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I came across this question recently:

For which compound acid catalyzed dehydration rate is highest? enter image description here

Now, they all are secondary alcohols so their dehydration should follow E1 mechanism and the stability of the carbocation formed in the RDS should be considered while comparing their rates.

Following this rule, I think, the answer should be (B) as ketone group shows (-)R effect in (C) and (-)I effect in (A) and (B), all which destabilize the carbocation formed.

But the actual answer is (A). How can that be? Is there some other mechanism to follow? I can’t understand what am I doing wrong.

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  • $\begingroup$ Basically, B and C have no difference when come to dehydration. So, according to your argument, B and D have similar rates. $\endgroup$
    – Mathew Mahindaratne
    Commented May 24, 2021 at 19:06
  • $\begingroup$ Driving force here to getting faster rate of A is gaining conjugation in the product. $\endgroup$
    – Mathew Mahindaratne
    Commented May 24, 2021 at 19:07

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Notice, that the first compound (A)can show tautomerism, after drawing the enol form we find that the carbocation lies at the allylic position, which means that it is stabilised through resonance forms, hence the compound A has highest rate of reaction.

Hope it helps.

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    $\begingroup$ Oh. I see it now. So due to allyl alcohol formation, the enol form is more stable than keto form, which is quite rare. And thus the enol form undergoes dehydration and not keto form. Is that the correct deduction from your answer? $\endgroup$
    – WhySee
    Commented May 24, 2021 at 17:48
  • $\begingroup$ @ApoorvSharma, in this particular compound the most stable is the keto form only, however the stable form is always the least reactive one under normal conditions, hence in this case enol form being less stable reacts faster as compared to that with keto form. $\endgroup$
    – green_32
    Commented May 24, 2021 at 22:22

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