A) I, II, VI, VII. VIII
B) I, II, III, IV, V, VI
C) III, IV, V, VI, IX
D) All of these
Correct Answer: B
Solution :
Halides or tosylates which are unreactive in \[{{S}_{N}}2\] cannot be used. I to VI cannot be prepared but VII to IX can be prepared. I. It cannot be prepared since \[M{{e}_{3}}C-X\] is a \[3{}^\circ \] halide. II. It cannot be prepared since \[\,M{{e}_{3}}C-C{{H}_{2}}-X\] is neopentyl halide and does not undergo \[{{S}_{N}}2\] reaction due to steric hindrance. III. It cannot be prepared since \[2{}^\circ \] amine gives poor yield due to elimination. IV. It cannot be prepared since arylhatide is not activated by electron donating group (Me group). \[Ar{{S}_{N}}\] reaction takes place only when the benzene ring is activated by electron withdrawing group present at o- and p-positions. V. It cannot be prepared, since electron withdrawing group (\[-N{{O}_{2}}\] group) is present m-position. \[Ar{{S}_{N}}\] reaction does not take place. VI. It cannot be prepared, since vinyl halides \[(C{{H}_{2}} = CM -X)\] are unreactive in \[{{S}_{N}}2\] reaction. VII. It can be prepared, since ?benzene ring is activated by electron withdrawing group \[\left( -N{{O}_{2}} \right)\] at p-position and hence \[Ar{{S}_{N}}\] reaction takes place. VIII. It can be prepared since benzene ring, is activated by electron withdrawing group\[\left( -N{{O}_{2}} \right)\] at o-position and hence \[Ar{{S}_{N}}\] reaction takes place. IX. It can be prepared since allyl halides \[\left( C{{H}_{2}} = CHC{{H}_{2}}X \right)\] can undergo \[{{S}_{N}}2\] reaction.You need to login to perform this action.
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