NaOt-Bu + chloroethane

Question

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?

Answer 1

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

Answer 2

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:

$$1°>2°>3°$$

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

$$3°>2°>1°$$

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):

While the reaction we're dealing with here is:

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:

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%:

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:

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.