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Question
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
Solution
Hex-2-ene is represented as:
\[\ce{CH3 - CH = CH - CH2 - CH2 - CH3}\]
Geometrical isomers of hex-2-ene are:
\[\begin{array}{cc}
\phantom{.............}\ce{H3C}\phantom{.........}\ce{CH2 - CH2 - CH3}\\
\backslash\phantom{.......}/\\
\ce{C = C}\\
/\phantom{.......}\backslash\\
\ce{H}\phantom{.........}\ce{H}\end{array}\]
cis-isomer
\[\begin{array}{cc}
\phantom{.}\ce{H3C}\phantom{......}\ce{H}\\
\backslash\phantom{.......}/\\
\ce{C = C}\\
/\phantom{.......}\backslash\\
\phantom{..............}\ce{H}\phantom{.......}\ce{CH2 - CH2 - CH3}\end{array}\]
trans-isomer
The dipole moment of cis-compound is a sum of the dipole moments of C–CH3 and C–CH2CH2CH3 bonds acting in the same direction.
The dipole moment of trans-compound is the resultant of the dipole moments of C–CH3 and C–CH2CH2CH3 bonds acting in opposite directions.
Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipole-dipole interaction and the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point than trans-isomer.
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