Additional Reading Materials

Chapter 7: Noncovalent Interactions and Functional Groups

Ch7.1 Forces Between Molecules

Consider these two aspects of the molecular-level environments in solid, liquid, and gaseous matter:

  • Particles in a solid are tightly packed together and often arranged in a regular pattern; in a liquid, they are close together with no regular arrangement; in a gas, they are far apart with no regular arrangement.
  • Particles in a solid vibrate about fixed positions and do not generally move in relation to one another; in a liquid, they move past each other but remain in essentially constant contact; in a gas, they move independently of one another except when they collide.

The differences in the properties of a solid, liquid, or gas reflect the strengths of the attractive forces between the atoms, molecules, or ions that make up each phase. The phase in which a substance exists depends on the relative extents of its intermolecular forces (IMFs) and the kinetic energies (KE) of its molecules. IMFs are the various forces of attraction that may exist between the molecules of a substance which serve to hold particles close together. The particles’ KE provides the energy required to overcome the attractive forces and thus increase the distance between particles. Figure 1 illustrates how changes in physical state may be induced by changing the temperature, hence, the average KE, of a given substance.

Three sealed flasks are labeled, “Crystalline solid,” “Liquid,” and “Gas,” from left to right. The first flask holds a cube composed of small spheres sitting on the bottom while the second flask shows a lot of small spheres in the bottom that are spaced a small distance apart from one another and have lines around them to indicate motion. The third flask shows a few spheres spread far from one another with larger lines to indicate motion. There is a right-facing arrow that spans the top of all three flasks. The arrow is labeled, “Increasing K E ( temperature ).” There is a left-facing arrow that spans the bottom of all three flasks. The arrow is labeled, “Increasing I M F.”
Figure 1. Transitions between solid, liquid, and gaseous states of a substance occur when conditions of temperature or pressure favor the associated changes in intermolecular forces. (Note: The space between particles in the gas phase is much greater than shown.)

As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between H2O molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid H2O.

Under appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, for example, the covalent bonds between the atoms. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. Figure 2 illustrates these different molecular forces. The strengths of these attractive forces can vary widely, although for mall molecules, the IMFs are usually weaker than intramolecular forces. For example, to overcome the IMFs in liquid HCl and convert it into gaseous HCl requires ~17 kJ/mol. To break the H-Cl covalent bond requires ~25 times more energy—430 kJ/mol.

An image is shown in which two molecules composed of a green sphere labeled “C l” connected on the right to a white sphere labeled “H” are near one another with a dotted line labeled “Intermolecular force ( weak )” drawn between them. A line connects the two spheres in each molecule and the line is labeled “Intramolecular force ( strong ).”
Figure 2. Intramolecular forces keep a molecule intact. Intermolecular forces hold multiple molecules together and determine many of a substance’s properties.

We will often use values such as boiling or enthalpies of vaporization as indicators of the relative strengths of IMFs present within substances, as we explore three major types of IMFs below.

Ch7.2 Dispersion Forces

One of the IMFs that is present in all condensed phases, regardless of the nature of the molecules composing the substance, is called the London dispersion force (in honor of German-born American physicist Fritz London who, in 1928, first explained it). This force is often referred to simply as the dispersion force. Because the electrons of a molecule are in constant motion at any moment in time, a molecule can develop an instantaneous dipole when its electrons are temporarily distributed asymmetrically. The presence of this instantaneous dipole can, in turn, distort the electron distribution of a neighboring molecule, producing an induced dipole. These two rapidly fluctuating, temporary dipoles result in an electrostatic attraction between the molecules—a so-called dispersion force as illustrated in Figure 3.

Two pairs of molecules are shown where each molecule has one larger blue side labeled “delta sign, negative sign” and a smaller red side labeled “delta sign, positive sign.” Toward the middle of the both molecules, but still on each distinct side, is a black dot. Between the two images is a dotted line labeled, “Attractive force.” In the first image, the red and blue sides are labeled, “Unequal distribution of electrons.” Below both images are brackets. The brackets are labeled, “Temporary dipoles.”
Figure 3. Dispersion forces result from the formation of temporary dipoles, as illustrated here for two nonpolar diatomic molecules.

Dispersion forces between molecules can attract the two molecules to each other. However, the forces are weak at long distance, and become significant only when the molecules are very close—it is proportional to [latex]\frac{1}{R^6}[/latex] where R is the intermolecular distance. Molecules with more electrons exhibit stronger dispersion forces than those with less electrons. For example, F2 and Cl2 are gases at room temperature (reflecting weaker attractive forces); Br2 is a liquid, and I2 is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table 1.

Halogen Number of e– Atomic Radius Melting Point Boiling Point
fluorine, F2 18 72 pm 53 K 85 K
chlorine, Cl2 34 99 pm 172 K 238 K
bromine, Br2 70 114 pm 266 K 332 K
iodine, I2 106 133 pm 387 K 457 K
astatine, At2 170 150 pm 575 K 610 K
Table 1. Melting and Boiling Points of the Halogens

The increase in melting and boiling points with increasing molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution is known as polarizability. A molecule that has a charge cloud that is more easily distorted is more polarizable and will have larger dispersion forces.

Example 1

London Dispersion Forces and Their Effects
Order the following compounds from lowest to highest boiling point: CH4, SiH4, GeH4, and SnH4. Explain your reasoning.

Solution
All of these molecules have a symmetric tetrahedral molecular geometry, and therefore are nonpolar. These compounds experience only dispersion forces. Hence, the molecule with less electrons will be less polarizable and have weaker dispersion forces. The number of electrons in CH4, SiH4, GeH4, and SnH4 are 10, 18, 36, and 54, respectively. Therefore, the order from lowest to highest boiling point is expected to be CH4 < SiH4 < GeH4 < SnH4.

A line graph, titled “Carbon Family,” is shown where the y-axis is labeled “Temperature, ( degree sign C )” and has values of “negative 200” to “negative 40” from bottom to top in increments of 20. The x-axis is labeled “Period” and has values of “0” to “5” in increments of 1. The first point on the graph is labeled “C H subscript 4” and is at point “2, negative 160.” The second point on the graph is labeled “S i H subscript 4” and is at point “3, negative 120” while the third point on the graph is labeled “G e H subscript 4” and is at point “4, negative 100.” The fourth point on the graph is labeled “S n H subscript 4” and is at point “5, negative 60.”

A graph of the actual boiling points of these compounds versus the period of the group 14 element shows this prediction to be correct:

Check Your Learning
Order the following hydrocarbons from lowest to highest boiling point: ethane (C2H6), propane (C3H8), and n-butane (C4H10).

Answer:

C2H6 < C3H8 < C4H10. All of these compounds are essentially nonpolar and only have London dispersion forces, therefore the molecule with more electrons (here, the longer chain linear alkane) will have stronger dispersion forces and the higher boiling point.

