactions, a beta-hydrogen and the leaving group are eliminated from an alkyl halide in reaction with a strong base to form an alkene.
• 7.14: Zaitsev's Rule
Zaitsev's Rule can be used to predict the regiochemistry of elimination reactions. Regiochemistry describes the orientation of reactions about carbon-carbon double bonds (C=C).
• 7.15: Characteristics of the E1 Reaction
The unimolecular E1 mechanism is a first order elimination reaction in which carbocation formation and stability are the primary factors for determining reaction pathway(s) and product(s).
• 7.16: E2 Regiochemistry and Cyclohexane Conformations
Cyclohexyl halides provides the perfect opportunity to learn and understand the regiochemistry of the E2 reaction and why Zaitsev's Rule does not always apply. The anti-coplanar orientation of the E2 mechanism can also be see with certain carbon chain diastereomers.
• 7.17: The E2 Reaction and the Deuterium Isotope Effect
The bimolecular transition state of the E2 reaction illustrates the effects of bond strength on reaction rates when studying the kinetic isotope effect of deuterium.
• 7.18: Comparison of E1 and E2 Reactions
The strength of the base is the primary consideration when distinguishing between the E1 and E2 pathways. The reaction solvent is a secondary consideration.
• 7.19: Comparing Substitution and Elimination Reactions
Chemical reactivity patterns can help us determine the most favorable pathway among the closely related SN1, SN2, E1, and E2 mechanisms.
• 7.20: Biological Substitution Reactions
A few examples of biochemical SN1 and SN2 mechanisms are introduced with an emphasis on the effects of the leaving group.
• 7.21: Biological Elimination Reactions
A few examples of biochemical E1 and E2 mechanisms are introduced.
• 7.22: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 7.23: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
07: Alkyl Halides- Nucleophilic Substitution and Elimination
Learning Objective
• classify alkyl halides
• predict relative boiling points and solubility of alkyl halides
Introduction
Alkyl halides are also known as haloalkanes. Alkyl halides are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). We will only look at compounds containing one halogen atom like th compounds below.
Alkyl halides fall into different classes depending on how the halogen atom is positioned on the chain of carbon atoms. Alkyl halides can be classified as primary, secondary, or tertiary. The chemical reactivity of alkyl halides is frequently discussed using alkyl halide classifications to help discern patterns and trends. Because the neutral bonding pattern for halogens is one bond and three lone pairs, the carbon and halogen always share a single bond. Alkyl halide classification is determined by the bonding pattern of the carbon atom bonded to the halogen as shown in the diagram below.
Primary alkyl halides
In a primary (1°) haloalkane, the carbon bonded to the halogen atom is only attached to one other alkyl group. Some examples of primary alkyl halides include thecompounds below.
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 halogen. There is an exception to this: CH3Br and the other methyl halides are often counted as primary alkyl halides even though there are no alkyl groups attached to the carbon with the halogen on it.
Secondary alkyl halides
In a secondary (2°) haloalkane, the carbon bonded with the halogen atom is joined directly to two other alkyl groups that can be the same or different. Some examples of secondary alkyl halides include thecompounds below.
Tertiary alkyl halides
In a tertiary (3°) halogenoalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different. Some examples of tertiary alkyl halides include thecompounds below.
Common Names
Many organic compounds are closely related to the alkanes and this similarity is incorporated into many common names. The reactions of alkanes with halogens produce halogenated hydrocarbons, compounds in which one or more hydrogen atoms of a hydrocarbon have been replaced by halogen atoms: The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane). The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide.
Examples
Give the common and IUPAC names for each compound.
1. CH3CH2CH2Br
2. (CH3)2CHCl
3. Give the IUPAC name for each compound.
a)
b)
SolutionS
1. The alkyl group (CH3CH2CH2–) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane.
2. The alkyl group [(CH3)2CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro-) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane.
3. a) The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
b) The parent alkane is hexane. Methyl (CH3) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.
Exercise
1. Give common and IUPAC names for each compound.
1. CH3CH2I
2. CH3CH2CH2CH2F
2. Give the IUPAC name for each compound.
a)
b)
Answer
1. a) ethyl iodide and iodoethane, respectively; Note the IUPAC name does not need a locator number because there is only one possible structure with two carbons and one iodine.
b) butyl fluoride and 1-fluorobutane
2. a) 2-chloro-2-methylbutane
b) 1-bromo-2-chloro-4-methylpentane
Halogens and the Character of the Carbon-Halogen Bond
With respect to electronegativity, halogens are more electronegative than carbons. This results in a carbon-halogen bond that is polarized. As shown in the image below, carbon atom has a partial positive charge, while the halogen has a partial negative charge.
The following image shows the relationship between the halogens and electronegativity. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
Haloalkanes Have Higher Boiling Points than Alkanes
When comparing alkanes and haloalkanes, we will see that haloalkanes have higher boiling points than alkanes containing the same number of carbons. London dispersion forces are the first of two types of forces that contribute to this physical property. You might recall from general chemistry that London dispersion forces increase with molecular surface area. In comparing haloalkanes with alkanes, haloalkanes exhibit an increase in surface area due to the substitution of a halogen for hydrogen. The incease in surface area leads to an increase in London dispersion forces, which then results in a higher boiling point.
Dipole-dipole interaction is the second type of force that contributes to a higher boiling point. As you may recall, this type of interaction is a coulombic attraction between the partial positive and partial negative charges that exist between carbon-halogen bonds on separate haloalkane molecules. Similar to London dispersion forces, dipole-dipole interactions establish a higher boiling point for haloalkanes in comparison to alkanes with the same number of carbons.
The table below illustrates how boiling points are affected by some of these properties. Notice that the boiling point increases when hydrogen is replaced by a halogen, a consequence of the increase in molecular size, as well as an increase in both London dispersion forces and dipole-dipole attractions. The boiling point also increases as a result of increasing the size of the halogen, as well as increasing the size of the carbon chain.
Solubility
Solubility in water
Alkyl halides have little to no solubility in water in spite of the polar carbon-halogen bond. The attraction between the alkyl halide molecules is stronger than the attraction between the alkyl halide and water. Alkyl halides have little to no solubility in water, but be aware of densities. Polyhalogenated alkanes such as dichloromethane can have densities greater than water.
Solubility in organic solvents
Alkyl halides are soluble in most organic solvents. The London Dispersion forces play a dominant role in solubility.
Exercises
Exercise
3. Classify (primary, secondary, tertiary, vicinal, or geminal) and give the IUPAC name for the following organohalides:
4. Classify (primary, secondary, tertiary, vicinal, or geminal) and draw the bond-line structures of the following compounds:
a) 2-Chloro-3,3-dimethylpentane
b) 1,1-Dichloro-4-isopropylcyclohexane
c) 3-bromo-3-ethylhexane
5. Arrange the following alkyl halides in order of decreasing boiling point.
6. Predict the solvent with great alkyl halide solubility.
a) water or hexane
b) water or 1-octanol
c) water or benzene
d) water or acetone
Solutions
3.
a) secondary; 5-ethyl-4-iodo-3methyl-octane
b) primary; 1-bromo-2,3,4-trimethyl-pentane
c) vicinal dihalide; 4-bromo-5-chloro-2-methyl-heptane
4. (A) secondary; (B) geminal dichloride (C) tertiary
5. A > B > C
6.
a) hexane
b) benzene
c) 1-octanol
d) acetone
Alkyl halides have little to no solubility in water, but be aware of densities. Polyhalogenated alkanes can have densities greater than water. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.01%3A_Alkyl_Halides_-_Structure_and_Physica.txt |
Learning Objective
• Discuss the common uses of alkyl halides
Halogen containing organic compounds are relatively rare in terrestrial plants and animals. The thyroid hormones T3 and T4 are exceptions; as is fluoroacetate, the toxic agent in the South African shrub Dichapetalum cymosum, known as "gifblaar". However, the halogen rich environment of the ocean has produced many interesting natural products incorporating large amounts of halogen. Some examples are shown below.
The ocean is the largest known source for atmospheric methyl bromide and methyl iodide. Furthermore, the ocean is also estimated to supply 10-20% of atmospheric methyl chloride, with other significant contributions coming from biomass burning, salt marshes and wood-rotting fungi. Many subsequent chemical and biological processes produce poly-halogenated methanes.
Synthetic organic halogen compounds are readily available by direct halogenation of hydrocarbons and by addition reactions to alkenes and alkynes. Many of these have proven useful as intermediates in traditional synthetic processes. Some halogen compounds, shown in the box. have been used as pesticides, but their persistence in the environment, once applied, has led to restrictions, including banning, of their use in developed countries. Because DDT is a cheap and effective mosquito control agent, underdeveloped countries in Africa and Latin America have experienced a dramatic increase in malaria deaths following its removal, and arguments are made for returning it to limited use. 2,4,5-T and 2,4-D are common herbicides that are sold by most garden stores. Other organic halogen compounds that have been implicated in environmental damage include the polychloro- and polybromo-biphenyls (PCBs and PBBs), used as heat transfer fluids and fire retardants; and freons (e.g. CCl2F2 and other chlorofluorocarbons) used as refrigeration gases and fire extinguishing agents.
Alkyl halides provide nice examples for learning about two very important organic reaction mechanism types: nucleophilic substitution and beta-elimination. In learning about these mechanisms in the context of alkyl halide reactivity, we will also learn some very fundamental ideas about three main players in many organic reactions: nucleophiles, electrophiles, and leaving groups. We'll start with an overview of the substitution and elimination reactions which alkyl halides undergo.
7.03: Preparation of Alkyl Halides
Learning Objective
• specify the reagents for the most efficient synthesis of alkyl halides using free-radical halogenation of alkanes (Chapter 5) or allylic halogenation of alkenes with NBS
Free radical halogenation of alkanes
Free radical halogenation of alkanes is the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. Light is required to initiate the radical formation and is a good example of a photochemical reaction. The simplest example is shown below for methane reacting with chlorine in the presence of light to form chloromethane and hydrogen chloride gas.
\[\ce{CH4 + Cl2 + energy → CH3Cl + HCl}\]
Free radical halogenation of alkanes has been thoroughly explained in chapter 5. The structure of the alkane is evaluated to choose between the high reactivity of chlorine (Cl2) and the high selectivity of bromine (Br2).
Allylic Bromination
When halogens are in the presence of unsaturated molecules such as alkenes, the expected reaction is addition to the double bond carbons resulting in a vicinal dihalide (halogens on adjacent carbons). The reaction is studied in a later chapter. To avoid halogen reactions at the alkene the halogen concentration is kept low enough that a substitution reaction occurs at the allylic position rather than addition at the double bond. The product is an allylic halide (halogen on carbon next to double bond carbons), which is acquired through a radical chain mechanism.
Why Substitution of Allylic Hydrogens?
As the table below shows, the dissociation energy for the allylic C-H bond is lower than the dissociation energies for the C-H bonds at the vinylic and alkylic positions. This is because the radical formed when the allylic hydrogen is removed is resonance-stabilized. Hence, given that the halogen concentration is low, substitution at the allylic position is favored over competing reactions. However, when the halogen concentration is high, addition at the double bond is favored because a polar reaction out competes the radical chain reaction.
Radical Allylic Bromination using NBS and light
Preparation of Bromine (low concentration)
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: Propagation
One bromine radical produced by homolytic cleavage in the initiation step removes an allylic hydrogen of the alkene molecule. A radical intermediate is generated, which is stabilized by resonance. The stability provided by delocalization of the radical in the alkene intermediate is the reason that substitution at the allylic position is favored over competing reactions such as addition at the double bond.
The intermediate radical then reacts with a Br2 molecule to generate the allylic bromide product and regenerate the bromine radical, which continues the radical chain mechanism. If the alkene reactant is asymmetric, two distinct product isomers are formed.
Step 3: Termination
The radical chain mechanism of allylic bromination can be terminated by any of the possible steps shown below.
Radical Allylic Chlorination
Like bromination, chlorination at the allylic position of an alkene is achieved when low concentrations of Cl2 are present. The reaction is run at high temperatures to achieve the desired results.
Industrial Uses
Allylic chlorination has important practical applications in industry. Since chlorine is inexpensive, allylic chlorinations of alkenes have been used in the industrial production of valuable products. For example, 3-chloropropene, which is necessary for the synthesis of products such as epoxy resin, is acquired through radical allylic chlorination (shown below).
Exercises
Exercises
1. Predict the two products of the allylic chlorination reaction of 1-heptene.
2. What conditions are required for allylic halogenation to occur? Why does this reaction outcompete other possible reactions such as addition when these conditions are met?
3. Predict the product of the allylic bromination reaction of 2-phenylheptane. (Hint: How are benzylic hydrogens similar to allylic hydrogens?)
4. The reactant 5-methyl-1-hexene generates the products 3-bromo-5-methyl-1-hexene and 1-bromo-5-methyl-2-hexene. What reagents were used in this reaction?
5. Predict the products of the following reactions:
Solutions
1. 3-chloro-1-heptene and 1-chloro-2-heptene
2. A low concentration of halide radical is sufficient for reaction at the allylic carbon without creating a reactivty environment for the pi bond of the alkene.
3. 2-bromo-2-phenylheptane
4. NBS with light
5. The product (A) is a 1° halogen which is more stable product even though the (B) had a better transition state with a 2° radical.
6.
7.04: Reactions of Alkyl Halides- Substitut
Learning Objective
• apply the alpha and beta labels to alkyl halides for substitution and elimination reactions - refer to section 7.4
Alkyl Halide Structure and Reaction Language
The carbon bonded to a halide is called the alpha-carbon. The carbons bonded to the alpha-carbon are called beta-carbons. Carbon atoms further removed from the alpha carbon are named by continuing the Greek alphabet (alpha, beta, gamma, delta, etc). In discussing the reactions of alkyl halides, it can be effective to use the alpha- and beta- labels. The structure for 2-bromopropane is used below to illustrate the application of these terms.
The Reactions - Nucleophilic Substitution and Elimination
Alkyl halides can undergo two major types of reactions - substitution and/or elimination. The substitution reaction is called a Nucleophilic Substitution reaction because the electrophilic alkyl halide forms a new bond with the nucleophile which substitutes for (replaces) the halogen at the alpha-carbon. Because carbon can only form four bonds, the halogen must leave and is called the "Leaving Group". Alkyl halides are excellent electrophiles because halogens share a polar bond with carbon, are polarizable, and form relatively stable leaving groups as halide anions. In the example below, 2-bromopropane is converted into propan-2-ol in a substitution reaction.
Allkyl halides can also undergo elimination reactions in the presence of strong bases. The elimination of a beta-hydrogen (hydrogen on a carbon vicinal to the alkyl halide carbon) and the halide produces a carbon-carbon double bond to form an alkene. In the example below, 2-bromopropane has undergone an elimination reaction to give an alkene - propene.
What decides whether you get substitution or elimination?
In the examples above, the reagents were the same for both substitution and elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. The product distribution depends on a number of factors. These factors will be explored in the remaining sections of this chapter. Depending on the structure of the alkyl halide, reagent type, reaction conditions, some reactions will only undergo only one pathway - substitution or elimination. While other alkyl halides will always produce a mixture of substitution and elimination products like the example above. The goal of efficient multiple-step synthetic pathways is to maximize the formation of a single product during each step. The reaction conditions explored in this chapter will be useful for future reactions we will study and learn.
Exercise
1. Classify the following reactions as "Substitutions" or "Eliminations".
Answer
1. a) substitution
b) elimination
c) elimination
d) substitution | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.02%3A_Common_Uses_of_Alkyl_Halides.txt |
Learning Objectives
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN2 reactions
• propose mechanisms for SN2 reactions
• draw and interpret Reaction Energy Diagrams for SN2 reactions
Introduction
In many ways, the proton transfer process of a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon.
In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) reacts with an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base.
In the next few sections, we are going to be discussing some general aspects of nucleophilic substitution reactions, and in doing so it will simplify things greatly if we can use some abbreviations and generalizations before we dive into real examples.
What is a nucleophile (Nu)?
Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. Nucleophiles can be negatively charged and some that are neutral with lone pair electrons. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates.
More specifically in laboratory reactions, halide and azide (N3-) anions are commonly seen acting as nucleophiles.
When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Some confusion in distinguishing basicity (base strength) and nucleophilicity (nucleophile strength) is inevitable. Since basicity is a less troublesome concept; it is convenient to start with it. Basicity refers to the ability of a base to accept a proton. Basicity may be related to the pKa of the corresponding conjugate acid, as shown below. The strongest bases have the weakest conjugate acids and vice versa. The range of basicities included in the following table is remarkable, covering over fifty powers of ten!
In an acid-base equilibrium the weakest acid and the weakest base will predominate (they will necessarily be on the same side of the equilibrium). Learning the pKa values for common compounds provides a useful foundation on which to build an understanding of acid-base factors in reaction mechanisms.
Base I (–) Cl (–) H2O CH3CO2(–) RS(–) CN(–) RO(–) NH2(–) CH3(–)
Conj. Acid HI HCl H3O(+) CH3CO2H RSH HCN ROH NH3 CH4
pKa -9 -7 -1.7 4.8 8 9.1 16 33 48
Nucleophilicity is a more complex property. It commonly refers to the rate of substitution reactions at the halogen-bearing carbon atom of a reference alkyl halide, such as CH3-Br. Thus the nucleophilicity of the Nu:(–) reactant in the following substitution reaction varies as shown in the chart below:
Nucleophilicity: CH3CO2 (–) < Cl(–) < Br(–) < N3(–) < CH3O(–) < CN(–) < I(–) < CH3S(–)
What is a Leaving Group (X or LG)?
In a similar fashion, we will call the leaving group 'X' for halogens as is customary. For other reactions, it will be more accurate to abbreviate the leaving group as "LG". The context of the reaction will dictate the abbreviation. Leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. Therefore, in this general picture we will not include a charge designation on the 'X' or 'LG' species. In referring to the comparison between acid-base chemistry and substitution reactions, the stability of the leaving group is evaluated the same way we evaluate the stability of conjugate bases.
When comparing the reactivity of electrophiles that vary only in their leaving groups, then leaving group stability plays a dominant role. The electrophile with the more stable leaving group will be favored. The lower the electron density of the leaving group, the more stable it is. Neutral leaving groups are favoring over charged leaving groups. When comparing charged leaving groups, apply the concepts used to determine the relative stability of conjugate bases:
1) identity or identities of the atom(s) holding the charge
2) delocalization of the charge via resonance
3) inductive effects
4) orbital hybridization
What is an Electrophile (E)?
An electrophile accepts electrons analogous to a Lewis acid. Electrophiles (E) are sometimes protonated and sometimes neutral. Electrophiles can also be called "Substrates". Since nucleophiles, leaving groups, and electrons may be charged or neutral, we will not include charges on 'Nu' or 'X' (or 'LG') or 'E'.
We will generalize the three other groups bonded on the electrophilic alpha-carbon as R1, R2, and R3: these symbols could represent hydrogens as well as alkyl groups. Finally, in order to keep figures from becoming too crowded, we will use in most cases the line structure convention in which the central, electrophilic carbon is not drawn out as a 'C'.
Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction:
To recognize neutral electrophiles, we will need to identify polarity and/or resonance with compounds to create a partial positive charge to attract the nucleophile. The electrophilicity of alkyl halides comes from the polar carbon-halogen bond.
The common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing on the right. Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is covalent bond strength. The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, alkyl fluorides and fluorocarbons in general are chemically and thermodynamically quite stable, and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative stability of the corresponding halide anions, which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pKa = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI.
Exercise
1. Since everything is relative in chemistry, one reaction's nucleophile can be another reaction's leaving group. Some functional groups can only react as a nuclephile or electrophile, while other functional groups can react as either a nuclephile or electrophile depending on the reaction conditions. Classify the following compounds as nucleophiles, electrophiles, or leaving groups. More than one answer may be possible.
a) bromoethane
b) hydroxide
c) water
d) chlorocyclohexane
e) ethanol
f) bromide
Answer
a) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
b) strong nucleophile
c) weak nucleophile and good leaving group
d) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
e) weak nucleophile, a poor electrophile without clever chemistry (stay tuned for future chapters), good leaving group
f) good nucleophile and a good leaving group
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution, SN2 and SN1. The SN2 reaction takes place in a single step with bond-forming and bond-breaking occurring simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
This is called an 'SN2' mechanism. In the term SN2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that it is a bimolecular reaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must react with the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the electron rich, leaving group blocks the way with electrostatic repulsion and steric hindrance.
The result of this backside penetration is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the SN2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile.
Exercise
2. Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
Solution
2.
SN2 Reactions Occur at sp3 Carbons with a Leaving Group
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons bonded to a leaving group. SN2 reactions cannot occur where the leaving group is attached to an sp2-hybridized carbon:
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
For future reference when discerning between substitution and elimination reactions, evaluating the structure of the electrophile can eliminate possible products. If the electrophilic carbon has no beta-hydrogens, then only substitution reactions can occur and elimination reactions are not possible (of course carbocation rearrangements may need to be considered). The first four halides shown on the left below do not give elimination reactions on treatment with base, because they have no β-hydrogens. The two halides on the right do not normally undergo such reactions because the potential elimination products have highly strained double or triple bonds. It is also worth noting that sp2 hybridized C–X compounds, such as the three on the right, do not normally undergo nucleophilic substitution reactions, unless other functional groups perturb the double bond(s).
Exercise
3. Predict which alkyl halides can undergo a SN2 reaction.
a) C6H5Br
b) CH3CH2CH2Br
c) CH2CHBr
d) CH3CH2CH2CHBrCH3
Solutions
3.
a) No, aryl halide.
b) Yes, primary alkyl halide
c) No, vinyl halide
d) Yes, secondary alkyl halide
SN2 Reaction Kinetics
In the term SN2, the S stands for substitution, the N stands for nucleophilic, and the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane.
If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate.
If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate.
The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics
Exercise
4. The reaction below follows the SN2 mechanism.
a) Write the rate law for this reaction.
b) Determine the value of the rate coefficient, k, if the initial concentrations are 0.01 M CH3Cl, 0.01 M NaOH, and the initial reaction rate is 6 x 10-10 M/s.
c) Calculate the initial reaction rate if the initial reactant concentrations are changed to 0.02 M CH3Cl and 0.0005 M NaOH.
Solutions
4.
a) rate = k [CH3Cl] [OH-]
b) substitute the data into the rate expression above and apply algebra to solve for k
k = 6 x 10-6 Lmol-1s-1
c) Using the rate law above, substitute the value for k from the previous question along with the new concentrations to determine the new initial rate.
rate = 6 x 10-10 M/s | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.05%3A_The_S2_Reaction.txt |
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN2 reactions
• predict the products and specify the reagents for SN2 reactions with stereochemistry
• propose mechanisms for SN2 reactions
• draw and interpret Reaction Energy Diagrams for SN2 reactions
Introduction
To understand why some combinations of alkyl halides and nucleophiles give a substitution reaction, whereas other combinations give elimination, and still others give no observable reaction, we must investigate systematically the way in which changes in reaction variables perturb the course of the reaction. The following general equation summarizes the factors that will be important in such an investigation where X represents the leaving group (a halide for this chapter).
In order of decreasing importance, the factors impacting SN2 reaction pathways are
1) structure of the alkyl halide
2) strength of the nucleophile
3) stability of the leaving group
4) type of solvent.
The bimolecular transition state of the SN2 pathway means that sterics are a primary consideration. The orbitals of the nucleophile must be able to penetrate through the reaction solution and create orbital overlap with the orbitals of the electrophilic carbon. The sterics of this mechanism can be determined by applying the bonding theories for individual compounds and ions to the interaction of the nucleophile and electrophile. The strength of the nucleophile will also influence the reaction along with the stability of the leaving group. Solvents can have a subtle yet measurable effect on SN2 pathway. Solvation may be defined as the interaction between molecules of solvent and particles of solute. The result of solvation is to stabilize (i.e., lower the energy of) the solute particles. Solvents with lone pairs of electrons are good at solvating cations. Protic (i.e., hydroxylic) solvents are able to solvate anions through hydrogen bonding. As water has two lone pairs of electrons and is also protic, it is good at solvating both anions and cations. The role of the solvent is often misunderstood and consequently given way too much importance. Do not drown in the solvent. Solvation effects are less significant than the structure of the alkyl halide, the reactivity of the nucleophile, and the stability of the leaving group.
The following variables and observables can be used to study the SN2 mechanism.
Variables
R change α-carbon from 1º to 2º to 3º
if the α-carbon is a
chiral center, set as (R) or (S)
X change from Cl to Br to I (F is relatively unreactive)
Nu: change from anion to neutral; change basicity; change
polarizability
Solvent polar vs. non-polar; protic vs. non-protic
Observables
Products substitution, elimination, no reaction.
Stereospecificity if the α-carbon is a chiral center what happens to its configuration?
Reaction Rate measure as a function of reactant concentration.
When several reaction variables may be changed, it is important to isolate the effects of each during the course of study. In other words: only one variable should be changed at a time, the others being held as constant as possible. For example, we can examine the effect of changing the halogen substituent from Cl to Br to I, using ethyl as a common R–group, cyanide anion as a common nucleophile, and ethanol as a common solvent. We would find a common substitution product, C2H5–CN, in all cases, but the speed or rate of the reaction would increase in the order: Cl < Br < I. This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class.
Bimolecular Nucleophilic Substitution Reactions Are Concerted
Bimolecular nucleophilic substitution (SN2) reactions are concerted, meaning they are a one step process. The bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electophilic carbon and the halogen.
The potential energy diagram for an SN2 reaction is shown below. The one-step mechanism means that only a single transition state is formed. A transition state, unlike a reaction intermediate, is a very short-lived species that cannot be isolated or directly observed.
Alkyl halide (Substrate) Structure and SN2 Reaction Rates
Now that we have discussed the effects that the leaving group, nucleophile, and solvent have on biomolecular nucleophilic substitution (SN2) reactions, it's time to turn our attention to how the substrate affects the reaction. Although the substrate, in the case of nucleophilic substitution of haloalkanes, is considered to be the entire molecule circled below, we will be paying particular attention to the alkyl portion of the substrate. In other words, we are most interested in the electrophilic center that bears the leaving group.
The SN2 transition state is very crowded with a total of five groups around the electrophilic center, the nucleophile, the leaving group, and three substituents.
If each of the three substituents in this transition state were small hydrogen atoms, as illustrated in the first example below, there would be little steric repulsion between the incoming nucleophile and the electrophilic center, thereby increasing the ease at which the nucleophilic substitution reaction can occur. Remember, for the SN2 reaction to occur, the nucleophile must be able to overlap orbitals with the electrophilic carbon center, resulting in the expulsion of the leaving group. If one of the hydrogens, however, were replaced with an R group, such as a methyl or ethyl group, there would be an increase in steric repulsion with the incoming nucleophile. If two of the hydrogens were replaced by R groups, there would be an even greater increase in steric repulsion with the incoming nucleophile.
How does steric hindrance affect the rate at which an SN2 reaction will occur? As each hydrogen is replaced by an R group, the rate of reaction is significantly diminished. This is because the addition of one or two R groups shields the backside of the electrophilic carbon impeding nucleophilic penetration.
The diagram below illustrates this concept, showing that electrophilic carbons attached to three hydrogen atoms results in faster nucleophilic substitution reactions, in comparison to primary and secondary haloalkanes, which result in nucleophilic substitution reactions that occur at slower or much slower rates, respectively. Notice that a tertiary haloalkane, that which has three R groups attached, does not undergo nucleophilic substitution reactions at all. The addition of a third R group to this molecule creates a carbon that is entirely blocked.
Substitutes on Neighboring Carbons Slow Nucleophilic Substitution Reactions
Previously we learned that adding R groups to the electrophilic carbon results in nucleophilic substitution reactions that occur at a slower rate. What if R groups are added to neighboring carbons? It turns out that the addition of substitutes on neighboring carbons will slow nucleophilic substitution reactions as well.
In the example below, 2-methyl-1-bromopropane differs from 1-bromopropane in that it has a methyl group attached to the carbon that neighbors the electrophilic carbon. The addition of this methyl group results in a significant decrease in the rate of a nucleophilic substitution reaction.
If R groups were added to carbons farther away from the electrophilic carbon, we would still see a decrease in the reaction rate. However, branching at carbons farther away from the electrophilic carbon would have a much smaller effect.
Nucleophilicity
There are predictable periodic trends in nucleophilicity. Moving horizontally across the second row of the table, the trend in nucleophilicity parallels the trend in basicity:
The reasoning behind the horizontal nucleophilicity trend is the same as the reasoning behind the basicity trend: more electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond. This horizontal trend also tells us that amines are more nucleophilic than alcohols, although both groups commonly act as nucleophiles in both laboratory and biochemical reactions.
Recall that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger area, which tends to increase stability (and thus reduce basicity).
The vertical periodic trend for nucleophilicity is somewhat more complicated because the solvent can influence the nucleophilicity trend in either direction. Let's take the simple example of the SN2 reaction below:
. . .where Nu- is one of the halide ions: fluoride, chloride, bromide, or iodide, and the leaving group I* is a radioactive isotope of iodine (which allows us to distinguish the leaving group from the nucleophile when both are iodide). If this reaction is occurring in a protic solvent (that is, a solvent that has a hydrogen bonded to an oxygen or nitrogen - water, methanol and ethanol are the most important examples), then the reaction will go fastest when iodide is the nucleophile, and slowest when fluoride is the nucleophile, reflecting the relative strength of the nucleophile.
Relative nucleophilicity in a protic solvent
This of course, is opposite that of the vertical periodic trend for basicity, where iodide is the least basic. What is going on here? Shouldn't the stronger base, with its more reactive unbonded valence electrons, also be the stronger nucleophile?
As mentioned above, it all has to do with the solvent. Remember, we are talking now about the reaction occurring in a protic solvent like ethanol. Protic solvent molecules form very strong ion-dipole interactions with the negatively-charged nucleophile, essentially creating a 'solvent cage' around the nucleophile:
In order for the nucleophile to react with the electrophile, it must break free, at least in part, from its solvent cage. The lone pair electrons on the larger, less basic iodide ion interact less tightly with the protons on the protic solvent molecules resulting in weaker solvation - thus the iodide nucleophile is better able to break free from its solvent cage compared the smaller, more basic fluoride ion, whose lone pair electrons are bound more tightly to the protons of the solvent cage.
