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Solutions are all around us. Air, for example, is a solution. If you live near a lake, a river, or an ocean, that body of water is not pure H2O but most probably a solution. Much of what we drink—for example, soda, coffee, tea, and milk—is at least in part a solution. Solutions are a large part of everyday life. A lot of the chemistry occurring around us happens in solution. In fact, much of the chemistry that occurs in our own bodies takes place in solution, and many solutions—such as the Ringer’s lactate IV solution—are important for our health. In our understanding of chemistry, we need to understand a little bit about solutions. In this chapter, you will learn about the special characteristics of solutions, how solutions are characterized, and some of their properties.
• 9.0: Prelude to Solutions
• 9.1: Solutions
Solutions form because a solute and a solvent experience similar intermolecular interactions.
• 9.2: Concentration
Various concentration units are used to express the amounts of solute in a solution. Concentration units can be used as conversion factors in stoichiometry problems. New concentrations can be easily calculated if a solution is diluted.
• 9.3: The Dissolution Process
When a solute dissolves, its individual particles are surrounded by solvent molecules and are separated from each other.
• 9.4: Properties of Solutions
Certain properties of solutions differ from those of pure solvents in predictable ways.
• 9.5: Chemical Equilibrium
Chemical equilibrium can be attained whether the reaction begins with all reactants and no products, all products and no reactants, or some of both. It may be tempting to think that once equilibrium has been reached, the reaction stops. Chemical equilibrium is a dynamic process. The forward and reverse reactions continue to occur even after equilibrium has been reached. Because the rates of the reactions are the same, there is no change in the relative concentrations of reactants and products.
• 9.6: Le Chatelier's Principle
The description of how a system responds to a stress to equilibrium has become known as Le Châtelier's principle: When a chemical system that is at equilibrium is disturbed by a stress, the system will respond in order to relieve the stress. Stresses to a chemical system involve changes in the concentrations of reactants or products, changes in the temperature of the system, or changes in the pressure of the system.
• 9.7: Osmosis and Diffusion
Fish cells, like all cells, have semipermeable membranes. Eventually, the concentration of "stuff" on either side of them will even out. A fish that lives in salt water will have somewhat salty water inside itself. Put it in freshwater, and the freshwater will, through osmosis, enter the fish, causing its cells to swell, and the fish will die. What will happen to a freshwater fish in the ocean?
• 9.E: Solutions (Exercises)
Problems and select solutions to this chapter.
• 9.S: Solutions (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
09: Solutions
If you watch any of the medical dramas on television, you may have heard a doctor (actually an actor) call for an intravenous solution of “Ringer’s lactate” (or “lactated Ringer’s” or "Ri-Lac"). So what is Ringer’s lactate?
Intravenous (IV) solutions are administered for two main reasons:
1. to introduce necessary substances into the bloodstream, such as ions for proper body function, sugar and other food substances for energy, or drugs to treat a medical condition, and
2. to increase the volume of the bloodstream.
Many people with acute or long-term medical conditions have received some type of an IV solution.
One basic IV solution, called a normal saline solution, is simply a dilute solution of NaCl dissolved in water. Normal saline is 9.0 g of NaCl dissolved in each liter of solution. Ringer’s lactate is a normal saline solution that also has small amounts of potassium and calcium ions mixed in. In addition, it contains about 2.5 g of lactate ions (C3H5O3) per liter of solution. The liver metabolizes lactate ions into bicarbonate (HCO3) ions, which help maintain the acid-base balance of blood. Many medical problems, such as heart attacks and shock, affect the acid-base balance of blood, and the presence of lactate in the IV solution eases problems caused by this imbalance.
Physicians can select from a range of premade IV solutions, in accordance with a patient’s particular needs. Ringer’s lactate is commonly used when a patient’s blood volume must be increased quickly. Another frequently used IV solution, called D5W, is a 5% solution of dextrose (a form of sugar) in water. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.00%3A_Prelude_to_Solutions.txt |
Learning Objectives
• To understand what causes solutions to form.
A solution is another name for a homogeneous mixture. A mixture as a material composed of two or more substances. In a solution, the combination is so intimate that the different substances cannot be differentiated by sight, even with a microscope. Compare, for example, a mixture of salt and pepper and another mixture consisting of salt and water. In the first mixture, we can readily see individual grains of salt and the flecks of pepper. A mixture of salt and pepper is not a solution. However, in the second mixture, no matter how carefully we look, we cannot see two different substances. Salt dissolved in water is a solution.
The major component of a solution, called the solvent, is typically the same phase as the solution itself. Each minor component of a solution (and there may be more than one) is called the solute. In most of the solutions we will describe in this textbook, there will be no ambiguity about whether a component is the solvent or the solute. For example, in a solution of salt in water, the solute is salt, and solvent is water.
Solutions come in all phases, and the solvent and the solute do not have to be in the same phase to form a solution (such as salt and water). For example, air is a gaseous solution of about 80% nitrogen and about 20% oxygen, with some other gases present in much smaller amounts. An alloy is a solid solution consisting of a metal (like iron) with some other metals or nonmetals dissolved in it. Steel, an alloy of iron and carbon and small amounts of other metals, is an example of a solid solution. Table $1$ lists some common types of solutions, with examples of each.
Table $1$: Types of Solutions
Solvent Phase Solute Phase Example
gas gas air
liquid gas carbonated beverages
liquid liquid ethanol (C2H5OH) in H2O (alcoholic beverages)
liquid solid saltwater
solid gas H2 gas absorbed by Pd metal
solid liquid Hg(ℓ) in dental fillings
solid solid steel alloys
What causes a solution to form? The simple answer is that the solvent and the solute must have similar intermolecular interactions. When this is the case, the individual particles of solvent and solute can easily mix so intimately that each particle of solute is surrounded by particles of solute, forming a solution. However, if two substances have very different intermolecular interactions, large amounts of energy are required to force their individual particles to mix intimately, so a solution does not form. Thus two alkanes like n-heptane, C7H16, and n-hexane, C6H14, are completely miscible in all proportions. The C7H16 and C6H14 molecules are so similar (recall Section 4.6) that there are only negligible differences in intermolecular forces.
For a similar reason, methanol, CH3OH, is completely miscible with water. In this case both molecules are polar and can form hydrogen bonds among themselves, and so there are strong intermolecular attractions within each liquid. However, CH3OH dipoles can align with H2O dipoles, and CH3OH molecules can hydrogen bond to H2O molecules, and so the attractions among unlike molecules in the solution are similar to those among like molecules in each pure liquid.
This process leads to a simple rule of thumb: like dissolves like. Solvents that are very polar will dissolve solutes that are very polar or even ionic. Solvents that are nonpolar will dissolve nonpolar solutes. Thus water, being polar, is a good solvent for ionic compounds and polar solutes like ethanol (C2H5OH). However, water does not dissolve nonpolar solutes, such as many oils and greases (Figure $1$).
We use the word soluble to describe a solute that dissolves in a particular solvent, and the word insoluble for a solute that does not dissolve in a solvent. Thus, we say that sodium chloride is soluble in water but insoluble in hexane (C6H14). If the solute and the solvent are both liquids and soluble in any proportion, we use the word miscible, and the word immiscible if they are not.
Example $1$
Water is considered a polar solvent. Which substances should dissolve in water?
1. methanol (CH3OH)
2. sodium sulfate (Na2SO4)
3. octane (C8H18)
Solution
Because water is polar, substances that are polar or ionic will dissolve in it.
1. Because of the OH group in methanol, we expect its molecules to be polar. Thus, we expect it to be soluble in water. As both water and methanol are liquids, the word miscible can be used in place of soluble.
2. Sodium sulfate is an ionic compound, so we expect it to be soluble in water.
3. Like other hydrocarbons, octane is nonpolar, so we expect that it would not be soluble in water.
Exercise $1$
Toluene (C6H5CH3) is widely used in industry as a nonpolar solvent. Which substances should dissolve in toluene?
1. water (H2O)
2. sodium sulfate (Na2SO4)
3. octane (C8H18)
Answer
Octane only.
Example $2$
Predict which of the following compounds will be most soluble in water:
1. $\underset{\text{Ethanol}}{\mathop{\text{CH}_{\text{3}}\text{CH}_{\text{2}}\text{OH}}}\,$
2. $\underset{\text{Hexanol}}{\mathop{\text{CH}_{\text{3}}\text{CH}_{\text{2}}\text{CH}_{\text{2}}\text{CH}_{\text{2}}\text{CH}_{\text{2}}\text{CH}_{\text{2}}\text{OH}}}\,$
Solution
Since ethanol contains an OH group, it can hydrogen bond to water. Although the same is true of hexanol, the OH group is found only at one end of a fairly large molecule. The rest of the molecule can be expected to behave much as though it were a nonpolar alkane. This substance should thus be much less soluble than the first. Experimentally we find that ethanol is completely miscible with water, while only 0.6 g hexanol dissolves in 100 g water.
Exercise $2$
Would I2 be more soluble in CCl4 or H2O?
Answer
I2 is nonpolar. Of the two solvents, CCl4 is nonpolar and H2O is polar, so I2 would be expected to be more soluble in CCl4.
Key Takeaway
• Solutions form because a solute and a solvent experience similar intermolecular interactions. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.01%3A_Solutions.txt |
Learning Objectives
• Calculate percentage concentration (m/m, v/v, m/v), ppm and ppb.
• Calculate the molarity of a solution.
• Use concentration units to calculate the amount of solute in a solution.
• Use molarity to determine quantities in chemical reactions.
• Determine the resulting concentration of a diluted solution.
To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms whose meanings depend on various factors.
Solubility
There is usually a limit to how much solute will dissolve in a given amount of solvent. This limit is called the solubility of the solute. Some solutes have a very small solubility, while other solutes are soluble in all proportions. Table $1$ lists the solubilities of various solutes in water. Solubilities vary with temperature, so Table $1$ includes the temperature at which the solubility was determined.
Table $1$: Solubilities of Various Solutes in Water at 25°C (Except as Noted)
Substance Solubility (g in 100 mL of H2O)
AgCl(s) 0.019
C6H6(ℓ) (benzene) 0.178
CH4(g) 0.0023
CO2(g) 0.150
CaCO3(s) 0.058
CaF2(s) 0.0016
Ca(NO3)2(s) 143.9
C6H12O6 (glucose) 120.3 (at 30°C)
KBr(s) 67.8
MgCO3(s) 2.20
NaCl(s) 36.0
NaHCO3(s) 8.41
C12H22O11 (sucrose) 204.0 (at 20°C)
If a solution contains so much solute that its solubility limit is reached, the solution is said to be saturated, and its concentration is known from information contained in Table $1$. If a solution contains less solute than the solubility limit, it is unsaturated. Under special circumstances, more solute can be dissolved even after the normal solubility limit is reached; such solutions are called supersaturated and are not stable. If the solute is solid, excess solute can easily recrystallize. If the solute is a gas, it can bubble out of solution uncontrollably, like what happens when you shake a soda can and then immediately open it.
Precipitation from Supersaturated Solutions
Recrystallization of excess solute from a supersaturated solution usually gives off energy as heat. Commercial heat packs containing supersaturated sodium acetate (NaC2H3O2) take advantage of this phenomenon. You can probably find them at your local drugstore.
Video $1$: Watered-down sodium acetate trihydrate. Needle crystal is truly wonderful structures
Most solutions we encounter are unsaturated, so knowing the solubility of the solute does not accurately express the amount of solute in these solutions. There are several common ways of specifying the concentration of a solution.
Percent Composition
There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100:
$\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%} \nonumber$
If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration.
Example $1$
A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution?
Solution
We can substitute the quantities given in the equation for mass/mass percent:
$\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}$
Exercise $1$
A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution?
Answer
7.90%
For gases and liquids, volumes are relatively easy to measure, so the concentration of a liquid or a gas solution can be expressed as a volume/volume percent (% v/v): the volume of a solute divided by the volume of a solution times 100:
$\mathrm{\%\: v/v = \dfrac{volume\: of\: solute}{volume\: of\: solution}\times100\%} \nonumber$
Again, the units of the solute and the solution must be the same. A hybrid concentration unit, mass/volume percent (% m/v), is commonly used for intravenous (IV) fluids (Figure $1$). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100:
$\mathrm{\%\: m/v = \dfrac{mass\: of\: solute\: (g)}{volume\: of\: solution\: (mL)}\times100\%} \nonumber$
Using Percent Composition in Calculations
The percent concentration can be used to produce a conversion factor between the amount of solute and the amount of solution. As such, concentrations can be useful in a variety of stoichiometry problems as discussed in Chapter 6. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor.
As an example, if the given concentration is 5% v/v solution of alcohol, this means that there are 5 mL of alcohol dissolved in every 100 mL solution.
5 mL alcohol = 100 mL solution
The two possible conversion factors are written as follows:
$\mathrm{\dfrac{5\: mL\: alcohol}{100\: mL\: solution}}$ or $\mathrm{\dfrac{100\: mL\: solution}{5\: mL\: alcohol}}$
Use the first conversion factor to convert from a given amount of solution to amount of solute. The second conversion factor is used to convert from a given amount of solute to amount of solution. Given any two quantities in any percent composition, the third quantity can be calculated, as the following example illustrates.
Example $2$
A sample of 45.0% v/v solution of ethanol (C2H5OH) in water has a volume of 115 mL. What volume of ethanol solute does the sample contain?
Solution
A percentage concentration is simply the number of parts of solute per 100 parts of solution. Thus, the percent concentration of 45.0% v/v implies the following:
$\mathrm{45.0\%\: v/v \rightarrow \dfrac{45\: mL\: C_2H_5OH}{100\: mL\: solution}}$
That is, there are 45 mL of C2H5OH for every 100 mL of solution. We can use this fraction as a conversion factor to determine the amount of C2H5OH in 115 mL of solution:
$\mathrm{115\: mL\: solution\times\dfrac{45\: mL\: C_2H_5OH}{100\: mL\: solution}=51.8\: mL\: C_2H_5OH}$
Exercise $2$
What volume of a 12.75% m/v solution of glucose (C6H12O6) in water is needed to obtain 50.0 g of C6H12O6?
Answer
$\mathrm{50.0\: g\: C_6H_12O_6\times\dfrac{100\: mL\: solution}{12.75\: g\: C_6H_12O_6}=392\: mL\: solution}$
Example $3$
A normal saline IV solution contains 9.0 g of NaCl in every liter of solution. What is the mass/volume percent of normal saline?
Solution
We can use the definition of mass/volume percent, but first we have to express the volume in milliliter units:
1 L = 1,000 mL
Because this is an exact relationship, it does not affect the significant figures of our result.
$\mathrm{\%\: m/v = \dfrac{9.0\: g\: NaCl}{1,000\: mL\: solution}\times100\%=0.90\%\: m/v}$
Exercise $3$
The chlorine bleach that you might find in your laundry room is typically composed of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mL of solution. What is the mass/volume percent of the bleach?
Answer
$\mathrm{\%\: m/v = \dfrac{27.0\: g\: NaOCl}{500.0\: mL\: solution}\times100\%=5.40\%\: m/v}$
In addition to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm) and parts per billion (ppb). Both of these units are mass based and are defined as follows:
$\mathrm{ppm=\dfrac{mass\: of\: solute}{mass\: of\: solution}\times1,000,000} \nonumber$
$\mathrm{ppb=\dfrac{mass\: of\: solute}{mass\: of\: solution}\times1,000,000,000} \nonumber$
Similar to parts per million and parts per billion, related units include parts per thousand (ppth) and parts per trillion (ppt).
Concentrations of trace elements in the body—elements that are present in extremely low concentrations but are nonetheless necessary for life—are commonly expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon’s Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm.
In aqueous solutions, 1 ppm is essentially equal to 1 mg/L, and 1 ppb is equivalent to 1 µg/L.
Example $4$
If the concentration of cobalt in a human body is 21 ppb, what mass in grams of Co is present in a body having a mass of 70.0 kg?
Solution
A concentration of 21 ppb means “21 g of solute per 1,000,000,000 g of solution.” Written as a conversion factor, this concentration of Co is as follows:
$\mathrm{21\: ppb\: Co \rightarrow \dfrac{21\: g\: Co}{1,000,000,000\: g\: solution}}$
We can use this as a conversion factor, but first we must convert 70.0 kg to gram units:
$\mathrm{70.0\: kg\times\dfrac{1,000\: g}{1\: kg}=7.00\times10^4\: g}$
Now we determine the amount of Co:
$\mathrm{7.00\times10^4\: g\: solution\times\dfrac{21\: g\: Co}{1,000,000,000\: g\: solution}=0.0015\: g\: Co}$
This is only 1.5 mg.
Exercise $4$
An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in parts per million?
Answer
0.14 ppm
Molarity
Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.
Molarity is defined as the number of moles of a solute dissolved per liter of solution:
$\mathrm{molarity=\dfrac{number\: of\: moles\: of\: solute}{number\: of\: liters\: of\: solution}} \nonumber$
Molarity is abbreviated M (often referred to as “molar”), and the units are often abbreviated as mol/L. It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore
$\mathrm{\dfrac{1.5\: mol\: NaCl}{0.500\: L\: solution}=3.0\: M\: NaCl} \nonumber$
which is read as “three point oh molar sodium chloride.” Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl(aq).”
Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters.
If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L?
Step 1: convert the mass of solute to moles using the molar mass of HCl (36.46 g/mol):
$22.4\cancel{gHCl}\times \dfrac{1\: mol\: HCl}{36.46\cancel{gHCl}}=0.614\, mol\; HCl \nonumber$
Step 2: use the definition of molarity to determine the concentration:
$M \: =\: \dfrac{0.614\: mol\: HCl}{1.56L\: solution}=0.394\, M HCl \nonumber$
Example $5$
What is the molarity of an aqueous solution of 25.0 g of NaOH in 750 mL?
Solution
Before we substitute these quantities into the definition of molarity, we must convert them to the proper units. The mass of NaOH must be converted to moles of NaOH. The molar mass of NaOH is 40.00 g/mol:
$\mathrm{25.0\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}=0.625\: mol\: NaOH}$
Next, we convert the volume units from milliliters to liters:
$\mathrm{750\: mL\times\dfrac{1\: L}{1,000\: mL}=0.750\: L}$
Now that the quantities are expressed in the proper units, we can substitute them into the definition of molarity:
$\mathrm{M=\dfrac{0.625\: mol\: NaOH}{0.750\: L}=0.833\: M\: NaOH}$
Exercise $5$
If a 350 mL cup of coffee contains 0.150 g of caffeine (C8H10N4O2), what is the molarity of this caffeine solution?
Answer
0.00221 M
Using Molarity in Calculations
The definition of molarity can also be used to calculate a needed volume of solution, given its concentration and the number of moles desired, or the number of moles of solute (and subsequently, the mass of the solute), given its concentration and volume. As in the percent concentration, molarity can also be expressed as a conversion factor.
Molarity is defined as moles solute per liter solution. There is an understood 1 in the denominator of the conversion factor. For example, a 3.0 M solution of sucrose means that there are three moles of sucrose dissolved in every liter of solution. Mathematically, this is stated as follows:
3.0 moles sucrose = 1 L solution
Dividing both sides of this expression by either side, we generate two possible conversion factors:
$\mathrm{\dfrac{3.0\: mol\: sucrose}{1\: L\: solution}}$ or $\mathrm{\dfrac{1\: L\: solution}{3.0\: mol\: sucrose}}$
The first conversion factor can be used to convert from volume (L) of solution to moles solute, and the second converts from moles of solute to volume (L) of solution.
For example, suppose we are asked how many moles of sucrose are present in 0.108 L of a 3.0 M sucrose solution. The given volume (0.108 L) is multiplied by the first conversion factor to cancel the L units, and find that 0.32 moles of sucrose are present.
$0.108\cancel{L\, solution}\times \dfrac{3.0\, mol\, sucrose}{\cancel{1L\, solution}}=0.32\, mol\, sucrose \nonumber$
How many liters of 3.0 M sucrose solution are needed to obtain 4.88 mol of sucrose? In such a conversion, we multiply the given (4.88 moles sucrose) with the second conversion factor. This cancels the moles units and converts it to liters of solution.
$4.88\cancel{mol\, sucrose}\times \dfrac{1\, L\, solution}{\cancel{3.0\, mol\, sucrose}}=1.63\, L\, solution \nonumber$
Example $6$
1. What volume of a 0.0753 M solution of dimethylamine [(CH3)2NH] is needed to obtain 0.450 mol of the compound?
2. Ethylene glycol (C2H6O2) is mixed with water to make auto engine coolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueous solution?
Solution
1. To solve for the volume, multiply the "given" (0.450 mol of dimethylamine) with the molarity conversion factor (0.0753 M). Use the proper conversion factor to cancel the unit "mol" and get the unit volume (L) of solution:
$\mathrm{0.450\: mol\: dimethylamine\times\dfrac{1\: L\: solution}{0.0753\: mol\: dimethylamine}=5.98\: L\: solution}$
2. The strategy in solving this problem is to convert the given volume (5.00 L) using the 6.00 M (conversion factor) to solve for moles of ethylene glycol, which can then be converted to grams.
Step 1: Convert the given volume (5.00 L) to moles ethylene glycol.
$\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}=30.0\: mol\: C_2H_6O_2}$
Step 2: Convert 30.0 mols C2H6O2 to grams C2H6O2. Molar mass of C2H6O2= 62.08 g/mol
$\mathrm{30.0\: mol\: C_2H_6O_2\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}$
The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:
$\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}$
The final answer is rounded off to 3 significant figures. Thus, there are 1,860 g of C2H6O2 in the specified amount of engine coolant.
Note: Dimethylamine has a “fishy” odor. In fact, organic compounds called amines cause the odor of decaying fish.
Example $\PageIndex{6A}$
1. What volume of a 0.0753 M solution of dimethylamine [(CH3)2NH] is needed to obtain 0.450 mol of the compound?
2. Ethylene glycol (C2H6O2) is mixed with water to make auto engine coolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueous solution?
Solution
This is an alternative method in case you don't want to use the conversion factor for molarity. In both parts, we will use the definition of molarity to solve for the desired quantity.
1. $\mathrm{0.0753\: M=\dfrac{0.450\: mol\: (CH_3)_2NH}{volume\: of\: solution}}$
To solve for the volume of solution, we multiply both sides by volume of solution and divide both sides by the molarity value to isolate the volume of solution on one side of the equation:
$\mathrm{volume\:of\:solution = \dfrac{0.450\:mol\:(CH_3)_2NH}{0.0753\:M}=5.98\:L}$
Note that because the definition of molarity is mol/L, the division of mol by M yields L, a unit of volume.
1. The molar mass of C2H6O2 is 62.08 g/mol., so
$\mathrm{6.00\: M=\dfrac{moles\: of\: solute}{5.00\: L}}$
To solve for the number of moles of solute, we multiply both sides by the volume:
moles of solute = (6.00 M)(5.00 L) = 30.0 mol
Note that because the definition of molarity is mol/L, the product M × L gives mol, a unit of amount. Now, using the molar mass of C3H8O3, we convert mol to g:
$\mathrm{30.0\: mol\times\dfrac{62.08\: g}{mol}=1,860\: g}$
Thus, there are 1,860 g of C2H6O2 in the specified amount of engine coolant.
Exercise $6$
1. What volume of a 0.0902 M solution of formic acid (HCOOH) is needed to obtain 0.888 mol of HCOOH?
2. Acetic acid (HC2H3O2) is the acid in vinegar. How many grams of HC2H3O2 are in 0.565 L of a 0.955 M solution?
Answer
a. 9.84 L
b. 32.4 g
Using Molarity in Stoichiometry Problems
Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit. Consider the following chemical equation:
HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq)
Suppose we want to know how many liters of aqueous HCl solution will react with a given mass of NaOH. A typical approach to answering this question is as follows:
Figure $2$: Typical approach to solving Molarity problems
In itself, each step is a straightforward conversion. It is the combination of the steps that is a powerful quantitative tool for problem solving.
Example $7$
How many milliliters of a 2.75 M HCl solution are needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows:
HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq)
Solution
We will follow the flowchart to answer this question. First, we convert the mass of NaOH to moles of NaOH using its molar mass, 40.00 g/mol:
$\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}=4.63\: mol\: NaOH}$
Using the balanced chemical equation, we see that there is a one-to-one ratio of moles of HCl to moles of NaOH. We use this to determine the number of moles of HCl needed to react with the given amount of NaOH:
$\mathrm{4.63\: mol\: NaOH\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}=4.63\: mol\: HCl}$
Finally, we use the definition of molarity to determine the volume of 2.75 M HCl needed:
$\mathrm{2.75\: M\: HCl=\dfrac{4.63\: mol\: HCl}{volume\: of\: HCl\: solution}}$
$\mathrm{volume\: of\: HCl=\dfrac{4.63\: mol\: HCl}{2.75\: M\: HCl}=1.68\: L\times\dfrac{1,000\: mL}{1\: L}=1,680\: mL}$
We need 1,680 mL of 2.75 M HCl to react with the NaOH.
The same multi-step problem can also be worked out in a single line, rather than as separate steps, as follows:
$\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}\times\dfrac{1\: L\: HCl\: solution}{2.75\: mol\: HCl}\times\dfrac{1000\: mL\: HCl\: solution}{1\: L\: HCl\: solution}=1,680\: mL\: HCl\: solution}$
Our final answer (rounded off to three significant figures) is 1,680 mL HCl solution.
Exercise $7$
How many milliliters of a 1.04 M H2SO4 solution are needed to react with 98.5 g of Ca(OH)2? The balanced chemical equation for the reaction is as follows:
$H_2SO_{4(aq)} + Ca(OH)_{2(s)} \rightarrow 2H_2O_{(ℓ)} + CaSO_{4(aq)} \nonumber$
Answer
1,280 mL
The general steps for performing stoichiometry problems such as this are shown in Figure $3$. You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure $3$ indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end.
Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins. Table $2$ lists the typical concentrations of some of these solutes.
Table $2$: Approximate Concentrations of Various Solutes in Some Solutions in the Body*
Solution Solute Concentration (M)
blood plasma Na+ 0.138
K+ 0.005
Ca2+ 0.004
Mg2+ 0.003
Cl 0.110
HCO3 0.030
stomach acid HCl 0.10
urine NaCl 0.15
PO43 0.05
NH2CONH2 (urea) 0.30
*Note: Concentrations are approximate and can vary widely.
Looking Closer: The Dose Makes the Poison
Why is it that we can drink 1 qt of water when we are thirsty and not be harmed, but if we ingest 0.5 g of arsenic, we might die? There is an old saying: the dose makes the poison. This means that what may be dangerous in some amounts may not be dangerous in other amounts.
Take arsenic, for example. Some studies show that arsenic deprivation limits the growth of animals such as chickens, goats, and pigs, suggesting that arsenic is actually an essential trace element in the diet. Humans are constantly exposed to tiny amounts of arsenic from the environment, so studies of completely arsenic-free humans are not available; if arsenic is an essential trace mineral in human diets, it is probably required on the order of 50 ppb or less. A toxic dose of arsenic corresponds to about 7,000 ppb and higher, which is over 140 times the trace amount that may be required by the body. Thus, arsenic is not poisonous in and of itself. Rather, it is the amount that is dangerous: the dose makes the poison.
Similarly, as much as water is needed to keep us alive, too much of it is also risky to our health. Drinking too much water too fast can lead to a condition called water intoxication, which may be fatal. The danger in water intoxication is not that water itself becomes toxic. It is that the ingestion of too much water too fast dilutes sodium ions, potassium ions, and other salts in the bloodstream to concentrations that are not high enough to support brain, muscle, and heart functions. Military personnel, endurance athletes, and even desert hikers are susceptible to water intoxication if they drink water but do not replenish the salts lost in sweat. As this example shows, even the right substances in the wrong amounts can be dangerous!
Equivalents
Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq). One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol/L of Na+(aq) is also 1 Eq/L because sodium has a 1+ charge. A 1 mol/L solution of Ca2+(aq) ions has a concentration of 2 Eq/L because calcium has a 2+ charge. Dilute solutions may be expressed in milliequivalents (mEq)—for example, human blood plasma has a total concentration of about 150 mEq/L. (For more information about the ions present in blood plasma, see Chapter 3, Section 3.3.)
Dilutions
When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of moles of the solute does not change, while the volume of the solution increases. We can set up an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2).
$\text{mol}_1 = \text{mol}_2 \nonumber$
Since the moles of solute in a solution is equal to the molarity multiplied by the liters, we can set those equal.
$M_1 \times L_1 = M_2 \times L_2 \nonumber$
Finally, because the two sides of the equation are set equal to one another, the volume can be in any units we choose, as long as that unit is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes:
$M_1 \times V_1 = M_2 \times V_2 \nonumber$
Suppose that you have $100. \: \text{mL}$ of a $2.0 \: \text{M}$ solution of $\ce{HCl}$. You dilute the solution by adding enough water to make the solution volume $500. \: \text{mL}$. The new molarity can easily be calculated by using the above equation and solving for $M_2$.
$M_2 = \frac{M_1 \times V_1}{V_2} = \frac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl} \nonumber$
The solution has been diluted by one-fifth since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value.
Dilution Equations
Any units of concentration and volume can be used, as long as both concentrations and both volumes have the same unit. For example, if we are using M (molarity), then we can express the equation as follows:
Molarityinitial × volumeinitial = Molarityfinal × volumefinal or
M1V1 = M2V2
If we are using percent, the dilution equation is as follows:
%initial × volumeinitial = %final × volumefinal or
%1V1 = %2V2
Example $8$
A 125 mL sample of 0.900 M NaCl is diluted to 1,125 mL. What is the final concentration of the diluted solution?
Solution
Because the volume units are the same, and we are looking for the molarity of the final solution, we can use (concentration × volume)initial = (concentration × volume)final:
(0.900 M × 125 mL) = (concentration × 1,125 mL)
We solve by isolating the unknown concentration by itself on one side of the equation. Dividing by 1,125 mL gives
$\mathrm{concentration = \dfrac{0.900\: M\times125\: mL}{1,125\: mL}=0.100\: M}$
as the final concentration.
Exercise $8$
a. A nurse uses a syringe to inject 5.00 mL of 0.550 M heparin solution (heparin is an anticoagulant drug) into a 250 mL IV bag, for a final volume of 255 mL. What is the concentration of the resulting heparin solution?
b. A 0.885 M solution of KBr whose initial volume is 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution?
Answer
a. 0.0108 M
b. 135.4 mL
Preparing IV Solutions
In a hospital emergency room, a physician orders an intravenous (IV) delivery of 100 mL of 0.5% KCl for a patient suffering from hypokalemia (low potassium levels). Does an aide run to a supply cabinet and take out an IV bag containing this concentration of KCl?
Not likely. It is more probable that the aide must make the proper solution from an IV bag of sterile solution and a more concentrated, sterile solution, called a stock solution, of KCl. The aide is expected to use a syringe to draw up some stock solution and inject it into the waiting IV bag and dilute it to the proper concentration. Thus the aide must perform a dilution calculation.
Figure $5$:Preparing IV Solution © Thinkstock. Medical personnel commonly must perform dilutions for IV solutions.
If the stock solution is 10.0% KCl and the final volume and concentration need to be 100 mL and 0.50%, respectively, then it is an easy calculation to determine how much stock solution to use:
(10%)V1 = (0.50%)(100 mL)
V1 = 5 mL
Of course, the addition of the stock solution affects the total volume of the diluted solution, but the final concentration is likely close enough even for medical purposes.
Medical and pharmaceutical personnel are constantly dealing with dosages that require concentration measurements and dilutions. It is an important responsibility: calculating the wrong dose can be useless, harmful, or even fatal!
Key Takeaways
• Various concentration units are used to express the amounts of solute in a solution.
• Concentration units can be used as conversion factors in stoichiometry problems.
• New concentrations can be easily calculated if a solution is diluted. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.02%3A_Concentration.txt |
Learning Objectives
• To describe the dissolution process at the molecular level
The Dissolution Process
What occurs at the molecular level to cause a solute to dissolve in a solvent? The answer depends in part on the solute, but there are some similarities common to all solutes.
Recall the rule that like dissolves like. This means that substances must have similar intermolecular forces to form solutions. When a soluble solute is introduced into a solvent, the particles of solute can interact with the particles of solvent. In the case of a solid or liquid solute, the interactions between the solute particles and the solvent particles are so strong that the individual solute particles separate from each other and, surrounded by solvent molecules, enter the solution. (Gaseous solutes already have their constituent particles separated, but the concept of being surrounded by solvent particles still applies.) This process is called solvation and is illustrated in Figure \(1\). When the solvent is water, the word hydration, rather than solvation, is used.
Ionic Compounds and Covalent Compounds as Solutes
In the case of molecular solutes like glucose, the solute particles are individual molecules. However, if the solute is ionic, the individual ions separate from each other and become surrounded by solvent particles. That is, the cations and anions of an ionic solute separate when the solute dissolves. This process is referred to as dissociation (Figure \(1\)).
The dissociation of soluble ionic compounds gives solutions of these compounds an interesting property: they conduct electricity. Because of this property, soluble ionic compounds are referred to as electrolytes. Many ionic compounds dissociate completely and are therefore called strong electrolytes. Sodium chloride is an example of a strong electrolyte. Some compounds dissolve but dissociate only partially, and solutions of such solutes may conduct electricity only weakly. These solutes are called weak electrolytes. Acetic acid (CH3COOH), the compound in vinegar, is a weak electrolyte. Solutes that dissolve into individual neutral molecules without dissociation do not impart additional electrical conductivity to their solutions and are called nonelectrolytes. Table sugar (C12H22O11) is an example of a nonelectrolyte.
The term electrolyte is used in medicine to mean any of the important ions that are dissolved in aqueous solution in the body. Important physiological electrolytes include Na+, K+, Ca2+, Mg2+, and Cl.
Example \(1\)
The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte.
1. potassium chloride (KCl)
2. fructose (C6H12O6)
3. isopropyl alcohol [CH3CH(OH)CH3]
4. magnesium hydroxide [Mg(OH)2]
Solution
Each substance can be classified as an ionic solute or a nonionic solute. Ionic solutes are electrolytes, and nonionic solutes are nonelectrolytes.
1. Potassium chloride is an ionic compound; therefore, when it dissolves, its ions separate, making it an electrolyte.
2. Fructose is a sugar similar to glucose. (In fact, it has the same molecular formula as glucose.) Because it is a molecular compound, we expect it to be a nonelectrolyte.
3. Isopropyl alcohol is an organic molecule containing the alcohol functional group. The bonding in the compound is all covalent, so when isopropyl alcohol dissolves, it separates into individual molecules but not ions. Thus, it is a nonelectrolyte
4. Magnesium hydroxide is an ionic compound, so when it dissolves it dissociates. Thus, magnesium hydroxide is an electrolyte.
Exercise \(1\)
The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte.
1. acetone (CH3COCH3)
2. iron(III) nitrate [Fe(NO3)3]
3. elemental bromine (Br2)
4. sodium hydroxide (NaOH)
Answer
a. nonelectrolyte
b. electrolyte
c. nonelectrolyte
d. electrolyte
Electrolytes in Body Fluids
Our body fluids are solutions of electrolytes and many other things. The combination of blood and the circulatory system is the river of life, because it coordinates all the life functions. When the heart stops pumping in a heart attack, the life ends quickly. Getting the heart restarted as soon as one can is crucial in order to maintain life.
The primary electrolytes required in the body fluid are cations (of calcium, potassium, sodium, and magnesium) and anions (of chloride, carbonates, aminoacetates, phosphates, and iodide). These are nutritionally called macrominerals.
Electrolyte balance is crucial to many body functions. Here's some extreme examples of what can happen with an imbalance of electrolytes: elevated potassium levels may result in cardiac arrhythmias; decreased extracellular potassium produces paralysis; excessive extracellular sodium causes fluid retention; and decreased plasma calcium and magnesium can produce muscle spasms of the extremities.
When a patient is dehydrated, a carefully prepared (commercially available) electrolyte solution is required to maintain health and well being. In terms of child health, oral electrolyte is given when a child is dehydrated due to diarrhea. The use of oral electrolyte maintenance solutions, which is responsible for saving millions of lives worldwide over the last 25 years, is one of the most important medical advances in protecting the health of children in the century, explains Juilus G.K. Goepp, MD, assistant director of the Pediatric Emergency Department of the Children's Center at Johns Hopkins Hospital. If a parent provides an oral electrolyte maintenance solution at the very start of the illness, dehydration can be prevented. The functionality of electrolyte solutions is related to their properties, and interest in electrolyte solutions goes far beyond chemistry.
Sports drinks are designed to rehydrate the body after excessive fluid depletion. Electrolytes in particular promote normal rehydration to prevent fatigue during physical exertion. Are they a good choice for achieving the recommended fluid intake? Are they performance and endurance enhancers like they claim? Who should drink them?
Typically, eight ounces of a sports drink provides between fifty and eighty calories and 14 to 17 grams of carbohydrate, mostly in the form of simple sugars. Sodium and potassium are the most commonly included electrolytes in sports drinks, with the levels of these in sports drinks being highly variable. The American College of Sports Medicine says a sports drink should contain 125 milligrams of sodium per 8 ounces as it is helpful in replenishing some of the sodium lost in sweat and promotes fluid uptake in the small intestine, improving hydration.
Gatorade
In the summer of 1965, the assistant football coach of the University of Florida Gators requested scientists affiliated with the university study why the withering heat of Florida caused so many heat-related illnesses in football players and provide a solution to increase athletic performance and recovery post-training or game. The discovery was that inadequate replenishment of fluids, carbohydrates, and electrolytes was the reason for the “wilting” of their football players. Based on their research, the scientists concocted a drink for the football players containing water, carbohydrates, and electrolytes and called it “Gatorade.” In the next football season the Gators were nine and two and won the Orange Bowl. The Gators’ success launched the sports-drink industry, which is now a multibillion-dollar industry that is still dominated by Gatorade.
Key Takeaway
• When a solute dissolves, its individual particles are surrounded by solvent molecules and are separated from each other. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.03%3A_The_Dissolution_Process.txt |
Learning Objectives
• To describe how the properties of solutions differ from those of pure solvents.
Solutions are likely to have properties similar to those of their major component—usually the solvent. However, some solution properties differ significantly from those of the solvent. Here, we will focus on liquid solutions that have a solid solute, but many of the effects we will discuss in this section are applicable to all solutions.
Colligative Properties
Solutes affect some properties of solutions that depend only on the concentration of the dissolved particles. These properties are called colligative properties. Four important colligative properties that we will examine here are vapor pressure depression, boiling point elevation, freezing point depression, and osmotic pressure.
Molecular compounds separate into individual molecules when they are dissolved, so for every 1 mol of molecules dissolved, we get 1 mol of particles. In contrast, ionic compounds separate into their constituent ions when they dissolve, so 1 mol of an ionic compound will produce more than 1 mol of dissolved particles. For example, every mole of NaCl that dissolves yields 1 mol of Na+ ions and 1 mol of Cl ions, for a total of 2 mol of particles in solution. Thus, the effect on a solution’s properties by dissolving NaCl may be twice as large as the effect of dissolving the same amount of moles of glucose (C6H12O6).
Vapor Pressure Depression
All liquids evaporate. In fact, given enough volume, a liquid will turn completely into a vapor. If enough volume is not present, a liquid will evaporate only to the point where the rate of evaporation equals the rate of vapor condensing back into a liquid. The pressure of the vapor at this point is called the vapor pressure of the liquid.
The presence of a dissolved solid lowers the characteristic vapor pressure of a liquid so that it evaporates more slowly. (The exceptions to this statement are if the solute itself is a liquid or a gas, in which case the solute will also contribute something to the evaporation process. We will not discuss such solutions here.) This property is called vapor pressure depression and is depicted in Figure $1$.
Boiling Point and Freezing Point Effects
A related property of solutions is that their boiling points are higher than the boiling point of the pure solvent. Because the presence of solute particles decreases the vapor pressure of the liquid solvent, a higher temperature is needed to reach the boiling point. This phenomenon is called boiling point elevation. For every mole of particles dissolved in a liter of water, the boiling point of water increases by about 0.5°C. The addition of one mole of sucrose (molecular compound) in one liter of water will raise the boiling point from 1000C to 100.50C but the addition of one mole of NaCl in one liter of water will raise the boiling point by 2 x 0.50C = 10C. Furthermore, the addition of one mole of $\ce{CaCl2}$ in one liter of water will raise the boiling point by 3 x 0.50C = 1.50C.
Some people argue that putting a pinch or two of salt in water used to cook spaghetti or other pasta makes a solution that has a higher boiling point, so the pasta cooks faster. In actuality, the amount of solute is so small that the boiling point of the water is practically unchanged.
The presence of solute particles has the opposite effect on the freezing point of a solution. When a solution freezes, only the solvent particles come together to form a solid phase, and the presence of solute particles interferes with that process. Therefore, for the liquid solvent to freeze, more energy must be removed from the solution, which lowers the temperature. Thus, solutions have lower freezing points than pure solvents do. This phenomenon is called freezing point depression. For every mole of particles in a liter of water, the freezing point decreases by about 1.9°C.
Both boiling point elevation and freezing point depression have practical uses. For example, solutions of water and ethylene glycol (C2H6O2) are used as coolants in automobile engines because the boiling point of such a solution is greater than 100°C, the normal boiling point of water. In winter, salts like NaCl and $\ce{CaCl2}$ are sprinkled on the ground to melt ice or keep ice from forming on roads and sidewalks (Figure $2$). This is because the solution made by dissolving sodium chloride or calcium chloride in water has a lower freezing point than pure water, so the formation of ice is inhibited.
Example $1$
Which solution’s freezing point deviates more from that of pure water—a 1 M solution of NaCl or a 1 M solution of $\ce{CaCl2}$?
Solution
Colligative properties depend on the number of dissolved particles, so the solution with the greater number of particles in solution will show the greatest deviation. When NaCl dissolves, it separates into two ions, Na+ and Cl. But when $\ce{CaCl2}$ dissolves, it separates into three ions—one Ca2+ ion and two Cl ions. Thus, mole for mole, $\ce{CaCl2}$ will have 50% more impact on freezing point depression than NaCl.
Exercise $1$
Which solution’s boiling point deviates more from that of pure water—a 1 M solution of $\ce{CaCl2}$ or a 1 M solution of MgSO4?
Answer
$\ce{CaCl2}$
Example $2$
Estimate the boiling point of 0.2 M $\ce{CaCl2}$ solution.
Solution
The boiling point increases 0.50C for every mole of solute per liter of water. For this estimation, let's assume that 1 liter of solution is roughly the same volume as 1 liter of water. A 0.2 M $\ce{CaCl2}$ solution contains 0.2 moles of $\ce{CaCl2}$ solution formula units per liter of solution. Each $\ce{CaCl2}$ unit separates into three ions.
$\mathrm{0.2\: mol\: CaCl_2\times\dfrac{3\: mol\: ions}{1\: mol\: CaCl_2}\times\dfrac{0.5\: deg\: C}{1\: mol\: ion}=0.3\: deg\: C} \nonumber$
The normal boiling point of water is 1000C, so the boiling point of the solution is raised to 100.30C.
Exercise $2$
Estimate the freezing point of 0.3 M $\ce{CaCl2}$ solution.
Answer
minus 1.70C
Osmotic Pressure
The last colligative property of solutions we will consider is a very important one for biological systems. It involves osmosis, the process by which solvent molecules can pass through certain membranes but solute particles cannot. When two solutions of different concentration are present on either side of these membranes (called semipermeable membranes), there is a tendency for solvent molecules to move from the more dilute solution to the more concentrated solution until the concentrations of the two solutions are equal. This tendency is called osmotic pressure. External pressure can be exerted on a solution to counter the flow of solvent; the pressure required to halt the osmosis of a solvent is equal to the osmotic pressure of the solution.
Osmolarity (osmol) is a way of reporting the total number of particles in a solution to determine osmotic pressure. It is defined as the molarity of a solute times the number of particles a formula unit of the solute makes when it dissolves (represented by $i$):
$osmol = M \times i\label{Eq1}$
If more than one solute is present in a solution, the individual osmolarities are additive to get the total osmolarity of the solution. Solutions that have the same osmolarity have the same osmotic pressure. If solutions of differing osmolarities are present on opposite sides of a semipermeable membrane, solvent will transfer from the lower-osmolarity solution to the higher-osmolarity solution. Counterpressure exerted on the high-osmolarity solution will reduce or halt the solvent transfer. An even higher pressure can be exerted to force solvent from the high-osmolarity solution to the low-osmolarity solution, a process called reverse osmosis. Reverse osmosis is used to make potable water from saltwater where sources of fresh water are scarce.
Example $3$
A 0.50 M NaCl aqueous solution and a 0.30 M Ca(NO3)2 aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow.
Solution
The solvent will flow into the solution of higher osmolarity. The NaCl solute separates into two ions—Na+ and Cl—when it dissolves, so its osmolarity is as follows:
osmol (NaCl) = 0.50 M × 2 = 1.0 osmol
The Ca(NO3)2 solute separates into three ions—one Ca2+ and two NO3—when it dissolves, so its osmolarity is as follows:
osmol [Ca(NO3)2] = 0.30 M × 3 = 0.90 osmol
The osmolarity of the Ca(NO3)2 solution is lower than that of the NaCl solution, so water will transfer through the membrane from the Ca(NO3)2 solution to the NaCl solution.
Exercise $3$
A 1.5 M C6H12O6 aqueous solution and a 0.40 M Al(NO3)3 aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow.
Answer
osmol C6H12O6 = 1.5; osmol Al(NO3)3 = 1.6
The solvent flows from C6H12O6 solution (lower osmolarity) to $\ce{Al(NO3)3}$ solution (higher osmolarity).
To Your Health: Dialysis
The main function of the kidneys is to filter the blood to remove wastes and extra water, which are then expelled from the body as urine. Some diseases rob the kidneys of their ability to perform this function, causing a buildup of waste materials in the bloodstream. If a kidney transplant is not available or desirable, a procedure called dialysis can be used to remove waste materials and excess water from the blood.
In one form of dialysis, called hemodialysis, a patient’s blood is passed though a length of tubing that travels through an artificial kidney machine (also called a dialysis machine). A section of tubing composed of a semipermeable membrane is immersed in a solution of sterile water, glucose, amino acids, and certain electrolytes. The osmotic pressure of the blood forces waste molecules and excess water through the membrane into the sterile solution. Red and white blood cells are too large to pass through the membrane, so they remain in the blood. After being cleansed in this way, the blood is returned to the body.
Dialysis is a continuous process, as the osmosis of waste materials and excess water takes time. Typically, 5–10 lb of waste-containing fluid is removed in each dialysis session, which can last 2–8 hours and must be performed several times a week. Although some patients have been on dialysis for 30 or more years, dialysis is always a temporary solution because waste materials are constantly building up in the bloodstream. A more permanent solution is a kidney transplant.
Cell walls are semipermeable membranes, so the osmotic pressures of the body’s fluids have important biological consequences. If solutions of different osmolarity exist on either side of the cells, solvent (water) may pass into or out of the cells, sometimes with disastrous results. Consider what happens if red blood cells are placed in a hypotonic solution, meaning a solution of lower osmolarity than the liquid inside the cells. The cells swell up as water enters them, disrupting cellular activity and eventually causing the cells to burst. This process is called hemolysis. If red blood cells are placed in a hypertonic solution, meaning one having a higher osmolarity than exists inside the cells, water leaves the cells to dilute the external solution, and the red blood cells shrivel and die. This process is called crenation. Only if red blood cells are placed in isotonic solutions that have the same osmolarity as exists inside the cells are they unaffected by negative effects of osmotic pressure. Glucose solutions of about 0.31 M, or sodium chloride solutions of about 0.16 M, are isotonic with blood plasma.
The concentration of an isotonic sodium chloride (NaCl) solution is only half that of an isotonic glucose (C6H12O6) solution because NaCl produces two ions when a formula unit dissolves, while molecular C6H12O6 produces only one particle when a formula unit dissolves. The osmolarities are therefore the same even though the concentrations of the two solutions are different.
Osmotic pressure explains why you should not drink seawater if you are abandoned in a life raft in the middle of the ocean. Its osmolarity is about three times higher than most bodily fluids. You would actually become thirstier as water from your cells was drawn out to dilute the salty ocean water you ingested. Our bodies do a better job coping with hypotonic solutions than with hypertonic ones. The excess water is collected by our kidneys and excreted.
Osmotic pressure effects are used in the food industry to make pickles from cucumbers and other vegetables and in brining meat to make corned beef. It is also a factor in the mechanism of getting water from the roots to the tops of trees!
Career Focus: Perfusionist
A perfusionist is a medical technician trained to assist during any medical procedure in which a patient’s circulatory or breathing functions require support. The use of perfusionists has grown rapidly since the advent of open-heart surgery in 1953.
Most perfusionists work in operating rooms, where their main responsibility is to operate heart-lung machines. During many heart surgeries, the heart itself must be stopped. In these situations, a heart-lung machine keeps the patient alive by aerating the blood with oxygen and removing carbon dioxide. The perfusionist monitors both the machine and the status of the blood, notifying the surgeon and the anesthetist of any concerns and taking corrective action if the status of the blood becomes abnormal.
Despite the narrow parameters of their specialty, perfusionists must be highly trained. Certified perfusion education programs require a student to learn anatomy, physiology, pathology, chemistry, pharmacology, math, and physics. A college degree is usually required. Some perfusionists work with other external artificial organs, such as hemodialysis machines and artificial livers. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.04%3A_Properties_of_Solutions.txt |
Learning Outcomes
• Explain chemical equilibrium.
• Write expression for calculating $K$.
• Calculate and compare Q and K values.
• Predict relative amounts of reactants and products based on equilibrium constant $K$.
Hydrogen and iodine gases react to form hydrogen iodide according to the following reaction:
$\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightleftharpoons 2 \ce{HI} \left( g \right) \nonumber$
\begin{align} &\text{Forward reaction:} \: \: \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right) \ &\text{Reverse reaction:} \: \: 2 \ce{HI} \left( g \right) \rightarrow \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \end{align} \nonumber
Initially, only the forward reaction occurs because no $\ce{HI}$ is present. As soon as some $\ce{HI}$ has formed, it begins to decompose back into $\ce{H_2}$ and $\ce{I_2}$. Gradually, the rate of the forward reaction decreases while the rate of the reverse reaction increases. Eventually the rate of combination of $\ce{H_2}$ and $\ce{I_2}$ to produce $\ce{HI}$ becomes equal to the rate of decomposition of $\ce{HI}$ into $\ce{H_2}$ and $\ce{I_2}$. When the rates of the forward and reverse reactions have become equal to one another, the reaction has achieved a state of balance. Chemical equilibrium is the state of a system in which the rate of the forward reaction is equal to the rate of the reverse reaction.
Chemical equilibrium can be attained whether the reaction begins with all reactants and no products, all products and no reactants, or some of both. The figure below shows changes in concentration of $\ce{H_2}$, $\ce{I_2}$, and $\ce{HI}$ for two different reactions. In the reaction depicted by the graph on the left (A), the reaction begins with only $\ce{H_2}$ and $\ce{I_2}$ present. There is no $\ce{HI}$ initially. As the reaction proceeds towards equilibrium, the concentrations of the $\ce{H_2}$ and $\ce{I_2}$ gradually decrease, while the concentration of the $\ce{HI}$ gradually increases. When the curve levels out and the concentrations all become constant, equilibrium has been reached. At equilibrium, concentrations of all substances are constant.
In reaction B, the process begins with only $\ce{HI}$ and no $\ce{H_2}$ or $\ce{I_2}$. In this case, the concentration of $\ce{HI}$ gradually decreases while the concentrations of $\ce{H_2}$ and $\ce{I_2}$ gradually increase until equilibrium is again reached. Notice that in both cases, the relative position of equilibrium is the same, as shown by the relative concentrations of reactants and products. The concentration of $\ce{HI}$ at equilibrium is significantly higher than the concentrations of $\ce{H_2}$ and $\ce{I_2}$. This is true whether the reaction began with all reactants or all products. The position of equilibrium is a property of the particular reversible reaction and does not depend upon how equilibrium was achieved.
Conditions for Equilibrium and Types of Equilibrium
It may be tempting to think that once equilibrium has been reached, the reaction stops. Chemical equilibrium is a dynamic process. The forward and reverse reactions continue to occur even after equilibrium has been reached. However, because the rates of the reactions are the same, there is no change in the relative concentrations of reactants and products for a reaction that is at equilibrium. The conditions and properties of a system at equilibrium are summarized below.
1. The system must be closed, meaning no substances can enter or leave the system.
2. Equilibrium is a dynamic process. Even though we don't necessarily see the reactions, both forward and reverse are taking place.
3. The rates of the forward and reverse reactions must be equal.
4. The amount of reactants and products do not have to be equal. However, after equilibrium is attained, the amounts of reactants and products will be constant.
The description of equilibrium in this concept refers primarily to equilibrium between reactants and products in a chemical reaction. Other types of equilibrium include phase equilibrium and solution equilibrium. A phase equilibrium occurs when a substance is in equilibrium between two states. For example, a stoppered flask of water attains equilibrium when the rate of evaporation is equal to the rate of condensation. A solution equilibrium occurs when a solid substance is in a saturated solution. At this point, the rate of dissolution is equal to the rate of recrystallization. Although these are all different types of transformations, most of the rules regarding equilibrium apply to any situation in which a process occurs reversibly.
Red blood cells transport oxygen to the tissues so they can function. In the absence of oxygen, cells cannot carry out their biochemical responsibilities. Oxygen moves to the cells attached to hemoglobin, a protein found in the red cells. In cases of carbon monoxide poisoning, $\ce{CO}$ binds much more strongly to the hemoglobin, blocking oxygen attachment and lowering the amount of oxygen reaching the cells. Treatment involves the patient breathing pure oxygen to displace the carbon monoxide. The equilibrium reaction shown below illustrates the shift toward the right when excess oxygen is added to the system:
$\ce{Hb(CO)_4} \left( aq \right) + 4 \ce{O_2} \left( g \right) \rightleftharpoons \ce{Hb(O_2)_4} \left( aq \right) + 4 \ce{CO} \left( g \right) \nonumber$
Equilibrium Constant
Consider the hypothetical reversible reaction in which reactants $\ce{A}$ and $\ce{B}$ react to form products $\ce{C}$ and $\ce{D}$. This equilibrium can be shown below, where the lowercase letters represent the coefficients of each substance.
$a \ce{A} + b \ce{B} \rightleftharpoons c \ce{C} + d \ce{D} \nonumber$
As we have established, the rates of the forward and reverse reactions are the same at equilibrium, and so the concentrations of all of the substances are constant. Since that is the case, it stands to reason that a ratio of the concentration for any given reaction at equilibrium maintains a constant value. The equilibrium constant $\left( K_\text{eq} \right)$ is the ratio of the mathematical product of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. For the general reaction above, the equilibrium constant expression is written as follows:
$K_\text{eq} = \frac{\left[ \ce{C} \right]^c \left[ \ce{D} \right]^d}{\left[ \ce{A} \right]^a \left[ \ce{B} \right]^b} \nonumber$
The concentrations of each substance, indicated by the square brackets around the formula, are measured in molarity units $\left( \text{mol/L} \right)$.
The value of the equilibrium constant for any reaction is only determined by experiment. As detailed in the above section, the position of equilibrium for a given reaction does not depend on the starting concentrations and so the value of the equilibrium constant is truly constant. It does, however, depend on the temperature of the reaction. This is because equilibrium is defined as a condition resulting from the rates of forward and reverse reactions being equal. If the temperature changes, the corresponding change in those reaction rates will alter the equilibrium constant. For any reaction in which a $K_\text{eq}$ is given, the temperature should be specified.
The equilibrium constant can vary over a wide range of values. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than $10^3$ indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between $H_2$ and $Cl_2$ to produce $HCl$, which has an equilibrium constant of $1.6 \times 10^{33}$ at 300 K. Because $H_2$ is a good reducing agent and $Cl_2$ is a good oxidizing agent, the reaction proceeds essentially to completion.
$H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)}$ $K$=$1.6 \times 10^{33}$ at 300K
In contrast, values of $K$ less than $10^{-3}$ indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants. An example is the decomposition of water. No wonder you won’t find water a very good source of oxygen gas and hydrogen gas at ordinary temperatures!
$H_2O_{(g)} \rightleftharpoons H_{2(g)} + ½ O_{2(g)}$ $K$=$8 \times 10^{-41}$ at 25°C
Figure $4$ summarizes the relationship between the magnitude of K and the relative concentrations of reactants and products at equilibrium for a general reaction, written as reactants $\rightleftharpoons$ products.
A large value of the equilibrium constant $K$ means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium.
• When $K$ is a very large number, the reaction is essentially irreversible, products predominate at equilibrium.
• When $K$ is between 1 and 1000, more products than reactants are present at equilibrium
• When $K$ is between 1 and 0.001, more reactants than products are present at equilibrium.
• When $K$ is a very small number, the reaction produces almost no products; only reactants are present at equilibrium.
Example $1$
Write the expression for the equilibrium constant K for the following reaction:
N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g)
Solution
The equilibrium constant equation is the ratio of the concentration of the products (NH3) to the concentration of the reactants (N2 and H2), each raised to the power of its coefficient in the balanced chemical equation.
$K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}$
Exercise $1$
Write the expression of the equilibrium constant K for the following reaction:.
2 NOCl (g) ⇄ 2 NO (g) + Cl2 (g)
Answer
$K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}$
Example $2$
Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants.
1. $H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\;\;\; K_{(700K)}=54$
2. $2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}\;\;\; K_{(1200K)}=3.1 \times 10^{−18}$
3. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)}+Cl_{2(g)}\;\;\; K_{(613K)}=97$
4. $2O_{3(g)} \rightleftharpoons 3O_{2(g)} \;\;\; K_{(298 K)}=5.9 \times 10^{55}$
Solution
Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both.
1. Only system 4 has $K \gg 10^3$, so at equilibrium it will consist of essentially only products.
2. System 2 has $K \ll 10^{−3}$, so the reactants have little tendency to form products under the conditions specified; thus, at equilibrium the system will contain essentially only reactants.
3. Both systems 1 and 3 have equilibrium constants in the range $10^3 \ge K \ge 10^{−3}$, indicating that the equilibrium mixtures will contain appreciable amounts of both products and reactants.
Exercise $2$
Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation:
$3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)} \nonumber$
Values of the equilibrium constant at various temperatures were reported as
• $K_{25°C} = 3.3 \times 10^8$,
• $K_{177°C} = 2.6 \times 10^3$, and
• $K_{327°C} = 4.1$.
1. At which temperature would you expect to find the highest proportion of $H_2$ and $N_2$ in the equilibrium mixture?
2. At which temperature would you expect to find the highest proportion of ammonia in the equilibrium mixture?
Answer
a. 327°C, where $K$ is smallest
b. 25°C
Reaction Quotient
The reaction quotient, $Q$, is used when questioning if we are at equilibrium. The calculation for $Q$ is exactly the same as for $K$ but we can only use $K$ when we know we are at equilibrium. Comparing $Q$ and $K$ allows the direction of the reaction to be predicted.
• $Q$ = $K$ equilibrium
• $Q$ < $K$ reaction proceeds to the right to form more products and decrease amount of reactants so value of $Q$ will increase
• $Q$ > $K$ reaction proceeds to the left to form more reactants and decrease amount of products so value of $Q$ will decrease
Key Takeaway
• As a chemical change proceeds, the quantities of the reactants/products will decrease, and those of the products/reactants will increase. Eventually the reaction slows down and the composition of the system stops changing. At this point the reaction is in its equilibrium state, and no further change in composition will occur as long as the system is left undisturbed.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.05%3A_Chemical_Equilibrium.txt |
Learning Outcomes
• Define Le Chatelier's principle.
• Predict how the change in amounts of substances, temperature, or pressure will affect amounts of reactants and products present at equilibrium.
Le Chatelier's Principle
Chemical equilibrium was studied by the French chemist Henri Le Chatelier (1850 - 1936) and his description of how a system responds to a stress to equilibrium has become known as Le Chatelier's principle: When a chemical system that is at equilibrium is disturbed by a stress, the system will respond in order to relieve the stress. Stresses to a chemical system involve changes in the concentrations of reactants or products, changes in the temperature of the system, or changes in the pressure of the system. We will discuss each of these stresses separately. The change to the equilibrium position in every case is either a favoring of the forward reaction or a favoring of the reverse reaction. When the forward reaction is favored, the concentrations of products increase, while the concentrations of reactants decrease. When the reverse reaction is favored, the concentrations of the products decrease, while the concentrations of reactants increase.
$\begin{array}{lll} \textbf{Original Equilibrium} & \textbf{Favored Reaction} & \textbf{Result} \ \ce{A} \rightleftharpoons \ce{B} & \text{Forward:} \: \ce{A} \rightarrow \ce{B} & \left[ \ce{A} \right] \: \text{decreases}; \: \left[ \ce{B} \right] \: \text{increases} \ \ce{A} \rightleftharpoons \ce{B} & \text{Reverse:} \: \ce{A} \leftarrow \ce{B} & \left[ \ce{A} \right] \: \text{increases}; \: \left[ \ce{B} \right] \: \text{decreases} \end{array} \nonumber$
Effect of Concentration
A change in concentration of one of the substances in an equilibrium system typically involves either the addition or the removal of one of the reactants or products. Consider the Haber-Bosch process for the industrial production of ammonia from nitrogen and hydrogen gases.
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right) \nonumber$
If the concentration of one substance in a system is increased, the system will respond by favoring the reaction that removes that substance. When more $\ce{N_2}$ is added, the forward reaction will be favored because the forward reaction uses up $\ce{N_2}$ and converts it to $\ce{NH_3}$. The forward reaction speeds up temporarily as a result of the addition of a reactant. The position of equilibrium shifts as more $\ce{NH_3}$ is produced. The concentration of $\ce{NH_3}$ increases, while the concentrations of $\ce{N_2}$ and $\ce{H_2}$ decrease. After some time passes, equilibrium is reestablished with new concentrations of all three substance. As can be seen in the figure below, if more $\ce{N_2}$ is added, a new equilibrium is achieved by the system. The new concentration of $\ce{NH_3}$ is higher because of the favoring of the forward reaction. The new concentration of the $\ce{H_2}$ is lower .The concentration of $\ce{N_2}$ is higher than in the original equilibrium, but went down slightly following the addition of the $\ce{N_2}$ that disturbed the original equilibrium. By responding in this way, the value of the equilibrium constant for the reaction, $K_\text{eq}$, does not change as a result of the stress to the system.
In other words, the amount of each substance is different but the ratio of the amount of each remains the same.
If more $\ce{NH_3}$ were added, the reverse reaction would be favored. This "favoring" of a reaction means temporarily speeding up the reaction in that direction until equilibrium is reestablished. Recall that once equilibrium is reestablished, the rates of the forward and reverse reactions are again equal. The addition of $\ce{NH_3}$ would result in increased formation of the reactants, $\ce{N_2}$ and $\ce{H_2}$.
An equilibrium can also be disrupted by the removal of one of the substances. If the concentration of a substance is decreased, the system will respond by favoring the reaction that replaces that substance. In the industrial Haber-Bosch process, $\ce{NH_3}$ is removed from the equilibrium system as the reaction proceeds. As a result, the forward reaction is favored so that more $\ce{NH_3}$ is produced. The concentrations of $\ce{N_2}$ and $\ce{H_2}$ decrease. Continued removal of $\ce{NH_3}$ will eventually force the reaction to go to completion until all of the reactants are used up. If either $\ce{N_2}$ or $\ce{H_2}$ were removed from the equilibrium system, the reverse reaction would be favored and the concentration of $\ce{NH_3}$ would decrease.
The effect of changes in concentration on an equilibrium system according to Le Chatelier's principle is summarized in the table below.
Table 9.6.1
Stress Response
addition of reactant forward reaction favored
addition of product reverse reaction favored
removal of reactant reverse reaction favored
removal of product forward reaction favored
Example $1$
Given this reaction at equilibrium:
$N_{2}+3H_{2}\rightleftharpoons 2NH_{3} \nonumber$
How will it affect the reaction if the equilibrium is stressed by each change?
1. H2 is added.
2. NH3 is added.
3. NH3 is removed.
Solution
1. If H2 is added, there is now more reactant, so the reaction will shift to the right (toward products) to reduce the added H2.
2. If NH3 is added, there is now more product, so the reaction will shift to the left (toward reactants) to reduce the added NH3.
3. If NH3 is removed, there is now less product, so the reaction will shift to the right (toward products) to replace the product removed.
Exercise $1$
Given this reaction at equilibrium:
$CO(g)+Br_{2}(g)\rightleftharpoons COBr_{2}(g) \nonumber$
How will it affect the reaction if the equilibrium is stressed by each change?
1. Br2 is removed.
2. COBr2 is added.
Answer
1. shift to the left (toward reactants)
2. shift to the left (toward reactants)
Effect of Temperature
Increasing or decreasing the temperature of a system at equilibrium is also a stress to the system. The equation for the Haber-Bosch process is written again below, as a thermochemical equation (i.e. it contains information about the energy gained or lost when the reaction occurs).
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right) + 91 \: \text{kJ} \nonumber$
The forward reaction is the exothermic direction: the formation of $\ce{NH_3}$ releases heat which is why that is shown as a product. The reverse reaction is the endothermic direction: as $\ce{NH_3}$ decomposes to $\ce{N_2}$ and $\ce{H_2}$, heat is absorbed. An increase in the temperature for this is like adding a product because heat is being released by the reaction. If we add a product then the reaction proceeds towards the formation of more reactants. Reducing the temperature for this system would be similar to removing a product which would favor the formation of more products. The amount of $\ce{NH_3}$ will increase and the amount of $\ce{N_2}$ and $\ce{H_2}$ will decrease.
For changes in concentration, the system responds in such a way that the value of the equilibrium constant, $K_\text{eq}$ is unchanged. However, a change in temperature shifts the equilibrium and the $K_\text{eq}$ value either increases or decreases. As discussed in the previous section, values of $K_\text{eq}$ are dependent on the temperature. When the temperature of the system for the Haber-Bosch process is increased, the resultant shift in equilibrium towards the reactants means that the $K_\text{eq}$ value decreases. When the temperature is decreased, the shift in equilibrium towards the products means that the $K_\text{eq}$ value increases.
Le Chatelier's principle as related to temperature changes can be illustrated easily be the reaction in which dinitrogen tetroxide is in equilibrium with nitrogen dioxide.
$\ce{N_2O_4} \left( g \right) + \text{heat} \rightleftharpoons 2 \ce{NO_2} \left( g \right) \nonumber$
Dinitrogen tetroxide $\left( \ce{N_2O_4} \right)$ is colorless, while nitrogen dioxide $\left( \ce{NO_2} \right)$ is dark brown in color. When $\ce{N_2O_4}$ breaks down into $\ce{NO_2}$, heat is absorbed (endothermic) according to the forward reaction above. Therefore, an increase in temperature (adding heat) of the system will favor the forward reaction. Conversely, a decrease in temperature (removing heat) will favor the reverse reaction.
Example $2$
Predict the effect of increasing the temperature on this equilibrium.
$PCl_{3}+Cl_{2}\rightleftharpoons PCl_{5}+60kJ \nonumber$
Solution
Because energy is listed as a product, it is being produced, so the reaction is exothermic. If the temperature is increasing, a product is being added to the equilibrium, so the equilibrium shifts to minimize the addition of extra product: it shifts to the left (back toward reactants).
Exercise $2$
Predict the effect of decreasing the temperature on this equilibrium.
$N_{2}O_{4}+57kJ\rightleftharpoons 2NO_{2} \nonumber$
Answer
Equilibrium shifts to the left (toward reactants).
Effect of Pressure
Changing the pressure of an equilibrium system in which gases are involved is also a stress to the system. A change in the pressure on a liquid or a solid has a negligible effect. We will return again the equilibrium for the Haber-Bosch process. Imagine the gases are contained in a closed system in which the volume of the system is controlled by an adjustable piston as shown in the figure below.
On the far left, the reaction system contains primarily $\ce{N_2}$ and $\ce{H_2}$, with only one molecule of $\ce{NH_3}$ present. As the piston is pushed inwards, the pressure of the system increases according to Boyle's law. This is a stress to the equilibrium. In the middle image, the same number of molecules is now confined in a smaller space and so the pressure has increased. According to Le Chatelier's principle, the system responds in order to relieve the stress. In the image on the right, the forward reaction has been favored and more $\ce{NH_3}$ is produced. The overall result is a decrease in the number of gas molecules in the entire system. This in turn decreases the pressure and provides a relief to the original stress of a pressure increase. An increase in pressure on an equilibrium system favors the reaction which products fewer total moles of gas. In this case, it is the forward reaction that is favored.
A decrease in pressure on the above system could be achieved by pulling the piston outward, increasing the container volume. The equilibrium would respond by favoring the reverse reaction in which $\ce{NH_3}$ decomposes to $\ce{N_2}$ and $\ce{H_2}$. This is because the overall number of gas molecules would increase and so would the pressure. A decrease in pressure on an equilibrium system favors the reaction which produces more total moles of gas. This is summarized in the table below.
Table 9.6.2
Stress Response
pressure increase reaction produces fewer gas molecules
pressure decrease reaction produces more gas molecules
Like changes in concentration, the $K_\text{eq}$ value for a given reaction is unchanged by a change in pressure. The amounts of each substance will change but the ratio will not. It is important to remember when analyzing the effect of a pressure change on equilibrium that only gases are affected. If a certain reaction involves liquids or solids, they should be ignored. For example, calcium carbonate decomposes according to the equilibrium reaction:
$\ce{CaCO_3} \left( s \right) \rightleftharpoons \ce{CaO} \left( s \right) + \ce{O_2} \left( g \right) \nonumber$
Oxygen is the only gas in the system. An increase in the pressure of the system slows the rate of decomposition of $\ce{CaCO_3}$ because the reverse reaction is favored. When a system contains equal moles of gas on both sides of the equation, pressure has no effect on the equilibrium position, as in the formation of $\ce{HCl}$ from $\ce{H_2}$ and $\ce{Cl_2}$.
$\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightleftharpoons 2 \ce{HCl} \left( g \right) \nonumber$
Example $3$
What is the effect on this equilibrium if pressure is increased?
$N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g) \nonumber$
Solution
According to Le Chatelier's principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. This particular reaction shows a total of 4 mol of gas as reactants and 2 mol of gas as products, so the reaction shifts to the right (toward the products side).
Exercise $3$
What is the effect on this equilibrium if pressure is decreased?
$3O_{2}(g)\rightleftharpoons 2O_{3}(g) \nonumber$
Answer
Reaction shifts to the left (toward reactants).
Application of Le Chatelier's Principle
Oxygen transport by the blood
In aerobic respiration, oxygen is transported to the cells where it is combined with glucose and metabolized to carbon dioxide, which then moves back to the lungs from which it is expelled.
hemoglobin + O2 oxyhemoglobin
The partial pressure of O2 in the air is 0.2 atm, sufficient to allow these molecules to be taken up by hemoglobin (the red pigment of blood) in which it becomes loosely bound in a complex known as oxyhemoglobin. At the ends of the capillaries which deliver the blood to the tissues, the O2 concentration is reduced by about 50% owing to its consumption by the cells. This shifts the equilibrium to the left, releasing the oxygen so it can diffuse into the cells.
Key Takeaways
• In a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, and the introduction of more reactants will lead to the formation of more products, but the ratio of Products/Reactants (equilibrium constant), K is unchanged.
• If temperature is changed, the numeric value K will change. If a reaction is exothermic (releases heat), an increase in the temperature will force the equilibrium to the left, causing the system to absorb heat and thus partially offsetting the rise in temperature. The opposite effect occurs for endothermic reactions, which are shifted to the right by rising temperature.
• The effect of pressure on an equilibrium is significant only for reactions which involve different numbers of moles of gases on the two sides of the equation. An increase in the total pressure will shift to the side with fewer moles of gas. A decrease in pressure will shift to the side with more moles of gas.
Contributors and Attributions
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky) | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.06%3A_Le_Chatelier%27s_Principle.txt |
Learning Outcomes
• Define osmosis and diffusion.
• Distinguish among hypotonic, hypertonic, and isotonic solutions.
• Describe a semipermeable membrane.
• Predict behavior of blood cells in different solution types.
• Describe flow of solvent molecules across a membrane.
• Identify the polar and nonpolar regions of a cell membrane.
• Explain the components present in a phospholipid.
Fish cells, like all cells, have semipermeable membranes. Eventually, the concentration of "stuff" on either side of them will even out. A fish that lives in salt water will have somewhat salty water inside itself. Put it in freshwater, and the freshwater will, through osmosis, enter the fish, causing its cells to swell, and the fish will die. What will happen to a freshwater fish in the ocean?
Osmosis
Imagine you have a cup that has $100 \: \text{mL}$ water, and you add $15 \: \text{g}$ of table sugar to the water. The sugar dissolves and the mixture that is now in the cup is made up of a solute (the sugar) that is dissolved in the solvent (the water). The mixture of a solute in a solvent is called a solution.
Imagine now that you have a second cup with $100 \: \text{mL}$ of water, and you add $45 \: \text{g}$ of table sugar to the water. Just like the first cup, the sugar is the solute, and the water is the solvent. But now you have two mixtures of different solute concentrations. In comparing two solutions of unequal solute concentration, the solution with the higher solute concentration is hypertonic, and the solution with the lower solute concentration is hypotonic. Solutions of equal solute concentration are isotonic. The first sugar solution is hypotonic to the second solution. The second sugar solution is hypertonic to the first.
You now add the two solutions to a beaker that has been divided by a semipermeable membrane, with pores that are too small for the sugar molecules to pass through, but are big enough for the water molecules to pass through. The hypertonic solution is one side of the membrane and the hypotonic solution on the other. The hypertonic solution has a lower water concentration than the hypotonic solution, so a concentration gradient of water now exists across the membrane. Water molecules will move from the side of higher water concentration to the side of lower concentration until both solutions are isotonic. At this point, equilibrium is reached.
Red blood cells behave the same way (see figure below). When red blood cells are in a hypertonic (higher concentration) solution, water flows out of the cell faster than it comes in. This results in crenation (shriveling) of the blood cell. On the other extreme, a red blood cell that is hypotonic (lower concentration outside the cell) will result in more water flowing into the cell than out. This results in swelling of the cell and potential hemolysis (bursting) of the cell. In an isotonic solution, the flow of water in and out of the cell is happening at the same rate.
Osmosis is the diffusion of water molecules across a semipermeable membrane from an area of lower concentration solution (i.e., higher concentration of water) to an area of higher concentration solution (i.e., lower concentration of water). Water moves into and out of cells by osmosis.
• If a cell is in a hypertonic solution, the solution has a lower water concentration than the cell cytosol, and water moves out of the cell until both solutions are isotonic.
• Cells placed in a hypotonic solution will take in water across their membranes until both the external solution and the cytosol are isotonic.
A red blood cell will swell and undergo hemolysis (burst) when placed in a hypotonic solution. When placed in a hypertonic solution, a red blood cell will lose water and undergo crenation (shrivel). Animal cells tend to do best in an isotonic environment, where the flow of water in and out of the cell is occurring at equal rates.
Diffusion
Passive transport is a way that small molecules or ions move across the cell membrane without input of energy by the cell. The three main kinds of passive transport are diffusion (or simple diffusion), osmosis, and facilitated diffusion. Simple diffusion and osmosis do not involve transport proteins. Facilitated diffusion requires the assistance of proteins.
Diffusion is the movement of molecules from an area of high concentration of the molecules to an area with a lower concentration. For cell transport, diffusion is the movement of small molecules across the cell membrane. The difference in the concentrations of the molecules in the two areas is called the concentration gradient. The kinetic energy of the molecules results in random motion, causing diffusion. In simple diffusion, this process proceeds without the aid of a transport protein. It is the random motion of the molecules that causes them to move from an area of high concentration to an area with a lower concentration.
Diffusion will continue until the concentration gradient has been eliminated. Since diffusion moves materials from an area of higher concentration to the lower, it is described as moving solutes "down the concentration gradient". The end result is an equal concentration, or equilibrium, of molecules on both sides of the membrane. At equilibrium, movement of molecules does not stop. At equilibrium, there is equal movement of materials in both directions.
Not everything can make it into your cells. Your cells have a plasma membrane that helps to guard your cells from unwanted intruders.
The Plasma Membrane and Cytosol
If the outside environment of a cell is water-based, and the inside of the cell is also mostly water, something has to make sure the cell stays intact in this environment. What would happen if a cell dissolved in water, like sugar does? Obviously, the cell could not survive in such an environment. So something must protect the cell and allow it to survive in its water-based environment. All cells have a barrier around them that separates them from the environment and from other cells. This barrier is called the plasma membrane, or cell membrane.
The Plasma Membrane
The plasma membrane (see figure below) is made of a double layer of special lipids, known as phospholipids. The phospholipid is a lipid molecule with a hydrophilic ("water-loving") head and two hydrophobic ("water-hating") tails. Because of the hydrophilic and hydrophobic nature of the phospholipid, the molecule must be arranged in a specific pattern as only certain parts of the molecule can physically be in contact with water. Remember that there is water outside the cell, and the cytoplasm inside the cell is mostly water as well. So the phospholipids are arranged in a double layer (a bilayer) to keep the cell separate from its environment. Lipids do not mix with water (recall that oil is a lipid), so the phospholipid bilayer of the cell membrane acts as a barrier, keeping water out of the cell, and keeping the cytoplasm inside the cell. The cell membrane allows the cell to stay structurally intact in its water-based environment.
The function of the plasma membrane is to control what goes in and out of the cell. Some molecules can go through the cell membrane to enter and leave the cell, but some cannot. The cell is therefore not completely permeable. "Permeable" means that anything can cross a barrier. An open door is completely permeable to anything that wants to enter or exit through the door. The plasma membrane is semipermeable, meaning that some things can enter the cell, and some things cannot.
Molecules that cannot easily pass through the bilayer include ions and small hydrophilic molecules, such as glucose, and macromolecules, including proteins and RNA. Examples of molecules that can easily diffuse across the plasma membrane include carbon dioxide and oxygen gas. These molecules diffuse freely in and out of the cell, along their concentration gradient. Though water is a polar molecule, it can also diffuse through the plasma membrane.
Cytosol
The inside of all cells also contain a jelly-like substance called cytosol. Cytosol is composed of water and other molecules, including enzymes, which are proteins that speed up the cell's chemical reactions. Everything in the cell sits in the cytosol, like fruit in a Jell-o mold. The term cytoplasm refers to the cytosol and all of the organelles, the specialized compartments of the cell. The cytoplasm does not include the nucleus. As a prokaryotic cell does not have a nucleus, the DNA is in the cytoplasm.
