Chemistry Answers Chapter 11

Answers Chapter 11 Exercises 1 The smallest division is 1 so the uncertainty is ±0.5. 2 The missing diamond has a mass of between 9.87 and 9.97 g. The found diamond has a mass between 9.9 and 10.3 g. As the ranges overlap, it could be the missing diamond. 3 (a) 4 × 10−2 g (c) 3.0 × 10−2 g (b) 2.22 × 102 cm3 (d) 3 × 10 or 3.0 × 10 °C (unspecified) 4 (a) 4 (b) unspecified (c) 3 (d) 4 5 A 6 A 7 D 9 A 10 D 11 C 12 8 A The average value = 49.0 s The uncertainty in the measurements is given as ±0.1 s but the results show that there is 52 10:01 additional uncertainty, suggesting that the value could be anywhere between 48.8 and 49.2 s. So the value could be quoted as 49.0 s ± 0.2 s. D 14 D 15 B 17 C 18 A 19 C 20 Number of moles = concentration × volume/1000 13 16 B 29 A (the spectrum on the left) corresponds to CH3CH2CHO B C 27 28 C B (the spectrum on the right) corresponds to CH3COCH3 Similarities Both have a molecular ion corresponding to 58. = 1.00 × 10.0/1000 = 0.0100 mol Differences % uncertainty in concentration = (0.05/1.00) × 100 = 5% A has peaks corresponding to 29 (CH3CH2+) and 28 (loss of CH3CH2). % uncertainty in volume = (0.1/10.0) × 100 = 1% B has a peak corresponding to 43 (loss of CH3). % uncertainty in number of moles = 5% + 1% = 6% 21 26 30 (a) Mass / charge Fragment 15 Absolute uncertainty in number of moles = (6/100) × 0.0100 = 0.0006 CH3 + 29 C 2H 5 + Number of moles = 0.0100 ± 0.0006 mol 43 C3H7 + (loss of CH3) (a) ∆T = 43.2 − 21.2°C = 22.0 °C 58 C4H10 + absolute uncertainty = (±)0.2°C (b) % uncertainty = 0.2/22.0 × 100% ≈ 1% (b) CH3CH2CH2CH3 Molecule 31 Corresponding saturated non-cyclic molecule (c) ∆H = −4.18 × 22.0/0.500 = −184 kJ mol−1 C 6H 6 C6H14 4 CH3COCH3 C 7 H 6 O2 C2H3Cl C 4H 9N C6H12O6 C 3H 8O C7H16O C2H5Cl C4H9NH2 C6H14O6 2 5 1 1 1 (d) 1% (e) absolute uncertainty = 1/100 × 184 = (±) 2 kJ mol−1 (f) experimental value for ∆H = −184 (±) 2 kJ mol−1 The literature value is outside this range. The random errors involved in reading the thermometer do not account for this difference. There are systematic errors. The assumptions on which the calculation is based are not strictly valid. Some of the heat of reaction passes into the surroundings and the other uncertainties in the measurements cannot be ignored. It should also be noted that the standard value for ΔH refers to standard conditions of 298 K and 100 kPa. 22 B 25 Concentration of chromium (from graph for absorbance of 0.215) = 3.34 µg dm−3 23 A 24 IHD 32 B D 35 (a) Empirical formula CH2O. Molecular formula C2H4O2. 33 34 B (b) IHD = 1 (c) O CH3 C OH 36 symmetric stretch IR active B 37 asymmetric stretch IR active symmetric bend IR active B 53 10:01 Answers 38 The polarity (of bond or molecule) changes as the bonds are bent or stretched. 39 Hex-1-ene shows an absorption in the range 1610–1680 cm−1 due to the presence of the C=C bond. 40 C–H bond 41 CH3OCH3 42 C 44 (a) 2 (b) 1 (c) 1 (d) 2 43 A 45 The H atoms are in three different environments. There are three peaks in the 1H NMR spectrum. 46 (a) CH3COCH2CH3 (b) 49 X-ray crystallography. 50 Monochromatic means all the X-rays have the same wavelength. The angle of diffraction depends on the wavelength. If the X-rays have different wavelengths, different diffraction angles/pattern would be obtained. It would be impossible to match the angles with the wavelengths. 51 Hydrogen atoms have a low electron density. 52 The atoms must have a regular arrangement if an ordered diffraction pattern is to be produced. 