OCR 2023 Chemistry A H032/02: Depth in chemistry AS Level Question Paper & Mark Scheme (Merged) Q5 g] A polyester is formed from 200 molecules of 4-hydroxybenzoic acid. What is the relative molecular mass, Mr , of the polyester ? - ANSWER -1.

OCR 2023 Chemistry A H032/02: Depth in chemistry AS Level Question Paper & Mark Scheme (Merged) Q5 g] A polyester is formed from 200 molecules of 4-hydroxybenzoic acid. What is the relative molecular mass, Mr , of the polyester ? - ANSWER -1. Structure of 4 - hydroxybenzoic acid 2. Mr of 4 - hydroxybenzoic acid: 138 3. 138 x 200 = 27600 4. There are [n-1] molecules of water lost, so there are 199 molecules of water lost 5. 27600 - [199 x18] =24018 *The Mr of the polyester is 24018* Q6 a] A student intends to synthesise ester C. [refer to the A LEVEL CHEMISTRY OCR A PAPER 2 PMT QP for the structure of ester C] Plan a two-stage synthesis to prepare 12.75 g of ester C, starting from the compound 2- methylpropanal, [CH3]2CHCHO. Assume that the overall percentage yield of ester C from the starting compound 2-methylpropanal is 40%. In your answer include the mass of 2-methylpropanal required, reagents, conditions and equations where appropriate. Purification details are not required [6 marks + QWC] - ANSWER -Preparation: - Reflux 2-methylpropanal with acidified potassium dichromate - The colour of the solution will turn from orange to dark green - Equation: [CH3]2CHCHO + [O] → [CH3]2CHCOOH - The intermediate product will be 2-methylpropanoic acid - Reflux 2-methylpropanoic acid with dilute HCl and methanoic acid - Equation: [CH3]2CHCOOH + CH3OH [CH3]2CHCOOCH3 + H2O - The end product will be ester C: methyl 2-methylpropanoate Calculation for the mass of 2-methylpropanal required: - The Mr of ester C is: 102 - mass / Mr = moles, so moles of C are: 0.125 - moles of C = 0.125 - 1 : 1 molar ratio [supposed to be] so moles of 2-methylpropanal needed are: 0.125 for 100% yield but 40% yield - 0.125 x [100 / 40] = 0.3125 - Mr of 2-methylpropanal = 72 - moles x Mr = mass so mass of 2-methylpropanal = 72 x 0.3125 - 3s.f. = 22.5 g *22.5g of 2-methylpropanal required* Q6 b] The mass spectrum of ester C is shown below [refer to the A LEVEL CHEMISTRY OCR A PAPER 2 PMT QP for the mass spectrum of ester C] Suggest possible structures for the species responsible for peaks Y and Z in the mass spectrum. - ANSWER -Peak Y = a fragment ion of m / z 43 = *[CH3]2CH+* Peak Z = a fragment ion of m / z 71 = *[CH3]2CHCO+* Q5 f] Compound B is a structural isomer of ester A. • Compound B reacts with aqueous sodium carbonate. • The 13C NMR spectrum of B has 4 peaks. Draw a possible structure for compound B. - ANSWER -- 4 different carbon environments for B - Is a carboxylic acid - structural formula: CH2BrCH[CH3]2COOH - name: 2,2-dimethyl-5-bromopentanoic acid Q7 a] This question is about benzene. Over time, the Kekulé and delocalised models have been used to describe the bonding and structure of a benzene molecule. Describe, in terms of orbital overlap, the similarities and differences between the bonding in the Kekulé model and the delocalised model of benzene [3 marks] - ANSWER -Similarities: - both structures have overlapping p orbitals and pi bonds Differences: - the delocalised model has a delocalised ring of electrons above and below the plane of the carbon ring while the Kekulé model has localised electrons around the C = C pi bonds Q7 b] Experimental evidence led to the general acceptance of the delocalised model over the Kekulé model. Describe two pieces of evidence to support the delocalised model of benzene [2 marks] - ANSWER -- X ray diffraction: this showed that all the C - C bonds in benzene were of an intermediate length between the C - C and the C = C bonds - Adding bromine water to benzene produced no reaction or colour change, this shows that the electron density of benzene is too low for it to contain C = C bonds Q8 [info] - ANSWER -Benzene is first reacted with ethanoyl chloride, CH3COCl, to form phenylethanone, shown below. [refer to the A LEVEL CHEMISTRY OCR A PAPER 2 PMT QP for the structure of phenylethanone] The reaction takes place in the presence of aluminium chloride, Al Cl 3, which acts as a catalyst. In the mechanism for this reaction, • ethanoyl chloride first reacts with aluminium chloride to form the CH3-C+=O cation • the CH3-C+=O cation then behaves as an electrophile. Q1 [info] - ANSWER -This question is about unsaturated aldehydes and alcohols. 