Minggu, 11 Oktober 2009

Delocalized Bonding I

Introduction

Numerous studies of Sn1 and E1 reactions have shown that the stabilities of carbocations increase in the order methyl <1o <>o <>o. See Figure 1.

Figure 1

Carbocation Stabilities

These trends, and many others, arise from extended orbital overlap. Specifically, the overlap involves a p orbital with a sigma bond. This type of s-p overlap is referred to as hyperconjugation.

s-p Conjugation

Carbocation Stabilities

The best way to gain an appreciation of the structural requirements for hyperconjugation is to inspect molecular models. Once you have examined several structures in which hyperconjugation is possible, you will be in a position to understand drawings like those in Figures 2 and 3, which depict various views of the orbital arrangements in the methyl and ethyl carbocations.

Figure 2

No Hyperconjugation Here

Figure 3

Hyperconjugation: Lending a Helping Hand

As the positive charge develops during a chemical reaction, the orbital overlap indicated by the arrow in Figure 3 provides a means for electron density to flow from the sigma bond toward the vacant p orbital of the carbocation. This is a form of electron delocalization, a type of resonance. Free rotation allows each of the C-H bonds to contribute electron density to the vacant p orbital . This charge delocalization reduces the energy of the ethyl cation relative to that of the methyl cation where no such orbital overlap is possible. As View 3 in Figure 3 indicates, the C-H bond is angled away from the p orbital. This reduces the overlap between these orbitals in comparison the the side-by-side overlap of two p orbitals in a pi bond, but beggars can't be choosers, and the carbocation gladly accepts all the help it can get.


Exercise 1 Draw structural diagrams similar to those in Figure 3 for the isopropyl cation, +CH(CH3)2. How many C-H bonds can adopt a conformation where the dihedral angle between the p orbital and the C-H bond is 0o?

Exercise 2 Draw structural diagrams similar to those in Figure 3 for the t-butyl cation, +C(CH3)3. How many C-H bonds can adopt a conformation where the dihedral angle between the p orbital and the C-H bond is 0o?

Exercise 3 Does the overlap shown in Figure 3 increase or decrease the electron density in the C-H bond?

Exercise 4 Given your answer to Exercise 3, would you expect the methyl C-H bond strength in propene, CH3CH=CH2, to be greater than or less than the methyl C-H bond strength in propane, CH3CH2CH3?


Alkene Stabilities

Orbital overlap of the type shown in Figure 3 also accounts for the increase in stability of alkenes that attends increasing the number of alkyl groups attached to the double bond: ethene <>

Figure 4

Alkene Stabilities

Figure 5

It's Still the Same Old Story...


Exercise 5 Draw structural diagrams similar to those in Figure 5 for the 2-methylpropene, (CH3)2C=CH2. How many C-H bonds can adopt a conformation where the dihedral angle between the p orbital on C2 and the C-H bond is 0o?

Exercise 6 A C-C sigma bond may overlap with an adjacent pi orbital in the same way that a C-H bond does. Draw structural diagrams similar to those in Figure 5 for that conformation of 1-butene, CH3CH2CH=CH2 where the dihedral angle between the p orbital on C2 and the C3-C4 bond is 0o. How many sigmas bonds can adopt a conformation where the dihedral angle between the p orbital on C2 and the sigma bond is 0o?

Exercise 7 How many C-H bonds in 2,3-dimethyl-2-butene, (CH3)2C=C(CH3)2 can adopt a conformation where the dihedral angle between a p orbital and the C-H bond is 0o?


Measuring Alkene Stabilities

How do we know that a monosubstituted alkene is more stable than a disubstituted alkene? What experimental data allows us to draw such a conclusion? We will consider two lines of experimental evidence, heats of hydrogenation and heats of combustion. Both of these approaches compare the heat released when isomeric alkenes are converted into the same product(s). Figure 6 illustrates the idea for the hydrogenation and combustion of 1-butene and (Z)-2-butene.

Figure 6

We're Gettin' Hot Now

Hydrogenation of 1-butene produces butane. The reaction releases 1.7 kcal/mol more energy than does hydrogenation of (Z)-2-butene, which produces the same product. Combustion analysis yields the same result: 1-butene releases 1.7 kcal/mol more energy than (Z)-2-butene. The fact that (Z)-2-butene releases less heat than the isomeric 1-butene must mean that it contains less energy. In other words, (Z)-2-butene is more stable than 1-butene.


