A) Propane
B) Propene
C) Propan-1-o1
D) PropyI amine
E) (e) None of these above
Correct Answer: E
Solution :
(e) Propane can be made from 1-bromopropane in a single step by reduction. \[LiAl{{H}_{4}}+4C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow{{}}\]\[4C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}+AlB{{r}_{3}}+LiBr\] \[LiAl{{H}_{4}}\] is a hydride donoar (\[{{H}^{-}}\]donor). Note \[{{3}^{o}}\] alkyl halides give alkenes. Because they undergo dehydrohalogenation. \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow[\Delta ]{Alcoholic\text{ }KOH}\]\[C{{H}_{3}}CH=C{{H}_{2}}+KCl+{{H}_{2}}O\] (elimination) \[\overset{propene}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}}}\,Br\xrightarrow[37.3\,K]{aq\,.\,\,KOH}\]\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH+KBr\] (substitution) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow[in\,\,presence\text{ }of\,{{C}_{2}}{{H}_{5}}OH]{H-N{{H}_{2}},\,\,\Delta }\]\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}N{{H}_{2}}\] (substitution)
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