Part 3. The chemistry of HALOGENOALKANES

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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3.3 Introduction to the nucleophilic substitution reactions of halogenoalkanes, the substitution reaction and hydrolysis with water and experiments with silver nitrate solution to investigate reactivity trends and introducing S_N1 and S_N2 mechanistic pathways

Hydrolysis of haloalkanes with NaOH/KOH is described in Part 3.4

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

3.3.1 Why do halogenoalkanes undergo nucleophilic substitution reactions?, why are haloalkanes more reactive than alkanes?, what are their reactivity trends? and note on sub-classification

3.3.2 The hydrolysis reaction between a tertiary halogenoalkane and pure water - an S_N1 mechanism

3.3.3 Hydrolysis of halogenoalkanes with pure water - investigating relative reactivity trends with AgNO₃ and a note on testing to identify which halogen atom is in the haloalkane molecules

3.3.4 The hydrolysis reaction of a primary haloalkane with pure water - an S_N2 mechanism

3.3.5 The hydrolysis of secondary halogenoalkanes with pure water

3.3.6 S_N1 and S_N2 water hydrolysis mechanisms connection with rate expressions, orders of reaction

See also

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

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).

3.3.1 Why do halogenoalkanes (RX) undergo nucleophilic substitution reactions?

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

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

e.g. hydrolysis with water or alkali to form an alcohol:

RX  +  H₂O  ===>  ROH  +  X^-  +  H⁺

RX  +  OH-  ===>  ROH  +  X^-

Why are haloalkanes more reactive than alkanes?

and what are their reactivity trends, plus a note on sub-classification reactivity and mechanism mode

Structure and reactivity:

R₃C-X = halogenoalkane (haloalkane, alkyl halide, halogenated alkane etc.)

R = H, alkyl or aryl  and X = group 7/17 halogen F, Cl, Br or I

Three trends need to be discussed before getting into the detailed chemistry of nucleophilic substitution reactions of halogenoalkanes

BUT first, a view of where we are going in terms of the mechanism of nucleophilic attack and subsequent substitution of the halogen atom by another atom/group.

Consider mechanism diagram 82a above in the context of  halogenoalkanes owe their reactivity, especially compared to the relatively unreactive alkanes, to two principal reasons:

[box (i)] The initial attack by the electron pair donating nucleophile

The C-X bond is polar (C^δ+-X^δ-), giving the carbon atom a partial positive charge and a region of lower electron density susceptible to attack by an electron pair donor - known as a nucleophile.

Alkanes do NOT have a polar bond.

[box (ii)] Apart from C-F, the C-Cl, C-Br and C-I bonds are weaker than C-H or C-C bonds, therefore bond scission will have a lower activation in a crucial mechanism step.

Trend (i) Bond polarity and reactivity trend - but take care, its not what you might think!

As already pointed out, the carbon-halogen bond is polar, C^δ+-X^δ- due to the difference in electronegativity between carbon and the halogen.

The C^δ+ carbon is then susceptible to nucleophilic attack by electron pair donor neutral molecules e.g. the nucleophiles :NH₃  and  H₂O: or nucleophile negative ions like. :OH^-  and  :CN^-.

This is the principal reason why halogenoalkanes undergo nucleophilic substitution reactions.

The order of bond polarity is: C-F  >  C-Cl  >  C-Br  >  C-I  because of the decreasing electronegativity of the halogen atom down group 7/17, hence the decrease in electronegativity difference between the carbon and halogen atom of the C-X bond.

BUT, contrary to what you might think, the reactivity trend towards nucleophilic attack is the complete opposite, and generally speaking, for the same alkyl group R, the reactivity order is:  R-I  >  R-Br  >  R-Cl  >  RF  because the trend in decreasing bond enthalpy overrides bond polarity trend - explained in the next section (ii) below.

Trend (ii) Bond enthalpies and reactivity trend

Apart from carbon-fluorine, the carbon-halogen bond is usually the weakest bond in the molecule and significantly weaker than the carbon-carbon or carbon-hydrogen bonds.

Average bond enthalpies/kJmol^-1: C-C  348, C-H  412, both requiring relatively high activation energies for reaction.

Average bond enthalpies/kJmol^-1: C-F 484, C-Cl  338, C-Br  276, C-I  238, generally lower than C-C or C-H bonds resulting in lower activation energies.

