Bulky Bases in Elimination Reactions
Last updated: November 5th, 2020 |
Elimination Reactions Using “Bulky Bases” – When The Zaitsev Product Is Minor
We’ve recently talked about Zaitsev’s rule in elimination reactions, and how the transition state leading to the more substituted alkene is lower in energy. This post covers reactions involving “bulky bases” where less of the Zaitsev product is obtained.
Table of Contents
- “Normal” E2 Reactions Follow Zaitsev’s Rule, Giving The “More Substituted” Alkene
- “Bulky Bases” Tend To Give A Higher Proportion Of “Non-Zaitsev” Products
- Bulky Bases Give More “Non-Zaitsev” Products Due To Steric Interactions With The Alkyl Halide
- Two Common Bulky Bases Are The t-Butoxide Ion And Lithium Di-Isopropyl Amide (LDA)
- (Advanced) References and Further Reading
Most elimination reactions follow Zaitsev’s rule : you should expect that the “more substituted” alkene will be formed if at all possible. Like in the elimination reaction below, for instance, we get 80% of the tetrasubstituted alkene [“Zaitsev” – more substituted because there are 4 carbons attached to the alkene] and 20% of the disubstituted “non-Zaitsev” product.
However, today we’ll talk about one interesting exception to this “rule” and how under certain conditions we actually end up with the “non-Zaitsev” alkene product instead.
For instance, instead of using sodium methoxide, (NaOCH3) if you use the base NaOC(CH3)3 [or KOC(CH3)3, changing sodium for potassium doesn’t really matter here] you end up with an interesting reversal of products!
So what’s going on here? Why might we get less of the Zaitsev product here and more of the “non-Zaitsev” product?
Well, if we draw out what the structure of the reactants might look like in their transition state, we can start to see why. [Note: this is not technically a transition state since we’re not drawing partial bonds, but you can at least see how the reactants are assembled]. The base in this instance – potassium t-butoxide – is an extremely bulky base, and the proton we remove to form the Zaitsev product is on a tertiary carbon. As the oxygen from the base draws nearer to this proton, a steric clash occurs. In essence the electron clouds around the methyl groups are interacting with each other, and the repulsion between these clouds will raise the energy of the transition state [remember – opposite charges attract, like charges repel]. This will slow down the reaction.
Looking at the reactant assembly that produces the non-Zaitsev product, the bulky base is removing a proton from a primary carbon. Steric clash is considerably reduced in comparison to that for the Zaitsev product. Elimination is faster, and we therefore end up with the less substituted alkene as our major product. Note: for an excellent set of 3D models that show the differences in excellent detail, visit this page on Reusch’s site.
This is one example of a reaction where the more thermodynamically stable product is not formed [recall that alkene stability increases with the number of carbons directly attached to the alkene].
So the bottom line for this post is that when performing an E2 reaction, using a bulky base will produce a greater proportion of non-Zaitsev alkene products relative to a less bulky base.
As far as we’ll see, the most common “bulky base” we need to consider is the t-butoxide ion, which can be drawn in many forms [see diagram]; occasionally you might see lithium di-isopropyl amide (LDA) used as well. For our purposes this completes the roster of bulky bases.
In the next post we’ll talk about an interesting observation we can make during certain E1 reactions.
- For an example where a bulky leaving group can lead to “non-Zaitsev” (aka “Hofmann”) products, see this post on the Hofmann Elimination.
- Mechanism of elimination reactions. Part X. Kinetics of olefin elimination from isopropyl, sec.-butyl, 2-n-amyl, and 3-n-amyl bromides in acidic and alkaline alcoholic media
M. L. Dhar, E. D. Hughes, and C. K. Ingold
J. Chem. Soc. 1948, 2058-2065
Table I in this paper shows that solvolysis of 2-bromobutane with 1 M NaOEt in ethanol gives 82% yield of alkene at 25 °C, but similar solvolysis at 80 °C gives 91.4% yield of alkene.
