Animated Solution for Chemistry - Organic Chemistry: In the following reactions, P, Q, R, and S are the major products. The correct statement about P, Q, R, and S is

This problem is a beautiful, comprehensive journey through several classic organic chemistry reaction pathways. It tests your ability to track carbon skeletons, identify functional group transformations, and recall specific name reactions. Let's break down each pathway systematically.

The very first step is to correctly identify the starting material from its skeletal structure. We see a chlorine atom attached to a primary carbon, which is in turn attached to a carbon bearing two methyl groups. This is the classic isobutyl arrangement. Therefore, our starting material is isobutyl chloride (or 1-chloro-2-methylpropane), which contains exactly 4 carbon atoms.

The Grignard Quench

In the first reaction sequence, isobutyl chloride is treated with magnesium in dry ether to form the Grignard reagent, isobutylmagnesium chloride.

This is followed by the addition of water. Water acts as a mild acid, quenching the Grignard reagent to form an alkane.

The resulting product P is isobutane, which is an alkane, not an alcohol. Thus, statement (A) is incorrect.

Carbonation and Kolbe's Electrolysis

Here, the Grignard reagent is reacted with carbon dioxide. This is a standard method for synthesizing carboxylic acids, effectively extending the carbon chain by one.

The resulting acid is isovaleric acid (5 carbons). Subsequent treatment with sodium hydroxide yields its sodium salt, Q (sodium 3-methylbutanoate).

The problem then asks about Kolbe's electrolysis of Q. In Kolbe's electrolysis, the carboxylate undergoes oxidative decarboxylation to form an alkyl radical, which then dimerizes.

Since our alkyl group $\text{R}$ is the isobutyl group (4 carbons), the dimerization product $\text{R-R}$ will be 2,5-dimethylhexane, which contains exactly $4+4=8$ carbons. Therefore, statement (B) is correct.

Oxidation and Cannizzaro Check

In the third sequence, the Grignard reagent reacts with acetaldehyde (${\text{CH}}_3\text{CHO}$). This nucleophilic addition forms a secondary alcohol after aqueous workup.

This secondary alcohol is then oxidized by chromium trioxide (${\text{CrO}}_3$) to yield a ketone, R (4-methylpentan-2-one).

The Cannizzaro reaction is strictly for aldehydes that lack $\alpha$-hydrogens. Since product R is a ketone and possesses $\alpha$-hydrogens, it will not undergo the Cannizzaro reaction. Thus, statement (C) is incorrect.

The Nitrogen Journey

The final pathway begins with a nucleophilic substitution using ethanolic NaCN, replacing the chlorine with a cyanide group and extending the carbon chain to 5 carbons.

Catalytic hydrogenation (${\text{H}}_2/\text{Ni}$) reduces the nitrile to a primary amine (isoamylamine).

Next, the carbylamine reaction (using ${\text{CHCl}}_3$ and KOH) converts the primary amine into an isocyanide.

Finally, reduction of the isocyanide with ${\text{LiAlH}}_4$ yields a secondary amine, S (N-methyl-3-methylbutan-1-amine), which has 6 carbons.

Because S is a secondary amine, statement (D) is incorrect.

In conclusion, only statement (B) accurately describes the chemical transformations.