# Hydride shift and Ring Expansion

Below is a problem I got stuck at in a test .

The first compund in this chain was easy to figure out , it was something like this :

Then I had this intermediate compound for B.(I didn't add the bromide anion because this is where my confusion starts)

Now ,here it shouldn't be correct to do a hydride shift outside the ring as it would result in a much less stable carbocation . But if I proceed using that idea and solving further for part D and E , I am getting option (b) as my answer and the correct answer is option (d). Although if I do a hydride shift there then that opens a way for a ring expansion and hence a correct answer (according to the answer key) .

• Consider if D is the primary alcohol from HCHO addition to the tert Grignard. The HI conditions could give A or a primary cation that would undergo hydride shift to the methylcyclohexane cation i.e the precursor to D – Waylander Jul 8 '20 at 10:48

Ring expansion takes place in the step $$\ce{D->E}$$.

You were going right till the intermediate. Your intermediate will be attacked by $$\ce{Br-}$$ anions (as temp. is normal, therefore $$\ce{S_N1}$$ mechanism). Product will be B.

Then, it will form grignard reagent as C, which will react with aldehyde to produce alcohol D.

After that, it will react with HI to form a carbocation intermediate, which will rearrange to give product E.

The whole mechanism is as follows,

• Looks correct, but E is the iodide not the bromide – Waylander Jul 8 '20 at 11:01
• @Waylander: Thanks, I've edited that. – Rahul Verma Jul 8 '20 at 11:06

Your answer A is correct. For B, I don't understand why you want to make stable carbocation to relatively high energy carbocation. Nonetheless, B is a compound, not a intermediate. Sure, dehydroxylation of A by $$\ce{HBr}$$ would give $$3^\circ$$-carbocation as indicated. But that intermediate would further proceed to react with remaining $$\ce{Br-}$$ to give B (1-bromomethylcyclopentane) as the product.

B reacts with $$\ce{Mg}$$ in ether to give the Grignard reagent C (methylcyclopentyl magnesium bromide). This react with $$\ce{HCHO}$$ (formaldehyde) to give the $$1^\circ$$-alcohol D (1-methylcyclopentylmethanol; the structure in answer $$(A)$$ with $$\ce{OH}$$ in place of $$\ce{I}$$).

Protonation of D by $$\ce{HI}$$ would give a $$1^\circ$$-alkylhydroxonium ion, which further subjected to dehydrate with concomitant ring expansion to give $$3^\circ$$-methylcyclohexyl carbocation. This carbocation would further react with remaining $$\ce{I-}$$ to give E (1-iodomethylcyclohexane; the structure in option $$(D)$$) as the final product.

• I think you mean D dehydrates to give the cation, not decarboxylates. – Waylander Jul 8 '20 at 11:57
• @Waylander: Yes, in deed. Thank you for pointing out that. – Mathew Mahindaratne Jul 8 '20 at 12:02