Question

Arrange the following compounds in increasing order of their boiling points:

• \(\mathrm{C_2H_5OH}\) (Ethanol)
• \(\mathrm{CH_3CHO}\) (Acetaldehyde)
• \(\mathrm{CH_3CH_2CH_3}\) (Propane)
• \(\mathrm{CH_3OCH_3}\) (Dimethyl ether)

• A. CH\(_3\)CH\(_2\)CH\(_3\) \(<\) CH\(_3\)OCH\(_3\) \(<\) CH\(_3\)CHO \(<\) C\(_2\)H\(_5\)OH
• B. C\(_2\)H\(_5\)OH \(<\) CH\(_3\)CHO \(<\) CH\(_3\)OCH\(_3\) \(<\) CH\(_3\)CH\(_2\)CH\(_3\)
• C. CH\(_3\)OCH\(_3\) \(<\) CH\(_3\)CH\(_2\)CH\(_3\) \(<\) CH\(_3\)CHO \(<\) C\(_2\)H\(_5\)OH
• D. CH\(_3\)CHO \(<\) C\(_2\)H\(_5\)OH \(<\) CH\(_3\)CH\(_2\)CH\(_3\) \(<\) CH\(_3\)OCH\(_3\)

Hint:

When comparing boiling points of organic compounds with similar molecular weights, prioritize intermolecular forces: hydrogen bonding > dipole-dipole interactions > London dispersion forces.

Answer 1

The Correct Option is A

Step 1: Identify the compounds and their molecular weights.

C\(_2\)H\(_5\)OH (Ethanol): An alcohol. Molecular Weight (MW) = 2(12) + 6(1) + 1(16) = 46 g/mol.
CH\(_3\)CHO (Acetaldehyde): An aldehyde. MW = 2(12) + 4(1) + 1(16) = 44 g/mol.
CH\(_3\)CH\(_2\)CH\(_3\) (Propane): An alkane. MW = 3(12) + 8(1) = 44 g/mol.
CH\(_3\)OCH\(_3\) (Dimethyl ether): An ether. MW = 2(12) + 6(1) + 1(16) = 46 g/mol. All four compounds have very similar molecular weights, so the boiling point will primarily depend on the strength of their intermolecular forces (IMFs). Step 2: Determine the primary intermolecular forces for each compound.
C\(_2\)H\(_5\)OH (Ethanol): Contains a hydroxyl (-OH) group, which allows for strong hydrogen bonding between molecules. It also has dipole-dipole interactions and London dispersion forces. Hydrogen bonding is the strongest IMF present.
CH\(_3\)CHO (Acetaldehyde): Contains a polar carbonyl (C=O) group. It exhibits significant dipole-dipole interactions due to the polarity of the C=O bond, in addition to London dispersion forces. It does not form hydrogen bonds as hydrogen is not directly bonded to oxygen.
CH\(_3\)CH\(_2\)CH\(_3\) (Propane): This is a nonpolar hydrocarbon. It only exhibits weak London dispersion forces.
CH\(_3\)OCH\(_3\) (Dimethyl ether): Contains a polar C-O-C bond. It exhibits dipole-dipole interactions due to the overall dipole moment, in addition to London dispersion forces. It does not form hydrogen bonds as hydrogen is not directly bonded to oxygen. Step 3: Compare the strengths of intermolecular forces and predict the boiling point order.
The general order of strength for these intermolecular forces is: Hydrogen Bonding > Dipole-Dipole Interactions > London Dispersion Forces. Applying this to the compounds: (A) Ethanol (C\(_2\)H\(_5\)OH) will have the highest boiling point due to strong hydrogen bonding. (B) Acetaldehyde (CH\(_3\)CHO) will have a higher boiling point than dimethyl ether because the dipole moment of the carbonyl group (C=O) in aldehydes typically leads to stronger dipole-dipole interactions compared to the less concentrated dipole in ethers (C-O-C). (C) Dimethyl ether (CH\(_3\)OCH\(_3\)) will have a higher boiling point than propane due to the presence of dipole-dipole interactions, which are stronger than purely London dispersion forces. (D) Propane (CH\(_3\)CH\(_2\)CH\(_3\)) will have the lowest boiling point as it only exhibits weak London dispersion forces. Therefore, the increasing order of boiling points is: CH\(_3\)CH\(_2\)CH\(_3\) (lowest) < CH\(_3\)OCH\(_3\) < CH\(_3\)CHO < C\(_2\)H\(_5\)OH (highest) Step 4: Verify with actual boiling points (for reference):
Propane (CH\(_3\)CH\(_2\)CH\(_3\)): Approx. -42 \(^{\circ}\)C
Dimethyl ether (CH\(_3\)OCH\(_3\)): Approx. -24 \(^{\circ}\)C
Acetaldehyde (CH\(_3\)CHO): Approx. 20 \(^{\circ}\)C
Ethanol (C\(_2\)H\(_5\)OH): Approx. 78 \(^{\circ}\)C The actual boiling points confirm the predicted order.