Question

Which of the following compounds is not formed in the iodoform reaction of acetone?
(A) ${CH}_{3}CO{CH}_{2}I$
(B) ${ICH}_{2}COC{H}_2$
(C) ${CH}_{3}COCH{I}_2$
(D) ${CH}_{3}COC{I}_3$

Answer 1

Hint: Iodoform reaction of acetone will lead to the products - acetic acid (${CH}_{3}COOH$) (sodium acetate) and iodoform ($CH{I}_3$). On writing the mechanism of the reaction, we can find certain stable intermediates are also formed. None of these intermediates contains a C=C bond. So, among the options given, the compound with C=C is the answer.

Complete Step by Step Solution:
-Structure of acetone is given below:

-Iodoform reaction is used to identify the presence of carbonyl compounds in compounds.
-Here in the iodoform reaction of acetone, the acetone will be reacted by $I_2$ in the presence of NaOH to form acetic acid (sodium acetate) and iodoform.
-When we look at the mechanism of this reaction, we can see many intermediates formed in between.
-Given below is the diagrammatic representation of the mechanism of iodoform reaction of acetone.-From the mechanism, we can see that the iodoform reaction of acetone leads to the products - iodoform ($CH{I}_3$) and acetic acid (${CH}_{3}COOH$ ).
-Apart from the final products, there are certain stable intermediates formed in the reaction - i.e, on addition of iodine on acetate ion, it leads to the formation of 1-iodoacetic acid (${CH}_{3}CO{CH}_{2}I$) and on successive addition of iodine , two more compounds are formed which are 1,1-di iodoacetic acid (${CH}_{3}COCH{I}_2$ ) and 1,1,1- tri iodoacetic acid (${CH}_{3}COC{I}_3$)
-So, these compounds ${CH}_{3}CO{CH}_{2}I$,${CH}_{3}COCH{I}_2$ , ${CH}_{3}COC{I}_3$are formed in the iodoform reaction of acetone, which are satisfied by Option (A), option (C) and option (D).
-In the question, it has been asked which among the options are not formed in the iodoform reaction of acetone and that compound is Option (B)${ICH}_{2}COC{H}_2$.

Therefore, the correct answer here is Option (B)${ICH}_{2}COC{H}_2$.

Note: Iodoform reaction occurs when there is an $\alpha$- hydrogen in that compound, and this $\alpha$ - hydrogen will be substituted with iodine leading to the formation of iodoform. In the case of acetone, there are three $\alpha$ - hydrogens present in the compound, so all the three $\alpha$ - hydrogens are replaced by iodine successively and thereby these stable intermediates are formed.