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

    http://filestore.aqa.org.uk/subjects...4-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!
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    (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!
    you made a mistake with part 2. the hydrogen falls of the COOH functional group and the Sodium attaches on instead.
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    (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!
    How come it reacts with COOH?
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    (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?
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    (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.
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    (Original post by Super199)
    How come it reacts with COOH?
    Think acid and base
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    (Original post by Serine Soul)
    Think acid and base
    precisely, also lets not forget that amino acids can exist as zwitterions
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    (Original post by 713Wave)
    precisely, also lets not forget that amino acids can exist as zwitterions
    :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 )
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    (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-?
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    (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 )
    yes, as far as I know the COOH group donates a H+ ion to the amine group :cool:.
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    (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+
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    (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+
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    in addition to part one, you have write NH3+ for BOTH amine groups on lysine.
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    (Original post by Serine Soul)
    You essentially get [amino acid]+ Cl-

    You write the Cl- near to the NH3+
    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?Name:  Amino acids.png
Views: 49
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    (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!
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    (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?Name:  Amino acids.png
Views: 49
Size:  4.8 KB
    Sure, just give me a mo. Need to get on my laptop
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    (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?Name:  Amino acids.png
Views: 49
Size:  4.8 KB
    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):
    Name:  peptide bond.png
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    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
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    (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):
    Name:  peptide bond.png
Views: 48
Size:  3.2 KB
    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
    What happens to the hydrogen on alanine?
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    (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

    Name:  dipeptide bond.png
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    Excuse my awful paint skills
 
 
 
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