Chlorobenzene reacts with trichloroacetaldehy | vedclass

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(C) In the presence of concentrated $H_2SO_4$,two moles of chlorobenzene react with one mole of chloral $(CCl_3CHO)$ through a condensation reaction.

Vedclass · Accepted Answer

$(Cl-C_6H_4)_2CH-CCl_3$ $(DDT)$

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(C) In the presence of concentrated $H_2SO_4$,two moles of chlorobenzene react with one mole of chloral $(CCl_3CHO)$ through a condensation reaction. The oxygen atom from the aldehyde group of chloral and the hydrogen atoms from the para-positions of the two chlorobenzene rings are removed as a water molecule $(H_2O)$. This results in the formation of $p,p'$-dichlorodiphenyltrichloroethane,commonly known as $DDT$. The reaction is: $2 C_6H_5Cl + CCl_3CHO \xrightarrow{H_2SO_4} (Cl-C_6H_4)_2CH-CCl_3 + H_2O$

Chlorobenzene reacts with trichloroacetaldehyde (chloral) in the presence of $H_2SO_4$.
$2 C_6H_5Cl + CCl_3CHO \xrightarrow{H_2SO_4} \text{Major Product}$
The major product formed is:

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Chlorobenzene reacts with trichloroacetaldehyde (chloral) in the presence of $H_2SO_4$. $2 C_6H_5Cl + CCl_3CHO \xrightarrow{H_2SO_4} \text{Major Product}$ The major product formed is: