Question

Which of the following compounds give(s) iodoform test ?

• A. CH2=CH-CHO
• B. CH3COCH3
• C. CH3COCH2COOC2H5
• D. CH3-CH(OH)-CH2-CH(OH)CH3

Answer 1

Answer: (B), (C), (D)

(B), (C), (D)

Answer 2

The iodoform test is given by compounds containing the methyl ketone group ($CH_3CO-$) or a methyl secondary alcohol group ($CH_3CH(OH)-$). Under the basic conditions of the test, the methyl secondary alcohol group is oxidized to the methyl ketone group.

Let's examine each compound:

(A) $CH_2=CH-CHO$ (Acrolein): This compound is an $\alpha,\beta$-unsaturated aldehyde. It does not contain a $CH_3CO-$ group or a $CH_3CH(OH)-$ group. Therefore, it does not give the iodoform test.

(B) $CH_3COCH_3$ (Acetone): This compound contains a $CH_3CO-$ group. Therefore, it gives the iodoform test. $CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 \downarrow + CH_3COONa + 3NaI + 3H_2O$

(C) $CH_3COCH_2COOC_2H_5$ (Ethyl acetoacetate): This compound contains a $CH_3CO-$ group. Therefore, it gives the iodoform test. $CH_3COCH_2COOC_2H_5 + 3I_2 + 4NaOH \rightarrow CHI_3 \downarrow + CH_3COONa + CH_2(Na)COOC_2H_5 + 3NaI + 3H_2O$

(D) $CH_3-CH(OH)-CH_2-CH(OH)CH_3$ (Pentane-2,4-diol): This compound contains two $CH_3CH(OH)-$ groups. Under the reaction conditions, these are oxidized to $CH_3CO-$ groups, forming pentane-2,4-dione ($CH_3COCH_2COCH_3$). Pentane-2,4-dione contains two $CH_3CO-$ groups and gives the iodoform test. $CH_3CH(OH)CH_2CH(OH)CH_3 \xrightarrow{I_2, OH^-} CH_3COCH_2COCH_3$ $CH_3COCH_2COCH_3 + 6I_2 + 8NaOH \rightarrow 2CHI_3 \downarrow + NaOCOCH_2COONa + 6NaI + 6H_2O$

Therefore, compounds (B), (C), and (D) give the iodoform test.