The shapes of molecules also affect the magnitudes of the dispersion forces between them. For example, boiling points for the isomers n-pentane, isopentane, and neopentane (Figure 4) are 36 °C, 27 °C, and 9.5 °C, respectively, even though these molecules have the same chemical formula, C5H12, and therefore same number of electrons. The difference in boiling points suggests that dispersion forces in the liquid phase are greatest for n-pentane and least for neopentane. In n-pentane, the elongated shape provides a greater surface area between n-pentane molecules when they come in contact, resulting in stronger dispersion forces. Neopentane molecules are the most compact of the three, offering the least available surface area for intermolecular contact and, hence, the weakest dispersion forces.

Three images of molecules are shown. The first shows a cluster of large, gray spheres each bonded together and to several smaller, white spheres. There is a gray, jagged line and then the mirror image of the first cluster of spheres is shown. Above these two clusters is the label, “Small contact area, weakest attraction,” and below is the label, “neopentane boiling point: 9.5 degrees C.” The second shows a chain of three gray spheres bonded by the middle sphere to a fourth gray sphere. Each gray sphere is bonded to several smaller, white spheres. There is a jagged, gray line and then the mirror image of the first chain appears. Above these two chains is the label, “Less surface area, less attraction,” and below is the label, “isopentane boiling point: 27 degrees C.” The third image shows a chain of five gray spheres bonded together and to several smaller, white spheres. There is a jagged gray line and then the mirror image of the first chain appears. Above these chains is the label, “Large contact area, strong attraction,” and below is the label, “n-pentane boiling point 36 degrees C.”
Figure 4. The strength of the dispersion forces increases with larger contact area between molecules, as demonstrated by the boiling points of these pentane isomers.

Geckos and Intermolecular Forces

Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos’ feet to behave this way.

Geckos’ toes are covered with hundreds of thousands of tiny hairs known as setae, with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae. The huge numbers of spatulae on its setae provide a gecko with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces—intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. The total attraction over millions of spatulae is large enough to support many times the gecko’s weight.

In 2014, two scientists developed a model to explain how geckos can rapidly transition from “sticky” to “non-sticky.” Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos’ feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Further investigations may eventually lead to the development of better adhesives and other applications.

Three figures are shown. The first is a photo of the bottom of a gecko’s foot. The second is bigger version which shows the setae. The third is a bigger version of the setae and shows the spatulae. (credit photo: modification of work by “JC*+A!”/Flickr)[/caption]

Ch7.3 Dipole-Dipole Attractions

Polar molecules have a partial positive charge on one side and a partial negative charge on the other side—a separation of charge called a dipole. The attractive force between the positive end of one dipole and the negative end of another dipole is called a dipole-dipole attraction. Consider a polar molecule such as hydrogen chloride, HCl, where the more electronegative Cl atom bears the partial negative charge and the less electronegative H atom bears the partial positive charge. In addition to dispersion forces, there are also dipole-dipole attraction between HCl molecules, as illustrated in Figure 5.

Two pairs of molecules are shown where each molecule has one larger blue side labeled “delta sign, negative sign” and a smaller red side labeled “delta sign, positive sign. In the first pair, the red sides of the two molecules both face to the left and the blue side to the right. A horizontal dotted line lies in between the two. In the second pair, the molecules face up and down, with the red and blue ends aligning. A horizontal dotted line lies between the red and blue ends facing upward and another lies between the red and blue ends facing downward.
Figure 5. This image shows two arrangements of polar molecules, such as HCl, that allow an attraction between the partial negative end of one molecule and the partial positive end of another.

The effect of dipole-dipole attraction is apparent when we compare the properties of HCl to the nonpolar F2. Both HCl and F2 consist of 18 electrons and have approximately the same shape. At a temperature of 150 K, molecules of both substances would have the same average KE. However, HCl is a liquid at that temperature, whereas F2, with only dispersion forces, is gaseous. The higher boiling point of HCl (188 K) compared to F2 (85 K) is a reflection of the additional dipole-dipole attractions between HCl molecules.

The strength of dipole-dipole attractions is dependent on the dipole moment of a molecule, which measures the extent of net charge separation in the molecule as a whole. Dipole moment has a unit of Debye (D). We determine the dipole moment by summing all the individual bond dipole moments while taking into account the 3D molecular structure.

If the bonds in a molecule are arranged such that their bond dipole moments cancel (vector sum equals zero), then the molecule is nonpolar. For example, in the linear CO2 molecule (Figure 6), each of the bonds is polar, but the molecule as a whole is nonpolar. This is because the bond dipole moments of the polar C=O bonds on opposite sides of the carbon atom cancel since they are equal in magnitude and are pointed in opposite directions. Other nonpolar molecules include BF3 (trigonal planar), CH4 (tetrahedral), PF5 (trigonal bipyramidal), and SF6 (octahedral), in which all the polar bonds are identical and are arranged such that their dipoles cancel.

Two images are shown and labeled, “a” and “b.” Image a shows a carbon atom bonded to two oxygen atoms in a ball-and-stick representation. Two arrows face away from the center of the molecule in opposite directions and are drawn horizontally like the molecule. These arrows are labeled, “Bond moments,” and the image is labeled, “Overall dipole moment equals 0.” Image b shows an oxygen atom bonded to two hydrogen atoms in a downward-facing v-shaped arrangement. An upward-facing, vertical arrow is drawn below the molecule while two upward and inward facing arrows are drawn above the molecule. The upper arrows are labeled, “Bond moments,” while the image is labeled, “Overall dipole moment.”
Figure 6. The overall dipole moment of a molecule depends on the individual bond dipole moments and how they are arranged. (a) Each CO bond has a bond dipole moment, but they point in opposite directions so that the net CO2 molecule is nonpolar. (b) In contrast, water is polar because the OH bond moments do not cancel out.

On the other hand, the water molecule (Figure 6) is bent because of the lone pairs on O, and the two O-H bond dipole moments do not cancel. Therefore, a water molecule has a net molecular dipole moment and is a polar molecule.

The OCS molecule has a structure similar to CO2, but a sulfur atom has replaced one of the oxygen atoms. To determine if this molecule is polar, we draw the molecular structure, which is linear:

An image shows a carbon atom double bonded to a sulfur atom and an oxygen atom which are arranged in a horizontal plane. Two arrows face away from the center of the molecule in opposite directions and are drawn horizontally like the molecule. The left-facing arrow is larger than the right-facing arrow. These arrows are labeled, “Bond moments,” and a left-facing arrow below the molecule is labeled, “Overall dipole moment.”

The C=O bond is considerably polar. Although C and S have very similar electronegativity values, S is slightly more electronegative than C, and so the C=S bond is just slightly polar. Hence OCS is a polar molecule, and because oxygen is more electronegative than sulfur, the oxygen end of the molecule is the negative end.

