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 Formaldehdye with Ethylmagnesium bromide followed by hydrolysis
(D) Acid catalyzed hydration of propene

Choose the correct answer from the options given below:

(A), (B) and (C) only

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

n-Propanol ($CH_{3}CH_{2}CH_{2}OH$) is a primary alcohol. Each reaction must be checked to see whether it forms this compound.

Analysis of the Statements

(A) Hydroboration–oxidation of propene

Hydroboration–oxidation proceeds through anti-Markovnikov addition, where the $OH$ group attaches to the less substituted carbon of the double bond.

Reaction:

$CH_{3}-CH=CH_{2} \rightarrow CH_{3}-CH_{2}-CH_{2}OH$

Product formed: n-Propanol

Thus, statement (A) is correct.

(B) Reduction of propanal

Propanal is an aldehyde. Reduction of aldehydes using hydrogen with catalysts such as Pd, Pt, or Ni converts them into primary alcohols.

Reaction:

$CH_3CH_2CHO \rightarrow CH_3CH_2CH_2OH$

Product formed: n-Propanol

Thus, statement (B) is correct.

(C) Reaction of formaldehyde with ethylmagnesium bromide

Grignard reagents react with formaldehyde to produce primary alcohols with one additional carbon atom.

Reaction:

$HCHO + C_2H_5MgBr \rightarrow CH_3CH_2CH_2OH$ (after hydrolysis)

Product formed: n-Propanol

Thus, statement (C) is correct.

(D) Acid catalyzed hydration of propene

Acid-catalyzed hydration follows Markovnikov addition. The OH group attaches to the more substituted carbon, producing:

$CH_3-CHOH-CH_3$

Product formed: 2-propanol, not n-propanol.

Thus, statement (D) is incorrect.