Discussion Forum

Identify Z in the following reaction  sequence

Answer:

Explanation:

$CH_{3}CH_{2}CH_{2}OH$    $\xrightarrow[{160^{0}-180^{0}C}]{conc.H_{2}SO_{4}}$       $CH_{3}CH_{}=CH_{2}$  $\underrightarrow{Br_{2}} \underrightarrow{alc.KOH}$         $CH_{3}CH_{}Br-CH_{2}Br$ 

                         (Y)

$[CH_{3}C(Br)=CH_{2}+CH_{3}CH=CHBr]$   $\xrightarrow[{-HBr}]{NaNH_{2}}$      $CH_{3}C\equiv CH$

    (A)    (B)                                                                                        (Z)

 Alcoholic KOH brings about dehydrobromination of Y and give a mixture of vinyl bromide (A and B) while 
NaNH₂ being a strong base than alc . KOH readily  brings about dehydrobromination of less reactive vinyl bromide to give propyne  $CH_{3}C\equiv CH$  i.e, (Z)