What are the different types of JC Chemistry tuition classes available at your centre?

We have 3 modes of learning for students to choose from: weekly physical classes at Bishan; weekly online lessons via Zoom; and on-demand video lessons.

Are you the only Chemistry tutor conducting the classes or do you engage other tutors to do that?

Please be rest assured that I conduct all my classes personally and I do not engage other tutors to teach at my centre.

What is your class size at Bishan?

Usually the class size is around 10 to 12 students. Max class size is 16 pax.

Is there makeup lesson if I miss a class?

Yes makeup lesson is available in the same week subject to availability. If makeup class is not possible, the student can view the recorded video lessons at Google Classroom.

Do you have trial lesson available? How much is it?

Yes we do. Fees for trial lesson are for Bishan class and for Zoom online class.

Does your centre offer O Level chemistry tuition? Do you teach IB / IP chemistry tuition? How about H1 Chemistry?

Chemistry Guru specialises in H2 Chemistry for JC1 and JC2 or IP Year 5 and IP Year 6 students only. Find out all the information about A Level H2 Chemistry and the detailed comparison between H1 and H2 Chemistry. We do not conduct lessons for H1 Chemistry, O Level Chemistry, IB Chemistry or IP Chemistry at IP Year 3 and IP Year 4.

2024 A Level H2 Chemistry Paper 1 Solutions - Questions 21 to 30

Question 21

2024p1 q21

Answer: C

Topic: Phenol / Phenylamine

Explanation:

By comparing molecular formula for Y (C₇H₇NO₂) and product (C₇H₄NO₂Br₃) we can deduce there are 3 H substituted by 3 Br due to the presence of highly activating -OH or -NH₂ group to benzene.

Hence we can identify the highly activating group and determine which one will have tri-substitution.

2024p1 q21 explain

Question 22

2024p1 q22

Answer: D

Topic: Organic Reactions

Explanation:

For each option we determine the product and deduce which is the required compound.

Type of Reactions:
A - elimination of water to form alkene
B - oxidation to form ketone
C - alkaline hydrolysis to form salt of acid
D - oxidation to form acid

2024p1 q22 explain

Question 23

2024p1 q23

Answer: C

Topic: Acidity of Alcohols and Phenol

Explanation:

Comparing phenol, water and ethanol, phenol is the most acidic (largest Ka, smallest pKa) while ethanol is the least acidic (smallest Ka, largest pKa).

Acidity of organic compounds is explained via stability of conjugate base.

Phenoxide is stabilised by resonance when lone pair of oxygen interacts with delocalised pi system of benzene, hence phenol is more acidic than water.

Ethoxide is destabilised as electron donating ethyl group intensifies negative charge on oxygen, hence ethanol is less acidic than water.

2024p1 q23 explain

Question 24

2024p1 q24

Answer: B

Topic: Organic Reactions

Explanation:

LiAlH₄ is a source of hydride [H^-] which is negatively charged. Therefore it will react with carbonyl carbon which is partial positive charge but not alkene which is electron rich.

2024p1 q24 explain

Statement 1 is true since C=O bond is polar and carbonyl carbon is partial positive.

Statement 2 is false as alkenes should react with electrophiles in first step of addition.

Statement 3 is false as ketones should be reduced to alcohols instead.

Question 25

2024p1 q25

Answer: D

Topic: Amide Hydrolysis

Explanation:

Amides will undergo alkaline hydrolysis to form amine and salt of acid.

2024p1 q25 explain

Question 26

2024p1 q26

Answer: A

Topic: Amino Acids

Explanation:

With excess hydroxide ions, the alpha acid will react to form carboxylate which is negatively charged. Hence the overall species should be negatively charged.

2024p1 q26 explain

Question 27

2024p1 q27

Answer: C

Topic: Electrolysis

Explanation:

We can use the Faraday's equations to calculate charge and moles of electrons.

Then we can find moles and mass of copper reduced at the cathode.

2024p1 q27 explain

Question 28

2024p1 q28

Answer: B

Topic: Electrochemistry

Explanation:

We can pull the half equations and E values for Br₂/Br^- and O₂/H₂O from the data booklet and calculate Ecell to be positive. This means reaction between O₂ and Br^- in acidic medium should be feasible under standard conditions.

2024p1 q28 explain1

Statement 1 is correct. In air the partial pressure of oxygen is smaller than 1 atm. Position of equilibrium for O₂/H₂O half equation shifts left and Er becomes less positive. This will give a less positive Ecell and make the reaction less feasible.

2024p1 q28 explain2

Statement 2 is likely correct. If activation energy is high, the rate of the reaction will be very slow and kinetically not feasible.

Statement 3 is incorrect. A higher concentration of Br^- will shift position of equilibrium for Br₂/Br^- half equation to the left, Eo becomes less positive and Ecell becomes more positive. This means the reaction should become more feasible instead.

2024p1 q28 explain3

Question 29

2024p1 q29

Answer: C

Topic: Transition Elements

Explanation:

The d-d splitting pattern for octahedral complexes is 2 on top and 3 at the bottom.

For each cation we determine the number of d-electrons, fill up the electron in box diagram and determine the number of electrons in the higher energy 3d orbitals.

2024p1 q29 explain

Question 30

2024p1 q30

Answer: B

Topic: Electrochemistry

Explanation:

SO₂ is oxidised to sulfate hence can reduce VO₂⁺ to a certain extent.

For each possible redox reaction between SO₂ and vanadium species, we calculate its Ecell. If positive, reaction is feasible and vanadium species will be reduced.

We can then consider the redox reaction between SO₂ and this new vanadium species and determine its feasibility.

Final oxidation state of vanadium is determined when the Ecell of the reaction is negative, reaction is not feasible and hence no more change in oxidation state for vanadium.

2024p1 q30 explain