Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

Below is the structure of Ace-Ala-Nme. I have labelled just four of the carbon atoms (1-4 in blue). I need to identify (i) the hybridization state, and (ii) the number of electron withdrawing groups for each.

enter image description here

I believe the amine (-NH) groups are electron donating while the carbonyls (-CO) are electron withdrawing. I therefore suspect that all four labelled carbon atoms are sp3 with 0 withdrawing groups. But this is assuming that the donating/withdrawing tendency of the amine and carbonyls cancel each other out. Can anyone confirm this to be the case?

share|improve this question
    
This is a neutral, un-ionized (i.e. non-zwitterionic form) form of Ala-Ala correct? Also, are you asking about the electron configuration of this dipeptide or hybridization of various atoms in this molecule? –  LordStryker Jul 29 at 14:26
    
@LordStryker It is indeed neutral and I'm interested in the individual atoms. Thank you. –  lemon Jul 29 at 14:29
    
This illustration must be just a fragment of an Alanine chain (polypeptide). The un-ionized Alanine dipeptide would intially begin with NH2 and terminate with COOH. Your picture does not indicate this. You may want to provide a proper image, re-label your atoms and then attempt to provide an adjusted answer. (See Figure 3 on page 5 of this article: iopscience.iop.org/0004-637X/765/2/111/pdf/…) –  LordStryker Jul 29 at 14:34
    
@LordStryker I have updated my question. I have seen this molecule referred to as "Ace-Ala-Nme" although it's mostly referred to as "alanine dipeptide" in the papers I've read (then again, these are papers by physicists rather than chemists). –  lemon Jul 29 at 14:53
2  
I think an amine (-NH) is electron donating but I am not an organic chemist. EDIT: The consensus in my lab is that -NH is an electron donating group. –  LordStryker Jul 29 at 15:06

1 Answer 1

up vote 3 down vote accepted

1: sp3 with 1 electron withdrawing group

2: sp3 with 2 electron withdrawing groups

3: sp3 with 0 electron withdrawing groups

4: sp3 with 1 electron withdrawing group

I would say your earlier analysis was correct. If you look at the resonance structures that can be drawn for an amide linkage, you find there is considerable positive charge both on carbon and nitrogen. This should serve to make them both electron withdrawing, relative to an attached carbon, in your structure.

enter image description here

EDIT: To clarify this answer with regards to some of the comments posted above, I'd like to add that typically when an amino nitrogen is situated on a double bond or aromatic ring it can donate electrons through resonance. In fact, that is what it is doing in the resonance structure on the right - it is donating its lone pair of electrons via resonance. However, the amino group is also electron withdrawing in an inductive sense relative to carbon. Further the $\ce{-NH3^+}$ group can no longer interact via resonance and is simply a strong electron withdrawing group. The resonance structure on the right is analogous to the $\ce{-NH3^+}$ situation and since it is a significant contributor to the overall description of the amide (cf. there is significant double bond character in the amide $\ce{C-N}$ bond) it acts inductively as a strong electron withdrawing group for the attached $\ce{sp^3}$ carbon.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.