Consider the above reaction and identify the Product P :
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Primary alcohols under acidic dehydration often undergo
carbocation rearrangements. Always check for hydride or alkyl shifts
before predicting the alkene product.
Step 1: Acid-catalysed dehydration
Cyclohexylmethanol is a primary alcohol.
In the presence of hot $H_3PO_4$ ($120^\circ$C), dehydration occurs via
protonation of the –OH group followed by loss of water.
A primary carbocation formed initially is unstable and undergoes
a 1,2-hydride shift from the cyclohexyl ring, producing a
more stable tertiary carbocation.
Elimination of a proton from this carbocation gives the
more substituted alkene (Zaitsev product):
\[
\text{A = 1-methylcyclohex-1-ene}
\]
Step 2: Hydroboration–oxidation
The alkene undergoes hydroboration–oxidation
\[
(BH_3)_2 \; \xrightarrow{} \; H_2O_2/\,OH^-
\]
This reaction proceeds via:
anti-Markovnikov addition
syn addition (stereochemistry not required here)
Hence, the –OH group attaches to the less substituted carbon
of the double bond (C-2).
Final Product:
\[
\text{2-methylcyclohexanol}
\]
\[
\boxed{\text{Option (C)}}
\]
