TBR OChem Chapter 1, Passage 1, Number 8, Page 58

[thekman786]

Hi guys,
The question asks:
The greatest amount of energy is required to break which of the following carbon-carbon bonds?
A) H3C--CH3
B) (H3C)3C--C(CH3)3
C) H2C=CH2
D) (H3C)2=C(CH3)2

The solution states the answer is D.

How I went about thinking about this question:
Double bonds are stronger than single bonds. This eliminates A and B.
Between C and D, the difference is substitution of the carbons to which the bond is attached.
Substituted carbons create weaker bonds. My justification for this: think of a primary carbon-hydrogen, and a tertiary carbon-hydrogen bond. With a homolytic (radical) cleavage of both bonds, the tertiary carbon product would be more stable than the primary carbon product, due to the fact that the tertiary carbon has more carbon substituents that are electron donating and can stabilize the radical. Thus since the tertiary carbon product is more stable, its bond with hydrogen is weaker.
Using the same logic to this problem, in C you have a primary carbon breaking a double bond, and in D, you have a tertiary carbon breaking a double bond. The tertiary carbon's bond should be weaker right?
I'm worried about this because I've completely thought out my logic and still gotten to the wrong answer.
Please help clarify this for me.
Thank you,

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[DocMC]

Hi guys,
The question asks:
The greatest amount of energy is required to break which of the following carbon-carbon bonds?
A) H3C--CH3
B) (H3C)3C--C(CH3)3
C) H2C=CH2
D) (H3C)2=C(CH3)2

The solution states the answer is D.

How I went about thinking about this question:
Double bonds are stronger than single bonds. This eliminates A and B.
Between C and D, the difference is substitution of the carbons to which the bond is attached.
Substituted carbons create weaker bonds. My justification for this: think of a primary carbon-hydrogen, and a tertiary carbon-hydrogen bond. With a homolytic (radical) cleavage of both bonds, the tertiary carbon product would be more stable than the primary carbon product, due to the fact that the tertiary carbon has more carbon substituents that are electron donating and can stabilize the radical. Thus since the tertiary carbon product is more stable, its bond with hydrogen is weaker.
Using the same logic to this problem, in C you have a primary carbon breaking a double bond, and in D, you have a tertiary carbon breaking a double bond. The tertiary carbon's bond should be weaker right?
I'm worried about this because I've completely thought out my logic and still gotten to the wrong answer.
Please help clarify this for me.
Thank you,

When you are looking at a C-C bond, the one that is less substituted is hardest to break (ie H2RC-CRH2 > BDE than R3C-C3R. When comparing two alkenes, the more substituted C=C bond has a higher BDE. Why? I don't exactly know.

 

[Vurday]

When you are looking at a C-C bond, the one that is less substituted is hardest to break (ie H2RC-CRH2 > BDE than R3C-C3R. When comparing two alkenes, the more substituted C=C bond has a higher BDE. Why? I don't exactly know.

From what I remember, this is because for an alkane, the more alkyl substituents you have, the more steric repulsion you get making substituted single bonds less stable than unsubstituted alkanes.

For alkenes though, the pi bond can be viewed as a combination of two adjacent radicals. Since radicals are electron deficient, they are stabilized by Electron Donating Groups. Alkyl groups are electron donating groups ==> double bonds are stabilized by substitution

Sterics still has a bit of a factor with alkenes as well. This is best seen by looking at the stability of disubstituted alkenes. If you look at the trend, you'll find that for alkenes with two substituents, stability follows the trend of trans > geminal > cis. This makes sense since the trans will have your bulky alkyl groups as far apart as possible, with cis having them as close as they can go.

Hope that helps!