Supplemental Resources
• The Plasma Membrane: www.youtube.com/watch?v=moPJkCbKjBs
Key Takeaways
• Water moves into and out of cells by osmosis.
• Water (solvent) moves from an area of lower concentration solution (i.e., higher concentration of water) to an area of higher concentration solution (i.e., lower concentration of water). | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.07%3A_Osmosis_and_Diffusion.txt |
Concept Review Exercises
1. What causes a solution to form?
2. How does the phrase like dissolves like relate to solutions?
Answers
1. Solutions form because a solute and a solvent have similar intermolecular interactions.
2. It means that substances with similar intermolecular interactions will dissolve in each other.
Exercises
1. Define solution.
2. Give several examples of solutions.
3. What is the difference between a solvent and a solute?
4. Can a solution have more than one solute in it? Can you give an example?
5. Does a solution have to be a liquid? Give several examples to support your answer.
6. Give at least two examples of solutions found in the human body.
7. Which substances will probably be soluble in water, a very polar solvent?
1. sodium nitrate (NaNO3)
2. hexane (C6H14)
3. isopropyl alcohol [(CH3)2CHOH]
4. benzene (C6H6)
8. Which substances will probably be soluble in toluene (C6H5CH3), a nonpolar solvent?
1. sodium nitrate (NaNO3)
2. hexane (C6H14)
3. isopropyl alcohol [(CH3)2CHOH]
4. benzene (C6H6)
9. The solubility of alcohols in water varies with the length of carbon chain. For example, ethanol (CH3CH2OH) is soluble in water in any ratio, while only 0.0008 mL of heptanol (CH3CH2CH2CH2CH2CH2CH2OH) will dissolve in 100 mL of water. Propose an explanation for this behavior.
10. Dimethyl sulfoxide [(CH3)2SO] is a polar liquid. Based on the information in Exercise 9, which do you think will be more soluble in it—ethanol or heptanol?
Answers
1. a homogeneous mixture
2. vinegar, dextrose IV, saline IV, coffee, tea, wine
1. A solvent is the majority component of a solution; a solute is the minority component of a solution.
4. yes. Coke or Pepsi has sugar, caffeine and carbon dioxide as solutes.
1. A solution does not have to be liquid; air is a gaseous solution, while some alloys are solid solutions (answers will vary).
6. Urine, plasma
1. probably soluble
2. probably not soluble
3. probably soluble
4. probably not soluble
8.
1. probably not soluble
2. probably soluble
3. probably not soluble
4. probably soluble
9. Small alcohol molecules have strong polar intermolecular interactions, so they dissolve in water. In large alcohol molecules, the nonpolar end overwhelms the polar end, so they do not dissolve very well in water.
10. Ethanol is a smaller molecule. It will be more soluble in water than heptanol.
Concept Review Exercises
1. What are some of the units used to express concentration?
2. Distinguish between the terms solubility and concentration.
Answers
1. % m/m, % m/v, ppm, ppb, molarity, and Eq/L (answers will vary)
2. Solubility is typically a limit to how much solute can dissolve in a given amount of solvent. Concentration is the quantitative amount of solute dissolved at any concentration in a solvent.
Exercises
1. Define solubility. Do all solutes have the same solubility?
2. Explain why the terms dilute or concentrated are of limited usefulness in describing the concentration of solutions.
3. If the solubility of sodium chloride (NaCl) is 30.6 g/100 mL of H2O at a given temperature, how many grams of NaCl can be dissolved in 250.0 mL of H2O?
4. If the solubility of glucose (C6H12O6) is 120.3 g/100 mL of H2O at a given temperature, how many grams of C6H12O6 can be dissolved in 75.0 mL of H2O?
5. How many grams of sodium bicarbonate (NaHCO3) can a 25.0°C saturated solution have if 150.0 mL of H2O is used as the solvent?
6. If 75.0 g of potassium bromide (KBr) are dissolved in 125 mL of H2O, is the solution saturated, unsaturated, or supersaturated?
7. Calculate the mass/mass percent of a saturated solution of NaCl. Use the data from Table $1$ "Solubilities of Various Solutes in Water at 25°C (Except as Noted)", assume that masses of the solute and the solvent are additive, and use the density of H2O (1.00 g/mL) as a conversion factor.
8. Calculate the mass/mass percent of a saturated solution of MgCO3 Use the data from Table $1$ "Solubilities of Various Solutes in Water at 25°C (Except as Noted)", assume that masses of the solute and the solvent are additive, and use the density of H2O (1.00 g/mL) as a conversion factor.
9. Only 0.203 mL of C6H6 will dissolve in 100.000 mL of H2O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of benzene in water.
10. Only 35 mL of aniline (C6H5NH2) will dissolve in 1,000 mL of H2O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of aniline in water.
11. A solution of ethyl alcohol (C2H5OH) in water has a concentration of 20.56% v/v. What volume of C2H5OH is present in 255 mL of solution?
12. What mass of KCl is present in 475 mL of a 1.09% m/v aqueous solution?
13. The average human body contains 5,830 g of blood. What mass of arsenic is present in the body if the amount in blood is 0.55 ppm?
14. The Occupational Safety and Health Administration has set a limit of 200 ppm as the maximum safe exposure level for carbon monoxide (CO). If an average breath has a mass of 1.286 g, what is the maximum mass of CO that can be inhaled at that maximum safe exposure level?
15. Which concentration is greater—15 ppm or 1,500 ppb?
16. Express the concentration 7,580 ppm in parts per billion.
17. What is the molarity of 0.500 L of a potassium chromate solution containing 0.0650 mol of K2CrO4?
18. What is the molarity of 4.50 L of a solution containing 0.206 mol of urea [(NH2)2CO]?
19. What is the molarity of a 2.66 L aqueous solution containing 56.9 g of NaBr?
20. If 3.08 g of Ca(OH)2 is dissolved in enough water to make 0.875 L of solution, what is the molarity of the Ca(OH)2?
21. What mass of HCl is present in 825 mL of a 1.25 M solution?
22. What mass of isopropyl alcohol (C3H8O) is dissolved in 2.050 L of a 4.45 M aqueous C3H8O solution?
23. What volume of 0.345 M NaCl solution is needed to obtain 10.0 g of NaCl?
24. How many milliliters of a 0.0015 M cocaine hydrochloride (C17H22ClNO4) solution is needed to obtain 0.010 g of the solute?
25. Aqueous calcium chloride reacts with aqueous silver nitrate according to the following balanced chemical equation:
CaCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Ca(NO3)2(aq)
How many moles of AgCl(s) are made if 0.557 L of 0.235 M CaCl2 react with excess AgNO3? How many grams of AgCl are made?
26. Sodium bicarbonate (NaHCO3) is used to react with acid spills. The reaction with sulfuric acid (H2SO4) is as follows:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2H2O(ℓ) + 2CO2(g)
If 27.6 mL of a 6.25 M H2SO4 solution were spilled, how many moles of NaHCO3 would be needed to react with the acid? How many grams of NaHCO3 is this?
27. The fermentation of glucose to make ethanol and carbon dioxide has the following overall chemical equation:
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)
If 1.00 L of a 0.567 M solution of C6H12O6 were completely fermented, what would be the resulting concentration of the C2H5OH solution? How many moles of CO2 would be formed? How many grams is this? If each mole of CO2 had a volume of 24.5 L, what volume of CO2 is produced?
28. Aqueous sodium bisulfite gives off sulfur dioxide gas when heated:
2NaHSO3(aq) → Na2SO3(aq) + H2O(ℓ) + SO2(g)
If 567 mL of a 1.005 M NaHSO3 solution were heated until all the NaHSO3 had reacted, what would be the resulting concentration of the Na2SO3 solution? How many moles of SO2 would be formed? How many grams of SO2 would be formed? If each mole of SO2 had a volume of 25.78 L, what volume of SO2 would be produced?
29. What is the concentration of a 1.0 M solution of K+(aq) ions in equivalents/liter?
30. What is the concentration of a 1.0 M solution of SO42(aq) ions in equivalents/liter?
31. A solution having initial concentration of 0.445 M and initial volume of 45.0 mL is diluted to 100.0 mL. What is its final concentration?
32. A 50.0 mL sample of saltwater that is 3.0% m/v is diluted to 950 mL. What is its final mass/volume percent?
Answers
1. Solubility is the amount of a solute that can dissolve in a given amount of solute, typically 100 mL. The solubility of solutes varies widely.
2. The term dilute means relatively less solute and the term concentrated implies relatively more solute. Both are of limited usefulness because these are not accurate.
3. 76.5 g
4. 90.2 g
5. 12.6 g
6. unsaturated
7. 26.5%
8. 2.15%
9. 0.203%
10. 3.4%
11. 52.4 mL
12. 5.18 g
13. 0.00321 g
14. 2.57 x 10-4 g
15. 15 ppm
16. 7,580,000 ppb
17. 0.130 M
18. 0.0458 M
19. 0.208 M
20. 0.0475 M
21. 37.6 g
22. 548 g
23. 0.496 L
24. 20 mL
25. 0.262 mol; 37.5 g
26. 0.345 mol; 29.0 g
27. 1.13 M C2H5OH; 1.13 mol of CO2; 49.7 g of CO2; 27.7 L of CO2
28. 0.503 M Na2SO3; 0.285 mol SO2; 18.3 g SO2; 471 L SO2
29. 1.0 Eq/L
30. 2.0 Eq/L
31. 0.200 M
32. 0.16 % m/v
Concept Review Exercise
1. Explain how the solvation process describes the dissolution of a solute in a solvent.
Answer
1. Each particle of the solute is surrounded by particles of the solvent, carrying the solute from its original phase.
Exercises
1. Describe what happens when an ionic solute like Na2SO4 dissolves in a polar solvent.
2. Describe what happens when a molecular solute like sucrose (C12H22O11) dissolves in a polar solvent.
3. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H2O to some extent.
1. NH4NO3
2. CO2
3. NH2CONH2
4. HCl
4. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H2O to some extent.
1. CH3CH2CH2OH
2. Ca(CH3CO2)2
3. I2
4. KOH
5. Will solutions of each solute conduct electricity when dissolved?
1. AgNO3
2. CHCl3
3. BaCl2
4. Li2O
6. Will solutions of each solute conduct electricity when dissolved?
1. CH3COCH3
2. N(CH3)3
3. CH3CO2C2H5
4. FeCl2
Answers
1. Each ion of the ionic solute is surrounded by particles of solvent, carrying the ion from its associated crystal.
2. Each sucrose molecule is surrounded by solvent molecules (attracted to each other via intermolecular forces of attraction).
1. electrolyte
2. nonelectrolyte
3. nonelectrolyte
4. electrolyte
4.
• nonelectrolyte
• electrolyte
• nonelectrolyte
• electrolyte
5.
1. yes
2. no
3. yes
4. yes
6.
a. no
b. no
c. no
d. yes
Concept Review Exercises
1. What are the colligative properties of solutions?
2. Explain how the following properties of solutions differ from those of the pure solvent: vapor pressure, boiling point, freezing point, and osmotic pressure.
Answers
1. Colligative properties are characteristics that a solution has that depend on the number, not the identity, of solute particles.
2. In solutions, the vapor pressure is lower, the boiling point is higher, the freezing point is lower, and the osmotic pressure is higher.
Exercises
1. In each pair of aqueous systems, which will have the lower vapor pressure?
1. pure water or 1.0 M NaCl
2. 1.0 M NaCl or 1.0 M C6H12O6
3. 1.0 M $\ce{CaCl2}$ or 1.0 M (NH4)3PO4
2. In each pair of aqueous systems, which will have the lower vapor pressure?
1. 0.50 M Ca(NO3)2 or 1.0 M KBr
2. 1.5 M C12H22O11 or 0.75 M Ca(OH)2
3. 0.10 M Cu(NO3)2 or pure water
3. In each pair of aqueous systems, which will have the higher boiling point?
1. pure water or a 1.0 M NaCl
2. 1.0 M NaCl or 1.0 M C6H12O6
3. 1.0 M $\ce{CaCl2}$ or 1.0 M (NH4)3PO4
4. In each pair of aqueous systems, which will have the higher boiling point?
1. 0.50 M Ca(NO3)2 or 1.0 M KBr
2. 1.5 M C12H22O11 or 0.75 M Ca(OH)2
3. 0.10 M Cu(NO3)2 or pure water
5. Estimate the boiling point of each aqueous solution. The boiling point of pure water is 100.0°C.
1. 0.50 M NaCl
2. 1.5 M Na2SO4
3. 2.0 M C6H12O6
6. Estimate the freezing point of each aqueous solution. The freezing point of pure water is 0.0°C.
1. 0.50 M NaCl
2. 1.5 M Na2SO4
3. 2.0 M C6H12O6
7. Explain why salt (NaCl) is spread on roads and sidewalks to inhibit ice formation in cold weather.
8. Salt (NaCl) and calcium chloride ($\ce{CaCl2}$) are used widely in some areas to minimize the formation of ice on sidewalks and roads. One of these ionic compounds is better, mole for mole, at inhibiting ice formation. Which is that likely to be? Why?
9. What is the osmolarity of each aqueous solution?
1. 0.500 M NH2CONH2
2. 0.500 M NaBr
3. 0.500 M Ca(NO3)2
10. What is the osmolarity of each aqueous solution?
1. 0.150 M KCl
2. 0.450 M (CH3)2CHOH
3. 0.500 M Ca3(PO4)2
11. A 1.0 M solution of an unknown soluble salt has an osmolarity of 3.0 osmol. What can you conclude about the salt?
12. A 1.5 M NaCl solution and a 0.75 M Al(NO3)3 solution exist on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and the direction of solvent flow, if any, across the membrane.
Answers
1. 1.0 M NaCl
2. 1.0 M NaCl
3. 1.0 M (NH4)3PO4
2.
• 1.0 M KBr
• 0.75 M Ca(OH)2
• 0.10 M Cu(NO3)2
3.
1. 1.0 M NaCl
2. 1.0 M NaCl
3. 1.0 M (NH4)3PO4
4.
• 1.0 M KBr
• 0.75 M Ca(OH)2
• 0.10 M Cu(NO3)2
5.
1. 100.5°C
2. 102.3°C
3. 101°C
6.
1. -1.9°C
2. -8.6°C
3. -3.8°C
7. NaCl lowers the freezing point of water, so it needs to be colder for the water to freeze.
8. $\ce{CaCl2}$ splits up into 3 ions while NaCl splits up into 2 ions only. $\ce{CaCl2}$ will be more effective.
9.
1. 0.500 osmol
2. 1.000 osmol
3. 1.500 osmol
10.
1. 0.300 osmol
2. 0.450 osmol
3. 2.50 osmol
11. It must separate into three ions when it dissolves.
12. Both NaCl and Al(NO3)3 have 3.0 osmol. There will be no net difference in the solvent flow.
Concept Review Exercises
1. What is chemical equilibrium?
2. What does the equilibrium constant tell us?
Answers
1. The rate of the forward reaction equals the rate of the reverse reaction.
2. The ratio of products and reactants when the system is at equilibrium.
EXERCISES
1. If the reaction H2 + I2 ⇌2HI is at equilibrium, do the concentrations of HI, H2, and I2 have to be equal?
2. Do the concentrations at equilibrium depend upon how the equilibrium was reached?
3. What does it mean if the Keq is > 1?
4. What does it mean if the Keq is < 1?
5. Does the equilibrium state depend on the starting concentrations?
6. Write an expression for the equilibrium constant K equation.
a. PCl5 (g) ⇄ PCl3 (g) + Cl2 (g)
b. $2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)$
7. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: 3C2H2(g)⟶C6H6(g). Which value of K would make this reaction most useful commercially? Explain your answer.
a. K ≈ 0.01
b. K ≈ 1
c. K ≈ 10.
8. Tell whether the reactants or the products are favored at equilibrium:
a. 2NH3(g) ⇌ N2(g) + 3H2(g) K = 172
b. 2SO3(g) ⇌ 2SO2(g) + O2(g) K = 0.230
c. 2NO(g) + Cl2(g) ⇌ 2NOCl(g) K = 4.6×104
d. N2(g) + O2(g) ⇌ 2NO(g) K = 0.050
AnswerS
1. No, the concentrations are constant but the concentrations do not have to be equal.
2. No.
3. More products than reactants are present at equilibrium.
4. More reactants than products present at equilibrium.
5. No. The equilibrium ratio does not depend on the initial concentrations.
6.
a. $K=\dfrac{[PCl_3][Cl_2]}{[PCl_5]}$
b. $K=\dfrac{[O_2]^3}{[O_3]^2}$
7. The answer is c. K ≈ 10.
Since $K=\dfrac{[C_6H_6]^2}{[C_2H_2]^3}$ (K≈10), this means that C6H6 predominates over C2H2. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.
8.
a. products
b. reactants
c. products
d. reactants
Concept Review Exercises
1. Define Le Chatelier’s principle.
2. List the three factors types of changes that can disturb the equilibrium of a system.
Answers
1. Le Chatelier’s principle states that a system at equilibrium is disturbed, it will respond in a way to minimize the disturbance.
2. temperature, change in amount of substance, change in pressure through change in volume
EXERCISES
1. How will each change affect the reaction?
PCl5(g) + heat ⇌PCl3(g) + Cl2(g)
1. Addition of PCl5
2. Addition of Cl2
3. Removal of PCl3
4. Increasing temperature
5. Decreasing temperature
6. Decreasing volume
2. How will each change affect the reaction?
HNO2(aq) ⇌H+(aq) + NO2(aq)
1. Removal of HNO2
2. Addition of HCl (i.e. adding more H+)
3. Increasing volume
4. Decreasing volume
5. Removal of NO2
6. Addition of OH (which will react with and remove H+)
3. How will each change affect the reaction?
CO2(g) + C(s) ⇌2CO(g) ΔH=172.5kJ
1. Addition of CO2
2. Removal of CO2
3. Increasing temperature
4. Decreasing temperature
5. Increasing volume
6. Addition of CO
4. How will each change affect the reaction?
H2(g) + I2(g) ⇌2HI(g) ΔH=−9.48kJ
1. Addition of H2
2. Removal of H2
3. Increasing temperature
4. Decreasing temperature
5. Increasing volume
6. Decreasing volume
1.
1. shift right
2. shift left
3. shift right
4. shift right
5. shift left
6. shift left
2.
1. shift left
2. shift left
3. no effect
4. no effect
5. shift right
6. shift right
3.
1. shift right
2. shift left
3. shift right
4. shift left
5. shift right
6. shift left
4.
1. shift right
2. shift left
3. shift left
4. shift right
5. no effect
6. no effect
Concept Review Exercises
1. What are some of the features of a semipermeable membrane?
2. What do the prefixes hyper, hypo, and iso mean?
Answers
1. A semipermeable membrane allows some substances to pass through but not others.
2. hyper – higher; hypo – lower; iso - same
EXERCISES
1. Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
1. Which one has a higher concentration?
2. Which way will water molecules flow?
3. Which volume will increase?
4. Which volume will decrease?
5. What will happen to the concentration of solution A?
6. What will happen to the concentration of solution B?
2. Two solutions with different concentrations and compositions are separated by a semipermeable membrane. The left-hand solution is a .50 M solution of MgSO4, while the right-hand solution contains CaCl2 at a concentration of .40 M. Determine the direction of the flow of solvent, left or right.
3. Given the following situations, wherein two tanks of different solutions are separated by a semipermeable membrane, determine the direction of the flow of solvent (water).
a. Solution A contains a 0.40 M concentration of CaCl2, while Solution B contains a 0.45 M concentration of KI
b. Solution A contains a 1.00 M concentration of NH4Cl, while Solution B contains a 1.00 M concentration of CH2O
4. Cells are placed in a solution and the cells then undergo hemolysis. What can be said about the relative concentrations of solute in the cell and the solution?
5. Describe the relative concentrations inside and outside a red blood cell when crenation occurs.
6. A saltwater fish is placed in a freshwater tank. What will happen to the fish? Describe the flow of water molecules to explain the outcome.
7. What makes up the "head" region of a phospholipid? Is it hydrophobic or hydrophilic?
8. What makes up the "tail" region of a phospholipid? Is it hydrophobic or hydrophilic?
Answers
1. Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.
1. Solution B
2. A →→ B
3. B
4. A
5. increase
6. decrease
2. Water (solvent) flows from left to right.
3.
a. Water flows from Solution B to Solution A.
b. Water flows from Solution B to Solution A.
4. Cells contain fluid with higher concentration than solution outside the cell.
5. Cells contain fluid with a lower concentration than the solution outside the cell.
6. Water molecules will flow from the tank water into the fish because the fish has a higher concentration of salt. If the fish absorbs too much water, it will die.
7. The "head" region is a phosphate group and it is hydrophilic.
8. The "tail" is a hydrocarbon tail and it is hydrophobic.
Contributors
• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
Additional Exercises
1. Calcium nitrate reacts with sodium carbonate to precipitate solid calcium carbonate:
$Ca(NO_3)_{2(aq)} + Na_2CO_{3(aq)} \rightarrow CaCO_{3(s)} + NaNO_{3(aq)}$
1. Balance the chemical equation.
2. How many grams of Na2CO3 are needed to react with 50.0 mL of 0.450 M Ca(NO3)2?
3. Assuming that the Na2CO3 has a negligible effect on the volume of the solution, find the osmolarity of the NaNO3 solution remaining after the CaCO3 precipitates from solution.
2. The compound HCl reacts with sodium carbonate to generate carbon dioxide gas:
$HCl_{(aq)} + Na_2CO_{3(aq)} \rightarrow H_2O_{(ℓ)} + CO_{2(g)} + NaCl_{(aq)}$
1. Balance the chemical equation.
2. How many grams of Na2CO3 are needed to react with 250.0 mL of 0.755 M HCl?
3. Assuming that the Na2CO3 has a negligible effect on the volume of the solution, find the osmolarity of the NaCl solution remaining after the reaction is complete.
3. Estimate the freezing point of concentrated aqueous HCl, which is usually sold as a 12 M solution. Assume complete ionization into H+ and Cl ions.
4. Estimate the boiling point of concentrated aqueous H2SO4, which is usually sold as an 18 M solution. Assume complete ionization into H+ and HSO4 ions.
5. Seawater can be approximated by a 3.0% m/m solution of NaCl in water. Determine the molarity and osmolarity of seawater. Assume a density of 1.0 g/mL.
6. Human blood can be approximated by a 0.90% m/m solution of NaCl in water. Determine the molarity and osmolarity of blood. Assume a density of 1.0 g/mL.
7. How much water must be added to 25.0 mL of a 1.00 M NaCl solution to make a resulting solution that has a concentration of 0.250 M?
8. Sports drinks like Gatorade are advertised as capable of resupplying the body with electrolytes lost by vigorous exercise. Find a label from a sports drink container and identify the electrolytes it contains. You should be able to identify several simple ionic compounds in the ingredients list.
9. Occasionally we hear a sensational news story about people stranded in a lifeboat on the ocean who had to drink their own urine to survive. While distasteful, this act was probably necessary for survival. Why not simply drink the ocean water? (Hint: See Exercise 5 and Exercise 6 above. What would happen if the two solutions in these exercises were on opposite sides of a semipermeable membrane, as we would find in our cell walls?)
Answers
1.
1. Ca(NO3)2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaNO3(aq)
2. 2.39 g
3. 1.80 osmol
2.
a. 2HCl (aq) + Na2CO3(aq) → H2O(ℓ) + CO2(g) + 2NaCl (aq)
b. 10.0 g
c. 1.51 M
1. −45.6°C
4. 118°C
1. 0.513 M; 1.026 osmol
6. molarity = 0.15 M; osmolarity = 0.31 M
7. 75.0 mL
8. magnesium chloride, calcium chloride (answers may vary)
1. The osmotic pressure of seawater is too high. Drinking seawater would cause water to go from inside our cells into the more concentrated seawater, ultimately killing the cells. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.E%3A_Solutions_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
A solution is a homogeneous mixture. The major component is the solvent, while the minor component is the solute. Solutions can have any phase; for example, an alloy is a solid solution. Solutes are soluble or insoluble, meaning they dissolve or do not dissolve in a particular solvent. The terms miscible and immiscible, instead of soluble and insoluble, are used for liquid solutes and solvents. The statement like dissolves like is a useful guide to predicting whether a solute will dissolve in a given solvent.
The amount of solute in a solution is represented by the concentration of the solution. The maximum amount of solute that will dissolve in a given amount of solvent is called the solubility of the solute. Such solutions are saturated. Solutions that have less than the maximum amount are unsaturated. Most solutions are unsaturated, and there are various ways of stating their concentrations. Mass/mass percent, volume/volume percent, and mass/volume percent indicate the percentage of the overall solution that is solute. Parts per million (ppm) and parts per billion (ppb) are used to describe very small concentrations of a solute. Molarity, defined as the number of moles of solute per liter of solution, is a common concentration unit in the chemistry laboratory. Equivalents express concentrations in terms of moles of charge on ions. When a solution is diluted, we use the fact that the amount of solute remains constant to be able to determine the volume or concentration of the final diluted solution.
Dissolving occurs by solvation, the process in which particles of a solvent surround the individual particles of a solute, separating them to make a solution. For water solutions, the word hydration is used. If the solute is molecular, it dissolves into individual molecules. If the solute is ionic, the individual ions separate from each other, forming a solution that conducts electricity. Such solutions are called electrolytes. If the dissociation of ions is complete, the solution is a strong electrolyte. If the dissociation is only partial, the solution is a weak electrolyte. Solutions of molecules do not conduct electricity and are called nonelectrolytes.
Solutions have properties that differ from those of the pure solvent. Some of these are colligative properties, which are due to the number of solute particles dissolved, not the chemical identity of the solute. Colligative properties include vapor pressure depression, boiling point elevation, freezing point depression, and osmotic pressure. Osmotic pressure is particularly important in biological systems. It is caused by osmosis, the passage of solvents through certain membranes like cell walls. The osmolarity of a solution is the product of a solution’s molarity and the number of particles a solute separates into when it dissolves. Osmosis can be reversed by the application of pressure; this reverse osmosis is used to make fresh water from saltwater in some parts of the world. Because of osmosis, red blood cells placed in hypotonic or hypertonic solutions lose function through either hemolysis or crenation. If they are placed in isotonic solutions, however, the cells are unaffected because osmotic pressure is equal on either side of the cell membrane. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.S%3A_Solutions_%28Summary%29.txt |
Many of us are familiar with the group of chemicals called acids. But do you know what it takes for a compound to be an acid? Actually, there are several different definitions of acid that chemistry uses, and each definition is appropriate under different circumstances. Less familiar—but just as important to chemistry and ultimately to us—is the group of chemicals known as bases. Both acids and bases are important enough that we devote an entire chapter to them—their properties and their reactions.
• 10.0: Prelude to Acids and Bases
One of the most concentrated acids in the body is stomach acid, which can be approximated as a 0.05 M hydrochloric acid solution. Special cells in the stomach wall secrete this acid, along with special enzymes, as part of the digestion process. In a laboratory, a 0.05 M solution of hydrochloric acid would dissolve some metals. How does the stomach survive the presence of such a reactive acid?
• 10.1: Arrhenius Definition of Acids and Bases
Arrhenius acid: a compound that increases the concentration of hydrogen ion (H+) in aqueous solution; Arrhenius base: a compound that increases the concentration of hydroxide ion (OH−) in aqueous solution. the reaction of an acid and a base
• 10.2: Brønsted-Lowry Definition of Acids and Bases
A Brønsted-Lowry acid is a proton donor, and a Brønsted-Lowry base is a proton acceptor. Brønsted-Lowry acid-base reactions are essentially proton transfer reactions.
• 10.3: Water - Both an Acid and a Base
Water molecules can act as both an acid and a base, depending on the conditions.
• 10.4: The Strengths of Acids and Bases
Acids and bases can be strong or weak depending on the extent of ionization in solution. Most chemical reactions reach equilibrium at which point there is no net change. The pH scale is used to succinctly communicate the acidity or basicity of a solution.
• 10.5: Buffers
A buffer is a solution that resists sudden changes in pH.
• 10.E: Acids and Bases (Exercises)
Problems and select solutions to the chapter.
• 10.S: Acids and Bases (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
10: Acids and Bases
One of the most concentrated acids in the body is stomach acid, which can be approximated as a 0.05 M hydrochloric acid solution. Special cells in the stomach wall secrete this acid, along with special enzymes, as part of the digestion process. In a laboratory, a 0.05 M solution of hydrochloric acid would dissolve some metals. How does the stomach survive the presence of such a reactive acid?
Actually, the stomach has several mechanisms for withstanding this chemical onslaught. First, the lining of the stomach is coated with a thin layer of mucus that contains some bicarbonate ions (HCO3). These react with the hydrochloric acid to produce water, carbon dioxide, and harmless chloride ions. If any acid penetrates through the mucus, it can attack the surface layer of stomach cells, called the gastric epithelium. Cells in the gastric epithelium are being constantly shed, so damaged cells are quickly removed and replaced with healthy cells.
However, if the gastric epithelium is destroyed faster than it can be replaced, the acid may reach the wall of the stomach, resulting in ulcers. If an ulcer grows large enough, it can expose blood vessels in the stomach wall, causing bleeding. In extreme situations, the loss of blood through a severe ulcer can threaten a person’s health.
Ulcers can also result from the presence of a certain bacterium—Helicobacter pylori—in the stomach. The mechanism for this ulcer formation is not the same as that for ulcers caused by stomach acid and is not completely understood. However, there are two main treatments for ulcers: (1) antacids to react chemically with excess hydrochloric acid in the stomach and (2) antibiotics to destroy the H. pylori bacteria in the stomach. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/10%3A_Acids_and_Bases/10.00%3A_Prelude_to_Acids_and_Bases.txt |
Learning Objectives
• To recognize a compound as an Arrhenius acid or an Arrhenius base.
• To describe characteristics of acids and bases.
• To write equations of neutralization reactions.
One way to define a class of compounds is by describing the various characteristics its members have in common. In the case of the compounds known as acids, the common characteristics include a sour taste, the ability to change the color of the vegetable dye litmus to red, and the ability to dissolve certain metals and simultaneously produce hydrogen gas. For the compounds called bases, the common characteristics are a slippery texture, a bitter taste, and the ability to change the color of litmus to blue. Acids and bases also react with each other to form compounds generally known as salts.
Although we include their tastes among the common characteristics of acids and bases, we never advocate tasting an unknown chemical!
Chemists prefer, however, to have definitions for acids and bases in chemical terms. The Swedish chemist Svante Arrhenius developed the first chemical definitions of acids and bases in the late 1800s. Arrhenius defined an acid as a compound that increases the concentration of hydrogen ion (H+) in aqueous solution. Many acids are simple compounds that release a hydrogen cation into solution when they dissolve. Similarly, Arrhenius defined a base as a compound that increases the concentration of hydroxide ion (OH) in aqueous solution. Many bases are ionic compounds that have the hydroxide ion as their anion, which is released when the base dissolves in water.
Table $1$: Formulas and Names for Some Acids and Bases
Acids Bases
Formula Name Formula Name
HCl(aq) hydrochloric acid NaOH(aq) sodium hydroxide
HBr(aq) hydrobromic acid KOH(aq) potassium hydroxide
HI(aq) hydriodic acid Mg(OH)2(aq) magnesium hydroxide
H2S(aq) hydrosulfuric acid Ca(OH)2(aq) calcium hydroxide
HC2H3O2(aq) acetic acid NH3(aq) ammonia
HNO3(aq) nitric acid NaHCO3 (aq) sodium bicarbonate
HNO2(aq) nitrous acid CaCO3 (aq) calcium carbonate
H2SO4(aq) sulfuric acid
H2SO3(aq) sulfurous acid
HClO3(aq) chloric acid
HClO4(aq) perchloric acid
HClO2(aq) chlorous acid
H3PO4(aq) phosphoric acid
H3PO3(aq) phosphorous acid
H2CO3(aq) carbonic acid
Many bases and their aqueous solutions are named using the normal rules of ionic compounds that were presented previously; that is, they are named as hydroxide compounds. For example, the base sodium hydroxide (NaOH) is both an ionic compound and an aqueous solution. However, aqueous solutions of acids have their own naming rules. The names of binary acids (compounds with hydrogen and one other element in their formula) are based on the root of the name of the other element preceded by the prefix hydro- and followed by the suffix -ic acid. Thus, an aqueous solution of HCl [designated “HCl(aq)”] is called hydrochloric acid, H2S(aq) is called hydrosulfuric acid, and so forth. Acids composed of more than two elements (typically hydrogen and oxygen and some other element) have names based on the name of the other element, followed by the suffix -ic acid or -ous acid, depending on the number of oxygen atoms in the acid’s formula. Other prefixes, like per- and hypo-, also appear in the names for some acids. Unfortunately, there is no strict rule for the number of oxygen atoms that are associated with the -ic acid suffix; the names of these acids are best memorized. Table $1$ lists some acids and bases and their names. Note that acids have hydrogen written first, as if it were the cation, while most bases have the negative hydroxide ion, if it appears in the formula, written last.
The name oxygen comes from the Latin meaning “acid producer” because its discoverer, Antoine Lavoisier, thought it was the essential element in acids. Lavoisier was wrong, but it is too late to change the name now.
Example $1$
Name each substance.
1. HF(aq)
2. Sr(OH)2(aq)
Solution
1. This acid has only two elements in its formula, so its name includes the hydro- prefix. The stem of the other element’s name, fluorine, is fluor, and we must also include the -ic acid ending. Its name is hydrofluoric acid.
2. This base is named as an ionic compound between the strontium ion and the hydroxide ion: strontium hydroxide.
Exercise $1$
Name each substance.
1. H2Se(aq)
2. Ba(OH)2(aq)
Answer
a. hydroselenic acid
b. barium hydroxide
Notice that one base listed in Table $1$—ammonia—does not have hydroxide as part of its formula. How does this compound increase the amount of hydroxide ion in aqueous solution? Instead of dissociating into hydroxide ions, ammonia molecules react with water molecules by taking a hydrogen ion from the water molecule to produce an ammonium ion and a hydroxide ion:
$NH_{3(aq)} + H_2O_{(ℓ)} \rightarrow NH^+_{4(aq)} + OH^−_{(aq)} \label{Eq1}$
Because this reaction of ammonia with water causes an increase in the concentration of hydroxide ions in solution, ammonia satisfies the Arrhenius definition of a base. Many other nitrogen-containing compounds are bases because they too react with water to produce hydroxide ions in aqueous solution.
Neutralization
As we noted previously, acids and bases react chemically with each other to form salts. A salt is a general chemical term for any ionic compound formed from an acid and a base. In reactions where the acid is a hydrogen ion containing compound and the base is a hydroxide ion containing compound, water is also a product. The general reaction is as follows:
acid + base → water + salt
The reaction of acid and base to make water and a salt is called neutralization. Like any chemical equation, a neutralization chemical equation must be properly balanced. For example, the neutralization reaction between sodium hydroxide and hydrochloric acid is as follows:
$NaOH{(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(ℓ)} \label{Eq2}$
with coefficients all understood to be one. The neutralization reaction between sodium hydroxide and sulfuric acid is as follows:
$2NaOH_{(aq)} + H_2SO_{4(aq)} \rightarrow Na_2SO_{4(aq)} + 2H_2O_{(ℓ)} \label{Eq3}$
Once a neutralization reaction is properly balanced, we can use it to perform stoichiometry calculations, such as the ones we practiced earlier.
There are a number of examples of acid-base chemistry in everyday life. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO3 is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter “rises.” Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.
$HCO_3^- (aq) + H^+ (aq) \rightarrow H_2 CO_3 (aq) \label{4.3.19}$
$H_2 CO_3 (aq) \rightarrow CO_2 (g) + H_2 O(l) \nonumber$
Example $2$
Nitric acid [HNO3(aq)] can be neutralized by calcium hydroxide [Ca(OH)2(aq)].
1. Write a balanced chemical equation for the reaction between these two compounds and identify the salt it produces.
2. For one reaction, 16.8 g of HNO3 is present initially. How many grams of Ca(OH)2 are needed to neutralize that much HNO3?
3. In a second reaction, 805 mL of 0.672 M Ca(OH)2 is present initially. What volume of 0.432 M HNO3 solution is necessary to neutralize the Ca(OH)2 solution?
Solution
1. Because there are two OH ions in the formula for Ca(OH)2, we need two moles of HNO3 to provide H+ ions. The balanced chemical equation is as follows: Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(ℓ)
The salt formed is calcium nitrate.
1. This calculation is much like the calculations we did in Chapter 6. First we convert the mass of HNO3 to moles using its molar mass of 1.01 + 14.01 + 3(16.00) = 63.02 g/mol; then we use the balanced chemical equation to determine the related number of moles of Ca(OH)2 needed to neutralize it; and then we convert that number of moles of Ca(OH)2 to the mass of Ca(OH)2 using its molar mass of 40.08 + 2(1.01) + 2(16.00) = 74.10 g/mol.
$\mathrm{16.8\: g\: HNO_3 \times \dfrac{1\: mol\: HNO_3}{63.02\: g\ HNO_3} \times \dfrac{1\: mol\: Ca(OH)_2}{2\: mol\: HNO_3} \times \dfrac{74.10\: g\: Ca(OH)_2}{1\: mol\: Ca(OH)_2}=9.88\: g\: Ca(OH)_2\: needed}$
1. Having concentration information allows us to employ the skills we developed in Chapter 9. We have two alternative solutions: the multi-step process and the combined one-line process (found at the bottom).
First, we use the concentration and volume data to determine the number of moles of Ca(OH)2 present. Recognizing that 805 mL = 0.805 L,
$\mathrm{0.672\: M\: Ca(OH)_2=\dfrac{mol\: Ca(OH)_2}{0.805\: L\: soln}}$
0.672 M CaOH)2 × (0.805 L soln) = mol Ca(OH)2 = 0.541 mol Ca(OH)2
We combine this information with the proper ratio from the balanced chemical equation to determine the number of moles of HNO3 needed:
$\mathrm{0.541\: mol\: Ca(OH)_2 \times \dfrac{2\: mol\: HNO_3}{1\: mol\: Ca(OH)_2}=1.08\: mol\: HNO_3}$
Now, using the definition of molarity one more time, we determine the volume of acid solution needed:
$\mathrm{0.432\: M\: HNO_3=\dfrac{1.08\: mol\: HNO_3}{volume\: of\: HNO_3}}$
$\mathrm{volume\: of\: HNO_3=\dfrac{1.08\: mol\: HNO_3}{0.432\: M\: HNO_3}=2.50\: L=2.50\times10^3\: mL\: HNO_3}$
$\mathrm{0.805\: L\: Ca(OH)_2\: soln \times \dfrac{0.672\: mol\: Ca(OH)_2 }{1\: L\ Ca(OH)_2\: soln } \times \dfrac{2\: mol\: HNO_3}{1\: mol\: Ca(OH)_2} \times \dfrac{1\: L\: HNO_3\: soln}{0.432\: mol\: HNO_3}=2.50\: L\: HNO_3\: soln\: needed}$
Exercise $2$
Hydrocyanic acid [HCN(aq)] can be neutralized by potassium hydroxide [KOH(aq)].
1. Write a balanced chemical equation for the reaction between these two compounds and identify the salt it produces.
2. For one reaction, 37.5 g of HCN is present initially. How many grams of KOH are needed to neutralize that much HCN?
3. In a second reaction, 43.0 mL of 0.0663 M KOH is present initially. What volume of 0.107 M HCN solution is necessary to neutralize the KOH solution?
Answer
a. KOH(aq) + HCN(aq) → KCN(aq) + H2O(ℓ) KCN is the salt.
b. 77.8 g
c. 0.0266 L or 26.6 mL
Hydrocyanic acid (HCN) is one exception to the acid-naming rules that specify using the prefix hydro- for binary acids (acids composed of hydrogen and only one other element).
Stomach Antacids
Our stomachs contain a solution of roughly 0.03 M HCl, which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, CaCO3. The reaction,
$CaCO_3(s)+2HCl(aq)⇌CaCl_2(aq)+H_2O(l)+CO_2(g) \nonumber$
not only neutralizes stomach acid, it also produces CO2(g), which may result in a satisfying belch.
Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, Mg(OH)2. It works according to the reaction:
$Mg(OH)_2(s)⇌Mg^{2+}(aq)+2OH^-(aq) \nonumber$
The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that :
$H_3O^+ + OH^- ⇌ 2H_2O(l) \nonumber$
This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect. Several antacids have aluminum hydroxide, Al(OH)3, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.
Example $3$
Assume that the stomach of someone suffering from acid indigestion contains 75 mL of 0.20 M HCl. How many Tums tablets are required to neutralize 90% of the stomach acid, if each tablet contains 500 mg of CaCO3? (Neutralizing all of the stomach acid is not desirable because that would completely shut down digestion.)
Solution
1. Write the balanced chemical equation for the reaction and then decide whether the reaction will go to completion.
2. Calculate the number of moles of acid present. Multiply the number of moles by the percentage to obtain the quantity of acid that must be neutralized. Using mole ratios, calculate the number of moles of base required to neutralize the acid.
3. Calculate the number of moles of base contained in one tablet by dividing the mass of base by the corresponding molar mass. Calculate the number of tablets required by dividing the moles of base by the moles contained in one tablet.
A. We first write the balanced chemical equation for the reaction:
$2HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2CO_3(aq)$
Each carbonate ion can react with 2 mol of H+ to produce H2CO3, which rapidly decomposes to H2O and CO2. Because HCl is a strong acid and CO32 is a weak base, the reaction will go to completion.
B. Next we need to determine the number of moles of HCl present:
$75\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .20\: mol\: HCl} {\cancel{L}} \right) = 0. 015\: mol\: HCl$
Because we want to neutralize only 90% of the acid present, we multiply the number of moles of HCl by 0.90:
$(0.015\: mol\: HCl)(0.90) = 0.014\: mol\: HCl$
We know from the stoichiometry of the reaction that each mole of CaCO3 reacts with 2 mol of HCl, so we need
$moles\: CaCO_3 = 0 .014\: \cancel{mol\: HCl} \left( \dfrac{1\: mol\: CaCO_3}{2\: \cancel{mol\: HCl}} \right) = 0 .0070\: mol\: CaCO_3$
C. Each Tums tablet contains
$\left( \dfrac{500\: \cancel{mg\: CaCO_3}} {1\: Tums\: tablet} \right) \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg\: CaCO_3}} \right) \left( \dfrac{1\: mol\: CaCO_3} {100 .1\: \cancel{g}} \right) = 0 .00500\: mol\: CaCO_ 3$
Thus we need $\dfrac{0.0070\: \cancel{mol\: CaCO_3}}{0.00500\: \cancel{mol\: CaCO_3}}= 1.4$ Tums tablets.
Exercise $3$
Assume that as a result of overeating, a person’s stomach contains 300 mL of 0.25 M HCl. How many Rolaids tablets must be consumed to neutralize 95% of the acid, if each tablet contains 400 mg of NaAl(OH)2CO3? The neutralization reaction can be written as follows:
$NaAl(OH)_2CO_3(s) + 4HCl(aq) \rightarrow AlCl_3(aq) + NaCl(aq) + CO_2(g) + 3H_2O(l)$
Answer
6.4 tablets
Key Takeaway
• An Arrhenius acid increases the H+ ion concentration in water, while an Arrhenius base increases the OH ion concentration in water. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/10%3A_Acids_and_Bases/10.01%3A_Arrhenius_Definition_of_Acids_and_Bases.txt |
Learning Objectives
1. Recognize a compound as a Brønsted-Lowry acid or a Brønsted-Lowry base.
2. Illustrate the proton transfer process that defines a Brønsted-Lowry acid-base reaction.
Ammonia (NH3) increases the hydroxide ion concentration in aqueous solution by reacting with water rather than releasing hydroxide ions directly. In fact, the Arrhenius definitions of an acid and a base focus on hydrogen ions and hydroxide ions. Are there more fundamental definitions for acids and bases?
In 1923, the Danish scientist Johannes Brønsted and the English scientist Thomas Lowry independently proposed new definitions for acids and bases. Rather than considering both hydrogen and hydroxide ions, they focused on the hydrogen ion only. A Brønsted-Lowry acid is a compound that supplies a hydrogen ion in a reaction. A Brønsted-Lowry base, conversely, is a compound that accepts a hydrogen ion in a reaction. Thus, the Brønsted-Lowry definitions of an acid and a base focus on the movement of hydrogen ions in a reaction, rather than on the production of hydrogen ions and hydroxide ions in an aqueous solution.
Let us use the reaction of ammonia in water to demonstrate the Brønsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products:
$NH_{3}(aq) + H_2O(ℓ) \rightarrow NH^+_{4}(aq) + OH^−(aq) \label{Eq1}$
What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows:
Because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule—which accepts the hydrogen ion—is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense.
Is an Arrhenius acid like hydrochloric acid still an acid in the Brønsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen atom is a single proton surrounded by a single electron. To make the hydrogen ion, we remove the electron, leaving a bare proton. Do we really have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H+ ion attaches itself to H2O to make H3O+, which is called the hydronium ion. For most purposes, H+ and H3O+ represent the same species, but writing H3O+ instead of H+ shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules.
A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like H5O2+ or H9O4+ rather than H3O+. It is simpler, however, to use H3O+.
With this in mind, how do we define HCl as an acid in the Brønsted-Lowry sense? Consider what happens when HCl is dissolved in H2O:
$HCl(g) + H_2O(ℓ) \rightarrow H_3O^+(aq) + Cl^−(aq) \label{Eq2}$
We can depict this process using Lewis electron dot diagrams:
Now we see that a hydrogen ion is transferred from the HCl molecule to the H2O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Brønsted-Lowry acid; as a hydrogen ion acceptor, H2O is a Brønsted-Lowry base. So HCl is an acid not just in the Arrhenius sense but also in the Brønsted-Lowry sense. Moreover, by the Brønsted-Lowry definitions, H2O is a base in the formation of aqueous HCl. So the Brønsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base—although the Arrhenius definition would not have labeled H2O a base in this circumstance.
All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases.
Example $1$
Aniline (C6H5NH2) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base.
Solution
C6H5NH2 and H2O are the reactants. When C6H5NH2 accepts a proton from H2O, it gains an extra H and a positive charge and leaves an OH ion behind. The reaction is as follows:
C6H5NH2(aq) + H2O(ℓ) → C6H5NH3+(aq) + OH(aq)
Because C6H5NH2 accepts a proton, it is the Brønsted-Lowry base. The H2O molecule, because it donates a proton, is the Brønsted-Lowry acid.
Exercise $1$
Caffeine (C8H10N4O2) is a stimulant found in coffees and teas. When dissolved in water, it can accept a proton from a water molecule. Write the chemical equation for this process and identify the Brønsted-Lowry acid and base.
Answer
C8H10N4O2(aq) + H2O(ℓ) → C8H11N4O2+(aq) + OH(aq)
B-L base B-L acid
The Brønsted-Lowry definitions of an acid and a base can be applied to chemical reactions that occur in solvents other than water. The following example illustrates.
Example $2$
Sodium amide (NaNH2) dissolves in methanol (CH3OH) and separates into sodium ions and amide ions (NH2). The amide ions react with methanol to make ammonia and the methoxide ion (CH3O). Write a balanced chemical equation for this process and identify the Brønsted-Lowry acid and base.
Solution
The equation for the reaction is between NH2 and CH3OH to make NH3 and CH3O is as follows:
NH2(solv) + CH3OH(ℓ) → NH3(solv) + CH3O(solv)
The label (solv) indicates that the species are dissolved in some solvent, in contrast to (aq), which specifies an aqueous (H2O) solution. In this reaction, we see that the NH2 ion accepts a proton from a CH3OH molecule to make an NH3 molecule. Thus, as the proton acceptor, NH2 is the Brønsted-Lowry base. As the proton donor, CH3OH is the Brønsted-Lowry acid.
Exercise $2$
Pyridinium chloride (C5H5NHCl) dissolves in ethanol (C2H5OH) and separates into pyridinium ions (C5H5NH+) and chloride ions. The pyridinium ion can transfer a hydrogen ion to a solvent molecule. Write a balanced chemical equation for this process and identify the Brønsted-Lowry acid and base.
Answer
C5H5NH+(solv) + C2H5OH(ℓ) → C5H5N(solv) + C2H5OH2+(solv)
B-L acid B-L base
Application in Everyday Life
Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure $1$). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a “sour” taste that we seem to enjoy.
Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.
To Your Health: Brønsted-Lowry Acid-Base Reactions in Pharmaceuticals
There are many interesting applications of Brønsted-Lowry acid-base reactions in the pharmaceutical industry. For example, drugs often need to be water soluble for maximum effectiveness. However, many complex organic compounds are not soluble or are only slightly soluble in water. Fortunately, those drugs that contain proton-accepting nitrogen atoms (and there are a lot of them) can be reacted with dilute hydrochloric acid [HCl(aq)]. The nitrogen atoms—acting as Brønsted-Lowry bases—accept the hydrogen ions from the acid to make an ion, which is usually much more soluble in water. The modified drug molecules can then be isolated as chloride salts:
$\mathrm{RN(sl\: aq) + H^+(aq) \rightarrow RNH^+(aq) \xrightarrow{Cl^-(aq)} RNHCl(s)} \label{Eq3}$
where RN represents some organic compound containing nitrogen. The label (sl aq) means “slightly aqueous,” indicating that the compound RN is only slightly soluble. Drugs that are modified in this way are called hydrochloride salts. Examples include the powerful painkiller codeine, which is commonly administered as codeine hydrochloride. Acids other than hydrochloric acid are also used. Hydrobromic acid, for example, gives hydrobromide salts. Dextromethorphan, an ingredient in many cough medicines, is dispensed as dextromethorphan hydrobromide. The accompanying figure shows another medication (lidocaine) as a hydrochloride salt.
Conjugate Acid-Base Pairs
According to the Bronsted-Lowry theory of acids and bases, an acid is a proton donor and a base is a proton acceptor. Once an acid has given up a proton, the part that remains is called the acid's conjugate base. This species is a base because it can accept a proton (to re-form the acid). The conjugate base of HF (first example below) is fluoride ion, F-.
$\mathrm{{\color{Red} Acid} = H^+ + {\color{Blue} Conjugate\: base\: of\: Acid}^-}$
${\color{Red} \mathrm{HF}} \rightleftharpoons \mathrm{H^+} + {\color{Blue} \mathrm{F^-}}$
${\color{Red} \mathrm{H_2O}} \rightleftharpoons \mathrm{H^+} + {\color{Blue} \mathrm{OH^-}}$
${\color{Red} \mathrm{NH_4^+}} \rightleftharpoons \mathrm{H^+} + {\color{Blue} \mathrm{NH_3}}$
Similarly, the part of the base that remains after a base accepts a proton is called the base's conjugate acid. This species is an acid because it can give up a proton (and thus re-form the base). The conjugate acid of fluoride ion, F- (first example below) is HF.
$\mathrm{H^+ + {\color{Blue} Base} = {\color{Red} Conjugate\: acid\: of\: Base}^+}$
$\mathrm{H^+ + {\color{Blue} F^- } \rightleftharpoons {\color{Red} HF}}$
$\mathrm{H^+ + {\color{Blue} OH^- } \rightleftharpoons {\color{Red} H_2O}}$
$\mathrm{H^+ + {\color{Blue} H_2O } \rightleftharpoons {\color{Red} H_3O^+}}$
$\mathrm{H^+ + {\color{Blue} NH_3 } \rightleftharpoons {\color{Red} NH_4^+}}$
To summarize, the conjugate base of HF is fluoride ion, F-, and the conjugate acid of fluoride ion, F-, is HF. The HF/F- pair is referred to as a conjugate acid-base pair. The difference in the formulas of a conjugate acid-base pair (example: HF and F-) is H+. The table below lists conjugate acid-base pairs for your reference so that you can figure out the strategy of identifying them. For any given acid or base, you should be able to give its conjugate base or conjugate acid. The formula of an acid's conjugate base is generated by removing a proton (H+) from the acid formula. The formula of the base's conjugate acid is formed by adding a proton (H+) to the formula of the base.
Table $1$. Conjugate acid-base pairs.
Conjugate Acid Conjugate Base
$\ce{H3O+}$ $\ce{H2O}$
$\ce{H2O}$ $\ce{OH-}$
$\ce{H2SO4}$ $\ce{HSO4-}$
$\ce{HSO4-}$ $\ce{SO4^2-}$
$\ce{NH4+}$ $\ce{NH3}$
$\ce{NH3}$ $\ce{NH2-}$
$\ce{CH3COOH}$ $\ce{CH3COO-}$
$\ce{CH3NH3+}$ $\ce{CH3NH2}$
Example $3$ : Conjugate Pairs
Write the formula of the conjugate base of (a) HCl and (b) HCO3.
Write the formula of the conjugate acid of (c) CH3NH2 and (d) OH.
Solution:
A conjugate base is formed by removing a proton (H+). A conjugate acid is formed by accepting a proton (H+).
1. After HCl donates a proton, a Cl ion is produced, and so Cl is the conjugate base.
2. After hydrogen carbonate ion, HCO3, donates a proton, its conjugate base, CO32is produced.
3. After accepting a proton (H+), CH3NH2 is converted to CH3NH3+, its conjugate acid.
4. After accepting a proton (H+), OHis converted to H2O, its conjugate acid.
Exercise $3$: Conjugate Pairs
Write the formula of the conjugate base of (a) HNO2 and (b) H2CO3.
Write the formula of the conjugate acid of (c) C6H5NH2 and (d) HCO3.
Answer
a. NO2 is the conjugate base of HNO2.
b. HCO3 is the conjugate base of H2CO3
c. C6H5NH3+ is the conjugate acid of C6H5NH2.
d. H2CO3 is the conjugate acid of HCO3
In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are $NH_4^+/NH_3$ and $H_2O/OH^−$. This means that the conjugate acid of the base NH3 is NH4+ while the conjugate base of the acid NH4+ is NH3. Similarly, the conjugate base of the acid H2O is OH-, and the conjugate acid of the base OH- is H2O.
In the forward reaction, the parent acid is H2O and and the parent base is NH3 (shown in the illustration below). The acid H2O loses a proton (H+) to form its conjugate base OH-. The base NH3 gains a proton, to produce its conjugate acid NH4+. In the reverse reaction, the acid NH4+ loses a proton (H+) to form its conjugate base NH3. The base OH- gains a proton, to produce its conjugate acid H2O.
When hydrogen fluoride (HF) dissolves in water and ionizes, protons are transferred from hydrogen fluoride (parent acid) molecules to water (parent base) molecules, yielding hydronium ions (conjugate acid of water) and fluoride ions (conjugate base of HF):
${\color{Red} \mathrm{HF}} + {\color{Blue} H_2O } \rightleftharpoons {\color{Blue} H_3O^+} + {\color{Red} \mathrm{F^-}}$
Example $3$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{CH3CO2H + H2O <=> H3O^{+} + CH3CO2^{-}} \nonumber$
Solution
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, $H_3O^+$ is the acid that donates a proton to the acetate ion, which acts as the base.
Once again, we have two conjugate acid–base pairs:
• the parent acid and its conjugate base ($CH_3CO_2H/CH_3CO_2^−$) and
• the parent base and its conjugate acid ($H_3O^+/H_2O$).
Example $4$
Identify the conjugate acid-base pairs in this equilibrium.
$(CH_{3})_{3}N + H_{2}O\rightleftharpoons (CH_{3})_{3}NH^{+} + OH^{-} \nonumber$
Solution
One pair is H2O and OH, where H2O has one more H+ and is the conjugate acid, while OH has one less H+ and is the conjugate base.
The other pair consists of (CH3)3N and (CH3)3NH+, where (CH3)3NH+ is the conjugate acid (it has an additional proton) and (CH3)3N is the conjugate base.
Exercise $4$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{NH2^{-} + H2O\rightleftharpoons NH3 + OH^{-}} \nonumber$
Answer:
H2O (acid) and OH (base); NH2 (base) and NH3 (acid)
The use of conjugate acid-base pairs allows us to make a very simple statement about relative strengths of acids and bases. The stronger an acid, the weaker its conjugate base, and, conversely, the stronger a base, the weaker its conjugate acid.
Key Takeaways
• A Brønsted-Lowry acid is a proton donor, and a Brønsted-Lowry base is a proton acceptor.
• Brønsted-Lowry acid-base reactions are essentially proton transfer reactions.
10.03: Water - Both an Acid and a Base
Learning Objectives
• To write chemical equations for water acting as an acid and as a base.
Water (H2O) is an interesting compound in many respects. Here, we will consider its ability to behave as an acid or a base.
In some circumstances, a water molecule will accept a proton and thus act as a Brønsted-Lowry base. We saw an example in the dissolving of HCl in H2O:
$\rm{HCl + H_2O(ℓ) \rightarrow H_3O^+(aq) + Cl^−(aq)} \label{Eq1}$
In other circumstances, a water molecule can donate a proton and thus act as a Brønsted-Lowry acid. For example, in the presence of the amide ion (see Example 4 in Section 10.2), a water molecule donates a proton, making ammonia as a product:
$H_2O(ℓ) + NH^−_{2}(aq) \rightarrow OH^−(aq) + NH_{3}(aq) \label{Eq2}$
In this case, NH2 is a Brønsted-Lowry base (the proton acceptor).
So, depending on the circumstances, H2O can act as either a Brønsted-Lowry acid or a Brønsted-Lowry base. Water is not the only substance that can react as an acid in some cases or a base in others, but it is certainly the most common example—and the most important one. A substance that can either donate or accept a proton, depending on the circumstances, is called an amphiprotic compound.
A water molecule can act as an acid or a base even in a sample of pure water. About 6 in every 100 million (6 in 108) water molecules undergo the following reaction:
$H_2O(ℓ) + H_2O(ℓ) \rightarrow H_3O^+(aq) + OH^−(aq) \label{Eq3}$
This process is called the autoionization of water (Figure $1$) and occurs in every sample of water, whether it is pure or part of a solution. Autoionization occurs to some extent in any amphiprotic liquid. (For comparison, liquid ammonia undergoes autoionization as well, but only about 1 molecule in a million billion (1 in 1015) reacts with another ammonia molecule.)
Note
It is rare to truly have pure water. Water exposed to air will usually be slightly acidic because dissolved carbon dioxide gas, or carbonic acid, decreases the pH slightly below 7. Alternatively, dissolved minerals, like calcium carbonate (limestone), can make water slightly basic.
Example $1$
Identify water as either a Brønsted-Lowry acid or a Brønsted-Lowry base.
1. H2O(ℓ) + NO2(aq) → HNO2(aq) + OH(aq)
2. HC2H3O2(aq) + H2O(ℓ) → H3O+(aq) + C2H3O2(aq)
Solution
1. In this reaction, the water molecule donates a proton to the NO2 ion, making OH(aq). As the proton donor, H2O acts as a Brønsted-Lowry acid.
2. In this reaction, the water molecule accepts a proton from HC2H3O2, becoming H3O+(aq). As the proton acceptor, H2O is a Brønsted-Lowry base.
Exercise $2$
Identify water as either a Brønsted-Lowry acid or a Brønsted-Lowry base.
1. HCOOH(aq) + H2O(ℓ) → H3O+(aq) + HCOO(aq)
2. H2O(ℓ) + PO43(aq) → OH(aq) + HPO42(aq)
Answer
1. H2O acts as the proton acceptor (Brønsted-Lowry base)
2. H2O acts as the proton donor (Brønsted-Lowry acid)
Key Takeaway
• Water molecules can act as both an acid and a base, depending on the conditions. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/10%3A_Acids_and_Bases/10.02%3A_Brnsted-Lowry_Definition_of_Acids_and_Bases.txt |
Learning Objectives
• Describe the difference between strong and weak acids and bases.
• Describe how a chemical reaction reaches chemical equilibrium.
• Define the pH scale and use it to describe acids and bases.
Acids and bases do not all demonstrate the same degree of chemical activity in solution. Different acids and bases have different strengths.
Strong and Weak Acids
Let us consider the strengths of acids first. A small number of acids ionize completely in aqueous solution. For example, when HCl dissolves in water, every molecule of HCl separates into a hydronium ion and a chloride ion:
$\ce{HCl(g) + H2O(l) ->[\sim 100\%] H_3O^{+}(aq) + Cl^{-} (aq)} \label{Eq1}$
HCl(aq) is one example of a strong acid, which is a compound that is essentially 100% ionized in aqueous solution. There are very few strong acids. The important ones are listed in Table $1$.
Table $1$: Strong Acids and Bases (All in Aqueous Solution)
Acids Bases
HCl LiOH
HBr NaOH
HI KOH
HNO3 Mg(OH)2
H2SO4 Ca(OH)2
HClO4
By analogy, a strong base is a compound that is essentially 100% ionized in aqueous solution. As with acids, there are only a few strong bases, which are also listed in Table $1$.
If an acid is not listed in Table $1$, it is likely a weak acid, which is a compound that is not 100% ionized in aqueous solution. Similarly, a weak base is a compound that is not 100% ionized in aqueous solution. For example, acetic acid ($\ce{HC2H3O2}$) is a weak acid. The ionization reaction for acetic acid is as follows:
$\ce{HC2H3O2(aq) + H2O(ℓ) \rightarrow H3O^{+}(aq) + C2H3O^{−}2(aq)} \label{Eq2}$
Depending on the concentration of HC2H3O2, the ionization reaction may occur only for 1%–5% of the acetic acid molecules.
Looking Closer: Household Acids and Bases
Many household products are acids or bases. For example, the owner of a swimming pool may use muriatic acid to clean the pool. Muriatic acid is another name for hydrochloric acid [$\ce{HCl(aq)}$]. Vinegar has already been mentioned as a dilute solution of acetic acid [$\ce{HC2H3O2(aq)}$]. In a medicine chest, one may find a bottle of vitamin C tablets; the chemical name of vitamin C is ascorbic acid ($\ce{HC6H7O6}$).
One of the more familiar household bases is ammonia (NH3), which is found in numerous cleaning products. As we mentioned previously, ammonia is a base because it increases the hydroxide ion concentration by reacting with water:
$\ce{NH3(aq) + H2O(ℓ) \rightarrow NH^{+}4(aq) + OH^{−}(aq)} \label{Eq3}$
Many soaps are also slightly basic because they contain compounds that act as Brønsted-Lowry bases, accepting protons from water and forming excess hydroxide ions. This is one reason that soap solutions are slippery.
Chemical Equilibrium in Weak Acids and Bases
The behavior of weak acids and bases illustrates a key concept in chemistry. Does the chemical reaction describing the ionization of a weak acid or base just stop when the acid or base is done ionizing? Actually, no. Rather, the reverse process—the reformation of the molecular form of the acid or base—occurs, ultimately at the same rate as the ionization process. For example, the ionization of the weak acid $\ce{HC2H3O2 (aq)}$ is as follows:
$\ce{HC2H3O2(aq) + H2O(ℓ) \rightarrow H3O^{+}(aq) + C2H3O^{−}2(aq)} \label{Eq4}$
The reverse process also begins to occur:
$\ce{H3O^{+}(aq) + C2H3O^{−}2(aq) \rightarrow HC2H3O2(aq) + H2O(ℓ)} \label{Eq5}$
Eventually, there is a balance between the two opposing processes, and no additional change occurs. The chemical reaction is better represented at this point with a double arrow:
$\ce{HC2H3O2(aq) + H2O(ℓ) <=> H3O^{+}(aq) + C2H3O^{-}2(aq)} \label{Eq6}$
The $\rightleftharpoons$ implies that both the forward and reverse reactions are occurring, and their effects cancel each other out. A process at this point is considered to be at chemical equilibrium (or equilibrium). It is important to note that the processes do not stop. They balance out each other so that there is no further net change; that is, chemical equilibrium is a dynamic equilibrium.
Example $1$: Partial Ionization
Write the equilibrium chemical equation for the partial ionization of each weak acid or base.
1. HNO2(aq)
2. C5H5N(aq)
Solution
1. HNO2(aq) + H2O(ℓ) ⇆ NO2(aq) + H3O+(aq)
2. C5H5N(aq) + H2O(ℓ) ⇆ C5H5NH+(aq) + OH(aq)
Exercise $1$
Write the equilibrium chemical equation for the partial ionization of each weak acid or base.
1. $HF_{(aq)}$
2. $AgOH_{(aq)}$
3. CH3NH2(aq)
Answer
a. HF(aq) + H2O(ℓ) ⇆ F(aq) + H3O+(aq)
b. AgOH(aq) ⇆ Ag+(aq) + OH(aq)
c. CH3NH2(aq) + H2O(ℓ) ⇆ CH3NH3+(aq) + OH(aq)
Acid Ionization Constant, $K_\text{a}$
The ionization for a general weak acid, $\ce{HA}$, can be written as follows:
$\ce{HA} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{A^-} \left( aq \right) \nonumber$
Because the acid is weak, an equilibrium expression can be written. An acid ionization constant $\left( K_\text{a} \right)$ is the equilibrium constant for the ionization of an acid.
$K_\text{a} = \frac{\left[ \ce{H^+} \right] \left[ \ce{A^-} \right]}{\left[ \ce{HA} \right]} \nonumber$
The acid ionization represents the fraction of the original acid that has been ionized in solution. Therefore, the numerical value of $K_\text{a}$ is a reflection of the strength of the acid. Weak acids with relatively higher $K_\text{a}$ values are stronger than acids with relatively lower $K_\text{a}$ values. Because strong acids are essentially $100\%$ ionized, the concentration of the acid in the denominator is nearly zero and the $K_\text{a}$ value approaches infinity. For this reason, $K_\text{a}$ values are generally reported for weak acids only.
The table below is a listing of acid ionization constants for several acids. Note that polyprotic acids have a distinct ionization constant for each ionization step, with each successive ionization constant being smaller than the previous one.
Name of Acid Ionization Equation $K_\text{a}$
Table $2$: Acid Ionization Constants at $25^\text{o} \text{C}$
Sulfuric acid
$\ce{H_2SO_4} \rightleftharpoons \ce{H^+} + \ce{HSO_4^-}$
$\ce{HSO_4} \rightleftharpoons \ce{H^+} + \ce{SO_4^{2-}}$
very large
$1.3 \times 10^{-2}$
Hydrofluoric acid $\ce{HF} \rightleftharpoons \ce{H^+} + \ce{F^-}$ $7.1 \times 10^{-4}$
Nitrous acid $\ce{HNO_2} \rightleftharpoons \ce{H^+} + \ce{NO_2^-}$ $4.5 \times 10^{-4}$
Benzoic acid $\ce{C_6H_5COOH} \rightleftharpoons \ce{H^+} + \ce{C_6H_5COO^-}$ $6.5 \times 10^{-5}$
Acetic acid $\ce{CH_3COOH} \rightleftharpoons \ce{H^+} + \ce{CH_3COO^-}$ $1.8 \times 10^{-5}$
Carbonic acid
$\ce{H_2CO_3} \rightleftharpoons \ce{H^+} + \ce{HCO_3^-}$
$\ce{HCO_3^-} \rightleftharpoons \ce{H^+} + \ce{CO_3^{2-}}$
$4.2 \times 10^{-7}$
$4.8 \times 10^{-11}$
Hydrofluoric acid $HF_{(aq)}$ reacts directly with glass (very few chemicals react with glass). Hydrofluoric acid is used in glass etching.
Strong and Weak Bases and Base Ionization Constant, Kb
As with acids, bases can either be strong or weak, depending on their extent of ionization. A strong base is a base, which ionizes completely in an aqueous solution. The most common strong bases are soluble metal hydroxide compounds such as potassium hydroxide. Some metal hydroxides are not as strong simply because they are not as soluble. Calcium hydroxide is only slightly soluble in water, but the portion that does dissolve also dissociates into ions.
A weak base is a base that ionizes only slightly in an aqueous solution. Recall that a base can be defined as a substance, which accepts a hydrogen ion from another substance. When a weak base such as ammonia is dissolved in water, it accepts an $\ce{H^+}$ ion from water, forming the hydroxide ion and the conjugate acid of the base, the ammonium ion.
$\ce{NH_3} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{NH_4^+} \left( aq \right) + \ce{OH^-} \left( aq \right) \nonumber$
The equilibrium greatly favors the reactants and the extent of ionization of the ammonia molecule is very small.
An equilibrium expression can be written for the reactions of weak bases with water. Because the concentration of water is extremely large and virtually constant, the water is not included in the expression. A base ionization constant $\left( K_\text{b} \right)$ is the equilibrium constant for the ionization of a base. For ammonia the expression is:
$K_\text{b} = \frac{\left[ \ce{NH_4^+} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{NH_3} \right]} \nonumber$
The numerical value of $K_\text{b}$ is a reflection of the strength of the base. Weak bases with relatively higher $K_\text{b}$ values are stronger than bases with relatively lower $K_\text{b}$ values. Table $3$ is a listing of base ionization constants for several weak bases.
Table $3$: Base Ionization Constants at $25^\text{o} \text{C}$
Name of Base Ionization Equation $K_\text{b}$
Methylamine $\ce{CH_3NH_2} + \ce{H_2O} \rightleftharpoons \ce{CH_3NH_3^+} + \ce{OH^-}$ $5.6 \times 10^{-4}$
Ammonia $\ce{NH_3} + \ce{H_2O} \rightleftharpoons \ce{NH_4^+} + \ce{OH^-}$ $1.8 \times 10^{-5}$
Pyridine $\ce{C_5H_5N} + \ce{H_2O} \rightleftharpoons \ce{C_5H_5NH^+} + \ce{OH^-}$ $1.7 \times 10^{-9}$
Acetate ion $\ce{CH_3COO^-} + \ce{H_2O} \rightleftharpoons \ce{CH_3COOH} + \ce{OH^-}$ $5.6 \times 10^{-10}$
Fluoride ion $\ce{F^-} + \ce{H_2O} \rightleftharpoons \ce{HF} + \ce{OH^-}$ $1.4 \times 10^{-11}$
Urea $\ce{H_2NCONH_2} + \ce{H_2O} \rightleftharpoons \ce{H_2NCONH_3^+} + \ce{OH^-}$ $1.5 \times 10^{-14}$
The Ion-Product of Water
As we have already seen, H2O can act as an acid or a base. Within any given sample of water, some $\ce{H2O}$ molecules are acting as acids, and other $\ce{H2O}$ molecules are acting as bases. The chemical equation is as follows:
$\color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{<=> H3O^{+} + OH^{−}} \label{Auto}$
Similar to a weak acid, the autoionization of water is an equilibrium process, and is more properly written as follows:
$\ce{H2O(ℓ) + H2O(ℓ) <=> H3O^{+}(aq) + OH^{-}(aq)} \label{Eq7}$
We often use the simplified form of the reaction:
$\ce{H2O(l) <=> H+(aq) + OH−(aq)} \nonumber$
The equilibrium constant for the autoionization of water is referred to as the ion-product for water and is given the symbol Kw.
$K_w = [\ce{H^{+}}][\ce{OH^{-}}] \nonumber$
The ion-product of water (Kw) is the mathematical product of the concentration of hydrogen ions and hydroxide ions. Note that H2O is not included in the ion-product expression because it is a pure liquid. The value of Kw is very small, in accordance with a reaction that favors the reactants. At 25oC, the experimentally determined value of $K_w$ in pure water is 1.0×10−14.
$K_w = [\ce{H^{+}}][\ce{OH^{−}}] = 1.0 \times 10^{−14} \nonumber$
In a sample of pure water, the concentrations of hydrogen and hydroxide ions are equal to one another. Pure water or any other aqueous solution in which this ratio holds is said to be neutral. To find the molarity of each ion, the square root of Kw is taken.
[H+] = [OH] = 1.0×10−7
The product of these two concentrations is 1.0×10−14
$\color{red}{\ce{[H^{+}]}} \color{black}{\times} \color{blue}{\ce{[OH^{-}]}} \color{black} = (1.0 \times 10^{-7})( 1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \nonumber$
• For acids, the concentration of H+ or [H+]) is greater than 1.0×10−7 M
• For bases, the concentration of OH or [OH] is greater than 1.0×10−7 M.
Aqueous HCl is an example of acidic solution. Hydrogen chloride (HCl) ionizes to produce H+ and Cl ions upon dissolving in water. This increases the concentration of H+ ions in the solution. According to Le Chatelier's principle, the equilibrium represented by
$\ce{H2O(l) <=> H^{+}(aq) + OH^{−}(aq)} \nonumber$
$\ce{HCl(g) -> H^{+}(aq) + Cl^{−}(aq)} \nonumber$
is forced to the left, towards the reactant. As a result, the concentration of the hydroxide ion decreases.
Now, consider KOH (aq), a basic solution. Solid potassium hydroxide (KOH) dissociates in water to yield potassium ions and hydroxide ions.
KOH(s) → K+(aq) + OH(aq)
The increase in concentration of the OH ions will cause a decrease in the concentration of the H+ ions.
No matter whether the aqueous solution is an acid, a base, or neutral:and the ion-product of [H+][OH] remains constant.
• For acidic solutions, [H+]) is greater than [OH].
• For basic solutions, [OH−] is greater than [H+].
• For neutral solutions, [H3O+] = [OH−] = 1.0×10−7M
This means that if you know $\ce{[H^{+}]}$ for a solution, you can calculate what $\ce{[OH^{−}]}$) has to be for the product to equal $1.0 \times 10^{−14}$, or if you know $\ce{[OH^{−}]}$), you can calculate $\ce{[H^{+}]}$. This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of $K_w$.
$K_w = \color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \label{eq10}$
Example $2$
Hydrochloric acid (HCl) is a strong acid, meaning it is 100% ionized in solution. What is the [H+] and the [OH] in a solution of 2.0×10−3 M HCl?
Known
• [HCl] = 2.0×10−3 M
• Kw = 1.0×10−14
Unknown
• [H+]=?M
• [OH]=?M
Because HCl is 100% ionized, the concentration of H+ ions in solution will be equal to the original concentration of HCl. Each HCl molecule that was originally present ionizes into one H+ ion and one Cl− ion. The concentration of OH− can then be determined from the [H+] and Kw.
Step 2: Solve.
[H+]=2.0×10−3 M
Kw = [H+][OH] = 1.0×10−14
[OH] = Kw/[H+] = 1.0×10−14/2.0×10−3 = 5.0×10−12 M
Step 3: Think about your result.
The [H+] is much higher than the [OH] because the solution is acidic. As with other equilibrium constants, the unit for Kw is customarily omitted.
Exercise $2$
Sodium hydroxide (NaOH) is a strong base. What is the [H+] and the [OH] in a 0.001 M NaOH solution at 25 °C?
Answer
[OH] = 0.001M or 1 x 10-3M; [H+]=1×1011M.
The pH Scale
One qualitative measure of the strength of an acid or a base solution is the pH scale, which is based on the concentration of the hydronium (or hydrogen) ion in aqueous solution.
$pH = -\log[H^+] \nonumber$
or
$pH = -\log[H_3O^+] \nonumber$
Figure $3$ illustrates this relationship, along with some examples of various solutions. Because hydrogen ion concentrations are generally less than one (for example $1.3 \times 10^{-3}\,M$), the log of the number will be a negative number. To make pH even easier to work with, pH is defined as the negative log of $[H_3O^+]$, which will give a positive value for pH.
A neutral (neither acidic nor basic) solution has a pH of 7. A pH below 7 means that a solution is acidic, with lower values of pH corresponding to increasingly acidic solutions. A pH greater than 7 indicates a basic solution, with higher values of pH corresponding to increasingly basic solutions. Thus, given the pH of several solutions, you can state which ones are acidic, which ones are basic, and which are more acidic or basic than others. These are summarized in Table $4. Table \(4$: Acidic, Basic and Neutral pH Values
Acidic, Basic and Neutral pH Values
Classification Relative Ion Concentrations pH at 25 °C
acidic [H+] > [OH] pH < 7
neutral [H+] = [OH] pH = 7
basic [H+] < [OH] pH > 7
Example $3$
Find the pH, given the $[H^+]$ of the following:
1. 1 ×10-3 M
2. 2.5 ×10-11 M
3. 4.7 ×10-9 M
Solution
pH = - log [H3O+]
Substitute the known quantity into the equation and solve. Use a scientific calculator for b and c.
1. pH = - log [1 × 10−3 ] = 3.0 (1 decimal place since 1 has 1 significant figure)
2. pH = - log [2.5 ×10-11] = 10.60 (2 decimal places since 2.5 has 2 significant figures)
3. pH = - log [4.7 ×10-9] = 8.33 (2 decimal places since 4.7 has 2 significant figures)
Note on significant figures:
Because the number(s) before the decimal point in the pH value relate to the power on 10, the number of digits after the decimal point (underlined) is what determines the number of significant figures in the final answer.
Exercise $3$
Find the pH, given [H+] of the following:
1. 5.8 ×10-4 M
2. 1.0×10-7 M
Answer
a. 3.24
b. 7.00
Table $5$ lists the pH of several common solutions. The most acidic among the listed solutions is battery acid with the lowest pH value (0.3). The most basic is 1M NaOH solution with the highest pH value of 14.0. Notice that some biological fluids (stomach acid and urine) are nowhere near neutral. You may also notice that many food products are slightly acidic. They are acidic because they contain solutions of weak acids. If the acid components of these foods were strong acids, the food would likely be inedible.
Table $5$: The pH Values of Some Common Solutions
Solution pH
battery acid 0.3
stomach acid 1–2
lemon or lime juice 2.1
vinegar 2.8–3.0
Coca-Cola 3
wine 2.8–3.8
beer 4–5
coffee 5
milk 6
urine 6
pure H2O 7
(human) blood 7.3–7.5
sea water 8
antacid (milk of magnesia) 10.5
NH3 (1 M) 11.6
bleach 12.6
NaOH (1 M) 14.0
Example $4$
Label each solution as acidic, basic, or neutral based only on the stated $pH$.
1. milk of magnesia, pH = 10.5
2. pure water, pH = 7
3. wine, pH = 3.0
Solution
1. With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH)2.)
2. Pure water, with a pH of 7, is neutral.
3. With a pH of less than 7, wine is acidic.
Exercise $4$
Identify each substance as acidic, basic, or neutral based only on the stated $pH$.
1. human blood with $pH$ = 7.4
2. household ammonia with $pH$ = 11.0
3. cherries with $pH$ = 3.6
Answer
a. slightly basic
b. basic
c. acidic
Acid Rain
Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO2 which forms carbonic acid:
$\ce{H2O (l) + CO2(g) ⟶ H2CO3(aq)} \label{14}$
$\ce{H2CO3(aq) \rightleftharpoons H^+(aq) + HCO3^- (aq)} \label{15}$
Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:
$\ce{H2O (l) + SO3(g) ⟶ H2SO4(aq)} \label{16}$
$\ce{H2SO4(aq) ⟶ H^+(aq) + HSO4^- (aq)} \label{17}$
Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.
Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure $4$). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.
Key Takeaways
• Acids and bases can be strong or weak depending on the extent of ionization in solution.
• Most chemical reactions reach equilibrium at which point there is no net change.