The picture changes if we switch to a polar aprotic solvent, such as acetone, in which there is a molecular dipole but no hydrogens bound to oxygen or nitrogen. Now, fluoride is the best nucleophile, and iodide the weakest.
Relative nucleophilicity in a polar aprotic solvent
The reason for the reversal is that, with an aprotic solvent, the ion-dipole interactions between solvent and nucleophile are much weaker: the positive end of the solvent's dipole is hidden in the interior of the molecule, and thus it is shielded from the negative charge of the nucleophile.
A weaker solvent-nucleophile interaction means a weaker solvent cage for the nucleophile to break through, so the solvent effect is much less important, and the more basic fluoride ion is also the better nucleophile.
Why not use a completely nonpolar solvent, such as hexane, for this reaction, so that the solvent cage is eliminated completely? The answer to this is simple - the nucleophile needs to be in solution in order to react at an appreciable rate with the electrophile, and a solvent such as hexane will not solvate an a charged (or highly polar) nucleophile at all. That is why chemists use polar aprotic solvents for nucleophilic substitution reactions in the laboratory: they are polar enough to solvate the nucleophile, but not so polar as to lock it away in an impenetrable solvent cage. In addition to acetone, three other commonly used polar aprotic solvents are acetonitrile, dimethylformamide (DMF), and dimethyl sulfoxide (DMSO).
In biological chemistry, where the solvent is protic (water), the most important implication of the periodic trends in nucleophilicity is that thiols are more powerful nucleophiles than alcohols. The thiol group in a cysteine amino acid, for example, is a powerful nucleophile and often acts as a nucleophile in enzymatic reactions, and of course negatively-charged thiolates (RS-) are even more nucleophilic. This is not to say that the hydroxyl groups on serine, threonine, and tyrosine do not also act as nucleophiles - they do.
Resonance effects on nucleophilicity
Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity. If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance.
The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group.
The leaving group
The more stable the leaving group, the lower the transition state energy, the lower the activation energy, the faster the reaction rate. Evaluating leaving group stability is analogous to determining relative acidity by evaluating conjugate base stability. The considerations are the same: identity of the atom(s) and relative position on the periodic table, resonance delocalization, and electronegativity. Orbital hybridization is rarely relevant.
As Size Increases, Basicity Decreases, Leaving Group Stability Increases:In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons.
When evaluating halogens as leaving groups, the same trend is significant. Fluoride has the highest electron density and is considered the worst leaving group to the point of no reactivity. As move down the column, the leaving groups have lower electron density and greater stability with iodide considered an excellent leaving group.
Resonance Decreases Basicity and Increases Leaving Group Stability:The formation of a resonance stabilized structure delocalizes the electrons over two or more atoms lowering the electron density of the leaving group and increases its stability. For halides as leaving groups there are no applications for this consideration, so we will look briefly at carbonyl chemistry to illustrate this effect. When comparing the hydrolysis rates of anhydrides and esters, anhydrides react spontaneously with water and undergo hydrolysis to form a resonance stabilized carboxylate ion. Whereas, ester hydrolysis is a much slower reaction and requires a catalyst to overcome the alkoxides as poor leaving groups. The details of these two reactions will be studied in greater detail later in this text.
As Electronegativity Increases, Basicity Decreases and Leaving Group Stability Increases: In general, if we move from the left of the periodic table to the right of the periodic table as shown in the diagram below, electronegativity increases. As electronegativity increases, basicity will decrease, meaning a species will be less likely to act as base; that is, the species will be less likely to share its electrons.
The following diagram illustrates this concept, showing -CH3 to be the worst leaving group and F- to be the best leaving group. This particular example should only be used to facilitate your understanding of this concept. In real reaction mechanisms, these groups are not good leaving groups at all. For example, fluoride is such a poor leaving group that SN2 reactions of fluoroalkanes are rarely observed.
Leaving Groups Across a Period
Solvent Effects on an SN2 reaction
The rate of an SN2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of protic solvents (those, such as water or alcohols, with hydrogen-bond donating capability) decreases the power of the nucleophile through strong solvation. WE can view the nucleophile as being locked in a solvent cage through the strong hydrogen-bond interactions between solvent protons and the reactive lone pairs on the nucleophile. A less powerful nucleophile in turn means a slower SN2 reaction.
SN2 reactions are faster in polar, aprotic solvents: those that lack hydrogen-bond donating capability. Below are several polar aprotic solvents that are commonly used in the laboratory:
These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile. By using an aprotic solvent we can raise the reactivity of the nucleophile. This can sometimes have dramatic effects on the rate at which a nucleophilic substitution reaction can occur. For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol.
Example
In each pair (A and B) below, which electrophile would be expected to react more rapidly in an SN2 reaction with the thiol group of cysteine as the common nucleophile?
Explanations to explain differences in chemical reactivity need to discuss structural and/or electrostatic differences between the reactants
a) Cpd B b/c it has a more stable leaving group.
The larger atomic size of S relative to O means the sulfide (CH3S-) will have a lower electron density than the alkoxide (CH3O-).
b) Cpd A b/c it has a more stable leaving group.
The neutral leaving group, (CH3)2S, is more stable than the charged sulfie leaving group (CH3S-).
c) Cpd B b/c the leaving group is resonance stabilized delocalizing the negative charge over two oxygen atoms. d) Cpd B b/c the leaving group has inductive electron withdrawal stabilization from the three fluorine atoms in addition to the resonance stabilzation.
Exercise
1. What product(s) do you expect from the reaction of 1-bromopentane with each of the following reagents in an SN2 reaction?
a) KI
b) NaOH
c) CH3C≡C-Li
d) NH3
2. Which in the following pairs is a better nuceophile?
a) (CH3CH2)2N- or (CH3CH2)2NH
b) (CH3CH2)3N or (CH3CH2)3B
c) H2O or H2S
3. Order the following in increasing reactivity for an SN2 reaction.
CH3CH2Br CH3CH2OTos (CH3CH2)3CCl (CH3CH2)2CHCl
4. Solvents benzene, ether, chloroform are non-polar and not strongly polar solvents. What effects do these solvents have on an SN2 reaction?
Answer
1. (a) - (d)
2.
a) (CH3CH2)2N- as there is a charge present on the nitrogen.
b) (CH3CH2)3N because a lone pair of electrons is present.
c) H2O as oxygen is more electronegative.
3.
4. They will decrease the reactivity of the reaction. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.06%3A_Characteristics_of_the_S2_Reaction.txt |
Learning Objective
• predict the products and specify the reagents for SN2reactions with stereochemistry
• propose mechanisms for SN2 reactions
• draw and interpret Reaction Energy Diagrams for SN2reactions
SN2 Reactions Are Stereospecific
The SN2 reaction is stereospecific like other concerted reactions.. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. The nuclephile and electrophile must be correctly oriented for orbital overlap to occur and trigger chemical reactivity. Experimental observations show that all SN2 reactions proceed with inversion of configuration; that is, the nucleophile will always penetrate from the backside in SN2 reactions. To think about why this might be true, remember that the nucleophile has a lone pair of electrons to be shared with the electrophilic carbon center and the leaving group is going to take a lone pair of electrons with it upon leaving. Because like charges repel each other, the nucleophile will always proceed by a backside displacement mechanism.
• Frontside Orientation: In a frontside orientation, the nucleophile approaches the electrophilic center on the same side as the leaving group. With frontside orientation, the stereochemistry of the product remains the same; that is, we have retention of configuration.
• Backside Orientation: In a backside orientation, the nucleophile approachss the electrophilic center on the side that is opposite to the leaving group. With backside orientation, the stereochemistry of the product does not stay the same. There is inversion of configuration.
For example, if the substrate is an R enantiomer, a frontside nucleophilic orientation results in retention of configuration, and the formation of the R enantiomer. A backside nucleophilic orientation results in inversion of configuration, and the formation of the S enantiomer.
Conversely, if the substrate is an S enantiomer, a frontside nucleophilic orientation results in retention of configuration, and the formation of the S enantiomer. A backside nucleophilic orientation results in inversion of configuration, and the formation of the R enantiomer.
Empirically, SN2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept of retention and inversion of configuration can also be applied to substrates that can exist as geometric isomers (cis and trans). If the cis configuration is the substrate, the resulting product will be trans. Conversely, if the trans configuration is the substrate, the resulting product will be cis.
Exercise
1. Predict the product of a nucleophilic substitution of (S)-2-bromopentane reacting with CH3CO2-, Show stereochemistry.
Answer
1.
Contributors and Attributions
• Racheal Curtis (UCD)
7.08: The S1 Reaction
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN1 reactions
• predict the products and specify the reagents for SN1 reactions with stereochemistry
• propose mechanisms for SN1 reactions
• draw and interpret Reaction Energy Diagrams for SN1 reactions
The SN1 mechanism with Stereochemistry
A second model for a nucleophilic substitution reaction is called the 'dissociative' or 'SN1' mechanism. In many cases, the nucleophile is the solvent, so this mechanism can also be called "solvolysis".
Step1: In the SN1 mechanism, the carbocation forms when the C-X bond breaks first, before the nucleophile approaches
Th carbocation has a central carbon with only three bonds and bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
Step 2: The nucleophile reacts with the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry. Because of this trigonal planar geometry, the nucleophile can approach the carbocation from either lobe of the empty p orbital (aka either side of the carbocation). This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization of the product occurs during SN1 reactions if the electrophilic carbon is chiral. If the intermediate from a chiral alkyl halide survives long enough to encounter a random environment, the products are expected to be racemic (a 50:50 mixture of enantiomers). On the other hand, if the departing halide anion temporarily blocks the front side, or if a nucleophile is oriented selectively at one or the other face, then the substitution might occur with predominant inversion or even retention of configuration.
As an example, the tertiary alkyl bromide below, (S)-3-bromo-3-methylhexane, would be expected to form a racemic mix of R- and S-3-methyl-3-hexanol after an SN1 reaction with water as the nucleophile.
Exercise
1. Draw the structure of the intermediate in the two-step nucleophilic substitution reaction (SN1) above.
Solution
1.
The SN1 Reaction Energy Diagram
The SN1 reaction is an example of a two-step reaction with a reaction intermediate. Evaluating reactive intermediates is a very important skill in the study of organic reaction mechanisms. Many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise
2. Draw structures representing transition state 1 (TS1) and transition state 2 (TS2) in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Solution
2.
SN1 Reaction Kinetics
In the first step of an SN1 mechanism, two charged species are formed from a neutral molecule. This step is much the slower of the two steps, and is therefore rate-determining. In the reaction energy diagram, the activation energy for the first step is higher than that for the second step indicating that the SN1 reaction has first order kinetics because the rate determining step involves one molecule splitting apart, not two molecules colliding. It is important to remember that first order refers to the rate law expression where the generic term substrate is used to describe the alkyl halide.
rate = k [substrate]
Because an SN1 reaction is first order overall the concentration of the nucleophile does not affect the rate. The implication is that the nucleophile does not participate in the rate limiting step or any prior steps, which suggests that the first step is the rate limiting step. Since the nucleophile is not involved in the rate-limiting first step, the nature of the nucleophile does not affect the rate of an SN1 reaction.
Exercise
3. Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is SN2, and reaction B is SN1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
4. Give the products of the following SN1 reaction. Show stereochemistry.
Solution
3. For Reaction A, the rate law is rate = k[CH3I][CH3S-]. Therefore, if the concentration of the nucleophile, CH3S-, is doubled and the concentration of the alkyl halide remains the same, then the reaction rate will double.
For Reaction B, the rate law is rate = k[CH3)3Br]. Therefore, if the concentration of the nucleophile, CH3SH, is doubled and the concentration of the alkyl halide remains the same, then reaction rate stays the same.
4. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.07%3A_Stereochemistry_of_the_SN2_Reaction.txt |
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for SN1 reactions
• predict the products and specify the reagents for SN1 reactions with stereochemistry
• propose mechanisms for SN1 reactions
• draw and interpret Reaction Energy Diagrams for SN1 reactions
In order of decreasing importance, the factors impacting SN1 reaction pathways are
1. structure of the alkyl halide
2. stability of the leaving group
3. type of solvent.
The unimolecular transition state of the SN1 pathway means that structure of the alkyl halide and stability of the leaving group are the primary considerations. Alkyl halides that can ionize to form stable carbocations are more reactive via the SN1 mechanism. Because carbocation stability is the primary energetic consideration, stabilization of the carbocation via solvation is also an important consideration.
Alkyl Halide Structure and Carbocation Stability
The first order kinetics of SN1 reactions suggest a two-step mechanism in which the rate-determining step consists of carbocation formation from the ionization of the alkyl halide as shown in the diagram below. In this mechanism, the carbocation is a high-energy intermediate the bonds immediately to nearby nucleophiles. The only reactant that is undergoing change in the first (rate-determining) step is the alkyl halide, so we expect such reactions would be unimolecular and follow a first-order rate equation. Hence the name SN1 is applied to this mechanism.
The Hammond postulate suggests that the activation energy of the rate-determining first step will be inversely proportional to the stability of the carbocation intermediate: the more stable the carbocation, the lower the activation energy, the faster the reactivity. Therefore, carbocation stability is a primary consideration in SN1 reactions. Carbocations can be stabilized by delocalizing the charge via resonance and through inductive electron donation of alkyl groups. Carbocations can also stabilize by rearrangement via 1,2-hydride or 1,2-methyl shifts. Carbocation rearrangements are explained in a subsequent section of this chapter.
Benzyl Carbocation
The relative stability of carbocations is summarized below.
Carbocation Stability CH3(+) < CH3CH2(+) < (CH3)2CH(+) CH2=CH-CH2(+) < C6H5CH2(+) (CH3)3C(+)
Consequently, we expect that 3º-alkyl halides will be more reactive than their 2º and 1º-counterparts in reactions that follow an SN1 mechanism. This is opposite to the reactivity order observed for the SN2 mechanism. Allylic and benzylic halides are exceptionally reactive by either mechanism. This trend is summarized in the diagram below.
Effects of Leaving Group
An SN1 reaction speeds up with a good leaving group. This is because the leaving group is involved in the rate-determining step. A good leaving group wants to leave so it breaks the C-Leaving Group bond faster. Once the bond breaks, the carbocation is formed and the faster the carbocation is formed, the faster the nucleophile can come in and the faster the reaction will be completed.
A good leaving group is a weak base because weak bases can hold the charge. They're happy to leave with both electrons and in order for the leaving group to leave, it needs to be able to accept electrons. Strong bases, on the other hand, donate electrons which is why they can't be good leaving groups. As you go from left to right on the periodic table, electron donating ability decreases and thus ability to be a good leaving group increases. Halides are an example of a good leaving group whos leaving-group ability increases as you go down the column.
The two reactions below is the same reaction done with two different leaving groups. One is significantly faster than the other. This is because the better leaving group leaves faster and thus the reaction can proceed faster.
Methyl Sulfate Ion Mesylate Ion Triflate Ion Tosylate Ion
CH3SO42- CH3SO32- CF3SO32- CH3C6H4SO32-
Solvent Effects on the SN1 Reaction
To facilitate the charge separation of the ionization reaction in the first step, a good ionizing solvent is needed. Two solvent characteristics will be particularly important - the polarity and the solvating power. The dielectric constant, ε, measures polarity of solvent molecules and their ability to orient themselves between ions to attenuate (reduce) the electrostatic force one ion exerts on the other. The higher the dielectric constant the more polar the substance and in the case of SN1 reactions, the faster the rate. A dielectric constant below 15 is usually considered non-polar. Solvents having high dielectric constants, such as water (ε=81), formic acid (ε=58), dimethyl sulfoxide (ε=45) & acetonitrile (ε=39) are generally considered better ionizing solvents than are some common organic solvents such as ethanol (ε=25), acetone (ε=21), methylene chloride (ε=9) & ether (ε=4). Below is the same reaction conducted in two different solvents. The relative reaction rate in water (ε=81) is 150,000 times faster than in methanol (ε=33).
Solvation refers to the solvent's ability to stabilize ions by encasing them in a sheath of weakly bonded solvent molecules. Anions are solvated by partial positive charges of hydrogen-bonding solvents. Cations are often best solvated by the nucleophilic sites on a solvent molecule (e.g. oxygen & nitrogen atoms). The interaction of the carbocations with these nucleophilic solvents may be strong enough to form covalent bonds to carbon, thus converting the intermediate to a substitution product and creating the reaction name "solvolysis". When solvolysis occurs with water, the actions are called "hydrolysis reactions" as shown in the reaction below.
Polar Protic and Polar Aprotic Solvents
Protic solvents contain polarized hydrogen. Whereas, aprotic solvents do NOT contain polarized hydrogen. For SN2 reactions, solvation of the nucleophile by polar protic solvents slows the reaction rate. However, in SN1 reaction the nucleophile is not a part of the rate-determining step so this concern is not relevant. In fact, polar protic solvents actually speed up the rate of SN1 reactions because the polar solvent helps stabilize the transition state and carbocation intermediate. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction. Polar aprotic solvents have a dipole moment, but their hydrogen is not highly polarized. Polar aprotic solvents are not used in SN1 reactions because some of them can react with the carbocation intermediate and give an unwanted side-product. Rather, polar protic solvents are preferred for unimolecular substitution reactions.
Effects of Nucleophile
The strength of the nucleophile does not affect the reaction rate of SN1 because the nucleophile is not involved in the rate-determining step. Since nucleophiles only participate in the fast second step, their relative molar concentrations rather than their nucleophilicities should be the primary product-determining factor. If a nucleophilic solvent such as water is used, its high concentration will assure that alcohols are the major product. However, if you have more than one nucleophile competing to bond to the carbocation, the strengths and concentrations of those nucleophiles affects the distribution of products. For example, if t-butylchloride reacts with a mixture of water and formic acid where the water and formic acid are competing nucleophiles, two different products are formed: (CH3)3COH and (CH3)3COCOH. The relative yields of these products depends on the concentrations and relative reactivities of the nucleophiles. With a higher electron density, water is considered the stronger nucleophile and the tertiary alcohol will be the major product if there are equal concentrations of competing nucleophiles.
Exercises
1. Rank the following by increasing reactivity in an SN1 reaction.
2. 3-bromo-1-pentene and 1-bromo-2-pentene undergo SN1 reaction at almost the same rate, but one is a secondary halide while the other is a primary halide. Explain why this is.
3. Label the following reactions as most likely occuring by an SN1 or SN2 mechanism. Suggest why.
Answers
1. Consider the stability of the intermediate, the carbocation.
A < D < B < C (most reactive)
2. They have the same intermediates when you look at the resonance forms.
3. A – SN1 *poor leaving group, protic solvent, secondary cation intermediate
B – SN2 *good leaving group, polar solvent, primary position.
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)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.09%3A_Characteristics_of_the_S1_Reaction.txt |
Learning Objective
• predict carbocation rearrangements in 1st order reactions
Whenever reactants like alkyl halides form carbocations, the carbocations are subject to a phenomenon known as carbocation rearrangement. A carbocation is highly reactive and holds the positive charge on carbon with a sextet rather than an octet. There are two types of rearrangements: hydride shift and alkyl shift. Rearrangements occur to create more stable carbocations. Reviewing carbocation stability from chapter 5 is helpful in identifying carbocations that can undergo rearrangement.
Once rearranged, the molecules can also undergo further unimolecular substitution (SN1) or unimolecular elimination (E1). Nucleophilic reactions often produce two products, a major product and a minor product. The major product is typically the rearranged product that is more substituted (aka more stable). The minor product, in contract, is typically the normal product that is less substituted (aka less stable). Similarly, we will see in subsequent sections of this chapter that for the unimolecular elimination reaction, a more substituted alkene can form through carbocation rearrangements ("stay tuned for coming attractions").
Hydride Shift
The hydride shift can also be called the 1,2-Hydride Shift because rearrangements primarily occur between adjacent carbon atoms. The 1,2 are communicating that the carbons are vicinal (adjacent). These numbers have nothing to do with the nomenclature of the reactant. We can see the phenomenon of hydride shift in solvolysis (SN1) reactions like the example below.
As shown in the following mechanism, the polarized carbon-chlorine bonds is heterolytically broken to produce a chloride ion and carbocation. The secondary carbocation undergoes a 1,2 hydride shift to produce the more stable tertiary carbocation. The oxygen of a water molecule acts as the nucleophile and reacts with the carbocation to form a protonated alcohol. The intermediate is deprotonated to form the final product, an alcohol. The mechanism for hydride shift occurs in multiple steps that includes various intermediates and transition states.
Exercise
1. Draw the bond-line structure for the major solvolysis product of each reaction.
Answer
Alkyl Shift
Not all carbocations have suitable hydrogen atoms (either secondary or tertiary) that are on adjacent carbon atoms available for rearrangement. In this case, the reaction can undergo a different mode of rearrangement known as alkyl shift (or alkyl group migration). Alkyl Shift acts very similarily to that of hydride shift. Instead of the proton (H) that shifts with the nucleophile, we see an alkyl group that shifts with the nucleophile instead. The shifting group carries its electron pair with it to furnish a bond to the neighboring or adjacent carbocation. The shifted alkyl group and the positive charge of the carbocation switch positions on the molecule.
Exercise
2. Draw the bond-line structure for the major solvolysis product of each reaction.
Answer
Alkyl Halide Classification and Carbocation Rearrangements
Reactions of tertiary carbocations react much faster than that of secondary carbocations and will form the major product almost exclusively. Alkyl shifts from a secondary carbocation to tertiary carbocation in SN1 reactions occur by independent steps. When the alkyl halide is primary, then slight variations and differences between the two reaction mechanisms. In reaction #1, we see that we have a secondary substrate. This undergoes alkyl shift because it does not have a suitable hydrogen on the adjacent carbon. Once again, the reaction is similar to hydride shift. The only difference is that we shift an alkyl group rather than shift a proton, while still undergoing various intermediate steps to furnish its final product.
With reaction #2, on the other hand, we can say that it undergoes a concerted mechanism. In short, this means that everything happens in one step. This is because primary carbocations cannot be an intermediate and they are relatively difficult processes since they require higher temperatures and longer reaction times. After protonating the alcohol substrate to form the alkyloxonium ion, the water must leave at the same time as the alkyl group shifts from the adjacent carbon to skip the formation of the unstable primary carbocation.
Exercise
3. Draw the bond-line structure for the major solvolysis product of each reaction.
Answer
1,3-Hydride and Greater Shifts
Typically, hydride shifts can occur at low temperatures. However, by heating the solutionf of a cation, it can easily and readily speed the process of rearrangement. One way to account for a slight barrier is to propose a 1,3-hydride shift interchanging the functionality of two different kinds of methyls. Another possibility is 1,2 hydride shift in which you could yield a secondary carbocation intermediate. Then, a further 1,2 hydride shift would give the more stable rearranged tertiary cation.
More distant hydride shifts have been observed, such as 1,4 and 1,5 hydride shifts, but these arrangements are too fast to undergo secondary cation intermediates.
Analogy
Carbocation rearrangements happen very readily and often occur in many organic chemistry reactions. Yet, we typically neglect this step. Dr. Sarah Lievens, a Chemistry professor at the University of California, Davis once said carbocation rearrangements can be observed with various analogies to help her students remember this phenomenon. For hydride shifts: "The new friend (nucleophile) just joined a group (the organic molecule). Because he is new, he only made two new friends. However, the popular kid (the hydrogen) glady gave up his friends to the new friend so that he could have even more friends. Therefore, everyone won't be as lonely and we can all be friends." This analogy works for alkyl shifts in conjunction with hydride shift as well.
• Jeffrey Ma | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.10%3A_Rearrangements_of_the_Carbocation_and.txt |
Learning Objective
• explain and apply Hammond's Postulate to substitution reactions
Now, back to transition states. Chemists are often very interested in trying to learn about what the transition state for a given reaction looks like, but addressing this question requires an indirect approach because the transition state itself cannot be observed. In order to gain some insight into what a particular transition state looks like, chemists often invoke the Hammond postulate, which states that a transition state resembles the structure of the nearest stable species. For an exergonic reaction, therefore, the transition state resembles the reactants more than it does the products.
If we consider a hypothetical exergonic reaction between compounds A and B to form AB, the distance between A and B would be relatively large at the transition state, resembling the starting state where A and B are two isolated species. In the hypothetical endergonic reaction between C and D to form CD, however, the bond formation process would be much further along at the TS point, resembling the product.
The Hammond Postulate is a very simplistic idea, which relies on an assumption that potential energy surfaces are parabolic. Although such an assumption is not rigorously true, it is fairly reliable and allows chemists to make energetic arguments about transition states by employing arguments about the stability of a related species. Since the formation of a reactive intermediate is very reliably endergonic, arguments about the stability of reactive intermediates can serve as proxy arguments about transition state stability.
The Hammond Postulate and the SN1 Reaction
The Hammond postulate suggests that the activation energy of the rate-determining first step will be inversely proportional to the stability of the carbocation intermediate. The stability of carbocations is shown qualitatively below:
Carbocation Stability
\[\ce{CH3(+) < CH3CH2(+) < (CH3)2CH(+) ≈ CH2=CH-CH2(+) < C6H5CH2(+) ≈ (CH3)3C(+)}\]
Consequently, we expect that 3º-alkyl halides will be more reactive than their 2º and 1º-counterparts in reactions that follow an SN1 mechanism. This is opposite to the reactivity order observed for the SN2 mechanism. Allylic and benzylic halides are exceptionally reactive by either mechanism.
7.12: Comparison of SN1 and SN2 Reactions
Learning Objective
• distinguish 1st or 2nd order substitution reactions
Predicting SN1 vs. SN2 mechanisms
When considering whether a nucleophilic substitution is likely to occur via an SN1 or SN2 mechanism, we really need to consider three factors:
1) The electrophile: when the leaving group is attached to a methyl group or a primary carbon, an SN2 mechanism is favored (here the electrophile is unhindered by surrounded groups, and any carbocation intermediate would be high-energy and thus unlikely). When the leaving group is attached to a tertiary, allylic, or benzylic carbon, a carbocation intermediate will be relatively stable and thus an SN1 mechanism is favored. These patterns of reactivity of summarized below.
Alkyl Halide Structure Possible Substitution Reactions
methyl and primary SN2 only
secondary SN2 and SN1
tertiary SN1 only
primary and secondary benzylic and allylic SN2 and SN1
tertiary benzylic and allylic SN1 only
vinyl and aryl NO reaction
2) The nucleophile: powerful nucleophiles, especially those with negative charges, favor the SN2 mechanism. Weaker nucleophiles such as water or alcohols favor the SN1 mechanism.
3) The solvent: Polar aprotic solvents favor the SN2 mechanism by enhancing the reactivity of the nucleophile. Polar protic solvents favor the SN1 mechanism by stabilizing the transition state and carbocation intermediate. SN1 reactions are called solvolysis reactions when the solvent is the nucleophile.
These patterns of reactivity are summarized in the table below.
Comparison between SN2 and SN1 Reactions
Reaction Parameter SN2 SN1
alkyl halide structure methyl > primary > secondary >>>> tertiary tertiary > secodary >>>> primary > methyl
nucleophile high concentration of a strong nucleophile poor nucleophile (often the solvent)
mechanism 1-step 2-stp
rate limiting step bimolecular transition state carbocation formation
rate law rate = k[R-X][Nu] rate = k[R-X]
stereochemisty inversion of configuration mixed configuration
solvent polar aprotic polar protic
For example, the reaction below has a tertiary alkyl bromide as the electrophile, a weak nucleophile, and a polar protic solvent (we’ll assume that methanol is the solvent). Thus we’d confidently predict an SN1 reaction mechanism. Because substitution occurs at a chiral carbon, we can also predict that the reaction will proceed with racemization.
In the reaction below, on the other hand, the electrophile is a secondary alkyl bromide – with these, both SN1 and SN2 mechanisms are possible, depending on the nucleophile and the solvent. In this example, the nucleophile (a thiolate anion) is strong, and a polar protic solvent is used – so the SN2 mechanism is heavily favored. The reaction is expected to proceed with inversion of configuration.
Exercise
1. Determine whether each substitution reaction shown below is likely to proceed by an SN1 or SN2 mechanism and explain your reasoning.
Answer
a) SN2 b/c primary alkyl halide with a strong nucleophile in a polar aprotic solvent.
b) SN1 b/c tertiary alkyl halide with a weak nucleophile that is also the solvent (solvolysis).
c) SN2 b/c secondary alkyl halides favor this mechanism when reacted with a strong nucleophile (and weak base) in a polar aprotic solvent. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.11%3A_The_Hammond_Postulate_and_Transition_.txt |
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for E2 reactions
• predict the products and specify the reagents for E2 reactions with stereochemistry
• propose mechanisms for E2 reactions
• draw and interpret Reaction Energy Diagrams for E2 reactions
In order of decreasing importance, the factors impacting E2 reaction pathways are
1) structure of the alkyl halide
2) strength of the base
3) stability of the leaving group
4) type of solvent.
The bimolecular transition state of the E2 pathway means that orientation of the base and leaving group are a primary consideration. Both the base and leaving group are electron rich and electrostatically repel each other forcing an anti-coplanar orientation between the base and leaving group. The structure of the alkyl halide must assume the orientation for an anti-coplar transition state. The strength of the base will also influence the reaction along with the stability of the leaving group. Solvents play a very minor role in E2 pathway.
Introduction
E2 reactions are typically seen with secondary and tertiary alkyl halides, but a hindered base is necessary with a primary halide. The mechanism by which it occurs is a single step concerted reaction with one transition state. The rate at which this mechanism occurs is second order kinetics, and depends on both the base and alkyl halide. A good leaving group is required because it is involved in the rate determining step. The leaving groups must be coplanar in order to form a pi bond; carbons go from sp3 to sp2 hybridization states.
To get a clearer picture of the interplay of these factors involved in a a reaction between a nucleophile/base and an alkyl halide, consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In one pathway, a methanethiolate nucleophile substitutes for bromine in an SN2 reaction. In the other (bottom) pathway, methoxide ion acts as a base (rather than as a nucleophile) in an elimination reaction. As we will soon see, the mechanism of this reaction is single-step, and is referred to as the E2 mechanism.
