Which of the following reactions will give n-Propanol as a product?
(A) Hydroboration oxidation of Propene
(B) Reduction of propanal with Palladium as a catalyst
(C) Reaction of Formaldehyde with Ethylmagnesium bromide followed by hydrolysis
(D) Acid catalyzed hydration of propene
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
n-Propanol (propan-1-ol) is \( \ce{CH3CH2CH2OH} \), a primary alcohol with the –OH on the end carbon. Each route is checked for where the –OH ends up.
(A) Hydroboration oxidation of Propene — Correct. Boron adds to the end carbon (the one with more hydrogen atoms), and oxidation puts the –OH there, an anti-Markovnikov result:
\( \displaystyle \ce{CH3-CH=CH2} \xrightarrow[\text{(ii) }\ce{H2O2}/\ce{OH-}]{\text{(i) }\ce{B2H6}} \ce{CH3CH2CH2OH} \)
(B) Reduction of propanal with Palladium as a catalyst — Correct. Catalytic hydrogenation reduces an aldehyde to a primary alcohol with the same carbon chain:
\( \displaystyle \ce{CH3CH2CHO + H2} \xrightarrow{\ce{Pd}} \ce{CH3CH2CH2OH} \)
(C) Reaction of Formaldehyde with Ethylmagnesium bromide followed by hydrolysis — Correct. A Grignard reagent with methanal gives a primary alcohol with one carbon more than the alkyl group; the ethyl group adds to the carbonyl carbon:
\( \displaystyle \ce{HCHO + C2H5MgBr -> C2H5CH2OMgBr} \)
\( \displaystyle \ce{C2H5CH2OMgBr + H2O -> CH3CH2CH2OH + Mg(OH)Br} \)
(D) Acid catalyzed hydration of propene — Incorrect. Water adds by Markovnikov’s rule through the more stable secondary carbocation, so the –OH goes to the middle carbon and the product is propan-2-ol:
\( \displaystyle \ce{CH3-CH=CH2 + H2O} \xrightarrow{\ce{H+}} \ce{CH3-CH(OH)-CH3} \)