Problem 178 - A Complete Resource Book in Chemistry for JEE Main

Question

Compound (A), \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\), optically active was oxidized by \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\)to an optically active compound (B), \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3} .(\mathrm{A})\) on oxidation with \(\mathrm{CrO}_{3}\) in pyridine gives optically inactive compound (C) which on treating with \(\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}\) gave 3 -methylpentane. (A) on oxidation with \(\mathrm{H}_{2} \mathrm{CrO}_{4}\) gave optically inactive compound (D). Solution \(\mathrm{HOCH}_{2}-\mathrm{CH}_{2}-\mathrm{C}^{*} \mathrm{H}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2}-\mathrm{CHO}\) $$ \text { (A) } \mathrm{HO}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2}-\mathrm{COOH} $$ (B) \(\mathrm{HOCH}_{2}-\mathrm{CH}_{2}-\mathrm{C}^{*} \mathrm{H}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2}-\mathrm{CHO}\) \(\mathrm{OHC}-\mathrm{CH}_{2}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2}-\mathrm{CHO}\) (C) \(\mathrm{HOCH}_{2}-\mathrm{CH}_{2}-\mathrm{C}^{*} \mathrm{H}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2}-\mathrm{CHO}\) \(\mathrm{H}_{2} \mathrm{CrO}_{4}\) \(\mathrm{HOOC}-\mathrm{CH}_{2}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2} \mathrm{COOH}\) (D) \(\mathrm{OHC}-\mathrm{CH}_{2}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2}-\mathrm{CHO}_{-\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}}{\longrightarrow}\) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) \(3-\) methylpentane Compound (B) on treating with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gives (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2} \mathrm{CHO}\) (b) CC1CCC(=O)OC1 (c) CC1CCOC(=O)C1 (d) None of these

Step-by-Step Solution

Answer

The treatment of (B) with dilute \( \mathrm{H}_2 \mathrm{SO}_4 \) results in option (d) None of these.

Step 1: Analyze starting compound (A)

The molecular formula of compound (A) is \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_2 \). It is optically active, indicating the presence of a chiral center. The structure provided is \( \mathrm{HO}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}^*(\mathrm{CH}_3)-\mathrm{CH}_2-\mathrm{COOH} \).

Step 2: Understand oxidation with Ag(NH3)2+

Compound (A) undergoes oxidation with \( \mathrm{Ag}(\mathrm{NH}_3)_2^+ \) which primarily oxidizes aldehydes to carboxylic acids. The formula for (B) is \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3 \), suggesting the aldehyde of (A) is converted to a carboxylic acid.

Step 3: Determine compound (B)

The structure of compound (B) is \( \mathrm{HOCH}_2-\mathrm{CH}_2-\mathrm{C}^*(\mathrm{CH}_3)-\mathrm{CH}_2-\mathrm{COOH} \), illustrating that the aldehyde group in (A) was oxidized to a carboxylic acid.

Step 4: Examine further oxidation with CrO3

When compound (A) is oxidized with \( \mathrm{CrO}_3 \), it loses optical activity, indicating the chiral center is likely removed or racemized. This results in compound (C), which is \( \mathrm{OHC}-\mathrm{CH}_2-\mathrm{CH}(\mathrm{CH}_3)-\mathrm{CH}_2-\mathrm{COOH} \).

Step 5: Consider the reaction with Zn-Hg/HCl

Compound (C) reduces via the Clemmensen reduction (using Zn-Hg/HCl) to yield 3-methylpentane, supporting the removal or modification of oxygen functionalities leading back to a simple alkane chain.

Step 6: Confirm compound (D) structure

The oxidation of (B) by \( \mathrm{H}_2 \mathrm{CrO}_4 \) would yield \( \mathrm{HOOC}-\mathrm{CH}_2-\mathrm{CH}(\mathrm{CH}_3)-\mathrm{CH}_2-\mathrm{COOH} \), as all primary alcohols convert to carboxylic acids and thereby losing optical activity.

Step 7: Analyze treatment of (B) with H2SO4

Compound (B) subjected to dilute \( \mathrm{H}_2 \mathrm{SO}_4 \) can result in structural rearrangement or cyclization. Due to the provided options and smell structures, none match the described compounds, hence (d) None of these.

Key Concepts

Chiral Center

In organic chemistry, a chiral center is an atom that has four different groups attached to it, giving rise to molecules that are non-superimposable on their mirror images. These types of molecules are called enantiomers. A chiral center often leads to optical activity, where each enantiomer rotates plane-polarized light in different directions. In our given compound (A), this chiral center is represented by the carbon labeled as * in the structure \( \mathrm{HOCH}_2-\mathrm{CH}_2-\mathrm{C}^{*}\mathrm{H}\left(\mathrm{CH}_3\right)-\mathrm{CH}_2-\mathrm{CHO} \). The presence of this chiral center is what makes compound (A) optically active, which means it has the ability to rotate polarized light. Whenever you see such a center, you should check for its ability to maintain optical activity in reactions.

Optical Activity

Optical activity refers to the ability of a chiral substance to rotate the plane of polarized light passing through it. This property is exclusively associated with chiral compounds that are not mirrored images of one another. In our case, compound (A) is originally optically active due to its chiral center. However, during its oxidation and subsequent reactions, we notice changes in optical activity. For instance, when oxidized by \( \mathrm{CrO}_3 \), the resulting compound (C) becomes optically inactive. This loss of optical activity likely indicates that a symmetric, possibly achiral product is formed, possibly through racemization or removal of the chiral center. Understanding optical activity is valuable for determining the purity and behavior of chemical compounds, particularly in the synthesis of pharmaceuticals.

Oxidation Reactions

Oxidation reactions in organic chemistry involve the increase in the oxidation state of a molecule, often by removing electrons or adding oxygen. In the context of our problem, compound (A) undergoes several oxidation reactions. Reaction with \( \mathrm{Ag} (\mathrm{NH}_3)_2^+ \) results in the oxidation of an aldehyde group to a carboxylic acid, converting it into compound (B) \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3 \). Moreover, the use of \( \mathrm{CrO}_3 \) and \( \mathrm{H}_2 \mathrm{CrO}_4 \) further oxidizes the compound, leading to optically inactive structures by transforming alcohol groups to carboxylic acids, essentially stripping the molecule of its optical activity. Recognizing how oxidizing agents work lets you predict changes in the functional groups of compounds.

Chemical Structure Analysis

Chemical structure analysis in organic chemistry is vital for understanding the reactivity and properties of a substance. This involves analyzing the molecular formula, identifying functional groups, and deducing structural changes during reactions. For compound (A) whose formula is \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_2 \), we start by establishing its structure based on optical activity and possible reagents. Through oxidation and reduction processes, we track transformations such as the conversion of carbonyl groups to carboxylic acids and alkanes. Techniques like oxidation, Clemmensen reduction, and catalytic processes help in deciphering the chemical pathways and resultant products seen in transformations to compounds (B), (C), and (D). Understanding these transformations aids in anticipating the behavior and outcome of organic reactions.