On descending the Group 7/17 halogens (X), the atomic radius increases, C-X bond length increases and the strength of the bond decreases.

Even the lowering of the bond enthalpy by 10kJ from C-C to C-Cl, combined with the polarity of the C-Cl bond, makes all the difference when comparing alkane and halogenoalkane reactivity.

Note that even though the most polar bond, based on differences in electronegativity, is the carbon - fluorine bond (C^δ+-F^δ-), but the bond enthalpy is so high that fluoroalkanes have the lowest reactivity of all halogenoalkanes and are very reluctant to undergo nucleophilic substitution reactions.

So what you find in practice is the reactivity trend of R-I  >  R-Br  >  R-Cl  >  RF , that completely overrides the bond polarity trend ...

... entirely due to the bond enthalpy trend:  C-F  >  C-Cl  >  C-Br  >  C-I  ....

... and the resulting activation energy trend will be in the same order i.e. decreasing from C-F to C-I,

... the lower the activation energy, the faster the reaction under the same conditions of temperature, concentration, reagent and solvent.

e.g. if you take the following series of halopropanes, the reactivity trend towards nucleophilic reagents would be:

CH₃CH₂CH₂-I  >  CH₃CH₂CH₂-Br   >   CH₃CH₂CH₂-Cl   ~ CH₃CH₂CH₂-F

When doing this comparison it is important to keep the alkyl group constant because there is a 'structural' haloalkane classification trend too in the reactivity of halogenoalkanes towards nucleophiles see (3) next.

Trend (3) A note on the structural sub-classification of halogenoalkanes and its effect on reactivity

You should understand the sub-classification of halogenoalkanes before studying their reactions.

The sub-classification of halogenoalkanes is based on structural differences, which can have significant chemical consequences e.g. on the rate of the reaction, mechanism of the reaction or the products formed in the reaction.

Halogenoalkanes are classified according to the atoms/groups attached to the carbon of the halogen atom X.

Primary halogenoalkanes have the structure R-CH₂-X, R = H, alkyl, aryl etc.

Apart from the likes of the methyl haloalkanes e.g. chloromethane (CH₃Cl) they have two hydrogen atoms and one alkyl/aryl group attached to the C of the C-X functional group.

e.g. chloroethane CH₃CH₂Cl, and  1-bromopropane

C₆H₅CH₂Cl is an example of an aryl (aromatic) substituent attached to the carbon of the C-X bond, but it is classed as a primary halogenoalkane (as well as an aromatic compound, but not classed as an aryl halide because the chlorine atom is not directly attached to the benzene ring.)

Secondary halogenoalkanes have the structure R₂CH-X, R = alkyl or aryl etc.

They have one hydrogen and two alkyl/aryl groups attached to the C of the C-X functional group.

e.g. 2-iodobutane CH₃CHICH₂CH₃ , and 2-chloropropane

Tertiary halogenoalkanes have the structure R₃C-X, R = alkyl or aryl etc.

They have three alkyl/aryl groups attached to the C of the C-X group.

e.g. 2-chloro-2-methylpropane (CH₃)₃CI

There no hydrogen atoms on the C of the C-X functional group.

 

For the same halogen atom, generally speaking the reactivity trends for haloalkanes is:

tertiary  >  secondary  >  primary

e.g. if you take the structural isomers of C₄H₉Cl, the reactivity trend towards nucleophilic reagents would be:

  >    >     ~

For the reactivity trend: tertiary  >   secondary  >  primary ~ primary halogenoalkanes.

The explanation for this is covered in the next few sections on this page when discussing the reaction mechanisms.

 

I have also included brief descriptions of how to do experiments to investigate the two most important reactivity trends based on:

(i) The nature of the halogen X in the C-X bond, the bond enthalpy based reactivity trend.

(ii) The sub-classification of the structure of the original halogenoalkane compared to its reactivity for the same halogen X in the C-X bond.

 

I've often added the boiling point (bpt) so can see what is a liquid and could be hydrolysed in a school/college laboratory experiment.  Strictly speaking all the reactants and products should be suffixed by (aq) apart from water (l).

 

Remember when studying the reaction mechanisms in the next sections:

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

e.g. for the hydrolysis of halogenoalkanes in pure water to form alcohols:

 R-X_(l)  +  H₂O:_(l)  ===>  R-OH_(aq)  +  H⁺_(aq)  +  X^-_(aq)   or more correctly ...

