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MHTCET 11 Apr 2026 Shift 1 Paper

Question : 90 of 150

Marks: +1, -0

The molecule with a maximum boiling point is

$\mathrm{CH}_{3}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}$
$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Br}$
$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}$
$\mathrm{CH}_{3}-C\left(\mathrm{CH}_{3}\right)(\mathrm{Cl})-\mathrm{OH}$

Solution:

Concept:

Boiling point depends on the strength of intermolecular forces.

Explanation:

Option A has two

-\mathrm{OH}

groups, so it can form many intermolecular hydrogen bonds.

Stronger hydrogen bonding requires more energy to separate molecules, giving a higher boiling point.

Options B and C are alkyl bromides, which show only dipole-dipole and van der Waals forces, not hydrogen bonding.

Option D has one

-\mathrm{OH}

group but is branched, which reduces surface area and hydrogen bonding.

Therefore, option A has the strongest intermolecular attraction.

Answer:

Option A:

\mathrm{CH}_{3}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}

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