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Difficulty: MediumAlkanoic Acids, Esterification, Saponification, Fats, and Oils

Arrange the following organic compounds, each having a relative molecular mass of approximately 58−60 g/mol58-60\text{ g/mol}, in order of INCREASING boiling point (from lowest boiling point to highest boiling point).

  1. 1Butane (C4H10\text{C}_4\text{H}_{10})
  2. 2Methyl formate (HCOOCH3\text{HCOOCH}_3)
  3. 3Propan-1-ol (C3H7OH\text{C}_3\text{H}_7\text{OH})
  4. 4Ethanoic acid (CH3COOH\text{CH}_3\text{COOH})

Answer

The correct sequence in order of increasing boiling point is Butane (C4H10\text{C}_4\text{H}_{10}), followed by Methyl formate (HCOOCH3\text{HCOOCH}_3), then Propan-1-ol (C3H7OH\text{C}_3\text{H}_7\text{OH}), and finally Ethanoic acid (CH3COOH\text{CH}_3\text{COOH}).
Boiling points depend on the relative strength of intermolecular forces when molecular masses are comparable (~60 g/mol). Butane is non-polar and exhibits only weak dispersion forces (lowest boiling point). Methyl formate is polar and exhibits dipole-dipole attractions. Propan-1-ol forms strong hydrogen bonds via its hydroxyl group. Ethanoic acid forms even stronger hydrogen bonds and stable cyclic dimers, giving it the highest boiling point.

Step-by-Step Solution

1
Identify the primary type of intermolecular force present in each compound of similar molar mass (≈58−60 g/mol\approx 58-60\text{ g/mol}).
Butane has London dispersion forces; Methyl formate has dipole-dipole forces; Propan-1-ol has hydrogen bonding; Ethanoic acid has extensive hydrogen bonding and dimer formation.
Boiling point increases as the strength of intermolecular forces holding the liquid molecules together increases.
2
Compare the compounds without hydrogen bonding capabilities (Butane vs. Methyl formate).
Butane is non-polar and exhibits only weak dispersion forces. Methyl formate has a polar carbonyl group (C=O\text{C=O}) causing dipole-dipole attractions, making its boiling point higher than butane.
Permanent dipole-dipole attractions are stronger than instantaneous dispersion forces for molecules of similar size.
3
Compare the hydrogen-bonded compounds (Propan-1-ol vs. Ethanoic acid).
Propan-1-ol forms intermolecular hydrogen bonds through its single hydroxyl group. Ethanoic acid forms stronger hydrogen bonds using both its carbonyl oxygen and hydroxyl hydrogen to form stable cyclic dimers.
Dimerization in alkanoic acids effectively doubles the molecular interaction area, requiring significantly more thermal energy to break apart during vaporization.
4
Arrange the compounds in order of increasing boiling point.
Butane < Methyl formate < Propan-1-ol < Ethanoic acid.
The progression of intermolecular force strength directly dictates the trend in boiling points.

Key Concept

Intermolecular Forces and Boiling Point Trends in Alkanoic Acids, Esters, Alkanols, and Alkanes
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Topics

Alkanoic Acids, Esterification, Saponification, Fats, and Oils

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