1. Benzene

2. Imine

3. Aldehyde

1. Benzene

2. Imine

3. Aldehyde

The more stable the carbocation, the lower the activation energy for reaching that intermediate will be. The more substituted a carbocation is, the more stable it is. The carbocation bonded to three alkanes (tertiary carbocation) is the most stable, and thus the correct answer.

Secondary carbocations will require more energy than tertiary, and primary carbocations will require the most energy.

The more stable the carbocation, the lower the activation energy for reaching that intermediate will be. The more substituted a carbocation is, the more stable it is. The carbocation bonded to three alkanes (tertiary carbocation) is the most stable, and thus the correct answer.

Secondary carbocations will require more energy than tertiary, and primary carbocations will require the most energy.

An aldehyde is more electrophilic than a ketone, so to do chemistry on the ketone, we must protect the aldehyde. A common protecting group for aldehydes and ketones is ethane-1,2-diol, as it forms a meta-stable five-membered acetal, which can be hydrolyzed to produce the original aldehyde or ketone by applying heat and acid.

As shown in the scheme below, which corresponds to the correct answer choice, once the aldehyde is protected, then the ketone can be reacted with the Grignard MeMgBr reagent to add a methyl group at the carbonyl. An acid workup removes the protecting group to reveal the original aldehyde, and affords the desired tertiary alcohol.

The schemes below illustrate why each of the other answer choices is wrong, as no other sequence will produce the desired product:

An aldehyde is more electrophilic than a ketone, so to do chemistry on the ketone, we must protect the aldehyde. A common protecting group for aldehydes and ketones is ethane-1,2-diol, as it forms a meta-stable five-membered acetal, which can be hydrolyzed to produce the original aldehyde or ketone by applying heat and acid.

As shown in the scheme below, which corresponds to the correct answer choice, once the aldehyde is protected, then the ketone can be reacted with the Grignard MeMgBr reagent to add a methyl group at the carbonyl. An acid workup removes the protecting group to reveal the original aldehyde, and affords the desired tertiary alcohol.

The schemes below illustrate why each of the other answer choices is wrong, as no other sequence will produce the desired product:

Rearrangements generally require carbocations. Carbocations are only formed under mechanisms with unimolecular rate-limiting steps (i.e. carbocation formation). reactions are unimolecular, involve carbocation intermediates, and therefore can undergo rearrangement.

Rearrangements generally require carbocations. Carbocations are only formed under mechanisms with unimolecular rate-limiting steps (i.e. carbocation formation). reactions are unimolecular, involve carbocation intermediates, and therefore can undergo rearrangement.

Enolates are formed by an oxygen anion bound to an alkene carbon. Reactions II and III include an enolate intermediate, as shown in the mechanisms below, whereas reaction I is a simple SN2 reaction and does not include an enolate intermediate. Enolates are highlighted in red.

Enolates are formed by an oxygen anion bound to an alkene carbon. Reactions II and III include an enolate intermediate, as shown in the mechanisms below, whereas reaction I is a simple SN2 reaction and does not include an enolate intermediate. Enolates are highlighted in red.