3.4 The reaction between halogenoalkanes and sodium hydroxide - a hydrolysis reaction to give alcohols

Doc Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK KS5 A/AS GCE IB advanced level organic chemistry students US K12 grade 11 grade 12 organic chemistry

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Part 3. The chemistry of HALOGENOALKANES (haloalkanes, alkyl halides)

Halogenoalkanes were once known as 'haloalkanes' or 'alkyl halides', but the correct IUPAC nomenclature is based on calling halogenated alkanes halogenoalkanes.

However, it seems ok to refer to chloroalkanes, bromoalkanes and iodoalkanes.

I've written the equations for the reactions showing the formation of an alcohol from the halogenoalkane in multiple styles and added the nucleophilic substitution mechanisms where appropriate.

Sub-index for this page

3.4.1 The hydrolysis reaction between halogenoalkanes and sodium/potassium hydroxide

3.4.2 The unimolecular S_N1 mechanism of alkaline hydrolysis of haloalkanes

3.4.3 The bimolecular S_N2 mechanism of alkaline hydrolysis of haloalkanes

3.4.4 A comparison S_N1 and S_N2 mechanisms and the reactivity of halogenoalkanes

3.4.5 Using alkaline hydrolysis and silver nitrate to identify the halogen functional group

3.4.6 S_N1 and S_N2 hydroxide hydrolysis mechanisms, rate expressions, orders of reaction

See also

My original detailed discussion of the nucleophilic substitution mechanism between a halogenoalkane and water/hydroxide ion

3.4.1 Hydrolysis reaction between halogenoalkanes and sodium/potassium hydroxide

A nucleophilic substitution reaction in which one functional group (halo/halogen, C-X) is replaced by another functional group (alcohol/hydroxy, C-OH)

You must know the structures of primary, secondary and tertiary halogenoalkanes (haloalkanes)

I've often added the boiling point (bpt) so can see what is a liquid and could be hydrolysed in a school/college laboratory.

Strictly speaking all the reactants and products should be suffixed by (aq) apart fro water (l).

(a) Introduction to the reaction between a halogenoalkane and the hydroxide ion and reagents and method

Both these soluble strong bases (alkalis) are source of a powerful nucleophile, the hydroxide ion (OH^–).

This is a nucleophilic substitution reaction because the attacking reagent is a nucleophile and the halogen functional group/atom is replaced by the hydroxy functional group (alcohol): RX  +  OH^-  ==> ROH  +  X^-

The hydrolysis usually takes faster with an alkali than pure water because water is a weaker nucleophile than the hydroxide ion.

The hydroxide ion (:OH^-) is a much stronger nucleophile than water, since it carries a negative charge.

Because the hydroxide ion can act as a strong base, there is a chance of an elimination reaction to form an alkene.

R₂CX-CHR₂  +  MOH  ===>  R₂C=CR₂  +  MX  +  H₂O

where X = halogen, R = H, alkyl or aryl, M = Na, K etc. The elimination change is highlighted in blue.

Four factors favouring the haloalkane substitution reaction ...

primary  >  secondary >  tertiary  halogenoalkane

Using water or aqueous ethanol rather than pure ethanol.

Using sodium hydroxide NaOH, KOH is an even stronger base

Low concentration of the base.

Reverse the factors to favour elimination !

 

I've added the boiling point (bpt) so can see what is a liquid and could be hydrolysed in a school/college laboratory.

Strictly speaking all the reactants and products should be suffixed by (aq) apart from water (l).

The reaction is usually carried out by refluxing the halogenoalkane with aqueous sodium hydroxide (NaOH) and potassium hydroxide (KOH).

The cold water cooled Liebig vertical condenser prevents the loss of volatile molecules e.g. solvent or product.

The diagram is common to many textbooks, but they never say how you can conveniently separate the alcohol.

Fractional distillation?

When an organic compound is insoluble in water, the reaction with an aqueous reagent is very slow because of a much reduced chance of reactant particle interactions e.g. compared to a homogenous mixture where the haloalkane dissolves in the reagent solvent.

So, even under reflux, the hydrolysis reaction can be quite slow because halogenoalkanes are insoluble in water, but ethanol solvent can be added to increase their solubility and speed of the reaction, BUT ...

