Chemistry · Alcohols, Phenols And Ethers · NEET
It is a way to make ethers. You take a sodium alkoxide, written R-ONa (or the aryl version, an aryloxide like phenoxide), and mix it with an alkyl halide R'-X. The product is the ether R-O-R' plus the salt NaX. Full equation: R-ONa + R'-X gives R-O-R' + NaX. It works because the alkoxide oxygen is a good nucleophile and the halide (Cl, Br, I) is a good leaving group.
It is SN2 (bimolecular nucleophilic substitution). The negatively charged alkoxide oxygen attacks the carbon that holds the halogen from the back side. In one single step, the new O-C bond forms while the C-X bond breaks and X leaves. Because it is SN2, the reaction is fastest when the carbon is not crowded, i.e. when the alkyl halide is primary (methyl or primary carbon).
This is the main limitation NCERT stresses. Alkoxides are strong bases as well as nucleophiles. When the alkyl halide is secondary or tertiary, the carbon is crowded, so the alkoxide cannot easily reach it for SN2. Instead the alkoxide acts as a base and pulls off a beta-hydrogen. This is elimination (E2), which gives an alkene, not an ether. With a tertiary alkyl halide the alkene is the ONLY product and no ether forms.
Always put the tertiary (or bulky) group as the ALKOXIDE and the primary/methyl group as the HALIDE. For tert-butyl methyl ether, use sodium tert-butoxide (CH3)3C-ONa plus methyl iodide CH3-I. Do NOT do the reverse: CH3ONa plus (CH3)3C-Br gives only 2-methylpropene (an alkene) by elimination and no ether at all.
No. Williamson synthesis needs the halide carbon to undergo SN2, but aryl halides (halogen on a benzene ring) do not react by SN2. To make anisole (methyl phenyl ether), use sodium phenoxide C6H5-ONa (the aryl part becomes the alkoxide) plus methyl iodide CH3-I. The aromatic ring must come in as the aryloxide, never as the halide.
Every alkyl halide with a beta-hydrogen has two possible reactions: substitution (SN2, makes the ether) and elimination (E2, makes the alkene). Which one wins depends on how crowded the carbon is and how strong and bulky the base is. Alkoxides are strong bulky bases, so as the halide gets more crowded (secondary, then tertiary), the base can no longer do SN2 and switches to elimination.
The reaction R-ONa + R'-X -> R-O-R' + NaX can be classified as:
The compound A on treatment with Na gives B, and with PCl5 gives C. B and C react together to give diethyl ether. A, B and C are, in order,
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. It can make symmetrical ethers (both R groups same, like diethyl ether) and unsymmetrical ethers (different R groups, like ethyl methyl ether). For unsymmetrical ethers you must choose the alkoxide and halide carefully so the halide stays primary.
Since it is SN2, the order of reactivity is primary greater than secondary greater than tertiary. Primary (and methyl) halides give the best yield of ether. Tertiary halides give only alkene.
An alcohol R-OH is only weakly acidic, so you need a strong base to remove its O-H hydrogen and make the alkoxide R-ONa. Sodium metal or sodium hydride (NaH) does this: 2R-OH + 2Na -> 2R-ONa + H2. The alkoxide is the actual nucleophile that attacks the halide.
They can give some ether, but elimination competes strongly and the yield is poor. NCERT says 'better results are obtained if the alkyl halide is primary.' For NEET, treat secondary as risky and tertiary as giving only alkene.
Dehydration of alcohol (with sulphuric acid) only works well for making simple symmetrical ethers from primary alcohols. Williamson synthesis is more general and is the main method for unsymmetrical ethers, because you pick the two pieces separately.
