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Question
A hydrocarbon of molecular mass 72 g mol–1 gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon.
Solution
\[\ce{C5H12}\], pentane has molecular mass 72 g mol–1 i.e., isomer of pentane which yields single monochloro derivative should have all the 12 hydrogens equivalents.
The hydrocarbon is
\[\begin{array}{cc}
\phantom{........}\ce{CH3}\\
\phantom{.....}\phantom{}|\\
\phantom{.....}\ce{CH3 - C - CH3}\\ \phantom{.....}\phantom{}|\\
\phantom{........}\ce{CH3}
\end{array}\]
Monochloro derivative:
\[\begin{array}{cc}
\phantom{........}\ce{CH3}\\
\phantom{.....}\phantom{}|\\
\phantom{.......}\ce{CH3 - C - CH2Cl}\\ \phantom{.....}\phantom{}|\\
\phantom{........}\ce{CH3}
\end{array}\]
Dichloro derivatives:
(i)
\[\begin{array}{cc}
\phantom{........}\ce{CH3}\\
\phantom{.....}\phantom{}|\\
\phantom{.......}\ce{CH3 - C - CHCl2}\\ \phantom{.....}\phantom{}|\\
\phantom{........}\ce{CH3}
\end{array}\]
(ii)
\[\begin{array}{cc}
\phantom{..........}\ce{CH2Cl}\\
\phantom{.....}\phantom{}|\\
\phantom{.......}\ce{H3C - C - CH2Cl}\\ \phantom{.....}\phantom{}|\\
\phantom{........}\ce{CH3}
\end{array}\]