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When an atom or a group of atoms is substituted for an \(\mathrm{H}\) atom in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\), the boiling point changes. Explain the order of the following boiling points: \(\mathrm{C}_{6} \mathrm{H}_{6}\left(80^{\circ} \mathrm{C}\right), \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\) \(\left(132^{\circ} \mathrm{C}\right), \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\left(156^{\circ} \mathrm{C}\right), \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\left(182^{\circ} \mathrm{C}\right)\).

Short Answer

Expert verified
The order of boiling points for the substituted benzenes is determined by the strength of intermolecular forces present in each compound. Benzene has only London dispersion forces, chlorobenzene has London dispersion forces and dipole-dipole forces, bromobenzene has stronger London dispersion forces and dipole-dipole forces, and phenol has London dispersion forces, dipole-dipole forces, and hydrogen bonding. The stronger the intermolecular forces, the higher the boiling point, resulting in the order: Benzene (80°C) < Chlorobenzene (132°C) < Bromobenzene (156°C) < Phenol (182°C).

Step by step solution

01

Identify the different substituted benzenes

In the exercise, the following substituted benzenes are provided: - Benzene (C6H6) (boiling point: 80°C) - Chlorobenzene (C6H5Cl) (boiling point: 132°C) - Bromobenzene (C6H5Br) (boiling point: 156°C) - Phenol (C6H5OH) (boiling point: 182°C). Now, the task is to explain the order of these boiling points.
02

Analyze intermolecular forces

Boiling points are determined by the strength of intermolecular forces in a compound. The stronger the intermolecular forces, the higher the boiling point. There are different types of intermolecular forces, such as London dispersion forces, dipole-dipole forces, and hydrogen bonding. We will analyze the presence and strength of these forces in each of the compounds.
03

Evaluate the effect of substituents on boiling points

For each of the compounds, let's look at how the substituent may affect the overall intermolecular forces. 1. Benzene (C6H6) - It has London dispersion forces because it is a nonpolar molecule. 2. Chlorobenzene (C6H5Cl) - The Cl atom is more electronegative than C and H atoms, making the molecule polar. It experiences both London dispersion forces and dipole-dipole forces. 3. Bromobenzene (C6H5Br) - The Br atom is more electronegative than C and H atoms, making the molecule polar as well. It also experiences both London dispersion forces and dipole-dipole forces. However, Br atom is larger and heavier than Cl, resulting in stronger London dispersion forces. 4. Phenol (C6H5OH) - The OH group is highly polar and forms hydrogen bonds with other molecules. It experiences London dispersion forces, dipole-dipole forces, and hydrogen bonding, which is the strongest type of intermolecular forces.
04

