What is the major product formed by reaction of sodium tert-butoxide and chloroethane?

I thought NaOt-Bu is a bulky base and the major product will be ethene due to E2. But the answer says t-butyl ethyl ether (due to SN2). Can you please help?

  • 1
    $\begingroup$ I think the answer is wrong. I'm on my phone so I can't give references, but if you search "t-butoxide reaction with primary haloalkanes" you will find references that support this. $\endgroup$
    – Waylander
    Oct 21, 2022 at 12:23
  • 4
    $\begingroup$ Here is a link to what @Waylander mentioned. NaO-t-But is rarely used as a base but KO-t-But is the preferred base. This has nothing to do with their inherent properties but rather their ease of preparation. Sodium (mp 98) does not melt in boiling HO-t-But (bp 82) where as potassium (mp 63) does. Potassium reacts faster than sodium in the solid state with alcohols. Potassium has the added advantage in this instance of being in the liquid state with a larger surface area. $\endgroup$
    – user55119
    Oct 21, 2022 at 17:29
  • $\begingroup$ It seems this is an edge case where steric hindrance of base and alkene stability are opposing factors and perhaps both the products will be have significant yields. It'd be good if the reaction could've been performed for real. $\endgroup$
    – Shub
    Nov 6, 2022 at 7:52

2 Answers 2


According to this book chapter by Robert J. Ouellette and J. David Rawn quoted here, original here

If a primary haloalkane is treated with tert-butoxide ion instead of ethoxide, the amount of elimination product increases significantly. The tert-butoxide ion is not only more basic than the ethoxide ion, it is also much more sterically hindered. The combination of these two factors favors elimination by an E2 process over substitution by an SN2 process.

It goes on to offer the example of 1-bromobutane reacting with t-butoxide to give 10% of the t-butoxybutyl ether (SN2 product) and 90% but-1-ene (E2 product). I see no reason why chloroethane should behave any differently. and

  • $\begingroup$ According to this source, the preparation of an unsymmetrical ether (like ETBE, i.e. the product in this reaction assuming SN2 dominance) claims a primary alkyl halide and a tertiary alkoxide ion are the best reagents. $\endgroup$
    – Sam202
    Nov 5, 2022 at 20:08
  • $\begingroup$ As the data I referred to shows, it is possible to make tBuOR ethers using Williamson. It is just not a very good way of doing it as you get more elimination. All the industrial preparations are done using tBuOH under acid catalysis e.g. sciencedirect.com/science/article/abs/pii/S1381514899000929 $\endgroup$
    – Waylander
    Nov 5, 2022 at 22:00
  • $\begingroup$ That's not necessarily true. If R=methyl, preparation of MTBE with t-butoxide and chloromethane won't produce any E2 product since there's no second carbon atom in the substrate to form a double bond with. In the case we're dealing with here, R=ethyl, and the formation of ethene (assuming E2 domination) implies forming an alkene (ethene) with 0 substituents to stabilize it. Conversely, in the example you cited, 1-bromobutane produces a mono-substituted alkene (1-butene), which is more stable than a non-substituted one. $\endgroup$
    – Sam202
    Nov 5, 2022 at 22:24
  • 1
    $\begingroup$ Your example involves 3 different variables with respect to this problem: (1) Electronegativity (Cl > Br). (2) Etyl-X vs Butyl-X. (3) Alkene product stability (0 substituents vs 1 substituent). I think it's not an analogous example to draw a valid conclusion from. $\endgroup$
    – Sam202
    Nov 5, 2022 at 22:28
  • $\begingroup$ @Sam202 I disagree. In my opinion it is the best example available to us as the actual reaction is not reported in the literature (unless someone knows better). The principle quoted by Ouellette & Rawn accords with my own experience as a practising synthetic chemist. $\endgroup$
    – Waylander
    Nov 5, 2022 at 23:21

Firstly, it's important to identify which species is the nucleophile/base, and which species is the substrate.

In this case, $\ce{(CH3)3CO^-}$ is the nucleophile/base, while $\ce{CH3CH2Cl}$ is the substrate.

This substrate is a primary (1°) alkyl halide because the $\ce{C}$ atom the $\ce{Cl}$ atom is bonded to, is only bonded to one $\ce{C}$ atom.

For $SN2$ reactions, substrate reactivity in terms of number of carbon substituents around the $\ce{C-X}$ bond is:


For $E2$ reactions, substrate reactivity is the opposite:


This is explained by the following:

Methyl groups ($\ce{CH3 -}$) act as electron donating groups, which help stabilize the positive partial charge that is formed in the $\ce{C}$ atom from which an $\ce{H}$ is extracted by a base (in an E2 reaction).

In other words, a double bond that is formed with a higher number of neighboring methyl groups will be more stable than one formed with a lower number.

If you try to perform $E2$ between a 1° alkyl halide (substrate) with a base (bulky or not), the resulting transition state will be very unstable because there are no substituents providing stabilization to the double bond that would be later formed.

This lower stability, in terms of kinetics, translates to higher activation energy and therefore, lower reaction rate.