53 (a) C6H5CH3 (b) Hydrogen atoms do not appear because of their low electron density (c) The saturated non-cyclic compound is C7H16 Type of Chemical No. of H Splitting hydrogen atom shift / ppm atoms pattern CH3CO 2.2–2.7 3 IHD = 12(16 − 8) = 4 (the IHD of a benzene ring = 4) 1 COCH2CH3 2.2–2.7 2 4 CH2CH3 0.9–1.0 3 3 48 Number ofpeaks Chemical shift / ppm No. of H atoms Splitting pattern Challenge yourself 47 2 2.2–2.7 3 2 9.4–10.0 1 4 2.1 6 1 Compound CH3CHO CH3COCH3 1 1 Y1 ave = Y2ave = 3 (−2×(−2)) + (−1×(−1)) + 0 + (1 × 1) + (2 × 2) (−2)2 + (−1)2 + 02 + 12 + 22 =1 R= 2 (−2 × 2) + (−1 × 1) + 0 + (1×(−1)) + (2×(−2)) (−2)2 + (−1)2 + 02 + 12 + 22 = −1 R= (a) Possible structures: CH3CH2COOH, CH3COOCH3, HCOOCH2CH3. (b) The peak at 8.0 ppm corresponds to R–COOH. There is no splitting as there are no hydrogen atoms bonded to neighbouring carbon atoms. The peak at 1.3 ppm corresponds to a CH3 group. The peak is split into a triplet because there is a neighbouring CH2 group. The peak at 4.3 ppm corresponds to the R–CH2–COO group. The peak is split into a quartet as there is a neighbouring CH3 group. Molecular structure: CH3CH2COOH Y1 ave = Y2ave = 3 3 Y1 ave = Y2ave = 3 (−2×(−2)) + (−1 × 2) + 0 + (1 × 1) + (2 × (−1)) 22 + 12 + 02 + 12 + 22 4−2+1−2 = 0.10 = 10 R= 4 Saturated hydrocarbons have the general formula CnH2n+2 For CnHp: H atoms needed = 2n + 2 − p H2 molecules needed = IHD = 12(2n + 2 − p) For CnHpOq: 54 10:01 only this compound would give a peak in the 9.4–10 ppm region / OWTTE [2 max] Oxygen forms two covalent bonds. Comparing ethane, C2H6: C–H, to ethanol, C2H5OH: C–O–H, we see that the presence of O has no impact on the IHD: (b) 2.5 ppm peak CH3–CO–CH3 also has hydrogen atoms on a carbon next to the >C=O group [2] IHD = 21(2n + 2 − p) For CnHpOqNr: (c) (i) 1700–1750 cm−1 (>C= O) Nitrogen forms three covalent bonds. Comparing C–H to C–N–H, we see that the presence of one N increases the IHD by 1: [1] (d) C3H6O+ and m/z = 58 IHD = 21(2n + 2 − p + r) C2H5+ and m/z = 29 For CnHpOqNrXs: CHO+ and m/z = 29 A halogen, X, forms one bond, like hydrogen, so can be treated in the same way: CH3+ and m/z = 15 [2 max] Penalize missing + sign once only. IHD = 21(2n + 2 − p + r − s) 5 [1] (ii) 1610–1680 cm−1 (>C= C<) / 3200–3600 cm−1 (–O–H) 6 E = hf E = 6.63 × 10−34 J s × 3.0 × 1014 s−1 = 1.989 × 10−19 J The energy of one mole of photons = 6.02 × 1023 mol−1 × 1.989 × 10−19 J (a) (stretches/vibrations in) HBr involve change in bond dipole / (stretches/vibrations in) Br2 do not involve change in bond dipole [1] (b) (i) I: O–H II: C–H III: C=O = 120 kJ mol−1 [3] Award [2] for C–H for I and O–H for II. 6 1/λ = 2100 cm−1 = 210000 m−1 (ii) m/z 102: molecular ion peak / (CH3)3CCOOH+ / C5H10O+ / M+ λ = 1/210000 m = 4.76 × 10−6 m m/z 57: (CH3)3C+ / (M–COOH)+ / C4H9+ f = 3.00 × 108 × 210000 = 6.30 × 1013 s−1 7 m/z 45: COOH+ This is the Bragg equation. This is covered in more detail in Chapter 12. [3] Penalize missing + once only. (iii) (H of) COOH group [1] (iv) nine hydrogens in the same environment/ (CH3)3C– (group) Practice questions [1] CH3 O 1 5 B 2 C 3 C 4 A (a) Compound: CH3–CH2 –CHO Explanation: [1 max] only this compound would give 3 peaks / OWTTE (v) (CH3)3CCOOH / (CH3)3CCO2H / H3C C C C [1] OH CH3 (vi) no peak at 11.5 ppm in spectrum of isomer / different chemical shift values four peaks (instead of two) / different number of peaks; only this compound has H atoms in 3 different chemical environments / OWTTE Three of these peaks can be split in actual spectrum, so allow for this in answers if exactly four peaks is not stated. only this compound has protons in ratio 3:2:1 in each environment / OWTTE different integration