Consider the above reaction, the major product P formed is
(a) CH3-CH(Br)-C(=CH2)-COOCH3
(b) CH3-CH=

Q: Why does the H+ attack the OH group and not the double bond or the ester?

While the double bond and ester oxygen can also be protonated, protonation of the alcohol OH group leads to the formation of a good leaving group (water). This pathway has a lower activation energy for the subsequent steps, driving the reaction forward.

Q: Isn't a secondary carbocation more stable than a primary one? Why does it rearrange?

Normally, yes. However, in this specific molecule, the rearrangement creates a primary carbocation that is part of an extended conjugated system with the ester's carbonyl group. This conjugation provides massive thermodynamic stability, far outweighing the typical primary vs. secondary difference.

Q: Why doesn't the bromide ion attack the secondary carbocation before it rearranges?

Carbocation rearrangements, especially those involving resonance (allylic shifts), are extremely fast—often faster than the nucleophilic attack. The reaction is thermodynamically controlled, leading to the most stable product.

Q: What is the role of the ester group in this reaction?

The ester group plays a crucial role in stabilizing the rearranged carbocation through extended conjugation. Without it, the primary carbocation would not be stable enough to form, and the reaction might yield a different major product.

Q: Can this reaction occur with HCl instead of HBr?

Yes, a similar reaction would occur with concentrated HCl, yielding the corresponding chloride product. The mechanism involving protonation, water loss, resonance, and nucleophilic attack remains the same.

Question

Consider the above reaction, the major product P formed is

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Welcome to a fascinating journey into the world of organic reaction mechanisms! Today, we are looking at a very interesting molecule: an allylic alcohol that also features an ester group. We are treating this molecule with concentrated hydrobromic acid ($HBr$).

Whenever you see an alcohol in the presence of a strong acid, your first instinct should be to look for protonation. The hydroxyl ($-OH$) group is a poor leaving group, but it has lone pairs on the oxygen atom that are eager to grab a proton.

### The Protonation Step
The strong acid provides an abundance of $H^+$ ions. The oxygen atom of the hydroxyl group acts as a nucleophile and attacks a proton.

This simple step is incredibly powerful. It transforms the poor $-OH$ leaving group into an oxonium ion ($-O^+H_2$). Oxygen is highly electronegative and strongly dislikes bearing a positive charge. It begins to pull the electron density from the carbon-oxygen bond towards itself, weakening the bond.

### Formation of the Carbocation
Unable to sustain the positive charge, the water molecule departs, taking the bonding electrons with it. This leaves behind a carbon atom with an incomplete octet, creating a secondary carbocation.

Now, a golden rule of organic chemistry is to always check if a newly formed carbocation can rearrange to a more stable form. Our secondary carbocation is located right next to a carbon-carbon double bond. This makes it an **allylic carbocation**.

### The Power of Resonance and Conjugation
Because it is allylic, the carbocation can undergo resonance. The $\pi$ electrons from the adjacent double bond shift over to fill the empty p-orbital. This movement shifts the positive charge to the terminal carbon atom, creating a primary carbocation.

Wait a minute! Isn't a primary carbocation less stable than a secondary one? Normally, yes. However, we must look at the bigger picture. The new double bond formed by this shift is now in direct conjugation with the carbonyl group ($C=O$) of the ester.

This creates an $\alpha,\beta$-unsaturated ester system. The extended conjugation provides massive thermodynamic stability to the entire molecule. This stabilizing effect far outweighs the typical energy difference between primary and secondary carbocations, making this resonance structure highly favored.

### The Final Nucleophilic Attack
With our highly stable, rearranged carbocation in place, the stage is set for the final act. The bromide ion ($Br^-$), which was left behind when the acid donated its proton, is a strong nucleophile.

It attacks the positively charged terminal carbon atom. This yields our final major product: an $\alpha,\beta$-unsaturated ester with a bromine atom attached to the end of the chain.

This reaction beautifully illustrates how thermodynamic stability, specifically through extended conjugation, can drive a reaction pathway and dictate the final structure of the product. Always keep an eye out for these hidden stabilizing factors!

Animated Solution for Chemistry - Organic Chemistry: Consider the above reaction, the major product P formed is

Select Answer:

Visualized Solution

Analyzing the Reactant

Protonation of Alcohol

Formation of Oxonium Ion

Leaving Group Departs

Resonance Stabilization

Nucleophilic Attack

Key Takeaways

The Sigma Insight: Alcohols, Phenols, Ethers

Analyzing the Setup Welcome to a fascinating journey into the world of organic reaction mechanisms! Today, we are looking at a very interesting molecule: an allylic alcohol that also features an ester group

The Protonation Step The strong acid provides an abundance of $H^{+}$ ions

Formation of the Carbocation Unable to sustain the positive charge, the water molecule departs, taking the bonding electrons with it

The Power of Resonance and Conjugation Because it is allylic, the carbocation can undergo resonance

The Final Nucleophilic Attack With our highly stable, rearranged carbocation in place, the stage is set for the final act

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Animated Solution for Chemistry - Organic Chemistry: Consider the above reaction, the major product P formed is

Select Answer:

Visualized Solution

Analyzing the Reactant

Protonation of Alcohol

Formation of Oxonium Ion

Leaving Group Departs

Resonance Stabilization

Nucleophilic Attack

Key Takeaways

The Sigma Insight: Alcohols, Phenols, Ethers

Analyzing the Setup Welcome to a fascinating journey into the world of organic reaction mechanisms! Today, we are looking at a very interesting molecule: an allylic alcohol that also features an ester group

The Protonation Step The strong acid provides an abundance of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H+ ions

Formation of the Carbocation Unable to sustain the positive charge, the water molecule departs, taking the bonding electrons with it

The Power of Resonance and Conjugation Because it is allylic, the carbocation can undergo resonance

The Final Nucleophilic Attack With our highly stable, rearranged carbocation in place, the stage is set for the final act

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The Protonation Step The strong acid provides an abundance of $H^{+}$ ions