Question

Hydroboration oxidation of 4-methyl-octene would give

• A. 4-methyl octanol
• B. 2-methyl decane
• C. 4-methyl heptanol
• D. 4-methyl-2-octanone

Hint:

Hydroboration-oxidation gives anti-Markovnikov alcohol without carbocation rearrangement.

Answer 1

The Correct Option is A

Step 1: Recall hydroboration-oxidation reaction.
Hydroboration-oxidation converts an alkene into an alcohol by anti-Markovnikov addition.
Reagents:
\[ BH_3 \, / \, THF \quad \text{followed by} \quad H_2O_2/NaOH \]
Step 2: Nature of addition.
OH group attaches to the less substituted carbon of the double bond (anti-Markovnikov rule).
Step 3: Apply to 4-methyl-octene.
Since the alkene has 8 carbon chain with a methyl group at position 4, the product remains an octanol derivative.
So the product formed is 4-methyl octanol.
Final Answer:
\[ \boxed{\text{(A) 4-methyl octanol}} \]