• The ion-product of [H+][OH] in an aqueous solution remains constant.
• A pH value is simply the negative of the logarithm of the H+ ion concentration (-log[H+]).
• The pH scale is used to succinctly communicate the acidity or basicity of a solution.
• A solution is acidic if pH < 7.
• A solution is basic if pH > 7.
• A solution is neutral if pH = 7. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/10%3A_Acids_and_Bases/10.04%3A_The_Strengths_of_Acids_and_Bases.txt |
Learning Objectives
• To define buffer and describe how it reacts with an acid or a base.
Weak acids are relatively common, even in the foods we eat. But we occasionally come across a strong acid or base, such as stomach acid, that has a strongly acidic pH of 1–2. By definition, strong acids and bases can produce a relatively large amount of hydrogen or hydroxide ions and, as a consequence, have a marked chemical activity. In addition, very small amounts of strong acids and bases can change the pH of a solution very quickly. If 1 mL of stomach acid [which we will approximate as 0.05 M HCl(aq)] is added to the bloodstream, and if no correcting mechanism is present, the pH of the blood would go from about 7.4 to about 4.9—a pH that is not conducive to continued living. Fortunately, the body has a mechanism for minimizing such dramatic pH changes.
The mechanism involves a buffer, a solution that resists dramatic changes in pH. A buffer (or buffered) solution is one that resists a change in its pH when H+ or OH ions are added or removed owing to some other reaction taking place in the same solution. Buffers do so by being composed of certain pairs of solutes: either a weak acid plus its conjugate base or a weak base plus its conjugate acid.
For example, a buffer can be composed of dissolved acetic acid (HC2H3O2, a weak acid) and sodium acetate (NaC2H3O2). Sodium acetate is a salt that dissociates into sodium ions and acetate ions in solution. For as long as acetic acid and acetate ions are present in significant amounts a solution, this can resist dramatic pH changes. Another example of a buffer is a solution containing ammonia (NH3, a weak base) and ammonium chloride (NH4Cl). Ammonium acetate is also a salt that dissociates into ammonium ions and chloride ions in solution. The presence of ammonium ions with ammonia molecules satisfies the requisite condition for a buffer solution.
How Buffers Work
The essential component of a buffer system is a conjugate acid-base pair whose concentration is fairly high in relation to the concentrations of added H+ or OH it is expected to buffer against. Let us use an acetic acid–sodium acetate buffer to demonstrate how buffers work. If a strong base—a source of OH(aq) ions—is added to the buffer solution, those hydroxide ions will react with the acetic acid in an acid-base reaction:
$HC_2H_3O_{2(aq)} + OH^−_{(aq)} \rightarrow H_2O_{(ℓ)} + C_2H_3O^−_{2(aq)} \label{Eq1}$
Rather than changing the pH dramatically by making the solution basic, the added hydroxide ions react to make water, and the pH does not change much.
Many people are aware of the concept of buffers from buffered aspirin, which is aspirin that also has magnesium carbonate, calcium carbonate, magnesium oxide, or some other salt. The salt acts like a base, while aspirin is itself a weak acid.
If a strong acid—a source of H+ ions—is added to the buffer solution, the H+ ions will react with the anion from the salt. Because HC2H3O2 is a weak acid, it is not ionized much. This means that if lots of hydrogen ions and acetate ions (from sodium acetate) are present in the same solution, they will come together to make acetic acid:
$H^+_{(aq)} + C_2H_3O^−_{2(aq)} \rightarrow HC_2H_3O_{2(aq)} \label{Eq2}$
Rather than changing the pH dramatically and making the solution acidic, the added hydrogen ions react to make molecules of a weak acid. Figure $1$ illustrates both actions of a buffer.
A simple buffer system might be a 0.2 M solution of sodium acetate; the conjugate pair here is acetic acid HAc and its conjugate base, the acetate ion Ac. The idea is that this conjugate pair "pool" will be available to gobble up any small (≤ 10–3 M) addition of H+ or OH that may result from other processes going on in the solution.
Buffers work well only for limited amounts of added strong acid or base. Once either solute is all reacted, the solution is no longer a buffer, and rapid changes in pH may occur. We say that a buffer has a certain capacity. Buffers that have more solute dissolved in them to start with have larger capacities, as might be expected.
Buffers made from weak bases and salts of weak bases act similarly. For example, in a buffer containing NH3 and NH4Cl, ammonia molecules can react with any excess hydrogen ions introduced by strong acids:
$NH_{3(aq)} + H^+_{(aq)} \rightarrow NH^+_{4(aq)} \label{Eq3}$
while the ammonium ion [NH4+(aq)] can react with any hydroxide ions introduced by strong bases:
$NH^+_{4(aq)} + OH^−_{(aq)} \rightarrow NH_{3(aq)} + H_2O_{(ℓ)} \label{Eq4}$
Example $1$
Which solute combinations can make a buffer solution? Assume all are aqueous solutions.
1. HCHO2 and NaCHO2
2. HCl and NaCl
3. CH3NH2 and CH3NH3Cl
4. NH3 and NaOH
Solution
1. Formic acid (HCHO2) is a weak acid, while NaCHO2 is a salt supplying —formate ion (CHO2), the conjugate base of HCHO2. The combination of these two solutes would make a buffer solution.
2. Hydrochloric acid (HCl) is a strong acid, not a weak acid, so the combination of these two solutes would not make a buffer solution.
3. Methylamine (CH3NH2) is like ammonia, a weak base. The compound CH3NH3Cl is a salt supplying CH3NH3+, the conjugate acid of CH3NH2. The combination of these two solutes would make a buffer solution.
4. Ammonia (NH3) is a weak base, but NaOH is a strong base. The combination of these two solutes would not make a buffer solution.
Exercise $1$
Which solute combinations can make a buffer solution? Assume all are aqueous solutions.
1. NaHCO3 and NaCl
2. H3PO4 and NaH2PO4
3. NH3 and (NH4)3PO4
4. NaOH and NaCl
Answer
b. H3PO4 (weak acid) and H2PO4- (conjugate base of H3PO4) make a buffer.
c. NH3 (weak base) and NH4+ (conjugate acid of NH3) make a buffer
Food and Drink App: The Acid That Eases Pain
Although medicines are not exactly "food and drink," we do ingest them, so let's take a look at an acid that is probably the most common medicine: acetylsalicylic acid, also known as aspirin. Aspirin is well known as a pain reliever and antipyretic (fever reducer).
The structure of aspirin is shown in the accompanying figure. The acid part is circled; it is the H atom in that part that can be donated as aspirin acts as a Brønsted-Lowry acid. Because it is not given in Table 10.5.1, acetylsalicylic acid is a weak acid. However, it is still an acid, and given that some people consume relatively large amounts of aspirin daily, its acidic nature can cause problems in the stomach lining, despite the stomach's defenses against its own stomach acid.
Because the acid properties of aspirin may be problematic, many aspirin brands offer a "buffered aspirin" form of the medicine. In these cases, the aspirin also contains a buffering agent-usually MgO-that regulates the acidity of the aspirin to minimize its acidic side effects.
As useful and common as aspirin is, it was formally marketed as a drug starting in 1899. The US Food and Drug Administration (FDA), the governmental agency charged with overseeing and approving drugs in the United States, wasn't formed until 1906. Some have argued that if the FDA had been formed before aspirin was introduced, aspirin may never have gotten approval due to its potential for side effects-gastrointestinal bleeding, ringing in the ears, Reye's syndrome (a liver problem), and some allergic reactions. However, recently aspirin has been touted for its effects in lessening heart attacks and strokes, so it is likely that aspirin is here to stay.
Buffer solutions are essential components of all living organisms.
• Our blood is buffered to maintain a pH of 7.4 that must remain unchanged as metabolically-generated CO2 (carbonic acid) is added and then removed by our lungs.
• Buffers in the oceans, in natural waters such as lakes and streams, and within soils help maintain their environmental stability against acid rain and increases in atmospheric CO2.
• Many industrial processes, such as brewing, require buffer control, as do research studies in biochemistry and physiology that involve enzymes, are active only within certain pH ranges.
The pH in living systems (Figure $1) is maintained by buffer systems. Table 7.3.2: pH in Living Systems Compartment pH Gastric Acid 1 Lysosomes 4.5 Granules of Chromaffin Cells 5.5 Human Skin 5.5 Urine 6 Neutral H2O at 37 °C 6.81 Cytosol 7.2 Cerebrospinal Fluid 7.3 Blood 7.43-7.45 Mitochondrial Matrix 7.5 Pancreas Secretions 8.1 Human blood has a buffering system to minimize extreme changes in pH. One buffer in blood is based on the presence of HCO3 and H2CO3 [H2CO3 is another way to write CO2(aq)]. With this buffer present, even if some stomach acid were to find its way directly into the bloodstream, the change in the pH of blood would be minimal. Inside many of the body’s cells, there is a buffering system based on phosphate ions. Medicine: The Buffer System in Blood The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction: $\ce{CO2}(g)+\ce{2H2O}(l)⇌\ce{H2CO3}(aq)⇌\ce{HCO3-}(aq)+\ce{H3O+}(aq) \nonumber$ The concentration of carbonic acid, H2CO3 is approximately 0.0012 M, and the concentration of the hydrogen carbonate ion, \(\ce{HCO3-}$, is around 0.024 M. Using the Henderson-Hasselbalch equation and the pKa of carbonic acid at body temperature, we can calculate the pH of blood:
$\mathrm{pH=p\mathit{K}_a+\log\dfrac{[base]}{[acid]}=6.1+\log\dfrac{0.024}{0.0012}=7.4} \nonumber$
The fact that the H2CO3 concentration is significantly lower than that of the $\ce{HCO3-}$ ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.
Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the $\ce{HCO3-}$ ion, producing H2CO3. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO2 from the blood through the lungs driving the equilibrium reaction such that [H3O+] is lowered. If the blood is too alkaline, a lower breath rate increases CO2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H+] and restoring an appropriate pH.
Career Focus: Blood Bank Technology Specialist
At this point in this text, you should have the idea that the chemistry of blood is fairly complex. Because of this, people who work with blood must be specially trained to work with it properly.
A blood bank technology specialist is trained to perform routine and special tests on blood samples from blood banks or transfusion centers. This specialist measures the pH of blood, types it (according to the blood’s ABO+/− type, Rh factors, and other typing schemes), tests it for the presence or absence of various diseases, and uses the blood to determine if a patient has any of several medical problems, such as anemia. A blood bank technology specialist may also interview and prepare donors to give blood and may actually collect the blood donation.
Blood bank technology specialists are well trained. Typically, they require a college degree with at least a year of special training in blood biology and chemistry. In the United States, training must conform to standards established by the American Association of Blood Banks.
Key Takeaway
• A buffer is a solution that resists sudden changes in pH.
Concept Review Exercise
1. Explain how a buffer prevents large changes in pH.
Answer
1. A buffer has components that react with both strong acids and strong bases to resist sudden changes in pH.
Exercises
1. Describe a buffer. What two related chemical components are required to make a buffer?
2. Can a buffer be made by combining a strong acid with a strong base? Why or why not?
3. Which solute combinations can make a buffer? Assume all are aqueous solutions.
1. HCl and NaCl
2. HNO2 and NaNO2
3. NH4NO3 and HNO3
4. NH4NO3 and NH3
4. Which solute combinations can make a buffer? Assume all are aqueous solutions.
1. H3PO4 and Na3PO4
2. NaHCO3 and Na2CO3
3. NaNO3 and Ca(NO3)2
4. HN3 and NH3
5. For each combination in Exercise 3 that is a buffer, write the chemical equations for the reactions of the buffer components when a strong acid and a strong base is added.
6. For each combination in Exercise 4 that is a buffer, write the chemical equations for the reaction of the buffer components when a strong acid and a strong base is added.
7. The complete phosphate buffer system is based on four substances: H3PO4, H2PO4, HPO42, and PO43. What different buffer solutions can be made from these substances?
8. Explain why NaBr cannot be a component in either an acidic or a basic buffer.
9. Explain why Mg(NO3)2 cannot be a component in either an acidic or a basic buffer.
Answers
1. A buffer resists sudden changes in pH. It has a weak acid or base and a salt of that weak acid or base.
2. No. Combining a strong acid and a strong base will produce salt and water. Excess strong acid or strong base will not act as a buffer.
1. not a buffer
2. buffer
3. not a buffer
4. buffer
4. 1. not a buffer
2. buffer
3. not a buffer
4. not buffer
1. 3b: strong acid: H+ + NO2 → HNO2; strong base: OH + HNO2 → H2O + NO2; 3d: strong acid: H+ + NH3 → NH4+; strong base: OH + NH4+ → H2O + NH3
2. 4b: strong acid: H+ + CO32 → HCO3; strong base: OH + HCO3 → H2O + CO32;
1. Buffers can be made by combining H3PO4 and H2PO4, H2PO4 and HPO42, and HPO42 and PO43.
2. NaBr splits up into two ions in solution, Na+ and Br. Na+ will not react with any added base knowing that NaOH is a strong base. Br- will not react with any added acid knowing that HBr is a strong acid. Because NaBr will not react with any added base or acid, it does not resist change in pH and is not a buffer.
1. Mg(NO3)2 includes two types of ions, Mg2+ and NO3. Mg(OH)2 is strong base and completely dissociates (100% falls apart), so Mg2+ will not react with any added base (0% combines with OH). HNO3 is strong acid and completely dissociates (100% falls apart), so NO3 will not react with any added acid (0% combines with H+). Because Mg(NO3)2 will not react with any added base or acid, it does not resist change in pH and is not a buffer. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/10%3A_Acids_and_Bases/10.05%3A_Buffers.txt |
Concept Review Exercises
1. Give the Arrhenius definitions of an acid and a base.
2. What is neutralization?
Answers
1. Arrhenius acid: a compound that increases the concentration of hydrogen ion (H+) in aqueous solution; Arrhenius base: a compound that increases the concentration of hydroxide ion (OH) in aqueous solution.
2. the reaction of an acid and a base
Exercises
1. Give two examples of Arrhenius acids.
2. Give two examples of Arrhenius bases.
3. List the general properties of acids.
4. List the general properties of bases.
5. Name each compound. (For acids, look up the name in Table 10.1.1. For bases, use the rules for naming ionic compounds from Chapter 3.)
a. HBr(aq)
b. Ca(OH)2(aq)
c. HNO3(aq)
d. Fe(OH)3(aq)
6. Name each compound.
a. HI(aq)
b. Cu(OH)2(aq)
c. H3PO4(aq)
d. CsOH(aq)
7. Write a balanced chemical equation for the neutralization of Ba(OH)2(aq) with HNO3(aq).
8. Write a balanced chemical equation for the neutralization of H2SO4(aq) with Cr(OH)3(aq).
9. Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction.
10. Identify the salt produced in each acid-base reaction below. Then, balance the equation.
a. 2HCl + Sr(OH)2 2H2O + ??
b. KNO3; HNO3 + KOH ?? + H2O
c. HF + Ca(OH)2 ---> ?? + H2O
11. How many moles of sodium hydroxide (NaOH) are needed to neutralize 0.844 mol of acetic acid (HC2H3O2)? (Hint: begin by writing a balanced chemical equation for the process.)
12. How many moles of perchloric acid (HClO4) are needed to neutralize 0.052 mol of calcium hydroxide [Ca(OH)2]? (Hint: begin by writing a balanced chemical equation for the process
13. Hydrazoic acid (HN3) can be neutralized by a base.
a. Write the balanced chemical equation for the reaction between hydrazoic acid and calcium hydroxide.
b. How many milliliters of 0.0245 M Ca(OH)2 are needed to neutralize 0.564 g of HN3?
14. Citric acid (H3C6H5O7) has three hydrogen atoms that can form hydrogen ions in solution.
a. Write the balanced chemical equation for the reaction between citric acid and sodium hydroxide.
b. If an orange contains 0.0675 g of H3C6H5O7, how many milliliters of 0.00332 M NaOH solution are needed to neutralize the acid?
15. Magnesium hydroxide [Mg(OH)2] is an ingredient in some antacids. How many grams of Mg(OH)2 are needed to neutralize the acid in 158 mL of 0.106 M HCl(aq)? It might help to write the balanced chemical equation first.
16. Aluminum hydroxide [Al(OH)3] is an ingredient in some antacids. How many grams of Al(OH)3 are needed to neutralize the acid in 96.5 mL of 0.556 M H2SO4(aq)? It might help to write the balanced chemical equation first.
17. Write the balanced chemical equation for the reaction between HBr and Ca(OH)2. What volume of 0.010 M HBr solution is be required to neutralize 25 mL of a 0.0100M Ca(OH)2 solution?
18. Write the balanced chemical equation for the reaction between HNO3 and KOH. What volume of 0.5M HNO3 is required to neutralize 60 mL of 0.4M KOH solution?
Answers
1. HCl and HNO3 (answers will vary)
2. NaOH and Ca(OH)2 (answers will vary)
1. sour taste, react with metals, react with bases, and turn litmus red
2. bitter taste, feels slippery, react with acids and turn litmus blue
1. a. hydrobromic acid
b. calcium hydroxide
c. nitric acid
d. iron(III) hydroxide
6. a. hydroiodic acid
b. cupric hydroxide
c. phosphoric acid
d. cesium hydroxide
7. 2HNO3(aq) + Ba(OH)2(aq) → Ba(NO3)2(aq) + 2H2O
8. 3H2SO4(aq) + 2Cr(OH)3(aq) → Cr2(SO4)3(aq) + 6H2O
9. Mg(OH)2 + 2HCl --> MgCl2 + 2H2O
10. a. SrCl2; 2HCl + Sr(OH)2 2H2O + SrCl2
b. KNO3; HNO3 + KOH KNO3 + H2O
c. CaF2; 2HF + Ca(OH)2 CaF2 + 2H2O
11. 0.844 mol
12. 0.104 mol
13. Part 1: 2HN3(aq) + Ca(OH)2 → Ca(N3)2 + 2H2O
Part 2: 268 mL
14. Part 1: H3C6H5O7(aq) + 3NaOH(aq) → Na3C6H5O7(aq) + 3H2O
Part 2: 317.5 mL
15. 0.488 g
16. 2.79 g
17. 2HBr + Ca(OH)2 CaBr2 + 2H2O; 50 mL HBr
18. HNO3 + KOHKNO3 + H2O; 48 mL HNO3
Concept Review Exercise
1. Give the definitions of a Brønsted-Lowry acid and a Brønsted-Lowry base.
Answer
1. A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor.
Exercises
1. Label each reactant as a Brønsted-Lowry acid or a Brønsted-Lowry base.
HCl(aq) + NH3(aq) → NH4+(aq) + Cl(aq)
2. Label each reactant as a Brønsted-Lowry acid or a Brønsted-Lowry base.
H2O(ℓ) + N2H4(aq) → N2H5+(aq) + OH(aq)
3. Explain why a Brønsted-Lowry acid can be called a proton donor.
4. Explain why a Brønsted-Lowry base can be called a proton acceptor.
5. Write the chemical equation of the reaction of ammonia in water and label the Brønsted-Lowry acid and base.
6. Write the chemical equation of the reaction of methylamine (CH3NH2) in water and label the Brønsted-Lowry acid and base.
7. Demonstrate that the dissolution of HNO3 in water is actually a Brønsted-Lowry acid-base reaction by describing it with a chemical equation and labeling the Brønsted-Lowry acid and base.
8. Identify the Brønsted-Lowry acid and base in the following chemical equation:
C3H7NH2(aq) + H3O+(aq) → C3H7NH3+(aq) + H2O(ℓ)
9. Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in each of the following equations
1. \(\ce{NO2- + H2O ⟶ HNO2 + OH-}\)
2. \(\ce{HBr + H2O ⟶ H3O+ + Br-}\)
3. \(\ce{HS- + H2O ⟶ H2S + OH-}\)
4. \(\ce{H2PO4- + OH- ⟶HPO4^2- + H2O}\)
5. \(\ce{H2PO4- + HCl ⟶ H3PO4 + Cl-}\)
10. Write the chemical equation for the reaction that occurs when cocaine hydrochloride (C17H22ClNO4) dissolves in water and donates a proton to a water molecule. (When hydrochlorides dissolve in water, they separate into chloride ions and the appropriate cation.)
11. If codeine hydrobromide has the formula C18H22BrNO3, what is the formula of the parent compound codeine?
Answers
1. HCl: Brønsted-Lowry acid; NH3: Brønsted-Lowry base
2. H2O: Brønsted-Lowry acid; N2H4: Brønsted-Lowry base
3. A Brønsted-Lowry acid gives away an H+ ion—nominally, a proton—in an acid-base reaction.
4. A Brønsted-Lowry base accepts an H+ ion (a proton) in an acid-base reaction.
5. NH3 + H2O → NH4+ + OH (here NH3 = Brønsted-Lowry base; H2O = Brønsted-Lowry acid)
6. CH3NH2 + H2O → CH3NH3+ + OH (here CH3NH2 = Brønsted-Lowry base; H2O = Brønsted-Lowry acid)
7. HNO3 + H2O → H3O+ + NO3(here HNO3 = Brønsted-Lowry acid; H2O = Brønsted-Lowry base)
8. C3H7NH2(aq) + H3O+(aq) → C3H7NH3+(aq) + H2O(ℓ) (here H3O+ = Brønsted-Lowry acid; C3H7NH2 = Brønsted-Lowry base)
Concept Review Exercises
1. Explain how water can act as an acid.
2. Explain how water can act as a base.
Answers
1. Under the right conditions, H2O can donate a proton, making it a Brønsted-Lowry acid.
2. Under the right conditions, H2O can accept a proton, making it a Brønsted-Lowry base.
Exercises
1. Is H2O(ℓ) acting as an acid or a base?
H2O(ℓ) + NH4+(aq) → H3O+(aq) + NH3(aq)
2. Is H2O(ℓ) acting as an acid or a base?
CH3(aq) + H2O(ℓ) → CH4(aq) + OH(aq)
3. In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some C2H3O2 solutions, the following reaction can occur:
C2H3O2(aq) + H2O(ℓ) → HC2H3O2(aq) + OH(aq)
Is H2O acting as an acid or a base in this reaction?
4. In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some NH4+ solutions, the following reaction can occur:
NH4+(aq) + H2O → NH3(aq) + H3O+(aq)
Is H2O acting as an acid or a base in this reaction?
5. Why is pure water considered neutral?
Answers
1. base
2. acid
1. acid
2. base
5. When water ionizes, equal amounts of H+ (acid) and OH(base) are formed, so the solution is neither acidic nor basic: H2O(ℓ) → H+(aq) + OH(aq)
Concept Review Exercises
1. Explain the difference between a strong acid or base and a weak acid or base.
2. Explain what is occurring when a chemical reaction reaches equilibrium.
3. Define pH.
AnswerS
1. A strong acid or base is 100% ionized in aqueous solution; a weak acid or base is less than 100% ionized.
2. The overall reaction progress stops because the reverse process balances out the forward process.
3. pH is a measure of the hydrogen ion concentration.
Exercises
1. Name a strong acid and a weak acid. (Hint: use Table 10.4.1.)
2. Name a strong base and a weak base. (Hint: use Table 10.4.1.)
3. Is each compound a strong acid or a weak acid? Assume all are in aqueous solution. (Hint: use Table 10.4.1.)
1. HF
2. HC2H3O2
3. HCl
4. HClO4
4. Is each compound a strong acid or a weak acid? Assume all are in aqueous solution. (Hint: use Table 10.4.1.)
1. H2SO4
2. HSO4
3. HPO42
4. HNO3
5. Is each compound a strong base or a weak base? Assume all are in aqueous solution. (Hint: use Table 10.4.1.)
1. NH3
2. NaOH
3. Mg(OH)2
4. Cu(OH)2
6. Is each compound a strong base or a weak base? Assume all are in aqueous solution. (Hint: use Table 10.4.1.)
1. KOH
2. H2O
3. Fe(OH)2
4. Fe(OH)3
7. Write the chemical equation for the equilibrium process for each weak acid in Exercise 3.
8. Write the chemical equation for the equilibrium process for each weak acid in Exercise 4.
9. Write the chemical equation for the equilibrium process for each weak base in Exercise 5.
10. Write the chemical equation for the equilibrium process for each weak base in Exercise 6.
11. Which is the stronger acid—HCl(aq) or HF(aq)?
12. Which is the stronger base—KOH(aq) or Ni(OH)2(aq)?
13. Consider the two acids in Exercise 11. For solutions that have the same concentration, which one would you expect to have a lower pH?
14. Consider the two bases in Exercise 12. For solutions that have the same concentration, which one would you expect to have a higher pH?
15. Consider the list of substances in Table \PageIndex3\PageIndex3.2"The pH Values of Some Common Solutions". What is the most acidic substance on the list that you have encountered recently?
16. Consider the list of substances in Table \PageIndex3\PageIndex3.2"The pH Values of Some Common Solutions". What is the most basic substance on the list that you have encountered recently?
17. Indicate whether solutions with the following pH values are acidic, basic, or neutral:
1. pH = 9.4
2. pH = 7.0
3. pH = 1.2
4. pH = 6.5
Answers
1. strong acid: HCl; weak acid: HC2H3O2 (answers will vary)
2. strong base: NaOH; weak base: NH3 (answers will vary)
1. weak
2. weak
3. strong
4. strong
1. strong
2. weak
3. weak
4. strong
1. weak
2. strong
3. strong
4. weak
1. strong
2. weak
3. weak
4. weak
7. 3a: HF(aq) ⇆ H+(aq) + F(aq); 3b: HC2H3O2(aq) ⇆ H+(aq) + C2H3O2(aq)
8. 4b: HSO4(aq) ⇆ H+(aq) + SO42(aq); 4c: HPO42(aq) ⇆ H+(aq) + PO43(aq)
9. 5a: NH3(aq) + H2O ⇆ NH4+(aq) + OH(aq); 5d: Cu(OH)2(aq) ⇆ Cu2+(aq) + 2OH(aq)
10. 6b: H2O + H2O ⇆ H3O+(aq) + OH(aq); 6c: Fe(OH)2(aq) ⇆ Fe2+(aq) + 2OH(aq); 6d: Fe(OH)3(aq) ⇆ Fe3+(aq) + 3OH(aq)
11. HCl(aq)
12. KOH(aq)
13. HCl(aq)
14. KOH(aq)
15. (answers will vary)
16. (answers will vary)
17. 1. basic
2. neutral
3. acidic (strongly)
4. acidic (mildly)
Concept Review Exercise
1. Explain how a buffer prevents large changes in pH.
Answer
1. A buffer has components that react with both strong acids and strong bases to resist sudden changes in pH.
Exercises
1. Describe a buffer. What two related chemical components are required to make a buffer?
2. Can a buffer be made by combining a strong acid with a strong base? Why or why not?
3. Which solute combinations can make a buffer? Assume all are aqueous solutions.
1. HCl and NaCl
2. HNO2 and NaNO2
3. NH4NO3 and HNO3
4. NH4NO3 and NH3
4. Which solute combinations can make a buffer? Assume all are aqueous solutions.
1. H3PO4 and Na3PO4
2. NaHCO3 and Na2CO3
3. NaNO3 and Ca(NO3)2
4. HN3 and NH3
5. For each combination in Exercise 3 that is a buffer, write the chemical equations for the reactions of the buffer components when a strong acid and a strong base is added.
6. For each combination in Exercise 4 that is a buffer, write the chemical equations for the reaction of the buffer components when a strong acid and a strong base is added.
7. The complete phosphate buffer system is based on four substances: H3PO4, H2PO4, HPO42, and PO43. What different buffer solutions can be made from these substances?
8. Explain why NaBr cannot be a component in either an acidic or a basic buffer.
9. Explain why Mg(NO3)2 cannot be a component in either an acidic or a basic buffer.
Answers
1. A buffer resists sudden changes in pH. It has a weak acid or base and a salt of that weak acid or base.
2. No. Combining a strong acid and a strong base will produce salt and water. Excess strong acid or strong base will not act as a buffer.
1. not a buffer
2. buffer
3. not a buffer
4. buffer
4. 1. not a buffer
2. buffer
3. not a buffer
4. not buffer
1. 3b: strong acid: H+ + NO2 → HNO2; strong base: OH + HNO2 → H2O + NO2; 3d: strong acid: H+ + NH3 → NH4+; strong base: OH + NH4+ → H2O + NH3
2. 4b: strong acid: H+ + CO32 → HCO3; strong base: OH + HCO3 → H2O + CO32;
1. Buffers can be made by combining H3PO4 and H2PO4, H2PO4 and HPO42, and HPO42 and PO43.
2. NaBr splits up into two ions in solution, Na+ and Br. Na+ will not react with any added base knowing that NaOH is a strong base. Br- will not react with any added acid knowing that HBr is a strong acid. Because NaBr will not react with any added base or acid, it does not resist change in pH and is not a buffer.
1. Mg(NO3)2 includes two types of ions, Mg2+ and NO3. Mg(OH)2 is strong base and completely dissociates (100% falls apart), so Mg2+ will not react with any added base (0% combines with OH). HNO3 is strong acid and completely dissociates (100% falls apart), so NO3 will not react with any added acid (0% combines with H+). Because Mg(NO3)2 will not react with any added base or acid, it does not resist change in pH and is not a buffer.
Additional Exercises
1. The properties of a 1.0 M HCl solution and a 1.0 M HC2H3O2 solution are compared. Measurements show that the hydrochloric acid solution has a higher osmotic pressure than the acetic acid solution. Explain why.
2. Of a 0.50 M HNO3 solution and a 0.50 M HC2H3O2 solution, which should have the higher boiling point? Explain why.
3. The reaction of sulfuric acid [H2SO4(aq)] with sodium hydroxide [NaOH(aq)] can be represented by two separate steps, with only one hydrogen ion reacting in each step. Write the chemical equation for each step.
4. The reaction of aluminum hydroxide [Al(OH)3(aq)] with hydrochloric acid [HCl(aq)] can be represented by three separate steps, with only one hydroxide ion reacting in each step. Write the chemical equation for each step.
5. A friend brings you a small sample of an unknown chemical. Assuming that the chemical is soluble in water, how would you determine if the chemical is an acid or a base?
6. A neutral solution has a hydrogen ion concentration of about 1 × 10−7 M. What is the concentration of the hydroxide ion in a neutral solution?
7. The Lewis definitions of an acid and a base are based on electron pairs, not protons. A Lewis acid is an electron pair acceptor, while a Lewis base is an electron pair donor. Use Lewis diagrams to show that
H+(aq) + OH(aq) → H2O(ℓ)
is an acid-base reaction in the Lewis sense as well as in the Arrhenius and Brønsted-Lowry senses.
8. Given the chemical reaction
NH3(g) + BF3(g) → NH3—BF3(s)
show that the reaction illustrated by this equation is an acid-base reaction if we use the Lewis definitions of an acid and a base (see Exercise 7). The product contains a bond between the N and B atoms.
Answers
1. HCl is a strong acid and yields more ions in solution. HC2H3O2 is a weak acid and undergoes partial ionization in solution.
2. HNO3 is a strong acid while HC2H3O2 is a weak acid. HNO3 dissociates 100% and its solution contains more ions. The more ions the solution contains the lower is its vapor pressure; the higher temperature is required for it to boil.
1. H2SO4 + NaOH → NaHSO4 + H2O; NaHSO4 + NaOH → Na2SO4 + H2O
2. Al(OH)3 + HCl → Al(OH)2Cl + H2O; Al(OH)2Cl + HCl → Al(OH)Cl2 + H2O; Al(OH)Cl2 + HCl → AlCl3 + H2O
1. One way is to add it to NaHCO3; if it bubbles, it is an acid. Alternatively, add the sample to litmus and look for a characteristic color change (red for acid, blue for base).
2. In a neutral solution, [OH-] = [H+] = 1.0 x 10-7 M
1. The O atom is donating an electron pair to the H+ ion, making the base an electron pair donor and the acid an electron pair acceptor.
2. The N atom is donating a lone pair to B in BF3, Hence NH3 is the Lewis base and BF3 is the Lewis acid. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/10%3A_Acids_and_Bases/10.E%3A_Acids_and_Bases_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
The earliest chemical definition of an acid, the Arrhenius definition, says that an acid is a compound that increases the amount of hydrogen ion (H+) in aqueous solution. An Arrhenius base is a compound that increases the amount of hydroxide ion (OH) in aqueous solution. While most bases are named as ionic hydroxide compounds, aqueous acids have a naming system unique to acids. Acids and bases react together in a characteristic chemical reaction called neutralization, in which the products are water and a salt. The principles of stoichiometry, along with the balanced chemical equation for a reaction between an acid and a base, can be used to determine how much of one compound will react with a given amount of the other.
A Brønsted-Lowry acid is any substance that donates a proton to another substance. A Brønsted-Lowry base is any substance that accepts a proton from another substance. The reaction of ammonia with water to make ammonium ions and hydroxide ions can be used to illustrate Brønsted-Lowry acid and base behavior.
Some compounds can either donate or accept protons, depending on the circumstances. Such compounds are called amphiprotic. Water is one example of an amphiprotic compound. One result of water being amphiprotic is that a water molecule can donate a proton to another water molecule to make a hydronium ion and a hydroxide ion. This process is called the autoionization of water and occurs in any sample of water.
Not all acids and bases are equal in chemical strength. A strong acid is an acid whose molecules are all dissociated into ions in aqueous solution. Hydrochloric acid is an example of a strong acid. Similarly, a strong base is a base whose molecules are dissociated into ions in aqueous solution. Sodium hydroxide is an example of a strong base. Any acid or base whose molecules are not all dissociated into ions in aqueous solution is a weak acid or a weak base. Solutions of weak acids and weak bases reach a chemical equilibrium between the un-ionized form of the compound and the dissociated ions. It is a dynamic equilibrium because acid and base molecules are constantly dissociating into ions and reassociating into neutral molecules.
The pH scale is a scale used to express the concentration of hydrogen ions in solution. A neutral solution, neither acidic nor basic, has a pH of 7. Acidic solutions have a pH lower than 7, while basic solutions have a pH higher than 7.
Buffers are solutions that resist dramatic changes in pH when an acid or a base is added to them. They contain a weak acid and a salt of that weak acid, or a weak base and a salt of that weak base. When a buffer is present, any strong acid reacts with the anion of the salt, forming a weak acid and minimizing the presence of hydrogen ions in solution. Any strong base reacts with the weak acid, minimizing the amount of additional hydroxide ions in solution. However, buffers only have limited capacity; there is a limit to the amount of strong acid or strong base any given amount of buffer will react with. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/10%3A_Acids_and_Bases/10.S%3A_Acids_and_Bases_%28Summary%29.txt |
Most chemists pay little attention to the nucleus of an atom except to consider the number of protons it contains because that determines an element’s identity. However, in nuclear chemistry, the composition of the nucleus and the changes that occur there are very important. Applications of nuclear chemistry may be more widespread than you realize. Many people are aware of nuclear power plants and nuclear bombs, but nuclear chemistry also has applications ranging from smoke detectors to medicine, from the sterilization of food to the analysis of ancient artifacts. In this chapter, we will examine some of the basic concepts of nuclear chemistry and some of the nuclear reactions that are important in our everyday lives.
• 11.0: Prelude to Nuclear Chemistry
A typical smoke detector contains an electric circuit that includes two metal plates about 1 cm apart. A battery in the circuit creates a voltage between the plates. Next to the plates is a small disk containing a tiny amount (∼0.0002 g) of the radioactive element americium (Am). The radioactivity of the americium ionizes the air between the plates, causing a tiny current to constantly flow between them.
• 11.1: Radioactivity
Atoms are composed of subatomic particles—protons, neutrons, and electrons. Protons and neutrons are located in the nucleus and provide most of the mass of the atom, while electrons circle the nucleus in shells and subshells and account for an atom’s size. There are three main forms of radioactive emissions and are alpha particles, beta particles, and gamma rays.
• 11.2: Half-Life
Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively. The amount of material left over after a certain number of half-lives can be easily calculated.
• 11.3: Units of Radioactivity
Radioactivity can be expressed in a variety of units, including rems, rads, and curies.
• 11.4: Uses of Radioactive Isotopes
Radioactivity has several practical applications, including tracers, medical applications, dating once-living objects, and the preservation of food.
• 11.5: Nuclear Energy
Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy.
• 11.E: Nuclear Chemistry (Exercises)
Select problems and solutions.
• 11.S: Nuclear Chemistry (Summary)
Chapter summary
11: Nuclear Chemistry
Most of us may not be aware of a device in our homes that guards our safety and, at the same time, depends on radioactivity to operate properly. This device is a smoke detector.
A typical smoke detector contains an electric circuit that includes two metal plates about 1 cm apart. A battery in the circuit creates a voltage between the plates. Next to the plates is a small disk containing a tiny amount (∼0.0002 g) of the radioactive element americium (Am). The radioactivity of the americium ionizes the air between the plates, causing a tiny current to constantly flow between them. (This constant drain on the battery explains why the batteries in smoke detectors should be replaced on a regular basis, whether the alarm has been triggered or not.)
When particles of smoke from a fire enter the smoke detector, they interfere with the ions between the metal plates, interrupting the tiny flow of current. When the current drops beneath a set value, another circuit triggers a loud alarm, warning of the possible presence of fire.
Although radioactive, the americium in a smoke detector is embedded in plastic and is not harmful unless the plastic package is taken apart, which is unlikely. Although many people experience an unfounded fear of radioactivity, smoke detectors provide an application of radioactivity that saves thousands of lives every year. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/11%3A_Nuclear_Chemistry/11.00%3A_Prelude_to_Nuclear_Chemistry.txt |
Learning Objectives
• To define and give examples of the major types of radioactivity.
Atoms are composed of subatomic particles—protons, neutrons, and electrons. Protons and neutrons are located in the nucleus and provide most of the mass of the atom, while electrons circle the nucleus in shells and subshells and account for an atom’s size. Remember, the notation for succinctly representing an isotope of a particular atom:
$\ce{^{12}_{6}C} \label{Eq1}$
The element in this example, represented by the symbol C, is carbon. Its atomic number, 6, is the lower left subscript on the symbol and is the number of protons in the atom. The mass number, the superscript to the upper left of the symbol, is the sum of the number of protons and neutrons in the nucleus of this particular isotope. In this case, the mass number is 12, which means that the number of neutrons in the atom is 12 − 6 = 6 (that is, the mass number of the atom minus the number of protons in the nucleus equals the number of neutrons). Occasionally, the atomic number is omitted in this notation because the symbol of the element itself conveys its characteristic atomic number. The two isotopes of hydrogen, 2H and 3H, are given their own names: deuterium (D) and tritium (T), respectively. Another way of expressing a particular isotope is to list the mass number after the element name, like carbon-12 or hydrogen-3.
Atomic theory in the 19th century presumed that nuclei had fixed compositions. But in 1896, the French scientist Henri Becquerel found that a uranium compound placed near a photographic plate made an image on the plate, even if the compound was wrapped in black cloth. He reasoned that the uranium compound was emitting some kind of radiation that passed through the cloth to expose the photographic plate. Further investigations showed that the radiation was a combination of particles and electromagnetic rays, with its ultimate source as the atomic nucleus. These emanations were ultimately called, collectively, radioactivity.
There are three main forms of radioactive emissions. The first is called an alpha particle, which is symbolized by the Greek letter α. An alpha particle is composed of two protons and two neutrons, and so it is the same as a helium nucleus. (We often use $\ce{^{4}_{2}He}$ to represent an alpha particle.) It has a 2+ charge. When a radioactive atom emits an alpha particle, the original atom’s atomic number decreases by two (because of the loss of two protons), and its mass number decreases by four (because of the loss of four nuclear particles). We can represent the emission of an alpha particle with a chemical equation—for example, the alpha-particle emission of uranium-235 is as follows:
$\ce{^{235}_{92}U \rightarrow \,_2^4He + \, _{90}^{231}Th} \label{Eq2}$
How do we know that a product of the reaction is $\ce{^{231}_{90}Th}$? We use the law of conservation of matter, which says that matter cannot be created or destroyed. This means we must have the same number of protons and neutrons on both sides of the chemical equation. If our uranium nucleus loses 2 protons, there are 90 protons remaining, identifying the element as thorium. Moreover, if we lose 4 nuclear particles of the original 235, there are 231 remaining. Thus, we use subtraction to identify the isotope of the thorium atom—in this case, $\ce{^{231}_{90}Th}$.
Chemists often use the names parent isotope and daughter isotope to represent the original atom and the product other than the alpha particle. In the previous example, $\ce{^{235}_{92}U}$ is the parent isotope, and $\ce{^{231}_{90}Th}$ is the daughter isotope. When one element changes into another in this manner, it undergoes radioactive decay.
Example $1$
Write the nuclear equation that represents the radioactive decay of radon-222 by alpha particle emission and identify the daughter isotope.
Solution
Radon has an atomic number of 86, so the parent isotope is represented as $\ce{^{222}_{86}Rn}$. We represent the alpha particle as $\ce{^{4}_{2}He}$ and use subtraction (222 − 4 = 218 and 86 − 2 = 84) to identify the daughter isotope as an isotope of polonium, $\mathrm{^{218}_{84}Po}$:
$\ce{_{86}^{222}Rn\rightarrow \, _2^4He + \, _{84}^{218}Po}$
Exercise $1$
Write the nuclear equation that represents the radioactive decay of polonium-209 by alpha particle emission and identify the daughter isotope.
Answer
$\ce{_{84}^{209}Po\rightarrow \, _2^4He + \, _{82}^{205}Pb}$
The second major type of radioactive emission is called a beta particle, symbolized by the Greek letter β. A beta particle is an electron ejected from the nucleus (not from the shells of electrons about the nucleus) and has a 1− charge. We can also represent a beta particle as $\ce{^0_{-1}e}$ or β. The net effect of beta particle emission on a nucleus is that a neutron is converted to a proton. The overall mass number stays the same, but because the number of protons increases by one, the atomic number goes up by one. Carbon-14 decays by emitting a beta particle:
$\ce{_6^{14}C \rightarrow \, _7^{14}N + \, _{-1}^0e} \label{Eq3}$
Again, the sum of the atomic numbers is the same on both sides of the equation, as is the sum of the mass numbers. (Note that the electron is assigned an “atomic number” of 1−, equal to its charge.)
The third major type of radioactive emission is not a particle but rather a very energetic form of electromagnetic radiation called gamma rays, symbolized by the Greek letter γ. Gamma rays themselves do not carry an overall electrical charge, but they may knock electrons out of atoms in a sample of matter and make it electrically charged (for which gamma rays are termed ionizing radiation). For example, in the radioactive decay of radon-222, both alpha and gamma radiation are emitted, with the latter having an energy of 8.2 × 10−14 J per nucleus decayed:
$\ce{_{86}^{222}Rn\rightarrow \, _{84}^{218}Po + \, _2^4He + \gamma} \label{Eq4}$
This may not seem like much energy, but if 1 mol of radon atoms were to decay, the gamma ray energy would be 49 million kJ!
Example $2$
Write the nuclear equation that represents the radioactive decay of boron-12 by beta particle emission and identify the daughter isotope. A gamma ray is emitted simultaneously with the beta particle.
Solution
The parent isotope is $\ce{^{12}_{5}B}$ while one of the products is an electron, $\ce{^{0}_{-1}e}$. So that the mass and atomic numbers have the same value on both sides, the mass number of the daughter isotope must be 12, and its atomic number must be 6. The element having an atomic number of 6 is carbon. Thus, the complete nuclear equation is as follows:
$\ce{_5^{12}B\rightarrow \, _6^{12}C + \, _{-1}^0e + \gamma} \nonumber$
The daughter isotope is $\ce{^{12}_6 C}$.
Exercise $2$
Write the nuclear equation that represents the radioactive decay of iodine-131 by beta particle emission and identify the daughter isotope. A gamma ray is emitted simultaneously with the beta particle.
Answer
$\ce{_53^{131}I\rightarrow \, _54^{131}Xe + \, _{-1}^0e + \gamma} \nonumber$
Alpha, beta, and gamma emissions have different abilities to penetrate matter. The relatively large alpha particle is easily stopped by matter (although it may impart a significant amount of energy to the matter it contacts). Beta particles penetrate slightly into matter, perhaps a few centimeters at most. Gamma rays can penetrate deeply into matter and can impart a large amount of energy into the surrounding matter. Table $1$ summarizes the properties of the three main types of radioactive emissions.
Table $1$: The Three Main Forms of Radioactive Emissions
Characteristic Alpha Particles Beta Particles Gamma Rays
symbols α, $\mathrm{_{2}^{4}He}$ β, $\ce{^{0}_{-1} e}$ γ
identity helium nucleus electron electromagnetic radiation
charge 2+ 1− none
mass number 4 0 0
penetrating power minimal (will not penetrate skin) short (will penetrate skin and some tissues slightly) deep (will penetrate tissues deeply)
Occasionally, an atomic nucleus breaks apart into smaller pieces in a radioactive process called spontaneous fission (or fission). Typically, the daughter isotopes produced by fission are a varied mix of products, rather than a specific isotope as with alpha and beta particle emission. Often, fission produces excess neutrons that will sometimes be captured by other nuclei, possibly inducing additional radioactive events. Uranium-235 undergoes spontaneous fission to a small extent. One typical reaction is
$\ce{_{92}^{235}U \rightarrow \, _{56}^{139}Ba + \, _{36}^{94}Kr + 2^1_0n} \label{Eq5}$
where $\ce{_0^1n}$ is a neutron. As with any nuclear process, the sums of the atomic numbers and the mass numbers must be the same on both sides of the equation. Spontaneous fission is found only in large nuclei. The smallest nucleus that exhibits spontaneous fission is lead-208.
Fission is the radioactive process used in nuclear power plants and one type of nuclear bomb.
Key Takeaway
The major types of radioactivity include alpha particles, beta particles, and gamma rays. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/11%3A_Nuclear_Chemistry/11.01%3A_Radioactivity.txt |
Learning Objectives
• To define half-life.
• To determine the amount of radioactive substance remaining after a given number of half-lives.
Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one-half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.
Consider the following example. Suppose we have 100.0 g of 3H (tritium, a radioactive isotope of hydrogen). It has a half-life of 12.3 y. After 12.3 y, half of the sample will have decayed to 3He by emitting a beta particle, so that only 50.0 g of the original 3H remains. After another 12.3 y—making a total of 24.6 y—another half of the remaining 3H will have decayed, leaving 25.0 g of 3H. After another 12.3 y—now a total of 36.9 y—another half of the remaining 3H will have decayed, leaving 12.5 g of 3H. This sequence of events is illustrated in Figure $1$.
We can determine the amount of a radioactive isotope remaining after a given number half-lives by using the following expression:
$\mathrm{amount\: remaining=initial\: amount\times \left ( \dfrac{1}{2} \right )^n} \label{Eq1}$
where n is the number of half-lives. This expression works even if the number of half-lives is not a whole number.
Example $1$
The half-life of 20F is 11.0 s. If a sample initially contains 5.00 g of 20F, how much 20F remains after 44.0 s?
Solution
If we compare the time that has passed to the isotope’s half-life, we note that 44.0 s is exactly 4 half-lives, so we can use Equation \ref{E1} with $n = 4$. Substituting and solving results in the following:
\begin{align*} \mathrm{amount\: remaining} &=5.00\: g \times \left(\dfrac{1}{2}\right)^4 \[4pt] &=5.00\: g \times \dfrac{1}{16} \[4pt] &=0.313\: g \end{align*} \nonumber
Less than one-third of a gram of 20F remains.
Exercise $2$
The half-life of 44Ti is 60.0 y. A sample initially contains 0.600 g of 44Ti. How much 44Ti remains after 180.0 y?
Answer
0.075 g.
Half-lives of isotopes range from fractions of a microsecond to billions of years. Table $1$ lists the half-lives of some isotopes.
Table $1$: Half-Lives of Various Isotopes
Isotope Half-Life
3H 12.3 y
14C 5,730 y
40K 1.26 × 109 y
51Cr 27.70 d
90Sr 29.1 y
131I 8.04 d
222Rn 3.823 d
235U 7.04 × 108 y
238U 4.47 × 109 y
241Am 432.7 y
248Bk 23.7 h
260Sg 4 ms
Example $2$
The isotope $\ce{I}$-125 is used in certain laboratory procedures and has a half-life of 59.4 days. If the initial activity of a sample of $\ce{I}$-125 is $32,000 \: \text{counts per minute (cpm)}$, how much activity will be present in 178.2 days?
Solution
We begin by determining how many half-lives are represented by 178.2 days:
$\dfrac{178.2 \: \text{days}}{59.4 \: \text{days/half-life}} = 3 \: \text{half-lives} \nonumber$
Then we simply count activity:
\begin{align} \text{initial activity} \: \left( t_0 \right) &= 32,000 \: \text{cpm} \nonumber\ \text{after one half-life} \: &= 16,000 \: \text{cpm} \nonumber\ \text{after two half-lives} \: &= 8,000 \: \text{cpm} \nonumber\ \text{after three half-lives} &= 4,000 \: \text{cpm} \nonumber \end{align} \nonumber
Be sure to keep in mind that the initial count is at time zero $\left( t_0 \right)$ and we subtract from that count at the first half-life. The second half-life has an activity of half the previous count (not the initial count).
Equation $\ref{Eq1}$ can be used to calculate the amount of radioactivity remaining after a given time:
$N_t = N_0 \times \left( 0.5 \right)^\text{number of half-lives} \nonumber$
where $N_t =$ activity at time $t$ and $N_0 =$ initial activity at time $t = 0$.
If we have an initial activity of $42,000 \: \text{cpm}$, what will the activity be after four half-lives?
\begin{align} N_t &= N_0 \left( 0.5 \right)^4 \nonumber \ &= \left( 42,000 \right) \left( 0.5 \right) \left( 0.5 \right) \left( 0.5 \right) \left( 0.5 \right) \nonumber \ &= 2625 \: \text{cpm} \nonumber \end{align} \nonumber
Typical radioactive decay curve.
The graph above illustrates a typical decay curve for $\ce{I}-125$. The activity decreases by one-half during each succeeding half-life.
Exercise $2$
A sample of $\ce{Ac}$-225 originally contained 80 grams and after 50 days only 2.5 grams of the original $\ce{Ac}$-225 remain. What is the half life of $\ce{Ac}$-225?
Answer
10 days
Looking Closer: Half-Lives of Radioactive Elements
Many people think that the half-life of a radioactive element represents the amount of time an element is radioactive. In fact, it is the time required for half—not all—of the element to decay radioactively. Occasionally, however, the daughter element is also radioactive, so its radioactivity must also be considered.
The expected working life of an ionization-type smoke detector (described in the opening essay) is about 10 years. In that time, americium-241, which has a half-life of about 432 y, loses less than 4% of its radioactivity. A half-life of 432 y may seem long to us, but it is not very long as half-lives go. Uranium-238, the most common isotope of uranium, has a half-life of about 4.5 × 109 y, while thorium-232 has a half-life of 14 × 109 y.
On the other hand, some nuclei have extremely short half-lives, presenting challenges to the scientists who study them. The longest-lived isotope of lawrencium, 262Lr, has a half-life of 3.6 h, while the shortest-lived isotope of lawrencium, 252Lr, has a half-life of 0.36 s. As of this writing, the largest atom ever detected has atomic number 118, mass number 293, and a half-life of 120 ns. Can you imagine how quickly an experiment must be done to determine the properties of elements that exist for so short a time?
Key Takeaways
• Natural radioactive processes are characterized by a half-life, the time it takes for half of the material to decay radioactively.
• The amount of material left over after a certain number of half-lives can be easily calculated. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/11%3A_Nuclear_Chemistry/11.02%3A_Half-Life.txt |
Learning Objectives
• To express amounts of radioactivity in a variety of units.
Previously, we used mass to indicate the amount of radioactive substance present. This is only one of several units used to express amounts of radiation. Some units describe the number of radioactive events occurring per unit time, while others express the amount of a person’s exposure to radiation.
Perhaps the direct way of reporting radioactivity is the number of radioactive decays per second. One decay per second is called one becquerel (Bq). Even in a small mass of radioactive material, however, there are many thousands of decays or disintegrations per second. The unit curie (Ci), now defined as 3.7 × 1010 decays per second, was originally defined as the number of decays per second in 1 g of radium. Many radioactive samples have activities that are on the order of microcuries (µCi) or more. Both the becquerel and curie can be used in place of grams to describe quantities of radioactive material. As an example, the amount of americium in an average smoke detector has an activity of 0.9 µCi.
The unit becquerel is named after Henri Becquerel, who discovered radioactivity in 1896. The unit curie is named after Polish scientist Marie Curie, who performed some of the initial investigations into radioactive phenomena and discovered the elements, polonium (Po) and radium (Ra) in the early 1900s.
Example \(1\)
A sample of radium has an activity of 16.0 mCi (millicuries). If the half-life of radium is 1,600 y, how long before the sample’s activity is 1.0 mCi?
Solution
The following table shows the activity of the radium sample over multiple half-lives:
Solutions to Example 11.3.1
Time in Years Activity
0 16.0 mCi
1,600 8.0 mCi
3,200 4.0 mCi
4,800 2.0 mCi
6,400 1.0 mCi
Over a period of 4 half-lives, the activity of the radium will be halved four times, at which point its activity will be 1.0 mCi. Thus, it takes 4 half-lives, or 4 × 1,600 y = 6,400 y, for the activity to decrease to 1.0 mCi.
Exercise \(1\)
A sample of radon has an activity of 60,000 Bq. If the half-life of radon is 15 h, how long before the sample’s activity is 3,750 Bq?
Answer
60 hrs.
Other measures of radioactivity are based on the effects it has on living tissue. Radioactivity can transfer energy to tissues in two ways: through the kinetic energy of the particles hitting the tissue and through the electromagnetic energy of the gamma rays being absorbed by the tissue. Either way, the transferred energy—like thermal energy from boiling water—can damage the tissue.
The rad (an acronym for radiation absorbed dose) is a unit equivalent to a gram of tissue absorbing 0.01 J:
1 rad = 0.01 J/g
Another unit of radiation absorption is the gray (Gy):
1 Gy = 100 rad
The rad is more common. To get an idea of the amount of energy this represents, consider that the absorption of 1 rad by 70,000 g of H2O (approximately the same mass as a 150 lb person) would increase its temperature by only 0.002°C. This may not seem like a lot, but it is enough energy to break about 1 × 1021 molecular C–C bonds in a person’s body. That amount of damage would not be desirable.
Predicting the effects of radiation is complicated by the fact that various tissues are affected differently by different types of emissions. To quantify these effects, the unit rem (an acronym for roentgen equivalent, man) is defined as
rem = rad × RBE
where RBE is the relative biological effectiveness factor is a number greater than or equal to 1 that takes into account the type of radioactive emission and sometimes the type of tissue being exposed. For beta particles, RBE factor equals 1. For alpha particles striking most tissues, the factor is 10, but for eye tissue, the factor is 30. Most radioactive emissions that people are exposed to are on the order of a few dozen millirems (mrem) or less; a medical X ray is about 20 mrem. A sievert (Sv) is a related unit and is defined as 100 rem.
What is a person’s annual exposure to radioactivity and radiation? Table \(1\) lists the sources and annual amounts of radiation exposure. It may surprise you to learn that fully 82% of the radioactivity and radiation exposure we receive is from natural sources—sources we cannot avoid. Fully 10% of the exposure comes from our own bodies—largely from 14C and 40K.
Table \(1\): Average Annual Radiation Exposure (Approximate)
Source Amount (mrem)
radon gas 200
medical sources 53
radioactive atoms in the body naturally 39
terrestrial sources 28
cosmic sources 28
consumer products 10
nuclear energy 0.05
Total 358
Flying from New York City to San Francisco adds 5 mrem to your overall radiation exposure because the plane flies above much of the atmosphere, which protects us from most cosmic radiation.
The actual effects of radioactivity and radiation exposure on a person’s health depend on the type of radioactivity, the length of exposure, and the tissues exposed. Table \(2\) lists the potential threats to health at various amounts of exposure over short periods of time (hours or days).
Table \(2\): Effects of Short-Term Exposure to Radioactivity and Radiation
Exposure (rem) Effect
1 (over a full year) no detectable effect
∼20 increased risk of some cancers
∼100 damage to bone marrow and other tissues; possible internal bleeding; decrease in white blood cell count
200–300 visible “burns” on skin, nausea, vomiting, and fatigue
>300 loss of white blood cells; hair loss
∼600 death
One of the simplest ways of detecting radioactivity is by using a piece of photographic film embedded in a badge or a pen. On a regular basis, the film is developed and checked for exposure. A comparison of the exposure level of the film with a set of standard exposures indicates the amount of radiation a person was exposed to.
Another means of detecting radioactivity is an electrical device called a Geiger counter (Figure \(1\)). It contains a gas-filled chamber with a thin membrane on one end that allows radiation emitted from radioactive nuclei to enter the chamber and knock electrons off atoms of gas (usually argon). The presence of electrons and positively charged ions causes a small current, which is detected by the Geiger counter and converted to a signal on a meter or, commonly, an audio circuit to produce an audible “click.”
Key Takeaway
• Radioactivity can be expressed in a variety of units, including rems, rads, and curies. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/11%3A_Nuclear_Chemistry/11.03%3A_Units_of_Radioactivity.txt |
Learning Objectives
• To learn some applications of radioactivity.
Radioactive isotopes have a variety of applications. Generally, however, they are useful either because we can detect their radioactivity or we can use the energy they release.
Radioactive isotopes are effective tracers because their radioactivity is easy to detect. A tracer is a substance that can be used to follow the pathway of that substance through some structure. For instance, leaks in underground water pipes can be discovered by running some tritium-containing water through the pipes and then using a Geiger counter to locate any radioactive tritium subsequently present in the ground around the pipes. (Recall that tritium, 3H, is a radioactive isotope of hydrogen.)
Tracers can also be used to follow the steps of a complex chemical reaction. After incorporating radioactive atoms into reactant molecules, scientists can track where the atoms go by following their radioactivity. One excellent example of this is the use of radioactive carbon-14 to determine the steps involved in the photosynthesis in plants. We know these steps because researchers followed the progress of the radioactive carbon-14 throughout the process.
Radioactive isotopes are useful for establishing the ages of various objects. The half-life of radioactive isotopes is unaffected by any environmental factors, so the isotope acts like an internal clock. For example, if a rock is analyzed and is found to contain a certain amount of uranium-235 and a certain amount of its daughter isotope, we can conclude that a certain fraction of the original uranium-235 has radioactively decayed. If half of the uranium has decayed, then the rock has an age of one half-life of uranium-235, or about 4.5 × 109 y. Many analyses like this, using a wide variety of isotopes, have indicated that the age of Earth itself is over 4 × 109 y. In another interesting example of radioactive dating, 3H dating has been used to verify the stated vintages of some old fine wines.
Carbon-14 (half-life is 5,370 y) is particularly useful in determining the age of once-living artifacts (e.g., animal or plant matter). A tiny amount of carbon-14 is produced naturally in the upper reaches of the atmosphere, and living things incorporate some of it into their tissues, building up to a constant, although very low, level. Once a living thing dies, however, it no longer acquires carbon-14, and as time passes, the carbon-14 that was in the tissues decays. If a once-living artifact is discovered and analyzed many years after its death, with the remaining carbon-14 compared to the known constant level, an approximate age of the artifact can be determined. Using such methods, scientists determined that the age of the Shroud of Turin (made of linen, which comes from the flax plant, and purported by some to be the burial cloth of Jesus Christ; Figure \(1\)) is about 600–700 y, not 2,000 y as claimed by some. Scientists were also able to use radiocarbon dating to show that the age of a mummified body found in the ice of the Alps was 5,300 y.
The radiation emitted by some radioactive substances can be used to kill microorganisms on a variety of foodstuffs, which extends the shelf life of these products. Produce such as tomatoes, mushrooms, sprouts, and berries are irradiated with the emissions from coba . This exposure kills a lot of the bacteria that cause spoilage, so the produce stays fresh longer. Eggs and some meat, such as beef, pork, and poultry, can also be irradiated. Contrary to the belief of some people, irradiation of food does not make the food itself radioactive.
Radioactive isotopes have numerous medical applications—diagnosing and treating illnesses and diseases. One example of a diagnostic application is using radioactive iodine-131 to test for thyroid activity (Figure \(2\)). The thyroid gland in the neck is one of the few places in the body with a significant concentration of iodine. To evaluate thyroid activity, a measured dose of iodine-131 is administered to a patient, and the next day a scanner is used to measure the amount of radioactivity in the thyroid gland. The amount of radioactive iodine that collects there is directly related to the activity of the thyroid, allowing trained physicians to diagnose both hyperthyroidism and hypothyroidism. Iodine-131 has a half-life of only 8 d, so the potential for damage due to exposure is minimal. Technetium-99 can also be used to test thyroid function. Bones, the heart, the brain, the liver, the lungs, and many other organs can be imaged in similar ways by using the appropriate radioactive isotope.
Very little radioactive material is needed in these diagnostic techniques because the radiation emitted is so easy to detect. However, therapeutic applications usually require much larger doses because their purpose is to preferentially kill diseased tissues. For example, if a thyroid tumor is detected, a much larger infusion (thousands of rem, as opposed to a diagnostic dose of less then 40 rem) of iodine-131 could help destroy the tumor cells. Similarly, radioactive strontium is used to not only detect but also ease the pain of bone cancers. Table \(1\) lists several radioactive isotopes and their medical uses.
Table \(1\): Some Radioactive Isotopes That Have Medical Applications
Isotope Use
32P cancer detection and treatment, especially in eyes and skin
59Fe anemia diagnosis
60Co gamma ray irradiation of tumors
99mTc brain, thyroid, liver, bone marrow, lung, heart, and intestinal scanning; blood volume determination
131I diagnosis and treatment of thyroid function
133Xe lung imaging
198Au liver disease diagnosis
In addition to the direct application of radioactive isotopes to diseased tissue, the gamma ray emissions of some isotopes can be directed toward the tissue to be destroyed. Cobalt-60 is a useful isotope for this kind of procedure.
To Your Health: Positron Emission Tomography Scans
One relatively rare form of radioactivity is called positron emission. It is similar to beta particle emission, except that instead of emitting an electron, a nucleus emits a positively charged electron, called a positron. A positron is actually a piece of antimatter; therefore, when a positron encounters an electron, both particles are converted into high-energy gamma radiation.
Isotopes that emit positrons can be employed in a medical imaging technique called positron emission tomography (PET). A patient receives a compound containing a positron-emitting isotope, either intravenously or by ingestion. The radioactive compound travels throughout the body, and the patient is then pushed slowly through a ring of sensors that detect the gamma radiation given off by the annihilation of positrons and electrons. A computer connected to the sensors constructs a three-dimensional image of the interior of part or all of the patient’s body, allowing doctors to see organs or tumors or regulate the function of various organs (such as the brain or the heart) to diagnose the medical condition of the patient.
Two isotopes that undergo positron emission are carbon-11 and fluorine-18, with half-lives of 20.4 and 110 min, respectively. Both isotopes can be incorporated into sugar molecules and introduced into the body. Doctors can use the intensity of gamma ray emission to find tissues that metabolize the sugar faster than other tissues; fast-metabolizing tissue is one sign of a malignant (i.e., cancerous) tumor. Researchers use similar techniques to map areas of the brain that are most active during specific tasks, such as reading or speaking.
PET is one of many diagnostic and treatment methods that physicians use to improve the quality of our lives. It is one of the many positive uses of radioactivity in society.
Key Takeaway
• Radioactivity has several practical applications, including tracers, medical applications, dating once-living objects, and the preservation of food. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/11%3A_Nuclear_Chemistry/11.04%3A_Uses_of_Radioactive_Isotopes.txt |
Learning Objectives
• Explain where nuclear energy comes from.
• Describe the difference between fission and fusion.
Nuclear changes occur with a simultaneous release of energy. Where does this energy come from? If we could precisely measure the masses of the reactants and the products of a nuclear reaction, we would notice that the amount of mass drops slightly in the conversion from reactants to products. Consider the following nuclear reaction, in which the molar mass of each species is indicated to four decimal places:
$\underbrace{\ce{^{235}U}}_{235.0439} \rightarrow \underbrace{\ce{^{139}Ba}}_{138.9088} + \underbrace{\ce{^{94}Kr}}_{93.9343} + \underbrace{\ce{2^1n}}_{2 \times 1.0087} \nonumber$
If we compare the mass of the reactant (235.0439) to the masses of the products (sum = 234.8605), we notice a mass difference of −0.1834 g, or −0.0001834 kg. Where did this mass go?
According to Albert Einstein’s theory of relativity, energy (E) and mass (m) are related by the following equation:
$E = mc^2 \label{Eq2}$
where c is the speed of light, or 3.00 × 108 m/s. In the course of the uranium nuclear chemical reaction, the mass difference is converted to energy, which is given off by the reaction:
\begin{align*} E &= (−0.0001834 \;kg)(3.00 \times 10^8\; m/s)^2 \[4pt] &= −1.65 \times 10^{13}\; J \[4pt] &= −1.65 \times 10^{10}\; kJ \label{Eq3} \end{align*}
That is, 16.5 billion kJ of energy are given off every time 1 mol of uranium-235 undergoes this nuclear reaction. This is an extraordinary amount of energy. Compare it to combustion reactions of hydrocarbons, which give off about 650 kJ/mol of energy for every CH2 unit in the hydrocarbon—on the order of hundreds of kilojoules per mole. Nuclear reactions give off billions of kilojoules per mole.
If this energy could be properly harvested, it would be a significant source of energy for our society. Nuclear energy involves the controlled harvesting of energy from fission reactions. The reaction can be controlled because the fission of uranium-235 (and a few other isotopes, such as plutonium-239) can be artificially initiated by injecting a neutron into a uranium nucleus. The overall nuclear equation, with energy included as a product, is then as follows:
$\ce{^{235}U + ^{1}n -> ^{139}Ba + ^{94}Kr + 3^{1}n + energy} \nonumber$
Thus, by the careful addition of extra neutrons into a sample of uranium, we can control the fission process and obtain energy that can be used for other purposes.
The Curie Family
Artificial or induced radioactivity was first demonstrated in 1934 by Irène Joliot-Curie and Frédéric Joliot, the daughter and son-in-law of Marie Curie.
Example $1$
Plutonium-239 can absorb a neutron and undergo a fission reaction to produce an atom of gold-204 and an atom of phosphorus-31. Write the balanced nuclear equation for the process and determine the number of neutrons given off as part of the reaction.
Solution
Using the data given, we can write the following initial equation:
$\mathrm{_0^1n+ \, _{94}^{239}Pu\rightarrow \, _{79}^{204}Au+ \, _{15}^{31}P + \, ?_0^1n}$
In balanced nuclear equations, the sums of the subscripts on each sides of the equation are the same, as are the sums of the superscripts. The subscripts are already balanced: 0 + 94 = 94 and 79 + 15 = 94. The superscripts on the left equal 240 (1 + 239) but equal 235 (204 + 31) on the right. We need five more mass number units on the right. Five neutrons should be the products of the process for the mass numbers to balance. (Because the atomic number of a neutron is zero, including five neutrons on the right does not change the overall sum of the subscripts.) Thus, the balanced nuclear equation is as follows:
$\mathrm{_0^1n + \, _{94}^{239}Pu \rightarrow \, _{79}^{204}Au + \, _{15}^{31}P + 5_0^1n}$
We predict that the overall process will give off five neutrons.
Exercise $1$
Uranium-238 can absorb a neutron and undergo a fission reaction to produce an atom of cesium-135 and an atom of rubidium-96. Write the balanced nuclear equation for the process and determine the number of neutrons given off as part of the reaction.
Answer
$\mathrm{_0^1n + \, _{92}^{238}U \rightarrow \, _{55}^{135}Cs + \, _{37}^{96}Rb+ 8_0^1n}$
We predict that the overall process will give off eight neutrons.
A nuclear reactor is an apparatus designed to carefully control the progress of a nuclear reaction and extract the resulting energy for useful purposes. Figure $1$ shows a simplified diagram of a nuclear reactor. The energy from the controlled nuclear reaction converts liquid water into high-pressure steam, which is used to run turbines that generate electricity.
Notice that the fission of uranium produces two more free neutrons than were present to begin with. These neutrons can themselves stimulate other uranium nuclei to undergo fission, releasing yet more energy and even more neutrons, which can in turn induce even more uranium fission. A single neutron can thus begin a process that grows exponentially in a phenomenon called a chain reaction:
1 → 2 → 4 → 8 → 16 → 32 → 64 → 128 → 256 → 512 → 1,024 → 2,048 → 4,096 → 8,192 → 16,384 →…
Because energy is produced with each fission event, energy is also produced exponentially and in an uncontrolled fashion. The quick production of energy creates an explosion. This is the mechanism behind the atomic bomb.
The first controlled chain reaction was achieved on December 2, 1942, in an experiment supervised by Enrico Fermi in a laboratory underneath the football stadium at the University of Chicago.
Although fairly simple in theory, an atomic bomb is difficult to produce, in part because uranium-235, the isotope that undergoes fission, makes up only 0.7% of natural uranium; the rest is mostly uranium-238, which does not undergo fission. (Remember that the radioactive process that a nucleus undergoes is characteristic of the isotope.) To make uranium useful for nuclear reactors, the uranium in uranium-235 must be enriched to about 3%. Enrichment of uranium is a laborious and costly series of physical and chemical separations. To be useful in an atomic bomb, the uranium in uranium-235 must be enriched to 70% or more. At lesser concentrations, the chain reaction cannot sustain itself, so no explosion is produced.
Fusion is another nuclear process that can be used to produce energy. In this process, smaller nuclei are combined to make larger nuclei, with an accompanying release of energy. One example is the hydrogen fusion, which makes helium. While the steps of the process are complicated, the net reaction is:
$\ce{4^1H \rightarrow ^4He} + 2.58 \times 10^{12}\; J \nonumber$
Notice that the amount of energy given off per mole of reactant is only a fraction of the amount given off by the fission of 1 mol of uranium-235. On a mass (per gram) basis, however, the hydrogen fusion emits many times more energy than fission does. In addition, the product of fission is helium gas, not a wide range of isotopes (some of which are also radioactive) produced by fission.
The practical problem is that to perform fusion, extremely high pressures and temperatures are necessary. Currently, the only known stable systems undergoing fusion are the interiors of stars. The conditions necessary for fusion can be created using an atomic bomb, but the resulting fusion is uncontrollable (and the basis for another type of bomb, a hydrogen bomb). Currently, researchers are looking for safe, controlled ways of producing useful energy using fusion.
Career Focus: Nuclear Medicine Technologist
Generally speaking, a radiological technician deals with X ray equipment and procedures. A nuclear medicine technologist has similar responsibilities, using compounds containing radioactive isotopes to help diagnose and treat disease.
Nuclear medicine technologists administer the substances containing the radioactive isotope and subsequently operate the apparatus that detects the radiation produced by radioactive decay. The apparatus may be as simple as a piece of photographic film or as complex as a series of computer-controlled electronic detectors. The images obtained by the technologist are interpreted by a specially trained physician.
One of the chief responsibilities of a nuclear medicine technologist is safety. Improper exposure to radioactivity can be harmful to both patient and technologist alike. Therefore, the technologist must adhere to strict safety standards to keep unnecessary exposure as low as possible. The technologist must also know how to dispose of waste materials safely and appropriately.
Key Takeaways
• Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur.
• In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/11%3A_Nuclear_Chemistry/11.05%3A_Nuclear_Energy.txt |
11.1 Radioactivity
Concept Review Exercise
1. What are the major types of radioactivity? Write chemical equations demonstrating each type.
Answer
1. The major types of radioactivity are alpha decay, beta decay, and gamma ray emission; alpha decay with gamma emission: $\mathrm{_{86}^{222}Rn \rightarrow \, _{84}^{218}Po + \, ^4_2He + \gamma}$; beta decay: $\ce{_6^{14}C \rightarrow _7^{14}N + ^0_{-1}e}$ (answers will vary)
Exercises
1. Define radioactivity.
2. Give an example of a radioactive isotope.
3. How many protons and neutrons are in each isotope?
1. $\mathrm{^{11}_5B}$
2. $\mathrm{^{27}_{13}Al}$
3. 56Fe
4. 224Rn
4. How many protons and neutrons are in each isotope?
1. $\mathrm{^{2}_{1}H}$
2. $\mathrm{^{112}_{48}Cd}$
3. 252Es
4. 40K
5. Describe an alpha particle. What nucleus is it equivalent to?
6. Describe a beta particle. What subatomic particle is it equivalent to?
7. Explain what gamma rays are.
8. Explain why it is inappropriate to refer to gamma rays as gamma “particles.”
9. Plutonium has an atomic number of 94. Write the chemical equation for the alpha particle emission of 244Pu. What is the daughter isotope?
10. Francium has an atomic number of 87. Write the chemical equation for the alpha particle emission of 212Fr. What is the daughter isotope?
11. Tin has an atomic number of 50. Write the chemical equation for the beta particle emission of 121Sn. What is the daughter isotope?
12. Technetium has an atomic number of 43. Write the chemical equation for the beta particle emission of 99Tc. What is the daughter isotope?
13. Energies of gamma rays are typically expressed in units of megaelectron volts (MeV), where 1 MeV = 1.602 × 10−13 J. Using data provided in the text, calculate the energy, in megaelectron volts, of the gamma ray emitted when radon-222 decays.
14. The gamma ray emitted when oxygen-19 gives off a beta particle is 0.197 MeV. What is its energy in joules? (See Exercise 13 for the definition of a megaelectron volt.)
15. Which penetrates matter more deeply—alpha particles or beta particles? Suggest ways to protect yourself against both particles.
16. Which penetrates matter more deeply—alpha particles or gamma rays? Suggest ways to protect yourself against both emissions.
17. Define nuclear fission.
18. What general characteristic is typically necessary for a nucleus to undergo spontaneous fission?
Answers
1. Radioactivity is the spontaneous emission of particles and radiation from atomic nuclei.
2. C-14 or 14C is an example of radioactive isotope (answers may vary).
1.
1. 5 protons; 6 neutrons
2. 13 protons; 14 neutrons
3. 26 protons; 30 neutrons
4. 86 protons; 138 neutrons
4.
1. 1 proton; 1 neutron
2. 48 protons; 64 neutrons
3. 99 protons;153 neutrons
4. 19 protons; 21 neutrons
1. An alpha particle is a combination of two protons and two neutrons and is equivalent to a helium nucleus.
6. A beta particle is an electron.
1. Gamma rays are high-energy electromagnetic radiation given off in radioactive decay.
8. Gamma rays have no mass. Hence not a particle.
1. $\mathrm{^{244}_{94}Pu\rightarrow \, _2^4He +\, ^{240}_{92}U}$; the daughter isotope is $\mathrm{^{240}_{92}U}$, an atom of uranium.
10. $\mathrm{^{212}_{87}Fr\rightarrow \, _2^4He +\, ^{208}_{85}At}$; the daughter isotope is $\mathrm{^{208}_{85}At}$, an atom of astatine.
11. $\mathrm{_{50}^{121}Sn \rightarrow \, _{-1}^0e + \, _{51}^{121}Sb}$; the daughter isotope is $\mathrm{_{51}^{121}Sb}$, an atom of antimony.
12. $\mathrm{_{43}^{99}Tc \rightarrow \, _{-1}^0e + \, _{44}^{99}Mo}$; the daughter isotope is $\mathrm{_{44}^{99}Mo}$, an atom of antimony.
13. 0.512 MeV
14. 3.16 x 10-14 J
15. Beta particles; shielding of the appropriate thickness can protect against both alpha and beta particles.
16. Gamma rays; can be shielded by thick, dense material such as lead (Pb). Alpha particles has low energy; can shielded by a piece of paper.
17. Nuclear fission is when large nuclei break down into smaller nuclei.
18. A nucleus must be very large. Examples are Th-232 and U-235.
11.2 Half-Life
Concept Review Exercises
1. Define half-life.
2. Describe a way to determine the amount of radioactive isotope remaining after a given number of half-lives.
Answers
1. Half-life is the amount of time needed for half of a radioactive material to decay.
2. take half of the initial amount for each half-life of time elapsed
Exercises
1. Do all isotopes have a half-life? Explain.
2. Which is more radioactive—an isotope with a long half-life or an isotope with a short half-life?
3. What percent of a sample remains after one half-life? Three half-lives?
4. The half-life of polonium-218 is 3.0 min. How much of a 0.540 mg sample would remain after 9.0 minutes have passed?
5. The half-life of protactinium-234 is 6.69 hours. If a 0.812 mg sample of Pa-239 decays for 40.14 hours, what mass of the isotope remains?
6. How long does it take for 1.00 g of 103Pd to decay to 0.125 g if its half-life is 17.0 d?
7. How long does it take for 2.00 g of 94Nb to decay to 0.0625 g if its half-life is 20,000 y?
8. It took 75 y for 10.0 g of a radioactive isotope to decay to 1.25 g. What is the half-life of this isotope?
9. It took 49.2 s for 3.000 g of a radioactive isotope to decay to 0.1875 g. What is the half-life of this isotope?
Answers
1. Only radioactive isotopes have half-lives.
2. An isotope with a shorter half-life decay more rapidly is more radioactive.
3. 1 half-life: 50%; 3 half-lives: 12.5%
4. 9.0 min = 3 half-lives (make 3 arrows): 0.540 mg --> 0.270 mg --> 0.135 mg --> 0.0675 mg
5. 0.0127 mg
6. 51.0 d
7. 100 000 y
8. 25 y
9. 12.3 s
11.3 Units of Radioactivity
Concept Review Exercise
1. What units are used to quantify radioactivity?
Answer
1. the curie, the becquerel, the rad, the gray, the sievert, and the rem
Exercises
1. Define rad.
2. Define rem.
3. How does a becquerel differ from a curie?
4. How is the curie defined?
5. A sample of radon gas has an activity of 140.0 mCi. If the half-life of radon is 1,500 y, how long before the activity of the sample is 8.75 mCi?
6. A sample of curium has an activity of 1,600 Bq. If the half-life of curium is 24.0 s, how long before its activity is 25.0 Bq?
7. If a radioactive sample has an activity of 65 µCi, how many disintegrations per second are occurring?
8. If a radioactive sample has an activity of 7.55 × 105 Bq, how many disintegrations per second are occurring?
9. Describe how a radiation exposure in rems is determined.
10. Which contributes more to the rems of exposure—alpha or beta particles? Why?
11. Use Table 11.3.2 to determine which sources of radiation exposure are inescapable and which can be avoided. What percentage of radiation is unavoidable?
12. What percentage of the approximate annual radiation exposure comes from radioactive atoms that are in the body naturally?
13. Explain how a film badge works to detect radiation.
14. Explain how a Geiger counter works to detect radiation.
Answers
1. Known as the radiation absorbed dose, a rad is the absorption of 0.01 J/g of tissue.
2. Known as roentgen equivalent man, a rem is an absorption of one rad times a factor. The factor is variable depending on the type of emission and the type of irradiated tissue.
