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AQA A Level Chemistry Unofficial Markscheme Paper 2 11th June 2019

2.57x10^9 s^-1 for Arhennius
Order of P was 1 and Q was 2
182kg of reactant to produce paracetamol
19991 mol^-3dm^9s^-1 for rate constant
1.39g of aromatic compound reacted?
25.8% percentage yield of cyclohexane

Test between secondary and tertiary alcohol - acidified potassium dichromate turns from orange to green w/ secondary and no change with tertiary
CO2 and emissions a) C=O bond lies in different wavelength range
b)Talk about CO2 having a large IR absorption (shown on the diagram) and not having a peak where the carbonyl group is

ethyl?amine is a better base than ammonia because the alkyl group donates electron density pushing out the lone pair on ammonia. Therefore has a greater ability to accept a proton

The first equation was you'd add 4NH3 and form 2NH4Br alongside the diamine

NaBH4 doesn't attack the carbon carbon bond as it is not polar

Golf Ball not biodegradable as it does not contain polar bonds

Molecular formula was C5H8
You got 1 & 1.6 which you had to times by 5 to get the whole number

Reagent and Conditions
Ethanolic KOH
H2 and Nickel Catalyst/LiAlH4 in a dry ether (ethoxyethane)

Draw First 2 curly arrows
1) Hydride ion attacking the carbon
2)Double bond breaking

Tollens Reagent forms silver mirror
Fehlings forms brick red precipitate
(edited 4 years ago)

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this is question 1 how do you get the answer
question.PNG8.9KB
Reply 2
Avatar for Nopefu
Nopefu
2
change the title to AQA chemistry
agree with the first 4 answers but don't remember the others, think i got 43% for yeild
Original post by justaname2001
2.57x10^9 s^-1 for Arhennius
Order of P was 1 and Q was 2
182kg of reactant to produce paracetamol
19991 mol^-3dm^9s^-1 for rate constant
1.39g of aromatic compound reacted?
25.8% percentage yield of cyclohexane

Test between secondary and tertiary alcohol - acidified potassium dichromate turns from orange to green w/ secondary and no change with tertiary
CO2 and emissions a) C=O bond lies in different wavelength range
b)Talk about CO2 having a large IR absorption (shown on the diagram) and not having a peak where the carbonyl group is

ethyl?amine is a better base than ammonia because the alkyl group donates electron density pushing out the lone pair on ammonia. Therefore has a greater ability to accept a proton
Reply 4
Avatar for habib.r2
habib.r2
7
Order was 1 and 0
Original post by Manningggggggggg
agree with the first 4 answers but don't remember the others, think i got 43% for yeild
Reply 5
Avatar for habib.r2
habib.r2
7
yes alot ppp are saying 20000 instead of 19991 but it was 19991 as it didny ask for most accurate significant figure as it did with other questions....
Original post by justaname2001
2.57x10^9 s^-1 for Arhennius
Order of P was 1 and Q was 2
182kg of reactant to produce paracetamol
19991 mol^-3dm^9s^-1 for rate constant
1.39g of aromatic compound reacted?
25.8% percentage yield of cyclohexane

Test between secondary and tertiary alcohol - acidified potassium dichromate turns from orange to green w/ secondary and no change with tertiary
CO2 and emissions a) C=O bond lies in different wavelength range
b)Talk about CO2 having a large IR absorption (shown on the diagram) and not having a peak where the carbonyl group is

ethyl?amine is a better base than ammonia because the alkyl group donates electron density pushing out the lone pair on ammonia. Therefore has a greater ability to accept a proton
Generally good practice to give an answer to reasonable sf. If your data is to 3sf, you can't give an answer to 5sf. Would expect them to take a mark off if you didn't write 20000.
yeh thats what i got.
Original post by habib.r2
Order was 1 and 0
Reply 8
Avatar for Spiket
Spiket
4
Can anyone remember if the final mechanism only asked for the first step?
yikes... well 1 mark at least

Original post by habib.r2
Order was 1 and 0
The order was definitely 1 and 2. My teacher has already been through the paper
Original post by Manningggggggggg
yikes... well 1 mark at least
Phew, thought 2.57 billion was a bit odd.. do we all agree on arhenius ?
Original post by justaname2001
2.57x10^9 s^-1 for Arhennius
Order of P was 1 and Q was 2
182kg of reactant to produce paracetamol
19991 mol^-3dm^9s^-1 for rate constant
1.39g of aromatic compound reacted?
25.8% percentage yield of cyclohexane

Test between secondary and tertiary alcohol - acidified potassium dichromate turns from orange to green w/ secondary and no change with tertiary
CO2 and emissions a) C=O bond lies in different wavelength range
b)Talk about CO2 having a large IR absorption (shown on the diagram) and not having a peak where the carbonyl group is

ethyl?amine is a better base than ammonia because the alkyl group donates electron density pushing out the lone pair on ammonia. Therefore has a greater ability to accept a proton
dibromhexane +4nh3 = diaminohexane +2NH4Br
1st Order and 0 Order
Why Amino acid seperates from Aqueous layer= i put it has different density considering it being higher density
25.8% for Yield
Tollens reagent for last question = Silver mirror
Rate Constant was 19991 mol^-3dm^-9s-1
2Bromoetc + 2Nh3= 2aminoetc + 2HBr
Step 1 was KCN aqueous Methanol
Step 2 was Ni +H2 and heat
I got summat like 178Kg
1.39g
C3H5 somehow for molecular formula
Solution goes from Cloudy to clear
i put h20 for hydrolysis as using acid or alkali would Go over Amino acids isoelectric point ??
For working out “A” i cant remember but you needed to DeLog the answer to get value for“A”

Original post by justaname2001
2.57x10^9 s^-1 for Arhennius
Order of P was 1 and Q was 2
182kg of reactant to produce paracetamol
19991 mol^-3dm^9s^-1 for rate constant
1.39g of aromatic compound reacted?
25.8% percentage yield of cyclohexane

Test between secondary and tertiary alcohol - acidified potassium dichromate turns from orange to green w/ secondary and no change with tertiary
CO2 and emissions a) C=O bond lies in different wavelength range
b)Talk about CO2 having a large IR absorption (shown on the diagram) and not having a peak where the carbonyl group is

ethyl?amine is a better base than ammonia because the alkyl group donates electron density pushing out the lone pair on ammonia. Therefore has a greater ability to accept a proton
(edited 4 years ago)
Original post by Spiket
Can anyone remember if the final mechanism only asked for the first step?


yeh just draw two curly arrows
Original post by Spiket
Can anyone remember if the final mechanism only asked for the first step?


I think so - I put 2 curly arrows
Original post by bob43212
dibromhexane +4nh3 = diaminohexane +2NH4Br

so i got a mark for this
this.PNG2.8KB
Yep the Q said 'draw two curly arrows'
Original post by ASchem4
yeh just draw two curly arrows
yeah it did, ah bye i did the whole mechanism. will i lose marks ?
Original post by Spiket
Can anyone remember if the final mechanism only asked for the first step?
shouldnt do
Original post by nessa.enthalpy
yeah it did, ah bye i did the whole mechanism. will i lose marks ?

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