Rate of Nucleophilic Substitution Contest – My Response

If you recall, the winners of this contest were chosen by pulling cards out of a hat, an unusually large (size 7 3/4″) black top hat. The questions asked by the winners* were:

1. Fluffy C. – Are the low reactivity of bromocyclohexane and bromocycloheptane towards NaI due to the way the rings fold?
2. Ross Y. – What is the relationship between angle strain and S_N1 reactivity in cycloalkyl bromides?
3. Kelsey W – Why is 3-bromocyclohexene more reactive than bromocyclohexane towards NaI?

These are all great questions and none of them have obvious answers so its a good thing I wasn’t forced to answer them on the night of the contest.
Fluffy C. – Are the low reactivity of bromocyclohexane and bromocycloheptane towards NaI due to the way the rings fold?

Yes. The folds bring hydrogens near the leaving group and nucleophile, which forces them to adopt distorted, relatively unstable geometries in their transition states. Destabilizing the transition state makes for a slower reaction.

I built transition state models for four S_N2 reactions. To make life simple for me, I used Br as both the nucleophile and leaving group. To reveal distortions, I varied the alkyl portion from methyl (no distortions) to various cycloalkyl systems(cyclopentyl, cyclohexyl, cycloheptyl).

First, take a numerical look at the transition state geometries:

• methyl transition state – BrC distance = 2.241, 2.241 angstrom, BrCBr angle = 180^o
• cyclopentyl t. state – BrC distance = 2.368, 2.368 angstrom, BrCBr angle = 154^o
• cyclohexyl t. state – BrC distance = 2.364, 2.400 angstrom, BrCBr angle = 150^o
• cycloheptyl t. state – BrC distance = 2.386, 2.388 angstrom, BrCBr angle = 151^o

The methyl transition state is special; the BrC distances are fairly short and the BrCBr angle is linear. These factors improve BrC orbital overlap in the transition state and stabilize it. You could say the BrC bonds are only “partially broken” in the transition state and less bond-breaking lowers the transition state energy.

BrC orbital overlap is much worse in the cyclohexyl and cycloheptyl transition states. We can “see” this by looking at the BrC distances (longer) and the BrCBr angle (not linear; remember that C is sp² hybridized and must use a 2p orbital to create overlap with the nucleophile and leaving group simultaneously). These factors suggest “more broken” BrC bonds and relatively high energy transition states.

The following pictures may help you understand why the cycloalkyl transition states are distorted. The first picture shows all four transition states “edge on”, going clockwise from methyl (upper left) -> cyclopentyl -> cycloheptyl -> cyclohexyl. (Click for larger image)

Obviously, the BrCBr angle is small because the nucleophile (Br) and leaving group (Br) must minimize steric repulsion with spectator (ring) atoms. The next image offers another perspective on the cyclohexyl (left) and cycloheptyl (right) transition states. (Click for larger image):

If you look closely, you will see that the Br atoms in each transition state are closely attended by two H. The folding of the rings brings these H into close proximity with Br and creates the steric repulsion that leads to small BrCBr angles (and also elongated BrC distances).

Ross Y. – What is the relationship between angle strain and S_N1 reactivity in cycloalkyl bromides?

Like the previous question, some molecular models and numbers are helpful. I built models of various cycloalkanes (“RH”), the corresponding cycloalkyl carbocations (“R⁺“), and corrected the energies for an aqueous medium. Then I combined these energies to obtain energies for the following ionization reaction:

R-H_aq   ?   R⁺_aq   +   H^–_aq

This isn’t quite the same reaction as the one Ross asked me about, but it was easier to model and it reveals whether angle strain is related to the difficulty of making a carbocation. Here are my results:

• cyclopropane ? cyclopropyl cation ?H = 566 kJ/mol
• cyclobutane ? cyclobutyl cation ?H = 476
• cyclopentane ? cyclopentyl cation ?H = 435
• cyclohexane ? cyclohexyl cation ?H = 440

These are all endothermic reactions (it’s hard to make a carbocation!), but there is a trend. Angle strain is greatest at the top of the list and this is also the most endothermic reaction. As angle strain falls so does the difficulty of making the carbocation.

But I need to be more careful in making this argument. What matters is the change in angle strain upon going from the neutral compound (RH) to the carbocation (R⁺). In order to think about this change in strain, let’s watch the CCC angle at the carbon that loses the leaving group, i.e., the carbon that becomes positively charged. The hybridization of this carbon changes when the leaving group takes off. We expect the ideal CCC angle to change from 109^o to 120^o, an increase of 11^o, but according to the models, the CCC angles actually change as follows:

• 3-membered ring: 60^o ? 65^o
• 4-membered ring: 89^o ? 95^o
• 5-membered ring: 105^o ? 112^o
• 6-membered ring: 111^o ? 124^o

The 5 and 6-membered rings come closest to matching the ideal angles and the ideal increase. The smaller rings do not come close to matching the ideal angles, either in the neutral ring or in the carbocation, and the angles in these rings increase less than the ideal amount. In other words, angle strain actually increases when we ionize smaller rings. This explains the trend in reaction energies, but it would be interesting to test this idea with additional compounds.

Kelsey W. – Why is 3-bromocyclohexene more reactive than bromocyclohexane towards NaI?

If you followed my answer to Fluffy (first question), then you’ll have no trouble following my argument here. First, my models reveal that the transition state involving the unsaturated ring is much less distorted than the transition state for the saturated ring:

• methyl transition state – BrC distance = 2.241, 2.241 angstrom, BrCBr angle = 180^o
• unsaturated cyclohexenyl t. state – BrC distance = 2.344, 2.365 angstrom, BrCBr angle = 156^o
• saturated cyclohexyl t. state – BrC distance = 2.364, 2.400 angstrom, BrCBr angle = 150^o

The transition state for the unsaturated molecule is characterized by shorter BrC distances and a larger BrCBr angle. Both of these features enhance BrC orbital overlap and stabilize the unsaturated transition state. According to my model, the double bond does not participate in the reaction directly. The C=C distance is 1.343 angstrom in the transition state which is just about the same distance as in the starting material (1.338).

Again, pictures of models show us how the absence/presence of nearby H is responsible for distortions in nucleophile and leaving group position. (Click for larger image)

If you count the number of axial H in each transition state, you will see that the unsaturated molecule (left) has two fewer H than the saturated molecule (right). Therefore, both Br, the nucleophile and the leaving group, encounter less steric repulsion in the unsaturated transition state.