# Why is there a hydride shift in the hydration of 1-methyl-5-phenyl-1,2,3,4,5,6,7,8-octahydronaphthalene?

I was attempting a question here in which you had to find a product after the hydration of alkenes.

Can someone explain why was there a hydride shift in this reaction?

According to me, initially, the positive charge should be more stable as it is tertiary and has 4 alpha hydrogens for hyperconjugation effect whereas, after the shift, the positive charge occurs on phenyl which should be unstable.

after the shift, the positive charge occurs on phenyl which should be unstable.

Are you sure about that? Firstly, the positive charge isn't 'on' the phenyl, it's on the Carbon to which the phenyl ring is attached and hence it enables resonance.

Out of many of the factors we look for the stability of a carbocation (which include the ones you mentioned like positive inductive effect in a tertiary Carbocation, Hyperconjugation, )

Resonance takes a major chunk of the competition along with hyperconjugation of the neighbouring Carbons which is what Oscar pointed out as well.

In general, the Carbocation formed is highly stabilised due to resonance with the whole phenyl ring so there's no problem in forming such a carbocation.

When you'll try and draw the resonating structures, you'd realise that the delocalisation of the charge throughout the ring stabilises it.

Thus,

1. Resonance
2. Hyperconjugation with 3 alpha Hydrogens
3. Tertiary Carbocation

All these factors ensure high stability of this carbocation

• Is resonance effect better than hyperconjugation in terms of stbility of carbocation? – Ritvish Jun 2 at 11:28
• In general, yes. Resonance is usually the major stabilising factor followed by hyperconjugation. Especially when there's resonance with a phenyl group, the number of resonating structures and a great extent of delocalisation of charge highly stabilise the carbocation. – Prajwal Tiwari Jun 2 at 11:30
• Besides that, you don't really lose a lot of hyperconjugation. The rearranged carbocation is still tertiary as well as being benzylic. – Oscar Lanzi Jun 2 at 11:31
• Also could you please tell me if the positive charge was present directly on phenyl,would it be unstable? – Ritvish Jun 2 at 11:34
• Yes. For stability through resonance, You need conjugation of pi bonds, charges, etc. But when it's directly on the vinylic carbon, ( i.e. the double bonded carbon) there is no conjugation so resonance is ruled out. Also, the carbon here being $\ce{sp2}$ is more electronegative(since, more s character) which instead destabilises the carbocation by pulling electron density from an already electron deficient region – Prajwal Tiwari Jun 2 at 11:38