Question

The set(s) of reagents that will effect the following conversion is(are)

• A. i) \(\mathrm{Ph_3P{=}CHOMe}\); ii) \(\mathrm{H_3O^+}\)
• B. i) \(\mathrm{TsNHNH_2}\); ii) \(^{n}\mathrm{BuLi}\) (2 equiv) then \(\mathrm{DMF}\)
• C. i) 1,3-dithiane, \(^{n}\mathrm{BuLi}\); ii) \(\mathrm{HgSO_4}\), dil. \(\mathrm{H_2SO_4}\)
• D. i) \(\mathrm{Me_3SI}\), \(\mathrm{NaH}\); ii) \(\mathrm{BF_3\cdot OEt_2}\)

Hint:

Look for a one-carbon Wittig-type homologation (Ph3P=CHOMe/H3O+) or a sulfonium-ylide epoxidation followed by a BF3-catalyzed Meinwald (1,2-hydride shift) rearrangement.

Answer 1

The Correct Option is A, D

Instead of working through each mechanism in full, it helps to first count atoms: the product has exactly one more carbon than the ketone starting material, and that extra carbon carries the new $\mathrm{-CHO}$ group while the original carbonyl oxygen is gone. Any correct reagent set has to (1) add one carbon to the carbonyl carbon and (2) end up delivering that carbon as an aldehyde, with the original ring carbon left as a plain $\mathrm{CH}$.

1. (A) $\mathrm{Ph_3P{=}CHOMe}$, then $\mathrm{H_3O^+}$: this ylide is a one-carbon Wittig reagent carrying a masked aldehyde (as a methyl enol ether). Olefination puts a $\mathrm{{=}CHOMe}$ group directly on the old carbonyl carbon, and aqueous acid simply unmasks that enol ether to the free aldehyde. Carbon count and functional group both match the product, so this route works.
2. (B) $\mathrm{TsNHNH_2}$, then 2 eq $^{n}\mathrm{BuLi}$/$\mathrm{DMF}$: this is the Shapiro sequence, which turns the carbonyl carbon into an alkene carbon (vinyllithium), then formylates it. The product keeps the new ring double bond, so it is more unsaturated than the target and is not the right structure.
3. (C) 1,3-dithiane/$^{n}\mathrm{BuLi}$, then $\mathrm{HgSO_4}$/dil. $\mathrm{H_2SO_4}$: lithiodithiane simply adds to the ketone as a nucleophile, giving an alcohol with the dithiane hanging off the same carbon; hydrolyzing the dithiane later just converts that group into a new carbonyl, giving a hydroxy-ketone rather than the target aldehyde with a plain $\mathrm{CH}$ at the ring position.
4. (D) $\mathrm{Me_3SI}$/$\mathrm{NaH}$, then $\mathrm{BF_3\cdot OEt_2}$: the sulfoxonium ylide methylenates the carbonyl to a spiro-epoxide (again adding exactly one carbon), and $\mathrm{BF_3}$ then triggers a Meinwald rearrangement, a Lewis-acid-promoted 1,2-hydride shift that converts the epoxide straight into the homologated aldehyde. This also matches the target exactly.

Only (A) and (D) add the correct one carbon AND land on an aldehyde with the ring carbon as a plain $\mathrm{CH}$; (B) leaves an extra double bond and (C) leaves an extra hydroxyl, so both are structurally wrong.

So the reagent sets that work are (A) and (D).