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# AQA CHEM4 discussion/unofficial mark scheme Watch

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1. Does anyone know how to do this?

Metal ions Q2+ in acidified aqueous solution can be oxidised by aqueous potassiumdichromate(VI).In a titration, an acidified 25.0 cm3 sample of a 0.140 mol dm–3 solution of Q2+(aq)required 29.2 cm3 of a 0.040 mol dm–3 solution of potassium dichromate(VI) forcomplete reaction.Determine the oxidation state of the metal Q after reaction with the potassiumdichromate(VI)

This is the solution, it loses me at multiplying it by 6

moles of dichromate = (29.2/1000)×0.04 = 0.001168 or 0.00117 (1) moles of Q2+ = (25/1000)×0.140 = 0.0035(0) (1) each mole of dichromate needs 6 electrons or half equation with 6 e- (1) moles of electrons = 6×0.001168 = 0.007008 or moles Q2+:moles (1)dichromate = 3:1Moles of electrons per mole of Q = 0.007008/0.0035 = 2.002 = 2 (gets previous (1)mark also)Q(IV) or Q4+

2. I have no idea if this is a valid method to obtain the answer but it's how I interpret that question :/

It's a redox rxn, ratio of Q2+ to dichromate is 3:1

Q2+ Must have lost 2 electrons so final state is Q4+

Open to corrections or further discussion
(Original post by hopingmedicinae)
Does anyone know how to do this?

Metal ions Q2+ in acidified aqueous solution can be oxidised by aqueous potassiumdichromate(VI).In a titration, an acidified 25.0 cm3 sample of a 0.140 mol dm–3 solution of Q2+(aq)required 29.2 cm3 of a 0.040 mol dm–3 solution of potassium dichromate(VI) forcomplete reaction.Determine the oxidation state of the metal Q after reaction with the potassiumdichromate(VI)

This is the solution, it loses me at multiplying it by 6

moles of dichromate = (29.2/1000)×0.04 = 0.001168 or 0.00117 (1) moles of Q2+ = (25/1000)×0.140 = 0.0035(0) (1) each mole of dichromate needs 6 electrons or half equation with 6 e- (1) moles of electrons = 6×0.001168 = 0.007008 or moles Q2+:moles (1)dichromate = 3:1Moles of electrons per mole of Q = 0.007008/0.0035 = 2.002 = 2 (gets previous (1)mark also)Q(IV) or Q4+
3. (Original post by hopingmedicinae)
Does anyone know how to do this?

Metal ions Q2+ in acidified aqueous solution can be oxidised by aqueous potassiumdichromate(VI).In a titration, an acidified 25.0 cm3 sample of a 0.140 mol dm–3 solution of Q2+(aq)required 29.2 cm3 of a 0.040 mol dm–3 solution of potassium dichromate(VI) forcomplete reaction.Determine the oxidation state of the metal Q after reaction with the potassiumdichromate(VI)

This is the solution, it loses me at multiplying it by 6

moles of dichromate = (29.2/1000)×0.04 = 0.001168 or 0.00117 (1) moles of Q2+ = (25/1000)×0.140 = 0.0035(0) (1) each mole of dichromate needs 6 electrons or half equation with 6 e- (1) moles of electrons = 6×0.001168 = 0.007008 or moles Q2+:moles (1)dichromate = 3:1Moles of electrons per mole of Q = 0.007008/0.0035 = 2.002 = 2 (gets previous (1)mark also)Q(IV) or Q4+

The reason you multiply by 6 is because the dichromate is being reduced
6H+ + 6e- + Cr2O7 2- -----> 2CrO4- + 3H2O this gives the moles of dichromate
Q2+ is oxidised. Prior to oxidation with manganate it is
Q----> q2+ + 2e-
Ratio is 3:1
4. for the Q on amino acids H bonding with one another could you have done NH group to the O on the OH group or did it have to be two NH's?
5. (Original post by Suits101)
1. I don't know - their general principle is right + wrong = wrong
2. It should be 112.0 but in a past paper which asked the same but with a different compound they allowed two answers (hence 114.0)
3. I don't know - but on all of my revision sheets dilute is a must, I'm pretty sure they would only allow heat under reflux
4. I don't know - but I and many others mentioned biodegradable, normally in markschemes they just want the science so you should be fine
Thanks so much
6. (Original post by logicmaryam)
Heey guys, I think we can all agree this paper was one of the hardest CHEM4 papers ever.

Predicted Grade boundaries by my friend
90 RAW for full UMS maybe
A* 82
A 74
B 66
C 58

Q1 20 marks
a) Ethanoic acid in water
CH3COOH ⇌ CH3COO- + H+ (1)
b) Ethanoic acid as base with HNO3
HNO3 + CH3COOH = CH3CHOOH2+ + NO3- (1)
c) pH of acid? 2.08 (3)d) pH when alkali added to acid 2.32 (4)
e) [salt]/[acid] = 3.33 (3) or 1.20?

For the Kc - 9.09x10^3?
k= 3.12x10-5 mol-1 dm3 s-1

Q8
Condition for step 1 (3)NaOH heat under reflux?
use distillation or filtration to separate

test for acyl chloride Q and some chlorohaloalkane P
acidified AgNO3
P - NVC
Q - white ppt of AgCl Some people put H2O which would give fumes of HCl.

