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R3.4

Electron-pair sharing reactions

Reactivity 3 · Mechanisms of chemical change · SL and HL

🎯What you need to be able to do

  • Recognise nucleophiles and electrophiles in chemical reactions, both neutral and charged.
  • Deduce equations, with descriptions of the movement of electron pairs, for nucleophilic substitution reactions.
  • Explain, with equations, the formation of ions by heterolytic fission, using curly arrows for the movement of electron pairs.
  • Deduce equations for the reactions of alkenes with water, halogens and hydrogen halides.
  • AHL Apply Lewis acid–base theory, and draw Lewis formulas showing coordination bond formation.
  • AHL Deduce the charge on a complex ion from the metal ion and the ligands present.
  • AHL Describe and explain the SN1 and SN2 mechanisms, including the stereospecific nature of SN2, and predict relative rates for different halogenoalkanes.
  • AHL Describe and explain the mechanism of electrophilic addition to symmetrical alkenes, use carbocation stability to predict the major product with an unsymmetrical alkene, and describe electrophilic substitution in benzene.

📚The chemistry

Nucleophiles and electrophiles

Nucleophile
forms a bond by donating both bonding electrons
“nucleus-loving” — attracted to a \( \delta+ \) or positive centre
has a lone pair
e.g. \( \mathrm{OH^{-}} \), \( \mathrm{CN^{-}} \), \( \mathrm{NH_3} \), \( \mathrm{H_2O} \), \( \mathrm{Cl^{-}} \)
Electrophile
forms a bond by accepting both bonding electrons
“electron-loving” — attracted to a region of high electron density
is electron-deficient
e.g. \( \mathrm{H^{+}} \), \( \mathrm{NO_2^{+}} \), \( \mathrm{Br^{\delta+}} \), \( \mathrm{BF_3} \), a carbocation

Note that a nucleophile may be neutral or negatively charged, and an electrophile neutral or positively charged — the definition is about electron pairs, not charge.

Heterolytic fission and curly arrows

Heterolytic fission is the breaking of a covalent bond in which both bonding electrons remain with one of the two fragments, producing a positive ion and a negative ion. It happens readily in polar bonds, where the more electronegative atom already has the greater share.

Curly arrows represent the movement of an electron pair. They are double-barbed, in contrast with the single-barbed fish hooks of R3.3, and two rules govern them:

  • the arrow starts at the electron pair that moves — a lone pair, or the middle of a bond — never at an atom or at a positive charge;
  • the arrow ends where that pair goes — at the atom forming the new bond, or onto an atom as a new lone pair.

Nucleophilic substitution

A halogenoalkane has a polar C–X bond: the halogen is more electronegative, so the carbon carries a \( \delta+ \) charge. That \( \delta+ \) carbon is the site a nucleophile attacks.

\[ \mathrm{Nu^{-}} + \mathrm{R{-}X} \rightarrow \mathrm{R{-}Nu} + \mathrm{X^{-}} \]

The nucleophile donates an electron pair to the carbon to form a new bond, while the C–X bond breaks heterolytically and the halide departs as the leaving group. Standard examples: with \( \mathrm{OH^{-}} \) to give an alcohol; with \( \mathrm{CN^{-}} \) to give a nitrile (which lengthens the carbon chain); with \( \mathrm{NH_3} \) to give an amine. At SL the mechanistic detail is not required — the equation and the electron-pair description are.

Electrophilic addition to alkenes

The C=C double bond has a region of high electron density above and below the plane of the molecule — the \( \pi \) system. That electron density attracts electrophiles, and because the \( \pi \) bond is the weaker of the two, it breaks and the alkene undergoes addition.

with a halogen
\( \mathrm{CH_2{=}CH_2} + \mathrm{Br_2} \rightarrow \mathrm{CH_2BrCH_2Br} \)
the test for unsaturation: bromine water is decolourised
with a hydrogen halide
\( \mathrm{CH_2{=}CH_2} + \mathrm{HBr} \rightarrow \mathrm{CH_3CH_2Br} \)
with water
\( \mathrm{CH_2{=}CH_2} + \mathrm{H_2O} \rightarrow \mathrm{CH_3CH_2OH} \)
steam, high temperature and pressure, acid catalyst

The bromine-water test is worth understanding rather than memorising: bromine is non-polar, but as it approaches the electron-rich double bond the \( \pi \) electrons induce a dipole in it, making one bromine atom \( \delta+ \) — an electrophile. Alkanes have no such electron-rich region, so bromine water is not decolourised by an alkane in the dark. (In UV light it is, but by radical substitution, which is a different reaction — hence “in the dark”.)

