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).
What type of alcohol is compound “A”?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Secondary aliphatic
An alcohol is primary, secondary or tertiary according to the carbon that carries the –OH group: a carbon attached to one, two or three other carbon atoms respectively. It is aliphatic when the –OH is on an open-chain (non-aromatic) carbon.
Compound A is formed from propanal and methylmagnesium bromide. A Grignard reagent adds to an aldehyde (other than methanal) to give a secondary alcohol:
\( \displaystyle \ce{CH3CH2CHO + CH3MgBr} \xrightarrow{\ce{H2O}} \ce{CH3CH2-CH(OH)-CH3} \)
A is butan-2-ol. The carbon bearing –OH is attached to two carbon atoms (a \( \ce{CH3} \) and a \( \ce{C2H5} \) group), and there is no benzene ring. It is also the product of reducing a ketone (butan-2-one), which always gives a 2° alcohol. So A is a secondary aliphatic alcohol.
Why the other options are incorrect
Option (1) Primary aliphatic — Incorrect. A primary alcohol such as butan-1-ol would come from methanal with a Grignard reagent, or from reducing an aldehyde, not a ketone.
Option (3) Tertiary aliphatic — Incorrect. A tertiary alcohol such as 2-methylpropan-2-ol comes from a ketone and a Grignard reagent; it cannot be made by reducing a ketone.
Option (4) Secondary aromatic — Incorrect. A has no benzene ring, so it is not aromatic.
NCERT (2026-27 edition):
From Grignard Reagents