General Reaction
Below is a mechanistic diagram of an elimination reaction by the E2 pathway:
.
In this reaction, ethoxide (CH3CH2O-) represents the base and Br representents a leaving group, typically a halogen. There is one transition state that shows the concerted reaction for the base attracting the hydrogen and the halogen taking the electrons from the bond. The product be both eclipse and staggered depending on the transition states. Eclipsed products have a synperiplanar transition states, while staggered products have anticoplanar (antiperiplanar) transition states. Staggered conformation is usually the major product because of its lower energy confirmation.
An E2 reaction has certain requirements to proceed:
• A strong base is necessary especially necessary for primary alkyl halides. Secondary and tertirary primary halides will procede with E2 in the presence of a base (OH-, RO-, R2N-)
• Both leaving groups should be on the same plane, this allows the double bond to form in the reaction. In the reaction above you can see both leaving groups are in the plane of the carbons.
• Follows Zaitsev's rule, the most substituted alkene is usually the major product.
• Hoffman Rule, a strically hindered base will result in the least substituted product.
E2 Reaction Coordinate
In the reaction energy diagram below, the base is represented as Ba-. The bimolecular transition state determines the overall reaction rate. It is important to note the anti-coplanar orientation of the base and the leaving group. Both the base and leaving group are electron rich and electrostatically repel each other forcing an anti-coplanar orientation between the base and leaving group.
anti-coplanar transition state
The Leaving Group Effect in E2 Reactions
As Size Increases, The Electron Density Decrease, The Ability of the Leaving Group to Leave Increases: Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms.
Exercise
1. Ignoring the alkene stereochemistry show the elimination product(s) of the following compounds.
Answer
1.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• Layne A. Morsch (University of Illinois Springfield)
7.14: Zaitsev's Rule
Learning Objective
• use Zaitsev’s rule to predict major and minor products of elimination reactions
Zaisev's Rule and Regioselectivity
The prefix "regio" indicates the interaction of reactants during bond making and/or bond breaking occurs preferentially by one orientation. Because the beta-carbons of an alkyl halide may not be equivalent, there can be more than one possible elimination product. Zaitsev's Rule can be used to predict the regiochemistry of elimination reactions.
Zaitsev’s or Saytzev’s (anglicized spelling) rule is an empirical rule used to predict regioselectivity of beta-elimination reactions occurring via the E1 or E2 mechanisms. It states that in a regioselective E1 or E2 reaction the major product is the more stable alkene with the more highly substituted double bond as shown in the example below.
If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below.
By using the strongly basic hydroxide nucleophile, we direct these reactions toward elimination. In both cases there are two different sets of beta-hydrogens available to the elimination reaction (these are colored red and magenta and the alpha carbon is blue). If the rate of each possible elimination was the same, we might expect the amounts of the isomeric elimination products to reflect the number of hydrogens that could participate in that reaction. For example, since there are three 1º-hydrogens (red) and two 2º-hydrogens (magenta) on beta-carbons in 2-bromobutane, statistics would suggest a 3:2 ratio of 1-butene and 2-butene in the products. This is not observed, and the latter predominates by 4:1. This departure from statistical expectation is even more pronounced in the second example, where there are six 1º-beta-hydrogens compared with one 3º-hydrogen. These results point to a strong regioselectivity favoring the more highly substituted product double bond, an empirical statement generally called the Zaitsev Rule.
Exercise
1. Ignoring the alkene stereochemistry show the elimination product(s) of the following compounds:
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.13%3A__Characteristics_of_the_E2_Reaction.txt |
Learning Objective
• determine the rate law & predict the mechanism based on its rate equation or reaction data for E1 reactions
• predict the products and specify the reagents for E1 reactions with stereochemistry
• propose mechanisms for E1 reactions
• draw and interpret Reaction Energy Diagrams for E1 reactions
General Reaction
Unimolecular Elimination (E1) is a reaction in which loss of the leaving group followed by removal of he beta-hydrogen results in the formation of a double bond.
It is similar to a unimolecular nucleophilic substitution reaction (SN1) in various ways. One being the formation of a carbocation intermediate as the rate determining (slow) step, hence the name unimolecular. . Alkyl halides that can ionize to form stable carbocations are more reactive via the E1 mechanism. Because carbocation stability is the primary energetic consideration, stabilization of the carbocation via solvation is also an important consideration. Because carbocations are highly reactive, the strength of the base is not important and weak bases can be used. Since SN1 and E1 reactions behave similarly, they often compete against each other. Many times, both these reactions will occur simultaneously to form different products from a single reaction. However, one can be favored over another through thermodynamic control. Heating the reaction favors elimination over substitution.
In order of decreasing importance, the factors impacting E1 reaction pathways are
1) structure of the alkyl halide
2) stability of the carbocation
3) type of solvent
4) strength of the base.
Mechanism for Alkyl Halides
As can be seen in the E1 mechanism below, the preliminary step is the leaving group (LG) leaving on its own. Because it takes the electrons in the bond along with it, the carbon that was attached to it loses its electron, making it a carbocation. Once it becomes a carbocation, a Lewis Base (B-) deprotonates the intermediate carbocation at the beta position, which then donates its electrons to the neighboring C-C bond to form a double bond. Unlike E2 reactions, which require the proton to be anti to the leaving group, E1 reactions only require a neighboring hydrogen. This is due to the fact that the leaving group has already left the molecule. The final product is an alkene along with the HB byproduct and leaving group salt.
Once again, we see the two steps of the E1 mechanism.
1. A base deprotonates a beta carbon to form a pi bond.
In this case we see a mixture of products rather than one discrete one. This is the case because the carbocation has two nearby carbons that are capable of being deprotonated, but that only one forms a major product (more stable).
Reactivity
Due to the fact that E1 reactions create a carbocation intermediate, rules present in \(S_N1\) reactions still apply.
As expected, tertiary carbocations are favored over secondary, primary and methyl’s. This is due to the phenomena of hyperconjugation, which essentially allows a nearby C-C or C-H bond to interact with the p orbital of the carbon to bring the electrons down to a lower energy state. Thus, this has a stabilizing effect on the molecule as a whole. In general, primary and methyl carbocations do not proceed through the E1 pathway for this reason, unless there is a means of carbocation rearrangement to move the positive charge to a nearby carbon. Secondary and Tertiary carbons form more stable carbocations, thus this formation occurs quite rapidly.
Secondary carbocations can be subject to the E2 reaction pathway, but this generally occurs in the presence of a good / strong base. Adding a weak base to the reaction disfavors E2, essentially pushing towards the E1 pathway. In many instances, solvolysis occurs rather than using a base to deprotonate. This means heat is added to the solution, and the solvent itself deprotonates a hydrogen. The medium can effect the pathway of the reaction as well. Polar protic solvents may be used to hinder nucleophiles, thus disfavoring E2 / Sn2 from occurring.
Regiochemistry & Stereochemistry of the E1 Reaction
The E1 reaction is regiospecific because it follows Zaitsev's rule that states the more substituted alkene is the major product. This infers that the hydrogen on the most substituted carbon is the most probable to be deprotonated, thus allowing for the most substituted alkene to be formed.
Unlike E2 reactions, the E1 reaction is not stereospecific. Thus, a hydrogen is not required to be anti-coplanar to the leaving group because the leaving group is gone. In the mechanism below, we can see two possible pathways for the reaction. Either one leads to a plausible resultant product, however, only one forms a major product. As stated by Zaitsev's rule, deprotonation of the most substituted carbon results in the most substituted alkene. This then becomes the most stable product due to hyperconjugation, and is also more common than the minor product.
Exercises
1. Which of these steps is the rate determining step (A or B)?
What is the major product formed (C or D)?
2. In order to produce the most stable alkene product, from which carbon should the base deprotonate (A, B, or C)?
If the carbocation were to rearrange, on which carbon would the positive charge go onto without sacrificing stability (A, B, or C)?
3) Predict the major product of the following reaction.
4) (True or False) – There is no way of controlling the product ratio of E1 / Sn1 reactions.
5) Explain why the presence of a weak base / nucleophile favors E1 reactions over E2.
Answer
1. A , C
2. B, B
3.
4. False - They can be thermodynamically controlled to favor a certain product over another.
5. By definition, an E1 reaction is a Unimolecular Elimination reaction. This means the only rate determining step is that of the dissociation of the leaving group to form a carbocation. Since E2 is bimolecular and the nucleophilic attack is part of the rate determining step, a weak base/nucleophile disfavors it and ultimately allows E1 to dominate. (Don't forget about SN1 which still pertains to this reaction simaltaneously).
Outside Sources
1. McMurry, J., Simanek, E. Fundamentals of Organic Chemistry, 6th edition. Cengage Learning, 2007.
Contributors
• Satish Balasubramanian | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.15%3A_Characteristics_of_the_E1_Reaction.txt |
Learning Objective
• use Zaitsev’s rule to predict major and minor products of elimination reactions including halocyclohexanes
Cyclohexane Conformation & Anti-coplanar Orientation
The concerted mechanism of the E2 reaction requires that the base and leaving group are orientated anti-coplanar to each other. When the beta-hydrogen and leaving group (halide for this chapter) are located on a 6-membered ring, then Zaitsev's Rule may not be followed. The beta-hydrogen and leaving group must both be in the axial position for the E2 reaction to occur. Consequently, E2 reactions of certain cycloalkyl halides show unusual rates and regioselectivity that are not explained by the principles thus far discussed. For example, trans-2-methyl-1-chlorocyclohexane reacts with alcoholic KOH at a much slower rate than does its cis-isomer. Furthermore, the product from elimination of the trans-isomer is 3-methylcyclohexene (not predicted by the Zaitsev rule), whereas the cis-isomer gives the predicted 1-methylcyclohexene as the chief product. These differences are described by the first two equations in the following diagram.
Unlike open chain structures, cyclic compounds generally restrict the spatial orientation of ring substituents to relatively few arrangements. Consequently, reactions conducted on such substrates often provide us with information about the preferred orientation of reactant species in the transition state. Stereoisomers are particularly suitable in this respect, so the results shown here contain important information about the E2 transition state.
The most sensible interpretation of the elimination reactions of 2- and 4-substituted halocyclohexanes is that this reaction prefers an anti orientation of the halogen and the beta-hydrogen which is attacked by the base. These anti orientations are colored in red in the above equations. The compounds used here all have six-membered rings, so the anti orientation of groups requires that they assume a diaxial conformation. The observed differences in rate are the result of a steric preference for equatorial orientation of large substituents, which reduces the effective concentration of conformers having an axial halogen. In the case of the 1-bromo-4-tert-butylcyclohexane isomers, the tert-butyl group is so large that it will always assume an equatorial orientation, leaving the bromine to be axial in the cis-isomer and equatorial in the trans. Because of symmetry, the two axial beta-hydrogens in the cis-isomer react equally with base, resulting in rapid elimination to the same alkene (actually a racemic mixture). This reflects the fixed anti orientation of these hydrogens to the chlorine atom. To assume a conformation having an axial bromine the trans-isomer must tolerate serious crowding distortions. Such conformers are therefore present in extremely low concentration, and the rate of elimination is very slow. Indeed, substitution by hydroxide anion predominates.
A similar analysis of the 1-chloro-2-methylcyclohexane isomers explains both the rate and regioselectivity differences. Both the chlorine and methyl groups may assume an equatorial orientation in a chair conformation of the trans-isomer, as shown in the top equation. The axial chlorine needed for the E2 elimination is present only in the less stable alternative chair conformer, but this structure has only one axial beta-hydrogen (colored red), and the resulting elimination gives 3-methylcyclohexene. In the cis-isomer the smaller chlorine atom assumes an axial position in the more stable chair conformation, and here there are two axial beta hydrogens. The more stable 1-methylcyclohexene is therefore the predominant product, and the overall rate of elimination is relatively fast.
An orbital drawing of the anti-transition state is shown on the right. Note that the base attacks the alkyl halide from the side opposite the halogen, just as in the SN2 mechanism. In this drawing the α and β carbon atoms are undergoing a rehybridization from sp3 to sp2 and the developing π-bond is drawn as dashed light blue lines. The symbol R represents an alkyl group or hydrogen. Since both the base and the alkyl halide are present in this transition state, the reaction is bimolecular and should exhibit second order kinetics. We should note in passing that a syn-transition state would also provide good orbital overlap for elimination, and in some cases where an anti-orientation is prohibited by structural constraints syn-elimination has been observed.
Alkyl Halide Chains
Instead, in an E2 reaction, stereochemistry of the double bond -- that is, whether the E or Z isomer results -- is dictated by the stereochemistry of the starting material, if it is diastereomeric. In other words, if the carbon with the hydrogen and the carbon with the halogen are both chiral, then one diastereomer will lead to one product, and the other diastereomer will lead to the other product.
The following reactions of potassium ethoxide with dibromostilbene (1,2-dibromo-1,2-diphenylethane) both occurred via an E2 mechanism. Two different diastereomers were used. Two different stereoisomers (E vs. Z) resulted.
Exercise
1. Which of the following compounds will react faster in an E2 reaction; trans-1-bromo-2-isopropylcyclohexane or cis-1-bromo-2-isopropylcyclohexane?
Answer
1. The cis isomer will react faster than the trans. The cis isomer has two possible perpendicular hydrogen in which it can eliminate from. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.16%3A_E2_Regiochemistry_and_Cyclohexane_Con.txt |
Learning Objective
• explain how the kinetic isotope effect (KIE) can be used to elucidate reaction mechanisms
Kinetic Isotope Effects
Kinetic Isotope Effects (KIEs) are used to determine reaction mechanisms by determining rate limiting steps and transition states and are commonly measured using NMR to detect isotope location or GC/MS to detect mass changes. In a KIE experiment an atom is replaced by its isotope and the change in rate of the reaction is observed. A very common isotope substitution is when hydrogen is replaced by deuterium. This is known as a deuterium effect and is expressed by the ratio kH/kD (as explained above). Normal KIEs for the deuterium effect are around 1 to 7 or 8. Large effects are seen because the percentage mass change between hydrogen and deuterium is great. Heavy atom isotope effects involve the substitution of carbon, oxygen, nitrogen, sulfur, and bromine, with effects that are much smaller and are usually between 1.02 and 1.10. The difference in KIE magnitude is directly related to the percentage change in mass. Large effects are seen when hydrogen is replaced with deuterium because the percentage mass change is very large (mass is being doubled) while smaller percent mass changes are present when an atom like sulfur is replaced with its isotope (increased by two mass units).
Primary KIEs
Primary kinetic isotope effects are rate changes due to isotopic substitution at a site of bond breaking in the rate determining step of a reaction.
Example
Consider the bromination of acetone: kinetic studies have been performed that show the rate of this reaction is independent of the concentration of bromine. To determine the rate determining step and mechanism of this reaction the substitution of a deuterium for a hydrogen can be made.
When hydrogen was replaced with deuterium in this reaction a $k_H \over k_D$ of 7 was found. Therefore the rate determining step is the tautomerization of acetone and involves the breaking of a C-H bond. Since the breaking of a C-H bond is involved, a substantial isotope effect is expected.
7.18: Comparison of E1 and E2 Reactions
Learning Objective
• distinguish 1st or 2nd order elimination reactions
Elimination reactions of alkyl halides can occur via the bimolecular E2 mechanism or unimolecular E1 mechanism as shown in the diagram below.
Comparing E1 and E2 mechanisms
When considering whether an elimination reaction is likely to occur via an E1 or E2 mechanism, we really need to consider three factors:
1) The base: strong bases favor the E2 mechanism, whereas, E1 mechanisms only require a weak base.
2) The solvent: good ionizing xolvents (polar protic) favor the E1 mechanism by stabilizing the carbocation intermediate.
3) The alkyl halide: primary alkyl halides have the only structure useful in distinguishing between the E2 and E1 pathways. Since primary carbocations do not form, only the E2 mechanism is possible.
Reaction Parameter E2 E1
alkyl halide structure tertiary > secondary > primary tertiary > secondary >>>> primary
nucleophile high concentration of a strong base weak base
mechanism 1-step 2-step
rate limiting step anti-coplanar bimolecular transition state carbocation formation
rate law rate = k[R-X][Base] rate = k[R-X]
stereochemisty retained configuration mixed configuration
solvent not important polar protic
Exercises
1. Predict the dominant elimination mechanism (E1 or E2) for each reaction below. Explain your reasoning.
2. Which one of the following groups of compounds would eliminate \(\ce{HCl}\) most readily on reaction with potassium hydroxide? Explain your reasoning, draw the bond-line structure and give the IUPAC name of the product.
a) \(\ce{(CH_3)_3CCl}\) \(\ce{CH_3CH_2CH_2CH_2Cl}\) \(\ce{CH_3CH(Cl)CH_2CH_3}\)
b) \(\ce{(CH_3)_3CCH_2Cl}\) \(\ce{(CH_3)_2CHCH_2Cl}\)
c)
3. Specify the reaction conditions to favor the indicated elimination mechanism.
Answer
1.
2.
3. a) strong base, such as hydroxide, an alkoxide, or equivalent
b) water or alcohol or equivalent weak base with heat
c) strong base, such as hydroxide, an alkoxide, or equivalent
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)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.17%3A_The_E2_Reaction_and_the_Deuterium_Iso.txt |
Learning Objective
• predict the products and specify the reagents for SN1, SN2, E1 and E2 reactions with stereochemistry
• propose mechanisms for SN1, SN2, E1 and E2 reactions
• draw, interpret, and apply Reaction Energy Diagrams for SN1, SN2, E1 and E2 reactions
Summary of Reaction Patterns
Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, the most important being the structure of the alkyl group and the nature of the nucleophilic reactant.In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable. Strong nucleophile favor substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor elimination.
The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both SN2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).
The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. Note that halogens bonded to sp2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents.
Nucleophile
Anionic Nucleophiles
( Weak Bases: I, Br, SCN, N3,
CH3CO2 , RS, CN etc. )
pKa's from -9 to 10 (left to right)
Anionic Nucleophiles
( Strong Bases: HO, RO )
pKa's > 15
Neutral Nucleophiles
( H2O, ROH, RSH, R3N )
pKa's ranging from -2 to 11
Alkyl Group
Primary
RCH2
Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g.
ClCH2CH2Cl + KOH ——> CH2=CHCl
SN2 substitution. (N ≈ S >>O)
Secondary
R2CH–
SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O)
In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly.
Tertiary
R3C–
E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed.
Allyl
H2C=CHCH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Benzyl
C6H5CH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Experimental Observations
Experimental observations are reported for the following reactions. These reactions include a range of alkyl halide structures in a variety of reaction conditions to illustrate the reaction patterns summarized above. In describing these, it is useful to designate the halogen-bearing carbon as alpha and the carbon atom(s) adjacent to it as beta, as noted in the first four equations shown below. Replacement or substitution of the halogen on the α-carbon (colored maroon) by a nucleophilic reagent is a commonly observed reaction, as shown in equations 1, 2, 5, 6 & 7 below. Also, since the electrophilic character introduced by the halogen extends to the β-carbons, and since nucleophiles are also bases, the possibility of base induced H-X elimination must also be considered, as illustrated by equation 3. Finally, there are some combinations of alkyl halides and nucleophiles that fail to show any reaction over a 24 hour period, such as the example in equation 4. For consistency, alkyl bromides have been used in these examples. Similar reactions occur when alkyl chlorides or iodides are used, but the speed of the reactions and the exact distribution of products will change.
In order to understand why some combinations of alkyl halides and nucleophiles give a substitution reaction, whereas other combinations give elimination, and still others give no observable reaction, we must investigate systematically the way in which changes in reaction variables perturb the course of the reaction. The following general equation summarizes the factors that will be important in such an investigation.
Friendly Reminder: One conclusion, relating the structure of the R-group to possible products, should be immediately obvious. If R- has no beta-hydrogens an elimination reaction is not possible, unless a structural rearrangement occurs first. The first four halides shown on the left below do not give elimination reactions on treatment with base, because they have no β-hydrogens. The two halides on the right do not normally undergo such reactions because the potential elimination products have highly strained double or triple bonds. It is also worth noting that sp2 hybridized C–X compounds, such as the three on the right, do not normally undergo nucleophilic substitution reactions, unless other functional groups perturb the double bond(s).
Exercise
1. Identify the dominant reaction mechanism (SN1, SN2, E1, or E2) and predict the major product for the following reactions.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.19%3A_Comparing_Substitution_and_Eliminatio.txt |
Objective
• discuss the importance of leaving groups in biological substitution reactions
Leaving Groups in Biochemical Reactions
In biological reactions, we do not often see halides serving as leaving groups (in fact, outside of some marine organisms, halogens are fairly unusual in biological molecules). More common leaving groups in biochemical reactions are phosphates, water, alcohols, and thiols. In many cases, the leaving group is protonated by an acidic group on the enzyme as bond-breaking occurs. For example, hydroxide ion itself seldom acts as a leaving group – it is simply too high in energy (too basic). Rather, the hydroxide oxygen is generally protonated by an enzymatic acid before or during the bond-breaking event, resulting in a (very stable) water leaving group.
More often, however, the hydroxyl group of an alcohol is first converted enzymatically to a phosphate ester in order to create a better leaving group. This phosphate ester can take the form of a simple monophosphate (arrow 1 in the figure below), a diphosphate (arrow 2), or a nucleotide monophosphate (arrow 3).
Due to resonance delocalization of the developing negative charge, phosphates are excellent leaving groups.
Here’s a specific example (from DNA nucleotide biosynthesis):
Here, the OH group on ribofuranose is converted to a diphosphate, a much better leaving group. Ammonia is the nucleophile in the second step of this SN1-like reaction.
What is important for now is that in each case, an alcohol has been converted into a much better leaving group, and is now primed for a nucleophilic substitution reaction.
SAM Methyltransferases
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).
Example
Contributors | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.20%3A_Biological_Substitution_Reactions.txt |
Learning Objective
• discuss enzymatic elimination reactions of histidine
Enzymatic E1 and E2 reactions
While most biochemical b-elimination reactions are of the E1cb type, some enzymatic E2 and E1 reactions are known. Like the enzymatic SN2 and SN1 substitution mechanisms discussed in chapters 8 and 9, the E2 and E1 models represent two possible mechanistic extremes, and actual enzymatic elimination reactions may fall somewhere in between. In an E1/E2 hybrid elimination, for example, Cβ-X bond cleavage may be quite advanced (but not complete) before proton abstraction takes place - this would lead to the build-up of transient partial positive charge on Cβ, but a discreet carbocation intermediate would not form. The extent to which partial positive charge builds up determines whether we refer to the mechanism as 'E1-like' or 'E2-like'.
A reaction in the histidine biosynthetic pathway provides a good example of a biological E1-like elimination step (we're looking specifically here at the first, enol-forming step in the reaction below - the second step is simply a tautomerization from the enol to the ketone product.
Notice in this mechanism that an E1cb elimination is not possible - there is no electron-withdrawing group (like a carbonyl) to stabilize the carbanion intermediate that would form if the proton were abstracted first. There is, however, an electron-donating group (the lone pair on a nitrogen) that can stabilize a positively-charged intermediate that forms when the water leaves.
Another good example of a biological E1-like reaction is the elimination of phosphate in the formation of 5-enolpyruvylshikimate-3-phosphate (EPSP), an intermediate in the synthesis of aromatic amino acids.
Experimental evidence indicates that significant positive charge probably builds up on Cβ of the starting compound, implying that C-O bond cleavage is advanced before proton abstraction occurs (notice the parallels to the Cope elimination in the previous section):
The very next step in the aromatic acid biosynthesis pathway is also an elimination, this time a 1,6-conjugated elimination rather than a simple beta-elimination.
An E1-like mechanism (as illustrated above) has been proposed for this step, but other evidence suggests that a free-radical mechanism may be involved.
While most E1 and E2 reactions involve proton abstraction, eliminations can also incorporate a decarboxylation step.
Isopentenyl diphosphate, the 'building block' for all isoprenoid compounds, is formed from a decarboxylation-elimination reaction.
Phenylpyruvate, a precursor in the biosynthesis of phenylalanine, results from a conjugated 1,6 decarboxylation-elimination.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
7.22: Additional Exercises
SN2
7-1 Predict which compound in each pair would undergo the SN2 reaction faster.
a) or
b) or
c) or
d) or
7-2 Predict the products of these nucleophilic substitution reactions, including stereochemistry when appropriate.
a)
b)
c)
d)
7-3 Show how each compound might be synthesized using SN2 reaction.
a)
b)
c)
7-4 Show 2 ways to synthesize the ether below using SN2 reaction
7-5 Of the two ways of synthesizing the compound in previous question (7-4), which one would be the most efficient? Why? Show the mechanism of the reaction as part of your explanation.
7-6 Arrange the compounds below in increasing order of nucleophilicity.
H2O; NH2-; CH3O-; CH3CH2O-
SN1
7-7 List the following carbocations in order of increasing stability.
7-8 Give the solvolysis product expected when the compound is heated in methanol.
a)
b)
c)
d)
7-9 Predict with compound in each pair will undergo an SN1 reaction more quickly.
a) or
b) or
c) or
d) or
7-10 Show how the following carbocations would rearrange to become more stable. Draw the mechanism of the rearrangement.
a)
b)
c)
d)
SN2 vs SN1
7-11 Predict whether each compound below would be more likely to undergo a SN2 or SN1 reaction.
a)
b)
c)
d)
7-12 Predict the product of the following reactions.
a)
b)
c)
d)
7-13 Show how each compound may be synthesized using nucleophilic substitution reactions.
a)
b)
c)
d)
e)
f)
g)
E2 vs E1
7-14 Predict the major products of the following reactions.
a)
b)
c)
7-15 Draw the expected major product when each of the following compounds is treated with hydroxide to give an E2 reaction.
a)
b)
c)
7-16 Predict all the elimination products of the following reactions and label the major product.
a)
b)
c)
d)
Substitution vs Elimination
7-17 Identify the function of the following reagents. The reagents will be a strong/weak nucleophile and/or a strong/weak base.
a) Cl-
b) NaH
c) t-BuO-
d) OH-
e) H2O
f) HS-
g) MeOH
7-18 Identify which mechanism the following reactions would undergo.
a)
b)
c)
d)
e)
7-19 Identify all the products of the following reactions and specify the major product.
a)
b)
c)
d)
e)
7-20 The following reaction yields five different products. Give the mechanisms for how each is formed.
7.23: Solutions to Additional Exercises
SN2
7-1
a)
b)
c)
d)
7-2
a)
b)
c) No reaction
d)
7-3
a)
b)
c)
7-4
First method:
Second method:
7-5 The second method is more efficient since the alkyl halide is not sterically hindered
7-6 H2O < NH2- < CH3CH2O-< CH3O-
7-7
7-8
a)
b)
c) No reaction
d)
7-9
a)
b)
c)
d)
7-10
a)
b)
c)
d)
7-11
a) SN2
b) SN2
c) SN1
d) SN1
7-12
a)
b)
c)
d)
7-13
a)
b)
c)
d)
e)
f)
g)
7-14
a)
b)
c)
7-15
a)
b)
c)
7-16
a)
b)
c)
d)
Substitution vs Elimination
7-17
a) Cl- ; strong nucleophile
b) NaH ; strong base
c) t-BuO- ; strong base
d) OH- ; strong nucleophile ; strong base
e) H2O ; weak nucleophile ; weak base
f) HS- ; strong nucleophile
g) MeOH ; weak nucleophile ; weak base
7-18
a) E2, SN1
b) SN2, E2
c) SN2
d) SN1, E1
e) E2, SN1
7-19
a)
b)
c)
d)
e)
7-20 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/07%3A_Alkyl_Halides-_Nucleophilic_Substitution_and_Elimination/7.21%3A_Biological_Elimination_Reactions.txt |
Learning Objectives:
After reading the chapter and completing the exercises and homework, a student can be able to:
• describe the electronic structure of alkenes using Molecular Orbital (MO) Theory and Orbital Hybridization - refer to section 8.1
• memorize the common names for vinylic and allylic groups including isoprene and styrene refer to section 8.2
• predict the relative physical properties of alkenes - refer to section 8.2
• recognize and classify the stereochemistry of alkenes using the cis/trans and E/Z systems - refer to section 8.3
• calculate the Degrees of Unsaturation (DU) and apply it to alkene structure - refer to section 8.4
• give the IUPAC names for alkenes given their structure & vice versa including E/Z isomers - refer to section 8.5 and chapter 3
• use heats of hydrogenation to compare the stabilities of alkenes - refer to section 8.6
• interpret and draw reaction energy diagrams for dehydrohalogenation of R-X’s and alcohol dehydration reactions - refer to sections 8.7 and 8.8 respectively and chapter 7
• propose mechanisms for a dehydrohalogenation or dehydration reactions - refer to sections 8.7 and 8.8 respectively and chapter 7
• predict the products and specify the reagents for alkene synthesis from dehydrohalogenation of R-X’s and alcohol dehydration reactions - refer to sections 8.7 and 8.8 respectively
• predict and explain the stereochemistry of E2 eliminations to form alkenes, especially from cyclohexanes - refer to sections 8.7 and 8.8 and chapter 7
• discuss the uses and sources of alkenes including catalytic cracking - refer to section 8.9
• 8.1: Alkene Structure
Alkenes are a class of hydrocarbons (i.e., containing only carbon and hydrogen). They are unsaturated compounds with at least one carbon-to-carbon double bond.
• 8.2: Physical Properties and Important Common Names
Alkenes are non-polar hydrocarbons with physical properties similar to alkanes. At room temperature, alkenes exist in all three phases, solid, liquids, and gases. The stereochemistry of the geometric isomers (cis/trans) can influence the physical properties.
• 8.3: The Alkene Double Bond and Stereoisomerism
The two lobes of the pi bond in the alkenes prevent rotation and are responsible for their rigid nature. The lack of rotation creates the potential for geometric isomers (cis/trans).
• 8.4: Degrees of Unsaturation
Calculating the degrees of unsaturation (DU) can provide useful information about the chemical structure from the molecular formula. The DU indicates the presence of rings and π bonds, but cannot distinguish between them.
• 8.5: The E/Z System (when cis/trans does not work)
Some alkenes cannot be unambiguously named using the cis/trans system. The Cahn-Ingold-Prelog (CIP) rules were used to develop the E/Z system for naming the stereoisomers of alkenes.
• 8.6: Stability of Alkenes
The energy released during alkene hydrogenation is called the heat of hydrogenation and indicates the relative stability of the double bond in the molecule.
• 8.7: Alkene Synthesis by Elimination of Alkyl Halides
The alkyl halide elimination reactions (E1 and E2) to synthesize alkenes are briefly reviewed. Refer to chapter 7 sections 13 through 18 for a complete explanation.
• 8.8: Alkene Synthesis by Dehydration of Alcohols
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.
• 8.9: Uses and Sources of Alkenes
Among the most important and most abundant organic chemicals produced worldwide are the two simple alkenes, ethylene and propylene. Thermal cracking is briefly explained.
• 8.10: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 8.11: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
08: Structure and Synthesis of Alkenes
Learning Objective
• describe the electronic structure of alkenes using Molecular Orbital (MO) Theory and Orbital Hybridization
Alkenes are a class of hydrocarbons (i.e., containing only carbon and hydrogen). They are unsaturated compounds with at least one carbon-to-carbon double bond. The double bond makes alkenes more reactive than alkanes. Olefin is another term used to describe alkenes. The alkene group can also be called a vinyl group and the carbons sharing the double bond can be called vinyl carbons.
Alkenes
Nomenclature of alkenes is covered in chapter 3. Condensed structural formulas for the first eight alkenes are listed in Table $1$ along with some relevant physical propoerties. Thus, CH2=CH2 stands for
The double bond is shared by the two carbon atoms and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C2H4, whereas that for ethane is C2H6.
Structure of Ethene - the simplest alkene
Ethene is not a very complicated molecule. It is made up of four 1s1 hydrogen atoms and two 2s2 2$p_x$1 2$p_y$2 carbon atoms. These carbon atoms already have four electrons, but they each want to get four more so that they have a full eight in the valence shell. Having eight valence electrons around carbon gives the atom itself the same electron configuration as neon, a noble gas. Carbon wants to have the same configuration as Neon because when it has eight valence electrons carbon is at its most stable, lowest energy state, it has all of the electrons that it wants, so it is no longer reactive.
This forms a total of three bonds to each carbon atom, giving them an $sp^2$ hybridization. Since the carbon atom is forming three sigma bonds instead of the four that it can, it only needs to hybridize three of its outer orbitals, instead of four. It does this by using the $2s$ electron and two of the $2p$ electrons, leaving the other unchanged. This new orbital is called an $sp^2$ hybrid because that's exactly what it is, it is made from one s orbital and two p orbitals.
When atoms are an $sp^2$ hybrid they have a trigonal planar structure. These structures are very similar to a 'peace' sign, there is a central atom with three atoms around it, all on one plane. Trigonal planar molecules have an ideal bond angle of 120° on each side.
The H-C-H bond angle is 117°, which is very close to the ideal 120° of a carbon with $sp^2$ hybridization. The other two angles (H-C=C) are both 121.5°.
Rigidity in Ethene
There is rigidity in the ethene molecule due to the double-bonded carbons. A double bond consists of one sigma bond formed by overlap of sp2 hybrid orbitals and one pi bond formed by overlap of parallel 2 p orbitals. In ethene there is no free rotation about the carbon-carbon sigma bond because these two carbons also share a $\pi$ bond. A $\pi$ bond is only formed when there is adequate overlap between both top and bottom p-orbitals. Free rotation the p-orbitals cause them to be 90° from each other breaking the $\pi$ bond because there would be no overlap. Since the $\pi$ bond is essential to the structure of ethene it must not break, so there can be not free rotation about the carbon-carbon sigma bond. The two carbon atoms of a double bond and the four atoms attached to them lie in a plane, with bond angles of approximately 120° as shownn in the figure below
(a) The σ-bonded framework is formed by the overlap of two sets of singly occupied carbon sp2 hybrid orbitals and four singly occupied hydrogen 1s orbitals to form electron-pair bonds. This uses 10 of the 12 valence electrons to form a total of five σ bonds (four C–H bonds and one C–C bond).
(b) One singly occupied unhybridized 2pz orbital remains on each carbon atom to form a carbon–carbon π bond. (Note: by convention, in planar molecules the axis perpendicular to the molecular plane is the z-axis.)
The first two alkenes in Table $1$, ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure $1$). Ethylene is a major commercial chemical. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products.
Exercise
1. Although there is only one alkene with the formula C2H4 (ethene) and only one with the formula C3H6 (propene), there are several alkenes with the formula C4H8. Draw all of the possible bond line structures for alkenes with the formula C4H8 including all possible structural and stereoisomers.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.01%3A_Alkene_Structure.txt |
Learning Objectives
• memorize the common names for vinylic and allylic groups including isoprene and styrene
• predict the relative physical properties of alkenes
Common names
The carbon atoms sharing the double bond can be referred to as the "vinyl carbons". The carbon atoms adjacent to the vinyl carbon atoms are called "allylic carbons". These carbon atoms have unique reactivity because of the potential for interaction with the pi bond.
Overall, common names remove the -ane suffix and add -ylene. There are a couple of unique ones like ethenyl's common name is vinyl and 2-propenyl's common name is allyl that need to be memorized.
• vinyl substituent H2C=CH-
• allyl substituent H2C=CH-CH2-
• allene molecule H2C=C=CH2
• isoprene is shown below
Physical Properties of Selected Alkenes
Some representative alkenes—their names, structures, and physical properties—are given in the table below.
Physical Properties of Some Selected Alkenes
IUPAC Name Molecular Formula Condensed Structural Formula Melting Point (°C) Boiling Point (°C)
ethene C2H4 CH2=CH2 –169 –104
propene C3H6 CH2=CHCH3 –185 –47
1-butene C4H8 CH2=CHCH2CH3 –185 –6
1-pentene C5H10 CH2=CH(CH2)2CH3 –138 30
1-hexene C6H12 CH2=CH(CH2)3CH3 –140 63
1-heptene C7H14 CH2=CH(CH2)4CH3 –119 94
1-octene C8H16 CH2=CH(CH2)5CH3 –102 121
Polarity and Physical Properties
Alkenes are non-polar hydrocarbons. The dominant intermolecular forces shared by alkenes are the London dispersion forces. These interactions are weak and temporary, so they are easily disrupted.
Physical States: The physical states reflect the weak attractive forces between molecules. Ethene, propene, and butene exist as colorless gases. Alkenes with 5 to 14 carbons are liquids, and alkenes with 15 carbons or more are solids.
Density: Alkenes are less dense than water with most densities in the range of 0.6 to 0.7 g/mL. Alkenes float on top of water.
Solubility: Alkenes are virtually insoluble in water, but dissolve in organic solvents. The reasons for this are exactly the same as for the alkanes.
Boiling Points: The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Boiling points of alkenes depend on more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules. In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains.
Compound Boiling points (oC)
Ethene -104
Propene -47
Trans-2-Butene 0.9
Cis-2-butene 3.7
Trans 1,2-dichlorobutene 155
Cis 1,2-dichlorobutene 152
1-Pentene 30
Trans-2-Pentene 36
Cis-2-Pentene 37
1-Heptene 115
3-Octene 122
3-Nonene 147
5-Decene 170
Melting Points: Melting points of alkenes depends on the packaging of the molecules so the stereochemistry of the carbon-carbon double bond has a strong influence on the relative melting points. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers. This effect is notable when comparing the melting points of fats and oils. The differences in the melting points is strongly influenced by the long hydrocarbon tails. Oils have a greater number of cis double bonds and exist as liquids at room temperature. Whereas, fats are primarily saturated and exist as solids at room temperature.
Compound Melting Points (0C)
Ethene -169
Propene -185
Butene -138
1-Pentene -165
Trans-2-Pentene -135
Cis-2-Pentene -180
1-Heptene -119
3-Octene -101.9
3-Nonene -81.4
5-Decene -66.3
Exercise
1. Draw the bond-line structures for the following compounds in order of increasing boiling point: 2-methyl-2-pentene; 2-hexene; isoprene; 2-heptene.
2. Which phase will contain the most 3-octene?
a) water or hexane
b) water or benzene
c) methanol or 1-octanol
Answer
1. relative boiling points
2. a) hexane (Hydrocarbons are hydrophobic and lipophilic.)
b) benzene (Hydrocarbons are hydrophobic and lipophilic.)
c) 1-octanol (Hydrocarbons seek the solvent with the most carbons and fewest polar groups.)
Contributors
• Trung Nguyen
• Jim Clark (Chemguide.co.uk)
• Layne A. Morsch (University of Illinois Springfield) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.02%3A_Physical_Properties_and_Important_Common_Names.txt |
Learning Objective
• recognize and classify the stereochemistry of alkenes using the cis/trans system
Steroisomerism in Alkenes
There is free rotation about the carbon-to-carbon single bonds (C–C) in alkanes. For example, in 1,2-dichloroethane, there is free rotation about the C–C bond. The two models shown represent exactly the same molecule; they are not isomers. You can draw structural formulas that look different, but if you bear in mind the possibility of this free rotation about single bonds, you should recognize that these two structures represent the same molecule:
In contrast, the structure of alkenes requires that the carbon atoms of a double bond and the two atoms bonded to each carbon atom all lie in a single plane, and that each doubly bonded carbon atom lies in the center of a triangle. This part of the molecule’s structure is rigid; rotation about doubly bonded carbon atoms is not possible without rupturing the bond. In 1,2-dichloroethene, restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond become significant. This leads to a special kind of isomerism. The isomer in which the two chlorine (Cl) atoms lie on the same side of the molecule is called the cis isomer (Latin cis, meaning “on this side”) and is named cis-1,2-dichloroethene. The isomer with the two Cl atoms on opposite sides of the molecule is the trans isomer (Latin trans, meaning “across”) and is named trans-1,2-dichloroethene. These two compounds are cis-trans isomers (or geometric isomers), compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule.
The diagram below summarizes the differences between alkanes and alkenes with respect to rotation where the carbons are red, hydrogens are off-white, and the chlorines are green.
In 1,2-dichloroethane (a), free rotation about the C–C bond allows the two structures to be interconverted by a twist of one end relative to the other. In 1,2-dichloroethene (b), restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond are significant.
Geometric Isomers have Different Physical Properties
Consider the alkene with the condensed structural formula CH3CH=CHCH3. We could name it 2-butene, but there are actually two such compounds; the double bond results in cis-trans isomerism. Cis-2-butene has both methyl groups on the same side of the molecule. Trans-2-butene has the methyl groups on opposite sides of the molecule as shown in the diagram below.
Cis-2-butene and trans-2-butene are unique compounds with slightly different physical properties as shown below. For stereospecific reactions, these compounds produce different stereoisomeric products under the same reaction conditions. This phenonenom of reactivity will be explored more closely in the next chapter on alkene reactivity.
Example
Which compounds can exist as cis-trans (geometric) isomers? Draw them.
1. CHCl=CHBr
2. CH2=CBrCH3
3. (CH3)2C=CHCH2CH3
4. CH3CH=CHCH2CH3
Solutions Explained
All four structures have a double bond and thus meet rule 1 for cis-trans isomerism.
1. This compound meets rule 2; it has two nonidentical groups on each carbon atom (H and Cl on one and H and Br on the other). It exists as both cis and trans isomers:
2. This compound has two hydrogen atoms on one of its doubly bonded carbon atoms; it fails rule 2 and does not exist as cis and trans isomers.
3. This compound has two methyl (CH3) groups on one of its doubly bonded carbon atoms. It fails rule 2 and does not exist as cis and trans isomers.
4. This compound meets rule 2; it has two nonidentical groups on each carbon atom and exists as both cis and trans isomers:
Exercise
Exercise
1. If the compound below can exist as cis-trans isomers, then draw both bond-line structures.
1. CH2=CHCH2CH2CH3
2. CH3CH=CHCH2CH3
3. CH3CH2CH=CHCH2CH3
4. CH2C(CH3)CH2CH3
5. CH3C(CH3)CHCH3
2. Write out the condensed structure for ethene.
3. Draw the Kekulé (Lewis) structure for ethene.
4. Draw the bond-line structure for ethene.
5. Why is it that the carbons in ethene cannot freely rotate around the carbon-carbon double bond?
Answer
1.
2. CH2CH2 The carbon-carbon double bond is not shown, but can be recognized by knowing the neutral bonding patterns of carbon.
3.
4." - " Note how easy it would be to misinterpret this small dash, therefore, the structure for ethene is typically shown with a condensed or Kekule structure.
5. The carbons cannot freely rotate about the carbon-carbon double bond because rotating p-orbitals would have to pass through a 90° point where there would no longer be any overlap, so the $\pi$ bond would have to break for there to be free rotation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.03%3A_The_Alkene_Double_Bond_and_Stereoisomerism.txt |
Learning Objectives
• calculate the Degrees of Unsaturation (DU) and apply it to alkene structure
Saturated and Unsaturated Molecules
In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated.
CH3CH2CH3 1-methyoxypentane
Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s).
CH3CH=CHCH3 3-chloro-5-octyne
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, these techniques require expensive instrumentation and are not always readily available. Fortunately, calculating the degrees of unsaturation provides useful information about the structure. The degree of unsaturation indicates the total number of pi bonds and rings within a molecule which makes it easier for one to figure out the molecular structure.
DU = Degrees of Unsaturation = (number of pi bonds) + (number of rings)
Alkenes (R2C=CR2) and alkynes (R–C≡C–R) are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas:
Calculating The Degree of Unsaturation (DU)
If the molecular formula is given, plug in the numbers into this formula:
$DoU= \dfrac{2C+2+N-X-H}{2} \tag{7.2.1}$
• $C$ is the number of carbons
• $N$ is the number of nitrogens
• $X$ is the number of halogens (F, Cl, Br, I)
• $H$ is the number of hydrogens
The molecular formula of a hydrocarbon provides information about the possible structural types it may represent. A saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8
[2C+2=(2x3)+2=8.]
The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6. For example, consider compounds having the formula C5H8. The formula of the five-carbon alkane pentane is C5H12 so the difference in hydrogen content is 4. This difference suggests such compounds may have a triple bond, two double bonds, a ring plus a double bond, or two rings. Some examples are shown here, and there are at least fourteen others!
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
When the DU is 4 or greater, the presence of benzene rings is very likely.
DU
Possible combinations of rings/ bonds
# of rings
# of double bonds
# of triple bonds
1
1
0
0
0
1
0
2
2
0
0
0
2
0
0
0
1
1
1
0
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example: Benzene
What is the Degree of Unsaturation for Benzene?
Solution
The molecular formula for benzene is C6H6. Thus,
DU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DU of 4 and have the same molecular formula as benzene. However, these compounds are very rare, unlike benzene. We will learn more about the reasons for benzen's high stability when we studey aromaticity in later chapters.
Exercises
1. Are the following molecules saturated or unsaturated:
1. (b.) (c.) (d.) C10H6N4
2. Using the molecules from (1) above, give the degrees of unsaturation for each.
3. Calculate the degrees of unsaturation, classify the compound as saturated or unsaturated, and list all the ring/pi bond combination possible for the following molecular formulas: (a.) C9H20 (b.) C7H8 (c.) C5H7Cl (d.) C9H9NO4
4. Calculate degrees of unsaturation (DoU) for the following, and propose a structure for each.
a) C5H8
b) C4H4
5. Calculate the degree of unsaturation (DoU) for the following molecules
a) C5H5N
b) C5H5NO2
c) C5H5Br
6. The following molecule is caffeine (C8H10N4O2), determine the degrees of unsaturation (DoU).
Answer
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3.
(a.) DU =0 ; saturated (Remember-a saturated molecule only contains single bonds)
(b.) DU = 4; unsaturated The molecule can contain any of these combinations of rings and pi bonds that add up to 4, such as (i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) DU = 2; unsaturated (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) DU = 6; (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
4.
5. a) 4 b) 4 c) 3
6. DU = 6 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.04%3A_Degrees_of_Unsaturation.txt |
Learning Objective
• recognize and classify the stereochemistry of alkenes using the cis/trans and E/Z systems
• give the IUPAC names for alkenes given their structure & vice versa including E/Z isomers
E/Z Nomenclature in Alkenes
The traditional system for naming the geometric isomers of an alkene, in which the same groups are arranged differently, is to name them as cis or trans. However, it is easy to find examples where the cis-trans system is not easily applied. IUPAC has a more complete system for naming alkene isomers. The R-S system for chirality is based on a set of "priority rules", which allow you to rank any groups. The rigorous IUPAC system for naming alkene isomers, called the E-Z system, is based on the same priority rules.
The priority rules are often called the Cahn-Ingold-Prelog (CIP) rules, after the chemists who developed the system.
The general strategy of the E-Z system is to analyze the two groups at each end of the double bond. For each vinyl carbon, rank the two groups using the CIP priority rules. Determine whether the higher priority group at one end of the double bond and the higher priority group at the other end of the double bond are on the same side (Z, from German zusammen = together or "on Zee Zame Zide") or on opposite sides (E, from German entgegen = opposite) of the double bond.
Example \(1\)
The figure below shows the two isomers of 2-butene. You should recognize them as cis and trans. Let's analyze them to see whether they are E or Z. Start with the left hand structure (the cis isomer). On C2 (the left end of the double bond), the two atoms attached to the double bond are C and H. By the CIP priority rules, C is higher priority than H (higher atomic number). Now look at C3 (the right end of the double bond). Similarly, the atoms are C and H, with C being higher priority. We see that the higher priority group is "down" at C2 and "down" at C3. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is (Z)-2-butene.
Now look at the right hand structure (the trans isomer). In this case, the priority group is "down" on the left end of the double bond and "up" on the right end of the double bond. Since the two priority groups are on opposite sides of the double bond, they are entgegen = opposite. Therefore, this is (E)-2-butene.
E/Z will work -- even when cis/trans fails
In simple cases, such as 2-butene, Z corresponds to cis and E to trans. However, that is not a rule. This section and the following one illustrate some idiosyncrasies that happen when you try to compare the two systems. The real advantage of the E-Z system is that it will always work. In contrast, the cis-trans system breaks down with many ambiguous cases.
Example \(1\)
The following figure shows two isomers of an alkene with four different groups on the double bond, 1-bromo-2-chloro-2-fluoro-1-iodoethene.
It should be apparent that the two structures shown are distinct chemicals. However, it is impossible to name them as cis or trans. On the other hand, the E-Z system works fine... Consider the left hand structure. On C1 (the left end of the double bond), the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at C2. The atoms are Cl and F, with Cl being higher priority. We see that the higher priority group is "down" at C1 and "down" at C2. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the (Z) isomer. Similarly, the right hand structure is (E).
E/Z will work, but may not agree with cis/trans
There are also molecules for which the E/Z system will not agree with the cis/trans system. Let's use 2-bromo-2-butene to explore this option. Is this compound cis or trans? This molecule is clearly cis. The two methyl groups are on the same side. More rigorously, the "parent chain" is cis.
Is this compound E or Z? There is a methyl at each end of the double bond. On the left, the methyl is the high priority group because the other group is -H. On the right, the methyl is the low priority group because the other group is -Br. That is, the high priority groups are -CH3 (left) and -Br (right). Thus the two priority groups are on opposite sides = entgegen = E.
This example should convince you that cis and Z are not synonyms. Cis/trans and E,Z are determined by distinct criteria. There may seem to be a simple correspondence, but it is not a rule. Be sure to determine cis,trans or E,Z separately, as needed.
Multiple double bonds
If the compound contains more than one double bond, then each one is analyzed and declared to be E or Z.
Example \(3\)
The configuration at the left hand double bond is E; at the right hand double bond it is Z. Thus this compound is (1E,4Z)-1,5-dichloro-1,4-hexadiene.
The double-bond rule in determining priorities
Example \(4\)
Consider the compound below
This is 1-chloro-2-ethyl-1,3-butadiene -- ignoring, for the moment, the geometric isomerism. There is no geometric isomerism at the second double bond, at 3-4, because it has 2 H at its far end.
What about the first double bond, at 1-2? On the left hand end, there is H and Cl; Cl is higher priority (by atomic number). On the right hand end, there is -CH2-CH3 (an ethyl group) and -CH=CH2 (a vinyl or ethenyl group). Both of these groups have C as the first atom, so we have a tie so far and must look further. What is attached to this first C? For the ethyl group, the first C is attached to C, H, and H. For the ethenyl group, the first C is attached to a C twice, so we count it twice; therefore that C is attached to C, C, H. CCH is higher than CHH; therefore, the ethenyl group is higher priority. Since the priority groups, Cl and ethenyl, are on the same side of the double bond, this is the Z-isomer; the compound is (Z)-1-chloro-2-ethyl-1,3-butadiene.
The "first point of difference" rule
Which is higher priority, by the CIP rules: a C with an O and 2 H attached to it or a C with three C? The first C has one atom of high priority but also two atoms of low priority. How do these "balance out"? Answering this requires a clear understanding of how the ranking is done. The simple answer is that the first point of difference is what matters; the O wins.
To illustrate this, consider the molecule at the left. Is the double bond here E or Z? At the left end of the double bond, Br > H. But the right end of the double bond requires a careful analysis.
At the right hand end, the first atom attached to the double bond is a C at each position. A tie, so we look at what is attached to this first C. For the upper C, it is CCC (since the triple bond counts three times). For the lower C, it is OHH -- listed in order from high priority atom to low. OHH is higher priority than CCC, because of the first atom in the list. That is, the O of the lower group beats the C of the upper group. In other words, the O is the highest priority atom of any in this comparison; thus the O "wins".
Therefore, the high priority groups are "up" on the left end (the -Br) and "down" on the right end (the -CH2-O-CH3). This means that the isomer shown is opposite = entgegen = E. And what is the name? The "name" feature of ChemSketch says it is (2E)-2-(1-bromoethylidene)pent-3-ynyl methyl ether.
Example \(1\)
The configuration about double bonds is undoubtedly best specified by the cis-trans notation when there is no ambiguity involved. Unfortunately, many compounds cannot be described adequately by the cis-trans system. Consider, for example, configurational isomers of 1 -fluoro- 1 -chloro-2- bromo-2-iodo-ethene, 9 and 10. There is no obvious way in which the cis-trans system can
be used:
A system that is easy to use and which is based on the sequence rules already described for the R,S system works as follows:
1. An order of precedence is established for the two atoms or groups attached to each end of the double bond according to the sequence rules of Section 19-6. When these rules are applied to 1-fluoro- 1-chloro-2-bromo-2- iodoethene, the priority sequence is:
• at carbon atom 1, C1 > F
• at carbon atom 2, I > Br
1. Examination of the two configurations shows that the two priority groups- one on each end- are either on the same side of the double bond or on opposite sides:
The Z isomer is designated as the isomer in which the top priority groups are on the same side (Z is taken from the German word zusammen- together). The E isomer has these groups on opposite sides (E, German for entgegen across). Two further examples show how the nomenclature is used:
Exercises
1. Order the following in increasing priority.
A) –H, –Cl, –OH
B) –CH3, CH2OH, CH2CH3
C) –C≡CH, –CH=CH2, –CH=O
2. Label the following as E or Z conformations.
Answer
1. A) –H < OH < Cl (highest priority)
B) –CH3 < CH2CH3 < CH2OH (highest priority)
C) –CH=CH2 < C≡CH < CH=O (highest priority)
2. A) Z B) Z C) E
Contributors and Attributions
>Robert Bruner (http://bbruner.org)
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)/08%3A_Structure_and_Synthesis_of_Alkenes/8.05%3A_The_E_Z_System_%28when_cis_trans_does_not_work%29.txt |
Learning Objective
• use heats of hydrogenation to compare the stabilities of alkenes
Heats of Hydrogenation
The stability of an alkene can be determined by measuring the amount of energy associated with the hydrogenation of the molecule. Since the double bond is breaking in this reaction, the energy released in hydrogenation is proportional to the energy in the double bond of the molecule. This is a useful tool because heats of hydrogenation can be measured very accurately. The \(\Delta H^o\) is usually around -30 kcal/mol for alkenes. Stability is simply a measure of energy. Lower energy molecules are more stable than higher energy molecules. More substituted alkenes are more stable than less substituted ones due to hyperconjugation. They have a lower heat of hydrogenation. The following illustrates stability of alkenes with various substituents:
In disubstituted alkenes, trans isomers are more stable than cis isomers due to steric hindrance. Also, internal alkenes are more stable than terminal ones. See the following isomers of butene:
In cycloalkenes smaller than cyclooctene, the cis isomers are more stable than the trans as a result of ring strain.
Exercises
1. When looking at their heats of hydrogenation, is the cis or the trans isomer generally more stable?
2. Arrange the following alkenes in order of increasing stability: 2,3-dimethyl-2-butene; trans-2-hexene; 2-methyl-2-pentene; cis-2-hexene
3. Which is the more stable alkene in each pair?
Answer
1. Trans alkenes are more stable as demonstrated by the lower heats of hydrogenation when compared to their cis-isomers.
2. (least substituted and cis) cis-2-hexene < trans-2-hexene < 2-methyl-2-pentene < 2,3-dimethyl-2-butene (most substituted)
3. A) 2 b/c trans with same substitution at C=C B) 1 b/c the C=C is more substituted
8.07: Alkene Synthesis by Elimination of Alkyl Halides
Learning Objective
• interpret and draw reaction energy diagrams for dehydrohalogenation of R-X’s
• propose mechanisms for a dehydrohalogenation reactions -
• predict the products and specify the reagents for alkene synthesis from dehydrohalogenation of R-X’s
• predict and explain the stereochemistry of E2 eliminations to form alkenes, especially from cyclohexanes
Alkene Synthesis by Elimination of Alkyl Halides is discussed in detail in chapter 7 sections 13 - 18. The major learning objectives are summarized briefly in this section.
Alkene Synthesis by Elimination of Alkyl Halides
When considering whether an elimination reaction is likely to occur via an E1 or E2 mechanism, we really need to consider three factors:
1. 1) The base: strong bases favor the E2 mechanism, whereas, E1 mechanisms only require a weak base.
2. 2) The solvent: good ionizing xolvents (polar protic) favor the E1 mechanism by stabilizing the carbocation intermediate.
3. 3) The alkyl halide: primary alkyl halides have the only structure useful in distinguishing between the E2 and E1 pathways. Since primary carbocations do not form, only the E2 mechanism is possible.
Reaction Parameter E2 E1
alkyl halide structure tertiary > secondary > primary tertiary > secondary >>>> primary
nucleophile high concentration of a strong base weak base
mechanism 1-step 2-step
rate limiting step anti-coplanar bimolecular transition state carbocation formation
rate law rate = k[R-X][Base] rate = k[R-X]
stereochemisty retained configuration mixed configuration
solvent not important polar protic | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.06%3A_Stability_of_Alkenes.txt |
Learning Objective
• interpret and draw reaction energy diagrams for alcohol dehydration reactions
• propose mechanisms for dehydration reactions
• predict the products and specify the reagents for alkene synthesis from alcohol dehydration reactions
• predict and explain the stereochemistry of E2 eliminations to form alkenes, especially from cyclohexanes
Dehydration of Alcohols to Yield Alkenes
One way to synthesize alkenes is by dehydration of alcohols. Alcohols undergo E1 or E2 mechanisms to lose water and form a double bond. This mechanism is analogous to the alkyl halide mechanism. The only difference is that hydroxide is a very poor leaving group so an extra step is required. The hydroxyl group is protonated so that water is now the leaving group. Therefore, 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 reaction temperature range 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).
Mechanism for the Dehydration of Alcohols
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 to form an alkyloxonium ion. This ion acts as a very good leaving group for either the E1 or E2 mechanism. The deprotonated acid (the conjugate base) then reacts with one of the beta-hydrogens to form a double bond.
Primary Alcohols and the E2 Mechanism
Primary alcohols undergo bimolecular elimination (E2 mechanism) while secondary and tertiary alcohols undergo unimolecular elimination (E1 mechanism). The relative reactivity of alcohols in dehydration reaction is ranked as the following
Methanol < primary < secondary < tertiary
Primary alcohol dehydrates through the E2 mechanism. Oxygen donates two electrons to a proton from sulfuric acid H2SO4, forming an alkyloxonium ion. The resulting conjugate base (HSO4) approaches in an anti-coplanar orientation relative to the leaving group and reacts with one adjacent hydrogen while the alkyloxonium ion simultaneously leaves in a concerted process, making a double bond.
Secondary and Tertiary Alcohols and the E1 Mechanism
Secondary and tertiary alcohols dehydrate through the E1 mechanism. Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, 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, forming a double bond.
For example, 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 the general rule that more substituted alkenes are more stable than less substituted alkenes, and trans alkenes are more stable than cis alkenes. Thereore, the trans diastereomer of the 2-butene product is most abundant.
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.
Carbocation Rearrangements and the E1 Mechanism
Carbocation stability is always a driving force in E1 mechanisms. It is important to evaluate the structure of all carbocation intermediates to look for the possibility of 1,2-hydride or 1,2-methyl shifts to form more stable carbocation intermediates. Carbocation rearrangements are discussed more completely in chapter 7.
Exercise
1. Draw the bond-line structure(s) for the product(s) formed and specify the mechanism (E1 or E2) for each reaction below.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.08%3A_Alkene_Synthesis_by_Dehydration_of_Alcohols.txt |
Learning Objectives
• discuss the uses and sources of alkenes including catalytic cracking
Uses of ethene and propene
Produced from ethylene (ethene)
Chemical Uses
ethanol solvent; constituent of cleaning preparations; in synthesis of esters
acetaldehyde slug killer, in the form of methaldehyde (CH3CHO)4
acetic acid manufacture of vinyl acetate polymers, ethyl acetate solvent and cellulose acetate polymers
ethylene oxide “cellosolves” (industrial solvents)
ethylene glycol anti-freeze; production of DacronOR
ethylene dichloride solvent; production of vinyl chloride
vinyl chloride manufacture of poly (vinyl chloride)—PVC
vinyl acetate manufacture of poly (vinyl acetate) used in paint emulsions, plywood adhesives and textiles
polyethylene “plastic” bags; toys; packaging
Produced from propylene (propene)
Chemical Uses
isopropyl alcohol rubbing alcohol; cosmetics; synthesis of acetone
propylene oxide manufacture of polyurethanes; polyesters
cumene industrial preparation of phenol and acetone
polypropylene molded articles (e.g., kitchenware); fibres for indoor-outdoor carpeting
Catalytic Cracking to Form Ethylene
Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking.