trace / different areas under the peaks / integration trace 55 10:01 Answers 7 would have a 3:2:2:3 peak area ratio [2 max] presence of peak between 1700 and 1750 cm−1 / peak for C=O Do not award mark if incorrect peak area ratios are given for the structure drawn in (v). absence of peak between 1610 and 1680 [2 max] cm−1 / peak for C=C (b) change in bond length / bond stretching / asymmetric stretch H change in bond angle / bending (of molecule) (a) C3H8O Accept more detailed formula such as CH3CH2CH2OH+. For (a) and (b), if charge is missing penalize once only. (c) (A) CH3CH2CH2OH Accept more detailed formula. [2] Accept more detailed formula. Hydrogen(s) missing, penalize once only. Award [1] if both structures correct but the wrong way round. 9 H H (a) (i) (2-)methylpropan-2-ol the (H atoms in the three) – CH3 groups are responsible for the peak at 1.3 ppm the –OH hydrogen atom is responsible for the peak at 2.0 ppm Accept explanations with suitable diagram. [3] (ii) (2-)methylpropan-1-ol the first peak (at 0.9 ppm) is due to the (H atoms in the) two –CH3 groups (bonded to the second carbon atom) / (CH3)2CHCH2OH the peak at 3.4 ppm is due to the (H atoms in the) –CH2– group / (CH3)2CHCH2OH (a) A: O– H B: C=O C: C–O Award [2] for three correct, [1] for two correct. (b) m/z = 74: C2H5COOH / C3H6O2 + [2] + m/z = 29: C2H5+ Penalize missing + charge once only. Do not award mark for m/z = 29: CHO+. (c) –COOH [3] [1] (d) CH3CH2COOH / CH3CH2CO2H More detailed structural formula may be given. [1] (a) absence of peak between 3200 and 3600 cm−1 / above 3000 cm−1 / peak for OH Accept explanations with suitable diagram. [3] (b) (i) butan-1-ol and butan-2-ol 74: M+ / C4H10O+ / CH3CH2CH2CH2OH+ and CH3CH2CH(OH)CH3 + m/z = 45: COOH+ 10 H O C ECF if structure is incorrect only if its NMR spectrum contains three peaks. [2] 11 [1] (B) CH3CH(OH)CH3 C Ignore order. [1] (b) CH3O+ / CH2OH+ C 3:2:1 induces molecular polarity / dipole moment / OWTTE [3] 8 H Accept CH3CH2CHO. Allow [1 max] for only stating vibrations. + H 59: C3H7O+ / (M – CH3)+ / CH2CH2CH2OH+ and CH2CH(OH)CH3+ / CH3CH2CH(OH)+ 45: C2H5O+ / (M – C2H5)+ / CH2CH2OH+ and CH(OH)CH3+ Accept explained answers instead of formulas. [4] 56 10:01 (ii) butan-1-ol Do not accept answers with reference to 13 C/14C isotopes and peak at m/z = 61. CH2OH+ / (M – C3H7)+ Do not accept C2H3O2+ / C3H7O+ does not exist. [1] Penalize missing + signs once only in parts (b)(i) and (ii). [2] (b) (i) A: C=O and B: C–O (c) they all contain O–H No mark if two bonds are given for AorB. they all contain C–H they all contain C–O Award [1 max] for same functional groups/ bonds. [2 max] (c) (i) this is because there are 3 peaks / 3:2:3 ratio H explanation of splitting into a singlet, a triplet and a quartet O C C H Only penalize once for missing charge in (a)(i) and (ii). [1] (iii) C2H3O2 produced has no charge / fragment produced after loss of C2H5 from molecular ion has no charge C C H H H / (ii) Relative peak area (ii) CH3CH2+ /C2H5+ / CHO [2] + H CH2CO2CH2CH3 First 2.0 3 Second 4.1 2 Quartet 0.9–1.0 1 Triplet Do not award mark if units are given. C4H8O2+ O H Chemical shift / ppm (a) (i) 88 H Peak methyl propanoate / CH3CH2COOCH3 is correct isomer because of higher chemical shift value of singlet (3.6 instead of 2.0–2.5) [4] 13 [1] [1] Do not accept COO. D could be CH3CH2COOCH3 or CH3COOCH2CH3 Third Splitting pattern 12 [1] Ignore names if incorrect. (ii) ester [3] Singlet (iii) (quartet means) neighbouring C: has 3 H atoms [2] Accept fragment(s) too unstable, fragment breaks up etc. 57 Z08_CHE_SB_IBDIP_9755_ANS.indd 57 10:01