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प्रश्न
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
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उत्तर
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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\[\begin{array}{cc}
\ce{CH3 - CH - CH2 - CH3 and CH3 - CH2 - O - CH2 - CH3}\\|\phantom{...........................................}\\
\ce{OH}\phantom{.........................................}\end{array}\]
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\[\begin{array}{cc}\ce{^+OH}\\||\\
\ce{H - C - OH}\end{array}\]
\[\begin{array}{cc}\ce{OH}\phantom{.}\\|\phantom{...}\\
\ce{H - C^+ - OH}\end{array}\]
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| I. | \[\begin{array}{cc} \phantom{.......................}\ce{O}\\ \phantom{.......................}||\\ \ce{CH3 - CH2 - CH2 - CH2 - C - H} \end{array}\] |
| II. | \[\begin{array}{cc} \phantom{.................}\ce{O}\\ \phantom{.................}||\\ \ce{CH3 - CH2 - CH2 - C - H} \end{array}\] |
| III. | \[\begin{array}{cc} \ce{CH3 - CH2 - C - CH2 - CH3}\\ \phantom{}||\\ \phantom{}\ce{O} \end{array}\] |
| IV. | \[\begin{array}{cc} \ce{CH3 - CH - CH2 - C - H}\\ \phantom{...}|\phantom{............}||\phantom{}\\ \phantom{...}\ce{CH3}\phantom{.........}\ce{O}\phantom{} \end{array}\] |
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| I. | CH3 – CH2 – CH2 – CH2 – OH |
| II. | \[\begin{array}{cc} \ce{CH3 - CH2 - CH - CH3}\\ \phantom{.....}|\\ \phantom{.......}\ce{OH} \end{array}\] |
| III. | \[\begin{array}{cc} \phantom{...}\ce{CH3}\\ \phantom{}|\\ \ce{CH3 - C - CH3}\\ \phantom{}|\\ \phantom{..}\ce{OH} \end{array}\] |
| IV. | \[\begin{array}{cc} \ce{CH3 - CH - CH2 - OH}\\ |\phantom{........}\\ \ce{CH3}\phantom{......} \end{array}\] |
| V. | CH3 – CH2 – O – CH2 – CH3 |
| VI. | CH3 – O – CH2 – CH2 – CH3 |
| VII. | \[\begin{array}{cc} \ce{CH3 - O - CH - CH3}\\ \phantom{...}|\\ \phantom{......}\ce{CH3} \end{array}\] |
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| I. | CH3 – CH2 – CH2 – CH2 – OH |
| II. | \[\begin{array}{cc} \ce{CH3 - CH2 - CH - CH3}\\ \phantom{.....}|\\ \phantom{.......}\ce{OH} \end{array}\] |
| III. | \[\begin{array}{cc} \phantom{...}\ce{CH3}\\ \phantom{}|\\ \ce{CH3 - C - CH3}\\ \phantom{}|\\ \phantom{..}\ce{OH} \end{array}\] |
| IV. | \[\begin{array}{cc} \ce{CH3 - CH - CH2 - OH}\\ |\phantom{........}\\ \ce{CH3}\phantom{......} \end{array}\] |
| V. | CH3 – CH2 – O – CH2 – CH3 |
| VI. | CH3 – O – CH2 – CH2 – CH3 |
| VII. | \[\begin{array}{cc} \ce{CH3 - O - CH - CH3}\\ \phantom{...}|\\ \phantom{......}\ce{CH3} \end{array}\] |
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And
\[\begin{array}{cc}
\phantom{.....................}\ce{CH3}\\
\phantom{................}/\\
\phantom{}\ce{CH3 - S - CH}\\
\phantom{...............}\backslash\\
\phantom{....................}\ce{CH3}
\end{array}\]
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