Arrange the following compounds in increasing order of their solubility in water:
(A) n-butanol
(B) Butan-1-amine
(C) Methanamine
(D) Aniline
Answer & explanation
Correct answer: option 3
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
$\text{Aniline} < \text{n-butanol} < \text{Butan-1-amine} < \text{Methanamine}$
$(D) < (A) < (B) < (C)$