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15.1 Halogenoalkanes

Syllabus
9701–2028–2029
Topic
15.1
Level
AS

Learning objectives

15.1.1Preparation of halogenoalkanes• Recall preparation of halogenoalkanes- (a) the free-radical substitution of alkanes by Cl 2 or Br2 with ultraviolet light, as exemplified by the reactions of ethane- (b) electrophilic addition of an alkene with a halogen, X 2, or hydrogen halide, HX(g), at room temp.- (c) substitution of an alcohol, e.g. by reaction with HX(g); or with KCl and conc. H2SO4 or conc. H3PO4; or with PCl 3 and heat; or with PCl5; or with SOCl 215.1.2Classify halogenoalkanes into primary,• Classify halogenoalkanes into primary, secondary and tertiary15.1.3Nucleophilic substitution• Describe the following nucleophilic substitution reactions:- (a) the reaction with NaOH(aq) and heat to produce an alcohol- (b) the reaction with KCN in ethanol and heat to produce a nitrile- (c) the reaction with NH3 in ethanol heated under pressure to produce an amine- (d) the reaction with aq. silver nitrate in ethanol as a method of identifying the halogen present as exemplified by bromoethane15.1.4Elimination reaction with NaOH in ethanol• Describe the elimination reaction with NaOH in ethanol and heat to produce an alkene as exemplified by bromoethane15.1.5SN1/SN2 substitution mechanisms• Describe the SN1 and SN2 mechanisms of nucleophilic substitution in halogenoalkanes including the inductive effects of alkyl groups15.1.6SN2/SN1 trends in halogenoalkanes• Recall that primary halogenoalkanes tend to react via the SN2 mechanism; tertiary halogenoalkanes via the SN1 mechanism; and secondary halogenoalkanes by a mixture of the two, depending on structure15.1.7Rate of hydrolysis of halogenoalkanes• Describe/explain: the different reactivities of halogenoalkanes (with particular reference to the relative strengths of the C–X bonds as exemplified by the reactions of halogenoalkanes with aq. silver nitrates)

Choose a halogenoalkane preparation from the starting functional group

Starting material Reaction Reagent and conditions Product pattern
alkane free-radical substitution Cl₂ or Br₂, ultraviolet light; ethane is the exemplar one H replaced by Cl or Br, plus HX
alkene electrophilic addition X₂, room temperature two X atoms add across C=C
alkene electrophilic addition HX(g), room temperature H and X add across C=C
alcohol substitution HX(g); or KCl with concentrated H₂SO₄/H₃PO₄; or PCl₃ and heat; or PCl₅; or SOCl₂ –OH replaced by X

Identify the starting functional group before choosing conditions. An alkane needs radical initiation, an alkene loses its π bond by addition, and an alcohol keeps its carbon skeleton while –OH is substituted.

CX2HX6+BrX2→UVCX2HX5Br+HBr\ce{C2H6 + Br2 ->[UV] C2H5Br + HBr}

These routes can differ in selectivity: further radical substitution or more than one possible alkene-addition product may occur. Conditions are part of the recalled route, not optional labels.

Classify the carbon directly bonded to the halogen

Carbon neighbours of the C–X carbon Class Example
1 primary, 1° CH₃CH₂Br
2 secondary, 2° CH₃CHBrCH₃
3 tertiary, 3° (CH₃)₃CBr

Locate X, identify the one carbon directly bonded to it, then count only the carbon atoms directly bonded to that carbon. Hydrogens, the halogen and more distant carbon atoms are not included in the count.

The classification does not use the total number of carbons in the molecule and does not describe the halogen as primary, secondary or tertiary. It describes the local substitution of the C–X carbon.

The nucleophile determines the substitution product

Nucleophile / reagent Conditions Organic product Carbon-count effect
NaOH(aq) heat alcohol unchanged
KCN in ethanol heat nitrile increases by one because the C of CN becomes part of the chain
NH₃ in ethanol heat under pressure primary amine unchanged
H₂O in aqueous AgNO₃/ethanol observe AgX formation alcohol plus halide ion; AgX identifies X unchanged

CHX3CHX2Br+OHX−→CHX3CHX2OH+BrX−\ce{CH3CH2Br + OH- -> CH3CH2OH + Br-}

CHX3CHX2Br+CNX−→CHX3CHX2CN+BrX−\ce{CH3CH2Br + CN- -> CH3CH2CN + Br-}

In the silver-nitrate test, ethanol helps the organic halogenoalkane mix with the aqueous reagent. Hydrolysis releases X⁻, which gives white AgCl, cream AgBr or yellow AgI; bromoethane therefore gives a cream precipitate.

