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Amino acids

Can someone explain 6b to me all 3 parts.
I genuinely don't understand how to do it?

http://filestore.aqa.org.uk/subjects/AQA-CHEM4-QP-JUN14.PDF

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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! :redface:
(edited 8 years ago)
Reply 2
Original post by haj101
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! :redface:


you made a mistake with part 2. the hydrogen falls of the COOH functional group and the Sodium attaches on instead.
Reply 3
Original post by haj101
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! :redface:

How come it reacts with COOH?
Reply 4
Original post by 713Wave
you made a mistake with part 2. the hydrogen falls of the COOH functional group and the Sodium attaches on instead.


Sodium?
Reply 5
Original post by Super199
Sodium?


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.
Original post by Super199
How come it reacts with COOH?


Think acid and base
Reply 7
Original post by Serine Soul
Think acid and base


precisely, also lets not forget that amino acids can exist as zwitterions :smile:
Original post by 713Wave
precisely, also lets not forget that amino acids can exist as zwitterions :smile:


:yep:

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 :colonhash:)
Reply 9
Original post by 713Wave
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.


So with the first one what happens to the cl-?
Original post by Serine Soul
:yep:

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 :colonhash:)


yes, as far as I know the COOH group donates a H+ ion to the amine group :cool:.
Original post by Super199
So with the first one what happens to the cl-?


You essentially get [amino acid]+ Cl-

You write the Cl- near to the NH3+ :smile:
Original post by Super199
So with the first one what happens to the cl-?


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+
in addition to part one, you have write NH3+ for BOTH amine groups on lysine.
Reply 14
Original post by Serine Soul
You essentially get [amino acid]+ Cl-

You write the Cl- near to the NH3+ :smile:


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?Amino acids.png
Original post by 713Wave
you made a mistake with part 2. the hydrogen falls of the COOH functional group and the Sodium attaches on instead.


Ooops I didn't explain that properly sorry!
Original post by Super199
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?Amino acids.png

Sure, just give me a mo. Need to get on my laptop
Original post by Super199
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?Amino acids.png

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):
peptide bond.png
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 :smile:
Reply 18
Original post by Serine Soul
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):
peptide bond.png
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 :smile:

What happens to the hydrogen on alanine?
Original post by Super199
What happens to the hydrogen on alanine?


The one attached to the carbon group? Stays where it is:

A dipeptide looks like this, where R and R' are different side chains attached the central carbon atoms

dipeptide bond.png
Excuse my awful paint skills :redface:

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