Question

A student performed analysis of aliphatic organic compound 'X' which on analysis gave C = 61.01%, H = 15.25%, N = 23.74%. This compound, on treatment with HNO\(_2\)/H\(_2\)O produced another compound 'Y' which did not contain any nitrogen atom. However, the compound 'Y' upon controlled oxidation produced another compound 'Z' that responded to iodoform test. The structure of 'X' is:

• A. Ph—CH—NH\(_2\) (with CH\(_3\) substituent)
• B. CH\(_3\)—CH—NH\(_2\) (with CH\(_3\) substituent)
• C. CH\(_3\)—CH\(_2\)—CH—CH\(_3\) (with NH\(_2\) substituent)
• D. CH\(_3\)—CH\(_2\)—CH\(_2\)—NH\(_2\)

Hint:

For nitrous acid reactions: - Primary aliphatic amines → alcohols. - Secondary amines → nitroso compounds. - Tertiary amines → no reaction.

Answer 1

The Correct Option is B

To determine the structure of the compound 'X', we must analyze the given information systematically:

1. The composition of compound 'X' includes C = 61.01%, H = 15.25%, and N = 23.74%. This suggests an aliphatic amine because it contains a significant percentage of nitrogen, typical for amines.
2. Treatment of 'X' with HNO\(_2\)/H\(_2\)O results in the formation of compound 'Y' that does not contain nitrogen. This reaction suggests that 'X' is a primary amine (R—NH\(_2\)), as primary amines react with nitrous acid (HNO\(_2\)) to form alcohols (R—OH) through diazotization.
3. The resultant compound 'Y', upon controlled oxidation, gives another compound 'Z' which responds to the iodoform test. The iodoform test is a chemical reaction used to identify methyl ketones or secondary alcohols with at least one methyl group next to the carbonyl.

Given these observations, we can infer:

• The iodoform test indicates the presence of a structure that can form CH\(_3\)C=O (acetyl group) or \(\text{CH}_3\)-OH(CH\(_3\)—C=O\) structure. For the iodoform test to be positive, compound 'Y' must have been oxidized to a methyl ketone.

Now, let's evaluate the options:

• **Ph—CH—NH\(_2\) (with CH\(_3\) substituent):** This is incorrect because the presence of a phenyl group (Ph) implies aromaticity, not aliphatic. Furthermore, aromatic primary amines behave differently with HNO\(_2\).
• **CH\(_3\)—CH—NH\(_2\) (with CH\(_3\) substituent):** This structure is 2-Aminopropane. On reacting with HNO\(_2\), it forms propan-2-ol. When oxidized, it can become acetone (propanone), which gives a positive iodoform test.
• **CH\(_3\)—CH\(_2\)—CH—CH\(_3\) (with NH\(_2\) substituent):** This would produce alcohol on diazotization, but the resulting butanol cannot form a methyl ketone for iodoform test.
• **CH\(_3\)—CH\(_2\)—CH\(_2\)—NH\(_2\):** This produces propanol, which doesn't yield a methyl ketone on oxidation.

Hence, the correct structure of 'X' is CH\(_3\)—CH—NH\(_2\) (with CH\(_3\) substituent), as only this structure correctly matches the reaction pathway leading to a molecule that would yield a positive iodoform test after oxidation.