Chloromethane, CH3Cl, is another example of a polar molecule. Although the polar C–Cl and C–H bonds are arranged in a tetrahedral geometry, the C–Cl bond has a larger bond moment than the C–H bond. Additionally, all of the bond dipole moments have an upward component in the orientation shown, since carbon is more electronegative than hydrogen and less electronegative than chlorine:

An image shows a carbon atom single bonded to three hydrogen atoms and a chlorine atom. There are arrows with crossed ends pointing from the hydrogen to the carbon near each bond, and one pointing from the carbon to the chlorine along that bond. The carbon and chlorine arrow is longer. This image uses dashes and wedges to give it a three-dimensional appearance.

Example 2

Polarity Simulations
Open the molecule polarity simulation and select the “Three Atoms” tab at the top. This should display a molecule ABC with three electronegativity adjustors. You can display or hide the bond dipole moments, molecular dipole momentss, and partial charges at the right. Turning on the Electric Field will show whether the molecule moves when exposed to a field.

Use the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if:

(a) A and C are very electronegative and B is in the middle of the range.

(b) A is very electronegative, and B and C are not.

Solution
(a) Molecular dipole moment points immediately between A and C.

(b) Molecular dipole moment points along the A–B bond, toward A.

Check Your Learning
Determine the partial charges that will give the largest possible bond dipole moments.

Answer:

The largest bond dipole moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond dipole moment will not change based on whether B is the most electronegative or the least, the direction of the bond dipole moment will.

Example 3

Dipole-Dipole Forces and Their Effects
Predict which will have the higher boiling point: N2 or CO. Explain your reasoning.

Solution
CO and N2 are both diatomic molecules with 14 electrons, so they experience similar London dispersion forces. Because CO is a polar molecule, it also experiences dipole-dipole attractions. N2 is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The additional dipole-dipole attractions between CO molecules is expected to lead to a higher boiling point. [The boiling point of N2 is 77 K, the boiling point of CO is 82 K.]

Check Your Learning
Predict which will have the higher boiling point: ICl or Br2. Explain your reasoning.

Answer:

ICl. ICl and Br2 have the same number of electrons (70) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br2 is nonpolar and does not. The additional dipole-dipole attractions require additional energy to overcome, so ICl will have the higher boiling point.

Ch7.4 Hydrogen Bonding

Nitrosyl fluoride (ONF, 24 electrons, boiling point 201 K) is a gas at room temperature. Water (H2O, 10 electrons, boiling point 373 K) is a liquid. We clearly cannot attribute this difference between the two compounds to dispersion forces, since both molecules have about the same shape and ONF has more electrons. We also cannot attribute it to differences in molecular dipole moments, since both molecules are polar and exhibit comparable dipole moments. This large difference in boiling points is due to the presence of an additional IMF, called hydrogen bonding. Hydrogen bonding occurs between a hydrogen atom that is covalently bonded to F, O, or N (denoted as X) and an electron lone pair on F, O, or N (denoted as Z). These three atoms, F, O, and N, are among the most electronegative elements in the periodic table, as well as being some of the smallest elements. These characteristics give rise to the formation of hydrogen bond (denoted at dotted bond): X-H···Z. Figure 7 illustrates hydrogen bonding between water molecules.

Five water molecules are shown near one another, but not touching. A dotted line lies between many of the hydrogen atoms on one molecule and the oxygen atom on another molecule.
Figure 7. Water molecules participate in multiple hydrogen-bonding interactions with nearby water molecules.

Despite use of the word “bond,” hydrogen bonds are still intermolecular attractive forces. A hydrogen bond is weaker than a typical covalent bond, only about 5-10% as strong, but it is stronger than dispersion forces between small molecules and many dipole-dipole attractions. Hydrogen bond is a distinctive intermolecular force from dipole-dipole attraction—while all substances that exhibit hydrogen bonding are polar molecules, not all polar molecules exhibit hydrogen bonding. In addition, dipole-dipole interactions only accounts for a part of the overall strength found in hydrogen bonding.

For small molecules, hydrogen bonds have a pronounced effect on a substance’s properties in condensed phases (liquids and solids). For example, consider the trends in boiling points for the binary hydrides of group 15 (NH3, PH3, AsH3, and SbH3), group 16 (H2O, H2S, H2Se, and H2Te), and group 17 (HF, HCl, HBr, and HI). The boiling points of the three heavier hydrides for each group are plotted in Figure 8. As we progress down any of these groups, the polarities of the molecules decrease slightly, whereas the number of electrons increase substantially. The effect of increasingly stronger dispersion forces dominates that of slightly weaker dipole-dipole attractions, and the boiling points are observed to increase steadily.

A line graph is shown where the y-axis is labeled “Boiling point (, degree sign, C )” and has values of “ negative 150” to “150” from bottom to top in increments of 50. The x-axis is labeled “Period” and has values of “0” to “5” in increments of 1. Three lines are shown on the graph and are labeled in the legend. The red line is labeled as “halogen family,” the blue is “oxygen family” and the green is “nitrogen family.” The first point on the red line is labeled “question mark” and is at point “2, negative 120”. The second point on the line is labeled “H C l” and is at point “3, negative 80” while the third point on the line is labeled “H B r” and is at point “4, negative 60”. The fourth point on the line is labeled “H I” and is at point “5, negative 40.” The first point on the green line is labeled “question mark” and is at point “2, negative 125.” The second point on the line is labeled “P H, subscript 3” and is at point “3, negative 80” while the third point on the line is labeled “A s H, subscript 3” and is at point “4, negative 55.” The fourth point on the line is labeled “S b H, subscript 3” and is at point “5, negative 10.” The first point on the blue line is labeled “question mark” and is at point “2, negative 80.” The second point on the line is labeled “H, subscript 2, S” and is at point “3, negative 55” while the third point on the line is labeled “H, subscript 2, S e” and is at point “4, negative 45.” The fourth point on the line is labeled “H, subscript 2, T e” and is at point “5, negative 3.”
Figure 8. For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase as you go down a group.

If we use this trend to predict the boiling points for the lightest hydride for each group, we would expect NH3 to boil at about −120 °C, H2O to boil at about −80 °C, and HF to boil at about −110 °C. However, the experimentally meastured boiling points for these compounds are dramatically higher than the trends would predict, as shown in Figure 9. The stark contrast between our predictions and reality provides compelling evidence for the presence of hydrogen bonding.

