3-ethyl-2-methylpentane-2,3-diol with sulfuric acid I was studying pinacol pinacolone reaction when I came across this. I believe the hydrogen cation will attack the hydroxyl group of carbon attached to two ethyl groups (since this carbocation is more stable due to the inductive effect), then methyl group will migrate and hence we get 3-ethyl-3-methylpentan-2-one, but the book says the answer is 2-ethyl-2-methylpentan-3-one. Is the answer in the book wrong or I am making a conceptual mistake?
Edit: This answer is based on a typo in the original question referring to a different starting compound than the OP intended.

A quick analysis of this reaction gives me the following:



3-ethyl-2-methylpentan-1,2-diol can be protonated by acid to produce cation A or B. Either of these cations can lose water to form a carbocation


The carbocation formed from A is a tertiary carbocation C that might not be likely to rearrange, but we'll consider the possibility in a bit.
The carbocation formed from B is a primary(!) carbocation D that seems unlikely to form.



  Pinacol rearrangement from cation A




A pinacol rearrangement from cation A through carbocation C forms an aldehyde: 3-ethyl-2-methylpentanal. 


  Pinacol rearrangement from cation B




Having established that carbocation D is unlikely to form, we might suspect no rearrangement, but the rearrangement can occur in a concerted fashion with the loss of water. The end result is a ketone: 4-ethyl-3-hexanone. These kinds of concerted rearrangements to avoid primary cations are also known to occur in Friedel-Crafts reactions.

Neither product seems to match what you provided as the book's answer. Is it possible you miss-named the product?


  Which product is more likely the major product?


Rather than tell you, I will give you a hint:

One rearrangement produces a thermodynamically more stable cation than the other.