 R-X_(l)  +  2H₂O:_(l)  ===>  R-OH_(aq)  +  H₃O⁺_(aq)  +  X^-_(aq)

is a typical haloalkane nucleophilic substitution reaction, where water is the nucleophile,

and it is also described as a hydrolysis, because the organic molecule reacts with water to give at least two products.

Where R = alkyl, H₂O: 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 3.3.3  and  3.3.4

The hydroxide ion (:OH^-) is a much stronger nucleophile than water, since it carries a negative charge, but I've written up hydrolysis of halogenoalkanes with strong alkalis is in Part 3.4

3.4 The substitution reaction of halogenoalkanes (haloalkanes) with sodium/potassium hydroxide to give alcohols

See also Introduction to organic chemical reaction mechanisms & technical terms explained

and Nucleophilic substitution by water/hydroxide ion [S_N1 or S_N2, hydrolysis to give alcohols]

with extra notes on kinetics, rds, molecularity, rate expression, activated complex etc.

TOP OF PAGE and sub-index

3.3.2 Hydrolysis reaction between a tertiary halogenoalkane and water - S_N1 mechanism

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

2-chloro-2-methylpropane (bpt 51^oC) is an example of a halogenoalkane that readily hydrolyses in water to give the appropriate tertiary alcohol and dilute hydrochloric acid (HCl, but fully ionised!).

2-chloro-2-methylpropane  +  water  ===>  2-methylpropan-2-ol  +  hydrochloric acid

  +  H₂O      +  H⁺  +  Cl^–   (structured formula equation)

  +  H₂O      +  H⁺  +  Cl^–   (abbreviated structured formula equation)

  +  H₂O      +  H⁺  +  Cl^–   (skeletal formula equation)

Note that a tertiary halogenoalkane on hydrolysis gives a tertiary alcohol.

 

It is a similar reaction with 2-bromo-2-methylpropane (bpt 73^oC)

hydrolysis conversion:      (structured formulae change)

2-bromo-2-methylpropane  +  water  ===>  2-methylpropan-2-ol  +  hydrobromic acid

(CH₃)₃CBr  +  H₂O    (CH₃)₃COH  +  H⁺  +  Br^–

or more correctly: (CH₃)₃CBr  +  2H₂O    (CH₃)₃COH  +  H₃O⁺  +  Br^–

 

The general mechanism for this particular reaction is shown below in mechanism diagram 10 for Cl, but same for Br or I.

mechanism 10 - nucleophilic substitution of a halogenoalkane by water (S_N1 unimolecular via carbocation)

This is the typical S_N1 mechanism by which tertiary halogenoalkanes hydrolyse

What we mean by an S_N1 nucleophilic substitution reaction.

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 step and only one molecule involved - hence the phrase 'unimolecular'.

Step (1) Heterolytic bond fission to create a carbocation and a chloride ion.

Heterolytic bond fission means the bonding pair of electrons goes completely to one of the atoms of the original bond, therefore automatically creating a positive and negative ion.

Step (2) A water molecule combines with the carbocation to give a protonated alcohol molecule.

Step (3) A proton is lost from the protonated alcohol molecule to give the alcohol e.g. 2-methylpropan-2-ol.

If R = CH₃, then above is the mechanism for the hydrolysis of 2-chloro-2-methylpropane, a tertiary haloalkane.

Note that step (1) is reversible.

e.g. if you mix 2-methylpropan-2-ol with concentrated hydrochloric acid you form 2-chloro-2-methylpropane !!!

  +  H₂O      +  H⁺  +  Cl^–

 

The reaction progress profile for the hydrolysis of a tertiary halogenoalkane with water 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 unimolecular step 1 is the rate determining step.

 

The mechanism for the hydrolysis of 2-bromo-2-methylpropane is shown below in mechanism diagram 71b below:

Step (1) Heterolytic bond fission to create a tertiary carbocation and a bromide ion - the slow rate determining step.

Here, heterolytic bond fission of the 2-bromo-2-methylpropane means the bonding pair of electrons goes completely to the bromine atom, automatically creating a positive carbocation and a negative bromide ion.

