Q.Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
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Start your 14-day free trial to unlock the full solution →Wurtz reaction couples two alkyl halide molecules; using one alkyl halide gives only even-carbon alkanes, and using two different alkyl halides to reach an odd-carbon count instead produces an inseparable mixture of three coupling products.
Wurtz reaction: 2 R-X + 2 Na --(dry ether)--> R-R + 2 NaX
This reaction works by coupling two alkyl halide molecules through the loss of the two halogen atoms (as NaX), joining the two alkyl fragments (R and R) together.
Why it fails for odd-carbon alkanes:
If a SINGLE alkyl halide (R-X) is used, the product R-R necessarily has TWICE the number of carbons that R has — so the product's carbon count is always an EVEN number (2 x however many carbons are in R). There is no way to combine one R group with itself and get an odd total.
To obtain an alkane with an odd number of carbons, one would need to couple two DIFFERENT alkyl halides, R-X and R'-X, where R and R' have carbon counts that add to an odd total. But when a mixture of two different alkyl halides is treated with sodium, THREE possible coupling products form simultaneously (since both R-X and R'-X molecules are present and can pair randomly):
R-R (symmetrical), R'-R' (symmetrical), and R-R' (the desired unsymmetrical, odd-carbon product)
These three products usually have close boiling points and are difficult/impractical to separate by simple distillation, and the desired unsymmetrical product is obtained mixed with, and often in lower yield than, the two symmetrical by-products. This makes Wurtz reaction an unreliable/impractical method for preparing odd-carbon alkanes in pure form.
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