[habbababba]

I read that when pure ethanol is used as a solvent during the reaction between a halogenoalkane and sodium hydroxide (absence of water), elimination predominates to produce the alkene.

In the case when aqueous sodium hydroxide is used, for the same haloalkane, substitution predominates.

This information feels vague to me because I learned that when it comes to deciding which reaction predominates over the other, many factors need to be taken into account including nucleophilicity vs basicity, the type of halogenoalkane (primary, secondary...), the solvent, the temperature and the leaving group.

In this particular case, all the factors are kept constant and the spotlight is on the solvent only.

How does changing the solvent from water to ethanol tips the balance in favor of elimination?

Thanks in advance.

[orgopete]

This information feels vague to me because I learned that when it comes to deciding which reaction predominates over the other, many factors need to be taken into account including nucleophilicity vs basicity, the type of halogenoalkane (primary, secondary...), the solvent, the temperature and the leaving group.

I agree with the above. For example, a tertiary halide is not going to give substitution by changing solvent. Perhaps you could give more information (and actual data) regarding this solvent effect.

[habbababba]

I wish I had access to any actual data, that would have certainly clarified things up. However, before I posted here I searched the matter on the Web and I found a couple of links (below) in which this solvent effect is mentioned. If you follow the second link, you'll find that the topic has been discussed but since no actual data is given, I am a bit doubtful of the explanation provided. What do you think?

I apologize if I'm not allowed to post any links here. I have no intentions for advertising the content of the two links.

http://www.chemguide.co.uk/mechanisms/elim/elimvsubst.html
https://www.physicsforums.com/threads/elimination-vs-substitution-ethanol-as-a-solvent.387177/

[Guitarmaniac86]

I'm going to assume the substitution reaction is S_N1 and the elimination is E2 because it has to be if a base is involved in elimination.

What happens in an S_N1 reaction? How does the solvent behave? Which solvent would you expect to be better at stabilising the intermediate in the S_N1 reaction?

[habbababba]

S_N1 mechanism involves a carbocation intermediate. I can see how water can make the process of forming the carbocation easier. The water molecules, with their partially positive and negative ends, can help separate the leaving group from the carbon, lowering the activation energy.

On the other hand, I don't see how ethanol allows hydroxide to act more as a Lewis/Bronsted base rather than a nucleophile.

Furthermore, the explanation I just provided applies only if the substitution follows the S_N1 path. For a methyl or primary haloalkane, S_N2 is the mechanism. This further complicates the matter because hydroxide approaches the haloalkane in both S_N2 and E2, but attacks at different sites in each mechanism. It seems that the solvent has somehow an effect on the nucleophilicity and on the basicity of the hydroxide.

[Guitarmaniac86]

S_N1 mechanism involves a carbocation intermediate. I can see how water can make the process of forming the carbocation easier. The water molecules, with their partially positive and negative ends, can help separate the leaving group from the carbon, lowering the activation energy.

On the other hand, I don't see how ethanol allows hydroxide to act more as a Lewis/Bronsted base rather than a nucleophile.

Furthermore, the explanation I just provided applies only if the substitution follows the S_N1 path. For a methyl or primary haloalkane, S_N2 is the mechanism. This further complicates the matter because hydroxide approaches the haloalkane in both S_N2 and E2, but attacks at different sites in each mechanism. It seems that the solvent has somehow an effect on the nucleophilicity and on the basicity of the hydroxide.

You are right that the solvent effects nucleophilicity and basicity of the hydroxide but that is secondary to how the reactions proceed.

We know elimination is favoured in ethanol. We know that water is better able to separate ion pairs in S_N1 reactions. We know ethanol is not polar enough to separate ion pairs. For substitutions in ethanol, what can we say about the activation energy? Is it higher or lower than elimination? If it is higher, what reaction is favoured?

Remember, these are two competing reactions, one is more favoured than the other and the solvent stabilises intermdiates. Draw the intermediate for elimination. What solvent, water or ethanol, would you expect to be better at stabilising the intermediate? Why?

[habbababba]

For substitutions in ethanol, what can we say about the activation energy? Is it higher or lower than elimination? If it is higher, what reaction is favoured?
Draw the intermediate for elimination. What solvent, water or ethanol, would you expect to be better at stabilising the intermediate? Why?

There is no intermediate in an E2 elimination. There is a transition state. Nevertheless, the activation energy for the transition state depends on the following factors:

- The type of halogenoalkane.
- The strength of the Bronsted base.
- The leaving group.
- The solvent.

Assuming it's a secondary haloalkane, I think it would be less energetically favorable to form a secondary carbocation (to go down the S_N1 path) than losing a H to the Bronsted base (to follow the E2 path). However, this isn't always true. We must take the strength of the Bronsted base into consideration. To illustrate, cyanide ion is much less basic than hydroxide and reacts with 2-chlorooctane to give the corresponding alkyl cyanide as the major product of substitution. But in the case where sodium hydroxide is used in ethanol, hydroxide is a strong enough base to abstract the hydrogen, rendering the activation energy for elimination lower than for substitution.

