Problem 2
Question
When another atom or group of atoms is substituted for one of the hydrogen atoms in benzene, \(C_{6} H_{6},\) the boiling point changes. Explain the order of the following boiling points: \(\mathrm{C}_{6} \mathrm{H}_{6}, 80^{\circ} \mathrm{C} ; \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}, 132^{\circ} \mathrm{C}\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}, 156^{\circ} \mathrm{C} ; \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}, 182^{\circ} \mathrm{C}\)
Step-by-Step Solution
Verified Answer
The atoms or groups of atoms substituted in the benzene molecule increase the boiling point by increasing the intermolecular forces of attraction. The strongest such force, hydrogen bonding, is seen in \(C_{6}H_{5}OH\) resulting in its high boiling point of 182°C.
1Step 1: Identify the nature of the substances
Benzene (\(C_{6} H_{6}\)) is a hydrocarbon and its structure tends to be non-polar. When hydrogen atoms in benzene are substituted with \(Cl\) (chlorine), \(Br\) (bromine), or \(OH\) (hydroxyl group), the substances become \(C_{6}H_{5}Cl\), \(C_{6}H_{5}Br\), and \(C_{6}H_{5}OH\) respectively.
2Step 2: Evaluate the impact of substitutions on boiling point
The boiling point of a substance depends on the intermolecular forces of attraction. Chlorine and bromine being larger atoms, induce polarizability, increasing the London Dispersion forces in the molecule, which increases the boiling point. However, the \(OH\) group not only causes the same effect due to oxygen being a relatively larger atom than hydrogen, but also makes the molecule polar leading to unbalanced charge distribution. This leads to the creation of hydrogen bonds, which are stronger intermolecular forces than London Dispersion forces. Hydrogen bond is the reason why the boiling point of \(C_{6}H_{5}OH\) is the highest among these substances.
Key Concepts
Intermolecular ForcesBenzene SubstitutionHydrogen Bonding
Intermolecular Forces
Understanding why substances have different boiling points starts with intermolecular forces. These forces are the attractions that help hold molecules together in liquid form. When we heat a substance, these forces need to be overcome for the substance to turn into gas. The stronger the intermolecular forces, the higher the boiling point.
- **London Dispersion Forces:** These are the weakest type of intermolecular forces. They arise from temporary shifts in electron density in atoms and molecules leading to a momentary induced dipole. All molecules experience London dispersion forces, but their strength depends on molecular size and shape. Larger molecules have a greater chance of inducing these forces.
- **Dipole-Dipole Interactions:** These forces occur between polar molecules. When there is a difference in electronegativity between atoms, the positive end of one molecule attracts the negative end of another, generating an intermolecular attraction.
- **Hydrogen Bonds:** These are very strong compared to other forms. They occur when hydrogen is directly bonded to highly electronegative atoms like nitrogen, oxygen, or fluorine, resulting in a significant positive charge in hydrogen, which is attracted to another electronegative atom in a nearby molecule.
Benzene Substitution
Benzene, with the formula \(C_6H_6\), is a basic aromatic hydrocarbon characterized by a stable ring of alternating carbon-carbon bonds. In modifications involving benzene, substituting hydrogen with other groups like chlorine, bromine, or hydroxyl, noticeable changes occur in chemical properties including boiling points.
When hydrogen atoms in benzene are replaced:
When hydrogen atoms in benzene are replaced:
- **Chlorine Substitution (\(C_6H_5Cl\)):** Increases the polarizability due to the larger size of the chlorine atom compared to hydrogen. This leads to stronger London dispersion forces which increase the boiling point of the molecule.
- **Bromine Substitution (\(C_6H_5Br\)):** Similar to chlorine, bromine is even larger, resulting in a relatively higher increase in London dispersion forces and a higher boiling point compared to chlorine substitution.
- **Hydroxyl (\(OH\)) Substitution (\(C_6H_5OH\)):** Adds both dipole character and potential for hydrogen bonding. These factors significantly increase the boiling point to much higher than that of chlorinated or brominated benzene derivatives.
Hydrogen Bonding
Hydrogen bonding is a particularly strong form of dipole-dipole interaction and substantially affects the boiling points of molecules. For hydrogen bonds to form, hydrogen must be directly bonded to a highly electronegative atom.In this exercise, benzene substituted with a hydroxyl group \((C_6H_5OH)\), allows for hydrogen bonding. Here's why it matters:
- **Electronegative Influence:** The oxygen in the hydroxyl group is highly electronegative, creating a substantial dipole moment within the molecule. This results in strong partial charges, facilitating hydrogen bonds.
- **Hydrogen Bonds Formation:** Once the hydroxyl group is part of the molecule, these partial charges make it possible for hydrogen bonds to form between the hydroxyl group of one molecule and the oxygen of a nearby molecule.
- **Boiling Point Effects:** These hydrogen bonds are much stronger than other types of intermolecular forces, requiring more energy to break. As a result, \(C_6H_5OH\) has the highest boiling point among the examined benzene derivatives.
Other exercises in this chapter
Problem 1
For each of the following substances describe the importance of dispersion (London) forces, dipoledipole interactions, and hydrogen bonding: (a) \(HCl;\) (b) \(
View solution Problem 5
One of the following substances is a liquid at room temperature and the others are gaseous: \(\mathrm{CH}_{3} \mathrm{OH}\) \(\mathrm{C}_{3} \mathrm{H}_{8} ; \m
View solution Problem 6
In which of the following compounds do you think that intramolecular hydrogen bonding is an important factor: \((\mathrm{a}) \mathrm{CH}_{3} \mathrm{CH}_{2} \ma
View solution Problem 7
How many water molecules can hydrogen bond to methanol?
View solution