I strongly believe that a carbocation should not rearrange to another if there are no immediate benefits(like a Greedy Algorithm).

reaction and possible products

The doubt hit me while solving this question. The answer mentioned in the book answers (b) but my answer comes out to be (a).

Here is the mechanism:


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    $\begingroup$ Ring expansion step gives you secondary carbocation, which is in higher energy state than original tertiary carbocation. $\endgroup$ Mar 7 '19 at 16:25
  • $\begingroup$ @MathewMahindaratne I am talking about the first shift. $\endgroup$ Mar 7 '19 at 16:31
  • $\begingroup$ Ring expansion and release of some strain presumably outweighs going from a tertiary to secondary carbocation. $\endgroup$
    – user55119
    Mar 7 '19 at 21:06

Option B is the correct one. The process of dehydration of alcohols follow E1 elimination mechanism which involves two steps out of which the first one is slow ionisation of C-X bond( where X is any hetro atom) and the second step is fast removal of $H^+$ ion .Since the first step is slow it provides enough time for the carbocation to rearrange and form a stable carbocation . The mechanism itself is self explanatory. Firstly as Oxygen possess lone pairs of electrons so the $H^+$ ion attacks the OH group as a result the $H_2O $ molecular gets eliminated resulting in a positive charge there. Now hydride shift takes place as the so formed $ \textbf{Carbocation is more stable as it has 7 alpha hydrogens compared to the 5 alpha hydrogens of the previous carbocation}$. Now the C-C bond undergoes ionisation to form a three degree Carbocation and a carbanion . Now this carbanion forms a bond with the Carbocation formed earlier to form a six member ring . Now the resulting carbocation undergoes methyl shift which produces more stable three degree Carbocation. Now a $H^+$ ion from the neighbouring carbon atom is ejected as a result the pie bond is formed.


Yes b is the correct answer.

The carbocation will do a hydride shift to the tertiary carbon. It will then have a ring expansion. Then it will have a methyl shift. It will then lose its H+ ion to give option b.

Actually its forming a much stable product on hydride shift. Ergo it does shift to another tertiary carbon


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