1. A becquerel is smaller and equals 1 decay per second. A curie is 3.7 × 1010 Bq.
4. A curie is defined as 3.7 × 1010 decays per second.
1. 6000 y
6. 144 s
1. 2.41 × 106 disintegrations per second
8. 7.55 × 105 disintegrations per second
1. The radiation exposure is determined by the number of rads times the quality factor of the radiation.
10. Alpha contributes more than beta because of its bigger size and electrical charge.
11. At least 16% (terrestrial and cosmic sources) of radioactivity is unavoidable; the rest depends on what else a person is exposed to.
12. About 11% come from radioactive atoms that are in the body naturally.
13. A film badge uses film, which is exposed as it is subjected to radiation.
14. The Geiger counter consists of a tube with electrodes and is filled with an inert (argon) gas. Radiation entering the tube ionizes the gas, and the ions are attracted to the electrodes and produce an electric pulse (clicking sound).
11.4 Uses of Radioactive Isotopes
Concept Review Exercise
1. Describe some of the different ways that amounts of radioactivity are applied in society.
Answer
1. Radioactive isotopes are used in dating, as tracers, and in medicine as diagnostic and treatment tools.
Exercises
1. Define tracer is and give an example of how tracers work.
2. Name two isotopes that have been used as tracers.
3. Explain how radioactive dating works.
4. Name an isotope that has been used in radioactive dating.
5. The current disintegration rate for carbon-14 is 14.0 Bq. A sample of burnt wood discovered in an archaeological excavation is found to have a carbon-14 decay rate of 3.5 Bq. If the half-life of carbon-14 is 5,700 y, approximately how old is the wood sample?
6. A small asteroid crashes to Earth. After chemical analysis, it is found to contain 1 g of technetium-99 to every 3 g of ruthenium-99, its daughter isotope. If the half-life of technetium-99 is 210,000 y, approximately how old is the asteroid?
7. What do you think are some of the positive aspects of irradiation of food?
8. What do you think are some of the negative aspects of irradiation of food?
9. Describe how iodine-131 is used to both diagnose and treat thyroid problems.
10. List at least five organs that can be imaged using radioactive isotopes.
11. Which radioactive emissions can be used therapeutically?
12. Which isotope is used in therapeutics primarily for its gamma ray emissions?
13. What volume of a radioisotope should be given if a patient needs 125 mCi of a solution which contains 45 mCi in 5.0 mL?
14. Sodium-24 is used to treat leukemia. A 36-kg patient is prescribed 145 μCi/kg and it is supplied to the hospital in a vial containing 250 μCi/mL. What volume should be given to the patient?
15. Lead-212 is one of the radioisotopes used in the treatment of breast cancer. A patient needs a 15 μCi dose and it is supplied as a solution with a concentration of 2.5 μCi/mL. What volume does the patient need? Given the half-life of lead is 10.6 hours, what will be the radioactivity of the sample after approximately four days?
Answers
1. A tracer follows the path of a chemical or a physical process. One of the uses of a tracer is following the path of water underground (answers will vary).
2. Tritium (3H) and Carbon-14 (14C) (answers will vary)
1. Radioactive dating works by comparing the amounts of parent and daughter isotopes and calculating back to how long ago all of the material was just the parent isotope.
4. Carbon-14 (14C) and Uranium-235 (235U) (answers will vary)
1. about 11,400 y
6. about 420,000 y
1. increased shelf life (answers will vary)
8. reduction in the food's vitamin content and cost
1. Iodine-131 is preferentially absorbed by the thyroid gland and can be used to measure the gland’s activity or destroy bad cells in the gland.
10. brain, bone, heart, thyroid, lung (answers will vary)
11. gamma rays, beta particles, or alpha particles
12. cobalt-60
13. 125mCi x (5.0mL/45mCi)=14mL
14. 36kg x (145μCi/kg) x (1mL/250μCi)=21mL
15. Volume given: 15μCi x (1mL/2.5μCi) = 6.0mL
Elapsed time in hours: 4 days x (24 hr/day) = 96 hr
Number of half-lives: 96 hrs/10.6 hours = 9
Radioactivity remaining after 9 half-lives: 0.029 μCi
11.5 Nuclear Energy
Concept Review Exercises
1. How is nuclear energy produced?
2. What is the difference between fission and fusion?
Answers
1. Nuclear energy is produced by carefully controlling the speed of a fission reaction.
2. In fission, large nuclei break down into small ones; in fusion, small nuclei combine to make larger ones. In both cases, a lot of energy is emitted.
Exercises
1. In the spontaneous fission of uranium-233, the following reaction occurs:
233U + 1n → 142Ce + 82Se + 101n
For every mole of 233U that decays, 0.1355 g of mass is lost. How much energy is given off per mole of 233U reacted?
2. In the spontaneous fission of plutonium-241, the following reaction occurs:
241Pu + 1n → 104Ru + 124Sn + 141n
For every mole of 241Pu that decays, 0.1326 g of mass is lost. How much energy is given off per mole of 241Pu reacted?
3. The two rarer isotopes of hydrogen—deuterium and tritium—can also be fused to make helium by the following reaction:
2H + 3H → 4He + 1n
In the course of this reaction, 0.01888 g of mass is lost. How much energy is emitted in the reaction of 1 mol of deuterium and tritium?
4. A process called helium burning is thought to occur inside older stars, forming carbon:
34He → 12C
If the reaction proceeds with 0.00781 g of mass lost on a molar basis, how much energy is given off?
5. Briefly describe how a nuclear reactor generates electricity.
6. Briefly describe the difference between how a nuclear reactor works and how a nuclear bomb works.
7. What is a chain reaction?
8. Why must uranium be enriched to supply nuclear energy?
Answers
1. 1.22 × 1013 J
2. 1.19 × 1013 J
1. 1.70 × 1012 J
4. 7.03 × 1011 J
1. A nuclear reactor generates heat, which is used to generate steam that turns a turbine to generate electricity.
6. Both nuclear reactor and nuclear bomb are powered by fission reaction however, in a nuclear reactor, the fission is monitored and controlled to occur continuously for a much longer time. In a nuclear bomb, the reaction is uncontrolled to explode in one event.
7. A chain reaction is an ever-expanding series of processes that, if left unchecked, can cause a runaway reaction and possibly an explosion.
8. Natural uranium ores contain only 0.7% U-235. Most nuclear reactors require enriched U-235 for their fuel.
11.6: Chapter Summary
Additional Exercises
1. Given that many elements are metals, suggest why it would be unsafe to have radioactive materials in contact with acids.
2. Many alpha-emitting radioactive substances are relatively safe to handle, but inhaling radioactive dust can be very dangerous. Why?
3. Uranium can be separated from its daughter isotope thorium by dissolving a sample in acid and adding sodium iodide, which precipitates thorium(III) iodide:
Th3+(aq) + 3I(aq) → ThI3(s)
If 0.567 g of Th3+ were dissolved in solution, how many milliliters of 0.500 M NaI(aq) would have to be added to precipitate all the thorium?
4. Thorium oxide can be dissolved in an acidic solution:
ThO2(s) + 4H+ → Th4+(aq) + 2H2O(ℓ)
How many milliliters of 1.55 M HCl(aq) are needed to dissolve 10.65 g of ThO2?
5. Radioactive strontium is dangerous because it can chemically replace calcium in the human body. The bones are particularly susceptible to radiation damage. Write the nuclear equation for the beta emission of strontium-90.
6. Write the nuclear equation for the beta emission of iodine-131, the isotope used to diagnose and treat thyroid problems.
7. A common uranium compound is uranyl nitrate hexahydrate [UO2(NO3)2_6H2O]. What is the formula mass of this compound?
8. Plutonium forms three oxides: PuO, PuO2, and Pu2O3. What are the formula masses of these three compounds?
9. A banana contains 600 mg of potassium, 0.0117% of which is radioactive potassium-40. If 1 g of potassium-40 has an activity of 2.626 × 105 Bq, what is the activity of a banana?
10. Smoke detectors typically contain about 0.25 mg of americium-241 as part of the smoke detection mechanism. If the activity of 1 g of americium-241 is 1.26 × 1011 Bq, what is the activity of americium-241 in the smoke detector?
11. Uranium hexafluoride (UF6) reacts with water to make uranyl fluoride (UO2F2) and hydrogen fluoride (HF). Balance the following chemical equation:
UF6 + H2O → UO2F2 + HF
12. The cyclopentadienyl anion (C5H5) is an organic ion that can make ionic compounds with positive ions of radioactive elements, such as Np3+. Balance the following chemical equation:
NpCl3 + Be(C5H5)2 → Np(C5H5)3 + BeCl2
Answers
1. Acids can dissolve metals, making aqueous solutions.
2. Alpha rays are dangerous only when the alpha emitter is in direct contact with tissue cells inside the body.
1. 14.7 mL
4. 104 mL
1. $\mathrm{^{90}_{38}Sr\rightarrow \, _{-1}^{0}e + \, _{39}^{90}Y}$
6. $\mathrm{^{131}_{53}I\rightarrow \, _{-1}^{0}e + \, _{54}^{131}Xe}$
1. 502 g/mol
8. PuO = 260.06 g/mol; PuO2 = 276.06 g/mol; Pu2O3 = 536.12 g/mol
9. about 18 Bq
10. 3.15 x 107 Bq
11. UF6 + 2H2O → UO2F2 + 4HF
12. 2NpCl3 + 3Be(C5H5)2 → 2Np(C5H5)3 + 3BeCl2 | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/11%3A_Nuclear_Chemistry/11.E%3A_Nuclear_Chemistry_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Some atoms have unstable nuclei that emit particles and high-energy electromagnetic radiation to form new elements that are more stable. This emission of particles and electromagnetic radiation is called radioactivity. There are three main types of spontaneous radioactive emission: alpha particles, which are equivalent to helium nuclei; beta particles, which are electrons; and gamma radiation, which is high-energy electromagnetic radiation. Another type of radioactive process is spontaneous fission, in which large nuclei spontaneously break apart into smaller nuclei and, often, neutrons. In all forms of radioactivity, new elements are formed from the radioactive reactants.
Radioactive isotopes decay at different rates. The rate of an isotope’s decay is expressed as a half-life, which is the amount of time required for half of the original material to decay. The length of its half-life is a characteristic of the particular isotope and can range from less than microseconds to billions of years.
Amounts of radioactivity are measured in several different ways. A becquerel is equal to one radioactive decay per second. A curie represents 3.7 × 1010 decays per second. Other units describe the amount of energy absorbed by body tissues. One rad is equivalent to 0.01 joule of energy absorbed per gram of tissue. Different tissues react differently to different types of radioactivity. The rem unit takes into account not only the energy absorbed by the tissues, but also includes a numerical multiplication factor to account for the type of radioactivity and the type of tissue. The average annual radiation exposure of a person is less than 360 millirem, over 80% of which is from natural sources. Radioactivity can be detected using photographic film or other devices such as Geiger counters.
Radioactive isotopes have many useful applications. They can be used as tracers to follow the journey of a substance through a system, like an underground waterway or a metabolic pathway. Radioactive isotopes can be used to date objects, since the amount of parent and daughter isotopes can sometimes be measured very accurately. Radioactive emission can be used to sterilize food for a longer edible lifetime. There are also a number of diagnostic and therapeutic medical applications for radioactive isotopes.
Radioactive processes occur with simultaneous changes in energy. This nuclear energy can be used to generate power for human use. Nuclear reactors use the energy released by fission of large isotopes to generate electricity. When carefully controlled, fission can produce a chain reaction that facilitates the continuous production of energy. If not carefully controlled, a very quick production of energy can result, as in an atomic bomb. Natural uranium does not contain enough of the proper isotope of uranium to work in a nuclear reactor, so it must first be enriched in uranium-235. Forcing small nuclei together to make larger nuclei, a process called fusion, also gives off energy; however, scientists have yet to achieve a controlled fusion process. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/11%3A_Nuclear_Chemistry/11.S%3A_Nuclear_Chemistry_%28Summary%29.txt |
We begin our study of organic chemistry with the alkanes, compounds containing only two elements, carbon and hydrogen, and having only single bonds. There are several other kinds of hydrocarbons, distinguished by the types of bonding between carbon atoms and by the properties that result from that bonding. We will first examine hydrocarbons with double bonds, with triple bonds, and with a special kind of bonding called aromaticity. Then we will study some compounds considered to be derived from hydrocarbons by replacing one or more hydrogen atoms with an oxygen-containing group. Finally, we focuse on organic acids and bases, after which we will be ready to look at the chemistry of life itself—biochemistry—in the remaining five chapters.
12: Organic Chemistry - Alkanes and Halogenated Hydrocarbons
Hydrocarbons are the simplest organic compounds, but they have interesting physiological effects. These effects depend on the size of the hydrocarbon molecules and where on or in the body they are applied. Alkanes of low molar mass—those with from 1 to approximately 10 or so carbon atoms—are gases or light liquids that act as anesthetics. Inhaling (“sniffing”) these hydrocarbons in gasoline or aerosol propellants for their intoxicating effect is a major health problem that can lead to liver, kidney, or brain damage or to immediate death by asphyxiation by excluding oxygen.
Swallowed, liquid alkanes do little harm while in the stomach. In the lungs, however, they cause “chemical” pneumonia by dissolving fatlike molecules from cell membranes in the tiny air sacs (alveoli). The lungs become unable to expel fluids, just as in pneumonia caused by bacteria or viruses. People who swallow gasoline or other liquid alkane mixtures should not be made to vomit, as this would increase the chance of getting alkanes into the lungs. (There is no home-treatment antidote for gasoline poisoning; call a poison control center.)
Liquid alkanes with approximately 5–16 carbon atoms per molecule wash away natural skin oils and cause drying and chapping of the skin, while heavier liquid alkanes (those with approximately 17 or more carbon atoms per molecule) act as emollients (skin softeners). Such alkane mixtures as mineral oil and petroleum jelly can be applied as a protective film. Water and aqueous solutions such as urine will not dissolve such a film, which explains why petroleum jelly protects a baby’s tender skin from diaper rash.
12.01: Organic Chemistry
Learning Objectives
• To recognize the composition and properties typical of organic and inorganic compounds.
Scientists of the 18th and early 19th centuries studied compounds obtained from plants and animals and labeled them organic because they were isolated from “organized” (living) systems. Compounds isolated from nonliving systems, such as rocks and ores, the atmosphere, and the oceans, were labeled inorganic. For many years, scientists thought organic compounds could be made by only living organisms because they possessed a vital force found only in living systems. The vital force theory began to decline in 1828, when the German chemist Friedrich Wöhler synthesized urea from inorganic starting materials. He reacted silver cyanate ($\ce{AgOCN}$) and ammonium chloride ($\ce{NH4Cl}$), expecting to get ammonium cyanate ($\ce{NH4OCN}$). What he expected is described by the following equation.
$\ce{AgOCN + NH_4Cl \rightarrow AgCl + NH_4OCN} \label{Eq1}$
Instead, he found the product to be urea (NH2CONH2), a well-known organic material readily isolated from urine. This result led to a series of experiments in which a wide variety of organic compounds were made from inorganic starting materials. The vital force theory gradually went away as chemists learned that they could make many organic compounds in the laboratory.
Today organic chemistry is the study of the chemistry of the carbon compounds, and inorganic chemistry is the study of the chemistry of all other elements. It may seem strange that we divide chemistry into two branches—one that considers compounds of only one element and one that covers the 100-plus remaining elements. However, this division seems more reasonable when we consider that of tens of millions of compounds that have been characterized, the overwhelming majority are carbon compounds.
The word organic has different meanings. Organic fertilizer, such as cow manure, is organic in the original sense; it is derived from living organisms. Organic foods generally are foods grown without synthetic pesticides or fertilizers. Organic chemistry is the chemistry of compounds of carbon.
Carbon is unique among the other elements in that its atoms can form stable covalent bonds with each other and with atoms of other elements in a multitude of variations. The resulting molecules can contain from one to millions of carbon atoms. We previously surveyed organic chemistry by dividing its compounds into families based on functional groups. We begin with the simplest members of a family and then move on to molecules that are organic in the original sense—that is, they are made by and found in living organisms. These complex molecules (all containing carbon) determine the forms and functions of living systems and are the subject of biochemistry.
Organic compounds, like inorganic compounds, obey all the natural laws. Often there is no clear distinction in the chemical or physical properties among organic and inorganic molecules. Nevertheless, it is useful to compare typical members of each class, as in Table $1$.
Table $1$: General Contrasting Properties and Examples of Organic and Inorganic Compounds
Organic Hexane Inorganic NaCl
low melting points −95°C high melting points 801°C
low boiling points 69°C high boiling points 1,413°C
low solubility in water; high solubility in nonpolar solvents insoluble in water; soluble in gasoline greater solubility in water; low solubility in nonpolar solvents soluble in water; insoluble in gasoline
flammable highly flammable nonflammable nonflammable
aqueous solutions do not conduct electricity nonconductive aqueous solutions conduct electricity conductive in aqueous solution
exhibit covalent bonding covalent bonds exhibit ionic bonding ionic bonds
Keep in mind, however, that there are exceptions to every category in this table. To further illustrate typical differences among organic and inorganic compounds, Table $1$ also lists properties of the inorganic compound sodium chloride (common table salt, NaCl) and the organic compound hexane (C6H14), a solvent that is used to extract soybean oil from soybeans (among other uses). Many compounds can be classified as organic or inorganic by the presence or absence of certain typical properties, as illustrated in Table $1$. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/12%3A_Organic_Chemistry_-_Alkanes_and_Halogenated_Hydrocarbons/12.00%3A_Prelude_to_Organic_Chemistry_-_Alkanes_and_Halogenated_Hydroca.txt |
Learning Objectives
• To identify and name simple (straight-chain) alkanes given formulas and write formulas for straight-chain alkanes given their names.
We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). Saturated, in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules.
The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C).
We previously introduced the three simplest alkanes—methane (CH4), ethane (C2H6), and propane (C3H8) and they are shown again in Figure $1$.
The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure $2$).
Methane (CH4), ethane (C2H6), and propane (C3H8) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH2 unit. The first 10 members of this series are given in Table $1$.
Table $1$: The First 10 Straight-Chain Alkanes
Name Molecular Formula (CnH2n + 2) Condensed Structural Formula Number of Possible Isomers
methane CH4 CH4
ethane C2H6 CH3CH3
propane C3H8 CH3CH2CH3
butane C4H10 CH3CH2CH2CH3 2
pentane C5H12 CH3CH2CH2CH2CH3 3
hexane C6H14 CH3CH2CH2CH2CH2CH3 5
heptane C7H16 CH3CH2CH2CH2CH2CH2CH3 9
octane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 18
nonane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3 35
decane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 75
Consider the series in Figure $3$. The sequence starts with C3H8, and a CH2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH2 group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series.
The principle of homology allows us to write a general formula for alkanes: CnH2n + 2. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C8H(2 × 8) + 2 = C8H18.
Key Takeaway
• Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/12%3A_Organic_Chemistry_-_Alkanes_and_Halogenated_Hydrocarbons/12.02%3A_Structures_and_Names_of_Alkanes.txt |
Learning Objectives
• To learn how alkane molecules can have branched chains and recognize compounds that are isomers.
We can write the structure of butane (C4H10) by stringing four carbon atoms in a row,
–C–C–C–C–
and then adding enough hydrogen atoms to give each carbon atom four bonds:
The compound butane has this structure, but there is another way to put 4 carbon atoms and 10 hydrogen atoms together. Place 3 of the carbon atoms in a row and then branch the fourth one off the middle carbon atom:
Now we add enough hydrogen atoms to give each carbon four bonds.
Notice that C4H10 is depicted with a bent chain in Figure \(1\). The four-carbon chain may be bent in various ways because the groups can rotate freely about the C–C bonds. However, this rotation does not change the identity of the compound. It is important to realize that bending a chain does not change the identity of the compound; all of the following represent the same compound, butane:
The structure of isobutane shows a continuous chain of three carbon atoms only, with the fourth attached as a branch off the middle carbon atom of the continuous chain, which is different from the structures of butane (compare the two structures in Figure \(1\).
Unlike C4H10, the compounds methane (CH4), ethane (C2H6), and propane (C3H8) do not exist in isomeric forms because there is only one way to arrange the atoms in each formula so that each carbon atom has four bonds.
Next beyond C4H10 in the homologous series is pentane. Each compound has the same molecular formula: C5H12. (Table 12.1.1 from the previous section has a column identifying the number of possible isomers for the first 10 straight-chain alkanes.) The compound at the far left is pentane because it has all five carbon atoms in a continuous chain. The compound in the middle is isopentane; like isobutane, it has a one CH3 branch off the second carbon atom of the continuous chain. The compound at the far right, discovered after the other two, was named neopentane (from the Greek neos, meaning “new”). Although all three have the same molecular formula, they have different properties, including boiling points: pentane, 36.1°C; isopentane, 27.7°C; and neopentane, 9.5°C.
A continuous (unbranched) chain of carbon atoms is often called a straight chain even though the tetrahedral arrangement about each carbon gives it a zigzag shape. Straight-chain alkanes are sometimes called normal alkanes, and their names are given the prefix n-. For example, butane is called n-butane. We will not use that prefix here because it is not a part of the system established by the International Union of Pure and Applied Chemistry.
12.04: Condensed Structural and Line-Angle Formulas
Learning Objectives
• Write condensed structural formulas for alkanes given complete structural formulas.
• Draw line-angle formulas given structural formulas.
We use several kinds of formulas to describe organic compounds. A molecular formula shows only the kinds and numbers of atoms in a molecule. For example, the molecular formula C4H10 tells us there are 4 carbon atoms and 10 hydrogen atoms in a molecule, but it doesn’t distinguish between butane and isobutane. A structural formula shows all the carbon and hydrogen atoms and the bonds attaching them. Thus, structural formulas identify the specific isomers by showing the order of attachment of the various atoms.
Unfortunately, structural formulas are difficult to type/write and take up a lot of space. Chemists often use condensed structural formulas to alleviate these problems. The condensed formulas show hydrogen atoms right next to the carbon atoms to which they are attached, as illustrated for butane:
The ultimate condensed formula is a line-angle formula, in which carbon atoms are implied at the corners and ends of lines, and each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds. For example, we can represent pentane (CH3CH2CH2CH2CH3) and isopentane [(CH3)2CHCH2CH3] as follows:
Parentheses in condensed structural formulas indicate that the enclosed grouping of atoms is attached to the adjacent carbon atom.
Key Takeaways
• Condensed chemical formulas show the hydrogen atoms (or other atoms or groups) right next to the carbon atoms to which they are attached.
• Line-angle formulas imply a carbon atom at the corners and ends of lines. Each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/12%3A_Organic_Chemistry_-_Alkanes_and_Halogenated_Hydrocarbons/12.03%3A_Branched-Chain_Alkanes.txt |
Learning Objectives
• To name alkanes by the IUPAC system and write formulas for alkanes given IUPAC names
As noted in previously, the number of isomers increases rapidly as the number of carbon atoms increases. There are 3 pentanes, 5 hexanes, 9 heptanes, and 18 octanes. It would be difficult to assign unique individual names that we could remember. A systematic way of naming hydrocarbons and other organic compounds has been devised by the International Union of Pure and Applied Chemistry (IUPAC). These rules, used worldwide, are known as the IUPAC System of Nomenclature. (Some of the names we used earlier, such as isobutane, isopentane, and neopentane, do not follow these rules and are called common names.) A stem name (Table $1$) indicates the number of carbon atoms in the longest continuous chain (LCC). Atoms or groups attached to this carbon chain, called substituents, are then named, with their positions indicated by numbers. For now, we will consider only those substituents called alkyl groups.
Table $1$: Stems That Indicate the Number of Carbon Atoms in Organic Molecules
Stem Number
meth- 1
eth- 2
prop- 3
but- 4
pent- 5
hex- 6
hept- 7
oct- 8
non- 9
dec- 10
An alkyl group is a group of atoms that results when one hydrogen atom is removed from an alkane. The group is named by replacing the -ane suffix of the parent hydrocarbon with -yl. For example, the -CH3 group derived from methane (CH4) results from subtracting one hydrogen atom and is called a methyl group. The alkyl groups we will use most frequently are listed in Table $2$. Alkyl groups are not independent molecules; they are parts of molecules that we consider as a unit to name compounds systematically.
Table $2$: Common Alkyl Groups
Parent Alkane Alkyl Group Condensed Structural Formula
methane methyl CH3
ethane ethyl CH3CH2
propane propyl CH3CH2CH2
isopropyl (CH3)2CH–
butane butyl* CH3CH2CH2CH2
*There are four butyl groups, two derived from butane and two from isobutane. We will introduce the other three where appropriate.
Simplified IUPAC rules for naming alkanes are as follows (demonstrated in Example $1$).
1. Name alkanes according to the LCC (longest continuous chain) of carbon atoms in the molecule (rather than the total number of carbon atoms). This LCC, considered the parent chain, determines the base name, to which we add the suffix -ane to indicate that the molecule is an alkane.
2. If the hydrocarbon is branched, number the carbon atoms of the LCC. Numbers are assigned in the direction that gives the lowest numbers to the carbon atoms with attached substituents. Hyphens are used to separate numbers from the names of substituents; commas separate numbers from each other. (The LCC need not be written in a straight line; for example, the LCC in the following has five carbon atoms.)
3. Place the names of the substituent groups in alphabetical order before the name of the parent compound. If the same alkyl group appears more than once, the numbers of all the carbon atoms to which it is attached are expressed. If the same group appears more than once on the same carbon atom, the number of that carbon atom is repeated as many times as the group appears. Moreover, the number of identical groups is indicated by the Greek prefixes di-, tri-, tetra-, and so on. These prefixes are not considered in determining the alphabetical order of the substituents. For example, ethyl is listed before dimethyl; the di- is simply ignored. The last alkyl group named is prefixed to the name of the parent alkane to form one word.
When these rules are followed, every unique compound receives its own exclusive name. The rules enable us to not only name a compound from a given structure but also draw a structure from a given name. The best way to learn how to use the IUPAC system is to put it to work, not just memorize the rules. It’s easier than it looks.
Example $1$
Name each compound.
Solution
1. The LCC has five carbon atoms, and so the parent compound is pentane (rule 1). There is a methyl group (rule 2) attached to the second carbon atom of the pentane chain. The name is therefore 2-methylpentane.
2. The LCC has six carbon atoms, so the parent compound is hexane (rule 1). Methyl groups (rule 2) are attached to the second and fifth carbon atoms. The name is 2,5-dimethylhexane.
3. The LCC has eight carbon atoms, so the parent compound is octane (rule 1). There are methyl and ethyl groups (rule 2), both attached to the fourth carbon atom (counting from the right gives this carbon atom a lower number; rule 3). The correct name is thus 4-ethyl-4-methyloctane.
Exercise $1$
Name each compound.
Example $2$
Draw the structure for each compound.
1. 2,3-dimethylbutane
2. 4-ethyl-2-methylheptane
Solution
In drawing structures, always start with the parent chain.
1. The parent chain is butane, indicating four carbon atoms in the LCC.
Then add the groups at their proper positions. You can number the parent chain from either direction as long as you are consistent; just don’t change directions before the structure is done. The name indicates two methyl (CH3) groups, one on the second carbon atom and one on the third.
Finally, fill in all the hydrogen atoms, keeping in mind that each carbon atom must have four bonds.
• The parent chain is heptane in this case, indicating seven carbon atoms in the LCC. –C–C–C–C–C–C–C–
Adding the groups at their proper positions gives
Filling in all the hydrogen atoms gives the following condensed structural formulas:
Note that the bonds (dashes) can be shown or not; sometimes they are needed for spacing.
Exercise $2$
Draw the structure for each compound.
1. 4-ethyloctane
2. 3-ethyl-2-methylpentane
3. 3,3,5-trimethylheptane
Key Takeaway
• Alkanes have both common names and systematic names, specified by IUPAC. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/12%3A_Organic_Chemistry_-_Alkanes_and_Halogenated_Hydrocarbons/12.05%3A_IUPAC_Nomenclature.txt |
Learning Objectives
• To identify the physical properties of alkanes and describe trends in these properties.
Because alkanes have relatively predictable physical properties and undergo relatively few chemical reactions other than combustion, they serve as a basis of comparison for the properties of many other organic compound families. Let’s consider their physical properties first.
Table \(1\) describes some of the properties of some of the first 10 straight-chain alkanes. Because alkane molecules are nonpolar, they are insoluble in water, which is a polar solvent, but are soluble in nonpolar and slightly polar solvents. Consequently, alkanes themselves are commonly used as solvents for organic substances of low polarity, such as fats, oils, and waxes. Nearly all alkanes have densities less than 1.0 g/mL and are therefore less dense than water (the density of H2O is 1.00 g/mL at 20°C). These properties explain why oil and grease do not mix with water but rather float on its surface.
Table \(1\): Physical Properties of Some Alkanes
Molecular Name Formula Melting Point (°C) Boiling Point (°C) Density (20°C)* Physical State (at 20°C)
methane CH4 –182 –164 0.668 g/L gas
ethane C2H6 –183 –89 1.265 g/L gas
propane C3H8 –190 –42 1.867 g/L gas
butane C4H10 –138 –1 2.493 g/L gas
pentane C5H12 –130 36 0.626 g/mL liquid
hexane C6H14 –95 69 0.659 g/mL liquid
octane C8H18 –57 125 0.703 g/mL liquid
decane C10H22 –30 174 0.730 g/mL liquid
*Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure.
Looking Closer: Gas Densities and Fire Hazards
Table \(1\) indicates that the first four members of the alkane series are gases at ordinary temperatures. Natural gas is composed chiefly of methane, which has a density of about 0.67 g/L. The density of air is about 1.29 g/L. Because natural gas is less dense than air, it rises. When a natural-gas leak is detected and shut off in a room, the gas can be removed by opening an upper window. On the other hand, bottled gas can be either propane (density 1.88 g/L) or butanes (a mixture of butane and isobutane; density about 2.5 g/L). Both are much heavier than air (density 1.2 g/L). If bottled gas escapes into a building, it collects near the floor. This presents a much more serious fire hazard than a natural-gas leak because it is more difficult to rid the room of the heavier gas.
Also shown in Table \(1\) are the boiling points of the straight-chain alkanes increase with increasing molar mass. This general rule holds true for the straight-chain homologs of all organic compound families. Larger molecules have greater surface areas and consequently interact more strongly; more energy is therefore required to separate them. For a given molar mass, the boiling points of alkanes are relatively low because these nonpolar molecules have only weak dispersion forces to hold them together in the liquid state.
Looking Closer: An Alkane Basis for Properties of Other Compounds
An understanding of the physical properties of the alkanes is important in that petroleum and natural gas and the many products derived from them—gasoline, bottled gas, solvents, plastics, and more—are composed primarily of alkanes. This understanding is also vital because it is the basis for describing the properties of other organic and biological compound families. For example, large portions of the structures of lipids consist of nonpolar alkyl groups. Lipids include the dietary fats and fatlike compounds called phospholipids and sphingolipids that serve as structural components of living tissues. These compounds have both polar and nonpolar groups, enabling them to bridge the gap between water-soluble and water-insoluble phases. This characteristic is essential for the selective permeability of cell membranes.
Key Takeaway
• Alkanes are nonpolar compounds that are low boiling and insoluble in water.
12.07: Chemical Properties of Alkanes
Learning Objectives
• To identify the main chemical properties of alkanes.
Alkane molecules are nonpolar and therefore generally do not react with ionic compounds such as most laboratory acids, bases, oxidizing agents, or reducing agents. Consider butane as an example:
Neither positive ions nor negative ions are attracted to a nonpolar molecule. In fact, the alkanes undergo so few reactions that they are sometimes called paraffins, from the Latin parum affinis, meaning “little affinity.”
Two important reactions that the alkanes do undergo are combustion and halogenation. Nothing happens when alkanes are merely mixed with oxygen ($O_2$) at room temperature, but when a flame or spark provides the activation energy, a highly exothermic combustion reaction proceeds vigorously. For methane (CH4), the reaction is as follows:
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{heat} \label{12.7.1}$
If the reactants are adequately mixed and there is sufficient oxygen, the only products are carbon dioxide ($CO_2$), water ($H_2O$), and heat—heat for cooking foods, heating homes, and drying clothes. Because conditions are rarely ideal, however, other products are frequently formed. When the oxygen supply is limited, carbon monoxide ($CO$) is a by-product:
$2CH_4 + 3O_2 \rightarrow 2CO + 4H_2O\label{12.7.2}$
This reaction is responsible for dozens of deaths each year from unventilated or improperly adjusted gas heaters. (Similar reactions with similar results occur with kerosene heaters.)
Alkanes also react with the halogens chlorine ($Cl_2$) and bromine ($Br_2$) in the presence of ultraviolet light or at high temperatures to yield chlorinated and brominated alkanes. For example, chlorine reacts with excess methane ($CH_4$) to give methyl chloride ($CH_3Cl$).
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl\label{12.7.3}$
With more chlorine, a mixture of products is obtained: CH3Cl, CH2Cl2, CHCl3, and CCl4. Fluorine ($F_2$), the lightest halogen, combines explosively with most hydrocarbons. Iodine ($I_2$) is relatively unreactive. Fluorinated and iodinated alkanes are produced by indirect methods.
Key Takeaway
• Alkanes react with oxygen (combustion) and with halogens (halogenation). | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/12%3A_Organic_Chemistry_-_Alkanes_and_Halogenated_Hydrocarbons/12.06%3A_Physical_Properties_of_Alkanes.txt |
Learning Objectives
• To name halogenated hydrocarbons given formulas and write formulas for these compounds given names.
Many organic compounds are closely related to the alkanes. As we noted previously, alkanes react with halogens to produce halogenated hydrocarbons, the simplest of which have a single halogen atom substituted for a hydrogen atom of the alkane. Even more closely related are the cycloalkanes, compounds in which the carbon atoms are joined in a ring, or cyclic fashion.
The reactions of alkanes with halogens produce halogenated hydrocarbons, compounds in which one or more hydrogen atoms of a hydrocarbon have been replaced by halogen atoms:
The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane). The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide. The IUPAC system uses the name of the parent alkane with a prefix indicating the halogen substituents, preceded by number indicating the substituent’s location. The prefixes are fluoro-, chloro-, bromo-, and iodo-. Thus CH3CH2Cl has the common name ethyl chloride and the IUPAC name chloroethane. Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names.
Example \(1\)
Give the common and IUPAC names for each compound.
1. CH3CH2CH2Br
2. (CH3)2CHCl
Solution
1. The alkyl group (CH3CH2CH2–) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane.
2. The alkyl group [(CH3)2CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro-) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane.
Exercise \(1\)
Give common and IUPAC names for each compound.
1. CH3CH2I
2. CH3CH2CH2CH2F
Example \(2\)
Give the IUPAC name for each compound.
Solution
1. The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
2. The parent alkane is hexane. Methyl (CH3) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.
Exercise \(2\)
Give the IUPAC name for each compound.
A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH4) can react with chlorine (Cl2), replacing one, two, three, or all four hydrogen atoms with Cl atoms. Several halogenated products derived from methane and ethane (CH3CH3) are listed in Table \(1\), along with some of their uses.
Table \(1\): Some Halogenated Hydrocarbons
Formula Common Name IUPAC Name Some Important Uses
Derived from CH4
CH3Cl methyl chloride chloromethane refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber
CH2Cl2 methylene chloride dichloromethane laboratory and industrial solvent
CHCl3 chloroform trichloromethane industrial solvent
CCl4 carbon tetrachloride tetrachloromethane dry-cleaning solvent and fire extinguishers (but no longer recommended for use)
CBrF3 halon-1301 bromotrifluoromethane fire extinguisher systems
CCl3F chlorofluorocarbon-11 (CFC-11) trichlorofluoromethane foaming plastics
CCl2F2 chlorofluorocarbon-12 (CFC-12) dichlorodifluoromethane refrigerant
Derived from CH3CH3
CH3CH2Cl ethyl chloride chloroethane local anesthetic
ClCH2CH2Cl ethylene dichloride 1,2-dichloroethane solvent for rubber
CCl3CH3 methylchloroform 1,1,1-trichloroethane solvent for cleaning computer chips and molds for shaping plastics
To Your Health: Halogenated Hydrocarbons
Once widely used in consumer products, many chlorinated hydrocarbons are suspected carcinogens (cancer-causing substances) and also are known to cause severe liver damage. An example is carbon tetrachloride (CCl4), once used as a dry-cleaning solvent and in fire extinguishers but no longer recommended for either use. Even in small amounts, its vapor can cause serious illness if exposure is prolonged. Moreover, it reacts with water at high temperatures to form deadly phosgene (COCl2) gas, which makes the use of CCl4 in fire extinguishers particularly dangerous.
Ethyl chloride, in contrast, is used as an external local anesthetic. When sprayed on the skin, it evaporates quickly, cooling the area enough to make it insensitive to pain. It can also be used as an emergency general anesthetic.
Bromine-containing compounds are widely used in fire extinguishers and as fire retardants on clothing and other materials. Because they too are toxic and have adverse effects on the environment, scientists are engaged in designing safer substitutes for them, as for many other halogenated compounds.
To Your Health: Chlorofluorocarbons and the Ozone Layer
Alkanes substituted with both fluorine (F) and chlorine (Cl) atoms have been used as the dispersing gases in aerosol cans, as foaming agents for plastics, and as refrigerants. Two of the best known of these chlorofluorocarbons (CFCs) are listed in Table \(1\).
Chlorofluorocarbons contribute to the greenhouse effect in the lower atmosphere. They also diffuse into the stratosphere, where they are broken down by ultraviolet (UV) radiation to release Cl atoms. These in turn break down the ozone (O3) molecules that protect Earth from harmful UV radiation. Worldwide action has reduced the use of CFCs and related compounds. The CFCs and other Cl- or bromine (Br)-containing ozone-destroying compounds are being replaced with more benign substances. Hydrofluorocarbons (HFCs), such as CH2FCF3, which have no Cl or Br to form radicals, are one alternative. Another is hydrochlorofluorocarbons (HCFCs), such as CHCl2CF3. HCFC molecules break down more readily in the troposphere, and fewer ozone-destroying molecules reach the stratosphere.
Key Takeaway
• The replacement of an hydrogen atom on an alkane by a halogen atom—F, Cl, Br, or I—forms a halogenated compound.
12.09: Cycloalkanes
Learning Objectives
• To name cycloalkanes given their formulas and write formulas for these compounds given their names.
The hydrocarbons we have encountered so far have been composed of molecules with open-ended chains of carbon atoms. When a chain contains three or more carbon atoms, the atoms can join to form ring or cyclic structures. The simplest of these cyclic hydrocarbons has the formula C3H6. Each carbon atom has two hydrogen atoms attached (Figure \(1\)) and is called cyclopropane.
To Your Health: Cyclopropane as an Anesthetic
With its boiling point of −33°C, cyclopropane is a gas at room temperature. It is also a potent, quick-acting anesthetic with few undesirable side effects in the body. It is no longer used in surgery, however, because it forms explosive mixtures with air at nearly all concentrations.
The cycloalkanes—cyclic hydrocarbons with only single bonds—are named by adding the prefix cyclo- to the name of the open-chain compound having the same number of carbon atoms as there are in the ring. Thus the name for the cyclic compound C4H8 is cyclobutane. The carbon atoms in cyclic compounds can be represented by line-angle formulas that result in regular geometric figures. Keep in mind, however, that each corner of the geometric figure represents a carbon atom plus as many hydrogen atoms as needed to give each carbon atom four bonds.
Some cyclic compounds have substituent groups attached. Example \(1\) interprets the name of a cycloalkane with a single substituent group.
Example \(1\)
Draw the structure for each compound.
1. cyclopentane
2. methylcyclobutane
Solution
1. The name cyclopentane indicates a cyclic (cyclo) alkane with five (pent-) carbon atoms. It can be represented as a pentagon.
• The name methylcyclobutane indicates a cyclic alkane with four (but-) carbon atoms in the cyclic part. It can be represented as a square with a CH3 group attached.
Exercise \(1\)
Draw the structure for each compound.
1. cycloheptane
2. ethylcyclohexane
The properties of cyclic hydrocarbons are generally quite similar to those of the corresponding open-chain compounds. So cycloalkanes (with the exception of cyclopropane, which has a highly strained ring) act very much like noncyclic alkanes. Cyclic structures containing five or six carbon atoms, such as cyclopentane and cyclohexane, are particularly stable. We will see later that some carbohydrates (sugars) form five- or six-membered rings in solution.
The cyclopropane ring is strained because the C–C–C angles are 60°, and the preferred (tetrahedral) bond angle is 109.5°. (This strain is readily evident when you try to build a ball-and-stick model of cyclopropane; see Figure \(1\).) Cyclopentane and cyclohexane rings have little strain because the C–C–C angles are near the preferred angles.
Key Takeaway
• Many organic compounds have cyclic structures. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/12%3A_Organic_Chemistry_-_Alkanes_and_Halogenated_Hydrocarbons/12.08%3A_Halogenated_Hydrocarbons.txt |
12.1: Organic Chemistry
Concept Review Exercises
1. Classify each compound as organic or inorganic.
1. C3H8O
2. CaCl2
3. Cr(NH3)3Cl3
4. C30H48O3N
2. Which compound is likely organic and which is likely inorganic?
1. a flammable compound that boils at 80°C and is insoluble in water
2. a compound that does not burn, melts at 630°C, and is soluble in water
Answers
1.
1. organic
2. inorganic
3. inorganic
4. organic
2.
1. organic
2. inorganic
1. Classify each compound as organic or inorganic.
1. C6H10
2. CoCl2
3. C12H22O11
2. Classify each compound as organic or inorganic.
1. CH3NH2
2. NaNH2
3. Cu(NH3)6Cl2
3. Which member of each pair has a higher melting point?
1. CH3OH and NaOH
2. CH3Cl and KCl
4. Which member of each pair has a higher melting point?
1. C2H6 and CoCl2
2. CH4 and LiH
1.
1. organic
2. inorganic
3. organic
1.
1. NaOH
2. KCl
Concept Review Exercises
1. In the homologous series of alkanes, what is the molecular formula for the member just above C8H18?
2. Use the general formula for alkanes to write the molecular formula of the alkane with 12 carbon atoms.
1. C9H20
2. C12H26
Exercises
1. What compounds contain fewer carbon atoms than C3H8 and are its homologs?
2. What compounds contain five to eight carbon atoms and are homologs of C4H10?
Answer
1. CH4 and C2H6
Concept Review Exercises
1. In alkanes, can there be a two-carbon branch off the second carbon atom of a four-carbon chain? Explain.
2. A student is asked to write structural formulas for two different hydrocarbons having the molecular formula C5H12. She writes one formula with all five carbon atoms in a horizontal line and the other with four carbon atoms in a line, with a CH3 group extending down from the first attached to the third carbon atom. Do these structural formulas represent different molecular formulas? Explain why or why not.
Answers
1. No; the branch would make the longest continuous chain of five carbon atoms.
2. No; both are five-carbon continuous chains.
Key Takeaway
• Alkanes with four or more carbon atoms can exist in isomeric forms.
Exercises
1. Briefly identify the important distinctions between a straight-chain alkane and a branched-chain alkane.
2. How are butane and isobutane related? How do they differ?
3. Name each compound.
4. Write the structural formula for each compound.
1. hexane
2. octane
5. Indicate whether the structures in each set represent the same compound or isomers.
1. CH3CH2CH2CH3 and
2. CH3CH2CH2CH2CH3 and
Answers
1. Straight-chain alkanes and branched-chain alkanes have different properties as well as different structures.
1.
1. pentane
2. heptane
1.
1. no
2. yes
Exercises
1. Write the condensed structural formula for each structural formula.
2. A condensed structural formula for isohexane can be written as (CH3)2CHCH2CH2CH3. Draw the line-angle formula for isohexane.
3. Draw a line-angle formula for the compound CH3CH2CH(CH3)CH2CH2CH3.
4. Give the structural formula for the compound represented by this line-angle formula:
Answers
1.
1. CH3CH3
2. CH3CH2CH3
3. CH3CH2CH2CH2CH3
Concept Review Exercises
1. What is a CH3 group called when it is attached to a chain of carbon atoms—a substituent or a functional group?
2. Which type of name uses numbers to locate substituents—common names or IUPAC names?
1. substituent
2. IUPAC names
Exercises
1. Briefly identify the important distinctions between an alkane and an alkyl group.
2. How many carbon atoms are present in each molecule?
1. 2-methylbutane
2. 3-ethylpentane
3. How many carbon atoms are present in each molecule?
1. 2,3-dimethylbutane
2. 3-ethyl-2-methylheptane
4. Draw the structure for each compound.
1. 3-methylpentane
2. 2,2,5-trimethylhexane
3. 4-ethyl-3-methyloctane
5. Draw the structure for each compound.
1. 2-methylpentane
2. 4-ethyl-2-methylhexane
3. 2,2,3,3-tetramethylbutane
6. Name each compound according to the IUPAC system.
7. Name each compound according to the IUPAC system.
8. What is a substituent? How is the location of a substituent indicated in the IUPAC system?
9. Briefly identify the important distinctions between a common name and an IUPAC name.
Answers
1. An alkane is a molecule; an alkyl group is not an independent molecule but rather a part of a molecule that we consider as a unit.
1.
1. 6
2. 10
1.
1.
1. 2,2,4,4-tetramethylpentane
2. 3-ethylhexane
1. Common names are widely used but not very systematic; IUPAC names identify a parent compound and name other groups as substituents.
Concept Review Exercises
1. Without referring to a table, predict which has a higher boiling point—hexane or octane. Explain.
2. If 25 mL of hexane were added to 100 mL of water in a beaker, which of the following would you expect to happen? Explain.
1. Hexane would dissolve in water.
2. Hexane would not dissolve in water and would float on top.
3. Hexane would not dissolve in water and would sink to the bottom of the container.
Answers
1. octane because of its greater molar mass
2. b; hexane is insoluble in water and less dense than water.
Exercises
1. Without referring to a table or other reference, predict which member of each pair has the higher boiling point.
1. pentane or butane
2. heptane or nonane
2. For which member of each pair is hexane a good solvent?
1. pentane or water
2. sodium chloride or soybean oil
1. pentane
2. nonane
Concept Review Exercises
1. Why are alkanes sometimes called paraffins?
2. Which halogen reacts most readily with alkanes? Which reacts least readily?
Answers
1. Alkanes do not react with many common chemicals. They are sometimes called paraffins, from the Latin parum affinis, meaning “little affinity.”
2. most readily: $F_2$; least readily: $I_2$
Exercises
1. Why do alkanes usually not react with ionic compounds such as most laboratory acids, bases, oxidizing agents, or reducing agents?
2. Write an equation for the complete combustion of methane ($CH_4$), the main component of natural gas).
3. What is the most important reaction of alkanes?
4. Name some substances other than oxygen that react readily with alkanes.
Answers
1. Alkanes are nonpolar; they do not attract ions.
Concept Review Exercises
1. What is the IUPAC name for the HFC that has the formula CH2FCF3? (Hint: you must use a number to indicate the location of each substituent F atom.)
2. What is the IUPAC name for the HCFC that has the formula CHCl2CF3?
Answers
1. 1,1,1,2-tetrafluoroethane
2. 1,1,1-trifluoro-2,2-dichloroethane
Exercises
1. Write the condensed structural formula for each compound.
1. methyl chloride
2. chloroform
2. Write the condensed structural formula for each compound.
1. ethyl bromide
2. carbon tetrachloride
3. Write the condensed structural formulas for the two isomers that have the molecular formula C3H7Br. Give the common name and the IUPAC name of each.
4. Write the condensed structural formulas for the four isomers that have the molecular formula C4H9Br. Give the IUPAC name of each.
5. What is a CFC? How are CFCs involved in the destruction of the ozone layer?
6. Explain why each compound is less destructive to the ozone layer than are CFCs.
1. fluorocarbons
2. HCFCs
Answers
1. CH3Cl
2. CHCl3
1. CH3CH2CH2Br, propyl bromide, 1-bromopropane; CH3CHBrCH3, isopropyl bromide, 2-bromopropane
1. compounds containing Cl, F, and C; by releasing Cl atoms in the stratosphere
Concept Review Exercises
1. What is the molecular formula of cyclooctane?
2. What is the IUPAC name for this compound?
Answers
1. C8H16
2. ethylcyclopropane
Exercises
1. Draw the structure for each compound.
1. ethylcyclobutane
2. propylcyclopropane
2. Draw the structure for each compound.
1. methylcyclohexane
2. butylcyclobutane
3. Cycloalkyl groups can be derived from cycloalkanes in the same way that alkyl groups are derived from alkanes. These groups are named as cyclopropyl, cyclobutyl, and so on. Name each cycloalkyl halide.
4. Halogenated cycloalkanes can be named by the IUPAC system. As with alkyl derivatives, monosubstituted derivatives need no number to indicate the position of the halogen. To name disubstituted derivatives, the carbon atoms are numbered starting at the position of one substituent (C1) and proceeding to the second substituted atom by the shortest route. Name each compound.
Answers
1. cyclopentyl bromide
2. cyclohexyl chloride
Additional Exercises
1. You find an unlabeled jar containing a solid that melts at 48°C. It ignites readily and burns readily. The substance is insoluble in water and floats on the surface. Is the substance likely to be organic or inorganic?
2. Give the molecular formulas for methylcyclopentane, 2-methylpentane, and cyclohexane. Which are isomers?
3. What is wrong with each name? (Hint: first write the structure as if it were correct.) Give the correct name for each compound.
1. 2-dimethylpropane
2. 2,3,3-trimethylbutane
3. 2,4-diethylpentane
4. 3,4-dimethyl-5-propylhexane
4. What is the danger in swallowing a liquid alkane?
5. Distinguish between lighter and heavier liquid alkanes in terms of their effects on the skin.
6. Following is the line formula for an alkane. Draw its structure and give its name.
7. Write equations for the complete combustion of each compound.
1. propane (a bottled gas fuel)
2. octane (a typical hydrocarbon in gasoline).
8. The density of a gasoline sample is 0.690 g/mL. On the basis of the complete combustion of octane, calculate the amount in grams of carbon dioxide (CO2) and water (H2O) formed per gallon (3.78 L) of the gasoline when used in an automobile.
9. Draw the structures for the five isomeric hexanes (C6H14). Name each by the IUPAC system.
10. Indicate whether the structures in each set represent the same compound or isomers.
11. Consider the line-angle formulas shown here and answer the questions.
1. Which pair of formulas represents isomers? Draw each structure.
2. Which formula represents an alkyl halide? Name the compound and write its condensed structural formula.
3. Which formula represents a cyclic alkane? Name the compound and draw its structure.
4. What is the molecular formula of the compound represented by (i)?
Answers
1. organic
1.
1. Two numbers are needed to indicate two substituents; 2,2-dimethylpropane.
2. The lowest possible numbers were not used; 2,2,3-trimethylbutane.
3. An ethyl substituent is not possible on the second carbon atom; 3,5-dimethylheptane.
4. A propyl substituent is not possible on the fifth carbon atom; 3,4,5-trimethyloctane.
1. Lighter alkanes wash away protective skin oils; heavier alkanes form a protective layer.
1.
1. C3H8 + 5O2 → 3CO2 + 4H2O
2. 2C8H18 + 25O2 → 16CO2 + 18H2O
1. CH3CH2CH2CH2CH2CH3; hexane
1.
1. ii and iii; CH3CH2CH2CH2CH2CH2CH3 and
2. iv; 3-chloropentane; CH3CH2CHClCH2CH3
3. i; ethylcyclopentane;
4. C7H14
12.S: Organic Chemistry- Alkanes and Halogenated Hydrocarbons (Summ
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the summary and ask yourself how they relate to the topics in the chapter.
Organic chemistry is the chemistry of carbon compounds, and inorganic chemistry is the chemistry of all the other elements. Carbon atoms can form stable covalent bonds with other carbon atoms and with atoms of other elements, and this property allows the formation the tens of millions of organic compounds. Hydrocarbons contain only hydrogen and carbon atoms.
Hydrocarbons in which each carbon atom is bonded to four other atoms are called alkanes or saturated hydrocarbons. They have the general formula CnH2n + 2. Any given alkane differs from the next one in a series by a CH2 unit. Any family of compounds in which adjacent members differ from each other by a definite factor is called a homologous series.
Carbon atoms in alkanes can form straight chains or branched chains. Two or more compounds having the same molecular formula but different structural formulas are isomers of each other. There are no isomeric forms for the three smallest alkanes; beginning with C4H10, all other alkanes have isomeric forms.
A structural formula shows all the carbon and hydrogen atoms and how they are attached to one another. A condensed structural formula shows the hydrogen atoms right next to the carbon atoms to which they are attached. A line-angle formula is a formula in which carbon atoms are implied at the corners and ends of lines. Each carbon atom is understood to be attached to enough hydrogen atoms to give each carbon atom four bonds.
The IUPAC System of Nomenclature provides rules for naming organic compounds. An alkyl group is a unit formed by removing one hydrogen atom from an alkane.
The physical properties of alkanes reflect the fact that alkane molecules are nonpolar. Alkanes are insoluble in water and less dense than water.
Alkanes are generally unreactive toward laboratory acids, bases, oxidizing agents, and reducing agents. They do burn (undergo combustion reactions).
Alkanes react with halogens by substituting one or more halogen atoms for hydrogen atoms to form halogenated hydrocarbons. An alkyl halide (haloalkane) is a compound resulting from the replacement of a hydrogen atom of an alkane with a halogen atom.
Cycloalkanes are hydrocarbons whose molecules are closed rings rather than straight or branched chains. A cyclic hydrocarbon is a hydrocarbon with a ring of carbon atoms | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/12%3A_Organic_Chemistry_-_Alkanes_and_Halogenated_Hydrocarbons/12.E%3A_Organic_Chemistry-_Alkanes_and_Halogenated_Hydrocarbons_%28Exer.txt |
Our modern society is based to a large degree on the chemicals we discuss in this chapter. Most are made from petroleum. Alkanes—saturated hydrocarbons—have relatively few important chemical properties other than that they undergo combustion and react with halogens. Unsaturated hydrocarbons—hydrocarbons with double or triple bonds—on the other hand, are quite reactive. In fact, they serve as building blocks for many familiar plastics—polyethylene, vinyl plastics, acrylics—and other important synthetic materials (e.g., alcohols, antifreeze, and detergents). Aromatic hydrocarbons have formulas that can be drawn as cyclic alkenes, making them appear unsaturated, but their structure and properties are generally quite different, so they are not considered to be alkenes. Aromatic compounds serve as the basis for many drugs, antiseptics, explosives, solvents, and plastics (e.g., polyesters and polystyrene). The two simplest unsaturated compounds—ethylene (ethene) and acetylene (ethyne)—were once used as anesthetics and were introduced to the medical field in 1924. However, it was discovered that acetylene forms explosive mixtures with air, so its medical use was abandoned in 1925. Ethylene was thought to be safer, but it too was implicated in numerous lethal fires and explosions during anesthesia. Even so, it remained an important anesthetic into the 1960s, when it was replaced by nonflammable anesthetics such as halothane ($\mathrm{CHBrClCF_3}$).
• 13.0: Prelude to Unsaturated and Aromatic Hydrocarbons
The two simplest unsaturated compounds—ethylene (ethene) and acetylene (ethyne)—were once used as anesthetics and were introduced to the medical field in 1924. However, it was discovered that acetylene forms explosive mixtures with air, so its medical use was abandoned in 1925. Ethylene was thought to be safer, but it too was implicated in numerous lethal fires and explosions during anesthesia. Even so, it remained an important anesthetic into the 1960s, when it was replaced by nonflammable anes
• 13.1: Alkenes- Structures and Names
Alkenes are hydrocarbons with a carbon-to-carbon double bond.
• 13.2: Cis-Trans Isomers (Geometric Isomers)
Cis-trans (geometric) isomerism exists when there is restricted rotation in a molecule and there are two nonidentical groups on each doubly bonded carbon atom.
• 13.3: Physical Properties of Alkenes
The physical properties of alkenes are much like those of the alkanes: their boiling points increase with increasing molar mass, and they are insoluble in water.
• 13.4: Chemical Properties of Alkenes
Alkenes undergo addition reactions, adding such substances as hydrogen, bromine, and water across the carbon-to-carbon double bond.
• 13.5: Polymers
Molecules having carbon-to-carbon double bonds can undergo addition polymerization.
• 13.6: Alkynes
Alkynes are similar to alkenes in both physical and chemical properties. For example, alkynes undergo many of the typical addition reactions of alkenes. The International Union of Pure and Applied Chemistry (IUPAC) names for alkynes parallel those of alkenes, except that the family ending is -yne rather than -ene. The IUPAC name for acetylene is ethyne. The names of other alkynes are illustrated in the following exercises.
• 13.7: Aromatic Compounds- Benzene
Aromatic hydrocarbons appear to be unsaturated, but they have a special type of bonding and do not undergo addition reactions.
• 13.8: Structure and Nomenclature of Aromatic Compounds
Aromatic compounds contain a benzene ring or have certain benzene-like properties; for our purposes, you can recognize aromatic compounds by the presence of one or more benzene rings in their structure.
• 13.E: Unsaturated and Aromatic Hydrocarbons (Exercises)
Select problems and solutions for the chapter.
• 13.S: Unsaturated and Aromatic Hydrocarbons (Summary)
A brief summary of the chapter.
13: Unsaturated and Aromatic Hydrocarbons
Our modern society is based to a large degree on the chemicals we discuss in this chapter. Most are made from petroleum. Alkanes—saturated hydrocarbons—have relatively few important chemical properties other than that they undergo combustion and react with halogens. Unsaturated hydrocarbons—hydrocarbons with double or triple bonds—on the other hand, are quite reactive. In fact, they serve as building blocks for many familiar plastics—polyethylene, vinyl plastics, acrylics—and other important synthetic materials (e.g., alcohols, antifreeze, and detergents). Aromatic hydrocarbons have formulas that can be drawn as cyclic alkenes, making them appear unsaturated, but their structure and properties are generally quite different, so they are not considered to be alkenes. Aromatic compounds serve as the basis for many drugs, antiseptics, explosives, solvents, and plastics (e.g., polyesters and polystyrene).
The two simplest unsaturated compounds—ethylene (ethene) and acetylene (ethyne)—were once used as anesthetics and were introduced to the medical field in 1924. However, it was discovered that acetylene forms explosive mixtures with air, so its medical use was abandoned in 1925. Ethylene was thought to be safer, but it too was implicated in numerous lethal fires and explosions during anesthesia. Even so, it remained an important anesthetic into the 1960s, when it was replaced by nonflammable anesthetics such as halothane (\(\ce{CHBrClCF3}\)). | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.00%3A_Prelude_to_Unsaturated_and_Aromatic_Hydrocarbons.txt |
Learning Objectives
• To name alkenes given formulas and write formulas for alkenes given names.
As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R2C=CR2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas:
Some representative alkenes—their names, structures, and physical properties—are given in Table \(1\).
Table \(1\): Physical Properties of Some Selected Alkenes
IUPAC Name Molecular Formula Condensed Structural Formula Melting Point (°C) Boiling Point (°C)
ethene C2H4 CH2=CH2 –169 –104
propene C3H6 CH2=CHCH3 –185 –47
1-butene C4H8 CH2=CHCH2CH3 –185 –6
1-pentene C5H10 CH2=CH(CH2)2CH3 –138 30
1-hexene C6H12 CH2=CH(CH2)3CH3 –140 63
1-heptene C7H14 CH2=CH(CH2)4CH3 –119 94
1-octene C8H16 CH2=CH(CH2)5CH3 –102 121
We used only condensed structural formulas in Table \(1\). Thus, CH2=CH2 stands for
The double bond is shared by the two carbons and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C2H4, whereas that for ethane is C2H6.
The first two alkenes in Table \(1\), ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure \(1\)). Ethylene is a major commercial chemical. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products.
Although there is only one alkene with the formula C2H4 (ethene) and only one with the formula C3H6 (propene), there are several alkenes with the formula C4H8.
Here are some basic rules for naming alkenes from the International Union of Pure and Applied Chemistry (IUPAC):
1. The longest chain of carbon atoms containing the double bond is considered the parent chain. It is named using the same stem as the alkane having the same number of carbon atoms but ends in -ene to identify it as an alkene. Thus the compound CH2=CHCH3 is propene.
2. If there are four or more carbon atoms in a chain, we must indicate the position of the double bond. The carbons atoms are numbered so that the first of the two that are doubly bonded is given the lower of the two possible numbers.The compound CH3CH=CHCH2CH3, for example, has the double bond between the second and third carbon atoms. Its name is 2-pentene (not 3-pentene).
3. Substituent groups are named as with alkanes, and their position is indicated by a number. Thus, the structure below is 5-methyl-2-hexene. Note that the numbering of the parent chain is always done in such a way as to give the double bond the lowest number, even if that causes a substituent to have a higher number. The double bond always has priority in numbering.
Example \(1\)
Name each compound.
Solution
1. The longest chain containing the double bond has five carbon atoms, so the compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the fourth carbon atom (rule 3), so the compound’s name is 4-methyl-2-pentene.
2. The longest chain containing the double bond has five carbon atoms, so the parent compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the third carbon atom (rule 3), so the compound’s name is 3-methyl-2-pentene.
Exercise \(1\)
Name each compound.
1. CH3CH2CH2CH2CH2CH=CHCH3
Just as there are cycloalkanes, there are cycloalkenes. These compounds are named like alkenes, but with the prefix cyclo- attached to the beginning of the parent alkene name.
Example \(2\)
Draw the structure for each compound.
1. 3-methyl-2-pentene
2. cyclohexene
Solution
1. First write the parent chain of five carbon atoms: C–C–C–C–C. Then add the double bond between the second and third carbon atoms:
Now place the methyl group on the third carbon atom and add enough hydrogen atoms to give each carbon atom a total of four bonds.
• First, consider what each of the three parts of the name means. Cyclo means a ring compound, hex means 6 carbon atoms, and -ene means a double bond.
Exercise \(2\)
Draw the structure for each compound.
1. 2-ethyl-1-hexene
2. cyclopentene
Key Takeaway
• Alkenes are hydrocarbons with a carbon-to-carbon double bond. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.01%3A_Alkenes-_Structures_and_Names.txt |
Learning Objectives
• Recognize that alkenes that can exist as cis-trans isomers.
• Classify isomers as cis or trans.
• Draw structures for cis-trans isomers given their names.
There is free rotation about the carbon-to-carbon single bonds (C–C) in alkanes. In contrast, the structure of alkenes requires that the carbon atoms of a double bond and the two atoms bonded to each carbon atom all lie in a single plane, and that each doubly bonded carbon atom lies in the center of a triangle. This part of the molecule’s structure is rigid; rotation about doubly bonded carbon atoms is not possible without rupturing the bond. Look at the two chlorinated hydrocarbons in Figure \(1\).
In 1,2-dichloroethane (part (a) of Figure \(1\)), there is free rotation about the C–C bond. The two models shown represent exactly the same molecule; they are not isomers. You can draw structural formulas that look different, but if you bear in mind the possibility of this free rotation about single bonds, you should recognize that these two structures represent the same molecule:
In 1,2-dichloroethene (Figure \(\PageIndex{1b}\)), however, restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond become significant. This leads to a special kind of isomerism. The isomer in which the two chlorine (Cl) atoms lie on the same side of the molecule is called the cis isomer (Latin cis, meaning “on this side”) and is named cis-1,2-dichloroethene. The isomer with the two Cl atoms on opposite sides of the molecule is the trans isomer (Latin trans, meaning “across”) and is named trans-1,2-dichloroethene. These two compounds are cis-trans isomers (or geometric isomers), compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule.
Consider the alkene with the condensed structural formula CH3CH=CHCH3. We could name it 2-butene, but there are actually two such compounds; the double bond results in cis-trans isomerism (Figure \(2\)).
Cis-2-butene has both methyl groups on the same side of the molecule. Trans-2-butene has the methyl groups on opposite sides of the molecule. Their structural formulas are as follows:
Figure \(3\): Models of (left) Cis-2-Butene and (right) Trans-2-Butene.
Note, however, that the presence of a double bond does not necessarily lead to cis-trans isomerism (Figure \(4\)). We can draw two seemingly different propenes:
Figure \(4\): Different views of the propene molecule (flip vertically). These are not isomers.
However, these two structures are not really different from each other. If you could pick up either molecule from the page and flip it over top to bottom, you would see that the two formulas are identical. Thus there are two requirements for cis-trans isomerism:
1. Rotation must be restricted in the molecule.
2. There must be two nonidentical groups on each doubly bonded carbon atom.
In these propene structures, the second requirement for cis-trans isomerism is not fulfilled. One of the doubly bonded carbon atoms does have two different groups attached, but the rules require that both carbon atoms have two different groups. In general, the following statements hold true in cis-trans isomerism:
• Alkenes with a C=CH2 unit do not exist as cis-trans isomers.
• Alkenes with a C=CR2 unit, where the two R groups are the same, do not exist as cis-trans isomers.
• Alkenes of the type R–CH=CH–R can exist as cis and trans isomers; cis if the two R groups are on the same side of the carbon-to-carbon double bond, and trans if the two R groups are on opposite sides of the carbon-to-carbon double bond.
Advanced Note: E/Z Isomerization
If a molecule has a C=C bond with one non-hydrogen group attached to each of the carbons, cis/trans nomenclature descried above is enough to describe it. However, if you have three different groups (or four), then the cis/trans approach is insufficient to describe the different isomers, since we do not know which two of the three groups are being described. For example, if you have a C=C bond, with a methyl group and a bromine on one carbon , and an ethyl group on the other, it is neither trans nor cis, since it is not clear whether the ethyl group is trans to the bromine or the methyl. This is addressed with a more advanced E,Z Convention [E] [E] [E] discussed elsewhere.
Cis-trans isomerism also occurs in cyclic compounds. In ring structures, groups are unable to rotate about any of the ring carbon–carbon bonds. Therefore, groups can be either on the same side of the ring (cis) or on opposite sides of the ring (trans). For our purposes here, we represent all cycloalkanes as planar structures, and we indicate the positions of the groups, either above or below the plane of the ring.
Example \(1\)
Which compounds can exist as cis-trans (geometric) isomers? Draw them.
1. CHCl=CHBr
2. CH2=CBrCH3
3. (CH3)2C=CHCH2CH3
4. CH3CH=CHCH2CH3
Solution
All four structures have a double bond and thus meet rule 1 for cis-trans isomerism.
1. This compound meets rule 2; it has two nonidentical groups on each carbon atom (H and Cl on one and H and Br on the other). It exists as both cis and trans isomers:
2. This compound has two hydrogen atoms on one of its doubly bonded carbon atoms; it fails rule 2 and does not exist as cis and trans isomers.
3. This compound has two methyl (CH3) groups on one of its doubly bonded carbon atoms. It fails rule 2 and does not exist as cis and trans isomers.
4. This compound meets rule 2; it has two nonidentical groups on each carbon atom and exists as both cis and trans isomers:
Exercise \(1\)
Which compounds can exist as cis-trans isomers? Draw them.
1. CH2=CHCH2CH2CH3
2. CH3CH=CHCH2CH3
3. CH3CH2CH=CHCH2CH3
Key Takeaway
• Cis-trans (geometric) isomerism exists when there is restricted rotation in a molecule and there are two nonidentical groups on each doubly bonded carbon atom. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.02%3A_Cis-Trans_Isomers_%28Geometric_Isomers%29.txt |
Learning Objectives
• To identify the physical properties of alkenes and describe trends in these properties.
The physical properties of alkenes are similar to those of the alkanes. The table at the start of the chapter shows that the boiling points of straight-chain alkenes increase with increasing molar mass, just as with alkanes. For molecules with the same number of carbon atoms and the same general shape, the boiling points usually differ only slightly, just as we would expect for substances whose molar mass differs by only 2 u (equivalent to two hydrogen atoms). Like other hydrocarbons, the alkenes are insoluble in water but soluble in organic solvents.
Looking Closer: Environmental Note
Alkenes occur widely in nature. Ripening fruits and vegetables give off ethylene, which triggers further ripening. Fruit processors artificially introduce ethylene to hasten the ripening process; exposure to as little as 0.1 mg of ethylene for 24 h can ripen 1 kg of tomatoes. Unfortunately, this process does not exactly duplicate the ripening process, and tomatoes picked green and treated this way don’t taste much like vine-ripened tomatoes fresh from the garden.
The bright red color of tomatoes is due to lycopene—a polyene.
Other alkenes that occur in nature include 1-octene, a constituent of lemon oil, and octadecene (C18H36) found in fish liver. Dienes (two double bonds) and polyenes (three or more double bonds) are also common. Butadiene (CH2=CHCH=CH2) is found in coffee. Lycopene and the carotenes are isomeric polyenes (C40H56) that give the attractive red, orange, and yellow colors to watermelons, tomatoes, carrots, and other fruits and vegetables. Vitamin A, essential to good vision, is derived from a carotene. The world would be a much less colorful place without alkenes.
Key Takeaway
• The physical properties of alkenes are much like those of the alkanes: their boiling points increase with increasing molar mass, and they are insoluble in water.
13.04: Chemical Properties of Alkenes
Learning Objectives
• To write equations for the addition reactions of alkenes with hydrogen, halogens, and water
Alkenes are valued mainly for addition reactions, in which one of the bonds in the double bond is broken. Each of the carbon atoms in the bond can then attach another atom or group while remaining joined to each other by a single bond. Perhaps the simplest addition reaction is hydrogenation—a reaction with hydrogen (H2) in the presence of a catalyst such as nickel (Ni) or platinum (Pt).
The product is an alkane having the same carbon skeleton as the alkene.
Alkenes also readily undergo halogenation—the addition of halogens. Indeed, the reaction with bromine (Br2) can be used to test for alkenes. Bromine solutions are brownish red. When we add a Br2 solution to an alkene, the color of the solution disappears because the alkene reacts with the bromine:
Another important addition reaction is that between an alkene and water to form an alcohol. This reaction, called hydration, requires a catalyst—usually a strong acid, such as sulfuric acid (H2SO4):
The hydration reaction is discussed later, where we deal with this reaction in the synthesis of alcohols.
Example \(1\)
Write the equation for the reaction between CH3CH=CHCH3 and each substance.
1. H2 (Ni catalyst)
2. Br2
3. H2O (H2SO4 catalyst)
Solution
In each reaction, the reagent adds across the double bond.
Exercise \(1\)
Write the equation for each reaction.
1. CH3CH2CH=CH2 with H2 (Ni catalyst)
2. CH3CH=CH2 with Cl2
3. CH3CH2CH=CHCH2CH3 with H2O (H2SO4 catalyst)
Key Takeaway
• Alkenes undergo addition reactions, adding such substances as hydrogen, bromine, and water across the carbon-to-carbon double bond. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.03%3A_Physical_Properties_of_Alkenes.txt |
Learning Objectives
• To draw structures for monomers that can undergo addition polymerization and for four-monomer-unit sections of an addition polymer.
The most important commercial reactions of alkenes are polymerizations, reactions in which small molecules, referred to in general as monomers (from the Greek monos, meaning “one,” and meros, meaning “parts”), are assembled into giant molecules referred to as polymers (from the Greek poly, meaning “many,” and meros, meaning “parts”). A polymer is as different from its monomer as a long strand of spaghetti is from a tiny speck of flour. For example, polyethylene, the familiar waxy material used to make plastic bags, is made from the monomer ethylene—a gas.
There are two general types of polymerization reactions: addition polymerization and condensation polymerization. In addition polymerization, the monomers add to one another in such a way that the polymer contains all the atoms of the starting monomers. Ethylene molecules are joined together in long chains. The polymerization can be represented by the reaction of a few monomer units:
The bond lines extending at the ends in the formula of the product indicate that the structure extends for many units in each direction. Notice that all the atoms—two carbon atoms and four hydrogen atoms—of each monomer molecule are incorporated into the polymer structure. Because displays such as the one above are cumbersome, the polymerization is often abbreviated as follows:
nCH2=CH2 [ CH2CH2 ] n
Many natural materials—such as proteins, cellulose and starch, and complex silicate minerals—are polymers. Artificial fibers, films, plastics, semisolid resins, and rubbers are also polymers. More than half the compounds produced by the chemical industry are synthetic polymers.
Some common addition polymers are listed in Table \(1\). Note that all the monomers have carbon-to-carbon double bonds. Many polymers are mundane (e.g., plastic bags, food wrap, toys, and tableware), but there are also polymers that conduct electricity, have amazing adhesive properties, or are stronger than steel but much lighter in weight.
Table \(1\): Some Addition Polymers
Monomer Polymer Polymer Name Some Uses
CH2=CH2 ~CH2CH2CH2CH2CH2CH2~ polyethylene plastic bags, bottles, toys, electrical insulation
CH2=CHCH3 polypropylene carpeting, bottles, luggage, exercise clothing
CH2=CHCl polyvinyl chloride bags for intravenous solutions, pipes, tubing, floor coverings
CF2=CF2 ~CF2CF2CF2CF2CF2CF2~ polytetrafluoroethylene nonstick coatings, electrical insulation
Medical Uses of Polymers
An interesting use of polymers is the replacement of diseased, worn out, or missing parts in the body. For example, about a 250,000 hip joints and 500,000 knees are replaced in US hospitals each year. The artificial ball-and-socket hip joints are made of a special steel (the ball) and plastic (the socket). People crippled by arthritis or injuries gain freedom of movement and relief from pain. Patients with heart and circulatory problems can be helped by replacing worn out heart valves with parts based on synthetic polymers. These are only a few of the many biomedical uses of polymers.
Key Takeaway
• Molecules having carbon-to-carbon double bonds can undergo addition polymerization.
13.06: Alkynes
Learning Objectives
• Describe the general physical and chemical properties of alkynes.
• Name alkynes given formulas and write formulas for alkynes given names.
The simplest alkyne—a hydrocarbon with carbon-to-carbon triple bond—has the molecular formula C2H2 and is known by its common name—acetylene (Figure \(1\)). Its structure is H–C≡C–H.
Acetylene is used in oxyacetylene torches for cutting and welding metals. The flame from such a torch can be very hot. Most acetylene, however, is converted to chemical intermediates that are used to make vinyl and acrylic plastics, fibers, resins, and a variety of other products.
Alkynes are similar to alkenes in both physical and chemical properties. For example, alkynes undergo many of the typical addition reactions of alkenes. The International Union of Pure and Applied Chemistry (IUPAC) names for alkynes parallel those of alkenes, except that the family ending is -yne rather than -ene. The IUPAC name for acetylene is ethyne. The names of other alkynes are illustrated in the following exercises.
Key Takeaway
• Alkynes are hydrocarbons with carbon-to-carbon triple bonds and properties much like those of alkenes. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.05%3A_Polymers.txt |
Learning Objectives
• To describe the bonding in benzene and the way typical reactions of benzene differ from those of the alkenes.
Next we consider a class of hydrocarbons with molecular formulas like those of unsaturated hydrocarbons, but which, unlike the alkenes, do not readily undergo addition reactions. These compounds comprise a distinct class, called aromatic hydrocarbons, with unique structures and properties. We start with the simplest of these compounds. Benzene (C6H6) is of great commercial importance, but it also has noteworthy health effects.
The formula C6H6 seems to indicate that benzene has a high degree of unsaturation. (Hexane, the saturated hydrocarbon with six carbon atoms has the formula C6H14—eight more hydrogen atoms than benzene.) However, despite the seeming low level of saturation, benzene is rather unreactive. It does not, for example, react readily with bromine, which, is a test for unsaturation.
Benzene is a liquid that smells like gasoline, boils at 80°C, and freezes at 5.5°C. It is the aromatic hydrocarbon produced in the largest volume. It was formerly used to decaffeinate coffee and was a significant component of many consumer products, such as paint strippers, rubber cements, and home dry-cleaning spot removers. It was removed from many product formulations in the 1950s, but others continued to use benzene in products until the 1970s when it was associated with leukemia deaths. Benzene is still important in industry as a precursor in the production of plastics (such as Styrofoam and nylon), drugs, detergents, synthetic rubber, pesticides, and dyes. It is used as a solvent for such things as cleaning and maintaining printing equipment and for adhesives such as those used to attach soles to shoes. Benzene is a natural constituent of petroleum products, but because it is a known carcinogen, its use as an additive in gasoline is now limited.
To explain the surprising properties of benzene, chemists suppose the molecule has a cyclic, hexagonal, planar structure of six carbon atoms with one hydrogen atom bonded to each. We can write a structure with alternate single and double bonds, either as a full structural formula or as a line-angle formula:
However, these structures do not explain the unique properties of benzene. Furthermore, experimental evidence indicates that all the carbon-to-carbon bonds in benzene are equivalent, and the molecule is unusually stable. Chemists often represent benzene as a hexagon with an inscribed circle:
The inner circle indicates that the valence electrons are shared equally by all six carbon atoms (that is, the electrons are delocalized, or spread out, over all the carbon atoms). It is understood that each corner of the hexagon is occupied by one carbon atom, and each carbon atom has one hydrogen atom attached to it. Any other atom or groups of atoms substituted for a hydrogen atom must be shown bonded to a particular corner of the hexagon. We use this modern symbolism, but many scientists still use the earlier structure with alternate double and single bonds.
To Your Health: Benzene and Us
Most of the benzene used commercially comes from petroleum. It is employed as a starting material for the production of detergents, drugs, dyes, insecticides, and plastics. Once widely used as an organic solvent, benzene is now known to have both short- and long-term toxic effects. The inhalation of large concentrations can cause nausea and even death due to respiratory or heart failure, while repeated exposure leads to a progressive disease in which the ability of the bone marrow to make new blood cells is eventually destroyed. This results in a condition called aplastic anemia, in which there is a decrease in the numbers of both the red and white blood cells.
Key Takeaway
• Aromatic hydrocarbons appear to be unsaturated, but they have a special type of bonding and do not undergo addition reactions. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.07%3A_Aromatic_Compounds-_Benzene.txt |
Learning Objectives
• Recognize aromatic compounds from structural formulas.
• Name aromatic compounds given formulas.
• Write formulas for aromatic compounds given their names.
Historically, benzene-like substances were called aromatic hydrocarbons because they had distinctive aromas. Today, an aromatic compound is any compound that contains a benzene ring or has certain benzene-like properties (but not necessarily a strong aroma). You can recognize the aromatic compounds in this text by the presence of one or more benzene rings in their structure. Some representative aromatic compounds and their uses are listed in Table \(1\), where the benzene ring is represented as C6H5.
