Concept:
Haloform Reaction:
- Acetaldehyde and all methyl ketones can be oxidized by hypohalite solutions to yield a haloform (chloroform, bromoform, or iodoform) and the reaction is called haloform reaction.
- A yellow ppt. of iodoform is seen.
The general reaction scheme is given as follows
R-COCH3 + X2 / OH(hydroxide) → RCOO- + CHX3 (halo form)
X2 + 2NaOH → NaOX + NaX + H2O
\(\begin{array}{{20}{c}} {\rm ,,,,,,O}\ {,,,,,,||}\ {\rm C{H_3} - C - R } \end{array} \begin{matrix} {} \\ {} \\ \xrightarrow{\rm 3NaOH} \end{matrix} \ \ \begin{array}{{20}{c}} {\rm ,,,,,,O}\ {,,,,,,||}\ {\rm C{H_3} - C - R } \end{array} \begin{matrix} {} \\ {} \\ + \ \ \rm 3NaOH \end{matrix} \)
\begin{array}{*{20}{c}} {\rm ,,,,,,O}\ {,,,,,,||}\ {\rm C{X_3} - C - R } \end{array} \begin{matrix} {} \\ {} \\ \rm \xrightarrow{NaOH} \underset{haloform}{CHX_3} + RCOONa \end{matrix}
Explanation:
Criteria for giving positive haloform test:
- All compounds containing CH3 –C=O and CH3CH-(OH) groups are positive to the Haloform test.
- Also, the compounds which on oxidation with halogen and alkali yield CH3 -C=O groups will respond positively to Haloform Test.
- Common compounds that respond to this test are
- CH3CH2OH
- CH3CHO
- CH3COCH3
- CH3COC2H5
- CH3CHOHCH3.
Step I (oxidation step)
I. \(\begin{array}{{20}{c}} {\rm C{H_3} - CH - R + {I_2}}\ {|,,,,,,,,,,}\ {\rm OH,,,,,,} \end{array} \begin{matrix} \begin{array}{{20}{c}} {\rm \ \ \ \ \ O,,,,,,,}\ { \ \ \ \ \ ||,,,,,,,}\ {\longrightarrow \rm C{H_3} - C - R + 2HI} \end{array} \\ {} \\ {} \\ {} \\ {} \end{matrix}\)
- Halogenation.
\(\begin{array}{{20}{c}} {\rm O,,,,,,,}\ {||,,,,,,,}\ {\rm C{H_3} - C - R + 3{I_2}} \end{array} \begin{array}{{20}{c}} {\rm \ \ \ \ O,,,,,,,}\ { \ \ \ \ ||,,,,,,,}\ {\longrightarrow \rm C{I_3} - C - R + 3{HI}} \end{array} \)
- Hydrolysis
\(\begin{array}{{20}{c}} {\rm C{I_3} - C - R + {NaOH}}\ {||, \ \ \ \ \ \ \ \ \ \ ,,,}\ {\rm O,,,,,, \ \ \ \ \ \ \ \ } \end{array} \begin{matrix} \begin{array}{{20}{c}} {\rm \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ O,,,,,,,}\ { \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||,,,,,,,}\ {\longrightarrow \rm \underset{Iodoform}{C{HI_3}} + R - C - ONa } \end{array} \\ {} \\ {} \\ {} \\ {} \end{matrix}\)
- 2-degree alcohols on oxidation produce CH3C=O-- group as shown above and thus reacts positively to haloform test.
- Thus, both the groups \(\begin{array}{{20}{c}} {\rm O,, \ \ ,,,,,,,}\ {|| \ \ ,,,,,,,,,,}\ { - \rm C - C{H_3}} \end{array}\) and \(\begin{array}{{20}{c}} {\rm OH,,,,,, \ \ \ ,,}\ {|,,,,,,,,, \ \ \ \ ,,,,}\ { - \rm{CH - CH}_3} \end{array}\)will give the haloform test.
- Isopropyl alcohol, (CH3)2CHOH is two-degree alcohol and thus will give positive iodoform because of oxidation to a keto methyl group.
- Ethanal, CH3CH=O will respond positively to the Haloform test.
- Benzyl alcohol PhCH2OH has no keto methyl group and being a 1-degree alcohol, will not respond to Iodoform test.
- Hence, negative iodoform test is given by Benzyl alcohol.