The hydrocarbons are mixed with a very fine catalyst powder. These days the catalysts are zeolites (complex aluminosilicates) - these are more efficient than the older mixtures of aluminium oxide and silicon dioxide. The whole mixture is then blown rather like a liquid through a reaction chamber at a temperature of about 500°C. Because the mixture behaves like a liquid, this is known as fluid catalytic cracking (or fluidized catalytic cracking). Although the mixture of gas and fine solid behaves as a liquid, this is nevertheless an example of heterogeneous catalysis - the catalyst is in a different phase from the reactants. The catalyst is recovered afterwards, and the cracked mixture is separated by cooling and further fractional distillation.
There isnot any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:
Or, showing more clearly what happens to the various atoms and bonds:
This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. The octane is one of the molecules found in petrol (gasoline).
Ethene
Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking.
There is not any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:
Or, showing more clearly what happens to the various atoms and bonds:
This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. You will remember that during the polymeriation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene. The reaction is done at high pressures in the presence of a trace of oxygen as an initiator.
8.10: Additional Exercises
Physical Properties of Alkenes
8-1 Explain why cis-2-butene is less stable than trans-2-butene.
8-2 Order the following alkenes in increasing order of stability.
8-3 Why do more substituted alkenes experience more stability compared to less substituted alkenes?
8-4 Identify whether the following alkenes are mono-, di-, tri-, or tetra-substituted.
8-5 Place the following alkenes in order of increasing boiling points.
Elements of Unsaturation and the Orbital Description of Alkenes
8-6 Using the following equation, calculate the degrees of unsaturation for the following compounds.
8-7 How many of each type of bonds (sigma/pi) make up a double bond?
Alkene Synthesis by Elimination of Alkyl Halides
8-8 Identify the major product(s) of the following reactions. Include stereochemistry.
8-9 Identify the major products of the following reactions.
8-10 Give the product of the following reaction.
8-11 What is the IUPAC name of the product formed by the reaction in the previous problem (8-10)?
Alkene Synthesis by Dehydration of Alcohols
8-12 Identify the product of the following reaction.
8-13 Draw the mechanism for the reaction in previous problem (8-12).
8-14 Draw the intermediate compounds for the following reaction.
8-15 Identify the product of the following reaction.
8.11: Solutions to Additional Exercises
Physical Properties of Alkenes
8-1 When in the cis configuration, the methyl groups experience steric strain as they are in close proximity to each other. They avoid steric interactions when in the trans configuration as they are able to stay as far apart as possible.
8-2
8-3 Alkyl groups are able to stabilize their neighboring carbon atoms by donating electron density, which allows for the delocalization of electron density and an increase in stability.
8-4
a) Disubstituted
b) Trisubstituted
c) Monosubstituted
d) Tetrasubstituted
8-5
Elements of Unsaturation and the Orbital Description of Alkenes
8-6
a) 4
b) 1
c) 2
d) 2
e) 5
f) 6
8-7 One sigma and one pi bond together make a double bond.
Alkene Synthesis by Elimination of Alkyl Halides
8-8
8-9
8-10
8-11 (1E)-1-chlorobut-1-ene
8-12
8-13
8-14
8-15 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/08%3A_Structure_and_Synthesis_of_Alkenes/8.09%3A_Uses_and_Sources_of_Alkenes.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• draw the general Electrophilic Addition Reaction (EAR) mechanism for an alkene - refer to section 9.1
• predict the products/specify the reagents for EAR of hydrohalic acids (HX) with symmetrical alkenes - refer to section 9.2
• predict the products/specify the reagents for EAR of hydrohalic acids (HX) with asymmetrical alkenes using Markovnikov's Rule for Regioselectivity - refer to section 9.3
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes - refer to sections 9.3 - 9.14
• predict the products, specify the reagents, and discern most efficient reaction for hydration of alkenes (acid catalyzed hydration; or oxymercuration/demercuration; or hydroboration/oxidation) - refer to sections 9.4, 9.5, and 9.6 respectively
• discern the stereochemical differences between the EAR of chiral and achiral alkenes - refer to sections 9.7 and 9.8
• predict the products/specify the reagents for halogenation and hydrohalogenation of alkenes - refer to sections 9.9 and 9.10 respectively
• recognize organic oxidation and reduction reactions - refer to sections 9.11 and 9.12
• predict the products/specify the reagents for hydrogenation (reduction) of alkenes - refer to section 9.11
• predict the products/specify the reagents for epoxidation of alkenes - refer to section 9.12
• predict the products/specify the reagents for dihydroxylation of alkenes - refer to sections 9.13 and 9.14
• predict the products/specify the reagents for oxidative cleavage of alkenes - refer to section 9.15
• predict the products of carbene additions to alkenes - refer to section 9.16
• predict the polymer/specify the monomer for radical, chain -growth polymers of alkenes - refer to section 9.17
• discuss an example biological addition reactions - refer to section 9.18
• 9.1: Electrophilic Addition Reactions (EARs)
Electrophilic addition reactions can occur in compounds containing pi bonds like the alkenes. Depending on the structure of the alkene and the specific reagents, the reactions can be regioselective and/or stereoselective.
• 9.2: Addition of Hydrogen Halides to Symmetrical Alkenes
The regioselective reaction of the carbon-carbon double bond in alkenes with hydrohalogens (HX) is a controlled by carbocation stability. Consequently, the symmetry of the alkene must be considered for this mechanistic pathway.
• 9.3: Alkene Asymmetry and Markovnikov's Rule
The regioselectivity of electrophilic addition reactions is determined by carbocation stability and is summarized by Markovnikov's Rule.
• 9.4: Hydration- Acid Catalyzed Addition of Water
Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a carbocation is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane.
• 9.5: Hydration- Oxymercuration-Demercuration
Oxymercuration is a stereospecific, regioselective electrophilic addition reaction because there are no carbocation rearrangements due to stabilization of the reactive intermediate. The Markovnikov products are reliably synthesized by this pathway.
• 9.6: Hydration - Hydroboration-Oxidation
Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes that are the non-Markovnikov products for alkene hydration.
• 9.7: Stereochemistry of Reactions - Hydration of Achiral Alkenes
For achiral alkenes, the symmetrical trigonal planar geometry of the carbocation leads to equivalent synthesis of both R and S products giving a racemic (50/50).
• 9.8: Stereochemistry of Reactions - Hydration of Chiral Alkenes
Chiral alkenes form electrophilic addition products in non 50:50 ratios due to the differences in steric effects between the enantiomeric starting materials.
• 9.9: Addition of Halogens
Halogens can act as electrophiles due to polarizability of their covalent bond and react with the pi bond of alkenes. This electrophilic addition mechanism is stereospecific. The orientation of the electrophile during a stereospecific electrophilic addition reaction will determine the stereochemistry of the product(s).
• 9.10: Formation of Halohydrins
When the halogenation reaction of alkenes is performed in a nucleophilic solvent like water or alcohol, then the solvent becomes the nucleophile to give halohydrin or haloalkoxy products.
• 9.11: Reduction of Alkenes - Catalytic Hydrogenation
Catalytic hydrogenation of a carbon-carbon double bond is a reduction reaction. The alkene orientation required for interaction with the surface of the catalyst means that t his reaction is stereospecific.
• 9.12: Oxidation of Alkenes - Epoxidation
Oxidation of alkenes is introduced and the epoxidation of alkenes is discussed.
• 9.13: Dihydroxylation of Alkenes
Alkenes can react to produce glycols (two adjacent hydroxyl groups) through either an anti- or syn- addition mechanism that is stereospecific.
• 9.14: Opening of Epoxides - Acidic versus Basic Conditions
Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions.
• 9.15: Oxidative Cleavage of Alkenes
Oxidative cleavage of alkenes can occur by several different pathways. The most common reagents and pathways are discussed in this section.
• 9.16: Addition of Carbenes to Alkenes - Cyclopropane Synthesis
A carbene such as methlyene will react with an alkene breaking the pi bond to form a cyclopropane. Since methylene is highly reactive, it is prepared in situ immediately preceding the addition of the alkene.
• 9.17: Radical Chain-Growth Polymerization
All the monomers from which addition polymers are made are alkenes or functionally substituted alkenes. The free radical mechanism for chain growth polymers is explained.
• 9.18: Biological Additions of Radicals to Alkenes
Some electrophilic addition reactions that take place in nature and the role of enzymes in such processes are introduced.
• 9.19: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
• 9.20: Solutions to Additional Exercises
This section has the solutions to the additional exercises from the previous section.
09: Reactions of Alkenes
Learning Objective
• draw the general Electrophilic Addition Reaction (EAR) mechanism for an alkene
Introduction
We are going to start by looking at ethene, because it is the simplest molecule containing a carbon-carbon double bond. What is true of C=C in ethene will be equally true of C=C in more complicated alkenes.
Ethene , C2H4, is often modeled as shown above. The double bond between the carbon atoms is, of course, two pairs of shared electrons. What the diagram doesn't show is that the two pairs aren't the same as each other. One of the pairs of electrons is held on the line between the two carbon nuclei as you would expect, but the other is held in a molecular orbital above and below the plane of the molecule. A molecular orbital is a region of space within the molecule where there is a high probability of finding a particular pair of electrons.
In this diagram, the line between the two carbon atoms represents a normal sigma bond - the pair of shared electrons lies in a molecular orbital on the line between the two nuclei where you would expect them to be. The other pair of electrons is found somewhere in the shaded part above and below the plane of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move around anywhere in this shaded region and can move freely from one half to the other. The pi electrons are not as fully under the control of the carbon nuclei as the electrons in the sigma bond. Because the pi bond electrons lie exposed above and below the rest of the molecule, they are relatively open to reaction with other compounds.
Electrophilic Addition Reactions (EARs)
In a sense, the pi bond is an unnecessary bond. The structure would hold together perfectly well with a single bond rather than a double bond. The pi bond often breaks and the electrons in it are used to join other atoms (or groups of atoms) onto the alkene molecule. To continue with our example, ethene undergoes addition reactions. An addition reaction is a reaction in which two molecules join together to make a bigger one. Nothing is lost in the process. All the atoms in the original molecules are found in the bigger one. The pi bond is electron rich and takes the role of the nucleophile seeking out an electrophile with its full or partial positive charge.
Using a general molecule X-Y as the electrophile reacting with ethene, the atoms 'X' and 'Y' are added to the carbon chain across the vinylic carbons as shown below.
Understanding the Electrophilic Addition Mechanism
The mechanism for the reaction between ethene and a molecule X-Y begins by recognizing the electrophilic nature of X-Y. Trends in relative electronegativity help identify bonds to create partial positives within polar compounds. As shown below, we are going to assume that Y is more electronegative than X, so that the pair of electrons is pulled slightly towards the Y end of the bond. This polarity means that the X atom carries a slight positive charge (partial positive).
The slightly positive X atom is an electrophile and attracts the exposed pi electrons in the ethene. Now imagine what happens as they approach each other. The electrons in the half of the pi bond nearest the XY are attracted to the partial positive charge as shown below.
The two electrons in the pi bond move closer towards the X until a covalent bond is made. Simultaneously, the electrons in the X-Y bond are pushed entirely onto the Y to form the anion Y- ion and a carbocation as shown below.
EAR Mechanism
The movements of the electrons for the EAR mechanism are shown with curved arrows.
Regiochemistry and Regioselective Reactions
Regiochemistry is the orientation of the electrophile relative to the pi bond of the alkene. Electrophilic addition reactions of alkenes can be regioselective depending on the symmetry and structure of the alkene. The details will be discussed in a later sections of this chapter.
Stereoselective Reactions
Different stereoisomeric reactants produce different stereoisomeric products. Electrophilic addition reactions of alkenes can be stereoselective depending on the symmetry and structure of the alkene. The details will be discussed in a later sections of this chapter.
Exercise \(1\)
1. In the next section, we will apply the Electrophilic Addition mechanism to actual compounds. Draw the complete mechanism when hydrogen bromide gas is bubbled through a solution of cyclopentene.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.01%3A_Electrophilic_Addition_Reactions_%28EARs%29.txt |
Learning Objective
• predict the products/specify the reagents for EAR of hydrohalic acids (HX) with symmetrical alkenes
This section looks at the reaction of symmetrical alkenes (like ethene, but-2-ene or cyclohexene) with hydrogen halides such as hydrogen chloride and hydrogen bromide. Since these alkenes have identical groups attached to each end of the carbon-carbon double bond, regioselectivity does not apply.
Addition to symmetrical alkenes
All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other.
For example, with ethene and hydrogen chloride, you get chloroethane:
With but-2-ene you get 2-chlorobutane:
What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product. The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space. That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.
Mechanism
The addition of hydrogen halides is one of the easiest electrophilic addition reactions because it uses the simplest electrophile: the proton. Hydrogen halides provide both a electrophile (proton) and a nucleophile (halide). First, the electrophile will attack the double bond and take up a set of π electrons, attaching it to the molecule (1). This is basically the reverse of the last step in the E1 reaction (deprotonation step). The resulting molecule will have a single carbon- carbon bond with a positive charge on one of them (carbocation). The next step is when the nucleophile (halide) bonds to the carbocation, producing a new molecule with both the original hydrogen and halide attached to the organic reactant (2). The second step will only occur if a good nucleophile is used.
Mechanism of Electrophilic Addition of Hydrogen Chloride to But-2-ene
All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals)s.
Reaction rates
Variation of rates when you change the halogen
Reaction rates increase in the order HF < HCl < HBr < HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions.
When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.
Variation of rates when you change the alkene
This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be.
Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond. For example:
There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.
Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this.
Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes.
The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.
The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:
The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.
Exercise
1. Draw the bond-line structures for the products of the following reactions.
Answer
1.
9.03: Alkene Asymmetry and Markovnikov's Rule
Learning Objective
• predict the products/specify the reagents for EAR of hydrohalic acids (HX) with asymmetrical alkenes using Markovnikov's Rule for Regioselectivity
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes
Addition to unsymmetrical alkenes
In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition of the hydrogen and the halogen across the double bond.
Markovnikov's Rule
If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product according to Markovnikov's Rule.
Markovnikov's Rule: When HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.
Applying Markovnikov's Rule to the reaction above, the hydrogen bonds with the CH2 group, because the CH2 group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count.
Regioselectivity - a closer look
If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if one product forms in greater amounts than the others, the overall reaction is said to be regioselective.
Say three reactions could occur between the hypothetical reactants A and B under the same conditions giving the constitutionally isomeric products C, D, and E.
There are two possibilities:
1. The three products form in equal amounts, i.e., of the total product 33% is C, another 33% D, the remaining 33% E. (These percentages are called relative yields of the products.)
If this is what is observed, the overall reaction between A and B is not regioselective.
2. One product forms in greater amounts than the others. Say, for example, the relative yields of C, D, and E are 25%, 50%, and 25%, respectively.
If this is what is observed, the overall reaction between A and B is regioselective.
eg:
Experimentally, 2 is the major product; 3 is the minor product. Thus, the overall reaction between 1 and HBr is regioselective toward 2.
If more than one reaction could occur between a set of reactants under the same conditions giving products that are constitutional isomers and if only one product is observed, the overall reaction is said to be 100% regioselective or regiospecific.
eg:
The only observed product is 5. (Relative yields of 5 and 6 are 100% and 0%, respectively.) Thus the overall reaction between 4 and HBr is regiospecific toward 5.
Regiospecificity is merely the limiting case of regioselectivity. All regiospecific reactions are regioselective, but not all regioselective reactions are regiospecific.
Exercises
1. Predict the product(s) for the following reactions:
2. In each case, suggest an alkene that would give the product shown.
Answers
1.
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.02%3A_Addition_of_Hydrogen_Halides_to_Symmetrical_Alkenes.txt |
Learning Objective
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes
• predict the products, specify the reagents, and discern most efficient reaction for hydration of alkenes (acid catalyzed hydration; or oxymercuration/demercuration; or hydroboration/oxidation)
What Is Electrophilic Hydration?
Electrophilic hydration is the addition of hydrogen and a hydroxyl group across the two carbons of a double bond. Electrophilic hydration is the reverse of dehydration of alcohols and so begins the circular nature of organic chemistry. Alcohols can be dehydrated to form alkenes and alkenes can undergo electrophilic addition reactions to form alcohols. Electrophilic hydrogen is essentially a proton: a hydrogen atom stripped of its electrons. Electrophilic hydrogen is commonly used to help break double bonds or restore catalysts.
Electrophilic hydration of alkenes has practical applications in making alcohols for fuels and reagents for other reactions. The basic reaction under certain temperatures (given below) is the following:
In later sections, we will learn that mercury (II) sulfate and borane are also electrophiles that can react with alkenes to form hydration products. Each reaction pathway has its own regio- and stereochemical considerations. In the example below, we see that the same alkene produces different hydration products depending on the hydration pathway.
Mechanism for Acid-catalyzed Hydration of Alkenes
Temperatures for Types of Alcohol Synthesis
Heat is used to catalyze electrophilic hydration; because the reaction is in equilibrium with the dehydration of an alcohol, which requires higher temperatures to form an alkene, lower temperatures are required to form an alcohol. The exact temperatures used are highly variable and depend on the product being formed.
• Primary Alcohol: Less than 170ºC
• Secondary Alcohol: Less than 100ºC
• Tertiary Alcohol: Less than 25ºC
But...Why Does Electrophilic Hydration Work?
• An alkene placed in an aqueous non-nucleophilic strong acid immediately "reaches out" with its double bond and attacks one of the acid's hydrogen atoms (meanwhile, the bond between oxygen and hydrogen performs heterolytic cleavage toward the oxygen—in other words, both electrons from the oxygen/hydrogen single bond move onto the oxygen atom).
• A carbocation is formed on the original alkene (now alkane) in the more-substituted position, where the oxygen end of water attacks with its 4 non-bonded valence electrons (oxygen has 6 total valence electrons because it is found in Group 6 on the periodic table and the second row down: two electrons in a 2s-orbital and four in 2p-orbitals. Oxygen donates one valence electron to each bond it forms, leaving four 4 non-bonded valence electrons).
• After the blue oxygen atom forms its third bond with the more-substituted carbon, it develops a positive charge (3 bonds and 2 valence electrons give the blue oxygen atom a formal charge of +1).
• The bond between the green hydrogen and the blue oxygen undergoes heterolytic cleavage, and both the electrons from the bond move onto the blue oxygen. The now negatively-charged strong acid picks up the green electrophilic hydrogen.
• Now that the reaction is complete, the non-nucleophilic strong acid is regenerated as a catalyst and an alcohol forms on the most substituted carbon of the current alkane. At lower temperatures, more alcohol product can be formed.
What is Regiochemistry and How Does It Apply?
Regiochemistry deals with where the substituent bonds on the product. Zaitsev's and Markovnikov's rules address regiochemistry, but Zaitsev's rule applies when synthesizing an alkene while Markovnikov's rule describes where the substituent bonds onto the product. In the case of electrophilic hydration, Markovnikov's rule is the only rule that directly applies. See the following for an in-depth explanation of regiochemistry Markovnikov explanation: Radical Additions--Anti-Markovnikov Product Formation
In the mechanism for a 3º alcohol shown above, the red H is added to the least-substituted carbon connected to the nucleophilic double bonds (it has less carbons attached to it). This means that the carbocation forms on the 3º carbon, causing it to be highly stabilized by hyperconjugationelectrons in nearby sigma (single) bonds help fill the empty p-orbital of the carbocation, which lessens the positive charge. More substitution on a carbon means more sigma bonds are available to "help out" (by using overlap) with the positive charge, which creates greater carbocation stability. In other words, carbocations form on the most substituted carbon connected to the double bond. Carbocations are also stabilized by resonance, but resonance is not a large factor in this case because any carbon-carbon double bonds are used to initiate the reaction, and other double bonded molecules can cause a completely different reaction. If the carbocation does originally form on the less substituted part of the alkene, carbocation rearrangements occur to form more substituted products.
Carbocation Rearrangements - a review
• Hydride shifts: a hydrogen atom bonded to a carbon atom next to the carbocation leaves that carbon to bond with the carbocation (after the hydrogen has taken both electrons from the single bond, it is known as a hydride). This changes the once neighboring carbon to a carbocation, and the former carbocation becomes a neighboring carbon atom.
In a more complex case, when alkenes undergo hydration, we also observe hydride shift. Below is the reaction of 3-methyl-1-butene with H3O+ that furnishes to make 2-methyl-2-butanol:
Once again, we see multiple products. In this case, however, we see two minor products and one major product. We observe the major product because the -OH substitutent is attached to the more substituted carbon. When the reactant undergoes hydration, the proton attaches to carbon #2. The carbocation is therefore on carbon #2. Hydride shift now occurs when the hydrogen on the adjacent carbon formally switch places with the carbocation. The carbocation is now ready to be attacked by H2O to furnish an alkyloxonium ion because of stability and hyperconjugation. The final step can be observed by another water molecule attacking the proton on the alkyloxonium ion to furnish an alcohol. We see this mechanism below:
• Alkyl shifts: if no hydrogen atoms are available for a hydride shift, an entire methyl group performs the same shift.
The nucleophile attacks the positive charge formed on the most substituted carbon connected to the double bond, because the nucleophile is seeking that positive charge. In the mechanism for a 3º alcohol shown above, water is the nucleophile. When the green H is removed from the water molecule, the alcohol attached to the most substituted carbon. Hence, electrophilic hydration follows Markovnikov's rule.
Stereochemistry of Acid-catalyzed Hydration
Stereochemistry deals with how the substituent bonds on the product directionally. Dashes and wedges denote stereochemistry by showing whether the molecule or atom is going into or out of the plane of the board. Whenever the bond is a simple single straight line, the molecule that is bonded is equally likely to be found going into the plane of the board as it is out of the plane of the board. This indicates that the product is a racemic mixture.
There is no stereochemical control in acid-catalyzed hydration reactions. The carbocation intermediate has the trigonal planar geometry of sp2 hybridization which allows the subsequent reaction with water from either orientation.
Electrophilic hydration adopts a stereochemistry wherein the substituent is equally likely to bond pointing into the plane of the board as it is pointing out of the plane of the board. The 3º alcohol product could look like either of the following products:
Note: Whenever a straight line is used along with dashes and wedges on the same molecule, it could be denoting that the straight line bond is in the same plane as the board. Practice with a molecular model kit and attempting the practice problems at the end can help eliminate any ambiguity.
Is this a Reversible Synthesis?
Electrophilic hydration is reversible because an alkene in water is in equilibrium with the alcohol product. To sway the equilibrium one way or another, the temperature or the concentration of the non-nucleophilic strong acid can be changed. For example:
• Less sulfuric or phosphoric acid and an excess of water help synthesize more alcohol product.
• Lower temperatures help synthesize more alcohol product.
Is There a Better Way to Add Water to Synthesize an Alcohol From an Alkene?
A more efficient pathway does exist: see Oxymercuration - Demercuration: A Special Electrophilic Addition. Oxymercuration does not allow for rearrangements, but it does require the use of mercury, which is highly toxic. Detractions for using electrophilic hydration to make alcohols include:
• Allowing for carbocation rearrangements
• Poor yields due to the reactants and products being in equilibrium
• Allowing for product mixtures (such as an (R)-enantiomer and an (S)-enantiomer)
• Using sulfuric or phosphoric acid
Exercise
1. Draw the bond-line structure for the major product of each reaction.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.04%3A_Hydration-_Acid_Catalyzed_Addition_of_Water.txt |
Learning Objective
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes
• predict the products, specify the reagents, and discern most efficient reaction for hydration of alkenes (acid catalyzed hydration; or oxymercuration/demercuration; or hydroboration/oxidation)
Introduction
Acid-catalyzed hydration of alkenes is limited by carbocation stability. Carbocation rearrangement can occur to form a more stable ion as shown in the example below.
Alkene hydration using the oxymercuration-demercuration reaction pathway reliably produces the Markovnikov product without carbocation rearrangment as shown in the example below.
Oxymercuration-Demercuration is a two step pathway used to produce alcohols.
Oxymercuration-Demercuration Mechanism
This mechanism is similar to the previous electrophilic addition reactions. 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, a water 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 water 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 oxymercuration - demercuration mechanism follows Markovnikov's Regioselectivity with the OH 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
1. Show how to prepare 3-methyl2-pentanol from 3-methyl-1-pentene.
Note: Questions 2-5 have not shown the water present in the sulfuric acid solution and have indicated a second neutralization step. Some authors simply write H+/H2O as a single step.
2. Draw the bond-line structure for the product.
3. Draw the bond-line structure for the product. How does the cyclopropane group affect the reaction?
4. Draw the bond-line structure for the product. (Hint: What is different about this problem?)
5. Draw the bond-line structure for the product(s). Indicate any shifts as well as the major product:
6. In each case, predict the product(s) of these reactants of oxymercuration.
7. Propose the alkene that was the reactant for each of these products of oxymercuration.
Answer
1.
2. This reaction is electrophilic hydration.
3. The answer is additional side products, but the major product formed is still the same (the product shown). Depending on the temperatures used, the cyclopropane may open up into a straight chain, which makes it unlikely that the major product will form (after the reaction, it is unlikely that the 3º carbon will remain as such).
4. A hydride shift actually occurs from the top of the 1-methylcyclopentane to where the carbocation had formed.
5. In the first picture shown below, an alkyl shift occurs but a hydride shift (which occurs faster) is possible. Why doesn't a hydride shift occur? The answer is because the alkyl shift leads to a more stable product. There is a noticeable amount of side product that forms where the two methyl groups are, but the major product shown below is still the most significant due to the hyperconjugation that occurs by being in between the two cyclohexanes.
6.
7. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.05%3A_Hydration-_Oxymercuration-Demercuration.txt |
Learning Objectives
• apply the principles of regioselectivity and stereoselectivity to the addition reactions of alkenes
• predict the products, specify the reagents, and discern most efficient reaction for hydration of alkenes (acid catalyzed hydration; or oxymercuration/demercuration; or hydroboration/oxidation)
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an anti-Markovnikov manner, where the hydrogen (from $\ce{BH3}$ or $\ce{BHR2}$) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene double bond. Furthermore, the borane acts as the electrophile 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.
The Borane Complex
It is very important to understand the structure and properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula $\ce{B2H6}$ (diborane).
$\ce{BH3→B2H6} \nonumber$
Since diborane dimer ignites spontaneously in air, it commercially distributed in ether or tetrahydrofuran (THF) solutions. In these solutions, the borane can exist as a Lewis acid-base complex which allows boron to have an octet of electrons.
The Mechanism
Step #1: Hydroboration of the alkene
The addition of the borane to the alkene is initiated and proceeds as a concerted reaction because bond breaking and bond formation occur at the same time. The vacant 2p orbital of the boron takes the role of electrophile and accepts the pi electrons from the nucleophilic alkene. The boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon. Both, the boron and the hydrogen add simultaneously on the same face of the double bond (syn addition). With a concerted mechanism, there is no carbocation formation.
Transition state
* Note that a carbocation is not formed. Therefore, no rearrangement takes place.
It is important to note that reaction continues two more times until all three hydrogens on the borane have reacted with alkenes to create the trialkylborane intermediate R3B.
Step #2: Oxidation of the Trialkylborane by Hydrogen Peroxide
The hydrogen peroxide ($\ce{HOOH}$) is the nucleophile in this reaction because it is the electron donor to the newly formed trialkylborane that resulted from the previous hydroboration.
$\ce{HOOH + OH^{-} -> HOO^{-} + HOH}$
In this second part of the mechanism, a rearrangement of an R group with its pair of bonding electrons to an adjacent oxygen results in the loss of a hydroxide ion.
Two more of these reactions with hydroperoxide will occur in order give a trialkylborate
In the final step of the oxidation process, the trialkylborate reacts with aqueous $\ce{NaOH}$ to give the alcohol and sodium borate (\ce{Na3BO3}\).
$\ce{(RO3)B + 3NaOH -> 3OH + Na3BO3}$
If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided at the end of this section.
Stereochemistry of the Hydroboration Step
The hydroboration reaction is among the few simple addition reactions that proceed cleanly in a syn fashion. As noted above, this is a single-step reaction. Since the bonding of the double bond carbons to boron and hydrogen is concerted, it follows that the geometry of this addition must be syn. Furthermore, rearrangements are unlikely inasmuch as a discrete carbocation intermediate is never formed. These features are illustrated for the hydroboration of α-pinene.
Since the hydroboration procedure is most commonly used to hydrate alkenes in an anti-Markovnikov fashion, we also need to know the stereoselectivity of the second oxidation reaction, which substitutes a hydroxyl group for the boron atom. Independent study has shown this reaction takes place with retention of configuration so the overall addition of water is also syn.
The hydroboration of α-pinene also provides a nice example of steric hindrance control in a chemical reaction. In the less complex alkenes used in earlier examples the plane of the double bond was often a plane of symmetry, and addition reagents could approach with equal ease from either side. In this case, one of the methyl groups bonded to C-6 (colored blue in the equation) covers one face of the double bond, blocking any approach from that side. All reagents that add to this double bond must therefore approach from the side opposite this methyl.
Exercises
1. Draw the bond-line structure of the product(s) for these following reactions?
a)
b)
c)
2. Draw the structural formulas for the alcohols that result from hydroboration-oxidation of the alkenes shown.
a)
b) (E)-3-methyl-2-pentene
3. Write out the reagents or products (A–D) shown in the following reaction schemes.
Answer
1.
a)
b)
c)
2.
a)
b)
3. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.06%3A_Hydration_-_Hydroboration-Oxidation.txt |
Learning Objective
• discern the stereochemical differences between the EAR of chiral and achiral alkenes
Stereochemistry and the Subtle Details
Organic reactions in the laboratory or in living systems can produce chiral centres. Consider reaction of 1-pentene with water (acid catalyzed). Markovnikov regiochemistry occurs and the OH adds to the second carbon. However, both R and S products occur giving a racemic (50/50) mixture of 2‑butanol. How does this occur? The proton addition to 1‑pentene results in a planar carbocation intermediate. A molecule of water is then equally likely to react from the top or the bottom of this cation to produce either (S)‑2‑pentanol or (R)‑2‑pentanol, respectively, as shown in the mechanism below.