Step (2) A water molecule combines with the carbocation to give protonated 2-methylpropan-2-ol

Step (3) A proton is lost from the protonated alcohol to a water molecule leaving the free alcohol 2-methylpropan-2-ol.

Overall reaction: (CH₃)₃CBr  +  2H₂O    (CH₃)₃COH  +  H₃O⁺  +  Br^–

TOP OF PAGE and sub-index

3.3.3 Hydrolysis of halogenoalkanes with pure water - investigating relative reactivity trends

AND also how to identify

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

Primary halogenoalkanes are quite slow to hydrolyse in pure water (see Part 3.3.4 for S_N2 mechanism).

To add some more skeletal formulae equations

REMINDER from previous courses

Extra observations to sort out which halogen atom is present in the haloalkane molecule

(i) The silver chloride precipitate dissolves in dilute ammonia solution.

(ii) The silver bromide precipitate only dissolves in concentrated ammonia solution.

(iii) The silver iodide precipitate does NOT dissolve in ammonia at all.

Note that these tests and comments only apply to when the halogenoalkane has been hydrolysed or reacted with silver nitrate in some way.

 

Two qualitative organic functional group tests for halogenoalkanes (haloalkanes) These haloalkane/halide ion tests can be adapted to qualitatively investigate the reactivity of haloalkanes
Halogenoalkanes (haloalkanes) chemical test R–X where R = alkyl, X = Cl, Br or I The halide is covalently bound (C–X bond), so the halogen X cannot react with the silver ion to form the ionic Ag+X–(s) precipitate until it is converted to the 'free' X– ionic form.	(i) Warm a few drops of the haloalkane with aqueous ethanolic silver nitrate solution, the ethanol increases the solubility of the immiscible haloalkanes. (ii) Gently simmering a few drops with aqueous NaOH (may need to add ethanol to increase solubility and reaction rate). Add dilute nitric acid followed by aqueous silver nitrate solution.	(i) Observe colour of precipitate and the effect of  ammonia solution on it (for rest of details see the diagram above) (ii) see the (i) notes as above for more details.	(i) AgNO3(aq) + RX(aq)  ==> R–NO3(aq)? +  AgX(s) (ii) The sodium hydroxide converts the halogen atom into the ionic halide ion in a hydrolysis reaction. RX(aq)  +  NaOH(aq)  ===>  ROH(aq)  +  NaX(aq) RX(aq)  +  OH-(aq)  ===>  ROH(aq)  +  X-(aq) then Ag+(aq)  +  X–(aq)  ===> AgX(s) The addition of dilute nitric acid prevents the precipitation of other silver salts or silver oxide (e.g. Ag2O forms if solution alkaline).

 

Trend 1 The relative reactivity of the C-X bond, where X = Cl, Br and I

You make up equal concentration mixtures of the halogenoalkane, ethanol and aqueous silver nitrate.

As the halogenoalkane hydrolyses, the freed halide ion forms a precipitate with silver nitrate.

You may need to warm all the solutions to 50-60^oC in a thermostated water bath to effect all three reactions.

You note the time the precipitate is first clearly visible from the reaction: Ag⁺_(aq)  +  X^-_(aq)  ===>  AgX_(s)

For a 'fair test' all the haloalkanes should be all of the same class i.e. in this case, all primary haloalkanes.

(a) 1-chlorobutane (bpt 79^oC)  +  water  ===>  butan-1-ol  +  hydrochloric acid

  +  H₂O    CH₃CH₂CH₂CH₂OH  +  H⁺  +  Cl^–  (structured formula equation)

  +  H₂O      +  H⁺  +  Cl^–   (skeletal formula equation)

This reaction is very slow because of the strong C-Cl bond, but eventually a white precipitate of silver chloride forms.

Ag⁺_(aq)  +  Cl^-_(aq)  ===>  AgCl_(s)

 

(b) 1-bromobutane (bpt 101^oC)  +  water  ===>  butan-1-ol  +  hydrobromic acid

CH₃CH₂CH₂CH₂Br  +  H₂O    CH₃CH₂CH₂CH₂OH  +  H⁺  +  Br^–

This reaction is faster, but still takes some tine, but eventually a cream precipitate of silver bromide forms.

