Animated Solution for Chemistry - Organic Chemistry: Match List-I with List-II. List-I (Chemical reaction) A. ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5⟶{\text{C}}_2{\text{H}}_5\text{OH}$ B. ${\text{CH}}_3{\text{COOCH}}_3⟶{\text{CH}}_3\text{CHO}$ C. ${\text{CH}}_3\text{C}\equiv \text{N}⟶{\text{CH}}_3\text{CHO}$ D. ${\text{CH}}_3\text{C}\equiv \text{N}⟶{\text{CH}}_3{\text{COCH}}_3$ List-II (Reagent used) 1. ${\text{CH}}_3\text{MgBr}/{\text{H}}_3{\text{O}}^{+}$ (1 equivalent) 2. ${\text{H}}_2{\text{SO}}_4/{\text{H}}_2\text{O}$ 3. $\text{DIBAL-H}/{\text{H}}_2\text{O}$ 4. ${\text{SnCl}}_2,\text{HCl}/{\text{H}}_2\text{O}$ Choose the most appropriate option given below.

The problem asks us to match four distinct chemical transformations with their appropriate reagents. This is a classic test of your knowledge of functional group interconversions in organic chemistry. Let's break down each reaction to understand the underlying chemistry.

Acidic Hydrolysis of Esters The first reaction shows ethyl acetate (${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5$) converting into ethanol (${\text{C}}_2{\text{H}}_5\text{OH}$). This process is the cleavage of an ester bond to yield a carboxylic acid and an alcohol.

To achieve this, we perform an acidic hydrolysis. The ester is treated with water in the presence of a strong acid catalyst, typically dilute sulfuric acid (${\text{H}}_2{\text{SO}}_4/{\text{H}}_2\text{O}$). The acid protonates the carbonyl oxygen, making the carbonyl carbon more electrophilic and susceptible to attack by water. Thus, A matches with 2.

Partial Reduction with DIBAL-H In the second reaction, methyl acetate (${\text{CH}}_3{\text{COOCH}}_3$) is transformed into acetaldehyde (${\text{CH}}_3\text{CHO}$). Notice that the ester is reduced, but the reduction stops at the aldehyde stage rather than proceeding all the way to a primary alcohol.

If we used a strong reducing agent like ${\text{LiAlH}}_4$, it would reduce the ester completely to ethanol. To stop at the aldehyde, we need a milder, sterically hindered reducing agent. DIBAL-H (Diisobutylaluminium hydride) is the perfect choice. At low temperatures, it forms a stable tetrahedral intermediate that only breaks down into the aldehyde upon aqueous workup. Thus, B matches with 3.

The Stephen Reduction The third reaction involves the conversion of acetonitrile (${\text{CH}}_3\text{C}\equiv \text{N}$) to acetaldehyde (${\text{CH}}_3\text{CHO}$). This is a classic named reaction known as the Stephen reduction.

In this reaction, the nitrile is treated with stannous chloride (${\text{SnCl}}_2$) and hydrochloric acid ($\text{HCl}$). The ${\text{SnCl}}_2$ reduces the nitrile to an iminium salt intermediate (${\text{CH}}_3\text{CH}={\text{NH}}_2^{+}{\text{Cl}}^{-}$). Subsequent hydrolysis of this salt yields the corresponding aldehyde. Thus, C matches with 4.

Grignard Synthesis of Ketones The final reaction shows acetonitrile (${\text{CH}}_3\text{C}\equiv \text{N}$) converting into acetone (${\text{CH}}_3{\text{COCH}}_3$). Here, a new carbon-carbon bond is formed, and the nitrogen is replaced by an oxygen atom.

This transformation is achieved using a Grignard reagent. Methylmagnesium bromide (${\text{CH}}_3\text{MgBr}$) acts as a strong nucleophile, attacking the electrophilic carbon of the nitrile to form an imine salt. Upon acidic hydrolysis (${\text{H}}_3{\text{O}}^{+}$), the imine salt is converted into a ketone. Thus, D matches with 1.

By systematically analyzing each transformation, we have determined the correct matches: A $\to$ 2 B $\to$ 3 C $\to$ 4 D $\to$ 1

This corresponds to option (c). Mastering these specific reagent functions is crucial for tackling complex organic synthesis problems!