... NaOH, and in particular KOH, are very strong bases, and if aqueous ethanol, and, in particular, pure ethanol is used as the solvent, you can get significant quantities of an alkene via an elimination reaction.

The elimination reaction is more likely in the order: tertiary > secondary > primary haloalkanes

See Part 3.7 The elimination reactions of halogenoalkanes (haloalkanes) with potassium hydroxide to give alkenes

(b) bromoethane (bpt 38^oC)  +  sodium hydroxide  ===>  ethanol  +  sodium bromide

  +  NaOH      +  NaBr

since hydroxide and bromide are free ions the equations are better written as ...

bromoethane  +  hydroxide ion  ===>  ethanol  +  bromide ion

  +  OH^–      +  Br^–   (displayed formula equation)

  +  OH^–      +  Br^–   (structured formula ionic equation)

  +  OH^–      +  Br^–   (skeletal formula equation)

(c) 1-bromopropane  +  sodium hydroxide  ===>  propan-1-ol  +  sodium bromide

  +  NaOH      +  NaBr

since hydroxide and bromide are free ions the equations are better written as ...

1-bromopropane  +  hydroxide ion  ===>  propan-1-ol  +  bromide ion

  +  OH^–      +  Br^–

  +  OH^–      +  Br^–

 

(d) bromocyclohexane  +  hydroxide ion  ===>  cyclohexanol  +  bromide ion

  +  OH^–    +  Br^–

Remember when studying the reaction mechanisms in the next few sections:

A neutral or negative nucleophile, Nuc: or Nuc:^-, is an electron pair donor that can attack an electron deficient partially/wholly positive carbon atom to form a new C-Nuc bond.

e.g. for the hydrolysis of halogenoalkanes with sodium/potassium hydroxide to form alcohols:

 R-X_(l)  +  :O-H^-_(aq)  ===>  R-O-H_(aq)  +  X^-_(aq)

is a typical haloalkane nucleophilic substitution reaction with strong bases.

Where R = alkyl, :OH^- is the nucleophile - electron pair donor (: on the O), X = halogen replaced and X^- is the displaced atom/group, in this case a halide ion, which is sometimes referred to as the 'leaving group'.

The mechanisms are discussed here in Part 3.4.2, 3.4.2  and  3.4.4

See also 3.3 Reactivity trends of halogenoalkanes - introduction to their nucleophilic substitution reactions, the substitution reaction between halogenoalkanes (haloalkanes) & water, experiments with silver nitrate solution

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3.4.2 The unimolecular S_N1 mechanism of alkaline hydrolysis with NaOH

The halogen atom gains the bonding pair of electrons from the original C-X bond to form the halide ion X^-.

There are two reactivity trends you should know:

(i) Always iodo (C-I)  >  bromo  (C-Br)  > chloro (C-Cl)  >  fluoro (C-F)

This trend is irrespective of whether the haloalkanes are all primary, secondary or tertiary.

This trend is determined by bond enthalpy and activation energy trend (see section 3.3.1)

This trend is irrespective of the mechanism.

(ii) Generally speaking tertiary  >  secondary  >  primary  haloalkanes

but things are complicated because there are two types of mechanism and two conflicting trends.

e.g. tertiary  >  secondary  >  primary via the S_N1 mechanism,

and  primary  >  secondary  >  tertiary  haloalkanes via the S_N2 mechanism.

so things are bit more complicated here and you can't appreciate these trend complexities until both mechanisms have been studied.

 

Mechanism diagram 1 - nucleophilic substitution of a halogenoalkane by hydroxide ion (S_N1 'unimolecular' via carbocation)

S_N1 unimolecular, a two step ionic mechanism via carbocation formation [mechanism 1 above]

In step (1) the C^δ+-X^δ- polar bond of the halogenoalkane splits heterolytically to form a carbocation and a free halide ion (X^-, e.g. chloride or bromide) and this is a reversible reaction.

The C-Cl bond is most likely to break because it is a weaker bond than the C-C or C-H bond AND breaks heterolytically because of the big difference in electronegativity between carbon (2.5) and chlorine (3.0) giving a polar C^δ+-Cl^δ- bond.

