AQA A-Level Chemistry Paper 2 - SOLUTION BANK (Q+A)

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÷by0
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Building a solution bank of all the questions/answers.

If you can remember a specific question that has not been contributed, please add it down below. Thank you in advance :five:

Question 1
1. This question is about amines.
1.1. Give an equation for the preparation of 1,6-diaminohexane by the reaction of 1,6-dibromohexane with an excess of ammonia. (2 marks)
Spoiler:
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C6H12Br2 + 2 NH3 —-> C6H12(NH3)2 + 2 HBr
or
C6H12Br2 + 4 NH3 —> 2 NH2(CH2)6NH2 + 2NH4Br (this is probably the correct equation but not sure)

1.2. Complete the mechanism for the reaction of ammonia with 6-bromohexylamine to form 1,6-diaminohexane. (2 marks)
Spoiler:
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Nucleophilic Sub. ammonia attacks Br.

Suggest the structure of a cyclic secondary amine that can be formed as a by-product in this reaction. (1 mark)
Spoiler:
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amino group attacks the Br on its own molecule.

1.3. 1,6-diaminohexane can also be formed in a two-stage synthesis, starting from 1,4-dibromobutane.
Suggest a reagent and a condition for each stage in the alternative synthesis.

Stage 1
Spoiler:
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Reagent: aqueous ethanol KCN
Condition: heat under reflux

Stage 2:
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Reagent: LiAlH4 in ether
Condition: don't know

Question 2
deduce formula of carboxylic acid, high res mass spec.

Question 3
golf balls.

Question 4
4. Substances P and Q react in a solution at a constant temperature.
The initial rate of reaction was studied in three experiments by measuring the change in concentration of P over the first 5 seconds of the reaction.
The data obtained are shown in table 1.
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4.1. Complete Table 2 to show the initial rate of reaction of P for each experiment.
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4.2. Calculate the order of reaction with respect to P and Q.
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P order 1
Q order 2

Question 5
The rate constant, k = Ae-Ea/RT
For this reaction at 25ºC, k = 3.46x10-8 s-1
The activation energy Ea = 96.2 kJ mol-1
The gas constant R = 8.31 J K mol-1

Calculate a value for the Arrhenius constant A for this reaction. Give the units for A.
Spoiler:
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A = 2.57x109 s-1

Question 6
6. Compounds A and B both have the molecular formula C4H8Br2 . A has a singlet, a triplet and a quartet in its H1 NMR spectrum. B has two singlets in its H1 NMR spectrum

6.1. Draw the structure of A and B.

Question 7
7.
Isomer X Isomer Y
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7.1. Give the IUPAC name of isomer X. (1 mark)
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cyclopentanone

7.2. Explain how and why isomers X and Y can be distinguished by comparing each of their:
    • Boiling points
    • C13 NMR spectra
    • IR spectra
Use data from Tables A and C in the data booklet in your answer. (6 marks)

Spoiler:
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Strongest intermolecular force: in X - permanent dipole, in y - hydrogen bonding.
Hydrogen bonding is the strongest intermolecular force, so requires more energy to overcome, thus Y will have a higher boiling point.

X will have C13 NMR peak in the range 190-220 ppm, due to the C=O ketone group. Y will not. However, Y will have a peak in range 50-90 ppm from -OH group, and a peak in 90-150 ppm range from the C=C. X will not have either of these.

The IR spectrum for Y will have a broad peak in the 3230-3550 range form the -OH (alcoholic) group. It will also have a sharp peak in the 1620-1680 cm-1 range from the C=C. X will not have this. X will have a sharp peak from the C=O group in the 1680-1750 cm-1 range.

Question 8
8. Paracetamol is a medicine commonly used to relieve mild pain traditionally paracetamol has been made in industry in a 3-step process.

8.1. Name the mechanism of the reaction in step 1[/sup][/sup]
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Electrophillic Substitution

8.2. Complete the equation for the reaction in step 2.
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aromatic + 6[H] => aromatic +2 H2O

In step 3, other aromatic products are formed as well as paracetamol
8.3. Draw the structure of one of these other aromatic products.
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phenyl ethanoate - this is a complete guess
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Chemists have recently developed a two step process to produce paracetamol from phenol. In the first step phenol is oxidised to Hydroquinone.
Spoiler:
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Phenol + H2O2 => Hydroquinone + H2O


In the second step hydroquinone reacts with ammonium to form paracetamol. complete the equation for the second step.
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Hydroquinone + H2NCOCH3 => paracetamol + H2O

Question 9

Question 10

Question 11

Question 12
12. This question is about chromatography.
12.1. Suggest why 2 solvents were used. (2 marks)
Spoiler:
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Polar species are only mobile in polar solvents, and likewise for non-polar. If you don't know the polarity of the constituents of the sample, using a both solvents separately will ensure all species are mobilised.

12.2 Deduce the minimum number of amino acids present. (1 mark)
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7 (7 dots after both solvents)

12.3. How can the spots be located?
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Locating agent (ninhydrin) / visible under UV light.

