- Identifying nucleophilic and electrophilic centers
- Curly arrow conventions in organic chemistry
- Intro to organic mechanisms
- Alkyl halide nomenclature and classification
- Sn1 mechanism: kinetics and substrate
- Sn1 mechanism: stereochemistry
- Carbocation stability and rearrangement introduction
- Carbocation rearrangement practice
- Sn1 mechanism: carbocation rearrangement
- Sn1 carbocation rearrangement (advanced)
- Sn2 mechanism: kinetics and substrate
- Sn2 mechanism: stereospecificity
- Sn1 and Sn2: leaving group
- Sn1 vs Sn2: Solvent effects
- Sn1 vs Sn2: Summary
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?(12 votes)
- Memorizing it. A tip is to make connections between the leaving group in the organic molecule and the anion in an acid. If you can remember the strength of the acid, you can know if the conjugated base is stable.(1 vote)
- So with the last reaction, was that a redox reaction followed by an SN1 reaction?5:27(3 votes)
- 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.(5 votes)
- Why is the pKa of water 15.7? I thought it was 14.(3 votes)
- For water at 25°C pKw = 14.
Why the pKa is listed as 15.7 is a good question – the short answer appears to be that organic chemists got this wrong!
You can read more about this here:
- 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?(2 votes)
- 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(2 votes)
- 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?(2 votes)
- 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.(3 votes)
- at5:27what is being done in the mechanism to protonate or deprotonate the substrate(2 votes)
- This may be late for you, but I think youd have to assume that the molecule is in a solvent with protons lying around. the oxygen forms a bond with a proton (sharing that lone pair), and you can see where that goes from there(2 votes)
- Why is ethanol such a week acid? at3:30(2 votes)
- 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.(3 votes)
- 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)
- [Instructor] The S N 1 and S N 2 reactions involve leaving groups. Let's look at this pKa table to study leaving groups in more detail. On the left, we have the acid. For example, hydroiodic acid, HI, with an approximate pKa of negative 11. Remember, the lower the pKa value, the stronger the acid. So on this table with the pKa value of negative 11, hydroiodic acid is the strongest acid and the stronger the acid, the more stable the conjugate base. So the conjugate base to HI is I minus the iodide anion. And since this is the conjugate based to the strongest acid, this is the most stable base. Let me write that down here. This is the most stable base on the table, which means that the iodide anion is an excellent leaving group because it is very stable. Next, we have hydrobromic acid, approximate pKa of negative nine. So the conjugate base would be the bromide anion, so also a stable conjugate base. So therefore, a good leaving group. For HCl, it's the chloride anion, also a good leaving group. So you see these halide anions as leaving groups all the time in organic mechanisms. Let me write this down here. So these are all examples of good leaving groups. Next, let's look at this acid on the left here. This is p-Toluenesulfonic acid with a pKa value of negative three, so it's still pretty acidic. The conjugate base to this is on the right here and we call this anion a tosylate group. Let me write this down. This is called a tosylate group. And since it's kind of a bulky group, instead of drawing this out all the time, you often see OTs written. So, OTs, like that, and you could put a negative charge on the oxygen here if you wanted to. So you'll see the tosylate group function as a leaving group in many reactions. Let's look at an example of another acid. So if I move down here to H3O+, the hydronium ion, with a pKa value of negative two. The conjugate base to H3O+ is H2O and water is also a good leaving group. So let's go back up here to the topic and we can see that all the acids that we talked about have negative pKa values, so negative 11, negative nine, negative seven, negative three, and negative two. And notice all of the conjugate bases are good leaving groups. So you can say that if an acid has a negative value for the pKa, the conjugate base will be a good leaving group. Let's look at another example of an acid. So, water. Water's pKa value is positive 15.7, so it's not a very strong acid. The conjugate base to water is the hydroxide anion, OH-, and this is a bad leaving group. So hydroxide ion is a bad leaving group and that's because water is not a strong acid. Look at this value for the pKa, positive 15.7. So if we look at ethanol, similar story here. So ethanol has a pKa value of positive 16. So the ethoxide anion is not a good leaving group, so this pKa values are in the positive and these conjugate bases must not be very stable which means they are bad leaving groups. Let me write that down here. So these are examples of bad leaving groups. Both S N 1 and S N 2 reactions need good leaving groups. However, the S N 1 reaction is even more sensitive. Let's look at tert-Butyl chloride. Let's say it's reacting via an S N 1 mechanism. The first step should be loss of leaving group. So these electrons come off onto the chlorine. We would form the chloride anion which has a negative one formal charge. We just saw on our pKa table that the chloride anion is a stable conjugate base. So therefore, this is a good leaving group. We're taking a bond away from the carbon in red, so the carbon in red gets a plus one formal charge and we form a tertiary carbocation as well. Since this is the rate determining step of our S N 1 mechanism, the formation of our stable anion, this formation of a good leaving group helps the S N 1 mechanism occur. Next, let's look at this alcohol here. If we approach it the same way as we did in the previous problem and we said, "Okay, first step is loss of the leaving group "and these electrons come off onto the oxygen." Think about what leaving group that is. That would be the hydroxide ion which we know from our pKa table is not a good leaving group. So the hydroxide ion is not as stable of an anion as the chloride anion. So the chloride anion is a good leaving group. The hydroxide anion is a bad leaving group. So that's not the first step of this mechanism. We need to make a better leaving group and you can do that by having a proton source. Let's say we have a source of protons, an acid in solution, so let's say there's an H+ here. The first step would be to protonate our alcohol, so our alcohol is gonna act as a base and pick up a proton. Let's draw the results of that. We have our ring. Let's put in that methyl group. And now our oxygen is bonded to two hydrogens. There's still a lone pair of electrons on this oxygen which give the oxygen a plus, which gives the oxygen a plus one formal charge. So the electrons here in magenta, let's say, pick up this proton to form this bond. Now we're ready for loss of a leaving group because if these electrons come off onto the oxygen now, we form water as a leaving group. Let me draw that in here. So here is the water molecule. Let me highlight those electrons in blue. These electrons come off onto the oxygen then we form water and we know from our pKa table that water is a good leaving group. We're taking a bond away from this carbon in red, so we're also gonna form a tertiary carbocation. Let me draw that in here. Here's our ring. Here's our methyl group. A plus one formal charge on the carbon in red. So by thinking about your pKa values, you can determine the stability of the conjugate base and therefore, if a leaving group is a good leaving group or a bad leaving group and that helps you out when you're drawing mechanisms.