SME
Organic Chemistry II
Enolates and alpha substitution
Use this for “Cyclohexanone (pKₐ 18) is treated with one equivalent of sodium ethoxide; ethanol’s pKₐ is 16. What percent is enolate?”
Diagram
ketone (19.3) gives its proton; HB (16) forms.
The arrow points to the weaker acid, the reactants.
log K = pKₐ(right) − pKₐ(left) = 16 − 19.3 = −3.3, K = 5.01 × 10⁻⁴.
This is an example. Type your own values to start a new problem.
pKₐ(ketone)
pKₐ of the ketone’s α hydrogen
example
pKₐ(HB)
pKₐ of the base’s conjugate acid
example
ΔpKₐ
Difference in pKₐ (equals log K)
calculated
K
Equilibrium constant
calculated
%
Percent enolate
calculated · %
Formulas
ΔpKₐ = pKₐ(HB) − pKₐ(ketone)
−3.3 = 16 − 19.3
K
=
10
ΔpKₐ
5.01 ×
10
-4
=
10
−3.3
%
=
100 ×
√
K
1 +
√
K
2.19
=
100 ×
√
5.01 ×
10
-4
1 +
√
5.01 ×
10
-4
Assumptions
•
The base takes an α hydrogen: ketone + B⁻ ⇌ enolate + HB, so log K = pKₐ(HB) − pKₐ(ketone), and K > 1 when HB is the weaker acid.
•
Equal amounts of ketone and base: enolate and HB form in equal amounts, so enolate ÷ ketone = √K and the percent enolate is 100√K ÷ (1 + √K).
•
Both pKₐ values are on one scale (water-like estimates, not DMSO values) and the enolate forms reversibly, so the percent is a guide.
•
The picture places the ketone and HB on the pKₐ ladder, with the arrow toward the weaker acid.
Step-by-step
WE KNOW
pKₐ(ketone) = 19.3, pKₐ(HB) = 16
FIND
difference in pKₐ (equals log K) (ΔpKₐ), equilibrium constant (K), percent enolate (%)
Step 1 · Find difference in pKₐ (equals log K): ΔpKₐ
Use ΔpKₐ = pKₐ(HB) − pKₐ(ketone)
Take the ketone’s pKₐ from the pKₐ of the base’s conjugate acid.
ΔpKₐ = 16 − 19.3
ΔpKₐ = −3.3
Step 2 · Find equilibrium constant (K)
Use K = 10ΔpKₐ
Undo the log: raise 10 to ΔpKₐ.
K
=
10
−3.3
K
=
5.01 ×
10
-4
Step 3 · Find percent enolate (%)
Use % = (100 × √(K)) ÷ (1 + √(K))
Enolate
over
ketone
is
√
K
,
so
the
enolate
is
√
K
parts
of
1
+
√
K
.
%
=
100 ×
√
5.01 ×
10
-4
1 +
√
5.01 ×
10
-4
%
=
100
×
√
5.01 ×
10
-4
1 + 0.022383
%
=
100
×
√
5.01 ×
10
-4
1.0224
%
=
100
×
0.022383
1.0224
%
=
2.2383
1.0224
% = 2.19% → Ketone and enolate are both present, so the enolate can also add to the ketone left over (aldol).
Check
Put the numbers back in. Both sides should match.
−3.3 = 16 − 19.3 ✓
5.01 ×
10
-4
=
10
−3.3
✓
2.19
=
100 ×
√
5.01 ×
10
-4
1 +
√
5.01 ×
10
-4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.