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Chapter 11: Problem 25

(a) What atoms must a molecule contain to participate in hydrogen bonding with other molecules of the same kind? (b) Which of the following molecules can form hydrogen bonds with other molecules of the same kind: \(\mathrm{CH}_{3} \mathrm{~F}, \mathrm{CH}_{3} \mathrm{NH}_{2}, \mathrm{CH}_{3} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{Br} ?\)

Short Answer

Expert verified
A molecule must have a hydrogen atom bonded to nitrogen (N), oxygen (O), or fluorine (F) to participate in hydrogen bonding with other molecules of the same kind. Out of the given molecules, \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) and \(\mathrm{CH}_{3}\mathrm{OH}\) can form hydrogen bonds with other molecules of the same kind, while \(\mathrm{CH}_{3}\mathrm{F}\) and \(\mathrm{CH}_{3}\mathrm{Br}\) cannot.

Step by step solution

01

Identify conditions for hydrogen bonding

To participate in hydrogen bonding, a molecule must have a hydrogen atom bonded to a highly electronegative element, usually nitrogen (N), oxygen (O), or fluorine (F). This creates a highly polar bond, enabling the positive hydrogen end to attract the negative end of another molecule containing N, O, or F.
02

Examine each molecule

Now, let's examine each of the given molecules and determine if they can form hydrogen bonds with other molecules of the same kind. 1. \(\mathrm{CH}_{3}\mathrm{F}\): This molecule has a hydrogen atom, but it is not bonded to the highly electronegative fluorine (F) atom. The hydrogen atoms are bonded to carbon (C), which is less electronegative. Therefore, this molecule cannot form hydrogen bonds with other molecules of the same kind. 2. \(\mathrm{CH}_{3}\mathrm{NH}_{2}\): Here, the hydrogen atoms are bonded to a highly electronegative nitrogen (N) atom. As a result, this molecule can participate in hydrogen bonding with other molecules of the same kind. 3. \(\mathrm{CH}_{3}\mathrm{OH}\): In this case, the hydrogen atom is bonded to a highly electronegative oxygen (O) atom, allowing this molecule to form hydrogen bonds with other molecules of the same kind. 4. \(\mathrm{CH}_{3}\mathrm{Br}\): This molecule has a hydrogen atom, but it is not bonded to a highly electronegative atom. The hydrogen atoms are bonded to carbon (C), and bromine (Br) is less electronegative than N, O, or F. Therefore, this molecule cannot form hydrogen bonds with other molecules of the same kind.
03

Conclusion

In summary, a molecule must have a hydrogen atom bonded to nitrogen (N), oxygen (O), or fluorine (F) to participate in hydrogen bonding with other molecules of the same kind. Out of the given molecules, \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) and \(\mathrm{CH}_{3}\mathrm{OH}\) can form hydrogen bonds with other molecules of the same kind, while \(\mathrm{CH}_{3}\mathrm{F}\) and \(\mathrm{CH}_{3}\mathrm{Br}\) cannot.

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Key Concepts

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

Electronegativity
Electronegativity is a fundamental concept in chemistry that refers to the ability of an atom to attract electrons towards itself in a chemical bond. This property can vary significantly across different elements. Generally, electronegativity increases across a period from left to right in the periodic table and decreases down a group.

When it comes to hydrogen bonding, the key elements involved are nitrogen (N), oxygen (O), and fluorine (F). These atoms are highly electronegative:
  • Fluorine is the most electronegative element, which makes it a strong hydrogen bond acceptor.
  • Oxygen comes next, known for its strong electron-pulling capabilities, often found in water molecules.
  • Nitrogen, while slightly less electronegative than oxygen, still readily participates in hydrogen bonds.
Understanding electronegativity helps us predict the behavior of molecules and their ability to form hydrogen bonds, particularly by assessing which atoms in a molecule might "pull" electrons and create polar bonds.
Hydrogen Bond Donor
A hydrogen bond donor is an atom within a molecule that provides the hydrogen atom needed for hydrogen bonding. For this to occur, the hydrogen must be bonded to an electronegative atom, creating a scenario where the hydrogen has a partial positive charge.