A line graph is shown where the y-axis is labeled “Boiling point, ( degree sign, C )” and has values of “negative 150” to “150” from bottom to top in increments of 50. The x-axis is labeled “Period” and has values of “0” to “5” in increments of 1. Three lines are shown on the graph and are labeled in the legend. The red line is labeled as “halogen family,” the blue is “oxygen family” and the green is “nitrogen family.” The first point on the red line is labeled “H F” and is at point “2, 25.” The second point on the line is labeled “H C l” and is at point “3, negative 80” while the third point on the line is labeled “H B r” and is at point “4, negative 60.” The fourth point on the line is labeled “H I” and is at point “5, negative 40.” The first point on the green line is labeled “N H, subscript 3” and is at point “2, negative 40.” The second point on the line is labeled “P H, subscript 3” and is at point “3, negative 80” while the third point on the line is labeled “A s H, subscript 3” and is at point “4, negative 55.” The fourth point on the line is labeled “S b H, subscript 3” and is at point “5, negative 10.” The first point on the blue line is labeled “H, subscript 2, O” and is at point “2, 100.” The second point on the line is labeled “H, subscript 2, S” and is at point “3, negative 55” while the third point on the line is labeled “H, subscript 2, S e” and is at point “4, negative 45.” The fourth point on the line is labeled “H, subscript 2, T e” and is at point “5, negative 3.”
Figure 9. In comparison to periods 3−5, the hydrides of period 2 elements in groups 17, 16 and 15 (F, O and N, respectively) exhibit anomalously high boiling points due to hydrogen bonding.

Example 4

Effect of Hydrogen Bonding on Boiling Points
Consider the compounds dimethylether (CH3OCH3), ethanol (CH3CH2OH), and propane (CH3CH2CH3). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning.

Solution
The 3D geometry of CH3OCH3, CH3CH2OH, and CH3CH2CH3 are similar, and they all have 26 electrons, so they will exhibit similar dispersion forces. Since CH3CH2CH3 is nonpolar, it exhibit only dispersion forces. Because CH3OCH3 is polar, it will also experience dipole-dipole attractions. Finally, CH3CH2OH has an −OH group, and so it will experience dispersion forces, dipole-dipole attraction and hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is CH3CH2CH3 < CH3OCH3 < CH3CH2OH. The boiling point of propane is −42.1 °C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.4 °C.

Check Your Learning
Would your predict the boiling point of methylamine (CH3NH2) to be greater, smaller, or about the same as ethane (CH3CH3, boiling point −89 °C). Explain your reasoning.

Answer:

The boiling point for methylamine is predicted to be significantly greater than that of ethane. CH3CH3 and CH3NH2 are similar in molecular geometry and number of electrons, but methylamine possesses an −NH group and therefore has dipole-dipole attraction and hydrogen bonding. This greatly increases its IMFs, and therefore its boiling point. (the known boiling point is −6 °C.)

Ch7.5 Alkanes/Petroleum

Each carbon atoms in an alkane has sp3 hybrid orbitals and is bonded to four other atoms, each of which is either carbon or hydrogen. Because of the sp3 hybridization, the bond angles in carbon chains are close to 109.5°, giving such chains in an alkane a zigzag shape.

The number of carbon atoms present in an alkane has no limit. Greater numbers of atoms in the molecules will lead to stronger intermolecular dispersion forces and correspondingly different physical properties. Properties such as melting point and boiling point (Table 2) usually vary smoothly and predictably as a function of the number of carbon and hydrogen atoms in a linear alkane molecule.

Alkane* Molecular Formula Melting Point (°C) Boiling Point (°C) Phase at STP** Number of Constitutional Isomers
methane CH4 –182.5 –161.5 gas 1
ethane C2H6 –183.3 –88.6 gas 1
propane C3H8 –187.7 –42.1 gas 1
butane C4H10 –138.3 –0.5 gas 2
pentane C5H12 –129.7 36.1 liquid 3
hexane C6H14 –95.3 68.7 liquid 5
heptane C7H16 –90.6 98.4 liquid 9
octane C8H18 –56.8 125.7 liquid 18
nonane C9H20 –53.6 150.8 liquid 35
decane C10H22 –29.7 174.0 liquid 75
tetradecane C14H30 5.9 253.5 solid 1858
octadecane C18H38 28.2 316.1 solid 60,523
Table 2. Properties of Some Alkanes

*Physical properties for C4H10 and heavier molecules are those of the normal isomer, n-butane, n-pentane, etc.

**STP indicates a temperature of 0 °C and a pressure of 1 atm.

As was discussed earlier, molecular shapes also affect the magnitudes of the dispersion forces. For example, 2-methylpropane (or isobutane) has the same C4H10 molecular formula as n-butane, but has a lower boiling point of -11.7 °C.

Alkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of C–H or C–C single bonds. Combustion is one such reaction:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Alkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, CH4, is the principal component of natural gas. Butane, C4H10, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of the differences in the boiling points of the various components of the mixture (see Figure 10), which arises from the differences in their intermolecular interactions.

This figure contains a photo of a refinery, showing large columnar structures. A diagram of a fractional distillation column is also shown. Near the bottom of the column, an arrow pointing into the column from the left shows a point of entry for heated crude oil. The column contains several layers at which different components are removed. At the very bottom, residue materials are removed through a pipe as indicated by an arrow out of the column. At each successive level, different materials are removed through pipes proceeding from the bottom to the top of the column. In order from bottom to top, these materials are fuel oil, followed by diesel oil, kerosene, naptha, gasoline, and refinery gas at the very top. To the right of the column diagram, a double sided arrow is shown that is blue at the top and gradually changes color to red moving downward. The blue top of the arrow is labeled, “Small molecules: low boiling point, very volatile, flows easily, ignites easily.” The red bottom of the arrow is labeled, “Large molecules: high boiling point, not very volatile, does not flow easily, does not ignite easily.”
Figure 10. In a column for the fractional distillation of crude oil, oil heated to about 425 °C in the furnace vaporizes when it enters the base of the tower. The vapors rise through bubble caps in a series of trays in the tower. As the vapors gradually cool, fractions of higher, then of lower, boiling points condense to liquids and are drawn off. (credit left: modification of work by Luigi Chiesa)

Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking.

There is not any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:

crackeqtn.gif

Or, showing more clearly what happens to the various atoms and bonds:

crackmodel.gif

This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. The octane is one of the molecules found in petrol (gasoline).

Ch7.6 Alcohols

Alcohols are derivatives of hydrocarbons in which an –OH group replaces a hydrogen atom. Although all alcohols have one or more hydroxyl (–OH) functional groups, they do not behave like bases such as NaOH. NaOH is an ionic compound that contains the OH– ions. Alcohols are covalent molecules; the hydroxyl group in an alcohol molecule is covalently bonded to a carbon atom.

Ethanol, CH3CH2OH, also called ethyl alcohol, is an important alcohol for human use. Ethanol is the alcohol produced by some species of yeast, and has long been prepared by humans via harnessing the metabolic efforts of these yeasts in fermenting various sugars:
This figure shows the reaction of glucose to produce ethanol and C O subscript 2. The reaction shows C subscript 6 H subscript 12 O subscript 6 ( a q ) arrow labeled “yeast” 2 C subscript 2 H subscript 5 O H (a q) plus 2 C O subscript 2 ( g ). The O H in ethanol is shown in red.