AHL Lewis acids and bases

A Lewis acid is an electron-pair acceptor. A Lewis base is an electron-pair donor.

This is a broader definition than Brønsted–Lowry (R3.1): every Brønsted–Lowry base is a Lewis base, since accepting a proton means donating a pair to it, but there are Lewis acids such as \( \mathrm{BF_3} \) and \( \mathrm{Al^{3+}} \) that contain no proton to donate and so are not Brønsted–Lowry acids at all.

The vocabulary maps directly onto the mechanisms: nucleophiles are Lewis bases and electrophiles are Lewis acids. When they react, the pair donated becomes a coordination bond (S2.2), which is why \( \mathrm{NH_3} + \mathrm{BF_3} \rightarrow \mathrm{H_3N{\rightarrow}BF_3} \) is a Lewis acid–base reaction.

Complex ions are the inorganic case: ligands are Lewis bases that donate lone pairs to a transition element cation, the Lewis acid. To deduce the overall charge, add the metal’s oxidation state to the sum of the ligand charges — so \( \mathrm{[Fe(CN)_6]^{4-}} \) contains \( \mathrm{Fe^{2+}} \), since \( +2 + 6(-1) = -4 \).

AHL SN1 and SN2

SN2 — primary halogenoalkanes
one concerted step: the nucleophile attacks the \( \delta+ \) carbon from the side opposite the leaving group, passing through a transition state with five groups partially bonded
rate \( = k[\mathrm{RX}][\mathrm{Nu^{-}}] \) — second order
stereospecific: the configuration is inverted, like an umbrella in the wind
SN1 — tertiary halogenoalkanes
two steps: the C–X bond breaks first to give a carbocation intermediate (slow, rate-determining), then the nucleophile attacks it (fast)
rate \( = k[\mathrm{RX}] \) — first order, independent of nucleophile concentration
the planar carbocation is attacked from either side, so a racemic mixture results

Secondary halogenoalkanes react by both mechanisms. Why the split? Two competing effects. A tertiary carbon is surrounded by three bulky alkyl groups, which sterically hinder attack from behind; and the tertiary carbocation is stabilised by the electron-releasing effect of those three alkyl groups, making the SN1 route viable. A primary carbon is unhindered, and its carbocation would be very unstable, so SN2 is the only option.

The leaving group controls the rate for a given mechanism, and the order is RI > RBr > RCl — iodoalkanes react fastest. The reason is the C–X bond enthalpy: C–I is much the weakest (about 228 kJ mol−1) and C–Cl the strongest (about 324), so the bond that must break in the rate-determining step breaks most easily for iodine. Note this runs opposite to bond polarity — C–Cl is the most polar — and bond strength wins. The roles of the solvent and the mechanism on rate are not assessed.

AHL Electrophilic addition mechanism

For a symmetrical alkene with HBr:

  1. The \( \pi \) electrons of the C=C are attracted to the \( \delta+ \) hydrogen of HBr. A curly arrow runs from the C=C to that hydrogen, and a second from the H–Br bond to the bromine, which leaves as \( \mathrm{Br^{-}} \).
  2. A carbocation is formed.
  3. A curly arrow runs from a lone pair on \( \mathrm{Br^{-}} \) to the positively charged carbon, forming the product.

With a non-polar electrophile such as \( \mathrm{Br_2} \), an extra sentence is needed at step 1: the approaching \( \pi \) electrons induce a dipole in the \( \mathrm{Br_2} \) molecule, so that the nearer atom becomes \( \delta+ \).

AHL Carbocation stability and the major product

With an unsymmetrical alkene there are two possible carbocations, and the reaction goes predominantly through the more stable one.

tertiary carbocation > secondary > primary   (most to least stable)

The reason is that alkyl groups are electron-releasing: each one pushes electron density towards the positive carbon, spreading the charge and lowering the energy. A tertiary carbocation has three such groups; a primary one has only one.

So propene with HBr gives predominantly 2-bromopropane, because the hydrogen adds to the carbon with more hydrogens already, generating a secondary carbocation rather than a primary one. That empirical pattern is Markovnikov’s rule, and the carbocation argument is the explanation behind it — give the explanation, not the rule.

AHL Electrophilic substitution in benzene

Benzene is highly unsaturated yet does not undergo addition. Addition would destroy the delocalised \( \pi \) system and lose the associated stabilisation energy (S2.2), which is energetically unfavourable. Instead benzene undergoes substitution, which preserves the ring.