Lidocaine 9 peaks is a or b more likely to be protonated, why?
B because ethyl group is electron releasing so the lone electron pair on N is more available because benzene ring is withdrawing the pair at A.

percentage mass of H = 9.4% because 22/233
why isn’t Ka used in tables for strong acids? (2)
Strong acids fully dissociates
Ka = [H+]??
Relationship between stereoisomers (1 or 2)
mirror images and non-superimposable?

Rate equation when propanone or something is increased
rate = [H+]
because propanone is in excess, H+ is limiting factor.

Why is “X” zero order?
not involved in the RDS and rate equation

with knowledge of carbonyls, deduce name of mechanism
haemacetol

2-Chloroethanoylchloride

some other compound was 2-amino-4-methylpentanoic acid

why is lidocaine hydrochloride used in medicine?
More soluble as it forms stronger hydrogen bonds? the other one was polar too but lidocaine had N-H+

Organic synthesis question (4)?
KCN aqueous and ethanolHydrogen and Ni catalyst high pressure or LiAlH4 in ether

orders of reaction (2)
first, 1
second, 2

Why are “X” sutures used in medicine instead of polythene one? (2 or 3)
Because X has polar bonds so is susceptible to attack by nucleophiles such as H2O in body so no need to remove it as it will break down over time. other one was non-polar

Benzene one was just acylation
electrophile CH3CO+
[ClCH2COCl]+* -> [COCl+](37Cl) + [ClCH2*](35 Cl)

Q9
C6H1402

61% Carbon
11.9% Hydrogen
27.1% Oxygen (100-11.9-61)

my method:
C : H : O
0.61/12 0.119/1 0.271/16
Then divide each answer by the answer of O to get C3H7O1 = Emperical. so X2 to get C6H14O2

potassium dichromate turned green so must be primary or secondary HO-CH2-C(CH3)2-O-CH3??

This one is by bluhbluhbluh
Equation representing ethanoic acid as a base with nitric acid:

CH3COOH + HNO3 -> CH3CO+NO3- + H2O

Identify what reagents are needed to distinguish between the two chlorine molecules, incl. observation. (Acyl chloride vs haloalkane (?))
Reagent: water/AgNO3 (?)/named carbonate
Observation: HCl fumes/White ppt./Effervescence -for the acyl chloride.

Which Nitrogen acts as a stronger base a vs. b (where a is attached to the benzene, and b is a tertiary amine) and why?
Nitrogen b due to the positive inductive effect of the carbons on the nitrogen, therefore making the lone pairs more available. Consequently nitrogen a is joined to a benzene so the lone pair delocalises into the benzene ring.

Order of reactions:
D 1
E 2

What is a racemic mixture?
Mixture containing equal quantities of each enantiomer.

How to distinguish between the two:
Expose to plane polarised light, rotates in opposite directions

Chiral Carbon relationship:
They are both mirror images of each other/enantiomers.

4.
Electrophile structure: CH3C(+)O

Reaction mechanism:
Arrow from benzene ring to C+ in CH3C(+)O, then arrow from hydrogen attached to the same carbon as the CH3CO back into the incomplete benzene ring (from carbon 1-4 i believe).

5. Can't remember exact questions, let me know i'll update ! =)
Notes:
Zwitterions have R-N(+)H3 and a COO-
Hydrogen bond between two amino acids: presumably show any partial bond from one hydrogen on one amino acid to either the nitrogen or oxygen on the other.
Dipeptides- essentially amides formed from 2 amino acids.

Advantage of using the polyester suture rather than the polyalkene?
the ester bond in polyester can hydrolyse whereas polyalkene C-C, C-H bonds are too strong thus cannot. Therefore the former can dissolve naturally whereas the former would require manual removal.

7. Conc. Of propanone increases by 100, What is the new rate equation and explain pls:
rate = k[H+]Due to the large excess of propanone, it is no longer affecting the rate as only the H+ concentration limits the speed at which the reaction proceeds.

Why is the initial rate of the reaction used to determine the order of the reactants rather than at any other point during the reaction? Or something
Initial rate takes into account the initial concentration of the reactants which is more accurate since, as the reaction proceeds the concentrations of the different species change. (?)

How do you calculate the initial rate of the reaction from a graph of concentration against time?
Calculate the gradient of the tangent to the slope. (?)

8. Organic synthesis to produce the branched primary amine:
Structure: Same as previous however Br is substituted with CN
Step 3: KCN w/ ethanol (?)
Step 4: H2/Ni, LiAlH4

Question 9 Notes
Potassium dichromate turns green, either a primary or secondary alcohol, or aldehyde
O-H absorption in the Mass spec
O-H in NMR spectrum, singlet obvs
CH3-O present in NMR spectrum, singlet
RCH2 in NMR spectrum, triplet ?
R(CH3)2 in NMR spectrum, doublet ?
Is this every question?
7. I haven't stopped thinking about this damn exam. It's going to haunt me all summer ffs.
8. (Original post by Abc321zxc)
Same...
9. does anyone know where to access this paper+ unit 2 paper???????????/
10. (Original post by Dudeyoufugly)
does anyone know where to access this paper+ unit 2 paper???????????/
Ask your teacher or you will have to wait until May/June for it to be officially released...

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