Answer
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Hint: We will draw the diagram of a monhalo compound, based on their placement in phenyl. For the second part we will use the mechanism of \[{S_N}2\] reaction and electron affinity to find the faster reacting compound.
Complete step by step answer:
a. A compound is classified as a monohalo compound when the hydrogen atom on carbon atom is replaced by the halogen. Depending upon how the carbon hybridized the compound will form a monohalo structure. Similarly in the given reaction we can see that hydrogen atom is replaced by chlorine from $PC{l_5}$ and resulted in hydrochloric acid production as well as the oxygen atom is replaced which resulted in $POC{l_3}$.
In this reaction the double bond is broken, and halogen is added on carbon.
b. \[{S_N}2\] reaction is a nucleophilic substitution reaction in which the reactions are bimolecular with simultaneous bond formation and breakage. It doesn’t require an intermediate to facilitate the reaction. The final product produced after \[{S_N}2\] reactions are stereochemistry inversion at their centre and have steric effect.
i. Between \[C{H_3}Br\] or \[C{H_3}I\], \[C{H_3}I\] will react faster to the \[{S_N}2\] reaction because Iodine present in the compound will leave easily due to large atomic size compared to atomic size of bromine.
ii. Between \[{(C{H_3})_3}C - Cl\] or \[C{H_3} - Cl\], \[C{H_3} - Cl\] will react faster to the \[{S_N}2\] reaction because \[C{H_3} - Cl\] is a primary halide and \[{(C{H_3})_3}C - Cl\] is tertiary halide and we know that primary halide reacts faster to \[{S_N}2\] reaction.
Note:
Student will make mistakes while making the monohalo diagrams of the compounds and they will also make mistakes in selecting the properties for elements which react faster in the \[{S_N}2\] reaction.
Complete step by step answer:
a. A compound is classified as a monohalo compound when the hydrogen atom on carbon atom is replaced by the halogen. Depending upon how the carbon hybridized the compound will form a monohalo structure. Similarly in the given reaction we can see that hydrogen atom is replaced by chlorine from $PC{l_5}$ and resulted in hydrochloric acid production as well as the oxygen atom is replaced which resulted in $POC{l_3}$.
In this reaction the double bond is broken, and halogen is added on carbon.
b. \[{S_N}2\] reaction is a nucleophilic substitution reaction in which the reactions are bimolecular with simultaneous bond formation and breakage. It doesn’t require an intermediate to facilitate the reaction. The final product produced after \[{S_N}2\] reactions are stereochemistry inversion at their centre and have steric effect.
i. Between \[C{H_3}Br\] or \[C{H_3}I\], \[C{H_3}I\] will react faster to the \[{S_N}2\] reaction because Iodine present in the compound will leave easily due to large atomic size compared to atomic size of bromine.
ii. Between \[{(C{H_3})_3}C - Cl\] or \[C{H_3} - Cl\], \[C{H_3} - Cl\] will react faster to the \[{S_N}2\] reaction because \[C{H_3} - Cl\] is a primary halide and \[{(C{H_3})_3}C - Cl\] is tertiary halide and we know that primary halide reacts faster to \[{S_N}2\] reaction.
Note:
Student will make mistakes while making the monohalo diagrams of the compounds and they will also make mistakes in selecting the properties for elements which react faster in the \[{S_N}2\] reaction.
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