Halogenalkanes

Structure of halogenalkanes

In halogenalkanes, the carbon-halogen (C-X) bond is formed by sp³-a hybridized carbon orbital and the p-orbital of the halogen. The larger the size of the halogen, the larger the size of the p-orbital, therefore, the larger the halogen, the longer the bond length, the overlap of the orbitals is smaller and the bond, respectively, is weaker.

Due to the fact that the electronegativity of halogens is greater than the electronegativity of carbon, the C-X bond is polar. The electron cloud is shifted towards the halogen, thereby giving the halogen a large negative charge, the carbon atom becomes less electronegative. As a result, the carbon atom becomes more susceptible to nucleophiles, i.e. to Lewis bases, halogen atoms - to Lewis acids.

Properties of halogenalkanes

The boiling point of haloalkanes is higher than the boiling point of the corresponding alkanes. Due to the low polarity, haloalkanes dissolve in nonpolar solutions and are insoluble in water.

Methyl chloride, methyl bromide, ethyl chloride - gases, methyl iodide and most other halogen derivatives are liquids with a sweet smell.

Reactions of halogenalkanes

Aliphatic nucleophilic substitution

As already mentioned, due to the features of the structure of haloalkane, carbon in the C-X bond is depleted of electrons, so the carbon atom is attacked by electron-rich substances called nucleophiles. The nucleophiles push out the halogen, which takes the electron pair with it. The nucleophile can be an anion, or a neutral molecule with an undivided electron pair. In the first case, the reaction product will be neutral, in the second case it will have a positive charge.

Reaction of nucleophile anion and haloalkane
Nu:^- + R^δ+-X^δ- → [R-Nu] + X^-
Reaction of a nucleophile with an undivided electron pair and a haloalkane
Nu: + R^δ+-X^δ- → [R-Nu]⁺ + X^-

The table below shows the reactions of aliphatic nucleophilic substitution of haloalkanes:

Bimolecular nucleophilic substitution

In the reaction of bimolecular nucleophilic substitution in haloalkanes, the nucleophile approaches the carbon atom on the reverse side of the C-X bond, forming a bond with carbon and thereby "pushing out" the halogen. Most of the energy, necessary for breaking the carbon-halogen bond, it comes from the released energy of the formation of a carbon-nucleophile bond. The reaction mechanism is as follows: there is a transitional stage when the nucleophile and the halogen are partially bound to the carbon atom, this stage lasts 10^-12 s - one period of molecular oscillation. Thus, the reaction occurs in one stage without forming intermediates.

Monomolecular nucleophilic substitution

The reaction of monomolecular nucleophilic substitution occurs in three stages.

The first stage is the heterolytic break of the R-X bond, with the formation of the carbation R⁺ and the halogen anion X^-. Such a rupture, as a rule, is an endothermic reaction and requires a large amount of energy, so it occurs slowly.

The second stage is the nucleophilic attack of the carbation R⁺. In the case of the nucleophile-anion reaction Nu^-, carbcation forms a bond with the R-Nu anion. If the reaction occurs with a neutral substance, Nu, then in the formed bond the nucleophile will have a positive charge of R-Nu⁺.

At the last stage, a cation is split off from the molecule.

Thus, for example, it is possible to obtain alcohol from haloalkane by hydrolysis of tert-butyl chloride:

1. Ionization of tert-butyl chloride
(CH₃)₃CCl → (CH₃)₃C⁺ + Cl^-
2. Water connection
(CH₃)₃C⁺ + H₂O → (CH₃)₃COH₂⁺
3. Proton cleavage
(CH₃)₃COH₂⁺ → (CH₃)₃COH + H⁺

Oxidation of halogenalkanes

Haloalkanes are split into carbon dioxide, water and hydrogen halides during oxidation, for example:

C₂H₅Cl + 3O₂ → 2CO₂ + 2H₂O + HCl

Elimination reactions

The elimination reaction in a haloalkane does not occur with carbon with a halogen, but with a neighboring atom, taking away one atom of hydrogen. As a result, the carbon atom becomes more charged, causing the halogen to split off from the neighboring atom and forming a double bond - alkene.

For asymmetric halogenalkanes, the Zaitsev rule works: hydrogen is split off to a large extent from the named hydrogenated atom. The reaction proceeds both by monomolecular elimination and by bimolecular elimination. elimination.

Obtaining halogenalkanes

From alcohols

To obtain halogenalkanes from alcohols, by replacing the OH group, the following reagents are used: HBr, HI, HCl + ZnCl₂, PBr₃, PI₃, PCl₅ and SOCl₂. Mainly, thionyl chloride (SOCl₂) is used, since by-products, in this case, there will be gases that are easily removed from the mixture.

3CH₃CH₂OH + PBr₃ → 3CH₃CH₂Br + H₃PO₃

From alkanes

The reaction mechanism is described in the section alkane halogenation. As a result of the reaction several products are formed, so this method is used quite rarely.

High temperature and ultraviolet:
CH₃-CH₂-CH₃ + Cl₂ → CH₃-CH₂-CH₂Cl (45%) + CH₃-CHCl-CH₃ (55%)

Halogen exchange

The exchange of halogens in halogenalkanes is used for iodine derivatives, an example of a reaction:

In acetone:
R-Cl + Na⁺I^- → R-I + Na-Cl

This reaction is very convenient, since sodium chloride is insoluble in acetone and precipitates.

From alkenes

The reaction mechanism is described in the article alkene halogenation, example reactions:

In the dark, in the presence of carbon tetrachloride:
C₂H₄ + Br₂ → C₂H₄Br₂

Qualitative reaction to haloalkanes

Halogens, nitrogen and sulfur are attached by a covalent bond to organic elements, therefore, to determine their presence it is necessary to convert the compounds to ionic. This can be achieved by gorenje organic compounds with metallic sodium. When an organic compound is heated with sodium, any halogens, nitrogen and sulfur are converted into inorganic sodium salts, such as sodium halide (for halides), sodium cyanide (for nitrogen), sodium sulfide (for sulfur) and sodium thiocyanate (for sulfur and nitrogen). Then, the amount of salt can be checked in an aqueous solution.