Chemistry Answers Chapter 10

Answers  Chapter 10 Exercises 1 (a) carboxylic acid; butanoic acid (b) halogenoalkane; 1,1-dichloropropane Cl (c) ketone; butanone (d) ester; methyl ethanoate (e) ether; methoxyethane Cl (f) ester; ethyl pentanoate 2 (a) CH3(CH2)4COOH (b) CH3CH2CH2CHO Cl (c) CH2CH(CH2)2CH3 (d) CH2BrCH(CH3)C2H5 or CH2BrCH(CH3)CH2CH3 (e) HCOOCH2CH3 Cl (f) CH3OCH2CH2CH3 (g) CH3C≡CCH3 3 5 A Cl 4 D Cl Cl H C C C Cl Cl H H 1,1,1,2,2-pentachloropropane Cl Cl Cl C C C Cl H H Cl H Cl C C C Cl H H Cl Cl Cl C C C H Cl H Cl Cl Cl C C C H H H H 1,1,1,2,3-pentachloropropane Cl 1,1,1,3,3-pentachloropropane H 1,1,2,2,3-pentachloropropane Cl 1,1,2,3,3-pentachloropropane 6 B 7 Benzene is a cyclic molecule with a planar framework of single bonds between the six carbon atoms and six hydrogen atoms. The carbon atoms are also bonded to each other by a delocalized cloud of electrons which forms a symmetrical region of electron density above 46 10:01 and below the plane of the ring. This is a very stable arrangement, so benzene has much lower energy than would be expected. 8 (a) Similar molar mass will mean molecules have approximately equal London (dispersion) forces and so differences in boiling point can be attributed to differences in dipole–dipole or hydrogen bonding. (b) C2H5COOH(aq) + C4H9OH(aq) → C2H5COOC4H9(aq) + H2O(l) 14 (c) no reaction; no colour change 15 (b) Solubility in hexane will increase with increasing chain length as the non-polar part of the molecule makes a larger contribution to its structure. 9 (a) C5H14(l) + 6O2(g) → 5CO(g) + 7H2O(l) Nucleophilic substitution involves an electronrich species (e.g. OH−) attacking an electrondeficient carbon atom (e.g. in chloroethane), leading to substitution of the halogen functional group by the nucleophile. C2H5Cl + OH− → C2H5OH + Cl− 16 Benzene has a very stable structure as a result of its symmetrical ring of delocalized electrons. Addition reactions would involve breaking this ring and therefore decreasing its stability. Substitution reactions in which one or more hydrogen atoms of the ring are replaced by other atoms or groups preserves the aromatic ring structure and therefore its stability. 17 (a) CH3CH2CH2CH2Br (b) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l) (c) C3H4(g) + O2(g) → 3C(s) + 2H2O(l) 10 (a) butanone; orange → green (b) methanal; orange → green Bromine + ethane initiation propagation primary CH3CH2CHBrCH3 secondary C(CH3)3Br tertiary (b) The tertiary halogenoalkane reacts by an SN1 mechanism. termination S = substitution; N = nucleophilic; 1 = unimolecular (c) RBr → R+ + Br− Overall, these reactions show how a mixture of products is formed. 11 12 (a) CH3CH2CH2CH3 butane (b) CH3CH2CH(OH)CH3 butan-2-ol (c) CH3CH2CHBrCH3 2-bromobutane (a) No observable change. (b) Burns with very smoky flame. (c) The bromine water changes from brown to colourless. 13 (a) C2H5OH(l) + 3O2 → 2CO2(g) + 3H2O(l) 2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(l) 18 C 19 (a) The carbon–halogen bond breaks more easily in the iodo- and bromo- derivatives than in the chloro- derivatives, so these compounds more readily undergo substitution reactions. (b) The substitution reaction of OH for Cl occurs in both these compounds, displacing Cl− and forming the white precipitate of AgCl, which darkens on exposure to air. The tertiary halogenoalkane C(CH3)3Cl isomer reacts more quickly than the primary isomer CH3CH2CH2CH2Cl because it undergoes an SN1 mechanism, which is faster. 47 10:01 Answers 20 Alkenes have a double bond which is an electrondense region and so is susceptible to attack by electrophiles which are themselves electron deficient. They undergo addition reactions because they are unsaturated; one of the bonds in the double bond breaks and incoming groups can add to the two carbon atoms. When bromine approaches but-2-ene, it is polarized by the electron density in the double bond. Electrons in the bromine–bromine bond are repelled away from the double bond, leading to the heterolytic fission of the bromine molecule. The Br+ product now attaches itself to one of the carbon atoms as the carbon–carbon bond breaks. This produces an unstable carbocation which then rapidly reacts with the Br− ion. The product is 2,3-dibromobutane. H H H H H C C C C H H H C 21 C 1 H H 23 C C Br H H H H H C C C C H Br Br H H C C C Br2 H H H C C 1 C H Cl2 H H H H C C C I Cl H H Concentrated H2SO4 and concentrated HNO3. (a) Use LiAlH4 in dry ether and heat. The acid is reduced first to the aldehyde and then to the alcohol. [+H] CH3CH2COOH C2H5CH2OH (b) Nitrobenzene is heated under reflux with tin and concentrated HCl, and the product is reacted