[Donaldson Tan]

this question comes from my department's organic chemistry examination last year. I study chemical engineering, so I don't study organic chemistry in depth as most of you do.  Exams are coming and I needed to be sure that I am right. Feel free to point my mistakes. Sometimes, I am not sure how to do them.

Trans-1-chloro-3-isopropylcycleohexane gives two main reaction products when heated with potassim tert-butoxide in 2-methyl-propan-2-ol.
a. draw the structure of the expected products and give them IUPAC names
b. explain the mechanism of this reaction; in what ratio will the two products be formed?
c. which rate law would you propose to describne the kinetics of this reaction (give the units of the rate constant)? what rate law would you propose if this reaction were carried out in methyl-propan-2-ol without the base?
d. Cis-1-chloro-3-isoproplycyclohexane undergoes reaction with much more slowly than the trans isomer under the same condition. provide an explaination for this observation and draw a schematic potential energy diagram for the reaction's progress.

firstly, i had to consider what reaction would happen. I have a strong base inside a protic solvent that contains a 2^o halogenoalkane. At first, I thought it should be an elimination reaction. Protic solvents make nucleophiles bulky, thus enhancing its basic character relatively.

There are two isomers of Trans-1-chloro-3-isopropylcycleohexane, namely Trans-1,R-chloro-3,R-isopropylcycleohexane and Trans-1,S-chloro-3,S-isopropylcycleohexane.

If the reaction proceed via elimination, then Trans-1,R-chloro-3,R-isopropylcycleohexane will produce 3,R-isopropyl-cyclohexene and 4,R-isopropyl-cyclohexene while Trans-1,S-chloro-3,S-isopropylcycleohexane will produce 3,S-isopropyl-cyclohexene and 4,S-isopropyl-cyclohexene. These are 4 reaction products, not two as specified in the question.

I always find it hard to decide if a 2^o halogenoalkane should proceed via SN1 or SN2.

However, considering SN2 mechanism, Trans-1-chloro-3-isopropylcycleohexane will produce R-tert-butanoxy-(3,S-isopropyl)-cyclohexane and S-tert-butanoxy-(3,R-isopropyl)-cyclohexane. 2 products are produced and it's in-line with the exam question.

I don't understand why my initial prediction for elimination to occur doesnt agree with the exam question. Hope someone can enlighten me,

(a) and (b) are manageable but (c) can be tough. The rate law I suggest is:
rate = k[Trans-1-chloro-3-isopropylcycleohexane][potassim tert-butoxide] with reference to the SN2 mechanism. However, without the potassium tert-butanoxide present, nothing will react with the chloro-isopropyl-cyclohexane. Heating the reaction mixture thus must result in re-arrangement of 2-methyl-propan-2-ol to form 2-methyl-propanol. I listed the mechanism for the alcohol rearrangement below:

1. (CH₃)₃C-OH <-> (CH₃)₃C⁺ + OH^-
2. (CH₃)₃C⁺ <-> H3C-CH(CH3)-CH₂⁺
3. H3C-CH(CH3)-CH₂⁺ + OH^- -> H3C-CH(CH3)-CH₂OH

Given (1) is highly favoured because a tertiary carbocation is formed, i believe (2) should be the rate determining step. (2) is the 1,2-hydride shift, which results in the transfer of the +ve charge from the 2nd carbon to the first carbon. I am not sure how to reflect an equilibrium step as the rate law. If the rate determining step is 3, it would be rather easy. The rate determining step seems to be (3) too because there would be a very low concentration of H3C-CH(CH3)-CH₂⁺. I hope someone can advise me on this.

I can't answer (d). Assuming the SN2 mechanism, it would need the nucleophile to attack the carbon attached to chlorine in the different plane with chlorine isn't there. In the case of trans, I would expect the isopropyl group to provide steric hindrance to the attack site of the nucleophile because it's in the same side of the ring as where the nucleophile should attack. In the case of cis, the isopropyl group is not in the same side of the ring as the attack site of the nucleophile. I actually expect the reaction for cis- to be faster. Please advice.

[Donaldson Tan]

i begin to think that i will have 2 elimination product if I assume I have an optically pure sample of Trans-1-chloro-3-isopropylcycleohexane to begin with. I am not sure if this is a valid assumption.

if the reaction is indeed E2, then I can account for (d) saying that in the cis-case, the chloro-group is sterically hindered to leave by the isopropyl group when tert-butanoxide to act as a base to abstract H.

Moreover, I am beginning to be more unsure of the alcohol rearrangement because the 1,2-hydride shift i propose leads to the formation of a less stable carbocation. Would anyone provide more insight?

