Below is one way that a simple 1,2‑diol can “capture’’ the two carbonyl groups of a diketone to give a **bicyclic diol‑diketone**.
(The exact product you obtain will depend on where the two hydroxyls and the two carbonyls are situated, but the logic of the mechanism and the retention of the four oxygens is the same.)
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1. The product that is obtained
O O
/ \ / \
O - C1‑C2‑C3‑C4‑C5‑C6–O
| | |
H H H
*The product is a six‑membered ring that has two fused rings:*
- A diol ring (the two –OH groups now act as an O‑bridge between C1 and C2).
- A diketone ring (the two carbonyl groups at C3 and C4 are still C=O).
A systematic name is **1,2‑dihydroxy‑3,4‑diketo‑cyclohexane** (sometimes written *1,2‑dihydroxy‑3,4‑dioxo‑cyclohexane*).
The key point is that the molecule still contains **four oxygen atoms** – two from the original diol and two from the original diketone – but all four are still present in the bicyclic skeleton.
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2. Step‑by‑step mechanism
> **Step A – First nucleophilic attack**
The lone‑pair on the OH of C1 attacks the carbonyl carbon of the diketone (C3). A tetrahedral intermediate is formed, and the C=O double bond is temporarily converted to an alkoxide.
O–C1 + O=C3–C4
/ \ \
H H H
↗
O–C1‑C3
| |
H O–C4
> **Step B – Collapse to give a hemiacetal**
The alkoxide (the oxygen that was originally part of the diol) collapses, breaking the C–O bond to the water that will leave. This gives a **hemiacetal** at C3 and releases a molecule of water (or proton‑assisted elimination if you wish to be rigorous).
O–C1‑C3
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H O–C4
> **Step C – Second nucleophilic attack**
Now the second hydroxyl group (on C2) attacks the remaining carbonyl carbon (C4). Again a tetrahedral intermediate is formed and collapses, yielding a second hemiacetal.
O–C1‑C3
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H O–C4
|
H
> **Step D – Cyclization / dehydration**
The two hemiacetals are now close enough that one can dehydrate the other, forming an **acyclic bridging ether** that connects the two rings together. In effect the two –OH groups have become ether bridges that fuse the two rings, while the two original C=O groups are left intact as carbonyls.
O–C1–C3
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H O
|
O–C2–C4
The final structure is a **bicyclic acetal/ketal** in which all four oxygen atoms of the starting materials are still present:
- two of them are the ether‑bridge oxygens that came from the diol,
- two of them are the carbonyl oxygens that came from the diketone.
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3. Why does the product still have four oxygens?
Because the reaction is a **concerted intramolecular nucleophilic attack followed by elimination of water** – it does not involve any oxidation, reduction, or oxidation‑dehydration that would remove an oxygen atom.
Both the diol’s two oxygens and the diketone’s two carbonyl oxygens are simply reorganized into new σ‑bonds (ether bridges and the remaining carbonyls) while all four atoms of oxygen survive intact.
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Bottom line:
*The bicyclic diol‑diketone is formed by intramolecular nucleophilic addition of the two hydroxyls of the diol to the two carbonyl carbons of the diketone, followed by loss of water. All four oxygen atoms of the starting diol (two O’s) and the diketone (two O’s) are retained in the final bicyclic skeleton.*
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