Animated Solution for Chemistry - Organic Chemistry: The correct reaction/reaction sequence that would produce a dicarboxylic acid as the major product is

Finding the Dicarboxylic Acid

Welcome to a classic organic chemistry puzzle! Our mission is straightforward but requires a sharp memory: we need to identify which of the given reaction sequences produces a dicarboxylic acid as its major product. A dicarboxylic acid is simply a molecule that boasts exactly two carboxyl groups ($-COOH$). To find the winner, we must act as chemical detectives and trace the transformation of each reactant step-by-step.

The Cyanide Substitution and Hydrolysis

Let's start by examining option (A). We are given a chlorohydrin, specifically $HO-C{H}_2-C{H}_2-Cl$. The first reagent is $NaCN$. This is a classic setup for a nucleophilic substitution reaction ($S_{N}2$). The cyanide ion ($C{N}^{-}$) acts as a nucleophile and kicks out the chloride leaving group, forming $HO-C{H}_2-C{H}_2-CN$.

The next step involves aqueous hydroxide followed by acidic hydrolysis ($H_3{O}^{+}$). This sequence completely hydrolyzes the nitrile group ($-CN$) into a carboxylic acid group ($-COOH$). The final product is 3-hydroxypropanoic acid ($HO-C{H}_2-C{H}_2-COOH$). Since this molecule contains only one $-COOH$ group, it is a monocarboxylic acid. Option A is out!

The Mild Oxidation of Glucose

Moving on to option (B), we encounter a familiar biomolecule: Glucose ($CHO-(CHOH{)}_4-C{H}_{2}OH$). The reagent here is bromine water ($B{r}_2/H_{2}O$).

Bromine water is known as a mild oxidizing agent. It is powerful enough to oxidize the highly reactive aldehyde group ($-CHO$) at the top of the chain into a carboxylic acid, but it leaves the secondary and primary alcohol groups completely untouched. The resulting product is gluconic acid ($COOH-(CHOH{)}_4-C{H}_{2}OH$). Once again, we only have a single $-COOH$ group. Option B is incorrect.

The Strong Oxidation of a Primary Alcohol

Let's skip to option (D) for a moment. Here, we have a complex molecule containing both a ketone and a primary alcohol group. The reagent is chromic acid ($H_{2}Cr{O}_4$), often associated with the Jones oxidation.

Chromic acid is a strong oxidizing agent. It will aggressively oxidize the primary alcohol ($-C{H}_{2}OH$) all the way to a carboxylic acid ($-COOH$). However, ketones are generally resistant to further oxidation under these conditions. The final product is a keto-carboxylic acid. Because it only possesses one $-COOH$ group, option D is also off the table.

The Two-Step Masterpiece

Finally, we arrive at option (C). The reactant is bromocyclohexane, a simple cyclic haloalkane. The reaction happens in two distinct phases.

Phase 1: Elimination The first reagent is alcoholic $KOH$. This is a crucial distinction! While aqueous $KOH$ favors substitution to form an alcohol, alcoholic $KOH$ provides strongly basic alkoxide ions that favor an $E2$ elimination reaction. A molecule of $HBr$ is removed, creating a double bond within the ring. The product of this first step is cyclohexene.

Phase 2: Oxidative Cleavage Now, the cyclohexene is subjected to hot, acidic $KMn{O}_4$. This is a favorite concept in competitive exams! Hot $KMn{O}_4$ is a vigorous oxidizing agent that doesn't just add oxygen; it completely shatters the carbon-carbon double bond. This process is known as oxidative cleavage.

When the double bond in the cyclohexene ring breaks, the ring opens up into a straight chain. The two carbons that originally formed the double bond are fully oxidized into carboxylic acid groups. The resulting molecule is a six-carbon chain with a $-COOH$ group at both ends: $HOOC-C{H}_2-C{H}_2-C{H}_2-C{H}_2-COOH$.

This molecule is adipic acid (hexanedioic acid). Because it contains exactly two carboxyl groups, it is a dicarboxylic acid! We have found our winner. Option (C) is the correct answer.