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प्रश्न
Explain haloform reaction with suitable example.
Write a note on the Haloform reaction.
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उत्तर
- This reaction is given by acetaldehyde, all methyl ketones (CH3–CO–R), and all alcohols containing CH3(CHOH)– group.
- When an alcohol or methyl ketone is warmed with sodium hydroxide and iodine, a yellow precipitate is formed. Here the reagent sodium hypoiodite is produced in situ.
- During the reaction, the sodium salt of the carboxylic acid is formed which contains one carbon atom less than the substrate.
- The methyl group is converted into haloform (CHX3).
e.g. Acetone is oxidized by sodium hypoiodite to give sodium salt of acetic acid and yellow precipitate of iodoform.
\[\begin{array}{cc}
\phantom{..}\ce{O}\phantom{.......................................}\ce{O}\phantom{..........................}\\
\phantom{..}||\phantom{.......................................}||\phantom{..........................}\\
\ce{\underset{\text{Acetone}}{CH3 - C - CH3} + \underset{\text{Sodium hypoiodite}}{3NaOI} ->[NaOH/I2][\Delta] \underset{\text{Sodium acetate}}{CH3 - C - O- Na+} + \underset{\text{Iodoform}}{CHI3 ↓} + 2NaOH}
\end{array}\]
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\[\begin{array}{cc} \phantom{...}\ce{R}\phantom{....}\ce{OR'}\phantom{}\\ \phantom{}\backslash\phantom{..}/\\ \ce{C}\\ \phantom{}/\phantom{..}\backslash\\ \phantom{...}\ce{H}\phantom{....}\ce{OR'}\phantom{} \end{array}\]
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\[\begin{array}{cc}
\ce{CH3 - CH2 - CH2 - C - CH3 ->[HO - CH2 - CH2 - CH2 - OH][dry HCl] ?}\\
||\phantom{.........}\\
\ce{O}\phantom{.........}
\end{array}\]
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\[\begin{array}{cc}
\ce{O}\phantom{........}\\
||\phantom{........}\\
\ce{\underset{(Acetone)}{CH3 - C - CH3}->[CrO3] B}
\end{array}\]
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