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In a set of reactions, ethyl benzene yielded a product D.
$C_6H_5CH_2CH_3 \xrightarrow[KOH]{KMnO_4}$ B $\xrightarrow[FeCl_3]{Br_2}$ C $\xrightarrow[H^+]{C_2H_5OH}$ D
'D' would be
$C_6H_5CH_2CH_3 \xrightarrow[KOH]{KMnO_4}$ B $\xrightarrow[FeCl_3]{Br_2}$ C $\xrightarrow[H^+]{C_2H_5OH}$ D
'D' would be
A
Ethyl m-bromobenzoate, m-$BrC_6H_4COOC_2H_5$
B
$C_6H_5-CH_2-CH(Br)-COOC_2H_5$
C
A dibromo-substituted $C_6H_3Br_2-CH_2COOC_2H_5$
D
m-Ethoxybenzoic acid, m-$C_2H_5O-C_6H_4-COOH$
Detailed Solution
Step 1: alkaline $KMnO_4$ oxidises the whole alkyl side chain of an alkylbenzene to –COOH, whatever its length: $C_6H_5CH_2CH_3 \rightarrow C_6H_5COOH$. So B is benzoic acid.
Step 2: $Br_2$/$FeCl_3$ brings about electrophilic substitution in the ring. The –COOH group is deactivating and meta directing, so bromine enters the meta position: C is m-bromobenzoic acid.
Step 3: with ethanol and acid the carboxylic acid is esterified (Fischer esterification): m-$BrC_6H_4COOH + C_2H_5OH \xrightarrow{H^+}$ m-$BrC_6H_4COOC_2H_5 + H_2O$
Hence D is ethyl m-bromobenzoate.
Step 2: $Br_2$/$FeCl_3$ brings about electrophilic substitution in the ring. The –COOH group is deactivating and meta directing, so bromine enters the meta position: C is m-bromobenzoic acid.
Step 3: with ethanol and acid the carboxylic acid is esterified (Fischer esterification): m-$BrC_6H_4COOH + C_2H_5OH \xrightarrow{H^+}$ m-$BrC_6H_4COOC_2H_5 + H_2O$
Hence D is ethyl m-bromobenzoate.