Question

Haloforms are trihalogen derivatives of
A. Ethane
B. Methane
C. Propane
D. Benzene

Answer 1

Hint: Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom shows positive haloform test. The methyl group is converted to haloform.

Complete step-by-step answer:In order to know that the halogen reaction, we can see that polyhalogenation, which is followed by the cleavage of the methyl group, takes place when the bromine halogen is applied to the methyl ketone in an aqueous sodium hydroxide solution.
The carboxylate and the necessary haloform, tribromomethane, are the reaction's end products. Chloroform, bromoform, and iodoform were previously produced using this procedure in an industrial setting.
As it is well known, a chemical is known as a trihalogen derivative when three hydrogen atoms are substituted by three halogen atoms.
Haloform compounds with the formula\[CH{X_3}\] ​in which \[X\] is a halogen atom, it is the trihalogen derivatives of methane.
The equation of the trihalogen derivate of methane is:
\[C{H_4}\xrightarrow[{ + 3X}]{{ - 3H}}CH{X_3}\,\,\,\,\,\,\,(X = Cl,Br,I)\]

Option ‘B’ is correct

Note:It should be noted that only ethanol is a main alcohol capable of producing the triiodomethane reaction. A surplus of methyl ketone is halogenated throughout the three-step haloform reaction process, which also yields carboxylate ions and the necessary haloform precipitate. Secondary alcohols that can be converted into methyl ketones, acetaldehyde, and ethanol are utilised as substrates in this process.