2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields

I. C2H5CH2C(CH3)2-OCH

Q: Why doesn't $S_N2$ happen here?

The alkyl halide is tertiary ($3^\circ$). The carbon bonded to the chlorine is surrounded by three bulky alkyl groups, creating massive steric hindrance that completely blocks the incoming nucleophile from attacking directly.

Q: Why is elimination the major product?

Sodium methoxide is a strong base. When a strong base reacts with a tertiary halide, the activation energy for abstracting a proton (elimination) is much lower than waiting for the carbocation to form or trying to bypass steric hindrance. Thus, $E2$ elimination dominates.

Q: What is the difference between Hofmann and Zaitsev products?

The Zaitsev product is the more substituted alkene, formed by removing a proton from the $\beta$-carbon with fewer hydrogens. The Hofmann product is the less substituted alkene, formed by removing a proton from the $\beta$-carbon with more hydrogens.

Q: Is the substitution product formed via $S_N1$ or $S_N2$?

It is formed via the $S_N1$ mechanism. The tertiary halide first loses the chloride ion to form a stable tertiary carbocation, which is then attacked by the methoxide nucleophile.

Q: Why do we get all three products instead of just the major one?

Chemical reactions are statistical processes. While the Zaitsev elimination might have the lowest activation energy and form the major product, the other pathways (Hofmann elimination and $S_N1$ substitution) still have accessible activation energies under these conditions, leading to a mixture of all three.

Question

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields

I. $C_2H_5CH_2\underset{\overset{|}{CH_3}}{\overset{\underset{|}{CH_3}}{C}}-OCH_3$

II. $C_2H_5CH_2\underset{\overset{|}{CH_3}}{C}=CH_2$

III. $C_2H_5CH=\underset{\overset{|}{CH_3}}{C}-CH_3$

Options:

(A) Both I and III

(B) Only III

(C) Both I and II

(D) All of the above

Sigma Percentile · Accepted Answer

### The Setup: A Classic Battleground

Imagine you are a tiny molecule of **2-chloro-2-methylpentane**. You are a tertiary alkyl halide, meaning the carbon atom holding your chlorine is surrounded by three other bulky carbon groups.

You are swimming in a flask of methanol, and suddenly, you encounter a fierce opponent: the **methoxide ion** ($CH_3O^-$).

This ion is not just a strong nucleophile looking for a positive center to attack; it is also a very strong base, hungry for protons. When a tertiary halide meets a strong base, a classic chemical battle begins: **Substitution vs. Elimination**.

### The Substitution Pathway ($S_N1$)

Our methoxide ion wants to attack the carbon holding the chlorine to perform an $S_N2$ substitution. But there is a massive problem: **steric hindrance**.

The three bulky alkyl groups act like bouncers at a club, completely blocking the methoxide from attacking from the back. Direct $S_N2$ is impossible!

However, the chlorine atom can leave on its own, taking its electrons to form a stable tertiary carbocation.

$$CH_3-C^+(CH_3)-CH_2CH_2CH_3$$

Once this stable carbocation is formed, the methoxide ion can easily swoop in and attach itself, forming an ether. This is the $S_N1$ pathway, and it gives us our first product: **2-methoxy-2-methylpentane** (Product I).

### The Elimination Pathway ($E2$)

But wait, the methoxide ion is a strong base! It doesn't just want to wait around for a carbocation to form. It wants action *now*.

Instead of attacking the crowded carbon, it looks at the neighboring carbons (the $\beta$-carbons) and spots their hydrogens ($\beta$-hydrogens).

By snatching a $\beta$-hydrogen, the methoxide can force the chlorine to leave simultaneously, creating a double bond. This is the $E2$ elimination mechanism.

Let's hunt for these $\beta$-hydrogens. We have two distinct types in our molecule.

### Hofmann vs. Zaitsev: The Tale of Two Alkenes

First, we have the primary $\beta$-hydrogens located on the terminal methyl groups.

If the methoxide base abstracts one of these easily accessible protons, the double bond forms at the very edge of the carbon chain. This results in a less substituted alkene, famously known as the **Hofmann product**.

This gives us Product II: **2-methylpent-1-ene**.

On the flip side, we have secondary $\beta$-hydrogens located on the internal $CH_2$ group.

If the base snatches one of these protons, the double bond forms in the middle of the chain. This creates a more substituted, highly stable alkene, known as the **Zaitsev product**.

This gives us Product III: **2-methylpent-2-ene**.

### The Final Verdict

So, what is the final outcome of this fierce battle?

Because we are using a strong base, the $E2$ elimination pathway will dominate, making the alkenes the major products (with the Zaitsev product usually being the most abundant due to its stability).

However, the $S_N1$ substitution pathway still occurs to a lesser extent, yielding the ether.

Chemical reactions are statistical processes, and under these conditions, all three pathways have accessible activation energies. Therefore, the reaction yields **all three products**.

The correct answer is **All of the above**!

Animated Solution for Chemistry - Organic Chemistry: 2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields I. $C_{2} H_{5} C H_{2} C H_{3} ∣ C ∣ C H_{3} - O C H_{3}$ II. $C_{2} H_{5} C H_{2} C H_{3} ∣ C = C H_{2}$ III. $C_{2} H_{5} C H = C H_{3} ∣ C - C H_{3}$

Select Answer:

Visualized Solution

$2$ -chloro- $2$ -methylpentane

Reagent Analysis

Substitution Pathway ( $S_{N} 1$ )

Identifying $β$ -Hydrogens

Hofmann Elimination

Zaitsev Elimination

Final Conclusion

The Sigma Insight: Haloalkanes & Haloarenes

The Setup

The Substitution Pathway ( $S_{N} 1$ )

The Elimination Pathway ( $E 2$ )

Hofmann vs

The Final Verdict

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