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Answer the following · Q1

Q.Identify A and B in the following reactions.
C6H5CH2Br --alco. KCN--> A --Na/ethanol--> B.

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Step 1. Form A: benzyl bromide + alcoholic KCN. This is the same alkyl-halide-to-nitrile substitution used to build up the starting nitrile in section 13.3.3's own worked Problem 13.1: C6H5-CH2-Br + KCN gives C6H5-CH2-CN (phenylacetonitrile, benzyl cyanide) + KBr -- one carbon (the nitrile carbon) has been added to the benzyl group.

Step 2. Form B: Mendius reduction of A (section 13.3.3). C6H5-CH2-CN, reduced with sodium in ethanol, gives C6H5-CH2-CH2-NH2, a primary amine (2-phenylethanamine, also known as phenethylamine) with the SAME carbon count as the nitrile A.

Step 3. Confirm the overall step-up pattern. Exactly as in section 13.3.3's own Problem 13.1 (methyl bromide to ethylamine), benzyl bromide (7 carbons: 6 ring + 1 CH2) is converted, via the one-carbon-longer nitrile A (8 carbons), to the amine B (also 8 carbons) -- an overall carbon-count-increasing, step-up sequence.

✓Final answer

A = C6H5-CH2-CN (phenylacetonitrile); B = C6H5-CH2-CH2-NH2 (2-phenylethanamine/phenethylamine).

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