Re: methlymalonic acid reaction with Na OH

"Panther" <black@xxxxxxx> wrote in message
news:dmicr9$ftv$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> i have tried this problem for a while but i cannot get it
> i think the formula for methylmalonic acid is this:
.....(CH3)HC(CO2H)2
>
(CH3)2>C<(CO2H)2 would Dimethylmalonic acid
>
> if i react with it sodium hydroxide (NaOH) what will the products be?
>
> please help :| i know H2O is in there, and i swear i have tried. i know it loses a
> H from this: (CH3)2 C (COOH)2
>
a) (CH3)2>C<(COOH)2 + NaOH = CH3)2>C< (COOH) (COO-)Na+ H2O

> and i think the ans is this:
>
> (CH3)2>C< (COOH) (COO-)Na+
>
> but it's diprotic so how come it only looses 1 H?
I don't know what you are driving at:
b) (CH3)2>C< (COOH) (COO-)Na+ + NaOH = (CH3)2 >C< (COONa)2 +H2O

(a+b) =
(CH3)2 >C< (COOH)2 + 2 NaOH = (CH3)2 >C< (COONa)2 + 2 H2O


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