Exercise 8 If you assume that the 1.7 kcal/mol difference in stability between 1-butene and (Z)-2-butene is due to the three additional hyperconjugative interactions in (Z)-2-butene, then each hyperconjugative interaction stabilizes the pi bond by 1.7/3= 0.57 kcal/mol. Estimate to the nearest 0.1 kcal/mol the heat of hydrogenation of 2-methyl-2-butene, CH3CH=C(CH3)2.
Exercise 9 Compounds A and B are isomeric disubstituted trans-alkenes. Which one is more stable? A B


p-p and n-p Conjugation

In our introduction to resonance theory, we outlined three situations where resonance interactions between a pi system and a pair of electrons on a substituent atom were possible. Figure 7 reiterates those three situations. We have just considered two sets of experimental data which led to the idea that p-s overlap leads to a more stable system. Now we will take a look at experimental evidence that led to similar postulates regarding p-p and p-n overlap.

Figure 7

Interacting Orbitals

Consider the solvolysis reactions shown in Equation 1, where G represents H, CH3, H2C=CH, C6H5, or CH3O.

Table 1 summarizes the relative rates of solvolysis under reaction conditions where all of the substrates display unimolecular kinetics. These values must be considered approximate since the solvent composition varied.

Table 1

Substituent Effects on Sn1 Solvolysis Rates

Entry
Common Name
G
krel
1
methyl chloride
H
1
2
ethyl chloride
CH3
1.7
3
allyl choride
H2C=CH
58
4
benzyl chloride
C6H5
654
5
methyl chlroromethyl ether
CH3O
1014


Exercise 10 Rewrite Equation 1 for each of the groups identified in Table 1.
We'll examine the effect of the methoxy group first since it is the most dramatic. Comparing entries 1 and 5 suggests that the carbocation generated during the solvolysis of methyl chloromethyl ether is much more stable than the one formed during the reaction of methyl chloride. Apparently there is an interaction between the CH3O group attached to the reaction center and the positive charge that develops at the reaction center as the C-Cl bond ionizes. Scheme 1 suggests how this might occur.

Scheme 1

Rationalizing Reaction Results

In the rate determining step, the chlorine atom departs from the reaction center to produce a carbocation. This is labeled "resonance contributor A" in the scheme. As the positive charge develops, electron density is drawn from the oxygen atom towards the electron deficient carbon as indicated by the red arrow. This generates "resonance contributor B". Note that donation of electron density from the oxygen to the carbon involves the overlap of a non-bonding orbital with a p orbital, i.e. it is an n-p interaction like that shown in the middle panel of Figure 7.

The structure of the actual intermediate formed in the reaction is described as a hybrid of resonance contributors A and B. Notice that in each resonance contributor the formal positive charge is assigned to a specific atom, while in the resonance hybrid, it is assigned to the entire molecule. In other words, the charge is localized in structures A and B, and delocalized in the hybrid structure. Although this hybrid is a "blend" of resonance contributors A and B, these two structures do not make equal contributions to the blend. Resonance contributor B makes a greater contribution. It does so for two reasons: 1. There is one more bond in this structure than in structure A. 2. All of the atoms in structure B have a filled valence shell. In structure A the carbon atom has only 6 electrons in its valence shell.

The reaction continues when a solvent molecule attacks the electron deficient carbon as shown at the bottom of Scheme 1. Formation of the C-O bond (arrow 1) results in the regeneration of the lone pair of electrons on the oxygen atom (arrow 2). The product is finally formed when a chloride ion removes the hydrogen atom from the positively charged oxygen.


Exercise 11 Consider the alternative reaction pathway shown below, then draw the structure of the intermediate that would be formed.

a. In this intermediate, what would the charge on the oxygen atom that came from the solvent be?

b. What would the charge on the other oxygen atom be?

c. What would the charge on the methylene carbon atom be?


Finally, let's examine two situations where the p orbital of a carbocation interacts with the p system of one or more multiple bonds. First we'll consider the p system of an alkene, then we'll look at the p system of an aromatic ring.

Solvolysis of allyl chloride occurs approximately 58 times faster than methyl chloride. Following the line of reasoning we have been developing, this suggests that the carbocation involved in the former reaction is more stable than that generated during the latter. Scheme 2 offers a rationalization that should be familiar to you by now.

Scheme 2

Rationalizing Reaction Results (Repeated)

Ionization of the C-Cl bond produces an allylic carbocation. As the red arrow indicates, electron density is then drawn from the p system of the adjacent double bond toward the positively charged center . This generates resonance contributor D. This resonance interaction indicates that the electron density is delocalized in this intermediate in a way that is not possible when G = H.

