Part 3. The chemistry of HALOGENOALKANES

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3.6 The reaction between halogenoalkanes and ammonia or amines - more nucleophilic substitution reactions

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 amines from the halogenoalkane in multiple styles and added the nucleophilic substitution mechanisms where appropriate.

Sub-index for this page

3.6.1 Introduction to the reaction between halogenoalkanes and ammonia  (study before 3.6.3)

3.6.2 Introduction to the reaction between halogenoalkanes and an amine  (study before 3.6.4)

3.6.3 The reaction mechanisms for halogenoalkanes reacting with ammonia (S_N1 and S_N2)

3.6.4 The reaction mechanisms for halogenoalkanes reacting with an amine (S_N1 and S_N2)

3.6.1 Introduction to the reaction between halogenoalkanes & ammonia to form a primary aliphatic amine

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

You must know the structures of primary, secondary and tertiary amines, and quaternary ammonium salts

The basic reaction is: R-X  +  2NH₃  ===> R-NH₂  +  NH₄X  (R = alky, X = Cl, Br, I)

Note the use of excess ammonia to free the aliphatic amine, which is itself, a base like ammonia AND can also act as a nucleophile.

In equations use two of the base, never say HCl is a product, it will always be neutralised by any base present to give the halide salt.

Ammonia (:NH₃) is the nucleophile, donating its lone pair of electrons to the δ+ carbon atom of the C-X bond to form the new C-N bond. So this is another nucleophilic substitution reaction.

Ammonia is an alkaline gas that is very soluble in water.

Ammonia is weaker nucleophile than the hydroxide ion since it doesn't carry an overall negative charge.

Unfortunately, you cannot heat the mixture of conc. ammonia and halogenoalkane under reflux because the ammonia gas would boil off i.e. ammonia would not be condensed by the reflux condenser.

Therefore, in industry (not school laboratory), the conc. ammonia and haloalkane must be heated in a sealed container and ethanol can be used as a solvent too.

However, leaving a mixture of conc. ammonia in aqueous ethanol with a halogenoalkane at room temperature for a 'long time' will result in amine formation - its just a rather slow in most cases.

With ammonia as the nucleophile, the initial product from a halogenoalkane, is always a primary amine.

 

(a) The reaction between ammonia and bromoethane  (a primary halogenoalkane)

Strictly speaking all the reactants and products should be suffixed by (aq)

The primary halogenoalkane bromoethane gives the primary aliphatic amine ethylamine/aminoethane.

bromoethane  +  ammonia  ===>  ethylamine (aminoethane)  +  hydrogen bromide

  +  2NH₃ +   NH₄Br 

(displayed formula equation showing the formation of a primary amine)

  +  2NH₃   +  NH₄Br

(structured formula equation)

  +  2NH₃ +  NH₄Br

(abbreviated structured formula equation)

  +  2NH₃ +  NH₄Br 

(skeletal formula equation)

CH₃CH₂Br  +  2NH₃ CH₃CH₂NH₂  + NH₄⁺  +  Br^-

(ionic equation) In terms of equations, I think that covers all bases !!!

 

You employ excess ammonia to combine with the hydrogen bromide which is effectively formed e.g.

(i) if no excess ammonia you get: CH₃CH₂Br  +  NH₃  CH₃CH₂NH₂  + HBr    CH₃CH₂NH₃⁺ + Br^–

There is an equilibrium between the free amine and its bromide salt, but best to use the equation 'style' below.

(ii) with excess ammonia you get: CH₃CH₂Br  +  2NH₃ CH₃CH₂NH₂  + NH₄Br

or ionically:   CH₃CH₂Br  +  2NH₃ CH₃CH₂NH₂  + NH₄⁺  +  Br^-

so you get the ammonium bromide salt and the free amine and not the amine's bromide salt.

 

Note with a large excess of the halogenoalkane you will get the secondary amine formed because the primary amine formed, is itself a nucleophile

e.g. CH₃CH₂Br  +  2CH₃CH₂NH₂ (CH₃CH₂)₂NH  +  CH₃CH₂NH₃⁺Br^-

The secondary amine is also a nucleophile, so a tertiary amine is formed

e.g. CH₃CH₂Br  +  2(CH₃CH₂)₂NH (CH₃CH₂)₃N  +  (CH₃CH₂)₂NH₂⁺Br^-

Finally. a quaternary ammonium salt can be formed

e.g. CH₃CH₂Br  +  (CH₃CH₂)₃N (CH₃CH₂)₄N⁺Br^-

More on these reactions and other examples are described further down the page.

 

(b) Reacting the secondary halogenoalkane 2-iodobutane with ammonia gives the primary 2-aminobutane

CH₃CHICH₂CH₃  +  2NH₃  ===>  CH₃CH(NH₂)CH₂CH₃  +   +  NH₄⁺  +  I^-

,  the structure of 2-aminobutane

 

(c) Adding the tertiary halogenoalkane  2-chloro-2-methylpropane to excess conc. ammonia at room temperature gives a primary amine.

(CH₃)₃C-Cl  +  2NH₃  ===>  (CH₃)₃C-NH₂  +  NH₄⁺  +  Cl^-

 

The industrial manufacture of aliphatic amines

Chloroalkanes would be most used in industry because they are much cheaper than bromoalkanes or iodoalkanes.

The ammonia/amine and halogenoalkane are heated under pressure in a sealed cylinder.

 

Chloroalkanes are readily manufactured by the free radical chlorination of alkanes from oil, see ....

Industrial manufacture of halogenoalkanes  and

Free radical chlorination/bromination to give halogenoalkanes (haloalkanes, alkyl halides)

TOP OF PAGE and sub-index

3.6.2 Introduction to the reaction between halogenoalkanes and amines to form secondary and tertiary aliphatic amines

You must know the structures of primary, secondary and tertiary amines, and quaternary ammonium salts

I'll outline the series of reactions that can happen starting with 1-chloropropane and ammonia.

The main new point to understand is the amine formed with ammonia and the haloalkane can also act as a nucleophile so you can get further reactions with the haloalkane to give secondary amines, tertiary amines and eventually a quaternary ammonium salt.

e.g. starting with 1-chloropropane and ammonia making a succession of further amine substitution products

(i) The initial product is a primary amine, 1-aminopropane (propylamine):

CH₃CH₂CH₂Cl  +  2NH₃  ===>  CH₃CH₂CH₂NH₂  +  NH₄⁺Cl^-

1-aminopropane (propylamine) structure  

(ii) Then the primary amine formed acts as a nucleophile and undergoes a substitution reaction with the original haloalkane (1-chloropropane) leading to the formation of a secondary amine:

CH₃CH₂CH₂Cl  +  2CH₃CH₂CH₂NH₂  ===>  (CH₃CH₂CH₂)₂NH  +  CH₃CH₂CH₂NH₃⁺Cl^-

The product is dipropylamine ,  .

(iii) Then the secondary amine formed acts as a nucleophile and undergoes a substitutes with the original haloalkane (1-chloropropane) leading to the formation of a tertiary amine:

CH₃CH₂CH₂Cl  +  2(CH₃CH₂CH₂)₂NH  ===>  (CH₃CH₂CH₂)₃N  +  (CH₃CH₂CH₂)₂NH₂⁺Cl^-

The product is tripropylamine

(iv) Finally the tertiary amine, again, acting as a nucleophile can undergo one more substitution reaction to form a quaternary ammonium salt.

CH₃CH₂CH₂Cl  + (CH₃CH₂CH₂)₃N  ===>  (CH₃CH₂CH₂)₄N⁺Cl^-

The product is tetrapropylammonium chloride

 

2nd example starting with bromoethane and ammonia

(i)  CH₃CH₂Br  +  2NH₃  ===>  CH₃CH₂NH₂  +  NH₄⁺Br^-

The product is the primary amine ethylamine (aminoethane)

(ii)  CH₃CH₂Br  +  2CH₃CH₂NH₂  ===>  (CH₃CH₂)₂NH  +  CH₃CH₂NH₃⁺Br^-

The product is the secondary amine diethylamine

(iii)  CH₃CH₂Br  +  2(CH₃CH₂)₂NH  ===>  (CH₃CH₂)₃N  +  (CH₃CH₂)₂NH₂⁺Br^-

The product is the tertiary amine triethylamine

(iv)  CH₃CH₂Br  + (CH₃CH₂)₃N  ===>  (CH₃CH₂)₄N⁺Br^-

The product is the quaternary ammonium salt tetraethylammonium bromide

 

By adjusting the ratio of the reactants, and the reaction conditions in the sealed reaction container, you can control which products you would like to predominate.

This is only one method to manufacture amines, and little used these days, because there lots of other more economic and 'greener' ways and you tend to get quite a mixture of products.

Quaternary ammonia salts find applications as surface active agents ('surfactants')

They act as cationic surfactants (opposite of fatty acid anionic surfactants).

They have a relatively small positive ionic hydrophilic end that solvates (bonds) with water and a much larger-longer hydrophobic end that can intermolecular bond with oil/fat molecules. Their applications include:

(i) in detergents to aid cleaning surfaces of fabrics to aid dyeing them

(ii) they kill bacteria by destroying their cell walls, so used in disinfectants, mouthwashes and toothpaste

TOP OF PAGE and sub-index

3.6.3 The reaction mechanisms for halogenoalkanes reacting with ammonia (S_N1 and S_N2 pathways)

Mechanism diagram 37 illustrates the general S_N1 unimolecular mechanism as haloalkane undergoes a nucleophilic substitution reaction with ammonia (shown for Br, but same for Cl or I).

Mechanism diagram 73b illustrates the S_N1 unimolecular mechanism as 2-bromo-2-methylpropane undergoes a nucleophilic substitution reaction with ammonia (shown for Br, but same for Cl or I).

Step 1: Heterolytic bond fission of the 2-bromo-2-methylpropane to give the tertiary carbocation and a bromide ion, C-Br bond pair shifts onto the Br atom - slowest unimolecular rate determining step, highest activation energy.

