Question:

Write the major product(s) in the following reactions: 
KMnO and KOH,

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In oxidation reactions with KMnO₄, alkyl groups are converted into carboxylic acids, and in electrophilic aromatic substitution, bromination leads to substitution at the positions ortho and para to the existing functional group.
Updated On: Feb 19, 2025
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Solution and Explanation

(1) The reaction involves oxidation using KMnO₄ and KOH, which leads to the formation of a carboxylic acid. The product is Benzoic acid (C₆H₅COOH). \[ \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \xrightarrow{\text{KMnO}_4, \text{KOH}} \text{C}_6\text{H}_5\text{COOH} \] (2) The reaction between benzaldehyde and acetone in the presence of dilute NaOH leads to the formation of Aldol product (4-hydroxy-4-phenylpent-3-en-2-one), followed by dehydration, which forms Cinnamon aldehyde. \[ \text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{COCH}_3 \xrightarrow{\text{dil NaOH}} \text{C}_6\text{H}_5\text{CH}= \text{CHCOCH}_3 \] (3) Bromination of benzoic acid in the presence of FeBr₃ leads to the formation of 2,4,6-Tribromobenzoic acid. \[ \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{Br}_2 / \text{FeBr}_3} \text{C}_6\text{H}_2\text{(Br)}_3\text{COOH} \]
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