Iodoform test can differentiate between
A. Methanol and Ethanol
B. $\mathrm{CH}_{3} \mathrm{COOH}$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$
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 given by compounds containing the methyl ketone group ($CH_3CO-$) or alcohols that can be oxidized to methyl ketones ($CH_3CH(OH)-$).
A. Methanol ($CH_3OH$) does not give the test, while Ethanol ($CH_3CH_2OH$) gives a positive iodoform test because it contains the $CH_3CH(OH)-$ group. Thus, they can be differentiated.
B. Acetic acid ($CH_3COOH$) and Propanoic acid ($CH_3CH_2COOH$) 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 ($CH_3CH_2OCH_2CH_3$) and Pentan-3-one ($CH_3CH_2COCH_2CH_3$) both do not give the iodoform test. Thus, they cannot be differentiated.
E. Anisole ($C_6H_5OCH_3$) does not give the test, while Acetone ($CH_3COCH_3$) 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.