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The more s character in the orbital holding a conjugate base's lone pair, the more stable the base: ethyne (sp, pKa 25) is far more acidic than ethene (44) and ethane (50), and only a base with a weaker conjugate acid can deprotonate it.
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You will give a hybrid's s character, rank C–H acids by hybridization, choose a base strong enough to deprotonate ethyne, and apply the same rule to nitrogen bases.
From lesson 1 you know that more s character holds electrons closer to the nucleus. From this unit you know a stable conjugate base means a strong acid. Put together, they explain why a C–H on a triple bond is acidic.
s character is the share of a hybrid that comes from the s orbital: sp 1/2, sp2 1/3, sp3 1/4. A carbanion is a carbon bearing a negative charge. The acetylide ion is HC≡C−, ethyne's conjugate base.
Remove a proton from a C–H and the carbon keeps the pair of electrons, as a lone pair in the hybrid orbital that made the bond.
| Hydrocarbon | Base's lone pair | s character | pKa |
|---|---|---|---|
| ethane, CH3CH3 | sp3 | 1/4 | 50 |
| ethene, CH2=CH2 | sp2 | 1/3 | 44 |
| ethyne, HC≡CH | sp | 1/2 | 25 |
An s orbital holds its electrons closer to the nucleus than a p orbital, so a lone pair in an sp orbital is held most tightly: the acetylide ion is the most stable of the three carbanions, and ethyne the most acidic hydrocarbon by twenty-five pKa units. Nothing about resonance or induction differs between the three; only the orbital.
Another way: steps
To compare by hybridization:
A base removes a proton only if its own conjugate acid is weaker (higher pKa) than the acid it attacks. Hydroxide cannot deprotonate ethyne, because water (16) is a stronger acid than ethyne (25). Sodium amide can, because ammonia is weaker (38). That is how chemists make acetylide ions to build carbon chains.
The rule works for any atom's lone pair. An sp3 amine nitrogen holds its pair loosely and is a good base; an sp2 pyridine nitrogen less so; an sp nitrile nitrogen holds it so tightly it is barely basic.
All carbon-bound hydrogens have equal acidity. They span 25 pKa units by hybridization alone.
More s character makes a base stronger. It makes the base more stable, and so weaker.
Any strong base removes any C–H. Its conjugate acid must be weaker.
Hybridization acidity is resonance. Nothing is delocalized; only the orbital changes.
Propyne's terminal C–H has pKa about 25; methanol, methoxide's conjugate acid, about 16.
Compare conjugate acids.
Methanol is the stronger acid, so the equilibrium lies back towards propyne.
Proton returns.
So methoxide cannot do it; a base like NH2− is needed.
Conjugate acid must be weaker.
What orbital holds each conjugate base's lone pair?
sp2 for benzene, sp3 for cyclohexane.
Which has more s character?
sp2, a third against a quarter: benzene's C–H is more acidic.
What fraction of an sp2 hybrid orbital is s character? (A fraction is fine.)
Answer:
Rank ethane, ethene and ethyne from most acidic C–H to least.
Number the steps in order (write the number in the box):
Ethyne's pKa is about 25 and ethane's about 50. How many pKa units more acidic is ethyne?
Answer:
To make the acetylide ion, HC≡C−, a chemist needs a base that removes ethyne's proton (pKa 25). Which base will do it? Conjugate acids: water pKa 16, ethanol 16, ammonia 38.
Nitrogen lone pairs follow the same rule. Match each nitrogen to how strong a base it is: the conjugate acids have pKa about 10.6, 5 and −10.
| strongest base: conjugate acid pKa about 10.6 | middling base: conjugate acid pKa about 5 | very weak base: conjugate acid pKa about −10 | |
|---|---|---|---|
| the sp3 nitrogen of ethanamine | |||
| the sp2 nitrogen of pyridine | |||
| the sp nitrogen of ethanenitrile |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For each hydrocarbon, give the hybridization of the carbon holding the lone pair in its conjugate base, and its approximate pKa.
| hybridization | pKa | |
|---|---|---|
| ethyne | ||
| ethene | ||
| ethane |
You can use hybridization to compare acidity. Tell someone why hydroxide cannot deprotonate ethyne. Next: predicting which way any proton transfer goes.
8. Your turn: which is more acidic, the C–H of benzene (sp2) or of cyclohexane (sp3)?, step 3