SN2 substitution and E2 eliminations are often competing reactions.
Upon treatment with basic nucleophiles, alkyl (pseudo)halides that do not undergo SN2 substitutions generally give either only E2 elimination products
when having beta-hydrogens (example: tert-alkyl halides) or SN1 substitution products via the reaction with solvent or the base when having no
beta-hydrogens (example: trityl halides). Other electrophiles which can not undergo SN2 substitions efficiently and are also unable to undergo E2
eliminations (example: neopentyl halides) commonly give rearrangement products.
Secondary alkyl (pseudo)halides are poor electrophiles in SN2 substitutions, but easily undergo E2 elimination, hence these are particularly prone to
the competition among the two types of the reaction. Whether there will be more SN2 or more E2 depends on the nucleophile, its basicity and steric
factors, solvent polarity and nature, and the temperature. But even primary alkyl (pseudo)halides, which are otherwise better electrophiles in SN2,
can undergo E2 elimination as the main route when the nucleophilicity to basicity ratio is poor in the nucleophile used (for example if using
t-butoxide as the nucleophile/base), or if this ratio is reduced by the solvent or reaction conditions.
Quote: Originally posted by cnidocyte  | | I read that size of the alkoxide and availability of its alpha carbons hydrogen are the main factors determining whether the elimination or
substitution reaction will occur. |
Well, yes though it is about sterics, not about hydrogen atoms per se. Hydrogen atoms are small, so the more alpha-hydrogens there are on the alkoxide
the less stericaly crowded this is. The less stericaly crowded, the more nucleophilic it is. The more nucleophilic it is, the higher its
nucleophilicity/basicity ratio is. The higher this ratio is, the better the SN2 substitution vs. E2 elimination ratio you get. Also, the more
alpha-alkyl groups (thus less alpha-hydrogens) the more basic the alkoxide is.
| Quote: Originally posted by cnidocyte | | Does that mean reacting a simple alkoxide like sodium ethoxide with a simple haloalkane like chloroethane would produce mainly diethyl ether rather
than ethylene? |
Yes, in ethanol as solvent, the main product would be diethyl ether, because ethyl chloride is a primary alkyl halide and ethoxide is a good
nucleophile even though quite basic. In aprotic solvents, like THF, where the basicity of the ethoxide increases dramatically (no solvation) while
nucleophilicity does not follow, there would be the E2 elimination reaction competing.
UTFSE for more information and please next time start beginer's and/or referenceless threads only in the Beginnings section.
[Edited on 25/7/2010 by Nicodem]
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