As the name implies a halogenoalkane is any molecule which is made up of the main alkane chain and a halogen group attached to the alkane chain. Three examples of halogenoalkanes are the following:

halo 1

With the naming being similar to that of alkanes, with side groups being placed in alphabetical order and using the smallest numbers possible.


Although halogenoalkane have got a halogen attached to a Carbon, which would produce a small charge difference due to the electronegativity of the halogen this is still not enough to make these soluble in water.

Boiling Point

The boiling pattern is similar to that of alkanes, but due to the fact that the halogens are electronegative this would mean that the attraction between separate molecules would be bigger then that of alkanes, and thus these would be liquid at lower temperatures.

Primary, Secondary and Tertiary Carbons

The words primary, secondary and tertiary are showing the number of Carbons attached to the α Carbon. The α Carbon is the Carbon which is attached to the halogen and therefore a primary carbon would have 1 Carbon atom attached to the α Carbon, a secondary Carbon would have two Carbons attached to the α Carbon and tertiary Carbon would have three Carbon atoms attached to the α Carbon. It must be noted that primary and tertiary halogenolakanes react with different mechanisms, and these would be analysed in more detail. Secondary Carbons would react with a mixture of the two mechanisms.

halo 2


It must be noted that the most common reaction for a halogenoalkane is the Substitution reaction and for this to occur there are two steps that must take place:

  • The molecule loses the halogen to provide a carbocation and a halide
  • The molecule attracts a partially negative which would displace the halogen.

When seeing these two steps one would think that a fluorohalogenoalkane would be the most reactive halogenoalkane, due to the fact that it would produce the highest positive charge on the Carbon due to its high electronegativity. This is not the case since the most important aspect of the reaction is the breaking of the bond between the Carbon and the halogen, and thus it would be easier to break a weak bond rather than a strong bond, and thus the reactivity of the halogens is as follows:

C-I > C-Br > C-Cl > C-F

seeing that the strength of the bonds is as follows:

halo 3

It must be noted that for most substitution reactions to take place reflux must be performed.



The name of the mechanism is nucleophilic substitution where the 1 indicates that there is only one reactant involved in the rate-determining step, giving a rate equation of

r = k *

halo 4

The first step in the mechanism is the loss of the halogenoalkane to produce a carbocation. A nucleophile would then attack the carbocation to produce a Carbon – Nucleophile bond and thus a new compound. The mechanism is seen below:


The name of the mechanism is nucleophilic substitution where the 2 indicates that there are two reactants involved in the rate-determining step, giving a rate equation of

r = k *

halo 5

The two mechanisms can occur on any halogenoalkane, but it must be noted that certain halogenoalkanes are more prone to a certain mechanism, and therefore analysis of the two mechanisms has to be performed.

In SN1 mechanism a carbocation is produced, and as seen in alkanes the higher the number of carbon groups attached to the α Carbon would mean that the carbocation is stabilised the most. This would mean that tertiary halogenoalkanes can stabilise the resultant carbocation more than a primary halogenoalkane.

In SN2 mechanism the reaction occurs in a single step where the reactants come in contact to form a 5 membered Carbon as an Intermediate. In order for this to occur the α Carbon should not have many groups surrounding it so that there would be the place available from where the nucleophile can attack. This would thus mean that primary halogenoalakanes are more prone to be attacked by a nucleophile then a tertiary halogenoalkane.

After analysis of the mechanisms, it can thus be said that primary halogenoalkanes react in an SN2 mechanism while tertiary halogenoalkanes react in an SN1 mechanism.  Secondary halogenoalkanes can react with both mechanisms.

Preparation of halogenoalkanes

From alcohols

halo 6

From alkenes

From alkenes, there are three types of halogenoalkanes that can be produced, mainly the dihalogenoalkane, halogenoalkane and halogenoalcohol. These can be produced according to the reagents used.

If an acid is used then a halogenoalkane is used:


If the pure halogen is used then the dihalogenoalkane is produced:

CH2=CHCH3 + Br2(l) à CH2BrCHBrCH3

If the halogen used is in aqueous conditions than the  halogenoalcohol is produced:

CH2=CHCH3 + Br2(aq) à CH2BrCH(OH)CH3

Reactions of Halogenoalkanes

With Hydroxide ion

There are two possible reactions that can take place, either substitution to form an alcohol or an elimination to form an alkene. It is of utter importance to know the concentration used, since high concentrations tend to produce elimination reaction, while a lower concentration would tend to produce the alcohol. Another important factor is the solvent used, for the preparation of a substitution reaction the solvent should be water while for an elimination reaction the solvent should be alcohol.

CH3CH2I + NaOH(aq) à CH3CH2OH + NaI (reflux)

CH3CH2I + NaOH(ethanol) à CH2=CH2 + NaI                (reflux)

Producing an alkane

Halogenoalkanes can be used to produce alkanes, mainly in two simple ways.

The first way is to produce Hydrogen atoms, which are very active, and these would displace the halogen. the hydrogen atoms would be prepared insitu, and before having the chance to react with another Hydrogen atom to form a Hydrogen molecule these would react with the halogenoalkane forming an alkane. This method can only be sued for iodoalkanes.


If sodium is reacted directly with sodium then the carbon chain would become double in length, with the reaction being as follows:

2CH3CH2I +2Na à CH3CH2CH2CH3 + 2NaI


Increasing Carbon chain

Increasing the Carbon chain is one of the most difficult processes in organic chemistry, but due to the fact that halogenoalkanes can react with the cyanide ion, this is one of the ways to do it. The solvent used must not be water since the hydroxide ion would react instead of the cyanide.

CH3CH2Br + KCN(propanone) à CH3CH2CN + KBr (reflux)

Producing a Grignard Reagent

Grignard Reagents are very reactive products that can be used to increase the Carbon chain. It must be noted that these are explosive in water, and thus these should never be introduced to a solution where water is present, and thus the solvent used is dry ether.

CH3CH2Cl + Mg(dry ether) à CH3CH2MgCl (reflux)

Finding the type of Halogenoalkane

The test for halides does not work on halogenoalkanes, mainly due to the fact the halogen is not readily available to bond with the silver ion. In order to obtain a halide ion one would have to first react the halogenoalkane with sodium, and obtain a common salt, NaX.

Two tests will then be used, the addition of silver nitrate followed by the addition of ammonium hydroxide.

Ion present Observation
Cl White precipitate
Br Very pale cream precipitate
I Very pale precipitate
Original precipitate Observation
AgCl Precipitate dissolved to give a colourless solution
AgBr Precipitate is almost unchanged using dilute ammonia solution, but dissolves in concentrated ammonia solution to give a colourless solution
AgI Precipitate is insoluble in ammonia solution of any concentration