Why E2 Favors Secondary Halides: Exploring 2-Chlorobutane Reactivity in Ethoxide

  • Thread starter silversurf
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In summary, KCN is a strong nucleophile that can deprotonate a secondary carbon to form a carbon-carbon double bond.
  • #1
silversurf
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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.


Homework Equations





The Attempt at a Solution

 
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  • #2
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?
 
  • #3
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?
 
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  • #4
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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  • #5
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.
 
  • #6
Actually never mind, I just realized the reason why is because it's a weak base.
 

1. Why does E2 reaction favor secondary halides over primary halides?

The E2 reaction is a type of elimination reaction where a molecule undergoes the loss of a leaving group and a proton simultaneously. In the case of 2-chlorobutane, the leaving group is the chlorine atom. Secondary halides have a larger number of electron-donating alkyl groups attached to the carbon with the leaving group, making the leaving group more stable. This results in a lower energy barrier for the E2 reaction to occur, making it more favorable compared to primary halides.

2. What is the role of ethoxide in the E2 reaction?

Ethoxide is the base used in the E2 reaction. It acts as a nucleophile, attacking the carbon atom adjacent to the leaving group. This initiates the elimination of the leaving group and a proton, resulting in the formation of an alkene. Ethoxide is a strong base, making it an ideal choice for promoting the E2 reaction.

3. How does the reactivity of 2-chlorobutane change with different concentrations of ethoxide?

The reactivity of 2-chlorobutane increases with higher concentrations of ethoxide. This is because the concentration of ethoxide directly affects the rate of the E2 reaction. As the concentration of ethoxide increases, the rate of the reaction also increases, resulting in a higher yield of the desired product, i.e. the alkene.

4. Can the E2 reaction be used for the synthesis of other compounds besides alkenes?

Yes, the E2 reaction can also be used for the synthesis of other compounds such as alkynes and dienes. This is because the E2 reaction follows the same mechanism of elimination of a leaving group and a proton, regardless of the specific functional group present. However, the choice of the base may vary depending on the desired product.

5. How does the reactivity of 2-chlorobutane in the E2 reaction compare to other halides?

The reactivity of 2-chlorobutane in the E2 reaction is higher compared to primary halides but lower compared to tertiary halides. This is because tertiary halides have even more electron-donating alkyl groups attached to the carbon with the leaving group, making the leaving group even more stable. Therefore, tertiary halides have the lowest energy barrier for the E2 reaction to occur and are the most reactive in this type of reaction.

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