A) X = dilute aqueous \[NaOH\] \[Y=HBr\]/acetic acid
B) X = concentrated alcoholic\[NaOH\]\[Y=HBr\]/acetic acid
C) X = dilute aqueous \[NaOH\] \[Y=B{{r}_{2}}/CHC{{l}_{3}}\]
D) X = concentrated aqueous \[NaOH\]\[Y=B{{r}_{2}}/CHC{{l}_{3}}\]
Correct Answer: B
Solution :
| \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Br\] |
| \[\xrightarrow[80{}^\circ C]{Alc\cdot NaOH(X)}C{{H}_{3}}-CH=C{{H}_{2}}\] |
| \[\xrightarrow{HBr(Y)}C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}\] |
| Step 1 Haloalkanes on reaction with alcoholic \[NaOH\]undergoes, elimination reaction to give alkenes. |
| Step 2 The formed alkene then reacts with \[HBr\]to give monobrominated product. This product is in accordance markonwnikoff addition. |
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