NCERT Solutions
Class 12 Chemistry
Haloalkanes and Haloarenes
Q.11
How will you bring about the following conversions?
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to1nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane (x) Benzene to biphenyl.
(i) Ethanol to but-1-yne
Since the conversion of ethanol to but-1-yne involves the addition the two extra carbons that is why the conversion is carried out in 3 steps
1. In the first step ethanol is treated with thionyl chloride (SOCl2) in the presence of pyridine to give chloroethane.
2. In the second step the acetylene is treated with sodamide (NaNH2) in the presence of liquid ammonia to give sodium acetylide.
3. The last step involves the reaction between chloroethane and sodium acetylide to give the final product as the But-1-yne and NaCl as the by-product.
(ii) Ethane to bromoethene
The Bromination of ethane (Br2 in the presence of light at 520-670K) gives bromoethane which on further treatment with alcoholic KOH results in the alkene called ethane which again on Bromination with Br2/CCl4 gives
Dibromide(vicinal bromide) called 1,2-dibromoethane which on treatment with alcoholic KOH gives bromoethene as the final product.
(iii) Propene on treatment with HBr in the presence of peroxide (antiMarkownikoff’s rule which states the negative part i.e. Br- goes to the carbon which has more number of hydrogens and positive part i.e. H+ goes to the carbon which has lesser number of hydrogens) gives 1-bromopropane which on treatment with silver nitrite in the presence of alcohol or water gives 1-nitropropane as the final product.
(iv) Chlorination of toluene (Cl2 at 773K) gives benzyl chloride with the
removal of HCl. Benzyl Chloride on further treatment with dilute KOH in the presence of heat gives benzyl alcohol as the final
product.(dilute KOH removes Cl from the benzyl chloride and replaces it by OH)
(v) Propene is treated with Br2/CCl4 i.e. bromination of alkene takes
place to give dibromides, in this case 1,2-dibromopropane which on further treatment with alcoholic KOH gives propyne.(two times removal of KBr)
(vi) Ethanol in treatment with thionyl chloride (SOCl2) in the presence of
pyridine gives ethyl chloride with the evolution of SO2 gas. The ethyl chloride on treatment with mercury fluoride gives ethyl fluoride (the Cl of ethyl chloride is being replaced by F from Hg2F2).
(vii) Bromoethane on treatment with alcoholic KCN forms Acetonitrile with the removal of KBr. CN is added in the case where there is a need to increase the number of carbons. Further the acetonitrile is treated with CH3MgBr (Grignard reagent) in the presence of ether, the intermediate so formed is hydrolysed, and the product formation takes place i.e. Propanone and NH3 is obtained as the by-product.
(viii) CH3-CH2-CH=CH2 +HBr---> CH3-CH2-CH(Br)-CH3
This reaction happens according to Markownikoff’s rule
Now reacting this with alcoholic KOH gives alkenes
CH3-CH2-CH (Br)-CH3 + KOH (alcohol)——-> CH3CH=CHCH3 + KBr + H2O
(ix) The conversion of 1-chlorobutane to n-octane in the presence of Na
metal and dry ether is called Wurtz reaction. Two equivalent chlorobutane reacts to give n-octane as a final product and NaCl as the by-product.
(xx) The conversion of benzene to biphenyl using Na and dry ether is
called FITTIG reaction.
Firstly, the benzene is treated with Br2/FeBr3 results in the formation of bromobenzene. It is an electrophilic substitution. Further the bromobenzene is treated with Na metal in presence of dry ether to form biphenyl as the final product and NaBr as the by-product. If instead of Na metal, bromobenzene is treated with Cu metal then the reaction is called Ullmann’s reaction. The final product obtained in this reaction will also be biphenyl.
