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Below is a problem I got stuck at in a test . enter image description here

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

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

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) .

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    $\begingroup$ 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 $\endgroup$
    – Waylander
    Jul 8 '20 at 10:48
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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,

enter image description here

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

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

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