Exam

JEE Main

Year

2025

Date

2 April

Shift

Evening Shift

Subject

Chemistry

Chapter

Carboxylic Acid Derivatives

JEE Main 2025 · 2 April · Evening Shift · Chemistry · Carboxylic Acid Derivatives

Consider the following reactions. From these reactions which...

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:

Step-by-step solution

(A) $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text{ mild condition }]{\text{ (i) } \mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}} \mathrm{R}-$$\mathrm{CONH}_2$$<br />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.<br />(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}-$$\mathrm{CHO}$$<br />(D)$$\mathrm{R}$ \cdot $\mathrm{CH}_2$ \cdot $\mathrm{OH}$ \xrightarrow{$\mathrm{PCC}$} $\mathrm{R}-$$\mathrm{CHO}$
(E)