[Pirlo61]

I need help figuring out the explanation for this question:

The reaction obtained by treating 3-bromo-2-methylpentane with sodium tert-butoxide in ethanol is:
A. Regioselective
B. Regiospecific
C. Stereoselective
D. Stereospecific
E. Exergonic

(N.B. only one answer is allowed)

thank you in advance

[Babcock_Hall]

Per forum rules, you must show your attempt before we can help you.  I would start by refreshing my memory regarding what each term means.  Then I would write down the structure of the product(s).

[Pirlo61]

Yep. So I thought of it like this: the reaction that in this case the 3-bromo-2-methylpentane (racemic mixture) gives with the tert-butoxide is obviously one of elimination with the E2 mechanism, in particular given that the base is strong, but it is bulky (t-BuOK is bulky like LDA) it mainly provides a Hofmann product by providing the least substituted alkene (4-methyl-2-pentene); the reaction, however, in addition to being regioselective, is also stereoselective and stereospecific because the (R)-3-bromo-2-methyl-pentane leads to trans-4-methyl-2-pentene, while the (S) enantiomer generates the cis-4-methyl-2-pentene. If I consider the minority product of the reaction (the one that provides the substituted alkene, i.e. 2-methyl-2-pentene) this product does not present stereogenic centers and does not provide diastereoisomerism. Therefore the correct answer should be both answer A, C and D at the same time; however I can only mark one answer: so I was thinking of marking E "exergonic reaction" as the only answer.

[clarkstill]

A few issues here:

... stereoselective and stereospecific because the (R)-3-bromo-2-methyl-pentane leads to trans-4-methyl-2-pentene, while the (S) enantiomer generates the cis-4-methyl-2-pentene.

If it were the Hofmann product formed then this isnt true - both (R) and (S) will give the same major (E)-isomeric product (see attachment).

Typically alkyl bromides will form the Zaitsev product, and I suspect this is the case here. The presence of the 2-methyl group means that the Zaitzev elimination product doesnt possess E/Z stereochemistry, so the reaction can't be described as either stereoselective or stereospecific. It certainly isn't regiospecific, but the fact it has a choice between either Hofmann or Zaitsev and chooses one means it is regioselective. I wouldn't have known off the top of my head whether the reaction is exergonic or endergonic, but assuming there isn't an error in the question I guess it must be endergonic, since (A) is certainly true.

[Babcock_Hall]

It is possible that answer E was intended to be "exothermic," which would probably be incorrect.  A model elimination in McMurry's textbook is endothermic but favored entropically.

[Pirlo61]

A few issues here:

... stereoselective and stereospecific because the (R)-3-bromo-2-methyl-pentane leads to trans-4-methyl-2-pentene, while the (S) enantiomer generates the cis-4-methyl-2-pentene.

If it were the Hofmann product formed then this isnt true - both (R) and (S) will give the same major (E)-isomeric product (see attachment).

Typically alkyl bromides will form the Zaitsev product, and I suspect this is the case here. The presence of the 2-methyl group means that the Zaitzev elimination product doesnt possess E/Z stereochemistry, so the reaction can't be described as either stereoselective or stereospecific. It certainly isn't regiospecific, but the fact it has a choice between either Hofmann or Zaitsev and chooses one means it is regioselective. I wouldn't have known off the top of my head whether the reaction is exergonic or endergonic, but assuming there isn't an error in the question I guess it must be endergonic, since (A) is certainly true.

Assuming that the reaction is regioselective and assuming that it provides e2 elimination according to Hoffman; Wouldn't the reaction also be stereoselective? Because both the (S) enantiomer and the (R) enantiomer should form both the E diastereoisomer and the Z diastomer (with E prevailing), I add a file of how I thought about it.

[rolnor]

Would not the conversion be exothermic in total? The formation of NaBr and t-BuOH is very exothermic

[clarkstill]

A few issues here:

... stereoselective and stereospecific because the (R)-3-bromo-2-methyl-pentane leads to trans-4-methyl-2-pentene, while the (S) enantiomer generates the cis-4-methyl-2-pentene.

If it were the Hofmann product formed then this isnt true - both (R) and (S) will give the same major (E)-isomeric product (see attachment).

Typically alkyl bromides will form the Zaitsev product, and I suspect this is the case here. The presence of the 2-methyl group means that the Zaitzev elimination product doesnt possess E/Z stereochemistry, so the reaction can't be described as either stereoselective or stereospecific. It certainly isn't regiospecific, but the fact it has a choice between either Hofmann or Zaitsev and chooses one means it is regioselective. I wouldn't have known off the top of my head whether the reaction is exergonic or endergonic, but assuming there isn't an error in the question I guess it must be endergonic, since (A) is certainly true.

Assuming that the reaction is regioselective and assuming that it provides e2 elimination according to Hoffman; Wouldn't the reaction also be stereoselective? Because both the (S) enantiomer and the (R) enantiomer should form both the E diastereoisomer and the Z diastomer (with E prevailing), I add a file of how I thought about it.

If it produced the Hofmann product then yes, there would be a stereoselectivity issue. However, my suggestion is that the Zaitsev (more substituted) product is formed, which doesnt have any E or Z isomers. I think the Hofmann product is often preferred in systems where the leaving group is a trialkylammonium because the reaction has an early (anion-like) transition state. When the leaving group is "good" such as Br, the TS is late (alkene-like) and the more substituted product is favoured.

[clarkstill]

Would not the conversion be exothermic in total? The formation of NaBr and t-BuOH is very exothermic

No idea - someone could do a Hess cycle here maybe. I guess you lose a C-H and C-Br bond and only gain a pi bond in the substrate. It probably depends on other stuff like solvation effect too. I find it odd that the question expects you to simply know the answer though.