So the C-Cl bonding pair of electrons leaves with the Cl atom as the Cl^- ion.

In step (2) the hydroxide ion is a negative electron pair donor and rapidly combines with the carbocation, forming the C-O bond in the alcohol product. The hydroxide ion is the nucleophile.

 Step (1) is the rate determining step with the much larger activation energy (see reaction profile diagram 45 below)

This mechanism is most likely with tertiary halogenoalkanes.

i.e. the mechanism is the same for tertiary bromoalkanes or iodoalkanes.

Primary halogenoalkanes tend to react by the S_N2 mechanism NOT involving a carbocation.

Secondary halogenoalkanes react via both mechanisms at the same time! The relative rate order for the S_N1 mechanism is tert > sec > prim i.e. following the pattern of decreasing stability of the carbocation.

The term 'unimolecular' indicates only one 'active species' involved in the rate determining step - the heterolytic bond fission.

Extra stereochemistry note

If the original haloalkane is a R/S isomer, i.e. one of the enantiomers, the product is a racemic mixture because the nucleophile can attack the carbocation with equal probability on each side of the trigonal bond system of the positive carbon atom (diagram above).

See Organic mechanism notes  and R/S isomerism notes.

 

The reaction progress profile for the hydrolysis of a tertiary halogenoalkane with hydroxide ion via the S_N1 carbocation mechanism - unimolecular rate determining step

Generalised reaction profiles showing the formation of the intermediate carbocation.

The heterolytic bond fission to generate the carbocation has by far the largest activation energy E_a1, so the unimolecular step 1 is the rate determining step.

 

Mechanism diagram 70b shows the SN1 mechanism for the hydrolysis of 2-bromo-2-methylpropane with sodium hydroxide - the sodium ion is not shown because it is a spectator ion - not involved!

S_N1 unimolecular, a two step ionic mechanism via carbocation formation [mechanism 1 above]

In step (1) the C^δ+-Br^δ- polar bond of the halogenoalkane splits heterolytically to form a carbocation and the free bromide ion.

The C-Br bond is most likely to break because it is a weaker bond than the C-C or C-H bond AND breaks heterolytically because of the difference in electronegativity between carbon  and bromine resulting in the polar C^δ+-Br^δ- bond.

So the C-Br bonding pair of electrons leaves with the Br atom as the Br^- ion.

In step (2) the hydroxide ion is an electron pair donor and rapidly combines with the carbocation, forming the C-O bond in the alcohol product.

The hydroxide ion is the nucleophile and the final product is 2-methylpropan-2-ol.

 Step (1) is the rate determining step with the much larger activation energy.

This mechanism is most likely with tertiary halogenoalkanes.

 

The relative ease of formation of carbocations determines the reactivity trend for the S_N1 mechanism.

The more alkyl groups attached to the carbon of the C-X bond (X = halogen), the greater the electron cloud shift to stabilise the carbocation - known as the inductive effect (+I effect).

A methyl group (-CH₃) and to smaller extent a -CH₂- group, have a tendency to push electron charge (+I, + inductive effect) from the sigma bonds (C-H, C-C).

This helps the release (substitution) of the halide atom (X) by shifting electron charge towards it.

As you go from primary to tertiary, you increase the alkyl groups attached to the C of the C-X functional group, hence increase this electron shift effect that in turn helps the halogen atom to gain the C-X bonding pair of electrons and leave to form the halide ion.

The more easily the carbocation is formed (lower activation energy), so the more likely the chance of the C-X bond breaking heterolytically to form the carbocation.

The greater the chance of carbocation formation, the greater the chance of it combining with the hydroxide ion to give the final alcohol product.

Hence, with reference to the carbocation diagram above, for the same halogen X, the reactivity of the halogenoalkane is: tertiary  >  secondary  >  primary

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3.4.3 The bimolecular S_N2 mechanism of alkaline hydrolysis by hydroxide ion

Remember: A neutral or negative nucleophile, Nuc: or Nuc:^-, is an electron pair donor that can attack an electron deficient partially/wholly (δ+/+) positive carbon atom to form a new C-Nuc (C:Nuc) bond and displace an atom/group in the process.

