The dehydration of CH$_3$CH$_2$CH$_2$CH$_2$OH will give?

Analysis

The question asks for the product formed when butan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$) undergoes dehydration. Dehydration is the removal of a water molecule ($\text{H}_2\text{O}$) from an alcohol to form an alkene.

Concept Explanation

Dehydration of alcohols typically occurs in the presence of an acid catalyst (like concentrated $\text{H}_2\text{SO}_4$) and heat. According to Zaitsev’s Rule, when an alcohol is dehydrated, the hydrogen atom is removed from the $\beta$-carbon (the carbon adjacent to the one holding the hydroxyl group) that has the fewest hydrogen atoms, leading to the formation of the most substituted alkene (the most stable product).

Step-by-Step Solution

1. Identify the structure: The reactant is butan-1-ol: $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}$.
2. Identify the carbons: • The carbon attached to the $-\text{OH}$ group is the $\alpha$-carbon. • The carbon next to it ($\text{CH}_2$) is the $\beta$-carbon.
3. Remove water: To form the alkene, we remove the $-\text{OH}$ group from the $\alpha$-carbon and one hydrogen atom from the $\beta$-carbon.
4. Rearrangement: In the presence of strong acid, the carbocation formed can undergo a 1,2-hydride shift to form a more stable secondary carbocation. • Initial: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+$ (primary) • Rearranged: $\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3$ (secondary)
5. Formation of the double bond: The loss of a proton from the secondary carbocation results in the formation of but-2-ene ($\text{CH}_3\text{CH=CHCH}_3$).

Evaluation of Options

• A and C are alkynes (triple bonds), which are not produced by simple dehydration.
• B represents but-1-ene ($\text{CH}_2\text{=CHCH}_2\text{CH}_3$), which is a minor product.
• D represents but-2-ene ($\text{CH}_3\text{CH=CHCH}_3$), which is the major product due to its higher stability.

Exam Tip

When dealing with dehydration of alcohols with four or more carbons, always look for the "most substituted" alkene as your major product. Remember the mantra: "The rich get richer" (for Markovnikov's addition) and "The poor get poorer" (for Zaitsev's elimination).