--%>

How haloalkanes are prepared from hydrocarbons?

Alkyl halides can be prepared from alkanes through substitution and from alkenes through addition of halogen acids or through allylic substitution.
    
From alkanes

When alkanes are treated with halogens, chlorine or bromine, in the presence of light or heat, they undergo free radical substitution and a mixture of mono- and poly- substituted products are obtained.

2242_alkyl halide.png 

Although, the substitution beyond monohalogenation may be suppressed by using alkane in excess yet the method is not of much practical use because of the difficulties of separation of such a mixture.

In case of higher alkanes, different isomeric products are formed even when mono-substitution is carried out.

1702_alkyl halide1.png 

In general, the ease of substitution of different types of hydrogen atoms is:

Benzylic, allylic > tertiary > secondary > primary > vinylic, aryl

The iodination of alkanes is reversible and is done by heating with iodine in the presence of oxidising agents like conc. HNO3, HIO4 orHIO3. The function of using such agents is to oxidize HI formed during the reaction to iodine, and hence shift the equilibrium in the forward direction.

264_alkyl halide2.png 

Due to formation of polysubstituted products and isomeric products, this method is not suitable for the laboratory preparation of pure haloalkanes. However, this method can be used for laboratory preparation of certain specific alkyl halides as given below:
    
When all the hydrogen atoms in the alkane are equivalent are equivalent, then it can form only one product on monosubstitution. In such cases this method may be applied.

924_alkyl halide3.png 
    
Allylic and benzylic halides can be prepared from alkenes and arenes respectively by this method because allylic and benzylic hydrogen atoms are substituted much more readily than vinylic and aryl hydrogen atoms.

260_alkyl halide4.png 

In such cases vinylic aryl hydrogens being less reactive do not participate in free radical substitution.

Allylic and benzylic hydrogen atoms are substituted very easily because their substitution proceeds via allylic and benzylic free radicals as intermediates. These intermediates are stabilized by resonance and hence being stable are formed at faster rate.
    
By halide exchange

Iodoalkanes can be obtained by treating bromo or chloroalkanes with a solution of sodium iodine in acetone or methanol. For example,

1778_alkyl halide5.png 

The reaction is known as Finkelstein reaction. This reaction is based on the fact that NaI is soluble in acetone but NaBr and NaCl are not. As a result, equilibrium in the above reaction is very much in favour of forward reaction. The reaction gives best result with primary halides.

Fluoroalkanes are difficult to prepare directly by the action of alkanes with fluorine. It is because fluoride has gor a high reactivity towards the hydrogen. It extracts all the hydrogen atoms from hydrocarbon molecule.

CH4 + 2F2  71_potassium permangnate3.png  4HF + C

However, Fluoroalkanes can be obtained by treating alkyl halides with salts like AgF, Hg2F2, CoF3 or SbF3. This reaction is known as Swarts reaction.

CH3Br + AgF 71_potassium permangnate3.png CH3F + AgBr

2CH3CH2Cl + Hg2F2  71_potassium permangnate3.png  2CH3CH2F + Hg2Cl
2

For replacement of two or three halogen atoms at the same carbon CoF3 or SbF3 is used.

67_alkyl halide6.png

   Related Questions in Chemistry

  • Q : Question based on relative lowering of

    Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre (d) Concentratio

  • Q : Calculation of concentration of the

    Choose the right answer from following. 200ml of a solution contains 5.85 dissolved sodium chloride. The concentration of the solution will be(Na= 23: cl = 35.5 ) (a) 1 molar (b) 2 molar (c) 0.5 molar (d) 0.25 molar

  • Q : Molarity of cane sugar solution 171 g

    171 g of cane sugar (C12H22O11)  is dissolved in one litre of water. Find the molarity of the solution: (i) 2.0 M (ii) 1.0 M (iii) 0.5 M (iv) 0.25 M Choose the right answer from above.

  • Q : Problem on equilibrium constant Ethanol

    Ethanol is manufactured from carbon monoxide and hydrogen at 600 K and 20 bars according to the reaction2 C0(g) + 4 H2(g) ↔ C2H5OH(g) + H2O (g)The feed stream contains 60 mol% H2, 20 m

  • Q : Analytical chemistry 37% weight of HCl

    37% weight of HCl and density is 1.1g/ml. find molarity of HCl

  • Q : Infrared Adsorption The adsorption of

    The adsorption of infrared radiation by diatomic molecules increases the vibrational energy fo molecules and gives information about the force constant for the "spring" of the molecule.;The molecular motion that has the next larger energy level spacing aft

  • Q : Meaning of Molar solution Molar

    Molar solution signifies 1 mole of solute present/existed in: (i) 1000g of solvent (ii) 1 litre of solvent (iii) 1 litre of solution (iv) 1000g of solution

  • Q : Finding Normality Can someone please

    Can someone please help me in getting through this problem. Concentrated H2SO4 has a density of 1.98 gm/ml and is 98% H2SO4 by weight. The normality is: (a) 2 N  (b) 19.8 N  (c) 39.6 N  (d) 98

  • Q : Relative lowering of vapour pressure

    Which of the following solutions will have a lower vapour pressure and why? a) A 5% aqueous solution of cane sugar. b) A 5% aqueous solution of urea.

  • Q : Molal elevation constant of water The

    The boiling point of 0.1 molal aqueous solution of urea is 100.18oC  at 1 atm. The molal elevation constant of water is: (a) 1.8    (b) 0.18   (c) 18    (d) 18.6Answer: (a) Kb