Question

Consider the above reaction and identify the Product P :

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Acid-catalysed dehydration of the primary alcohol proceeds through a rearranged tertiary carbocation to give 1-methylcyclohexene; hydroboration-oxidation then adds H and OH anti-Markovnikov.

Answer 1

The Correct Option is D

Step 1 (cyclohexylmethanol, \(\mathrm{H_3PO_4}\), \(120^\circ\mathrm{C}\)): protonation and loss of water gives a primary carbocation \(\mathrm{CH_2^+}\) attached to the ring, which rearranges by ring-carbon hydride shift to the more stable tertiary 1-methylcyclohexyl cation. Loss of a proton gives the most substituted alkene (Saytzeff): A = 1-methylcyclohexene.
Step 2 (\(\mathrm{(BH_3)_2}\); then \(\mathrm{H_2O_2/OH^-}\), \(\mathrm{H_2O}\)): hydroboration-oxidation adds H and OH across the double bond in an anti-Markovnikov, syn manner, so OH goes to the less substituted carbon (the ring carbon next to the one bearing \(\mathrm{CH_3}\)).
P = 2-methylcyclohexan-1-ol, i.e. a cyclohexane ring with \(\mathrm{OH}\) and \(\mathrm{CH_3}\) on adjacent carbons, which is option (4).
Answer: (4) 2-methylcyclohexanol (ring with adjacent CH3 and OH)