Heating of 2-chloro-1-phenyl butane with EtOK/EtOH gives X as the major product. Reaction of X with

Q: Why doesn't the base abstract a proton from the CH2 group on the right?

Abstracting a proton from the right CH2 group would form 1-phenylbut-2-ene. While possible, abstracting the benzylic proton forms 1-phenylbut-1-ene, which is conjugated with the benzene ring. Conjugated alkenes are significantly more stable, making this the major product.

Q: Why doesn't Oxymercuration involve carbocation rearrangement?

Instead of a free carbocation, oxymercuration forms a cyclic mercurinium ion intermediate. This three-membered ring prevents the carbon skeleton from rearranging, ensuring the addition happens exactly where the double bond was.

Q: How does Markovnikov's rule apply to the mercurinium ion?

The cyclic mercurinium ion is asymmetric. The carbon that can better stabilize a positive charge (in this case, the benzylic carbon) will bear a larger share of the partial positive charge ($\delta+$). The nucleophile (water) attacks this more electrophilic carbon.

Q: What is the role of NaBH4 in the final step?

NaBH4 is a reducing agent. It replaces the mercury-containing group ($-HgOAc$) with a hydrogen atom via a radical mechanism, completing the addition of water across the double bond.

Q: Could we just use dilute H2SO4 to get the same product?

In this specific molecule, acid-catalyzed hydration (dilute H2SO4) would form a benzylic carbocation. Since the benzylic carbocation is already highly stable, it might not rearrange, and you could get the same product. However, Oxymercuration is generally preferred in synthesis to absolutely guarantee no rearrangements occur.

Question

Heating of 2-chloro-1-phenyl butane with EtOK/EtOH gives $X$ as the major product. Reaction of $X$ with $	ext{Hg(OAc)}_2 / 	ext{H}_2	ext{O}$ followed by $	ext{NaBH}_4$ gives $Y$ as the major product. $Y$ is

Sigma Percentile · Accepted Answer

### The Setup: E2 Elimination

We begin our journey with **2-chloro-1-phenylbutane**. When this molecule is subjected to heat in the presence of a strong base like potassium ethoxide in ethanol ($	ext{EtOK}/	ext{EtOH}$), it undergoes a classic dehydrohalogenation via the **E2 mechanism**.

The critical question here is regioselectivity: which proton will the base abstract? The base has a choice between the protons on the $C_1$ (benzylic) position and the $C_3$ position. Abstracting a proton from $C_3$ would yield 1-phenylbut-2-ene. However, abstracting the benzylic proton from $C_1$ yields **1-phenylbut-1-ene**.

Why is the latter preferred? The double bond in 1-phenylbut-1-ene is in direct **conjugation** with the $\pi$-system of the benzene ring. This extended delocalization provides immense thermodynamic stability, making it the overwhelmingly favored major product, which we call **$X$**.

### The Transformation: Oxymercuration-Demercuration

Now that we have our conjugated alkene $X$, we subject it to $	ext{Hg(OAc)}_2$ and $	ext{H}_2	ext{O}$. This initiates the **Oxymercuration** sequence. Unlike acid-catalyzed hydration which forms a free carbocation, the electrophilic mercury species adds across the double bond to form a three-membered cyclic **mercurinium ion**.

This cyclic intermediate is the secret to why this reaction never undergoes carbocation rearrangements. However, the ring is not perfectly symmetrical. The carbon atom that can better stabilize a positive charge will bear a larger fraction of the partial positive charge ($\delta+$).

In our molecule, the benzylic carbon ($C_1$) is adjacent to the phenyl ring, which stabilizes the $\delta+$ through resonance. Consequently, the nucleophilic water molecule attacks this highly electrophilic benzylic carbon, popping the three-membered ring open.

### The Finale: Demercuration

With the hydroxyl group now firmly attached to the benzylic position, we are left with an organomercury intermediate. The final step is **Demercuration**, achieved by adding the reducing agent sodium borohydride ($	ext{NaBH}_4$).

$	ext{NaBH}_4$ replaces the $-	ext{HgOAc}$ group with a simple hydrogen atom via a radical mechanism. The net result of this entire two-step sequence is the Markovnikov addition of water across the double bond without any risk of skeletal rearrangement.

Our final major product **$Y$** is **1-phenylbutan-1-ol**, perfectly matching option (c).

Animated Solution for Chemistry - Organic Chemistry: Heating of 2-chloro-1-phenyl butane with EtOK/EtOH gives $X$ as the major product. Reaction of $X$ with $Hg(OAc)_{2} / H_{2} O$ followed by $NaBH_{4}$ gives $Y$ as the major product. $Y$ is

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Visualized Solution

$A na l y z in g t h e R e a c t an t$

$E 2 E l imina t i o n$

$F or ma t i o n o f A l k e n e X$

$O x y m er c u r a t i o n$

$R e g i ose l ec t i v i t y o f A tt a c k$

$D e m er c u r a t i o n t o P r o d u c t Y$

$Summary\ &\ Variations$

The Sigma Insight: Haloalkanes & Haloarenes

The Setup

The Transformation

The Finale

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