Question

The following reaction takes place:

• A.
• B.
• C.
• D.

Hint:

When a primary alcohol reacts with PBr\(_3\), it undergoes substitution, followed by elimination with alcoholic KOH to form alkenes.

Answer 1

The Correct Option is A

Conversion of Cyclohexanol to Cyclohexene

Step 1: Alcohol to Alkyl Bromide (PBr₃ Reaction)

Cyclohexanol + PBr₃ → Bromocyclohexane + H₃PO₃

• Mechanism: Nucleophilic substitution (S_N2-like)
  1. Oxygen lone pair attacks phosphorus in PBr₃
  2. Br⁻ leaves as leaving group
  3. New O-P bond forms while O-H bond breaks
  4. Br⁻ attacks backside, displacing the -OPBr₂ group
• Key Features:
  • Converts -OH (poor leaving group) to -Br (excellent leaving group)
  • Proceeds with inversion of configuration at carbon
  • Phosphorous byproduct is phosphorous acid (H₃PO₃)

Step 2: Elimination to Form Alkene (alc. KOH Reaction)

Bromocyclohexane + KOH (alc.) → Cyclohexene + KBr + H₂O

• Mechanism: Bimolecular Elimination (E2)
  1. Strong base (OH⁻) abstracts β-hydrogen
  2. C-Br bond breaks simultaneously
  3. π-bond forms between α- and β-carbons
• Key Features:
  • Requires anti-periplanar arrangement of H and Br
  • Favored in alcoholic solution (prevents competing substitution)
  • Follows Zaitsev's rule - forms more substituted alkene when possible

Why Other Products Don't Form

Possible Product	Why Not Formed
B (Cyclohexane)	Would require reduction conditions, not elimination
C (1,3-Cyclohexadiene)	Double elimination doesn't occur under these conditions
D (Bromocyclohexene)	Not a typical product of this reaction sequence

Critical Observation: This two-step process (alcohol → alkyl halide → alkene) is a classic method for dehydrating alcohols under mild conditions.

Therefore, the correct final product is A (Cyclohexene).

Answer 2

Step 1: \(PBr_3\) Reaction

Reagent: \(PBr_3\) (phosphorus tribromide)

Function: Converts alcohols (R-OH) into alkyl bromides (R-Br). It's a common way to introduce a good leaving group (bromide) into a molecule.

Step 2: alc. KOH Reaction

Reagent: Alcoholic KOH (potassium hydroxide in ethanol or another alcohol)

Function: A strong base under alcoholic conditions favors elimination reactions (specifically, E2 elimination). E2 elimination removes a proton from a carbon adjacent to the carbon bearing the leaving group (in this case, bromine), forming a double bond.

Overall Sequence

Cyclohexanol (the starting material) is converted to bromocyclohexane.

Bromocyclohexane undergoes E2 elimination, forming cyclohexene.

Therefore, the correct answer is A.