Problem 3 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An organic compound (A) (molecular formula \(\left.\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\right)\) was hydrolysed with dil. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to give a carboxylic acid (B) and an alochol (C). 'C' gives white turbidity immediately when treated with anhydrous \(\mathrm{ZnCl}_{2}\) and conc. HCl. The organic compound (A) is :

Step-by-Step Solution

Answer

Compound (A) is tert-butyl acetate (CH₃COOtC₄H₉).

Step 1: Understanding the Problem

We are given an organic compound \( A \) with the molecular formula \( C_6H_{12}O_2 \). It undergoes hydrolysis to produce a carboxylic acid \( B \) and an alcohol \( C \). Compound \( C \) produces white turbidity when treated with anhydrous \( ZnCl_2 \) and conc. HCl, known as the Lucas test. This indicates that \( C \) is a tertiary alcohol.

Step 2: Analyzing the Reaction

The hydrolysis of compound \( A \) using dilute \( H_2SO_4 \) suggests that \( A \) could be an ester, as esters are known to undergo hydrolysis to produce a carboxylic acid and an alcohol. The general reaction for ester hydrolysis is:\[ RCOOR' + H_2O \xrightarrow{H^+} RCOOH + R'OH \]where \( R'OH \) is the alcohol.

Step 3: Identifying Alcohol \( C \)

Given that \( C \) gives an immediate white turbidity in the Lucas test, \( C \) must be a tertiary alcohol. The simplest tertiary alcohol possible with part of the formula \( C_6H_{12}O_2 \) is tert-butyl alcohol, \( (CH_3)_3COH \).

Step 4: Determining Carboxylic Acid \( B \)

Since \( (CH_3)_3COH \) contributes \( C_4H_9O \) to the molecular formula, the remainder \( C_2H_3O_2 \) must correspond to the carboxylic acid. The only viable candidate is acetic acid, \( CH_3COOH \).

Step 5: Predicting Organic Compound \( A \)

Combining the identified carboxylic acid \( CH_3COOH \) and the alcohol \( (CH_3)_3COH \), we can construct the ester \( CH_3COO-tC_4H_9 \), which is tert-butyl acetate. Thus, compound \( A \) is tert-butyl acetate.

Key Concepts

Hydrolysis

Hydrolysis is a fascinating chemical reaction involving the breaking down of a compound due to its reaction with water. In organic chemistry, it's particularly significant because it helps transform molecules into simpler substances. In the described exercise, we explore hydrolysis on a compound with the molecular formula \( C_6H_{12}O_2 \).

• Hydrolysis requires a catalyst, often an acid or a base, to proceed effectively. In this case, dilute sulfuric acid \( (H_2SO_4) \) is used.
• The process breaks the bond within the compound \( A \), producing a carboxylic acid \( B \) and an alcohol \( C \).
• Hydrolysis is often applied to esters, where the reaction is fundamental in creating simpler molecules such as carboxylic acids and alcohols.

Understanding hydrolysis is crucial, not just for solving textbook problems, but for appreciating how complex biological systems break down nutrients and other compounds. By understanding this, we learn how crucial reactions transform our world at a molecular level.

Carboxylic Acid

Carboxylic acids are a class of organic acids characterized by the presence of at least one carboxyl group \( (-COOH) \). They are a significant product of hydrolysis reactions, as seen in this exercise where compound \( A \) breaks down into a carboxylic acid \( B \).

• The carboxyl group is a combination of a carbonyl \( (C=O) \) and a hydroxyl group \( (OH) \).
• These acids are known for their acidic nature and ability to donate a proton \( (H^+) \).
• In the solution to the exercise, the carboxylic acid formed is acetic acid \( (CH_3COOH) \), a simple yet vital organic acid.

Carboxylic acids play a fundamental role in organic chemistry, being key components in synthesis and metabolism. Their simple structure belies a complexity that's crucial in the diverse world of organic reactions.

Tertiary Alcohol

Tertiary alcohols are a specific type of alcohol where the hydroxyl group \( (OH) \) is attached to a carbon atom that is itself connected to three other carbon atoms. They stand out due to their unique structural characteristics and reactivity.

• The Lucas test is a simple method to distinguish tertiary alcohols from primary and secondary alcohols. When an alcohol is mixed with Lucas reagent \( (ZnCl_2 + HCl) \), tertiary alcohols like the one mentioned in the original exercise immediately form an insoluble layer (white turbidity).
• The test is indicative because tertiary alcohols react the fastest due to the stability of the resultant carbocation.
• In this particular exercise, the alcohol \( C \) formed post-hydrolysis is tert-butyl alcohol \( (CH_3)_3COH \).

Understanding tertiary alcohols is not only academically interesting but also practical, as these compounds are prevalent in numerous chemical applications. They underline the fascinating interplay between chemical structure and reactivity.