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I need to find the bond angles for carbamic acid (as a self-given practice). As I understand it, the structure looks like this:

        O
        ||
H - N - C - O - H
    |
    H

I'm trying to find the O-C-O bond angle, but I'm not sure how the - H on the end will factor in to the answer. Everything I read about it is dense and fairly hard to understand.

I've tried finding the number of electron domains around the C atom (I think it's 4), but I'm not sure where to go from there.

What is the general procedure for doing this? How would it look if I were trying to find a different angle (like H-N-H)?

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A quick google of 'carbamic acid bond angles' reveals many papers in the literature measuring/computing the bond angles in this and similar molecules. See here, for example.

That said, chemists are often not concerned with the exact bond angle, instead, we often look at the structure and make predictions based upon bonding and the substituents around the atoms.

In carbamic acid, the simplest carbamate, we can consider the central carbonyl to be sp2 hybridised, giving it a planar structure with bond angles of 120. Due to conjugation with the nitrogen lone pair, the N can also be considered to be sp2 hybridised, and also have bond angles of around 120.

Although this is quite a crude analysis, it holds up quite well for sp2 centres, with very little variation (except in cases of exceptional steric bulk forcing groups to move closer/further apart, or in strained ring systems). The paper referenced confirms that in this system (carbamic acid), the deviation is as little as 4 degrees.

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  • $\begingroup$ Wow, that made a lot of sense. So the O-C-O bond angle would be the same even if the rightmost H atom was (somehow) left out and the rest of the molecule was otherwise the same? $\endgroup$ – Kaylee Nov 30 '15 at 20:13
  • $\begingroup$ @Kaylee Not exactly, but anion also has this type of geometry. $\endgroup$ – Mithoron Dec 1 '15 at 2:02

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