9.08: Stereochemistry of Reactions - Hydration of Chiral Alkenes
Learning Objective
• discern the stereochemical differences between the EAR of chiral and achiral alkenes
Stereochemistry - the Subtle Details
In the previous section, the addition of water to the achiral alkene produced a racemic mixture of two enantiomeric alcohols. They are produced in equal amounts so the mixture is optically inactive. What would occur if we carried out a similar reaction on a chiral alkene? Consider (S)-3-methyl-1-pentene reacting with water (acid catalyzed). Proton addition produces a carbocation intermediate that is chiral (* denotes stereogenic centre). That intermediate does not have a plane of symmetry and therefore attack by water is not equal from the top and bottom. This ultimately produces R and S products in a non 50:50 ratio as shown in the mechanism below.
Exercise
1. Predict the products of the following reaction showing stereochemistry.
Answer
1. The products are diastereomers of one another.
9.09: Addition of Halogens
Learning Objective
• predict the products/specify the reagents for halogenation of alkenes
Introduction
As the halogen molecule, for example Br2, approaches the double bond of an alkene, electrons in the double bond repel electrons in the bromine molecule causing polarization of the halogen bond. This interaction induces a dipole moment in the halogen molecule bond allowing one of the halogens to gain a partial positive charge and take the role of electrophile. The nucleophilic pi electrons form a bond to the electrophilic halogen while the halogen molecular bond heterolytically breaks to release bromide as a leaving group. The halogen addition is not regioselective but stereoselective. Stereochemistry of this addition is analogous to the oxymercuration mechanism. In this reaction, a bromonium (halogenium) ion froms as the intermediate. The bromonium ion formation stabilizes the positive charge and prevents carbocation rearrangement. In the second step, the bromide released from the first step takes the role of thenucleophile and reacts with the cyclic bromonium ion with back side orientation. Therefore, the stereochemistry of the product is a vicinial dihalides through anti-addition.
$\ce{R_2C=CR_2 + X_2 \rightarrow R_2CX-CR_2X} \label{8.2.1}$
Halogens that are commonly used in this type of the reaction are: Br2 and Cl2. In thermodynamical terms I2 is too slow for this reaction because of the size of its atom, and F2 is too vigorous and explosive. Solvents that are used for this type of electrophilic halogenation are inert (e.g., CCl4) can be used in this reaction.
Because halogen with negative charge can attack any carbon from the opposite side of the cycle it creates a mixture of steric products.Optically inactive starting material produce optically inactive achiral products (meso) or a racemic mixture.
Electrophilic addition mechanism consists of two steps.
Before constructing the mechanism let us summarize conditions for this reaction. We will use Br2 in our example for halogenation of ethylene. Halogens can act as electrophiles due to polarizability of their covalent bond. Addition of halogens is stereospecific and produces vicinial dihalides with anti-addition. Cis starting materials will give a mixture of enantiomers and trans starting materials produce a meso compound.
Nucleophile pi electrons of alkene double bond
Electrophile halogen (Cl2 or Br2)
Regiochemistry none
Stereochemistry anti-additon
Step 1: The addition the Br-Br bond polarizes, heterolytic cleavage occurs and Br with the positive charge forms a intermediate cycle with the double bond.
Step 2: The bromide anion reacts with either carbon of the bridged bromonium ion from the back side of the ring. The ring opens up and the two halogens are in the anti-position relative to each other.
Exercise
1. What is the mechanism of adding Cl2 to the cyclohexene?
2. A reaction of Br2 molecule in an inert solvent with alkene follows?
a) syn addition
b) anti addition
c) Morkovnikov rule
3.
4.
5. Predict the product of the product of 1,2-dimethylcyclopentene reacting with Br2 with proper stereochemistry.
6. Predict the products for 1,2-dimethylcyclpentene reacting with HCl, give the proper stereochemistry. What is the relationship between the two products?
Answer
1.
2. b
3.enantiomer
4.
5.
6.
These compounds are enantiomers. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.07%3A_Stereochemistry_of_Reactions_-_Hydration_of_Achiral_Alkenes.txt |
Learning Objective
• predict the products/specify the reagents for hydrohalogenation of alkenes
Introduction
The proton is not the only electrophilic species that initiates addition reactions to the double bond of alkenes. Lewis acids like the halogens, boron hydrides and certain transition metal ions are able to accept the alkene pi-electrons. The resulting positively charged intermediates attract nucleophiles to give addition products.The electrophilic character of the halogens is well known. Fluorine adds uncontrollably with alkenes, and the addition of iodine is unfavorable, so these are not useful preparative methods. Chlorine (Cl2) and bromine (Br2) react selectively with the double bond of alkenes, so we will focus on these reactions.
The addition of chlorine and bromine to alkenes, as shown below, produces vicinal dihalo-compounds. In this reaction, we can assume that the solvent was something that is not nucleophilic, such as tetrahydrofuran (THF).
R2C=CR2 + X2 ——> R2CX-CR2X
If this same reaction is performed in a nucleophilic solvent like water or an alcohol, then the solvent becomes the nucleophile in the second step and reacts with the bromonium (or chloronium) ion to form a halohydrin as shown below.
R2C=CR2 + X2 + H2O ——> R2COH-CR2X
There are also other halogen-containing reagents that add to double bonds, such as hypohalous acids, HOX, and sulfenyl chlorides, RSCl. These reagents are unsymmetrical, so their addition to unsymmetrical double bonds may in principle take place in two ways. In practice, these addition reactions are regioselective, with one of the two possible constitutionally isomeric products being favored. The electrophilic moiety in both of these reagents is the halogen.
(CH3)2C=CH2 + HOBr ——> (CH3)2COH-CH2Br
(CH3)2C=CH2 + C6H5SCl ——> (CH3)2CCl-CH2SC6H5
Mechanisms explain the Regioselectivity
X2/H2O or X2/ROH: The regioselectivity of halohydrin formation from an alkene reaction with a halogen in a nucleophilic solvent is analogous to the oxymercuration-demercuration pathway. The halogen molecule takes the role of electrophile accepting nucleophilic pi electrons from the alkene while simultaneously forming a bond with the other vinyl carbon to create a bromonium (or chloroium) ion. The bromonium (or chloronium) ion formation stabilizes the positive charge and prevents carbocation rearrangement. The solvent takes the role of the nucleophile because it is present is a much greater percentage than the leaving group and reacts with the most substituted carbon of the cyclic bromonium (or chloronium) ion to create regiochemistry. The stereochemistry of this reaction is anti-addition because the solvent approaches the bromonium ion with back side orientation to produce the addition product. However, since the interaction of the halogen with the alkene can occur from above or below, there is no stereochemical control in this reaction and a mixture of enantiomers will be produced when applicable. The final step of this mechanism is a proton transfer to a solvent water molecule to neutralize the addition product.
HOX or RSCl: The regioselectivity of the hypohalous acids and sulfenyl chloride reactions may be explained by the same mechanism we used to rationalize the Markovnikov rule. Bonding of an electrophilic species to the double bond of an alkene forms preferentially to produce the more stable (more highly substituted) carbocation. This intermediate should then combine rapidly with a nucleophilic species to produce the addition product.
To apply this mechanism we need to determine the electrophilic moiety in each of the reagents. By using electronegativity differences we can dissect common addition reagents into electrophilic and nucleophilic moieties, as shown on the right. In the case of hypochlorous and hypobromous acids (HOX), these weak Brønsted acids (pKa's ca. 8) do not react as proton donors; and since oxygen is more electronegative than chlorine or bromine, the electrophile will be a halide cation. The nucleophilic species that bonds to the intermediate carbocation is then hydroxide ion, or more likely water (the usual solvent for these reagents), and the products are called halohydrins. Sulfenyl chlorides add in the opposite manner because the electrophile is a sulfur cation, RS(+), whereas the nucleophilic moiety is chloride anion (chlorine is more electronegative than sulfur).
Below are some examples illustrating the addition of various electrophilic halogen reagents to alkene groups. Notice the specific regiochemistry of the products, as explained above.
Exercise
1. Predict the product of the following reaction:
2. When butene is treated with NBS in the presence of water, the product shows that the bromine is on the least substituted carbon, is this Markovnikov or anti-Markovnikov?
Answer
1.
2. Since the bromine is the first addition to the alkene, this addition would be an anti-Markovnikov addition. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.10%3A_Formation_of_Halohydrins.txt |
Learning Objective
• recognize organic oxidation and reduction reactions
• predict the products/specify the reagents for hydrogenation (reduction) of alkenes
Introduction
Addition of hydrogen to a carbon-carbon double bond to form an alkane is a reduction reaction that is also called catalytic hydrogenation. Hydrogenation of a double bond is a thermodynamically favorable reaction because it forms a more stable (lower energy) product. In other words, the energy of the product is lower than the energy of the reactant; thus it is exothermic (heat is released). The heat released is called the heat of hydrogenation, which is an indicator of a molecule’s stability. Regioselectivity is not an issue because the same group (a hydrogen atom) is bonded to each of the vinyl carbons. The simplest source of two hydrogen atoms is molecular hydrogen (H2), but mixing alkenes with hydrogen does not result in any discernible reaction. Although the overall hydrogenation reaction is exothermic, a high activation energy prevents it from taking place under normal conditions. This restriction may be circumvented by the use of a catalyst, as shown in the reactions below.
The O-chem View of Oxidation and Reduction
For inorganic chemistry, the flow of electrons is easily counted with the change in oxidation numbers of the metals and non-metals. The expressions "LEO says GER" for "Loss of Electrons is Oxidation and Gain of Electrons is Reduction" or "OIL RIG" for "Oxidation Is Loss and Reduction Is Gain" can be useful guides to recognizing oxidation and reduction reactions for inorganic chemistry. However for organic chemistry, most of the reactants and products are neutral so the electron flow is more difficult to track. For organic compounds, oxidation and reduction reactions can be recognized at least three different ways.
1) Oxidation is an increase in the number of carbon to oxygen bonds or a decrease in the number of carbon to hydrogen bonds.
2) Reduction is the opposite of oxidation so it is a decrease in the number of carbon to oxygen bonds or an increase in the number of carbon to hydrogen bonds.
3) For reactions that do not involve a change in the bonding of carbon with oxygen and hydrogen, then we need to look at the differences in electronegativity. The shared electrons are assigned to the more electronegative element to determine the oxidation numbers.
The Catalyst
The reaction between hydrogen (H2) gas and an alkene (a carbon-carbon double bond) requires an active metal catalyst. A catalyst increases the reaction rate by lowering the activation energy of the reaction. Although the catalyst is not consumed in the reaction, it is required to accelerate the reaction sufficiently to be observed in a reasonable amount of time. Catalysts commonly used in alkene hydrogenation are: platinum, palladium, and nickel. The metal catalyst acts as a surface on which the reaction takes place. This increases the rate by putting the reactants in close proximity to each other, facilitating interactions between them. With this catalyst present, the sigma bond of H2 breaks, and the two hydrogen atoms instead bind to the metal (see #2 in the figure below). The $\pi$ bond of the alkene weakens as it also interacts with the metal as shown in step #3 of the diagram below.
Since both the reactants are bound to the metal catalyst, the hydrogen atoms can easily add, one at a time, to the previously double-bonded carbons as shown in steps #4 and #5 above. The position of both of the reactants bound to the catalyst makes it so the hydrogen atoms are only exposed to one side of the alkene. This explains why the hydrogen atoms add to same side of the molecule, called syn-addition.
Alkene Stability and Catalytic Hydrogenation
As shown in the reaction energy diagram below, the hydrogenation of alkenes is exothermic, and heat is released corresponding to the ΔE (colored green).
This heat of reaction can be used to evaluate the thermodynamic stability of alkenes having different numbers of alkyl substituents on the double bond. For example, the following table lists the heats of hydrogenation for three C5H10 alkenes which give the same alkane product (2-methylbutane). Since a large heat of reaction indicates a high energy reactant, these heats are inversely proportional to the stabilities of the alkene isomers. To a rough approximation, we see that each alkyl substituent on a double bond stabilizes this functional group by a bit more than 1 kcal/mole.
Alkene Isomer (CH3)2CHCH=CH2
3-methyl-1-butene
CH2=C(CH3)CH2CH3
2-methyl-1-butene
(CH3)2C=CHCH3
2-methyl-2-butene
Heat of Reaction
( ΔHº )
–30.3 kcal/mole –28.5 kcal/mole –26.9 kcal/mole
Stereochemistry of Catalytic Hydrogenation
From the mechanism shown below, we expect the addition of hydrogen to occur with syn-stereoselectivity since both reactants approach the same side of the catalyst's surface.
For example, 1,2-dimethylcyclopentene is reduced to 1,2-dimethylcyclopentane during catalytic hydrogenation.
Exercises
1. Use the catalytic hydrogenation of ethene with platinum oxide to answer the following questions.
1. 0.500 mol of ethene reacts with _______ mol of hydrogen.
2. Ethene is being _______; while _______ is being oxidized.
3. The oxidation number of carbon in ethene is _______; in ethane it is _______.
2. When 1.000 g of a certain triglyceride (fat) is treated with hydrogen gas in the presence of Adams’ catalyst, it is found that the volume of hydrogen gas consumed at 99.8 kPa and 25.0°C is 162 mL. A separate experiment indicates that the molar mass of the fat is 914 g mol−1. How many carbon-carbon double bonds does the compound contain?
3. Bromobutene reacts with hydrogen gas in the presence of a platinum catalyst. What is the name of the product?
4. Cyclohexene reacts with hydrogen gas in the presence of a palladium catalyst. What is the name of the product?
5. What is the stereochemistry of an alkene hydrogenation reaction?
6. When looking at their heats of hydrogenation, is the cis or the trans isomer generally more stable?
7. 2-chloro-4-ethyl-3methylcyclohexene reacts with hydrogen gas in the presence of a platinum catalyst. What is the name of the product?
8. Predict the products if the following alkenes were reacted with catalytic hydrogen.
Answer
1. a. 0.500 mole of hydrogen gas
b. Ethene is being reduced; while hydrogen is being oxidized.
c. The oxidation number of carbon in ethene is −2; in ethane it is −3.
2. Amount of hydrogen consumed $\begin{array}{l}=n\text{\hspace{0.17em}}\text{mol}\ \text{=}\frac{PV}{RT}\ =\frac{99.8\text{\hspace{0.17em}}\text{kPa}×0.162\text{\hspace{0.17em}}\text{L}}{8.31\text{\hspace{0.17em}}\text{kPa}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}×298\text{\hspace{0.17em}}\text{K}}\ =6.53×{10}^{-3}\text{\hspace{0.17em}}\text{mol}\text{\hspace{0.17em}}{\text{H}}_{2}\end{array}$
Amount of fat used $\begin{array}{l}=\frac{\left(1.000\text{\hspace{0.17em}}\text{g}\right)×\left(1\text{\hspace{0.17em}}\text{mol}\right)}{\left(914\text{\hspace{0.17em}}\text{g}\right)}\ =1.09×{10}^{-3}\text{\hspace{0.17em}}\text{mol}\text{\hspace{0.17em}}\text{fat}\end{array}$
Ratio of moles of hydrogen consumed to moles of fat $\begin{array}{l}=6.53×{10}^{-3}:1.09×{10}^{-3}\ =6:1\end{array}$
Thus, the fat contains six carbon-carbon double bonds per molecule.
3. Bromobutane
4. Cyclohexane
5. Syn-addition
6. Trans
7. 2-chloro-4-ethyl-3methylcyclohexane
8.
9. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.11%3A_Reduction_of_Alkenes_-_Catalytic_Hydrogenation.txt |
Learning Objective
• recognize organic oxidation and reduction reactions
• predict the products/specify the reagents for epoxidation of alkenes
Oxidation - a closer look
There are a variety of oxidative reagents that can react with alkenes. These reagents oxidize the alkene to different degrees and have different synthetic applications. It can be helpful to describe the relative oxidative strength of the reagents. Some reagents are so strong that the carbon chain will be cleaved at the alkene. This reactivity can also be a useful distinction. Before we explore the specific details of these different reaction pathways, let's look at the overall patterns of functional group reactivity.
The are four levels of oxidation for alkenes. The gentlest and least oxidative is epoxide (oxacyclopropane) formation in which the vinyl carbons share a single oxygen atom as a three membered ring. Moderate oxidation will convert the alkene into a vicinal diol in which each vinyl carbon is bonded to an independent oxygen atom. The stronger oxidative reactions cleave the carbon chain at the alkene. While the overall chemical process is an oxidation reaction, the work-up (second step) of the reaction can be performed under reductive or gentle conditions or a strong, oxidative cleavage reaction can occur with the strongest reagents. These four reaction pathways are summarized below.
Epoxide (Oxacyclopropane) Synthesis by Peroxycarboxylic Acid
Oxacyclopropane rings, also called epoxide rings, are useful reagents that may be opened by further reaction to form anti vicinal diols. One way to synthesize oxacyclopropane rings is through the reaction of an alkene with a peroxycarboxylic acid, such as MCPBA (m-chloroperoxybenzoic acid). 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.
Mechanism
The mechanism is a concerted reaction between the alkene and peroxyacid. As seen with other concerted reactions, it is stereospecific: a cis-alkene will produce a cis-epoxide and a trans alkene will produce a trans-epoxide.
Peroxycarboxylic acids are generally unstable. An exception is MCPBA, shown in the mechanism above. Often abbreviated MCPBA, it is a stable crystalline solid. Consequently, MCPBA is popular for laboratory use. However, MCPBA can be explosive under some conditions. Peroxycarboxylic acids are sometimes replaced in industrial applications by monoperphthalic acid, or the monoperoxyphthalate ion bound to magnesium, which gives magnesium monoperoxyphthalate (MMPP). In either case, a nonaqueous solvent such as chloroform, ether, acetone, or dioxane is used. This is because in an aqueous medium with any acid or base catalyst present, the epoxide ring is hydrolyzed to form a vicinal diol, a molecule with two OH groups on neighboring carbons. (For more explanation of how this reaction leads to vicinal diols, see below.) However, in a nonaqueous solvent, the hydrolysis is prevented and the epoxide ring can be isolated as the product. Reaction yields from this reaction are usually about 75%. The reaction rate is affected by the nature of the alkene, with more nucleophilic double bonds resulting in faster reactions.
Example \(1\)
Since the transfer of oxygen is to the same side of the double bond, the resulting oxacyclopropane ring will have the same stereochemistry as the starting alkene. A good way to think of this is that the alkene is rotated so that some constituents are coming forward and some are behind. Then, the oxygen is inserted on top. (See the product of the above reaction.) One way the epoxide ring can be opened is by an acid catalyzed oxidation-hydrolysis. Oxidation-hydrolysis gives a vicinal diol, a molecule with OH groups on neighboring carbons. For this reaction, the dihydroxylation is anti since, due to steric hindrance, the ring is attacked from the side opposite the existing oxygen atom. Thus, if the starting alkene is trans, the resulting vicinal diol will have one S and one R stereocenter. But, if the starting alkene is cis, the resulting vicinal diol will have a racemic mixture of S, S and R, R enantiomers.
Exercise \(1\)
1. Predict the product of the reaction of cis-2-hexene with MCPBA (meta-chloroperoxybenzoic acid)
a) in acetone solvent.
b) in an aqueous medium with acid or base catalyst present.
2. Predict the product of the reaction of trans-2-pentene with magnesium monoperoxyphthalate (MMPP) in a chloroform solvent.
3. Predict the product of the reaction of trans-3-hexene with MCPBA in ether solvent.
4. Predict the reaction of propene with MCPBA.
a) in acetone solvent
b) after aqueous work-up.
5. Predict the reaction of cis-2-butene in chloroform solvent.
Answer
1. a) Cis-2-methyl-3-propyloxacyclopropane
b) Racemic (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol
2. Trans-3-ethyl-2-methyloxacyclopropane.
3. Trans-3,4-diethyloxacyclopropane.
4. a) 1-ethyl-oxacyclopropane
b) Racemic (2S)-1,2-propandiol and (2R)-1,2-propanediol
5. Cis-2,3-dimethyloxacyclopropane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.12%3A_Oxidation_of_Alkenes_-_Epoxidation.txt |
Learning Objective
• predict the products/specify the reagents for dihydroxylation of alkenes
Dihydroxylation of alkenes
Alkenes can be dihydroxylated by two different stereochemical pathways: anti-dihydroxylation or syn-dihydroxylation. The opening of epoxides follows the anti-dihydroxylation mechanism, while potassium permanganate or osmium tetroxide produce the syn-dihydroxylated products. The osmium tertroxide reaction can also take place by a two-step process: 1) OsO4 in pyridine followed by 2) H2S or NaHSO3. It is important to note that different professors will emphasize different reagent systems to accomplish the same chemical reaction. In these situations, it can be helpful to recognize the role of each reagent to discern patterns.
Anti Dihydroxylation
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. The mechanism for the ring opening of epoxides depends on the reaction conditions and is discussed in more detail in the next section of this chapter.
Syn Dihydroxylation
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.
Dihydroxylated products (glycols) are obtained by reaction with aqueous potassium permanganate (pH > 8) or osmium tetroxide in pyridine solution. Both reactions appear to proceed by the same mechanism (shown below); the metallocyclic intermediate may be isolated in the osmium reaction. In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. From the mechanism shown here we would expect syn-stereoselectivity in the bonding to oxygen, and regioselectivity is not an issue.
When viewed in context with the previously discussed addition reactions, the hydroxylation reaction might seem implausible. Permanganate and osmium tetroxide have similar configurations, in which the metal atom occupies the center of a tetrahedral grouping of negatively charged oxygen atoms. How, then, would such a species interact with the nucleophilic pi-electrons of a double bond? A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum. Back-bonding of the nucleophilic oxygens to the antibonding π*-orbital completes this interaction. The result is formation of a metallocyclic intermediate, as shown above.
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.
$OsO_4$ is formed slowly when osmium powder reacts with gasoues $O_2$ at ambient temperature. Reaction of bulk solid requires heating to 400 °C:
$Os_{(s)} + 2O_{2\;(g)} \rightarrow OS_4$
Since Osmium tetroxide is expensive and highly toxic, the reaction with alkenes has been modified. Catalytic amounts of OsO4 and stoichiometric amounts of an oxidizing agent such as hydrogen peroxide are now used to eliminate some hazards. Also, an older reagent that was used instead of OsO4 was potassium permanganate, $KMnO_4$. Although syn diols will result from the reaction of KMnO4 and an alkene, potassium permanganate is less useful since it gives poor yields of the product because of overoxidation.
Chemical Highlight
Antitumor drugs have been formed by using dihydroxylation. This method has been applied to the enantioselective synthesis of ovalicin, which is a class of fungal-derived products called antiangiogenesis agents. These antitumor products can cut off the blood supply to solid tumors. A derivative of ovalicin, TNP-470, is chemically stable, nontoxic, and noninflammatory. TNP-470 has been used in research to determine its effectiveness in treating cancer of the breast, brain, cervix, liver, and prostate.
Exercise
1. Give the major product.
2. What is the product in the dihydroxylation of (Z)-3-hexene?
3. What is the product in the dihydroxylation of (E)-3-hexene?
4. Draw the intermediate of this reaction.
5. Fill in the missing reactants, reagents, and product.
Answer
1. A syn-1,2-ethanediol is formed. There is no stereocenter in this particular reaction. The OH groups are on the same side.
2. Meso-3,4-hexanediol is formed. There are 2 stereocenters in this reaction.
3. A racemic mixture of 3,4-hexanediol is formed. There are 2 stereocenters in both products.
4. A cyclic osmic ester is formed.
5. The Diels-Alder cycloaddition reaction is needed in the first box to form the cyclohexene. The second box needs a reagent to reduce the intermediate cyclic ester (not shown). The third box has the product: 1,2-cyclohexanediol.
• Shivam Nand | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.13%3A_Dihydroxylation_of_Alkenes.txt |
Learning Objective
• predict the products/specify the reagents for dihydroxylation of alkenes
Epoxide ring-opening reactions - SN1 vs. SN2, regioselectivity, and stereoselectivity
The nonenzymatic ring-opening reactions of epoxides provides an oppourtunity to review the nucelophilic substitution mechansims. Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions. 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 reacts with the nucleophile under steric considerations and produces product B in the example below.
Conversely, when solvolysis occurs in acidic methanol, the reaction occurs by a mechanism with substantial SN1 character, and the more substituted carbon reacts with the nucleophile under electrostatic considerations and produces product A in the example below.
These are both 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).
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 potent: a 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, nucleophilic reactions take place with backside orientation relative to the leaving group, resulting in inversion at the electrophilic carbon.
Probably the best way to depict the acid-catalyzed epoxide ring-opening reaction is as a hybrid, or cross, between an SN2 and SN1 mechanism. First, the oxygen is protonated, creating a good leaving group (step 1 below) . Electrostatic considerations have greater importance with a protonated intermediate. As the carbon-oxygen bond begins to break (step 2), positive charge builds on the more substituted carbon with greater carbocation stability.
Unlike in an SN1 reaction, the nucleophile reacts with the electrophilic carbon (step 3) before a complete carbocation intermediate has a chance to form.
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, the regiochemical outcome is different from the base-catalyzed reaction. In the acid-catalyzed process, the nucleophile reacts with the more substituted carbon because it is this carbon that holds a greater degree of positive charge and electrostatics (carbocation stability) take a dominant role in determining the mechanism.
Example \(1\)
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!
Solution
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 attack the less substituted carbon and an SN2 like reaction. However, if one of the epoxide carbons is tertiary, the halogen anion will primarily attack the tertialy cabon in a SN1 like reaction.
Example \(1\)
Exercise
1. Given the following, predict the product assuming only the epoxide is affected. (Remember stereochemistry)
2. Predict the product of the following, similar to above but a different nucleophile is used and not in acidic conditions. (Remember stereochemistry)
3. 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
1.
Note that the stereochemistry has been inverted
2,
3.
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)/09%3A_Reactions_of_Alkenes/9.14%3A_Opening_of_Epoxides_-_Acidic_versus_Basic_Conditions.txt |
Learning Objective
• predict the products/specify the reagents for oxidative cleavage of alkenes
Overview
Oxidative cleavage can occur by several different reaction pathways. The cleavage can be strong or gentle depending on the reaction conditions and/or the work-up of the initial reaction product. Both alkenes and alkynes can undergo cleavage reactions. This section will focus on alkenes.
Gentle cleavage of alkenes occurs by two primary reaction pathways: ozolysis with a reductive work-up or syn-dihydroxylation followed by oxidation with perioidc acid. Gentle cleavage will leave terminal carbons partially oxidized to aldehydes. Strong cleavage of alkenes will fully oxidize terminal carbons to carboxylic acids. Internal carbons become ketones by either reaction pathway.
Why so many reactions?
At first glance, it may seem silly to have more than more reaction pathway for the same functional group conversion. As this course proceeds, it will become necessary to target reactions to a single functional group of an organic compound with multiple functional groups. A particular reaction pathway can be advantageous when the reactivity of the entire molecule is considered, not just a single functional group.
Gentle Cleavage: Syn-Dihydroxylation followed by Periodic Acid
Alkenes can also be gently cleavage in a two-step reaction sequence in which the alkene first undergoes syn-dihydroxylation using cold, slightly basic KMnO4 or OsO4/H2O2 followed by oxidation with periodic acid (HIO4). Both reaction sequences are shown below using 1-methylcyclohexene as an example.
Gentle Cleavage: Ozonolysis with a Reduction Work-up
Ozonolysis is a method of oxidatively cleaving alkenes or alkynes using ozone (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes by breaking them down into smaller, more easily identifiable pieces. Ozonolysis also occurs naturally and would break down repeated units used in rubber and other polymers. On an industrial scale, azelaic acid and pelargonic acids are produced from ozonolysis.
Ozonolysis Reaction Mechanism
The gaseous ozone is first passed through the desired alkene solution in either methanol or dichloromethane. The first intermediate product is an ozonide molecule which is then further reduced to carbonyl products. This results in the breaking of the carbon-carbon double bond and is replaced by a carbon-oxygen double bond instead.
Step 1:
The first step in the mechanism of ozonolysis is the initial electrophilic addition of ozone to the carbon-carbon double bond to form the molozonide intermediate. Due to low stablility of molozonide, it continues reacting and breaks apart to form a carbonyl and a carbonyl oxide molecule.
Step 2:
The electrons of the carbonyl and the carbonyl oxide form the stable ozonide intermediate which can then undergo an oxidative or reductive work-up to form the products of interest. A reductive workup converts the ozonide molecule into the desired carbonyl products with aldehyde on terminal carbons. An oxidative workup converts the ozonide molecule into the desired carbonyl products with carboxylic acids on terminal carbons. The two reaction workup conditions are summarized below.
Exercises
Answers
Strong Cleavage from Strong Oxidative Reactions
When the reaction conditions for potassium permanganate are warm, then the oxidation reaction is stronger and cleavage occurs at the alkene with any terminal carbons fully oxidizing to carboxylic acids. When the ozonolysis reaction is followed by an oxidative work-up, then any terminal carbons will oxidize fully to the carboxylic acid as shown in the example below. The example for 1-methylcycohexene is shown below.
Exercise
5. What would you expect the products to be from the reaction of cis-2-pentene with m-chloro-peroxybenzoic acid? Show the stereochemistry of the final product.
6. Give a reaction scheme with starting alkenes and required reagents to produce the following compounds.
Answer
5.
6.
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)/09%3A_Reactions_of_Alkenes/9.15%3A_Oxidative_Cleavage_of_Alkenes.txt |
Learning Objective
• predict the products of carbene additions to alkenes
Introduction
Carbenes were once only thought of as short lived intermediates. The reactions of this section only deal with these short lived carbenes which are mostly prepared in situ, in conjunction with the main reaction. However, there do exist so called persistent carbenes. These persistent carbenes are stabilized by a variety of methods often including aromatic rings or transition metals. In general a carbene is neutral and has 6 valence electrons, 2 of which are non bonding. These electrons can either occupy the same sp2 hybridized orbital to form a singlet carbene (with paired electrons), or two different sp2 orbitals to from a triplet carbene (with unpaired electrons). The chemistry of triplet and singlet carbenes is quite different but can be oversimplified to the statement: singlet carbenes usually retain stereochemistry while triplet carbenes do not. The carbenes discussed in this section are singlet and thus retain stereochemistry.
