1. Adrenalin is a hormone which raises blood pressure, increases the depth of breathing and delays fatigue in muscles, thus allowing people to show great strength under stress. Benzedrine is a pharmaceutical which stimulates the central nervous system in a similar manner to adrenalin. HO H ( ) NH HO (OH) N Benzedrine Adrenalin (a) (i) On the structure for benzedrine mark with a ( * ) any asymmetric carbon atom that causes chirality. Suggest why adrenalin is more soluble in water than is benzedrine.......... () (b) Give the structural formulae of the organic products obtained when benzedrine reacts with: (i) an aqueous acid such as dilute hydrochloric acid; ethanoyl chloride in the absence of a catalyst; (iii) excess ethanoyl chloride in the presence of the catalyst anhydrous aluminium chloride. () Maltby Academy 1
(d) It is possible to eliminate a molecule of water from adrenalin which for the purpose of this question may be represented as R (OH) NH. Draw the structural formulae of the two stereoisomers produced. () (e) The mass spectra of both benzedrine and adrenalin have a peak at a mass/charge ratio of 44. Draw the structure of the species which give these peaks. (i) in benzedrine; in adrenalin. (Total 11 marks). This question concerns the three isomers A, B and C, each of which has a relative molecular mass of 14. O C C O H A B OH C Maltby Academy
(a) The mass spectrum of substance A is shown below. Identify the species responsible for the peaks labelled 1, and. Peak 1... Peak... Peak... () (b) The infra-red spectra of two of these substances were also measured. (i) Use the table and the spectra below to identify which spectrum is that of substance C. Bond Wavenumber / cm 1 Bond Wavenumber / cm 1 C H (arenes) 000-100 O H (hydrogen bonded) 00-570 C H (alkanes) 850-000 O H (not hydrogen bonded) 580-650 C==O 1680-1750 C==C (arenes) 1450-1600 The spectrum of substance C is spectrum number... Give one reason for your choice.... (iii) Give one other reason why the other spectrum could not be that of substance C....... Maltby Academy
Transmittance Transmittance Relative intensity 100 80 60 40 1 0 0 5 50 75 100 15 m/e 100 50 0 4000 000 000 1500 1000 Wavenumber/cm 1 100 50 0 4000 000 000 1500 1000 Wavenumber/cm 1 (c) State which of the substances A, B and C will react with the following reagents and state what would be observed. Maltby Academy 4
(i) Bromine dissolved in hexane. Substance(s)... Observation... () A warm ammoniacal solution of silver nitrate. Substance(s)... Observation... () (iii),4-dinitrophenylhydrazine solution. Substance(s)... Observation... () (iv) Give the structural formula of the organic product(s) obtained in (c)(i). Maltby Academy 5
(v) Give the structural formula of the organic product(s) obtained in (c). (Total 18 marks). The drug ibuprofen can be synthesised from benzene by the route shown below. H C H C H C H C Step A Step B Step C CO/Pd catalyst C O H C OH H C CO H Ibuprofen (a) Name the type and mechanism of the reaction in Step A, and suggest a suitable reagent and catalyst. Type and mechanism... Name of the reagent for Step A... Catalyst... () Maltby Academy 6
(b) Step C is a reduction. Give ONE reason why lithium tetrahydridoaluminate, LiAlH 4, is preferred to hydrogen as a reducing agent in this reaction.......... () (c) A sample of the final product was analysed by combustion. 1.00 g was burnt in oxygen. It produced.78 g carbon dioxide and 0.786 g water. State the molecular formula of ibuprofen and show that these results are consistent with it. (4) Maltby Academy 7
