Write about the oxidation reaction of methyl ketone compounds $(R-CO-CH_3)$ by the haloform reaction or write a note on the Haloform test.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Methyl ketones $(R-CO-CH_3)$ are oxidized by sodium hypohalite $(NaOX)$ to the sodium salts of corresponding carboxylic acids having one carbon atom less than that of the carbonyl compound. The methyl group is converted to haloform $(CHX_3)$. This oxidation does not affect a carbon-carbon double bond $(C=C)$,if present in the molecule.
$R-CO-CH_3 + 3NaOX \rightarrow R-COONa + CHX_3 + 2NaOH$
The reaction proceeds in two steps:
$(i)$ $R-CO-CH_3 + 3NaOX \rightarrow R-CO-CX_3 + 3NaOH$
(ii) $R-CO-CX_3 + NaOH \rightarrow R-COONa + CHX_3$
Iodoform reaction with sodium hypoiodite $(NaOI)$ is also used for the detection of the $CH_3CO-$ group or $CH_3CH(OH)-$ group,which produces iodoform ($CHI_3$,yellow precipitate) on oxidation.

Explore More

Similar Questions

In the following reaction,
$CH_3-C \equiv CH$ $\xrightarrow[H_3O^+]{Hg^{2+}} X$ $\xrightarrow[PhCHO]{dil. NaOH} Y$
$X$ and $Y$,respectively,are

Pyrolysis of acetone gives $CH_2=C=O$ called

Arrange the following compounds in increasing order of their enol content:
$1. CH_3CHO$
$2. CH_3COCH_3$
$3. CH_3COCH_2CHO$
$4. CH_3COCH_2COCH_3$

An organic compound $A$ has the molecular formula $C_3H_6O$. It gives the iodoform test. When it is saturated with $HCl$,it gives a compound $B$ with the molecular formula $C_9H_{14}O$. What are $A$ and $B$ respectively?

$C_6H_5-CH=CHCHO \xrightarrow{X} C_6H_5-CH=CHCH_2OH$. In the above sequence,$X$ can be:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo