Why is the behavior of sodium ethoxide inconsistent?

In a reaction with 1-bromobutane, the anionic part of sodium ethoxide substitutes for $\ce{Br}$:

$$ \ce{CH3CH2CH2CH2Br + C2H5ONa -> CH3CH2CH2CH2-O-CH2CH3}$$

Conversely, in a reaction with 2-bromobutane, the anionic part eliminates $\ce{Br}$:

$$\ce{CH3CH2CHBrCH3 + C2H5ONa -> CH3CH2CH=CH2}$$

I simply don't see how changing the position of $\ce{Br}$ changes the type of reaction. I believe sodium ethoxide, being bulky, must always do elimination, consistent with Rule #3 here.

Given the above: can sodium ethoxide be forced to one type of reaction with a given reactant, by enforcing some reaction conditions?

  • $\begingroup$ I'm sure there are already several questions covering when a reagent causes E2 and when it causes SN2, did you have a look at them already? Thank you. $\endgroup$ Commented Mar 29, 2018 at 14:12
  • 1
    $\begingroup$ Think about the difference in steric hindrance in the two halides.(primary and secondary). Also, in the second reaction its not that only elimination occurs. $\endgroup$
    – MollyCooL
    Commented Mar 29, 2018 at 14:15
  • $\begingroup$ @GaurangTandon I except sodium ethoxide to show E2 only when a hallide is present. I've seen examples where even the primary hallide undergoes E2 $\endgroup$ Commented Mar 29, 2018 at 14:29
  • $\begingroup$ On a side note, every single opening or closing parenthesis in your formula lines is redundant. $\endgroup$ Commented Mar 29, 2018 at 14:38
  • $\begingroup$ @IvanNeretin I think it adds readability. Although they are redundant. $\endgroup$ Commented Mar 29, 2018 at 14:58

2 Answers 2


The above reactions belong to Williamson Synthesis of Ethers, probably the best of the alternatives for prepartion of ethers!

The Williamson Ether synthesis reactions follow SN2 mechanism. Since the SN2 mechanism proceeds through a single step where the nucleophile performs a “backside attack” on the alkyl halide, the only thing stopping this, is steric hindrance.

Methyl and primary alkyl halides are excellent substrates for this synthesis, as they provide no/very less hindrance for the approaching nucleophile.

Since alkoxides are strong bases, competition with elimination (E2) becomes a concern once the alkyl halide becomes more sterically hindered. This is exactly why, the first reaction that you’ve mentioned, proceeds by SN2 but in the second both elimination(major) and substitution (minor) takes place.

Also note that, when tertiary alkyl halides are used, there is no hope for substitution because of the bulky groups present around carbon. Hence, only elimination takes place in this case.

So the reactions are,

  • With primary alkyl halide, $$ \ce{CH3CH2CH2CH2Br + C2H5ONa -> CH3CH2CH2CH2-O-CH2CH3}$$
  • With secondary alkyl halide, $$\ce{CH3CH2CHBrCH3 + C2H5ONa -> CH3CH2CH=CH2(major)}$$

Moving onto conditions to favour a particular type of reaction:

  • One way to attempt to get the SN2 to be favoured over elimination is to use a polar aprotic solvent (such as acetonitrile or DMSO) that will increase the nucleophilicity of the alkoxide.
  • Increasing the temperature favours elimination over substitution.

Source for the above conditions: chem.ucalgary

I believe sodium ethoxide being bulky should always do elimination

In most cases, rather than the incoming nucleophile, the hindrance in the the alkyl halide plays a more important role.

The rule #3 mentioned in the link you stated clearly says that the elimination takes place for a tertiary carbon in presence of a strong base like $\ce{CH3CH2-O-Na}$.

  • $\begingroup$ Nice Answer! If I use polar aprotic solvent and the hallide is $2^0$ will it show SN2 or E2? $\endgroup$ Commented Mar 30, 2018 at 4:49
  • $\begingroup$ @SmarthBansal Thanks:). Majorily SN2 because of the aforesaid reasons. $\endgroup$
    – MollyCooL
    Commented Mar 30, 2018 at 4:59
  • $\begingroup$ So using aprotic guarantees it's SN2? $\endgroup$ Commented Mar 30, 2018 at 5:00
  • $\begingroup$ As per the link, SN2 will be favoured. But you cannot say that it will only be substitution. I can guarantee that it surely won’t be Elimination only. ** Note: Substitution and elimination reactions are strongly influenced by many experimental factors** $\endgroup$
    – MollyCooL
    Commented Mar 30, 2018 at 5:03

You chose a good link but look at the not very helpful rule:

Your mentioned link clearly stated followings rules for the reactions of strong base/nucleophile with a primary, secondary, and tertiary halides, which are well acceptable:

Rule 1: A reaction of strong base/nucleophile with a primary halide (and methyl halide) will almost certainly be SN2.

Exceptions: (a) Bulky bases such as $\ce {(CH3)3CO-}$ (note that $\ce {CH3CH2O-}$ is not considered to be 'bulky') gives E2 product predominantly; and (b) some primary halides such as allyl and benzyl halides can form relatively stable carbocations (by resonance stabilization, for example) may proceed through either of SN2, SN1, or E1 pathway, based on other conditions such as solvent and temperature.

Rule 3: A strong base/nucleophile usually favors SN2/E2 pathways (Neither SN1 nor E1 should be considered).

Exceptions: Both Rule 1 exceptions apply here. Apart from them, tertiary halides give E2 product predominantly because of steric hindrance present in tertiary carbon prevents backward attack by the base/nucleophile, regardless of its size. The elimination products are, however, depend on the size of the base/nucleophile: (a) Bulky bases such as $\ce {(CH3)3CO-}$ would favor to give less substituted alkene ("Hoffmann" Rule), while (b) not so "bulky" bases such as $\ce {CH3CH2O-}$ would favor to give most substituted (hence most stable) alkene ("Zaitsev" Rule).

Rule 5: If all else being equal, a reaction of strong base/nucleophile with a secondary halide is based on other conditions such as solvent: For example, (a) polar aprotic solvents favor SN2 (substitution) pathway, while (b) polar protic solvents favor E2 (elimination) pathway.


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