(d) Ibuprofen can be analysed by instrumental methods. The infrared spectra of ibuprofen and two other drugs, aspirin and paracetamol, not necessarily in that order, are shown opposite. H C Ibuprofen has the formula H C CO H Aspirin has the formula CO H O C O OH Paracetamol has the formula H N C O (i) Explain, referring to the structure of each molecule, why infrared spectroscopy is not a good technique to distinguish aspirin from ibuprofen. Maltby Academy 8
Transmittance (%) Transmittance (%) Transmittance (%) Deduce which of X, Y or Z is the infrared spectrum of paracetamol, giving a piece of evidence from the spectrum you select. Spectrum X 100 50 Spectrum Y 0 4000 000 000 Wavenumber/cm 1 1500 1000 100 50 Spectrum Z 0 4000 000 000 Wavenumber/cm 1 1500 1000 100 50 0 4000 000 000 1500 1000 Wavenumber/cm 1 Maltby Academy 9
(iii) Ibuprofen and aspirin can be distinguished using their mass spectra. A line at mass/charge ratio 57 occurs only in the mass spectrum of ibuprofen. Give the formula of the ion which produces this line. Suggest the mass/charge ratio of one line which occurs in the mass spectrum of aspirin but not ibuprofen, and the formula of the species which produces it. () (Total 14 marks) 4. This question concerns the following compounds containing four carbon atoms. A Butanoic acid, COOH B Butanone, CO C Propyl methanoate, HCOO D Butanoyl chloride, COCl Select, from A to D, the compound that (a) can be made by the oxidation of a primary alcohol. A B C D Maltby Academy 10
(b) would be expected to react most rapidly with ethanol. A B C D (c) would have 4 different chemical shifts in its nmr spectrum and a broad absorption between 500 00 cm 1 in its infrared spectrum. A B C D (Total marks) 5. In moths a pheromone, P, acts as an attractant for the opposite sex. P has the molecular formula C 7 H 1 O. What can be deduced about the structure of P from the following information? (a) (i) 1 mole of P reacts with 1 mole of Br molecules to form a compound with the formula C 7 H 1 OBr. When lithium tetrahydridoaluminate is reacted with P a compound with the formula C 7 H 14 O is formed. Maltby Academy 11
(iii) P forms an orange precipitate with,4-dinitrophenylhydrazine. (iv) When P is heated with Fehling s or Benedict s solution, the solution remains blue. (v) P is a Z-isomer. (b) What does the following physical data tell you about the structure of P? Use your Data booklet where necessary. (i) The infrared spectrum of P has the following absorptions at wavenumbers above 1600 cm 1. 060 cm 1 90 cm 1 1690 cm 1 1660 cm 1 () Maltby Academy 1
The nmr spectrum does not have a peak corresponding to a chemical shift,, of between 9 and 10. (iii) The mass spectrum showed the presence of peaks at mass/charge ratios of 15 and 9, but no peak at 4. () (c) Given that P has a straight chain of carbon atoms in its formula, use the information you have deduced above to suggest a displayed formula for the pheromone P. () Maltby Academy 1
(d) How could you use a purified sample of the orange precipitate in (a)(iii) to confirm the formula of P?............... () (Total 16 marks) 6. (a) (i) *( ) 1 (the three) OH groups allow adrenalin to form more hydrogen bonds with water (than does benzedrine) (b) (i) ( )NH (Cl ) + 1 Can use R in place of C 6 H 5 ( ) in both (i) and ( )N C ( 1) H O 1 (iii) C ( ) N C O O substitution(s) in ring at any position(s) production of amide H (d) R NH H C C C C H R H H NH Maltby Academy 14
