In the following reactions, P, Q, R,and S are the major products.

The correct statement about P, Q, R,and S is:

Correct Statement: Q undergoes Kolbe’s electrolysis to give an eight-carbon product.

Let us analyze each reaction step-by-step using the starting material, 1-chloro-2-methylpropane (isobutyl chloride, (CH₃)₂CHCH₂Cl), as shown in the images provided:

1. Formation of Product P:

Reaction steps:

(i) Isobutyl chloride reacts with Mg in dry ether to form the Grignard reagent, isobutylmagnesium chloride:
(CH₃)₂CHCH₂Cl + Mg → (CH₃)₂CHCH₂MgCl

(ii) Protonation with H₂O gives isobutane (2-methylpropane):
(CH₃)₂CHCH₂MgCl + H₂O → (CH₃)₂CHCH₃ + Mg(OH)Cl

Thus, P is an alkane (isobutane), not an alcohol. Statement 1 is incorrect.

2. Formation of Product Q:

Reaction steps:

(i) Isobutyl chloride with Mg in dry ether forms (CH₃)₂CHCH₂MgCl.

(ii) Reaction with CO₂ followed by (iii) acid hydrolysis (H₃O⁺) yields 3-methylbutanoic acid:
(CH₃)₂CHCH₂COOH (a 5-carbon carboxylic acid).

(iv) Treatment with NaOH yields sodium 3-methylbutanoate:
Q = (CH₃)₂CHCH₂COONa

Kolbe's electrolysis of a sodium salt of carboxylate R-COONa proceeds via free radical dimerization at the anode to form R-R:

$2{\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2{\text{COO}}^{-}\to {\left({\text{CH}}_3\right)}_2{\text{CHCH}}_2-{\text{CH}}_2\text{CH}{\left({\text{CH}}_3\right)}_2+2{\text{CO}}_2+2{\text{e}}^{-}$

The resulting hydrocarbon product is 2,5-dimethylhexane, which contains exactly 8 carbon atoms (4 carbons from each alkyl fragment). Thus, Statement 2 is correct.

3. Formation of Product R:

Reaction steps:

(i) & (ii) Grignard reagent (CH₃)₂CHCH₂MgCl reacts with acetaldehyde (CH₃CHO) followed by hydrolysis to yield 4-methylpentan-2-ol (a secondary alcohol with 6 carbons).

(iii) Oxidation with CrO₃ converts the secondary alcohol to a ketone, 4-methylpentan-2-one:
R = (CH₃)₂CHCH₂COCH₃

Since R is a ketone (and possesses α-hydrogens), it does not undergo the Cannizzaro reaction. Statement 3 is incorrect.

4. Formation of Product S:

Reaction steps:

(i) Nucleophilic substitution with ethanolic NaCN gives 3-methylbutanenitrile: (CH₃)₂CHCH₂CN.

(ii) Reduction with H₂/Ni converts the nitrile to a 1° amine, 3-methylbutan-1-amine: (CH₃)₂CHCH₂CH₂NH₂ (5 carbons).

(iii) Carbylamine reaction (CHCl₃/KOH, Δ) converts the 1° amine into an isocyanide: (CH₃)₂CHCH₂CH₂NC.

(iv) Reduction of the isocyanide with LiAlH₄ yields a secondary amine, N-methyl-3-methylbutan-1-amine:
S = (CH₃)₂CHCH₂CH₂NHCH₃ (6 carbons)

Thus, S is a secondary amine, not a primary amine. Statement 4 is incorrect.

Conclusion:

The only correct statement is that Q undergoes Kolbe’s electrolysis to give an eight-carbon product.