Identify the product Y in the following scheme C6H6+CH3Cl AlCl3 X KMnO4 / H2SO4 Y
Correct answer: C. Benzoic acid.
- A. Toluene
- B. Xylene
- C. Benzoic acid.
- D. Acetophenone
Explanation
The introduction of an alkyl group in the benzene ring in the presence of an alkyl halide and a catalyst AICI3 is called Friedel-Crafts alkylation or Alkylation. In the first step alkyl group is added to benzene resulting in the formation of toluene, hence X is toluene.In the second step toluene is oxidised in the presence of KMnO4 which is called 'side chain oxidation'. It results in formation of benzoic acid. So, Y is benzoic acid.
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About Benzene and Aromatic Compounds
Benzene has a planar ring of delocalised π electrons, giving aromatic stability and equal carbon-carbon bond lengths. The topic covers resonance, electrophilic substitution reactions such as nitration, halogenation, sulphonation and Friedel-Crafts reactions, along with substituent effects and the distinction between substitution in benzene and addition in ordinary alkenes.
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