The reactivity of a singlet carbene is concerted and similar to that of electrophilic or nucleophilic addition (although, triplet carbenes react like biradicals, explaining why sterochemistry is not retained). The highly reactive nature of carbenes leads to very fast reactions in which the rate determining step is generally carbene formation.
Preparation of methylene
The preparation of methylene starts with the yellow gas diazomethane, CH2N2. Diazomethane can be exposed to light, heat or copper to facilitate the loss of nitrogen gas and the formation of the simplest carbene methylene. The process is driven by the formation of the nitrogen gas which is a very stable molecule.
Carbene reaction with alkenes
A carbene such as methlyene will react with an alkene which will break the double bond and result with a cyclopropane. The reaction will usually leave stereochemistry of the double bond unchanged. As stated before, carbenes are generally formed along with the main reaction; hence the starting material is diazomethane not methylene.
In the above case cis-2-butene is converted to cis-1,2-dimethylcyclopropane. Likewise, below the trans configuration is maintained.
Additional Types of Carbenes and Carbenoids
In addition to the general carbene with formula R2C there exist a number of other compounds that behave in much the same way as carbenes in the synthesis of cyclopropane. Halogenated carbenes are formed from halomethanes. An example is dicholorcarbene, Cl2C. These halogenated carbenes will form cyclopropanes in the same manner as methylene but with the interesting presence of two halogen atoms in place of the hydrogen atoms.
Carbenoids are substances that form cyclopropanes like carbenes but are not technically carbenes. One common example is the stereospecific Simmon-Smith reaction which utilizes the carbenoid ICH2ZnI. The carbenoid is formed in situ via the mixing of a Zn-Cu couple with CH2I2.Since this reacts thesame as a carbene, the same methods can be applied to determine the product. An example of this is given as problem 5.
Exercise \(1\)
1. Knowing that cycloalkenes react much the same as regular alkenes what would be the expected structure of the product of cyclohexene and diazomethane facilitated by copper metal?
2. What would be the result of a Simmons-Smith reaction that used trans-3-pentene as a reagent?
3. What starting material could be used to form cis-1,2-diethylcyclopropane?
4. What would the following reaction yield?
5. Draw the product of this reaction. What type of reaction is this?
6. Predict the following products. Will they be the same product?
Answer
1. The product will be a bicyclic ring, Bicyclo[4.1.0]heptane.
2. The stereochemistry will be retained making a cyclopropane with trans methyl and ethyl groups. Trans-1-ethyl-2-methylcyclopropane
3. The cis configuration will be maintained from reagent to product so we would want to start with cis-3-hexene. A Simmons Smith reagent, or methylene could be used as the carbene or carbenoid.
4. The halogenated carbene will react the same as methylene yielding, cis-1,1-dichloro-2,3dimethylcyclopropane.
5. This is a Simmons-Smith reaction which uses the carbenoid formed by the CH2I2 and Zu-Cu. The reaction results in the same product as if methylene was used and retains stereospecificity. Iodine metal and the Zn-Cu are not part of the product. The product is trans-1,2-ethyl-methylcyclopropane.
6. No they will not be the same product, they will be isomers of each other. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.16%3A_Addition_of_Carbenes_to_Alkenes_-_Cyclopropane_Synthesis.txt |
Learning Objective
• predict the polymer/specify the monomer for radical, chain -growth polymers of alkenes
Introduction
All the monomers from which addition polymers are made are alkenes or functionally substituted alkenes. The most common and thermodynamically favored chemical transformations of alkenes are addition reactions. Many of these addition reactions are known to proceed in a stepwise fashion by way of reactive intermediates, and this is the mechanism followed by most polymerizations. A general diagram illustrating this assembly of linear macromolecules, which supports the name chain growth polymers, is presented here. Since a pi-bond in the monomer is converted to a sigma-bond in the polymer, the polymerization reaction is usually exothermic by 8 to 20 kcal/mol. Indeed, cases of explosively uncontrolled polymerizations have been reported.
It is useful to distinguish four polymerization procedures fitting this general description.
• Radical Polymerization The initiator is a radical, and the propagating site of reactivity (*) is a carbon radical.
• Cationic Polymerization The initiator is an acid, and the propagating site of reactivity (*) is a carbocation.
• Anionic Polymerization The initiator is a nucleophile, and the propagating site of reactivity (*) is a carbanion.
• Coordination Catalytic Polymerization The initiator is a transition metal complex, and the propagating site of reactivity (*) is a terminal catalytic complex.
Radical Chain-Growth Polymerization
Virtually all of the monomers described above are subject to radical polymerization. Since this can be initiated by traces of oxygen or other minor impurities, pure samples of these compounds are often "stabilized" by small amounts of radical inhibitors to avoid unwanted reaction. When radical polymerization is desired, it must be started by using a radical initiator, such as a peroxide or certain azo compounds. The formulas of some common initiators, and equations showing the formation of radical species from these initiators are presented below.
By using small amounts of initiators, a wide variety of monomers can be polymerized. One example of this radical polymerization is the conversion of styrene to polystyrene, shown in the following diagram. The first two equations illustrate the initiation process, and the last two equations are examples of chain propagation. Each monomer unit adds to the growing chain in a manner that generates the most stable radical. Since carbon radicals are stabilized by substituents of many kinds, the preference for head-to-tail regioselectivity in most addition polymerizations is understandable. Because radicals are tolerant of many functional groups and solvents (including water), radical polymerizations are widely used in the chemical industry.
In principle, once started a radical polymerization might be expected to continue unchecked, producing a few extremely long chain polymers. In practice, larger numbers of moderately sized chains are formed, indicating that chain-terminating reactions must be taking place. The most common termination processes are Radical Combination and Disproportionation. These reactions are illustrated by the following equations. The growing polymer chains are colored blue and red, and the hydrogen atom transferred in disproportionation is colored green. Note that in both types of termination two reactive radical sites are removed by simultaneous conversion to stable product(s). Since the concentration of radical species in a polymerization reaction is small relative to other reactants (e.g. monomers, solvents and terminated chains), the rate at which these radical-radical termination reactions occurs is very small, and most growing chains achieve moderate length before termination.
The relative importance of these terminations varies with the nature of the monomer undergoing polymerization. For acrylonitrile and styrene combination is the major process. However, methyl methacrylate and vinyl acetate are terminated chiefly by disproportionation.
Another reaction that diverts radical chain-growth polymerizations from producing linear macromolecules is called chain transfer. As the name implies, this reaction moves a carbon radical from one location to another by an intermolecular or intramolecular hydrogen atom transfer (colored green). These possibilities are demonstrated by the following equations
Chain transfer reactions are especially prevalent in the high pressure radical polymerization of ethylene, which is the method used to make LDPE (low density polyethylene). The 1º-radical at the end of a growing chain is converted to a more stable 2º-radical by hydrogen atom transfer. Further polymerization at the new radical site generates a side chain radical, and this may in turn lead to creation of other side chains by chain transfer reactions. As a result, the morphology of LDPE is an amorphous network of highly branched macromolecules.
Exercise
1. Propose the monomer units in the following polymers:
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.17%3A__Radical_Chain-Growth_Polymerization.txt |
Learning Objective
• discuss an example biological addition reactions
Radical mechanisms for flavin-dependent reactions
Flavin coenzymes, like their nicotinamide adenine dinucleotide counterparts, can act as hydride acceptors and donors. In these redox reactions, two electrons are transferred together in the form of a hydride ion. Flavin, however, is also capable of mediating chemical steps in which a single unpaired electron is transferred - in other words, radical chemistry. This is due to the ability of the flavin system to form a stabilized radical intermediate called a semiquinone, formed when FADH2 (or FMNH2) donates a single electron, or when FAD (or FMN) accepts a single electron.
This single-electron transfer capability of flavins is critical to their metabolic role as the entry point of electrons into the electron transport phase of respiration. Electrons 'harvested' from the oxidation of fuel molecules are channeled, one by one, by FMNH2 into the electron transport chain, where they eventually reduce molecular oxygen. NADH is incapable of single electron transfer - all it can do is transfer two electrons, in the form of a hydride, to FMN; the regenerated FMNH2 is then able to continue sending single electrons into the transport chain.
You will learn more details about this process in a biochemistry class.
Because flavins are capable of single-electron as well as two-electron chemistry, the relevant mechanisms of flavoenzyme-catalyzed reactions are often more difficult to determine. Recall the dehydrogenation reaction catalyzed by acyl-CoA dehydrogenase (section 16.5C) - it involves the transfer of two electrons and two protons (ie. a hydrogen molecule) to FAD. Both electrons could be transferred together, with the FAD coenzyme simply acting as a hydride acceptor (this is the mechanism we considered previously). However, because the oxidizing coenzyme being used is FAD rather than NAD+, it is also possible that the reaction could proceed by a single-electron, radical intermediate process. In the alternate radical mechanism proposed below, for example, the enolate intermediate first donates a single electron to FAD, forming a radical semiquinone intermediate (step 2). The second electron is transferred when the semiquinone intermediate abstracts a hydrogen from Cb in a homolytic fashion (step 3).
Scientists are still not sure which mechanism - the hydride transfer mechanism that we saw in section 16.5B or the single electron transfer detailed above - more accurately depicts what is going on in this reaction.
The conjugated elimination catalyzed by chorismate synthase (section 14.3B) is another example of a reaction where the participation of flavin throws doubt on the question of what is the relevant mechanism. This could simply be a conjugated E1' reaction, with formation of an allylic carbocation intermediate. The question plaguing researchers studying this enzyme, however, is why FADH2 is required. This is not a redox reaction, and correspondingly, FADH2 is not used up in the course of the transformation - it just needs to be bound in the active site in order for the reaction to proceed. Given that flavins generally participate in single-electron chemistry, this is an indication that radical intermediates may be involved. Recently an alternative mechanism, involving a flavin semiquinone intermediate, has been proposed (J. Biol. Chem 2004, 279, 9451). Notice that a single electron is transferred from substrate to coenzyme in step 2, then transferred back in step 4. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.18%3A_Biological_Additions_of_Radicals_to__Alkenes.txt |
Addition of Hydrogen Halides to Alkenes
9-1 Give the IUPAC name for the product of the following reaction.
9-2 Draw the reaction mechanism of the previous problem (9-1).
9-3 Identify the product of the following reactions.
9-4 Identify the products of the following reactions.
Addition of Water: Hydration of Alkenes
9-5 Identify the product of the following reactions.
9-6 Identify the product of the following reaction.
9-7 Propose a plausible route of synthesis for the following product starting with 1-butanol.
9-8 Which of the following alkenes can be used to obtain 3,4-dimethylpentan-2-ol through a hydration reaction using dilute acid?
a) (2Z)-3,4-dimethylpent-2-ene
b) 3,4-dimethylpent-1-ene
c) 2,3,4-trimethylpent-2-ene
d) 2,4-dimethyl-3-methylidenepentane
Hydration by Oxymercuration-Demercuration
9-9 Explain why hydration of an alkene by Oxymercuration-Demercuration gives the Markovnikov product.
9-10 Identify the products of the following reactions.
9-11 Identify the product of the following reaction.
9-12 Propose a possible route of synthesis for the following ether starting with 2-ethylbutan-1-ol.
9-13 Give the IUPAC name of the product of the following reaction.
a) 2-methylpropan-2-ol
b) 2,2-dimethylbutane
c) 2-methoxy-2-methylpropane
d) 1,1-dimethylcyclopropane
Hydroboration of Alkenes
9-14 Identify the products of the following reactions.
9-15 Identify the products of the following reactions.
9-16 Give the IUPAC name for the product of the following reaction.
Addition of Halogens to Alkenes
9-17 Identify the products of the following reactions.
9-18 Identify the product of the following reaction.
9-19 What is the product of the following reaction?
a) (2Z)-2-bromo-3-methylpent-2-ene
b) 2-bromo-3-methylpentane
c) 2,3-dibromo-3-methylpentane
d) (2E)-4-bromo-3-methylpent-2-ene
9-20 What reagents can be used in each step to obtain the following products?
9-21 Explain why you do not obtain a mixture of cis- and trans-brominated products when you react Br2/CCl4 with cyclopentene.
Formation of Halohydrins
9-22 Identify the product of the following reaction, making sure to include stereochemistry.
9-23 Draw the mechanism of the reaction in the previous problem (9-22).
9-24 Give the IUPAC name for the product of the following reaction.
9-25 Identify the product of the following reaction.
Catalytic Hydrogenation of Alkenes
9-26 Identify the products of the following reactions.
9-27 Suggest a possible route of synthesis, that includes a catalytic hydrogenation step, to obtain the following product.
9-28 Identify the product of the following reaction.
9-29 Identify the product of the following reaction, making sure to include stereochemistry.
Addition of Carbenes to Alkenes
9-30 Identify the product of the following reaction.
9-31 Identify the product(s) of the following reaction.
9-32 Identify the product of the reaction when (2E)-3-methylhex-2-ene reacts with the carbene product from the previous problem (9-31), then reacts with Br2 and hv.
9-33 Propose a possible route of synthesis for the following reaction.
9-34 Identify the product of the following reaction.
Epoxidation of Alkenes and Acid-Catalyzed Opening of Epoxides
9-35 Identify the product of the following reaction.
9-36 Identify the product of the following reactions, specifying stereochemistry where appropriate.
9-37 Identify the products of the following reaction, including stereochemistry.
9-38 What is the product of the following reaction?
a) benzoic acid
b) 1-cyclohexylethan-1-ol
c) 1-cyclohexylethan-1-one
d) cyclohexanol
9-39 What is the product of the following reaction?
a) 4-chlorobutane-1,3-diol
b) 3-chlorobutan-1-ol
c) 2-chlorobutane-1,4-diol
d) 2,4-dichlorobutan-1-ol
9-40 Draw the arrows for the following epoxidation reaction to show the movement of electrons.
Syn Dihydroxylation of Alkenes
9-41 Identify the product of the following reaction, including stereochemistry.
9-42 Give the IUPAC name for the product(s) of the following reaction. Include stereochemistry.
9-43 Identify the product of the following reaction.
9-44 Suggest a possible route of synthesis for the following compound starting with cyclopentanol.
Oxidative Cleavage of Alkenes
9-45 Identify the products of the following reactions.
9-46 Identify the products of the following reactions.
9-47 Identify the product of the following reaction.
Polymerization of Alkenes
9-48 Identify the alkene monomer that composes the following polymer.
9-49 Draw the mechanism for the acid catalyzed formation of the polymer in the previous problem (9-48).
9-50 Draw the resulting polymer of the following reaction. Draw the chain four monomers in length.
9.20: Solutions to Additional Exercises
9-1
9-2
9-3
9-4
9-5
9-6
9-7
9-8 B.
Hydration by Oxymercuration-Demercuration
9-9 When an alkene is going through oxymercuration, it proceeds through a three-membered mercurinium ion intermediate. This does not allow for rearrangement as no carbocation is formed. In order to open the intermediate ring, the water molecule will attack the most substituted carbon, thus giving the Markovnikov regioselectivity in the final product.
9-10
9-11
9-12
9-13 C.
Hydroboration of Alkenes
9-14
9-15
9-16 2-chloro-3-methylbutane
Addition of Halogens to Alkenes
9-17
9-18
9-19 B.
9-20
A: HBr, ROOR
B: NaNH2
9-21 You do not obtain a mixture of cis- and trans-brominated products from this reaction (only trans products) due to the intermediate the reaction goes through. No carbocation is formed, which would allow the Br- to attack from two possible sides of the carbocation. Instead, a bromonium ion is formed and in order to add the second Br, it needs to attack one side of the bromonium ion intermediate, causing the product to always have a trans configuration.
Formation of Halohydrins
9-22
9-23
9-24 Buta-1,3-diene
9-25
9-26
9-27
9-28
9-29
9-30
9-31
9-32
9-33
9-34
9-35
9-36
9-37
9-38 C.
9-39 A.
9-40
9-41
9-42
9-43
9-44
9-45
9-46
9-47
9-48
9-49
9-50 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/09%3A_Reactions_of_Alkenes/9.19%3A__Additional_Exercises.txt |
learning objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• apply bonding theories to the structure of alkynes and distinguish between internal and terminal triple bonds - refer to section 10.1
• predict relative physical properties of alkynes, such as relative boiling points and solubilities - refer to section 10.1
• predict the products and specify the reagents for the synthesis of alkynes from the double elimination of dihaloalkanes refer to section 10.2
• predict the products and specify the reagents for the Electrophilic Addition Reactions (EARs) of alkynes with HX and X2 - refer to section 10.3
• predict the products and specify the reagents for the Markovnikov-products of alkyne hydration - refer to section 10.4
• predict the products and specify the reagents for the anti-Markovnikov-products of alkyne hydration - refer to section 10.5
• predict the products and specify the reagents for the full or partial reduction of alkynes - refer to section 10.6
• predict the products and specify the reagents for the oxidation of alkynes - refer to section 10.7
• explain why alkynes are more acidic than alkanes and alkenes - refer to section 10.8
• predict the products and specify the reagents to generate nucleophilic acetylide ions and heavy metal acetylides - refer to section 10.8
• predict the products and specify the reagents to synthesize larger alkynes with acetylide ions - refer to section 10.9
• use retrosynthetic analysis to design a multi-step synthesis with correct regiochemistry and stereochemistry using the reactions studied to date - refer to section 10.10
Please note: IUPAC nomenclature and important common names of alkynes were explained in Chapter 3.
10: Alkynes
Learning Objective
• apply bonding theories to the structure of alkynes and distinguish between internal and terminal triple bonds
Alkynes: Terminal vs Internal
Alkynes are organic molecules with carbon-carbon triple bonds. They are unsaturated hydrocarbons with the empirical formula of CnH2n-2. The simplest alkyne is ethyne which has the common name acetylene. Acetylene is a common name to memorize.
It is important to distinguish between terminal and internal alkynes because they can undergo different patterns of reactivity.
Electronic Structure
The sp hybridization of the carbon-carbon triple bond results in the perpendicular orientation of the sigma bond and two pi bonds. The close proximity of the electrons in this geometry orientation creates molecules with less stability. The structure of the carbon-carbon triple bond strongly influences the chemical reactivity of alkynes and the acidity of terminal alkynes. Because of its linear configuration (the bond angle of a sp-hybridized carbon is 180º), a ten-membered carbon ring is the smallest that can accommodate this function without excessive strain.
Physical Properties
Alkynes are nonpolar, unsaturated hydrocarbons with physical properties similar to alkanes and alkenes. Alkynes dissolve in organic solvents, have slight solubility in polar solvents, and are insoluble in water. Compared to alkanes and alkenes, alkynes have slightly higher boiling points. For example, ethane has a boiling point of -88.6 C, while ethene is -103.7 C and ethyne has a higher boiling point of -84.0 ?C.
Exercise
1. Arrange ethane, ethene, and acetylene in order of decreasing carbon-carbon length.
2. How many pi bonds and sigma bonds are involved in the structure of ethyne?
3. What contribute to the weakness of the pi bonds in an alkyne?
4. Arrange the following hydrocarbons in order of decreasing boiling point: 1-heptyne, 1-hexyne, 2-methyl-1-hexyne.
5. Predict the solvent with greater 2-butyne solubility. a) water or 1-octanol? b) water or acetone? c) ethanol or hexane?
Answer
1. relative carbon-carbon bond length: ethane < ethene < acetylene
2. There are three sigma bonds and two pi bonds.
3. The sigma bond and two pi bonds are all perpendicular to each other in the triple bond creating electron repulsion between the three pairs of bonding electrons in the triple bond.
4. 1-heptyne (99.7C) > 2-methyl-1-hexyne (91C) > 1-hexyne (71C)
5. a) 1-octanol b) acetone c) hexane
Outside links
• www.ucc.ie/academic/chem/dolc...t/alkynes.html
• www.cliffsnotes.com/WileyCDA/...eId-22631.html
Contributors and Attributions
• Bao Kha Nguyen, Garrett M. Chin | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/10%3A_Alkynes/10.01%3A_Structure_and_Physical_Properties.txt |
Introduction
To synthesize alkynes from dihaloalkanes we use dehydrohalogenation. The majority of these reactions take place using alkoxide bases (other strong bases can also be used) with high temperatures. This combination results in the majority of the product being from the E2 mechanism. Recall that the E2 mechanism is a concerted reaction (occurs in 1 step). However, in this 1 step there are 3 different changes in the molecule. This is the reaction between 2-bromo-2-methylpropane and sodium hydroxide.
Now, if we apply this concept using 2 halides on vicinal or geminal carbons, the E2 reaction will take place twice resulting in the formation of 2 pi bonds and an alkyne as shown in the examples below where the strong base is symbolized B-.
Double E2 of a Vicinal Dihalide Double E2 of a Geminal Dihalilde
It is important to note that the reaction of terminal haloalkanes requires 3 equivalents of base instead of 2 because of the relative acidity of alkynes that is discussed in a later section of this chapter.
The mechanism of a reaction between 2,3-dibromopentane with sodium amide in liquid ammonia is shown below where liquid ammonia is not part of the reaction, but is used as a solvent.
Notice the intermediate of the alkyne synthesis. It is stereospecifically in its anti form. Because the second proton and halogen are pulled off the molecule this is unimportant to the synthesis of alkynes.
Preparation of Alkynes from Alkenes
Lastly, we will briefly look at how to prepare alkynes from alkenes. This is a simple process using first halogenation of the alkene bond to form the dihaloalkane, and next, using the double elimination process form the alkyne.
This first process is gone over in much greater detail in the page on halogenation of an alkene. In general, chlorine or bromine is used with an inert halogenated solvent like chloromethane to create a vicinal dihalide from an alkene. The vicinal dihalide formed is then reacted with a strong base and heated to produce an alkyne. The two-step reaction pathway is shown below.
In The Lab
Due to the strong base and high temperatures needed for this reaction to take place, the triple bond may change positions. An example of this is when reactants that should form a terminal alkyne, form a 2-alkyne instead. The use of NaNH2 in liquid NH3 is used in order to prevent this from happening due to its lower reacting temperature. Even so, most chemists will prefer to use nucleophilic substitution instead of elimination when trying to form a terminal alkyne.
Exercise \(1\)
Question 1: Why would we need 3 bases for every terminal dihaloalkane instead of 2 in order to form an alkyne?
Question 2: What are the major products of the following reactions:
a.) 1,2-Dibromopentane with sodium amide in liquid ammonia
b.) 1-Pentene first with Br2 and chloromethane, followed by sodium ethoxide (Na+ -O-CH2CH3)
Question 3: What would be good starting molecules for the synthesis of the following molecules:
Question 4: Use a 6 carbon diene to synthesize a 6 carbon molecule with 2 terminal alkynes.
Answer
Answer 1: Remember that hydrogen atoms on terminal alkynes make the alkyne acidic. One of the base molecules will pull off the terminal hydrogen instead of one of the halides like we want.
Answer 2:
a.) 1-Pentyne
b.) 1-Pentyne
Answer 3:
Answer 4: Bromine or chlorine can be used with different inert solvents for the halogenation. This can be done using many different bases. Liquid ammonia is used as a solvent and needs to be followed by an aqueous work-up. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/10%3A_Alkynes/10.02%3A_10.2_Synthesis_of_Alkynes_-_Elimination_Reactions_of_Dihalides.txt |
Learning Objective
• predict the products and specify the reagents for the Electrophilic Addition Reactions (EARs) of alkynes with HX and X2
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Since the most common chemical transformation of a carbon-carbon double bond is an addition reaction, we might expect the same to be true for carbon-carbon triple bonds. Indeed, most of the alkene addition reactions also take place with alkynes with similar regio- and stereoselectivity.
When the addition reactions of electrophilic reagents, such as strong Brønsted acids and halogens, to alkynes are studied we find a curious paradox. The reactions are even more exothermic than the additions to alkenes, and yet the rate of addition to alkynes is slower by a factor of 100 to 1000 than addition to equivalently substituted alkenes. The reaction of one equivalent of bromine with 1-penten-4-yne, for example, gave 4,5-dibromo-1-pentyne as the chief product.
HC≡C-CH2-CH=CH2 + Br2 → HC≡C-CH2-CHBrCH2Br
Although these electrophilic additions to alkynes are sluggish, they do take place and generally display Markovnikov Rule regioselectivity and anti-stereoselectivity. One problem, of course, is that the products of these additions are themselves substituted alkenes and can therefore undergo further addition. Because of their high electronegativity, halogen substituents on a double bond act to reduce its nucleophilicity, and thereby decrease the rate of electrophilic addition reactions. Consequently, there is a delicate balance as to whether the product of an initial addition to an alkyne will suffer further addition to a saturated product. Although the initial alkene products can often be isolated and identified, they are commonly present in mixtures of products and may not be obtained in high yield. The following reactions illustrate many of these features. In the last example, 1,2-diodoethene does not suffer further addition inasmuch as vicinal-diiodoalkanes are relatively unstable.
As a rule, electrophilic addition reactions to alkenes and alkynes proceed by initial formation of a pi-complex, in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond. Such complexes are formed reversibly and may then reorganize to a reactive intermediate in a slower, rate-determining step. Reactions with alkynes are more sensitive to solvent changes and catalytic influences than are equivalent alkenes.
Why are the reactions of alkynes with electrophilic reagents more sluggish than the corresponding reactions of alkenes? After all, addition reactions to alkynes are generally more exothermic than additions to alkenes, and there would seem to be a higher π-electron density about the triple bond ( two π-bonds versus one ). Two factors are significant in explaining this apparent paradox. First, although there are more π-electrons associated with the triple bond, the sp-hybridized carbons exert a strong attraction for these π-electrons, which are consequently bound more tightly to the functional group than are the π-electrons of a double bond. This is seen in the ionization potentials of ethylene and acetylene.
Acetylene HC≡CH + Energy → [HC≡CH •(+) + e(–) ΔH = +264 kcal/mole
Ethylene H2C=CH2 + Energy → [H2C=CH2] •(+) + e(–) ΔH = +244 kcal/mole
Ethane H3C–CH3 + Energy → [H3C–CH3] •(+) + e(–) ΔH = +296 kcal/mole
As defined by the preceding equations, an ionization potential is the minimum energy required to remove an electron from a molecule of a compound. Since pi-electrons are less tightly held than sigma-electrons, we expect the ionization potentials of ethylene and acetylene to be lower than that of ethane, as is the case. Gas-phase proton affinities show the same order, with ethylene being more basic than acetylene, and ethane being less basic than either. Since the initial interaction between an electrophile and an alkene or alkyne is the formation of a pi-complex, in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond, the relatively slower reactions of alkynes becomes understandable.
A second factor is presumed to be the stability of the carbocation intermediate generated by sigma-bonding of a proton or other electrophile to one of the triple bond carbon atoms. This intermediate has its positive charge localized on an unsaturated carbon, and such vinyl cations are less stable than their saturated analogs. Indeed, we can modify our earlier ordering of carbocation stability to include these vinyl cations in the manner shown below. It is possible that vinyl cations stabilized by conjugation with an aryl substituent are intermediates in HX addition to alkynes of the type Ar-C≡C-R, but such intermediates are not formed in all alkyne addition reactions.
Carbocation
Stability
CH3(+) RCH=CH(+) < RCH2(+) RCH=CR(+) < R2CH(+) CH2=CH-CH2(+) < C6H5CH2(+) R3C(+)
Methyl 1°-Vinyl 2°-Vinyl 1°-Allyl 1°-Benzyl
Application of the Hammond postulate indicates that the activation energy for the generation of a vinyl cation intermediate would be higher than that for a lower energy intermediate. This is illustrated for alkenes versus alkynes by the following energy diagrams.
Despite these differences, electrophilic additions to alkynes have emerged as exceptionally useful synthetic transforms.
Addition of Hydrogen Halide to an Alkyne
Summary: Reactivity order of hydrogen halides: HI > HB r> HCl > HF.
Follows Markovnikov’s rule:
• Hydrogen adds to the carbon with the greatest number of hydrogens, the halogen adds to the carbon with fewest hydrogens.
• Protination occurs on the more stable carbocation. With the addition of HX, haloalkenes form.
• With the addition of excess HX, you get anti addition forming a geminal dihaloalkane.
Addition of a HX to an Internal Alkyne
As shown in Figure 2 below, the $\pi$ electrons react with the hydrogen of the HBr and because the alkyne carbons are equivalent it does not matter which carbon adds the hydrogen. Once the hydrogen is covalently bonded to one of the carbons, the bromide will react with the carbocation intermediate to form a vinyl halide as shown in the example of forming 2-bromobutene from 2-butyne reacting with HBr. The reaction below assumes a 1:1 mole ratio of the alkyne and HBr.
Now, what happens if there is excess HBr?
Addition due to excess HX yields a geminal dihaloalkane
Here, the electrophilic addition proceeds with the same steps used to achieve the product in Addition of a HX to an Internal Alkyne. The $\pi$ electrons react with the hydrogen (shown in blue) adding it to the carbon on the left because the lone pair electrons of the bromine can help stabilize the carbocation intermediate that reacts with the bromide ions to form a geminal dihalide.
Addition of HX to Terminal Alkyne
For terminal alkynes, the carbon atoms sharing the triple bond are not equivalent. The addition of HX to terminal alkynes occurs in a Markovnikov-manner in which the halide attaches to the most substituted carbon. The pi electrons react with the hydrogen and it bonds to the terminal carbon. The bromide reacts with the resulting carbocation intermediate to form the vinyl halide. The overall reaction and mechanism are shown below.
Addition due to excess HBr present
Similar to the addition of excess HBr to internal alkynes, both halides will add to the same carbon to form a geminal dihalide.
HBr Addition With Radical Mechanism
Most hydrogen halide reactions with terminal alkynes occur in a Markovnikov-manner in which the halide attaches to the most substituted carbon since it is the most positively polarized. However, there are two specific reactions among alkynes where anti-Markovnikov reactions take place: the radical addition of HBr and Hydroboration Oxidation reactions. For alkynes, an anti-Markovnikov addition takes place for terminal alkynes.
The Br of the Hydrogen Bromide (H-Br) attaches to the less substituted 1-carbon of the terminal alkyne shown below in an anti-Markovnikov manner while the Hydrogen proton attaches to the second carbon. As mentioned above, the first carbon is the less substituted carbon since it has fewer bonds attached to carbons and other substituents. The H-Br reagent must also be reacted with heat or some other radicial initiator such as a peroxide in order for this reaction to proceed in this manner. This presence of the radical or heat leads to the anti-Markovnikov addition since it produces the most stable reaction.