Ag⁺_(aq)  +  Br^-_(aq)  ===>  AgBr_(s)

(c) 1-iodobutane (bpt 130^oC)  +  water  ===>  butan-1-ol  +  hydroiodic acid

CH₃CH₂CH₂CH₂I  +  H₂O    CH₃CH₂CH₂CH₂OH  +  H⁺  +  I^–

This reaction the fastest, but still takes time, and a pale yellow precipitate of silver iodide forms.

Ag⁺_(aq)  +  I^-_(aq)  ===>  AgI_(s)

Although a very simple experiment, reactions (a) to (c) clearly demonstrate the difference in reactivity of the carbon - halogen bond

 i.e. the reactivity order of C-I  >  C-Br  >  C-I,

and from which you can infer the relative reactivity of C-F and C-At haloalkanes.

The explanation based on bond enthalpy trend is discussed in the first section 3.3.1

If you can do the experiment with the equivalent secondary haloalkanes, things go a bit faster (*).

e.g. reactivity order for 2-haloalkanes:  CH₃CH₂CHICH₃  >  CH₃CH₂CHBrCH₃  >  CH₃CH₂CHClCH₃

For a fair comparison test, you should try to keep the structures as similar as possible e.g. all primary secondary or tertiary, and just vary the halogen atom.

(*) Note that trend 2 experiments described below, investigate the reactivity trend tert  >  sec  >  prim by keeping the halogen atom constant.

 

Trend 2 The relative reactivity of primary, secondary and tertiary halogenoalkanes

You can do a similar comparison experiment to determine the relative reactivity of isomeric (ideally) primary, secondary and tertiary halogenoalkanes, but here you keep the halogen atom constant.

Again, you make up equal concentration mixtures of the halogenoalkane, ethanol and aqueous silver nitrate.

As the halogenoalkane hydrolyses, the freed halide ion forms a precipitate with silver nitrate.

You may need to warm all the solutions to 50-60^oC in a thermostated water bath to effect all three reactions.

Again, you note the time the precipitate is first clearly visible for the reaction: Ag⁺_(aq)  +  X^-_(aq)  ===>  AgX_(s)

The following equations involve the four isomers of the halogenoalkane general formula C₄H₉Br, (same for Cl or I)

For a fair comparison test, you should just vary the structures and keep the halogen atom constant.

(d) A primary halogenoalkane giving a primary alcohol on hydrolysis with water

hydrolysis conversion:      (displayed formulae change)

1-bromobutane (bpt 101^oC)  +  water  ===>  butan-1-ol  +  hydrobromic acid

CH₃CH₂CH₂CH₂Br  +  H₂O    CH₃CH₂CH₂CH₂OH  +  H⁺  +  Br^–

The silver bromide precipitate is relatively very slow to form: Ag⁺_(aq)  +  Br^-_(aq)  ===>  AgBr_(s)

 

(e) Another isomeric primary halogenoalkane giving a primary alcohol on hydrolysis with water

hydrolysis conversion:    

1-bromo-2-methylpropane (bpt 91^oC)  +  water  ===>  2-methylpropan-1-ol  +  hydrobromic acid

(CH₃)₂CHCH₂Br  +  H₂O    (CH₃)₂CHCH₂OH  +  H⁺  +  Br^–

Again, the silver bromide precipitate is relatively very slow to form: Ag⁺_(aq)  +  Br^-_(aq)  ===>  AgBr_(s)

 

(f) A secondary halogenoalkane giving a secondary alcohol on hydrolysis with water

hydrolysis conversion:    (structured formulae change)

2-bromobutane (bpt 91^oC)  +  water  ===>  butan-2-ol  +  hydrobromic acid

CH₃CHBrCH₂CH₃  +  H₂O    CH₃CH(OH)CH₂CH₃  +  H⁺  +  Br^–

Here, the silver bromide precipitate forms a bit faster: Ag⁺_(aq)  +  Br^-_(aq)  ===>  AgBr_(s)

(g) A tertiary halogenoalkane giving a tertiary alcohol on hydrolysis with water

hydrolysis conversion:      (structured formulae change)

2-bromo-2-methylpropane (bpt 73^oC)  +  water  ===>  2-methylpropan-2-ol  +  hydrobromic acid

(CH₃)₃C-Br  +  H₂O    (CH₃)₃C-OH  +  H⁺  +  Br^–

Here, the silver bromide precipitate forms the faster: Ag⁺_(aq)  +  Br^-_(aq)  ===>  AgBr_(s)

You can repeat the experiment with a series of chloroalkanes or iodoalkanes (preferably isomers) and you should get the same reactivity order which is ...

