2022 A Level H2 Chemistry Paper 1 Solutions - Questions 21 to 30

Question 21

2022 p1 21 question

Answer: A

Topic: Alkene


When HBr is added to both alkenes, we can use Markovnikov rule (H adds to C with more H) to deduce the major product 2,5-dibromo-2,5-dimethylhexane.

2022 p1 21 explain

Question 22

2022 p1 22 question

Answer: B

Topic: Halogenoalkane


The product is a racemic mixture hence reaction is via SN1 mechanism.

In step 1 the C-Cl bond breaks heterolytically to form a carbocation intermediate which is trigonal planar with a plane of symmetry.

2022 p1 22 explain

Question 23

2022 p1 23 question

Answer: B

Topic: Halogenoalkane


Only halogenoalkane will undergo nucleophilic substitution with NaOH.

Halogenobenzene will not react as C-X bond is stabilised by resonance.

Therefore only 3 of the compounds with halogenoalkanes will give aqueous halides that can be precipitated with AgNO3.

When shaken with ammonia solution only AgCl will dissolve.

Therefore only 2 of the compounds will have AgI ppt remaining after reaction with ammonia solution.

2022 p1 23 explain

Question 24

2022 p1 24 question

Answer: B

Topic: Intro to Organic Chem


Compound 1 has chiral carbon hence optically active.

For the other compounds we can rotate C-C bond between carbons 3 and 4 so that it's easier to determine if carbons 3 and 4 are mirror images of each other.

For compound 2 there is no internal mirror plane hence it is optically active.

For compounds 3 and 4, there is an internal mirror plane hence this is considered a meso compound.

The 2 chiral carbons will cancel out each other's optical activity and hence optically inactive.

2022 p1 24 explain

Question 25

2022 p1 25 question

Answer: D

Topic: Organic Synthesis


Options A and B are wrong since carbonyl compounds do not react with conc H2SO4 or NaOH(aq).

Option C will give a final product very close to the required answer, salt of conjugate base instead of acid group.

Option D is the best answer since last step with conc H2SO4 will eliminate alcohol to alkene and protonate conjugate base to carboxylic acid and give us the required final product.

2022 p1 25 explain

Question 26

2022 p1 26 question

Answer: D

Topic: Carboxylic Acid


By considering stability of conjugate base, HCOOH is more acidic and has a bigger Ka value than CH3COOH, hence option A is wrong.

CH2ClCOOH is more acidic and has a bigger Ka value than CH3COOH hence option B is wrong.

Option C is wrong since acids of different strength will dissociate to different extent to form different concentrations of H+ and conjugate base.

Option D is correct since benzoic acid is weaker than 4-chlorobenzoic acid with a smaller Ka value, hence need a higher concentration of benzoic acid to dissociate same concentration of H+ to give same pH.

2022 p1 26 explain

Question 27

2022 p1 27 question

Answer: A

Topic: Nitrogen Compound


Amine is basic and will react with acid HCl, while amide is neutral and will not react with HCl under cold conditions.

2022 p1 27 explain

Question 28

2022 p1 28 question

Answer: A

Topic: Carbonyl Compound


Both Y and Z can be oxidised by K2Cr2O7 hence primary alcohol, secondary alcohol or aldehyde FG present.

Y gives positive iodoform test hence will have CH3-CO-R or CH3-CHOH-R structure.

Only Z reacts with sodium hence OH group present, Z = CH2OHCH2OH

Y does not have OH group, Y = CH3COCHO

Since Y contains aldehyde, it will react with both Fehling's solution and 2,4-DNPH.

2022 p1 28 explain

Question 29

2022 p1 29 question

Answer: D

Topic: Electrochemistry


With Zn we can calculate Ecell for reaction between Zn + VO2+, Zn + VO2+ and Zn + V3+ to show that all Ecell are positive and reactions are all feasible.

Hence VO2+ will be reduced all the way to V2+, final colour is violet.

With Sn, only Ecell between Sn + V3+ is negative and reaction not feasible.

Hence VO2+ will be reduced to V3+ only, final colour is green.

2022 p1 29 explain

Question 30

2022 p1 30 question

Answer: D

Topic: Transition Element


For transition elements the additional electrons are added to inner 3d subshell which will shield valence 4s electrons.

The increase in nuclear charge and shielding effect will cancel out and effective nuclear charge will be very similar or invariant.

This causes transition elements to have very similar physical and chemical properties such as ionisation energies, ionic radii, density, charge density and so on.

Option A and C are wrong as they are suggesting the valence 4s electrons are causing the shielding effect.

Option B is wrong since there are no electrons in 4p subshell for transition elements.

2022 p1 30 explain

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