The reaction is completed by attack of a solvent molecule at the positively charge center. This produces an oxonium ion, which is then deprotonated by chloride ion to yield the final product.


Exercise 12 Draw the structure of the hybrid of resonance contributors C and D in Scheme 2.

Exercise 13 Which resonance structure would you expect to make the more important contribution to the resonance hybrid? C D They would contribute equally.

Exercise 14 Draw the structure of the product that would be formed by attack of the solvent on the positively charged carbon in resonance contributor D.


Before we consider the final example in this topic, you should be aware of an assumption that underlies the arguments presented here. Reactions rates are determined by activation energies. In the case of an Sn1 or E1 reaction, this is the difference in energy between the reactants and the transition state leading to the formation of a carbocationic intermediate. The underlying assumption is that any factor which reduces the energy of the intermediate will also reduce the energy of the transition state leading to that intermediate, thus making the compound react faster. Consider the solvolyses of the isomeric alkyl chlorides shown in reactions 2 and 3.


Exercise 15 1-chloro-2-butene ismonosubstituteddisubstituted trisubstitutedtetrasubstituted

Exercise 16 4-chloro-1-butene ismonosubstituteddisubstituted trisubstitutedtetrasubstituted

Exercise 17 Which alkene is more stable? 1-chloro-2-butene4-chloro-1-butene


The fact of the matter is that reaction 2 proceeds faster than reaction 3. Yet 1-chloro-2-butene is more stable than 4-chloro-1-butene. How can this be? Consider the two reaction coordinate diagrams shown in Figure 8. They represent the first step in the solvolysis of each of these isomeric alkyl halides.

Figure 8

Stability vs Reactivity

The smallest red bar at the right of the diagram indicates that 1-chloro-2-butene is slightly more stable than 4-chloro-1-butene (approximately 1.7 kcal/mol according to the data in Figure 6). The longest red bar approximates the difference in stability between the two carbocationic intermediates, I1 and I2. The p bond stabilizes I2 in a way that it cannot stabilize I1. (The sp3 hybridized carbon between the p orbitals on C-2 and C-4 insulates the carbocation from the p bond in I1. No resonance interaction is possible. See Exercise 9.) In other words, the charge in I2 is delocalized while in I1 it is not. Since the positive charge that is present in I2 is partially developed in transition state TS2, the p bond must afford some stabilization of TS2 that it does not provide TS1. This is indicated by the red bar of intermediate length in Figure 8. Thus, while 1-chloro-2-butene is more stable than 4-chloro-1-butene, it is also more reactive. Now let's turn our attention one final example.

Table 1 shows that benzyl chloride reacts more than 10 times faster than allyl chloride. Apparently the carbocationic intermediate formed in this reaction is even more stable than the allylic carbocation. This suggests that the extended orbital overlap is greater in the benzylic cation than in the allylic cation. Scheme 3 shows how.

Scheme 3

Further Extensions

In this case, ionization of the C-Cl bond produces resonance contributor E. Formation of the carbocation induces electron density to flow from the p system of the aromatic ring toward the positively charged benzylic carbon as shown by the arrow labeled 1. This produces resonance contributor F. (Compare the resonance interaction E <---> F to the C <---> D interaction shown in Scheme 2.) Now, in resonance contributor F, there is a double bond adjacent to the positively charged carbon, and the p electrons of that bond are drawn toward the positive charge as indicated by arrow 2. This interaction generates resonance contributor G. A similar situation applies in G, and, as arrow 3 shows, electron density is drawn from the adjacent p system to produce resonance contributor H.


Exercise 15 Use the formalism shown in Scheme 3 to depict the resonance interaction that converts structure H back to structure E.

Exercise 16 Draw the hybrid of resonance contributors E-H.


The interpretation of Scheme 3 is that the positive charge is delocalized over all the atoms in the intermediate; the charge distribution is not equal at every carbon, rather the electron density is lowest at those atoms that have a formal charge in a given resonance contributor. There is a pattern of charge distribution that is repeated throughout resonance theory: in drawing a series of resonance contributors, the formal charge will be localized on the 1st, 3rd, 5th atom, etc.

Resonance contributors E, F, G, and H do not contribute equally to the resonance hybrid. Structure E makes the most important contribution because it is the only structure in which the aromatic p system is in tact. We have seen that the aromatic ring is classified as a separate functional group, different from an alkene. This is because the p system of benzene and related molecules is much more stable than the p system of a simple alkene. In structures F, G, and H the aromatic p system is disrupted.

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