Large quantities of ethanol can be synthesized from the addition reaction of water with ethylene using an acid as a catalyst:
This reaction shows two carbons connected by a double bond, each with two bonded H atoms plus H O H arrow labeled “H subscript 3 O superscript plus” followed by two carbon atoms connected with a single bond with 5 bonded H atoms and an O H group shown in red at the right end of the molecule. The O of this group is shown with 2 pairs of electron dots.

Alcohols containing two or more hydroxyl groups can be made. Examples include 1,2-ethanediol (ethylene glycol, used in antifreeze) and 1,2,3-propanetriol (glycerine, used as a solvent for cosmetics and medicines):
Structural formulas for 1 comma 2 dash ethanediol and 1 comma 2 comma 3 dash propanetriol are shown. The first structure has a two C atom hydrocarbon chain with an O H group attached to each carbon. The O H groups are shown in red an each O atom has two sets of electron dots. Each C atom also has two H atoms bonded to it. The second structure shows a three C atom hydrocarbon chain with an O H group bonded to each carbon. The O H groups are shown in red, and each O atom has two sets of electron dots. The first C atom has two H atoms bonded to it. The second C atom has one H atom bonded to it. The third C atom has two H atoms bonded to it.

Ethylene glycol is the principal component of engine coolant for automobiles and is also used to manufacture polyester fibers. In 2005, nearly 1.8 × 1010 kg was produced worldwide. Glycerin is used as a lubricant and in the manufacture of explosives:

When nitroglycerin is mixed with a solid material such as nitrocellulose (which is made by treating cotton or wood pulp with nitric acid), the product is a form of dynamite.

Intermolecular Forces

The presence of the -OH functional group in alcohol makes alcohol molecules polar and capable of hydrogen bonding. For example, the ethylene glycol (1,2-ethanediol) molecule has two -OH groups which can participate in hydrogen bonding, and the glycerin (1,2,3-propanetriol) molecule has three. Both substances have rather high boiling points (198°C for ethylene glycol and 290°C for glycerin) and are syrupy, viscous liquids at room temperature. Their resistance to flowing freely is due to the network of hydrogen bonds that links each molecule to several of its fellows, making it more difficult for them to slide past one another. This highlight again the effect of hydrogen bonding on intermolecular forces and physical properties.

The IMFs present in alcohol are dispersion forces, dipole-dipole interaction, and hydrogen bonding. Such molecules will always have higher boiling points than similarly sized molecules without hydrogen bonding, i.e. more heat is necessary to separate them. For example, ethanol and dimethylether (methoxymethane) both have the same molecular formula, C2H6O.

They have the same number of electrons, and a similar length to the molecule. The dispersion forces and dipole-dipole attractions in each will be much the same. However, ethanol has a hydrogen atom attached directly to an oxygen – and that oxygen still has exactly the same two lone pairs as in a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one O-H bond in each ethanol molecule.

In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens aren’t sufficiently δ+ (no O-H bond) for hydrogen bonds to form. The effect due to the presence of additional hydrogen bonding in ethanol is reflected in the boiling points of ethanol and methoxymethane:

 

ethanol (with hydrogen bonding) image 78.5°C
methoxymethane (without hydrogen bonding) image -24.8°C

It is important to realize that dispersion forces and dipole-dipole interactions are also present in alcohol molecules. For example, all the following molecules contain the same number of electrons, and the first two are much the same length. The higher boiling point of the 1-butanol (butan-1-ol) compared to pentane is due to the additional hydrogen bonding as well as dipole-dipole interactions.

Comparing the two alcohols, both boiling points are high but they are not the same. The boiling point of the isobutanol (2-methylpropan-1-ol) isn’t as high as the butan-1-ol because the branching in the molecule lowers the strength of its dispersion forces.

We know that some liquids mix with each other in all proportions; in other words, they have infinite mutual solubility and are said to be miscible. Liquids that mix with water in all proportions are usually polar substances or substances that form hydrogen bonds. For such liquids, the dipole-dipole interactions and/or hydrogen bonding of the solute molecules with the water solvent molecules are at least as strong as those between molecules in the pure solute or in the pure solvent. Hence, the two kinds of molecules mix easily.

Likewise, nonpolar liquids are miscible with each other because there is no appreciable difference in the strengths of solute-solute, solvent-solvent, and solute-solvent intermolecular attractions. The solubility of polar molecules in polar solvents and of nonpolar molecules in nonpolar solvents is an illustration of the chemical axiom “like dissolves like.”

Two liquids that do not mix to an appreciable extent are called immiscible. Layers are formed when we pour immiscible liquids into the same container. Gasoline, benzene, carbon tetrachloride, and many other nonpolar liquids are immiscible with water. The attraction between the molecules of such nonpolar liquids and polar water molecules is ineffectively weak. The only strong attractions in such a mixture are between the water molecules, so they effectively squeeze out the molecules of the nonpolar liquid. The distinction between immiscibility and miscibility is really one of degrees, such that miscible liquids are of infinite mutual solubility, while liquids said to be immiscible are of very low (though not zero) mutual solubility.

Two liquids, such as bromine and water, that are of moderate mutual solubility are said to be partially miscible. Two partially miscible liquids usually form two layers when mixed. In the case of the bromine and water mixture, the upper layer is water, saturated with bromine, and the lower layer is bromine saturated with water. Since bromine is nonpolar, and, thus, not very soluble in water, the water layer is only slightly discolored by the bright orange bromine dissolved in it. Since the solubility of water in bromine is very low, there is no noticeable effect on the dark color of the bromine layer (Figure 11).

This figure shows three test tubes. The first test tube holds a dark orange-brown substance. The second test tube holds a clear substance. The amount of substance in both test tubes is the same. The third test tube holds a dark orange-brown substance on the bottom with a lighter orange substance on top. The amount of substance in the third test tube is almost double of the first two.
Figure 11. Bromine (the deep orange liquid on the left) and water (the clear liquid in the middle) are partially miscible. The top layer in the mixture on the right is a saturated solution of bromine in water; the bottom layer is a saturated solution of water in bromine. (credit: Paul Flowers)

For alcohol molecules, the -OH end is polar and capable of hydrogen bonding, making it hydrophilic (“water-loving”), capable of strong intermolecular interactions with water molecules. On the other hand, the hydrocarbon (alkyl) end of an alcohol molecule is nonpolar and is hydrophobic (“water-fearing”), where the only interaction with water molecules are dispersion forces. Hence, although the alkyl part of the molecule does not influence the reactivity of alcohol molecules, it does have a significant impact on the solubility of alcohols in water. Alcohols with small alkyl groups, e.g. methanol and ethanol, are completely miscible in water, while larger alcohol molecules, such as 1-octanol, are immiscible in water.

