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Chapter 12: Problem 2

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: \(\mathrm{C}_{6} \mathrm{H}_{6}, 80^{\circ} \mathrm{C} ; \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}, 132^{\circ} \mathrm{C}\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}, 156^{\circ} \mathrm{C} ; \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}, 182^{\circ} \mathrm{C}\)

Short Answer

Expert verified
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.

Step by step solution

01

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.
02

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.

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Most popular questions from this chapter

All solids contain defects or imperfections of structure or composition. Defects are important because they influence properties, such as mechanical strength. Two common types of defects are a missing ion in an otherwise perfect lattice, and the slipping of an ion from its normal site to a hole in the lattice. The holes discussed in this chapter are often called interstitial sites, since the holes are in fact interstices in the array of spheres. The two types of defects described here are called point de kcts because they occur within specific sites. In the 1930 s, two solidstate physicists, W. Schottky and J. Fraenkel, studied the two types of point defects: A Schottky defect corresponds to a missing ion in a lattice, while a Fraenkel defect corresponds to an ion that is displaced into an interstitial site. (a) An example of a Schottky defect is the absence of a \(\mathrm{Na}^{+}\) ion in the NaCl structure. The absence of a \(\mathrm{Na}^{+}\) ion means that a \(\mathrm{Cl}^{-}\) ion must also be absent to preserve electrical neutrality. If one NaCl unit is missing per unit cell, does the overall stoichiometry change, and what is the change in density? (b) An example of a Fraenkel defect is the movement of \(a \mathrm{Ag}^{+}\) ion to a tetrahedral interstitial site from its normal octahedral site in \(\mathrm{AgCl}\), which has a structure like \(\mathrm{NaCl}\). Does the overall stoichiometry of the compound change, and do you expect the density to change? (c) Titanium monoxide (TiO) has a sodium chloridelike structure. X-ray diffraction data show that the edge length of the unit cell is \(418 \mathrm{pm}\). The density of the crystal is \(4.92 \mathrm{g} / \mathrm{cm}^{3}\) Do the data indicate the presence of vacancies? If so, what type of vacancies?

If the triple point pressure of a substance is greater than 1 atm, which two of the following conclusions are valid? (a) The solid and liquid states of the substance cannot coexist at equilibrium. (b) The melting point and boiling point of the substance are identical. (c) The liquid state of the substance cannot exist. (d) The liquid state cannot be maintained in a beaker open to air at 1 atm pressure. (e) The melting point of the solid must be greater than \(0^{\circ} \mathrm{C}\) (f) The gaseous state at 1 atm pressure cannot be condensed to the solid at the triple point temperature.

As a liquid evaporated from an open container, its temperature was observed to remain roughly constant. When the same liquid evaporated from a thermally insulated container (a vacuum bottle or Dewar flask), its temperature was observed to drop. How would you account for this difference?

Because solid \(p\) -dichlorobenzene, \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2},\) sublimes rather easily, it has been used as a moth repellent. From the data given, estimate the sublimation pressure of \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}(\mathrm{s})\) at \(25^{\circ} \mathrm{C} .\) For \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2} ; \mathrm{mp}=\) \(53.1^{\circ} \mathrm{C} ;\) vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}(1)\) at \(54.8^{\circ} \mathrm{C}\) is \(10.0 \mathrm{mmHg} ; \Delta H_{\text {fus }}=17.88 \mathrm{kJ} \mathrm{mol}^{-1} ; \Delta H_{\text {vap }}=\) \(72.22 \mathrm{k}] \mathrm{mol}^{-1}\)

A crystalline solid contains three types of ions, \(\mathrm{Na}^{+}, \mathrm{O}^{2-},\) and \(\mathrm{Cl}^{-}\). The solid is made up of cubic unit cells that have \(\mathrm{O}^{2-}\) ions at each corner, \(\mathrm{Na}^{+}\) ions at the center of each face, and \(\mathrm{Cl}^{-}\) ions at the center of the cells. What is the chemical formula of the compound? What are the coordination numbers for the \(\mathrm{O}^{2-}\) and \(\mathrm{Cl}^{-}\) ions? If the length of one edge of the unit cell is \(a,\) what is the shortest distance from the center of a \(\mathrm{Na}^{+}\) ion to the center of an \(\mathrm{O}^{2-}\) ion? Similarly, what is the shortest distance from the center of a \(\mathrm{Cl}^{-}\) ion to the center of an \(\mathrm{O}^{2-}\) ion?
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