Iodoform test can differentiate between

A. Methanol and Ethanol
B. $$CH_{3}COOH$$ and $$CH_{3}CH_{2}COOH$$
C. Cyclohexene and cyclohexanone
D. Diethyl ether and Pentan-3-one
E. Anisole and acetone

Choose the correct answer from the options given below:

The iodoform test is used to detect the presence of a methyl group attached to a carbonyl group ($$CH_3CO-$$) or a secondary alcohol group ($$CH_3CH(OH)-$$), as well as ethanol and acetaldehyde. We need to identify which pairs can be differentiated by this test.

Key Principle: The iodoform test is positive for compounds containing the $$CH_3CO-$$ group (or structures that can be oxidized to it). Specifically, it is positive for:

- Acetaldehyde ($$CH_3CHO$$)

- Methyl ketones ($$CH_3COR$$)

- Ethanol ($$CH_3CH_2OH$$, oxidized to $$CH_3CHO$$)

- Secondary alcohols of the form $$CH_3CH(OH)R$$

Pair A: Methanol and Ethanol

- Methanol ($$CH_3OH$$): Does NOT give iodoform test (no $$CH_3CO$$ or $$CH_3CHOH$$ group)

- Ethanol ($$CH_3CH_2OH$$): Gives POSITIVE iodoform test (oxidized to acetaldehyde)

Can be differentiated.

Pair B: $$CH_3COOH$$ and $$CH_3CH_2COOH$$

- Both are carboxylic acids. Carboxylic acids do not give the iodoform test (the $$CH_3CO-$$ in acetic acid is $$CH_3COOH$$, not a ketone or aldehyde). Both are negative.

Cannot be differentiated.

Pair C: Cyclohexene and Cyclohexanone

- Cyclohexene: No $$CH_3CO$$ group. Negative.

- Cyclohexanone: It is a cyclic ketone but does NOT have a $$CH_3CO-$$ group (the carbonyl is flanked by $$CH_2$$ groups). Negative.

Cannot be differentiated.

Pair D: Diethyl ether and Pentan-3-one

- Diethyl ether ($$C_2H_5OC_2H_5$$): Negative.

- Pentan-3-one ($$CH_3CH_2COCH_2CH_3$$): No $$CH_3CO$$ group (the carbonyl is flanked by $$CH_2$$ groups). Negative.

Cannot be differentiated.

Pair E: Anisole and Acetone

- Anisole ($$C_6H_5OCH_3$$): Negative.

- Acetone ($$CH_3COCH_3$$): Has $$CH_3CO$$ group. Positive.

Can be differentiated.

The pairs that can be differentiated are A and E.

The correct answer is Option 3: A & E Only.