1. What are the products to the reaction below:
NH3+CH3CH2CH2Br-->?
2. Why does oxaloacetic acid has a higher Ka compared to succinic acid?
3. Which hydrogen atoms will malic lose if it ionizes completely?
No ideas on any of these questions. Please help!
Thank you!
1.
RX + NH3 ==> RNH3*X (the amine salt)
What does *represent?
RCOOH compounds are organic acids. The end H atom is acidic.
I'm sure you had the ionization of acetic acid, HC2H3O2, in general chemistry. That ionizes as HC2H3O3 ==> H^+ + C2H3O2^- BUT if we wrote that in terms of its structure (instead of just the empirical formula of HC2H3O3) it would be
CH3COOH and that is the end H that is ionizing. Since malic acid has TWO COOH groups, it can lose both H atoms of both COOH groups. In fact it can lose them in steps (and does), as
HOOC-R-COOH ==> (-OOC-R-COOH)^- + H^+ and
(-OOC-R-COOH) ==> (-OOC-R-COO-)^-2 + H^+ where R stands for everything between the two end COOH groups.
I wrote the * just to show that it is a salt, actually consisting of
RNH3^+ + Cl^- but it can be written as RNH3Cl as well. The * takes the place of a period in the middle of the line which I can't write on the computer. Often it is written that way, too, as RNH3(period in the middle of the line)Cl.
For #1, will the products be CH3CH2CH2NH3 and Br2?
I don't really get this....
No.
CH3CH2CH2Br + NH3 ==> CH3CH2CH2NH3Br.
I took a shortcut by using R to stand for CH3CH2CH2 and X to stand for Br to make
RX + NH3 ==> RNH3Br.
If you treat the CH3CH2CH2NH3Br with base (like NaOH), you will get the
CH3CH2CH2NH2 + H2O + NaBr (in other words by treating with NaOH you convert the amine salt to the free amine, the extra H of the RNH3 combines with the OH of the base to make water and the Br^- then combines with the Na of the NaOH to make NaBr.)
So it would be CH3CH2CH2NH2 and HBr?
Thanks
No. I gave you the products in my previous response. You do NOT get the free amine (CH3CH2CH2NH2) unless you add NaOH to the salt.
The reaction of CH3CH2CH2Br with NH3 produces JUST the salt, CH3CH2CH2NH3Br.
But there is no such a choice...
Actually,this question is a multiple choice question, I'll type out the whole question again. Sorry for not typing out the choices at the first time...
The products of this reaction are ___
CH3CH2CH2Br+NH3-->
a. CH3CH2CH2NH2Br and H2
b. CH3CH2CH2NH3 and Br2
c. CH3CH2CH2NH2 and HBr
d. CH3CH2CH3 and NH2Br
I wish you had, also.
Check your answers in the last post to make sure all of them are right (no charges omitted, etc). If all of them are correct answers from your original, then a can't possibly be right, b can't be right UNLESS there is a + charge on the CH3CH2CH2NH3 and c can't be right. That leaves c as the least of the evils and I would go with that. That probably is the correct one. But I might like to point out to your prof (and to you as to why I was so persistent with my initial answer) that CH3CH2CH2NH2 is a base and HBr is an acid. Just as NH3 + HCl gives NH4Cl (base + acid = salt), then CH3CH2CH2NH2(base) + HBr(acid) = salt (which is CH3CH2CH2NH3Br). So the two CAN'T be in the same container without reacting to form, guess what, the salt.