Why in this the nonbonding electrons of N group does not attack C2 as I think it is more electropositive ?
Then how pyruvic acid is converted to alanine.
0
$\begingroup$
$\endgroup$
-
3$\begingroup$ Firstly your numbering is the wrong way round. Carboxylic acid should have higher priority. In any case the carboxylic acid carbon is less electronegative (mesomeric donation of OH oxygen lone pair into C=O $\pi^*$) and this is especially so when you consider that with ammonia, the carboxylic acid should be mostly deprotonated. $\endgroup$ – orthocresol♦ Oct 16 '16 at 19:31
-
$\begingroup$ @orthocresol then how pyruvic acid is converted to alanine $\endgroup$ – Koolman Oct 17 '16 at 8:43
-
$\begingroup$ It is a process called reductive amination, look it up. $\endgroup$ – orthocresol♦ Oct 17 '16 at 8:49
-
1$\begingroup$ @orthocresol so here reductive amination takes place or not $\endgroup$ – Koolman Oct 17 '16 at 10:54
-
$\begingroup$ Did you consider reduction to lactic acid (A) and subsequent formation of lactide (C) upon heating? $\endgroup$ – Klaus-Dieter Warzecha Oct 17 '16 at 14:30
1
$\begingroup$
$\endgroup$
The keto group is more reactive compared to the ester group due to the mesomeric effect (+M effect) caused by the hydroxyl group in the ester functionality, as already pointed out by orthocresol. See also https://en.wikipedia.org/wiki/Carbonyl_reduction. Generally: $$ \mathrm{acylchloride} > \mathrm{anhydride} > \mathrm{aldehyde} > \mathrm{ketone} > \mathrm{ester} \approx \mathrm{carboxyl} > \mathrm{amide} $$
A very detailed discussion on the subject is given for example in Warren Organic Synthesis: Strategy and Control or in Francis A. Carey Advanced Organic Chemistry - Part A.
answered Oct 16 '16 at 20:20
-
$\begingroup$ Does here reductive amination takes place or not $\endgroup$ – Koolman Oct 17 '16 at 14:11
