Hydration
Reactions
Master acid-catalyzed hydration and oxymercuration-demercuration: two essential methods for adding water across alkene double bonds.
Organic Chemistry 1
What Is Hydration?
A hydration reaction adds \(\text{H}_2\text{O}\) across a carbon-carbon double bond (\(\text{C}=\text{C}\)), converting an alkene into an alcohol. The \(\pi\) bond breaks, and an \(\text{-OH}\) group and an \(\text{-H}\) are added to the two carbons.
This transformation is fundamental in organic synthesis because alcohols are versatile intermediates for building complex molecules.
Why Two Methods?
Acid-Catalyzed
Uses \(\text{H}_3\text{O}^+\) or dilute \(\text{H}_2\text{SO}_4\). Follows Markovnikov's rule. Risk of carbocation rearrangements.
Oxymercuration
Uses \(\text{Hg(OAc)}_2\), then \(\text{NaBH}_4, \text{NaOH}\). Markovnikov product without rearrangement. Two-step process.
Markovnikov's Rule Refresher
Acid-Catalyzed Hydration: The Mechanism
1
Protonation of the alkene: The \(\pi\) bond acts as a nucleophile and attacks \(\text{H}^+\) from the acid, forming a carbocation on the more substituted carbon (Markovnikov).
2
Nucleophilic attack by water: \(\text{H}_2\text{O}\) attacks the carbocation, forming an oxonium ion (protonated alcohol).
3
Deprotonation: A base (water or conjugate base) removes a proton to yield the neutral alcohol and regenerate \(\text{H}^+\).
Mechanism Diagram: Acid-Catalyzed
Oxymercuration-Demercuration
This two-step sequence avoids carbocation rearrangements entirely. It uses a mercuric acetate reagent in Step 1, followed by reduction with sodium borohydride in Step 2.
1
Oxymercuration: \(\text{Hg(OAc)}_2\) in \(\text{H}_2\text{O/THF}\) forms a mercurinium ion bridge across the double bond. Water attacks the more substituted carbon (Markovnikov).
2
Demercuration: \(\text{NaBH}_4, \text{NaOH}\) replaces the \(\text{-HgOAc}\) group with \(\text{-H}\), giving the final Markovnikov alcohol.
The Mercurinium Ion Bridge
Key Distinction: Carbocation vs. Bridge
Acid-Catalyzed
Forms a free carbocation. The positive charge sits on one carbon, which can rearrange via hydride or methyl shifts to a more stable carbocation.
Oxymercuration
Forms a mercurinium bridge. The positive charge is distributed across the Hg and two carbons, preventing rearrangement. The nucleophile (\(\text{H}_2\text{O}\)) attacks the more substituted carbon, leading to an overall Markovnikov addition of \(\text{-H}\) and \(\text{-OH}\).
How to Choose Your Method
1
Draw the alkene and identify the more substituted carbon of the \(\text{C}=\text{C}\).
2
Check for rearrangement risk: Would the carbocation formed by protonation be adjacent to a more stable position? If yes, consider oxymercuration.
3
Place \(\text{-OH}\) on the more substituted carbon (Markovnikov) for both methods.
4
Write reagents: Acid-catalyzed: \(\text{H}_2\text{O}, \text{H}_2\text{SO}_4\). Oxymercuration: (1) \(\text{Hg(OAc)}_2, \text{H}_2\text{O/THF}\), (2) \(\text{NaBH}_4, \text{NaOH}\).
5
Verify product: For oxymercuration, confirm no rearrangement occurred. For acid-catalyzed hydration, check if a more stable carbocation could form via a shift, and draw the rearranged product if so.
Example 1: Hydration of Propene
Example 2: When Rearrangement Strikes
Consider 3,3-dimethyl-1-butene with acid-catalyzed hydration:
Example 3: Oxymercuration Saves the Day
Same substrate: 3,3-dimethyl-1-butene. Now using oxymercuration-demercuration:
A Note on Stereochemistry
Acid-Catalyzed
The carbocation intermediate is planar (\(\text{sp}^2\) hybridized). Water can attack from either face, so the product is a racemic mixture (not stereospecific).
Oxymercuration
Water attacks anti to the mercurinium bridge (backside attack). However, demercuration is not stereospecific, so the overall result is also a mixture.
Common Mistakes
✕ Anti-Markovnikov Placement
Placing the \(\text{-OH}\) on the less substituted carbon. Both methods give Markovnikov products. Anti-Markovnikov requires hydroboration-oxidation.
✕ Ignoring Rearrangements
Using acid-catalyzed hydration on substrates prone to shifts (neopentyl, cyclopropylmethyl) and predicting the unrearranged product.
✕ Wrong Reagents
Confusing oxymercuration reagents with hydroboration: \(\text{Hg(OAc)}_2\) is NOT the same as \(\text{BH}_3\). Remember: Hg = Markovnikov, B = anti-Markovnikov.
Comparison Summary
| Acid-Catalyzed | Oxymercuration | |
|---|---|---|
| Reagents | \(\text{H}_2\text{O}, \text{H}^+\) | (1) \(\text{Hg(OAc)}_2\), \(\text{H}_2\text{O}\) (2) \(\text{NaBH}_4, \text{NaOH}\) |
| Regioselectivity | Markovnikov | Markovnikov |
| Intermediate | Carbocation | Mercurinium ion |
| Rearrangements | Yes (possible) | No |
| Stereochemistry | Racemic mixture | Mixture (not stereospecific) |
| Steps | One-step | Two-step |
| Best for | Simple alkenes | Rearrangement-prone substrates |
Question 1 of 5
What is the major product of acid-catalyzed hydration of 1-methylcyclohexene?
Markovnikov's rule places the \(\text{-OH}\) on the more substituted carbon (C1, which already bears the methyl group), giving 1-methylcyclohexanol.
Question 2 of 5
Why does oxymercuration-demercuration avoid carbocation rearrangements?
The mercurinium ion is a three-membered ring bridging both carbons. Because no free carbocation exists, there is no driving force for hydride or methyl shifts.
Question 3 of 5
Acid-catalyzed hydration of 3,3-dimethyl-1-butene gives an unexpected product. What is it?
The initial \(2^\circ\) carbocation at C2 undergoes a methyl shift to form a more stable \(3^\circ\) carbocation at C3. Water then attacks C3 to give 2,3-dimethyl-2-butanol.
Question 4 of 5
Which reagent set gives a Markovnikov alcohol without rearrangement?
Choice A is hydroboration-oxidation (anti-Markovnikov). Choice C is acid-catalyzed (Markovnikov but can rearrange). Choice D is dihydroxylation. Only oxymercuration-demercuration (Choice B) gives Markovnikov without rearrangement.
Question 5 of 5
In acid-catalyzed hydration, \(\text{H}^+\) is described as a catalyst. Why?
A catalyst speeds up a reaction without being consumed overall. \(\text{H}^+\) protonates the alkene (consumed), then is released during deprotonation of the oxonium ion (regenerated). The net reaction is simply \(\text{alkene} + \text{H}_2\text{O} \rightarrow \text{alcohol}\).