Oxidation Reactions

A carbon atom typically forms four bonds. Therefore, in an alcohol where carbon is bonded to an -OH group, there can be up to three carbon atoms directly bonded to the carbon atom bonded to the oxygen in -OH. As such, alcohols can be classified according to the number of alkyl groups attached to the carbon with the hydroxyl group. If one alkyl group is attached to that carbon, the alcohol is a primary (1º) alcohol. If two alkyl groups are attached, the alcohol is a secondary (2º) alcohol. If three alkyl groups are attached, the alcohol is a tertiary (3º) alcohol (Figure 12).

Figure 12. Generic forms of primary, secondary, and tertiary alcohol; R, R’, and R” designate alkyl groups that can be different. In orange are an example molecule of each type of alcohol drawn in skeletal structure.

All alcohols can be completely oxidized to carbon dioxide and water by oxygen in the air; that is, all alcohols are combustible. Like hydrocarbons, combustion is an important reaction of alcohols, but controlled oxidation is even more important because it can convert alcohols into other compounds that are chemically useful. The ease with which an alcohol can be oxidized and the extent of the oxidation depends on whether the alcohol is primary, secondary, or tertiary.

For primary alcohols, controlled, stepwise oxidation first yields compounds called aldehydes, then aldehydes can be further oxidized to carboxylic acids.

Oxidation of an organic compound can usually be recognized because either an oxygen atom is added to a molecule or two hydrogen atoms are lost from a molecule. For example, stepwise oxidation of ethanol first produces the aldehyde ethanal (commonly called acetaldehyde); further oxidation produces the carboxylic acid, ethanoic acid (commonly called acetic acid). An aldehyde has the functional group –CHO, where the carbon atom is double-bonded to an oxygen atom. A carboxylic acid has the functional group –COOH, in which the carbon atom is double-bonded to an oxygen atom and single-bonded to an oxygen atom in an OH group.

Note that acetaldehyde differs from ethanol by loss of one H atom from the oxygen atom and one H atom from the carbon on the right. Acetic acid differs from acetaldehyde by having an addition O atom on the right-hand carbon atom.

Controlled oxidation can be carried out in the laboratory using an aqueous solution of potassium permanganate (KMnO4(aq)) or an aqueous solution of potassium dichromate (K2Cr2O7(aq)).

When a similar controlled oxidation is applied to a secondary alcohol, such as 2-propanol, the oxidized molecule contains a C=O group that has two other carbon atoms attached to the C atom, this functional group is called a ketone. The ketone formed from 2-propanol is called propanone (commonly called acetone).

Again, note that in the ketone the number of hydrogen atoms is fewer by two than the number in the secondary alcohol. The oxidation corresponds with loss of two hydrogen atoms. Ketones are difficult to oxidize further, because there is no way to add another oxygen atom to the carbon atom in the C=O group, nor is there a way to remove hydrogen atoms from the C and O atoms in the C=O group.

Tertiary alcohols, which have no hydrogen atoms attached to the carbon that is bonded to the –OH group, are difficult to oxidize. If a primary alcohol, a secondary alcohol, and a tertiary alcohol are dissolved in water in three beakers and then treated with either potassium permanganate or potassium dichromate, only the primary and secondary alcohols will react. (The reaction can be observed because both permanganate ions and dichromate ions are colored (purple and orange, respectively). Thus for the primary and secondary alcohols, the color will disappear, but for tertiary alcohols there will be no color change.

Ch7.7 Carboxylic Acids

The odor of vinegar is caused by the presence of acetic acid, a carboxylic acid, in the vinegar. Carboxylic acids contain a carbonyl group (C=O) and a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. The second oxygen atom also bonds to a hydrogen atom. The names for carboxylic acids include prefixes that denote the lengths of the carbon chains in the molecules and are derived following nomenclature rules similar to those for inorganic acids and salts (The functional group for a carboxylic acid is shown in red on the left):
Two structures are shown. The first structure is labeled, “ethanoic acid,” and, “acetic acid.” This structure indicates a C atom to which H atoms are bonded above, below and to the left. To the right of this in red is a bonded group comprised of a C atom to which an O atom is double bonded above. To the right of the red C atom, an O atom is bonded which has an H atom bonded to its right. Both O atoms have two sets of electron dots. The second structure is labeled, “methyl ethanoate,” and, “methyl acetate.” This structure indicates a C atom to which H atoms are bonded above, below and to the left. In red, bonded to the right is a C atom with a double bonded O atom above and a single bonded O atom to the right. To the right of this last O atom in black is another C atom to which H atoms are bonded above, below and to the right. Both O atoms have two pairs of electron dots.

The simplest carboxylic acid is formic acid, HCOOH, known since 1670. Its name comes from the Latin word formicus, which means “ant”; it was first isolated by the distillation of red ants. It is partially responsible for the pain and irritation of ant and wasp stings, and is responsible for a characteristic odor of ants that can be sometimes detected in their nests.

Acetic acid, CH3COOH, constitutes 3–6% of vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present. Yeast cells present in the juice carry out the fermentation reactions, which change the sugar present in the juice to ethanol, then to acetic acid. Pure acetic acid has a penetrating odor and produces painful burns. It is an excellent solvent for many organic and some inorganic compounds, and it is essential in the production of cellulose acetate, a component of many synthetic fibers such as rayon.

Butyric acid, CH3CH2CH2-COOH, a component of rancid butter and Limburger cheese, has a vile odor. Adipic acid (IUPAC name: hexanedioic acid) is an example of a dicarboxylic acid—it has two -COOH functional groups, one on each end—and is used to make nylon.

Since the carboxyl group contains a highly polar C=O as well as an O-H group, hydrogen bonding is extensive among molecules of the carboxylic acids. Pure acetic acid is called glacial acetic acid because its melting point of 16.6 °C is high enough that it can freeze in a cold laboratory. It is also quite thick and syrupy because of extensive hydrogen bonding. The hydrogen bonding interaction strength between carboxylic acid molecules are generally stronger than that between alcohol molecules. For example, the boiling point of acetic acid is 118 °C, while that of 1-propanol is 97 °C and 2-propanol (isopropyl alcohol) is 82.6 °C.

The hydrogen atom in the -COOH group is acidic and will react with a base to form an ionic salt:
A chemical reaction is shown. On the left, a structure of propionic acid is indicated. This structure includes a 2 carbon hydrocarbon group on the left end in black. Above, below, and to the left, H atoms are bonded. This group is bonded to a red group comprised of a C atom to which an O atom is double bonded above. To the right of the red C atom, an O atom is connected with a single bond. To the right of the O atom, an H atom is bonded. To the right of this structure appears a plus and N a O H. Following the reaction arrow, the propionate ion is shown. This structure is in brackets. Appearing inside the brackets, is a 2 carbon hydrocarbon group on the left end. Above, below, and to the left, H atoms are bonded. To the right of this group, a group in red is attached comprised of a C atom to which an O atom is double bonded above and a second O atom is single bonded to the right. Outside the brackets appears a superscript minus symbol. This is followed by a plus sign, N a superscript plus another plus sign and H subscript 2 O. The singly bonded O atom in the propionate ion structure has 3 pairs of electron dots. All other O atoms have two pairs of electron dots.

