Looking for classes? Ksquare Career Institute, Bengaluru →
Carboxylic acids
Concepts tested here
- Side-chain oxidation, bromination, esterification
All Questions
2010 AIPMT-PRE 1 question
-
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 beStep 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.