3.4 Electron-pair sharing reactions

Learning objectives

3.4.1Nucleophiles• Donate electron pair to electrophile• Recognize nucleophiles as electron-rich species 3.4.2Nucleophilic substitution• Nucleophile donates pair, forms new bond• Another bond breaks, leaving group departs• Deduce substitution products from nucleophile and leaving group 3.4.3Heterolytic fission• Both electrons stay with one fragment• Forms ions• Use curly arrows to show electron-pair movement 3.4.4Electrophiles• Accept electron pair from nucleophile• Recognize electrophiles as electron-deficient species 3.4.5Electrophilic addition to alkenes• High electron density at C=C bond• Reactions: + H₂O, + halogens, + hydrogen halides• Deduce alkene addition equations; SL mechanisms are not assessed 3.4.6Lewis acid-base theory• Lewis acid = electron-pair acceptor• Lewis base = electron-pair donor• Classify nucleophiles/electrophiles using Lewis acid-base theory 3.4.7Coordination bond formation• Lewis base + Lewis acid• Nucleophiles = Lewis bases• Electrophiles = Lewis acids• Explain coordination bond formation with electron-pair donation 3.4.8Complex ions• Ligands donate electron pairs to transition metal cations• Identify ligands and central transition metal cations in complex ions 3.4.9(HL)—Nucleophilic substitution mechanisms• Primary halogenoalkanes: SN2• Tertiary halogenoalkanes: SN1• Distinguish one-step SN2 and two-step SN1; include stereospecific SN2 3.4.10(HL)—Leaving group effects• Identity influences rate• Link carbon-halogen bond enthalpy and polarity to reaction rate 3.4.11(HL)—Electrophilic addition mechanisms• Symmetrical alkenes + halogens/water/HX• Describe mechanisms with curly arrows and carbocation/halonium intermediates as appropriate 3.4.12(HL)—Carbocation stability• Unsymmetrical alkenes + HX or H₂O• Major product from more stable carbocation• Predict and explain major products from carbocation stability 3.4.13(HL)—Electrophilic substitution in benzene• Mechanism with charged electrophile E⁺• Formation of the electrophile is not assessed

A nucleophile is an electron-rich species that donates an electron pair to an electron-deficient centre.

both curved arrows begin at the nucleophile electron pair and end at E; Nu- gives a negatively charged adduct while neutral Nu gives a neutral adduct; the coordinate bond is shown as electron-pair donation from Nu to E.

Look for an available electron pair: OH⁻ and CN⁻ use a negative charge and lone pair, while NH₃ uses a lone pair without being an anion. A curly arrow must start at that pair and point toward the atom where the new bond forms.

1 mark

Identify a nucleophile which could be used for this reaction.

The nucleophile donates a pair to carbon while the leaving group departs with its bonding pair. Deduce the product by replacing the leaving group with the nucleophile.

the hydroxide arrow starts at oxygen and ends at the carbon bearing chlorine; the leaving-group arrow starts at the C-Cl bond and ends at chlorine; carbon is delta positive and chlorine delta negative; CH3CH2OH and Cl- conserve every atom and charge.

For CH₃CH₂Br + OH⁻, the C–O bond forms as the C–Br bond breaks, giving CH₃CH₂OH + Br⁻. Account for charge and every atom in the product; the leaving group takes the bonding pair rather than departing as a neutral bromine atom.

4 marks

Explain the mechanism of the reaction, using curly arrows to represent the movement of electron pairs.

In heterolytic fission both bonding electrons remain with one fragment, producing ions. Curly arrows show movement of an electron pair.

a full two-electron curly arrow begins at the A-B bond and ends at B; both bonding electrons are transferred to B; the products are exactly A plus and B minus.

Place the curly-arrow tail on the bond being broken and its head on the fragment receiving both electrons. Then assign charges from electron ownership: heterolysis creates ions, unlike homolysis, which gives one electron to each radical.

2 marks

Contrast homolytic and heterolytic fission.

Homolytic fission:

Heterolytic fission:

An electrophile is an electron-deficient species that accepts an electron pair from a nucleophile.

BF3 is trigonal planar with exactly three B-F single bonds; boron is labelled delta positive and each fluorine delta negative; no lone pair, formal charge, or fourth ligand is added to boron.

the skeletal structure is butanoic acid with four carbons; the C=O double bond and O-H bond are distinct; the carbonyl carbon is delta positive and carbonyl oxygen delta negative.

Identify the electron-poor atom, not merely a positive-looking formula. H⁺ and carbocations are electrophiles, and the δ⁺ carbon in a polar C–X bond can also accept a pair. The incoming curly arrow ends at this acceptor.

