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Chapter 11: Problem 24

Propyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) and isopropyl alcohol \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH}\right],\) whose space- filling models are shown, have boiling points of \(97.2^{\circ} \mathrm{C}\) and \(82.5^{\circ} \mathrm{C}\), respectively. Explain why the boiling point of propyl alcohol is higher, even though both have the molecular formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\).

Short Answer

Expert verified
The boiling point of propyl alcohol (\(97.2^{\circ} \mathrm{C}\)) is higher than that of isopropyl alcohol (\(82.5^{\circ} \mathrm{C}\)) due to stronger intermolecular forces in propyl alcohol. Both molecules can form hydrogen bonds due to their -OH groups, but propyl alcohol's straight-chain structure leads to stronger dispersion forces compared to isopropyl alcohol's branched structure. The cumulative effect of the hydrogen bonding and stronger dispersion forces makes the overall intermolecular forces in propyl alcohol stronger, resulting in a higher boiling point.

Step by step solution

01

Identify the molecular structure of each compound

: Propyl alcohol has the chemical formula \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}\) and has a straight-chain structure. Isopropyl alcohol has the chemical formula \(\left(\mathrm{CH}_3\right)_2\mathrm{CHOH}\) and has a branched structure.
02

Determine the types of intermolecular forces present in each compound

: Both propyl alcohol and isopropyl alcohol have polar -OH (hydroxyl) groups, which make both molecules polar. For the -OH hydroxyl group, hydrogen bond is the dominant intermolecular force. Due to its straight-chain structure, propyl alcohol has stronger dispersion forces between its molecules compared to the branched structure of isopropyl alcohol.
03

Relate boiling point to intermolecular forces

: The boiling point of a substance is related to the strength of its intermolecular forces. The stronger the intermolecular forces, the higher the boiling point, as more energy is required to overcome these forces and transform the substance into the gas phase.
04

Compare the intermolecular forces in propyl alcohol and isopropyl alcohol

: While both propyl alcohol and isopropyl alcohol can form hydrogen bonds due to their -OH groups, the straight-chain structure of propyl alcohol leads to stronger dispersion forces between its molecules compared to isopropyl alcohol's branched structure. The cumulative effect of the hydrogen bonding and stronger dispersion forces makes the overall intermolecular forces in propyl alcohol stronger than those in isopropyl alcohol.
05

Conclude and explain the difference in boiling points

: The stronger intermolecular forces in propyl alcohol are responsible for its higher boiling point (\(97.2^{\circ} \mathrm{C}\)) compared to isopropyl alcohol (\(82.5^{\circ} \mathrm{C}\)), even though both compounds have the same molecular formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\).

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Most popular questions from this chapter

True or false: (a) For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable. (b) For the noble gases the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance dipole- dipole interactions, when present, are always larger than dispersion forces. (d) All other factors being the same, dispersion forces between linear molecules are greater than dispersion forces between molecules whose shapes are nearly spherical.

The following quote about ammonia \(\left(\mathrm{NH}_{3}\right)\) is from a textbook of inorganic chemistry: "It is estimated that \(26 \%\) of the hydrogen bonding in \(\mathrm{NH}_{3}\) breaks down on melting, \(7 \%\) on warming from the melting to the boiling point, and the final \(67 \%\) on transfer to the gas phase at the boiling point." From the standpoint of the kinetic energy of the molecules, explain (a) why there is a decrease of hydrogen-bonding energy on melting and (b) why most of the loss in hydrogen bonding occurs in the transition from the liquid to the vapor state.

Ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) is the major component of antifreeze. It is a slightly viscous liquid, not very volatile at room temperature, with a boiling point of \(198^{\circ} \mathrm{C}\). Pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right),\) which has about the same molecular weight, is a nonviscous liquid that is highly volatile at room temperature and whose boiling point is \(36.1^{\circ} \mathrm{C}\). Explain the differences in the physical properties of the two substances.

The following data present the temperatures at which certain vapor pressures are achieved for dichloromethane \(\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)\) and methyl iodide \(\left(\mathrm{CH}_{3} \mathrm{I}\right)\) : $$ \begin{array}{lllll} \text { Vapor Pressure } & & & & \\ \text { (torr): } & 10.0 & 40.0 & 100.0 & 400.0 \\ \hline T \text { for } \mathrm{CH}_{2} \mathrm{Cl}_{2}\left({ }^{\circ} \mathrm{C}\right): & -43.3 & -22.3 & -6.3 & 24.1 \\ T \text { for } \mathrm{CH}_{3} \mathrm{I}\left({ }^{\circ} \mathrm{C}\right): & -45.8 & -24.2 & -7.0 & 25.3 \end{array} $$ (a) Which of the two substances is expected to have the greater dipole-dipole forces? Which is expected to have the greater dispersion forces? Based on your answers, explain why it is difficult to predict which compound would be more volatile. (b) Which compound would you expect to have the higher boiling point? Check your answer in a reference book such as the CRC Handbook of Chemistry and Physics. (c) The order of volatility of these two substances changes as the temperature is increased. What quantity must be different for the two substances in order for this phenomenon to occur? (d) Substantiate your answer for part (c) by drawing an appropriate graph.

Ethyl chloride \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) boils at \(12^{\circ} \mathrm{C}\). When liquid \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) under pressure is sprayed on a room-temperature \(\left(25^{\circ} \mathrm{C}\right)\) surface in air, the surface is cooled considerably. (a) What does this observation tell us about the specific heat of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(g)\) as compared with \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(l) ?\) (b) Assume that the heat lost by the surface is gained by ethyl chloride. What enthalpies must you consider if you were to calculate the final temperature of the surface?
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