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busupshot83

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On page 362 of Kaplan Blue Book, it states the following about the melting point of alkenes:

Trans-alkenes have higher melting points than cis-alkenes due to higher symmetry.

I thought a cis molecule would have more symmetry than a trans molecule. Does this have anything to do with dipoles? If so, how?
 

MrDreamWeaver

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Trans molecules pack together more efficiently = greater intermolecular forces = more energy needed to melt the molecule = higher mp
 

Lonely Sol

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To answer you question about dipoles, the reason the cis configuration has a higher boiling point as compare to trans is because cis has a net dipole moment while trans do not have dipole moment (they cancel out). A net dipole moment means it has high boiling point. Hope that helps!!
 
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CyborgNinjaX

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Huh what? In what regard? Somebody please tell me what's right! I read through that section yesterday and I'm trying to make sure everything sticks together perfectly. (hopes that streetwolf replies)
 

busupshot83

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Huh what? In what regard? Somebody please tell me what's right! I read through that section yesterday and I'm trying to make sure everything sticks together perfectly. (hopes that streetwolf replies)

If Kaplan really is wrong, then I guess the best way to think of trans having a higher melting point than cis is based off of its tightly-packed structure, not its symmetry.
 

Streetwolf

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If Kaplan really is wrong, then I guess the best way to think of trans having a higher melting point than cis is based off of its tightly-packed structure, not its symmetry.

Yes, since it's tightly-packed it will want to remain in the solid state moreso than its cis counterpart. It's symmetrical around the double bond. While you might think cis is the more symmetrical molecule, it doesn't help for what we're talking about. It's just symmetrical across the double bond - not around it.
 
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