AQA A-level Chemistry paper 2 7405/2 Unofficial Markscheme

U mans know the drill by now, if ur answersmatch, thumbs up the post ygm:

- 9.07g (first question mass thing)
- 0.04 (t=0 tangent initial rate)
- 0.65
- 0.0281 mol^-1 dm^3 s^-1 (rate constant, k)
- 50.7 kjmol^-1 (the activation energy)
- 0.62 (rearrange kc and work out moles of suttin)
- 1.45x10^-4 (took an L on this question, correct answer is 145mg)
- 3 peaks in carbon 13 nmr
- 2 peaks in proton nmr
- There was a nucleophilic addition-elimination reaction somewhere and the product was N-ethylpropanamide

- the final 2 questions, basically for the first one, you gotta talk about inductive effect and benzene ring causing lone pair of electron on n of the ammonia, to become partially delocalised onto the ring, making it less available, the one where the ammonia was furthest away from the benzene ring with an alkyl group before it (ch2ch2) was the strognest, due to positive inductive effect. the weakest base was the one where the ch2ch2 was at the end, and ammonia was inbetween the benzene ring and the alkyl group, due to negative inductive effect, and lone pair ultimately fully delocalising onto the ring), the middle one was just the ammonia attached to the ring.
(I think it was FEG or suttin)

- for the final one though, for the intermediate you should get a chloro or bromo group attached to the ch2, and it is ultraviolet light + cl2 or br2
then for the second intermediate, you should have the cl or br replaced by a CN, and reagent = kcn, conditions is aqueous & ethanol as solvent)

then the final stage (dont know if u had to talk about it), but basically heat the second intermediate under reflux with conc hcl/sn, to form the final product.

ayt safe

(edited 6 years ago)

Reply 1

Aliff

8

That paper was a blessing! :biggrin:

Reply 2

tsr120

6

Original post by brumtown0121

U mans know the drill by now, if ur answersmatch, thumbs up the post ygm:

- 9.07g (first question mass thing)
- 0.04 (t=0 tangent initial rate)
- 0.65
- 0.0281
- 50.7 kjmol^-1 (the activation energy)
- 0.62
- 1.45x10^-4 (i know my answers wrong so holla the right one)
- 3 peaks in carbon 13 nmr
- 2 peaks in proton nmr
- the final 2 questions, basically for the first one, you gotta talk about inductive effect and benzene ring causing lone pair of electron on n of the ammonia, to become partially delocalised onto the ring, making it less available, the one where the ammonia was furthest away from the benzene ring with an alkyl group before it (ch2ch2) was the strognest, due to positive inductive effect. the weakest base was the one where the ch2ch2 was at the end, and ammonia was inbetween the benzene ring and the alkyl group, due to negative inductive effect, and lone pair ultimately fully delocalising onto the ring), the middle one was just the ammonia attached to the ring.
(I think it was FEG or suttin)

- for the final one though, for the intermediate you should get a chloro or bromo group attached to the ch2, and it is ultraviolet light + cl2 or br2
then for the second intermediate, you should have the cl or br replaced by a CN, and reagent = kcn, conditions is aqueous & ethanol as solvent)

then the final stage (dont know if u had to talk about it), but basically heat the second intermediate under reflux with conc hcl/sn, to form the final product.

ayt safe

Thanks for this!! I don't think hcl/sn is right though because that's only for reducing no2 on benzene ring. Should be h2/nickel or LiAlh4

Reply 3

RKM21

3

There was no question on the back page right?

Reply 4

Khanman123

16

Jeeez, I messed up that last question man. 145mg for that other one. I got 15.8 or suttin like that for Ea.

