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Question: Phthalimide + KOH$_{(alc)}$ → [A] [A]+ CH$_3$CH$_2$CH$_2$ Br$\overset{\triangle}{\rightarrow}$[B] [...

Phthalimide + KOH(alc)_{(alc)} → [A]

[A]+ CH3_3CH2_2CH2_2 Br→△\overset{\triangle}{\rightarrow}[B] [B]+H2_2O, OH→△\overset{\triangle}{\rightarrow}[C] [D]

The final products [C] and [D] in the sequence of the above reactions are.

A

CH3_3NH C2_2H5_5 ,

B

CH3_3CH = CH2_2,

C

(CH3_3)3_3N ,

D

CH3_3CH2_2CH2_2NH2_2,

Answer

CH3_3CH2_2CH2_2NH2_2, and a benzene ring is shown with two carboxylate groups (–COO⁻K⁺) attached to adjacent carbon atoms on the ring.

Explanation

Solution

The reaction sequence involves the Gabriel synthesis.

  1. Formation of [A]: Phthalimide reacts with alcoholic KOH to form potassium phthalimide.

  2. Formation of [B]: Potassium phthalimide reacts with CH3_3CH2_2CH2_2Br (an alkyl halide) to yield N-propylphthalimide.

  3. Formation of [C] and [D]: Hydrolysis of N-propylphthalimide yields n-propylamine ([C], CH3_3CH2_2CH2_2NH2_2) and potassium phthalate ([D], a benzene ring with two -COO−^-K+^+ groups at adjacent positions).