[webassignbuddy]

(a) Select the anionic nucleophile which will have the largest K, equilibrium constant.

(b) Select the anionic nucleophile which will have the smallest K, equilibrium constant.

Hint: the most basic anion will give the largest K while the least basic anion will give the smallest K.
^
What does that mean???!!!!!

Here is MY reasoning:
1) Large Keq means more products (the product side handles situation better).
2) Br would be the most stable by itself because it's the biggest atom (bigger atom = more spread out outer eectrons are from positively charged nucleus = more stability of atom by itself)

Also, F < O < S < Br (F is the smallest and most electronegative while Br is the largest and the least electronegative)

3) Most favored product that would form would be with F (which is really unstable before it attacks the epoxide and then becomes somewhat more stable whereas the Br would become unstable)?

*delete me*!

[Babcock_Hall]

The strongest base is associated with the weakest conjugate acid.  The weakest base is associated with the strongest conjugate acid.  How do we compare the relative strengths of two acids?

[webassignbuddy]

The strongest base is associated with the weakest conjugate acid.  The weakest base is associated with the strongest conjugate acid.  How do we compare the relative strengths of two acids?

pKa?
But I don't know those by heart!

Is there another way to go about figuring out the answer to this question?
All I remember my teacher saying in lecture was that the bigger the atom, the better.
But I don't see how that relates to this

[webassignbuddy]

The strongest base is associated with the weakest conjugate acid.  The weakest base is associated with the strongest conjugate acid.  How do we compare the relative strengths of two acids?

Here are 2 slides that kind of talk about what you're talking about but I don't understand how they relate to K equilibrium.

[webassignbuddy]

[webassignbuddy]

The strongest base is associated with the weakest conjugate acid.  The weakest base is associated with the strongest conjugate acid.  How do we compare the relative strengths of two acids?

You still there?

[TwistedConf]

Hint: the most basic anion will give the largest K while the least basic anion will give the smallest K.
^
What does that mean???!!!!!

It means that this is an extremely poor question.

Presumably, the question aims to ask you to identify the best nucleophile (which is often, but not always, the strongest base).  However, nucleophilicity is measured kinetically, and thus asking the question in the context of equilibrium constants is not proper or correct.

[Babcock_Hall]

I am a little confused by the question.  Are we looking in the direction of forming the epoxide as a product, or are we looking in the direction of opening the epoxide?

[discodermolide]

To the original question, I think that S^- is at least as good as Br^-. And F^- is definitely the worst.
Perhaps you should look at the Hard and Soft discussions of this forum and this web page it may have relevance to your problem?

http://www.chemicalforums.com/index.php?topic=6054.0
http://en.wikipedia.org/wiki/HSAB_theory

[webassignbuddy]

I am a little confused by the question.  Are we looking in the direction of forming the epoxide as a product, or are we looking in the direction of opening the epoxide?

I'm confused too

[Babcock_Hall]

@OP, Your original post looks like opening of an epoxide by a nucleophile.  The slides you showed in the fourth and fifth comment look like formation of an epoxide.  If that is the case, then we are comparing different leaving groups, which is a different kettle of fish.

[orgopete]

(a) Select the anionic nucleophile which will have the largest K, equilibrium constant.

(b) Select the anionic nucleophile which will have the smallest K, equilibrium constant.

Hint: the most basic anion will give the largest K while the least basic anion will give the smallest K.

I thought this was self explanatory. I agreed with the hint as far as predicting the equilibrium (unless of course someone has some data to the contrary). I didn't think bromide would open an epoxide and if it did, the reverse reaction is quite favored. It seems the least effective at forming bromo-alcohol. I am assuming the conditions are as indicated though.

HBr + H2O  H3O(+) + Br(-)

This could also be interpreted as Br(-) and would be the most effective. I assume this is not meant.

[webassignbuddy]

So I have the answers:

Nucleophile A (which is the oxygen) has the LARGEST K, equilibrium constant
Nucleophile D (which is the bromine) has the SMALLEST K, equilibrium constant...

I understand why Bromine would have the smallest K. Bromine is the biggest and the bulkiest and can cope with being negative. And since Br is good at dealing with a negative charge and is stable compared to F, it makes a very weak nucleophile. And since it is the wakest nucleophile, it will have the smallest K equil. constant (products side of the reaction is not so favored/the reaction isn't pushed to the right because reactants are favored).

HOWEVER, I still don't understand why O has the largest K equilibrium constant compared to F and S.

It seems like what the teacher did was divided up the 4 nucleophiles into groups (based on atomic sizes) to answer each question.

Nucleophiles with long carbon chains (O vs. S): (smaller atom; less stable; more EN) O < S (bigger atom; more stable; less EN)
Nucleophiles as single elements (F vs. Br): (smaller atom; less stable; more EN) F < Br (bigger atom; more stable; less EN)

But I still don't understnad how she came to the conclusion that O had the largest K

[orgopete]

Slide 1 of reply 3 shows the equilibria with hydroxide. Water and alcohol are nearly equal while HBr greatly favors bromide formation. Slide 2 shows the equilibria of the reaction in question, but from the reverse side. That is all you need to answer the question. It shows bromide opening of an expoxide is an uphill reaction, and more uphill than sulfide or alkoxide. Nucleophile A will have the largest amount of ring opening. This is indicated by Slide 2.

Certainly, mercaptides react to open epoxides and it would be difficult to know where an equilibrium might lie. However, it can also be difficult to know whether the product are also influenced by the reaction conditions. For example, the intent of reacting a sulfide with an epoxide might be presumed to form its product and avoid its reverse reaction. If it's reaction were carried out in a protic solvent, protonation of the alkoxide intermediate will stop the reverse reaction. Since the reaction only requires a catalytic amount of base, the mercaptan is the most acidic component and either directly or indirectly will form more mercaptide. This is different than the equilibrium shown on Slide 2. It shows a reaction of an alkoxide to form an epoxide plus mercaptide, the reverse reaction. Simply by formation of the conjugate base of a stronger acid, this reaction should be favored, however, steric strain can increase the energy level of the epoxide and reduce the equilibrium.

The other slides tell you about factors in the reaction, such as relief of strain. Fluoride was not on the slides, but it should have predictable properties. Acidity is data. It should tell you HBr is a stronger acid than HF. The arguments about atom size are an arguments to correct for the erroneous prediction that HF should be a stronger acid from electronegativity theory.