- Beiträge zur Kenntniss der flüchtigen organischen Basen
Aug. Wilk. von Hofmann
Just. Lieb. Ann. Chem. 1851, 78 (3), 253-286
Early paper on Hofmann eliminations by its discoverer. Eliminations of quaternary ammonium salts favor loss of ethylene over larger groups.
- —The nature of the alternating effect in carbon chains. Part XVIII. Mechanism of exhaustive methylation and its relation to anomalous hydrolysis
Walther Hanhart and Christopher Kelk Ingold
J. Chem. Soc. 1927, 997-1020
Prof. Ingold mentions in this paper, “It follows from the basic hypothesis that the ease of removal of the b-proton (reaction A) depends (a) on its vulnerability, (b) on the proton-avidity of the attacking anion”
- Steric Effects in Elimination Reactions. VII. The Effect of the Steric Requirements of Alkoxide Bases on the Direction of Bimolecular Elimination
Herbert C. Brown, Ichiro Moritani, and Y. Okamoto
Journal of the American Chemical Society 1956, 78 (10), 2193-2197
As Nobel Laureate Prof. H. C. Brown (Purdue) states in this paper, “[…] with increasing basic strength of the alkoxide bases (C2H5O– < (CH3)3– < CO–) there is observed not a decrease, but rather an increase in the selectivity of the reagent. It may be concluded therefore that the increase in base strength does not play any major role in altering the isomer distribution in the present reaction.”
Check out Table IV. For 2-bromobutane, 1.0 M KOtBu gives a 53:47 ratio of 1-butene to 2-butene. For 2-bromopentene, it’s 66:34 for 1-pentene vs 2-pentene. The ratio of “anti-Zaitsev” alkenes gets higher when the bromide is tertiary; for 2-bromo-2-methylbutane and 1.0 M KOtBu the ratio is 72:28 for the “anti-Zaitsev”. [30:70 when the base is KOEt].
If you’re really interested see this chart (thanks, Ben!) Olefin Product Distribution
- Steric Effects in Elimination Reactions. IX. The Effect of the Steric Requirements of the Leaving Group on the Direction of Bimolecular Elimination in 2-Pentyl Derivatives
Herbert C. Brown and Owen H. Wheeler
Journal of the American Chemical Society 1956, 78 (10), 2199-2202
In this paper, Prof. Brown shows that, all things being equal, bulkier leaving groups also lead to formation of less substituted olefins. 2-bromopentane gives 31% yield of 1-pentene upon solvolysis with KOEt, but 2-(trimethylammonium)-pentane gives 98% of yield of 1-pentene.
- Steric Effects in Elimination Reactions. X. Steric Strains as a Factor in Controlling the Direction of Bimolecular Eliminations. The Hofmann Rule as a Manifestation of Steric Strain
Herbert C. Brown and I. Moritani
Journal of the American Chemical Society 1956, 78 (10), 2203-2210
This is Prof. Brown’s paper wrapping up the topic of steric effects in E2 eliminations. Prof. Brown used various pyridine bases varying in steric hindrance around the nitrogen (pyridine, 2-methylpyridine (2-picoline), and 2,6-dimethylpyridine (2,6-lutidine). Increasing the steric bulk of the base does increase the yield of the less substituted olefin.
- Mechanism of elimination reactions. Part XIX. Kinetics and steric course of elimination from isomeric menthyl chlorides
E. D. Hughes, C. K. Ingold, and J. B. Rose
J. Chem. Soc. 1953, 3839-3845
This is an example of the Zaitsev rule in a cyclohexane system. Neomenthyl chloride gives 78% 3-menthene and 22% 2-menthene with EtO– in ethanol.
- Eliminations in Cyclic cis‐trans‐Isomers
Dr. W. Hückel and Priv.‐Doz. Dr. M. Hanack
Angew. Chem. Int. Ed. 1967, 6 (6), 534-544
Study on E1 and E2 eliminations in cyclic systems. E2 eliminations will give the “Hofmann” product if anti-arrangement of H and leaving group is not possible, whereas E1 will give Zaitsev regardless.