The boiling point of trans-but-2-ene is lower than that of its cis isomer but the melting point of the former is higher than the later. Why is it not following the same order? Is there any factor of structural symmetry coming playing a role here?
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asked Mar 3, 2015 at 16:39
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Yes . You are right that structural symmetry comes into play .
Boiling point depends upon intermolecular interactions which over here is more in cis due to its net dipole moment . The dipole moment enables electronic interactions which hold molecules together . This shows some general factors of boiling point . Also the below link to the google book covers the structural aspect of a molecule towards b.p on pg47.
For melting points , let me quote Joseph Hornback from Chapter 2.6 Organic Chemistry pg46 (google books) :
The shape of a compound is very important and has a dramatic effect on its melting point . More symmetrical molecules pack into the crystal lattice better , allowing closer approach and larger attractive forces , resulting in higher melting points .
Melting points of alkenes depends on the packaging of the molecules. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers.(source)
Trans 2-Butene has more symmetry than its cis isomer which results in better packing and hence higher melting point .
For further reading I suggest you visit 'Molecular Shape' over here.
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$\begingroup$ What about the boiling point ? Why is it not following the same pattern ? Can you include that in the answer too please. Thanks $\endgroup$ Mar 5, 2015 at 3:09
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$\begingroup$ @DhrubaBanerjee The second sentence refers to the b.p , also pg46,47 of the google book can be viewed for details . Anyway , I will edit to add a few more references . $\endgroup$– Del PateMar 6, 2015 at 16:43