Table \(1\): Some Representative Aromatic Compounds
Name Structure Typical Uses
aniline C6H5–NH2 starting material for the synthesis of dyes, drugs, resins, varnishes, perfumes; solvent; vulcanizing rubber
benzoic acid C6H5–COOH food preservative; starting material for the synthesis of dyes and other organic compounds; curing of tobacco
bromobenzene C6H5–Br starting material for the synthesis of many other aromatic compounds; solvent; motor oil additive
nitrobenzene C6H5–NO2 starting material for the synthesis of aniline; solvent for cellulose nitrate; in soaps and shoe polish
phenol C6H5–OH disinfectant; starting material for the synthesis of resins, drugs, and other organic compounds
toluene C6H5–CH3 solvent; gasoline octane booster; starting material for the synthesis of benzoic acid, benzaldehyde, and many other organic compounds
Example \(1\)
Which compounds are aromatic?
Solution
1. The compound has a benzene ring (with a chlorine atom substituted for one of the hydrogen atoms); it is aromatic.
2. The compound is cyclic, but it does not have a benzene ring; it is not aromatic.
3. The compound has a benzene ring (with a propyl group substituted for one of the hydrogen atoms); it is aromatic.
4. The compound is cyclic, but it does not have a benzene ring; it is not aromatic.
Exercise \(1\)
Which compounds are aromatic?
In the International Union of Pure and Applied Chemistry (IUPAC) system, aromatic hydrocarbons are named as derivatives of benzene. Figure \(1\) shows four examples. In these structures, it is immaterial whether the single substituent is written at the top, side, or bottom of the ring: a hexagon is symmetrical, and therefore all positions are equivalent.
Although some compounds are referred to exclusively by IUPAC names, some are more frequently denoted by common names, as is indicated in Table \(1\).
When there is more than one substituent, the corners of the hexagon are no longer equivalent, so we must designate the relative positions. There are three possible disubstituted benzenes, and we can use numbers to distinguish them (Figure \(2\)). We start numbering at the carbon atom to which one of the groups is attached and count toward the carbon atom that bears the other substituent group by the shortest path.
In Figure \(2\), common names are also used: the prefix ortho (o-) for 1,2-disubstitution, meta (m-) for 1,3-disubstitution, and para (p-) for 1,4-disubstitution. The substituent names are listed in alphabetical order. The first substituent is given the lowest number. When a common name is used, the carbon atom that bears the group responsible for the name is given the number 1:
Example \(2\)
Name each compound using both the common name and the IUPAC name.
Solution
1. The benzene ring has two chlorine atoms (dichloro) in the first and second positions. The compound is o-dichlorobenzene or 1,2-dichlorobenzene.
2. The benzene ring has a methyl (CH3) group. The compound is therefore named as a derivative of toluene. The bromine atom is on the fourth carbon atom, counting from the methyl group. The compound is p-bromotoluene or 4-bromotoluene.
3. The benzene ring has two nitro (NO2) groups in the first and third positions. It is m-dinitrobenzene or 1,3-dinitrobenzene.
4. Note: The nitro (NO2) group is a common substituent in aromatic compounds. Many nitro compounds are explosive, most notably 2,4,6-trinitrotoluene (TNT).
Exercise \(2\)
Name each compound using both the common name and the IUPAC name.
• Sometimes an aromatic group is found as a substituent bonded to a nonaromatic entity or to another aromatic ring. The group of atoms remaining when a hydrogen atom is removed from an aromatic compound is called an aryl group. The most common aryl group is derived from benzene (C6H6) by removing one hydrogen atom (C6H5) and is called a phenyl group, from pheno, an old name for benzene.
Polycyclic Aromatic Hydrocarbons
Some common aromatic hydrocarbons consist of fused benzene rings—rings that share a common side. These compounds are called polycyclic aromatic hydrocarbons (PAHs).
The three examples shown here are colorless, crystalline solids generally obtained from coal tar. Naphthalene has a pungent odor and is used in mothballs. Anthracene is used in the manufacture of certain dyes. Steroids, a large group of naturally occurring substances, contain the phenanthrene structure.
To Your Health: Polycyclic Aromatic Hydrocarbons and Cancer
The intense heating required for distilling coal tar results in the formation of PAHs. For many years, it has been known that workers in coal-tar refineries are susceptible to a type of skin cancer known as tar cancer. Investigations have shown that a number of PAHs are carcinogens. One of the most active carcinogenic compounds, benzopyrene, occurs in coal tar and has also been isolated from cigarette smoke, automobile exhaust gases, and charcoal-broiled steaks. It is estimated that more than 1,000 t of benzopyrene are emitted into the air over the United States each year. Only a few milligrams of benzopyrene per kilogram of body weight are required to induce cancer in experimental animals.
Biologically Important Compounds with Benzene Rings
Substances containing the benzene ring are common in both animals and plants, although they are more abundant in the latter. Plants can synthesize the benzene ring from carbon dioxide, water, and inorganic materials. Animals cannot synthesize it, but they are dependent on certain aromatic compounds for survival and therefore must obtain them from food. Phenylalanine, tyrosine, and tryptophan (essential amino acids) and vitamins K, B2 (riboflavin), and B9 (folic acid) all contain the benzene ring. Many important drugs, a few of which are shown in Table \(2\), also feature a benzene ring.
So far we have studied only aromatic compounds with carbon-containing rings. However, many cyclic compounds have an element other than carbon atoms in the ring. These compounds, called heterocyclic compounds, are discussed later. Some of these are heterocyclic aromatic compounds.
Table \(2\): Some Drugs That Contain a Benzene Ring
Name Structure
aspirin
acetaminophen
ibuprofen
amphetamine
sulfanilamide
Key Takeaway
• Aromatic compounds contain a benzene ring or have certain benzene-like properties; for our purposes, you can recognize aromatic compounds by the presence of one or more benzene rings in their structure. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.08%3A_Structure_and_Nomenclature_of_Aromatic_Compounds.txt |
Concept Review Exercises
1. Briefly identify the important distinctions between a saturated hydrocarbon and an unsaturated hydrocarbon.
2. Briefly identify the important distinctions between an alkene and an alkane.
3. Classify each compound as saturated or unsaturated. Identify each as an alkane, an alkene, or an alkyne.
1. CH3CH2C≡CCH3
Answers
1. Unsaturated hydrocarbons have double or triple bonds and are quite reactive; saturated hydrocarbons have only single bonds and are rather unreactive.
2. An alkene has a double bond; an alkane has single bonds only.
1. saturated; alkane
2. unsaturated; alkyne
3. unsaturated; alkene
Exercises
1. Draw the structure for each compound.
1. 2-methyl-2-pentene
2. 2,3-dimethyl-1-butene
3. cyclohexene
2. Draw the structure for each compound.
1. 5-methyl-1-hexene
2. 3-ethyl-2-pentene
3. 4-methyl-2-hexene
3. Name each compound according to the IUPAC system.
4. Name each compound according to the IUPAC system.
Answers
1. 2-methyl-1-pentene
2. 2-methyl-2-pentene
3. 2,5-dimethyl-2-hexene
Concept Review Exercises
1. What are cis-trans (geometric) isomers? What two types of compounds can exhibit cis-trans isomerism?
2. Classify each compound as a cis isomer, a trans isomer, or neither.
Answers
1. Cis-trans isomers are compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule. Alkenes and cyclic compounds can exhibit cis-trans isomerism.
1. trans (the two hydrogen atoms are on opposite sides)
2. cis (the two hydrogen atoms are on the same side, as are the two ethyl groups)
3. cis (the two ethyl groups are on the same side)
4. neither (flipping the bond does not change the molecule. There are no isomers for this molecule)
Exercises
1. Draw the structures of the cis-trans isomers for each compound. Label them cis and trans. If no cis-trans isomers exist, write none.
1. 2-bromo-2-pentene
2. 3-hexene
3. 4-methyl-2-pentene
4. 1,1-dibromo-1-butene
5. 2-butenoic acid (CH3CH=CHCOOH)
2. Draw the structures of the cis-trans isomers for each compound. Label them cis and trans. If no cis-trans isomers exist, write none.
1. 2,3-dimethyl-2-pentene
2. 1,1-dimethyl-2-ethylcyclopropane
3. 1,2-dimethylcyclohexane
4. 5-methyl-2-hexene
5. 1,2,3-trimethylcyclopropane
Answer
1. a: none. There are two distinct geometric isomers, but since there are there are four different groups off the double bond, these are both cis/trans isomers (they are technically E/Z isomers discussed elsewhere).
b:
c:
d:
e:
Concept Review Exercises
1. Briefly describe the physical properties of alkenes. How do these properties compare to those of the alkanes?
2. Without consulting tables, arrange the following alkenes in order of increasing boiling point: 1-butene, ethene, 1-hexene, and propene.
Answers
1. Alkenes have physical properties (low boiling points, insoluble in water) quite similar to those of their corresponding alkanes.
2. ethene < propene < 1-butene < 1-hexene
Exercises
1. Without referring to a table or other reference, predict which member of each pair has the higher boiling point.
1. 1-pentene or 1-butene
2. 3-heptene or 3-nonene
2. Which is a good solvent for cyclohexene, pentane or water?
1. 1-pentene
2. 3-nonene
Concept Review Exercises
1. What is the principal difference in properties between alkenes and alkanes? How are they alike?
2. If C12H24 reacts with HBr in an addition reaction, what is the molecular formula of the product?
Answers
1. Alkenes undergo addition reactions; alkanes do not. Both burn.
2. C12H24Br2
Exercises
1. Complete each equation.
1. (CH3) 2C=CH2 + Br2
2. $\mathrm{CH_2\textrm{=C}(CH_3)CH_2CH_3 + H_2 \xrightarrow{Ni}}$
2. Complete each equation.
1. $\mathrm{CH_2\textrm{=CHCH=C}H_2 + 2H_2\xrightarrow{Ni}}$
2. $\mathrm{(CH_3)_2\textrm{C=C}(CH_3)_2 + H_2O \xrightarrow{H_2SO_4}}$
Answer
1. (CH3)2CBrCH2Br
2. CH3CH(CH3)CH2CH3
Concept Review Exercises
1. What is a monomer? What is a polymer? How do polymer molecules differ from the molecules we have discussed in earlier sections of this chapter?
2. What is addition polymerization? What structural feature usually characterizes molecules used as monomers in addition polymerization?
3. What is the molecular formula of a polymer molecule formed by the addition polymerization of 175 molecules of vinyl chloride (CH2=CHCl)?
Answers
1. Monomers are small molecules that can be assembled into giant molecules referred to as polymers, which are much larger than the molecules we discussed earlier in this chapter.
2. In addition polymerization, the monomers add to one another in such a way that the polymer contains all the atoms of the starting monomers.
3. C350H525Cl175
Exercises
1. Write the condensed structural formula of the monomer from which Saran is formed. A segment of the Saran molecule has the following structure: CH2CCl2CH2CCl2CH2CCl2CH2CCl2.
2. Write the condensed structural formula for the section of a molecule formed from four units of the monomer CH2=CHF.
1. H2C=CCl2
Concept Review Exercises
1. Briefly identify the important differences between an alkene and an alkyne. How are they similar?
2. The alkene (CH3)2CHCH2CH=CH2 is named 4-methyl-1-pentene. What is the name of (CH3)2CHCH2C≡CH?
3. Do alkynes show cis-trans isomerism? Explain.
Answers
1. Alkenes have double bonds; alkynes have triple bonds. Both undergo addition reactions.
2. 4-methyl-1-pentyne
3. No; a triply bonded carbon atom can form only one other bond. It would have to have two groups attached to show cis-trans isomerism.
Exercises
1. Draw the structure for each compound.
1. acetylene
2. 3-methyl-1-hexyne
2. Draw the structure for each compound.
1. 4-methyl-2-hexyne
2. 3-octyne
3. Name each alkyne.
1. CH3CH2CH2C≡CH
2. CH3CH2CH2C≡CCH3
1. H–C≡C–H
1. 1-pentyne
2. 2-hexyne
Concept Review Exercises
1. How do the typical reactions of benzene differ from those of the alkenes?
2. Briefly describe the bonding in benzene.
3. What does the circle mean in the chemist’s representation of benzene?
Answers
1. Benzene is rather unreactive toward addition reactions compared to an alkene.
2. Valence electrons are shared equally by all six carbon atoms (that is, the electrons are delocalized).
3. The six electrons are shared equally by all six carbon atoms.
Exercises
1. Draw the structure of benzene as if it had alternate single and double bonds.
2. Draw the structure of benzene as chemists usually represent it today.
Concept Review Exercises
1. Briefly identify the important characteristics of an aromatic compound.
2. What is meant by the prefixes meta, ortho, or para? Give the name and draw the structure for a compound that illustrates each.
3. What is a phenyl group? Give the structure for 3-phenyloctane.
Answers
1. An aromatic compound is any compound that contains a benzene ring or has certain benzene-like properties.
2. meta = 1,3 disubstitution; (answers will vary)
ortho = 1,2 disubstitution
para = 1,4 disubstitution or 1-bromo-4-chlorobenzene
3. phenyl group: C6H5 or
3-phenyloctane:
Exercises
1. Is each compound aromatic?
2. Is each compound aromatic?
3. Draw the structure for each compound.
1. toluene
2. m-diethylbenzene
3. 3,5-dinitrotoluene
4. Draw the structure for each compound.
1. p-dichlorobenzene
2. naphthalene
3. 1,2,4-trimethylbenzene
5. Name each compound with its IUPAC name.
6. Name each compound with its IUPAC name.
Answers
1. yes
2. no
1. ethylbenzene
2. isopropylbenzene
3. o-bromotoluene
4. 3,5-dichlorotoluene
Additional Exercises
1. Classify each compound as saturated or unsaturated.
1. CH3C≡CCH3
2. Classify each compound as saturated or unsaturated.
3. Give the molecular formula for each compound.
4. When three isomeric pentenes—X, Y, and Z—are hydrogenated, all three form 2-methylbutane. The addition of Cl2 to Y gives 1,2-dichloro-3-methylbutane, and the addition of Cl2 to Z gives 1,2-dichloro-2-methylbutane. Draw the original structures for X, Y, and Z.
5. Pentane and 1-pentene are both colorless, low-boiling liquids. Describe a simple test that distinguishes the two compounds. Indicate what you would observe.
6. Draw and name all the alkene cis-trans isomers corresponding to the molecular formula C5H10. (Hint: there are only two.)
7. The complete combustion of benzene forms carbon dioxide and water:
C6H6 + O2 → CO2 + H2O
Balance the equation. What mass, in grams, of carbon dioxide is formed by the complete combustion of 39.0 g of benzene?
8. Describe a physiological effect of some PAHs.
9. What are some of the hazards associated with the use of benzene?
10. What is wrong with each name? Draw the structure and give the correct name for each compound.
1. 2-methyl-4-heptene
2. 2-ethyl-2-hexene
3. 2,2-dimethyl-3-pentene
11. What is wrong with each name?
1. 2-bromobenzene
2. 3,3-dichlorotoluene
3. 1,4-dimethylnitrobenzene
12. Following are line-angle formulas for three compounds. Draw the structure and give the name for each.
13. Following are ball-and-stick molecular models for three compounds (blue balls represent H atoms; red balls are C atoms). Write the condensed structural formula and give the name for each.
Answers
1.
1. unsaturated
2. unsaturated
1.
1. C6H10
2. C4H8
1. Add bromine solution (reddish-brown) to each. Pentane will not react, and the reddish-brown color persists; 1-pentene will react, leaving a colorless solution.
1. 2C6H6 + 15O2 → 12CO2 + 6H2O; 132 g
1. carcinogenic, flammable
1.
1. number not needed
2. can’t have two groups on one carbon atom on a benzene ring
3. can’t have a substituent on the same carbon atom as the nitro group
1.
1. CH3CH=CHCH2CH2CH3; 2-hexene
13.S: Unsaturated and Aromatic Hydrocarbons (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Any hydrocarbon containing either a double or triple bond is an unsaturated hydrocarbon. Alkenes have a carbon-to-carbon double bond. The general formula for alkenes with one double bond is CnH2n. Alkenes can be straight chain, branched chain, or cyclic. Simple alkenes often have common names, but all alkenes can be named by the system of the International Union of Pure and Applied Chemistry.
Cis-trans isomers (or geometric isomers) are characterized by molecules that differ only in their configuration around a rigid part of the structure, such as a carbon–to-carbon double bond or a ring. The molecule having two identical (or closely related) atoms or groups on the same side is the cis isomer; the one having the two groups on opposite sides is the trans isomer.
The physical properties of alkenes are quite similar to those of alkanes. Like other hydrocarbons, alkenes are insoluble in water but soluble in organic solvents.
More reactive than alkanes, alkenes undergo addition reactions across the double bond:
• Addition of hydrogen (hydrogenation):
CH2=CH2 + H2 → CH3CH3
• Addition of halogen (halogenation):
CH2=CH2 + X2XCH2CH2X
where X = F, Cl, Br, or I.
• Addition of water (hydration):
CH2=CH2 + HOH → HCH2CH2OH
Alkenes also undergo addition polymerization, molecules joining together to form long-chain molecules.
…CH2=CH2 + CH2=CH2 + CH2=CH2 +…→…CH2CH2–CH2CH2–CH2CH2–…
The reactant units are monomers, and the product is a polymer.
Alkynes have a carbon-to-carbon triple bond. The general formula for alkynes is CnH2n − 2. The properties of alkynes are quite similar to those of alkenes. They are named much like alkenes but with the ending -yne.
The cyclic hydrocarbon benzene (C6H6) has a ring of carbon atoms. The molecule seems to be unsaturated, but it does not undergo the typical reactions expected of alkenes. The electrons that might be fixed in three double bonds are instead delocalized over all six carbon atoms.
A hydrocarbon containing one or more benzene rings (or other similarly stable electron arrangements) is an aromatic hydrocarbon, and any related substance is an aromatic compound. One or more of the hydrogen atoms on a benzene ring can be replaced by other atoms. When two hydrogen atoms are replaced, the product name is based on the relative position of the replacement atoms (or atom groups). A 1,2-disubstituted benzene is designated as an ortho (o-) isomer; 1,3-, a meta (m-) isomer; and 1,4-, a para (p-) isomer. An aromatic group as a substituent is called an aryl group.
A polycyclic aromatic hydrocarbon (PAH) has fused benzene rings sharing a common side. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/13%3A_Unsaturated_and_Aromatic_Hydrocarbons/13.E%3A_Unsaturated_and_Aromatic_Hydrocarbons_%28Exercises%29.txt |
Ethanol and resveratrol, a phenol, are representatives of two of the families of oxygen-containing compounds that we consider in this chapter. Two other classes, aldehydes and ketones, are formed by the oxidation of alcohols. Ethers, another class, are made by the dehydration of alcohols.
• 14.0: Prelude to Organic Compounds of Oxygen
One of the more familiar chemical compounds on Earth is ethyl alcohol (ethanol). As the intoxicant in alcoholic beverages, ethanol is often simply called alcohol. If ethanol is diluted, as it is in wine, beer, or mixed drinks with about 1 oz of liquor, and if it is consumed in small quantities, it is relatively safe. In excess—four or more drinks in a few hours—it causes intoxication, which is characterized by a loss of coordination, nausea and vomiting, and memory blackouts.
• 14.1: Organic Compounds with Functional Groups
The functional group, a structural arrangement of atoms and/or bonds, is largely responsible for the properties of organic compound families.
• 14.2: Alcohols - Nomenclature and Classification
In the IUPAC system, alcohols are named by changing the ending of the parent alkane name to -ol. Alcohols are classified according to the number of carbon atoms attached to the carbon atom that is attached to the OH group.
• 14.3: Physical Properties of Alcohols
Alcohols have higher boiling points than do ethers and alkanes of similar molar masses because the OH group allows alcohol molecules to engage in hydrogen bonding. Alcohols of four or fewer carbon atoms are soluble in water because the alcohol molecules engage in hydrogen bonding with water molecules; comparable alkane molecules cannot engage in hydrogen bonding.
• 14.4: Reactions that Form Alcohols
Many alcohols are made by the hydration of alkenes. Ethanol can be made by the fermentation of sugars or starch from various sources.
• 14.5: Reactions of Alcohols
Alcohols can be dehydrated to form either alkenes (higher temperature, excess acid) or ethers (lower temperature, excess alcohol). Primary alcohols are oxidized to form aldehydes. Secondary alcohols are oxidized to form ketones. Tertiary alcohols are not readily oxidized.
• 14.6: Glycols and Glycerol
Glycols are alcohols with two OH groups on adjacent carbon atoms. Glycerol is the most important trihydroxy alcohol.
• 14.7: Phenols
Phenols are compounds in which an OH group is attached directly to an aromatic ring. Many phenols are used as antiseptics.
• 14.8: Ethers
To give ethers common names, simply name the groups attached to the oxygen atom, followed by the generic name ether. If both groups are the same, the group name should be preceded by the prefix di-. Ether molecules have no OH group and thus no intermolecular hydrogen bonding. Ethers therefore have quite low boiling points for a given molar mass. Ether molecules have an oxygen atom and can engage in hydrogen bonding with water molecules. An ether molecule has about the same solubility in water as
• 14.9: Aldehydes and Ketones- Structure and Names
The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The common names of ketones, like those of ethers, consist of the names of the groups attached to the carbonyl group, followed by the word ketone. Stem names of aldehydes and ketones are derived from those of the parent alkanes, using an -al ending for an aldehydes and an -one ending for a ketone.
• 14.10: Properties of Aldehydes and Ketones
The polar carbon-to-oxygen double bond causes aldehydes and ketones to have higher boiling points than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols that engage in intermolecular hydrogen bonding. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation.
• 14.11: Organic Sulfur Compounds
Thiols, thioethers, and disulfides are common in biological compounds.
• 14.E: Organic Compounds of Oxygen (Exercises)
• 14.S: Organic Compounds of Oxygen (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the summary and ask yourself how they relate to the topics in the chapter.
14: Organic Compounds of Oxygen
One of the more familiar chemical compounds on Earth is ethyl alcohol (ethanol). As the intoxicant in alcoholic beverages, ethanol is often simply called alcohol. If ethanol is diluted, as it is in wine, beer, or mixed drinks with about 1 oz of liquor, and if it is consumed in small quantities, it is relatively safe. In excess—four or more drinks in a few hours—it causes intoxication, which is characterized by a loss of coordination, nausea and vomiting, and memory blackouts.
Excessive ingestion of ethanol over a long period of time leads to cirrhosis of the liver, alteration of brain cell function, nerve damage, and strong physiological addiction. Alcoholism—an addiction to ethanol—is the most serious drug problem in the United States. Heavy drinking shortens a person’s life span by contributing to diseases of the liver, the cardiovascular system, and virtually every other organ of the body.
In small quantities—one or two drinks a day—ethanol might promote health. In addition to the possible benefits of modest amounts of ethanol, a chemical in red wines, resveratrol, is thought to lower the risk of heart disease. Resveratrol, found in red grapes, is an antioxidant. It inhibits the oxidation of cholesterol and subsequent clogging of the arteries. One need not drink wine to get the benefits of resveratrol, however. It can be obtained by eating the grapes or drinking red grape juice. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.00%3A_Prelude_to_Organic_Compounds_of_Oxygen.txt |
Learning Objectives
• to describe functional groups and explain why they are useful in the study of organic chemistry.
Previously, we considered several kinds of hydrocarbons. Now we examine some of the many organic compounds that contain functional groups. We first introduced the idea of the functional group, a specific structural arrangement of atoms or bonds that imparts a characteristic chemical reactivity to the molecule. If you understand the behavior of a particular functional group, you will know a great deal about the general properties of that class of compounds. In this chapter, we make a brief yet systematic study of some of organic compound families. Each family is based on a common, simple functional group that contains an oxygen atom or a nitrogen atom. Some common functional groups are listed in Table \(1\).
Table \(1\): Selected Organic Functional Groups
Name of Family General Formula Functional Group Suffix*
alkane RH none -ane
alkene R2C=CR2 -ene
alkyne RC≡CR –C≡C– -yne
alcohol ROH –OH -ol
thiol RSH –SH -thiol
ether ROR –O– ether
aldehyde -al
ketone -one
carboxylic acid -oic acid
*Ethers do not have a suffix in their common name; all ethers end with the word ether.
Summary
The functional group, a structural arrangement of atoms and/or bonds, is largely responsible for the properties of organic compound families.
14.02: Alcohols - Nomenclature and Classification
Learning Objectives
• Identify the general structure for an alcohol.
• Identify the structural feature that classifies alcohols as primary, secondary, or tertiary.
• Name alcohols with both common names and IUPAC names
An alcohol is an organic compound with a hydroxyl (OH) functional group on an aliphatic carbon atom. Because OH is the functional group of all alcohols, we often represent alcohols by the general formula ROH, where R is an alkyl group. Alcohols are common in nature. Most people are familiar with ethyl alcohol (ethanol), the active ingredient in alcoholic beverages, but this compound is only one of a family of organic compounds known as alcohols. The family also includes such familiar substances as cholesterol and the carbohydrates. Methanol (CH3OH) and ethanol (CH3CH2OH) are the first two members of the homologous series of alcohols.
Nomenclature of Alcohols
Alcohols with one to four carbon atoms are frequently called by common names, in which the name of the alkyl group is followed by the word alcohol:
According to the International Union of Pure and Applied Chemistry (IUPAC), alcohols are named by changing the ending of the parent alkane name to -ol. Here are some basic IUPAC rules for naming alcohols:
1. The longest continuous chain (LCC) of carbon atoms containing the OH group is taken as the parent compound—an alkane with the same number of carbon atoms. The chain is numbered from the end nearest the OH group.
2. The number that indicates the position of the OH group is prefixed to the name of the parent hydrocarbon, and the -e ending of the parent alkane is replaced by the suffix -ol. (In cyclic alcohols, the carbon atom bearing the OH group is designated C1, but the 1 is not used in the name.) Substituents are named and numbered as in alkanes.
3. If more than one OH group appears in the same molecule (polyhydroxy alcohols), suffixes such as -diol and -triol are used. In these cases, the -e ending of the parent alkane is retained.
Figure \(1\) shows some examples of the application of these rules.
Example \(1\)
Give the IUPAC name for each compound.
• HOCH2CH2CH2CH2CH2OH
Solution
1. Ten carbon atoms in the LCC makes the compound a derivative of decane (rule 1), and the OH on the third carbon atom makes it a 3-decanol (rule 2).
The carbon atoms are numbered from the end closest to the OH group. That fixes the two methyl (CH3) groups at the sixth and eighth positions. The name is 6,8-dimethyl-3-decanol (not 3,5-dimethyl-8-decanol).
2. Five carbon atoms in the LCC make the compound a derivative of pentane. Two OH groups on the first and fifth carbon atoms make the compound a diol and give the name 1,5-pentanediol (rule 3).
Exercise \(1\)
Give the IUPAC name for each compound.
Example \(2\)
Draw the structure for each compound.
1. 2-hexanol
2. 3-methyl-2-pentanol
Solution
1. The ending -ol indicates an alcohol (the OH functional group), and the hex- stem tells us that there are six carbon atoms in the LCC. We start by drawing a chain of six carbon atoms: –C–C–C–C–C–C–.
The 2 indicates that the OH group is attached to the second carbon atom.
Finally, we add enough hydrogen atoms to give each carbon atom four bonds.
• The numbers indicate that there is a methyl (CH3) group on the third carbon atom and an OH group on the second carbon atom.
Exercise \(2\)
Draw the structure for each compound.
1. 3-heptanol
• 2-methyl-3-hexanol
Classification of Alcohols
Some of the properties of alcohols depend on the number of carbon atoms attached to the specific carbon atom that is attached to the OH group. Alcohols can be grouped into three classes on this basis.
• A primary (1°) alcohol is one in which the carbon atom (in red) with the OH group is attached to one other carbon atom (in blue). Its general formula is RCH2OH.
• A secondary (2°) alcohol is one in which the carbon atom (in red) with the OH group is attached to two other carbon atoms (in blue). Its general formula is R2CHOH.
• A tertiary (3°) alcohol is one in which the carbon atom (in red) with the OH group is attached to three other carbon atoms (in blue). Its general formula is R3COH.
Table \(1\) names and classifies some of the simpler alcohols. Some of the common names reflect a compound’s classification as secondary (sec-) or tertiary (tert-). These designations are not used in the IUPAC nomenclature system for alcohols. Note that there are four butyl alcohols in the table, corresponding to the four butyl groups: the butyl group (CH3CH2CH2CH2) discussed before, and three others:
Table \(1\): Classification and Nomenclature of Some Alcohols
Condensed Structural Formula Class of Alcohol Common Name IUPAC Name
CH3OH methyl alcohol methanol
CH3CH2OH primary ethyl alcohol ethanol
CH3CH2CH2OH primary propyl alcohol 1-propanol
(CH3)2CHOH secondary isopropyl alcohol 2-propanol
CH3CH2CH2CH2OH primary butyl alcohol 1-butanol
CH3CH2CHOHCH3 secondary sec-butyl alcohol 2-butanol
(CH3)2CHCH2OH primary isobutyl alcohol 2-methyl-1-propanol
(CH3)3COH tertiary tert-butyl alcohol 2-methyl-2-propanol
secondary cyclohexyl alcohol cyclohexanol
Summary
In the IUPAC system, alcohols are named by changing the ending of the parent alkane name to -ol. Alcohols are classified according to the number of carbon atoms attached to the carbon atom that is attached to the OH group. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.01%3A_Organic_Compounds_with_Functional_Groups.txt |
Learning Objectives
• Explain why the boiling points of alcohols are higher than those of ethers and alkanes of similar molar masses.
• Explain why alcohols and ethers of four or fewer carbon atoms are soluble in water while comparable alkanes are not soluble.
Alcohols can be considered derivatives of water (H2O; also written as HOH).
Like the H–O–H bond in water, the R–O–H bond is bent, and alcohol molecules are polar. This relationship is particularly apparent in small molecules and reflected in the physical and chemical properties of alcohols with low molar mass. Replacing a hydrogen atom from an alkane with an OH group allows the molecules to associate through hydrogen bonding (Figure \(1\)).
Recall that physical properties are determined to a large extent by the type of intermolecular forces. Table \(1\) lists the molar masses and the boiling points of some common compounds. The table shows that substances with similar molar masses can have quite different boiling points.
Table \(1\): Comparison of Boiling Points and Molar Masses
Formula Name Molar Mass Boiling Point (°C)
CH4 methane 16 –164
HOH water 18 100
C2H6 ethane 30 –89
CH3OH methanol 32 65
C3H8 propane 44 –42
CH3CH2OH ethanol 46 78
C4H10 butane 58 –1
CH3CH2CH2OH 1-propanol 60 97
Alkanes are nonpolar and are thus associated only through relatively weak dispersion forces. Alkanes with one to four carbon atoms are gases at room temperature. In contrast, even methanol (with one carbon atom) is a liquid at room temperature. Hydrogen bonding greatly increases the boiling points of alcohols compared to hydrocarbons of comparable molar mass. The boiling point is a rough measure of the amount of energy necessary to separate a liquid molecule from its nearest neighbors. If the molecules interact through hydrogen bonding, a relatively large quantity of energy must be supplied to break those intermolecular attractions. Only then can the molecule escape from the liquid into the gaseous state.
Alcohols can also engage in hydrogen bonding with water molecules (Figure \(2\)). Thus, whereas the hydrocarbons are insoluble in water, alcohols with one to three carbon atoms are completely soluble. As the length of the chain increases, however, the solubility of alcohols in water decreases; the molecules become more like hydrocarbons and less like water. The alcohol 1-decanol (CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2OH) is essentially insoluble in water. We frequently find that the borderline of solubility in a family of organic compounds occurs at four or five carbon atoms.
Summary
Alcohols have higher boiling points than do ethers and alkanes of similar molar masses because the OH group allows alcohol molecules to engage in hydrogen bonding. Alcohols of four or fewer carbon atoms are soluble in water because the alcohol molecules engage in hydrogen bonding with water molecules; comparable alkane molecules cannot engage in hydrogen bonding.
14.04: Reactions that Form Alcohols
Learning Objectives
• To describe how to prepare alcohols from alkenes
Methanol is prepared by combining hydrogen gas and carbon monoxide at high temperatures and pressures in the presence of a catalyst composed of zinc oxide (ZnO) and chromium oxide (Cr2O3) catalyst:
$\mathrm{2H_2 + CO \underset{ZnO,\: Cr_2O_3}{\xrightarrow{200\: atm,\: 350^\circ C}} CH_3OH} \nonumber$
Methanol is an important solvent and is used as an automotive fuel, either as the pure liquid—as in some racing cars—or as an additive in gasoline.
Nearly 2 billion gallons of methanol are produced each year in the United States by the catalytic reduction of carbon monoxide with hydrogen gas.
Many simple alcohols are made by the hydration of alkenes. Ethanol is made by the hydration of ethylene in the presence of a catalyst such as sulfuric acid (H2SO4).
In a similar manner, isopropyl alcohol is produced by the addition of water to propene (propylene).
Additional Exercise $1$ describes how to use a generalization called Markovnikov’s rule to predict the results when the addition of water to an alcohol has two possible products.
Example $1$
Write the equation for the reaction of 2-butene with water to form 2-butanol. Indicate that sulfuric acid is used as a catalyst.
Solution
First write the condensed structural formula of 2-butene and indicate that it reacts with water. Then write the condensed structural formula of 2-butanol after the reaction arrow to indicate that it is the product. Finally, write the formula for the catalyst above the arrow.
Exercise $1$
Write the equation for the reaction of cyclopentene with water to form cyclopentanol. Indicate that phosphoric acid (H3PO4) is used as a catalyst.
Many OH compounds in living systems are formed by alkene hydration. Here is an example that occurs in the Krebs cycle: fumarate is hydrated to form malate.
In addition to its preparation from ethylene, ethanol is made by the fermentation of sugars or starch from various sources (potatoes, corn, wheat, rice, etc.). Fermentation is catalyzed by enzymes found in yeast and proceeds by an elaborate multistep mechanism. We can represent the overall process as follows:
To Your Health: The Physiological Effects of Alcohols
Methanol is quite poisonous to humans. Ingestion of as little as 15 mL of methanol can cause blindness, and 30 mL (1 oz) can cause death. However, the usual fatal dose is 100 to 150 mL. The main reason for methanol’s toxicity is that we have liver enzymes that catalyze its oxidation to formaldehyde, the simplest member of the aldehyde family:
Formaldehyde reacts rapidly with the components of cells, coagulating proteins in much the same way that cooking coagulates an egg. This property of formaldehyde accounts for much of the toxicity of methanol.
Organic and biochemical equations are frequently written showing only the organic reactants and products. In this way, we focus attention on the organic starting material and product, rather than on balancing complicated equations.
Ethanol is oxidized in the liver to acetaldehyde:
The acetaldehyde is in turn oxidized to acetic acid (HC2H3O2), a normal constituent of cells, which is then oxidized to carbon dioxide and water. Even so, ethanol is potentially toxic to humans. The rapid ingestion of 1 pt (about 500 mL) of pure ethanol would kill most people, and acute ethanol poisoning kills several hundred people each year—often those engaged in some sort of drinking contest. Ethanol freely crosses into the brain, where it depresses the respiratory control center, resulting in failure of the respiratory muscles in the lungs and hence suffocation. Ethanol is believed to act on nerve cell membranes, causing a diminution in speech, thought, cognition, and judgment.
Rubbing alcohol is usually a 70% aqueous solution of isopropyl alcohol. It has a high vapor pressure, and its rapid evaporation from the skin produces a cooling effect. It is toxic when ingested but, compared to methanol, is less readily absorbed through the skin.
Summary
Many alcohols are made by the hydration of alkenes. Ethanol can be made by the fermentation of sugars or starch from various sources. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.03%3A_Physical_Properties_of_Alcohols.txt |
Learning Objectives
1. Give two major types of reactions of alcohols.
2. Describe the result of the oxidation of a primary alcohol.
3. Describe the result of the oxidation of a secondary alcohol.
Chemical reactions in alcohols occur mainly at the functional group, but some involve hydrogen atoms attached to the OH-bearing carbon atom or to an adjacent carbon atom. Of the three major kinds of alcohol reactions, which are summarized in Figure $1$, two—dehydration and oxidation—are considered here. The third reaction type—esterification—is covered elsewhere.
Dehydration
As noted in Figure $1$, an alcohol undergoes dehydration in the presence of a catalyst to form an alkene and water. The reaction removes the OH group from the alcohol carbon atom and a hydrogen atom from an adjacent carbon atom in the same molecule:
Under the proper conditions, it is possible for the dehydration to occur between two alcohol molecules. The entire OH group of one molecule and only the hydrogen atom of the OH group of the second molecule are removed. The two ethyl groups attached to an oxygen atom form an ether molecule.
(Ethers are discussed in elsewhere) Thus, depending on conditions, one can prepare either alkenes or ethers by the dehydration of alcohols.
Both dehydration and hydration reactions occur continuously in cellular metabolism, with enzymes serving as catalysts and at a temperature of about 37°C. The following reaction occurs in the "Embden–Meyerhof" pathway
Although the participating compounds are complex, the reaction is the same: elimination of water from the starting material. The idea is that if you know the chemistry of a particular functional group, you know the chemistry of hundreds of different compounds.
Oxidation
Primary and secondary alcohols are readily oxidized. We saw earlier how methanol and ethanol are oxidized by liver enzymes to form aldehydes. Because a variety of oxidizing agents can bring about oxidation, we can indicate an oxidizing agent without specifying a particular one by writing an equation with the symbol [O] above the arrow. For example, we write the oxidation of ethanol—a primary alcohol—to form acetaldehyde—an aldehyde—as follows:
We shall see that aldehydes are even more easily oxidized than alcohols and yield carboxylic acids. Secondary alcohols are oxidized to ketones. The oxidation of isopropyl alcohol by potassium dichromate ($\ce{K2Cr2O7}$) gives acetone, the simplest ketone:
Unlike aldehydes, ketones are relatively resistant to further oxidation, so no special precautions are required to isolate them as they form. Note that in oxidation of both primary (RCH2OH) and secondary (R2CHOH) alcohols, two hydrogen atoms are removed from the alcohol molecule, one from the OH group and other from the carbon atom that bears the OH group.
These reactions can also be carried out in the laboratory with chemical oxidizing agents. One such oxidizing agent is potassium dichromate. The balanced equation (showing only the species involved in the reaction) in this case is as follows:
$\ce{8H^{=} + Cr2O7^{2-} + 3CH3CH2OH -> 3CH3CHO + 2Cr^{3+} + 7H2O} \nonumber$
Alcohol oxidation is important in living organisms. Enzyme-controlled oxidation reactions provide the energy cells need to do useful work. One step in the metabolism of carbohydrates involves the oxidation of the secondary alcohol group in isocitric acid to a ketone group:
The overall type of reaction is the same as that in the conversion of isopropyl alcohol to acetone.
Tertiary alcohols (R3COH) are resistant to oxidation because the carbon atom that carries the OH group does not have a hydrogen atom attached but is instead bonded to other carbon atoms. The oxidation reactions we have described involve the formation of a carbon-to-oxygen double bond. Thus, the carbon atom bearing the OH group must be able to release one of its attached atoms to form the double bond. The carbon-to-hydrogen bonding is easily broken under oxidative conditions, but carbon-to-carbon bonds are not. Therefore tertiary alcohols are not easily oxidized.
Example $1$
Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow.
1. CH3CH2CH2CH2CH2OH
Solution
The first step is to recognize the class of each alcohol as primary, secondary, or tertiary.
1. This alcohol has the OH group on a carbon atom that is attached to only one other carbon atom, so it is a primary alcohol. Oxidation forms first an aldehyde and further oxidation forms a carboxylic acid.
2. This alcohol has the OH group on a carbon atom that is attached to three other carbon atoms, so it is a tertiary alcohol. No reaction occurs.
3. This alcohol has the OH group on a carbon atom that is attached to two other carbon atoms, so it is a secondary alcohol; oxidation gives a ketone.
Exercise $1$
Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow.
Summary
Alcohols can be dehydrated to form either alkenes (higher temperature, excess acid) or ethers (lower temperature, excess alcohol). Primary alcohols are oxidized to form aldehydes. Secondary alcohols are oxidized to form ketones. Tertiary alcohols are not readily oxidized. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.05%3A_Reactions_of_Alcohols.txt |
Learning Objectives
• To describe the structure and uses of some common polyhydric alcohols.
Alcohols with two OH groups on adjacent carbon atoms are commonly known as glycols. The most important of these is 1,2-ethanediol (the common name is ethylene glycol), a sweet, colorless, somewhat viscous liquid.
Another common glycol, 1,2-propanediol, is commonly called propylene glycol. Its physical properties are quite similar to those of ethylene glycol.
Commonly called glycerol or glycerin, 1,2,3-propanetriol is the most important trihydroxy alcohol. Like the two glycols, it is a sweet, syrupy liquid. Glycerol is a product of the hydrolysis of fats and oils.
Ethylene glycol is the main ingredient in many antifreeze mixtures for automobile radiators. The two OH groups lead to extensive intermolecular hydrogen bonding. This results in a high boiling point—198°C; thus ethylene glycol does not boil away when it is used as an antifreeze. It is also completely miscible with water. A solution of 60% ethylene glycol in water freezes at −49°C (−56°F) and thus protects an automobile radiator down to that temperature. Ethylene glycol is also used in the manufacture of polyester fiber and magnetic film used in tapes for recorders and computers.
To Your Health: Glycols and Human Health
Ethylene glycol is quite toxic. Because it is sweet, pets often lap up spills of leaked antifreeze from a garage floor or driveway. Sometimes people, especially children, drink it. As with methanol, its toxicity is due to a metabolite. Liver enzymes oxidize ethylene glycol to oxalate ion.
In the kidneys, the oxalate ion combines with the calcium ($Ca^{2+}$) ion, precipitating as calcium oxalate ($CaC_2O_4$).
$\ce{Ca^{2+}(aq) + C_2O_4^{2−}(aq) \rightarrow CaC_2O_4(s)} \nonumber$
These crystals cause renal damage and can lead to kidney failure and death.
Although propylene glycol has physical properties much like those of ethylene glycol, its physiological properties are quite different. Propylene glycol is essentially nontoxic, and it can be used as a solvent for drugs and as a moisturizing agent for foods. Like other alcohols, propylene glycol is oxidized by liver enzymes.
In this case, however, the product is pyruvate ion, a normal intermediate in carbohydrate metabolism. Glycerol, a product of fat metabolism, is essentially nontoxic.
Summary
Glycols are alcohols with two OH groups on adjacent carbon atoms. Glycerol is the most important trihydroxy alcohol.
14.07: Phenols
Learning Objectives
• To describe the structure and uses of some phenols
Compounds in which an OH group is attached directly to an aromatic ring are designated ArOH and called phenols. Phenols differ from alcohols in that they are slightly acidic in water. They react with aqueous sodium hydroxide (NaOH) to form salts.
$\ce{ArOH (aq) + NaOH (aq) \rightarrow ArONa (aq) + H_2O} \nonumber$
The parent compound, C6H5OH, is itself called phenol. (An old name, emphasizing its slight acidity, was carbolic acid.) Phenol is a white crystalline compound that has a distinctive (“hospital smell”) odor.
To Your Health: Phenols and Us
Phenols are widely used as antiseptics (substances that kill microorganisms on living tissue) and as disinfectants (substances intended to kill microorganisms on inanimate objects such as furniture or floors). The first widely used antiseptic was phenol. Joseph Lister used it for antiseptic surgery in 1867. Phenol is toxic to humans, however, and can cause severe burns when applied to the skin. In the bloodstream, it is a systemic poison—that is, one that is carried to and affects all parts of the body. Its severe side effects led to searches for safer antiseptics, a number of which have been found.
An operation in 1753, painted by Gaspare Traversi, of a surgery before antiseptics were used.
One safer phenolic antiseptic is 4-hexylresorcinol (4-hexyl-1,3-dihydroxybenzene; resorcinol is the common name for 1,3-dihydroxybenzene, and 4-hexylresorcinol has a hexyl group on the fourth carbon atom of the resorcinol ring). It is much more powerful than phenol as a germicide and has fewer undesirable side effects. Indeed, it is safe enough to be used as the active ingredient in some mouthwashes and throat lozenges.
The compound 4-hexylresorcinol is mild enough to be used as the active ingredient in antiseptic preparations for use on the skin.
Summary
Phenols are compounds in which an OH group is attached directly to an aromatic ring. Many phenols are used as antiseptics. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.06%3A_Glycols_and_Glycerol.txt |
Learning Objectives
• Describe the structural difference between an alcohol and an ether that affects physical characteristics and reactivity of each.
• Name simple ethers.
• Describe the structure and uses of some ethers.
With the general formula ROR′, an ether may be considered a derivative of water in which both hydrogen atoms are replaced by alkyl or aryl groups. It may also be considered a derivative of an alcohol (ROH) in which the hydrogen atom of the OH group is been replaced by a second alkyl or aryl group:
$\mathrm{HOH\underset{H\: atoms}{\xrightarrow{replace\: both}}ROR'\underset{of\: OH\: group}{\xleftarrow{replace\: H\: atom}}ROH} \nonumber$
Simple ethers have simple common names, formed from the names of the groups attached to oxygen atom, followed by the generic name ether. For example, CH3–O–CH2CH2CH3 is methyl propyl ether. If both groups are the same, the group name should be preceded by the prefix di-, as in dimethyl ether (CH3–O–CH3) and diethyl ether CH3CH2–O–CH2CH3.
Ether molecules have no hydrogen atom on the oxygen atom (that is, no OH group). Therefore there is no intermolecular hydrogen bonding between ether molecules, and ethers therefore have quite low boiling points for a given molar mass. Indeed, ethers have boiling points about the same as those of alkanes of comparable molar mass and much lower than those of the corresponding alcohols (Table $1$).
Table $1$: Comparison of Boiling Points of Alkanes, Alcohols, and Ethers
Condensed Structural Formula Name Molar Mass Boiling Point (°C) Intermolecular Hydrogen Bonding in Pure Liquid?
CH3CH2CH3 propane 44 –42 no
CH3OCH3 dimethyl ether 46 –25 no
CH3CH2OH ethyl alcohol 46 78 yes
CH3CH2CH2CH2CH3 pentane 72 36 no
CH3CH2OCH2CH3 diethyl ether 74 35 no
CH3CH2CH2CH2OH butyl alcohol 74 117 yes
Ether molecules do have an oxygen atom, however, and engage in hydrogen bonding with water molecules. Consequently, an ether has about the same solubility in water as the alcohol that is isomeric with it. For example, dimethyl ether and ethanol (both having the molecular formula C2H6O) are completely soluble in water, whereas diethyl ether and 1-butanol (both C4H10O) are barely soluble in water (8 g/100 mL of water).
Example $1$
What is the common name for each ether?
1. CH3CH2CH2OCH2CH2CH3
Solution
1. The carbon groups on either side of the oxygen atom are propyl (CH3CH2CH2) groups, so the compound is dipropyl ether.
2. The three-carbon group is attached by the middle carbon atom, so it is an isopropyl group. The one-carbon group is a methyl group. The compound is isopropyl methyl ether.
Exercise $1$
What is the common name for each ether?
1. CH3CH2CH2CH2OCH2CH2CH2CH3
To Your Health: Ethers as General Anesthetics
A general anesthetic acts on the brain to produce unconsciousness and a general insensitivity to feeling or pain. Diethyl ether (CH3CH2OCH2CH3) was the first general anesthetic to be used.
Diethyl ether is relatively safe because there is a fairly wide gap between the dose that produces an effective level of anesthesia and the lethal dose. However, because it is highly flammable and has the added disadvantage of causing nausea, it has been replaced by newer inhalant anesthetics, including the fluorine-containing compounds halothane, enflurane, and isoflurane. Unfortunately, the safety of these compounds for operating room personnel has been questioned. For example, female operating room workers exposed to halothane suffer a higher rate of miscarriages than women in the general population.
These three modern, inhalant, halogen-containing, anesthetic compounds are less flammable than diethyl ether.
Summary
To give ethers common names, simply name the groups attached to the oxygen atom, followed by the generic name ether. If both groups are the same, the group name should be preceded by the prefix di-. Ether molecules have no OH group and thus no intermolecular hydrogen bonding. Ethers therefore have quite low boiling points for a given molar mass. Ether molecules have an oxygen atom and can engage in hydrogen bonding with water molecules. An ether molecule has about the same solubility in water as the alcohol that is isomeric with it. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.08%3A_Ethers.txt |
Learning Objectives
• Identify the general structure for an aldehyde and a ketone.
• Use common names to name aldehydes and ketones.
• Use the IUPAC system to name aldehydes and ketones.
The next functional group we consider, the carbonyl group, has a carbon-to-oxygen double bond.
Carbonyl groups define two related families of organic compounds: the aldehydes and the ketones.
The carbonyl group is ubiquitous in biological compounds. It is found in carbohydrates, fats, proteins, nucleic acids, hormones, and vitamins—organic compounds critical to living systems.
In a ketone, two carbon groups are attached to the carbonyl carbon atom. The following general formulas, in which R represents an alkyl group and Ar stands for an aryl group, represent ketones.
In an aldehyde, at least one of the attached groups must be a hydrogen atom. The following compounds are aldehydes:
In condensed formulas, we use CHO to identify an aldehyde rather than COH, which might be confused with an alcohol. This follows the general rule that in condensed structural formulas H comes after the atom it is attached to (usually C, N, or O).
The carbon-to-oxygen double bond is not shown but understood to be present. Because they contain the same functional group, aldehydes and ketones share many common properties, but they still differ enough to warrant their classification into two families.
Naming Aldehydes and Ketones
Both common and International Union of Pure and Applied Chemistry (IUPAC) names are frequently used for aldehydes and ketones, with common names predominating for the lower homologs. The common names of aldehydes are taken from the names of the acids into which the aldehydes can be converted by oxidation.
The stems for the common names of the first four aldehydes are as follows:
• 1 carbon atom: form-
• 2 carbon atoms: acet-
• 3 carbon atoms: propion-
• 4 carbon atoms: butyr-
Because the carbonyl group in a ketone must be attached to two carbon groups, the simplest ketone has three carbon atoms. It is widely known as acetone, a unique name unrelated to other common names for ketones.
Generally, the common names of ketones consist of the names of the groups attached to the carbonyl group, followed by the word ketone. (Note the similarity to the naming of ethers.) Another name for acetone, then, is dimethyl ketone. The ketone with four carbon atoms is ethyl methyl ketone.
Example \(1\)
Classify each compound as an aldehyde or a ketone. Give the common name for each ketone.
Solution
1. This compound has the carbonyl group on an end carbon atom, so it is an aldehyde.
2. This compound has the carbonyl group on an interior carbon atom, so it is a ketone. Both alkyl groups are propyl groups. The name is therefore dipropyl ketone.
3. This compound has the carbonyl group between two alkyl groups, so it is a ketone. One alkyl group has three carbon atoms and is attached by the middle carbon atom; it is an isopropyl group. A group with one carbon atom is a methyl group. The name is therefore isopropyl methyl ketone.
Exercise \(1\)
Classify each compound as an aldehyde or a ketone. Give the common name for each ketone.
Here are some simple IUPAC rules for naming aldehydes and ketones:
• The stem names of aldehydes and ketones are derived from those of the parent alkanes, defined by the longest continuous chain (LCC) of carbon atoms that contains the functional group.
• For an aldehyde, drop the -e from the alkane name and add the ending -al. Methanal is the IUPAC name for formaldehyde, and ethanal is the name for acetaldehyde.
• For a ketone, drop the -e from the alkane name and add the ending -one. Propanone is the IUPAC name for acetone, and butanone is the name for ethyl methyl ketone.
• To indicate the position of a substituent on an aldehyde, the carbonyl carbon atom is always considered to be C1; it is unnecessary to designate this group by number.
• To indicate the position of a substituent on a ketone, number the chain in the manner that gives the carbonyl carbon atom the lowest possible number. In cyclic ketones, it is understood that the carbonyl carbon atom is C1.
Example \(2\)
Give the IUPAC name for each compound.
Solution
1. There are five carbon atoms in the LCC. The methyl group (CH3) is a substituent on the second carbon atom of the chain; the aldehyde carbon atom is always C1. The name is derived from pentane. Dropping the -e and adding the ending -al gives pentanal. The methyl group on the second carbon atom makes the name 2-methylpentanal.
2. There are five carbon atoms in the LCC. The carbonyl carbon atom is C3, and there are methyl groups on C2 and C4. The IUPAC name is 2,4-dimethyl-3-pentanone.
3. There are six carbon atoms in the ring. The compound is cyclohexanone. No number is needed to indicate the position of the carbonyl group because all six carbon atoms are equivalent.
Exercise
Give the IUPAC name for each compound.
Example \(3\)
Draw the structure for each compound.
1. 7-chlorooctanal
2. 4-methyl–3-hexanone
Solution
1. The octan- part of the name tells us that the LCC has eight carbon atoms. There is a chlorine (Cl) atom on the seventh carbon atom; numbering from the carbonyl group and counting the carbonyl carbon atom as C1, we place the Cl atom on the seventh carbon atom.
2. The hexan- part of the name tells us that the LCC has six carbon atoms. The 3 means that the carbonyl carbon atom is C3 in this chain, and the 4 tells us that there is a methyl (CH3) group at C4:
Exercise \(3\)
Draw the structure for each compound.
1. 5-bromo-3-iodoheptanal
2. 5-bromo-4-ethyl-2-heptanone
Summary
The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The common names of ketones, like those of ethers, consist of the names of the groups attached to the carbonyl group, followed by the word ketone. Stem names of aldehydes and ketones are derived from those of the parent alkanes, using an -al ending for an aldehydes and an -one ending for a ketone. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.09%3A_Aldehydes_and_Ketones-_Structure_and_Names.txt |
Learning Objectives
• Explain why the boiling points of aldehydes and ketones are higher than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols.
• Compare the solubilities in water of aldehydes and ketones of four or fewer carbon atoms with the solubilities of comparable alkanes and alcohols.
• Name the typical reactions take place with aldehydes and ketones.
• Describe some of the uses of common aldehydes and ketones.
The carbon-to-oxygen double bond is quite polar, more polar than a carbon-to-oxygen single bond. The electronegative oxygen atom has a much greater attraction for the bonding electron pairs than does the carbon atom. The carbon atom has a partial positive charge, and the oxygen atom has a partial negative charge:
In aldehydes and ketones, this charge separation leads to dipole-dipole interactions that are great enough to significantly affect the boiling points. Table $1$ shows that the polar single bonds in ethers have little such effect, whereas hydrogen bonding between alcohol molecules is even stronger.
Table $1$: Boiling Points of Compounds Having Similar Molar Masses but Different Types of Intermolecular Forces
Compound Family Molar Mass Type of Intermolecular Forces Boiling Point (°C)
CH3CH2CH2CH3 alkane 58 dispersion only –1
CH3OCH2CH3 ether 60 weak dipole 6
CH3CH2CHO aldehyde 58 strong dipole 49
CH3CH2CH2OH alcohol 60 hydrogen bonding 97
Formaldehyde is a gas at room temperature. Acetaldehyde boils at 20°C; in an open vessel, it boils away in a warm room. Most other common aldehydes are liquids at room temperature.
Although the lower members of the homologous series have pungent odors, many higher aldehydes have pleasant odors and are used in perfumes and artificial flavorings. As for the ketones, acetone has a pleasant odor, but most of the higher homologs have rather bland odors.
The oxygen atom of the carbonyl group engages in hydrogen bonding with a water molecule.
The solubility of aldehydes is therefore about the same as that of alcohols and ethers. Formaldehyde, acetaldehyde, and acetone are soluble in water. As the carbon chain increases in length, solubility in water decreases. The borderline of solubility occurs at about four carbon atoms per oxygen atom. All aldehydes and ketones are soluble in organic solvents and, in general, are less dense than water.
Oxidation of Aldehydes and Ketones
Aldehydes and ketones are much alike in many of their reactions, owing to the presence of the carbonyl functional group in both. They differ greatly, however, in one most important type of reaction: oxidation. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation.
The aldehydes are, in fact, among the most easily oxidized of organic compounds. They are oxidized by oxygen ($\ce{O2}$) in air to carboxylic acids.
$\ce{2RCHO + O_2 -> 2RCOOH} \nonumber$
The ease of oxidation helps chemists identify aldehydes. A sufficiently mild oxidizing agent can distinguish aldehydes not only from ketones but also from alcohols. Tollens’ reagent, for example, is an alkaline solution of silver ($\ce{Ag^{+}}$) ion complexed with ammonia (NH3), which keeps the $\ce{Ag^{+}}$ ion in solution.
$\ce{H_3N—Ag^{+}—NH_3} \nonumber$
When Tollens’ reagent oxidizes an aldehyde, the $\ce{Ag^{+}}$ ion is reduced to free silver ($\ce{Ag}$).
$\underbrace{\ce{RCHO(aq)}}_{\text{an aldehyde}} + \ce{2Ag(NH3)2^{+}(aq) -> RCOO^{-} +} \underbrace{\ce{2Ag(s)}}_{\text{free silver}} + \ce{4NH3(aq) + 2H2O} \nonumber$
Deposited on a clean glass surface, the silver produces a mirror (Figure $1$). Ordinary ketones do not react with Tollens’ reagent.
Although ketones resist oxidation by ordinary laboratory oxidizing agents, they undergo combustion, as do aldehydes.
Some Common Carbonyl Compounds
Formaldehyde has an irritating odor. Because of its reactivity, it is difficult to handle in the gaseous state. For many uses, it is therefore dissolved in water and sold as a 37% to 40% aqueous solution called formalin. Formaldehyde denatures proteins, rendering them insoluble in water and resistant to bacterial decay. For this reason, formalin is used in embalming solutions and in preserving biological specimens.
Aldehydes are the active components in many other familiar substances. Large quantities of formaldehyde are used to make phenol-formaldehyde resins for gluing the wood sheets in plywood and as adhesives in other building materials. Sometimes the formaldehyde escapes from the materials and causes health problems in some people. While some people seem unaffected, others experience coughing, wheezing, eye irritation, and other symptoms.
The odor of green leaves is due in part to a carbonyl compound, cis-3-hexenal, which with related compounds is used to impart a “green” herbal odor to shampoos and other products.
Acetaldehyde is an extremely volatile, colorless liquid. It is a starting material for the preparation of many other organic compounds. Acetaldehyde is formed as a metabolite in the fermentation of sugars and in the detoxification of alcohol in the liver. Aldehydes are the active components of many other familiar materials (Figure $2$).
Acetone is the simplest and most important ketone. Because it is miscible with water as well as with most organic solvents, its chief use is as an industrial solvent (for example, for paints and lacquers). It is also the chief ingredient in some brands of nail polish remover.
To Your Health: Acetone in Blood, Urine, and Breath
Acetone is formed in the human body as a by-product of lipid metabolism. Normally, acetone does not accumulate to an appreciable extent because it is oxidized to carbon dioxide and water. The normal concentration of acetone in the human body is less than 1 mg/100 mL of blood. In certain disease states, such as uncontrolled diabetes mellitus, the acetone concentration rises to higher levels. It is then excreted in the urine, where it is easily detected. In severe cases, its odor can be noted on the breath.
Ketones are also the active components of other familiar substances, some of which are noted in the accompanying figure.
Certain steroid hormones have the ketone functional group as a part of their structure. Two examples are progesterone, a hormone secreted by the ovaries that stimulates the growth of cells in the uterine wall and prepares it for attachment of a fertilized egg, and testosterone, the main male sex hormone. These and other sex hormones affect our development and our lives in fundamental ways.
Summary
The polar carbon-to-oxygen double bond causes aldehydes and ketones to have higher boiling points than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols that engage in intermolecular hydrogen bonding. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.10%3A_Properties_of_Aldehydes_and_Ketones.txt |
Learning Objectives
• Identify thiols (mercaptans) by the presence of an SH group.
• The mild oxidation of thiols gives disulfides.
Because sulfur is in the same group (6A) of the periodic table as oxygen, the two elements have some similar properties. We might expect sulfur to form organic compounds related to those of oxygen, and indeed it does. Thiols (also called mercaptans), which are sulfur analogs of alcohols, have the general formula RSH. Methanethiol (also called methyl mercaptan), has the formula CH3SH. Ethanethiol (ethyl mercaptan) is the most common odorant for liquid propane (LP) gas.
The mild oxidation of thiols gives compounds called disulfides.
$\mathrm{2 RSH \xrightarrow{\,[O]\,} RSSR} \nonumber$
The amino acids cysteine [HSCH2CH(NH2)COOH] and methionine [CH3SCH2CH2CH(NH2)COOH] contain sulfur atoms, as do all proteins that contain these amino acids. Disulfide linkages (–S–S–) between protein chains are extremely important in protein structure.
Thioethers, which are sulfur analogs of ethers, have the form general formula RSR′. An example is dimethylsulfide (CH3SCH3), which is responsible for the sometimes unpleasant odor of cooking cabbage and related vegetables. Note that methionine has a thioether functional group.
Career Focus: Paramedic
Paramedics are highly trained experts at providing emergency medical treatment. Their critical duties often include rescue work and emergency medical procedures in a wide variety of settings, sometimes under extremely harsh and difficult conditions. Like other science-based professions, their work requires knowledge, ingenuity, and complex thinking, as well as a great deal of technical skill. The recommended courses for preparation in this field include anatomy, physiology, medical terminology, and—not surprisingly—chemistry. An understanding of basic principles of organic chemistry, for example, is useful when paramedics have to deal with such traumas as burns from fuel (hydrocarbons) or solvent (alcohols, ethers, esters, and so on) fires and alcohol and drug overdoses.
To become a paramedic requires 2–4 y of training and usually includes a stint as an emergency medical technician (EMT). An EMT provides basic care, can administer certain medications and treatments, such as oxygen for respiratory problems and epinephrine (adrenalin) for allergic reactions, and has some knowledge of common medical conditions. A paramedic, in contrast, must have extensive knowledge of common medical problems and be trained to administer a wide variety of emergency drugs.
Paramedics usually work under the direction of a medical doctor with a title such as “medical director.” Some paramedics are employed by fire departments and may work from a fire engine that carries medical equipment as well as fire-fighting gear. Some work from hospital-sponsored ambulances and continue to care for their patients after reaching the hospital emergency room. Still other paramedics work for a government department responsible for emergency health care in a specific geographical area. Finally, some work for private companies that contract to provide service for a government body.
An experienced paramedic has a broad range of employment options, including training for mountain or ocean rescue, working with police department special weapons and tactics (SWAT) teams, or working in isolated settings such as on oil rigs. With their expertise at treating and stabilizing patients before quickly moving them to a hospital, paramedics often provide the first critical steps in saving an endangered life. The following quotation, inscribed on the Arlington National Cemetery headstone of Army Lieutenant R. Adams Cowley, who is often called the “father” of shock trauma medicine, serves as the motto for many paramedic units: “Next to creating a life the finest thing a man can do is save one.” —Abraham Lincoln
Summary
Thiols, thioethers, and disulfides are common in biological compounds. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.11%3A_Organic_Sulfur_Compounds.txt |
Concept Review Exercises
1. What is the functional group of an alkene? An alkyne?
2. Does CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3 have a functional group? Explain.
Answers
1. carbon-to-carbon double bond; carbon-to-carbon triple bond
2. No; it has nothing but carbon and hydrogen atoms and all single bonds.
Exercises
1. What is the functional group of 1-butanol (CH3CH2CH2CH2OH)?
2. What is the functional group of butyl bromide, CH3CH2CH2CH2Br?
1. OH
Concept Review Exercises
1. Is isobutyl alcohol primary, secondary, or tertiary? Explain.
2. What is the LCC in 2-ethyl-1-hexanol? What is taken as the LCC in naming the compound? Explain.
Answers
1. primary; the carbon atom bearing the OH group is attached to only one other carbon atom
2. 7 carbon atoms; the 6-atom chain includes the carbon atom bearing the OH group
Exercises
1. Name each alcohol and classify it as primary, secondary, or tertiary.
1. CH3CH2CH2CH2CH2CH2OH
2. Name each alcohol and classify it as primary, secondary, or tertiary.
3. Draw the structure for each alcohol.
1. 3-hexanol
2. 3,3-dimethyl-2-butanol
3. cyclobutanol
4. Draw the structure for each alcohol.
1. cyclopentanol
2. 4-methyl-2-hexanol
3. 4,5-dimethyl-3-heptanol
Answers
1. 1-hexanol; primary
2. 3-hexanol; secondary
3. 3,3-dibromo-2-methyl-2-butanol; tertiary
Concept Review Exercises
1. Why is ethanol more soluble in water than 1-hexanol?
2. Why does 1-butanol have a lower boiling point than 1-hexanol?
Answers
1. Ethanol has an OH group and only 2 carbon atoms; 1-hexanol has one OH group for 6 carbon atoms and is thus more like a (nonpolar) hydrocarbon than ethanol is.
2. The molar mass of 1-hexanol is greater than that of 1-butanol.
Exercises
Answer the following exercises without consulting tables in the text.
1. Arrange these alcohols in order of increasing boiling point: ethanol, methanol, and 1-propanol.
2. Which has the higher boiling point—butane or 1-propanol?
3. Arrange these alcohols in order of increasing solubility in water: 1-butanol, methanol, and 1-octanol.
4. Arrange these compounds in order of increasing solubility in water: 1-butanol, ethanol, and pentane.
Answers
1. methanol < ethanol < 1-propanol
1. 1-octanol < 1-butanol < methanol
Concept Review Exercises
1. Why is methanol more toxic than ethanol?
2. How does rubbing alcohol cool a feverish patient?
Answers
1. Methanol is oxidized to formaldehyde, which destroys tissue; ethanol is oxidized to acetaldehyde and then acetic acid, a normal metabolite.
2. Evaporation removes heat.
Exercises
1. From what alkene is ethanol made? Draw its condensed structural formula.
2. Can methanol be made from an alkene? Explain.
Answer
1. ethylene; CH2=CH2
14.5: Reactions of Alcohols
Conceptual Questions
1. In a reaction, compound W with the molecular formula C4H10O is converted to compound X with the formula C4H8O. Is W oxidized, reduced, dehydrated, or none of these? Explain.
2. In a reaction, 2 mol of compound Y with the molecular formula C4H10O is converted to 1 mol of compound Z with the formula C8H18O. Is Y oxidized, reduced, or neither? Explain.
Answers
1. oxidized; H is removed
2. neither; water is removed
Exercises
1. Name the three major types of chemical reactions of alcohols.
2. Why do tertiary alcohols not undergo oxidation? Can a tertiary alcohol undergo dehydration?
3. Draw the structure of the product for each reaction.
4. Draw the structure of the product for each reaction.
5. Write an equation for the dehydration of 2-propanol to yield each compound type.
1. an alkene
2. an ether
6. Draw the structure of the alkene formed by the dehydration of cyclohexanol.
Answers
1. dehydration, oxidation, and esterification
1. $\mathrm{CH_3CHOHCH_3\underset{180^\circ C,\: excess\: acid}{\xrightarrow{conc\: H_2SO_4}} CH_3COCH_3+H_2O}$
2. $\mathrm{2CH_3CHOHCH_3 \underset{180^\circ C,\: excess\: acid}{\xrightarrow{conc\: H_2SO_4}}(CH_3)_2CHOCH(CH_3)_2+H_2O}$
Concept Review Exercises
1. In the oxidation of propylene glycol to pyruvic acid, what functional groups in the reactant are involved? What new functional groups appear in the product?
2. Oxalate ion is formed by the oxidation of ethylene glycol. In what kind of reaction is the oxalate ion involved?
Answers
1. two OH groups; a ketone group and a carboxylic acid group
2. precipitation
Exercises
1. What is a glycol?
2. Why is ethylene glycol so much more toxic to humans than propylene glycol?
3. Draw the structure for each compound.
1. 1,5-pentanediol
2. propylene glycol
4. Draw the structure for each compound.
1. 1,3-hexandiol
2. glycerol
Answers
1. an alcohol with two OH groups on adjacent carbon atoms
1. HOCH2CH2CH2CH2CH2OH
Concept Review Exercises
1. How do phenols differ from alcohols in terms of structure and properties?
2. How do phenols differ in properties from aromatic hydrocarbons?
Answers
1. Phenols have an OH group attached directly to an aromatic ring. Phenols are weakly acidic.
2. Phenols have an OH group and are somewhat soluble in water.
Exercises
1. Name each compound.
2. Name each compound.
3. Draw the structure for each compound.
1. m-iodophenol
2. p-methylphenol (p-cresol)
4. Draw the structure for each compound.
1. 2,4,6-trinitrophenol (picric acid)
2. 3,5-diethylphenol
Answers
1. o-nitrophenol
2. p-bromophenol
Concept Review Exercises
1. Why does diethyl ether (CH3CH2OCH2CH3) have a much lower boiling point than 1-butanol (CH3CH2CH2CH2OH)?
2. Which is more soluble in water—ethyl methyl ether (CH3CH2OCH3) or 1-butanol (CH3CH2CH2CH2OH)? Explain.
Answers
1. Diethyl ether has no intermolecular hydrogen bonding because there is no OH group; 1-butanol has an OH and engages in intermolecular hydrogen bonding.
2. Ethyl methyl ether (three carbon atoms, one oxygen atom) is more soluble in water than 1-butanol (four carbon atoms, one oxygen atom), even though both can engage in hydrogen bonding with water.
Exercises
1. How can ethanol give two different products when heated with sulfuric acid? Name these products.
2. Which of these ethers is isomeric with ethanol—CH3CH2OCH2CH3, CH3OCH2CH3, or CH3OCH3?
3. Name each compound.
1. CH3OCH2CH2CH3
4. Name each compound.
1. CH3CH2CH2CH2OCH3
2. CH3CH2OCH2CH2CH3
5. Draw the structure for each compound.
1. methyl ethyl ether
2. tert-butyl ethyl ether
6. Draw the structure for each compound.
1. diisopropyl ether
2. cyclopropyl propyl ether
Answers
1. Intramolecular (both the H and the OH come from the same molecule) dehydration gives ethylene; intermolecular (the H comes from one molecule and the OH comes from another molecule) dehydration gives diethyl ether.
1. methyl propyl ether
2. ethyl isopropyl ether
1. CH3OCH2CH3
Concept Review Exercises
1. Give the structure and IUPAC name for the compound that has the common name m-bromobenzaldehyde.
2. Give the IUPAC name for glyceraldehyde, (HOCH2CHOHCHO). (Hint: as a substituent, the OH group is named hydroxy.)
Answers
1. 3-bromobenzaldehyde
2. 2,3-dihydroxypropanal
Answers
1. The stem names of aldehydes and ketones are derived from those of the parent alkanes, defined by the longest continuous chain (LCC) of carbon atoms that contains the functional group.
2. For an aldehyde, drop the -e from the alkane name and add the ending -al. Methanal is the IUPAC name for formaldehyde, and ethanal is the name for acetaldehyde.
3. For a ketone, drop the -e from the alkane name and add the ending -one. Propanone is the IUPAC name for acetone, and butanone is the name for ethyl methyl ketone.
4. To indicate the position of a substituent on an aldehyde, the carbonyl carbon atom is always considered to be C1; it is unnecessary to designate this group by number.
5. To indicate the position of a substituent on a ketone, number the chain in the manner that gives the carbonyl carbon atom the lowest possible number. In cyclic ketones, it is understood that the carbonyl carbon atom is C1.
6. Give the structure and IUPAC name for the compound that has the common name m-bromobenzaldehyde.
7. Give the IUPAC name for glyceraldehyde, (HOCH2CHOHCHO). (Hint: as a substituent, the OH group is named hydroxy.)
8. 3-bromobenzaldehyde
9. 2,3-dihydroxypropanal
10. Name each compound.
11. Name each compound.
1. CH3CH2CH2CH2CH2CHO
12. Draw the structure for each compound.
1. butyraldehyde
2. 2-hexanone
3. p-nitrobenzaldehyde
13. Draw the structure for each compound.
1. 5-ethyloctanal
2. 2-chloropropanal
3. 2-hydroxy-3-pentanone
1. propanal or propionaldehyde
2. butanal or butyraldehyde
3. 3-pentanone or diethyl ketone
4. benzaldehyde
1. CH3CH2CH2CHO
Concept Review Exercises
1. What feature of their structure makes aldehydes easier to oxidize than ketones?
2. How does the carbon-to-oxygen bond of aldehydes and ketones differ from the carbon-to-carbon bond of alkenes?
Answers
1. the H on the carbonyl carbon atom
2. The carbon-to-oxygen double bond is polar; the carbon-to-carbon double bond is nonpolar.
Exercises
1. Which compound in each pair has the higher boiling point?
1. acetone or 2-propanol
2. dimethyl ether or acetaldehyde
2. Which compound in each pair has the higher boiling point?
1. butanal or 1-butanol
2. acetone or isobutane
3. Draw the structure of the alcohol that could be oxidized to each compound.
1. cyclohexanone
2. 2-methyl-1-propanal
4. Draw the structure of the alcohol that could be oxidized to each compound.
1. 2-pentanone
2. o-methylbenzaldehyde
5. Acetaldehyde is treated with each substance.
1. Ag+(aq)—What inorganic product, if any, is formed?
2. K2Cr2O7 in an acid solution—What organic product, if any, is formed?
6. Acetone is treated with each substance.
1. Ag+(aq) —What inorganic product, if any, is formed?
2. K2Cr2O7 in an acid solution—What organic product, if any, is formed?
Answers
1. 2-propanol
2. acetaldehyde
1. silver metal (Ag)
2. acetic acid (CH3COOH)
Concept Review Exercises
1. What is the functional group of a thiol? Write the condensed structural formula for ethanethiol (ethyl mercaptan).
2. What is the functional group of a disulfide? Write the condensed structural formula for dipropyl disulfide.
Answers
1. SH; CH3CH2SH
2. –S–S–; CH3CH2CH2SSCH2CH2CH3
Exercises
1. A common natural gas odorant is tert-butyl mercaptan. What is its condensed structural formula?
2. Write the equation for the oxidation of ethanethiol to diethyl disulfide.
1. (CH3)3CSH
Extra Exercises
1. Describe two ways that ethanol can be prepared. Which method is used to produce alcoholic beverages?
2. Give the structure of the alkene from which isopropyl alcohol is made by reaction with water in an acidic solution.
3. Ethanol is used as a solvent for some drugs that are not soluble in water. Why is methanol not used in medicines?
4. Give the structure of the alkene that is made from tert-butyl alcohol [(CH3)3COH] by reaction with water in an acidic solution.
5. Classify each conversion as oxidation, dehydration, or hydration (only the organic starting material and product are shown):
1. CH3OH → HCHO
2. CH3CHOHCH3 → CH3CH=CH2
3. CH2=CHCH2CH3 → CH3CHOHCH2CH3
6. Classify each conversion as oxidation, dehydration, or hydration (only the organic starting material and product are shown.):
1. CH3CHOHCH3 → CH3COCH3
2. HOOCCH=CHCOOH → HOOCCH2CHOHCOOH
3. 2 CH3OH → CH3OCH3
7. Why is methanol so much more toxic to humans than ethanol?
8. Each of the four isomeric butyl alcohols is treated with potassium dichromate (K2Cr2O7) in acid. Give the product (if any) expected from each reaction.
9. Draw the structures and give IUPAC names for the four isomeric aldehydes having the formula C5H10O.
10. Write an equation for the reaction of phenol with aqueous NaOH.
11. Write an equation for the ionization of phenol in water.
12. Draw the structures and give the common and IUPAC names for the three isomeric ketones having the formula C5H10O.
13. As we shall see in Chapter 16 "Carbohydrates", 2,3-dihydroxypropanal and 1,3-dihydroxyacetone are important carbohydrates. Draw their structures.
14. Glutaraldehyde (pentanedial) is a germicide that is replacing formaldehyde as a sterilizing agent. It is less irritating to the eyes, the nose, and skin. Write the condensed structural formula of glutaraldehyde.
15. Why does the oxidation of isopropyl alcohol give a ketone, whereas the oxidation of isobutyl alcohol gives an aldehyde?
16. Identify each compound as an alcohol, a phenol, or an ether. Classify any alcohols as primary (1°), secondary (2°), or tertiary (3°).
1. CH3CH2CH2OH
17. Identify each compound as an alcohol, a phenol, or an ether. Classify any alcohols as primary, secondary, or tertiary.
1. CH3CH2OCH2CH3
18. Tell whether each compound forms an acidic, a basic, or a neutral solution in water.
19. When water is added to ethylene in the presence of an acid catalyst, only one product—ethanol—is possible. However, when water is added to propylene, two products are possible—1-propanol and 2-propanol—but only 2-propanol is formed. In 1870, the Russian chemist Vladimir V. Markovnikov proposed a rule to predict the products of such reactions: Considering water to be HOH, the hydrogen atom of water goes on the carbon atom (of the two involved in the double bond) that has the most hydrogen atoms already bonded to it. The OH group goes on the carbon atom with fewer hydrogen atoms. Use Markovnikov’s rule to predict the product of the addition of water to each compound.
1. 2-methylpropene
2. 1-butene
3. 2-methyl-1-pentene
4. 2-methyl-2-pentene
20. Ethyl alcohol, like rubbing alcohol (isopropyl alcohol), is often used for sponge baths. What property of alcohols makes them useful for this purpose?
21. In addition to ethanol, the fermentation of grain produces other organic compounds collectively called fusel oils (FO). The four principal FO components are 1-propanol, isobutyl alcohol, 3-methyl-1-butanol, and 2-methyl-1-butanol. Draw a structure for each. (FO is quite toxic and accounts in part for hangovers.)
22. Draw and name the isomeric ethers that have the formula C5H12O.
23. Menthol is an ingredient in mentholated cough drops and nasal sprays. It produces a cooling, refreshing sensation when rubbed on the skin and so is used in shaving lotions and cosmetics. Thymol, the aromatic equivalent of menthol, is the flavoring constituent of thyme.
1. To what class of compounds does each belong?
2. Give an alternate name for thymol.
24. Write the equation for the production of ethanol by the addition of water to ethylene. How much ethanol can be made from 14.0 kg of ethylene?
25. Methanol is not particularly toxic to rats. If methanol were newly discovered and tested for toxicity in laboratory animals, what would you conclude about its safety for human consumption?
26. The amino acid cysteine has the formula HSCH2CH(NH2)COOH. What is the sulfur-containing functional group in the cysteine molecule?
27. The amino acid methionine has the formula CH3SCH2CH2CH(NH2)COOH. What functional groups are in methionine?
28. Tetrahydrocannabinol is the principal active ingredient in marijuana. What functional groups are present in this molecule?
Answers
1. addition of water to ethylene; fermentation (for beverages)
1. Methanol is too toxic.
1.
1. oxidation
2. dehydration
3. hydration
1. Methanol is oxidized in the body to toxic formaldehyde; ethanol is oxidized to the less toxic acetaldehyde.
1. C6H5OH + H2O → C6H5O + H3O+
1. Isopropyl alcohol is a secondary alcohol, whereas isobutyl alcohol is a primary alcohol.
1.
