IB Chemistry R3.4: Electron-pair sharing reactions worked solutions

\ce{CN-} has a lone pair to donate. \ce{NH4+} and \ce{CH4} have none, and \ce{BF3} accepts a pair.

\ce{OH-} replaces Br, giving the alcohol butan-1-ol. Butanenitrile needs \ce{CN-} and butylamine needs ammonia.

In heterolytic fission both bonding electrons stay with one atom, so a cation and an anion form. Option A and C describe homolytic fission.

Hex-1-ene has a C=C pi bond that polarizes \ce{Br2} and adds it. The other three do not react in the dark.

A. Electrophilic addition puts one bromine atom on each carbon of the double bond: 1,2-dibromopropane. 2-Bromopropane would need hydrogen bromide, and 2,2- or 1,3-dibromopropane are not addition products.

Chlorine is more electronegative than carbon, so the C– Cl bond is polar and carbon is \delta+ (1). The nucleophile is attracted to this electron-poor carbon and donates its lone pair to it (1).

x+4(0)+2(-1)=+1, so x=+3. Neutral ammonia adds no charge; each chloride adds -1.

\ce{AlCl3} has six electrons around aluminium and accepts a lone pair. The other three each have a lone pair to donate.

C. A ligand donates a lone pair to the metal ion. \ce{CN-} has one on carbon. \ce{K+} is a counter-ion, \ce{Fe^3+} is the central ion and the whole bracket is the complex ion.

C. \ce{BF3} accepts a lone pair but has no proton to donate. \ce{CH3COOH}, \ce{NH4+} and \ce{HNO3} all donate \ce{H+}, so they are Brønsted–Lowry acids.

A. \ce{NO2+} is electron-poor and accepts a pair from the delocalized ring. \ce{NO2-} and \ce{HSO4-} are anions with lone pairs, and \ce{NH4+} is not the nitrating species.

(a) Nucleophilic substitution: hydroxide replaces bromide (1). (b) The first arrow starts at a lone pair on the oxygen of \ce{OH-} and ends at the \delta+ carbon (1). The second starts at the C– Br bond and ends at the Br atom, which leaves as \ce{Br-} (1). (c) \ce{CN-} replaces Br, adding one carbon: \ce{CH3CH2CN}, propanenitrile (1).

(a) \ce{CH3CH2CH2C^+HCH3} is secondary (1) and \ce{CH3CH2CH2CH2C^+H2} is primary (1). (b) Alkyl groups donate electron density towards the positive carbon, so the secondary cation, with two, is more stable (1). It forms by the lower-energy pathway and so is favoured (1). (c) Bromide attacks the secondary cation: 2-bromopentane, \ce{CH3CH2CH2CHBrCH3} (1).

(a) The bond is a coordinate (dative covalent) bond (1). Both electrons of the pair come from the lone pair on nitrogen, which donates to the silver ion (Lewis base to Lewis acid) (1). (b) Two ammonia ligands give coordination number 2 (the mark is for both statements): charge =+1+2(0)=1+ (1). (c) Ammonia donates a lone pair, which is what a Lewis base does (1). The carbon accepts the pair, so it is the electrophile and a Lewis acid (1).

(a) 1-Bromobutane is primary: SN2, one concerted step (1). 2-Bromo-2-methylpropane is tertiary: SN1, two steps (1). (b) The reactants are neutral \ce{RBr} and \ce{OH-}, so the transition state [\ce{HO\cdots C\cdots Br}] has charge 0+(-1)=1- (1). (c) The SN1 reaction of the tertiary compound forms the carbocation (1). It is an intermediate: it lies in an energy minimum between two transition states and can in principle be detected, whereas a transition state is a maximum (1).

(a) 1-Iodobutane is fastest, then 1-bromobutane, then 1-chlorobutane (1). (b) \ce{C-I}=228, \ce{C-Br}=285, \ce{C-Cl}=324\,\mathrm{kJ\,mol^{-1}} (1). (c) The carbon–halogen bond is broken in the rate-determining step (1). The weaker the bond, the less energy is needed and the faster the reaction, so bond strength outweighs polarity and the strong C–Cl bond makes the chloroalkane slowest (1).

(a) The \pi electrons of the alkene repel the electrons in \ce{Br-Br}, inducing a \delta+ on the nearer bromine, which then accepts a pair (1). (b) Arrow 1 starts at the C=C bond and ends at the nearer \delta+ Br; arrow 2 starts at the Br–Br bond and ends on the far Br, which leaves as \ce{Br-} (1). The intermediate is a carbocation with charge 1+ (1). Arrow 3 starts at a lone pair on \ce{Br-} and ends at the positive carbon (1). (c) 2,3-dibromobutane, \ce{CH3CHBrCHBrCH3} (1).

(a) Electrophilic substitution: a ring hydrogen is replaced by an electron-pair acceptor (1). (b) \ce{AlCl3} is a Lewis acid, accepting a lone pair from \ce{Cl2} (1). (c) Free-radical substitution (1), attacked by the chlorine radical, \ce{Cl.} (1).

(a) Step 1 is electrophilic addition, giving X, 2-bromobutane (1); step 2 is nucleophilic substitution (1). (b) Ammonia replaces Br: \ce{CH3CH2CH(NH2)CH3}, butan-2-amine (1). (c) Both bonding electrons go to bromine, so \ce{H+} and \ce{Br-} form (1).

