Question: Find number of $\alpha$ - hydrogen in the major product A....

Question

Find number of $\alpha$ - hydrogen in the major product A.

Visual representation for Organic Chemistry

Answer 1

Solution

The given reaction is the acid-catalyzed dehydration of an alcohol.

1. Structure of the starting material: The starting material is 3,3-dimethylbutan-2-ol. Its structure is:

      CH₃
      |
CH₃ - C - CH - CH₃
      |   |
      CH₃ OH

Or, more clearly: $\text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}(\text{OH})-\text{CH}_3$

2. Mechanism of Acid-Catalyzed Dehydration:

Step 1: Protonation of the hydroxyl group. The oxygen atom of the hydroxyl group is protonated by $\text{H}_2\text{SO}_4$ (a strong acid) to form an oxonium ion, which is a good leaving group ( $\text{H}_2\text{O}$ ). $\text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}(\text{OH})-\text{CH}_3 + \text{H}^+ \rightleftharpoons \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}(\text{OH}_2^+)-\text{CH}_3$
Step 2: Loss of water to form a carbocation. The protonated alcohol loses a water molecule, forming a secondary carbocation. $\text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}(\text{OH}_2^+)-\text{CH}_3 \rightarrow \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}^+-\text{CH}_3 + \text{H}_2\text{O}$ The carbocation formed is a secondary carbocation: 3,3-dimethylbutan-2-yl cation.
```
      CH₃
      |
CH₃ - C - CH⁺ - CH₃
      |
      CH₃
```
Step 3: Carbocation Rearrangement (1,2-Methyl Shift). The secondary carbocation is adjacent to a quaternary carbon (the carbon with two methyl groups). A 1,2-methyl shift occurs from the quaternary carbon to the secondary carbocation, forming a more stable tertiary carbocation. This rearrangement occurs to achieve greater stability.
```
      CH₃                      CH₃
      |                        |
CH₃ - C - CH⁺ - CH₃   --1,2-methyl shift-->   CH₃ - C⁺ - CH - CH₃
      |                                        |
      CH₃                                      CH₃
```
The new carbocation is 2,3-dimethylbutan-2-yl cation. This is a tertiary carbocation, which is more stable than the secondary carbocation.
Step 4: Elimination of a proton to form the major alkene product. The tertiary carbocation will lose a $\beta$ -hydrogen (a hydrogen from an adjacent carbon) to form an alkene. According to Saytzeff's Rule, the major product will be the most substituted alkene (the one with the most $\alpha$ -hydrogens).

The tertiary carbocation is:
```
      CH₃
      |
CH₃ - C⁺ - CH - CH₃
      |
      CH₃
```
Let's identify the $\beta$ -hydrogens (hydrogens on carbons adjacent to the carbon with the positive charge):
1. Two $\text{CH}_3$ groups directly attached to the $\text{C}^+$ (on C2 in 2,3-dimethylbutan-2-yl cation). Each has 3 $\beta$ -hydrogens. (Total 6 $\beta$ -H)
2. One $\text{CH}$ group (on C3) which is part of the isopropyl group. This has 1 $\beta$ -hydrogen. (Total 1 $\beta$ -H)
- Possibility 1: Elimination from one of the $\text{CH}_3$ groups (6 $\beta$ -H available). This leads to 2,3-dimethylbut-1-ene: $\text{CH}_2=\text{C}(\text{CH}_3)-\text{CH}(\text{CH}_3)_2$ $\alpha$ -hydrogens for 2,3-dimethylbut-1-ene:
  - One $\text{CH}_3$ group attached to the double bond carbon (3 $\alpha$ -H).
  - One $\text{CH}$ group attached to the double bond carbon (1 $\alpha$ -H). Total $\alpha$ -hydrogens = 3 + 1 = 4.
- Possibility 2: Elimination from the $\text{CH}$ group (1 $\beta$ -H available). This leads to 2,3-dimethylbut-2-ene: $\text{CH}_3-\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)-\text{CH}_3$ $\alpha$ -hydrogens for 2,3-dimethylbut-2-ene:
  - Two $\text{CH}_3$ groups attached to one double bond carbon (3 + 3 = 6 $\alpha$ -H).
  - Two $\text{CH}_3$ groups attached to the other double bond carbon (3 + 3 = 6 $\alpha$ -H). Total $\alpha$ -hydrogens = 6 + 6 = 12.

3. Major Product A: According to Saytzeff's Rule, the major product is the most substituted alkene, which is the one with the highest number of $\alpha$ -hydrogens. Comparing the two products:

2,3-dimethylbut-1-ene has 4 $\alpha$ -hydrogens.
2,3-dimethylbut-2-ene has 12 $\alpha$ -hydrogens.

Therefore, the major product A is 2,3-dimethylbut-2-ene.

4. Number of $\alpha$ -hydrogens in Major Product A: The major product A is 2,3-dimethylbut-2-ene. Structure: $\text{CH}_3-\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)-\text{CH}_3$ The $\alpha$ -carbons are the carbons directly attached to the double bond carbons. In this molecule, all four methyl groups are $\alpha$ -carbons. Each methyl group has 3 hydrogens. Total number of $\alpha$ -hydrogens = 4 (methyl groups) $\times$ 3 (hydrogens per methyl group) = 12.

The final answer is $\boxed{12}$ .

Explanation of the solution: The reaction is acid-catalyzed dehydration of 3,3-dimethylbutan-2-ol.

Protonation of $-\text{OH}$ group forms $-\text{OH}_2^+$ .
Loss of $\text{H}_2\text{O}$ forms a secondary carbocation (3,3-dimethylbutan-2-yl cation).
A 1,2-methyl shift occurs to form a more stable tertiary carbocation (2,3-dimethylbutan-2-yl cation).
Elimination of a $\beta$ $β$ -hydrogen from the tertiary carbocation follows Saytzeff's rule, yielding the most substituted alkene.
- Elimination of $\text{H}$ from a $\text{CH}_3$ group gives 2,3-dimethylbut-1-ene (4 $\alpha$ -hydrogens).
- Elimination of $\text{H}$ from the $\text{CH}$ group gives 2,3-dimethylbut-2-ene (12 $\alpha$ -hydrogens).
The major product A is 2,3-dimethylbut-2-ene, which has 12 $\alpha$ -hydrogens.