Question

Propene on hydroboration-oxidation gives ________.

• A. 1-propanol
• B. 2-propanol
• C. propanal
• D. propanone
• E. ethanoic acid

Hint:

Hydration/Acid-catalyzed: Markovnikov (2-propanol); Hydroboration: Anti-Markovnikov (1-propanol).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Hydroboration-oxidation is a two-step organic reaction that converts an alkene into an alcohol. The key feature of this reaction is its regioselectivity, which is opposite to that of many other addition reactions to alkenes.
Step 2: Key Formula or Approach
The hydroboration-oxidation of an alkene results in the net anti-Markovnikov addition of water across the double bond. - Markovnikov's Rule: In the addition of H-X to an alkene, the H atom adds to the double-bonded carbon that has more H atoms. - Anti-Markovnikov's Rule: The H atom adds to the double-bonded carbon that has fewer H atoms. In the context of adding H-OH, hydroboration-oxidation places the -OH group on the less substituted carbon atom of the double bond.
Step 3: Detailed Explanation
1. Identify the Alkene and its Double Bond Carbons:
The starting material is propene: \(CH_3-CH=CH_2\). The double bond is between carbon-1 (\(=CH_2\)) and carbon-2 (\(-CH=\)). - Carbon-1 has two hydrogen atoms. - Carbon-2 has one hydrogen atom. Carbon-1 is the less substituted carbon. 2. Apply the Anti-Markovnikov Rule:
The hydroboration-oxidation sequence adds an -OH group to the less substituted carbon atom and an H atom to the more substituted carbon atom. - The -OH group will add to carbon-1. - The H atom will add to carbon-2. 3. Determine the Product:
\[ \underbrace{CH_3-CH=CH_2}_{\text{Propene}} \xrightarrow{\text{1. } BH_3 \text{ 2. } H_2O_2, OH^-} \underbrace{CH_3-CH_2-CH_2-OH}_{\text{1-Propanol}} \] The product is 1-propanol. In contrast, an acid-catalyzed hydration would follow Markovnikov's rule to give 2-propanol.
Step 4: Final Answer
Propene on hydroboration-oxidation gives 1-propanol.