Reply 5

eza11

2

i got 0.074 for tangent but its dependent on the line u draw got like 0.38/ 5.0

Reply 6

Aliff

8

Original post by brumtown0121

U mans know the drill by now, if ur answersmatch, thumbs up the post ygm:

- 9.07g (first question mass thing)
- 0.04 (t=0 tangent initial rate)
- 0.65
- 0.0281
- 50.7 kjmol^-1 (the activation energy)
- 0.62
- 1.45x10^-4 (i know my answers wrong so holla the right one)
- 3 peaks in carbon 13 nmr
- 2 peaks in proton nmr
- the final 2 questions, basically for the first one, you gotta talk about inductive effect and benzene ring causing lone pair of electron on n of the ammonia, to become partially delocalised onto the ring, making it less available, the one where the ammonia was furthest away from the benzene ring with an alkyl group before it (ch2ch2) was the strognest, due to positive inductive effect. the weakest base was the one where the ch2ch2 was at the end, and ammonia was inbetween the benzene ring and the alkyl group, due to negative inductive effect, and lone pair ultimately fully delocalising onto the ring), the middle one was just the ammonia attached to the ring.
(I think it was FEG or suttin)

- for the final one though, for the intermediate you should get a chloro or bromo group attached to the ch2, and it is ultraviolet light + cl2 or br2
then for the second intermediate, you should have the cl or br replaced by a CN, and reagent = kcn, conditions is aqueous & ethanol as solvent)

then the final stage (dont know if u had to talk about it), but basically heat the second intermediate under reflux with conc hcl/sn, to form the final product.

ayt safe

The last step in the final qs is wrong, because that step is to convert nitrobenze to phenylamine. You should have instead react it with Ni and H2 gas or LiAlH3 in dry ether.

Reply 7

Swissblade

8

Used H₂/Ni catalyst for the last one.

Anyone else get this?

6 carbons, 4 ¹³C peaks and peaks at C=O and acid O-H.

(edited 6 years ago)

Reply 8

Khanman123

16

I got EGF for the benzene ring bit

Reply 9

CaiusMartius

12

Original post by brumtown0121

U mans know the drill by now, if ur answersmatch, thumbs up the post ygm:

- 9.07g (first question mass thing)
- 0.04 (t=0 tangent initial rate)
- 0.65
- 0.0281 mol^-1 dm^3 s^-1 (rate constant, k)
- 50.7 kjmol^-1 (the activation energy)
- 0.62
- 1.45x10^-4 (i know my answers wrong so holla the right one)
- 3 peaks in carbon 13 nmr
- 2 peaks in proton nmr
- the final 2 questions, basically for the first one, you gotta talk about inductive effect and benzene ring causing lone pair of electron on n of the ammonia, to become partially delocalised onto the ring, making it less available, the one where the ammonia was furthest away from the benzene ring with an alkyl group before it (ch2ch2) was the strognest, due to positive inductive effect. the weakest base was the one where the ch2ch2 was at the end, and ammonia was inbetween the benzene ring and the alkyl group, due to negative inductive effect, and lone pair ultimately fully delocalising onto the ring), the middle one was just the ammonia attached to the ring.
(I think it was FEG or suttin)

- for the final one though, for the intermediate you should get a chloro or bromo group attached to the ch2, and it is ultraviolet light + cl2 or br2
then for the second intermediate, you should have the cl or br replaced by a CN, and reagent = kcn, conditions is aqueous & ethanol as solvent)

then the final stage (dont know if u had to talk about it), but basically heat the second intermediate under reflux with conc hcl/sn, to form the final product.

ayt safe

I got the same as you for the base ammonia one, I can't tell if everyone has got the same or different tbh looool

Reply 10

sername

3

Number of moles = 0.789 (maybe for the Kc question?)
rate constant k = 0.028
H2/Ni Catalyst or LiAlH4
The methylbenze should be in excess (in order for 1 substitution only)
I got EGF (not sure)
To distinguish between KLMN, add tollen's reagent to get N (only one with silver mirror), NaHCO3 to the remaining KLM to get the one without COOH (L I think, only one WITHOUT effervesence) to KM add K2Cr2O7/c.H2SO4 to identify K (only one with orange to green) and therefore M (by elimination)
145 mg for P (I think)

Correct me if I'm wrong

(edited 6 years ago)

Reply 11

Hassankhan2405

2

Did anyone get 870Mg for one of the questions? The Ideal Gas Equation?