# Does a tertiary carbocation rearrange to another tertiary carbocation?

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

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:

• Ring expansion step gives you secondary carbocation, which is in higher energy state than original tertiary carbocation. – Mathew Mahindaratne Mar 7 '19 at 16:25
• @MathewMahindaratne I am talking about the first shift. – harshit54 Mar 7 '19 at 16:31
• Ring expansion and release of some strain presumably outweighs going from a tertiary to secondary carbocation. – 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.