Silver nitrate is not the nucleophile that replaces X. Water hydrolyses the C–X bond, then Ag⁺ traps the released halide ion as AgX.

Hot ethanolic hydroxide eliminates HX to form C=C

With NaOH in ethanol and heat, a base removes H from a carbon adjacent to the C–X carbon while X leaves. A new C=C forms between those two carbon atoms: this is elimination.

CHX3CHX2Br+OHX−→ethanol,  heatCHX2=CHX2+HX2O+BrX−\ce{CH3CH2Br + OH- ->[ethanol,\ heat] CH2=CH2 + H2O + Br-}

Bromoethane has only one adjacent carbon position, so it gives ethene. A longer unsymmetrical halogenoalkane may have H atoms on different adjacent carbons and can therefore form more than one positional alkene product.

Solvent changes the dominant pathway: aqueous NaOH and heat is used for nucleophilic substitution to an alcohol; NaOH in ethanol and heat is used for elimination to an alkene.

SN2 is concerted; SN1 separates ionisation from attack

Feature SN2 SN1
steps one concerted step two stages
first electron movement nucleophile lone pair → C as C–X pair → X C–X pair → X, forming a carbocation
carbon access favoured when the C–X carbon is less crowded nucleophile attacks after the planar carbocation forms
alkyl inductive effect crowding by extra alkyl groups hinders direct attack extra alkyl groups donate electron density and stabilise the carbocation

In SN2, draw a curly arrow from the nucleophile's lone pair to the C–X carbon and another from the C–X bond to X. Bond formation and bond breaking occur together; there is no carbocation intermediate.

In SN1, first draw the C–X electron pair moving to X to form X⁻ and a carbocation. In the second stage, draw the nucleophile's lone pair to the positive carbon. If a neutral nucleophile such as water attacks, a later proton transfer gives the neutral product.

The subscripts distinguish the molecularity of the rate-determining process, not the number of arrows or products. Full curly arrows represent electron-pair movement and must begin at a bond or lone pair.

Carbon substitution predicts the usual SN1/SN2 pathway

Halogenoalkane class Usual pathway Structural reason
primary SN2 C–X carbon is accessible; a primary carbocation is poorly stabilised
secondary mixture of SN1 and SN2 intermediate crowding and carbocation stabilisation make conditions and detailed structure important
tertiary SN1 direct attack is crowded; three alkyl groups stabilise the tertiary carbocation by induction

Classify the C–X carbon first, then apply the trend. The trend links two competing effects: steric access controls direct SN2 attack, while alkyl electron donation controls whether the SN1 carbocation can form.

This is a tendency, not a new definition of primary, secondary or tertiary. A secondary halogenoalkane cannot be assigned one universal mechanism without considering its structure and reaction conditions.

Weaker C–X bonds give faster halogenoalkane hydrolysis

Bond Relative bond strength Expected hydrolysis reactivity AgX observation after X⁻ is released
C–F strongest slowest AgF is soluble, so no AgF precipitate
C–Cl strong slower white AgCl
C–Br weaker faster cream AgBr
C–I weakest fastest yellow AgI

Hydrolysis requires the C–X bond to break. Down Group 17 the bond becomes longer and weaker, so less energy is needed for C–I cleavage than for C–Br or C–Cl cleavage; iodoalkanes therefore release halide ions fastest under matched conditions.

Warm comparable halogenoalkanes with aqueous silver nitrate in ethanol using equal concentrations, volumes and temperature. Time the first precipitate: a shorter time means faster halide release and therefore faster hydrolysis under that controlled comparison.

Do not predict the order from C–X polarity alone: C–F is highly polar but very strong. Substrate class also affects mechanism, so compare like-for-like carbon skeletons when using the test to infer the halogen trend.