December 23, 2024, 07:29:34 PM
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Topic: "Substitution" refers to the replacing of H. Double bond placement matters?  (Read 2581 times)

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Offline Lo.Lee.Ta.

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It is obvious that something like (CH3)3CH is more highly substituted than CH4.

But what about double bonds placed differently in 2 molecules.

I ask this because I am working on understanding Zaitsev's Rule.
It says that the more highly substituted alkene product predominates.

Between CH3CH=CHCH3   and    CH3CH2CH=CH2, why is it that the CH3CH=CHCH3 is more highly substituted?
Both of these molecules have a double bond that replaced 2 hydrogens, and both have 8 H's in the molecule.
And I thought substitution was just about replacing H's.
So why is CH3CH=CHCH3 more highly substituted than CHCH2CH=CH2?

Thanks! :)

Offline blackzoid

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Zaitsev's Rule states that the most stable alkene is formed in an elimination, which usually happens to be the most subtituted. Though both possible outcomes are equally substituted, you get primarily one product since it is far easier to remove a hydrogen from a secondary carbon than a primary carbon. Therefore when considering Zaitsev's rule it would be better to think about which carbon will lose a hydrogen more readily.

Offline Lo.Lee.Ta.

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I'm wondering if I can think of it like this:

In CH3CH=CHCH3, the double bond connects a CH and CH3 on the left and on the right, connects a CH and CH3.

In CH3CH2CH=CH2, the double bond connects a CH3CH2 and a CH on the left and on the right, connects a CH and an H.

Since the double bond in CH3CH=CHCH3 has more carbons it is directly bonded to, it means it is more highly substituted.

Offline orgopete

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This is how I think of it. A double bond is electron deficient, hence lower pKa. A triple bond is more so, and an even lower pKa. Carbons are electron donors and therefore stabilize a double bond. Double bonds can have 4, 3, 2, 1, or 0 carbons attached. to them. You can find a table in your textbook showing you the stability order. In the two compounds being compared, one is mono substituted and the other disubstituted. The disubstituted is the preferred product.

I don't agree with blackzoid. Under SN1-like conditions (polar protic solvents), then the hydrogen of the more substituted carbon become more acidic (because of the electrons being attracted to the carbocation). This is like a hyperconjugation effect. As far as acidity is concerned, then methane>primary>secondary>tertiary. This is also consistent with carbon being a better electron donor.
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