2003/Paper 2/Qn 4
Liquid CO2 is not cheap. It is expensive to keep it in the
liquid state. The correct reason is that it is the most volatile
and therefore most easily separated after the extraction is
complete. An additional reason is that it is non-toxic, compared to
benzene.
Polar vs non-polar, isn't a binary 1 or 0. It is a spectrum.
Caffeine is partially polar, partially non-polar, as is carbon
dioxide. CO2 may be overall non-polar as it's dipoles cancel out,
but the central C is permanently partial +ve and the O atoms are
permanently partial -ve, and therefore can have dipole-dipole
interactions with the polar groups of caffeine.
2005/Paper 2/Qn 5
COOH groups also react with PCl5, not just OH groups.
Y is oxidized by K2Cr2O7, and therefore Y is either a primary or
secondary alcohol, and Z is either a carboxylic acid or ketone.
Z has a carboxylic acid COOH group, instead of an alcoholic OH
group, since it is an oxidation product of Y.
Y is 2-hydroxypropanoic acid, exists as optical isomers.
Z is 2-oxopropanoic acid.
arxoz posted :
Didn’t know where to post my chem qns.
I am a private As student so I don’t have the privilege of clearing
my doubts with a teacher so banking all my hopes on receiving gd
solutions.
Both Qns are from A level papers.
2003/Paper 2/Qn 4
Decaffeinated coffee and tea are made by extracting the caffeine
from solid coffee and tea using a solvent.
Suggest 2 reasons which of the following industrial solvents would
be the most suitable
-benzene
-a hydrocarbon,such as cyclohexane
-liquid CO2
My Answer- Liquid CO2. Readily available therefore
cheap. -Organic compounds may contaminate the coffee or tea.
Are my answers right? How can the solvents given (all non-polar
with weak id-id) can dissolve the caffeine (polar).
2005/Paper 2/Qn 5
An organic compound Y is found in the urine of patients
suffering from diabetes. Compound Y can be converted into compound
Z. The following tests are carried out in the order given. In each
case, state all deductions.
(i)With aq alkaline iodine, Y gives a yellow ppt.
My answer- Y is either a primary/secondary alcohol with a methyl
sidechain or a carbonyl compound with a methyl sidechain.
(ii)Y gives white fumes when treated with PCl5.
My answer-Y has a hydroxyl functional group.
(iii)On heating under reflux with acidified K2Cr2O7, Y gives a
green-blue solution from which an organic compound Z may be
obtained.
My answer- Z may be either alcohol or ketone. It may also be
ester.
(iv)Compound Z is a carbonyl compound. Therefore Y is a carbonyl
compound with methyl sidechain and also an alcohol with methyl
sidechain.
(v) On treatment with PCL5, Z gives white fumes .
Compound Z has hydroxyl functional group. Compound Y has also
tertiary alcohol groups.
From part b of the qn, formula of Y can be deduced. =C6H12O6
.
What is the displayed formula of Y and Z.
My answer- Y=CH3CHC(OH)2C(OH)2COCH3
For Y, I can also come up with alternative answers which I think
is not possible.
Are my deductions right?
Thanks In Advance.