Halogenoalkane reactivity trend: tertiary  >  secondary  >  primary

Explanation

Tertiary haloalkanes hydrolyse via S_N1 unimolecular mechanism (see section 3.3.2), which has a lower activation energy than the S_N2 mechanism by which primary haloalkanes hydrolyse (see section 3.3.4).

Secondary haloalkanes can hydrolyse via both mechanistic pathways and so are intermediate in reactivity.

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).

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

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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3.3.4 The mechanism of hydrolysis of primary haloalkanes by water (S_N2 mechanism pathway)

mechanism 35 nucleophilic substitution of a halogenoalkane by water (S_N2 bimolecular)

This is the typical S_N2 mechanism by which primary halogenoalkanes hydrolyse, doesn't involve a carbocation

General scheme for the S_N2 nucleophilic substitution reaction of e.g. a primary halogenoalkane undergoing direct hydrolysis with water. RCH₂X where R = H or alkyl and X is the halogen atom Cl, Br or I).

Step (1): The C^δ+-X^δ- bond is polar because of the difference in electronegativity between carbon (2.5) and chlorine (3.0), so the electron rich nucleophile, the water molecule, attacks the partially positive carbon.

The nucleophilic water acts as an electron pair donor (Lewis base) to bond with the 'delta positive' carbon to give the C-O bond of the protonated alcohol.

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

Step (1) is the rate determining step (rds) and the rate effectively only depends on the halogenoalkane concentration.

BUT the intermediate is [R₃COH₂]⁺ NOT an R₃C⁺ carbocation.

In step (2) another water molecule rapidly accepts the proton from the protonated alcohol to leave the free alcohol product.

This mechanism is most likely with primary halogenoalkanes, but very slow if at all with water, much faster with the hydroxide ion, a more powerful nucleophile (negative ion as well as an electron pair donor).

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 water 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 highest activation energy, E_a1, is by far the largest of the two and so the bimolecular step 1 is the rate determining step. The H⁺ product is actually a H₃O⁺ via a 2nd water molecule.

 

Mechanism diagrams 71a and 71c show the hydrolysis of bromoethane by water.

Note the concept of the transition state in the reaction profile *, 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 H⁺ product is actually a H₃O⁺ via a 2nd water molecule.

The [transition state], sometimes called the 'activation complex' is neutral because the nucleophile water is also neutral.

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

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3.3.5 The hydrolysis of secondary halogenoalkanes

Secondary halogenoalkanes can hydrolyse by either an S_N1 or S_N2 mechanism.

The S_N1 mechanism is faster than the S_N2 pathway, and so is the dominant mechanism hence rate of hydrolysis order is tert  >  sec  >  prim

e.g. quoting from section 3.3.3 on reactivity trends

The tertiary halogenoalkane, 2-chloro-2-methylpropane, produces an almost instant white precipitate of silver chloride when mixed with ethanolic silver nitrate solution at room temperature.

This contrasts with the very slow hydrolysis of isomeric primary halogenoalkane 1-chlorobutane, that only produces a white precipitate after warmed to ~60^oC for some time.

In exams you should be given equal credit for describing and S_N1 or S_N2 mechanism for the hydrolysis of a secondary halogenoalkane.

TOP OF PAGE and sub-index

3.3.6 S_N1 and S_N2 water hydrolysis mechanisms - their connection with rate expressions and orders of reaction - a kinetic discussion

(when first encountering this, you may not have done advanced 'rates of reaction' studies, so return to section 3.3.6 later in your course)

Reminder !

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

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 10 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 water 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.

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][H₂O], k₂ = 2nd order rate constant, [RX] = concentration of haloalkane,

and [H₂O] = concentration of water, in the rate expression.

Since water or aqueous ethanol is the solvent, the concentration of water is so high, it is effectively constant, so the concentration of the halogenoalkane controls the rate.

The kinetics simplifies to: rate = k[RX]

However, this is not the case for the hydrolysis using sodium hydroxide solution (Part 3.4).

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

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.

 

The order of reaction can only be determined by experiment - see

Obtaining rate data, interpreting data, deducing orders of reaction and rate expressions

 

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