Pre-requisite Reading- Type of reactions
According to the IUPAC nomenclature, a halogen derivative of an alkane is called a Haloalkane, and the common or the trivial name is Alkyl Halide. Enclosed in the common name lies the mnemonic to remember few of the common preparation methods.
Mnemonic : AAAlkyl Halide (Alkane, Alkene, Alcohol, Halogen transfer)
AAAlkyl Halide represents the preparation of a haloalkane (alkyl halide) from an alkane, alkene, alcohol and replacing an existing halogen with the other in halogen transfer reactions.
A) Preparation of Haloalkane from Alkanes- Free Radical Halogenation
Pre-requisite reading- Free radical reaction and mechanism, Properties of Halogen, Types of equivalent hydrogen, Free radical stability, Identification of allylic and benzylic position, Identification of 1o, 2o and 3o alkyl groups
Alkanes are very stable, inert molecules that undergo only a few reactions - cracking (breaking down a larger carbon chain into smaller carbon fragments), combustion (oxidation of an alkane to give carbon dioxide and water molecules ) and free radical halogenation reactions (replacing Hydrogen of an alkane with halogen).
For a free radical halogenation, an alkane is reacted with the halogen (Cl2 or Br2) in the presence of UV light or sunlight as the catalyst. In absence of a light source, a very high temperature of 520 K -670 K can be employed.
The order of the halogens from most reactive to the least is- F2 > Cl2 > Br2 > I2 . Most reactions are done using Cl2 or Br2 as halogenation with F2 is highly reactive, explosive and is hard to control. Reactions with I2, on the other hand, are very sluggish and reversible. It is due to the formation of the byproduct HI along with the desired iodoalkane. In such situations, the reaction is done using I2 along with an oxidizing agent such as - HNO3, HgO, HI that destroys the HI formed by converting it back to I2. According to the Le Chatelier principle, increase in I2 concentration pushes the reaction to obtain more of iodoalkane.
The outcome of a halogenation reaction using Cl2 or Br2 is mono, di and polysubstituted haloalkanes depending on the number and type of replaceable hydrogens. For example, for methane the number of Hydrogens are four from which it can form four halogenated compounds.
But for three carbon molecule like propane the maximum number of replaceable hydrogen is eight so one can obtain twenty two combinations of halogenated products in presence of excess halogen.
But if the ratio of alkane to the halogen is tweaked to have less halogen per molecule of alkane (1: 0.8), then excess halogen will not be available for forming polyhalogen species, and the primary product would be a monohalogen compound. The issue does not resolve at the formation of the monohalogen compound. From a molecule like propane, it can give two types of monohalogenated products due to the presence of two types of the hydrogens.
The major product is the one where the radical before the attachment of the halogen is most stable. A radical being an electron deficient species follows the stability order similar to the carbocations- 3o> 2o> 1o. Therefore, 2o radical formed gives 2-chloropropane as the major product over 1-chloropropane that is obtained from a less stable primary radical. The 2-chloropropane and 1-chloropropane are positional isomers with a very similar boiling point that would be hard to separate.
Another example of the type of the hydrogen and most stable product formed is shown for isopentane molecule -
Selective halogenation at the desired position is not possible for an alkane without giving side products. But similar to the carbocations, a radical is stable next to an allylic or benzylic position, and would selectively give the monohalogenated major product.
Halogenation reactions are hard to control at the desired monosubstituted stage and several times gives polysubstituted products making isolation of the compound a tedious process. It is, therefore, not a preferred method to halogenate an alkane. One other way is to halogenate of an alkene after which it can be reduced to give a haloalkane. For making a fluoro or an iodoalkane, halogen transfer reactions are preferred.
Note that bromination is more selective than chlorination in deciding that the major product always is of the most stable radical formed. The number of hydrogens (1o vs 2o vs 3o) does not control the major and minor product ratio.
a) Identify the number of possible monobromo structural isomers expected to be formed on free radical monobromination of the following compounds (quiz)