Draw the structure of the organic product formed when alanine NH2-CH(CH3)-COOH reacts with CH3Cl in a 1:1 mole ratio.
The answer is CH3-NH-CH(CH3)-COOH, but I don't know how to get to it. A little help, please?!
I'm guessing the amine group acts as a nucleophile, so its similar to nucleophilic substitution with ammonia, for the mechanism; You draw an aroow from the lone pair on N atom to the C in CH3Cl, then an arrow from the C-Cl bond to the Cl. The N will have a positive charge, so you'd draw an arrow from an N-H bond to the N atom. Hope you sort of understood that
I'm guessing the amine group acts as a nucleophile, so its similar to nucleophilic substitution with ammonia, for the mechanism; You draw an aroow from the lone pair on N atom to the C in CH3Cl, then an arrow from the C-Cl bond to the Cl. The N will have a positive charge, so you'd draw an arrow from an N-H bond to the N atom. Hope you sort of understood that
Ooh, I see what you mean — thanks!
Did you not convert alanine to a zwitterion first, though? That's what I'm a bit confused about. I thought you had to draw the zwitterion out and 'apply the changes' subject to the change in its environment (e.g. low/high pH). If I did exactly what you said with the zwitterion I'd end up with NH2/COO- either side. Can you explain why you don't change alanine to its zwitterion first?
I'm guessing the amine group acts as a nucleophile, so its similar to nucleophilic substitution with ammonia, for the mechanism; You draw an aroow from the lone pair on N atom to the C in CH3Cl, then an arrow from the C-Cl bond to the Cl. The N will have a positive charge, so you'd draw an arrow from an N-H bond to the N atom. Hope you sort of understood that
Think this is how you do it? Not sure if it would be a neucleoplilic addition elimination though?
That answer is the one they're looking for... but their answer is wrong. A 1:1 ratio of primary amine and haloalkane will give you 0.25:0.75 quaternary amine salt:unreacted primary amine.
Also you need to consider the solvent used. As you are reacting with CH3Cl it will be in a non polar solvent, and hence the alanine will not exist as the zwitterion.
That answer is the one they're looking for... but their answer is wrong. A 1:1 ratio of primary amine and haloalkane will give you 0.25:0.75 quaternary amine salt:unreacted primary amine.
Also you need to consider the solvent used. As you are reacting with CH3Cl it will be in a non polar solvent, and hence the alanine will not exist as the zwitterion.
The answer I've provided is from the mark scheme? I've learnt amino acids exist as zwitterions in solution and in solid state, so I just assumed...
The answer I've provided is from the mark scheme? I've learnt amino acids exist as zwitterions in solution and in solid state, so I just assumed...
Well if they were mixed in aqueous solution there would be a few issues:
-CH3Cl not soluble in water -Zwitterion not a good nucleophile
Hence I presume that this reaction would be taking place in a non-polar solvent where the charges on the zwitterion would not be stabilised and hence the amino acid will exist in its non-zwitterion form, where the nitrogen is nucleophilic.
If the answer is in your mark scheme then learn that this is the case. Still doesn't make it correct though! Another example of the A level courses lying to you.