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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?

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    $\begingroup$ The name ‘2-ethyl-2-methylpentan-3-one’ is not correct. The systematic name is 4,4-dimethylhexan-3-one. $\endgroup$ – Loong Jun 29 '15 at 18:49
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With the correct starting compound, 3-ethyl-2-methylpentan-2,3-diol, predicting the product is tough! Here is the reaction scheme for this compound:

enter image description here

  1. Alkonium cations A and D both equally likely because they are both tertiary. Thus both alcohols should be equally basic.
  2. Carbocations B and E are both tertiary carbocations that seem equally likely to form.
  3. The rearrangement from B to C involves an ethyl migration, which seems less likely than the methyl migration involved in the rearrangement from E to F.

Compound F is 3-ethyl-3-methylpentan-2-one, which is what you chose as the major product. Compound C is 4,4-dimethylhexan-3-one, which is what the book says is the major product.

Without experimental evidence, I would have trouble choosing between C and F. If I had to choose, my first instinct would be F because it involved the methyl shift instead of the ethyl shift. Methyl shifts happen faster (methyl is smaller) than ethyl shifts.

My reasoning (and probably yours) is wrong, and here is why. We can only consider methyl > ethyl from the same carbocation. For example, in the following rearrangement:

enter image description here

In the rearrangment step at the end of the sequence, the methyl group is more likely to move than the ethyl group. We can make this comparison here because the pathway is identical up until the rearrangement.

In your reaction, the pathways toward the generation of the carbocations are not equivalent.

In other words, if carbocation B forms preferentially over carbocation E, then it might not matter that the methyl group migrates faster. We get more ethyl group migration because we have more of that carbocation formed. If carbocation B forms preferentially, one of my two assumptions about A, B, D, and E above must be incorrect. I will restate them below for convenience:

  1. Alkonium cations A and D both equally likely because they are both tertiary. Thus both alcohols should be equally basic.
  2. Carbocations B and E are both tertiary carbocations that seem equally likely to form.

I'll leave it to you now. Which assumption is not correct? Why is that assumption incorrect? Why do cations A and B form preferentially over D and E?

I'll give you a hint: Have you considered steric factors?

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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:

enter image description here

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

enter image description here

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

Pinacol rearrangement from cation B

enter image description here

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.

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  • $\begingroup$ i made a mistake in the naming sorry.please see now $\endgroup$ – Swapnava Chaudhuri Jul 7 '15 at 7:27
  • $\begingroup$ Thanks for checking. Rather than delete this answer, I am going to add a new one since the two are very different. This answer is still a good analysis for the wrong starting compounding. $\endgroup$ – Ben Norris Jul 7 '15 at 12:13

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