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# Sn1 and Sn2: leaving group

Using pKa table to determine leaving group ability for Sn1 and Sn2 reactions.

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• How can you determine a leaving group without pkas being listed? • So with the last reaction, was that a redox reaction followed by an SN1 reaction? • The first part is an acid-base reaction, but it is not a redox reaction. You can know this by observing that no electrons are being transferred. You could also calculate the oxidation state for the oxygen before and after protonation and see that it is -2 in each case.

You may have been confused by the fact that the oxygen acquires a positive formal charge, but this does not indicate that a redox reaction has occurred.
• Why is the pKa of water 15.7? I thought it was 14. • I thought that since Ka of the hydronium ion is 10 to the power of minus seven, thus pKa of the h3o+ should be -7? • Consider the origin of the 1 x 10ˉ⁷ value - it comes from the experimentally determined equilibrium constant for the autoionization of pure water, represented by the equation:

H₂O (l) + H₂O (l) = H₃O⁺ (aq) + OH⁻ (aq)

whose equilibrium constant Kw is 1 x 10ˉ¹⁴

Kw = [ H₃O⁺] • [OH⁻] = 1 x 10ˉ¹⁴

and since H₃O⁺ and OH⁻ are produced in equal quantity it follows

[H₃O⁺] = 1 x 10ˉ⁷ and [OH⁻] = 1 x 10ˉ⁷

thus the 1 x 10ˉ⁷ simply refers to the (very low) concentration of hydronium and hydroxide ions present in pure neutral water and not to the acidity constant of the hydronium ion in water, i.e. to the equilibrium constant for the reaction:

H₃O⁺ (aq) + H₂O (l) = H₂O (l) + H₃O⁺ (aq)

Also, note that the pKa expression has a sneaky negative sign, pKa = - log (Ka), so
- log (1 x 10ˉ⁷) would have been -(-7) = +7.

Remember that the relationships:

Ka • Kb = 1 x 10ˉ¹⁴

pKa + pKb = 14

refer to conjugate acid-base pairs in water as the solvent -

e.g. acetic acid/acetate ion (AcOH/AcO⁻) in water -

Dissolving acetic acid in water gives:

AcOH (aq) + H₂O (l) = H₃O⁺ (aq) + AcO⁻ (aq)

Ka (AcOH) = [H₃O⁺] • [AcO⁻] / [AcOH] = 1.74 x 10ˉ⁵

pKa (AcOH) = -log (1.74 x 10ˉ⁵) = 4.76

the Kb and hence pKb of the acetate ion is not simply the K value for the reverse of this reaction, but rather refers to the equilibrium obtained by dissolving the conjugate base of acetic acid in water (e.g. dissolving sodium acetate in water) which gives:

AcO⁻ (aq) + H₂O (l) = OH⁻(aq) + AcOH (aq)

Kb (AcO⁻) = [OH⁻] • [AcOH] / [AcO⁻] = 5.75 x 10ˉ¹⁰

pKb (AcO⁻) = -log (5.75 x 10ˉ¹⁰) = 9.24

since the pKa of acetic acid is pKa (AcOH) = 4.76, we could have used pKa + pKb = 14
to obtain the pKb of the acetate ion:

4.76 + pKb (AcO⁻) = 14

pKb (AcO⁻) = 14 - 4.76 = 9.24

Going back to the original question the correct values with water as the solvent are:

pKa (H₂O) = 14

pKa (H₃O⁺) = 0

the values given in the video lessons, 15.7 and -2 respectively, derive from what some would argue as faulty reasoning when dealing with the equilibrium expressions. I've posted a bit more regarding this below if you are interested in the extra info.

Thermodynamic equilibrium constants should really be expressed in terms of activities. Activity is a measure the 'effective concentration' of each reagent or product in the balanced equation that takes into account deviations from ideal solution behavior which are a result of intermolecular forces. In some ways this is a similar concept to using the Van der Waals equation in place of the ideal gas law when dealing with non-ideal gases.

In simple terms the activity (a), of a species (X), can be related to its concentration [X] through the expression a(X) = γ[X] where γ is an activity coefficient. So for the reaction:

A + B = C + D

the equilibrium expression is in the (familiar) form:

K = a(C) • a(D) / a(A) • a(B)

In the equilibrium expressions presented in the video lessons, where concentration is being used in place of activities, the approximation is being made that for dilute solutions the activity coefficient for each species is equal to one (γ = 1) and so activity can be approximated to concentration a(X) ≈ [X]. Another assumption being made is that the activity of the solvent can be approximated to the activity of the pure liquid which then assumes a value of 1 - i.e. a(solvent) = 1. This is the true origin of being able to exclude the solvent from the equilibrium expression - it really is there but it has a value of 1 !

It's easier to get your head around it with an example so consider the question of the pKₐ of water with water as the solvent:

H₂O (l) + H₂O (l) = H₃O⁺ (aq) + OH⁻ (aq)

Here we have water acting as both Bronsted acid (the bit we are interested in) and Bronsted base. Writing the equilibrium expression in terms of activities, and noting that the activity of the solvent is equal to 1 gives:

Ka (H₂O) = a(H₃O⁺) • a(OH⁻) / a(H₂O) • a(H₂O)

Ka (H₂O) = a(H₃O⁺) • a(OH⁻) / 1 • 1

now approximating activities with concentration:

Ka (H₂O) = [H₃O⁺] • [OH⁻] / 1 • 1

Ka (H₂O) = [H₃O⁺] • [OH⁻]

this equation should now look very familiar - it's just the autoionization constant of water, with the experimentally determined value of 1 x 10ˉ¹⁴, so Ka (H₂O) = Kw (H₂O) = 1 x 10ˉ¹⁴ and

pKa (H₂O) = - log (1 x 10ˉ¹⁴) = 14

it follows that the Kb of water is also 1 x 10ˉ¹⁴, so the pKb of water is also 14, pKb (H₂O) = 14.

