Question

When a tertiary alcohol ($C_4​H_{10}​O$) reacts with 20% $H_3​PO_4$​ at 358 K, it gives a compound ($C_4​H_8​$) as the major product. The IUPAC name of this compound is:

• A. But-1-ene
• B. But-2-ene
• C. Cyclobutane
• D. 2-Methylpropene

Hint:

The tertiary alcohol dehydrates under acidic conditions to yield 2-methylpropene () as the major product.

Answer 1

The Correct Option is D

The problem deals with the dehydration of a tertiary alcohol under acidic conditions to form an alkene. Here is the step-by-step reasoning to identify the correct product and its IUPAC name:

1. The given alcohol is a tertiary alcohol with the molecular formula C_4H_{10}O. The most common tertiary alcohol with this formula is 2-methyl-2-propanol, also known as t-butanol.
2. When t-butanol is treated with 20% phosphoric acid (H_3PO_4) at 358 K, it undergoes dehydration. Dehydration involves the removal of a water molecule, resulting in the formation of an alkene.
3. Tertiary alcohols dehydrate to form alkenes via an E1 elimination mechanism, which proceeds through the formation of a carbocation intermediate. In this case, the carbocation is formed at the 2-position, leading to a more stable tertiary carbocation.
4. The IUPAC name for this compound, which is the major product of the reaction, is 2-Methylpropene.
5. Let's evaluate the other options:
   • But-1-ene: It requires deprotonation from the terminal carbon, which is less favorable.
   • But-2-ene: Not formed as it would require rearrangement that is not favorable given the carbocation stability.
   • Cyclobutane: No cyclization occurs given the reaction conditions and structure of the alcohol.
6. Therefore, the correct answer is 2-Methylpropene, which matches the reaction mechanism and expected outcome.