Problem 113 - A Complete Resource Book in Chemistry for JEE Main

Question

Predict the product 'B' in the sequence of reaction \(\mathrm{HC} \equiv \mathrm{CH} \frac{30 \% \mathrm{H}_{2} \mathrm{SO}_{4}}{\mathrm{HgSO}_{4}} \mathrm{~A} \stackrel{\mathrm{NaOH}}{\longrightarrow} \mathrm{B}\) (a) \(\mathrm{CH}_{3} \mathrm{CHO}\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}\) (c) \(\mathrm{CH}_{3} \mathrm{COONa}\) CC(O)CC=O (d)

Step-by-Step Solution

Answer

Product 'B' is \(\mathrm{CH}_{3}\mathrm{COONa}\) (option c).

Step 1: Identify the Reaction Type

The reaction involves acetylene \(\mathrm{HC} \equiv \mathrm{CH}\) reacting with 30% \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the presence of \(\mathrm{HgSO}_{4}\). This mixture is known to catalyze the hydration of acetylene to form an aldehyde.

Step 2: Predict Product 'A'

The result of the hydration of acetylene would transform \(\mathrm{HC} \equiv \mathrm{CH}\) into acetaldehyde (ethanal), \(\mathrm{CH}_{3}\mathrm{CHO}\), which is product 'A'.

Step 3: Identify the Second Reaction

In the second part of the sequence, product 'A', \(\mathrm{CH}_{3}\mathrm{CHO}\), reacts with \(\mathrm{NaOH}\). This reaction involves a common aldol condensation reaction mechanism or conversion to an alcohol or salt.

Step 4: Predict Product 'B'

The reaction of aldehyde (here, acetaldehyde) with \(\mathrm{NaOH}\) typically forms the sodium salt of the corresponding carboxylic acid due to the Cannizzaro reaction, resulting in \(\mathrm{CH}_{3}\mathrm{COONa}\). Thus, product 'B' is acetyl chloride salt \(\mathrm{CH}_{3}\mathrm{COONa}\).

Key Concepts

Hydration Reaction

A hydration reaction is a chemical process where a water molecule is added to a substrate. This type of reaction is quite significant in organic chemistry as it often transforms alkynes and alkenes into alcohols or aldehydes. In the context of the given exercise, acetylene, \(\mathrm{HC} \equiv \mathrm{CH}\), undergoes a hydration reaction to form acetaldehyde, an aldehyde. Here’s how this happens:

• First, the acetylene interacts with sulfuric acid \(\mathrm{H}_{2} \mathrm{SO}_{4}\) mixed with mercuric sulfate, \(\mathrm{HgSO}_{4}\).
• This mixture catalyzes the reaction, effectively adding a water molecule across the triple bond of acetylene.
• The final product of this reaction is acetaldehyde, \(\mathrm{CH}_{3}\mathrm{CHO}\).

What makes this reaction particularly fascinating is its ability to convert simple hydrocarbons into more complex alcohols or carbonyl compounds, expanding the range of substances that can be synthesized from basic materials.

Cannizzaro Reaction

The Cannizzaro reaction is an important reaction in organic chemistry, particularly for carbonyl compounds that cannot easily undergo keto-enol tautomerism. When applied to aldehydes like acetaldehyde (\(\mathrm{CH}_3\mathrm{CHO}\)), the Cannizzaro reaction occurs in the presence of a strong base such as sodium hydroxide (\(\mathrm{NaOH}\)). Here is the process explained:

• The aldehyde, which lacks alpha hydrogen atoms, can undergo this self-redox reaction where one molecule is oxidized and another is reduced.
• For acetaldehyde, one molecule is oxidized to form a carboxylate ion while another is reduced to form an alcohol.
• Ultimately, this reaction results in the formation of a carboxylic acid salt for those aldehydes involved. Specifically, acetaldehyde forms the sodium salt of acetic acid, \(\mathrm{CH}_3\mathrm{COONa}\).

This reaction serves as a powerful example of how chemical processes can be manipulated to yield various useful compounds through internal electron transfer mechanisms within molecules.

Sodium Salt Formation

Sodium salt formation, particularly within the Cannizzaro reaction context, involves transforming an aldehyde or alcohol into a salt form under basic conditions. Here is how it fits into the specified exercise:

• When acetaldehyde \(\mathrm{CH}_3\mathrm{CHO}\) is treated with \(\mathrm{NaOH}\), the resulting reaction leads to the formation of the sodium salt of a carboxylic acid.
• This conversion process involves the aldehyde undergoing a reaction where one molecule is reduced while another is oxidized, forming the sodium acetate salt \(\mathrm{CH}_3\mathrm{COONa}\).
• The sodium ion from the \(\mathrm{NaOH}\) binds with the carboxylate group, stabilizing the compound as a salt, which is typically more soluble or easier to handle in various chemical processes.

Understanding the formation of sodium salts is crucial because these compounds are often more stable and can be manipulated in different synthetic pathways, offering versatility in chemical manufacturing and research.