6b i- acid is being added therefore it will react with the base (NH2) donating a proton to NH2 Therefore it will turn to NH3 (with a positive charge as an extra proton has a positive charge (ii) NaOH will react with COOH accepting a proton from COOH therefore it will turn to COO (negatively charged as a proton has been taken) (Iii) esterification- COOCH3 The rest of the structure remains the same I hope you understood!
6b i- acid is being added therefore it will react with the base (NH2) donating a proton to NH2 Therefore it will turn to NH3 (with a positive charge as an extra proton has a positive charge (ii) NaOH will react with COOH accepting a proton from COOH therefore it will turn to COO (negatively charged as a proton has been taken) (Iii) esterification- COOCH3 The rest of the structure remains the same I hope you understood!
you made a mistake with part 2. the hydrogen falls of the COOH functional group and the Sodium attaches on instead.
6b i- acid is being added therefore it will react with the base (NH2) donating a proton to NH2 Therefore it will turn to NH3 (with a positive charge as an extra proton has a positive charge (ii) NaOH will react with COOH accepting a proton from COOH therefore it will turn to COO (negatively charged as a proton has been taken) (Iii) esterification- COOCH3 The rest of the structure remains the same I hope you understood!
yes, NaOH will split into Na+ and OH-. the hydrogen on On COOH falls off and Na+ bonds with the COO-. the hydrogen that fell off (H+ ion) reacts with OH- to make water.
precisely, also lets not forget that amino acids can exist as zwitterions
I also recently found out that, during esterification in the presence of an acid of an amino acid, the amine group becomes NH3+ rather than staying NH2 (dropped a mark because of it in my mock )
yes, NaOH will split into Na+ and OH-. the hydrogen on On COOH falls off and Na+ bonds with the COO-. the hydrogen that fell off (H+ ion) reacts with OH- to make water.
I also recently found out that, during esterification in the presence of an acid of an amino acid, the amine group becomes NH3+ rather than staying NH2 (dropped a mark because of it in my mock )
yes, as far as I know the COOH group donates a H+ ion to the amine group .
Nothing, the cl- will stay diluted in solution as 2cl-. Also for the sodium, you cannot show the bond between the oxygen and the sodium, you have to write O(minus sign)Na+
Do you mind helping me with 6d as well. Do you get rid of a OH from lysine and a H from alanine? As that forms water? The bit I don't get it is, where did the other hydrogen go from the NH2 in alanine when you combine them?
Do you mind helping me with 6d as well. Do you get rid of a OH from lysine and a H from alanine? As that forms water? The bit I don't get it is, where did the other hydrogen go from the NH2 in alanine when you combine them?
Do you mind helping me with 6d as well. Do you get rid of a OH from lysine and a H from alanine? As that forms water? The bit I don't get it is, where did the other hydrogen go from the NH2 in alanine when you combine them?
Why isn't it what I have done?
Okay, have you seen what a peptide bond looks like? It's essentially this, and looks like this between any two amino acids (the C=O and N-H can be flipped around though):
What happens is that the NH2 of amino acid 1 and the COOH of amino acid 2 react to form a bond, and this involves the removal of water, as the -OH of the COOH reacts with a H from the NH2, and only from an NH2 group
Okay, have you seen what a peptide bond looks like? It's essentially this, and looks like this between any two amino acids (the C=O and N-H can be flipped around though):
What happens is that the NH2 of amino acid 1 and the COOH of amino acid 2 react to form a bond, and this involves the removal of water, as the -OH of the COOH reacts with a H from the NH2, and only from an NH2 group