A) \[{{C}_{6}}{{H}_{5}}Br+C{{H}_{2}}=CH-C{{H}_{2}}ONa\]
B) \[C{{H}_{2}}=CHC{{H}_{2}}Br+{{C}_{6}}{{H}_{5}}ONa\]
C) \[{{C}_{6}}{{H}_{5}}CH=CHBr+C{{H}_{3}}ONa\]
D) \[C{{H}_{2}}=CHBr+{{C}_{6}}{{H}_{5}}C{{H}_{2}}ONa\]
Correct Answer: B
Solution :
In 'a' bromobenzene (aryl halide), in 'c' and 'd' vinyl halide are less reactive towards \[{{S}_{N}}2\] reaction. 'b' is the correct option. \[C{{H}_{2}}=CH-C{{H}_{2}}-Br+{{C}_{6}}{{H}_{5}}ONa\]\[\xrightarrow[-NaBr]{}C{{H}_{2}}=CH-C{{H}_{2}}-O-{{C}_{6}}{{H}_{5}}\]
You need to login to perform this action.
You will be redirected in
3 sec