According to S. Bailey and A. Bailey:

When alkyl halides are treated with a nucleophile, they can undergo either substitution or elimination. The most important factor in determining which will occur is the stability of the alkene that could be formed by elimination should it predominate.

The steric hindrance you're referring to (bulky reactant) as a way of discriminating between $E2$ vs $SN2$ dominance, is more impactful when the substrate is what's bulky, not the nucleophile/base itself.

For example, if we switched things around and had a small nucleophile/base like $\ce{CH3O-}$ react with a bulky substrate like $\ce{(CH3)3C-Cl}$,$\;$ $E2$ would dominate over $SN2$, which is consistent with the conclusion you reached.

I could not find an explicit source with porcentual yield estimates for the competing products in this reaction by $SN2$ and $E2$ (ETBE and ethene, respectively), but I will try to make the case for why I believe, based on what I've researched so far, that $SN2$ would slightly dominate in this case.

First, I will use a similar reaction with known product yields as a reference point.

According to Yurkanis Bruice, the reaction between t-butoxide and 1-bromopentane yields 15% of the $SN2$ product (1-tert-butoxypentane), and 85% of the $E2$ product (1-pentene):

enter image description here

While the reaction we're dealing with here is:

enter image description here

There's 3 important differences that would promote $SN2$ and not $E2$ when comparing both reactions:

(1) $R$ group size in substrate:

The R group in the reference reaction is pentyl, while in our reaction it's ethyl, which is a difference of 3 carbon groups.

According to Yurkanis Bruice, in similar cases, a difference of just 1 carbon group implies more than a two-fold increase in $SN2$ rate:

The rate of an SN2 reaction depends not only on the number of alkyl groups attached to the carbon that is undergoing nucleophilic attack but also on their size. For example, bromoethane and 1-bromopropane are both primary alkyl halides, but bromoethane is more than twice as reactive in an SN2 reaction, because the bulkier alkyl group on the carbon undergoing nucleophilic attack in 1-bromopropane provides greater steric hindrance to back-side attack.

So, we can expect that the absence of 3 carbon groups in the substrate of our reaction will boost the $SN2$ rate by a significant factor.

(2) Polarity of C-X bond in substrate:

$X$ is bromine in the reference reaction, while it is chlorine in our reaction.

The dipole moment $\mu$ of the C-Cl bond is higher than C-Br and even C-F. Their value according to L.G. Wade is:

enter image description here

This means that the carbon atom in C-Cl is more electrophilic (has a higher positive partial charge) than the carbon atom in C-Br.

In other words, the attraction between the nucleophile (t-butoxide) and such carbon atom is higher in our reaction, which also promotes $SN2$.

(3) Alkene product stability:

The $E2$ product in the reference reaction is a mono-substituted alkene (1-pentene), while it is a non-subsituted alkene (ethene) in our reaction. As mentioned before, substituted alkenes are more stable than non-substituted ones.

As a secondary reference point to estimate how important alkene stability is for $E2$, Marc Loudon claims the yield of ethene when bromoethane reacts with ethoxide is only 1%:

enter image description here

Furthermore, the addition of just 1 carbon group to the substrate R chain results in a 10-fold increase in yield for the resulting mono-substituted alkene:

enter image description here

I think it's reasonable to believe that the combination of these 3 differentiating factors (in this very particular case) offset the factor of steric hindrance in t-butoxide and lead to the $SN2$ product being the major product (although probably by a small margin), as claimed by OP's book and according to S. Bailey's alkene stability trend.

As a final note, Yurkanis Bruice claims Williamson ether synthesis (i.e. the particular $SN2$ reaction taking place in our case) is significantly more efficient when bulky alkoxides are used compared to when bulky substrates are used:

Consequently, a Williamson ether synthesis should be designed in such a way that the less hindered alkyl group is provided by the alkyl halide and the more hindered alkyl group comes from the alkoxide ion.

  • 3
    $\begingroup$ Do you have any citations to support your claim that the substrate steric bulk is significantly more impactful than the nucleophile bulk? $\endgroup$
    – Waylander
    Oct 22, 2022 at 8:13
  • $\begingroup$ @Waylander "When alkyl halides are treated with a nucleophile, they can undergo either substitution or elimination. The most important factor in determining which will occur is the stability of the alkene that could be formed by elimination should it predominate". Organic Chemistry by Bailey Jr. (page 223). Book can be consulted here: archive.org/details/organicchemistry0000bail/page/n551/mode/2up $\endgroup$
    – Sam202
    Oct 22, 2022 at 19:39
  • $\begingroup$ that quote says nothing about the effect of nucleophile sterics. See the reference I quote in my answer. $\endgroup$
    – Waylander
    Nov 5, 2022 at 14:22
  • $\begingroup$ In the context the quote is made, it says that when weighing in all the factors previously discussed in terms of S vs E selectivity (i.e. solvent, steric hindrance, alkene stability, etc.), alkene stability is the most important one. $\endgroup$
    – Sam202
    Nov 5, 2022 at 20:20

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