"Reaction of compound A with sodium amide and ammonia followed by bromomethane yielded compound B.
Compound C is produced when B is isolated and treated with aqueous acid. Subsequent treatment of C with 1,2-ethanethiol in toluene, catalysed by boron trifluoride, and followed by a Raney nickel reduction produced compound D.
D was found to be a symmetrical compound as well as a flammable gas. The addition of bromine water to A and B, independently, turned the solution from red to clear. Analysis of C revealed a molecular formula of C4H8O"
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My reasoning:
(C) is 2-butanone, (D) is butane
I think (A) is acetone - NaNH2 removes the alpha proton, the alpha carbon attacks the electrophile CH3-I.
Then (B) is 2-butanone also? Not sure what the acid is doing, perhaps it's there to react with Br-?
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Is this viable? Or am I on the wrong track?
Thanks