The alternative value often quoted, pKa (H₂O) = 15.7, comes from assuming that we somehow have two different types of water molecules - the 'solvent' ones, whose activity is equal to 1, and 'solute' ones, i.e. the ones that undergo ionization, whose activity is then approximated by the molar concentration of pure water i.e. 55.3 mol/L. This is a strange distinction to make because it is obvious that all the water molecules are identical, and therefore their activities should also be identical and equal to 1. However, continuing with this idea leads to:

Ka (H₂O) = [H₃O⁺] • [OH⁻] / [H₂O] • 1

Ka (H₂O) = 1 x 10ˉ¹⁴ / 55.3 = 1.8 x 10ˉ¹⁶

pKa (H₂O) = - log (1.8 x 10ˉ¹⁶) = 15.74

likewise, the pKb of water is then also reasoned to be 15.74, pKb (H₂O) = 15.74.

It is then argued that since H₃O⁺ and H₂O form a conjugate pair, and pKb (H₂O) = 15.74, then using the relation pKa + pKb = 14 one can obtain the pKa of the hydronium ion in water, pKa (H₃O⁺):

pKa (H₃O⁺) + pKb (H₂O) = 14

pKa (H₃O⁺) + 15.74 = 14

pKa (H₃O⁺) = - 1.74

whereas if we use the correct value for the pKb of water, pKb (H₂O) = 14 we get

pKa (H₃O⁺) + pKb (H₂O) = 14

pKa (H₃O⁺) + 14 = 14

pKa (H₃O⁺) = 0

This same result is also be obtained from:

H₃O⁺ (aq) + H₂O (l) = H₂O (l) + H₃O⁺ (aq)

Ka (H₃O⁺) = a(H₂O) • a(H₃O⁺) / a(H₃O⁺) • a(H₂O)

Ka (H₃O⁺) = 1 • [H₃O⁺] / [H₃O⁺] • 1

Ka (H₃O⁺) = 1

pKa (H₃O⁺) = - log (1) = 0
• I'm not sure I've fully grasped the concept of a good vs bad leaving group. Could anyone give a bit more context to the terms "good" and "bad"? Does a "bad leaving group" refer to a conjugate base that is simply too unstable to detach from the molecule and therefore must be protonated to enable detachment for substitution? Does a "good leaving group" refer to a LG that has a low activation energy or something along those lines? Do I just need to refresh my understanding of pKa values and the answer will become clear? • A leaving group is a fragment of a molecule (with a pair of electrons) which departs from the original molecule. Being good in this sense means that it is stable on its own and can exist separately from the original molecule easily. And bad would mean that it is unstable when separated from the original molecule.

Keep in mind that leaving groups exist on a spectrum of good versus bad; it's not an either/or situation. So something like an iodide anion is considered a good leaving group, as is a chloride anion, but the iodide is a better leaving group compared to the chloride.

Additionally leaving groups don't have to be bases, but those are the most commonly observed ones.

If the leaving group is a base, then generally weaker bases are better leaving groups. The reasoning here is that the reaction of the leaving group departing is in equilibrium with the leaving group reattaching itself back to the original molecule. A strong base would favor the reactants while a weak base would favor the products. This is why hydroxide (a strong base) is considered a bad leaving group, but water (a weak base) is considered a good leaving group.

Resonance is also a common contributor to leaving group stability. If a leaving group can more evenly distribute the additional negative charge acquired from the pair of electron then this will make it more stable on its own.

Hope that helps.
• at what is being done in the mechanism to protonate or deprotonate the substrate • Why is ethanol such a week acid? at • I thought you said water is a poor leaving group because it has a pKa of 15, why is it a good LG in the second example?
(1 vote) • Water's pKa value being high means hydroxide would be the bad leaving group, not water itself. Water is the conjugate base of the hydronium ion where the Pka is decently low at -2. When thinking about water as the leaving group we're looking at the Pka value of -2, not 15.

If that's too abstract, a simple way to explain if something is a good leaving group is if it is stable on its own after it has left a substrate. Water is of course perfectly stable on it's own, but hydroxide is a strong base which reacts easily (particularly with the substrate recreating the original reactant). Therefore making water a good leaving group, and hydroxide a bad leaving group. Hope that helps.
• Can we do SN2 reaction with tosylate group? is it considered very bulky due to the benzylic ring and the S with the double bonds to the oxygens?
(1 vote) • How do you know when to use a SN1 reaction and when to use a SN2 reaction? How do I know if which one will be just one step and which one would have two?
(1 vote) • If the leaving group in an Sn1 reaction is not a good leaving group (not a stable ion) then look for a way to make the leaving group better. That will require another component for the reaction so you would need to be concerned with the concentration of that atom/molecule in the solution making it an Sn2 process.

That was how I understood the video at least.
(1 vote)