A) propanal
B) propane
C) propyne
D) propene
Correct Answer: D
Solution :
\[ROH\xrightarrow[\begin{smallmatrix} P{{X}_{3}}(X=\,Cl,Br\,I)or \\ ZnC{{l}_{2}}/HCl \end{smallmatrix}]{PC{{l}_{5}}\,or\,SOC{{l}_{2}}\,or}RCl\] \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\xrightarrow[-HCl]{PC{{l}_{5}}}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl\] \[\xrightarrow[(dehydrohalogenation)]{alc.\,KOH}\underset{Alkene}{\mathop{C{{H}_{3}}CH=C{{H}_{2}}}}\,\] B is an alkene (propene)You need to login to perform this action.
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