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Step-by-Step Solution
Step 1: Identify the final product (ethylamine)
The question states that upon treatment with Br2 and KOH, the compound “C” gives ethylamine (C2H5NH2). This is a characteristic outcome of the Hoffmann bromamide degradation reaction, which converts an amide (with “n” carbons) into a primary amine having one fewer carbon (with “n−1” carbons).
Step 2: Relate ethylamine (C2H5NH2) to the amide “C”
If ethylamine has 2 carbons, the corresponding amide “C” must have 3 carbons (because Hoffmann bromamide reaction removes one carbon). Thus, “C” is propionamide, which has the formula CH3CH2CONH2.
Step 3: Determine how “C” (propionamide) is formed from “B”
The question indicates that compound “B,” on heating, gives “C.” Typically, when an ammonium salt of a carboxylic acid is heated, it dehydrates to form the corresponding amide. Hence, “B” must be the ammonium salt of a 3-carbon carboxylic acid. Specifically, it is the ammonium salt of propionic acid: CH3CH2COO−NH4+.
Step 4: Identify the original acid "A"
To form this ammonium salt “B,” compound “A” reacts with NH3. Since “B” is the ammonium salt of propionic acid, “A” must be propionic acid, whose formula is CH3CH2COOH.
Step 5: Conclude the structure of “A”
Therefore, the organic compound “A” that eventually yields ethylamine via the sequence of reactions described is CH3CH2COOH (propionic acid).