12.4 peptide chain hydrolysis.

Question 13

13.
13.1 Add H bonds to the diagram
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5 in total:
3 bonds between C and G
2 bonds between A and T


13.2
13.3 Explain what is meant my the term complimentary in the context of DNA. (2 marks)
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DNA bases on one strand always form hydrogen bonds with their complimentary base pair on the other strand. Guanine always hydrogen bonds to cytosine, and adenine always binds to thymine.
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Fahad9
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Hopefully it will work out by tonight 🤞
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jamesctu
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This question is about isomers
Give a reagent and observations for a test – tube reaction to distingujish between 2- methylbutan 1 ol and 2-methylbutan-2-ol
Reagent
Observation with 2-methylbutan-1-ol
Observation eith 2-methylbutan-2-ol
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÷by0
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Reagent is acidified potassium dichromate. It goes from orange to green with the 2º alcohol, because is oxidises the alcohol (to a ketone). With 3º alcohol there is no change / remains orange. 3º alcohols cannot be oxidised.
(Original post by jamesctu)
This question is about isomers
Give a reagent and observations for a test – tube reaction to distingujish between 2- methylbutan 1 ol and 2-methylbutan-2-ol
Reagent
Observation with 2-methylbutan-1-ol
Observation eith 2-methylbutan-2-ol
Do you know what the numbers of the questions are so i can create a solution bank?
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jamesctu
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(Original post by ÷by0)
Reagent is acidified potassium dichromate. It goes from orange to green with the 2º alcohol, because is oxidises the alcohol (to a ketone). With 3º alcohol there is no change / remains orange. 3º alcohols cannot be oxidised.

Do you know what the numbers of the questions are so i can create a solution bank?
i think that question 6
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jamesctu
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Compounds A and B both have the molecular formula c4h8Br2
A has a singlet,a triplet and a quartet in its 1H NMR spectrum.
B has only two singlets in its 1H NMR spectrum
Draw the structure for each A and B
A B
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÷by0
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(Original post by jamesctu)
Compounds A and B both have the molecular formula c4h8Br2
A has a singlet,a triplet and a quartet in its 1H NMR spectrum.
B has only two singlets in its 1H NMR spectrum
Draw the structure for each A and B
A B
for A:
Singlet (S)=> no neighbouring H environments
Triplet (T)=> 2 H adjacent
quartet (Q) => 3 adjacent

CH2-CBr2-CH2-CH3
(S) (Q) (T)

For B:

Singlets mean all Hydrogens have no hydrogenates on their neighbouring carbon
=> (BrH2C)-CBr-(CH3)2
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jamesctu
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(Original post by ÷by0)
for A:
Singlet (S)=> no neighbouring H environments
Triplet (T)=> 2 H adjacent
quartet (Q) => 3 adjacent

CH2-CBr2-CH2-CH3
(S) (Q) (T)

For B:

Singlets mean all Hydrogens have no hydrogenates on their neighbouring carbon
=> (BrH2C)-CBr-(CH3)2
this is still all in the 6 question i think
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÷by0
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thanks
(Original post by jamesctu)
this is still all in the 6 question i think
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phosphene
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DNA -

Part one was the hydrogen bond drawings

Then you had to circle the nucleotides (vague wording lol)

Explain why strands complementary
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jamesctu
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q8
calculate the mass in kg of hydroquinone (mr=110.0) needed to produce 250 kg of parecetamol
(Original post by ÷by0)
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phosphene
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Not 100% sure but I think 12 was the TLC question

Minimum number of amino acids (8 or 7 can’t recall)
How to see them? Something about UV
How to hydrolyse them (hcl probs)
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jamesctu
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(Original post by phosphene)
Not 100% sure but I think 12 was the TLC question

Minimum number of amino acids (8 or 7 can’t recall)
How to see them? Something about UV
How to hydrolyse them (hcl probs)
what did u get for the wavelength
i got
F
G
E
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÷by0
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i don't remember letter orders, but it was ester, the alcohol, then carboxylic acid i think.
(Original post by jamesctu)
what did u get for the wavelength
i got
F
G
E
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jamesctu
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(Original post by ÷by0)
i don't remember letter orders, but it was ester, the alcohol, then carboxylic acid i think.
this was the displayed formula of each one what was the order
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E, G, F, i think. The bottom two graphs had broad OH peaks.
(Original post by jamesctu)
this was the displayed formula of each one what was the order
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jamesctu
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how do you answer this
(Original post by jamesctu)
this was the displayed formula of each one what was the order
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jamesctu
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this is q8
how do you answer it
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justme2001
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#19
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The reaction was actually with ammonium ethanoate not ammonia so the formula would have been CH3CONH4

Which would produce the correct thing on the right hand side
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Dreametuber
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one of the questions was draw structures of c6h3br3 which have 2 peaks and 4 peaks in carbon nmr
also there was a question on draw the structure of the vegetable oil (c17H31 x 2 and C17H33 on the molecule)
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