The correct reaction/reaction sequence that would produce a dicarboxylic acid as the major product is

The correct option is C.

To determine which reaction sequence produces a dicarboxylic acid as the major product, we analyze the reaction steps for each option:

Option A:
The starting material is 3-chloropropan-1-ol:
$\text{HO-CH}$₂-CH₂-CH₂-Cl
1. Nucleophilic substitution with cyanide ion (${\text{CN}}^{-}$) from $\text{NaCN}$ replaces the chloride leaving group, yielding a hydroxynitrile:
$\text{HO-CH}$₂-CH₂-CH₂-CN
2. Hydrolysis of the nitrile group ($\text{-CN}$) under basic conditions (${\text{OH}}^{-}/{\text{H}}_2\text{O}$) followed by acidic workup (${\text{H}}_3{\text{O}}^{+}$) converts the nitrile group to a carboxylic acid group ($\text{-COOH}$):
$\text{HO-CH}$₂-CH₂-CH₂-COOH
This product is 4-hydroxybutanoic acid, which contains only one carboxylic acid group (a monocarboxylic acid). Therefore, Option A is incorrect.

Option B:
The starting material is D-glucose:
$\text{CHO-(CHOH)}$₄-CH₂OH
When treated with bromine water (${\text{Br}}_2/{\text{H}}_2\text{O}$), which is a mild oxidizing agent, only the terminal aldehyde group ($\text{-CHO}$) is oxidized to a carboxylic acid group ($\text{-COOH}$). The primary alcohol group ($\text{-CH}$₂OH) remains unaffected:
$\text{HOOC-(CHOH)}$₄-CH₂OH
This reaction yields gluconic acid, which is a monocarboxylic acid. Therefore, Option B is incorrect.

Option C:
The starting material is bromocyclohexane:
1. Treatment with potassium hydroxide in ethanol ($\text{KOH / EtOH}$) under heating promotes a dehydrohalogenation (E2 elimination) reaction, eliminating $\text{HBr}$ to produce cyclohexene:
${\text{C}}_6{\text{H}}_{11}\text{Br}\stackrel{\text{KOH / EtOH, Δ}}{\to }{\text{C}}_6{\text{H}}_{10}\text{(cyclohexene)}$
2. Treating cyclohexene with hot, acidic potassium permanganate (${\text{KMnO}}_4/{\text{H}}_2{\text{SO}}_4,\Delta$) causes oxidative cleavage of the carbon-carbon double bond ($\text{C=C}$). Both alkene carbons are oxidized to carboxylic acid groups:
${\text{C}}_6{\text{H}}_{10}\stackrel{{\text{KMnO}}_4/{\text{H}}_2{\text{SO}}_4,\Delta }{\to }\text{HOOC-(CH}$₂)₄-COOH
This opens the ring and yields adipic acid (hexanedioic acid), which contains two carboxylic acid groups (a dicarboxylic acid). Therefore, Option C is correct.

Option D:
The starting material represents a cyclic ketone. The reagents do not lead to the ring opening or oxidative functionalization required to generate a dicarboxylic acid as the major product. Therefore, Option D is incorrect.