OCR B Salters - F335 Exam - 15 June 2011

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Are the reaction conditions for diazotisation < 5°C, NaNO2, HCl and phenylamine?

Then this can couple with phenylamine or phenol which is electrophillic substitution?

Reply 201

I'mBadAtMaths

Original post by JackWoot

Are the reaction conditions for diazotisation < 5°C, NaNO2, HCl and phenylamine?

Then this can couple with phenylamine or phenol which is electrophillic substitution?

Diazotisation is with dilute HCl.

Do you know the conditions for the coupling reraction?

Reply 202

ABCDemily

Original post by I'mBadAtMaths

Diazotisation is with dilute HCl.

Do you know the conditions for the coupling reraction?

Kept below 10C to prevent a phenol forming. Also, the reagents have to be created in situ as they are very volatile and would otherwise bugger off and now work.

Also, when you couple the diazonium salt with the phenol, has to be done over iceee.

Reply 203

RipperRoo

Original post by ABCDemily

Is it not also < 5C? I remember that being in a mark scheme. Also it said alkaline conditions. However I can't find that in the revision book.

Reply 204

pearlover

Original post by pav

won't a good colourfast dye be the one that forms covalent bonds? ( since that is the strongest bond )

yep they're fibre reactive dyes which are the most permanent type of dye as they form strong covalent bonds.. the one i said before also has a strong colourfast bt yea the covalent bond one will be stronger so an even better colourfast than that

Reply 205

pearlover

when arenes undergo electrophilic substitution for example the nitronium ions, the process in the cgp book is explained in three simple steps:
-nitronium ion attacks benzene
-an unstable intermediate is formed
-H+ ion is lost

how can I elaborate on that to get a better answer that if needed to can get me more marks? cz thats abit too simple to write the exam right?

Reply 206

RipperRoo

Original post by pearlover

nitronium ion attacks electron rich benzene, and the H+ ion is lost in order to regain the delocalised system and restore it's original stability.

Reply 207

pav

can anyone exaplian why CH2(COOH)-C(OH)(COOH)-CH2(COOH) has proton environments in the ration 1:4:3. is there a symmetry here?

Also, if the question says that 15g of ethanol produced 10.6 of pure ethanal. what is the percentage yield of ethanal? why is the answer 73.9%??

Reply 208

RipperRoo

Original post by pav

can anyone exaplian why CH2(COOH)-C(OH)(COOH)-CH2(COOH) has proton environments in the ration 1:4:3. is there a symmetry here?

Also, if the question says that 15g of ethanol produced 10.6g of pure ethanal. what is the percentage yield of ethanal? why is the answer 73.9%??

Well I can answer the second Q, CH3CH2OH ----> CH3CHO + H2

So 1 mole of ethanol makes 1 mole of ethanal because of the stoichiometry.

moles of ethanol = 15/46 = 0.326

0.326 * 44 (Mr of ethanal) = 14.344g

10.6/14.344 * 100 = 73.9%

Hope that makes sense

(edited 12 years ago)

Reply 209

zahre

Original post by pav

For the percentage yield question the mr of ehanol is 46 and the mr of ethanal is 44. The question says that 15g of ETHANOL produced 10.6g of ETHANAL. so you divide the mr of ethanal by that of ethanol and times by 15. So:

44/46*15=14.34782609g this is the maximum amount of ethanal that can be produced from 15g of ethanol. So you then do 10.6/14.34782609*100=73.9% to 1dp.
:biggrin:

Hope it helps.

Reply 210

zahre

Original post by pav

For the nmr one, I would have thought there are 4 different proton environments in the ratio 1:4:2:1 :s-smilie:

Reply 211

MyJunkIsYou

Do we need to know about calculating solubilities, I know Kow isn't in the specification, are the solubility questions related to that so they are also no longer included?

Reply 212

zahre

Original post by MyJunkIsYou

Do we need to know about calculating solubilities, I know Kow isn't in the specification, are the solubility questions related to that so they are also no longer included?

No we dont need to know about calculating solubilities. Its not in the spec anymore.

Reply 213

beckyxo

Right at the beginning of AI in the revision book (the salters one not the cgp one) it says
'you need to be aware of risks/benefits of a given chemical process in terms of hazards associated with raw materials, reactants, prodcuts and by-products, explosions, acidic gases, flammable gases and toxic emissions (where appropriate).'
I don't think this has come up much before has it? I've done most of the past papers now and still dont really know what this is referring to in the exam :/
however i just got a motivation email from my teacher which made me smile :smile:

simply said:
Becky
Keep up the good work
Alister

Reply 214

RipperRoo

Original post by beckyxo

simply said:
Becky
Keep up the good work
Alister

Well I remember one question being "What is a safety hazard about this reaction". It produced nitric acid but you had to say it was 'corrosive/toxic' to get a mark.

Reply 215

beckyxo

Original post by JackWoot

Well I remember one question being "What is a safety hazard about this reaction". It produced nitric acid but you had to say it was 'corrosive/toxic' to get a mark.

oh, well thats easier than I thought haha

Reply 216

pav

Original post by JackWoot

your brilliant!..Thanks a lot

Reply 217

pav

Original post by zahre

For the nmr one, I would have thought there are 4 different proton environments in the ratio 1:4:2:1 :s-smilie:

that's what i thought too but apprently it's not. this question is from chemical ideas so im hoping it's a lot harder then what we would get in the exam, but there is no guarantee :frown: