Question

Which of the following compounds undergo haloform reaction?

A.${ CH }_{ 3 }{ COCH }_{ 3 }$
B.${ HCHO }$
C.${ CH }_{ 3 }{ CH }_{ 2 }{ Br }$
D.${ CH }_{ 3 }{ OCH }_{ 3 }$

Answer 1

Hint:
Haloform reaction: Compounds having ${ CH }_{ 3 }{ C=O }$ group or ${ CH }_{ 3 }{ OH }$ group react with sodium hypohalite to give haloform. This method is used for the detection of the methyl keto group.

Complete step-by-step answer:
The Iodoform test is given by aldehydes and ketones having alpha-methyl group ${ CH }_{ 3 }{ C=O }$. Even alcohols having alpha-methyl group show iodoform test.
${ CH }_{ 3 }{ COCH }_{ 3 }$: In acetone, ${ CH }_{ 3 }{ C=O }$ The group is present, so it shows the iodoform test.
${ HCHO }$: There is no beta hydrogen in formaldehyde to deprotonate, so they will not deprotonate and hence, do not show iodoform test.
${ CH }_{ 3 }{ CH }_{ 2 }{ Br }$: As there is no ${ CH }_{ 3 }{ C=O }$ or ${ CH }_{ 3 }{ OH }$ in ethyl bromide, so it will not give a haloform reaction.
${ CH }_{ 3 }{ OCH }_{ 3 }$: Methoxy methane does not show iodoform reaction as there is no presence of the ${ C=O }$ group.
   The correct option is A.

Additional Information:
In the Iodoform test, carbonyl compound (aldehyde or ketone) or a compound containing ${ CH }_{ 3 }{ C=O }$ group, reacts with ${ I }_{ 2 }$, in presence of ${ NaOH }$, to give ${ CH }_{ 3 }{ I }$ (yellow precipitate).
The iodoform test is a very useful method to identify the presence of these methyl ketones or acetaldehyde in an unknown compound.
 The reaction of iodine, a base, and a methyl ketone gives a yellow precipitate along with an “antiseptic” smell.

Note:
The possibility to make a mistake is that you may choose option B. As I discussed above, carbonyl compounds give the iodoform test but only those compounds having an alpha-methyl group and there is no alpha carbon in formaldehyde, so it does not give iodoform test.