(e) (i) (( ) NH ) + 1 ( NH ) + / ((OH) ) + max 1 for (e) if no charges shown must show some structure in answers ie. C H 5 N(0) 1 [11] + 7. (a) Peak 1 m/e 77: C 6 H 5 Peak m/e 105: C 6 H 5 CO + + Peak m/e 14: C 6 H 5 CO if + left off penalize only once (b) (i) IR spectrum number 1 Because of the broad peak at 0cm 1 caused by OH / because peak at 0 is OH / because it does not have a peak at 1750 and is only one without C O 1 (iii) The other has a peak at approx. 1700 cm 1 caused by C=O 1 (c) (i) C 1 (brown) goes colourless 1 B 1 silver mirror / black precipitate 1 (iii) A and B red/orange/yellow ppt 1 (iv) C 6 H 5 Br Br OH 1 (v) C 6 H 5 COO but allow the carboxylic acid as product consequential 1 [15] Maltby Academy 15
8. (a) Electrophilic substitution IGNORE extras eg Friedel Craft, alkylation UNLESS contradictory 1-chloro-()-methylpropane IGNORE punctuation Accept ()-methyl-1-chloropropane Accept ( ) Cl/( ) Cl Accept Bromo / iodo for chloro Reject 1-methyl--chloropropane Reject missing 1 from position of Cl in name Catalyst AlCl /aluminium chloride Accept Al Cl 6, AlBr, FeBr (b) LiAlH 4 is a source of H / hydride ion Hydrogen might reduce/attack benzene ring/ H won t attack region of negative charge/ H can attack (δ + ) C in keto group Reject comments on conditions or safety eg temperature, pressure Reject LiAlH 4 /H is a more powerful reducing agent Reject H is a nucleophile/a stronger nucleophile Reject any mention of attack on carboxylate ion (for nd mark) Maltby Academy 16
(c) Note: although many candidates have calculated the empirical formula, this is not required. Molecular formula of ibuprofen = C 1 H 18 O Allow if given at end Allow marks for masses and number of moles if answers are rounded to SF in OR but method is correct. EITHER M r = 06 1 1 g = mol = 4.854 10 mol 16 mass CO produced from 1 C = 1 44 4.854 10 =.78 g mass H O from 18 H = 9 18 4.854 10 = 0.787 g OR Mass C = Mass H = (.78 1) 44 (0.786) 9 = 0.758g = 0.087g (0.758) Moles C = = 0.06 1 Moles H = 0.087 Ratio C:H = 0.06: 0.087 = 1:18 4 (d) (i) (Aspirin and ibuprofen) both contain same (types of) bond(s)(so absorb at same frequency/wavenumber) 1 Accept list of at least 4 bonds which are present in both Reject groups for bonds Data is required for mark Y = paracetamol Peak at 500 00 (N H) IGNORE mention of amine OR 500 140 (N H or amide) OR 750 00 ((phenolic) O H) OR Only Y has peaks above 000 cm 1 (so must contain different type of bond to X and Z) 1 Reject C H in arene = 00 as present in both Reject 1700 160 (amide) Maltby Academy 17
(iii) 57 in Ibuprofen C 4 H 9 + / ( ) + /( ) + OR C O H + /CCO H + Accept structural or displayed formulae Do not allow lines at 15 ( + ) 76 (C 6 H 4 + ) 4 (C H 7 + or CO + ) 45 (COOH + ) as present in both Aspirin 59 OCO + /C H O + OR 11 C 6 H 4 CO H + OR 180 C 9 H 8 O 4 + (parent ion) OR 17 C 6 H 4 (CO H)O + Penalise no/wrong charges once only [14] 9. (a) A 1 (b) D 1 (c) A 1 [] 10. (a) (i) contains one carbon-carbon double bond Accept alkene 1 is a carbonyl compound / C=O group reduced (to (OH)) Accept aldehyde or ketone 1 (iii) is a carbonyl compound Accept aldehyde or ketone 1 Maltby Academy 18
(iv) is a ketone / P is not an aldehyde 1 Reject aldehyde (v) has two groups on the same side of a C=C Accept cis isomer 1 (b) (i) QWC 060 alkene (C H stretching) 90 alkane (C H stretching) 1690 ketones (C=O stretching) 1660 alkene (C=C stretching) 4 Correct marks Correct marks Correct 1 mark not an aldehyde 1 (iii) QWC 15 group 9 C H 5 group 4 no C H 7 group (c) C H 5 C C H H CO ketone and Z rest of molecule Accept Fully displayed (d) Measure its melting temperature And compare with data book values [16] Maltby Academy 19