Explain the order of boiling points

To recap, here are the types of intermolecular forces present in each compound: - Benzene: London dispersion forces - Chlorobenzene: London dispersion forces and dipole-dipole forces - Bromobenzene: London dispersion forces (stronger than in chlorobenzene) and dipole-dipole forces - Phenol: London dispersion forces, dipole-dipole forces, and hydrogen bonding. As we established earlier, molecules with stronger intermolecular forces have higher boiling points. Therefore, the boiling point order can be explained: Benzene < Chlorobenzene < Bromobenzene < Phenol This is consistent with the given boiling points (80°C, 132°C, 156°C, and 182°C).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intermolecular Forces
Intermolecular forces are the forces of attraction between molecules. They play a crucial role in determining the physical properties of substances, such as boiling points. The primary types of intermolecular forces include London dispersion forces, dipole-dipole forces, and hydrogen bonds. Understanding these forces helps us explain why different substances have varying boiling points.
  • London Dispersion Forces: These are the weakest intermolecular forces. They occur due to temporary fluctuations in the electron distribution within atoms and molecules, leading to momentary dipoles. All molecules, regardless of being polar or nonpolar, exhibit London dispersion forces. However, they are stronger in larger and heavier atoms or molecules due to their greater electron cloud.
  • Dipole-Dipole Forces: These exist in polar molecules, where there is a permanent dipole moment due to differences in electronegativity between the bonded atoms. The positive end of one polar molecule attracts the negative end of another, leading to a stronger intermolecular force compared to London dispersion forces.
  • Hydrogen Bonding: This is a special type of dipole-dipole interaction that occurs when hydrogen is directly bonded to a highly electronegative atom such as nitrogen, oxygen, or fluorine. Hydrogen bonds are the strongest intermolecular forces and significantly increase the boiling points of substances.
The presence and strength of these forces determine how much energy is required to convert a liquid to a gas, which is observed as the boiling point.
Substituted Benzene
Substituted benzenes are formed when one or more hydrogen atoms in benzene are replaced by another atom or groups of atoms. This substitution affects the molecular properties, including the boiling point of the compounds. Let's take a closer look at some common substituted benzenes and how their substituents affect boiling points.
Benzene is a simple aromatic compound and is nonpolar, exhibiting only London dispersion forces. When hydrogen atoms in benzene are replaced with atoms such as chlorine or bromine, the polarity of the molecule increases due to the electronegativity of these atoms.
  • Chlorobenzene (C extsubscript{6}H extsubscript{5}Cl): Here, a chlorine atom substitutes for a hydrogen atom. Chlorine is more electronegative than carbon, creating a dipole that adds dipole-dipole forces to the intermolecular attractions. This increases the boiling point from that of regular benzene.
  • Bromobenzene (C extsubscript{6}H extsubscript{5}Br): Similar to chlorobenzene, bromine adds polarity to the benzene ring. However, bromine is larger and heavier, enhancing the London dispersion forces, thus increasing the boiling point even further compared to chlorobenzene.
Each substituent alters the molecule's structure and forces, impacting the boiling point. This is why the order of boiling points is benzene < chlorobenzene < bromobenzene.
Hydrogen Bonding
Hydrogen bonding is a particularly strong type of intermolecular force that occurs only in specific conditions. It's key to understanding why some organic compounds exhibit much higher boiling points compared to others. In the context of substituted benzene, let's consider phenol (C extsubscript{6}H extsubscript{5}OH).
  • Phenol: Phenol has an -OH group bonded to the benzene ring. The oxygen in the hydroxyl group is highly electronegative, drawing electron density away from the hydrogen and creating a large dipole moment. This allows the hydrogen to engage in hydrogen bonding with nearby molecules. These bonds are more robust than typical dipole-dipole interactions.
  • Impact on Boiling Point: The strong hydrogen bonds in phenol mean more energy is required to break these interactions and transition it to a gaseous state. Consequently, phenol has a significantly higher boiling point compared to other substituted benzenes like chlorobenzene and bromobenzene.
Overall, hydrogen bonds are like intermolecular super-glue, drastically raising boiling points where they are present.

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Most popular questions from this chapter

For a given substance, the liquid crystalline phase tends to be more viscous than the liquid phase. Why?

(a) What is the relationship between surface tension and temperature? (b) What is the relationship between viscosity and temperature? (c) Why do substances with high surface tension also tend to have high viscosities?

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 greater than dispersion forces. (d) All other factors being the same, dispersion forces between linear molecules are greater than those between molecules whose shapes are nearly spherical.

The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temperature and pressure. In an experiment, \(5.00 \mathrm{~L}\) of \(\mathrm{N}_{2}\) gas is passed through \(7.2146 \mathrm{~g}\) of liquid benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(26.0^{\circ} \mathrm{C}\). The liquid remaining after the experiment weighs \(5.1493 \mathrm{~g}\). Assuming that the gas becomes saturated with benzene vapor and that the total gas volume and temperature remain constant, what is the vapor pressure of the benzene in torr?

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 \(\left(\mathrm{CH}_{3} \mathrm{I}\right)\) : (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 for this phenomenon to occur? (d) Substantiate your answer for part (c) by drawing an appropriate graph.

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