Chemistry Preparation of Haloalkanes and Haloarenes

### Topics Covered :

● Nature of C-X Bond
● Methods of Preparation
● From Alcohols
● From Hydrocarbons
● By Free Radical Halogenation
● By Electrophilic Substitution
● Sandmeyer's Reaction
● From Alkenes
● Halogen Exchange

### Nature of C-X Bond :

=> Since halogen atoms are more electronegative than carbon, the carbon-halogen bond of alkyl halide is polarised; the carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge as shown in fig.

=> 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 color{red}(C—F) to color{red}(C—l).

=> Some typical bond lengths, bond enthalpies and dipole moments are given in Table 10.2.

### Methods of Preparation :

Haloalkanes and haloarenes are prepared as follow :

### From Alcohols :

=> Alkyl halides are best prepared from alcohols, which are easily accessible.

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

● Thionyl chloride is preferred because the other two products are escapable gases.

● Hence the reaction gives pure alkyl halides.

=> 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.

=> The preparation of alkyl chloride is carried out either by passing dry hydrogen chloride gas through a solution of alcohol or by heating a solution of alcohol in concentrated aqueous acid.

color{red}(R - OH + HX overset(ZnCl_2)→ R- X + H_2O)

color{red}(R - OH +NaBr + H_2SO_4 → R - Br + NaHSO_4 +H_2O)

color{red}(3R - OH +PX_3 → 3R - X +H_3PO_3 \ \ \ \ \ (X = Cl , Br))

color{red}(R - OH +PCl_5 → R - Cl +POCl_3+HCl)

color{red}(R- OH underset(red P//X_3) overset(X_a = Br_2 , I_2)→ R- X)

color{red}(R - OH + SOCl_2 → R - Cl +SO_2+HCl)

=> The reactions of primary and secondary alcohols with color{red}(HX) require the presence of a catalyst, color{red}(ZnCl_2).

● With tertiary alcohols, the reaction is conducted by simply shaking with concentrated color{red}(HCl) at room temperature.

● Constant boiling with color{red}(HBr) (48%) is used for preparing alkyl bromide.

● Good yields of color{red}(R—I) may be obtained by heating alcohols with sodium or potassium iodide in 95% phosphoric acid.

● The order of reactivity of alcohols with a given haloacid is 3^° > 2^° > 1^°.

● The above method is not applicable for the preparation of aryl halides because the carbon-oxygen bond in phenols has a partial double bond character and is difficult to break being stronger than a single bond.

### From Hydrocarbons :

From hydrocarbons, these are prepared by the following methods :

### By free radical halogenation :

Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric mono- and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the yield of any one compound is low.

color{red}(CH_3CH_2CH_2CH_3 undersettext(or heat) overset(Cl_2//UV light)→ CH_3CH_2CH_2CH_2Cl + CH_3CH_2CHClCH_3)

Q 3022156031

Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of (CH_3)_2CHCH_2CH_3.

Solution:

In the given molecule, there are four different types of hydrogen atoms. Replacement of these hydrogen atoms will give the following

(CH_3)_2CHCH_2CH_2Cl \ \ \ \ \ \ \ \ \ \ \ \ (CH_3)_2CHCH(Cl)CH_3
(CH_3)_2C(Cl)CH_2CH_3 \ \ \ \ \ \ \ \ \ \ \ \ \ CH_3CH(CH_2Cl)CH_2CH_3

### By electrophilic substitution :

=> Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts like iron or iron(III) chloride.

=> The ortho and para isomers can be easily separated due to large difference in their melting points.

=> Reactions with iodine are reversible in nature and require the presence of an oxidising agent (color{red}(HNO_3, HIO_4)) to oxidise the color{red}(HI) formed during iodination.

=> Fluoro compounds are not prepared by this method due to high reactivity of fluorine.

### Sandmeyer’s reaction

=> When a primary aromatic amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed.

=> Mixing the solution of freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by color{red}(–Cl) or color{red}(–Br). See fig.1.

=> 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. See fig.2.

### From alkenes :

(i) color{green}("Addition of Hydrogen Halides ") : An alkene is converted to corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide or hydrogen iodide. See fig.1

● Propene yields two products, however only one predominates as per Markovnikov’s rule.

color{red}(CH_3CH = CH_2+H- I → undersettext(minor)(CH_3CH_2CH_2I) +undersettext(major)(CH_3CHICH_3))

(ii) color{green}("Addition of Halogens ") : In the laboratory, addition of bromine in color{red}(C Cl_4) to an alkene resulting in discharge of reddish brown colour of bromine constitutes an important method for the detection of double bond in a molecule.

● The addition results in the synthesis of vic-dibromides, which are colourless. See fig.2.
Q 2616478379

Write the products of the following reactions:

Solution:

### Halogen Exchange :

=> Alkyl iodides are often prepared by the reaction of alkyl chlorides/ bromides with color{red}(NaI) in dry acetone.

● This reaction is known as text(Finkelstein reaction).

color{red}(R - X + NaI → R - I + NaX)

color{red}(X = Cl , Br)

● color{red}(NaCl) or color{red}(NaBr) thus formed is precipitated in dry acetone.

● It facilitates the forward reaction according to Le Chatelier’s Principle.

color{green}("Swarts Reaction ") : The synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as color{red}(AgF, Hg_2F_2, CoF_2) or color{red}(SbF_3) is termed as Swarts reaction.

color{red}(H_3C - Br +AgF → H_3 C - F + AgBr)