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).
Which of the isomers of the compound “A” has maximum boiling point?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1-Butanol
Alcohols have high boiling points because their molecules are held together by intermolecular hydrogen bonds. Among isomeric alcohols (same number of carbon atoms), the boiling point is decided by the van der Waals forces between the carbon chains: branching makes the molecule more compact, reduces its surface area and weakens these forces, so it lowers the boiling point. A straight chain with the –OH at the end, where hydrogen bonding is least hindered, boils highest.
A (\( \ce{C4H10O} \)) is butan-2-ol. The alcohols with this formula and their boiling points are:
\( \displaystyle \text{butan-1-ol (391 K)} > \text{2-methylpropan-1-ol (381 K)} > \text{butan-2-ol (373 K)} > \text{2-methylpropan-2-ol (355 K)} \)
So 1-butanol, the unbranched primary alcohol, has the maximum boiling point.
Why the other options are incorrect
Option (2) 2-Butanol — Incorrect. The –OH is on an inner carbon, which makes the molecule more compact; it boils at 373 K.
Option (3) 2-methyl-propanol — Incorrect. It has a branched chain; 2-methylpropan-1-ol boils at 381 K and 2-methylpropan-2-ol at 355 K, both below 1-butanol.
Option (4) 2,2-dimethylpropanol — Incorrect. It is not an isomer of A (it is C5H12O); even so, its compact, highly branched chain gives it a boiling point of about 386 K, below 1-butanol.
Note: It seems that there is a mistake in the original NTA paper:
Option (4) prints 2,2-dimethylpropanol, (CH3)3C–CH2OH, which is C5H12O and not an isomer of A (C4H10O).
Option (3) prints 2-methyl-propanol without the position of the –OH group, so it could mean 2-methylpropan-1-ol or 2-methylpropan-2-ol.
NCERT (2026-27 edition):
Boiling Points
