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glycine with an NH3Cl where the NH2 is...
NaOH
NaOH
if you want to react glycine hydrochloride with NaOH all that will happen is that glycine will be generated in the amine form and you will get the sodium salt of the carboxylic acid end formed as ions.
HO2CCH2NH3Cl + 2NaOH --> Na+[O2CCH2NH2]- + NaCl + 2H2O
HO2CCH2NH3Cl + 2NaOH --> Na+[O2CCH2NH2]- + NaCl + 2H2O
charco
if you want to react glycine hydrochloride with NaOH all that will happen is that glycine will be generated in the amine form and you will get the sodium salt of the carboxylic acid end formed as ions.
HO2CCH2NH3Cl + 2NaOH --> Na+[O2CCH2NH2]- + NaCl + 2H2O
HO2CCH2NH3Cl + 2NaOH --> Na+[O2CCH2NH2]- + NaCl + 2H2O
Why do you have sodium chloride as your product? I think you meant hydrochloric acid.
no I don't - I mean sodium chloride, that's why I wrote it!
When an acid reacts wth a base they do so in the required stoichiometric amounts.
For example in the reaction between sulphuric acid and sodium hydroxide, 2 moles of NaOH are required for complete neutralisation.
H2SO4 + 2NaOH --> Na2SO4`+ 2H2O
It would be possible to prepare sodium hydrogen sulphate by adding only 1 mole of NaOH to 1 mole of H2SO4
H2SO4 + NaOH --> NaHSO4`+ H2O
But the OP did not say that the NaOH was restricted in quantity in any way.
NaOH will react with both ends of the amino acid salt. Therefore 2 moles of NaOH are required, as in the equation I posted.
The first mole of NaOH removes the HCl of the amine salt making NaCl and H2O, and then the second NaOH reacts with the acidic -COOH group making the sodium salt of the carboxylic acid group + H2O
For example in the reaction between sulphuric acid and sodium hydroxide, 2 moles of NaOH are required for complete neutralisation.
H2SO4 + 2NaOH --> Na2SO4`+ 2H2O
It would be possible to prepare sodium hydrogen sulphate by adding only 1 mole of NaOH to 1 mole of H2SO4
H2SO4 + NaOH --> NaHSO4`+ H2O
But the OP did not say that the NaOH was restricted in quantity in any way.
NaOH will react with both ends of the amino acid salt. Therefore 2 moles of NaOH are required, as in the equation I posted.
The first mole of NaOH removes the HCl of the amine salt making NaCl and H2O, and then the second NaOH reacts with the acidic -COOH group making the sodium salt of the carboxylic acid group + H2O
Original post by Umeh
I mean glycine hydrochloride with sodium hydroxide solution
Necrophilia Award 1st Class .... 9 years and 3 days later!
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