Aromatic nitriles \[(ArCN)\] are not prepared by reaction
(A) $ArX + KCN$
(B) $Ar{N_2}Cl + CuCN$
(C) $ArCON{H_2} + {P_2}{O_5}$
(D) $ArCON{H_2} + SOC{l_2}$
Answer
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Hint: In the lab, aromatic nitriles are frequently made from aniline using diazonium compounds. The Sandmeyer response is as follows. Transition metal cyanides are necessary. To know whether aromatic nitrile will be formed by a reaction or not depends on the reagents and the bond that the element forms with the aromatic ring.
Complete Step by Step Solution:
\[Ar{N_2}Cl\] is an aryl i.e., it has a phenyl group and has \[{N_2}Cl\] on it so it is a diazonium chloride. Now, \[{N_2}^ + \] is a good leaving group so when it is reacted with\[CuCN\] , \[CN\] attacks \[{N_2}\] bond of the diazonium chloride and we get \[ArCN\] . Therefore we obtain an aromatic nitrile on reaction between \[Ar{N_2}Cl\] with \[CuCN\] .
Now the reactions in options C. and D. are similar with the only difference that their reagents are not the same.
\[ArCON{H_2}\] is an amide group compound. When it is reacted with \[{P_2}{O_5}\] or\[SOC{l_2}\] , dehydration occurs i.e., \[{H_2}O\] leave. Now both carbon and nitrogen form double bonds with oxygen and hydrogen respectively. As \[{H_2}O\] leaves carbon and nitrogen both lose their double bond. As a result they make a double bond with each other in addition to an already existing bond and thus form \[ArCN\] i.e., an aromatic nitrile.
Now when \[ArX\] (an aromatic halide) is made to react with \[KCN\] , \[CN\] cannot attack the bond of X in the aromatic halide because of the partial double bond character of this halide group with this aromatic ring. Hence no nucleophilic substitution takes place. And hence we do not get an aromatic nitrile.
Hence, option A. is the answer.
Note: Due to the halogen's -I and +M effects, which cause its electron to delocalize in the benzene ring through resonance and bind to it, it cannot be substituted by \[CN\] and exhibits inertness toward \[KCN\] whereas the alternative produces aromatic nitrile.
Complete Step by Step Solution:
\[Ar{N_2}Cl\] is an aryl i.e., it has a phenyl group and has \[{N_2}Cl\] on it so it is a diazonium chloride. Now, \[{N_2}^ + \] is a good leaving group so when it is reacted with\[CuCN\] , \[CN\] attacks \[{N_2}\] bond of the diazonium chloride and we get \[ArCN\] . Therefore we obtain an aromatic nitrile on reaction between \[Ar{N_2}Cl\] with \[CuCN\] .
Now the reactions in options C. and D. are similar with the only difference that their reagents are not the same.
\[ArCON{H_2}\] is an amide group compound. When it is reacted with \[{P_2}{O_5}\] or\[SOC{l_2}\] , dehydration occurs i.e., \[{H_2}O\] leave. Now both carbon and nitrogen form double bonds with oxygen and hydrogen respectively. As \[{H_2}O\] leaves carbon and nitrogen both lose their double bond. As a result they make a double bond with each other in addition to an already existing bond and thus form \[ArCN\] i.e., an aromatic nitrile.
Now when \[ArX\] (an aromatic halide) is made to react with \[KCN\] , \[CN\] cannot attack the bond of X in the aromatic halide because of the partial double bond character of this halide group with this aromatic ring. Hence no nucleophilic substitution takes place. And hence we do not get an aromatic nitrile.
Hence, option A. is the answer.
Note: Due to the halogen's -I and +M effects, which cause its electron to delocalize in the benzene ring through resonance and bind to it, it cannot be substituted by \[CN\] and exhibits inertness toward \[KCN\] whereas the alternative produces aromatic nitrile.
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