[cyberdot38]

I have been working on this equation since 1pm Eastern. Now I have a huge headache.

NH2CH2COOH + NH2CHCOOH  w/heat  ?
                            |
-------------------CH3


I know that the OH from one, plus and O from the other forms the H2O.  Then after loosing to form the water, the two join together where the OH is removed. What confuses me is the leftover "O"!  I tried using a subscript 2 in front, but I think that is totally wrong. If you can direct me in the right direction perhaps I can figure out what to do next.

A big thank you to the person who lends me a hand.

[Sev]

I know that the OH from one, plus and O from the other forms the H2O. 

Do you mean H?  Not sure if I understand your question, looks like a classic dehydration giving amide bond.
http://en.wikipedia.org/wiki/Peptide_bond

[cyberdot38]

This is the general equation we were given:

-COOH + NH3  using heat results is H2O + -C double bond O, N <--attach the N to the C though.

If I could type in the equations as I have them written, perhaps they would make more sense.  Here is a different equation that I already figured out. Perhaps this one will show you what I am trying to do.

                                     
CH3COOH + CH3NHCH3   -heat-   H2O  +  CH3C double bond O above the C, then NCH3 
                                                                                                              |
                                                                                                            CH3

Can you understand any of this? 
       

[Sev]

Is that what you mean?

[cyberdot38]

Yes, the only difference is the radicals have been replaced with CH3 and so on.  I was hoping someone would see what I was talking about, so they can put my previous equation together. Are you able to do so?

[agrobert]

This is a generalized amide bond formation as shown above.  Those are not radicals, they are R groups meaning that they can be any substituent that is reasonable, including dimethyl amine.

[Sev]

RCOOH +H₂NR → R(CO)NHR + H₂O.

If that isn't what you mean, I don't know.

[agrobert]

acetic acid + dimethyl amine --> N,N-dimethylacetamide + water

CH₃COOH + HN(CH₃)₂ --> CH₃C(O)N(CH₃)₂ + H₂O

[cyberdot38]

NH2CH2COOH + NH2CHCOOH  w/heat  ?
                            |
-------------------CH3

This is the equation.  Is this the solution?

CH3COOH + HN(CH3)2 --> CH3C(O)N(CH3)2 + H2O

They don't look the same to me.

[agrobert]

Ok I get it now.

Glycine + Alanine --> GLY-ALA + H₂O

H₂NCH₂COOH + H₂NCH(CH₃)COOH

-->

^-O(O)CCH₂NHC(O)CH(CH₃)NH₃⁺  OR  HO(O)CCH₂NHC(O)CH(CH₃)NH₂ + H₂O

But you really should be able to figure this out from your text.

This is similar

http://homepages.ius.edu/GKIRCHNE/peptide.jpg

[cyberdot38]

O(O)CCH2NHC(O)CH(CH3)NH3+

Thanks for helping, would it be possible for you to explain how you got this answer?

[agrobert]

The acid group (-COOH) easily donates a proton (Bronsted Acid) to become (-COO^-) and the amide group (-NH₂) easily accepts this proton (Bronsted Base) and the salt is formed in solution to beome (-NH₃⁺)

Check this out

http://en.wikipedia.org/wiki/Zwitterion

[cyberdot38]

I SEE!!!   That's so great....Thanks for your brain picking.  You are awsome.

[ultrashogun]

This reaction can procede to form longer chains.

[AWK]

First of all, the reaction can proceed in two main ways:
Ala + Gly = Ala-Gly
or Gly + Ala = Gly-Ala
but also such side reactions are possible:
Ala + Ala = Ala-Ala
Gly + Gly = Gly-Gly
During this reaction proceeded without protective groups the next step is a cyclization rather then polymerization -
both Ala-Gly and Gly-Ala form diketopiperazines cyclo-Ala-Gly