1. ether
2. tertiary alcohol
3. phenol
4. secondary alcohol
1.
1. tert-butyl alcohol
2. 2-butanol
3. 2-methyl-2-pentanol
4. 2-methyl-2-pentanol
1.
1. menthol: alcohol; thymol: phenol
2. 2-isopropyl-5-methylphenol
1. It might be ruled safe until tested on humans.
1. sulfide, amino, and carboxylic acid | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/14%3A_Organic_Compounds_of_Oxygen/14.E%3A_Organic_Compounds_of_Oxygen_%28Exercises%29.txt |
Organic acids have been known for ages. Prehistoric people likely made acetic acid when their fermentation reactions went awry and produced vinegar instead of wine. The Sumerians (2900–1800 BCE) used vinegar as a condiment, a preservative, an antibiotic, and a detergent. Citric acid was discovered by an Islamic alchemist, Jabir Ibn Hayyan (also known as Geber), in the 8th century, and crystalline citric acid was first isolated from lemon juice in 1784 by the Swedish chemist Carl Wilhelm Scheele. Medieval scholars in Europe were aware that the crisp, tart flavor of citrus fruits is caused by citric acid. Naturalists of the 17th century knew that the sting of a red ant’s bite was due to an organic acid that the ant injected into the wound. The acetic acid of vinegar, the formic acid of red ants, and the citric acid of fruits all belong to the same family of compounds—carboxylic acids. Soaps are salts of long-chain carboxylic acids. Prehistoric people also knew about organic bases—by smell if not by name; amines are the organic bases produced when animal tissue decays. The organic compounds that we consider in this chapter are organic acids and bases. We will also consider two derivatives of carboxylic acids: esters and amides. An ester is derived from a carboxylic acid and an alcohol. Fats and oils are esters, as are many important fragrances and flavors. An amide is derived from a carboxylic acid and either ammonia or an amine. Proteins, often called “the stuff of life,” are polyamides.
• 15.0: Prelude to Organic Acids and Bases and Some of Their Derivatives
Organic acids have been known for ages. Prehistoric people likely made acetic acid when their fermentation reactions went awry and produced vinegar instead of wine. The Sumerians (2900–1800 BCE) used vinegar as a condiment, a preservative, an antibiotic, and a detergent.
• 15.1: Carboxylic Acids - Structures and Names
Simple carboxylic acids are best known by common names based on Latin and Greek words that describe their source (e.g., formic acid, Latin formica, meaning “ant”). Greek letters, not numbers, designate the position of substituted acids in the common naming convention. IUPAC names are derived from the LCC of the parent hydrocarbon with the -e ending of the parent alkane replaced by the suffix -oic and the word acid.
• 15.2: The Formation of Carboxylic Acids
Whether in the laboratory or in the body, the oxidation of aldehydes or primary alcohols forms carboxylic acids.
• 15.3: Physical Properties of Carboxylic Acids
Carboxylic acids have high boiling points compared to other substances of comparable molar mass. Boiling points increase with molar mass. Carboxylic acids having one to four carbon atoms are completely miscible with water. Solubility decreases with molar mass.
• 15.4: Chemical Properties of Carboxylic Acids- Ionization and Neutralization
Soluble carboxylic acids are weak acids in aqueous solutions. Carboxylic acids neutralize bases to form salts.
• 15.5: Esters - Structures and Names
An ester has an OR group attached to the carbon atom of a carbonyl group.
• 15.6: Physical Properties of Esters
Esters have polar bonds but do not engage in hydrogen bonding and are therefore intermediate in boiling points between the nonpolar alkanes and the alcohols, which engage in hydrogen bonding. Ester molecules can engage in hydrogen bonding with water, so esters of low molar mass are therefore somewhat soluble in water.
• 15.7: Preparation of Esters
Esters are made by the reaction of a carboxylic acid with an alcohol, a process that is called esterification.
• 15.8: Hydrolysis of Esters
Hydrolysis is a most important reaction of esters. Acidic hydrolysis of an ester gives a carboxylic acid and an alcohol. Basic hydrolysis of an ester gives a carboxylate salt and an alcohol.
• 15.9: Esters of Phosphoric Acid
Inorganic acids such as \(H_3PO_4\) form esters. The esters of phosphoric acid are especially important in biochemistry.
• 15.10: Amines - Structures and Names
An amine is a derivative of ammonia in which one, two, or all three hydrogen atoms are replaced by hydrocarbon groups. Amines are classified as primary, secondary, or tertiary by the number of hydrocarbon groups attached to the nitrogen atom. Amines are named by naming the alkyl groups attached to the nitrogen atom, followed by the suffix -amine.
• 15.11: Physical Properties of Amines
Primary and secondary amines have higher boiling points than those of alkanes or ethers of similar molar mass because they can engage in intermolecular hydrogen bonding. Their boiling points are lower than those of alcohols because alcohol molecules have hydrogen atoms bonded to an oxygen atom, which is more electronegative. The boiling points of tertiary amines, which cannot engage in hydrogen bonding because they have no hydrogen atom on the nitrogen atom.
• 15.12: Amines as Bases
Amines are bases; they react with acids to form salts. Salts of aniline are properly named as anilinium compounds, but an older system is used to name drugs: the salts of amine drugs and hydrochloric acid are called “hydrochlorides.” Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring.
• 15.13: Amides- Structures and Names
Amides have a general structure in which a nitrogen atom is bonded to a carbonyl carbon atom. In names for amides, the -ic acid of the common name or the -oic ending of the IUPAC for the corresponding carboxylic acid is replaced by -amide.
• 15.14: Physical Properties of Amides
Most amides are solids at room temperature; the boiling points of amides are much higher than those of alcohols of similar molar mass. Amides of five or fewer carbon atoms are soluble in water.
• 15.15: Formation of Amides
Amides are prepared by the reaction of a carboxylic acid with ammonia or an amine.
• 15.16: Chemical Properties of Amides- Hydrolysis
The hydrolysis of an amide produces a carboxylic acid and ammonia or an amine.
• 15.S: Organic Acids and Bases and Some of Their Derivatives (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the summary and ask yourself how they relate to the topics in the chapter.
15: Organic Acids and Bases and Some of Their Derivatives
Organic acids have been known for ages. Prehistoric people likely made acetic acid when their fermentation reactions went awry and produced vinegar instead of wine. The Sumerians (2900–1800 BCE) used vinegar as a condiment, a preservative, an antibiotic, and a detergent. Citric acid was discovered by an Islamic alchemist, Jabir Ibn Hayyan (also known as Geber), in the 8th century, and crystalline citric acid was first isolated from lemon juice in 1784 by the Swedish chemist Carl Wilhelm Scheele. Medieval scholars in Europe were aware that the crisp, tart flavor of citrus fruits is caused by citric acid. Naturalists of the 17th century knew that the sting of a red ant’s bite was due to an organic acid that the ant injected into the wound. The acetic acid of vinegar, the formic acid of red ants, and the citric acid of fruits all belong to the same family of compounds—carboxylic acids. Soaps are salts of long-chain carboxylic acids.
Prehistoric people also knew about organic bases—by smell if not by name; amines are the organic bases produced when animal tissue decays. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.00%3A_Prelude_to_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives.txt |
Learning Objectives
• Name carboxylic acids with common names.
• Name carboxylic acids according to IUPAC nomenclature.
Carboxylic acids occur widely in nature, often combined with alcohols or other functional groups, as in fats, oils, and waxes. They are components of many foods, medicines, and household products (Figure \(1\)). Not surprisingly, many of them are best known by common names based on Latin and Greek words that describe their source.
The simplest carboxylic acid, formic acid (HCOOH), was first obtained by the distillation of ants (Latin formica, meaning “ant”). The bites of some ants inject formic acid, and the stings of wasps and bees contain formic acid (as well as other poisonous materials).
The next higher homolog is acetic acid, which is made by fermenting cider and honey in the presence of oxygen. This fermentation produces vinegar, a solution containing 4%–10% acetic acid, plus a number of other compounds that add to its flavor. Acetic acid is probably the most familiar weak acid used in educational and industrial chemistry laboratories.
Pure acetic acid solidifies at 16.6°C, only slightly below normal room temperature. In the poorly heated laboratories of the late 19th and early 20th centuries in northern North America and Europe, acetic acid often “froze” on the storage shelf. For that reason, pure acetic acid (sometimes called concentrated acetic acid) came to be known as glacial acetic acid, a name that survives to this day.
The third homolog, propionic acid (CH3CH2COOH), is seldom encountered in everyday life. The fourth homolog, butyric acid (CH3CH2CH2COOH), is one of the most foul-smelling substances imaginable. It is found in rancid butter and is one of the ingredients of body odor. By recognizing extremely small amounts of this and other chemicals, bloodhounds are able to track fugitives. Models of the first four carboxylic acids are shown in Figure \(2\).
The acid with the carboxyl group attached directly to a benzene ring is called benzoic acid (C6H5COOH).
The common names of carboxylic acids use Greek letters (α, β, γ, δ, and so forth), not numbers, to designate the position of substituent groups in acids. These letters refer to the position of the carbon atom in relation to the carboxyl carbon atom.
In the nomenclature system of the International Union of Pure and Applied Chemistry (IUPAC), the parent hydrocarbon is the one that corresponds to the longest continuous chain (LCC) containing the carboxyl group. The -e ending of the parent alkane is replaced by the suffix -oic and the word acid. For example, the carboxylic acid derived from pentane is pentanoic acid (CH3CH2CH2CH2COOH). As with aldehydes, the carboxyl carbon atom is counted first; numbers are used to indicate any substituted carbon atoms in the parent chain.
Greek letters are used with common names; numbers are used with IUPAC names.
Example \(1\)
Give the common and IUPAC names for each compound.
1. ClCH2CH2CH2COOH
Solution
1. The LCC contains four carbon atoms; the compound is therefore named as a substituted butyric (or butanoic) acid.
The chlorine atom is attached to the γ-carbon in the common system or C4 in the IUPAC system. The compound is γ-chlorobutyric acid or 4-chlorobutanoic acid.
2. The LCC contains four carbon atoms; the compound is therefore named as a substituted butyric (or butanoic) acid.
The bromine (Br) atom is at the α-carbon in the common system or C2 in the IUPAC system. The compound is α-bromobutyric acid or 2-bromobutanoic acid.
Exercise \(1\)
Give the IUPAC name for each compound.
1. ClCH2CH2CH2CH2COOH
2. (CH3)2CHCH2CHBrCOOH
Example \(2\)
Write the condensed structural formula for β-chloropropionic acid.
Solution
Propionic acid has three carbon atoms: C–C–COOH. Attach a chlorine (Cl) atom to the parent chain at the beta carbon atom, the second one from the carboxyl group: Cl–C–C–COOH. Then add enough hydrogen atoms to give each carbon atom four bonds: ClCH2CH2COOH.
Exercise \(2\)
Write the condensed structural formula for 4-bromo-5-methylhexanoic acid.
Key Takeaways
• Simple carboxylic acids are best known by common names based on Latin and Greek words that describe their source (e.g., formic acid, Latin formica, meaning “ant”).
• Greek letters, not numbers, designate the position of substituted acids in the common naming convention.
• IUPAC names are derived from the LCC of the parent hydrocarbon with the -e ending of the parent alkane replaced by the suffix -oic and the word acid.
15.02: The Formation of Carboxylic Acids
Learning Objectives
• To describe the preparation of carboxylic acids.
As we noted previously, the oxidation of aldehydes or primary alcohols forms carboxylic acids:
In the presence of an oxidizing agent, ethanol is oxidized to acetaldehyde, which is then oxidized to acetic acid.
This process also occurs in the liver, where enzymes catalyze the oxidation of ethanol to acetic acid.
$\mathrm{CH_3CH_2OH \underset{oxidizing\: agent}{\xrightarrow{alcohol\: dehydrogenase}} CH_3CHO \underset{oxidizing\: agent}{\xrightarrow{alcohol\: dehydrogenase}} CH_3COOH} \nonumber$
Acetic acid can be further oxidized to carbon dioxide and water.
Summary
Whether in the laboratory or in the body, the oxidation of aldehydes or primary alcohols forms carboxylic acids.
15.03: Physical Properties of Carboxylic Acids
Learning Objectives
• Compare the boiling points of carboxylic acids with alcohols of similar molar mass.
• Compare the solubilities of carboxylic acids in water with the solubilities of comparable alkanes and alcohols in water.
Many carboxylic acids are colorless liquids with disagreeable odors. The carboxylic acids with 5 to 10 carbon atoms all have “goaty” odors (explaining the odor of Limburger cheese). These acids are also produced by the action of skin bacteria on human sebum (skin oils), which accounts for the odor of poorly ventilated locker rooms. The acids with more than 10 carbon atoms are waxlike solids, and their odor diminishes with increasing molar mass and resultant decreasing volatility.
Carboxylic acids exhibit strong hydrogen bonding between molecules. They therefore have high boiling points compared to other substances of comparable molar mass.
The carboxyl group readily engages in hydrogen bonding with water molecules (Figure \(1\)). The acids with one to four carbon atoms are completely miscible with water. Solubility decreases as the carbon chain length increases because dipole forces become less important and dispersion forces become more predominant. Hexanoic acid [CH3(CH2)4COOH] is barely soluble in water (about 1.0 g/100 g of water). Palmitic acid [CH3(CH2)14COOH], with its large nonpolar hydrocarbon component, is essentially insoluble in water. The carboxylic acids generally are soluble in such organic solvents as ethanol, toluene, and diethyl ether.
Table 15.4.1 lists some physical properties for selected carboxylic acids. The first six are homologs. Notice that the boiling points increase with increasing molar mass, but the melting points show no regular pattern.
Table \(1\): Physical Constants of Carboxylic Acids
Condensed Structural Formula Name of Acid Melting Point (°C) Boiling Point (°C) Solubility (g/100 g of Water)
HCOOH formic acid 8 100 miscible
CH3COOH acetic acid 17 118 miscible
CH3CH2COOH propionic acid –22 141 miscible
CH3(CH2)2COOH butyric acid –5 163 miscible
CH3(CH2)3COOH valeric acid –35 187 5
CH3(CH2)4COOH caproic acid –3 205 1.1
C6H5COOH benzoic acid 122 249 0.29
Key Takeaways
• Carboxylic acids have high boiling points compared to other substances of comparable molar mass. Boiling points increase with molar mass.
• Carboxylic acids having one to four carbon atoms are completely miscible with water. Solubility decreases with molar mass. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.01%3A_Carboxylic_Acids_-_Structures_and_Names.txt |
Learning Objectives
• Name the typical reactions that take place with carboxylic acids.
• Describe how carboxylic acids react with basic compounds.
Water-soluble carboxylic acids ionize slightly in water to form moderately acidic solutions.
$\mathrm{RCOOH + H_2O\rightleftharpoons RCOO^{-}+H_3O^+} \nonumber$
Their aqueous solutions exhibit the typical properties of acids, such as changing litmus from blue to red.
The anion formed when a carboxylic acid dissociates is called the carboxylate anion (RCOO).
Whether soluble in water or not, carboxylic acids react with aqueous solutions of sodium hydroxide (NaOH), sodium carbonate (Na2CO3), and sodium bicarbonate (NaHCO3) to form salts:
RCOOH + NaOH(aq) → RCOONa+(aq) + H2O
2RCOOH + Na2CO3(aq) → 2RCOONa+(aq) + H2O + CO2(g)
RCOOH + NaHCO3(aq) → RCOONa+(aq) + H2O + CO2(g)
In these reactions, the carboxylic acids act like inorganic acids: they neutralize basic compounds. With solutions of carbonate ($CO_3^{2–}$) and bicarbonate ($HCO_3^{–}$) ions, they also form carbon dioxide gas.
Carboxylic acid salts are named in the same manner as inorganic salts: the name of the cation is followed by the name of the organic anion. The name of the anion is obtained by dropping the -ic ending of the acid name and replacing it with the suffix -ate. This rule applies whether we are using common names or International Union of Pure and Applied Chemistry (IUPAC) names:
The salts of long-chain carboxylic acids are called soaps. We discuss the chemistry of soaps elsewhere.
Example $1$
Write an equation for each reaction.
1. the ionization of propionic acid in water (H2O)
2. the neutralization of propionic acid with aqueous sodium hydroxide (NaOH)
Solution
Propionic acid has three carbon atoms, so its formula is CH2CH2COOH.
1. Propionic acid ionizes in water to form a propionate ion and a hydronium (H3O+) ion. CH3CH2COOH(aq) + H2O(ℓ) → CH3CH2COO(aq) + H3O+(aq)
2. Propionic acid reacts with NaOH(aq) to form sodium propionate and water. CH3CH2COOH(aq) + NaOH(aq) → CH3CH2COONa+(aq) + H2O(ℓ)
Exercise $1$
Write an equation for each reaction.
1. the ionization of formic acid in water
2. the ionization of p-chlorobenzoic acid in water
Example $2$
Write an equation for the reaction of decanoic acid with each compound.
1. aqueous sodium hydoxide (NaOH)
2. aqueous sodium bicarbonate (NaHCO3)
Solution
1. Decanoic acid has 10 carbon atoms. It reacts with NaOH to form a salt and water (H2O). CH3(CH2)8COOH + NaOH(aq) → CH3(CH2)8COONa+(aq) + H2O(ℓ)
2. With NaHCO3, the products are a salt, H2O, and carbon dioxide (CO2). CH3(CH2)8COOH + NaHCO3(aq) → CH3(CH2)8COONa+(aq) + H2O(ℓ) + CO2(g)
Exercise $3$
Write an equation for the reaction of benzoic acid with each compound.
1. aqueous sodium hydroxide (NaOH)
2. aqueous sodium bicarbonate (NaHCO3)
To Your Health: Organic Salts as Preservatives
Some organic salts are used as preservatives in food products. They prevent spoilage by inhibiting the growth of bacteria and fungi. Calcium and sodium propionate, for example, are added to processed cheese and bakery goods; sodium benzoate is added to cider, jellies, pickles, and syrups; and sodium sorbate and potassium sorbate are added to fruit juices, sauerkraut, soft drinks, and wine. Look for them on ingredient labels the next time you shop for groceries.
Key Takeaways
• Soluble carboxylic acids are weak acids in aqueous solutions.
• Carboxylic acids neutralize bases to form salts.
15.05: Esters - Structures and Names
Learning Objectives
• Identify the general structure for an ester.
• Use common names to name esters.
• Name esters according to the IUPAC system.
Esters have the general formula RCOOR′, where R may be a hydrogen atom, an alkyl group, or an aryl group, and R′ may be an alkyl group or an aryl group but not a hydrogen atom. (If it were hydrogen atom, the compound would be a carboxylic acid.) Figure \(1\) shows models for two common esters.
Esters occur widely in nature. Unlike carboxylic acids, esters generally have pleasant odors and are often responsible for the characteristic fragrances of fruits and flowers. Once a flower or fruit has been chemically analyzed, flavor chemists can attempt to duplicate the natural odor or taste. Both natural and synthetic esters are used in perfumes and as flavoring agents.
Fats and vegetable oils are esters of long-chain fatty acids and glycerol. Esters of phosphoric acid are of the utmost importance to life.
Names of Esters
Although esters are covalent compounds and salts are ionic, esters are named in a manner similar to that used for naming salts. The group name of the alkyl or aryl portion is given first and is followed by the name of the acid portion. In both common and International Union of Pure and Applied Chemistry (IUPAC) nomenclature, the -ic ending of the parent acid is replaced by the suffix -ate (Table \(1\)).
Table \(1\): Nomenclature of Esters
Condensed Structural Formula Common Name IUPAC Name
HCOOCH3 methyl formate methyl methanoate
CH3COOCH3 methyl acetate methyl ethanoate
CH3COOCH2CH3 ethyl acetate ethyl ethanoate
CH3CH2COOCH2CH3 ethyl propionate ethyl propanoate
CH3CH2CH2COOCH(CH3)2 isopropyl butyrate isopropyl butanoate
ethyl benzoate ethyl benzoate
Example \(1\)
Give the common and IUPAC names for each compound.
Solution
The alkyl group attached directly to the oxygen atom is a butyl group (in green).
The part of the molecule derived from the carboxylic acid (in red) has three carbon atoms. It is called propionate (common) or propanoate (IUPAC). The ester is therefore butyl propionate or butyl propanoate.
1. An alkyl group (in green) is attached directly to the oxygen atom by its middle carbon atom; it is an isopropyl group. The part derived from the acid (that is, the benzene ring and the carbonyl group, in red) is benzoate. The ester is therefore isopropyl benzoate (both the common name and the IUPAC name).
Exercise \(1\)
Give the common and IUPAC names for each compound.
Example \(2\)
Draw the structure for ethyl pentanoate.
Solution
Start with the portion from the acid. Draw the pentanoate (five carbon atoms) group first; keeping in mind that the last carbon atom is a part of the carboxyl group.
Then attach the ethyl group to the bond that ordinarily holds the hydrogen atom in the carboxyl group.
Exercise \(2\)
Draw the structure for phenyl pentanoate.
Key Takeaway
• An ester has an OR group attached to the carbon atom of a carbonyl group. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.04%3A_Chemical_Properties_of_Carboxylic_Acids-_Ionization_and_Neutraliz.txt |
Learning Objectives
• Compare the boiling points of esters with alcohols of similar molar mass.
• Compare the solubilities of esters in water with the solubilities of comparable alkanes and alcohols in water.
Ester molecules are polar but have no hydrogen atom attached directly to an oxygen atom. They are therefore incapable of engaging in intermolecular hydrogen bonding with one another and thus have considerably lower boiling points than their isomeric carboxylic acids counterparts. Because ester molecules can engage in hydrogen bonding with water molecules, however, esters of low molar mass are somewhat soluble in water. Borderline solubility occurs in those molecules that have three to five carbon atoms. Table \(1\) lists the physical properties of some common esters.
Esters are common solvents. Ethyl acetate is used to extract organic solutes from aqueous solutions—for example, to remove caffeine from coffee. It also is used to remove nail polish and paint. Cellulose nitrate is dissolved in ethyl acetate and butyl acetate to form lacquers. The solvent evaporates as the lacquer “dries,” leaving a thin film on the surface. High boiling esters are used as softeners (plasticizers) for brittle plastics.
Table \(1\): Physical Properties of Some Esters
Condensed Structural Formula Name Molar Mass Melting Point (°C) Boiling Point (°C) Aroma
HCOOCH3 methyl formate 60 −99 32
HCOOCH2CH3 ethyl formate 74 −80 54 rum
CH3COOCH3 methyl acetate 74 −98 57
CH3COOCH2CH3 ethyl acetate 88 −84 77
CH3CH2CH2COOCH3 methyl butyrate 102 −85 102 apple
CH3CH2CH2COOCH2CH3 ethyl butyrate 116 −101 121 pineapple
CH3COO(CH2)4CH3 pentyl acetate 130 −71 148 pear
CH3COOCH2CH2CH(CH3)2 isopentyl acetate 130 −79 142 banana
CH3COOCH2C6H5 benzyl acetate 150 −51 215 jasmine
CH3CH2CH2COO(CH2)4CH3 pentyl butyrate 158 −73 185 apricot
CH3COO(CH2)7CH3 octyl acetate 172 −39 210 orange
Summary
Esters have polar bonds but do not engage in hydrogen bonding and are therefore intermediate in boiling points between the nonpolar alkanes and the alcohols, which engage in hydrogen bonding. Ester molecules can engage in hydrogen bonding with water, so esters of low molar mass are therefore somewhat soluble in water.
15.07: Preparation of Esters
Learning Objectives
• To identify and describe the substances from which most esters are prepared.
Some esters can be prepared by esterification, a reaction in which a carboxylic acid and an alcohol, heated in the presence of a mineral acid catalyst, form an ester and water:
The reaction is reversible. As a specific example of an esterification reaction, butyl acetate can be made from acetic acid and 1-butanol.
A Closer Look: Condensation Polymers
A commercially important esterification reaction is condensation polymerization, in which a reaction occurs between a dicarboxylic acid and a dihydric alcohol (diol), with the elimination of water. Such a reaction yields an ester that contains a free (unreacted) carboxyl group at one end and a free alcohol group at the other end. Further condensation reactions then occur, producing polyester polymers.
The most important polyester, polyethylene terephthalate (PET), is made from terephthalic acid and ethylene glycol monomers:
Polyester molecules make excellent fibers and are used in many fabrics. A knitted polyester tube, which is biologically inert, can be used in surgery to repair or replace diseased sections of blood vessels. PET is used to make bottles for soda pop and other beverages. It is also formed into films called Mylar. When magnetically coated, Mylar tape is used in audio- and videocassettes. Synthetic arteries can be made from PET, polytetrafluoroethylene, and other polymers.
Summary
Esters are made by the reaction of a carboxylic acid with an alcohol, a process that is called esterification. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.06%3A_Physical_Properties_of_Esters.txt |
Learning Objectives
• Describe the typical reaction that takes place with esters.
• Identify the products of an acidic hydrolysis of an ester.
• Identify the products of a basic hydrolysis of an ester.
Esters are neutral compounds, unlike the acids from which they are formed. In typical reactions, the alkoxy (OR′) group of an ester is replaced by another group. One such reaction is hydrolysis, literally “splitting with water.” The hydrolysis of esters is catalyzed by either an acid or a base.
Acidic hydrolysis is simply the reverse of esterification. The ester is heated with a large excess of water containing a strong-acid catalyst. Like esterification, the reaction is reversible and does not go to completion.
As a specific example, butyl acetate and water react to form acetic acid and 1-butanol. The reaction is reversible and does not go to completion.
Example \(1\)
Write an equation for the acidic hydrolysis of ethyl butyrate (CH3CH2CH2COOCH2CH3) and name the products.
Solution
Remember that in acidic hydrolysis, water (HOH) splits the ester bond. The H of HOH joins to the oxygen atom in the OR part of the original ester, and the OH of HOH joins to the carbonyl carbon atom:
The products are butyric acid (butanoic acid) and ethanol.
Exercise \(1\)
Write an equation for the acidic hydrolysis of methyl butanoate and name the products.
When a base (such as sodium hydroxide [NaOH] or potassium hydroxide [KOH]) is used to hydrolyze an ester, the products are a carboxylate salt and an alcohol. Because soaps are prepared by the alkaline hydrolysis of fats and oils, alkaline hydrolysis of esters is called saponification (Latin sapon, meaning “soap,” and facere, meaning “to make”). In a saponification reaction, the base is a reactant, not simply a catalyst. The reaction goes to completion:
As a specific example, ethyl acetate and NaOH react to form sodium acetate and ethanol:
Example \(2\)
Write an equation for the hydrolysis of methyl benzoate in a potassium hydroxide solution.
Solution
In basic hydrolysis, the molecule of the base splits the ester linkage. The acid portion of the ester ends up as the salt of the acid (in this case, the potassium salt). The alcohol portion of the ester ends up as the free alcohol.
Exercise \(2\)
Write the equation for the hydrolysis of ethyl propanoate in a sodium hydroxide solution.
Summary
Hydrolysis is a most important reaction of esters. Acidic hydrolysis of an ester gives a carboxylic acid and an alcohol. Basic hydrolysis of an ester gives a carboxylate salt and an alcohol.
15.09: Esters of Phosphoric Acid
do, inorganic acids such as nitric acid (HNO), sulfuric acid (HSO), and phosphoric acid (HPO) also form esters. The esters of phosphoric acid are especially important in . A phosphoric acid can form a monoalkyl, a dialkyl, or a trialkyl by reaction with one, two, or three molecules of an . . . The bonds between phosphate units in adenosine triphosphate (ATP) are called . These are high- bonds that store from the metabolism of foods. of ATP releases as it is needed for biochemical processes (for instance, for muscle contraction). Phosphate esters are also important structural constituents of phospholipids and nucleic acids. . | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.08%3A_Hydrolysis_of_Esters.txt |
Learning Objectives
• Identify the general structure for an amine.
• Identify the functional group for amines.
• Determine the structural feature that classifies amines as primary, secondary, or tertiary.
• Use nomenclature systems to name amines.
Amines are classified according to the number of carbon atoms bonded directly to the nitrogen atom. A primary (1°) amine has one alkyl (or aryl) group on the nitrogen atom, a secondary (2°) amine has two, and a tertiary (3°) amine has three (Figure \(1\)).
To classify alcohols, we look at the number of carbon atoms bonded to the carbon atom bearing the OH group, not the oxygen atom itself. Thus, although isopropylamine looks similar to isopropyl alcohol, the former is a primary amine, while the latter is a secondary alcohol.
The common names for simple aliphatic amines consist of an alphabetic list of alkyl groups attached to the nitrogen atom, followed by the suffix -amine. (Systematic names are often used by some chemists.) The amino group (NH2) is named as a substituent in more complicated amines, such as those that incorporate other functional groups or in which the alkyl groups cannot be simply named.
Example \(1\)
Name and classify each compound.
1. CH3CH2CH2NH2
2. CH3CH2NHCH2CH3
3. CH3CH2CH2NHCH3
Solution
1. There is only one alkyl group attached to the nitrogen atom, so the amine is primary. A group of three carbon atoms (a propyl group) is attached to the NH2 group through an end carbon atom, so the name is propylamine.
2. There are two methyl groups and one ethyl group on the nitrogen atom. The compound is ethyldimethylamine, a tertiary amine.
3. There are two ethyl groups attached to the nitrogen atom; the amine is secondary, so the compound is diethylamine.
4. The nitrogen atom has a methyl group and a propyl group, so the compound is methylpropylamine, a secondary amine.
Exercise \(1\)
Name and classify each compound.
1. CH3CH2CH2CH2NH2
2. CH3CH2CH2NHCH2CH2 CH3
Example \(2\)
Draw the structure for each compound and classify.
1. isopropyldimethylamine
2. dipropylamine
Solution
1. The name indicates that there are an isopropyl group (in red) and two methyl groups (in green) attached to the nitrogen atom; the amine is tertiary.
2. The name indicates that there are two propyl groups attached to the nitrogen atom; the amine is secondary. (The third bond on the nitrogen atom goes to a hydrogen atom.) CH3CH2CH2NHCH2CH2CH3
Exercise \(2\)
Draw the structure for each compound and classify.
1. ethylisopropylamine
2. diethylpropylamine
The primary amine in which the nitrogen atom is attached directly to a benzene ring has a special name—aniline. Aryl amines are named as derivatives of aniline.
Example \(3\)
Name this compound.
Solution
The benzene ring with an amino (NH2) group is aniline. The compound is named as a derivative of aniline: 3-bromoaniline or m-bromoaniline.
Exercise \(3\)
Name this compound.
Example \(4\)
Draw the structure for p-ethylaniline and classify.
Solution
The compound is a derivative of aniline. It is a primary amine having an ethyl group located para to the amino (NH2) group.
Exercise \(4\)
Draw the structure for p-isopropylaniline and classify.
Example \(5\)
Draw the structure for 2-amino-3-methylpentane.
Solution
Always start with the parent compound: draw the pentane chain. Then attach a methyl group at the third carbon atom and an amino group at the second carbon atom.
Exercise \(5\)
Draw the structure for 2-amino-3-ethyl-1-chloroheptane.
Ammonium (NH4+) ions, in which one or more hydrogen atoms are replaced with alkyl groups, are named in a manner analogous to that used for simple amines. The alkyl groups are named as substituents, and the parent species is regarded as the NH4+ ion. For example, CH3NH3+ is the methylammonium ion. The ion formed from aniline (C6H5NH3+) is called the anilinium ion.
Example \(6\)
Name each ion.
1. CH3NH3+
2. (CH3)2NH2+
3. (CH3)3NH+
4. (CH3)4N+
Solution
The ions have one, two, three, and four methyl (CH3) groups attached to a nitrogen atom. Their names are as follows:
1. methylammonium ion
2. dimethylammonium ion
3. trimethylammonium ion
4. tetramethylammonium ion
Exercise \(6\)
Name each ion.
1. CH3CH2NH3+
2. (CH3CH2)3NH+
3. (CH3CH2CH2)2NH2+
4. (CH3CH2CH2CH2)4N+
Summary
An amine is a derivative of ammonia in which one, two, or all three hydrogen atoms are replaced by hydrocarbon groups. The amine functional group is as follows:
Amines are classified as primary, secondary, or tertiary by the number of hydrocarbon groups attached to the nitrogen atom. Amines are named by naming the alkyl groups attached to the nitrogen atom, followed by the suffix -amine.
15.11: Physical Properties of Amines
Learning Objectives
• Explain why the boiling points of primary and secondary amines are higher than those of alkanes or ethers of similar molar mass but are lower than those of alcohols.
• Compare the boiling points of tertiary amines with alcohols, alkanes, and ethers of similar molar mass.
• Compare the solubilities in water of amines of five or fewer carbon atoms with the solubilities of comparable alkanes and alcohols in water.
Primary and secondary amines have hydrogen atoms bonded to an nitrogen atom and are therefore capable of hydrogen bonding (part (a) of Figure $1$), although not as strongly as alcohol molecules (which have hydrogen atoms bonded to an oxygen atom, which is more electronegative than nitrogen). These amines boil at higher temperatures than alkanes but at lower temperatures than alcohols of comparable molar mass. For example, compare the boiling point of methylamine (CH3NH2; −6°C) with those of ethane (CH3CH3; −89°C) and methanol (CH3OH; 65°C). Tertiary amines have no hydrogen atom bonded to the nitrogen atom and so cannot participate in intermolecular hydrogen bonding. They have boiling points comparable to those of ethers (Table $1$).
Table $1$: Physical Properties of Some Amines and Comparable Oxygen-Containing Compounds
Name Condensed Structural Formula Class Molar Mass Boiling Point (°C) Solubility at 25°C (g/100 g Water)
butylamine CH3CH2CH2CH2NH2 73 78 miscible
diethylamine (CH3CH2)2NH 73 55 miscible
butyl alcohol CH3CH2CH2CH2OH 74 118 8
dipropylamine (CH3CH2CH2)2NH 101 111 4
triethylamine (CH3CH2)3N 101 90 14
dipropyl ether (CH3CH2CH2)2O 102 91 0.25
All three classes of amines can engage in hydrogen bonding with water (Figure $\PageIndex{1b}$). Amines of low molar mass are quite soluble in water; the borderline of solubility in water is at five or six carbon atoms.
To Your Health: Amines in Death and Life
Amines have “interesting” odors. The simple ones smell very much like ammonia. Higher aliphatic amines smell like decaying fish. Or perhaps we should put it the other way around: Decaying fish give off odorous amines. The stench of rotting fish is due in part to two diamines: putrescine and cadaverine. They arise from the decarboxylation of ornithine and lysine, respectively, amino acids that are found in animal cells.
$\ce{HOCH2CH2OH} \nonumber$
Aromatic amines generally are quite toxic. They are readily absorbed through the skin, and workers must exercise caution when handling these compounds. Several aromatic amines, including β-naphthylamine, are potent carcinogens.
Key Takeaways
• Primary and secondary amines have higher boiling points than those of alkanes or ethers of similar molar mass because they can engage in intermolecular hydrogen bonding. Their boiling points are lower than those of alcohols because alcohol molecules have hydrogen atoms bonded to an oxygen atom, which is more electronegative.
• The boiling points of tertiary amines, which cannot engage in hydrogen bonding because they have no hydrogen atom on the nitrogen atom, are comparable to those of alkanes and ethers of similar molar mass.
• Because all three classes of amines can engage in hydrogen bonding with water, amines of low molar mass are quite soluble in water. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.10%3A_Amines_-_Structures_and_Names.txt |
Learning Objectives
• Name the typical reactions that take place with amines.
• Describe heterocyclic amines.
Recall that ammonia (NH3) acts as a base because the nitrogen atom has a lone pair of electrons that can accept a proton. Amines also have a lone electron pair on their nitrogen atoms and can accept a proton from water to form substituted ammonium (NH4+) ions and hydroxide (OH) ions:
As a specific example, methylamine reacts with water to form the methylammonium ion and the OH ion.
Nearly all amines, including those that are not very soluble in water, will react with strong acids to form salts soluble in water.
Amine salts are named like other salts: the name of the cation is followed by the name of the anion.
Example $1$
What are the formulas of the acid and base that react to form [CH3NH2CH2CH3]+CH3COO?
Solution
The cation has two groups—methyl and ethyl—attached to the nitrogen atom. It comes from ethylmethylamine (CH3NHCH2CH3). The anion is the acetate ion. It comes from acetic acid (CH3COOH).
Exercise $1$
What are the formulas of the acid and base that react to form $\ce{(CH3CH2CH2)3NH^{+}I^{−}}$?
To Your Health: Amine Salts as Drugs
Salts of aniline are properly named as anilinium compounds, but an older system, still in use for naming drugs, identifies the salt of aniline and hydrochloric acid as “aniline hydrochloride.” These compounds are ionic—they are salts—and the properties of the compounds (solubility, for example) are those characteristic of salts. Many drugs that are amines are converted to hydrochloride salts to increase their solubility in aqueous solution.
Heterocyclic Amines
Looking back at the various cyclic hydrocarbons discussed previously, we see that all the atoms in the rings of these compounds are carbon atoms. In other cyclic compounds, called heterocyclic compounds (Greek heteros, meaning “other”), nitrogen, oxygen, sulfur, or some other atom is incorporated in the ring. Many heterocyclic compounds are important in medicine and biochemistry. Some compose part of the structure of the nucleic acids, which in turn compose the genetic material of cells and direct protein synthesis.
Many heterocyclic amines occur naturally in plants. Like other amines, these compounds are basic. Such a compound is an alkaloid, a name that means “like alkalis.” Many alkaloids are physiologically active, including the familiar drugs caffeine, nicotine, and cocaine.
To Your Health: Three Well-Known Alkaloids
Caffeine is a stimulant found in coffee, tea, and some soft drinks. Its mechanism of action is not well understood, but it is thought to block the activity of adenosine, a heterocyclic base that acts as a neurotransmitter, a substance that carries messages across a tiny gap (synapse) from one nerve cell (neuron) to another cell. The effective dose of caffeine is about 200 mg, corresponding to about two cups of strong coffee or tea.
Nicotine acts as a stimulant by a different mechanism; it probably mimics the action of the neurotransmitter acetylcholine. People ingest this drug by smoking or chewing tobacco. Its stimulant effect seems transient, as this initial response is followed by depression. Nicotine is highly toxic to animals. It is especially deadly when injected; the lethal dose for a human is estimated to be about 50 mg. Nicotine has also been used in agriculture as a contact insecticide.
Cocaine acts as a stimulant by preventing nerve cells from taking up dopamine, another neurotransmitter, from the synapse. High levels of dopamine are therefore available to stimulate the pleasure centers of the brain. The enhancement of dopamine action is thought to be responsible for cocaine’s “high” and its addictive properties. After the binge, dopamine is depleted in less than an hour. This leaves the user in a pleasureless state and (often) craving more cocaine.
Cocaine is used as the salt cocaine hydrochloride and in the form of broken lumps of the free (unneutralized) base, which is called crack cocaine.
$\underbrace{\ce{C17H21O4N}}_{\text{cocaine (freebase)}} + \ce{HCl ->} \underbrace{\ce{C17H21O4NH^{+}Cl^{-}}}_{\text{cocaine hydrochloride}} \nonumber$
Because it is soluble in water, cocaine hydrochloride is readily absorbed through the watery mucous membranes of the nose when it is snorted. Crack cocaine is more volatile than cocaine hydrochloride. It vaporizes at the temperature of a burning cigarette. When smoked, cocaine reaches the brain in 15 s.
Summary
Amines are bases; they react with acids to form salts. Salts of aniline are properly named as anilinium compounds, but an older system is used to name drugs: the salts of amine drugs and hydrochloric acid are called “hydrochlorides.” Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.12%3A_Amines_as_Bases.txt |
Learning Objectives
• Identify the general structure for an amide.
• Identify the functional group for an amide.
• Names amides with common names.
• Name amides according to the IUPAC system.
The amide functional group has an nitrogen atom attached to a carbonyl carbon atom. If the two remaining bonds on the nitrogen atom are attached to hydrogen atoms, the compound is a simple amide. If one or both of the two remaining bonds on the atom are attached to alkyl or aryl groups, the compound is a substituted amide.
The carbonyl carbon-to-nitrogen bond is called an amide linkage. This bond is quite stable and is found in the repeating units of protein molecules, where it is called a peptide linkage.
Simple amides are named as derivatives of carboxylic acids. The -ic ending of the common name or the -oic ending of the International Union of Pure and Applied Chemistry (IUPAC) name of the carboxylic acid is replaced with the suffix -amide.
Example \(1\)
Name each compound with the common name, the IUPAC name, or both.
Solution
1. This amide has two carbon atoms and is thus derived from acetic acid. The OH of acetic acid is replaced by an NH2 group. The -ic from acetic (or -oic from ethanoic) is dropped, and -amide is added to give acetamide (or ethanamide in the IUPAC system).
2. This amide is derived from benzoic acid. The -oic is dropped, and -amide is added to give benzamide.
Exercise \(1\)
Name each compound with the common name, the IUPAC name, or both.
Key Takeaways
• Amides have a general structure in which a nitrogen atom is bonded to a carbonyl carbon atom.
• The functional group for an amide is as follows:
• In names for amides, the -ic acid of the common name or the -oic ending of the IUPAC for the corresponding carboxylic acid is replaced by -amide.
15.14: Physical Properties of Amides
Learning Objectives
• Compare the boiling points of amides with alcohols of similar molar mass.
• Compare the solubilities in water of amides of five or fewer carbon atoms with the solubilities of comparable alkanes and alcohols in water.
With the exception of formamide (HCONH2), which is a liquid, all simple amides are solids (Table \(1\)). The lower members of the series are soluble in water, with borderline solubility occurring in those that have five or six carbon atoms. Like the esters, solutions of amides in water usually are neutral—neither acidic nor basic.
Table \(1\): Physical Constants of Some Unsubstituted Amides
Condensed Structural Formula Name Melting Point (°C) Boiling Point (°C) Solubility in Water
HCONH2 formamide 2 193 soluble
CH3CONH2 acetamide 82 222 soluble
CH3CH2CONH2 propionamide 81 213 soluble
CH3CH2CH2CONH2 butyramide 115 216 soluble
C6H5CONH2 benzamide 132 290 slightly soluble
The amides generally have high boiling points and melting points. These characteristics and their solubility in water result from the polar nature of the amide group and hydrogen bonding (Figure \(1\)). (Similar hydrogen bonding plays a critical role in determining the structure and properties of proteins, deoxyribonucleic acid [DNA], ribonucleic acid [RNA], and other giant molecules so important to life processes.
Key Takeaways
• Most amides are solids at room temperature; the boiling points of amides are much higher than those of alcohols of similar molar mass.
• Amides of five or fewer carbon atoms are soluble in water?. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.13%3A_Amides-_Structures_and_Names.txt |
Learning Objectives
• To describe the preparation procedure for amides.
The addition of ammonia (NH3) to a carboxylic acid forms an amide, but the reaction is very slow in the laboratory at room temperature. Water molecules are split out, and a bond is formed between the nitrogen atom and the carbonyl carbon atom.
In living cells, amide formation is catalyzed by enzymes. Proteins are polyamides; they are formed by joining amino acids into long chains. In proteins, the amide functional group is called a peptide bond.
Polyamides
Just as the reaction of a diol and a diacid forms a polyester, the reaction of a diacid and a diamine yields a polyamide. The two difunctional monomers often employed are adipic acid and 1,6-hexanediamine. The monomers condense by splitting out water to form a new product, which is still difunctional and thus can react further to yield a polyamide polymer.
Some polyamides are known as nylons. Nylons are among the most widely used synthetic fibers—for example, they are used in ropes, sails, carpets, clothing, tires, brushes, and parachutes. They also can be molded into blocks for use in electrical equipment, gears, bearings, and valves.
Key Takeaway
• Amides are prepared by the reaction of a carboxylic acid with ammonia or an amine.
15.16: Chemical Properties of Amides- Hydrolysis
Learning Objectives
• To identify the typical reaction that amides undergo.
Generally, amides resist hydrolysis in plain water, even after prolonged heating. In the presence of added acid or base, however, hydrolysis proceeds at a moderate rate. In living cells, amide hydrolysis is catalyzed by enzymes. Amide hydrolysis is illustrated in the following example:
Hydrolysis of an amide in acid solution actually gives a carboxylic acid and the salt of ammonia or an amine (the ammonia or amine initially formed is neutralized by the acid). Basic hydrolysis gives a salt of the carboxylic acid and ammonia or an amine.
Example \(1\)
Write the equation for the hydrolysis of each compound.
1. butyramide
2. benzamide
Solution
1. The hydrolysis of a simple amide produces an organic acid and ammonia. Butyramide thus yields butyric acid and ammonia.
• The hydrolysis of an amide produces an organic acid and ammonia. Benzamide thus yields benzoic acid and ammonia.
Exercise \(1\)
Write the equation for the hydrolysis of each compound.
1. propionamide (propanamide)
2. hexanamide
Career Focus: Athletic Trainer
Athletic training is an allied health-care profession recognized by the American Medical Association. The athletic trainer’s role is to recognize, evaluate, and provide immediate care for athletic injuries; prevent athletic injuries by taping, bandaging, and bracing vulnerable body parts; make referrals to medical doctors when necessary; and rehabilitate injured athletes. Athletic trainers work in high schools, colleges, and other organizations where athletics programs are found. Athletic trainers usually have a degree from an accredited athletic training program whose curriculum includes such basic science courses as biology, chemistry, and physics. These studies provide the necessary background for more applied courses, such as anatomy and physiology, exercise physiology, kinesiology, and nutrition. Knowledge of chemistry is necessary for understanding pharmacological and medical terminology. For example, athletic trainers must understand the action of numerous drugs, many of which are esters, amines, or amides like those mentioned in this chapter.
Athletic trainers may have administrative duties, such as the responsibility for ordering supplies. They also need to be able to evaluate nutritional supplements because providing the wrong one can get an athlete banned from competition and may bring sanctions against a school. In short, the athletic trainer is responsible for the overall health and well-being of the athletes in his or her charge.
Key Takeaway
• The hydrolysis of an amide produces a carboxylic acid and ammonia or an amine. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.15%3A_Formation_of_Amides.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the summary and ask yourself how they relate to the topics in the chapter.
A carboxylic acid (RCOOH) contains the functional group COOH, called the carboxyl group, which has an OH group attached to a carbonyl carbon atom. An ester (RCOOR′) has an OR′ group attached to a carbonyl carbon atom. An amine is derived from ammonia (NH3), with one, two, or all three of the hydrogen atoms of NH3 replaced by an alkyl (or an aryl) group. The amide functional group has a carbonyl group joined to a nitrogen atom from NH3 or an amine.
There are many familiar carboxylic acids. The R group may be a hydrogen atom (as in formic acid, HCOOH), an alkyl group (as in acetic acid, CH2COOH), or an aryl group (as in benzoic acid, C6H5COOH). The location of substituents along the carbon chain is indicated by a Greek letter (for common names) or a number (for names from the International Union of Pure and Applied Chemistry).
A carboxylic acid is formed by the oxidation of an aldehyde with the same number of carbon atoms. Because aldehydes are formed from primary alcohols, these alcohols are also a starting material for carboxylic acids.
Carboxylic acids have strong, often disagreeable, odors. They are highly polar molecules and readily engage in hydrogen bonding, so they have relatively high boiling points.
Carboxylic acids are weak acids. They react with bases to form salts and with carbonates and bicarbonates to form carbon dioxide gas and the salt of the acid.
Esters are pleasant-smelling compounds that are responsible for the fragrances of flowers and fruits. They have lower boiling points than comparable carboxylic acids because, even though ester molecules are somewhat polar, they cannot engage in hydrogen bonding. However, with water, esters can engage in hydrogen bonding; consequently, the low molar mass esters are soluble in water. Esters can be synthesized by esterification, in which a carboxylic acid and an alcohol are combined under acidic conditions. Esters are neutral compounds that undergo hydrolysis, a reaction with water. Under acidic conditions, hydrolysis is essentially the reverse of esterification. When carried out under basic conditions, the process is called saponification.
Inorganic acids also react with alcohols to form esters. Some of the most important esters in biochemistry are those formed from phosphoric acid.
Amines are nitrogen-containing organic molecules derived from ammonia (NH3). A primary (1°) amine (RNH2) has one organic group bonded to the nitrogen atom, a secondary (2°) amine (R2NH) has two organic groups bonded to the nitrogen atom, and a tertiary (3°) amine (R3N) has three organic groups bonded to the nitrogen atom. Amines are basic compounds that react with strong acids to produce ammonium (NH4+) salts. A cyclic compound in which the ring contains one or more noncarbon atoms is called a heterocyclic compound. There are many heterocyclic amines, including many physiologically important ones. Alkaloids are heterocyclic amines found in many plants. Caffeine, nicotine, and cocaine are familiar alkaloids.
Organic compounds containing a carbonyl group bonded to a nitrogen atom are amides, and the carbon-to-nitrogen bond is an amide linkage (or a peptide linkage). Most amides are colorless and odorless, and the lighter ones are soluble in water. Because they are polar molecules, amides have comparatively high boiling points and melting points. Amides are synthesized from carboxylic acids and NH3 or amines. Amides are neutral compounds. They resist hydrolysis in water, but acids, bases, and enzymes catalyze the reaction. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.S%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives__%28Summary%.txt |
Glucose is one of the carbohydrates you will learn about in this chapter as we begin the study of biochemistry—the chemistry of molecules found in living organisms. Later we will study the other three major types of macromolecules found in living organisms: lipids, proteins, and nucleic acids.
• 16.0: Prelude to Carbohydrates
People with diabetes are impaired in their ability to metabolize glucose, a sugar needed by the body for energy; as a result, excessive quantities of glucose accumulate in the blood and the urine. The characteristic symptoms of diabetes are weight loss, constant hunger, extreme thirst, and frequent urination (the kidneys excrete large amounts of water in an attempt to remove the excess sugar from the blood).
• 16.1: Carbohydrates
Carbohydrates are an important group of biological molecules that includes sugars and starches. Photosynthesis is the process by which plants use energy from sunlight to synthesize carbohydrates. A monosaccharide is the simplest carbohydrate and cannot be hydrolyzed to produce a smaller carbohydrate molecule. Disaccharides contain two monosaccharide units, and polysaccharides contain many monosaccharide units.
• 16.2: Classes of Monosaccharides
Monosaccharides can be classified by the number of carbon atoms in the structure and/or the type of carbonyl group they contain (aldose or ketose). Most monosaccharides contain at least one chiral carbon and can form stereoisomers. Enantiomers are a specific type of stereoisomers that are mirror images of each other.
• 16.3: Important Hexoses
Three abundant hexoses in living organisms are the aldohexoses D-glucose and D-galactose and the ketohexose D-fructose.
• 16.4: Cyclic Structures of Monosaccharides
Monosaccharides that contain five or more carbons atoms form cyclic structures in aqueous solution. Two cyclic stereoisomers can form from each straight-chain monosaccharide; these are known as anomers. In an aqueous solution, an equilibrium mixture forms between the two anomers and the straight-chain structure of a monosaccharide in a process known as mutarotation.
• 16.5: Properties of Monosaccharides
Monosaccharides are crystalline solids at room temperature and quite soluble in water. Monosaccharides are reducing sugars; they reduce mild oxidizing agents, such as Tollens’ or Benedict’s reagents.
• 16.6: Disaccharides
Maltose is composed of two molecules of glucose joined by an α-1,4-glycosidic linkage. It is a reducing sugar that is found in sprouting grain. Lactose is composed of a molecule of galactose joined to a molecule of glucose by a β-1,4-glycosidic linkage. It is a reducing sugar that is found in milk. Sucrose is composed of a molecule of glucose joined to a molecule of fructose by an α-1,β-2-glycosidic linkage. It is a nonreducing sugar that is found in sugar cane and sugar beets.
• 16.7: Polysaccharides
Starch is a storage form of energy in plants. It contains two polymers composed of glucose units: amylose (linear) and amylopectin (branched). Glycogen is a storage form of energy in animals. It is a branched polymer composed of glucose units. It is more highly branched than amylopectin. Cellulose is a structural polymer of glucose units found in plants. It is a linear polymer with the glucose units linked through β-1,4-glycosidic bonds.
• 16.S: Carbohydrates (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
16: Carbohydrates
In the United States, 17.9 million people have been diagnosed with diabetes, and experts estimate that at least another 5.7 million people have the disease but have not been diagnosed. In 2006, diabetes was the seventh leading cause of death, listed on 72,507 death certificates. Moreover, it was a contributing factor in over 200,000 deaths in which the cause was listed as something else, such as heart or kidney disease.
People with diabetes are impaired in their ability to metabolize glucose, a sugar needed by the body for energy; as a result, excessive quantities of glucose accumulate in the blood and the urine. The characteristic symptoms of diabetes are weight loss, constant hunger, extreme thirst, and frequent urination (the kidneys excrete large amounts of water in an attempt to remove the excess sugar from the blood).
An important diagnostic test for diabetes is the oral glucose tolerance test, which measures the level of glucose in blood plasma. A first measurement is made after a fast of at least 8 h, followed by another measurement 2 h after the person drinks a flavored solution of 75 g of glucose dissolved in water. At the second measurement, the glucose plasma level should be no higher than 139 mg/dL. Individuals with a value between 140 and 199 mg/dL are diagnosed with prediabetes, while those with a value of 200 mg/dL or above are diagnosed with diabetes. Following a diagnosis of diabetes a person will need to monitor his or her blood glucose levels daily (or more often) using a glucose meter. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/16%3A_Carbohydrates/16.00%3A_Prelude_to_Carbohydrates.txt |
Learning Objectives
• To recognize carbohydrates and classify them as mono-, di-, or polysaccharides.
All carbohydrates consist of carbon, hydrogen, and oxygen atoms and are polyhydroxy aldehydes or ketones or are compounds that can be broken down to form such compounds. Examples of carbohydrates include starch, fiber, the sweet-tasting compounds called sugars, and structural materials such as cellulose. The term carbohydrate had its origin in a misinterpretation of the molecular formulas of many of these substances. For example, because its formula is C6H12O6, glucose was once thought to be a “carbon hydrate” with the structure C6·6H2O.
Example $1$
Which compounds would be classified as carbohydrates?
Solution
1. This is a carbohydrate because the molecule contains an aldehyde functional group with OH groups on the other two carbon atoms.
2. This is not a carbohydrate because the molecule does not contain an aldehyde or a ketone functional group.
3. This is a carbohydrate because the molecule contains a ketone functional group with OH groups on the other two carbon atoms.
4. This is not a carbohydrate; although it has a ketone functional group, one of the other carbons atoms does not have an OH group attached.
Exercise $1$
Which compounds would be classified as carbohydrates?
Green plants are capable of synthesizing glucose (C6H12O6) from carbon dioxide (CO2) and water (H2O) by using solar energy in the process known as photosynthesis:
$\ce{6CO_2 + 6H_2O} + \text{686 kcal} \rightarrow \ce{C_6H_{12}O_6 + 6O_2} \label{$1$}$
(The 686 kcal come from solar energy.) Plants can use the glucose for energy or convert it to larger carbohydrates, such as starch or cellulose. Starch provides energy for later use, perhaps as nourishment for a plant’s seeds, while cellulose is the structural material of plants. We can gather and eat the parts of a plant that store energy—seeds, roots, tubers, and fruits—and use some of that energy ourselves. Carbohydrates are also needed for the synthesis of nucleic acids and many proteins and lipids.
Animals, including humans, cannot synthesize carbohydrates from carbon dioxide and water and are therefore dependent on the plant kingdom to provide these vital compounds. We use carbohydrates not only for food (about 60%–65% by mass of the average diet) but also for clothing (cotton, linen, rayon), shelter (wood), fuel (wood), and paper (wood).
The simplest carbohydrates—those that cannot be hydrolyzed to produce even smaller carbohydrates—are called monosaccharides. Two or more monosaccharides can link together to form chains that contain from two to several hundred or thousand monosaccharide units. Prefixes are used to indicate the number of such units in the chains. Disaccharide molecules have two monosaccharide units, trisaccharide molecules have three units, and so on. Chains with many monosaccharide units joined together are called polysaccharides. All these so-called higher saccharides can be hydrolyzed back to their constituent monosaccharides.
Compounds that cannot be hydrolyzed will not react with water to form two or more smaller compounds.
Summary
Carbohydrates are an important group of biological molecules that includes sugars and starches. Photosynthesis is the process by which plants use energy from sunlight to synthesize carbohydrates. A monosaccharide is the simplest carbohydrate and cannot be hydrolyzed to produce a smaller carbohydrate molecule. Disaccharides contain two monosaccharide units, and polysaccharides contain many monosaccharide units. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/16%3A_Carbohydrates/16.01%3A_Carbohydrates.txt |
Learning Objectives
• Classify monosaccharides as aldoses or ketoses and as trioses, tetroses, pentoses, or hexoses.
• Distinguish between a D sugar and an L sugar.
The naturally occurring monosaccharides contain three to seven carbon atoms per molecule. Monosaccharides of specific sizes may be indicated by names composed of a stem denoting the number of carbon atoms and the suffix -ose. For example, the terms triose, tetrose, pentose, and hexose signify monosaccharides with, respectively, three, four, five, and six carbon atoms. Monosaccharides are also classified as aldoses or ketoses. Those monosaccharides that contain an aldehyde functional group are called aldoses; those containing a ketone functional group on the second carbon atom are ketoses. Combining these classification systems gives general names that indicate both the type of carbonyl group and the number of carbon atoms in a molecule. Thus, monosaccharides are described as aldotetroses, aldopentoses, ketopentoses, ketoheptoses, and so forth. Glucose and fructose are specific examples of an aldohexose and a ketohexose, respectively.
Example \(1\)
Draw an example of each type of compound.
1. a ketopentose
2. an aldotetrose
Solution
1. The structure must have five carbon atoms with the second carbon atom being a carbonyl group and the other four carbon atoms each having an OH group attached. Several structures are possible, but one example is shown.
• The structure must have four carbon atoms with the first carbon atom part of the aldehyde functional group. The other three carbon atoms each have an OH group attached. Several structures are possible, but one example is shown.
Exercise \(1\)
Draw an example of each type of compound.
1. an aldohexose
2. a ketotetrose
The simplest sugars are the trioses. The possible trioses are shown in part (a) of Figure \(1\); glyceraldehyde is an aldotriose, while dihydroxyacetone is a ketotriose. Notice that two structures are shown for glyceraldehyde. These structures are stereoisomers, and hence are isomers having the same structural formula but differing in the arrangement of atoms or groups of atoms in three-dimensional space. If you make models of the two stereoisomers of glyceraldehyde, you will find that you cannot place one model on top of the other and have each functional group point in the same direction. However, if you place one of the models in front of a mirror, the image in the mirror will be identical to the second stereoisomer in part (b) of Figure \(1\). Molecules that are nonsuperimposable (nonidentical) mirror images of each other are a type of stereoisomer called enantiomers (Greek enantios, meaning “opposite”).
Note
These are another type of stereoisomers than the cis-trans (geometric) isomers previously discussed.
A key characteristic of enantiomers is that they have a carbon atom to which four different groups are attached. Note, for example, the four different groups attached to the central carbon atom of glyceraldehyde (part (a) of Figure \(1\)). A carbon atom that has four different groups attached is a chiral carbon. If a molecule contains one or more chiral carbons, it is likely to exist as two or more stereoisomers. Dihydroxyacetone does not contain a chiral carbon and thus does not exist as a pair of stereoisomers. Glyceraldehyde, however, has a chiral carbon and exists as a pair of enantiomers. Except for the direction in which each enantiomer rotates plane-polarized light, these two molecules have identical physical properties. One enantiomer has a specific rotation of +8.7°, while the other has a specific rotation of −8.7°.
H. Emil Fischer, a German chemist, developed the convention commonly used for writing two-dimensional representations of the monosaccharides, such as those in part (a) of Figure \(1\). In these structural formulas, the aldehyde group is written at the top, and the hydrogen atoms and OH groups that are attached to each chiral carbon are written to the right or left. (If the monosaccharide is a ketose, the ketone functional group is the second carbon atom.) Vertical lines represent bonds pointing away from you, while horizontal lines represent bonds coming toward you. The formulas of chiral molecules represented in this manner are referred to as Fischer projections.
The two enantiomers of glyceraldehyde are especially important because monosaccharides with more than three carbon atoms can be considered as being derived from them. Thus, D- and L-glyceraldehyde provide reference points for designating and drawing all other monosaccharides. Sugars whose Fischer projections terminate in the same configuration as D-glyceraldehyde are designated as D sugars; those derived from L-glyceraldehyde are designated as L sugars.
By convention, the penultimate (next-to-last) carbon atom has been chosen as the carbon atom that determines if a sugar is D or L. It is the chiral carbon farthest from the aldehyde or ketone functional group.
Looking Closer: Polarized Light
A beam of ordinary light can be pictured as a bundle of waves; some move up and down, some sideways, and others at all other conceivable angles. When a beam of light has been polarized, however, the waves in the bundle all vibrate in a single plane. Light altered in this way is called plane-polarized light. Much of what chemists know about stereoisomers comes from studying the effects they have on plane-polarized light. In this illustration, the light on the left is not polarized, while that on the right is polarized.
Sunlight, in general, is not polarized; light from an ordinary light bulb or an ordinary flashlight is not polarized. One way to polarize ordinary light is to pass it through Polaroid sheets, special plastic sheets containing carefully oriented organic compounds that permit only light vibrating in a single plane to pass through. To the eye, polarized light doesn’t “look” any different from nonpolarized light. We can detect polarized light, however, by using a second sheet of polarizing material, as shown here.
In the photo on the left, two Polaroid sheets are aligned in the same direction; plane-polarized light from the first Polaroid sheet can pass through the second sheet. In the photo on the right, the top Polaroid sheet has been rotated 90° and now blocks the plane-polarized light that comes through the first Polaroid sheet.
Certain substances act on polarized light by rotating the plane of vibration. Such substances are said to be optically active. The extent of optical activity is measured by a polarimeter, an instrument that contains two polarizing lenses separated by a sample tube, as shown in the accompanying figure. With the sample tube empty, maximum light reaches the observer’s eye when the two lenses are aligned so that both pass light vibrating in the same plane. When an optically active substance is placed in the sample tube, that substance rotates the plane of polarization of the light passing through it, so that the polarized light emerging from the sample tube is vibrating in a different direction than when it entered the tube. To see the maximum amount of light when the sample is in place, the observer must rotate one lens to accommodate the change in the plane of polarization.
Some optically active substances rotate the plane of polarized light to the right (clockwise) from the observer’s point of view. These compounds are said to be dextrorotatory; substances that rotate light to the left (counterclockwise) are levorotatory. To denote the direction of rotation, a positive sign (+) is given to dextrorotatory substances, and a negative sign (−) is given to levorotatory substances.
Summary
Monosaccharides can be classified by the number of carbon atoms in the structure and/or the type of carbonyl group they contain (aldose or ketose). Most monosaccharides contain at least one chiral carbon and can form stereoisomers. Enantiomers are a specific type of stereoisomers that are mirror images of each other. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/16%3A_Carbohydrates/16.02%3A_Classes_of_Monosaccharides.txt |
Learning Objectives
• To identify the structures of D-glucose, D-galactose, and D-fructose and describe how they differ from each other.
Although a variety of monosaccharides are found in living organisms, three hexoses are particularly abundant: D-glucose, D-galactose, and D-fructose (Figure \(1\)). Glucose and galactose are both aldohexoses, while fructose is a ketohexose.
Glucose
D-Glucose, generally referred to as simply glucose, is the most abundant sugar found in nature; most of the carbohydrates we eat are eventually converted to it in a series of biochemical reactions that produce energy for our cells. It is also known by three other names: dextrose, from the fact that it rotates plane-polarized light in a clockwise (dextrorotatory) direction; corn sugar because in the United States cornstarch is used in the commercial process that produces glucose from the hydrolysis of starch; and blood sugar because it is the carbohydrate found in the circulatory system of animals. Normal blood sugar values range from 70 to 105 mg glucose/dL plasma, and normal urine may contain anywhere from a trace to 20 mg glucose/dL urine.
The Fischer projection of D-glucose is given in Figure \(2\). Glucose is a D sugar because the OH group on the fifth carbon atom (the chiral center farthest from the carbonyl group) is on the right. In fact, all the OH groups except the one on the third carbon atom are to the right.
Galactose
D-Galactose does not occur in nature in the uncombined state. It is released when lactose, a disaccharide found in milk, is hydrolyzed. The galactose needed by the human body for the synthesis of lactose is obtained by the metabolic conversion of D-glucose to D-galactose. Galactose is also an important constituent of the glycolipids that occur in the brain and the myelin sheath of nerve cells. For this reason it is also known as brain sugar. The structure of D-galactose is shown in Figure \(1\). Notice that the configuration differs from that of glucose only at the fourth carbon atom.
Fructose
D-Fructose, also shown in Figure \(1\), is the most abundant ketohexose. Note that from the third through the sixth carbon atoms, its structure is the same as that of glucose. It occurs, along with glucose and sucrose, in honey (which is 40% fructose) and sweet fruits. Fructose (from the Latin fructus, meaning “fruit”) is also referred to as levulose because it has a specific rotation that is strongly levorotatory (−92.4°). It is the sweetest sugar, being 1.7 times sweeter than sucrose, although many nonsugars are several hundred or several thousand times as sweet (Table \(1\)).
Table \(1\): The Relative Sweetness of Some Compounds (Sucrose = 100)
Compound Relative Sweetness
lactose 16
maltose 32
glucose 74
sucrose 100
fructose 173
aspartame 18,000
acesulfame K 20,000
saccharin 30,000
sucralose 60,000
Looking Closer: Artificial Sweeteners
Although sweetness is commonly associated with mono- and disaccharides, it is not a property found only in sugars. Several other kinds of organic compounds have been synthesized that are far superior as sweetening agents. These so-called high-intensity or artificial sweeteners are useful for people with diabetes or other medical conditions that require them to control their carbohydrate intake. The synthetic compounds are noncaloric or used in such small quantities that they do not add significantly to the caloric value of food.
The first artificial sweetener—saccharin—was discovered by accident in 1879. It is 300 times sweeter than sucrose, but it passes through the body unchanged and thus adds no calories to the diet. After its discovery, saccharin was used until it was banned in the early 1900s. However, during the sugar-short years of World War I, the ban was lifted and was not reinstated at the war’s end. One drawback to the use of saccharin is its bitter, metallic aftertaste. The initial solution to this problem was to combine saccharin with cyclamate, a second artificial sweetener discovered in 1937.
In the 1960s and 1970s, several clinical tests with laboratory animals implicated both cyclamate and saccharin as carcinogenic (cancer-causing) substances. The results from the cyclamate tests were completed first, and cyclamate was banned in the United States in 1969. Then a major study was released in Canada in 1977 indicating that saccharin increased the incidence of bladder cancer in rats. The US Food and Drug Administration (FDA) proposed a ban on saccharin that raised immediate public opposition because saccharin was the only artificial sweetener still available. In response, Congress passed the Saccharin Study and Labeling Act in 1977, permitting the use of saccharin as long as any product containing it was labeled with a consumer warning regarding the possible elevation of the risk of bladder cancer. Today this warning is no longer required; moreover, the FDA is currently reviewing the ban on cyclamate, as 75 additional studies and years of usage in other countries, such as Canada, have failed to show that it has any carcinogenic effect.
A third artificial sweetener, aspartame, was discovered in 1965. This white crystalline compound is about 180 times sweeter than sucrose and has no aftertaste. It was approved for use in 1981 and is used to sweeten a wide variety of foods because it blends well with other food flavors. Aspartame is not used in baked goods, however, because it is not heat stable.
In the body (or when heated), aspartame is initially hydrolyzed to three molecules: the amino acids aspartic acid and phenylalanine and an alcohol methanol. Repeated controversy regarding the safety of aspartame arises partly from the fact that the body metabolizes the released methanol to formaldehyde. It should be noted, though, that a glass of tomato juice has six times as much methanol as a similar amount of a diet soda containing aspartame. The only documented risk connected to aspartame use is for individuals with the genetic disease phenylketonuria (PKU); these individuals lack the enzyme needed to metabolize the phenylalanine released when aspartame is broken down by the body. Because of the danger to people with PKU, all products containing aspartame must carry a warning label.
Acesulfame K, discovered just two years after aspartame (1967), was approved for use in the United States in 1988. It is 200 times sweeter than sugar and, unlike aspartame, is heat stable. It has no lingering aftertaste.
One of the newest artificial sweeteners to gain FDA approval (April 1998) for use in the United States is sucralose, a white crystalline solid approximately 600 times sweeter than sucrose. Sucralose is synthesized from sucrose and has three chlorine atoms substituted for three OH groups. It is noncaloric because it passes through the body unchanged. It can be used in baking because it is heat stable.
All of the extensive clinical studies completed to date have indicated that these artificial sweeteners approved for use in the United States are safe for consumption by healthy individuals in moderate amounts.
Summary
Three abundant hexoses in living organisms are the aldohexoses D-glucose and D-galactose and the ketohexose D-fructose. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/16%3A_Carbohydrates/16.03%3A_Important_Hexoses.txt |
Learning Objectives
• Define what is meant by anomers and describe how they are formed.
• Explain what is meant by mutarotation.
So far we have represented monosaccharides as linear molecules, but many of them also adopt cyclic structures. This conversion occurs because of the ability of aldehydes and ketones to react with alcohols:
You might wonder why the aldehyde reacts with the OH group on the fifth carbon atom rather than the OH group on the second carbon atom next to it. Recall that cyclic alkanes containing five or six carbon atoms in the ring are the most stable. The same is true for monosaccharides that form cyclic structures: rings consisting of five or six carbon atoms are the most stable.
When a straight-chain monosaccharide, such as any of the structures shown in Figure \(1\), forms a cyclic structure, the carbonyl oxygen atom may be pushed either up or down, giving rise to two stereoisomers, as shown in Figure \(2\). The structure shown on the left side of Figure \(2\), with the OH group on the first carbon atom projected downward, represent what is called the alpha (α) form. The structures on the right side, with the OH group on the first carbon atom pointed upward, is the beta (β) form. These two stereoisomers of a cyclic monosaccharide are known as anomers; they differ in structure around the anomeric carbon—that is, the carbon atom that was the carbonyl carbon atom in the straight-chain form.
It is possible to obtain a sample of crystalline glucose in which all the molecules have the α structure or all have the β structure. The α form melts at 146°C and has a specific rotation of +112°, while the β form melts at 150°C and has a specific rotation of +18.7°. When the sample is dissolved in water, however, a mixture is soon produced containing both anomers as well as the straight-chain form, in dynamic equilibrium (part (a) of Figure \(2\)). You can start with a pure crystalline sample of glucose consisting entirely of either anomer, but as soon as the molecules dissolve in water, they open to form the carbonyl group and then reclose to form either the α or the β anomer. The opening and closing repeats continuously in an ongoing interconversion between anomeric forms and is referred to as mutarotation (Latin mutare, meaning “to change”). At equilibrium, the mixture consists of about 36% α-D-glucose, 64% β-D-glucose, and less than 0.02% of the open-chain aldehyde form. The observed rotation of this solution is +52.7°.
Even though only a small percentage of the molecules are in the open-chain aldehyde form at any time, the solution will nevertheless exhibit the characteristic reactions of an aldehyde. As the small amount of free aldehyde is used up in a reaction, there is a shift in the equilibrium to yield more aldehyde. Thus, all the molecules may eventually react, even though very little free aldehyde is present at a time.
Commonly, (e.g., in Figures \(1\) and \(2\)) the cyclic forms of sugars are depicted using a convention first suggested by Walter N. Haworth, an English chemist. The molecules are drawn as planar hexagons with a darkened edge representing the side facing toward the viewer. The structure is simplified to show only the functional groups attached to the carbon atoms. Any group written to the right in a Fischer projection appears below the plane of the ring in a Haworth projection, and any group written to the left in a Fischer projection appears above the plane in a Haworth projection.
The difference between the α and the β forms of sugars may seem trivial, but such structural differences are often crucial in biochemical reactions. This explains why we can get energy from the starch in potatoes and other plants but not from cellulose, even though both starch and cellulose are polysaccharides composed of glucose molecules linked together.
Summary
Monosaccharides that contain five or more carbons atoms form cyclic structures in aqueous solution. Two cyclic stereoisomers can form from each straight-chain monosaccharide; these are known as anomers. In an aqueous solution, an equilibrium mixture forms between the two anomers and the straight-chain structure of a monosaccharide in a process known as mutarotation.
16.05: Properties of Monosaccharides
Learning Objectives
• To identify the physical and chemical properties of monosaccharides.
Monosaccharides such as glucose and fructose are crystalline solids at room temperature, but they are quite soluble in water, each molecule having several OH groups that readily engage in hydrogen bonding. The chemical behavior of these monosaccharides is likewise determined by their functional groups.
An important reaction of monosaccharides is the oxidation of the aldehyde group, one of the most easily oxidized organic functional groups. Aldehyde oxidation can be accomplished with any mild oxidizing agent, such as Tollens’ reagent or Benedict’s reagent. With the latter, complexed copper(II) ions are reduced to copper(I) ions that form a brick-red precipitate [copper(I) oxide; Figure \(1\)].
Any carbohydrate capable of reducing either Tollens’ or Benedict’s reagents without first undergoing hydrolysis is said to be a reducing sugar. Because both the Tollens’ and Benedict’s reagents are basic solutions, ketoses (such as fructose) also give positive tests due to an equilibrium that exists between ketoses and aldoses in a reaction known as tautomerism.
These reactions have been used as simple and rapid diagnostic tests for the presence of glucose in blood or urine. For example, Clinitest tablets, which are used to test for sugar in the urine, contain copper(II) ions and are based on Benedict’s test. A green color indicates very little sugar, whereas a brick-red color indicates sugar in excess of 2 g/100 mL of urine.
Summary
Monosaccharides are crystalline solids at room temperature and quite soluble in water. Monosaccharides are reducing sugars; they reduce mild oxidizing agents, such as Tollens’ or Benedict’s reagents. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/16%3A_Carbohydrates/16.04%3A_Cyclic_Structures_of_Monosaccharides.txt |
Learning Objectives
• Identify the structures of sucrose, lactose, and maltose.
• Identify the monosaccharides that are needed to form sucrose, lactose, and maltose
Previously, you learned that monosaccharides can form cyclic structures by the reaction of the carbonyl group with an OH group. These cyclic molecules can in turn react with another alcohol. Disaccharides (C12H22O11) are sugars composed of two monosaccharide units that are joined by a carbon–oxygen-carbon linkage known as a glycosidic linkage. This linkage is formed from the reaction of the anomeric carbon of one cyclic monosaccharide with the OH group of a second monosaccharide.
The disaccharides differ from one another in their monosaccharide constituents and in the specific type of glycosidic linkage connecting them. There are three common disaccharides: maltose, lactose, and sucrose. All three are white crystalline solids at room temperature and are soluble in water. We’ll consider each sugar in more detail.
Maltose
Maltose occurs to a limited extent in sprouting grain. It is formed most often by the partial hydrolysis of starch and glycogen. In the manufacture of beer, maltose is liberated by the action of malt (germinating barley) on starch; for this reason, it is often referred to as malt sugar. Maltose is about 30% as sweet as sucrose. The human body is unable to metabolize maltose or any other disaccharide directly from the diet because the molecules are too large to pass through the cell membranes of the intestinal wall. Therefore, an ingested disaccharide must first be broken down by hydrolysis into its two constituent monosaccharide units.
In the body, such hydrolysis reactions are catalyzed by enzymes such as maltase. The same reactions can be carried out in the laboratory with dilute acid as a catalyst, although in that case the rate is much slower, and high temperatures are required. Whether it occurs in the body or a glass beaker, the hydrolysis of maltose produces two molecules of D-glucose.
$\mathrm{maltose \xrightarrow{H^+\: or\: maltase} \textrm{2 D-glucose}} \nonumber$
Maltose is a reducing sugar. Thus, its two glucose molecules must be linked in such a way as to leave one anomeric carbon that can open to form an aldehyde group. The glucose units in maltose are joined in a head-to-tail fashion through an α-linkage from the first carbon atom of one glucose molecule to the fourth carbon atom of the second glucose molecule (that is, an α-1,4-glycosidic linkage; see Figure $1$). The bond from the anomeric carbon of the first monosaccharide unit is directed downward, which is why this is known as an α-glycosidic linkage. The OH group on the anomeric carbon of the second glucose can be in either the α or the β position, as shown in Figure $1$.
Lactose
Lactose is known as milk sugar because it occurs in the milk of humans, cows, and other mammals. In fact, the natural synthesis of lactose occurs only in mammary tissue, whereas most other carbohydrates are plant products. Human milk contains about 7.5% lactose, and cow’s milk contains about 4.5%. This sugar is one of the lowest ranking in terms of sweetness, being about one-sixth as sweet as sucrose. Lactose is produced commercially from whey, a by-product in the manufacture of cheese. It is important as an infant food and in the production of penicillin.
Lactose is a reducing sugar composed of one molecule of D-galactose and one molecule of D-glucose joined by a β-1,4-glycosidic bond (the bond from the anomeric carbon of the first monosaccharide unit being directed upward). The two monosaccharides are obtained from lactose by acid hydrolysis or the catalytic action of the enzyme lactase:
Many adults and some children suffer from a deficiency of lactase. These individuals are said to be lactose intolerant because they cannot digest the lactose found in milk. A more serious problem is the genetic disease galactosemia, which results from the absence of an enzyme needed to convert galactose to glucose. Certain bacteria can metabolize lactose, forming lactic acid as one of the products. This reaction is responsible for the “souring” of milk.
Example $1$
For this trisaccharide, indicate whether each glycosidic linkage is α or β.
Solution
The glycosidic linkage between sugars 1 and 2 is β because the bond is directed up from the anomeric carbon. The glycosidic linkage between sugars 2 and 3 is α because the bond is directed down from the anomeric carbon.
Exercise $1$
For this trisaccharide, indicate whether each glycosidic linkage is α or β.
To Your Health: Lactose Intolerance and Galactosemia
Lactose makes up about 40% of an infant’s diet during the first year of life. Infants and small children have one form of the enzyme lactase in their small intestines and can digest the sugar easily; however, adults usually have a less active form of the enzyme, and about 70% of the world’s adult population has some deficiency in its production. As a result, many adults experience a reduction in the ability to hydrolyze lactose to galactose and glucose in their small intestine. For some people the inability to synthesize sufficient enzyme increases with age. Up to 20% of the US population suffers some degree of lactose intolerance.
