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Chemistry question help

I was wondering how you work out what the product of 2 amino 3 hydroxy propanoic acid and bromomethane will be? How do you work this out?

Also, for question 5ci.) on this paper, how do you work this out? https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Past-Papers/AQA-Old/Unit-4/June%202011%20QP%20-%20Unit%204%20AQA%20Chemistry%20A-level.pdf
As for your first question - you have a primary amine (R-NH2) reacting with a haloalkane (R-X), and this will produce a secondary amine (R2-NH).
Typically, these reactions are done using excess ethanolic amine, as well as under basic conditions i.e. using NaOH. This is to deprotonate the amine group (forming water), and also to get rid of any X- which will be lying around (forming an NaX salt).
Essentially, one of the hydrogens in the R-NH2 group will be replaced by the R group in R-X. This will turn R-NH2 into R2-NH.
Use the above reasoning to work out which product will be formed in your case.

As for your second question - this looks like it'd be an anhydride + alcohol →\rightarrow ester + carboxylic acid reaction.
However, there's a catch... The anhydride is cyclic, and the ester formed has two R groups.
The problem with this is that for one, you need to remember to include an excess of alcohol due to the two R groups, and for two, you will no longer produce a carboxylic acid. In this case, the alcohol would be ethanol.
What basically happens is:

•

A molecule of ethanol will lose OH- and join onto the anhydride group's central O atom. This breaks one of the C-O bonds, forming a new COO-CH2CH3 bond whilst leaving a C=O bond bare,

•

A second molecule will come along and lose H+, so that it can join onto the bare C=O group. This will form another COO-CH2CH3 bond. Now, the ester has been made,

•

The lost H+ and the OH- will combine to produce one mole of water.


Hopefully you can visualise what happens for that reaction, as it's hard to describe it using words and not diagrams :smile:
(edited 5 years ago)
Original post by Kian Stevens
As for your first question - you have a primary amine (R-NH2) reacting with a haloalkane (R-X), and this will produce a secondary amine (R2-NH).
Typically, these reactions are done using excess ethanolic amine, as well as under basic conditions i.e. using NaOH. This is to deprotonate the amine group (forming water), and also to get rid of any X- which will be lying around (forming an NaX salt).
Essentially, one of the hydrogens in the R-NH2 group will be replaced by the R group in R-X. This will turn R-NH2 into R2-NH.
Use the above reasoning to work out which product will be formed in your case.

As for your second question - this looks like it'd be an anhydride + alcohol →\rightarrow ester + carboxylic acid reaction.
However, there's a catch... The anhydride is cyclic, and the ester formed has two R groups.
The problem with this is that for one, you need to remember to include an excess of alcohol due to the two R groups, and for two, you will no longer produce a carboxylic acid. In this case, the alcohol would be ethanol.
What basically happens is:

•

A molecule of ethanol will lose OH- and join onto the anhydride group's central O atom. This breaks one of the C-O bonds, forming a new COO-CH2CH3 bond whilst leaving a C=O bond bare,

•

A second molecule will come along and lose H+, so that it can join onto the bare C=O group. This will form another COO-CH2CH3 bond. Now, the ester has been made,

•

The lost H+ and the OH- will combine to produce one mole of water.


Hopefully you can visualise what happens for that reaction, as it's hard to describe it using words and not diagrams :smile:

Thank you!

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