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General Methods of Preparation of Alkanes


Table of Content


Decarboxylation refers to the process of removal of CO2 from the molecules having -COOH group. Saturated monocarboxylic acid salt of sodium potassium on dry distillation with soda lime gives alkane.

 RCOONa\xrightarrow[-CO^{_{2}}]{Soda Lime} R-H              


RCOONa +NaOH\overset{CaO}{\rightarrow} R-H + Na_{2}CO_{3}

The alkane formed by decarboxylation process always contains one carbon atom less than the original acid.The yield is good in case of lower members but poor for higher members.

Soda lime is prepared by soaking quick lime CaO in caustic soda solution and then drying the products. It is generally written as NaOH + CaO. Its reaction is milder than caustic soda. Otherwise the reaction will occur violently. Also CaO used alongwith NaOH keeps it dry (NaOH is hygroscopic) to aid fusion.

The decarboxylation of sodium formate yields H2.

HCOONa + NaOH cao H2 + Na2CO3

CH3COONa + NaOH cao CH4 + Na2CO3                        

Wurtz Reaction

In wourtz reaction  a solution of alkyl halide in ether on heating with sodium gives alkane.

R-X + 2Na + X-R \overset{Dry Ether}{\rightarrow} R-R + 2NaX

An alkyl halide on Wurtz reaction leads to the formation of symmetrical alkane having an even number of carbon atoms. Two different alkyl halides, on Wurtz reaction give all possible alkanes.

CH3X + Na + C2H5X → CH3CH2CH3 + CH3CH3 + CH3CH2CH2CH3

The different involved are steps are:

CH3X + 2Na + C2H5X → CH3CH2CH3 + 2NaX

CH3X + 2Na + C2H5X →CH3CH3 + 2NaX

C2H5X + 2Na + C2H5X → C2H5C2H5 + 2NaX

The separation of mixture into individual members is not easy because their boiling points are near to each other and thus Wurtz reaction is not suitable for the synthesis of alkanes containing odd number of carbon atoms.

When  Zn is used in Wurtz reaction  in place of Na, the reaction is named as Frankland method.

Limitations of Wurtz reaction :

a.Methane can not be obtained by this method

b.The reaction fails in case of tertiary halides

Mechanism of Wurtz reaction : 

The mechanism of Wurtz reaction is although not clear however two mechanisms are proposed for this reaction.

The first proposed mechanism of Wurtz reaction involved  formation of an  Intermediate organometallic compound:

 RX + 2Na →   [RNa]  + NaX


RX + [RNa] → R-R + NaX

Another proposed mechanizm of Wurtz reaction involved  formation of Intermediate  free radicals:
 RX + Na → R. + Nax

                 Free radicals

 R. + R. → R-R

By the Reduction of Alkyl Halides

Alkyl halides on reduction with nascent hydrogen form alkanes.  R-X  + 2[H]→R-H + HX. The nascent hydrogen may be obtained by any one of the following

  1. Zn + HCI                              

  2. Zn + CH3COOH

  3. Zn-Cu couple in ethanol

  4. Red P + HI

  5. Al-Hg + ethanol

Alkyl halides can also be reduced catalytically to alkane by H2/Pd or LiAIH4 or by H2/Ni. The yields are generally high and the hydrocarbons formed are pure.

Note : Zn-Cu couple is prepared by adding Zn granules in aqueous CuSO4 solution where copper is deposited on the Zn pieces.

By Hydrogenation of Alkenes((>C=C<) : Sabatier and Senderen's Method 

Alkenes and alkynes on catatlytic hydrogenation give alkanes

CH2 = CH2 + H2 catalyst-ni CH3-CH3

CH≡CH + 2H2 catalyst-ni CH3-CH3

Catalyst Ni is used in finely divided form. If Pt or Pd are used as catalyst, reaction occurs at normal temperature. Also some times Raney nickel is used as catalyst. It is obtained by boiling Ni-AI alloy with NaOH, when AI dissolves leaving Ni in finely divided state. The filtered, washed and died Ni is known as Raney Nickel. Raney Ni is effective at room temperature and atmospheric pressure.

Kolbe's Electrolysis Method

Alkanes are formed, on electrolysis of concentrated aqueous solution of sodium or potassium salt of saturated mono carboxylic acids


Electrolysis of an acid salt gives symmetrical alkane. However, in case of mixture of carboxylic acid salts, all probable alkanes are formed.

R1COOK + R2COOK electrolysis  R1-R2 + 2CO2 + H2 + 2NaOH

                                                      (R1-R1 and R2-R2 are also formed).

By Grignard Reagents

Organic compounds in which a metal atom is directly linked to carbon atom are known as organometallic compound. e.g. HC≡CNa, (C2H5)4 Pb, (C2H5)2 Zn

Alkyl or aryl magnesium halide (R-MgX) are also called Grignard reagents or organometallic compounds.Grignard reagent on double decomposition with water or with other compounds having active H(the hydrogen attached on O, N, F or triple bonded carbon atom are known as active hydrogen) give alkane.





By Reduction of Alcohols, Aldehydes, Ketones or Fatty Acids and their Derivatives

The reduction of either of the above in presence of red P & HI gives corresponding alkane.














By Reduction of Carbonyl Compounds

The reduction of carbonyl compounds by amalgamated zinc and conc. HCI also yields alkanes. This is Clemmensen reduction.

CH3CHO + 2H2  chemical-3 CH3CH3 + H2O

CH3COOH + 2H2 chemical-3 CH3CH2CH3 + H2O

Carbonyl compounds may also be reduced to alkanes by Wolf Kishner reaction


By the Hydrolysis of AI or Be Carbides

Only CH4 can be obtained by the hydrolysis of Be or Al carbides.

AI4C3 + 12H2delta 4AI(OH)3 + 3CH4

Be2C + 4H2delta  2Be(OH)2 + CH4

Note :

1.   Calcium carbide reacts with water to give acetylene.

2.   Magnesium carbide, Mg2C2 reacts with water to give propyne.

3.   CH4 can be obtained by passing a mixture of H2S and CS2 through red and Cu tube

CS2 + 2H2chemical-cu  CH4 + 4Cu2S

By Hydroboration of Alkenes

Alkenes on hydroboration give trialkyl borane as a result of addition of diborane on olefinic bond. This trialkyl borane on treatment with acetic acid or propanoic acid yields alkane.

2R - CH=CH2 chemical-1 2(RCH2CH2)3B chemical-2 2RCH2CH3

By Corey- House Synthesis

Alkyl chloride say chloroethane reacts with lithium in presence of ether to give lithium alkyl then reacts with CuI to give lithium dialkyl cuprate. This lithium dialkyl cuprate now again reacts with alkyl chloride to given alkane.

CH3CH2CI + 2Li ether  CH3CH2Li + LiCl

2CH3CH2Li + CuI → Li(CH3CH2)2 Cu + LiL

Li(CH3CH2)2Cu + CH3CH2CI → CH3CH2CH2CH3 + CH3CH2Cu + LiCl

Refer the below mentioned links to get an immediate solution to all queries on Organic chemistry:

JEE Organic Chemistry Syllabus

Organic Chemistry Revision Notes

Reference books of Organic Chemistry

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