Solveeit Logo
Login

Question

Question: Here C is: \( C{H_3}COOK\xrightarrow{{electrolysis}}A\xrightarrow[\Delta ]{{{{1000}^0}C}}\,B\xrig...

Here C is:
CH3COOK→electrolysisA→Δ10000C B→alkalineKMnO4 CC{H_3}COOK\xrightarrow{{electrolysis}}A\xrightarrow[\Delta ]{{{{1000}^0}C}}\,B\xrightarrow[{alkaline}]{{KMn{O_4}\,}}C
(A) CH3COOHC{H_3}COOH
(B) CH2OH−CH2OHC{H_2}OH - C{H_2}OH
(C) CHO−CHOCHO - CHO
(D) COOH−COOHCOOH - COOH

Explanation

Solution

Hint : Electrolysis of carboxylic salts yields an alkane along with the liberation of carbon dioxide at the anode. This alkane on further decomposition at higher temperature yields an alkene and upon further oxidation of alkene, it will get converted into an alcohol.

Complete step by step answer
Decarboxylation is a reaction in which a carboxyl group is eliminated from the chain along with the liberation of carbon dioxide gas. It is mostly done in the reaction of carboxylic acids. It is a reversible process. You can produce alkane from the carboxylic acid by removing the CO2C{O_2} . The Kolbe decarboxylation reaction involves the conversion of a carboxylic salt to an alkane with the liberation of CO2C{O_2} at the anode. The salt could be a sodium salt or a potassium salt. In the given reaction, the salt is of potassium so when we treat it with H2O{H_2}O . It will produce alkane with CO2C{O_2} and NaOHNaOH and H2{H_2} . The reaction will be:
CH3COOK→H2Oelectrolysis CH3−CH3+2KOH+H2+CO2C{H_3}COOK\xrightarrow[{{H_2}O}]{{electrolysis}}\,C{H_3} - C{H_3} + 2KOH + {H_2} + C{O_2}
Hence AA is Ethane. This ethane formed upon the catalytic cracking will get converted into an alkene at a high temperature. The reaction will be:
CH3−CH3→Δ10000CCH2−CH2+H2C{H_3} - C{H_3}\xrightarrow[\Delta ]{{{{1000}^0}C}}C{H_2} - C{H_2} + {H_2}
Hence BB is ethane. This ethene produced in reaction with cold alkaline KMnO4KMn{O_4} will undergo oxidation in which the double bonds between the carbon atoms will break and be replaced by an oxygen atom. In this way, alkene will get converted into ethylene glycol. This reaction occurs via the formation of an intermediate. The reaction will be;
CH2−CH2→KMnO4AlkalineCH2OH−CH2OHC{H_2} - C{H_2}\xrightarrow[{KMn{O_4}}]{{Alkaline}}C{H_2}OH - C{H_2}OH
Hence CC is CH2OH−CH2OHC{H_2}OH - C{H_2}OH . Therefore option (B) is correct.

Note
Upon decarboxylation reaction, the carboxylic salt gets converted into alkane and when this alkane is treated at high temperature in the range of 750−10000C750 - {1000^0}C , it gets converted into alkene and when you treat an alkene with alkaline potassium permanganate it will yield a diol.