Chemistry · Hydrocarbons · NEET
In the Wurtz reaction two alkyl halide molecules join end to end: 2 R-X + 2 Na give R-R + 2 NaX. You are simply adding one carbon chain to another identical carbon chain. If each piece has n carbons, the product has n + n = 2n carbons. Doubling any whole number always gives an EVEN number. So the clean product can never have an odd carbon count. This is why NCERT says Wurtz is used to prepare higher alkanes with an EVEN number of carbon atoms.
7 is an ODD number, so it cannot be split into two equal identical pieces. To reach 7 carbons you would have to couple two DIFFERENT halides, for example a 3-carbon and a 4-carbon halide (CH3CH2CH2-X + CH3CH2CH2CH2-X). But sodium couples them randomly, so you also get the two symmetrical by-products (6-carbon n-hexane and 8-carbon n-octane) at the same time. The result is a mixture of three alkanes that are hard to separate, so n-heptane is formed in poor yield. This exact idea was asked in NEET 2020.
Using two different halides R-X and R'-X gives THREE alkanes at once: R-R (both same first piece), R'-R' (both same second piece), and R-R' (the cross product you wanted). Because all three form together, the wanted unsymmetrical alkane R-R' is only a fraction of the mixture, and separating them is difficult. That is why Wurtz is good ONLY for symmetrical alkanes made from two identical halides.
Methane has only 1 carbon. The Wurtz reaction always at least DOUBLES the carbon count, so the smallest alkane it can give is ethane (C2H6, from 2 CH3-X). There is no way to get a 1-carbon product by joining two carbon chains. So methane (and any odd-carbon or single-carbon alkane) cannot come out of a clean Wurtz reaction.
Only symmetrical alkanes with an even number of carbons, because they split into two identical halves. Examples: ethane (2 CH3-X), n-butane (2 C2H5-X), n-hexane (2 n-C3H7-X), and 2,3-dimethylbutane (2 isopropyl-X). Each uses the SAME halide twice, so you get one main product in good yield. For NEET, spot the alkane whose chain divides into two equal identical parts.
Which of the following alkane cannot be made in good yield by Wurtz reaction?
Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which, by Wurtz reaction, is converted to a gaseous hydrocarbon containing less than four carbon atoms. (A) is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. It joins two alkyl halides, so if each has n carbons the product has 2n carbons. This is why only even-carbon alkanes come out cleanly and why the smallest possible product is ethane, not methane.
Yes, if the branched alkane is symmetrical. For example 2,3-dimethylbutane forms cleanly from two isopropyl halides. What matters is that the molecule splits into two identical halves, not whether it is straight or branched.
Sodium couples the radicals randomly. With R-X and R'-X you get R-R, R'-R' and R-R' all together. Only R-R' is the unsymmetrical alkane you wanted, so its yield is low and separation is hard.
Wurtz-Fittig couples an alkyl halide with an aryl halide to make an alkylarene (like ethylbenzene). It works because the two partners are chosen to give a wanted cross product, but the plain Wurtz reaction with two different ALKYL halides still suffers the mixture problem.
Check the carbon count. If it is ODD, Wurtz fails (needs two different halides). If it is EVEN, check that it can split into two equal identical pieces. Odd-carbon options are almost always the answer to 'cannot be made by Wurtz'.
