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

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Question 21 • Answer: A • Topic: Alkene • Explanation: • 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. • Question 22 • Answer: B • Topic: Halogenoalkane • Explanation: • 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. • Question 23 • Answer: B • Topic: Halogenoalkane • Explanation: • 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. • Question 24 • Answer: B • Topic: Intro to Organic Chem • Explanation: • 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. • Question 25 • Answer: D • Topic: Organic Synthesis • Explanation: • 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. • Question 26 • Answer: D • Topic: Carboxylic Acid • Explanation: • 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. • Question 27 • Answer: A • Topic: Nitrogen Compound • Explanation: • Amine is basic and will react with acid HCl, while amide is neutral and will not react with HCl under cold conditions. • Question 28 • Answer: A • Topic: Carbonyl Compound • Explanation: • 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. • Question 29 • Answer: D • Topic: Electrochemistry • Explanation: • With Zn we can calculate Ecell for reaction between Zn + VO2^+, Zn + VO^2+ 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. • Question 30 • Answer: D • Topic: Transition Element • Explanation: • 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. • Subscribe to our channel for more A Level Chemistry video lessons! • View the full video with description at my website https://chemistryguru.com.sg/2022-p1-... • We are conducting weekly classes for A Level H2 Chem at Bishan, Singapore and online webinar. Find out more at https://chemistryguru.com.sg/ • 00:00 Introduction • 00:13 Question 21 • 03:54 Question 22 • 07:55 Question 23 • 12:43 Question 24 • 18:05 Question 25 • 23:05 Question 26 • 31:42 Question 27 • 33:32 Question 28 • 39:57 Question 29 • 45:10 Question 30

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