with NaOH. H C6H5NO2 [+H] C6H5NH2 (c) Ethanal is heated with NaBH4(aq). But-1-ene + HBr → 2-bromobutane ICl is polarized: Iδ+ Clδ− owing to the greater electronegativity of Cl than I. So when it undergoes heterolytic fission it will form I+ and Cl−. By application of Markovnikov’s rule, the I+ will attach to the terminal carbon, while Cl− will add to carbon 2. The product is therefore 1-iodo-2-chloropropane. I1 H The stronger acid H2SO4 protonates the HNO3, leading to production of the nitronium ion NO2+. This is a strong electrophile which reacts with the π electrons of the benzene ring, substituting for H. 24 H H I Br1 Application of Markovnikov’s rule enables us to predict that the electrophile H+ will add to the terminal carbon-forming a secondary carbocation, as this is stabilized by the positive inductive effect of the alkyl groups. Br− will then add to carbon 2, forming 2-bromobutane. 22 H Brδ2 H H H Brδ1 H H H H H CH3CHO 25 [+H] CH3CH2OH Start with ethanol. Take one portion and oxidize it using acidified potassium(VI) dichromate solution and heat under reflux to allow the reaction to go to completion. C2H5OH [+O] CH3COOH The product is ethanoic acid. React the ethanoic acid product with another portion of the ethanol by warming it in the presence of some concentrated H2SO4. The esterification reaction yields ethyl ethanoate. CH3COOH + C2H5OH → CH3COOC2H5 48 10:01 26 React the 1-chlorobutane with NaOH in warm aqueous solution to convert it into butan-1-ol. Challenge yourself C4H9Cl + NaOH → C4H9OH + NaCl Oxidize the butan-1-ol using acidified potassium(VI) dichromate solution and heat under reflux to allow the reaction to go to completion. [+O] C4H9OH 1 All four C atoms in the molecule are sp3 hybridized because they form four single bonds which are tetrahedrally arranged. The nitrogen atom is also sp3 hybridized, as its four electron domains are also tetrahedrally arranged. Note that here the hybridization also includes the lone pair on the nitrogen atom. 2 Complete combustion: 2C2H6 + 7O2 → 4CO2 + 6H2O C3H7COOH 27 C 28 3-methylhexane: CH3CH2CH(CH3)CH2CH2CH3 H C: −3 → +4 C C 2H 5 29 (a) H CH3 Incomplete combustion: 2C2H6 + 5O2 → 4CO + 6H2O C 3H 7 C: −3 → +2 H H H H H C C C C C H H H H C C C H H H H 3 Heterolytic describes breaking of the bond, producing two different products. The products are ions, and the reaction mechanism involves attraction of the electron density of the C=C double bond to the positive ion. 4 The repeating unit in polystyrene is —CH(C6H5)–CH2— 5 The cyanide ion, CN−, and ammonia, NH3, are nucleophiles that react with halogenoalkanes in substitution reactions. They act as ligands with transition metal ions, forming complexes such as [Cu(NH3)4(H2O)2]2+ and [Cu(CN)4]3−. They act as Lewis bases by donating a lone pair of electrons. For example: H Z-pent-2-ene H H H C C H H E-pent-2-ene (b) H H Cl Cl H C C C C H H H NH3 + BCl3 → NH3BCl3 Z-2,3-dichlorobut-2-ene 6 H H Cl C C H H C C Cl H E-2,3-dichlorobut-2-ene H The order of the reaction with respect to each reactant can be deduced from experiments in which the concentration of each reactant in turn is changed, and the initial rate of the reaction then measured. If, for example, the concentration of halogenoalkane is doubled while the concentration of OH− remains constant, and the rate is found to have doubled, then it indicates that the reaction is first order with respect to halogenoalkane. Examples of this type of experiment and the interpretation of the results are given in Chapter 6. 49 10:01 Answers 7 8 With bromine water, the water can also take part in the second part of the reaction because of its lone pairs. The carbocation is attacked by water in competition with Br−, and the major product is 2-bromoethanol, CH2BrCH2OH. The bromine water is decolorized from brown. The relative concentration of bromoethanol and 1,2-dibromoethane depends on the strength of the bromine water used. The –NH2 group in phenylamine is electron donating due to conjugation of the lone pair of electrons on N with the ring electrons. As a result, the electron density of the ring is increased, making it more susceptible to electrophilic attack. In contrast, the –NO2 group in nitrobenzene is electron withdrawing due to the electronegativity of the nitrogen and oxygen atoms. d2 O d2 O Also, the electrons in its double bond conjugate with the π electrons in the ring, causing the electron density of the ring to be decreased, making it less susceptible to electrophilic attack. In both square planar and octahedral compounds, geometric isomers can arise due to groups having the possibility of being in adjacent (cis) or in across (trans) positions. In tetrahedral compounds, all positions are adjacent to each other, so these isomers are not possible. H 10 Practice questions 1 C 2 C 3 D 4 A 5 A 6 A 7 B 8 C 9 B 10 B 11 A 12 C 13 A 14 D 15 (a) A: 1-bromobutane B: 2-bromobutane C: 2-bromo-2-methylpropane D: 1-bromo-2-methylpropane N 9 than the trans isomer, and its density is less. cis-Butenedioic acid is a stronger acid because when H+ is lost, the cis anion is more stable than the trans form. (b) (i) C / 2-bromo-2-methylpropane; unimolecular nucleophilic substitution [2] (ii) RBr → R+ + Br− (iii) A / 1-bromobutane / D / 1-bromo-2methylpropane OH H H − HO C CH3CH2CH2 O C Br C H O H O Br C CH3CH2CH2 H C H [1] Allow use of 2-bromo-2-methylpropane instead of RBr. H C [4] Penalize incorrect punctuation, e.g. commas for hyphens, only once. Accept 2-bromomethylpropane and 1-bromomethylpropane for C and D respectively. O H C HO OH + Br H − − H CH2CH2CH3 (CH3)2CH Br H cis-butenedioic acid melting point 139 °C The cis isomer, maleic acid, has a lower melting point as it forms fewer intermolecular bonds. The cis isomer is much more soluble in water (CH3)2CH H − H HO C Br C HO H + Br − H CH(CH3)2 50 10:01 curly arrow going from lone pair/negative charge on O in OH– to C No mark for change to clear, or for decolorized with no reference to original colour. (b) Chloroethene: Do not allow curly arrow originating on H in OH−. H curly arrow showing Br leaving C Accept curly arrow either going from bond between C and Br to Br in 1-bromobutane or in the transition state. representation of transition state showing negative charge, square brackets and partial bonds Cl No mark if the lone pairs are missing on Cl. Accept lines, dots or crosses for e– pairs. Poly(chloroethene): [ Do not penalize if HO and Br are not at 180° to each other. CH2 (b) (iii) rate doubles as the rate is proportional to [OH–] / OH– appears in the ratedetermining / slow step / first order with respect to OH– [2] Accept other commercial applications. 17 (a) A: H (a) Colour change from yellow / orange / rust colour / red / brown to colourless [1] H H H C C C Cl H H H C H H B: H Accept ‘turn’ instead of ‘rotate’ but not ‘bend’/‘reflect’. 16 [2] n (d) rate of 1-bromobutane is faster; C–Br bond is weaker / breaks more easily than C–Cl bond [2] Do not accept ‘similar’ in place of ‘identical’. ] (c) (hydration of ethene for the manufacture of) ethanol / C2H4 + H2O → C2H5OH; (synthesis of) CH3COOH / ethanoic / acetic acid; (synthesis of) ethylene glycol / 1,2-ethanediol / ethane-1,2-diol; (synthesis of) drugs / pesticides; (hydrogenation of unsaturated oils in the manufacture of) margarine [2 max] (c) (b) (i) no change as [OH–] does not appear in the rate equation / in the rate determining step Physical properties identical (apart from effect on plane-polarized light); chemical properties are identical (except with other chiral compounds) [5] CHCl n and square brackets are not required. Continuation bonds must be shown. Do not award fourth mark if OH–C bond is represented. (e) 2-bromobutane / B; (plane-) polarized light shone through; enantiomers rotate plane of plane-polarized light to left or right / opposite directions (by same amount) C H [4] Award [1] if correctly predicts no rate change for SN1 and doubling of rate for SN2 without suitable explanation. H H H H C C C OH H H H C H H C: H H H C C H O C H C H H H 51 10:01 Answers of A and D. M2 (for B) and M5 (for E) may also be scored for substitution product if primary chloroalkane used. Penalize missing hydrogens once only in Q7. H D: H H H C H C C Cl C H (b) CH3CH2COOH + CH3OH → CH3CH2COOCH3 + H2O H H [1] for reactants and [1] for products. (concentrated) sulfuric acid / H2SO4 H E: Do not accept just H+ or acid. methyl propanoate H H H H C C 18 H C OH C H Award [1 max] if A and D are other way round (and nothing else correct). Award [2 max] if A and D are other way round but one substitution product B or E is correct based on initial choice of A and D. Award [3 max] if A and D are other way round but both substitution products B and E are correct based on initial choice [1] (b) +6 [1] (c) Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O H Accept condensed formulas. (a) (the solution changes) from orange to green Do not accept 6, 6+ or the use of Roman numerals unless they have already been penalized in 2(a). H H [4] [5] [1] (d) CH3CH2OH → CH3CHO + 2H+ + 2e– Cr2O72– + 3CH3CH2OH + 8H+ → 2Cr3+ + 3CH3CHO + 7H2O [3] For second equation award [1] for correct reactants and products and [1] for correct balancing. (e) H+ is a reactant / OWTTE [1] (f) ethanoic acid / CH3COOH / acid [1] Accept acetic acid. 10:01