3-Methylbut-2-enal, shown below, is used as a food flavouring. 3-Methylbut-2-enal is reacted with hydrogen bromide, forming a mixture of two organic products. [refer to the A LEVEL CHEMISTRY OCR A PAPER 2 PMT QP for the structure of 3-Methylbut-2-enal] Q1 a] One of the organic products forms in a much greater quantity than the other organic product. Outline the reaction mechanism for the formation of one of the organic products. Include curly arrows and relevant dipoles. - ANSWER -Mechanism: Start product: 3-methylbut-2-enal - curly arrow from centre of the C = C bond to the [negative dipole] Br atom of the H - Br bond - curly arrow from the centre of the H - Br bond to the [positive dipole] H atom on it - carbocation intermediate with a C - Br bond on the adjacent carbon - Curly arrow from H to the + on the carbocation intermediate End product: 3-methyl-2-bromobut-2-anal [tertiary carbocation is the most stable carbocation] Q1 b] Explain why one of the organic products forms in a much greater quantity than the other organic product. [2 marks] - ANSWER -- The tertiary carbocation is the most stable carbocation, - The major product forms from the tertiary carbocation Q2 a] Geraniol and citronellal, shown below, are isomers present in 'citronella oil', used as an insect repellent. Geraniol and citronellal are structural isomers of each other. • They also show stereoisomerism. Describe how the observations from a chemical test would distinguish between geraniol and citronellal [2 marks] - ANSWER -- Add Tollen's reagent to test tubes of citronellal and geraniol - The test tube with citronellal in it would have a silver mirror on it cause citronellal is an aldehyde, while nothing would happen in the test tube with geraniol Q2 b] What is the molecular formula of geraniol? [1 mark] - ANSWER -C10H18O Q2 c] Explain why geraniol and citronellal are structural isomers of each other [1 mark] - ANSWER -- They both have the same molecular formula [C10H18O] but different structural formulae Q3 a] Explain the term stereoisomerism [1 mark] - ANSWER -Same structural and molecular formulae but different arrangements of atoms in space Q3 b] The structures of geraniol and citronellal are repeated below with the carbon atoms numbered. [refer to the A LEVEL CHEMISTRY OCR A PAPER 2 PMT QP for the structures of geraniol and citronellal] Explain the types of stereoisomerism shown by geraniol and citronellal. In your answer, refer to the numbered carbon atoms in the structures above [4 marks] - ANSWER -Geraniol stereoisomers: - on the C = C bond on carbon 2, there is E / Z isomerism - the highest priority groups are the CH2OH and the CH2CH2CHC[CH3]2 groups - E isomer has the highest priority groups on opposite sides of the C = C bond - Z isomer has the highest priority groups on the same side of the C = C bond Citronellal stereoisomers: - carbon 3 is a chiral centre with 4 different groups around carbon 3 - the groups are: CH2CHO, H, CH2CH2CHC[CH3]2, CH3 - it has two optical isomers that are non - superimposible mirror images of each other Q4 a] This question is about α-amino acids, RCH(NH2)COOH. Table 17.1 shows the R groups in four amino acids [refer to the A LEVEL CHEMISTRY OCR A PAPER 2 PMT QP for the Fig. 17.1 and the flowchart] In the boxes, draw the organic products for the reactions of serine shown below. - ANSWER -The up arrow: the product of