A) Benzyl chloride
B) Ethyl chloride
C) Chlorobenzene
D) Isopropyl chloride
Correct Answer: A
Solution :
Aliphatic \[{{S}_{N}}1\] reaction is carried out in two steps. In form of slow step Step (i) carbonium (carbocation) ion is formed and its formation is based upon the stability Stability order of carbocation \[{{C}_{6}}\overset{+}{\mathop{C}}\,{{H}_{2}}>C{{H}_{3}}-\overset{+}{\mathop{C}}\,H-C{{H}_{3}}>C{{H}_{3}}-\overset{+}{\mathop{C}}\,{{H}_{2}}\] and in step (ii) nucleophile is attracted towards the carbonium ion in form of fast step to give final product. Hence, in benzyl chloride, ethyl chloride and isopropyl chloride order of \[{{S}_{N}}1\] reaction is benzyl chloride > isopropyl chloride > ethyl chloride In chlorobenzene, mechanism of \[{{S}_{N}}1\] reaction is differ to aliphatic alkyl halide. The aryl halides are much less reactive as compared to alkyl halides, towards nucleophilic reagents in either \[{{S}_{N}}1\] or \[{{S}_{N}}2\] reaction. The carbon-halogen bond in the aryl halide is quite strong and only forcing conditions can break up this bond. Hence, Step (i) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}-Cl\xrightarrow{Slow}{{C}_{6}}{{H}_{5}}\overset{+}{\mathop{C}}\,{{H}_{2}}+C{{l}^{-}}\] Rate \[\propto \,\,[{{C}_{6}}{{H}_{5}}C{{H}_{2}}Cl]\] Step (ii) \[{{C}_{6}}{{H}_{5}}-\overset{+}{\mathop{C}}\,{{H}_{2}}+N{{u}^{-}}\xrightarrow{Fast}{{C}_{6}}{{H}_{5}}C{{H}_{2}}Nu\]
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