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प्रश्न
HCl is added to a hydrocarbon ‘A’ \[\ce{(C4H8)}\] to give a compound ‘B’ which on hydrolysis with aqueous alkali forms tertiary alcohol ‘C’ \[\ce{(C4H10O)}\]. Identify ‘A’ ,‘B’ and ‘C’.
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उत्तर
\[\begin{array}{cc}
\phantom{...}\ce{CH3}\phantom{..............................}\ce{CH3}\phantom{..............................}\ce{CH3}\\
|\phantom{..................................}|\phantom{.................................}|\\
\ce{\underset{(A)}{\underset{(C4H8)}{\underset{2-Methylprop-1-ene}{H3C - C = CH2 + HCl}}} ->[Markovnikov][rule] H3C - C - CH3 ->[KOH (aq_.)Hydrolysis][\Delta] H3C - C - CH3}\\
\phantom{...................................}|\phantom{.................................}|\\
\phantom{.....................................}\ce{\underset{(B)}{\underset{2-Chloro-2-methylpropane}{Cl}}\phantom{...............}\ce{\underset{(C)}{\underset{(C4H10O)}{\underset{2-Hydroxy-2-methylpropane}{OH}}}}}
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3}\phantom{................}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether] A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{.................}
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3}\phantom{................}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl->[Na/ dry ether]A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{................}\\
\end{array}\]
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\[\begin{array}{cc}
\ce{CH3}\phantom{................}\\
|\phantom{...................}\\
\ce{CH3 - C - CH2 - Cl ->[Na/dry ether] A}\\
|\phantom{...................}\\
\ce{CH3}\phantom{................}\\
\end{array}\]
