The final product formed by distilling ethyl alcohol with excess of \[C{l_2}\] and \[Ca{(OH)_2}\] is
A. \[C{H_3}CHO\]
B. \[CC{l_3}CHO\]
C. \[CHC{l_3}\]
D. \[{(C{H_3})_2}O\]

Hint: We know that ethyl alcohol is an organic substance, and its chemical formula is \[{C_2}{H_5}OH\]. Ethyl alcohol can be oxidized to give carbonyl compounds. Calcium hydroxide and chlorine can be used in haloform tests.

Complete step-by-step answer:Acetaldehyde is created when ethyl alcohol gets oxidized in the presence of \[C{l_2}\] and \[Ca{(OH)_2}\]. The reaction is shown below.
\[C{l_2}+C{H_3} - C{H_2} - OH \xrightarrow {Ca{(OH)_2}} C{H_3}CHO +2HCl\]

Acetaldehyde has methyl carbonyl group and can give positive haloform test in the presence of halogen and base such as sodium hydroxide and calcium hydroxide. Chloroform is the end product of the reaction between chlorine and ethyl alcohol when \[Ca{(OH)_2}\] is present. The reaction is shown below.
$ 2C{H_3}CHO+3C{l_2} \xrightarrow {Ca{(OH)_2}} 2CHC{l_3} + Ca{(HCOO)_2}$

Therefore, \[CHC{l_3}\] is the end result of distilling excess \[C{l_2}\] and \[Ca{(OH)_2}\] into ethyl alcohol.

Option ‘C’ is correct

Note: Haloform test is used to detect methyl carbonyl compounds. In this test, halogen molecules and a base such as sodium hydroxide, calcium hydroxide are added to the unknown solution. If the unknown solution contains any organic compound that either have methyl carbonyl group or can be converted into methyl carbonyl compounds give positive result for the test.