When neopentyl alcohol is heated with an acid, it slowly converted into an 85 : 15 mixture of alkene

Q: Why doesn't a hydride shift happen instead of a methyl shift?

A hydride shift requires a hydrogen atom on the adjacent carbon. In the neopentyl carbocation, the adjacent carbon is a quaternary carbon bonded to three methyl groups and zero hydrogens. Therefore, a methyl shift is the only possible rearrangement.

Q: Why is the tertiary carbocation so much more stable than the primary one?

Tertiary carbocations are stabilized by two main factors: the inductive effect (alkyl groups push electron density toward the positive charge) and hyperconjugation (overlap of adjacent C-H or C-C sigma bonds with the empty p-orbital of the carbocation). A tertiary carbocation has more of these stabilizing interactions than a primary one.

Q: What exactly is Saytzeff's rule?

Saytzeff's rule (or Zaitsev's rule) states that in an elimination reaction, the most substituted alkene will be the major product. 'More substituted' means there are more alkyl groups attached to the carbons of the double bond. These alkyl groups stabilize the double bond through hyperconjugation.

Q: Could the primary carbocation just eliminate a proton directly without rearranging?

Technically, yes, but the activation energy for that direct elimination is much higher than the activation energy for the rearrangement. Because the rearrangement leads to a significantly more stable intermediate, it happens much faster, making the rearranged products the overwhelmingly dominant ones.

Q: How do I know which product is A and which is B?

The problem states the mixture is 85:15 of A and B. This means A is the major product and B is the minor product. Based on Saytzeff's rule, the more substituted alkene (2-methylbut-2-ene) is the major product (A), and the less substituted one (2-methylbut-1-ene) is the minor product (B).

Question

When neopentyl alcohol is heated with an acid, it slowly converted into an 85 : 15 mixture of alkenes A and B, respectively. What are these alkenes?

Sigma Percentile · Accepted Answer

### The Dehydration of Neopentyl Alcohol: A Tale of Rearrangement

When we heat an alcohol with a strong acid like $H_2SO_4$, we are setting the stage for an acid-catalyzed dehydration reaction. The goal is to remove a molecule of water and form a double bond, creating an alkene. However, when our starting material is **neopentyl alcohol** (2,2-dimethylpropan-1-ol), the reaction takes a fascinating detour before reaching its final destination.

### The Initial Setup: Protonation and Departure

The reaction begins with the protonation of the hydroxyl ($-OH$) group. The oxygen atom uses its lone pair to grab a proton ($H^+$) from the acid, transforming into $-OH_2^+$. This is a crucial step because the hydroxide ion is a terrible leaving group, but water is an excellent, stable leaving group.

Once protonated, the water molecule departs, taking its bonding electrons with it. This leaves behind a **primary ($1^\circ$) carbocation**: $(CH_3)_3C-CH_2^+$.

### The Plot Twist: Carbocation Rearrangement

Nature abhors high-energy states, and a primary carbocation is highly unstable due to the lack of sufficient hyperconjugation and inductive stabilization. The molecule desperately seeks a way to lower its energy.

Notice the adjacent carbon atom—it is a quaternary carbon bonded to three methyl groups. To achieve stability, the molecule undergoes a **1,2-methyl shift**. An entire methyl group ($-CH_3$), along with its bonding pair of electrons, migrates from the adjacent carbon to the positively charged primary carbon.

This elegant internal rearrangement transforms the unstable primary carbocation into a highly stable **tertiary ($3^\circ$) carbocation**: $CH_3-C^+(CH_3)-CH_2-CH_3$. The positive charge is now stabilized by the electron-donating inductive effect of three alkyl groups and extensive hyperconjugation.

### The Finale: Elimination and Saytzeff's Rule

With a stable tertiary carbocation formed, the final step is the elimination of a proton ($H^+$) from an adjacent carbon to form the carbon-carbon double bond. Looking at our tertiary carbocation, there are two types of adjacent carbons that can lose a proton:

1. **The $CH_2$ group:** Removing a proton from here yields $CH_3-C(CH_3)=CH-CH_3$ (2-methylbut-2-ene). This is a highly substituted (trisubstituted) alkene.
2. **The $CH_3$ groups:** Removing a proton from one of the two equivalent methyl groups on the left yields $CH_2=C(CH_3)-CH_2-CH_3$ (2-methylbut-1-ene). This is a less substituted (disubstituted) alkene.

According to **Saytzeff's Rule**, elimination reactions under thermodynamic control (like heating with acid) will favor the formation of the more highly substituted, and therefore more stable, alkene.

Thus, the trisubstituted alkene, **2-methylbut-2-ene**, is the major product (Product A, 85%). The disubstituted alkene, **2-methylbut-1-ene**, is the minor product (Product B, 15%).

Animated Solution for Chemistry - Organic Chemistry: When neopentyl alcohol is heated with an acid, it slowly converted into an 85 : 15 mixture of alkenes A and B, respectively. What are these alkenes?

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Visualized Solution

$(CH_{3})_{3} C - CH_{2} OH$

$Acid-Catalyzed Dehydration$

$Formation of 1^{\circ} Carbocation$

$Carbocation Rearrangement$

$Elimination (Saytzeff Product)$

$Elimination (Hofmann Product)$

$Identifying A and B$

$Thermodynamic vs Kinetic Control$

The Sigma Insight: Hydrocarbons

The Dehydration of Neopentyl Alcohol

The Initial Setup

The Plot Twist

The Finale

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