Haloform reaction is not given by

C H_{3} C O C H_{3}

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C H_{3} C O C_{2} H_{5}

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C_{6} H_{5} C O C_{2} H_{5}

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C H_{3} C H (O H) C H_{3}

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Solution

The correct option is C $C_{6} H_{5} C O C_{2} H_{5}$
The haloform is a chemical reaction, where a haloform $C H X_{3}$ (X is a halogen) is produced by the exhaustive halogenation of a methyl ketone (a molecule containing the $R C O C H_{3}$ group) in the presence of a base. R may be alkyl or aryl. The reaction can be used to produce chloroform $(C H C l_{3}),$ bromoform $(C H B r_{3}),$ or iodoform $(C H I_{3}) .$

Substrates that successfully undergo the halform reaction are methyl ketones and secondary alcohols oxidizable to methyl ketones, such as isopropanol. The only primary alcohol and aldehyde to undergo this reaction are ethanol and ethanal, respectively.

For haloform reaction, 3 hydrogens on alpha carbon are required.
Therefore, the compound $C_{6} H_{5} C O C_{2} H_{5}$ can not undergo haloform reaction. Hence, the correct answer is option (c).

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