# Reaction of 1-(1-chloroethyl)-1-methylcyclobutane under E1 Conditions

What is the major product of the following reaction?

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It looks to me that the $$\ce{Cl}$$ group leaves then there's an alkyl shift to where the $$\ce{Cl}$$ group leaves in order to form a tertiary carbocation. Then $$\ce{EtOH}$$ attacks the beta hydrogen and forms a double bond to give product B:

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However, the answer I came up with was incorrect. Any ideas why?

• It's a good attempt, but as I said there are quite a number of alkyl groups you could migrate - you might want to try to list all of them and see why each of them may or may not be so favoured. – orthocresol Nov 20 '15 at 21:12
• Before solving this problem, you should know the migrating aptitude of groups. en.wikipedia.org/wiki/Migratory_aptitude – Aditya Dev Nov 22 '15 at 1:32
• I've thought of another one I guess. I'm pretty confident nobody actually mentioned it as something that happens in my ochem class but could the ring expand with an alkyl shift? So it would create product A. Problem is why would A be favored over B? Both involve a tertiary carbocation... is the carbocation in a ring more stable or something? – John Nov 22 '15 at 6:16
• I would say it is because there is less ring strain in A than in B. – orthocresol Nov 22 '15 at 12:12

## 1 Answer

You are correct that an alkyl shift will take place but there are two possible options for which group moves. Your option is possible but there is another option which involves a ring expansion to generate product A.

This mechanism is favoured because it relieves the ring strain in the four membered ring. Since the reaction is run under reversible conditions, the thermodynamic product (with the least ring strain) will dominate.

• The migration of the primary alkyl group is also favoured over the methyl group since it can stabilise the positive charge in the TS better. – orthocresol May 27 '16 at 21:39