Alkyl halide (haloalkane) and aryl halide (haloarene)

The replacement of hydrogen atoms(s) in a hydrocarbon, aliphatic or aromatic, by halogen atom(s) results in the formation of alkyl halide (haloalkane) and aryl halide (haloarene), respectively.

Aliphatic hydrocarbon

Butane
Cyclobutane
Ethene

An aliphatic compound is a hydrocarbon compound containing carbon and hydrogen joined together in straight chains, branched trains or non-aromatic rings. Aliphatic compounds may be saturated (e.g., hexane and other alkanes) or unsaturated (e.g., hexene and other alkene, as well as alkynes).

Aromatic hydrocarbon

Aromatic hydrocarbons, also called aromatic compounds, are compounds that contain benzene as a part of their structure. Benzene is a cyclic hydrocarbon with the formula C6H6.

There are three major classes of halogenated organic compounds: the alkyl halides, the vinyl halides, and the aryl halides. An alkyl halide simply has a halogen atom bonded to one of the sp3 hybrid carbon atoms of an alkyl group. A vinyl halide has a halogen atom bonded to one of the sp2 hybrid carbon atoms of an alkene. An aryl halide has a halogen atom bonded to one of the sp2 hybrid carbon atoms of an aromatic ring. The chemistry of vinyl halides and aryl halides is different from that of alkyl halides because their bonding and hybridization are different.

Polar C-Cl bond

The carbon–halogen bond in an alkyl halide is polar because halogen atoms are more electronegative than carbon atoms. Most reactions of alkyl halides result from breaking this polarized bond. The electrostatic potential map of chloromethane (Figure 6-1) shows higher electron density (red) around the chlorine atom and relatively low electron density (blue) around the carbon and hydrogen atoms. The carbon atom has a partial positive charge, making it somewhat electrophilic. A nucleophile can attack this electrophilic carbon, and the halogen atom can leave as a halide ion, taking the bonding pair of electrons with it.

By serving as a leaving group, the halogen can be eliminated from the alkyl halide, or it can be replaced (substituted for) by a wide variety of functional groups. This versatility allows alkyl halides to serve as intermediates in the synthesis of many other functional groups.

Electrophiles and Nucleophiles

Whenever one compound uses its electrons to attack another compound, we call the attacker a nucleophile, and we call the compound being attacked an electrophile. It is very simple to tell the difference between an electrophile and a nucleophile. You just look at the arrows and see which compound is attacking the other. A nucleophile will always use a region of high electron density (either a lone pair or a bond) to attack the electrophile (which, by definition, has a region of low electron density that can be attacked). These are important terms, so let’s make sure we know how to identify nucleophiles and electrophiles.

On basis of number of halogen atoms

Haloalkanes and haloarenes may be classified as follows: These may be classified as mono, di, or polyhalogen (tri-,tetra-, etc.) compounds depending on whether they contain one, two or more halogen atoms in their structures. For example,

Compunds containing sp3C-X Bond(X=F, Cl, Br, I)

Compounds containing sp2C-X bond
Nomenclature

Which of the following naming is correct? Why?

1-bromo-4-methylhexane or 1-bromo-4-ethylpentane


Nature of C-X bond

Since the size of halogen atom increases as we go down the group in the periodic table, fluorine atom is the smallest and iodine atom, the largest. Consequently the carbon-halogen bond length also increases from C-F to C-I

Methods of Preparation

From Alcohols using halogen acids, phosphorus halides or thionyl chloride.

The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride.

Why thionyl chloride is preferred?

Thionyl chloride is preferred because the other two products are escapable gases. Hence the reaction gives pure alkyl halides.

How do we produce phosphorus tribromide and triiodide?

Phosphorus tribromide and triiodide are usually generated in situ (produced in the reaction mixture) by the reaction of red phosphorus with bromine and iodine respectively.

How can you prepare alkyl chloride using hyrogen chloride gas.?

1) By passing dry hydrogen chloride gas through a solution of alcohol.
2) By heating a solution of alcohol in concenctrated aqueous acid.

Free-Radical Halogenation

Free-radical halogenation is rarely an effective method for the synthesis of alkyl halides. It usually produces mixtures of products because these are differetn kinds of hydrogen atoms that can be abstracted. Also, more than one halogen atom may react, giving multiple substitutions.

For example, the chlorination of propane can give a messy mixture of products.

In industry, free-radical halogenation is sometimes useful because the reagents are cheap, the mixture of products can be separated by distillation, and each of the individual products is sold separately. In a laboratory, however, we need a good yield of one particular product. Free-radical halogenation rarely provides good selectivity and yield, so it is seldom used in the laboratory.

Laboratory syntheses using free-radical halogenation are generally limited to specialized compounds that give a single major product, such as the following examples.

Electrophilic Aromatic Substitution

Sandmeyer's reaction

Copper(I) salts (cuprous salts) have a special affinity for diazonium salts. Cuprous chloride bromide and cuprous cyanide react with arenediazonium salts to give aryl chlorides, aryl bromides and aryl cyanides. The use of cuprous salts to replace arenediazonium groups is called the Sandmeyer reaction. The Sandmeyer reaction (using cuprous cyanide) is also an excellent method of attaching another carbon substituent to an aromatic ring.

OR

Replacement of the diazonium group by iodine does not require the presence of cuprous halide and is done simply by shaking the diazonium salt with potassium iodide.

Markovnikov and anti-Markovnikov

See book

Example 10.4 Ncert

Finkelstein Reaction

Finkelstein reaction is a halogen exchange reaction between haloalkane and a salt of a different halogenide.

$$ \ce{R-CH2-Cl + KI -> R-CH2-I + KCl} $$

Le Chatelier's Principle

The driving force in this reaction is precipitation of KCl. It is important to find a solvent that would dissolve KI better than KCl.

Then solution will always have [I-]> [Cl-] (more iodide anions than cloride anions). As a result this equilibrium will be driven to $$\ce{R-CH2-I +KCl}$$. Acetone is exactly this type of a reagent.

For this case Le Chatelier's principle means that reaction will be driven to the direction in which one of the reagents gets outside of the solution. Since KCl keeps precipitating the reaction will be driven to the disired product.

Questions