Epoxide Opening

[2PacClone23]

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Why doesn't CH3 attack the more substituted carbon? I thought in acid, that's what it does.

It's certainly not in base, isn't H3O+ acid?

 

[RalphMouth]

The H3O+ is a separate step that protonates the opened ring.

The first step is simply a nucleophilic attack by the gringard reagent on the least substituted side causing the opening.

You are partially correct. If the acid catalyst was present in the first step, it would protonate the O, causing an opening and a carbocation to form on the more substituted carbon, leading to an addition there. This doesn't happen here (I think gringard are destroyed by acid in the same step anyway).

 

[2PacClone23]

The H3O+ is a separate step that protonates the opened ring.

The first step is simply a nucleophilic attack by the gringard reagent on the least substituted side causing the opening.

You are partially correct. If the acid catalyst was present in the first step, it would protonate the O, causing an opening and a carbocation to form on the more substituted carbon, leading to an addition there. This doesn't happen here (I think gringard are destroyed by acid in the same step anyway).

Ah Thanks!

 

[UCSD1984]

Ah Thanks!

Yea, you'll never see acid in the same step as the Grignard.