In the context of hydrogen bonding, typical hydrogen bond donors include:
  • Nitrogen (as in amines like CH3NH2)
  • Oxygen (such as in alcohols like CH3OH)
These atoms not only contribute to high polarity in the bond with hydrogen but also make the hydrogen available to interact with an electronegative atom from another molecule, thereby forming a strong intermolecular attraction known as a hydrogen bond. Without such donors, molecules cannot effectively participate in hydrogen bonding, as seen in less electronegative pairings like hydrogen bonded to carbon.
Intermolecular Forces
Intermolecular forces are forces of attraction or repulsion between neighboring particles (molecules, atoms, or ions). These forces are crucial in determining the physical properties of substances such as boiling points and solubility.

Hydrogen bonds are a specific type of intermolecular force. They are particularly strong dipole-dipole attractions that occur when a hydrogen atom bonded to a highly electronegative atom (like N, O, or F) attracts an electronegative atom on a neighboring molecule. This allows molecules to cling to each other more tightly:
  • Hydrogen bonds give water its unique properties, such as a high boiling point.
  • Such bonds also make hydrogen bonding participants, like CH3OH and CH3NH2, have higher boiling points compared to non-hydrogen bonding molecules.
Overall, understanding these interactions helps explain why some liquids, like water and alcohols, remain liquid at room temperature, while others, like methane, do not.

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Most popular questions from this chapter

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 and \(82.5^{\circ} \mathrm{C}\), respectively. Explain why the boiling point of propyl alcohol is higher, even though both have the molecular formula, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\).

True or false: (a) Molecules containing polar bonds must be polar molecules and have dipole-dipole forces. (b) For the halogen 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, the more polar bonds there are in a molecule, the stronger the dipole-dipole interaction. \(\mathbf{d}\) ) All other factors being the same, total attractive forces between linear molecules are greater than those between molecules whose shapes are nearly spherical. (e) The more electronegative the atom, the more polarizable it is.

Appendix \(\mathrm{B}\) lists the vapor pressure of water at various external pressures. (a) Plot the data in Appendix B,vapor pressure versus temperature \(\left({ }^{\circ} \mathrm{C}\right) .\) From your plot, estimate the vapor pressure of water at body temperature, \(37^{\circ} \mathrm{C}\). (b) Explain the significance of the data point at \(101.3 \mathrm{kPa}, 100^{\circ} \mathrm{C} .(\mathbf{c}) \mathrm{A}\) city at an altitude of \(1525 \mathrm{~m}\) above sea level has a barometric pressure of \(84.3 \mathrm{kPa}\). To what temperature would you have to heat water to boil it in this city? (d) A city at an altitude of \(150 \mathrm{~m}\) below sea level would have a barometric pressure of \(103.14 \mathrm{kPa}\). To what temperature would you have to heat water to boil it in this city?

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and the absolute temperatures at which they were measured, \(T_{1}\) and \(T_{2}\) : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a component of which is octane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)\). Octane has a vapor pressure of \(1.86 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(19.3 \mathrm{kPa}\) at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. \((\mathbf{c})\) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.81 . (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

Due to the environmental concern of fluorocarbons as refrigerants, a refrigerant based on a mixture of hydrocarbons was used as a replacement. It is a patented blend of ethane, propane, butane, and isobutane. Isobutane has a normal boiling point of \(-12^{\circ} \mathrm{C}\). The molar specific heat of liquid phase and gas phase isobutane are \(129.7 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) and \(95.2 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) respectively. The heat of vaporization for this compound is \(21.3 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(25.0 \mathrm{~g}\) of isobutane from a liquid at \(-50^{\circ} \mathrm{C}\) to a gas at \(40^{\circ} \mathrm{C}\).
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