Animated Solution for Chemistry - Organic Chemistry: The major product [C] of the following reaction sequence will be ${\text{CH}}_2=\text{CH}-\text{CHO}\underset{\text{(ii) }{\text{SOCl}}_2}{\overset{\text{(i) }{\text{NaBH}}_4}{\to }}[\text{A}]$ $\text{Benzene}+[\text{A}]\stackrel{\text{Anhy. }{\text{AlCl}}_3}{\to }[\text{B}]\stackrel{\text{DBr}}{\to }[\text{C}]$

From Reduction to Rearrangement

Organic chemistry is often a beautiful sequence of logical transformations, and this problem is a perfect showcase of that elegance. We are tasked with tracking a molecule through a multi-step journey involving reduction, substitution, electrophilic aromatic substitution, and finally, an electrophilic addition with a hidden trap.

Our starting material is an $\alpha ,\beta$-unsaturated aldehyde: ${\text{CH}}_2=\text{CH}-\text{CHO}$. The first reagent thrown at it is sodium borohydride (${\text{NaBH}}_4$).

${\text{NaBH}}_4$ is a mild and highly selective reducing agent. It acts as a source of hydride ions (${\text{H}}^{-}$), which are nucleophilic. Because the carbon-carbon double bond is also electron-rich, the hydride ion is repelled by it. However, the carbonyl carbon is highly electrophilic. Thus, ${\text{NaBH}}_4$ selectively reduces the aldehyde group to a primary alcohol, leaving the alkene untouched.

Following this, thionyl chloride (${\text{SOCl}}_2$) is used to substitute the newly formed hydroxyl group with a chlorine atom. This gives us our first major intermediate, allyl chloride (${\text{CH}}_2=\text{CH}-{\text{CH}}_2\text{Cl}$), labeled as [A].

Next, we introduce benzene and anhydrous aluminum chloride (${\text{AlCl}}_3$). This is the classic recipe for a Friedel-Crafts Alkylation.

Anhydrous ${\text{AlCl}}_3$ is a strong Lewis acid. It eagerly accepts the electron pair from the chlorine atom of allyl chloride, effectively ripping it off the molecule. This generates a highly reactive electrophile: the allyl carbocation (${\text{CH}}_2=\text{CH}-{\text{CH}}_2^{+}$).

This electrophile attacks the electron-rich $\pi$-cloud of the benzene ring. After the temporary loss of aromaticity, a proton is lost to restore the stable benzene ring, yielding allylbenzene ($\text{Ph}-{\text{CH}}_2-\text{CH}={\text{CH}}_2$). This is our intermediate [B].

The final step involves the addition of deuterium bromide ($\text{DBr}$) to allylbenzene. This is an electrophilic addition reaction that follows Markovnikov's rule.

The electrophile, ${\text{D}}^{+}$, attacks the double bond. It adds to the terminal carbon to generate the more stable secondary carbocation at the adjacent position:

Many students would immediately attach the bromide ion to this secondary carbocation. However, we must always look for opportunities to form a more stable carbocation. Right next to our secondary positive charge is a benzylic position.

By performing a $1,2-\text{hydride shift}$, a hydrogen atom (with its bonding electrons) migrates from the benzylic carbon to the secondary carbon. This shifts the positive charge to the benzylic position:

This new benzylic carbocation is exceptionally stable because the positive charge is delocalized over the entire benzene ring via resonance.

With the most stable carbocation now formed, the nucleophilic bromide ion (${\text{Br}}^{-}$) attacks the benzylic position.

The final major product [C] is $\text{Ph}-\text{CH}(\text{Br})-{\text{CH}}_2-{\text{CH}}_2\text{D}$. This perfectly matches option (c), showcasing how a deep understanding of reaction mechanisms and intermediate stability is crucial for conquering organic chemistry.