Question

In the given reaction, product A is:
$CHC{l_3}\xrightarrow[{excess}]{{^{KOH(aq)}}}A$
A.Formic acid
B.Potassium formate
C.Acetic acid
D.Potassium acetate

Answer 1

Hint: We have to remember that the polyhalogen compounds are carbon compounds with more than one halogen atom (elements of the modern periodic table's group $17$). Trichloromethane, also known as chloroform, is an organic compound with the formula $CHC{l_3}$. It is a trihalomethane and one of the four chloromethanes. Potassium hydroxide, also known as caustic potash, is an inorganic compound with the formula $KOH$. $KOH$ is a prototypical solid base, along with sodium hydroxide.

Complete answer:
We have to know that as chloroform reacts with aqueous $KOH$, the chlorines on the carbon atom are substituted one by one by $ - OH$ groups from $KOH$via a nucleophilic reaction (SN2). In theory, it forms \[CHC{l_2}\left( {OH} \right)\]\[CHCl{\left( {OH} \right)_2}\] and \[CH{\left( {OH} \right)_3}\]in that order, while removing $KCl$at each step. \[CHCl{\left( {OH} \right)_2}\] will spontaneously release one molecule of \[{H_2}O\]and produce \[HC\left( { = O} \right)Cl\] or formyl chloride, which will then be hydrolyzed to produce \[HCOOK\]or Potassium Formate because \[C{l^ - }\] is an outstanding leaving group.
We get \[CH{\left( {OH} \right)_3}\]from aq. $KOH$, which is less stable and loses \[{H_2}O\]to give \[HCOOH\], and then reacts with another mole of aq. $KOH$to give potassium formate.
Therefore, the correct option is option (B).
Sodium formate and sodium chloride can be generated by hydrolyzing chloroform with \[NaOH\]and tetrabutylammonium bromide. This formate can be protonated in the same way as formic acid.
Hence, option (A) is incorrect.
Acetic acid cannot be produced from chloroform but rather produced by the oxidation of the intermediates formed during the production of chloroform.
Hence, Option (C) is incorrect.
Potassium acetate is formed by the reaction of $KOH$with acetic acid and not chloroform.
Hence, option (D) is incorrect.

Note:
It must be noted that one mole of Chloroform will react with four moles of $KOH$ to produce one mole of \[HCOOK\] (potassium formate) along with three moles of \[KCl\]and two moles of \[{H_2}O\]. This reaction is only possible when excess of $KOH$ is present and the reaction is allowed to reach its completion. Also, this reaction is only for aqueous $KOH$; alcoholic $KOH$ gives a different result.