Question

Haloform reaction is not given by
a.) $\text{C}{{\text{H}}_{\text{3}}}\text{COC}{{\text{H}}_{\text{3}}}$
b.) $\text{C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}$
c.) ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{CO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}$
d.) $\text{C}{{\text{H}}_{\text{3}}}\text{CH(OH)C}{{\text{H}}_{\text{3}}}$

Answer 1

Hint: To know which among the mentioned compounds will perform the haloform reactions, we need to have an idea about the definition of haloform reaction. Once we see the haloform reaction's aim and meaning, the criteria for a compound to perform the haloform reaction will be known to us.

Complete step-by-step answer:
The haloform reaction is the reaction of a methyl ketone with chlorine, bromine, or iodine in the presence of Hydroxide ions to give a carboxylate ion and a haloform. There is one aldehyde that undergoes the haloform reaction, which is acetaldehyde.
Substrates that successfully undergo the haloform reaction are methyl ketones and secondary alcohols oxidizable to methyl ketone, such as isopropanol. The only primary alcohol and aldehyde to undergo this reaction are ethanol and ethanal, respectively.
To perform the haloform reaction, there should be three hydrogens on the alpha carbon.
The first option is $\text{C}{{\text{H}}_{\text{3}}}\text{COC}{{\text{H}}_{\text{3}}}$.

The name of the compound is Acetone. There are 6 alpha Hydrogen atoms in Acetone. So it can perform the Haloform reaction.

Now moving on to the second option is that is mentioned in the question is $\text{C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}$
The name of the compound is Methyl ethyl ketone. This compound does not have 3 alpha Hydrogen atoms and does not fulfil the criteria for the haloform reaction. So, it cannot undergo the haloform reaction. Moreover, it is a methyl ketone, which makes it very suitable for the haloform reaction.
The third compound that is mentioned among the options is ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{CO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}$.
The name of this compound is Propiophenone. From the formula of the compound it is clear that it will form an electrophile rather than being a nucleophile. So, this compound cannot take part in the haloform reaction.
The last compound that is mentioned as option is $\text{C}{{\text{H}}_{\text{3}}}\text{CH(OH)C}{{\text{H}}_{\text{3}}}$.
The name of the compound is Isopropyl alcohol.
When isopropyl alcohol reacts with KOBr, it oxidizes to methyl ketone. Thus, we can say it performs haloform.
Therefore, the correct answer is Option C.

Note: To distinguish between the compounds which can form a part of the haloform reaction and which cannot we need to know the following. In alcohols beta carbon must be a methyl group. In secondary alcohols only 2-Alkanol give haloform test. 2-Propanol, 2-Butanol, 2-Hexanol all form haloform reactions, but 3-Hexanol cannot.