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Two amides are given: X = CH3CONH2 (acetamide) and Y = CH3CONHCH3 (N-methylacetamide). X treated with LiAlH4 then H2O gives P. Y treated with LiAlH4 then H2O gives Q. Then P is treated sequentially with (1) CHCl3/KOH, (2) LiAlH4, (3) H2O. What is the final product?

1. Compound (X)
2. Compound (Y)
3. Compound (P)
4. Compound (Q)

Correct answer: Compound (Q)

Solution

P = ethanamine (CH3CH2NH2). CHCl3/KOH converts P to ethyl isocyanide (CH3CH2NC). LiAlH4 reduces isocyanide (R-NC) to N-methylamine (R-NH-CH3), giving CH3CH2NHCH3 = N-methylethanamine = Q.

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