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.
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.
Originally posted by UltimaOnline: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 4Decaffeinated 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 CO2My 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.
Originally posted by arxoz:Thanks for the Reply.
What do you mean by “Polar vs non-polar, isn’t a binary 1 or 0” ?
From now on, Arxoz, post your questions in this very thread, ok? Or if you prefer, you can always start a new thread on this forum, just label it as "[H2 Chemistry] - Organic Chem Qn" or whatever topic the question is on.
Are alcohols polar or non-polar. The correct answer is, they are partially polar and partially non-polar. The alkyl group of alcohols are non-polar, while the hydroxy group of alcohols are polar.
So it's not a binary black vs white or 1 (eg. polar) vs 0 (eg. non-polar), it is a relativity spectrum of "how polar is it? compared to..."
For instance, ethanol is more polar compared to propanol, but ethanol is less polar compared to methanol.