--%>
Assignment Help - homework help

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 + Hg2Cl2

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 : Vapour pressure over mercury Choose the

    Choose the right answer from following. At 300 K, when a solute is added to a solvent its vapour pressure over the mercury reduces from 50 mm to 45 mm. The value of mole fraction of solute will be: (a)0.005 (b)0.010 (c)0.100 (d)0.900

    		•
  • Q : Which is polar HCl or HF Which one is

    Which one is polar HCl or HF?

    		•
  • Q : Concentration of urea Help me to go

    Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<

    		•
  • Q : Adiabatic compression A lean natural

    A lean natural gas is available at 18oC and 65 bars and must be compressed for economical pipeline transportation. The gas is first adiabatically compressed to 200 bars and then isobarically (i.e. at constant pressure) cooled to 25°C. The gas, which is

    		•
  • Q : What is electrolysis? Explain with

    Passage of a current through a solution can produce an electrolysis reaction.Much additional information on the properties of the ions in an aqueous solution can be obtained from studies of the passage of a direct current (dc) through a cell containing a s

    		•
  • Q : Molality of Sulfuric acid Choose the

    Choose the right answer from following. The molality of 90% H2SO4 solution is: [density=1.8 gm/ml]  (a)1.8 (b) 48.4 (c) 9.18 (d) 94.6

    		•
  • Q : Concentration of Sodium chloride

    Provide solution of this question. If 25 ml of 0.25 M NaCl solution is diluted with water to a volume of 500ml the new concentration of the solution is : (a) 0.167 M (b) 0.0125 M (c) 0.833 M (d) 0.0167 M

    		•
  • Q : Mole fraction of solute The mole

    The mole fraction of the solute in 1 molal aqueous solution is: (a) 0.027 (b) 0.036 (c) 0.018 (d) 0.009What is the correct answer.

    		•
  • Q : Strength of Nacl in solution To 5.85gm

    To 5.85gm of Nacl one kg of water is added to prepare of solution. What is the strength of Nacl in this solution (mol. wt. of nacl = 58.5)? (a) 0.1 Normal (b) 0.1 Molal (c) 0.1 Molar (d) 0.1 FormalAnswer:

    		•
  • Q : Problem related to molality Help me to

    Help me to solve this problem. What is the molality of a solution which contains 18 g of glucose (C6,H12, O6) in 250 g of water:  (a) 4.0 m (b) 0.4 m (c) 4.2 m (d) 0.8 m

    		•