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:

Question

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:

Accepted Answer

Correct answer: C. (c) The iodoform test is given by compounds containing the methyl ketone group ( CH_3 CO- ) or alcohols that can be oxidized to methyl ketones ( CH_3 CH ( OH )- ). (A) Methanol ( CH_3 OH ) does not give the test, while Ethanol ( CH_3 CH_2 OH ) gives a positive iodoform test because it contains the CH_3 CH ( OH ) - group. Thus, they can be differentiated. (B) Acetic acid ( CH_3 COOH ) and Propanoic acid ( CH_3 CH_2 COOH ) both do not give the iodoform test. Carboxylic acids generally do not respond to this test. Thus, they cannot be differentiated. (C) Cyclohexene and cyclohexanone both do not give the iodoform test. Cyclohexanone lacks a methyl group adjacent to the carbonyl carbon. Thus, they cannot be differentiated. (D) Diethyl ether ( CH_3 CH_2 OCH_2 CH_3 ) and Pentan-3-one ( CH_3 CH_2 COCH_2 CH_3 ) both do not give the iodoform test. Thus, they cannot be differentiated. (E) Anisole ( C_6H_5 OCH_3) does not give the test, while Acetone ( CH_3 COCH_3) gives a positive iodoform test as it is a methyl ketone. Thus, they can be differentiated. Therefore, iodoform test can differentiate pairs A and E.

$Iodoform test can differentiate between(A) Methanol and Ethanol(B) CH_{3} COOH and CH_{3} CH_{2} COOH (C) Cyclohexene and cyclohexanone(D) Diethyl ether and Pentan-3-one(E) Anisole and acetone Choose the correct answer from the options given below:$

Solution

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