Question

An organic compound ‘X’ on treatment with PCC in dichloromethane gives the compound Y. Compound ‘Y’ reacts with I₂ and alkali to form yellow precipitate of triiodomethane. The compound X is

• A. CH₃CHO
• B. CH₃COCH₃
• C. CH₃CH₂OH
• D. CH₃COOH

Answer 1

The Correct Option is C

To solve the problem, we need to identify the organic compound 'X' that, upon treatment with PCC, forms a compound 'Y' which then reacts with iodine and alkali to produce a yellow precipitate of triiodomethane (iodoform). This reaction is known as the iodoform test.

1. Understanding the Iodoform Test:
The iodoform test is positive for compounds containing the $CH_3CO-$ group or compounds that can be oxidized to form this group. This means either methyl ketones ($RCOCH_3$) or secondary alcohols of the form $RCH(OH)CH_3$ will give a positive iodoform test.

2. PCC Oxidation:
PCC (pyridinium chlorochromate) is a mild oxidizing agent that converts primary alcohols to aldehydes and secondary alcohols to ketones.

3. Analyzing the Options:
Let's analyze each option to see if it fits the criteria:
(A) $CH_3CHO$ (Acetaldehyde): This is an aldehyde containing the $CH_3CO-$ group. It can directly give iodoform test without PCC oxidation.
(B) $CH_3COCH_3$ (Acetone): This is a ketone containing the $CH_3CO-$ group. It will directly give iodoform test without PCC oxidation. However, if 'X' were acetone, 'Y' would also be acetone.
(C) $CH_3CH_2OH$ (Ethanol): PCC will oxidize ethanol to acetaldehyde ($CH_3CHO$). Acetaldehyde contains the $CH_3CO-$ group and will give a positive iodoform test. So 'X' is ethanol, 'Y' is acetaldehyde. This is a valid pathway.
(D) $CH_3COOH$ (Acetic Acid): This is a carboxylic acid. PCC does not react with carboxylic acids.

Final Answer:
The compound X is $CH_3CH_2OH$.

Answer 2

PCC (Pyridinium chlorochromate) is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes. In this case, compound 'X' is a primary alcohol, CH₃CH₂OH (ethanol), which is oxidized by PCC to form acetaldehyde (CH₃CHO), compound 'Y'.

Next, compound 'Y' (acetaldehyde) reacts with iodine (I₂) in the presence of alkali, forming a yellow precipitate of triiodomethane (CHI₃) as a result of the iodoform reaction. The iodoform test is positive for compounds that have the structure R-CHOH- (where R is a methyl group or equivalent), such as acetaldehyde (CH₃CHO).

Thus, the correct compound 'X' is CH₃CH₂OH (ethanol).

The correct answer is (C) : CH₃CH₂OH.