Why is Wurtz reaction not good for tert. alkyl halide ?

ON WURTZ REACTION  THE PRODUCT FORMED IS =

nOW THERE IS too much steric crowding on the two carbons and thus hinders its +I effect.

And now chemistry says that it is unstable having crowding  on carbon bearing methyl grps.

Plz AAPROVE!!

DUE TO THE PRESENCE OF MORE STERIC HINDRENCE.

Hi Sujeet

Its a simple reaction which is carried by free radical mechanism, you can refer the link http://www.organic-chemistry.org/namedreactions/wurtz-reaction.shtm to understand the mechanism. As its a free radical mechanism 3-degree Carbon will be very stable and hence the reaction will move at a faster rate.

he mechanism of the reaction goes like this :

One electron from the metal is transferred to the halogen to produce a metal halide and an alkyl radical.

R-X + M → R• + M⁺X⁻

The alkyl radical then accepts an electron from another metal atom to form an alkyl anion and the metal becomes cationic. This intermediate has been isolated in a several cases.

R• + M → R⁻M⁺

The nucleophilic carbon of the alkyl anion then displaces the halide in another molecule through a S_N2 reaction, forming a new carbon-carbon covalent bond.

R⁻M⁺ + R-X → R-R + M⁺X⁻

(from wikipedia)

so in the third step, there is SN2 reaction. tertiary alkyl group are very large groups.  the alkyl anion will not be able to attack another molecule of the alkyl halide due to steric hindrance or repulsion between molecules. so tertiary alkyl halides do not give this reaction.