1 mark

Which species is the electrophile?

\mathrm{CH}_{3} \mathrm{Br}+\mathrm{OH}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{Br}^{-}

The electron-rich C=C attacks an electrophile. Deduce addition products with water, halogens or hydrogen halides within the SL mechanism boundary.

the C=C pi pair attacks Br delta positive and the Br-Br pair moves to Br delta negative; the textbook bromoethyl carbocation plus Br- intermediate is preserved; Br- attacks C+ and forms neutral 1,2-dibromoethane; two carbons, four hydrogens, and two bromines are conserved.

Treat the C=C as the reactive site and place the two added groups on its two carbon atoms. Bromine addition removes the double bond and forms a dibromoalkane; hydration forms an alcohol. At SL, deducing these products does not require a mechanism.

Reagent	Groups added across C=C	Product check
X₂ (for example Br₂)	X and X	vicinal dihalogenoalkane; C=C becomes C–C
HX	H and X	halogenoalkane; conserve the H and halogen from HX
H₂O/steam under acid-catalysed hydration conditions	H and OH	alcohol; conserve the carbon skeleton

At SL, use reagent and atom conservation to deduce products; curly-arrow mechanisms are not assessed in this card.

1 mark

Predict the product of the reaction between ethene and bromine.

A Lewis acid accepts an electron pair; a Lewis base donates an electron pair. Nucleophiles correspond to Lewis bases and electrophiles to Lewis acids.

In BF₃ + NH₃ → F₃B←NH₃, NH₃ donates the pair and is the Lewis base; BF₃ accepts it and is the Lewis acid. Classify the roles from electron-pair movement rather than from whether H⁺ appears.

1 mark

What is the role of the $\mathrm{CN}^{-}$ ion in the reaction of 1-chloropropane with excess KCN in ethanol?

\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{Cl}+\mathrm{KCN} \rightarrow \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{CN}+\mathrm{KCl}

A ligand acts as a Lewis base and donates an electron pair to a Lewis-acid transition-metal cation, forming a coordinate bond.

the full curly arrow starts at the nitrogen lone pair and ends at boron; reactants contain exactly BF3 and NH3; the product contains one B-N coordination bond with three F and three H; no proton transfer or extra ligand is implied.

Lewis base donates an electron pair to Lewis acid; nucleophile donates an electron pair to electrophile; ligand donates an electron pair to transition element cation; all three arrows run from donor on the left to acceptor on the right.

Show a coordination bond with an arrow from a ligand lone pair to the metal ion. The arrow records the origin of the shared pair; after formation the bond is not a different electrostatic species from other covalent bonds.

1 mark

Outline how ammonia acts as a Lewis base when it forms the complex ion

\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{2+}(\mathrm{aq}) .

Identify the central transition-metal cation and the surrounding ligands. Each ligand donates an electron pair to the metal centre.

the bracketed ion contains one Cu and exactly six neutral H2O ligands; oxygen is the donor atom directed toward Cu; the six-coordinate geometry is octahedral; the overall charge is exactly 2 plus.

Read [Cu(NH₃)₄]²⁺ as one Cu centre with four NH₃ ligands and coordination number 4. Use ligand charges and the overall bracket charge to deduce the metal oxidation state; do not confuse coordination number with oxidation state.

1 mark

Which statements are correct for the complex ion $\left[\mathrm{FeCl}_{4}\right]^{2-}$ ?

I. Chloride ions are behaving as ligands.
II. The oxidation state of iron is +3 .
III. Iron ion forms coordination bonds with chloride ions.

Mechanism	Typical substrate/context	Steps and stereochemical result
SN2	primary; some secondary	one concerted backside attack; inversion at a stereogenic centre
SN1	tertiary; some secondary	two steps through a planar carbocation; both configurations can form

hydroxide attacks opposite bromine in one concerted step; the transition state is bracketed, overall negative, and has partial HO-C and C-Br bonds; no carbocation intermediate is introduced; ethanol and Br- are both shown with conserved atoms and charge.

C-Cl heterolysis occurs before hydroxide attack; the tertiary carbocation has three CH3 substituents and an explicit positive charge; hydroxide donates its electron pair to C+ in the second step; Cl- appears once as the leaving-group product.

SN2 is a concerted backside attack: the nucleophile bonds as the carbon–halogen bond breaks. SN1 first forms a carbocation; the nucleophile then attacks the planar intermediate. Primary substrates generally favour SN2 and tertiary substrates generally favour SN1 because steric access and carbocation stability differ. Secondary substrates can support either mechanism, so use the conditions as well as the substrate class.

