Chemistry CUET UG 22-05-2023 Shift 3 Paper

When phenol is treated with chloroform ($\text{CHCl}_3$) in the presence of aqueous sodium hydroxide, an aldehyde group ($-\text{CHO}$) is introduced at the ortho position of the benzene ring, yielding salicylaldehyde ($o$-hydroxybenzaldehyde). This reaction proceeds via the intermediate dichlorocarbene ($:\text{CCl}_2$) electrophile and is universally known as the Reimer-Tiemann reaction.

Analysis of intermolecular forces and boiling points of comparable molecular weights:

• $n$-Butane (B): Non-polar hydrocarbon with only weak London dispersion forces $\rightarrow$ lowest boiling point ($309\text{ K}$ or below, $\text{b.p.} = -0.5^\circ\text{C} \approx 272.5\text{ K}$).
• Ethoxyethane (D): Weak dipole-dipole interactions without hydrogen bonding $\rightarrow \text{b.p.} = 307.6\text{ K}$.
• Pentan-1-al (C): Polar carbonyl group with strong dipole-dipole attractions $\rightarrow \text{b.p.} = 376\text{ K}$.
• Pentan-1-ol (A): Extensive intermolecular hydrogen bonding between hydroxyl groups $\rightarrow$ highest boiling point ($411\text{ K}$).

Reaction analysis:

• Acid-catalyzed hydration of propene follows Markovnikov's rule, adding $-\text{OH}$ to the secondary carbon to form propan-2-ol (compound A):
  $$\text{CH}_3-\text{CH}=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3-\text{CH(OH)}-\text{CH}_3$$
• Hydroboration-oxidation of propene with diborane followed by alkaline $\text{H}_2\text{O}_2$ yields anti-Markovnikov addition of water, producing propan-1-ol (compound C):
  $$\text{CH}_3-\text{CH}=\text{CH}_2 \xrightarrow{\text{B}_2\text{H}_6 / \text{H}_2\text{O}_2, \text{OH}^-} \text{CH}_3-\text{CH}_2-\text{CH}_2\text{OH}$$
• Both compounds have the identical molecular formula ($\text{C}_3\text{H}_8\text{O}$) and the same functional group ($-\text{OH}$), differing only in the position of the hydroxyl group along the carbon chain (carbon-2 vs carbon-1). Therefore, compounds A and C are positional isomers.