The product of a terminal alkyne that is reacted with a peroxide (or light) and H-Br is a 1-bromoalkene.
Regioselectivity: The Bromine can attach in a syn or anti manner which means the resulting alkene can be both cis and trans. Syn addition is when both Hydrogens attach to the same face or side of the double bond (i.e. cis) while the anti addition is when they attach on opposite sides of the bond (trans).
Halogenation of Alkynes
The additon of X2 to alkynes is analogous to the addition of X2 to alkenes. The halogen molecule becomes polarized by the approach of the nucleophilic alkyne. The pi electrons of the alkyne react with the bromine to form a carbon-bromine bond and cyclic halonium ion with halide as the leaving group. The formation of the cyclic halonium ion requires anti-addition of the nucleophilic halide to produce a vicinal dihalide alkene as shown in the reaction below.
Exercise
1. Draw the structure, and give the IUPAC name, of the product formed in each of the reactions listed below.
1. $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{HCl}}]}}$
2. $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{HCl}}]}}$
3. $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{Br}_2}]}}$
4. $\ce{\sf{CH3-C#C-CH3->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{Br}_2}]}}$
5. $\ce{\sf{CH3CH2-C#C-H->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{HCl}}]}}$
6. $\ce{\sf{CH3CH2-C#C-H->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{HCl}}]}}$
7. $\ce{\sf{CH3CH2CH2-C#C-H->[\displaystyle{\textrm{1 equiv}}][\displaystyle{\textrm{Br}_2}]}}$
8. $\ce{\sf{CH3CH2CH2-C#C-H->[\displaystyle{\textrm{excess}}][\displaystyle{\textrm{Br}_2}]}}$
Answer
1. (Z)-2-chloro-2-butene
2. 2,2-dichlorobutane
3. (E)-2,3-dibromo-2-butene
4. 2,2,3,3-tetrabromobutane
5. 2-chloro-1-butene
6. 2,2-dichlorobutane
7. (E)-1,2-dibromo-1-pentene
8. 1,1,2,2-tetrabromopentane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/10%3A_Alkynes/10.03%3A_Reactions_of_Alkynes_-_Addition_of_HX_and_X.txt |
Learning Objective
• predict the products and specify the reagents for the Markovnikov-products of alkyne hydration
Reaction: Hydration of Alkynes (Markovnikov's Rule)
Hydration of alkynes begins similar to the hydration of alkenes through the addition of the first water molecule. However, this first hydration reaction forms an enol, an alcohol bonded to a vinyl carbon. Enols immediately undergo a special type of isomerization reaction called tautomerization to form carbonyl groups - aldehydes or ketones. To keep things simple, this reaction is called "enol-keto" tautomerization with the understanding that aldehydes form on terminal alkyne carbons. As with alkenes, hydration (addition of water) to alkynes requires a strong acid, usually sulfuric acid with a mercuric sulfate catalyst as shown below.
Enol-Keto Tautomers
Tautomers are defined as rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) 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. The three examples shown below illustrate these reactions for different substitutions of the triple-bond. The tautomerization step is indicated by a red arrow. For terminal alkynes the addition of water follows the Markovnikov rule, as in the second example below, and the final product ia a methyl ketone ( except for acetylene, shown in the first example ). For internal alkynes ( the triple-bond is within a longer chain ) the addition of water is not regioselective. If the triple-bond is not symmetrically located ( i.e. if R & R' in the third equation are not the same ) two isomeric ketones will be formed.
With the addition of water, alkynes can be hydrated to form enols that spontaneously tautomerize to ketones. The reaction is catalyzed by mercury ions and follows Markovnikov’s Rule A useful functional group conversion for multiple -step syntheses is to hydrate terminal alkynes to produce methyl ketones.
Hydration of Alkyne Mechanism
The first step is an acid/base reaction where the π electrons of the triple bond acts as a Lewis base and reacts with the proton therefore protonating the carbon with the most hydrogen substituents as expected by Markovnikov's Rule. In the second step, the nucleophilic water molecule reacts with the electrophilic carbocation to produce an oxonium ion. The oxonium ion is deprotonated by a base to produce an enol which immediately tautomerizes into a ketone. The hydration reaction for propyne is shown below with its mechanism to illustrate the electron flow of the mechanism.
Exercise
1. Draw the structure of the product formed when each of the substances below is treated with H2O/H2SO4 in the presence of HgSO4.
1. CH3CHCCH
2. Draw the structure of the keto form of the compound shown below. Which form would you expect to be the most stable?
3. What alkyne would you start with to gain the following products, in an oxidation reaction? Keep in mind resonance.
4. Propose a reaction scheme for the following compound starting from the alkyne and showing required reagents and intermediates.
Answer
1.
2. ; The keto form should be the most stable.
3.
4.
10.05: Hydration of Alkynes for Anti-Markovnikov Products
Learning Objective
• predict the products and specify the reagents for the anti-Markovnikov-products of alkyne hydration
Introduction
The hydroboration-oxidation of alkynes is similar to the reaction with alkenes. However, there is one important difference. The alkyne has two pi bonds and both are capable of reacting with borane (BH3). To limit the reactivity to only one of the pi bonds within the alkyne, a dialkyl borane reagent (R2BH) is used. Replacing two of the hydrogens on the borane with alkyl groups also creates steric hindrance so that the hydroboration reaction produces the regioselective,
anti-Markovnikov product. Disiamylborane (Sia2BH) and 9-borabicyclo[3.3.1]nonane (9-BBN) are two common reagents for the hydroboration step. Their structures are shown below.
The oxidation reagents ( a basic hydrogen peroxide solution) are the same for both alkenes and alkynes, The overall reaction is shown below.
Mechanism
The hydroboration reaction of alkynes has the same stereo- and regiochemistry as the alkene reaction. The primary difference is the steric hindrance of the two isoamyl groups of the dialkyl borane creates anti-Markovnikov regeioselectivity. The hydrogen and boron bond with the same orientation to the alkyne carbon with syn-addition stereochemistry to form the enol. The enol immediately tautomerizes to the keto form which is an aldehyde for terminal alkynes. The hydration of 1-propyne is shown below along with the reaction mechanism.
Exercise
1. Draw the bond-line structure(s) for the product(s) of each reaction.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/10%3A_Alkynes/10.04%3A_Hydration_of_Alkynes_for_Markovnikov_Products.txt |
Learning Objective
• predict the products and specify the reagents for the full or partial reduction of alkynes
Introduction and Overview
Alkynes can undergo reduction reactions similar to alkenes. These reactions are also called hydrogenation reactions. With the presence of two pi bonds within the carbon-carbon triple bonds, the reduction reactions can be partial or complete depending on the reagents. Since partial reduction of an alkyne produces an alkene, the stereochemistry of the addition mechanism determines whether the cis- or trans- alkene is formed. The three most significant alkyne reduction reactions are summarized below.
Hydrogenation and the Relative Stability of Hydrocarbons
Like alkenes, alkynes readily undergo catalytic hydrogenation partially to cis- or trans- alkenes or fully to alkanes depending on the reaction employed.
The catalytic addition of hydrogen to 2-butyne provides heat of reaction data that reflect the relative thermodynamic stabilities of these hydrocarbons, as shown above. From the heats of hydrogenation, shown in blue in units of kcal/mole, it would appear that alkynes are thermodynamically less stable than alkenes to a greater degree than alkenes are less stable than alkanes. The standard bond energies for carbon-carbon bonds confirm this conclusion. Thus, a double bond is stronger than a single bond, but not twice as strong. The difference ( 63 kcal/mole ) may be regarded as the strength of the π-bond component. Similarly, a triple bond is stronger than a double bond, but not 50% stronger. Here the difference ( 54 kcal/mole ) may be taken as the strength of the second π-bond. The 9 kcal/mole weakening of this second π-bond is reflected in the heat of hydrogenation numbers ( 36.7 - 28.3 = 8.4 ).
Catalytic Hydrogenation of an Alkyne
Alkynes can be fully hydrogenated into alkanes with the help of a platinum, paladium, or nickel catalyst. Because the reaction is catalyzed on the surface of the metal, it is common for these catalysts to dispersed on carbon (Pd/C) or finely dispersed as nickel (Raney-Ni). The full reduction of 2-butyne is shown below as an example.
Hydrogenation of an Alkyne to a Cis-Alkene
Since alkynes are thermodynamically less stable than alkenes, we expect addition reactions of alkynes to be more exothermic and relatively faster than equivalent reactions of alkenes. For catalytic hydrogenation, the Pt, Pd, or Ni catalysts are so effective in promoting addition of hydrogen to both double and triple carbon-carbon bonds that the alkene intermediate formed by hydrogen addition to an alkyne cannot be isolated. A less efficient catalyst, Lindlar's catalyst permits alkynes to be converted to alkenes without further reduction to an alkane. Lindlar’s Catalyst transforms an alkyne to a cis-alkene because the hydrogenation reaction is occurring on the surface of the metal. Both hydrogen atoms are added to the same side of the alkyne as shown in the syn-addition mechanism for hydrogenation of alkenes in the previous chapter.
Lindlar's catalyst is prepared by deactivating (or poisoning) a conventional palladium catalyst. Lindlar’s catalyst has three components: palladium-calcium carbonate, lead acetate and quinoline. The quinoline serves to prevent complete hydrogenation of the alkyne to an alkane. This approach is similar to the one used for hydration of alkynes using a dialkyl borane for hydroboration. A strong reagent is modified into a less reactive form.
Hydrogenation of an Alkyne to a Trans-Alkene
Alkynes can be reduced to trans-alkenes with the use of sodium dissolved in an ammonia solvent. A sodium radical donates an electron to one of the p-orbitals in the carbon-carbon triple bond. This reaction forms an anion that can be protonated by a hydrogen atom in the ammonia solvent which prompts another sodium radical to donate an electron to the second p-orbital. The resulting anion is also protonated by a hydrogen from the ammonia solvent to produce a trans-alkene according to the mechanism shown below.
Mechanism for Hydrogenation of Alkynes to trans-Alkenes
Exercise
1. Using any alkyne how would you prepare the following compounds: pentane, trans-4-methyl-2-pentene, cis-4-methyl-2-pentene.
Answer
1. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/10%3A_Alkynes/10.06%3A_10.6_Reduction_of_Alkynes.txt |
Learning Objective
• predict the products and specify the reagents for the oxidation of alkynes
Alkynes, similar to alkenes, can be oxidized gently or strongly depending on the reaction environment. Since alkynes are less stable than alkenens, the reactions conditions can be gentler. For examples, alkynes form vicinal dicarbonyls in neutral permanganate solution. For the alkene reaction to vicinal dialcohols, the permanganate reaction requires a lightly basic environment for the reaction to occur. During strong oxidation with ozone or basic potassium permanganate, the alkyne is cleaved into two products. Because at least one of the reaction products is a carboxylic acid, it is important to consider the acid-base chemistry of the product in the reaction solution. Carboxylic acids are deprotonated in basic solutions to carboxylates. A second reaction step is required to protonate the carboxylate to the neutral form of the carboxylic acid. The generic reactions are summarized below for the different oxidative conditions - gentle or strong.
Gentle Alkyne Oxidation
Strong Alkyne Oxidation - Oxidative Cleavage
Exercise
1. Draw the bond-line structures for the product(s) of the following reactions.
Answer
1.
10.08: Acidity of Terminal Alkynes and Acetylide Ions
Learning Objectives
• explain why alkynes are more acidic than alkanes and alkenes
• predict the products and specify the reagents to generate nucleophilic acetylide ions and heavy metal acetylides
Acidity of Terminal Alkynes and Acetylilde Ion Formation
Terminal alkynes are much more acidic than most other hydrocarbons. Removal of the proton leads to the formation of an acetylide anion, RC=C:-. The origin of the enhanced acidity can be attributed to the stability of the acetylide anion, which has the unpaired electrons in an sp hybridized orbital. The stability results from occupying an orbital with a high degree of s-orbital character. There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The enhanced acidity with greater s-character occurs despite the fact that the homolytic C-H BDE is larger.
Table 9.7.1: Akynes
Compound Conjugate Base Hybridization "s Character" pKa C-H BDE (kJ/mol)
CH3CH3 CH3CH2- sp3 25% 50 410
CH2CH2 CH2CH- sp2 33% 44 473
HCCH HCC- sp 50% 25 523
Consequently, acetylide anions can be readily formed by deprotonation using a sufficiently strong base. Amide anion (NH2-), in the form of NaNH2 is commonly used for the formation of acetylide anions.
Exercise
1. Given that the pKa of water is 14.00, would you expect hydroxide ion to be capable of removing a proton from each of the substances listed below? Justify your answers, briefly.
1. ethanol (pKa = 16)
2. acetic acid (pKa = 4.72)
3. acetylene (pKa = 25)
Answer
Answers:
1.
1. No, The pKa of ethanol is similar to the pKa of water so proton exchange is comparable for both protonation and deprotonation between alcohols and water. Alcohols can be considered "alkylated water" and share many similarities in both physical properties and chemical reactivity.
2. Yes, very well. There is a difference of 11 pKa units between the pKa of water and the pKa of acetic acid. The equilibrium lies well to the right with acetate as the predominate form of the original acetic acid.
3. No, hardly at all. The hydroxide ion is too weak a base to remove a proton from acetylene. The equilibrium lies so far to the left that it is considered a "No Reaction".
10.09: Synthesis of Larger Alkynes from Acetylides
Learning Objective
• predict the products and specify the reagents to synthesize larger alkynes with acetylide ions
Nucleophilic Substitution Reactions of Acetylides
Acetylide anions are strong bases and strong nucleophiles. Therefore, they are able to displace halides and other leaving groups in substitution reactions. The product is a substituted alkyne with a longer, continuous carbon chain.
Because the ion is a very strong base, the substitution reaction follows the SN2 mechanism and is most efficient with methyl or primary halides without substitution near the reaction center.
Secondary, tertiary or even bulky primary substrates will give elimination by the E2 mechanism.
Nucleophilic Addition of Acetylides to Carbonyls
Acetylide anions will add to aldehydes and ketones to form alkoxides that are subsequently protonated to form propargyl alcohols.
With aldehydes and non-symmetric ketones, in the absence of chiral catalyst, the product will be a racemic mixture of the two enantiomers.
Exercise
1. The pKa of ammonia is 35. Estimate the equilibrium constant for the deprotonation of pent-1-yne by amide, as shown above.
2. Give the possible reactants for the following formations:
3. Propose a synthetic route to produce 2-pentene from propyne and an alkyl halide.
Answer
1. Assuming the pKa of pent-1-yne is about 25, then the difference in pKas is 10. Since pentyne is more acidic, the formation of the acetylide will be favored at equilibrium, so the equilibrium constant for the reaction is about 1010.
2.
3. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/10%3A_Alkynes/10.07%3A_Oxidation_of_Alkynes.txt |
Learning Objective
• use retrosynthetic analysis to design a multi-step synthesis with correct regiochemistry and stereochemistry using the reactions studied to date
Introduction
The study of organic chemistry introduces students to a wide range of interrelated reactions. Alkenes, for example, may be converted to structurally similar alkanes, alcohols, alkyl halides, epoxides, glycols and boranes; cleaved to smaller aldehydes, ketones and carboxylic acids; and enlarged by carbocation and radical additions as well as cycloadditions. Most of these reactions are shown in the Alkene Reaction Map below. All of these products may be subsequently transformed into a host of new compounds incorporating a wide variety of functional groups. Consequently, the logical conception of a multi-step synthesis for the construction of a designated compound from a specified starting material becomes one of the most challenging problems that may be posed. Functional group reaction maps like the one below for alkenes can be helpful in designing multi-step syntheses. It can be helpful to build and design your own reaction maps for each functional group studied.
Alkene Reaction Map
Please note: The reagents for each chemical transformation have been intentionally omitted so that this map can be used as a study tool. The answers are provided at the end of this section as part of the exercises.
Simple Multi-Step Syntheses
A one or two step sequence of simple reactions is not that difficult to deduce. For example, the synthesis of meso-3,4-hexanediol from 3-hexyne can occur by more than one multi-step pathway.
One approach would be to reduce the alkyne to cis or trans-3-hexene before undertaking glycol formation. Permanaganate or osmium tetroxide hydroxylation of cis-3-hexene would form the desired meso isomer.
From trans-3-hexene, it would be necessary to first epoxidize the alkene with a peracid followed by ring opening with acidic or basic hydrolysis.
Longer multi-step syntheses require careful analysis and thought, since many options need to be considered. Like an expert chess player evaluating the long range pros and cons of potential moves, the chemist must appraise the potential success of various possible reaction paths, focusing on the scope and limitations constraining each of the individual reactions being employed. The skill is acquired by practice, experience, and often trial and error.
Thinking it Through with 3 Examples
The following three examples illustrate strategies for developing multi-step syntheses from the reactions studied in the first ten chapters of this text. It is helpful to systematically look for structural changes beginning with the carbon chain and brainstorm relevant functional group conversion reactions. Retro-synthesis is the approach of working backwards from the product to the starting material.
In the first example, we are asked to synthesize 1-butanol from acetylene.
The carbon chain doubles in size indicating an acetylide SN2 reaction with an alkyl halide. Primary alcohol formation from an anti-Markovnikov alkene hydration reaction (hydroboration-oxidation) is more likely than a substitution reaction. Applying retro-synthesis, we work backwards from the alcohol to the alkene to the alkyne from an acetylide reaction that initially builds the carbon chain.
Retro-Synthesis
Working forwards, we specify the reagents needed for each transformation identified from the retro-synthesis. The ethylbromide must also be derived from acetylene so multiple reaction pathways are combined as shown below.
In the second example, we are asked to synthesize 1,2-dibromobutane from acetylene.
Once again there is an increase in the carbon chain length indicating an acetylide SN2 reaction with an alkyl halide similar to the first example. The hydrohalogenation can be subtle to discern because the hydrogen atoms are not shown in bond-line structures. Comparing the chemical formulas of 1-butyne with 1,2-dibromobutane, there is a difference of two H atoms and two Br atoms indicating hydrohalogenation and not halogenation. The addition of both bromine atoms to the same carbon atom also supports the idea that hydrohalogenation occurs on an alkyne and not an alkene. The formation of the geminal dihalide also indicates hydrohalogenation instead of halogenatioin because halogenation produces vicinal dihalides. With this insight, the retro-synthesis indicates the following series of chemical transformations.
Retro-Synthesis
Working forwards, we specify the reagents needed for each transformation.
In the third example, we are asked to produce 6-oxoheptanal from methylcyclohexane.
Counting the carbons, the starting material and product both contain seven carbon atoms and there is a cleavage reaction of an alkene under reductive conditions. One important missing aspect of this reaction is a good leaving group (LG). Alkanes are chemically quite boring. We can burn them as fuel or perform free-radical halogenation to create alkyl halides with excellent leaving groups. With these observations, the following retro-synthesis is reasonable.
Retro-Synthesis
Working forwards, we specify the reagents needed for each reaction. For the initial free-radical halogenation of the alkane, we have the option of chlorine (Cl2) or bromine (Br2). Because methylcyclohexane has several different classifications of carbons, the selectivity of Br2 is more important than the faster reactivity of Cl2. An strong base with heat can be used for the second step to follow an E2 mechanism and form 1-methylcyclohexene. The aldhyde group on the final product indicates gentle oxidative cleavage by any of several reaction pathways. These reactions can be combined in to the following multi-step synthesis.
Reaction Maps to Build Functional Group Conversion Mastery
After working through the examples above, we can see how important it is to memorize all of the functional group reactions studied in the first ten chapters. We can apply the knowledge of these reactions to the wisdom of multi-step syntheses.
Please note: The reagents for each chemical transformation have been intentionally omitted so that these maps can be used as a study tools. The answers are provided at the end of this section as part of the exercises.
Alkyne Reaction Map
Exercise
1. Starting at 3-hexyne predict synthetic routes to achieve:
a) trans-3-hexene
b) 3,4-dibromohexane
c) 3-hexanol.
2. Starting with acetylene and any alkyl halides propose a synthesis to make
a) pentanal
b) hexane.
Answer
1.
2.
10.11: Additional Exercises
Alkyne Reactions
10-1 Predict the product of these following reactions:
a)
b)
c)
d)
10-2 Using acetylene as the starting material, show how you would synthesize the following compounds
a)
b) but-2-yne
c)
d)
10-3 Identify the reagents needed to turn hex-1-yne into the following compounds
a) hexane
b) oct-3-yne
c) cis-hept-2-ene
d) trans-hept-2-ene
e) 2,2-dibromohexane
f) 1-bromohexene
10-4 Show how you would accomplish the following synthetic transformations.
a)
b)
c)
d)
e)
f)
10-5 Deduce the structure of each unknown from the information given.
a) Upon catalytic hydrogenation, unknown A yields pentane. Ozonolysis of A yields butanoic acid, HOOC(CH2)2CH3 and CO2. Draw the structure of compound A
b) Upon catalytic hydrogenation, unknown B yields pentane. Ozonolysis of B yields acetaldehyde, CH3CHO, and propionaldehyde, CH3CH2CHO.
10-6 Use compound A from the previous problem (10-5) and any additional reagents you may need to synthesize the following compound.
10.12: Solutions to Additional Exercises
10-1
a)
b)
c)
d)
10-2
a)
b)
c)
d)
10-3
a)
b)
c)
d)
e)
f)
10-4
a)
b)
c)
d)
e)
f)
10-5
a)
b)
10-6 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/10%3A_Alkynes/10.10%3A_An_Introduction_to_Multiple_Step_Synthesis.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• correlate regions of the electromagnetic spectrum to spectroscopic techniques - refer to section 11.1
• explain how an IR spectrometer works and the IR region interacts with organic compounds - refer to section 11.2
• explain the role of asymmetry in IR absorption - refer to section 11.3
• interpret IR spectra - refer to section 11.4, 11.5, and 11.6
• expalin how a mass spectrometer works - refer to section 11.7
• explain the source of the base peak and molecular ion in a mass spectrum - refer to section 11.7
• correlate bond strength to fragmentation patterns - refer to section 11.8
• use fragmentation patterns to elucidate structural features of organic compounds - refer to section 11.9
• explain how high-resolution mass can be used to determine chemical formulas - refer to section 11.10
• 11.1: The Electromagnetic Spectrum and Spectroscopy
Spectroscopy is an experimental method used by chemists to elucidate structural information. The interaction between a compound or sample and a selected region of the electromagnetic spectrum can be measured both qualitatively and quantitatively.
• 11.2: Infrared (IR) Spectroscopy
The infrared region of the electromagnetic spectrum causes asymmetric bonds to stretch, bend, and/or vibrate. This interaction can be measured to help elucidate chemical structures.
• 11.3: IR-Active and IR-Inactive Vibrations
Asymmetry and polarity increase the strength of IR absorption (infrared active). Symmetrical carbon-carbon double and triple bonds will not absorb IR light and are called "infrared inactive".
• 11.4: Interpretting IR Spectra
The analysis and interpretation of the IR spectra for several compounds are explained.
• 11.5: Infrared Spectra of Some Common Functional Groups
One of the most common applications of infrared spectroscopy is the identification of organic compounds. The IR spectra for the major classes of organic molecules are shown and discussed.
• 11.6: Summary and Tips to Distinguish between Carbonyl Functional Groups
This summary includes the minimum information that needs to be memorized to interpret IR spectra in first year organic chemistry along with some tips on how to distinguish between the different functional groups that all contain at least one carbonyl structural feature.
• 11.7: Mass Spectrometry - an introduction
Mass spectrometry is an analytic method that employs ionization and mass analysis of compounds in order to determine the mass, formula and structure of the compound being analyzed. A mass analyzer is the component of the mass spectrometer that takes ionized masses and separates them based on charge to mass ratios and outputs them to the detector where they are detected and later converted to a digital output.
• 11.8: Fragmentation Patterns in Mass Spectrometry
When interpreting fragmentation patterns, as you might expect, the weakest carbon-carbon bonds are the ones most likely to break.
• 11.9: Useful Patterns for Structure Elucidation
Pattern recognition is a chemistry student's best friend, especially when analyzing and interpreting mass spectra.
• 11.10: Determination of the Molecular Formula by High Resolution Mass Spectrometry
High resolution mass spectrometry can determine molecular formulas by distinguishing between the masses of compounds based on the isotopic distribution of each element in the compound.
11: Infrared Spectroscopy and Mass Spectrometry
Objectives
After completing this section, you should be able to
1. write a brief paragraph discussing the nature of electromagnetic radiation.
2. write the equations that relate energy to frequency, frequency to wavelength and energy to wavelength, and perform calculations using these relationships.
3. describe, in general terms, how absorption spectra are obtained.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• electromagnetic radiation
• electromagnetic spectrum
• hertz (Hz)
• infrared spectroscopy
• photon
• quantum
Study Notes
From your studies in general chemistry or physics, you should be familiar with the idea that electromagnetic radiation is a form of energy that possesses wave character and travels through space at a speed of 3.00 × 108m · s−1. However, such radiation also displays some of the properties of particles, and on occasion it is more convenient to think of electromagnetic radiation as consisting of a stream of particles called photons.
In spectroscopy, the frequency of the electromagnetic radiation being used is usually expressed in hertz (Hz), that is, cycles per second. Note that 1 Hz = s−1.
A quantum is a small, definite quantity of electromagnetic radiation whose energy is directly proportional to its frequency. (The plural is “quanta.”) If you wish, you can read about the properties of electromagnetic radiation and the relationships among wavelength, frequency and energy, or refer to your general chemistry textbook if you still have it.
Note also that in SI units, Planck’s constant is 6.626 × 10−34J · s.
The electromagnetic spectrum
Electromagnetic radiation, as you may recall from a previous chemistry or physics class, is composed of electrical and magnetic waves which oscillate on perpendicular planes. Visible light is electromagnetic radiation. So are the gamma rays that are emitted by spent nuclear fuel, the x-rays that a doctor uses to visualize your bones, the ultraviolet light that causes a painful sunburn when you forget to apply sun block, the infrared light that the army uses in night-vision goggles, the microwaves that you use to heat up your frozen burritos, and the radio-frequency waves that bring music to anybody who is old-fashioned enough to still listen to FM or AM radio.
Just like ocean waves, electromagnetic waves travel in a defined direction. While the speed of ocean waves can vary, however, the speed of electromagnetic waves – commonly referred to as the speed of light – is essentially a constant, approximately 300 million meters per second. This is true whether we are talking about gamma radiation or visible light. Obviously, there is a big difference between these two types of waves – we are surrounded by the latter for more than half of our time on earth, whereas we hopefully never become exposed to the former to any significant degree. The different properties of the various types of electromagnetic radiation are due to differences in their wavelengths, and the corresponding differences in their energies: shorter wavelengths correspond to higher energy.
High-energy radiation (such as gamma- and x-rays) is composed of very short waves – as short as 10-16 meter from crest to crest. Longer waves are far less energetic, and thus are less dangerous to living things. Visible light waves are in the range of 400 – 700 nm (nanometers, or 10-9 m), while radio waves can be several hundred meters in length.
The notion that electromagnetic radiation contains a quantifiable amount of energy can perhaps be better understood if we talk about light as a stream of particles, called photons, rather than as a wave. (Recall the concept known as ‘wave-particle duality’: at the quantum level, wave behavior and particle behavior become indistinguishable, and very small particles have an observable ‘wavelength’). If we describe light as a stream of photons, the energy of a particular wavelength can be expressed as:
$E = \dfrac{hc}{\lambda} \tag{12.5.1}$
where E is energy in kcal/mol, λ (the Greek letter lambda) is wavelength in meters, c is 3.00 x 108 m/s (the speed of light), and h is 9.537 x 10-14 kcal•s•mol-1, a number known as Planck’s constant.
Because electromagnetic radiation travels at a constant speed, each wavelength corresponds to a given frequency, which is the number of times per second that a crest passes a given point. Longer waves have lower frequencies, and shorter waves have higher frequencies. Frequency is commonly reported in hertz (Hz), meaning ‘cycles per second’, or ‘waves per second’. The standard unit for frequency is s-1.
When talking about electromagnetic waves, we can refer either to wavelength or to frequency - the two values are interconverted using the simple expression:
$\lamabda \nu = c \tag{12.5.2}$
where ν (the Greek letter ‘nu’) is frequency in s-1. Visible red light with a wavelength of 700 nm, for example, has a frequency of 4.29 x 1014 Hz, and an energy of 40.9 kcal per mole of photons. The full range of electromagnetic radiation wavelengths is referred to as the electromagnetic spectrum.
Notice in the figure above that visible light takes up just a narrow band of the full spectrum. White light from the sun or a light bulb is a mixture of all of the visible wavelengths. You see the visible region of the electromagnetic spectrum divided into its different wavelengths every time you see a rainbow: violet light has the shortest wavelength, and red light has the longest.
Example
Visible light has a wavelength range of about 400-700 nm. What is the corresponding frequency range? What is the corresponding energy range, in kcal/mol of photons?
Solution
Molecular spectroscopy – the basic idea
In a spectroscopy experiment, electromagnetic radiation of a specified range of wavelengths is allowed to pass through a sample containing a compound of interest. The sample molecules absorb energy from some of the wavelengths, and as a result jump from a low energy ‘ground state’ to some higher energy ‘excited state’. Other wavelengths are not absorbed by the sample molecule, so they pass on through. A detector on the other side of the sample records which wavelengths were absorbed, and to what extent they were absorbed.
Here is the key to molecular spectroscopy: a given molecule will specifically absorb only those wavelengths which have energies that correspond to the energy difference of the transition that is occurring. Thus, if the transition involves the molecule jumping from ground state A to excited state B, with an energy difference of ΔE, the molecule will specifically absorb radiation with wavelength that corresponds to ΔE, while allowing other wavelengths to pass through unabsorbed.
By observing which wavelengths a molecule absorbs, and to what extent it absorbs them, we can gain information about the nature of the energetic transitions that a molecule is able to undergo, and thus information about its structure. | 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/11.01%3A_The_Electromagnetic_Spectrum_and_Spectroscopy.txt |
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