A lone pair of electrons on the oxygen atom of the hydroxide ion (:OH^-) attacks the δ+ carbon atom of the carbon -halogen bond (C-X) to form a new C-O sigma bond (single covalent bond).

 

Mechanism diagram 2 - nucleophilic substitution of a halogenoalkane by hydroxide ion (S_N2 'bimolecular')

The simplest representation of the one step mechanism for the alkaline hydrolysis of a halogenoalkane e.g. a chloroalkane.

It is a bimolecular mechanism because two reactant molecules/ions collide in a single step mechanism.

 

Mechanism diagram 34 - nucleophilic substitution of a halogenoalkane by hydroxide ion (S_N2 'bimolecular')

In diagram 34 the intermediate 'transition state' or 'activated complex' is shown.

S_N2 'bimolecular', a one step bimolecular collision mechanism [mechanisms 34]

The C^δ+-Cl^δ- bond is polar because of the difference in electronegativity between carbon (2.5) and chlorine (3.0), so the electron rich nucleophile, the hydroxide ion, attacks the slightly positive carbon.

The nucleophile acts as an electron pair donor (Lewis base) to bond with the C^δ+ carbon to make the C-O bond in the newly formed C-OH alcohol group.

Simultaneously the chlorine atom is ejected, taking with it the C-X bond pair, so forming the chloride ion on expulsion.

This mechanism is most likely with primary halogenoalkanes.

i.e. the mechanism is the same for primary bromoalkanes or iodoalkanes.

Tertiary halogenoalkanes tend to react by the S_N1 mechanism involving a carbocation, secondary halogenoalkanes react via both mechanisms.

The reaction progress profile for the hydrolysis of a primary halogenoalkane with hydroxide ion via the S_N2 'transition state' mechanism - bimolecular rate determining step

Mechanism diagram 45b: Reaction progress profile for S_N2 bimolecular hydrolysis by water of halogenoalkanes.

Note the concept of the [transition state] in the reaction profile, for the blue dotted lines, think of a C-X bond half-broken and the C-O bond half-formed at the point in the mechanism - at the top of the potential energy hump!

The [transition state], sometimes called the 'activation complex' carries a negative charge because of the negative charge on the hydroxide ion.

The energy released by the formation of the C-O bond more than compensates for the breaking of the C-X bond

 

Mechanism diagram 70a shows the hydrolysis of bromoethane with sodium hydroxide, the hydroxide ion being the attacking nucleophile - the electron pair donor to the δ+ carbon.

The 2nd version shows all the non-bonding pairs of electrons involved,

The 3rd version shows the [transition state] (or activated complex), the point where the C-Br is half broken and the C-O bond is half formed.

On the right is the 'simple' reaction progress profile with top of the hump representing where the 'transition state' or 'activated complex' is momentarily formed.

Below is full 3D version of the nucleophilic substitution reaction between a chloroalkane and the hydroxide ion.

Mechanism 33 - nucleophilic substitution of a halogenoalkane by hydroxide ion (S_N2 'bimolecular')

The diagram shows the intermediate transition state ('activated complex') in a 3D representation - note the inversion of the crucial carbon atom (R₃C^δ+), if it is chiral (an asymmetric carbon atom), you don't produce the optical (R/S) stereoisomers you might expect! This carbon atom is chiral if all three 'R' groups are different (R, R', R" can be H, alkyl or aryl).

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3.4.4 A comparison S_N1 and S_N2 mechanisms and the reactivity of halogenoalkanes towards the hydroxide ion

mechanism 1 - nucleophilic substitution of a halogenoalkane by hydroxide ion

(S_N1 'unimolecular' hydrolysis via carbocation)

mechanism 34 - nucleophilic substitution of a halogenoalkane by hydroxide ion

(S_N2 'bimolecular' hydrolysis via a 'transition state')

The relative rates of hydrolysis via SN1 and SN2 mechanisms, for the different sub-classes of halogen, is shown in the graph below - mechanism diagram 75.

The sub-class trend for the same halogen X:  is tertiary  >  secondary  >  primary for the S_N1 mechanism via a carbocation.

