The table shows information about three carbon compounds: [table]
Compound | Structural formula | Molecular weight | Boiling point Y | | 58 | 0oC X | | 60 | 97oC Y | | 60 | 118oC Z | | 60 | ?
[/table] What is the best estimate for the boiling point of compound Z?

Question

BlackTom AI · Accepted Answer

Intermolecular forces — dispersion, dipole–dipole, and hydrogen bonding — control boiling points, so compare structures to predict Z.

1) Relevant knowledge points and concepts
- Molecules with similar molecular weight can have very different boiling points depending on intermolecular forces.  
- Nonpolar hydrocarbons: only London dispersion forces → low b.p.  
- Polar molecules with C=O and lone pairs: dipole–dipole interactions (moderate b.p.).  
- Molecules with O–H groups (able to donate and accept H-bonds): strong hydrogen bonding → high b.p. Carboxylic acids also form stable dimers, raising b.p. further.

2) Step-by-step reasoning using the table
- Compound with MW 58 and b.p. 0°C is a nonpolar hydrocarbon (butane) — dispersion only.  
- Compound X, MW 60, b.p. 97°C is an alcohol (propanol) — strong H-bond donors and acceptors → high b.p.  
- Compound Y, MW 60, b.p. 118°C is acetic acid (contains –COOH) — very strong H-bonding and dimer formation → even higher b.p.  
- Compound Z, MW 60, shows an ester/formate structure (no O–H) with a C=O and an –O– (i.e., methyl formate HCOOCH3). It can accept H-bonds but cannot donate; therefore it has dipole–dipole interactions but weaker overall intermolecular forces than alcohols or acids.

3) Calculations / intermediate steps
- All three MW = 60 species are compared: alcohol (97°C), carboxylic acid (118°C), and Z. Ranking by expected strength: acid > alcohol > ester/formate > nonpolar hydrocarbon. Given butane (58) b.p. 0°C and methyl formate’s intermediate polarity, Z’s b.p. should be modestly above 0°C and well below 97–118°C.

4) Why the correct answer is correct
- Methyl formate (HCOOCH3) has a known boiling point ≈ 31°C. This matches the expected intermediate b.p. for a polar ester that cannot form O–H hydrogen bonds.

5) Why other choices are incorrect
- 0°C: too low (matches nonpolar butane, not a polar ester).  
- 101°C, 114°C, 156°C: too high — these are in the range of alcohols/acids or much larger polar molecules, inconsistent with Z’s lack of O–H donation.

6) Conclusion / confirmation
- Therefore the best estimate for compound Z’s boiling point is 31°C, which matches the provided answer.

The table shows information about three carbon compounds: [table] Compound | Structural formula | Molecular weight | Boiling point Y | | 58 | 0oC X | | 60 | 97oC Y | | 60 | 118oC Z | | 60 | ? [/table] What is the best estimate for the boiling point of compound Z?Single choice

Similar Questions

The table shows information about three carbon compounds: [table] Compound | Structural formula | Molecular weight | Boiling point Y | | 58 | 0oC X | | 60 | 97oC Y | | 60 | 118oC Z | | 60 | ? [/table] What is the best estimate for the boiling point of compound Z?

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