the reaction of serine when H+ is added *HOCH2CH[+NH3]COOH* The diagonal arrow leading to the top right box: the product of the reaction of serine when [CH3]2CHOH/H2SO4 is added *HOCH2CH[NH2]COOCH[CH3]2 [ester]* The right arrow: the product of the reaction when excess CH3COCl is added *CH3COOCH2CH[CH3CONH]COOH* Q4 b] A student is provided with one of the four amino acids in Table 17.1. A student carries out a titration with a standard solution of hydrochloric acid to identify the amino acid. The student's method is outlined below. • The student dissolves 5.766 g of the amino acid in water and makes the solution up to 250.0 cm3 in a volumetric flask. • The student titrates this solution with 25.0 cm3 of 0.150 mol dm−3 hydrochloric acid. • 21.30 cm3 of the amino acid solution were required for complete neutralisation of the hydrochloric acid. Determine which amino acid the student used [4 marks] - ANSWER -mol = conc x vol and moles = mass / Mr 1. Vol of HCl: 0.025 dm3, conc of HCl: 0.15 mol dm-3, 2. so, moles of HCl: conc x vol = 3.75 x10^-3 NOTE: 1 : 1 molar ratio of reactants 3. moles of AA: 3.75 x10^-3, vol of AA sol: 0.0213 4. moles / vol = 0. mol dm^-3 in the solution 5. mol dm / 4 = 0. moles in 5.766g of AA 6. Mr of AA without R group = 74 Mrs of AA with R groups alanine: 89 [74 + 15] serine: 107 [74 + 31] leucine: 131 [74 + 57] glycine: 75 [74 + 1] 7. Moles = mass / Mr, so Mr = mass / moles 8. Mr is 131 g mol^-1 *The amino acid is leucine* Q5 a] This question is about esters. The structure of ester A is shown below. [refer to the A LEVEL CHEMISTRY OCR A PAPER 2 PMT QP for the structure of ester A] What is the systematic name of ester A? - ANSWER -ethyl 4-bromopropanoate Q5 b] In the boxes, draw the organic products for the reactions of the functional groups in ester A shown below. Each reaction forms two organic products. [refer to the A LEVEL CHEMISTRY OCR A PAPER 2 PMT QP for the flowchart] - ANSWER -The top arrow: the reaction of ethyl 4-bromopropanoate with H+[aq] Products: *ethanol* and *4-bromopropanoic acid* The bottom arrow: the reaction of ethyl 4-bromopropanoate with excess OH-[aq] Products: *ethanol* and *4-hydroxypropanoate* Q5 c] Name the type of reactions of ester A shown in b] - ANSWER -Acid hydrolysis [top arrow] and base hydrolysis [bottom arrow] Q5 e] The protons in ester A are in four different environments, which are labelled 1- 4 on the structure below [refer to the A LEVEL CHEMISTRY OCR A PAPER 2 PMT QP for the structure of ester A] Complete the table to predict the proton NMR spectrum of ester A. - ANSWER -Proton environment 1: HC - Br chemical shift of 1: between [delta] 3.0 - 4.3 ppm splitting pattern of 1: triplet with an area of 2 under the peak Proton environment 2: HC - CO - chemical shift of 2: between [delta] 2.0 - 3.0 ppm splitting pattern of 2: triplet with an area of 2 under the peak Proton environment 3: HC - O chemical shift of 3: between [delta] 3.0 - 4.3 ppm splitting pattern of 3: quartet with an area of 2 under the peak Proton environment 4: HC - R chemical shift of 4: between [delta] 0.5 -1.9 ppm splitting pattern of 4: triplet with an area of 3 under the peak Q8 a] Complete the mechanism for the reaction. Include equations to show the role of the AlCl3 catalyst, relevant curly arrows and the structure of the intermediate. - ANSWER -Mechanism: 1. curly arrow from the ring of electrons in benzene to the CH3CO+ electrophile 2. intermediate with both CH3CO+ and H attached to the arene ring. The arene ring is broken and goes up to C6 and C2 only in a horseshoe with a + charge in the middle of it. There is a curly arrow from the middle of the C - H bond into the centre of the arene ring 3. product: phenylethanone Breakdown of catalyst: AlCl3 + CH3COCl AlCl4- + CH3CO+ Reformation of catalyst: AlCl4- + H+ AlCl3 + HCl Q8 b] Complete the flowchart for the synthesis of compounds D and E from phenylethanone. - ANSWER -