[Donaldson Tan]

i think the alcohol rearrangement is not feasible in view that it forms a less stable carbocation. In absence of the strong base, the alcohol solvent acts as a nucleophile to attack the halogenocyclohexane. Ether will be formed as a result. The mechanism is SN2. Does anyone disagree?

[corey2]

tBuOK is a really powerful base, but an extremely low nucleophile, very good for elimination reactions.

[dexangeles]

tert-butoxide is a strong hindered base and will force an E2 elimination

the probable reason why they said 2 products, they where not being too specific especially with the R and S configurations, they probably just want to know if you know where the double bonds would occur.  Specially when they ask the ration, meaning they probably want you to determine the hindrance on one side cause by the iso substituent, thus you'll have more of the 4 and less of the 3 (all going thru E2 and no E1 because of the strong hindered base).

for D, the cis conformation actually gives both substituent in the equatorial plane which makes it more stable.  If stability is the judge that determines speed of reaction, then that's you answer, but I'm not 100percent sure about this.  Let me get back to you on it.

oh please don't forget that the compund is gonna be in a chair conformation.

[Donaldson Tan]

i did consider the boat-chair conformation. whether the chloro-alkyl-cyclohexane is boat or chair won't affect its R/S configuration because it's already specified we are dealing with trans chloro-cyclohexane. transition from boat to chair or vice versa won't change its optical configuration, ie. R remains R and S remains S.

Perhaps Ad Majorem Dei Gloriam is right that i think too much and should not consider enantiomerism at all.

[movies]

I think you all are on the right track.  I would expect an E2 mechanism with the two alkene positional isomers Corey2 drew.  I believe that the question assumes that racemic mixtures "count" as only one compound.  That's a fairly common assumption on organic exams, even though it isn't rigorously correct.

The key to part (d) is that the Cl has to occupy an axial position to have the proper orbital alignment for the elimination to occur (Anyone want to try and draw the orbitals for it?!  It's a good exercise!).  Along the lines of what AMDG said, for this to occur with the cis compound then both substituents would have to be axial, which means high energy.

[Donaldson Tan]

The key to part (d) is that the Cl has to occupy an axial position to have the proper orbital alignment for the elimination to occur (Anyone want to try and draw the orbitals for it?!  It's a good exercise!).  Along the lines of what AMDG said, for this to occur with the cis compound then both substituents would have to be axial, which means high energy.

hmm.. can someone elaborate more on this?

[movies]

Just draw out the two cyclohexane ring flips.

[dexangeles]

what do you guys use to draw those?

[movies]

ChemDraw.

Check it out:
http://products.cambridgesoft.com/family.cfm?FID=2

[Donaldson Tan]

if the reaction is indeed E2, then I can account for (d) saying that in the cis-case, the chloro-group is sterically hindered to leave by the isopropyl group when tert-butanoxide to act as a base to abstract H.

so i was right to say that Cl- was hindered sterically to leave during the SN2 process..

[movies]

Well, it's not so much that the Cl leaving is hidered but that the nucleophile approaching is hindered.

At least for the S_N2 process.

[Donaldson Tan]

for (d), we are considering the reaction between the cis isomer and tert-butanoxide in the alcohol solvent, so so it must be E2. LOL. I am obviously still pre-occupied with my initial blunder. LOL.

But still then, all I see is the Cl- is hindered sterically from leaving during the E2 process. I cant see how tert-butanoxide is hindered sterically to abstract H. Note tert-butanoxide can attack the cyclohexane ring at two sites for beta elimination. Those two sites seems free from any possible steric hindrance. Please advise. Thanks in advance.

[movies]

The size of the base has a bigger effect in the mixture of alkene isomers you get from the trans form (parts a and b) because the isopropyl group is closer in that case since it is equatorial.  I would therefore expect more of the product corresponding to elimination on the side away from the isopropyl group, so more of 4-isopropyl-cyclohexene.

I would expect that the product mixture from the cis case would actually be closer to 50:50.  The point is that the reaction is much, much slower for the cis case because it must first do a ring flip into the conformation where both substituents are axial before the elimination can occur.  Whether the Cl is hindered from leaving or not doesn't have a huge effect (it's probably negligeable in this case).  The key is that you need a lot of energy just to cause the ring flip to initiate the elimination reaction.

In the trans case, however, this ring flip doesn't take nearly as much energy.  In fact, the conformer with an axial Cl is probably lower in energy than the ring flip with the axial isopropyl group.