OCR Chemistry A F324 Rings, Polymers and Analysis Tue 19 June 2012

Oh nd i gave you a question before: it is what are the limitatins of gas chromotography

* Components may have similar retention times, peak shape and detector response.
* Unknown compounds have no retention times for comparison.
* Not all compounds will be separated and detected. Some could 'hide' beneath another substance.

Question:
Describe the use of GC-MS in forensics, environmental analysis, airport security and in space probes.

Reply 181

Robpattinsonxxx

11

Revision anyone? :smile:

Reply 182

Robpattinsonxxx

11

anyone wanna do some question and answer sessions? :smile:

Reply 183

otrivine

14

Original post by Robpattinsonxxx

anyone wanna do some question and answer sessions? :smile:

yes please :wink:

Reply 184

racheatworld

17

Original post by Robpattinsonxxx

anyone wanna do some question and answer sessions? :smile:

Explain how from benzene, you would synthesise an azo dye with phenol as the coupling agent. State all reagents and conditions and name mechanisms where appropriate.

Reply 185

Robpattinsonxxx

11

Original post by racheatworld

Explain how from benzene, you would synthesise an azo dye with phenol as the coupling agent. State all reagents and conditions and name mechanisms where appropriate.

So from NaNO2 and excess HCl you generate HNO2 (and NaCl). Add cold HNO2 to phenylamine to form the diazonium salt. The NH2 is replaced by the azo functional group and is positive. This occurs in situ and below 10 degrees Under alkaline conditions, add phenol to form the azo dye. One of the bonds of the azo group forms a bond with the 4th carbon on the phenol. Azo dye is brightly coloured.

Reply 186

LGrosvenor101

9

Original post by racheatworld

Explain how from benzene, you would synthesise an azo dye with phenol as the coupling agent. State all reagents and conditions and name mechanisms where appropriate.

Step 1: Add concentrated H2S04 and HN03 to Benzene, this forms C6H5N02. (electrophilic substitution)
Step 2: Need Tin (Sn) and concentrated HCl, this is to reduce the nitro group to form C6H5NH2.
Step 3: Add NaN02 and HCl (in situ?) <10 degrees to form the diazonium salt
Step 4: Add phenol (IN ALKALINE CONDITIONS!) also at < 10 degrees to form to azo dye! :smile:

Reply 187

racheatworld

17

Original post by LGrosvenor101

Step 1: Add concentrated H2S04 and HN03 to Benzene, this forms C6H5N02. (electrophilic substitution)
Step 2: Need Tin (Sn) and concentrated HCl, this is to reduce the nitro group to form C6H5NH2.
Step 3: Add NaN02 and HCl (in situ?) <10 degrees to form the diazonium salt
Step 4: Add phenol (IN ALKALINE CONDITIONS!) also at < 10 degrees to form to azo dye! :smile:

Perfect answer well done :biggrin:

Reply 188

kimmey

7

i was doing this past paper and i dont understand how why when in GC you have a alkene liquid as a mobile phase and you want to seperate an alkene, alcohol and ester, why is it the alkene takes the longest and the alcohol takes the fastest to pass through, how are we meant to know which things have the shortest retention times

Reply 189

Smiley Face :)

Do u think the grade boundaries will be as high as they were in jan 2012??

Reply 190

Robpattinsonxxx

11

What were they for an a in jan 2012?

Reply 191

racheatworld

17

Original post by kimmey

i was doing this past paper and i dont understand how why when in GC you have a alkene liquid as a mobile phase and you want to seperate an alkene, alcohol and ester, why is it the alkene takes the longest and the alcohol takes the fastest to pass through, how are we meant to know which things have the shortest retention times

The alkene takes the longest because it is non-polar like the mobile phase, and it will dissolve the most in the alkene mobile phase. The alcohol will be the fastest because the alcohol has a polar C-O bond and can form H bonds. The polar bond means that the alcohol does not dissolve as much in the non-polar alkene mobile phase. The esters have the 2nd longest retention times because although they have a polar C-O bond, they have a large hydrocarbon chain length which is non-polar and so this part of the ester will dissolve easily in the non-polar alkene mobile phase. However, it will be hard to distinguish the retention times between the 2 esters because they have a similar structure, therefore similar retention times.

Reply 192

Fatima0065

5

Hi does anyone know any good websites for sample answers on proton nmr and carbon 13 because i hate it :smile:

im always losing marks in past papers because of it...thankyou in advance 😊

Reply 193

tomkeys

2

Original post by Fatima0065

Hi does anyone know any good websites for sample answers on proton nmr and carbon 13 because i hate it :smile:

im always losing marks in past papers because of it...thankyou in advance 😊

I imagine looking at the marking points for the questions you've been doing would be quite helpful. Its not so much the structure determination part of the question that is the hard part, but the communication of your reasons for your choice of structure.

Reply 194

Fatima0065

5

Original post by tomkeys

I imagine looking at the marking points for the questions you've been doing would be quite helpful. Its not so much the structure determination part of the question that is the hard part, but the communication of your reasons for your choice of structure.

Thnks fro the tips :smile:

ill look at the markschems and see what im doing wrong :smile:

Reply 195

JMulvany

Original post by racheatworld

Perfect answer well done :biggrin:

what about the temperature in step one? should be 50 degrees C under reflux,
and step two is also under reflux

question was all conditions that is all

Reply 196

umair.khan

Hi everyone, erm how much marks do you need on this paper to get 28 UMS :biggrin:

Reply 197

Tateco

14

Original post by umair.khan

Hi everyone, erm how much marks do you need on this paper to get 28 UMS :biggrin:

Grade boundaries tend to come out after exams have taken place.

Reply 198

Is_Different_Good?

Original post by tomkeys

I imagine looking at the marking points for the questions you've been doing would be quite helpful. Its not so much the structure determination part of the question that is the hard part, but the communication of your reasons for your choice of structure.