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Question
- Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
- Classify the isomers of alcohols in the above question as primary, secondary and tertiary alcohols.
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Solution
- The isomeric alcohols with molecular formula C5H12O are:
- \[\ce{\underset{Pentanol (1^\circ)}{CH3 - CH2 - CH2 - CH2 - CH2 - OH}}\]
\[\begin{array}{cc}
\ce{CH3 - CH2 - CH2 - \overset{∗}{C}H - CH3}\\
\phantom{....................}|\\
\phantom{........................}\ce{\underset{Pentan-2-ol (2^\circ)}{OH}}\
\end{array}\]
\[\begin{array}{cc}
\ce{CH3 - CH2 - CH - CH2 - CH3}\\
|\phantom{..}\\
\phantom{..}\ce{\underset{Pentan-3-ol (2^\circ)}{OH}}\
\end{array}\]
\[\begin{array}{cc}
\ce{H3C - H2C - H\overset{∗}{C} - CH2OH}\\
\phantom{.........}|\\
\phantom{.............}\ce{\underset{2-Methylbutan-1-ol (1^\circ)}{CH3}}\
\end{array}\]
\[\begin{array}{cc}
\ce{CH3}\phantom{....................}\\
|\phantom{........................}\\
\ce{\underset{3-Methylbutan-1-ol (1^\circ)}{CH3 - CH - CH2 - CH2 - OH}}\\
\end{array}\]
\[\begin{array}{cc}
\ce{CH3}\phantom{.......}\\
|\phantom{...........}\\
\ce{CH3 - C - CH2 - CH3}\\
|\phantom{............}\\
\ce{\underset{2-Methylbutan-2-ol (3^\circ)}{OH}}\phantom{.........}\
\end{array}\]
\[\begin{array}{cc}
\ce{CH3}\phantom{.......}\\
|\phantom{..........}\\
\ce{CH3 - C - CH2 - OH}\\
|\phantom{..........}\\
\ce{\underset{2, 2-Dimethylpropan-1-ol (1^\circ)}{CH3}}\phantom{......}\
\end{array}\]
\[\begin{array}{cc}
\ce{CH3}\phantom{...}\ce{OH}\phantom{.}\\
|\phantom{.........}|\phantom{...}\\
\ce{\underset{3-Methylbutan-2-ol (2^\circ)}{CH3 - CH - \underset{∗}{C}H - CH3}}\
\end{array}\]
Isomers b, d and h contain chiral carbon atoms; thus, they exhibit enantiomerism.
- Isomers a, d, e, and g are primary alcohols.
Isomers b, c and h are secondary alcohols.
Isomer f is a tertiary alcohol.
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\[\begin{array}{cc}
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