Carboxylic acids are weak acids, meaning they are not 100% ionized in water. For example, about 1% of the acetic acid molecules dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution. However, when we look at comparable alcohols, such as ethanol, carboxylic acids are stronger acids by over ten powers of ten! Why should the presence of a carbonyl group adjacent to a hydroxyl group have such a profound effect on the acidity of the hydroxyl proton?

The acidity of a molecule is related to its behavior in acid-base equilibrium (we will learn this in much more detail later on in the semester), and equilibrium favors the thermodynamically more stable side. Consequently, anything that stabilizes the conjugate base (A–) of an acid will make that acid (H–A) more easily dissociate in water (i.e. a stronger acid).

The conjugate base of carboxylic acid, the carboxylate anion, is stabilized by resonance, as shown in Figure 13. The two contributing structures have equal weight in the hybrid, and the C–O bonds are of equal length (between a double and a single bond). Hence, the presence of carbonyl group make this stabilization possible, leading to a markedly increased acidity. In contrast, there’s no such resonance stabilization for the conjugate base of alcohol, the alkoxide anion.

Figure 13. Acetic acid is more than 1010 times more acidic than isopropanol, even though both molecules breaks an O-H bond in an acid-base reaction. This increased acidity is due to resonance stabilization of the excess electron in the acetate anion (the negative charge is shared by both oxygen atoms in acetate anion).

Ch7.8 Esters

The odor of ripe bananas and many other fruits is due to the presence of esters, compounds that can be prepared by the reaction of a carboxylic acid with an alcohol. Because esters do not have hydrogen bonds between molecules, they have lower vapor pressures than the alcohols and carboxylic acids from which they are derived (see Figure 14).

There are nine structures represented in this figure. The first is labeled, “raspberry,” and, “iso-butyl formate.” It shows an H atom with a line going up and to the right which then goes down and to the right. It goes up and to the right again and down and to the right and up and to the right. At the first peak is a double bond to an O atom. At the first trough is an O atom. At the second trough, there is a line going straight down. The second is labeled, “apple,” and, “butyl acetate.” There is a line that goes up and to the right, down and to the right, up and to the right, and down and to the right. At the second peak is a double bond to an O atom. At the end, on the right is O C H subscript 3. The third is labeled, “pineapple,” and, “ethyl butyrate.” It is a line that goes up and to the right, down and to the right, up and to the right, down and to the right, up and to the right, and down and to the right. At the second peak is a double bond to an O atom and at the second trough is an O atom. The fourth is labeled, “rum,” and “propyl isobutyrate.” It shows a line that goes down and to the right, up and to the right, down and to the right, up and to the right, down and to the right and up and to the right. The first complete peak has a double bond to an O atom and the second trough has an O atom. The fifth is labeled, “peach,” and “benzyl acetate.” It shows a line that goes up and to the right, down and to the right, up and to the right and down and to the right. This line connects to a hexagon with a circle inside it. The first peak has a double bond to an O atom and the first trough has an O atom. The sixth is labeled, “orange,” and, “octyl acetate.” It shows a line that goes up and to the right and down and to the right and up and to the right and down and to the right and up and to the right and down and to the right and up and to the right and down and to the right and up and to the right and down and to the right. The first peak has a double bond to an O atom and the first complete trough has and an O atom. The seventh is labeled, “wintergreen,” and “methyl salicylate.” It shows a hexagon with a circle inside of it. On the right, is a bond down and to the right to an O H group. On the right is a bond to a line that goes up and to the right and down and two the right and up and to the right. At the first peak is a double bond to an O atom, the next trough shows and O atom and at the end of the line is a C H subscript 3 group. The eighth is labeled, “honey,” and “methyl phenylacetate.” It shows a hexagon with a circle inside of it. It shows it connecting to a line on the right that goes down and to the right then up and to the right and down and to the right and up and to the right. At the first peak that is not part of the hexagon is a double bond to an O atom. At the last trough is an O atom. The ninth is labeled, “strawberry,” and “ethyl methylphenylglycidate.” This shows a hexagon with a circle inside of it. On the right, it connects to a line that goes up and to the right and down and to the right and up and to the right and down and to the right and up and to the right and down and to the right. At the first peak is a line that extends above and below. Below, it connects to an O atom. At the next trough, the line extends down and to the left to the same O atom. At the next peak is a double bond to an O atom and at the next trough is an O atom.
Figure 14. Esters are responsible for the odors associated with various plants and their fruits.

Similar to carboxylic acids, esters contain a carbonyl group (C=O) and a second oxygen atom bonded to the carbonyl carbon. In an ester, the second oxygen atom bonds to another carbon atom. The names for esters include prefixes that denote the lengths of the carbon chains in the molecules (the functional groups for an ester are shown in red on the right):
Two structures are shown. The first structure is labeled, “ethanoic acid,” and, “acetic acid.” This structure indicates a C atom to which H atoms are bonded above, below and to the left. To the right of this in red is a bonded group comprised of a C atom to which an O atom is double bonded above. To the right of the red C atom, an O atom is bonded which has an H atom bonded to its right. Both O atoms have two sets of electron dots. The second structure is labeled, “methyl ethanoate,” and, “methyl acetate.” This structure indicates a C atom to which H atoms are bonded above, below and to the left. In red, bonded to the right is a C atom with a double bonded O atom above and a single bonded O atom to the right. To the right of this last O atom in black is another C atom to which H atoms are bonded above, below and to the right. Both O atoms have two pairs of electron dots.

Esters are produced by the reaction of carboxylic acids with alcohols. For example, ethyl acetate, CH3COOCH2CH3, is formed when acetic acid reacts with ethanol:
A chemical reaction is shown. On the left, a C H subscript 3 group bonded to a red C atom. The C atom forms a double bond with an O atom which is also in red. The C atom is also bonded to an O atom which is bonded to an H atom, also in red. A plus sign is shown, which is followed by H O C H subscript 2 C H subscript 3. The H O group is in red. Following a reaction arrow, a C H subscript 3 group is shown which is bonded to a red C atom with a double bonded O atom and a single bonded O. To the right of this single bonded O atom, a C H subscript 2 C H subscript 3 group is attached and shown in black. This structure is followed by a plus sign and H subscript 2 O. The O atoms in the first structure on the left and the structure following the reaction arrow have two pairs of electron dots.

The distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits are due to the presence of one or more esters (Figure 15). Among the most important natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, C3H5(OH)3, with large carboxylic acids, such as palmitic acid, CH3(CH2)14COOH, stearic acid, CH3(CH2)16COOH, and oleic acid, CH3(CH2)7CH=CH(CH2)7COOH . Oleic acid is an unsaturated acid; it contains a C=C double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds.

This is a photo of a bright red strawberry being held in a human hand.
Figure 15. Over 350 different volatile molecules (many members of the ester family) have been identified in strawberries. (credit: Rebecca Siegel)

Ch7.9 Amines

Amines are molecules that contain carbon-nitrogen bonds. The nitrogen atom in an amine has a lone pair of electrons and three single bonds to other atoms, either carbon or hydrogen. Various nomenclatures are used to derive names for amines, but all involve the class-identifying suffix –ine as illustrated here for a few simple examples:
Three structures are shown, each with a red, central N atom which has a pair of electron dots indicated in red above the N atoms. The first structure is labeled methyl amine. To the left of the N, a C H subscript 3 group is bonded. H atoms are bonded to the right and bottom of the central N atom. The second structure is labeled dimethyl amine. This structure has C H subscript 3 groups bonded to the left and right of the N atom and a single H atom is bonded below. The third structure is labeled trimethyl amine, which has C H subscript 3 groups bonded to the left, right, and below the central N atom.

Similar to alcohols, you can group amines by the number of carbon atoms bonded to the nitrogen. In the above example, methyl amine is a primary amine, dimethyl amine is a secondary amine, and trimethyl amine is a tertiary amine.

Like ammonia, amines are weak bases due to the lone pair of electrons on their nitrogen atoms:
Two reactions are shown. In the first, ammonia reacts with H superscript plus. An unshared pair of electron dots sits above the N atom. To the left, right, and bottom, H atoms are bonded. This is followed by a plus symbol and an H atom with a superscript plus symbol. To the right of the reaction arrow, ammonium ion is shown in brackets with a superscript plus symbol outside. Inside the brackets, the N atom is shown with H atoms bonded on all four sides. In a very similar second reaction, methyl amine reacts with H superscript plus to yield methyl ammonium ion. The methyl amine structure is like ammonia except a C H subscript 3 group is attached in place of the left most H atom in the structure. Similarly, the resulting methyl ammonium ion is represented in brackets with a superscript plus symbol appearing outside. Inside, the structure is similar to that of methyl amine except that an H atom appears at the top of the N atom where the unshared electron pair was previously shown.

The basicity of an amine’s nitrogen atom plays an important role in much of the compound’s chemistry. Amine functional groups are found in a wide variety of compounds, including natural and synthetic dyes, polymers, vitamins, and medications such as penicillin and codeine. They are also found in many molecules essential to life, such as amino acids, hormones, neurotransmitters, and DNA.

Addictive Alkaloids

Since ancient times, plants have been used for medicinal purposes. One class of substances, called alkaloids, found in many of these plants has been isolated and found to contain cyclic molecules with an amine functional group. These amines are bases. They can react with H3O+ in a dilute acid to form an ammonium salt, and this property is used to extract them from the plant:

R3N + H3O+ + Cl– → [R3NH+]Cl– + H2O

The name alkaloid means “like an alkali.” Thus, an alkaloid reacts with acid. The free compound can be recovered after extraction by reaction with a base:

[R3NH+]Cl– + OH– → R3N + H2O + Cl–

The structures of many naturally occurring alkaloids have profound physiological and psychotropic effects in humans. Examples of these drugs include nicotine, morphine, codeine, and heroin. The plant produces these substances, collectively called secondary plant compounds, as chemical defenses against the numerous pests that attempt to feed on the plant:

Molecular structures of nicotine, morphine, codeine, and heroin are shown. These large structures share some common features, including rings. In the complex structures of morphine, codeine, and heroin, bonds to some O atoms in the structures are indicated with dashed wedges and bonds to some H atoms and N atoms are shown as solid wedges.

In these diagrams, as is common in representing structures of large organic compounds, carbon atoms in the rings and the hydrogen atoms bonded to them have been omitted for clarity. The solid wedges indicate bonds that extend out of the page. The dashed wedges indicate bonds that extend into the page. Notice that small changes to a part of the molecule change the properties of morphine, codeine, and heroin. Morphine, a strong narcotic used to relieve pain, contains two hydroxyl functional groups, located at the bottom of the molecule in this structural formula. Changing one of these hydroxyl groups to a methyl ether group forms codeine, a less potent drug used as a local anesthetic. If both hydroxyl groups are converted to esters of acetic acid, the powerfully addictive drug heroin results (Figure 16).

This is a photo of a field of red-orange poppies.
Figure 16. Poppies can be used in the production of opium, a plant latex that contains morphine from which other opiates, such as heroin, can be synthesized. (credit: Karen Roe)

Ch7.10 Amides

Amides are molecules that contain nitrogen atoms connected to the carbon atom of a carbonyl group. Like amines, various nomenclature rules may be used to name amides, but all include use of the class-specific suffix -amide:
This figure shows three structures. Two examples are provided. The basic structure has an H atom or R group bonded to a C atom which is double bonded to an O atom. The O atom as two sets of electron dots. The C atom is bonded to an N atom which in turn is bonded to two R groups or two H atoms. The N atom as one set of electron dots. The next structure includes acetamide, which has C H subscript 3 bonded to a C atom with a doubly bonded O atom. The second C atom is also bonded to N H subscript 2. Hexanamide has a hydrocarbon chain of length 6 involving all single bonds. The condensed structure is shown here. To the sixth C atom at the right end of the chain, an O atom is double bonded and an N H subscript 2 group is single bonded.

Amides can be produced when carboxylic acids react with amines or ammonia. A water molecule is eliminated from the reaction, and the C-N amide bond is formed from the remaining pieces of the carboxylic acid and the amine (note the similarity to formation of an ester from a carboxylic acid and an alcohol):
A chemical reaction is shown between a carboxylic acid and amine to form an amide and water. Structures are shown. The carboxylic acid is shown as a C H subscript 3 group bonded to a C H subscript 2 group bonded to a C atom with a double bonded O atom above and an O H group bonded to the right. There is a plus sign. The amine is shown as an N atom with two H atoms bonded to the bottom and left sides. A C H subscript 3 group is bonded to the right side of the N atom. To the right of an arrow, an amide is shown as a C H subscript 3 group bonded to a C H subscript 2 group bonded to a C atom which is double bonded to an O atom above and an N with an H atom bonded below. The N atom is bonded to a C H subscript 3 group. The final product indicated after a plus sign is water, H subscript 2 O.

The reaction between amines and carboxylic acids to form amides is biologically important. It is through this reaction that amino acids (molecules containing both an amine and a carboxylic acid functional group) link together in a polymer to form peptides and proteins.

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