[kmlchicago]

Hello, everyone -

My organic chemistry class was assigned a molecular modeling project. Using the computer program Spartan, I have built the molecules 1,1-dibromoethene and 1-1-difluoroethene. I have computationally determined the following about their respective bond angles:

Br-C-Br = 115°
F-C-F = 109.5°

I want to know why this difference is observed. Since the carbon is sp2 hybridized, shouldn't the angles both be 120°? Additionally, why is the F-C-F bond angle less than the Br-C-Br angle? I know F is the most electronegative element, and Br and F are both halogens with incomplete octets. Also, I know a pi bond exists across the C=C bond (an alkene), so perhaps these electrons are influencing the angles. Please let me know your thoughts, or perhaps point me to a resource that could help further.

Thank you very much for your help.

-Kyle Lee, Chicago

[orgo814]

Br is larger than F

[magician4]

F-C-F = 109.5°

tried to reproduce this value on my Spartan software, but failed , both with MM based and with Hartree-Fock based energy optimizing algorithms

sure this isn't a typo ? ( 119,5° would be just fine with me instead)


with respect to bond angels in Br-C-Br :

p- sp² - [itex]\sigma[/itex] bonds are a little shorter than p- sp³ - [itex]\sigma[/itex] bonds.
Br - atoms being slightly obese, they yell for distance from each other, and, being attached to the same carbon, hence for longer bonds Br - C  (as this is the only game in town)

  this repulsive effect makes the "clean" sp² hybridization not the most optimum energetic solution for the system.
Instead, a "broken" hybridization of the sp^(2+x) type ; 0 < x <1 is the optimum for the system

... and this increased p- character makes for different bond angels, more towards 109,5° ( i.e. towards sp³)


regards

Ingo

[opsomath]

Since the carbon is sp2 hybridized, shouldn't the angles both be 120°?

Ok, I wanted to deal with this first. 120 is the "ideal" bond angle for a sp2 hybridized atom. The only way you will have a perfect 120° angle is if each of the three things bonded to the carbon is exactly identical. If there is any lack of symmetry, the bulkier groups will push the less bulky groups away and distort the angle. The same thing is true for sp3 hybridized stuff.

Now, the other thing. Your value for the F-C-F bond angle is simply wrong. I don't know if you calculated it wrong, or if you just wrote it down wrong, but it's wrong. I'm guessing you calculated it wrong. I note that 109.5° is about right for an sp3 atom; perhaps you accidentally built the wrong structure. (In fact, I'd bet money that that's what happened.)

[magician4]

The only way you will have a perfect 120° angle is if each of the three things bonded to the carbon is exactly identical.

I tend to disagree: the group attached to the double bond doesn't need to be identical to those two groups attached to the singlebonds for 120° to occur.
Ethene has -H , - H and =CH₂ attached to the central carbon, respectively , and is 120° nonetheless

on the other hand, even if all 4 substituents attached to the central C=C are identical, still this ain't a guarantee for 120°: I₂C=CI₂ has a bond angle I-C- I of 114° evenso*⁾


regards

Ingo


*⁾ Bock et al., Ang.Chem.Int.Ed.Engl. 31, 550  (1992)

[opsomath]

the group attached to the double bond doesn't need to be identical to those two groups attached to the singlebonds for 120° to occur.

It certainly does. While two groups that are similar in size might coincidentally give a bond angle that is close to 120, the only way to get a perfect 120° is to have a completely symmetrical molecule. Ethene is actually about 117°.

http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Structure_and_Bonding_in_Ethene-The_Pi_Bond

even if all 4 substituents attached to the central C=C are identical, still this ain't a guarantee for 120°

Hmm? I never said it did. I mean the three things attached to the sp2 center must be identical. For instance, tert-butyl carbocation has a perfect 120° bond angle. Carbon tetrachloride has a perfect 109.5° bond angle.

[magician4]

It certainly does. While two groups that are similar in size might coincidentally give a bond angle that is close to 120, the only way to get a perfect 120° is to have a completely symmetrical molecule.

in my understanding, it is not necessary that all three of them are identical (though this is a solution of the problem , too), but only that the repulsive effects of the three substituents X₁-C , X₂ -C and C=Y would exactly cancel each other out

Ethene is actually about 117°.

didn't know that : thank you for the hint
always thought that those two hydrogens were small enough to be of no further importance for the structure.
but then: you never stop learning

again, thank you

regards

Ingo

[opsomath]

in my understanding, it is not necessary that all three of them are identical (though this is a solution of the problem , too), but only that the repulsive effects of the three substituents X1-C , X2 -C and C=Y would exactly cancel each other out

This is technically correct! The trouble is that unless the substituents are exactly the same, there will always be some slight difference. In cases where the substituents are very close in size, the difference can be negligible, though.

always thought that those two hydrogens were small enough to be of no further importance for the structure.

"No further importance" is kind of a misleading way to look at it. Every functional group takes up some space (think Van der Waals radius of the atoms involved in the group). A mere hydrogen takes up a lot less than an entire double-bonded =CH2 group. So, the double bonded group pushes a little harder, making the HCH bond angle smaller than the HCC.

[kmlchicago]

Thanks to everyone for your help, especially magician4 and opsomath! I understand the concept much better now, especially with regard to atomic size influencing bond angle. I appreciate your taking the time to explain.

[Corribus]

I'm curious.  Ignoring the specified bond angles for a moment, did your models predict that these molecules are planar?

Also: what is the predicted C=C bond length?

[opsomath]

I'm curious.  Ignoring the specified bond angles for a moment, did your models predict that these molecules are planar?

Also: what is the predicted C=C bond length?

If OP answers this question, it will be much easier to help explain the results.