Arrange the following compounds in increasing order of their solubility in water:

(A) n-butanol
(B) Butan-1-amine
(C) Methanamine
(D) Aniline

(D), (A), (B), (C)

The correct answer is Option (3) → (D), (A), (B), (C)

Solubility in water depends on:

• Ability to form hydrogen bonds
• Polarity
• Size of hydrophobic hydrocarbon part (bigger chain $\rightarrow$ lower solubility)
• Amines are generally more soluble than alcohols of similar size because they can be protonated in water.

Analysis of Each Compound

(D) Aniline ($C_6H_5NH_2$)

Large hydrophobic benzene ring reduces solubility. Although $-NH_2$ can H-bond, the aromatic ring dominates, making it poorly soluble.

Least soluble among the given.

(A) n-Butanol ($C_4H_9OH$)

Has one $-OH$ group (H-bonding possible) but a four-carbon chain reduces polarity. Moderately soluble but limited due to long alkyl chain.

(B) Butan-1-amine ($C_4H_9NH_2$)

Similar carbon chain as butanol, but amines are more soluble than alcohols because they form H-bonds and also partially ionize in water. More soluble than n-butanol.

(C) Methanamine ($CH_3NH_2$)

Very small molecule, highly polar, excellent hydrogen bonding, and readily forms ammonium ion in water. Most soluble.

Increasing Order of Solubility

$\text{Aniline} < \text{n-butanol} < \text{Butan-1-amine} < \text{Methanamine}$

$(D) < (A) < (B) < (C)$