In people with lactose intolerance, some of the unhydrolyzed lactose passes into the colon, where it tends to draw water from the interstitial fluid into the intestinal lumen by osmosis. At the same time, intestinal bacteria may act on the lactose to produce organic acids and gases. The buildup of water and bacterial decay products leads to abdominal distention, cramps, and diarrhea, which are symptoms of the condition.
The symptoms disappear if milk or other sources of lactose are excluded from the diet or consumed only sparingly. Alternatively, many food stores now carry special brands of milk that have been pretreated with lactase to hydrolyze the lactose. Cooking or fermenting milk causes at least partial hydrolysis of the lactose, so some people with lactose intolerance are still able to enjoy cheese, yogurt, or cooked foods containing milk. The most common treatment for lactose intolerance, however, is the use of lactase preparations (e.g., Lactaid), which are available in liquid and tablet form at drugstores and grocery stores. These are taken orally with dairy foods—or may be added to them directly—to assist in their digestion.
Galactosemia is a condition in which one of the enzymes needed to convert galactose to glucose is missing. Consequently, the blood galactose level is markedly elevated, and galactose is found in the urine. An infant with galactosemia experiences a lack of appetite, weight loss, diarrhea, and jaundice. The disease may result in impaired liver function, cataracts, mental retardation, and even death. If galactosemia is recognized in early infancy, its effects can be prevented by the exclusion of milk and all other sources of galactose from the diet. As a child with galactosemia grows older, he or she usually develops an alternate pathway for metabolizing galactose, so the need to restrict milk is not permanent. The incidence of galactosemia in the United States is 1 in every 65,000 newborn babies.
Sucrose
Sucrose, probably the largest-selling pure organic compound in the world, is known as beet sugar, cane sugar, table sugar, or simply sugar. Most of the sucrose sold commercially is obtained from sugar cane and sugar beets (whose juices are 14%–20% sucrose) by evaporation of the water and recrystallization. The dark brown liquid that remains after the recrystallization of sugar is sold as molasses.
The sucrose molecule is unique among the common disaccharides in having an α-1,β-2-glycosidic (head-to-head) linkage. Because this glycosidic linkage is formed by the OH group on the anomeric carbon of α-D-glucose and the OH group on the anomeric carbon of β-D-fructose, it ties up the anomeric carbons of both glucose and fructose.
This linkage gives sucrose certain properties that are quite different from those of maltose and lactose. As long as the sucrose molecule remains intact, neither monosaccharide “uncyclizes” to form an open-chain structure. Thus, sucrose is incapable of mutarotation and exists in only one form both in the solid state and in solution. In addition, sucrose does not undergo reactions that are typical of aldehydes and ketones. Therefore, sucrose is a nonreducing sugar.
The hydrolysis of sucrose in dilute acid or through the action of the enzyme sucrase (also known as invertase) gives an equimolar mixture of glucose and fructose. This 1:1 mixture is referred to as invert sugar because it rotates plane-polarized light in the opposite direction than sucrose. The hydrolysis reaction has several practical applications. Sucrose readily recrystallizes from a solution, but invert sugar has a much greater tendency to remain in solution. In the manufacture of jelly and candy and in the canning of fruit, the recrystallization of sugar is undesirable. Therefore, conditions leading to the hydrolysis of sucrose are employed in these processes. Moreover, because fructose is sweeter than sucrose, the hydrolysis adds to the sweetening effect. Bees carry out this reaction when they make honey.
The average American consumes more than 100 lb of sucrose every year. About two-thirds of this amount is ingested in soft drinks, presweetened cereals, and other highly processed foods. The widespread use of sucrose is a contributing factor to obesity and tooth decay. Carbohydrates such as sucrose, are converted to fat when the caloric intake exceeds the body’s requirements, and sucrose causes tooth decay by promoting the formation of plaque that sticks to teeth.
Summary
Maltose is composed of two molecules of glucose joined by an α-1,4-glycosidic linkage. It is a reducing sugar that is found in sprouting grain. Lactose is composed of a molecule of galactose joined to a molecule of glucose by a β-1,4-glycosidic linkage. It is a reducing sugar that is found in milk. Sucrose is composed of a molecule of glucose joined to a molecule of fructose by an α-1,β-2-glycosidic linkage. It is a nonreducing sugar that is found in sugar cane and sugar beets | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/16%3A_Carbohydrates/16.06%3A_Disaccharides.txt |
Learning Objectives
• To compare and contrast the structures and uses of starch, glycogen, and cellulose.
The polysaccharides are the most abundant carbohydrates in nature and serve a variety of functions, such as energy storage or as components of plant cell walls. Polysaccharides are very large polymers composed of tens to thousands of monosaccharides joined together by glycosidic linkages. The three most abundant polysaccharides are starch, glycogen, and cellulose. These three are referred to as homopolymers because each yields only one type of monosaccharide (glucose) after complete hydrolysis. Heteropolymers may contain sugar acids, amino sugars, or noncarbohydrate substances in addition to monosaccharides. Heteropolymers are common in nature (gums, pectins, and other substances) but will not be discussed further in this textbook. The polysaccharides are nonreducing carbohydrates, are not sweet tasting, and do not undergo mutarotation.
Starch
Starch is the most important source of carbohydrates in the human diet and accounts for more than 50% of our carbohydrate intake. It occurs in plants in the form of granules, and these are particularly abundant in seeds (especially the cereal grains) and tubers, where they serve as a storage form of carbohydrates. The breakdown of starch to glucose nourishes the plant during periods of reduced photosynthetic activity. We often think of potatoes as a “starchy” food, yet other plants contain a much greater percentage of starch (potatoes 15%, wheat 55%, corn 65%, and rice 75%). Commercial starch is a white powder.
Starch is a mixture of two polymers: amylose and amylopectin. Natural starches consist of about 10%–30% amylose and 70%–90% amylopectin. Amylose is a linear polysaccharide composed entirely of D-glucose units joined by the α-1,4-glycosidic linkages we saw in maltose (part (a) of Figure \(1\)). Experimental evidence indicates that amylose is not a straight chain of glucose units but instead is coiled like a spring, with six glucose monomers per turn (part (b) of Figure \(1\)). When coiled in this fashion, amylose has just enough room in its core to accommodate an iodine molecule. The characteristic blue-violet color that appears when starch is treated with iodine is due to the formation of the amylose-iodine complex. This color test is sensitive enough to detect even minute amounts of starch in solution.
Amylopectin is a branched-chain polysaccharide composed of glucose units linked primarily by α-1,4-glycosidic bonds but with occasional α-1,6-glycosidic bonds, which are responsible for the branching. A molecule of amylopectin may contain many thousands of glucose units with branch points occurring about every 25–30 units (Figure \(2\)). The helical structure of amylopectin is disrupted by the branching of the chain, so instead of the deep blue-violet color amylose gives with iodine, amylopectin produces a less intense reddish brown.
Dextrins are glucose polysaccharides of intermediate size. The shine and stiffness imparted to clothing by starch are due to the presence of dextrins formed when clothing is ironed. Because of their characteristic stickiness with wetting, dextrins are used as adhesives on stamps, envelopes, and labels; as binders to hold pills and tablets together; and as pastes. Dextrins are more easily digested than starch and are therefore used extensively in the commercial preparation of infant foods.
The complete hydrolysis of starch yields, in successive stages, glucose:
starch → dextrins → maltose → glucose
In the human body, several enzymes known collectively as amylases degrade starch sequentially into usable glucose units.
Glycogen
Glycogen is the energy reserve carbohydrate of animals. Practically all mammalian cells contain some stored carbohydrates in the form of glycogen, but it is especially abundant in the liver (4%–8% by weight of tissue) and in skeletal muscle cells (0.5%–1.0%). Like starch in plants, glycogen is found as granules in liver and muscle cells. When fasting, animals draw on these glycogen reserves during the first day without food to obtain the glucose needed to maintain metabolic balance.
Glycogen is structurally quite similar to amylopectin, although glycogen is more highly branched (8–12 glucose units between branches) and the branches are shorter. When treated with iodine, glycogen gives a reddish brown color. Glycogen can be broken down into its D-glucose subunits by acid hydrolysis or by the same enzymes that catalyze the breakdown of starch. In animals, the enzyme phosphorylase catalyzes the breakdown of glycogen to phosphate esters of glucose.
About 70% of the total glycogen in the body is stored in muscle cells. Although the percentage of glycogen (by weight) is higher in the liver, the much greater mass of skeletal muscle stores a greater total amount of glycogen.
Cellulose
Cellulose, a fibrous carbohydrate found in all plants, is the structural component of plant cell walls. Because the earth is covered with vegetation, cellulose is the most abundant of all carbohydrates, accounting for over 50% of all the carbon found in the vegetable kingdom. Cotton fibrils and filter paper are almost entirely cellulose (about 95%), wood is about 50% cellulose, and the dry weight of leaves is about 10%–20% cellulose. The largest use of cellulose is in the manufacture of paper and paper products. Although the use of noncellulose synthetic fibers is increasing, rayon (made from cellulose) and cotton still account for over 70% of textile production.
Like amylose, cellulose is a linear polymer of glucose. It differs, however, in that the glucose units are joined by β-1,4-glycosidic linkages, producing a more extended structure than amylose (part (a) of Figure \(3\)). This extreme linearity allows a great deal of hydrogen bonding between OH groups on adjacent chains, causing them to pack closely into fibers (part (b) of Figure \(3\)). As a result, cellulose exhibits little interaction with water or any other solvent. Cotton and wood, for example, are completely insoluble in water and have considerable mechanical strength. Because cellulose does not have a helical structure, it does not bind to iodine to form a colored product.
Cellulose yields D-glucose after complete acid hydrolysis, yet humans are unable to metabolize cellulose as a source of glucose. Our digestive juices lack enzymes that can hydrolyze the β-glycosidic linkages found in cellulose, so although we can eat potatoes, we cannot eat grass. However, certain microorganisms can digest cellulose because they make the enzyme cellulase, which catalyzes the hydrolysis of cellulose. The presence of these microorganisms in the digestive tracts of herbivorous animals (such as cows, horses, and sheep) allows these animals to degrade the cellulose from plant material into glucose for energy. Termites also contain cellulase-secreting microorganisms and thus can subsist on a wood diet. This example once again demonstrates the extreme stereospecificity of biochemical processes.
Career Focus: Certified Diabetes Educator
Certified diabetes educators come from a variety of health professions, such as nursing and dietetics, and specialize in the education and treatment of patients with diabetes. A diabetes educator will work with patients to manage their diabetes. This involves teaching the patient to monitor blood sugar levels, make good food choices, develop and maintain an exercise program, and take medication, if required.
Summary
Starch is a storage form of energy in plants. It contains two polymers composed of glucose units: amylose (linear) and amylopectin (branched). Glycogen is a storage form of energy in animals. It is a branched polymer composed of glucose units. It is more highly branched than amylopectin. Cellulose is a structural polymer of glucose units found in plants. It is a linear polymer with the glucose units linked through β-1,4-glycosidic bonds. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/16%3A_Carbohydrates/16.07%3A_Polysaccharides.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Carbohydrates, a large group of biological compounds containing carbon, hydrogen, and oxygen atoms, include sugars, starch, glycogen, and cellulose. All carbohydrates contain alcohol functional groups, and either an aldehyde or a ketone group (or a functional group that can be converted to an aldehyde or ketone). The simplest carbohydrates are monosaccharides. Those with two monosaccharide units are disaccharides, and those with many monosaccharide units are polysaccharides. Most sugars are either monosaccharides or disaccharides. Cellulose, glycogen, and starch are polysaccharides.
Many carbohydrates exist as stereoisomers, in which the three-dimensional spatial arrangement of the atoms in space is the only difference between the isomers. These particular stereoisomers contain at least one chiral carbon, a carbon atom that has four different groups bonded to it. A molecule containing a chiral carbon is nonsuperimposable on its mirror image, and two molecules that are nonsuperimposable mirror images of each other are a special type of stereoisomer called enantiomers. Enantiomers have the same physical properties, such as melting point, but differ in the direction they rotate polarized light.
A sugar is designated as being a D sugar or an L sugar according to how, in a Fischer projection of the molecule, the hydrogen atom and OH group are attached to the penultimate carbon atom, which is the carbon atom immediately before the terminal alcohol carbon atom. If the structure at this carbon atom is the same as that of D-glyceraldehyde (OH to the right), the sugar is a D sugar; if the configuration is the same as that of L-glyceraldehyde (OH to the left), the sugar is an L sugar.
Monosaccharides of five or more carbons atoms readily form cyclic structures when the carbonyl carbon atom reacts with an OH group on a carbon atom three or four carbon atoms distant. Consequently, glucose in solution exists as an equilibrium mixture of three forms, two of them cyclic (α- and β-) and one open chain. In Haworth projections, the alpha form is drawn with the OH group on the “former” carbonyl carbon atom (anomeric carbon) pointing downward; the beta form, with the OH group pointing upward; these two compounds are stereoisomers and are given the more specific term of anomers. Any solid sugar can be all alpha or all beta. Once the sample is dissolved in water, however, the ring opens up into the open-chain structure and then closes to form either the α- or the β-anomer. These interconversions occur back and forth until a dynamic equilibrium mixture is achieved in a process called mutarotation.
The carbonyl group present in monosaccharides is easily oxidized by Tollens’ or Benedict’s reagents (as well as others). Any mono- or disaccharide containing a free anomeric carbon is a reducing sugar. The disaccharide maltose contains two glucose units joined in an α-1,4-glycosidic linkage. The disaccharide lactose contains a galactose unit and a glucose unit joined by a β-1,4-glycosidic linkage. Both maltose and lactose contain a free anomeric carbon that can convert to an aldehyde functional group, so they are reducing sugars; they also undergo mutarotation. Many adults, and some children, have a deficiency of the enzyme lactase (which is needed to break down lactose) and are said to be lactose intolerant. A more serious problem is the genetic disease galactosemia, which results from the absence of an enzyme needed to convert galactose to glucose.
The disaccharide sucrose (table sugar) consists of a glucose unit and a fructose unit joined by a glycosidic linkage. The linkage is designated as an α-1,β-2-glycosidic linkage because it involves the OH group on the first carbon atom of glucose and the OH group on the second carbon atom of fructose. Sucrose is not a reducing sugar because it has no anomeric carbon that can reform a carbonyl group, and it cannot undergo mutarotation because of the restrictions imposed by this linkage.
Starch, the principal carbohydrate of plants, is composed of the polysaccharides amylose (10%–30%) and amylopectin (70%–90%). When ingested by humans and other animals, starch is hydrolyzed to glucose and becomes the body’s energy source. Glycogen is the polysaccharide animals use to store excess carbohydrates from their diets. Similar in structure to amylopectin, glycogen is hydrolyzed to glucose whenever an animal needs energy for a metabolic process. The polysaccharide cellulose provides structure for plant cells. It is a linear polymer of glucose units joined by β-1,4-glycosidic linkages. It is indigestible in the human body but digestible by many microorganisms, including microorganisms found in the digestive tracts of many herbivores. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/16%3A_Carbohydrates/16.S%3A_Carbohydrates_%28Summary%29.txt |
The lipids are a large and diverse group of naturally occurring organic compounds that are related by their solubility in nonpolar organic solvents (e.g., ether, chloroform, acetone and benzene) and general insolubility in water. There is great structural variety among the lipids, as will be demonstrated in the following sections.
• 17.0: Prelude to Lipids
Lipids are not defined by the presence of specific functional groups, as carbohydrates are, but by a physical property—solubility. Compounds isolated from body tissues are classified as lipids if they are more soluble in organic solvents, such as dichloromethane, than in water. Hence, the lipid category includes not only fats and oils, which are esters of the trihydroxy alcohol glycerol and fatty acids, but also compounds derived from phosphoric acid, carbohydrates, amino alcohols and steroids.
• 17.1: Fatty Acids
Fatty acids are carboxylic acids that are the structural components of many lipids. They may be saturated or unsaturated. Most fatty acids are unbranched and contain an even number of carbon atoms. Unsaturated fatty acids have lower melting points than saturated fatty acids containing the same number of carbon atoms.
• 17.2: Fats and Oils
Fats and oils are composed of molecules known as triglycerides, which are esters composed of three fatty acid units linked to glycerol. An increase in the percentage of shorter-chain fatty acids and/or unsaturated fatty acids lowers the melting point of a fat or oil. The hydrolysis of fats and oils in the presence of a base makes soap and is known as saponification. Double bonds present in unsaturated triglycerides can be hydrogenated to convert oils (liquid) into margarine (solid).
• 17.3: Membranes and Membrane Lipids
Lipids are important components of biological membranes. These lipids have dual characteristics: part of the molecule is hydrophilic, and part of the molecule is hydrophobic. Membrane lipids may be classified as phospholipids, glycolipids, and/or sphingolipids. Proteins are another important component of biological membranes. Integral proteins span the lipid bilayer, while peripheral proteins are more loosely associated with the surface of the membrane.
• 17.4: Steroids
Steroids have a four-fused-ring structure and have a variety of functions. Cholesterol is a steroid found in mammals that is needed for the formation of cell membranes, bile acids, and several hormones. Bile salts are secreted into the small intestine to aid in the digestion of fats.
• 17.E: Exercises
Problems and select solutions for the chapter.
• 17.S: Lipids (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
17: Lipids
On July 11, 2003, the Food and Drug Administration amended its food labeling regulations to require that manufacturers list the amount of trans fatty acids on Nutrition Facts labels of foods and dietary supplements, effective January 1, 2006. This amendment was a response to published studies demonstrating a link between the consumption of trans fatty acids and an increased risk of heart disease. Trans fatty acids are produced in the conversion of liquid oils to solid fats, as in the creation of many commercial margarines and shortenings. They have been shown to increase the levels of low-density lipoproteins (LDLs)—complexes that are often referred to as bad cholesterol—in the blood. In this chapter, you will learn about fatty acids and what is meant by a trans fatty acid, as well as the difference between fats and oils. You will also learn what cholesterol is and why it is an important molecule in the human body.
Fats and oils, found in many of the foods we eat, belong to a class of biomolecules known as lipids. Gram for gram, they pack more than twice the caloric content of carbohydrates: the oxidation of fats and oils supplies about 9 kcal of energy for every gram oxidized, whereas the oxidation of carbohydrates supplies only 4 kcal/g. Although the high caloric content of fats may be bad news for the dieter, it says something about the efficiency of nature’s designs. Our bodies use carbohydrates, primarily in the form of glucose, for our immediate energy needs. Our capacity for storing carbohydrates for later use is limited to tucking away a bit of glycogen in the liver or in muscle tissue. We store our reserve energy in lipid form, which requires far less space than the same amount of energy stored in carbohydrate form. Lipids have other biological functions besides energy storage. They are a major component of the membranes of the 10 trillion cells in our bodies. They serve as protective padding and insulation for vital organs. Furthermore, without lipids in our diets, we would be deficient in the fat-soluble vitamins A, D, E, and K.
Lipids are not defined by the presence of specific functional groups, as carbohydrates are, but by a physical property—solubility. Compounds isolated from body tissues are classified as lipids if they are more soluble in organic solvents, such as dichloromethane, than in water. By this criterion, the lipid category includes not only fats and oils, which are esters of the trihydroxy alcohol glycerol and fatty acids, but also compounds that incorporate functional groups derived from phosphoric acid, carbohydrates, or amino alcohols, as well as steroid compounds such as cholesterol (Figure \(1\) presents one scheme for classifying the various kinds of lipids). We will discuss the various kinds of lipids by considering one subclass at a time and pointing out structural similarities and differences as we go. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/17%3A_Lipids/17.00%3A_Prelude_to_Lipids.txt |
Learning Objectives
• To recognize the structures of common fatty acids and classify them as saturated, monounsaturated, or polyunsaturated.
Fatty acids are carboxylic acids that are structural components of fats, oils, and all other categories of lipids, except steroids. More than 70 have been identified in nature. They usually contain an even number of carbon atoms (typically 12–20), are generally unbranched, and can be classified by the presence and number of carbon-to-carbon double bonds. Thus, saturated fatty acids contain no carbon-to-carbon double bonds, monounsaturated fatty acids contain one carbon-to-carbon double bond, and polyunsaturated fatty acids contain two or more carbon-to-carbon double bonds.
Table \(1\) lists some common fatty acids and one important source for each. The atoms or groups around the double bonds in unsaturated fatty acids can be arranged in either the cis or trans isomeric form. Naturally occurring fatty acids are generally in the cis configuration.
Table \(1\): Some Common Fatty Acids Found in Natural Fats
Name Abbreviated Structural Formula Condensed Structural Formula Melting Point (°C) Source
lauric acid C11H23COOH CH3(CH2)10COOH 44 palm kernel oil
myristic acid C13H27COOH CH3(CH2)12COOH 58 oil of nutmeg
palmitic acid C15H31COOH CH3(CH2)14COOH 63 palm oil
palmitoleic acid C15H29COOH CH3(CH2)5CH=CH(CH2)7COOH 0.5 macadamia oil
stearic acid C17H35COOH CH3(CH2)16COOH 70 cocoa butter
oleic acid C17H33COOH CH3(CH2)7CH=CH(CH2)7COOH 16 olive oil
linoleic acid C17H31COOH CH3(CH2)3(CH2CH=CH)2(CH2)7COOH −5 canola oil
α-linolenic acid C17H29COOH CH3(CH2CH=CH)3(CH2)7COOH −11 flaxseed
arachidonic acid C19H31COOH CH3(CH2)4(CH2CH=CH)4(CH2)2COOH −50 liver
Two polyunsaturated fatty acids—linoleic and α-linolenic acids—are termed essential fatty acids because humans must obtain them from their diets. Both substances are required for normal growth and development, but the human body does not synthesize them. The body uses linoleic acid to synthesize many of the other unsaturated fatty acids, such as arachidonic acid, a precursor for the synthesis of prostaglandins. In addition, the essential fatty acids are necessary for the efficient transport and metabolism of cholesterol. The average daily diet should contain about 4–6 g of the essential fatty acids.
To Your Health: Prostaglandins
Prostaglandins are chemical messengers synthesized in the cells in which their physiological activity is expressed. They are unsaturated fatty acids containing 20 carbon atoms and are synthesized from arachidonic acid—a polyunsaturated fatty acid—when needed by a particular cell. They are called prostaglandins because they were originally isolated from semen found in the prostate gland. It is now known that they are synthesized in nearly all mammalian tissues and affect almost all organs in the body. The five major classes of prostaglandins are designated as PGA, PGB, PGE, PGF, and PGI. Subscripts are attached at the end of these abbreviations to denote the number of double bonds outside the five-carbon ring in a given prostaglandin.
The prostaglandins are among the most potent biological substances known. Slight structural differences give them highly distinct biological effects; however, all prostaglandins exhibit some ability to induce smooth muscle contraction, lower blood pressure, and contribute to the inflammatory response. Aspirin and other nonsteroidal anti-inflammatory agents, such as ibuprofen, obstruct the synthesis of prostaglandins by inhibiting cyclooxygenase, the enzyme needed for the initial step in the conversion of arachidonic acid to prostaglandins.
Their wide range of physiological activity has led to the synthesis of hundreds of prostaglandins and their analogs. Derivatives of PGE2 are now used in the United States to induce labor. Other prostaglandins have been employed clinically to lower or increase blood pressure, inhibit stomach secretions, relieve nasal congestion, relieve asthma, and prevent the formation of blood clots, which are associated with heart attacks and strokes.
Although we often draw the carbon atoms in a straight line, they actually have more of a zigzag configuration (Figure \(\PageIndex{2a}\)). Viewed as a whole, however, the saturated fatty acid molecule is relatively straight (Figure \(\PageIndex{2b}\)). Such molecules pack closely together into a crystal lattice, maximizing the strength of dispersion forces and causing fatty acids and the fats derived from them to have relatively high melting points. In contrast, each cis carbon-to-carbon double bond in an unsaturated fatty acid produces a pronounced bend in the molecule, so that these molecules do not stack neatly. As a result, the intermolecular attractions of unsaturated fatty acids (and unsaturated fats) are weaker, causing these substances to have lower melting points. Most are liquids at room temperature.
Waxes are esters formed from long-chain fatty acids and long-chain alcohols. Most natural waxes are mixtures of such esters. Plant waxes on the surfaces of leaves, stems, flowers, and fruits protect the plant from dehydration and invasion by harmful microorganisms. Carnauba wax, used extensively in floor waxes, automobile waxes, and furniture polish, is largely myricyl cerotate, obtained from the leaves of certain Brazilian palm trees. Animals also produce waxes that serve as protective coatings, keeping the surfaces of feathers, skin, and hair pliable and water repellent. In fact, if the waxy coating on the feathers of a water bird is dissolved as a result of the bird swimming in an oil slick, the feathers become wet and heavy, and the bird, unable to maintain its buoyancy, drowns.
Summary
Fatty acids are carboxylic acids that are the structural components of many lipids. They may be saturated or unsaturated. Most fatty acids are unbranched and contain an even number of carbon atoms. Unsaturated fatty acids have lower melting points than saturated fatty acids containing the same number of carbon atoms. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/17%3A_Lipids/17.01%3A_Fatty_Acids.txt |
Learning Objectives
• Explain why fats and oils are referred to as triglycerides.
• Explain how the fatty acid composition of the triglycerides determines whether a substance is a fat or oil.
• Describe the importance of key reactions of triglycerides, such as hydrolysis, hydrogenation, and oxidation.
Fats and oils are the most abundant lipids in nature. They provide energy for living organisms, insulate body organs, and transport fat-soluble vitamins through the blood.
Structures of Fats and Oils
Fats and oils are called triglycerides (or triacylcylgerols) because they are esters composed of three fatty acid units joined to glycerol, a trihydroxy alcohol:
If all three OH groups on the glycerol molecule are esterified with the same fatty acid, the resulting ester is called a simple triglyceride. Although simple triglycerides have been synthesized in the laboratory, they rarely occur in nature. Instead, a typical triglyceride obtained from naturally occurring fats and oils contains two or three different fatty acid components and is thus termed a mixed triglyceride.
Tristearin
a simple triglyceride
a mixed triglyceride
A triglyceride is called a fat if it is a solid at 25°C; it is called an oil if it is a liquid at that temperature. These differences in melting points reflect differences in the degree of unsaturation and number of carbon atoms in the constituent fatty acids. Triglycerides obtained from animal sources are usually solids, while those of plant origin are generally oils. Therefore, we commonly speak of animal fats and vegetable oils.
No single formula can be written to represent the naturally occurring fats and oils because they are highly complex mixtures of triglycerides in which many different fatty acids are represented. Table \(1\) shows the fatty acid compositions of some common fats and oils. The composition of any given fat or oil can vary depending on the plant or animal species it comes from as well as on dietetic and climatic factors. To cite just one example, lard from corn-fed hogs is more highly saturated than lard from peanut-fed hogs. Palmitic acid is the most abundant of the saturated fatty acids, while oleic acid is the most abundant unsaturated fatty acid.
Table \(1\): Average Fatty Acid Composition of Some Common Fats and Oils (%)*
Lauric Myristic Palmitic Stearic Oleic Linoleic Linolenic
Fats
butter (cow) 3 11 27 12 29 2 1
tallow 3 24 19 43 3 1
lard 2 26 14 44 10
Oils
canola oil 4 2 62 22 10
coconut oil 47 18 9 3 6 2
corn oil 11 2 28 58 1
olive oil 13 3 71 10 1
peanut oil 11 2 48 32
soybean oil 11 4 24 54 7
*Totals less than 100% indicate the presence of fatty acids with fewer than 12 carbon atoms or more than 18 carbon atoms.
Coconut oil is highly saturated. It contains an unusually high percentage of the low-melting C8, C10, and C12 saturated fatty acids.
Terms such as saturated fat or unsaturated oil are often used to describe the fats or oils obtained from foods. Saturated fats contain a high proportion of saturated fatty acids, while unsaturated oils contain a high proportion of unsaturated fatty acids. The high consumption of saturated fats is a factor, along with the high consumption of cholesterol, in increased risks of heart disease.
Physical Properties of Fats and Oils
Contrary to what you might expect, pure fats and oils are colorless, odorless, and tasteless. The characteristic colors, odors, and flavors that we associate with some of them are imparted by foreign substances that are lipid soluble and have been absorbed by these lipids. For example, the yellow color of butter is due to the presence of the pigment carotene; the taste of butter comes from two compounds—diacetyl and 3-hydroxy-2-butanone—produced by bacteria in the ripening cream from which the butter is made.
Fats and oils are lighter than water, having densities of about 0.8 g/cm3. They are poor conductors of heat and electricity and therefore serve as excellent insulators for the body, slowing the loss of heat through the skin.
Chemical Reactions of Fats and Oils
Fats and oils can participate in a variety of chemical reactions—for example, because triglycerides are esters, they can be hydrolyzed in the presence of an acid, a base, or specific enzymes known as lipases. The hydrolysis of fats and oils in the presence of a base is used to make soap and is called saponification. Today most soaps are prepared through the hydrolysis of triglycerides (often from tallow, coconut oil, or both) using water under high pressure and temperature [700 lb/in2 (∼50 atm or 5,000 kPa) and 200°C]. Sodium carbonate or sodium hydroxide is then used to convert the fatty acids to their sodium salts (soap molecules):
Looking Closer: Soaps
Ordinary soap is a mixture of the sodium salts of various fatty acids, produced in one of the oldest organic syntheses practiced by humans (second only to the fermentation of sugars to produce ethyl alcohol). Both the Phoenicians (600 BCE) and the Romans made soap from animal fat and wood ash. Even so, the widespread production of soap did not begin until the 1700s. Soap was traditionally made by treating molten lard or tallow with a slight excess of alkali in large open vats. The mixture was heated, and steam was bubbled through it. After saponification was completed, the soap was precipitated from the mixture by the addition of sodium chloride (NaCl), removed by filtration, and washed several times with water. It was then dissolved in water and reprecipitated by the addition of more NaCl. The glycerol produced in the reaction was also recovered from the aqueous wash solutions.
Pumice or sand is added to produce scouring soap, while ingredients such as perfumes or dyes are added to produce fragrant, colored soaps. Blowing air through molten soap produces a floating soap. Soft soaps, made with potassium salts, are more expensive but produce a finer lather and are more soluble. They are used in liquid soaps, shampoos, and shaving creams.
Dirt and grime usually adhere to skin, clothing, and other surfaces by combining with body oils, cooking fats, lubricating greases, and similar substances that act like glues. Because these substances are not miscible in water, washing with water alone does little to remove them. Soap removes them, however, because soap molecules have a dual nature. One end, called the head, carries an ionic charge (a carboxylate anion) and therefore dissolves in water; the other end, the tail, has a hydrocarbon structure and dissolves in oils. The hydrocarbon tails dissolve in the soil; the ionic heads remain in the aqueous phase, and the soap breaks the oil into tiny soap-enclosed droplets called micelles, which disperse throughout the solution. The droplets repel each other because of their charged surfaces and do not coalesce. With the oil no longer “gluing” the dirt to the soiled surface (skin, cloth, dish), the soap-enclosed dirt can easily be rinsed away.
The double bonds in fats and oils can undergo hydrogenation and also oxidation. The hydrogenation of vegetable oils to produce semisolid fats is an important process in the food industry. Chemically, it is essentially identical to the catalytic hydrogenation reaction described for alkenes.
In commercial processes, the number of double bonds that are hydrogenated is carefully controlled to produce fats with the desired consistency (soft and pliable). Inexpensive and abundant vegetable oils (canola, corn, soybean) are thus transformed into margarine and cooking fats. In the preparation of margarine, for example, partially hydrogenated oils are mixed with water, salt, and nonfat dry milk, along with flavoring agents, coloring agents, and vitamins A and D, which are added to approximate the look, taste, and nutrition of butter. (Preservatives and antioxidants are also added.) In most commercial peanut butter, the peanut oil has been partially hydrogenated to prevent it from separating out. Consumers could decrease the amount of saturated fat in their diet by using the original unprocessed oils on their foods, but most people would rather spread margarine on their toast than pour oil on it.
Many people have switched from butter to margarine or vegetable shortening because of concerns that saturated animal fats can raise blood cholesterol levels and result in clogged arteries. However, during the hydrogenation of vegetable oils, an isomerization reaction occurs that produces the trans fatty acids mentioned in the opening essay. However, studies have shown that trans fatty acids also raise cholesterol levels and increase the incidence of heart disease. Trans fatty acids do not have the bend in their structures, which occurs in cis fatty acids and thus pack closely together in the same way that the saturated fatty acids do. Consumers are now being advised to use polyunsaturated oils and soft or liquid margarine and reduce their total fat consumption to less than 30% of their total calorie intake each day.
Fats and oils that are in contact with moist air at room temperature eventually undergo oxidation and hydrolysis reactions that cause them to turn rancid, acquiring a characteristic disagreeable odor. One cause of the odor is the release of volatile fatty acids by hydrolysis of the ester bonds. Butter, for example, releases foul-smelling butyric, caprylic, and capric acids. Microorganisms present in the air furnish lipases that catalyze this process. Hydrolytic rancidity can easily be prevented by covering the fat or oil and keeping it in a refrigerator.
Another cause of volatile, odorous compounds is the oxidation of the unsaturated fatty acid components, particularly the readily oxidized structural unit
in polyunsaturated fatty acids, such as linoleic and linolenic acids. One particularly offensive product, formed by the oxidative cleavage of both double bonds in this unit, is a compound called malonaldehyde.
Rancidity is a major concern of the food industry, which is why food chemists are always seeking new and better antioxidants, substances added in very small amounts (0.001%–0.01%) to prevent oxidation and thus suppress rancidity. Antioxidants are compounds whose affinity for oxygen is greater than that of the lipids in the food; thus they function by preferentially depleting the supply of oxygen absorbed into the product. Because vitamin E has antioxidant properties, it helps reduce damage to lipids in the body, particularly to unsaturated fatty acids found in cell membrane lipids.
Summary
Fats and oils are composed of molecules known as triglycerides, which are esters composed of three fatty acid units linked to glycerol. An increase in the percentage of shorter-chain fatty acids and/or unsaturated fatty acids lowers the melting point of a fat or oil. The hydrolysis of fats and oils in the presence of a base makes soap and is known as saponification. Double bonds present in unsaturated triglycerides can be hydrogenated to convert oils (liquid) into margarine (solid). The oxidation of fatty acids can form compounds with disagreeable odors. This oxidation can be minimized by the addition of antioxidants. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/17%3A_Lipids/17.02%3A_Fats_and_Oils.txt |
Learning Objectives
• Identify the distinguishing characteristics of membrane lipids.
• Describe membrane components and how they are arranged.
All living cells are surrounded by a cell membrane. Plant cells (Figure \(\PageIndex{1A}\)) and animal cells (Figure \(\PageIndex{1B}\)) contain a cell nucleus that is also surrounded by a membrane and holds the genetic information for the cell. Everything between the cell membrane and the nuclear membrane—including intracellular fluids and various subcellular components such as the mitochondria and ribosomes—is called the cytoplasm. The membranes of all cells have a fundamentally similar structure, but membrane function varies tremendously from one organism to another and even from one cell to another within a single organism. This diversity arises mainly from the presence of different proteins and lipids in the membrane.
The lipids in cell membranes are highly polar but have dual characteristics: part of the lipid is ionic and therefore dissolves in water, whereas the rest has a hydrocarbon structure and therefore dissolves in nonpolar substances. Often, the ionic part is referred to as hydrophilic, meaning “water loving,” and the nonpolar part as hydrophobic, meaning “water fearing” (repelled by water). When allowed to float freely in water, polar lipids spontaneously cluster together in any one of three arrangements: micelles, monolayers, and bilayers (Figure \(2\)).
Micelles are aggregations in which the lipids’ hydrocarbon tails—being hydrophobic—are directed toward the center of the assemblage and away from the surrounding water while the hydrophilic heads are directed outward, in contact with the water. Each micelle may contain thousands of lipid molecules. Polar lipids may also form a monolayer, a layer one molecule thick on the surface of the water. The polar heads face into water, and the nonpolar tails stick up into the air. Bilayers are double layers of lipids arranged so that the hydrophobic tails are sandwiched between an inner surface and an outer surface consisting of hydrophilic heads. The hydrophilic heads are in contact with water on either side of the bilayer, whereas the tails, sequestered inside the bilayer, are prevented from having contact with the water. Bilayers like this make up every cell membrane (Figure \(3\)).
In the bilayer interior, the hydrophobic tails (that is, the fatty acid portions of lipid molecules) interact by means of dispersion forces. The interactions are weakened by the presence of unsaturated fatty acids. As a result, the membrane components are free to mill about to some extent, and the membrane is described as fluid.
The lipids found in cell membranes can be categorized in various ways. Phospholipids are lipids containing phosphorus. Glycolipids are sugar-containing lipids. The latter are found exclusively on the outer surface of the cell membrane, acting as distinguishing surface markers for the cell and thus serving in cellular recognition and cell-to-cell communication. Sphingolipids are phospholipids or glycolipids that contain the unsaturated amino alcohol sphingosine rather than glycerol. Diagrammatic structures of representative membrane lipids are presented in Figure \(4\).
Phosphoglycerides (also known as glycerophospholipids) are the most abundant phospholipids in cell membranes. They consist of a glycerol unit with fatty acids attached to the first two carbon atoms, while a phosphoric acid unit, esterified with an alcohol molecule (usually an amino alcohol, as in part (a) of Figure \(5\)) is attached to the third carbon atom of glycerol (part (b) of Figure \(5\)). Notice that the phosphoglyceride molecule is identical to a triglyceride up to the phosphoric acid unit (part (b) of Figure \(5\)).
There are two common types of phosphoglycerides. Phosphoglycerides containing ethanolamine as the amino alcohol are called phosphatidylethanolamines or cephalins. Cephalins are found in brain tissue and nerves and also have a role in blood clotting. Phosphoglycerides containing choline as the amino alcohol unit are called phosphatidylcholines or lecithins. Lecithins occur in all living organisms. Like cephalins, they are important constituents of nerve and brain tissue. Egg yolks are especially rich in lecithins. Commercial-grade lecithins isolated from soybeans are widely used in foods as emulsifying agents. An emulsifying agent is used to stabilize an emulsion—a dispersion of two liquids that do not normally mix, such as oil and water. Many foods are emulsions. Milk is an emulsion of butterfat in water. The emulsifying agent in milk is a protein called casein. Mayonnaise is an emulsion of salad oil in water, stabilized by lecithins present in egg yolk.
Sphingomyelins, the simplest sphingolipids, each contain a fatty acid, a phosphoric acid, sphingosine, and choline (Figure \(6\)). Because they contain phosphoric acid, they are also classified as phospholipids. Sphingomyelins are important constituents of the myelin sheath surrounding the axon of a nerve cell. Multiple sclerosis is one of several diseases resulting from damage to the myelin sheath.
Most animal cells contain sphingolipids called cerebrosides (Figure \(7\)). Cerebrosides are composed of sphingosine, a fatty acid, and galactose or glucose. They therefore resemble sphingomyelins but have a sugar unit in place of the choline phosphate group. Cerebrosides are important constituents of the membranes of nerve and brain cells.
The sphingolipids called gangliosides are more complex, usually containing a branched chain of three to eight monosaccharides and/or substituted sugars. Because of considerable variation in their sugar components, about 130 varieties of gangliosides have been identified. Most cell-to-cell recognition and communication processes (e.g., blood group antigens) depend on differences in the sequences of sugars in these compounds. Gangliosides are most prevalent in the outer membranes of nerve cells, although they also occur in smaller quantities in the outer membranes of most other cells. Because cerebrosides and gangliosides contain sugar groups, they are also classified as glycolipids.
Membrane Proteins
If membranes were composed only of lipids, very few ions or polar molecules could pass through their hydrophobic “sandwich filling” to enter or leave any cell. However, certain charged and polar species do cross the membrane, aided by proteins that move about in the lipid bilayer. The two major classes of proteins in the cell membrane are integral proteins, which span the hydrophobic interior of the bilayer, and peripheral proteins, which are more loosely associated with the surface of the lipid bilayer (Figure \(3\)). Peripheral proteins may be attached to integral proteins, to the polar head groups of phospholipids, or to both by hydrogen bonding and electrostatic forces.
Small ions and molecules soluble in water enter and leave the cell by way of channels through the integral proteins. Some proteins, called carrier proteins, facilitate the passage of certain molecules, such as hormones and neurotransmitters, by specific interactions between the protein and the molecule being transported.
Summary
Lipids are important components of biological membranes. These lipids have dual characteristics: part of the molecule is hydrophilic, and part of the molecule is hydrophobic. Membrane lipids may be classified as phospholipids, glycolipids, and/or sphingolipids. Proteins are another important component of biological membranes. Integral proteins span the lipid bilayer, while peripheral proteins are more loosely associated with the surface of the membrane. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/17%3A_Lipids/17.03%3A_Membranes_and_Membrane_Lipids.txt |
Learning Objectives
• To identify the functions of steroids produced in mammals.
All the lipids discussed so far are saponifiable, reacting with aqueous alkali to yield simpler components, such as glycerol, fatty acids, amino alcohols, and sugars. Lipid samples extracted from cellular material, however, also contain a small but important fraction that does not react with alkali. The most important nonsaponifiable lipids are the steroids. These compounds include the bile salts, cholesterol and related compounds, and certain hormones (such as cortisone and the sex hormones).
Steroids occur in plants, animals, yeasts, and molds but not in bacteria. They may exist in free form or combined with fatty acids or carbohydrates. All steroids have a characteristic structural component consisting of four fused rings. Chemists identify the rings by capital letters and number the carbon atoms as shown in Figure \(\PageIndex{1a}\). Slight variations in this structure or in the atoms or groups attached to it produce profound differences in biological activity.
Cholesterol
Cholesterol (Figure \(\PageIndex{1b}\)) does not occur in plants, but it is the most abundant steroid in the human body (240 g is a typical amount). Excess cholesterol is believed to be a primary factor in the development of atherosclerosis and heart disease, which are major health problems in the United States today. About half of the body’s cholesterol is interspersed in the lipid bilayer of cell membranes. Much of the rest is converted to cholic acid, which is used in the formation of bile salts. Cholesterol is also a precursor in the synthesis of sex hormones, adrenal hormones, and vitamin D.
Excess cholesterol not metabolized by the body is released from the liver and transported by the blood to the gallbladder. Normally, it stays in solution there until being secreted into the intestine (as a component of bile) to be eliminated. Sometimes, however, cholesterol in the gallbladder precipitates in the form of gallstones (Figure \(2\)). Indeed, the name cholesterol is derived from the Greek chole, meaning “bile,” and stereos, meaning “solid.”
To Your Health: Cholesterol and Heart Disease
Heart disease is the leading cause of death in the United States for both men and women. The Centers for Disease Control and Prevention reported that heart disease claimed 631,636 lives in the United States (26% of all reported deaths) in 2006.
Scientists agree that elevated cholesterol levels in the blood, as well as high blood pressure, obesity, diabetes, and cigarette smoking, are associated with an increased risk of heart disease. A long-term investigation by the National Institutes of Health showed that among men ages 30 to 49, the incidence of heart disease was five times greater for those whose cholesterol levels were above 260 mg/100 mL of serum than for those with cholesterol levels of 200 mg/100 mL or less. The cholesterol content of blood varies considerably with age, diet, and sex. Young adults average about 170 mg of cholesterol per 100 mL of blood, whereas males at age 55 may have cholesterol levels at 250 mg/100 mL or higher because the rate of cholesterol breakdown decreases with age. Females tend to have lower blood cholesterol levels than males.
To understand the link between heart disease and cholesterol levels, it is important to understand how cholesterol and other lipids are transported in the body. Lipids, such as cholesterol, are not soluble in water and therefore cannot be transported in the blood (an aqueous medium) unless they are complexed with proteins that are soluble in water, forming assemblages called lipoproteins. Lipoproteins are classified according to their density, which is dependent on the relative amounts of protein and lipid they contain. Lipids are less dense than proteins, so lipoproteins containing a greater proportion of lipid are less dense than those containing a greater proportion of protein.
Research on cholesterol and its role in heart disease has focused on serum levels of low-density lipoproteins (LDLs) and high-density lipoproteins (HDLs). One of the most fascinating discoveries is that high levels of HDLs reduce a person’s risk of developing heart disease, whereas high levels of LDLs increase that risk. Thus the serum LDL:HDL ratio is a better predictor of heart disease risk than the overall level of serum cholesterol. Persons who, because of hereditary or dietary factors, have high LDL:HDL ratios in their blood have a higher incidence of heart disease.
How do HDLs reduce the risk of developing heart disease? No one knows for sure, but one role of HDLs appears to be the transport of excess cholesterol to the liver, where it can be metabolized. Therefore, HDLs aid in removing cholesterol from blood and from the smooth muscle cells of the arterial wall.
Dietary modifications and increased physical activity can help lower total cholesterol and improve the LDL:HDL ratio. The average American consumes about 600 mg of cholesterol from animal products each day and also synthesizes approximately 1 g of cholesterol each day, mostly in the liver. The amount of cholesterol synthesized is controlled by the cholesterol level in the blood; when the blood cholesterol level exceeds 150 mg/100 mL, the rate of cholesterol biosynthesis is halved. Hence, if cholesterol is present in the diet, a feedback mechanism suppresses its synthesis in the liver. However, the ratio of suppression is not a 1:1 ratio; the reduction in biosynthesis does not equal the amount of cholesterol ingested. Thus, dietary substitutions of unsaturated fat for saturated fat, as well as a reduction in consumption of trans fatty acids, is recommended to help lower serum cholesterol and the risk of heart disease.
Steroid Hormones
Hormones are chemical messengers that are released in one tissue and transported through the circulatory system to one or more other tissues. One group of hormones is known as steroid hormones because these hormones are synthesized from cholesterol, which is also a steroid. There are two main groups of steroid hormones: adrenocortical hormones and sex hormones.
The adrenocortical hormones, such as aldosterone and cortisol (Table \(1\)), are produced by the adrenal gland, which is located adjacent to each kidney. Aldosterone acts on most cells in the body, but it is particularly effective at enhancing the rate of reabsorption of sodium ions in the kidney tubules and increasing the secretion of potassium ions and/or hydrogen ions by the tubules. Because the concentration of sodium ions is the major factor influencing water retention in tissues, aldosterone promotes water retention and reduces urine output. Cortisol regulates several key metabolic reactions (for example, increasing glucose production and mobilizing fatty acids and amino acids). It also inhibits the inflammatory response of tissue to injury or stress. Cortisol and its analogs are therefore used pharmacologically as immunosuppressants after transplant operations and in the treatment of severe skin allergies and autoimmune diseases, such as rheumatoid arthritis.
Table \(1\): Representative Steroid Hormones and Their Physiological Effects
Hormone Effect
regulates salt metabolism; stimulates kidneys to retain sodium and excrete potassium
stimulates the conversion of proteins to carbohydrates
regulates the menstrual cycle; maintains pregnancy
stimulates female sex characteristics; regulates changes during the menstrual cycle
stimulates and maintains male sex characteristics
The sex hormones are a class of steroid hormones secreted by the gonads (ovaries or testes), the placenta, and the adrenal glands. Testosterone and androstenedione are the primary male sex hormones, or androgens, controlling the primary sexual characteristics of males, or the development of the male genital organs and the continuous production of sperm. Androgens are also responsible for the development of secondary male characteristics, such as facial hair, deep voice, and muscle strength. Two kinds of sex hormones are of particular importance in females: progesterone, which prepares the uterus for pregnancy and prevents the further release of eggs from the ovaries during pregnancy, and the estrogens, which are mainly responsible for the development of female secondary sexual characteristics, such as breast development and increased deposition of fat tissue in the breasts, the buttocks, and the thighs. Both males and females produce androgens and estrogens, differing in the amounts of secreted hormones rather than in the presence or absence of one or the other.
Sex hormones, both natural and synthetic, are sometimes used therapeutically. For example, a woman who has had her ovaries removed may be given female hormones to compensate. Some of the earliest chemical compounds employed in cancer chemotherapy were sex hormones. For example, estrogens are one treatment option for prostate cancer because they block the release and activity of testosterone. Testosterone enhances prostate cancer growth. Sex hormones are also administered in preparation for sex-change operations, to promote the development of the proper secondary sexual characteristics. Oral contraceptives are synthetic derivatives of the female sex hormones; they work by preventing ovulation.
Bile Salts
Bile is a yellowish green liquid (pH 7.8–8.6) produced in the liver. The most important constituents of bile are bile salts, which are sodium salts of amidelike combinations of bile acids, such as cholic acid (part (a) of Figure \(3\)) and an amine such as the amino acid glycine (part (b) of Figure \(3\)). They are synthesized from cholesterol in the liver, stored in the gallbladder, and then secreted in bile into the small intestine. In the gallbladder, the composition of bile gradually changes as water is absorbed and the other components become more concentrated.
Because they contain both hydrophobic and hydrophilic groups, bile salts are highly effective detergents and emulsifying agents; they break down large fat globules into smaller ones and keep those smaller globules suspended in the aqueous digestive environment. Enzymes can then hydrolyze fat molecules more efficiently. Thus, the major function of bile salts is to aid in the digestion of dietary lipids.
Surgical removal is often advised for a gallbladder that becomes infected, inflamed, or perforated. This surgery does not seriously affect digestion because bile is still produced by the liver, but the liver’s bile is more dilute and its secretion into the small intestine is not as closely tied to the arrival of food.
Summary
Steroids have a four-fused-ring structure and have a variety of functions. Cholesterol is a steroid found in mammals that is needed for the formation of cell membranes, bile acids, and several hormones. Bile salts are secreted into the small intestine to aid in the digestion of fats. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/17%3A_Lipids/17.04%3A_Steroids.txt |
Additional Exercises
1. The melting point of elaidic acid is 52°C.
1. What trend is observed when comparing the melting points of elaidic acid, oleic acid, and stearic acid? Explain.
2. Would you expect the melting point of palmitelaidic acid to be lower or higher than that of elaidic acid? Explain.
2. Examine the labels on two brands of margarine and two brands of shortening and list the oils used in the various brands.
3. Draw a typical lecithin molecule that incorporates glycerol, palmitic acid, oleic acid, phosphoric acid, and choline. Circle all the ester bonds.
4. In cerebrosides, is the linkage between the fatty acid and sphingosine an amide bond or an ester bond? Justify your answer.
5. Serine is an amino acid that has the following structure. Draw the structure for a phosphatidylserine that contains a palmitic acid and a palmitoleic acid unit.
6. Explain whether each compound would be expected to diffuse through the lipid bilayer of a cell membrane.
1. potassium chloride
2. CH3CH2CH2CH2CH2CH3
3. fructose
7. Identify the role of each steroid hormone in the body.
1. progesterone
2. aldosterone
3. testosterone
4. cortisol
8. How does the structure of cholic acid differ from that of cholesterol? Which compound would you expect to be more polar? Why?
1. What fatty acid is the precursor for the prostaglandins?
2. Identify three biological effects of prostaglandins.
9. Why is it important to determine the ratio of LDLs to HDLs, rather than just the concentration of serum cholesterol?
Answers
1. Stearic acid has the highest melting point, followed by elaidic acid, and then oleic acid with the lowest melting point. Elaidic acid is a trans fatty acid, and the carbon chains can pack together almost as tightly as those of the saturated stearic acid. Oleic acid is a cis fatty acid, and the bend in the hydrocarbon chain keeps these carbon chains from packing as closely together; fewer interactions lead to a much lower melting point.
2. The melting point of palmitelaidic acid should be lower than that of elaidic acid because it has a shorter carbon chain (16, as compared to 18 for elaidic acid). The shorter the carbon chain, the lower the melting point due to a decrease in intermolecular interactions.
1. regulates the menstrual cycle and maintains pregnancy
2. regulates salt metabolism by stimulating the kidneys to retain sodium and excrete potassium
3. stimulates and maintains male sex characteristics
4. stimulates the conversion of proteins to carbohydrates
1. arachidonic acid
2. induce smooth muscle contraction, lower blood pressure, and contribute to the inflammatory response
17.S: Lipids (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
Lipids, found in the body tissues of all organisms, are compounds that are more soluble in organic solvents than in water. Many of them contain fatty acids, which are carboxylic acids that generally contain an even number of 4–20 carbon atoms in an unbranched chain. Saturated fatty acids have no carbon-to-carbon double bonds. Monounsaturated fatty acids have a single carbon-to-carbon double bond, while polyunsaturated fatty acids have more than one carbon-to-carbon double bond. Linoleic and linolenic acid are known as essential fatty acids because the human body cannot synthesize these polyunsaturated fatty acids. The lipids known as fats and oils are triacylglycerols, more commonly called triglycerides—esters composed of three fatty acids joined to the trihydroxy alcohol glycerol. Fats are triglycerides that are solid at room temperature, and oils are triglycerides that are liquid at room temperature. Fats are found mainly in animals, and oils found mainly in plants. Saturated triglycerides are those containing a higher proportion of saturated fatty acid chains (fewer carbon-to-carbon double bonds); unsaturated triglycerides contain a higher proportion of unsaturated fatty acid chains.
Saponification is the hydrolysis of a triglyceride in a basic solution to form glycerol and three carboxylate anions or soap molecules. Other important reactions are the hydrogenation and oxidation of double bonds in unsaturated fats and oils.
Phospholipids are lipids containing phosphorus. In phosphoglycerides, the phosphorus is joined to an amino alcohol unit. Some phosphoglycerides, like lecithins, are used to stabilize an emulsion—a dispersion of two liquids that do not normally mix, such as oil and water. Sphingolipids are lipids for which the precursor is the amino alcohol sphingosine, rather than glycerol. A glycolipid has a sugar substituted at one of the OH groups of either glycerol or sphingosine. All are highly polar lipids found in cell membranes.
Polar lipids have dual characteristics: one part of the molecule is ionic and dissolves in water; the rest has a hydrocarbon structure and dissolves in nonpolar substances. Often, the ionic part is referred to as hydrophilic (literally, “water loving”) and the nonpolar part as hydrophobic (“water fearing”). When placed in water, polar lipids disperse into any one of three arrangements: micelles, monolayers, and bilayers. Micelles are aggregations of molecules in which the hydrocarbon tails of the lipids, being hydrophobic, are directed inward (away from the surrounding water), and the hydrophilic heads that are directed outward into the water. Bilayers are double layers arranged so that the hydrophobic tails are sandwiched between the two layers of hydrophilic heads, which remain in contact with the water.
Every living cell is enclosed by a cell membrane composed of a lipid bilayer. In animal cells, the bilayer consists mainly of phospholipids, glycolipids, and the steroid cholesterol. Embedded in the bilayer are integral proteins, and peripheral proteins are loosely associated with the surface of the bilayer. Everything between the cell membrane and the membrane of the cell nucleus is called the cytoplasm.
Most lipids can be saponified, but some, such as steroids, cannot be saponified. The steroid cholesterol is found in animal cells but never in plant cells. It is a main component of all cell membranes and a precursor for hormones, vitamin D, and bile salts. Bile salts are the most important constituents of bile, which is a yellowish-green liquid secreted by the gallbladder into the small intestine and is needed for the proper digestion of lipids. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/17%3A_Lipids/17.E%3A_Exercises.txt |
Proteins may be defined as compounds of high molar mass consisting largely or entirely of chains of amino acids. Their masses range from several thousand to several million daltons (Da). In addition to carbon, hydrogen, and oxygen atoms, all proteins contain nitrogen and sulfur atoms, and many also contain phosphorus atoms and traces of other elements. Proteins serve a variety of roles in living organisms and are often classified by these biological roles. Muscle tissue is largely protein, as are skin and hair. Proteins are present in the blood, in the brain, and even in tooth enamel. Each type of cell in our bodies makes its own specialized proteins, as well as proteins common to all or most cells. We begin our study of proteins by looking at the properties and reactions of amino acids, which is followed by a discussion of how amino acids link covalently to form peptides and proteins. We end the chapter with a discussion of enzymes—the proteins that act as catalysts in the body.
• 18.0: Prelude to Amino Acids, Proteins, and Enzymes
Insulin is a hormone that is synthesized in the pancreas. Insulin stimulates the transport of glucose into cells throughout the body and the storage of glucose as glycogen. People with diabetes do not produce insulin or use it properly. The isolation of insulin in 1921 led to the first effective treatment for these individuals.
• 18.1: Properties of Amino Acids
Amino acids can be classified based on the characteristics of their distinctive side chains as nonpolar, polar but uncharged, negatively charged, or positively charged. The amino acids found in proteins are L-amino acids.
• 18.2: Reactions of Amino Acids
Amino acids can act as both an acid and a base due to the presence of the amino and carboxyl functional groups. The pH at which a given amino acid exists in solution as a zwitterion is called the isoelectric point (pI).
• 18.3: Peptides
The amino group of one amino acid can react with the carboxyl group on another amino acid to form a peptide bond that links the two amino acids together. Additional amino acids can be added on through the formation of addition peptide (amide) bonds. A sequence of amino acids in a peptide or protein is written with the N-terminal amino acid first and the C-terminal amino acid at the end (writing left to right).
• 18.4: Proteins
Proteins can be divided into two categories: fibrous, which tend to be insoluble in water, and globular, which are more soluble in water. A protein may have up to four levels of structure. The primary structure consists of the specific amino acid sequence. The peptide chain can form an α-helix or β-pleated sheet, which is known as secondary structure and are incorporated into the tertiary structure of the folded polypeptide. The quaternary structure describes the arrangements of subunits.
• 18.5: Enzymes
An enzyme is a biological catalyst, a substance that increases the rate of a chemical reaction without being changed or consumed in the reaction. A systematic process is used to name and classify enzymes.
• 18.6: Enzyme Action
A substrate binds to a specific region on an enzyme known as the active site, where the substrate can be converted to product. The substrate binds to the enzyme primarily through hydrogen bonding and other electrostatic interactions. The induced-fit model says that an enzyme can undergo a conformational change when binding a substrate. Enzymes exhibit varying degrees of substrate specificity.
• 18.7: Enzyme Activity
Initially, an increase in substrate concentration increases the rate of an enzyme-catalyzed reaction. As the enzyme molecules become saturated with substrate, this increase in reaction rate levels off. The rate of an enzyme-catalyzed reaction increases with an increase in the concentration of an enzyme. At low temperatures, an increase in temperature increases the rate of an enzyme-catalyzed reaction; at higher temperatures, the protein will denature. Enzymes have optimum pH ranges.
• 18.8: Enzyme Inhibition
An irreversible inhibitor inactivates an enzyme by bonding covalently to a particular group at the active site. A reversible inhibitor inactivates an enzyme through noncovalent, reversible interactions. A competitive inhibitor competes with the substrate for binding at the active site of the enzyme. A noncompetitive inhibitor binds at a site distinct from the active site.
• 18.9: Enzyme Cofactors and Vitamins
Vitamins are organic compounds that are essential in very small amounts for the maintenance of normal metabolism. Vitamins are divided into two broad categories: fat-soluble vitamins and water-soluble vitamins. Most water-soluble vitamins are needed for the formation of coenzymes, which are organic molecules needed by some enzymes for catalytic activity.
• 18.E: Amino Acids, Proteins, and Enzymes (Exercises)
Problems and select solutions for the chapter.
• 18.S: Amino Acids, Proteins, and Enzymes (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
18: Amino Acids Proteins and Enzymes
The 1923 Nobel Prize in Medicine or Physiology was awarded to Frederick Grant Banting and John James Richard Macleod for their discovery of the protein insulin. In 1958, the Nobel Prize in Chemistry was awarded to Frederick Sanger for his discoveries concerning the structure of proteins and, in particular, the structure of insulin. What is so important about insulin that two Nobel Prizes have been awarded for work on this protein?
Insulin is a hormone that is synthesized in the pancreas. Insulin stimulates the transport of glucose into cells throughout the body and the storage of glucose as glycogen. People with diabetes do not produce insulin or use it properly. The isolation of insulin in 1921 led to the first effective treatment for these individuals. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.00%3A_Prelude_to_Amino_Acids_Proteins_and_Enzymes.txt |
Learning Objectives
• To recognize amino acids and classify them based on the characteristics of their side chains.
The proteins in all living species, from bacteria to humans, are constructed from the same set of 20 amino acids, so called because each contains an amino group attached to a carboxylic acid. The amino acids in proteins are α-amino acids, which means the amino group is attached to the α-carbon of the carboxylic acid. Humans can synthesize only about half of the needed amino acids; the remainder must be obtained from the diet and are known as essential amino acids. However, two additional amino acids have been found in limited quantities in proteins: Selenocysteine was discovered in 1986, while pyrrolysine was discovered in 2002.
The amino acids are colorless, nonvolatile, crystalline solids, melting and decomposing at temperatures above 200°C. These melting temperatures are more like those of inorganic salts than those of amines or organic acids and indicate that the structures of the amino acids in the solid state and in neutral solution are best represented as having both a negatively charged group and a positively charged group. Such a species is known as a zwitterion.
Classification
In addition to the amino and carboxyl groups, amino acids have a side chain or R group attached to the α-carbon. Each amino acid has unique characteristics arising from the size, shape, solubility, and ionization properties of its R group. As a result, the side chains of amino acids exert a profound effect on the structure and biological activity of proteins. Although amino acids can be classified in various ways, one common approach is to classify them according to whether the functional group on the side chain at neutral pH is nonpolar, polar but uncharged, negatively charged, or positively charged. The structures and names of the 20 amino acids, their one- and three-letter abbreviations, and some of their distinctive features are given in Table \(1\).
Table \(1\): Common Amino Acids Found in Proteins
Common Name Abbreviation Structural Formula (at pH 6) Molar Mass Distinctive Feature
Amino acids with a nonpolar R group
glycine gly (G) 75 the only amino acid lacking a chiral carbon
alanine ala (A) 89
valine val (V) 117 a branched-chain amino acid
leucine leu (L) 131 a branched-chain amino acid
isoleucine ile (I) 131 an essential amino acid because most animals cannot synthesize branched-chain amino acids
phenylalanine phe (F) 165 also classified as an aromatic amino acid
tryptophan trp (W) 204 also classified as an aromatic amino acid
methionine met (M) 149 side chain functions as a methyl group donor
proline pro (P) 115 contains a secondary amine group; referred to as an α-imino acid
Amino acids with a polar but neutral R group
serine ser (S) 105 found at the active site of many enzymes
threonine thr (T) 119 named for its similarity to the sugar threose
cysteine cys (C) 121 oxidation of two cysteine molecules yields cystine
tyrosine tyr (Y) 181 also classified as an aromatic amino acid
asparagine asn (N) 132 the amide of aspartic acid
glutamine gln (Q) 146 the amide of glutamic acid
Amino acids with a negatively charged R group
aspartic acid asp (D) 132 carboxyl groups are ionized at physiological pH; also known as aspartate
glutamic acid glu (E) 146 carboxyl groups are ionized at physiological pH; also known as glutamate
Amino acids with a positively charged R group
histidine his (H) 155 the only amino acid whose R group has a pKa (6.0) near physiological pH
lysine lys (K) 147
arginine arg (R) 175 almost as strong a base as sodium hydroxide
The first amino acid to be isolated was asparagine in 1806. It was obtained from protein found in asparagus juice (hence the name). Glycine, the major amino acid found in gelatin, was named for its sweet taste (Greek glykys, meaning “sweet”). In some cases an amino acid found in a protein is actually a derivative of one of the common 20 amino acids (one such derivative is hydroxyproline). The modification occurs after the amino acid has been assembled into a protein.
Configuration
Notice in Table \(1\) that glycine is the only amino acid whose α-carbon is not chiral. Therefore, with the exception of glycine, the amino acids could theoretically exist in either the D- or the L-enantiomeric form and rotate plane-polarized light. As with sugars, chemists used L-glyceraldehyde as the reference compound for the assignment of absolute configuration to amino acids. Its structure closely resembles an amino acid structure except that in the latter, an amino group takes the place of the OH group on the chiral carbon of the L-glyceraldehyde and a carboxylic acid replaces the aldehyde. Modern stereochemistry assignments using the Cahn-Ingold-Prelog priority rules used ubiquitously in chemistry show that all of the naturally occurring chiral amino acids are S except Cys which is R.
We learned that all naturally occurring sugars belong to the D series. It is interesting, therefore, that nearly all known plant and animal proteins are composed entirely of L-amino acids. However, certain bacteria contain D-amino acids in their cell walls, and several antibiotics (e.g., actinomycin D and the gramicidins) contain varying amounts of D-leucine, D-phenylalanine, and D-valine.
Summary
Amino acids can be classified based on the characteristics of their distinctive side chains as nonpolar, polar but uncharged, negatively charged, or positively charged. The amino acids found in proteins are L-amino acids. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.01%3A_Properties_of_Amino_Acids.txt |
Learning Objectives
• To explain how an amino acid can act as both an acid and a base.
The structure of an amino acid allows it to act as both an acid and a base. An amino acid has this ability because at a certain pH value (different for each amino acid) nearly all the amino acid molecules exist as zwitterions. If acid is added to a solution containing the zwitterion, the carboxylate group captures a hydrogen (H+) ion, and the amino acid becomes positively charged. If base is added, ion removal of the H+ ion from the amino group of the zwitterion produces a negatively charged amino acid. In both circumstances, the amino acid acts to maintain the pH of the system—that is, to remove the added acid (H+) or base (OH) from solution.
Example \(1\)
1. Draw the structure for the anion formed when glycine (at neutral pH) reacts with a base.
2. Draw the structure for the cation formed when glycine (at neutral pH) reacts with an acid.
Solution
1. The base removes H+ from the protonated amine group.
• The acid adds H+ to the carboxylate group.
Exercise \(1\)
1. Draw the structure for the cation formed when valine (at neutral pH) reacts with an acid.
2. Draw the structure for the anion formed when valine (at neutral pH) reacts with a base.
The particular pH at which a given amino acid exists in solution as a zwitterion is called the isoelectric point (pI). At its pI, the positive and negative charges on the amino acid balance, and the molecule as a whole is electrically neutral. The amino acids whose side chains are always neutral have isoelectric points ranging from 5.0 to 6.5. The basic amino acids (which have positively charged side chains at neutral pH) have relatively high examples. Acidic amino acids (which have negatively charged side chains at neutral pH) have quite low examples (Table \(1\)).
Table \(1\): ExampIes of Some Representative Amino Acids
Amino Acid Classification pI
alanine nonpolar 6.0
valine nonpolar 6.0
serine polar, uncharged 5.7
threonine polar, uncharged 6.5
arginine positively charged (basic) 10.8
histidine positively charged (basic) 7.6
lysine positively charged (basic) 9.8
aspartic acid negatively charged (acidic) 3.0
glutamic acid negatively charged (acidic) 3.2
Amino acids undergo reactions characteristic of carboxylic acids and amines. The reactivity of these functional groups is particularly important in linking amino acids together to form peptides and proteins, as you will see later in this chapter. Simple chemical tests that are used to detect amino acids take advantage of the reactivity of these functional groups. An example is the ninhydrin test in which the amine functional group of α-amino acids reacts with ninhydrin to form purple-colored compounds. Ninhydrin is used to detect fingerprints because it reacts with amino acids from the proteins in skin cells transferred to the surface by the individual leaving the fingerprint.
Summary
Amino acids can act as both an acid and a base due to the presence of the amino and carboxyl functional groups. The pH at which a given amino acid exists in solution as a zwitterion is called the isoelectric point (pI). | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.02%3A_Reactions_of_Amino_Acids.txt |
Learning Objectives
• Explain how a peptide is formed from individual amino acids.
• Explain why the sequence of amino acids in a protein is important.
Two or more amino acids can join together into chains called peptides. Previously, we discussed the reaction between ammonia and a carboxylic acid to form an amide. In a similar reaction, the amino group on one amino acid molecule reacts with the carboxyl group on another, releasing a molecule of water and forming an amide linkage:
An amide bond joining two amino acid units is called a peptide bond. Note that the product molecule still has a reactive amino group on the left and a reactive carboxyl group on the right. These can react with additional amino acids to lengthen the peptide. The process can continue until thousands of units have joined, resulting in large proteins.
A chain consisting of only two amino acid units is called a dipeptide; a chain consisting of three is a tripeptide. By convention, peptide and protein structures are depicted with the amino acid whose amino group is free (the N-terminal end) on the left and the amino acid with a free carboxyl group (the C-terminal end) to the right.
The general term peptide refers to an amino acid chain of unspecified length. However, chains of about 50 amino acids or more are usually called proteins or polypeptides. In its physiologically active form, a protein may be composed of one or more polypeptide chains.
For peptides and proteins to be physiologically active, it is not enough that they incorporate certain amounts of specific amino acids. The order, or sequence, in which the amino acids are connected is also of critical importance. Bradykinin is a nine-amino acid peptide (Figure \(1\)) produced in the blood that has the following amino acid sequence:
arg-pro-pro-gly-phe-ser-pro-phe-arg
This peptide lowers blood pressure, stimulates smooth muscle tissue, increases capillary permeability, and causes pain. When the order of amino acids in bradykinin is reversed,
arg-phe-pro-ser-phe-gly-pro-pro-arg
the peptide resulting from this synthesis shows none of the activity of bradykinin.
Just as millions of different words are spelled with our 26-letter English alphabet, millions of different proteins are made with the 20 common amino acids. However, just as the English alphabet can be used to write gibberish, amino acids can be put together in the wrong sequence to produce nonfunctional proteins. Although the correct sequence is ordinarily of utmost importance, it is not always absolutely required. Just as you can sometimes make sense of incorrectly spelled English words, a protein with a small percentage of “incorrect” amino acids may continue to function. However, it rarely functions as well as a protein having the correct sequence. There are also instances in which seemingly minor errors of sequence have disastrous effects. For example, in some people, every molecule of hemoglobin (a protein in the blood that transports oxygen) has a single incorrect amino acid unit out of about 300 (a single valine replaces a glutamic acid). That “minor” error is responsible for sickle cell anemia, an inherited condition that usually is fatal.
Summary
The amino group of one amino acid can react with the carboxyl group on another amino acid to form a peptide bond that links the two amino acids together. Additional amino acids can be added on through the formation of addition peptide (amide) bonds. A sequence of amino acids in a peptide or protein is written with the N-terminal amino acid first and the C-terminal amino acid at the end (writing left to right). | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.03%3A_Peptides.txt |
Learning Objectives
• Describe the four levels of protein structure.
• Identify the types of attractive interactions that hold proteins in their most stable three-dimensional structure.
• Explain what happens when proteins are denatured.
• Identify how a protein can be denatured.
Each of the thousands of naturally occurring proteins has its own characteristic amino acid composition and sequence that result in a unique three-dimensional shape. Since the 1950s, scientists have determined the amino acid sequences and three-dimensional conformation of numerous proteins and thus obtained important clues on how each protein performs its specific function in the body.
Proteins are compounds of high molar mass consisting largely or entirely of chains of amino acids. Because of their great complexity, protein molecules cannot be classified on the basis of specific structural similarities, as carbohydrates and lipids are categorized. The two major structural classifications of proteins are based on far more general qualities: whether the protein is (1) fiberlike and insoluble or (2) globular and soluble. Some proteins, such as those that compose hair, skin, muscles, and connective tissue, are fiberlike. These fibrous proteins are insoluble in water and usually serve structural, connective, and protective functions. Examples of fibrous proteins are keratins, collagens, myosins, and elastins. Hair and the outer layer of skin are composed of keratin. Connective tissues contain collagen. Myosins are muscle proteins and are capable of contraction and extension. Elastins are found in ligaments and the elastic tissue of artery walls.
Globular proteins, the other major class, are soluble in aqueous media. In these proteins, the chains are folded so that the molecule as a whole is roughly spherical. Familiar examples include egg albumin from egg whites and serum albumin in blood. Serum albumin plays a major role in transporting fatty acids and maintaining a proper balance of osmotic pressures in the body. Hemoglobin and myoglobin, which are important for binding oxygen, are also globular proteins.
Levels of Protein Structure
The structure of proteins is generally described as having four organizational levels. The first of these is the primary structure, which is the number and sequence of amino acids in a protein’s polypeptide chain or chains, beginning with the free amino group and maintained by the peptide bonds connecting each amino acid to the next. The primary structure of insulin, composed of 51 amino acids, is shown in Figure \(1\).
A protein molecule is not a random tangle of polypeptide chains. Instead, the chains are arranged in unique but specific conformations. The term secondary structure refers to the fixed arrangement of the polypeptide backbone. On the basis of X ray studies, Linus Pauling and Robert Corey postulated that certain proteins or portions of proteins twist into a spiral or a helix. This helix is stabilized by intrachain hydrogen bonding between the carbonyl oxygen atom of one amino acid and the amide hydrogen atom four amino acids up the chain (located on the next turn of the helix) and is known as a right-handed α-helix. X ray data indicate that this helix makes one turn for every 3.6 amino acids, and the side chains of these amino acids project outward from the coiled backbone (Figure \(2\)). The α-keratins, found in hair and wool, are exclusively α-helical in conformation. Some proteins, such as gamma globulin, chymotrypsin, and cytochrome c, have little or no helical structure. Others, such as hemoglobin and myoglobin, are helical in certain regions but not in others.
Another common type of secondary structure, called the β-pleated sheet conformation, is a sheetlike arrangement in which two or more extended polypeptide chains (or separate regions on the same chain) are aligned side by side. The aligned segments can run either parallel or antiparallel—that is, the N-terminals can face in the same direction on adjacent chains or in different directions—and are connected by interchain hydrogen bonding (Figure \(3\)). The β-pleated sheet is particularly important in structural proteins, such as silk fibroin. It is also seen in portions of many enzymes, such as carboxypeptidase A and lysozyme.
Tertiary structure refers to the unique three-dimensional shape of the protein as a whole, which results from the folding and bending of the protein backbone. The tertiary structure is intimately tied to the proper biochemical functioning of the protein. Figure \(4\) shows a depiction of the three-dimensional structure of insulin.
Four major types of attractive interactions determine the shape and stability of the tertiary structure of proteins. You studied several of them previously.
1. Ionic bonding. Ionic bonds result from electrostatic attractions between positively and negatively charged side chains of amino acids. For example, the mutual attraction between an aspartic acid carboxylate ion and a lysine ammonium ion helps to maintain a particular folded area of a protein (part (a) of Figure \(5\)).
2. Hydrogen bonding. Hydrogen bonding forms between a highly electronegative oxygen atom or a nitrogen atom and a hydrogen atom attached to another oxygen atom or a nitrogen atom, such as those found in polar amino acid side chains. Hydrogen bonding (as well as ionic attractions) is extremely important in both the intra- and intermolecular interactions of proteins (part (b) of Figure \(5\)).
3. Disulfide linkages. Two cysteine amino acid units may be brought close together as the protein molecule folds. Subsequent oxidation and linkage of the sulfur atoms in the highly reactive sulfhydryl (SH) groups leads to the formation of cystine (part (c) of Figure \(5\)). Intrachain disulfide linkages are found in many proteins, including insulin (yellow bars in Figure \(1\)) and have a strong stabilizing effect on the tertiary structure.
1. Dispersion forces. Dispersion forces arise when a normally nonpolar atom becomes momentarily polar due to an uneven distribution of electrons, leading to an instantaneous dipole that induces a shift of electrons in a neighboring nonpolar atom. Dispersion forces are weak but can be important when other types of interactions are either missing or minimal (part (d) of Figure \(5\)). This is the case with fibroin, the major protein in silk, in which a high proportion of amino acids in the protein have nonpolar side chains. The term hydrophobic interaction is often misused as a synonym for dispersion forces. Hydrophobic interactions arise because water molecules engage in hydrogen bonding with other water molecules (or groups in proteins capable of hydrogen bonding). Because nonpolar groups cannot engage in hydrogen bonding, the protein folds in such a way that these groups are buried in the interior part of the protein structure, minimizing their contact with water.
When a protein contains more than one polypeptide chain, each chain is called a subunit. The arrangement of multiple subunits represents a fourth level of structure, the quaternary structure of a protein. Hemoglobin, with four polypeptide chains or subunits, is the most frequently cited example of a protein having quaternary structure (Figure \(6\)). The quaternary structure of a protein is produced and stabilized by the same kinds of interactions that produce and maintain the tertiary structure. A schematic representation of the four levels of protein structure is in Figure \(7\).
The primary structure consists of the specific amino acid sequence. The resulting peptide chain can twist into an α-helix, which is one type of secondary structure. This helical segment is incorporated into the tertiary structure of the folded polypeptide chain. The single polypeptide chain is a subunit that constitutes the quaternary structure of a protein, such as hemoglobin that has four polypeptide chains.
Denaturation of Proteins
The highly organized structures of proteins are truly masterworks of chemical architecture. But highly organized structures tend to have a certain delicacy, and this is true of proteins. Denaturation is the term used for any change in the three-dimensional structure of a protein that renders it incapable of performing its assigned function. A denatured protein cannot do its job. (Sometimes denaturation is equated with the precipitation or coagulation of a protein; our definition is a bit broader.) A wide variety of reagents and conditions, such as heat, organic compounds, pH changes, and heavy metal ions can cause protein denaturation (Figure \(1\)).
Figure \(1\): Protein Denaturation Methods
Method Effect on Protein Structure
Heat above 50°C or ultraviolet (UV) radiation Heat or UV radiation supplies kinetic energy to protein molecules, causing their atoms to vibrate more rapidly and disrupting relatively weak hydrogen bonding and dispersion forces.
Use of organic compounds, such as ethyl alcohol These compounds are capable of engaging in intermolecular hydrogen bonding with protein molecules, disrupting intramolecular hydrogen bonding within the protein.
Salts of heavy metal ions, such as mercury, silver, and lead These ions form strong bonds with the carboxylate anions of the acidic amino acids or SH groups of cysteine, disrupting ionic bonds and disulfide linkages.
Alkaloid reagents, such as tannic acid (used in tanning leather) These reagents combine with positively charged amino groups in proteins to disrupt ionic bonds.
Anyone who has fried an egg has observed denaturation. The clear egg white turns opaque as the albumin denatures and coagulates. No one has yet reversed that process. However, given the proper circumstances and enough time, a protein that has unfolded under sufficiently gentle conditions can refold and may again exhibit biological activity (Figure \(8\)). Such evidence suggests that, at least for these proteins, the primary structure determines the secondary and tertiary structure. A given sequence of amino acids seems to adopt its particular three-dimensional arrangement naturally if conditions are right.
The primary structures of proteins are quite sturdy. In general, fairly vigorous conditions are needed to hydrolyze peptide bonds. At the secondary through quaternary levels, however, proteins are quite vulnerable to attack, though they vary in their vulnerability to denaturation. The delicately folded globular proteins are much easier to denature than are the tough, fibrous proteins of hair and skin.
