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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
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)
1.2. Complete the mechanism for the reaction of ammonia with 6-bromohexylamine to form 1,6-diaminohexane. (2 marks)
Suggest the structure of a cyclic secondary amine that can be formed as a by-product in this reaction. (1 mark)
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
Stage 2:
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.
![Name: Q4 Table 1 .png
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4.1. Complete Table 2 to show the initial rate of reaction of P for each experiment.
4.2. Calculate the order of reaction with respect to P and Q.
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.
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
![Name: pentanone.png
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![Name: pentol.png
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7.1. Give the IUPAC name of isomer X. (1 mark)
7.2. Explain how and why isomers X and Y can be distinguished by comparing each of their:
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]
8.2. Complete the equation for the reaction in step 2.
In step 3, other aromatic products are formed as well as paracetamol
8.3. Draw the structure of one of these other aromatic products.
Chemists have recently developed a two step process to produce paracetamol from phenol. In the first step phenol is oxidised to Hydroquinone.
In the second step hydroquinone reacts with ammonium to form paracetamol. complete the equation for the second step.
Question 9
Question 10
Question 11
Question 12
12. This question is about chromatography.
12.1. Suggest why 2 solvents were used. (2 marks)
12.2 Deduce the minimum number of amino acids present. (1 mark)
12.3. How can the spots be located?
12.4 peptide chain hydrolysis.
Question 13
13.
13.1 Add H bonds to the diagram
13.2
13.3 Explain what is meant my the term complimentary in the context of DNA. (2 marks)
If you can remember a specific question that has not been contributed, please add it down below. Thank you in advance
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:
Show
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)
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:
Show
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:
Show
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:
Show
Reagent: aqueous ethanol KCN
Condition: heat under reflux
Condition: heat under reflux
Stage 2:
Spoiler:
Show
Reagent: LiAlH4 in ether
Condition: don't know
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.
4.1. Complete Table 2 to show the initial rate of reaction of P for each experiment.
Spoiler: ![Name: Table 2 qu4.png
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Show
4.2. Calculate the order of reaction with respect to P and Q.
Spoiler:
Show
P order 1
Q order 2
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:
Show
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
7.1. Give the IUPAC name of isomer X. (1 mark)
Spoiler:
Show
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
Spoiler:
Show
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.
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]
Spoiler:
Show
Electrophillic Substitution
8.2. Complete the equation for the reaction in step 2.
Spoiler:
Show
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.
Spoiler:
Show
phenyl ethanoate - this is a complete guess
![Name: Aromatic product.png
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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:
Show
Phenol + H2O2 => Hydroquinone + H2O
In the second step hydroquinone reacts with ammonium to form paracetamol. complete the equation for the second step.
Spoiler:
Show
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:
Show
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)
Spoiler:
Show
7 (7 dots after both solvents)
12.3. How can the spots be located?
Spoiler:
Show
Locating agent (ninhydrin) / visible under UV light.
12.4 peptide chain hydrolysis.
Question 13
13.
13.1 Add H bonds to the diagram
Spoiler:
Show
5 in total:
3 bonds between C and G
2 bonds between A and T
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)
Spoiler:
Show
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.
Last edited by ÷by0; 2 years ago
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Fahad9
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jamesctu
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#3
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
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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#4
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?
(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
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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jamesctu
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#5
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#5
(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?
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?
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jamesctu
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#6
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#6
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
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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#7
(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
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
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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#8
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#8
(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
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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phosphene
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#10
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#10
DNA -
Part one was the hydrogen bond drawings
Then you had to circle the nucleotides (vague wording lol)
Explain why strands complementary
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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#11
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#11
q8
calculate the mass in kg of hydroquinone (mr=110.0) needed to produce 250 kg of parecetamol
calculate the mass in kg of hydroquinone (mr=110.0) needed to produce 250 kg of parecetamol
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phosphene
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#12
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)
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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#13
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#13
(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)
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)
i got
F
G
E
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÷by0
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#14
i don't remember letter orders, but it was ester, the alcohol, then carboxylic acid i think.
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jamesctu
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#15
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#15
(Original post by ÷by0)
i don't remember letter orders, but it was ester, the alcohol, then carboxylic acid i think.
i don't remember letter orders, but it was ester, the alcohol, then carboxylic acid i think.
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÷by0
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#16
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
this was the displayed formula of each one what was the order
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jamesctu
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#17
how do you answer this
(Original post by jamesctu)
this was the displayed formula of each one what was the order
this was the displayed formula of each one what was the order
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justme2001
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#19
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
Which would produce the correct thing on the right hand side
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Dreametuber
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#20
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)
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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