
Find A ${C_6}{H_6} + A$ $\xrightarrow {AlC{l_3}} {C_6}{H_5}CON{H_2}$
A. $N{H_2}COHN{H_2}$
B. $ClCON{H_2}$
C. $C{H_3}CON{H_2}$
D. $C{H_2}\left( {Cl} \right)CON{H_2}$
Answer
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Hint: this reaction is an electrophilic reaction of benzene. This reaction takes place in the presence of Friedel craft reagent that is aluminium chloride. This reaction leads to the formation of benzamide.
Complete step by step answer:
This reaction is an example of Friedel craft acylation. This reaction involves the electrophilic substitution of the hydrogen of an arene with another electrophile. The electrophile is formed by the aluminium chloride. This reaction usually leads to only monoacetylated compounds.
This can be demonstrated using the option b that is carbamic chloride. Since the product formed contains one more carbon than benzene. Therefore, it cannot react with the compounds of the other options since the other options contain more than one carbon atom and thus reaction with them will not lead to formation of benzamide.
The reaction is following the steps given below. First is the formation of electrophile. This is shown below:
$ClCON{H_2} + AlC{l_3} \to CON{H_2}^ + + AlC{l_4}^ - $
Therefore, the electrophile will now react with benzene and undergoes the following reaction to form benzamide.
${C_6}{H_6} + CON{H_2}^ + \to {C_6}{H_5}CON{H_2} + {H^ + }$
The hydrogen ion now reacts with the nucleophile formed in the reaction to form the following products.
$AlC{l_4}^ - + {H^ + } \to AlC{l_3} + HCl$
Therefore, the complete reaction can be demonstrated below:
${C_6}{H_6} + CON{H_2}^ + + AlC{l_4}^ - \to {C_6}{H_5}CON{H_2} + HCl + AlC{l_3}$
Therefore, the compound that should be in the place of an unknown variable $A$ should be carbamic chloride given by the molecular formula $ClCON{H_2}$ .
So, the correct answer is Option B .
Note: Through electrophilic substitution reaction known as Friedel craft acylation, benzene can be converted into benzamide.
This reaction requires the replacing group to be an electrophile. Therefore the chlorine from the carbamic chloride is transferred to aluminium chloride.
Carbamic chloride is used because it has only one carbon atom and therefore, when it reacts with benzene it will form benzamide.
Complete step by step answer:
This reaction is an example of Friedel craft acylation. This reaction involves the electrophilic substitution of the hydrogen of an arene with another electrophile. The electrophile is formed by the aluminium chloride. This reaction usually leads to only monoacetylated compounds.
This can be demonstrated using the option b that is carbamic chloride. Since the product formed contains one more carbon than benzene. Therefore, it cannot react with the compounds of the other options since the other options contain more than one carbon atom and thus reaction with them will not lead to formation of benzamide.
The reaction is following the steps given below. First is the formation of electrophile. This is shown below:
$ClCON{H_2} + AlC{l_3} \to CON{H_2}^ + + AlC{l_4}^ - $
Therefore, the electrophile will now react with benzene and undergoes the following reaction to form benzamide.
${C_6}{H_6} + CON{H_2}^ + \to {C_6}{H_5}CON{H_2} + {H^ + }$
The hydrogen ion now reacts with the nucleophile formed in the reaction to form the following products.
$AlC{l_4}^ - + {H^ + } \to AlC{l_3} + HCl$
Therefore, the complete reaction can be demonstrated below:
${C_6}{H_6} + CON{H_2}^ + + AlC{l_4}^ - \to {C_6}{H_5}CON{H_2} + HCl + AlC{l_3}$
Therefore, the compound that should be in the place of an unknown variable $A$ should be carbamic chloride given by the molecular formula $ClCON{H_2}$ .
So, the correct answer is Option B .
Note: Through electrophilic substitution reaction known as Friedel craft acylation, benzene can be converted into benzamide.
This reaction requires the replacing group to be an electrophile. Therefore the chlorine from the carbamic chloride is transferred to aluminium chloride.
Carbamic chloride is used because it has only one carbon atom and therefore, when it reacts with benzene it will form benzamide.
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