1 of 31© boardworks ltd 2009. 2 of 31© boardworks ltd 2009 how are halogenoalkanes made? there are...

1 of 31 © Boardworks Ltd 2009


How are halogenoalkanes made?

There are several ways by which halogenoalkanes can be made, including:

free radical substitution of an alkane:

electrophilic addition of HX or X2 to an alkene:

CH4 + Cl2 CH3Cl + HCl

C2H4 + HBr C2H5Br

C2H4 + Br2 C2H4Br2


Free radical substitution: Cl2 + CH4


Other products of chain reactions

If an alkane is more than two carbons in length then any of the hydrogen atoms may be substituted, leading to a mixture of different isomers. For example:

The mixture of products is difficult to separate, and this is one reason why chain reactions are not a good method of preparing halogenoalkanes.

1-chloropropane 2-chloropropane


Further substitution in chain reactions

Some chloromethane molecules formed during free radical substitution between methane and chlorine will undergo further substitution to form dichloromethane. Further substitution can occur until all hydrogens are substituted.

The further substituted chloroalkanes are impurities that must be removed. The amount of these molecules can be decreased by reducing the proportion of chlorine in the reaction mixture.


Chain reactions and ozone


Free radical reactions: true or false?


Bonding in alkenes


Structure of alkenes summary


Double bonds and electrophiles

The double bond of an alkene is an area of high electron density, and therefore an area of high negative charge.

The negative charge of the double bond may be attacked by electron-deficient species, which will accept a lone pair of electrons.

Alkenes undergo addition reactions when attacked by electrophiles. This is called electrophilic addition.

These species have either a full positive charge or slight positive charge on one or more of their atoms. They are called electrophiles, meaning ‘electron loving’.


Electrophilic addition mechanism: 1

In the first stage of electrophilic addition, the positive charge on the electrophile is attracted to the electron density in the double bond.

As the electrophile approaches the double bond, electrons in the A–B bond are repelled towards B.

The pi bond breaks, and A bonds to the carbon, forming a carbocation – an ion with a positively-charge carbon atom. The two electrons in the A–B bond move to B forming a B- ion.

δ+ δ-


Electrophilic addition mechanism: 2

In the second stage of electrophilic addition, the B- ion acts as a nucleophile and attacks the carbocation.

The lone pair of electrons on the B- ion are attracted towards the positively-charged carbon in the carbocation, causing B to bond to it.

Because both electrons in the bond that joins B- to the carbocation ion come from B-, the bond is a co-ordinate bond.


More on the bromine water test

A simple equation for the bromine water test with ethene is:

However, because water is present in such a large amount, a water molecule (which has a lone pair) adds to one of the carbon atoms, followed by the loss of a H+ ion.

CH2=CH2 + Br2 + H2O CH2BrCH2Br + H2O

CH2=CH2 + Br2 + H2O CH2BrCH2OH + HBr

The major product of the test is not 1,2-dibromoethane (CH2BrCH2Br) but 2-bromoethan-1-ol (CH2BrCH2OH).


Electrophilic addition reactions


Addition to unsymmetrical alkenes

When an electrophile (e.g. HBr) attacks an alkene with three or more carbon atoms (e.g. propene), a mix of products is formed. This is because these alkenes are unsymmetrical.

Unequal amounts of each product are formed due to the relative stabilities of the carbocation intermediates.

minor product:1-bromopropane

major product:2-bromopropane

HBr


A chain of carbon atoms can be represented by R when drawing organic structures. This is an alkyl group (general formula CnH2n+1).

Primary (1°) carbocations have one alkyl group attached to the positively-charged carbon.

Secondary (2°) carbocations have two alkyl groups attached to the positively-charged carbon.

Tertiary (3°) carbocations have three alkyl groups attached to the positively-charged carbon.

Structure of carbocations


The stability of carbocations increases as the number of alkyl groups on the positively-charged carbon atom increases.

The stability increases because alkyl groups contain a greater electron density than hydrogen atoms. This density is attracted towards, and reduces, the positive charge on the carbon atom, which has a stabilizing effect.

Stability of carbocations

increasing stability

tertiaryprimary secondary


Structure of carbocations


Electrophiles: true or false?


Polar bonds and nucleophiles

The carbon–halogen bond in halogenoalkanes is polar because all halogens are more electronegative than carbon.

The polar bond means that the carbon atom has a small positive charge (δ+), which attracts substances with a lone pair of electrons. These are nucleophiles, meaning ‘nucleus (positive charge) loving’. Examples include:

δ+ δ- δ+ δ- δ+ δ- δ+ δ-

ammonia cyanide hydroxide


Nucleophiles (Nu-) attack the carbon of a carbon–halogen (C–X) bond, because the electron pair on the nucleophile is attracted towards the small positive charge on the carbon.

Reaction with nucleophiles

The electrons in the C–X bond are repelled as the Nu- approaches the carbon atom.

δ+ δ-

The Nu- bonds to the carbon and the C–X

bond breaks. The two electrons move to the halogen, forming a halide ion.

The halide is substituted, so this is a nucleophilic substitution reaction.


Nucleophilic substitution reactions


Rate of nucleophilic substitution

The rate of a nucleophilic substitution reaction depends on the strength of the carbon–halogen bond rather than the degree of polarization in the bond.

The C–I bond is the weakest and so most readily undergoes nucleophilic substitution. The rate of reactions involving iodoalkanes is the highest.

238C–I

276C–Br

338C–Cl

484C–F

Strength (kJ mol-1)Bond


Nucleophilic substitution


Elimination in halogenoalkanes

In the reaction with a strong base, halogenoalkanes will undergo not only nucleophilic substitution but also elimination reactions, forming alkenes and water.

The OH- acts as both a base and a nucleophile. When acting as a base, the OH- removes H+ from the halogenoalkane, which also results in the formation of a halide ion.

The reaction between a halogenoalkane and a strong base usually results in the formation of a mixture of substitution and elimination products.


Elimination mechanism


Mixture of elimination products

If the carbon chain is four or more carbons in length and the halogen is not attached to a terminal carbon, a mixture of positional isomers may be formed.

attack at Aattack at BA B

but-2-ene but-1-ene


Conditions are important

The conditions for the reaction that favour substitution or elimination are different.

Base strength: the stronger the base used, the more elimination is favoured. Sodium hydroxide in aqueous solution contains OH-, but when dissolved in ethanol, CH3CH2O- is also present, which is a stronger base.

Therefore elimination is favoured by NaOH in ethanolic solution, and substitution is favoured by NaOH in aqueous solution.

Temperature: elimination is favoured at hotter temperatures whereas substitution is favoured by warm conditions.


Primary, secondary or tertiary?

Primary halogenoalkanes favour substitution whereas tertiary halogenoalkanes favour elimination.

primary tertiarysecondary

elimination more likely

substitution more likely


Elimination or substitution?


Glossary


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Multiple-choice quiz

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