JEE MAIN - Chemistry (2025 - 2nd April Evening Shift - No. 14)
Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product ?
(A) $\quad \mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { mild condition }]{\text { (i) } \stackrel{+}{\mathrm{H}} / \mathrm{H}_2 \mathrm{O}}$
(B) $\quad \mathrm{R}-\mathrm{MgX} \xrightarrow[\text { (ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{CO}_2}$
(C) $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { (ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{SnCl}_2 / \mathrm{HCl}}$
(D) $\quad \mathrm{R} \cdot \mathrm{CH}_2 \cdot \mathrm{OH} \xrightarrow{\mathrm{PCC}}$
(E) _2nd_April_Evening_Shift_en_14_1.png)
Choose the correct answer from the options given below :
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 supplied then it gets convert in to $-$COOH .
(B) $\mathrm{R}-\mathrm{MgX} \xrightarrow[\text { (ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{CO}_2} \mathrm{R}-\mathrm{COOH}$
(C) $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { (ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{SnCl}_2 / \mathrm{HCl}} \mathrm{R}-\mathrm{CHO}$
(D) $\mathrm{R} \cdot \mathrm{CH}_2 \cdot \mathrm{OH} \xrightarrow{\mathrm{PCC}} \mathrm{R}-\mathrm{CHO}$
_2nd_April_Evening_Shift_en_14_2.png)
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