The argument for this trend based on the ease of formation and stability of carbocation is explained in 3.4.2 and shown in the diagram above - the stabilising effect of the electron shifts are shown in brown chevrons.

However, the reactivity trend for the S_N2 bimolecular reaction is primary > secondary > tertiary.

This trend can be explained by the effect of other alkyl groups attached to the carbon atom of the carbon-halogen bond.

The more alkyl groups attached to this carbon, the more it is shielded from attack by the nucleophile - the hydroxide ion.

This is an example of a stereochemical effect on the rate of a reaction.

Therefore the presence of alkyl groups here, has the complete opposite effect on the rate of the bimolecular S_N2 mechanism, as it does on the rate of the S_N1 mechanism.

Hence the two conflicting graph curves for the rates of S_N1 and S_N2 reactions.

Primary halogenoalkanes react predominantly by the S_N2 mechanism.

Secondary halogenoalkanes react by either the S_N1 or S_N2 mechanism.

Tertiary halogenoalkanes react predominantly by the S_N1 mechanism.

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3.4.5 Using alkaline hydrolysis of haloalkanes to identify the halogen functional group

Since halogenoalkanes do not dissolve in water AND some are very slow to hydrolyse in pure water, the best approach is to carefully heat a few drops of the compound with aqueous sodium hydroxide solution for a few minutes in a pyrex boiling tube.

This frees the halogen atom as the halide ion, so for R = alkyl, X = halogen Cl, Br or I:

RX_(l)  +  NaOH_(aq)  ===>  ROH_(aq)  +  NaX_(aq)

RX_(l)  +  OH^-_(aq)  ===>  ROH_(aq)  +  X^-_(aq)

After leaving to cool, the resulting solution is neutralised and made acid with excess dilute nitric acid - test with litmus.

The nitric acid prevents the precipitation of other insoluble silver salts.

A few drops of aqueous silver nitrate are then added to the solution and the colour of the precipitate noted and its solubility in dil. and conc. ammonia. See the table below the 'picture' on how to interpret the results.

add from e.g three structures

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3.4.6 S_N1 and S_N2 hydroxide hydrolysis mechanisms, rate expressions, orders of reaction

Don't worry if you haven't done rate expressions and orders of reaction yet, you will do them later in your course.

 

Reminder of what we mean by an S_N1 nucleophilic substitution reaction (e.g. unimolecular mechanism 1 above).

The S signifies substitution, N signifies nucleophilic (attack) and the 1 means a unimolecular step 1 that also determines the rate of the reaction because the heterolytic bond fission of the C-Cl bond is the slowest rate determining step and only one molecule involved - hence the phrase 'unimolecular'.

The rate of the reaction is controlled ONLY by the concentration of the haloalkane for a given solvent:

rate = k₁[RX]

k₁ = 1st order rate constant, [RX] = concentration of haloalkane in the rate expression.

The concentration of the hydroxide ion is irrelevant to the rate of the reaction.

If the hydrolysis kinetics show up as a 1st order rate expression, it is indicative of a S_N1 unimolecular carbocation mechanism.

 

Reminder of what an S_N2 nucleophilic substitution reaction is (e.g. mechanism 34 above).

S signifies substitution, N signifies nucleophilic and the 2 means a bimolecular step1 that also determines the rate of the reaction via a transition state - known as the rate determining step by this particular mechanism.

Don't worry if you haven't done rate expressions and orders of reaction yet, you can ignore the next paragraph until later in your course.

The rate of the reaction is controlled by the concentration of the haloalkane and the water:

rate = k₂[RX][OH^-]

k₂ = 2nd order rate constant, [RX] = concentration of haloalkane,

and [OH^-] = concentration of hydroxide ion, in the rate expression.

The term 'bimolecular' in this case refers to the two molecule collision in the rate determining step 1 - between the halogenoalkane and the hydroxide ion.

This is deduced from the equation, not from experiments.

The order of the reaction, is the sum of the powers to which the concentration terms are raised in the rate expression (e.g. here it is 1 + 1 = 2 = 2nd order rate expression).

If the hydrolysis kinetics show up as a 2nd order rate expression, it is indicative of a S_N2 bimolecular mechanism.