[snoopylinz]

Hi there,

I have an urgent question that I am completely stumped on.  I have attached the question below.  I understand this is an addition reaction whereby the H+ of the acid will add to the double bond, therefore removing it, creating a carbocation.  I believe this carbocation will rearrange to the more stable tertiary carbocation and OH- from water will attack as the nucleophile.  What I don't get is how this 4-membered ring becomes a 5-membered ring.

Any help would be much appreciated!

Thanks,
Lindsay

[snoopylinz]

I should have also added that the question is to show how this reaction occurs via its mechanism.

Thanks!

[fledarmus]

Can you draw those two steps of the mechanism, showing electron movements with arrows?

[snoopylinz]

This would be my understanding...

Thanks!

[Doc Oc]

You're on the right track with the alkyl shift.  Rather than moving the methyl group, can you move something else to create a tertiary carbocation?

[snoopylinz]

Hm, I'm not sure what else I would move, but I did realize that I drew my structure wrong (never actually moved the alkyl group) so here are my new structures, but I'm still not doing it right...

[discodermolide]

There are other C-C bonds you can migrate. Look at your product!

[snoopylinz]

Hmm, so what I have drawn is allowed? I just didn't know how that carbon-carbon bond would break to allow a shift of the methyl group...

[Doc Oc]

There are other C-C bonds you can migrate. Look at your product!

This.

The key is not to get too infatuated with specific substrates, but rather concepts.  So I know you've seen methyl migrations as a way to stabilize carbocations.  But fundamentally, what you're doing is breaking a carbon-carbon bond and shifting it to create a more stable carbocation, and what migrates can be just about anything, not just methyl.  So take a look at some other C-C bonds and see if you can move those to get what you want.

Edit: I see you've figured it out.  The reasons it rearranges this way is that you're relieving a lot of ring strain by opening the product from a 4 to a 5-membered ring.  If it were a 5 or 6-membered ring, you'd probably have methyl migration instead.

[snoopylinz]

So what I've done in the picture above - is that correct? It gives me the right answer and I'm not sure what other bonds there are that I could break...

[snoopylinz]

Ah, amazing!!

Thank you so much to everyone who helped here.

So just to clarify, the ring can only open in this case because of the formation of the carbocation, correct?  If there were no carbocation, this rearragement/ring opening would not be possible, right?

Thanks.

[discodermolide]

Ah, amazing!!

Thank you so much to everyone who helped here.

So just to clarify, the ring can only open in this case because of the formation of the carbocation, correct?  If there were no carbocation, this rearragement/ring opening would not be possible, right?

Thanks.

And release of ring strain!

[Arctic-Nation]

So what I've done in the picture above - is that correct? It gives me the right answer and I'm not sure what other bonds there are that I could break...

A minor remark: the way you've drawn your intermediate after the ring expansion suggests that the carbocation is still on the same carbon atom as before the rearrangement, while it obviously has shifted. The intermediate is of course symmetrical otherwise, and you can say you've only flipped your molecule, but it's bad practice to do that without good reason.
Numbering the atoms is a good way to avoid mistakes like these, especially in rearrangements where it can be difficult to keep track of where all the atoms end up at.