The mechanism with a charged electrophile \( \mathrm{E^{+}} \):

  1. A curly arrow runs from the delocalised \( \pi \) system to \( \mathrm{E^{+}} \), forming a C–E bond and giving a positively charged intermediate in which the delocalisation is partially broken.
  2. A curly arrow runs from the C–H bond of that same carbon back into the ring, restoring full delocalisation and releasing \( \mathrm{H^{+}} \).

The classic example is nitration, in which a mixture of concentrated nitric and sulfuric acids generates the electrophile \( \mathrm{NO_2^{+}} \). The formation of the electrophile will not be assessed, but a nice linking point is that in that mixture the sulfuric acid protonates the nitric acid — so \( \mathrm{HNO_3} \), normally a strong acid, is here acting as a base.

✏️Worked example

(a) Write an equation for the reaction of 1-bromopropane with aqueous sodium hydroxide, name the organic product, identify the nucleophile and the leaving group, and state the type of bond fission.
(b) AHL State and explain the mechanism 1-bromopropane follows, and contrast it with that of 2-bromo-2-methylpropane, including the rate equation for each.
(c) AHL Predict the major product of the reaction of propene with HBr, and explain the prediction.
(d) AHL State the type of reaction benzene undergoes with \( \mathrm{NO_2^{+}} \), and explain why it is not addition.

(a)

\[ \mathrm{CH_3CH_2CH_2Br} + \mathrm{OH^{-}} \rightarrow \mathrm{CH_3CH_2CH_2OH} + \mathrm{Br^{-}} \]

The organic product is propan-1-ol. The nucleophile is \( \mathrm{OH^{-}} \), which donates a lone pair to the \( \delta+ \) carbon; the leaving group is \( \mathrm{Br^{-}} \). The C–Br bond breaks heterolytically, both bonding electrons going to the bromine.

(b) 1-Bromopropane is a primary halogenoalkane, so it reacts by SN2. The carbon bearing the bromine has only one alkyl group, so it is sterically unhindered and the hydroxide can attack from the side opposite the bromine; and the primary carbocation that SN1 would require is too unstable to form. The reaction is a single concerted step passing through a transition state in which the incoming and leaving groups are both partially bonded. Since both species are involved in the rate-determining step:

\[ \text{rate} = k[\mathrm{CH_3CH_2CH_2Br}][\mathrm{OH^{-}}] \qquad \text{second order} \]

2-Bromo-2-methylpropane is tertiary, so it reacts by SN1. The three bulky methyl groups block backside attack, and the tertiary carbocation is stabilised by their electron-releasing effect, so the C–Br bond breaks first in a slow, rate-determining step, and the hydroxide then attacks the carbocation rapidly. Only the halogenoalkane appears in the rate-determining step:

\[ \text{rate} = k[\mathrm{(CH_3)_3CBr}] \qquad \text{first order} \]

A further contrast: SN2 is stereospecific, inverting the configuration at the carbon, whereas the planar SN1 carbocation is attacked from either face and gives a racemic mixture.

(c) Propene is \( \mathrm{CH_3CH{=}CH_2} \), an unsymmetrical alkene, so the \( \mathrm{H^{+}} \) can add to either double-bonded carbon.

  • Adding H to carbon 1 (the CH2 end) gives a secondary carbocation, \( \mathrm{CH_3\overset{+}{C}HCH_3} \).
  • Adding H to carbon 2 gives a primary carbocation, \( \mathrm{CH_3CH_2\overset{+}{C}H_2} \).

The secondary carbocation is more stable, because it has two electron-releasing alkyl groups spreading the positive charge rather than one. The reaction therefore proceeds predominantly through it, and the \( \mathrm{Br^{-}} \) attacks that carbon. The major product is 2-bromopropane, with 1-bromopropane as the minor product.

(d) Electrophilic substitution. Benzene has a delocalised \( \pi \) system spread over all six carbons, which makes it substantially more stable than a structure with three localised double bonds. An addition reaction would use up two of the delocalised electrons in forming \( \sigma \) bonds and would therefore destroy the delocalisation, losing that stabilisation energy — which is energetically unfavourable. Substitution replaces a hydrogen and restores the delocalised system in the second step, so the stabilisation is retained. This is why benzene does not decolourise bromine water in the dark, whereas an alkene does.