Do not treat SN1 and SN2 as labels that follow substrate class with no exceptions, or confuse inversion in SN2 with the mixture possible after planar SN1 attack.

4 marks

Sketch the mechanism of the reaction for step 1 in part (b), using curly arrows to show the movement of electron pairs.

Leaving-group identity affects substitution rate through carbon–halogen bond enthalpy and polarity: a weaker, more easily broken bond generally permits faster departure.

For comparable halogenoalkanes, the weaker C–I bond usually makes iodide a better leaving group and substitution faster than for chloride. Explain the trend with bond enthalpy and the actual rate-determining bond change, not polarity alone.

2 marks

Explain why $\mathrm{CH}_{3} \mathrm{CHIC}\left(\mathrm{CH}_{3}\right)_{3}$ reacts faster than $\mathrm{CH}_{3} \mathrm{CHBrC}\left(\mathrm{CH}_{3}\right)_{3}$ .

Use curly arrows from the alkene π bond to the electrophile, then from the intermediate to the nucleophile. Halogens may pass through a halonium intermediate; HX or water may pass through a carbocation.

the pi bond attacks H delta positive and the H-Br bond pair moves to Br; the intermediate is the valid secondary butyl carbocation plus Br-; the bromide lone pair attacks C+; the final condensed formula is 2-bromobutane with no duplicated hydrogens.

hex-3-ene protonation forms the central secondary carbocation; water attacks from an oxygen lone pair to C+; the oxonium intermediate is positively charged and deprotonates to neutral hexan-3-ol; H+ is regenerated and six carbons are conserved without duplicate hydrogens.

Begin with an arrow from the alkene π bond to the electrophile. HX or acid-catalysed hydration can form a carbocation, whereas halogen addition uses a bridged halonium ion; the next arrow must start from the nucleophile's pair.

3 marks

Describe the mechanism of this reaction, using curly arrows to represent the movement of electron pairs.

For HX or water addition to an unsymmetrical alkene, compare the possible carbocations and select the pathway through the more stable carbocation as the major product.

the upper route forms a valid secondary carbocation with one H on C+; the lower route forms a primary carbocation; 2-bromopropane is correctly labelled major and 1-bromopropane minor; three carbons are conserved in every structure.

For an unsymmetrical alkene, draw both protonation routes and compare the carbocations: tertiary is generally more stable than secondary, then primary. Choose the product from the more stable intermediate; do not apply this shortcut to a halonium pathway.

2 marks

Explain which of the two structural isomers is the major product.

Benzene attacks the charged electrophile E+, forms a charged intermediate, then loses H+ to restore aromaticity. The formation of E+ is outside this assessed mechanism boundary.

the attacking electron-pair arrow starts at the benzene pi system and ends at N+ of NO2+; the N=O pi-bond arrow ends on oxygen; the sigma-complex carbon bears both H and NO2; the incomplete dashed arc and ring plus sign show aromaticity loss and delocalized positive charge.

a water lone pair attacks the hydrogen on the sigma-complex carbon; the C–H bond electron pair returns to the ring; nitrobenzene with a restored aromatic circle and H3O+ are formed; NO2 remains attached to the same ring carbon.

The first step temporarily disrupts aromatic delocalization; loss of H⁺ then restores the ring and replaces H by E. An addition product would leave aromaticity broken, which explains why the assessed outcome is substitution.

4 marks

Explain the mechanism for the nitration of benzene, using curly arrows to indicate the movement of electron pairs.

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—IB ChemistryRead

3.4 Electron-pair sharing reactions

Learning objectives

Recognizing Nucleophiles

Nucleophilic Substitution

Heterolytic Fission

Recognizing Electrophiles

Electrophilic Addition to Alkenes

Lewis Acids and Bases

Coordination Bonds

Ligands and Complex Ions

SN1 and SN2 Substitution

Leaving Groups and Rate

Electrophilic-Addition Mechanisms

Major Products of Unsymmetrical Addition

Electrophilic Substitution of Benzene

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Identifying Nucleophiles

Deducing Substitution Products

Showing Electron-Pair Movement

Recognizing Electrophiles

Deducing Alkene-Addition Products

Classifying Lewis Acids and Bases

Explaining Coordinate-Bond Formation

Identifying Complex-Ion Components

Comparing SN1 and SN2

Explaining Leaving-Group Effects

Drawing Addition Mechanisms

Predicting the Major Addition Product

Drawing Benzene Electrophilic Substitution