JEE Mains · Chemistry · STD 12 - 8.2 Carboxylic acids and their derivative
Consider the following reactions. From these reactions which reaction will give carboxylic acid
as a major product ?
(A) \(\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { mild condition }]{\text { (i) } \mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}}\)
(B) \(\mathrm{R}-\mathrm{MgX} \xrightarrow[\text { (ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{CO}_2}\)
(C) \(\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { (ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{SnCl}_2 / \mathrm{HCl}}\)
(D) \(\mathrm{R} \cdot \mathrm{CH}_2 \cdot \mathrm{OH} \xrightarrow{\mathrm{PCC}}\)
(E)

Choose the correct answer from the options given below:
- A A and D only
- B A,B and E only
- C B,C and E only
- D B and E only
Answer & Solution
Correct Answer
(D) B and E only
Step-by-step Solution
Detailed explanation
(A) \(\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { mild condition }]{\text { (i) } \mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}} \mathrm{R}-\mathrm{CONH}_2\) Under mild condition amide is formed because this reaction is typically slow if further more heat will…
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