Check it. For mechanism questions, check the classification of the carbon first — count how many carbons are attached to the one bearing the functional group — because everything else follows from it: primary means SN2 means second order; tertiary means SN1 means first order. For carbocation questions, check that your major product corresponds to the more substituted carbocation, and confirm by counting alkyl groups on the positive carbon. For curly arrows, check that every arrow starts at an electron pair — a lone pair or a bond — and never at a positive charge or a bare atom.
Drawing the curly arrow from \( \mathrm{H^{+}} \) to the double bond. It goes the other way. Arrows show where the electrons move, and \( \mathrm{H^{+}} \) has none to give — the electron pair comes from the C=C and goes to the hydrogen. The same error appears as an arrow starting at a carbocation’s positive charge instead of at the nucleophile’s lone pair. Both lose the mark even when the products are right, because the mechanism is what is being examined. And in an SN2 transition state, draw partial bonds with dashed lines and the negative charge delocalised across both — not a carbon with five full bonds, which is impossible.

📝Practise

Work through these on paper, then reveal the answer.

1. Classify each as a nucleophile or an electrophile, and justify: \( \mathrm{CN^{-}} \), \( \mathrm{NO_2^{+}} \), \( \mathrm{NH_3} \), \( \mathrm{BF_3} \), \( \mathrm{H_2O} \).
\( \mathrm{CN^{-}} \) — nucleophile. It is negatively charged and has a lone pair on carbon which it donates to form a new bond. \( \mathrm{NO_2^{+}} \) — electrophile. It is positively charged and electron-deficient, so it accepts an electron pair; it is the species that attacks benzene in nitration. \( \mathrm{NH_3} \) — nucleophile, despite being neutral: nitrogen has a lone pair available to donate. \( \mathrm{BF_3} \) — electrophile, despite being neutral: boron has only six electrons in its valence shell, so it is electron-deficient and readily accepts a pair (it is the classic Lewis acid). \( \mathrm{H_2O} \) — nucleophile, again neutral: oxygen has two lone pairs, one of which it can donate, which is what happens in the hydrolysis of a halogenoalkane. The lesson is that charge does not determine the classification — the availability or deficiency of an electron pair does.
2. Write equations and name the organic products for the reaction of 1-chlorobutane with (a) aqueous KOH, (b) ethanolic KCN, (c) excess NH3. State what each nucleophile contributes.
(a) \( \mathrm{CH_3CH_2CH_2CH_2Cl} + \mathrm{OH^{-}} \rightarrow \mathrm{CH_3CH_2CH_2CH_2OH} + \mathrm{Cl^{-}} \) — butan-1-ol. The nucleophile \( \mathrm{OH^{-}} \) donates a lone pair from oxygen. (b) \( \mathrm{CH_3CH_2CH_2CH_2Cl} + \mathrm{CN^{-}} \rightarrow \mathrm{CH_3CH_2CH_2CH_2CN} + \mathrm{Cl^{-}} \) — pentanenitrile. Note that this reaction lengthens the carbon chain by one, which makes it synthetically valuable; the nitrile carbon becomes carbon 1 in the name. (c) \( \mathrm{CH_3CH_2CH_2CH_2Cl} + \mathrm{NH_3} \rightarrow \mathrm{CH_3CH_2CH_2CH_2NH_2} + \mathrm{HCl} \) — butan-1-amine. Ammonia donates the lone pair on nitrogen; excess ammonia is used because the amine product is itself a nucleophile and would otherwise react further. In every case the mechanism is nucleophilic substitution, the C–Cl bond breaks heterolytically, and \( \mathrm{Cl^{-}} \) is the leaving group.
3. Explain why bromine water is decolourised by ethene but not by ethane in the dark, and write the equation for the reaction that does occur.
Ethene decolourises it. The C=C double bond contains a \( \pi \) system with a high electron density above and below the plane of the molecule. As a bromine molecule approaches, those \( \pi \) electrons repel the electrons in the Br–Br bond and thereby induce a dipole, making the nearer bromine atom \( \delta+ \). That \( \delta+ \) atom is an electrophile and is attacked by the \( \pi \) electrons, and electrophilic addition follows: \( \mathrm{CH_2{=}CH_2} + \mathrm{Br_2} \rightarrow \mathrm{CH_2BrCH_2Br} \), 1,2-dibromoethane, which is colourless — hence the decolourisation. Ethane does not. It is saturated, with no \( \pi \) system and no region of high electron density, so it cannot induce a dipole in the bromine and there is nothing for an electrophile to attack. Its C–C and C–H bonds are also strong and essentially non-polar. Why “in the dark” matters: in ultraviolet light, ethane does decolourise bromine, but by radical substitution (R3.3), producing bromoethane and HBr — a completely different mechanism, which is why the test must be carried out in the dark.
