--%>

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 : Explain the molecular mass with respect

    During the formation of polymers, different macromolecules have different degree of polymerisation i.e. they have varied chain lengths. Thus, the molecular masses of the individual macromolecules in a particular sample of the polymer are different. Hence, an average value of the molecular mass is

  • Q : Problem based on molarity Select the

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Laws of Chemical Combination Laws of

    Laws of Chemical Combination- In order to understand the composition of the compounds, it is necessary to have a theory which accounts for both qualitative and quantitative observations during chem

  • Q : Molarity of acid solution If 20ml of

    If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is: (a) 0.1M (b) 0.2M  (c) 0.3M (d) 0.4M Choose the right answer fron above.

  • Q : Calculating amount of Sodium hydroxide

    Choose the right answer from following. The amount of NaOH in gms in 250cm3 of a0.100M NaOH solution would be : (a) 4 gm (b) 2 gm (c) 1 gm (d) 2.5 gm

  • Q : Problem based on lowering in vapour

    Help me to solve this problem. An aqueous solution of glucose was prepared by dissolving 18 g of glucose in 90 g of water. The relative lowering in vapour pressure is: (a) 0.02 (b)1 (c) 20 (d)180

  • 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 decinormal Select the right

    Select the right answer of the question. How much water is required to dilute 10 ml of 10 N hydrochloric acid to make it exactly decinormal (0.1 N): (a) 990 ml (b) 1000 ml (c) 1010 ml (d) 100 ml

  • Q : Molality of glucose Help me to go

    Help me to go through this problem. Molecular weight of glucose is 180. A solution of glucose which contains 18 gms per litre is : (a) 2 molal (b) 1 molal (c) 0.1 molal (d)18 molal

  • Q : Problem on mole fraction of glucose

    Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998