The major product of the following reaction is
[Reactant Image] - (H2SO4) - ?
(a) [Image]
(b) [Ima

Q: Why doesn't the $NO_2$ group react with the acid?

The nitro group ($-NO_2$) is strongly electron-withdrawing and highly deactivated. It does not readily undergo protonation or electrophilic attack under these conditions compared to the much more reactive $\pi$-electrons of the alkene.

Q: Why does the carbocation form at the secondary carbon instead of the primary one?

According to Markovnikov's rule, protonation occurs to yield the most stable carbocation. A secondary carbocation is stabilized by hyperconjugation from adjacent alkyl groups, whereas a primary carbocation lacks this stabilization.

Q: Why does the attack happen at the ortho position and not the para position?

The attack is intramolecular. The carbocation is tethered to the ring via a short alkyl chain, making the ortho position geometrically accessible. Reaching the para position would require an impossibly strained, long-bridged ring.

Q: Could a 6-membered ring form instead?

Yes, a 6-membered ring can form if the primary carbocation is generated. However, because the secondary carbocation is formed much faster and is more stable, the 5-membered ring derived from it is the major product.

Q: Why does the intermediate lose a proton at the end?

The intermediate (sigma complex) is non-aromatic and relatively high in energy. By losing a proton, the electrons return to the ring, restoring the highly stable, delocalized aromatic $\pi$-system.

Question

The major product of the following reaction is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to a beautiful organic chemistry puzzle! We are given a fascinating reactant: a benzene ring equipped with a strongly deactivating nitro group ($-NO_2$) and an electron-rich 3-butenyl side chain ($-CH_2-CH_2-CH=CH_2$) at the ortho position.

The reagent provided is sulfuric acid ($H_2SO_4$), which acts as a classic source of protons ($H^+$).

**Whenever you see an alkene in the presence of a strong acid, your first instinct should be electrophilic addition.** The setup is a perfect recipe for an intramolecular reaction, where the molecule will eventually react with itself!

### The Master Equation: Protonation and Carbocation Stability

The double bond in our side chain is packed with $\pi$-electrons, making it highly nucleophilic. It reaches out and attacks the $H^+$ ion from the acid.

But here is the critical question: which carbon of the double bond will take the hydrogen?

**According to Markovnikov's rule, the electrophile adds in a way that generates the most stable carbocation intermediate.**

If the hydrogen attaches to the inner carbon, we get a primary ($1^{\circ}$) carbocation. However, if it attaches to the terminal carbon, we create a secondary ($2^{\circ}$) carbocation on the inner carbon.

$$R-CH=CH_2 + H^+ \rightarrow R-\overset{+}{C}H-CH_3$$

**A secondary carbocation is significantly more stable than a primary one due to hyperconjugation and inductive effects.** Therefore, the $2^{\circ}$ carbocation is our major intermediate that will drive the rest of the mechanism.

### The Intramolecular Attack: Closing the Ring

Now, imagine the dynamic setup we have created. We have a highly reactive, electron-deficient carbocation dangling right next to an electron-rich benzene ring.

It is an intramolecular trap! The carbocation swoops in and attacks the $\pi$-electron cloud of the benzene ring.

It specifically targets the ortho position relative to the side chain. Conveniently, this position is also meta to the strongly deactivating $-NO_2$ group, making it the most favorable site for electrophilic aromatic substitution.

### Final Calculation: Restoring Aromaticity

Boom! The attack forms a brand new carbon-carbon single bond, elegantly closing the loop to create a five-membered ring fused to our benzene.

Notice how the terminal methyl group ($-CH_3$) is left hanging on the outside of this newly formed ring. This intermediate is known as a sigma complex (or arenium ion), but it has temporarily lost its aromaticity.

To regain that sweet, sweet aromatic stability, the intermediate rapidly sheds a proton ($H^+$).

**The major product is a fused five-membered ring with a methyl group, perfectly matching option (b).**

This is how you conquer complex organic mechanisms—step by logical step!

Animated Solution for Chemistry - Hydrocarbons: The major product of the following reaction is

Select Answer:

Visualized Solution

Analyzing the Reactant

Protonation of Alkene

Formation of $2^{\circ}$ Carbocation

Intramolecular Electrophilic Attack

Ring Closure

Restoring Aromaticity

The Sigma Insight: Aromatic Hydrocarbons

Analyzing the Setup

The Master Equation

The Intramolecular Attack

Final Calculation