Organic Compounds Containing Halogens Question 189
Question: In the following sequence of reactions $ CH_3CH_2CH_2Br\xrightarrow{KOH(alc)}(A)\xrightarrow{HBr}(B)\xrightarrow{KOH(aq.)}(C), $ The product is
[JIPMER 2001]
Options:
A) Propan - 2 - ol
B) Propan - l - ol
C) Propyne
D) Propene
Show Answer
Answer:
Correct Answer: A
Solution:
$ CH_3-CH_2-CH_2-Br\xrightarrow{alc\text{.}KOH}\underset{\text{(A)}}{\mathop{CH_3-CH=CH_2}} $
$ \underset{\text{Propan-2-ol}}{\mathop{\begin{matrix} CH_3-CH-CH_3\xleftarrow{aq.KOH} \\ \overset{|}{\mathop{OH}} \\ \end{matrix} }}\begin{matrix} CH_3-CH-CH_3 \\ \overset{|}{\mathop{Br}} \\ \end{matrix} $
BETA
