Alkyl Halides: Structure, Uses, and Synthesis

Alkyl Halides

Structure and Classification of Alkyl Halides

Alkyl halides are organic compounds in which a halogen atom (X) is bonded to an sp3-hybridized carbon atom. The nature of the carbon to which the halogen is attached determines the classification of the alkyl halide.

• Primary (1°) Alkyl Halide: The halogen is attached to a carbon bonded to one other carbon.

• Secondary (2°) Alkyl Halide: The halogen is attached to a carbon bonded to two other carbons.

• Tertiary (3°) Alkyl Halide: The halogen is attached to a carbon bonded to three other carbons.

• Common halogens: F, Cl, Br, I

The C–X bond is polar due to differences in electronegativity between carbon and the halogen.

• Example: Chloromethane (CH3Cl) is a simple alkyl halide.

Additional info: The electron density map (EPM) of chloromethane shows electron withdrawal by the halogen, making the carbon partially positive and the halogen partially negative.

Uses of Alkyl Halides

Solvent Properties and Reactivity

Some alkyl halides are used as aprotic, non-acidic solvents in organic chemistry. Their ability to dissolve a wide range of compounds makes them useful in laboratory and industrial settings.

• Common solvents:

  • Carbon tetrachloride (CCl4)

  • Chloroform (CHCl3)

  • Methylene chloride (CH2Cl2, also called dichloromethane or DCM)

Alkyl halides are also excellent electrophiles in substitution and elimination reactions:

• Substitution: A nucleophile (Nu-) replaces the halogen atom.

• Elimination: A base removes a proton, resulting in the formation of an alkene.

General equations:

• Substitution: $\ce{R-X + Nu^- \rightarrow R-Nu + X^-}$

• Elimination: $\ce{R-CH_2-CH_2-X + Base \rightarrow R-CH=CH_2 + X^- + HB^+}$

Synthesis of Alkyl Bromides

Free Radical Bromination

Alkyl bromides can be synthesized using free radical bromination of alkanes. This reaction is promoted by light (hv) and involves the addition of bromine (Br2) to a hydrocarbon.

• The bromine atom is preferentially added to the most substituted sp3 carbon (tertiary > secondary > primary).

• This selectivity is due to the stability of the intermediate free radical.

• Chlorination is less selective and thus less useful for complex molecules.

Example:

• Reaction of 2-methylpropane with Br2 and light yields the major product with Br on the tertiary carbon.

General equation:

• $\ce{R-H + Br_2 \xrightarrow{hv} R-Br + HBr}$

Additional info: In more complex molecules, bromination is highly regioselective, favoring the most substituted carbon.

Synthesis of Alkyl Halides: Benzylic and Allylic Bromination

Special Reactivity at Benzylic and Allylic Positions

Bromine reacts preferentially at the benzylic (adjacent to benzene) and allylic (adjacent to an alkene) positions. This is due to the stability of the benzylic and allylic radicals formed during the reaction.

• Benzylic bromination: Bromine adds to the carbon next to an aromatic ring.

• Allylic bromination: Bromine adds to the carbon next to a double bond.

• NBS (N-bromosuccinimide): Used as a brominating agent for allylic and benzylic positions, as it provides a controlled source of Br2.

Example:

• Allylic bromination of cyclohexene with NBS and light yields the allylic bromide.

General equation:

• $\ce{R-CH_2-CH=CH_2 + NBS \xrightarrow{hv} R-CH(Br)-CH=CH_2}$

Additional info: These reactions are covered in more detail in chapters on alkenes and aromatic compounds.

Practice Problem: Monobromination Product

Predicting the Major Product

When predicting the major product of a monobromination reaction, bromine is added to the most substituted carbon atom due to the stability of the resulting radical intermediate.

• Example: Bromination of methylcyclopentane with Br2 and light yields the major product with Br on the tertiary carbon.

General approach:

• Identify the most substituted carbon atom.

• Replace the hydrogen on that carbon with a bromine atom.

Additional info: This principle applies to most free radical halogenation reactions.