enter image description here

here,should I prefer hydride shift or a product with more $\alpha$ hydrogens?

my doubt is...as hydride shift is easier than methyl shift,the answer should be $a$ or $c$.if methyl shift takes place the alkene formed has more $\alpha$ hydrogens than $a$ and $c$.what should I prefer? The question is to find major product.also,irrespective of the answer I would like to know which is more stable $a$ or $c$?


Compound $a$ is stabler than compound $c$ because the alkene in $a$ has $7$ hyperconjugative structures due to $7 \alpha$-Hydrogen, whereas $c$ has only $5$ hyperconjugative structures. Hyperconjugation or Nathan-Baker Effect is a stabilizing phenomenon due to resonance-like charge delocalisation. Also, the reason why here hydride shift takes place is because it leads to a stabler carbocation. Hence the product should be $a$.

  • $\begingroup$ methyl shift also produces stable carbocation? $\endgroup$ – IITM May 12 at 7:30
  • $\begingroup$ True but hydride shift would be preferred, due to more hyperconjugation $\endgroup$ – Ritam_Dasgupta May 12 at 7:38
  • $\begingroup$ how do I know whether to consider stability of intermediate or product $\endgroup$ – IITM May 12 at 7:45
  • $\begingroup$ You consider stability of intermediate first. Once you know which carbocation it is, you start considering product. Please be advised though, try to avoid exam-help questions on this site, it is not in accordance with the guidelines. $\endgroup$ – Ritam_Dasgupta May 12 at 7:47
  • $\begingroup$ ok,and this is not exam-help question it is assignment-doubt question :) thanks for answering $\endgroup$ – IITM May 12 at 7:50

Dehydration of alcohols (in case of secondary and tertiary) proceeds through a carbocation intermediate, as you already have mentioned. You can refer this link in Master Organic Chemistry for the full mechanism.

In this case, the formation of carbocation is the slowest step, hence the rate determining step. So the stability of carbocation must be considered in the mechanism. After the formation of the initial carbocation, it can rearrange either by methyl shift or hydride shift. I could spot two reasons why hydride shift will be preferred -

  1. Hydride has more migratory aptitude than methyl
  2. After methyl shift, the new carbocation will have 6 alpha hydrogen for hyperconjugation. By hydride shift it will have 7. The latter is hence more stable and preferred.

After this step, when the new double bond is formed, it can give either A or C as you said. (A) will have more hyperconjugative structures than (C) so (A) is major.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.