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Halogenoalkanes

Written by tutor Heidi R.

Halogenoalkanes are an exciting class of compounds as they are highly reactive, compared to alkanes, when reacted with nucleophiles or bases. Recall halogens, in organic chemistry we often refer to as X, are F, Cl, Br and I. Halogenoalkanes are based on alkanes so they have all single bonds and are therefore SP3 hybridized. With the exception of the fluroalkane they are more reactive than alkanes principally due to two reasons:

1. C-X bond is polar due to the difference in electronegativity between carbon and X. The carbon which is now considered to be positive is now susceptible to nucleophilic attack by either a neutral nucleophile (H20, NH3, ROH etc) or by a negatively charged nucleophile (OH-, NH2-, H-, CN- etc). The halogen which is electronegative will accept the electrons from the broken C-X bond and leave.
2. C-X bonds are usually weaker than a C-C bond or a C-H bond. (C-F bonds are stronger hence they are less reactive). As the halogen increases in size the C-X bond becomes weaker and weaker. This is because the bond length is increasing as the halogen increases in size.

IUPAC Nomenclature

Halogenoalkanes are named based on the longest parent alkane in the structure, even if the halogen is not connected to any of the carbons. Use the lowest possible number for any halogens or alkyl chains. You may encounter some common names when dealing with halogenoalkanes especially when the solvents are concerned. For example dichoromethane (CH2Cl2) is often referred to as methylene chloride.

Halogenoalkanes Classification

In order to understand the reactions of the halogenoalkanes we need to be able to classify them. There are four different types.

1. Methyl halides. The halide is attached only to the methyl group.
2. Primary halides. The carbon where the halogen is bonded is only attached to one other carbon.
3. Secondary halides. The carbon where the halogen is bonded is only attached to two other carbons.
4. Tertiary halides. The carbon where the halogen is bonded is attached to three other carbons.

Preparation of Halogenoalkanes

Halogenoalkanes can be prepared by several different methods.

1. Free radical halogenation. This is a good lab synthesis for bromination as this is a highly selective reaction. Tertertiary bromination is more selective than secondary which is more selective than primary. Chlorination on the other hand is usually a poor lab synthesis as it is not selective.
2. Electrophilic addition of hydrogen halides across the double bond. This is one of the easiest electrophilic additions across a double bond. The hydrogen (proton) acts as the electrophile and the halide acts like the nucleophile. First the electrophile will receive the pair of electrons from the pie bond therefore attaching it the molecule. This will leave a positive charge (carbocation) on the other carbon where the double bond was. The next step in the nucleophile (halide) will donate its electrons to the carbocation and thus attaching itself to the other carbon. The resulting molecule will have all single bonds and the hydrogen and halide will be attached. This reactions is usually markovikov addition (Hydrogens add to the side with most hydrogens).

3. Free radical addition of hydrogen halides across a double bond. By adding peroxide into the reaction vessel with hydrogen halide we can force the reaction to go anti markovikov.

Reactions of Halogenolalkanes

1. Halogens are great leaving groups so they will undergo substitution reactions readily. There are two mechanisms available for substitution, unimolecular (SN1) and bimolecular (SN2). The exact mechanism depends on the halogenoalkane classification and also the nature of the nucleophile. No matter what the mechanism the resulting product would be the nucleophile displacing the halide.
2. Halogenoalkanes will also readily undergo elimination reactions. Just like in the substitution reactions there are two main reaction mechanisms governing elimination. (E1 and E2). Again the exact mechanism depends on the halogenoalkane classification and the nature of the base. An elimination reaction is the removal of two atoms, in this case a hydrogen and a halide, resulting in a double bond. If the alkyl halide has more than two carbons in its chain, and the carbon atoms next to the carbon atom with the halogen both have hydrogen atoms bonded to them, more than one product can be formed. The major product can usually be predicted by Zaitsev's rule. Usually the more substituted alkene will be the major product.

It should be noted that substitution and elimination reaction are in competition with other.

Physical Properties

As expected the boiling points of different length chain halogenoalkanes increases as the chain length increases due to more London dispersion forces. The boiling points of the same length chain halogenoalkane increases with increasing size of halogen due to more London dispersion forces. If there are isomers of the same molecular formula the compound that is the most branched will have the lowest boiling point.

Halogenoalkanes are usually soluble in non polar solvents such as hexane but have little solubility in water.

Halogenoalkanes Quiz

The most reactive halogenoalkane contains which halogen?

A. Fluorine
B. Chlorine
C. Bromine
D. Iodine
The correct answer here would be D.

Iodoalkanes are the most reactive because they have the weakest bond of C-I, thus they are most easily broken, making them most reactive.

A secondary halogenoalkane may be described as:

A. The halide is attached only to the methyl group
B. The carbon where the halogen is bonded is only attached to one other carbon.
C. The carbon where the halogen is bonded is attached to two other carbons.
D. The carbon where the halogen is bonded is attached to three other carbons.
The correct answer here would be C.

Halogenoalkanes can be prepared by

A. Nucleophilic addition to carboxylic acids
B. Electrophilic addition to aromatic rings
C. Electrophilic addition across double bonds
D. Elimination reactions
The correct answer here would be C.

Halogenoalkanes may undergo which of the following reactions?
i. elimination
ii. nucleophilic substitution
iii. electrophilic aromatic substitution
iv. nucleophilic addition
v. electrophilic addition

A. none of the above
B. all of the above
C. just i
D. just i and ii
E. just i and iv
The correct answer here would be D.
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