Alcohols
Alcohols are formed when a hydrogen atom in aliphatic hydrocarbon is replaced by –OH group. Alcohols are classified as primary, secondary and tertiary alcohols depending on the attachment of –OH group to primary, secondary and tertiary carbon atom, respectively. In alcohols, the oxygen of the –OH group is attached to carbon by a sigma (σ) bond formed by the overlap of a sp3 hybridised orbital of carbon with a sp3 hybridised orbital of oxygen. Alcohols consist of two parts, an alkyl group and a hydroxyl group. The properties of alcohols are chiefly due to the hydroxyl group. The nature of alkyl groups simply modify these properties. The boiling points of alcohols increase with increase in the number of carbon atoms (increase in van der Waals forces). In alcohols, the boiling points decrease with increase of branching in carbon chain (because of decrease in van der Waals forces with decrease in surface area). Alcohols undergo dehydration (removal of a molecule of water) to form alkenes on treating with a protic acid e.g., concentrated H2SO4 or H3PO4, or catalysts such as anhydrous zinc chloride or alumina. Alcohols are versatile compounds. They act both as nucleophiles and electrophiles. An alcohol with molecular formula C4H10O (A) is prepared by the reduction of corresponding ketone. A is also prepared by the reaction of propanal with methylmagnesium bromide followed by hydrolysis. Compound A on dehydration gives alkene (B).
The IUPAC name and structure of compound “A” is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Butan-2-ol (structure 1)
A Grignard reagent adds to the carbonyl carbon of an aldehyde or ketone; hydrolysis then gives an alcohol whose carbon skeleton is the carbonyl compound plus the alkyl group of the Grignard reagent. Methanal gives a 1° alcohol, other aldehydes give 2° alcohols, and ketones give 3° alcohols.
Propanal (3 carbons) and methylmagnesium bromide (1 carbon) give a four-carbon secondary alcohol:
\( \displaystyle \ce{CH3CH2CHO + CH3MgBr -> CH3CH2CH(OMgBr)CH3} \)
\( \displaystyle \ce{CH3CH2CH(OMgBr)CH3} \xrightarrow{\ce{H2O}} \ce{CH3CH2CH(OH)CH3} \)
For the IUPAC name, the longest chain has four carbons (butane) and the –OH is on C-2, so A is butan-2-ol. This fits the other information too: its formula is \( \ce{C4H10O} \), and it is obtained by reducing the ketone butan-2-one. Structure 1 shows this four-carbon chain with the –OH on the second carbon.
Why the other options are incorrect
Option (2) structure 2 (Butanol) — Incorrect. It shows butan-1-ol, a primary alcohol, which cannot be formed from propanal and \( \ce{CH3MgBr} \) or by reducing a ketone.
Option (3) structure 3 (2-methylpropan-1-ol) — Incorrect. It has a branched chain; the Grignard product here has an unbranched four-carbon chain.
Option (4) structure 4 (3-methylbutan-2-ol) — Incorrect. It has five carbon atoms (\( \ce{C5H12O} \)), not the formula \( \ce{C4H10O} \) of A.