BR Ochem Ch1 Passage 2

[deleted388502]

Hi,

I just want to clear up Question 12 in BR Ochem phase 1 passage 2.

It gives an amide and asks which correctly describes the geometry of the molecule:

A. Nitrogen has trigonal pyramidal geometry, so the two hydrogens are outside of the plane created by the four other atoms.
B. The nitrogen has tetrahedral geometry, so the two hydrogens are outside of the plane created by the other four atoms.
C. The carbon has tetrahedral geometry, so the carbon hydrogen is outside of the plane created by the other five atoms.
D. The six atoms are coplanar

So the answer is D and I would love to hear how someone else approaches this problem. My first thought was that the nitrogen was tetrahedral but I now realize it's between sp2 and sp3, but I don't understand the explanation BR gives. How can the nitrogen be entirely sp2 and trigonal pyramidal?

Also, in an amide, the passage has a bunch of questions that talk about how the oxygen is where it's protonated. Why is the oxygen more basic than the nitrogen in an amide? Question 14 says that the most stable hydrogen-bond between amides is from the carbonyl oxygen to the H on the nitrogen - in interpreting this question, why would you choose this answer over the carbonyl oxygen and H on the carbon?


Sorry for all the questions, and I'm greatly appreciative to anyone that can provide insight on this.

Thanks

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[Teleologist]

Right, amides are sp2 hybridized because they have three sigma bonds. This rules out B and C because sp2 hybridization does not correspond with a tetrahedral molecular geometry. Choice A is also wrong; nitrogen has 4 sigma bonds; the geometry should be tetrahedral about the nitrogen.

I'm not sure what you mean by the nitrogen is sp2 and trigonal pyramidal. Which nitrogen are you talking about?

Also the carbonyl oxygen is more basic because when it's protonated it can keep its 0 formal charge. A lone pair on the carbonyl oxygen makes the bond with a hydrogen proton and a bonding pair (remember it's C = O) can jump on top of the oxygen to maintain its 0 formal charge. This sorta mirrors the behavior of acetate ion in the presence of hydrogen protons. On the other hand if you protonated the nitrogen you would have a positive formal charge.

If you could post a picture of the passage and questions everything would be a lot clearer.

Note the abstraction of hydrogen protons above. Technically I should have drawn hydronium ions but the picture gets the point across.

 

[deleted388502]

Here are the pictures! Your oxygen explanation definitely makes sense, thank you!


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[DrknoSDN]

Oddly, someone else asked a question about a near identical problem on Thursday.

Answer Here:
http://forums.studentdoctor.net/thr...passage-iv-question-24.1071520/#post-15235851

 

[Chrisz]

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Here are the pictures! Your oxygen explanation definitely makes sense, thank you!


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The book gives a superficial explanation of why the nitrogen is planar with the rest of the atoms in the molecule. However, this is enough to answer the question that the second resonance structure shows that the nitrogen has some sp2 character. The proper explanation is that the lone pair of the nitrogen goes into the p orbital instead of an sp3 orbital. The reason lies in the fact that we see there is a pi bond between carbon and oxygen. If the lone pair resides in the p orbital of nitrogen, it can overlap with the p orbital of the carbon atom and form conjugation. Conjugation provides stability. An example of conjugation providing stability is 1,3 cyclohexanediene vs 1, 5 cyclohexadiene. Because there is conjugation between 1,3 cyclohexanediene, 1,3 cyclohexanediene is relatively more stable compared to 1, 5 cyclohexadiene. This is the reason why we see 1,3 cyclohexanediene release less heat upon hydrogenation to cyclohexane than 1,5 cyclohexanediene. 1,3 cyclohexanediene is relatively more stable compared to 1, 5 cyclohexadiene does.