OCR F324-June 2013 Chemistry

Can someone take a look at an interesting question 4b)ii) from Jan 12 F324

QP http://www.ocr.org.uk/Images/79471-question-paper-unit-f324-rings-polymers-and-analysis.pdf

MS http://www.ocr.org.uk/Images/61007-mark-scheme-unit-f324-rings-polymers-and-analysis-january.pdf

The question is about Gas chromatography, there's a liquid alkane stationary phase, a sample which has been contaminated with an alcohol and alkane, question is suggest how these would be seperated.

Perhaps someone can explain the answer below. Guessing: Is it because alcohols or more soluble componds turn to gas easier/quicker. or because since the alcohol has -OH groups it dissolves more in the carrier gas, travels faster. is it because the alkane and the liquid alkane stationary phase form van der waals forces between themselves which slows the alkane down, even though the alkane is a gas it can still form van der waals forces? you can see how confused i am. but i got 144/150 in jan f325. Please can everyone give there ideas about this question

Alcohol would have short retention time
AND
alkane would have long retention time

Transesterification is the process in which Biodiesel and Glycerol are formed from a Trigylceride and an alcohol such as Methanol or Ethanol when reacted with a H2SO4 catalyst.

I haven't done this paper so I just scrolled through quickly to see the question you were on about.

In chromatography in general you have a stationary phase and mobile phase. The sample is carried by the mobile phase, and it is slowed by the stationary phase.

In GC, the mobile phase is the inert gas H2 (g). This carries the components of the sample when they are in gaseous form. Therefore the more volatile a component is the shorter its retention time as it spends less time in the other form; dissolved into the stationary phase. Which in this case is a liquid supported on a solid. Ie the long chain alkane with a high boiling point coated (and therefore supported) on the inside of the capillary tube. When a component is not in its gaseous phase and moving through the column it is dissolved into the stationary phase due to its solubility.

Now, as the stationary phase is a long chain alkane, it cannot form hydrogen bonds, as it is not polar. This means -OH bonds are irrelevant in this case. It can only form VDW forces. The relevance of this is the principle; "Like dissolve like". A polar molecule such as H2O will dissolve another polar molecule such as an alcohol. A polar molecule such as H2O cannot dissolve a non polar molecule such as a long chain alkane.
Alternatively a long chain alkane cannot dissolve a polar molecule to the same extent it can dissolve a non polar molecule such as itself.

With this in mind we can say the alkane in the sample will have the strongest affinity for the stationary phase so have the shortest retention time as it will be slowed down the most/spend the most time dissolved into the stationary phase.

Next up lets look at the two esters, they are long molecules so their polar regions do not have as much significance as in the alcohol. These will have a retention time somewhere between the alcohol and the alkane. If you would like me to explain this more please ask.

Finally the alcohol is the most polar structure out of the 4 organic compounds, this causes it to have the least amount of interaction with the stationary phase so it will spend most of its time as a gas. Therefore it should have the shortest retention time.

This is my understanding of the question, if the MS says something different let me know. As I said I don't want to look at it as I need to do this paper to assess my ability and I'd rather not have cheated, thanks. :P

Transesterification is the process in which Biodiesel and Glycerol are formed from a Trigylceride and an alcohol such as Methanol or Ethanol when reacted with a H2SO4 catalyst.

I haven't done this paper so I just scrolled through quickly to see the question you were on about.

In chromatography in general you have a stationary phase and mobile phase. The sample is carried by the mobile phase, and it is slowed by the stationary phase.

In GC, the mobile phase is the inert gas H2 (g). This carries the components of the sample when they are in gaseous form. Therefore the more volatile a component is the shorter its retention time as it spends less time in the other form; dissolved into the stationary phase. Which in this case is a liquid supported on a solid. Ie the long chain alkane with a high boiling point coated (and therefore supported) on the inside of the capillary tube. When a component is not in its gaseous phase and moving through the column it is dissolved into the stationary phase due to its solubility.

Now, as the stationary phase is a long chain alkane, it cannot form hydrogen bonds, as it is not polar. This means -OH bonds are irrelevant in this case. It can only form VDW forces. The relevance of this is the principle; "Like dissolve like". A polar molecule such as H2O will dissolve another polar molecule such as an alcohol. A polar molecule such as H2O cannot dissolve a non polar molecule such as a long chain alkane.
Alternatively a long chain alkane cannot dissolve a polar molecule to the same extent it can dissolve a non polar molecule such as itself.

With this in mind we can say the alkane in the sample will have the strongest affinity for the stationary phase so have the shortest retention time as it will be slowed down the most/spend the most time dissolved into the stationary phase.

Next up lets look at the two esters, they are long molecules so their polar regions do not have as much significance as in the alcohol. These will have a retention time somewhere between the alcohol and the alkane. If you would like me to explain this more please ask.

Finally the alcohol is the most polar structure out of the 4 organic compounds, this causes it to have the least amount of interaction with the stationary phase so it will spend most of its time as a gas. Therefore it should have the shortest retention time.

This is my understanding of the question, if the MS says something different let me know. As I said I don't want to look at it as I need to do this paper to assess my ability and I'd rather not have cheated, thanks. :P

Great answer, thanks alot

good answer. Have some tasty and nutritious rep

hello, just wondering if anyone has the January 2013 past paper for f324 or knows where i can get it? thanks

If anyone's done the January 2013 paper, what did your think of it?

If anyone's done the January 2013 paper, what did your think of it?

do we need to know anything on the amide functional groups or the acyl chlorides? my teacher mentioned them but I haven't come across them in any book!

can someone please summarise the reactions we need to know?

thank you so much!

When amino acids undergo condensation polymerisation they form long chain molecules connected my amide (peptide) bonds. These molecules are called polyamides.
Don't worry about acyl chlorides I don't think they are on the spec. Acid anhydrides are you may want to practice them.


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does anyone have the 2013 paper ?

When describing phenol, is it okay to refer to the -OH as a hydroxyl group? Ie Phenol is a benzene molecule where one of the Carbon atoms is bonded to a hydroxyl group.

Yes, that's what the book says anyway

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Cool. I just did June 2011. 53/60, 44 was an A. Harshly marked. :P some tips:

When forming Azo dyes, if the temperature goes above 10 degrees Celsius when you have the diazonium ion; it reacts with water to form phenol.

C6H5N2+ + H2O ---> C6H5OH + N2 (g) + H+ (aq)

I wasn't sure so put H+ or 1/2 H2 (g). So didn't credit myself for the mark.

Artificial sugars that were used a few years ago have been removed as they had side effects.

Cosmetic products are tested before production to test if they cause side effects.

Condensation polymerisation is where monomers join together forming a polymer and a small molecule such as H2O.

I think that was about it.


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