(a) For X doubling [\ce{OH-}] changes nothing, so rate =k[\mathrm{X}] (1). For Y the rate doubles with [\ce{OH-}] and so depends on it: rate =k[\mathrm{Y}][\ce{OH-}] (1). (b) X: k=1.5\times10^{-5}/0.050=3.0\times10^{-4}\,\mathrm{s^{-1}} (1). Y: k=3.0\times10^{-6}/(0.050\times0.10)=6.0\times10^{-4}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}} (1). (c) X is tertiary with a rate independent of hydroxide: SN1. Y is primary with a rate depending on both: SN2 (1). (d) Slower: the C– Cl bond (324\,\mathrm{kJ\,mol^{-1}}) is stronger than C– Br (285), so the slow step is slower (1).

(a) \ce{NO2+} (1). (b) Arrow 1 starts at the delocalized pi electrons and ends at nitrogen of \ce{NO2+} (1); this gives a positively charged intermediate in which delocalization is partly lost (1). In step 2 the C– H bond pair moves back into the ring as \ce{H+} is removed by a base (1). (c) An addition product would have no delocalized ring and would give up its stabilization (about 152\,\mathrm{kJ\,mol^{-1}}) (1). Substitution restores the delocalized ring (1).

(a) SN2 (1). The carbon bonded to Br is primary, so it is little hindered and a primary carbocation would be unstable (1). Arrow 1 starts at a lone pair on the oxygen of \ce{OH-} and ends at the \delta+ carbon (1). Arrow 2 starts at the C–Br bond and ends at the bromine, which leaves as \ce{Br-} (1). (b) \ce{H+} adds to the terminal \ce{CH2}, giving the secondary cation \ce{CH3C^+HCH2CH2Br} (1). Two alkyl groups push electron density towards the positive carbon, so it is more stable than the primary alternative (1). Bromide then attacks it: \ce{CH3CHBrCH2CH2Br}, 1,3-dibromobutane (1).

(a) x+5(0)+(-1)=+2, so x=+3 (1). (b) Only the two chloride ions outside the bracket are free: n(\ce{AgCl})=2\times0.010=0.020\,\mathrm{mol} (1). The third chloride is a ligand bonded to cobalt by a coordinate bond, so it does not precipitate (1). (c) \ce{Co^3+} accepts lone pairs, so it is the Lewis acid; the nitrogen atoms of \ce{NH3} and the chloride ion donate lone pairs, so they are Lewis bases (1). Electron pairs, not protons, are transferred, so it is a Lewis reaction only (1).

(a) The orange bromine water becomes colourless (1); electrophilic addition (1). (b) The alkene is \ce{CH3CH=CHCH3}, but-2-ene, which is symmetrical, so H and Br give only 2-bromobutane (1). But-1-ene and 2-methylpropene are unsymmetrical, so each gives two different bromoalkanes, a major and a minor product (1). (c) Water adds as H and OH: \ce{CH3CH(OH)CH2CH3}, butan-2-ol (1).

(a) \ce{CH3CH2CH2CH2Br + OH- -> CH3CH2CH2CH2OH + Br-} (1); n=13.7/137.0=0.100\,\mathrm{mol} (1). (b) Theoretical mass =0.100\times74.1=7.41\,\mathrm{g} (1); yield =4.45/7.41\times100=60.1\% (1). (c) SN2, since the substrate is primary (1). The rate equation is rate =k[\mathrm{RBr}][\ce{OH-}], first order in hydroxide, so doubling [\ce{OH-}] doubles the rate (1).

(a) Only the halogenoalkane appears in the rate equation, so the nucleophile is not in the slow step: SN1 (1). In the slow step the C–Br bond breaks heterolytically, giving a carbocation and \ce{Br-} (1). (b) t_{1/2}=\ln2/k=0.693/(3.0\times10^{-4})=2.3\times10^3\,\mathrm{s} (2 marks: formula and value, 1 each). (c) \ln([\mathrm{R}]_0/[\mathrm{R}])=kt with ratio 10: t=\ln10/k=2.303/(3.0\times10^{-4})=7.7\times10^3\,\mathrm{s} (1 for the method, 1 for the value).

(a) \ce{CH3CH2Cl + 2NH3 -> CH3CH2NH2 + NH4Cl} (1 for the organic product, 1 for the balanced equation). (b) The first \ce{NH3} donates its lone pair to the \delta+ carbon, forming the C–N bond (1). The second accepts \ce{H+} from the intermediate \ce{CH3CH2NH3+} using its lone pair (1). Both roles are donation of the nitrogen lone pair; they differ in the atom that receives the pair, carbon or hydrogen (1). (c) [\ce{OH-}]=\sqrt{K_\mathrm{b}c}=\sqrt{4.5\times10^{-4}\times0.20}=9.5\times10^{-3}\,\mathrm{mol\,dm^{-3}} (1); \mathrm{pOH}=2.02, so \mathrm{pH}=14.00-2.02=11.98 (1).

(a) Bonds broken: \ce{C=C}+\ce{Br-Br}=614+193=807\,\mathrm{kJ\,mol^{-1}} (1). Bonds formed: \ce{C-C}+2\,\ce{C-Br}=346+2(285)=916. \Delta H=807-916=-109\,\mathrm{kJ\,mol^{-1}} (1). (b) The \pi electron density of ethene repels the electrons in \ce{Br-Br}, inducing a dipole (1). The nearer bromine becomes \delta+ and accepts an electron pair from the \pi bond, so bromine acts as an electrophile (1). (c) n=2.80/28.0=0.100\,\mathrm{mol} (1); energy released =0.100\times109=10.9\,\mathrm{kJ} (1).