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In the given reaction, product A is:
CHCl3excessKOH(aq)A
A.Formic acid
B.Potassium formate
C.Acetic acid
D.Potassium acetate

Answer
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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 CHCl3. 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 KOHvia a nucleophilic reaction (SN2). In theory, it forms CHCl2(OH)CHCl(OH)2 and CH(OH)3in that order, while removing KClat each step. CHCl(OH)2 will spontaneously release one molecule of H2Oand produce HC(=O)Cl or formyl chloride, which will then be hydrolyzed to produce HCOOKor Potassium Formate because Cl is an outstanding leaving group.
We get CH(OH)3from aq. KOH, which is less stable and loses H2Oto give HCOOH, and then reacts with another mole of aq. KOHto give potassium formate.
Therefore, the correct option is option (B).
Sodium formate and sodium chloride can be generated by hydrolyzing chloroform with NaOHand 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 KOHwith 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 KCland two moles of H2O. 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.
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