Step 2: The ammonia nucleophile (electron pair donor) bonds to the carbocation to form the protonated primary amine - fast.

Step 3: An excess ammonia molecule removes a proton from the protonated amine to leave the free primary amine - fast.

 

Mechanism diagram 9 illustrates the general S_N2 bimolecular mechanism as haloalkane undergoes a nucleophilic substitution reaction with ammonia (shown for Br, but same for Cl or I).

S_N2 'bimolecular', a two step mechanism [mechanism 9 above]

Step (1) the C^δ+-Br^δ- bond is polar, so the electron rich nucleophile, the ammonia molecule, attacks the slightly positive carbon (δ+).

The nucleophile acts as an electron pair donor (Lewis base) to bond with the 'positive' carbon.

Simultaneously the bromine atom is ejected, taking with it the C-Br bonding pair of electrons, so forming the bromide ion.

This is the slowest rate determining step.

In step (2) one of the excess ammonia molecules can remove a proton to leave the primary amine product (fast).

Mechanism diagram 73a shows the S_N2 bimolecular mechanism as bromoethane undergoes a nucleophilic substitution reaction with ammonia (shown for Br, but same for Cl or I).

In this case I've shown the 'transition state' or 'activated complex' plus the simple reaction progress profile.

Ammonia (:NH₃) is the nucleophile, its lone pair of electrons attack the δ+ carbon atom of the C-X bond, simultaneously the C-X bond breaks and the new C-N bond forms.

Initially a protonated amine forms, so another ammonia molecule removes the proton to form the ammonium ion - shown in the 1st version, but the 2nd version just treats it as a concerted process via the transition state - which is neutral since no charge on the attacking ammonia nucleophile.

You can insert the formation of the [transition state] in the 1st version of 73a if you so wish.

TOP OF PAGE and sub-index

3.6.4 The reaction mechanisms for halogenoalkanes reacting with an amine (S_N1 and S_N2 pathways)

Mechanism diagram 38 illustrates the general S_N1 unimolecular mechanism as haloalkane undergoes a nucleophilic substitution reaction with a primary amine (shown for Br, but same for Cl or I).

S_N1 unimolecular, a three step ionic mechanism via carbocation formation [mechanism 38 above]

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

In step (2) the nucleophilic electron pair donating primary amine molecule rapidly adds to the carbocation to give the protonated amine product R₃C-NH₂⁺.

The primary amine is now acting as the nucleophile.

In step (3) one of the excess amine molecules can remove a proton to leave the primary amine product.

Mechanism diagram 74b shows the S_N1 unimolecular mechanism as 2-bromo-2methylpropane undergoes a nucleophilic substitution reaction with the primary amine ethylamine (shown for Br, but same for Cl or I).

Step 1: Heterolytic bond fission of the 2-bromo-2-methylpropane to give the tertiary carbocation and a bromide ion, C-Br bond pair shifts onto Br atom - slowest unimolecular rate determining step, highest activation energy.

Step 2: The ethylamine nucleophile (electron pair donor) bonds to the carbocation to form the protonated secondary amine - fast.

Step 3: An excess amine molecule removes a proton from the protonated amine to leave the free secondary amine.

 

Mechanism diagram 11 illustrates the S_N2 bimolecular mechanism as haloalkane undergoes a nucleophilic substitution reaction with a primary amine (shown for Br, but same for Cl or I).

S_N2 'bimolecular', a two step mechanism [mechanism 11 above]

Step (1) the C^δ+-Br^δ- bond is polar, so the electron rich nucleophile, the ammonia molecule, attacks the slightly positive carbon. The nucleophile acts as an electron pair donor (Lewis base) to bond with the 'positive' carbon. Simultaneously the bromine atom is ejected, taking with it the C-Br bonding pair of electrons, so forming the bromide (halide) ion.

In step (2) one of the excess amine molecules can remove a proton to leave the primary amine product.

 

Mechanism diagram 74a illustrates the S_N2 bimolecular mechanism as bromoethane undergoes a nucleophilic substitution reaction with the primary amine ethylamine (shown for Br, but same for Cl or I).

Ethylamine, an electron pair donor, act as the nucleophile, its lone pair of electrons attack the δ+ carbon atom of the C-Br bond, simultaneously the C-Br bond breaks and the new C-N bond forms as the C-Br bond pair is shifted onto the bromine atom.

Initially a protonated amine forms, so another ammonia molecule removes the proton to form the ethylammonium ion - shown in the 1st version.

In the 2nd version (diagram 74c below) just treats the proton loss as a concerted process via the transition state - which is neutral since no charge on the attacking ammonia nucleophile.

You can insert the formation of the [transition state] in the 1st version of 74c (below) if you so wish.

 

Mechanism diagram 74c illustrates the S_N2 bimolecular mechanism as bromoethane undergoes a nucleophilic substitution reaction with the primary amine ethylamine (shown for Br, but same for Cl or I).

In this case I've shown the 'transition state' or 'activated complex' plus the simple reaction progress profile.

You can insert the formation of the [transition state] in the 1st version diagram 74a if you so wish.

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