Chapter 10: Problem 58

The pinacol reartangement is general for all glycols. In the rearrangeanent of piracol, a symmetrical diol, equivalent carbocations are formed no matter which - Ol1 becomes protonated and leaves. Studies of unsymmetrical vicinal diols reveal that the OH group that becomes protomated and leaves is the one that gives rise to the more stable carboration. For example, treatment of 2 -methyd 1,2 propaumediol wìth cold concentrated sulfuric acid gives a 3 " carbocation. Subsequerat migration of hydride ion (H: -) followed by transfer of a proton from the new cation to solvent gines 2-methylporopanil.

Short Answer

Expert verified

Answer: The major product formed in this reaction is 2-methylpropanal.

Step by step solution

Write down the given reactants and reagent

We are given 2-methyl-1,2-propanediol as the reactant and cold concentrated sulfuric acid as the reagent for the pinacol rearrangement.

Understand the pinacol rearrangement mechanism

In the pinacol rearrangement, an OH group of the vicinal diol becomes protonated and leaves, forming a carbocation. This is typically followed by the migration of a substituent group to form a new carbocation. Then, the new cation transfers a proton to the solvent, yielding the final product.

Identify the OH group that becomes protonated and leaves

For the given reactant (2-methyl-1,2-propanediol) we must evaluate which OH group will lead to a more stable carbocation when protonated. The choice will be between a 2° carbocation (from the OH bound to C1) and a 3° carbocation (from the OH bound to C2). The more stable carbocation is the 3° carbocation, hence the OH group on C2 will be protonated and will leave.

Predict the migration of the hydride ion and formation of a new cation

After the departure of the protonated OH group from C2, the adjacent hydride ion (H:-) from C1 will migrate to the C2 carbocation, forming a new carbocation at C1.

Determine the proton transfer to the solvent to yield the final product

Finally, the new carbocation at C1 will undergo deprotonation in the presence of the solvent (acting as a base), resulting in the formation of the final product, 2-methylpropanal.

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Short Answer

Step by step solution

Write down the given reactants and reagent

Understand the pinacol rearrangement mechanism

Identify the OH group that becomes protonated and leaves

Predict the migration of the hydride ion and formation of a new cation

Determine the proton transfer to the solvent to yield the final product

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