Which of the following methods are used for the synthesis of an alcoh…

Which of the following methods are used for the synthesis of an alcohol?

(A) Hydroboration–oxidation of alkenes

(B) By reduction of aldehydes and ketones

(D) Hydration of alkyne in the presence of Hg²⁺/H⁺

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 2

The correct answer is Option (2) → (A), (B) and (C) only

Alcohols are prepared from alkenes by adding water across the C=C bond, and from carbonyl compounds by adding hydrogen across the C=O bond. The addition of water to an alkyne behaves differently, because the product is an enol that rearranges to a carbonyl compound.

(A) Hydroboration–oxidation of alkenes — Correct. Diborane adds to the alkene, and the trialkylborane is oxidised by \( \ce{H2O2} \) in NaOH. The product is the alcohol with OH on the less substituted carbon (anti-Markovnikov).
\( \displaystyle \ce{CH3-CH=CH2} \xrightarrow[\text{(ii) }\ce{H2O2}/\ce{OH^-}]{\text{(i) }\ce{B2H6}} \ce{CH3-CH2-CH2OH} \)
(B) By reduction of aldehydes and ketones — Correct. With \( \ce{H2} \)/Pd, \( \ce{NaBH4} \) or \( \ce{LiAlH4} \), aldehydes give primary alcohols and ketones give secondary alcohols.
\( \displaystyle \ce{RCHO} \xrightarrow{\ce{NaBH4}} \ce{RCH2OH}\quad\quad \ce{RCOR'} \xrightarrow{\ce{NaBH4}} \ce{RCH(OH)R'} \)
(C) By acid catalysed hydration of alkenes — Correct. Water adds in the presence of an acid, following Markovnikov’s rule.
\( \displaystyle \ce{CH3-CH=CH2 + H2O} \xrightarrow{\ce{H+}} \ce{CH3-CH(OH)-CH3} \)
(D) Hydration of alkyne in the presence of Hg²⁺/H⁺ — Incorrect. The enol first formed is unstable and tautomerises to a carbonyl compound, so the product is an aldehyde or a ketone, not an alcohol.
\( \displaystyle \ce{HC#CH + H2O} \xrightarrow{\ce{Hg^2+}/\ce{H+}} \ce{CH3CHO} \)