4. AHL Explain why the hydrolysis of 1-iodobutane is faster than that of 1-chlorobutane, and why 2-bromo-2-methylpropane hydrolyses faster than 1-bromobutane.
Iodo against chloro: both are primary and both react by SN2, so the mechanism is the same and the difference must lie in the leaving group. The rate-determining step requires the C–X bond to break, and the C–I bond is much weaker than C–Cl (about 228 against 324 kJ mol−1), because iodine is a much larger atom so the shared pair is further from the nuclei and the overlap is poorer. Less energy is needed, so the reaction is faster. Note this is the opposite of what bond polarity alone would predict — C–Cl is the more polar bond — so bond strength is the deciding factor, and saying “C–I is more polar” is wrong. Tertiary against primary: here the mechanisms differ. 2-bromo-2-methylpropane is tertiary and reacts by SN1, whose rate-determining step is the formation of a tertiary carbocation that is well stabilised by the electron-releasing effect of three alkyl groups, so its activation energy is comparatively low. 1-bromobutane is primary and must react by SN2, which has a higher activation energy because the crowded transition state must form. The tertiary compound therefore hydrolyses faster.
5. AHL Describe the mechanism of the reaction of but-2-ene with HBr, giving the curly arrows in words, and explain why but-1-ene gives two products but but-2-ene gives only one.
Mechanism. Step 1: a curly arrow runs from the C=C \( \pi \) bond to the \( \delta+ \) hydrogen of HBr, forming a new C–H bond; simultaneously a second curly arrow runs from the H–Br bonding pair onto the bromine, which leaves as \( \mathrm{Br^{-}} \). This is heterolytic fission of the H–Br bond. A carbocation intermediate is produced. Step 2: a curly arrow runs from a lone pair on \( \mathrm{Br^{-}} \) to the positively charged carbon, forming the C–Br bond and giving the product. Why but-2-ene gives one product. \( \mathrm{CH_3CH{=}CHCH_3} \) is symmetrical about the double bond: adding H to either double-bonded carbon gives the same secondary carbocation, so there is only one possible product, 2-bromobutane. Why but-1-ene gives two. \( \mathrm{CH_2{=}CHCH_2CH_3} \) is unsymmetrical: adding H to carbon 1 gives a secondary carbocation and hence 2-bromobutane, while adding H to carbon 2 gives a primary carbocation and hence 1-bromobutane. The secondary carbocation is more stable, because two electron-releasing alkyl groups spread the positive charge rather than one, so 2-bromobutane is the major product and 1-bromobutane the minor one.
6. AHL (a) Deduce the oxidation state of the metal and the type of bonding in \( \mathrm{[Cr(H_2O)_6]^{3+}} \) and \( \mathrm{[CoCl_4]^{2-}} \). (b) Explain how Lewis acid–base theory relates to nucleophiles and electrophiles, and give one Lewis acid that is not a Brønsted–Lowry acid.
(a) \( \mathrm{[Cr(H_2O)_6]^{3+}} \): water is neutral, so the charge is carried entirely by the metal — chromium is +3. \( \mathrm{[CoCl_4]^{2-}} \): \( x + 4(-1) = -2 \), so cobalt is +2. In both, the ligands donate a lone pair to the metal ion, so the metal–ligand bonds are coordination (dative covalent) bonds, in which both electrons of the shared pair come from the ligand. (b) A Lewis base is an electron-pair donor and a Lewis acid is an electron-pair acceptor. That is exactly the definition of a nucleophile and an electrophile respectively, so every nucleophile is a Lewis base and every electrophile is a Lewis acid — the two vocabularies describe the same event, one in organic and one in general terms. Ligands are Lewis bases; transition element cations are Lewis acids. A Lewis acid that is not a Brønsted–Lowry acid: \( \mathrm{BF_3} \). It accepts an electron pair readily, because boron has only six valence electrons, but it contains no hydrogen at all and therefore cannot donate a proton. \( \mathrm{Al^{3+}} \) and \( \mathrm{AlCl_3} \) are equally good answers. This is why the Lewis definition is described as broader: it includes everything Brønsted–Lowry covers, and more.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Chemistry LibreTexts — SN1 and SN2 with animated stereochemistry, which makes the inversion of configuration in SN2 obvious in a way a static diagram never manages.
  • MolView or any 3D builder — build a tertiary halogenoalkane and rotate it, and the steric hindrance argument stops being an assertion.
  • Your data booklet — the bond enthalpy table, which is where the C–I against C–Cl leaving-group argument comes from, and which you should quote numbers from rather than asserting relative strengths.