CHEM2403 - biomolecules Prac report help - and yes i know uni has finished! (1 Viewer)

[shes_jinxed]

CHEM2403 - biomolecules HELP! with Prac report - and yes i know uni has finished!

Hey guys,
Long story short there was a mess up with one of my prac reports blah blah basically i have till this friday to hand in the report for prac 4 (the two part one with the enzyme and optical rotations jog any memories) anyway i was wondering if you could help me with the questions at the end. I passed theory and the exam but they will fail me if i dont get a good mark in this prac..........i hate the chem department. I love u long time for your help lol

I realise these questions are probably really easy, but because I couldnt do the second half of the lab I have no idea what they are going on about

Q1) If a smample has a measured specific optical rotation of +12, and the pure (R)-isomer has a sepcific rotation of -48, calculate the percentage of (R)- and (S)-enantiomers in the sample.

Ok i really have no idea about this one
%ee (optical purity) =( [a]D sample / [a]D pure enantiomer ) x100

note: [a]D = i think specific rotation
So i would guess %ee = ( 12/-48) x 100
= -25%
but that doesnt seem right and im not sure how thats going to give me the % of R and S im also given

%ee = { [(R) - (S)] / [(R) + (S) ] } x 100

i know this should be easy but i have no idea any help would be great

Q2. When ninhydrin reacts with proline. it gives a yello dye, rather than a blue/violet dye. Suggest why it acts differently to other aminoacids.
Its just that the side chain is cylic so it locks up the back bone of proline which makes it a unique amino acid therefore reacts differently to ninhydrin. Right?

Q3. The esters of amino acids are commerically sold as the hydrochloride salts. Why? (shint: identify the electrophilic and nucleophilic sites present in the hydrochloride salt and in the free amine)
Im guessing its because without the it as the hydrochloride salt that is
Cl-N3H+ as one R group stops the oxygen on the other R group: CO2CH3 donating its electrons to the N but why is this helpful

Q4. How would you convert N-acetyl-(s)-phenylalanine to pure (s)-phenylalanine?

I know everyone is in holiday mode, but it would be amazing if anyone could help.
Thanks a bunch and happy holidays

 

[tennille]

Re: CHEM2403 - biomolecules HELP! with Prac report - and yes i know uni has finished!

The reaction of ninhydrin with proline is "incomplete" because it is a secondary amine (not like the other amino acids with are primary). That's all I remember putting down.

For Q1, you work out your %ee by ([a]d sample/[a]d pure enantiomer) *100. Once you work that out, you can work out what (R-S)/(R+S) is (since that's equal to %ee). R+S is always 100, so then you have the answer for R-S. So since you know what R+S and R-S is, you can use simultaneous equations to work out how much there is of each. I think that's how you do it. I don't have my report on me.

I can't remember the answers for the rest. I hope that helps.

 

[Tabanacle]

Re: CHEM2403 - biomolecules HELP! with Prac report - and yes i know uni has finished!

i hope this is right for q1, but basically if the (R) enantiomer has an optical light rotation of -48 then the (S) enantiomer has an optical rotation of +48 (sorry to state the obvious but i'm trying to explain myself here). If the rotation was 0, the mixture would be a racemic (50-50) mixture of (R) & (S). Since the rotation is +12 there is clearly more S than R. What I did (not sure if this is right, i didn't actually do this subject) is instead of having a scale of -48 to +48, is the add 48 to everything and have a scale of 0 to 96, 0 corresponding to pure R, 96 to pure S. This means the optical rotation is now 12+48=60.

The amount of S is then simply 60/96*100 = 62.5%

This makes sense because optical rotation of R + optical rotation of S = (1-.625)*-48 + .625*48 = 12 as required

 

[tennille]

Re: CHEM2403 - biomolecules HELP! with Prac report - and yes i know uni has finished!

That would mean that R is 37.5%. But if that is the case, if you use %ee (enantiomeric excess) = (R-S)/(R+S), you'll end up with a negative percentage. I don't know anymore.

 

[Tabanacle]

Re: CHEM2403 - biomolecules HELP! with Prac report - and yes i know uni has finished!

again i haven't done the subject, but the %ee would be 25% i.e (S-R)/(S+R) since S is in excess

 

[shes_jinxed]

Re: CHEM2403 - biomolecules HELP! with Prac report - and yes i know uni has finished!

i hope this is right for q1, but basically if the (R) enantiomer has an optical light rotation of -48 then the (S) enantiomer has an optical rotation of +48 (sorry to state the obvious but i'm trying to explain myself here). If the rotation was 0, the mixture would be a racemic (50-50) mixture of (R) & (S). Since the rotation is +12 there is clearly more S than R. What I did (not sure if this is right, i didn't actually do this subject) is instead of having a scale of -48 to +48, is the add 48 to everything and have a scale of 0 to 96, 0 corresponding to pure R, 96 to pure S. This means the optical rotation is now 12+48=60.

The amount of S is then simply 60/96*100 = 62.5%

This makes sense because optical rotation of R + optical rotation of S = (1-.625)*-48 + .625*48 = 12 as required

thanks so much both you!!!!!!!! i checked with another friend of mine and they got the same answers 62.5 and 37.5
thanks for all your help happy holidays!