Why is E2 preferred over SN2 for 2-chlorobutane with sodium ethoxide?

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SUMMARY

The reaction of 2-chlorobutane with sodium ethoxide favors the E2 elimination mechanism over the SN2 substitution due to the strong basicity of ethoxide, which is a stronger base than acetate (pKa 4.8). The major product formed is trans-2-butene, as elimination reactions typically follow Zaitsev's rule, producing the most substituted double bond. The discussion clarifies that secondary alkyl halides require a strong base to deprotonate, and the geometric stability of the product is also considered in determining the major product. Understanding the relationship between basicity and nucleophilicity is essential for predicting reaction outcomes.

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  • Understanding of E2 and SN2 reaction mechanisms
  • Familiarity with pKa and pKb values
  • Knowledge of Zaitsev's rule in elimination reactions
  • Concept of geometric isomerism in alkenes
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  • Study the differences between E2 and SN2 mechanisms in detail
  • Learn about the application of Zaitsev's rule in various elimination reactions
  • Research the relationship between pKa, pKb, and basicity/nucleophilicity
  • Explore the effects of solvent on reaction mechanisms, particularly aprotic vs. protic solvents
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Chemistry students, organic chemists, and anyone interested in understanding reaction mechanisms involving alkyl halides and strong bases.

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Homework Statement


Why would this reaction favor E2 over SN2? Since its a secondary halide reacting with a strong base/nucleophile, and all other conditions being equal isn't SN2 favored? Or is ethoxide somehow a stronger base than nucleophile? If so, how can you tell that it's a better base than nucleophile? Below is the question.

What is the major product formed from the reaction of 2-chlorobutane in aqueous sodium ethoxide?
A.

2-ethoxybutane
B.

1-butene
C.

cis-2-butene
D.

trans-2-butene

Correct Answer
Explanation:
D. 2-Chlorobutane, a secondary alkyl halide, will react with a strong base, such as ethoxide, via an E2 mechanism. Choice A can be eliminated because it is the result of a substitution reaction. Elimination reactions generally follow Zaitsev's rule, which states that the product with the most substituted double bond is formed (provided an unhindered base is used). 1-Butene has a terminal double bond, which is less substituted than the double bond in 2-butene, so choice B can be eliminated. The most stable geometric isomer will have the two largest groups on opposite sides of the double bond, so choice C can be eliminated.


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The Attempt at a Solution

 
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For secondary alkyl halides, any nucleophile that is a base stronger than acetate (pKa 4.8) will deprotonate rather than undergo substitution. What do you think this says about the acidity of secondary alkyl halides with alpha protons?
 
Hi Chemisttree,
I think I understand but I'm not sure. I'm confused on why you used pka of acetate, for comparison, to determine basicity. Are you just using this as a starting point converting to pkb and then looking at basicity? If so couldn't we just look at each pkb and pick the one with the smallest pkb to determine the stronger base. Is there a way to determine basicity, relative to nucleophilicity without using pkb/pka values? I read there's a trend where basicity is highest going towards the upper left hand side of the periodic table and a trend where nucleophilicity is highest going towards the bottom left hand side of the periodic table but they aren't explained and I can't find information on what these trends are based on, relative to each other. Would you happen to know what these trends are based on? About your question, this basically tells us that the alpha proton in this case is not very acidic so we need a strong base right?
 
Last edited:
silversurf said:
Hi Chemisttree,
I think I understand but I'm not sure. I'm confused on why you used pka of acetate, for comparison, to determine basicity. Are you just using this as a starting point converting to pkb and then looking at basicity? If so couldn't we just look at each pkb and pick the one with the smallest pkb to determine the stronger base. Is there a way to determine basicity, relative to nucleophilicity without using pkb/pka values?

We use either pKa or pKb as a measure of basisity. Of course when we use pKa, we assume the protonated form. pKa of acetate being 4.8 is actually that of the acid form... acetic acid. Kind of confusing to speak of base strength in terms of acidity but you will note that they are related by the formula pKa +pKb = 14.

Is there a way to determine basisity relative to nucleophilicity without using pKa/pKb? Not really. You see, even a fairly weak base like acetate is a pretty good nucleophile. Pretty good towards the right alkyl halide, that is. Good nucleophile for primary alkyl halides without beta carbon branching, not so good toward secondary, tertiary or primary with beta carbon branching.

About your question, this basically tells us that the alpha proton in this case is not very acidic so we need a strong base right?

Actually it was a bit of a trick question. If a base only a little stronger than acetate is capable of deprotonating alpha to the C-X bond you might be tempted to assume that something like chloroethane is a fairly strong acid and might exist in equilibrium as:
CH3-CH2-X <------> CH-2-CH2-X + H+

which doesn't happen... hence the "2" subscript for an E2 elimination. Plucking off that proton doesn't happen without the acetate and a particular arrangement of the C-Cl bond anti to the proton being removed. It is more than an acid/base thing.
 
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chemisttree said:
Is there a way to determine basisity relative to nucleophilicity without using pKa/pKb? Not really. You see, even a fairly weak base like acetate is a pretty good nucleophile. Pretty good towards the right alkyl halide, that is. Good nucleophile for primary alkyl halides without beta carbon branching, not so good toward secondary, tertiary or primary with beta carbon branching.

@hat about KCN reacting with a secondary carbon, in aprotic solution, this results in a nucleophilic attack sn2 major product, how would we know that the major product is SN2 in this case? since aprotic conditions favor sn2 and e2 and secondary carbons are in the middle favor ability for both too.
 
Actually never mind, I just realized the reason why is because it's a weak base.
 

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