All of the above has been said assuming that the substitution reaction follows the S_N1 path. Concerning the S_N2 path, it seems, on one hand, that steric hindrance makes this path energetically less favorable than the E2 mechanism. In other words, abstraction of H is more energetically favorable than nucleophilic attack. On the other hand, the hydroxide ion is more exposed in ethanol than in water (the water molecules 'cage' the hydroxide ion more than ethanol) and as a result one would expect that the hydroxide ion can become a better nucleophile in ethanol than in water! Since this isn't the case according to some sources, either my analysis is wrong or it's incomplete. Other than that, I don't see how the solvent affects the transition state in E2.

[Guitarmaniac86]

For substitutions in ethanol, what can we say about the activation energy? Is it higher or lower than elimination? If it is higher, what reaction is favoured?
Draw the intermediate for elimination. What solvent, water or ethanol, would you expect to be better at stabilising the intermediate? Why?

There is no intermediate in an E2 elimination. There is a transition state. Nevertheless, the activation energy for the transition state depends on the following factors:

- The type of halogenoalkane.
- The strength of the Bronsted base.
- The leaving group.
- The solvent.

Assuming it's a secondary haloalkane, I think it would be less energetically favorable to form a secondary carbocation (to go down the S_N1 path) than losing a H to the Bronsted base (to follow the E2 path). However, this isn't always true. We must take the strength of the Bronsted base into consideration. To illustrate, cyanide ion is much less basic than hydroxide and reacts with 2-chlorooctane to give the corresponding alkyl cyanide as the major product of substitution. But in the case where sodium hydroxide is used in ethanol, hydroxide is a strong enough base to abstract the hydrogen, rendering the activation energy for elimination lower than for substitution.

All of the above has been said assuming that the substitution reaction follows the S_N1 path. Concerning the S_N2 path, it seems, on one hand, that steric hindrance makes this path energetically less favorable than the E2 mechanism. In other words, abstraction of H is more energetically favorable than nucleophilic attack. On the other hand, the hydroxide ion is more exposed in ethanol than in water (the water molecules 'cage' the hydroxide ion more than ethanol) and as a result one would expect that the hydroxide ion can become a better nucleophile in ethanol than in water! Since this isn't the case according to some sources, either my analysis is wrong or it's incomplete. Other than that, I don't see how the solvent affects the transition state in E2.

I mixed my terms, I apologise. I was trying to get across that the transition state in E2 is more stable in ethanol than in water, and this is played out by the difference in activation energy which you have understood.

In your original post you asked how solvent tips the balance in favour of elimination, in this case ethanol is better at stabilising the transition state in elimination than it is at stabilising the carbocation in a substitution reaction. This is because there is a better interaction between ethanol and transition state.

In terms of S_N2 reactions, polar protic solvents like ethanol and water disfavour S_N2 reactions because there is an ion dipole interaction. This ion dipole interaction reduces the nucleophilicity of the hydroxide ion. This is the reason why elimination is favoured when all things are kept equal; because the hydroxide is unable to act as a nucleophile and instead acts as a base. This also explains why in water the S_N1 reaction predominates with low concentrations of base; the water attacks in the carbocation and the hydroxide ion deprotonates the C-OH₂⁺. I will be disagreed with on this point I know: the hydroxide is acting as a base and not a nucleophile.

In terms of elimination, the reaction rate is faster than the rate for S_N1, ethanol reduces the nucleophilicity of the hydroxide ion but increases its basicity and at he same time it stabilises the transition state in the elimination pathway but is unable to stabilise the carbocation adequately should it form. I think I am also missing the "silver bullet" trying to explain this. I hope I havent added to your confusion.

Also, I have deliberately avoided talking about hard and soft nucleophiles because I dont like this idea.

[habbababba]

in this case ethanol is better at stabilising the transition state in elimination than it is at stabilising the carbocation in a substitution reaction. This is because there is a better interaction between ethanol and transition state.

Could you please expand on what you mean by ethanol interacts better with the transition state?

Thanks.

[Guitarmaniac86]

in this case ethanol is better at stabilising the transition state in elimination than it is at stabilising the carbocation in a substitution reaction. This is because there is a better interaction between ethanol and transition state.

Could you please expand on what you mean by ethanol interacts better with the transition state?

Thanks.

In general terms the transition state does not have a charge so the interaction between it and the solvent is one of intermolecular attraction. The ethanol keeps the transition state in solution by essentially keeping it dissolved, which further reduces the activation energy. Water, being more polar than ethanol, is not as able to stabilise the transition state through intermolecular attraction as there is no charges for its polarised bonds to stabilise.

[phth]

The reaction also depends on the halide product.  Would an alkyl chloride, bromide or iodide be better, and why?  The best way by SN2 would involve formation of an ester so the % deprotonation decreases by a factor of 10[/sup]11[/sup] e.g. acetate versus hydroxide.  This is an example of pKa and basicity.  Water would work better, and in the correct solvent choice and time it will react with secondary alkyl halides: example http://www.organic-chemistry.org/abstracts/literature/724.shtm