Tuesday, January 29, 2008

Revision - Preparation of alkanes by Wurtz reaction

When an alkyl halide (usually bromide or iodide) is treated with sodium in dry ether, a symmetrical alkane containing both twice the number of carbon atoms of alkyl halide is obtained.

RX + 2Na + XR ---> R-R + 2NaX (catalyst in dry ether)

There is a possibility, in the reaction to use different alkyl halides instead of a single halide. If two different halides are taken with the aim of preparing an alkane with odd number of carbon atoms, a mixture of products is obtained in stead of a single alkane. This is because in this case three reactions takes place and three different products are obtained.

4 comments:

Aryya said...

what is the exact role of ether in the reaction?I saw in a book that !-chloro-3-bromocyclobutane undergoes Wurtz reaction in dry aliphatic ethers to give a bicyclo compound but in dry dioxane it gives 1-chloro-3(3-chlorocyclobutyl)cyclobutane.Why does this happen?

KVSSNrao said...

Please give the book, where you have read these statements. Then if not me, some other visitor will access that book and give you the clarification. As issues become more specific reference to the source becomes important to find out the topic and search for more detail.

Such a search will be useful to deepen the knolwedge. Your comment is appreciated.

Unknown said...

I AM A 11TH student and i am having tremendous difficulty in understanding these diff. rxns..
though i try to understand them but i get confused...........help....it would b rlly appreciated..........!plz

Unknown said...

kindly send d rply on kashishkapoor53@yahoo.com....plz