The following question was asked in IITJEE-Screening 2000,

Which of the following will be most readily dehydrated in acidic conditions?


I thought that the answer would simply be (c), as its intermediate would be stabilized by resonance with keto group and it is a sec-carbocation (which is common in all the options, so deosn't matter).

But my answer was wrong as per answer key. Am I missing something?

The correct answer as per key was,


  • 1
    $\begingroup$ The carbocation is not stabilised in c since the carbon connected to it is supposed to show a negative character. Alpha carbon to keto. Both (a) and (c) give the same resonance stabilised alkene at the end. so answer should be a $\endgroup$ Jul 24, 2020 at 12:53
  • $\begingroup$ @Safdar, what do you mean by "negative char"? charge? $\endgroup$ Jul 24, 2020 at 12:55
  • $\begingroup$ Alpha carbon to keto is electronegative in nature, an example of that would be in aldol condensation. $\endgroup$ Jul 24, 2020 at 12:56
  • 1
    $\begingroup$ All resonance is not stabilizing resonance. $\endgroup$ Jul 24, 2020 at 13:10
  • 1
    $\begingroup$ The reason that a) dehydrates faster than c) and d) is that a) dehydrates through the more substituted enol of the keto group. Effectively, the enol is now an allylic alcohol which, under acid conditions, loses water. $\endgroup$
    – user55119
    Jul 24, 2020 at 14:02

2 Answers 2


A carbonyl group normally destabilizes carbocations but stabilizes carbanions.

In this particular case, the rate of dehydration isn't guarded by carbocation stability, but by CH- acidity. In acidic conditions, OH group in alcohols is relatively easily protonated. Then, if there is a relatively acidic hydrogen in $\beta$-position it immediately dissociates, together with a water molecule.

Excercise: Try to draw a mechanism for dehydration of (a) in basic conditions. Hint: it does occur and produces the same product. It does require more heating, though.

  • $\begingroup$ Why is it guarded by "CH- acidity"? Just because found in experiments? $\endgroup$ Jul 24, 2020 at 13:21
  • $\begingroup$ @RobinSingh ... I can't answer this question in simple terms. At best, I can reccomend to find a good org.chem.book and read chapter on elimination reaction mechanism? $\endgroup$
    – permeakra
    Jul 24, 2020 at 13:38

The other answer by permeakra is sufficient and complete. But, I'd like to add an explanation on how

"A carbonyl group normally destabilizes carbocations but stabilizes carbanions"

For instance, someone can think that the carbocation would resonate with the carbonyl group as what we've learnt earlier, I'd also thought in that way, initially. But, from the following resonance forms,


We can see there is a positive charge on oxygen, which is highly unstable resonance form. So, in this way it destabilizes the intermediate (here carbocation).

Now, one can think what happens when we have a carbanion ;)

  • 1
    $\begingroup$ @Safdar It is possible and does happen on occasion. Electronegativity of an atom in a molecule and electronegativity of an atom on Polling scale are completely different things. For example, what is a better electron acceptor - carbon or chlorine? Probably chlorine. But one of the strongest electron-withdrawing groups in organic chemistry is $-CF_3$ with electron accepted by carbon. Similarly, $Mn$ in $KMnO_4$ Is a very strong oxidizer, capable of oxidizing chlorine in $HCl$ $\endgroup$
    – permeakra
    Jul 24, 2020 at 15:04
  • $\begingroup$ @Safdar It is and it happens. Carbocation with this structure WILL have electron on an pi orbital with contribution of p orbitals of two carbons. Vacant valence active atomic orbitals remain completely vacant only in exceptional or very specific circumstances, and this isn't the case. $\endgroup$
    – permeakra
    Jul 24, 2020 at 15:11

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