Question

What is the product formed when acetylene reacts with hypochlorous acid :
(A) $C{H_3}COCl$
(B) $ClC{H_2}CHO$
(C) $C{l_2}CHCHO$
(D) $ClC{H_2}COOH$

Answer 1

In order to solve this question we need to see the complete analysis of the reaction when acetylene reacts with what happens when acetylene reacts with hypochlorous acid. To form $HCC{l_2} - C{(OH)_2}H$ after this the dehydration of water results in the final product.

Complete step-by-step answer: For solving this question we need to have the knowledge of the reactants so first is acetylene whose formula is $HC \equiv CH$ and the second one is hypochlorous acid whose formula is $HOCl$ these two reacts with each other to form an intermediate whose formula is $HCC{l_2} - C{(OH)_2}H$ .
Now we will see it according to equation:
$HC \equiv CH + 2HOCl \to HCC{l_2} - C{(OH)_2}H$
This is the complete reaction in which a molecule of acetylene (IUPAC name ethyne) reacts with 2 mole of hypochlorous acid to form the intermediate.Now after the dehydration of this intermediate the final product is formed. So we will be seeing it by the reaction:
$HCC{l_2} - C{(OH)_2}H\xrightarrow{{ - {H_2}O}}HCC{l_2} - CHO$
So now we will be analyzing this final reaction and know what is the final product.So when this intermediate is dehydrated then our final product is gained which is di-chloro acetaldehyde.

Hence our final product is gained and the third option which is (C) is correct.

Note: In between the reaction when there is intermediate the reversible product mono acetaldehyde is also formed whose formula is $OHC - C{H_2}Cl$ .When the hypochlorous acid is attacking in the acetylene the –OH group attacks one carbon and joins in it and the Cl group is joined to another carbon and joins in that.So one bond is broken and the triple bond between two C is converted into a double bond.