The major product obtained in the following reaction is
Reactant + DIBAL-H - ?

Q: Why doesn't DIBAL-H reduce the double bond?

DIBAL-H is an electrophilic reducing agent that targets polar multiple bonds like C=O and C≡N. Isolated, non-polar C=C double bonds are generally unaffected.

Q: What if we used LiAlH4 instead of DIBAL-H?

Lithium aluminum hydride is a much stronger reducing agent. It would reduce the lactone completely to a diol, converting the carbonyl group into a primary alcohol (-CH2OH).

Q: How do we know which bond in the lactone breaks?

The reduction involves the delivery of a hydride to the electrophilic carbonyl carbon, followed by the collapse of the tetrahedral intermediate which expels the ring oxygen as a leaving group. Thus, the acyl C-O bond breaks.

Q: Is this a beta-lactone or a gamma-lactone?

By counting the atoms in the ring containing the oxygen and carbonyl, we see it has 4 atoms. This makes it a beta-lactone.

Q: Why doesn't the aldehyde get further reduced to an alcohol by DIBAL-H?

At low temperatures (typically -78°C), the tetrahedral intermediate formed after the first hydride addition is stable. It only collapses to form the aldehyde during the aqueous workup, after the DIBAL-H has been destroyed.

Question

The major product obtained in the following reaction is

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are looking at a fascinating molecular architecture: a bicyclic compound where a five-membered cyclopentene ring is fused to a four-membered lactone ring. This is our starting material. We are treating this molecule with **DIBAL-H** (Diisobutylaluminium hydride), a very specific and highly useful reducing agent in organic chemistry.

### The Master Reagent: DIBAL-H
What makes DIBAL-H so special? Unlike its aggressive cousin, Lithium Aluminum Hydride ($LiAlH_4$), DIBAL-H is a selective reducing agent. When used at low temperatures (typically $-78^\circ C$), it has the remarkable ability to reduce esters and lactones directly to **aldehydes**, stopping the reduction process right there without proceeding all the way to primary alcohols.

### Cracking Open the Lactone
Let's focus our attention on the four-membered lactone ring. A lactone is essentially a cyclic ester. During the reduction process, DIBAL-H delivers a hydride ion to the electrophilic carbonyl carbon. This forms a stable tetrahedral intermediate.

Upon aqueous workup, this intermediate collapses, leading to the cleavage of the acyl carbon-oxygen single bond.

The carbonyl carbon is transformed into an aldehyde group ($-CHO$), and the ring oxygen atom is protonated to become a hydroxyl group ($-OH$).

### Tracing the Atoms
If we trace the atoms in our specific bicyclic system, the top bridgehead carbon was attached to the ring oxygen. Therefore, after the bond breaks, this top bridgehead carbon will now hold the newly formed $-OH$ group.

Conversely, the bottom bridgehead carbon was part of the carbonyl group. After reduction, this carbon becomes the $-CHO$ group.

### The Fate of the Double Bond
But what about the double bond in the cyclopentene ring? This is where the selectivity of DIBAL-H shines again. DIBAL-H is an electrophilic reducing agent and generally does not interact with isolated, non-polar carbon-carbon double bonds.

Therefore, the $C=C$ double bond remains completely unaffected and stays exactly in its original position on the left side of the ring.

### Final Conclusion
Putting all these pieces together, our major product will have a hydroxyl group at the top bridgehead, an aldehyde group at the bottom bridgehead, and the double bond intact on the left side. This perfectly matches the structure shown in **Option (a)**.

Animated Solution for Chemistry - Aldehydes and Ketones: The major product obtained in the following reaction is

Select Answer:

Visualized Solution

Visual Anchor

Role of DIBAL-H

Lactone Cleavage

Product Formation

Double Bond Unaffected

Final Answer

The Way Forward

The Sigma Insight: Preparation of Aldehydes and Ketones

Analyzing the Setup Imagine you are looking at a fascinating molecular architecture: a bicyclic compound where a five-membered cyclopentene ring is fused to a four-membered lactone ring

The Master Reagent

Cracking Open the Lactone Let's focus our attention on the four-membered lactone ring

Tracing the Atoms If we trace the atoms in our specific bicyclic system, the top bridgehead carbon was attached to the ring oxygen

The Fate of the Double Bond But what about the double bond in the cyclopentene ring? This is where the selectivity of DIBAL-H shines again

Final Conclusion Putting all these pieces together, our major product will have a hydroxyl group at the top bridgehead, an aldehyde group at the bottom bridgehead, and the double bond intact on the left side

Similar Questions

Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in the presence of $HgSO_{4}$ and $H_{2} SO_{4}$ ?