Summary
Proteins can be divided into two categories: fibrous, which tend to be insoluble in water, and globular, which are more soluble in water. A protein may have up to four levels of structure. The primary structure consists of the specific amino acid sequence. The resulting peptide chain can form an α-helix or β-pleated sheet (or local structures not as easily categorized), which is known as secondary structure. These segments of secondary structure are incorporated into the tertiary structure of the folded polypeptide chain. The quaternary structure describes the arrangements of subunits in a protein that contains more than one subunit. Four major types of attractive interactions determine the shape and stability of the folded protein: ionic bonding, hydrogen bonding, disulfide linkages, and dispersion forces. A wide variety of reagents and conditions can cause a protein to unfold or denature. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.04%3A_Proteins.txt |
Learning Objectives
• Explain the functions of enzymes.
• Explain how enzymes are classified and named.
A catalyst is any substance that increases the rate or speed of a chemical reaction without being changed or consumed in the reaction. Enzymes are biological catalysts, and nearly all of them are proteins. The reaction rates attained by enzymes are truly amazing. In their presence, reactions occur at rates that are a million (106) or more times faster than would be attainable in their absence. What is even more amazing is that enzymes perform this function at body temperature (~37°C) and physiological pH (pH ~7), rather than at the conditions that are typically necessary to increase reaction rates (high temperature or pressure, the use of strong oxidizing or reducing agents or strong acids or bases, or a combination of any of these). In addition, enzymes are highly specific in their action; that is, each enzyme catalyzes only one type of reaction in only one compound or a group of structurally related compounds. The compound or compounds on which an enzyme acts are known as its substrates.
Hundreds of enzymes have been purified and studied in an effort to understand how they work so effectively and with such specificity. The resulting knowledge has been used to design drugs that inhibit or activate particular enzymes. An example is the intensive research to improve the treatment of or find a cure for acquired immunodeficiency syndrome (AIDS). AIDS is caused by the human immunodeficiency virus (HIV). Researchers are studying the enzymes produced by this virus and are developing drugs intended to block the action of those enzymes without interfering with enzymes produced by the human body. Several of these drugs have now been approved for use by AIDS patients.
Table \(1\): Classes of Enzymes
Class Type of Reaction Catalyzed Examples
oxidoreductases oxidation-reduction reactions Dehydrogenases catalyze oxidation-reduction reactions involving hydrogen and reductases catalyze reactions in which a substrate is reduced.
transferases transfer reactions of groups, such as methyl, amino, and acetyl Transaminases catalyze the transfer of amino group, and kinases catalyze the transfer of a phosphate group.
hydrolases hydrolysis reactions Lipases catalyze the hydrolysis of lipids, and proteases catalyze the hydrolysis of proteins
lyases reactions in which groups are removed without hydrolysis or addition of groups to a double bond Decarboxylases catalyze the removal of carboxyl groups.
isomerases reactions in which a compound is converted to its isomer Isomerases may catalyze the conversion of an aldose to a ketose, and mutases catalyze reactions in which a functional group is transferred from one atom in a substrate to another.
ligases reactions in which new bonds are formed between carbon and another atom; energy is required Synthetases catalyze reactions in which two smaller molecules are linked to form a larger one.
The first enzymes to be discovered were named according to their source or method of discovery. The enzyme pepsin, which aids in the hydrolysis of proteins, is found in the digestive juices of the stomach (Greek pepsis, meaning “digestion”). Papain, another enzyme that hydrolyzes protein (in fact, it is used in meat tenderizers), is isolated from papayas. As more enzymes were discovered, chemists recognized the need for a more systematic and chemically informative identification scheme. In the current numbering and naming scheme, under the oversight of the Nomenclature Commission of the International Union of Biochemistry, enzymes are arranged into six groups according to the general type of reaction they catalyze (Table \(1\)), with subgroups and secondary subgroups that specify the reaction more precisely.
Figure \(1\): Structure of the alcohol dehydrogenase protein (E.C.1.1.1.1) (EE ISOZYME) complexed wtih nicotinamide adenini dinulceotide (NAD) and zinc (PDB: 1CDO).
Each enzyme is assigned a four-digit number, preceded by the prefix EC—for enzyme classification—that indicates its group, subgroup, and so forth. This is demonstrated in Table \(2\) for alcohol dehydrogenase. Each enzyme is also given a name consisting of the root of the name of its substrate or substrates and the -ase suffix. Thus urease is the enzyme that catalyzes the hydrolysis of urea.
Table \(2\): Assignment of an Enzyme Classification Number
Alcohol Dehydrogenase: EC 1.1.1.1
The first digit indicates that this enzyme is an oxidoreductase; that is, an enzyme that catalyzes an oxidation-reduction reaction.
The second digit indicates that this oxidoreductase catalyzes a reaction involving a primary or secondary alcohol.
The third digit indicates that either the coenzyme NAD+ or NADP+ is required for this reaction.
The fourth digit indicates that this was the first enzyme isolated, characterized, and named using this system of nomenclature.
The systematic name for this enzyme is alcohol:NAD+ oxidoreductase, while the recommended or common name is alcohol dehydrogenase.
Reaction catalyzed:
Summary
An enzyme is a biological catalyst, a substance that increases the rate of a chemical reaction without being changed or consumed in the reaction. A systematic process is used to name and classify enzymes. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.05%3A_Enzymes.txt |
Learning Objectives
• To describe the interaction between an enzyme and its substrate.
Enzyme-catalyzed reactions occur in at least two steps. In the first step, an enzyme molecule (E) and the substrate molecule or molecules (S) collide and react to form an intermediate compound called the enzyme-substrate (E–S) complex. (This step is reversible because the complex can break apart into the original substrate or substrates and the free enzyme.) Once the E–S complex forms, the enzyme is able to catalyze the formation of product (P), which is then released from the enzyme surface:
$S + E \rightarrow E–S \tag{$1$}$
$E–S \rightarrow P + E \tag{$2$}$
Hydrogen bonding and other electrostatic interactions hold the enzyme and substrate together in the complex. The structural features or functional groups on the enzyme that participate in these interactions are located in a cleft or pocket on the enzyme surface. This pocket, where the enzyme combines with the substrate and transforms the substrate to product is called the active site of the enzyme (Figure $1$).
The active site of an enzyme possesses a unique conformation (including correctly positioned bonding groups) that is complementary to the structure of the substrate, so that the enzyme and substrate molecules fit together in much the same manner as a key fits into a tumbler lock. In fact, an early model describing the formation of the enzyme-substrate complex was called the lock-and-key model (Figure $2$). This model portrayed the enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site.
Working out the precise three-dimensional structures of numerous enzymes has enabled chemists to refine the original lock-and-key model of enzyme actions. They discovered that the binding of a substrate often leads to a large conformational change in the enzyme, as well as to changes in the structure of the substrate or substrates. The current theory, known as the induced-fit model, says that enzymes can undergo a change in conformation when they bind substrate molecules, and the active site has a shape complementary to that of the substrate only after the substrate is bound, as shown for hexokinase in Figure $3$. After catalysis, the enzyme resumes its original structure.
The structural changes that occur when an enzyme and a substrate join together bring specific parts of a substrate into alignment with specific parts of the enzyme’s active site. Amino acid side chains in or near the binding site can then act as acid or base catalysts, provide binding sites for the transfer of functional groups from one substrate to another or aid in the rearrangement of a substrate. The participating amino acids, which are usually widely separated in the primary sequence of the protein, are brought close together in the active site as a result of the folding and bending of the polypeptide chain or chains when the protein acquires its tertiary and quaternary structure. Binding to enzymes brings reactants close to each other and aligns them properly, which has the same effect as increasing the concentration of the reacting compounds.
Example $1$
1. What type of interaction would occur between an OH group present on a substrate molecule and a functional group in the active site of an enzyme?
2. Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified.
Solution
1. An OH group would most likely engage in hydrogen bonding with an appropriate functional group present in the active site of an enzyme.
2. Several amino acid side chains would be able to engage in hydrogen bonding with an OH group. One example would be asparagine, which has an amide functional group.
Exercise $1$
1. What type of interaction would occur between an COO group present on a substrate molecule and a functional group in the active site of an enzyme?
2. Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified.
One characteristic that distinguishes an enzyme from all other types of catalysts is its substrate specificity. An inorganic acid such as sulfuric acid can be used to increase the reaction rates of many different reactions, such as the hydrolysis of disaccharides, polysaccharides, lipids, and proteins, with complete impartiality. In contrast, enzymes are much more specific. Some enzymes act on a single substrate, while other enzymes act on any of a group of related molecules containing a similar functional group or chemical bond. Some enzymes even distinguish between D- and L-stereoisomers, binding one stereoisomer but not the other. Urease, for example, is an enzyme that catalyzes the hydrolysis of a single substrate—urea—but not the closely related compounds methyl urea, thiourea, or biuret. The enzyme carboxypeptidase, on the other hand, is far less specific. It catalyzes the removal of nearly any amino acid from the carboxyl end of any peptide or protein.
Enzyme specificity results from the uniqueness of the active site in each different enzyme because of the identity, charge, and spatial orientation of the functional groups located there. It regulates cell chemistry so that the proper reactions occur in the proper place at the proper time. Clearly, it is crucial to the proper functioning of the living cell.
Summary
A substrate binds to a specific region on an enzyme known as the active site, where the substrate can be converted to product. The substrate binds to the enzyme primarily through hydrogen bonding and other electrostatic interactions. The induced-fit model says that an enzyme can undergo a conformational change when binding a substrate. Enzymes exhibit varying degrees of substrate specificity. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.06%3A_Enzyme_Action.txt |
Learning Objectives
• To describe how pH, temperature, and the concentration of an enzyme and its substrate influence enzyme activity.
The single most important property of enzymes is the ability to increase the rates of reactions occurring in living organisms, a property known as catalytic activity. Because most enzymes are proteins, their activity is affected by factors that disrupt protein structure, as well as by factors that affect catalysts in general. Factors that disrupt protein structure include temperature and pH; factors that affect catalysts in general include reactant or substrate concentration and catalyst or enzyme concentration. The activity of an enzyme can be measured by monitoring either the rate at which a substrate disappears or the rate at which a product forms.
Concentration of Substrate
In the presence of a given amount of enzyme, the rate of an enzymatic reaction increases as the substrate concentration increases until a limiting rate is reached, after which further increase in the substrate concentration produces no significant change in the reaction rate (part (a) of Figure \(1\)). At this point, so much substrate is present that essentially all of the enzyme active sites have substrate bound to them. In other words, the enzyme molecules are saturated with substrate. The excess substrate molecules cannot react until the substrate already bound to the enzymes has reacted and been released (or been released without reacting).
Let’s consider an analogy. Ten taxis (enzyme molecules) are waiting at a taxi stand to take people (substrate) on a 10-minute trip to a concert hall, one passenger at a time. If only 5 people are present at the stand, the rate of their arrival at the concert hall is 5 people in 10 minutes. If the number of people at the stand is increased to 10, the rate increases to 10 arrivals in 10 minutes. With 20 people at the stand, the rate would still be 10 arrivals in 10 minutes. The taxis have been “saturated.” If the taxis could carry 2 or 3 passengers each, the same principle would apply. The rate would simply be higher (20 or 30 people in 10 minutes) before it leveled off.
Concentration of Enzyme
When the concentration of the enzyme is significantly lower than the concentration of the substrate (as when the number of taxis is far lower than the number of waiting passengers), the rate of an enzyme-catalyzed reaction is directly dependent on the enzyme concentration (part (b) of Figure \(1\)). This is true for any catalyst; the reaction rate increases as the concentration of the catalyst is increased.
Temperature
A general rule of thumb for most chemical reactions is that a temperature rise of 10°C approximately doubles the reaction rate. To some extent, this rule holds for all enzymatic reactions. After a certain point, however, an increase in temperature causes a decrease in the reaction rate, due to denaturation of the protein structure and disruption of the active site (part (a) of Figure \(2\)). For many proteins, denaturation occurs between 45°C and 55°C. Furthermore, even though an enzyme may appear to have a maximum reaction rate between 40°C and 50°C, most biochemical reactions are carried out at lower temperatures because enzymes are not stable at these higher temperatures and will denature after a few minutes.
At 0°C and 100°C, the rate of enzyme-catalyzed reactions is nearly zero. This fact has several practical applications. We sterilize objects by placing them in boiling water, which denatures the enzymes of any bacteria that may be in or on them. We preserve our food by refrigerating or freezing it, which slows enzyme activity. When animals go into hibernation in winter, their body temperature drops, decreasing the rates of their metabolic processes to levels that can be maintained by the amount of energy stored in the fat reserves in the animals’ tissues.
Hydrogen Ion Concentration (pH)
Because most enzymes are proteins, they are sensitive to changes in the hydrogen ion concentration or pH. Enzymes may be denatured by extreme levels of hydrogen ions (whether high or low); any change in pH, even a small one, alters the degree of ionization of an enzyme’s acidic and basic side groups and the substrate components as well. Ionizable side groups located in the active site must have a certain charge for the enzyme to bind its substrate. Neutralization of even one of these charges alters an enzyme’s catalytic activity.
An enzyme exhibits maximum activity over the narrow pH range in which a molecule exists in its properly charged form. The median value of this pH range is called the optimum pH of the enzyme (part (b) of Figure \(2\)). With the notable exception of gastric juice (the fluids secreted in the stomach), most body fluids have pH values between 6 and 8. Not surprisingly, most enzymes exhibit optimal activity in this pH range. However, a few enzymes have optimum pH values outside this range. For example, the optimum pH for pepsin, an enzyme that is active in the stomach, is 2.0.
Summary
Initially, an increase in substrate concentration leads to an increase in the rate of an enzyme-catalyzed reaction. As the enzyme molecules become saturated with substrate, this increase in reaction rate levels off. The rate of an enzyme-catalyzed reaction increases with an increase in the concentration of an enzyme. At low temperatures, an increase in temperature increases the rate of an enzyme-catalyzed reaction. At higher temperatures, the protein is denatured, and the rate of the reaction dramatically decreases. An enzyme has an optimum pH range in which it exhibits maximum activity. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.07%3A_Enzyme_Activity.txt |
Learning Objectives
• Explain what an enzyme inhibitor is.
• Distinguish between reversible and irreversible inhibitors.
• Distinguish between competitive and noncompetitive inhibitors.
Previously, we noted that enzymes are inactivated at high temperatures and by changes in pH. These are nonspecific factors that would inactivate any enzyme. The activity of enzymes can also be regulated by more specific inhibitors. Many compounds are poisons because they bind covalently to particular enzymes or kinds of enzymes and inactivate them (Table \(1\)).
Table \(1\): Poisons as Enzyme Inhibitors
Poison Formula Example of Enzyme Inhibited Action
arsenate \(\ce{AsO4^{3−}}\) glyceraldehyde 3-phosphate dehydrogenase substitutes for phosphate
iodoacetate \(\ce{ICH2COO^{−}}\) triose phosphate dehydrogenase binds to cysteine \(\ce{SH}\) group
diisopropylfluoro-phosphate (DIFP; a nerve poison) acetylcholinesterase binds to serine \(\ce{OH}\) group
Irreversible Inhibition: Poisons
An irreversible inhibitor inactivates an enzyme by bonding covalently to a particular group at the active site. The inhibitor-enzyme bond is so strong that the inhibition cannot be reversed by the addition of excess substrate. The nerve gases, especially Diisopropyl fluorophosphate (DIFP), irreversibly inhibit biological systems by forming an enzyme-inhibitor complex with a specific OH group of serine situated at the active sites of certain enzymes. The peptidases trypsin and chymotrypsin contain serine groups at the active site and are inhibited by DIFP.
Reversible Inhibition
A reversible inhibitor inactivates an enzyme through noncovalent, more easily reversed, interactions. Unlike an irreversible inhibitor, a reversible inhibitor can dissociate from the enzyme. Reversible inhibitors include competitive inhibitors and noncompetitive inhibitors. (There are additional types of reversible inhibitors.) A competitive inhibitor is any compound that bears a structural resemblance to a particular substrate and thus competes with that substrate for binding at the active site of an enzyme. The inhibitor is not acted on by the enzyme but does prevent the substrate from approaching the active site.
The degree to which a competitive inhibitor interferes with an enzyme’s activity depends on the relative concentrations of the substrate and the inhibitor. If the inhibitor is present in relatively large quantities, it will initially block most of the active sites. But because the binding is reversible, some substrate molecules will eventually bind to the active site and be converted to product. Increasing the substrate concentration promotes displacement of the inhibitor from the active site. Competitive inhibition can be completely reversed by adding substrate so that it reaches a much higher concentration than that of the inhibitor.
Studies of competitive inhibition have provided helpful information about certain enzyme-substrate complexes and the interactions of specific groups at the active sites. As a result, pharmaceutical companies have synthesized drugs that competitively inhibit metabolic processes in bacteria and certain cancer cells. Many drugs are competitive inhibitors of specific enzymes.
A classic example of competitive inhibition is the effect of malonate on the enzyme activity of succinate dehydrogenase (Figure \(1\)). Malonate and succinate are the anions of dicarboxylic acids and contain three and four carbon atoms, respectively. The malonate molecule binds to the active site because the spacing of its carboxyl groups is not greatly different from that of succinate. However, no catalytic reaction occurs because malonate does not have a CH2CH2 group to convert to CH=CH. This reaction will also be discussed in connection with the Krebs cycle and energy production.
To Your Health: Penicillin
Chemotherapy is the strategic use of chemicals (that is, drugs) to destroy infectious microorganisms or cancer cells without causing excessive damage to the other, healthy cells of the host. From bacteria to humans, the metabolic pathways of all living organisms are quite similar, so the search for safe and effective chemotherapeutic agents is a formidable task. Many well-established chemotherapeutic drugs function by inhibiting a critical enzyme in the cells of the invading organism.
An antibiotic is a compound that kills bacteria; it may come from a natural source such as molds or be synthesized with a structure analogous to a naturally occurring antibacterial compound. Antibiotics constitute no well-defined class of chemically related substances, but many of them work by effectively inhibiting a variety of enzymes essential to bacterial growth.
Penicillin, one of the most widely used antibiotics in the world, was fortuitously discovered by Alexander Fleming in 1928, when he noticed antibacterial properties in a mold growing on a bacterial culture plate. In 1938, Ernst Chain and Howard Florey began an intensive effort to isolate penicillin from the mold and study its properties. The large quantities of penicillin needed for this research became available through development of a corn-based nutrient medium that the mold loved and through the discovery of a higher-yielding strain of mold at a United States Department of Agriculture research center near Peoria, Illinois. Even so, it was not until 1944 that large quantities of penicillin were being produced and made available for the treatment of bacterial infections.
Penicillin functions by interfering with the synthesis of cell walls of reproducing bacteria. It does so by inhibiting an enzyme—transpeptidase—that catalyzes the last step in bacterial cell-wall biosynthesis. The defective walls cause bacterial cells to burst. Human cells are not affected because they have cell membranes, not cell walls.
Several naturally occurring penicillins have been isolated. They are distinguished by different R groups connected to a common structure: a four-member cyclic amide (called a lactam ring) fused to a five-member ring. The addition of appropriate organic compounds to the culture medium leads to the production of the different kinds of penicillin.
The penicillins are effective against gram-positive bacteria (bacteria capable of being stained by Gram’s stain) and a few gram-negative bacteria (including the intestinal bacterium Escherichia coli). They are effective in the treatment of diphtheria, gonorrhea, pneumonia, syphilis, many pus infections, and certain types of boils. Penicillin G was the earliest penicillin to be used on a wide scale. However, it cannot be administered orally because it is quite unstable; the acidic pH of the stomach converts it to an inactive derivative. The major oral penicillins—penicillin V, ampicillin, and amoxicillin—on the other hand, are acid stable.
Some strains of bacteria become resistant to penicillin through a mutation that allows them to synthesize an enzyme—penicillinase—that breaks the antibiotic down (by cleavage of the amide linkage in the lactam ring). To combat these strains, scientists have synthesized penicillin analogs (such as methicillin) that are not inactivated by penicillinase.
Some people (perhaps 5% of the population) are allergic to penicillin and therefore must be treated with other antibiotics. Their allergic reaction can be so severe that a fatal coma may occur if penicillin is inadvertently administered to them. Fortunately, several other antibiotics have been discovered. Most, including aureomycin and streptomycin, are the products of microbial synthesis. Others, such as the semisynthetic penicillins and tetracyclines, are made by chemical modifications of antibiotics; and some, like chloramphenicol, are manufactured entirely by chemical synthesis. They are as effective as penicillin in destroying infectious microorganisms. Many of these antibiotics exert their effects by blocking protein synthesis in microorganisms.
Initially, antibiotics were considered miracle drugs, substantially reducing the number of deaths from blood poisoning, pneumonia, and other infectious diseases. Some seven decades ago, a person with a major infection almost always died. Today, such deaths are rare. Seven decades ago, pneumonia was a dreaded killer of people of all ages. Today, it kills only the very old or those ill from other causes. Antibiotics have indeed worked miracles in our time, but even miracle drugs have limitations. Not long after the drugs were first used, disease organisms began to develop strains resistant to them. In a race to stay ahead of resistant bacterial strains, scientists continue to seek new antibiotics. The penicillins have now been partially displaced by related compounds, such as the cephalosporins and vancomycin. Unfortunately, some strains of bacteria have already shown resistance to these antibiotics.
Some reversible inhibitors are noncompetitive. A noncompetitive inhibitor can combine with either the free enzyme or the enzyme-substrate complex because its binding site on the enzyme is distinct from the active site. Binding of this kind of inhibitor alters the three-dimensional conformation of the enzyme, changing the configuration of the active site with one of two results. Either the enzyme-substrate complex does not form at its normal rate, or, once formed, it does not yield products at the normal rate. Because the inhibitor does not structurally resemble the substrate, the addition of excess substrate does not reverse the inhibitory effect.
Feedback inhibition is a normal biochemical process that makes use of noncompetitive inhibitors to control some enzymatic activity. In this process, the final product inhibits the enzyme that catalyzes the first step in a series of reactions. Feedback inhibition is used to regulate the synthesis of many amino acids. For example, bacteria synthesize isoleucine from threonine in a series of five enzyme-catalyzed steps. As the concentration of isoleucine increases, some of it binds as a noncompetitive inhibitor to the first enzyme of the series (threonine deaminase), thus bringing about a decrease in the amount of isoleucine being formed (Figure \(2\)).
Summary
An irreversible inhibitor inactivates an enzyme by bonding covalently to a particular group at the active site. A reversible inhibitor inactivates an enzyme through noncovalent, reversible interactions. A competitive inhibitor competes with the substrate for binding at the active site of the enzyme. A noncompetitive inhibitor binds at a site distinct from the active site. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.08%3A_Enzyme_Inhibition.txt |
Learning Objectives
• To explain why vitamins are necessary in the diet.
Many enzymes are simple proteins consisting entirely of one or more amino acid chains. Other enzymes contain a nonprotein component called a cofactor that is necessary for the enzyme’s proper functioning. There are two types of cofactors: inorganic ions [e.g., zinc or Cu(I) ions] and organic molecules known as coenzymes. Most coenzymes are vitamins or are derived from vitamins.
Vitamins are organic compounds that are essential in very small (trace) amounts for the maintenance of normal metabolism. They generally cannot be synthesized at adequate levels by the body and must be obtained from the diet. The absence or shortage of a vitamin may result in a vitamin-deficiency disease. In the first half of the 20th century, a major focus of biochemistry was the identification, isolation, and characterization of vitamins. Despite accumulating evidence that people needed more than just carbohydrates, fats, and proteins in their diets for normal growth and health, it was not until the early 1900s that research established the need for trace nutrients in the diet.
Vitamin Physiological Function Effect of Deficiency
Table \(1\): Fat-Soluble Vitamins and Physiological Functions
vitamin A (retinol) formation of vision pigments; differentiation of epithelial cells night blindness; continued deficiency leads to total blindness
vitamin D (cholecalciferol) increases the body’s ability to absorb calcium and phosphorus osteomalacia (softening of the bones); known as rickets in children
vitamin E (tocopherol) fat-soluble antioxidant damage to cell membranes
vitamin K (phylloquinone) formation of prothrombin, a key enzyme in the blood-clotting process increases the time required for blood to clot
Because organisms differ in their synthetic abilities, a substance that is a vitamin for one species may not be so for another. Over the past 100 years, scientists have identified and isolated 13 vitamins required in the human diet and have divided them into two broad categories: the fat-soluble vitamins, which include vitamins A, D, E, and K, and the water-soluble vitamins, which are the B complex vitamins and vitamin C. All fat-soluble vitamins contain a high proportion of hydrocarbon structural components. There are one or two oxygen atoms present, but the compounds as a whole are nonpolar. In contrast, water-soluble vitamins contain large numbers of electronegative oxygen and nitrogen atoms, which can engage in hydrogen bonding with water. Most water-soluble vitamins act as coenzymes or are required for the synthesis of coenzymes. The fat-soluble vitamins are important for a variety of physiological functions. The key vitamins and their functions are found in Tables \(1\) and \(2\).
Table \(2\): Water-Soluble Vitamins and Physiological Functions
Vitamin Coenzyme Coenzyme Function Deficiency Disease
vitamin B1 (thiamine) thiamine pyrophosphate decarboxylation reactions beri-beri
vitamin B2 (riboflavin) flavin mononucleotide or flavin adenine dinucleotide oxidation-reduction reactions involving two hydrogen atoms
vitamin B3 (niacin) nicotinamide adenine dinucleotide or nicotinamide adenine dinucleotide phosphate oxidation-reduction reactions involving the hydride ion (H) pellagra
vitamin B6 (pyridoxine) pyridoxal phosphate variety of reactions including the transfer of amino groups
vitamin B12 (cyanocobalamin) methylcobalamin or deoxyadenoxylcobalamin intramolecular rearrangement reactions pernicious anemia
biotin biotin carboxylation reactions
folic acid tetrahydrofolate carrier of one-carbon units such as the formyl group anemia
pantothenic Acid coenzyme A carrier of acyl groups
vitamin C (ascorbic acid) none antioxidant; formation of collagen, a protein found in tendons, ligaments, and bone scurvy
Vitamins C and E, as well as the provitamin β-carotene can act as antioxidants in the body. Antioxidants prevent damage from free radicals, which are molecules that are highly reactive because they have unpaired electrons. Free radicals are formed not only through metabolic reactions involving oxygen but also by such environmental factors as radiation and pollution.
β-carotene is known as a provitamin because it can be converted to vitamin A in the body.
Free radicals react most commonly react with lipoproteins and unsaturated fatty acids in cell membranes, removing an electron from those molecules and thus generating a new free radical. The process becomes a chain reaction that finally leads to the oxidative degradation of the affected compounds. Antioxidants react with free radicals to stop these chain reactions by forming a more stable molecule or, in the case of vitamin E, a free radical that is much less reactive (vitamin E is converted back to its original form through interaction with vitamin C).
Summary
Vitamins are organic compounds that are essential in very small amounts for the maintenance of normal metabolism. Vitamins are divided into two broad categories: fat-soluble vitamins and water-soluble vitamins. Most water-soluble vitamins are needed for the formation of coenzymes, which are organic molecules needed by some enzymes for catalytic activity. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.09%3A_Enzyme_Cofactors_and_Vitamins.txt |
Concept Review Exercises
1. What is the general structure of an α-amino acid?
2. Identify the amino acid that fits each description.
1. also known as aspartate
2. almost as strong a base as sodium hydroxide
3. does not have a chiral carbon
1. aspartic acid
2. arginine
3. glycine
3. Write the side chain of each amino acid.
1. serine
2. arginine
3. phenylalanine
4. Write the side chain of each amino acid.
1. aspartic acid
2. methionine
3. valine
5. Draw the structure for each amino acid.
1. alanine
2. cysteine
3. histidine
6. Draw the structure for each amino acid.
1. threonine
2. glutamic acid
3. leucine
7. Identify an amino acid whose side chain contains a(n)
1. amide functional group.
2. aromatic ring.
3. carboxyl group.
8. Identify an amino acid whose side chain contains a(n)
1. OH group
2. branched chain
3. amino group
1. CH2OH
1.
2.
1. asparagine or glutamine
2. phenylalanine, tyrosine, or tryptophan
3. aspartic acid or glutamic acid
Concept Review Exercises
1. Define each term.
1. zwitterion
2. isoelectric point
2. Draw the structure for the anion formed when alanine (at neutral pH) reacts with a base.
3. Draw the structure for the cation formed when alanine (at neutral pH) reacts with an acid.
Answers
1. an electrically neutral compound that contains both negatively and positively charged groups
2. the pH at which a given amino acid exists in solution as a zwitterion
Exercises
1. Draw the structure of leucine and determine the charge on the molecule in a(n)
1. acidic solution (pH = 1).
2. neutral solution (pH = 7).
3. a basic solution (pH = 11)
2. Draw the structure of isoleucine and determine the charge on the molecule in a(n)
1. acidic solution (pH = 1).
2. neutral solution (pH = 7).
3. basic solution (pH = 11).
Concept Review Exercises
1. Distinguish between the N-terminal amino acid and the C-terminal amino acid of a peptide or protein.
2. Describe the difference between an amino acid and a peptide.
3. Amino acid units in a protein are connected by peptide bonds. What is another name for the functional group linking the amino acids?
Answers
1. The N-terminal end is the end of a peptide or protein whose amino group is free (not involved in the formation of a peptide bond), while the C-terminal end has a free carboxyl group.
2. A peptide is composed of two or more amino acids. Amino acids are the building blocks of peptides.
3. amide bond
Exercises
1. Draw the structure for each peptide.
1. gly-val
2. val-gly
2. Draw the structure for cys-val-ala.
3. Identify the C- and N-terminal amino acids for the peptide lys-val-phe-gly-arg-cys.
4. Identify the C- and N-terminal amino acids for the peptide asp-arg-val-tyr-ile-his-pro-phe.
Answers
1. C-terminal amino acid: cys; N-terminal amino acid: lys
Concept Review Exercises
1. What is the predominant attractive force that stabilizes the formation of secondary structure in proteins?
2. Distinguish between the tertiary and quaternary levels of protein structure.
3. Briefly describe four ways in which a protein could be denatured.
Answers
1. hydrogen bonding
2. Tertiary structure refers to the unique three-dimensional shape of a single polypeptide chain, while quaternary structure describes the interaction between multiple polypeptide chains for proteins that have more than one polypeptide chain.
3. (1) heat a protein above 50°C or expose it to UV radiation; (2) add organic solvents, such as ethyl alcohol, to a protein solution; (3) add salts of heavy metal ions, such as mercury, silver, or lead; and (4) add alkaloid reagents such as tannic acid
Exercises
1. Classify each protein as fibrous or globular.
1. albumin
2. myosin
3. fibroin
2. Classify each protein as fibrous or globular.
1. hemoglobin
2. keratin
3. myoglobin
3. What name is given to the predominant secondary structure found in silk?
4. What name is given to the predominant secondary structure found in wool protein?
5. A protein has a tertiary structure formed by interactions between the side chains of the following pairs of amino acids. For each pair, identify the strongest type of interaction between these amino acids.
1. aspartic acid and lysine
2. phenylalanine and alanine
3. serine and lysine
4. two cysteines
6. A protein has a tertiary structure formed by interactions between the side chains of the following pairs of amino acids. For each pair, identify the strongest type of interaction between these amino acids.
1. valine and isoleucine
2. asparagine and serine
3. glutamic acid and arginine
4. tryptophan and methionine
7. What level(s) of protein structure is(are) ordinarily disrupted in denaturation? What level(s) is(are) not?
8. Which class of proteins is more easily denatured—fibrous or globular?
Answers
1. globular
2. fibrous
3. fibrous
1. β-pleated sheet
1. ionic bonding
2. dispersion forces
3. dispersion forces
4. disulfide linkage
1. Protein denaturation disrupts the secondary, tertiary, and quaternary levels of structure. Only primary structure is unaffected by denaturation.
Concept Review Exercise
In the small intestine, sucrose is hydrolyzed to form glucose and fructose in a reaction catalyzed by sucrase.
1. Identify the substrate in this reaction.
2. Name the enzyme.
1. sucrose
2. sucrase
Exercises
1. Identify the substrate catalyzed by each enzyme.
1. lactase
2. cellulase
3. peptidase
2. Identify the substrate catalyzed by each enzyme.
1. lipase
2. amylase
3. maltase
3. Identify each type of enzyme.
1. decarboxylase
2. protease
3. transaminase
4. Identify each type of enzyme.
1. dehydrogenase
2. isomerase
3. lipase
Answers
1. lactose
2. cellulose
3. peptides
1. lyase
2. hydrolase
3. transferase
Concept Review Exercises
1. Distinguish between the lock-and-key model and induced-fit model of enzyme action.
2. Which enzyme has greater specificity—urease or carboxypeptidase? Explain.
Answers
1. The lock-and-key model portrays an enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site. The induced fit model portrays the enzyme structure as more flexible and is complementary to the substrate only after the substrate is bound.
2. Urease has the greater specificity because it can bind only to a single substrate. Carboxypeptidase, on the other hand, can catalyze the removal of nearly any amino acid from the carboxyl end of a peptide or protein.
Exercises
1. What type of interaction would occur between each group present on a substrate molecule and a functional group of the active site in an enzyme?
1. COOH
2. NH3+
3. OH
4. CH(CH3)2
2. What type of interaction would occur between each group present on a substrate molecule and a functional group of the active site in an enzyme?
1. SH
2. NH2
3. C6H5
4. COO
3. For each functional group in Exercise 1, suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you identified.
4. For each functional group in Exercise 2, suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you identified.
Answers
1. hydrogen bonding
2. ionic bonding
3. hydrogen bonding
4. dispersion forces
1. The amino acid has a polar side chain capable of engaging in hydrogen bonding; serine (answers will vary).
2. The amino acid has a negatively charged side chain; aspartic acid (answers will vary).
3. The amino acid has a polar side chain capable of engaging in hydrogen bonding; asparagine (answers will vary).
4. The amino acid has a nonpolar side chain; isoleucine (answers will vary).
Concept Review Exercises
1. The concentration of substrate X is low. What happens to the rate of the enzyme-catalyzed reaction if the concentration of X is doubled?
2. What effect does an increase in the enzyme concentration have on the rate of an enzyme-catalyzed reaction?
Answers
1. If the concentration of the substrate is low, increasing its concentration will increase the rate of the reaction.
2. An increase in the amount of enzyme will increase the rate of the reaction (provided sufficient substrate is present).
Exercises
1. In non-enzyme-catalyzed reactions, the reaction rate increases as the concentration of reactant is increased. In an enzyme-catalyzed reaction, the reaction rate initially increases as the substrate concentration is increased but then begins to level off, so that the increase in reaction rate becomes less and less as the substrate concentration increases. Explain this difference.
2. Why do enzymes become inactive at very high temperatures?
3. An enzyme has an optimum pH of 7.4. What is most likely to happen to the activity of the enzyme if the pH drops to 6.3? Explain.
4. An enzyme has an optimum pH of 7.2. What is most likely to happen to the activity of the enzyme if the pH increases to 8.5? Explain.
Answers
1. In an enzyme-catalyzed reaction, the substrate binds to the enzyme to form an enzyme-substrate complex. If more substrate is present than enzyme, all of the enzyme binding sites will have substrate bound, and further increases in substrate concentration cannot increase the rate.
1. The activity will decrease; a pH of 6.3 is more acidic than 7.4, and one or more key groups in the active site may bind a hydrogen ion, changing the charge on that group.
Concept Review Exercises
1. What is the difference between a cofactor and a coenzyme?
2. How are vitamins related to coenzymes?
Answers
1. A coenzyme is one type of cofactor. Coenzymes are organic molecules required by some enzymes for activity. A cofactor can be either a coenzyme or an inorganic ion.
2. Coenzymes are synthesized from vitamins.
Exercises
1. Identify each vitamin as water soluble or fat soluble.
1. vitamin D
2. vitamin C
3. vitamin B12
2. Identify each vitamin as water soluble or fat soluble.
1. niacin
2. cholecalciferol
3. biotin
3. What vitamin is needed to form each coenzyme?
1. pyridoxal phosphate
2. flavin adenine dinucleotide
3. coenzyme A
4. nicotinamide adenine dinucleotide
4. What coenzyme is formed from each vitamin?
1. niacin
2. thiamine
3. cyanocobalamin
4. pantothenic acid
5. What is the function of each vitamin or coenzyme?
1. flavin adenine dinucleotide
2. vitamin A
3. biotin
6. What is the function of each vitamin or coenzyme?
1. vitamin K
2. pyridoxal phosphate
3. tetrahydrofolate
Answers
1. fat soluble
2. water soluble
3. water soluble
1. vitamin B6 or pyridoxine
2. vitamin B2 or riboflavin
3. pantothenic acid
4. vitamin B3 or niacin
1. needed by enzymes that catalyze oxidation-reduction reactions in which two hydrogen atoms are transferred
2. needed for the formation of vision pigments
3. needed by enzymes that catalyze carboxylation reactions
Concept Review Exercises
1. What are the characteristics of an irreversible inhibitor?
2. In what ways does a competitive inhibitor differ from a noncompetitive inhibitor?
Answers
1. It inactivates an enzyme by bonding covalently to a particular group at the active site.
2. A competitive inhibitor structurally resembles the substrate for a given enzyme and competes with the substrate for binding at the active site of the enzyme. A noncompetitive inhibitor binds at a site distinct from the active site and can bind to either the free enzyme or the enzyme-substrate complex.
Exercises
1. What amino acid is present in the active site of all enzymes that are irreversibly inhibited by nerve gases such as DIFP?
2. Oxaloacetate (OOCCH2COCOO) inhibits succinate dehydrogenase. Would you expect oxaloacetate to be a competitive or noncompetitive inhibitor? Explain.
1. serine
Additional Exercises
1. Draw the structure of the amino acid γ-aminobutyric acid (GABA). Would you expect to find GABA in the amino acid sequence of a protein? Explain.
2. Draw the structure of the amino acid homocysteine (R group = CH2CH2SH). Would you expect to find homocysteine in the amino acid sequence of a protein? Justify your answer.
3. Write equations to show how leucine can act as a buffer (that is, how it can neutralize added acid or base).
4. Write equations to show how isoleucine can act as a buffer (that is, how it can neutralize added acid or base).
5. Glutathione (γ-glutamylcysteinylglycine) is a tripeptide found in all cells of higher animals. It contains glutamic acid joined in an unusual peptide linkage involving the carboxyl group of the R group (known as γ-carboxyl group), rather than the usual carboxyl group (the α-carboxyl group). Draw the structure of glutathione.
6. Draw the structure of the pentapeptide whose sequence is arg-his-gly-leu-asp. Identify which of the amino acids have R groups that can donate or gain hydrogen ions.
7. Bradykinin is a peptide hormone composed of nine amino acids that lowers blood pressure. Its primary structure is arg-pro-pro-gly-phe-ser-pro-phe-arg. Would you expect bradykinin to be positively charged, negatively charged, or neutral at a pH of 6.0? Justify your answer.
8. One of the neurotransmitters involved in pain sensation is a peptide called substance P, which is composed of 11 amino acids and is released by nerve-cell terminals in response to pain. Its primary structure is arg-pro-lys-pro-gln-gln-phe-phe-gly-leu-met. Would you expect this peptide to be positively charged, negatively charged, or neutral at a pH of 6.0? Justify your answer.
9. Carbohydrates are incorporated into glycoproteins. Would you expect the incorporation of sugar units to increase or decrease the solubility of a protein? Justify your answer.
10. Some proteins have phosphate groups attached through an ester linkage to the OH groups of serine, threonine, or tyrosine residues to form phosphoproteins. Would you expect the incorporation of a phosphate group to increase or decrease the solubility of a protein? Justify your answer.
11. Refer to Table 18.5 and determine how each enzyme would be classified.
1. the enzyme that catalyzes the conversion of ethanol to acetaldehyde
2. the enzyme that catalyzes the breakdown of glucose 6-phosphate to glucose and inorganic phosphate ion (water is also a reactant in this reaction)
12. Refer to Table 18.5 and determine how each enzyme would be classified.
1. the enzyme that catalyzes the removal of a carboxyl group from pyruvate to form acetate
2. the enzyme that catalyzes the rearrangement of 3-phosphoglycerate to form 2-phosphoglycerate
13. The enzyme lysozyme has an aspartic acid residue in the active site. In acidic solution, the enzyme is inactive, but activity increases as the pH rises to around 6. Explain why.
14. The enzyme lysozyme has a glutamic acid residue in the active site. At neutral pH (6–7), the enzyme is active, but activity decreases as the pH rises. Explain why.
15. The activity of a purified enzyme is measured at a substrate concentration of 1.0 μM and found to convert 49 μmol of substrate to product in 1 min. The activity is measured at 2.0 μM substrate and found to convert 98 μmol of substrate to product/minute.
1. When the substrate concentration is 100 μM, how much substrate would you predict is converted to product in 1 min? What if the substrate concentration were increased to 1,000 μM (1.0 mM)?
2. The activities actually measured are 676 μmol product formed/minute at a substrate concentration of 100 μM and 698 μmol product formed/minute at 1,000 μM (1.0 mM) substrate. Is there any discrepancy between these values and those you predicted in Exercise 15a? Explain.
16. A patient has a fever of 39°C. Would you expect the activity of enzymes in the body to increase or decrease relative to their activity at normal body temperature (37°C)?
17. Using your knowledge of factors that influence enzyme activity, describe what happens when milk is pasteurized.
Answers
1. This amino acid would not be found in proteins because it is not an α-amino acid.
1. Bradykinin would be positively charged; all of the amino acids, except for arginine, have R groups that do not become either positively or negatively charged. The two arginines are R groups that are positively charged at neutral pH, so the peptide would have an overall positive charge.
1. Carbohydrates have many OH groups attached, which can engage in hydrogen bonding with water, which increases the solubility of the proteins.
1.
1. oxidoreductase
2. hydrolase
1. The enzyme is active when the carboxyl group in the R group of aspartic acid does not have the hydrogen attached (forming COO); the hydrogen is removed when the pH of the solution is around pH 6 or higher.
1.
1. at 100 μM, you would predict that the rate would increase 100 times to 4,900 μmol of substrate to product in 1 min; at 1.0 mM, you would predict an increase to 49,000 μmol of substrate to product in 1 min.
2. There is a great discrepancy between the predicted rates and actual rates; this occurs because the enzyme becomes saturated with substrate, preventing a further increase in the rate of the reaction (the reaction is no longer linear with respect to substrate concentration because it is at very low concentrations).
1. When milk is pasteurized, it is heated to high temperatures. These high temperatures denature the proteins in bacteria, so they cannot carry out needed functions to grow and multiply. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.E%3A_Amino_Acids_Proteins_and_Enzymes_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
A protein is a large biological polymer synthesized from amino acids, which are carboxylic acids containing an α-amino group. Proteins have a variety of important roles in living organisms, yet they are made from the same 20 L-amino acids. About half of these amino acids, the essential amino acids, cannot be synthesized by the human body and must be obtained from the diet. In the solid state and in neutral solutions, amino acids exist as zwitterions, species that are charged but electrically neutral. In this form, they behave much like inorganic salts. Each amino acid belongs to one of four classes depending on the characteristics of its R group or amino acid side chain: nonpolar, polar but neutral, positively charged, and negatively charged. Depending on the conditions, amino acids can act as either acids or bases, which means that proteins act as buffers. The pH at which an amino acid exists as the zwitterion is called the isoelectric point (pI).
The amino acids in a protein are linked together by peptide bonds. Protein chains containing 10 or fewer amino acids are usually referred to as peptides, with a prefix such as di- or tri- indicating the number of amino acids. Chains containing more than 50 amino acid units are referred to as proteins or polypeptides. Proteins are classified globular or fibrous, depending on their structure and resulting solubility in water. Globular proteins are nearly spherical and are soluble in water; fibrous proteins have elongated or fibrous structures and are not soluble in water.
Protein molecules can have as many as four levels of structure. The primary structure is the sequence of amino acids in the chain. The secondary structure is the arrangement of adjacent atoms in the peptide chain; the most common arrangements are α-helices or β-pleated sheets. The tertiary structure is the overall three-dimensional shape of the molecule that results from the way the chain bends and folds in on itself. Proteins that consist of more than one chain have quaternary structure, which is the way the multiple chains are packed together.
Four types of intramolecular and intermolecular forces contribute to secondary, tertiary, and quaternary structure: (1) hydrogen bonding between an oxygen or a nitrogen atom and a hydrogen atom bound to an oxygen atom or a nitrogen atom, either on the same chain or on a neighboring chain; (2) ionic bonding between one positively charged side chain and one negatively charged side chain; (3) disulfide linkages between cysteine units; and (4) dispersion forces between nonpolar side chains.
Because of their complexity, protein molecules are delicate and easy to disrupt. A denatured protein is one whose conformation has been changed, in a process called denaturation, so that it can no longer do its physiological job. A variety of conditions, such as heat, ultraviolet radiation, the addition of organic compounds, or changes in pH can denature a protein.
An enzyme is an organic catalyst produced by a living cell. Enzymes are such powerful catalysts that the reactions they promote occur rapidly at body temperature. Without the help of enzymes, these reactions would require high temperatures and long reaction times.
The molecule or molecules on which an enzyme acts are called its substrates. An enzyme has an active site where its substrate or substrates bind to form an enzyme-substrate complex. The reaction occurs, and product is released:
$E + S → E–S → E + P \nonumber$
The original lock-and-key model of enzyme and substrate binding pictured a rigid enzyme of unchanging configuration binding to the appropriate substrate. The newer induced-fit model describes the enzyme active site as changing its conformation after binding to the substrate.
Most enzymes have maximal activity in a narrow pH range centered on an optimum pH. In this pH range, the enzyme is correctly folded, and catalytic groups in the active site have the correct charge (positive, negative, or neutral). For most enzymes, the optimum pH is between 6 and 8.
Substances that interfere with enzyme function are called inhibitors. An irreversible inhibitor inactivates enzymes by forming covalent bonds to the enzyme, while a reversible inhibitor inactivates an enzyme by a weaker, noncovalent interaction that is easier to disrupt. A competitive inhibitor is a reversible inhibitor that is structurally similar to the substrate and binds to the active site. When the inhibitor is bound, the substrate is blocked from the active site and no reaction occurs. Because the binding of such an inhibitor is reversible, a high substrate concentration will overcome the inhibition because it increases the likelihood of the substrate binding. A noncompetitive inhibitor binds reversibly at a site distinct from the active site. Thus, it can bind to either the enzyme or the enzyme-substrate complex. The inhibitor changes the conformation of the active site so that the enzyme cannot function properly. Noncompetitive inhibitors are important in feedback inhibition, in which the amount of product produced by a series of reactions is carefully controlled. The final product in a series of reactions acts as a noncompetitive inhibitor of the initial enzyme.
Simple enzymes consist entirely of one or more amino acid chains. Complex enzymes are composed of one or more amino acid chains joined to cofactors—inorganic ions or organic coenzymes. Vitamins are organic compounds that are essential in very small amounts for the maintenance of normal metabolism and generally cannot be synthesized at adequate levels by the body. Vitamins are divided into two broad categories: fat-soluble vitamins and water-soluble vitamins. Many of the water-soluble vitamins are used for the synthesis of coenzymes. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/18%3A_Amino_Acids_Proteins_and_Enzymes/18.S%3A_Amino_Acids_Proteins_and_Enzymes_%28Summary%29.txt |
The blueprint for the reproduction and the maintenance of each organism is found in the nuclei of its cells, concentrated in elongated, threadlike structures called chromosomes. These complex structures, consisting of DNA and proteins, contain the basic units of heredity, called genes. The number of chromosomes (and genes) varies with each species. Human body cells have 23 pairs of chromosomes having 20,000–40,000 different genes.
Sperm and egg cells contain only a single copy of each chromosome; that is, they contain only one member of each chromosome pair. Thus, in sexual reproduction, the entire complement of chromosomes is achieved only when an egg and sperm combine. A new individual receives half its hereditary material from each parent. Calling the unit of heredity a “gene” merely gives it a name. But what really are genes and how is the information they contain expressed? One definition of a gene is that it is a segment of DNA that constitutes the code for a specific polypeptide. If genes are segments of DNA, we need to learn more about the structure and physiological function of DNA. We begin by looking at the small molecules needed to form DNA and RNA (ribonucleic acid)—the nucleotides.
• 19.0: Prelude to Nucleic Acids
In the 1970s, an intense research effort began that eventually led to the production of genetically engineered human insulin—the first genetically engineered product to be approved for medical use. To accomplish this feat, researchers first had to determine how insulin is made in the body and then find a way of causing the same process to occur in nonhuman organisms, such as bacteria or yeast cells. Many aspects of these discoveries are presented in this chapter on nucleic acids.
• 19.1: Nucleotides
Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose.
• 19.2: Nucleic Acid Structure
DNA is the nucleic acid that stores genetic information. RNA is the nucleic acid responsible for using the genetic information in DNA to produce proteins. Nucleotides are joined together to form nucleic acids through the phosphate group of one nucleotide connecting in an ester linkage to the OH group on the third carbon atom of the sugar unit of a second nucleotide. Nucleic acid sequences are written starting with the nucleotide having a free phosphate group (the 5′ end).
• 19.3: Replication and Expression of Genetic Information
In DNA replication, each strand of the original DNA serves as a template for the synthesis of a complementary strand. DNA polymerase is the primary enzyme needed for replication. In transcription, a segment of DNA serves as a template for the synthesis of an RNA sequence. RNA polymerase is the primary enzyme needed for transcription. Three types of RNA are formed during transcription: mRNA, rRNA, and tRNA.
• 19.4: Protein Synthesis and the Genetic Code
In translation, the information in mRNA directs the order of amino acids in protein synthesis. A set of three nucleotides (codon) codes for a specific amino acid.
• 19.5: Mutations and Genetic Diseases
The nucleotide sequence in DNA may be modified either spontaneously or from exposure to heat, radiation, or certain chemicals and can lead to mutations. Mutagens are the chemical or physical agents that cause mutations. Genetic diseases are hereditary diseases that occur because of a mutation in a critical gene.
• 19.6: Viruses
Viruses are very small infectious agents that contain either DNA or RNA as their genetic material. The human immunodeficiency virus (HIV) causes acquired immunodeficiency syndrome (AIDS).
• 19.E: Nucleic Acids (Exercises)
Problems and select solutions for the chapter.
• 19.S: Nucleic Acids (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
19: Nucleic Acids
Following the initial isolation of insulin in 1921, diabetic patients could be treated with insulin obtained from the pancreases of cattle and pigs. Unfortunately, some patients developed an allergic reaction to this insulin because its amino acid sequence was not identical to that of human insulin. In the 1970s, an intense research effort began that eventually led to the production of genetically engineered human insulin—the first genetically engineered product to be approved for medical use. To accomplish this feat, researchers first had to determine how insulin is made in the body and then find a way of causing the same process to occur in nonhuman organisms, such as bacteria or yeast cells. Many aspects of these discoveries are presented in this chapter on nucleic acids.
19.01: Nucleotides
Learning Objectives
• To identify the different molecules that combine to form nucleotides.
The repeating, or monomer, units that are linked together to form nucleic acids are known as nucleotides. The deoxyribonucleic acid (DNA) of a typical mammalian cell contains about 3 × 109 nucleotides. Nucleotides can be further broken down to phosphoric acid (H3PO4), a pentose sugar (a sugar with five carbon atoms), and a nitrogenous base (a base containing nitrogen atoms).
$\mathrm{nucleic\: acids \underset{down\: into}{\xrightarrow{can\: be\: broken}} nucleotides \underset{down\: into}{\xrightarrow{can\: be\: broken}} H_3PO_4 + nitrogen\: base + pentose\: sugar} \nonumber$
If the pentose sugar is ribose, the nucleotide is more specifically referred to as a ribonucleotide, and the resulting nucleic acid is ribonucleic acid (RNA). If the sugar is 2-deoxyribose, the nucleotide is a deoxyribonucleotide, and the nucleic acid is DNA.
The nitrogenous bases found in nucleotides are classified as pyrimidines or purines. Pyrimidines are heterocyclic amines with two nitrogen atoms in a six-member ring and include uracil, thymine, and cytosine. Purines are heterocyclic amines consisting of a pyrimidine ring fused to a five-member ring with two nitrogen atoms. Adenine and guanine are the major purines found in nucleic acids (Figure $1$).
The formation of a bond between C1′ of the pentose sugar and N1 of the pyrimidine base or N9 of the purine base joins the pentose sugar to the nitrogenous base. In the formation of this bond, a molecule of water is removed. Table $1$ summarizes the similarities and differences in the composition of nucleotides in DNA and RNA.
The numbering convention is that primed numbers designate the atoms of the pentose ring, and unprimed numbers designate the atoms of the purine or pyrimidine ring.
Table $1$: Composition of Nucleotides in DNA and RNA
Composition DNA RNA
purine bases adenine and guanine adenine and guanine
pyrimidine bases cytosine and thymine cytosine and uracil
pentose sugar 2-deoxyribose ribose
inorganic acid phosphoric acid (H3PO4) H3PO4
The names and structures of the major ribonucleotides and one of the deoxyribonucleotides are given in Figure $2$.
Apart from being the monomer units of DNA and RNA, the nucleotides and some of their derivatives have other functions as well. Adenosine diphosphate (ADP) and adenosine triphosphate (ATP), shown in Figure $3$, have a role in cell metabolism. Moreover, a number of coenzymes, including flavin adenine dinucleotide (FAD), nicotinamide adenine dinucleotide (NAD+), and coenzyme A, contain adenine nucleotides as structural components.
Summary
Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/19%3A_Nucleic_Acids/19.00%3A_Prelude_to_Nucleic_Acids.txt |
Skills to Develop
• Identify the two types of nucleic acids and the function of each type.
• Describe how nucleotides are linked together to form nucleic acids.
• Describe the secondary structure of DNA and the importance of complementary base pairing.
Nucleic acids are large polymers formed by linking nucleotides together and are found in every cell. Deoxyribonucleic acid (DNA) is the nucleic acid that stores genetic information. If all the DNA in a typical mammalian cell were stretched out end to end, it would extend more than 2 m. Ribonucleic acid (RNA) is the nucleic acid responsible for using the genetic information encoded in DNA to produce the thousands of proteins found in living organisms.
Primary Structure of Nucleic Acids
Nucleotides are joined together through the phosphate group of one nucleotide connecting in an ester linkage to the OH group on the third carbon atom of the sugar unit of a second nucleotide. This unit joins to a third nucleotide, and the process is repeated to produce a long nucleic acid chain (Figure $1$). The backbone of the chain consists of alternating phosphate and sugar units (2-deoxyribose in DNA and ribose in RNA). The purine and pyrimidine bases branch off this backbone.
Each phosphate group has one acidic hydrogen atom that is ionized at physiological pH. This is why these compounds are known as nucleic acids.
Figure $1$ Structure of a Segment of DNA. A similar segment of RNA would have OH groups on each C2′, and uracil would replace thymine.
Like proteins, nucleic acids have a primary structure that is defined as the sequence of their nucleotides. Unlike proteins, which have 20 different kinds of amino acids, there are only 4 different kinds of nucleotides in nucleic acids. For amino acid sequences in proteins, the convention is to write the amino acids in order starting with the N-terminal amino acid. In writing nucleotide sequences for nucleic acids, the convention is to write the nucleotides (usually using the one-letter abbreviations for the bases, shown in Figure $1$) starting with the nucleotide having a free phosphate group, which is known as the 5′ end, and indicate the nucleotides in order. For DNA, a lowercase d is often written in front of the sequence to indicate that the monomers are deoxyribonucleotides. The final nucleotide has a free OH group on the 3′ carbon atom and is called the 3′ end. The sequence of nucleotides in the DNA segment shown in Figure $1$ would be written 5′-dG-dT-dA-dC-3′, which is often further abbreviated to dGTAC or just GTAC.
Secondary Structure of DNA
The three-dimensional structure of DNA was the subject of an intensive research effort in the late 1940s to early 1950s. Initial work revealed that the polymer had a regular repeating structure. In 1950, Erwin Chargaff of Columbia University showed that the molar amount of adenine (A) in DNA was always equal to that of thymine (T). Similarly, he showed that the molar amount of guanine (G) was the same as that of cytosine (C). Chargaff drew no conclusions from his work, but others soon did.
At Cambridge University in 1953, James D. Watson and Francis Crick announced that they had a model for the secondary structure of DNA. Using the information from Chargaff’s experiments (as well as other experiments) and data from the X ray studies of Rosalind Franklin (which involved sophisticated chemistry, physics, and mathematics), Watson and Crick worked with models that were not unlike a child’s construction set and finally concluded that DNA is composed of two nucleic acid chains running antiparallel to one another—that is, side-by-side with the 5′ end of one chain next to the 3′ end of the other. Moreover, as their model showed, the two chains are twisted to form a double helix—a structure that can be compared to a spiral staircase, with the phosphate and sugar groups (the backbone of the nucleic acid polymer) representing the outside edges of the staircase. The purine and pyrimidine bases face the inside of the helix, with guanine always opposite cytosine and adenine always opposite thymine. These specific base pairs, referred to as complementary bases, are the steps, or treads, in our staircase analogy (Figure $2$).
Figure $2$ DNA Double Helix. (a) This represents a computer-generated model of the DNA double helix. (b) This represents a schematic representation of the double helix, showing the complementary bases.
The structure proposed by Watson and Crick provided clues to the mechanisms by which cells are able to divide into two identical, functioning daughter cells; how genetic data are passed to new generations; and even how proteins are built to required specifications. All these abilities depend on the pairing of complementary bases. Figure $3$ shows the two sets of base pairs and illustrates two things. First, a pyrimidine is paired with a purine in each case, so that the long dimensions of both pairs are identical (1.08 nm).
Figure $3$ Complementary Base Pairing. Complementary bases engage in hydrogen bonding with one another: (a) thymine and adenine; (b) cytosine and guanine.
If two pyrimidines were paired or two purines were paired, the two pyrimidines would take up less space than a purine and a pyrimidine, and the two purines would take up more space, as illustrated in Figure $4$. If these pairings were ever to occur, the structure of DNA would be like a staircase made with stairs of different widths. For the two strands of the double helix to fit neatly, a pyrimidine must always be paired with a purine. The second thing you should notice in Figure $3$ is that the correct pairing enables formation of three instances of hydrogen bonding between guanine and cytosine and two between adenine and thymine. The additive contribution of this hydrogen bonding imparts great stability to the DNA double helix.
Figure $4$ Difference in Widths of Possible Base Pairs
Summary
• DNA is the nucleic acid that stores genetic information. RNA is the nucleic acid responsible for using the genetic information in DNA to produce proteins.
• Nucleotides are joined together to form nucleic acids through the phosphate group of one nucleotide connecting in an ester linkage to the OH group on the third carbon atom of the sugar unit of a second nucleotide.
• Nucleic acid sequences are written starting with the nucleotide having a free phosphate group (the 5′ end).
• Two DNA strands link together in an antiparallel direction and are twisted to form a double helix. The nitrogenous bases face the inside of the helix. Guanine is always opposite cytosine, and adenine is always opposite thymine.
Concept Review Exercises
1.
1. Name the two kinds of nucleic acids.
2. Which type of nucleic acid stores genetic information in the cell?
2. What are complementary bases?
3. Why is it structurally important that a purine base always pair with a pyrimidine base in the DNA double helix?
Answers
1.
1. deoxyribonucleic acid (DNA) and ribonucleic acid (RNA)
2. DNA
2. the specific base pairings in the DNA double helix in which guanine is paired with cytosine and adenine is paired with thymine
3. The width of the DNA double helix is kept at a constant width, rather than narrowing (if two pyrimidines were across from each other) or widening (if two purines were across from each other).
Exercises
1. For this short RNA segment,
1. identify the 5′ end and the 3′ end of the molecule.
2. circle the atoms that comprise the backbone of the nucleic acid chain.
3. write the nucleotide sequence of this RNA segment.
2. For this short DNA segment,
1. identify the 5′ end and the 3′ end of the molecule.
2. circle the atoms that comprise the backbone of the nucleic acid chain.
3. write the nucleotide sequence of this DNA segment.
3. Which nitrogenous base in DNA pairs with each nitrogenous base?
1. cytosine
2. adenine
3. guanine
4. thymine
4. Which nitrogenous base in RNA pairs with each nitrogenous base?
1. cytosine
2. adenine
3. guanine
4. thymine
5. How many hydrogen bonds can form between the two strands in the short DNA segment shown below?
5′ ATGCGACTA 3′ 3′ TACGCTGAT 5′
6. How many hydrogen bonds can form between the two strands in the short DNA segment shown below?
5′ CGATGAGCC 3′ 3′ GCTACTCGG 5′
Answers
1. c. ACU
1.
1. guanine
2. thymine
3. cytosine
4. adenine
1. 22 (2 between each AT base pair and 3 between each GC base pair) | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/19%3A_Nucleic_Acids/19.02%3A_Nucleic_Acid_Structure.txt |
Learning Objectives
• Describe how a new copy of DNA is synthesized.
• Describe how RNA is synthesized from DNA.
• Identify the different types of RNA and the function of each type of RNA.
We previously stated that deoxyribonucleic acid (DNA) stores genetic information, while ribonucleic acid (RNA) is responsible for transmitting or expressing genetic information by directing the synthesis of thousands of proteins found in living organisms. But how do the nucleic acids perform these functions? Three processes are required: (1) replication, in which new copies of DNA are made; (2) transcription, in which a segment of DNA is used to produce RNA; and (3) translation, in which the information in RNA is translated into a protein sequence.
Replication
New cells are continuously forming in the body through the process of cell division. For this to happen, the DNA in a dividing cell must be copied in a process known as replication. The complementary base pairing of the double helix provides a ready model for how genetic replication occurs. If the two chains of the double helix are pulled apart, disrupting the hydrogen bonding between base pairs, each chain can act as a template, or pattern, for the synthesis of a new complementary DNA chain.
The nucleus contains all the necessary enzymes, proteins, and nucleotides required for this synthesis. A short segment of DNA is “unzipped,” so that the two strands in the segment are separated to serve as templates for new DNA. DNA polymerase, an enzyme, recognizes each base in a template strand and matches it to the complementary base in a free nucleotide. The enzyme then catalyzes the formation of an ester bond between the 5′ phosphate group of the nucleotide and the 3′ OH end of the new, growing DNA chain. In this way, each strand of the original DNA molecule is used to produce a duplicate of its former partner (Figure \(1\)). Whatever information was encoded in the original DNA double helix is now contained in each replicate helix. When the cell divides, each daughter cell gets one of these replicates and thus all of the information that was originally possessed by the parent cell.
Example \(1\)
A segment of one strand from a DNA molecule has the sequence 5′‑TCCATGAGTTGA‑3′. What is the sequence of nucleotides in the opposite, or complementary, DNA chain?
Solution
Knowing that the two strands are antiparallel and that T base pairs with A, while C base pairs with G, the sequence of the complementary strand will be 3′‑AGGTACTCAACT‑5′ (can also be written as TCAACTCATGGA).
Exercise \(1\)
A segment of one strand from a DNA molecule has the sequence 5′‑CCAGTGAATTGCCTAT‑3′. What is the sequence of nucleotides in the opposite, or complementary, DNA chain?
What do we mean when we say information is encoded in the DNA molecule? An organism’s DNA can be compared to a book containing directions for assembling a model airplane or for knitting a sweater. Letters of the alphabet are arranged into words, and these words direct the individual to perform certain operations with specific materials. If all the directions are followed correctly, a model airplane or sweater is produced.
In DNA, the particular sequences of nucleotides along the chains encode the directions for building an organism. Just as saw means one thing in English and was means another, the sequence of bases CGT means one thing, and TGC means something different. Although there are only four letters—the four nucleotides—in the genetic code of DNA, their sequencing along the DNA strands can vary so widely that information storage is essentially unlimited.
Transcription
For the hereditary information in DNA to be useful, it must be “expressed,” that is, used to direct the growth and functioning of an organism. The first step in the processes that constitute DNA expression is the synthesis of RNA, by a template mechanism that is in many ways analogous to DNA replication. Because the RNA that is synthesized is a complementary copy of information contained in DNA, RNA synthesis is referred to as transcription. There are three key differences between replication and transcription:
1. RNA molecules are much shorter than DNA molecules; only a portion of one DNA strand is copied or transcribed to make an RNA molecule.
2. RNA is built from ribonucleotides rather than deoxyribonucleotides.
3. The newly synthesized RNA strand does not remain associated with the DNA sequence it was transcribed from.
The DNA sequence that is transcribed to make RNA is called the template strand, while the complementary sequence on the other DNA strand is called the coding or informational strand. To initiate RNA synthesis, the two DNA strands unwind at specific sites along the DNA molecule. Ribonucleotides are attracted to the uncoiling region of the DNA molecule, beginning at the 3′ end of the template strand, according to the rules of base pairing. Thymine in DNA calls for adenine in RNA, cytosine specifies guanine, guanine calls for cytosine, and adenine requires uracil. RNA polymerase—an enzyme—binds the complementary ribonucleotide and catalyzes the formation of the ester linkage between ribonucleotides, a reaction very similar to that catalyzed by DNA polymerase (Figure \(2\)). Synthesis of the RNA strand takes place in the 5′ to 3′ direction, antiparallel to the template strand. Only a short segment of the RNA molecule is hydrogen-bonded to the template strand at any time during transcription. When transcription is completed, the RNA is released, and the DNA helix reforms. The nucleotide sequence of the RNA strand formed during transcription is identical to that of the corresponding coding strand of the DNA, except that U replaces T.
Example \(2\)
A portion of the template strand of a gene has the sequence 5′‑TCCATGAGTTGA‑3′. What is the sequence of nucleotides in the RNA that is formed from this template?
Solution
Four things must be remembered in answering this question: (1) the DNA strand and the RNA strand being synthesized are antiparallel; (2) RNA is synthesized in a 5′ to 3′ direction, so transcription begins at the 3′ end of the template strand; (3) ribonucleotides are used in place of deoxyribonucleotides; and (4) thymine (T) base pairs with adenine (A), A base pairs with uracil (U; in RNA), and cytosine (C) base pairs with guanine (G). The sequence is determined to be 3′‑AGGUACUCAACU‑5′ (can also be written as 5′‑UCAACUCAUGGA‑3′).
Exercise \(2\)
A portion of the template strand of a gene has the sequence 5′‑CCAGTGAATTGCCTAT‑3′. What is the sequence of nucleotides in the RNA that is formed from this template?
Three types of RNA are formed during transcription: messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA). These three types of RNA differ in function, size, and percentage of the total cell RNA (Table \(1\)). mRNA makes up only a small percent of the total amount of RNA within the cell, primarily because each molecule of mRNA exists for a relatively short time; it is continuously being degraded and resynthesized. The molecular dimensions of the mRNA molecule vary according to the amount of genetic information a given molecule contains. After transcription, which takes place in the nucleus, the mRNA passes into the cytoplasm, carrying the genetic message from DNA to the ribosomes, the sites of protein synthesis. Elsewhere, we shall see how mRNA directly determines the sequence of amino acids during protein synthesis.
Table \(1\): Properties of Cellular RNA in Escherichia coli
Type Function Approximate Number of Nucleotides Percentage of Total Cell RNA
mRNA codes for proteins 100–6,000 ~3
rRNA component of ribosomes 120–2900 83
tRNA adapter molecule that brings the amino acid to the ribosome 75–90 14
Ribosomes are cellular substructures where proteins are synthesized. They contain about 65% rRNA and 35% protein, held together by numerous noncovalent interactions, such as hydrogen bonding, in an overall structure consisting of two globular particles of unequal size.
Molecules of tRNA, which bring amino acids (one at a time) to the ribosomes for the construction of proteins, differ from one another in the kinds of amino acid each is specifically designed to carry. A set of three nucleotides, known as a codon, on the mRNA determines which kind of tRNA will add its amino acid to the growing chain. Each of the 20 amino acids found in proteins has at least one corresponding kind of tRNA, and most amino acids have more than one.
The two-dimensional structure of a tRNA molecule has three distinctive loops, reminiscent of a cloverleaf (Figure \(3\)). On one loop is a sequence of three nucleotides that varies for each kind of tRNA. This triplet, called the anticodon, is complementary to and pairs with the codon on the mRNA. At the opposite end of the molecule is the acceptor stem, where the amino acid is attached.
Summary
• In DNA replication, each strand of the original DNA serves as a template for the synthesis of a complementary strand.
• DNA polymerase is the primary enzyme needed for replication.
• In transcription, a segment of DNA serves as a template for the synthesis of an RNA sequence.
• RNA polymerase is the primary enzyme needed for transcription.
• Three types of RNA are formed during transcription: mRNA, rRNA, and tRNA. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/19%3A_Nucleic_Acids/19.03%3A_Replication_and_Expression_of_Genetic_Information.txt |
Learning Objectives
• describe the characteristics of the genetic code.
• describe how a protein is synthesized from mRNA.
One of the definitions of a gene is as follows: a segment of deoxyribonucleic acid (DNA) carrying the code for a specific polypeptide. Each molecule of messenger RNA (mRNA) is a transcribed copy of a gene that is used by a cell for synthesizing a polypeptide chain. If a protein contains two or more different polypeptide chains, each chain is coded by a different gene. We turn now to the question of how the sequence of nucleotides in a molecule of ribonucleic acid (RNA) is translated into an amino acid sequence.
How can a molecule containing just 4 different nucleotides specify the sequence of the 20 amino acids that occur in proteins? If each nucleotide coded for 1 amino acid, then obviously the nucleic acids could code for only 4 amino acids. What if amino acids were coded for by groups of 2 nucleotides? There are 42, or 16, different combinations of 2 nucleotides (AA, AU, AC, AG, UU, and so on). Such a code is more extensive but still not adequate to code for 20 amino acids. However, if the nucleotides are arranged in groups of 3, the number of different possible combinations is 43, or 64. Here we have a code that is extensive enough to direct the synthesis of the primary structure of a protein molecule.
Video: NDSU Virtual Cell Animations project animation "Translation". For more information, see VCell, NDSU(opens in new window) [vcell.ndsu.nodak.edu]
The genetic code can therefore be described as the identification of each group of three nucleotides and its particular amino acid. The sequence of these triplet groups in the mRNA dictates the sequence of the amino acids in the protein. Each individual three-nucleotide coding unit, as we have seen, is called a codon.
Protein synthesis is accomplished by orderly interactions between mRNA and the other ribonucleic acids (transfer RNA [tRNA] and ribosomal RNA [rRNA]), the ribosome, and more than 100 enzymes. The mRNA formed in the nucleus during transcription is transported across the nuclear membrane into the cytoplasm to the ribosomes—carrying with it the genetic instructions. The process in which the information encoded in the mRNA is used to direct the sequencing of amino acids and thus ultimately to synthesize a protein is referred to as translation.
Before an amino acid can be incorporated into a polypeptide chain, it must be attached to its unique tRNA. This crucial process requires an enzyme known as aminoacyl-tRNA synthetase (Figure \(1\)). There is a specific aminoacyl-tRNA synthetase for each amino acid. This high degree of specificity is vital to the incorporation of the correct amino acid into a protein. After the amino acid molecule has been bound to its tRNA carrier, protein synthesis can take place. Figure \(2\) depicts a schematic stepwise representation of this all-important process.
Early experimenters were faced with the task of determining which of the 64 possible codons stood for each of the 20 amino acids. The cracking of the genetic code was the joint accomplishment of several well-known geneticists—notably Har Khorana, Marshall Nirenberg, Philip Leder, and Severo Ochoa—from 1961 to 1964. The genetic dictionary they compiled, summarized in Figure \(3\), shows that 61 codons code for amino acids, and 3 codons serve as signals for the termination of polypeptide synthesis (much like the period at the end of a sentence). Notice that only methionine (AUG) and tryptophan (UGG) have single codons. All other amino acids have two or more codons.
Example \(1\): Using the Genetic Code
A portion of an mRNA molecule has the sequence 5′‑AUGCCACGAGUUGAC‑3′. What amino acid sequence does this code for?
Solution
Use Figure \(3\) to determine what amino acid each set of three nucleotides (codon) codes for. Remember that the sequence is read starting from the 5′ end and that a protein is synthesized starting with the N-terminal amino acid. The sequence 5′‑AUGCCACGAGUUGAC‑3′ codes for met-pro-arg-val-asp.
Exercise \(4\)
A portion of an RNA molecule has the sequence 5′‑AUGCUGAAUUGCGUAGGA‑3′. What amino acid sequence does this code for?
Further experimentation threw much light on the nature of the genetic code, as follows:
1. The code is virtually universal; animal, plant, and bacterial cells use the same codons to specify each amino acid (with a few exceptions).
2. The code is “degenerate”; in all but two cases (methionine and tryptophan), more than one triplet codes for a given amino acid.
3. The first two bases of each codon are most significant; the third base often varies. This suggests that a change in the third base by a mutation may still permit the correct incorporation of a given amino acid into a protein. The third base is sometimes called the “wobble” base.
4. The code is continuous and nonoverlapping; there are no nucleotides between codons, and adjacent codons do not overlap.
5. The three termination codons are read by special proteins called release factors, which signal the end of the translation process.
6. The codon AUG codes for methionine and is also the initiation codon. Thus methionine is the first amino acid in each newly synthesized polypeptide. This first amino acid is usually removed enzymatically before the polypeptide chain is completed; the vast majority of polypeptides do not begin with methionine.
Summary
In translation, the information in mRNA directs the order of amino acids in protein synthesis. A set of three nucleotides (codon) codes for a specific amino acid. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/19%3A_Nucleic_Acids/19.04%3A_Protein_Synthesis_and_the_Genetic_Code.txt |
Learning Objectives
• To describe the causes of genetic mutations and how they lead to genetic diseases.
We have seen that the sequence of nucleotides in a cell’s deoxyribonucleic acid (DNA) is what ultimately determines the sequence of amino acids in proteins made by the cell and thus is critical for the proper functioning of the cell. On rare occasions, however, the nucleotide sequence in DNA may be modified either spontaneously (by errors during replication, occurring approximately once for every 10 billion nucleotides) or from exposure to heat, radiation, or certain chemicals. Any chemical or physical change that alters the nucleotide sequence in DNA is called a mutation. When a mutation occurs in an egg or sperm cell that then produces a living organism, it will be inherited by all the offspring of that organism.
Common types of mutations include substitution (a different nucleotide is substituted), insertion (the addition of a new nucleotide), and deletion (the loss of a nucleotide). These changes within DNA are called point mutations because only one nucleotide is substituted, added, or deleted (Figure \(1\)). Because an insertion or deletion results in a frame-shift that changes the reading of subsequent codons and, therefore, alters the entire amino acid sequence that follows the mutation, insertions and deletions are usually more harmful than a substitution in which only a single amino acid is altered.
The chemical or physical agents that cause mutations are called mutagens. Examples of physical mutagens are ultraviolet (UV) and gamma radiation. Radiation exerts its mutagenic effect either directly or by creating free radicals that in turn have mutagenic effects. Radiation and free radicals can lead to the formation of bonds between nitrogenous bases in DNA. For example, exposure to UV light can result in the formation of a covalent bond between two adjacent thymines on a DNA strand, producing a thymine dimer (Figure \(2\)). If not repaired, the dimer prevents the formation of the double helix at the point where it occurs. The genetic disease xeroderma pigmentosum is caused by a lack of the enzyme that cuts out the thymine dimers in damaged DNA. Individuals affected by this condition are abnormally sensitive to light and are more prone to skin cancer than normal individuals.
Sometimes gene mutations are beneficial, but most of them are detrimental. For example, if a point mutation occurs at a crucial position in a DNA sequence, the affected protein will lack biological activity, perhaps resulting in the death of a cell. In such cases the altered DNA sequence is lost and will not be copied into daughter cells. Nonlethal mutations in an egg or sperm cell may lead to metabolic abnormalities or hereditary diseases. Such diseases are called inborn errors of metabolism or genetic diseases. A partial listing of genetic diseases is presented in Figure \(1\), and two specific diseases are discussed in the following sections. In most cases, the defective gene results in a failure to synthesize a particular enzyme.
Figure \(1\): Some Representative Genetic Diseases in Humans and the Protein or Enzyme Responsible
Disease Responsible Protein or Enzyme
alkaptonuria homogentisic acid oxidase
galactosemia galactose 1-phosphate uridyl transferase, galactokinase, or UDP galactose epimerase
Gaucher disease glucocerebrosidase
gout and Lesch-Nyhan syndrome hypoxanthine-guanine phosphoribosyl transferase
hemophilia antihemophilic factor (factor VIII) or Christmas factor (factor IX)
homocystinuria cystathionine synthetase
maple syrup urine disease branched chain α-keto acid dehydrogenase complex
McArdle syndrome muscle phosphorylase
Niemann-Pick disease sphingomyelinase
phenylketonuria (PKU) phenylalanine hydroxylase
sickle cell anemia hemoglobin
Tay-Sachs disease hexosaminidase A
tyrosinemia fumarylacetoacetate hydrolase or tyrosine aminotransferase
von Gierke disease glucose 6-phosphatase
Wilson disease Wilson disease protein
PKU results from the absence of the enzyme phenylalanine hydroxylase. Without this enzyme, a person cannot convert phenylalanine to tyrosine, which is the precursor of the neurotransmitters dopamine and norepinephrine as well as the skin pigment melanin.
When this reaction cannot occur, phenylalanine accumulates and is then converted to higher than normal quantities of phenylpyruvate. The disease acquired its name from the high levels of phenylpyruvate (a phenyl ketone) in urine. Excessive amounts of phenylpyruvate impair normal brain development, which causes severe mental retardation.
PKU may be diagnosed by assaying a sample of blood or urine for phenylalanine or one of its metabolites. Medical authorities recommend testing every newborn’s blood for phenylalanine within 24 h to 3 weeks after birth. If the condition is detected, mental retardation can be prevented by immediately placing the infant on a diet containing little or no phenylalanine. Because phenylalanine is plentiful in naturally produced proteins, the low-phenylalanine diet depends on a synthetic protein substitute plus very small measured amounts of naturally produced foods. Before dietary treatment was introduced in the early 1960s, severe mental retardation was a common outcome for children with PKU. Prior to the 1960s, 85% of patients with PKU had an intelligence quotient (IQ) less than 40, and 37% had IQ scores below 10. Since the introduction of dietary treatments, however, over 95% of children with PKU have developed normal or near-normal intelligence. The incidence of PKU in newborns is about 1 in 12,000 in North America.
Every state in the United States has mandated that screening for PKU be provided to all newborns.
Several genetic diseases are collectively categorized as lipid-storage diseases. Lipids are constantly being synthesized and broken down in the body, so if the enzymes that catalyze lipid degradation are missing, the lipids tend to accumulate and cause a variety of medical problems. When a genetic mutation occurs in the gene for the enzyme hexosaminidase A, for example, gangliosides cannot be degraded but accumulate in brain tissue, causing the ganglion cells of the brain to become greatly enlarged and nonfunctional. This genetic disease, known as Tay-Sachs disease, leads to a regression in development, dementia, paralysis, and blindness, with death usually occurring before the age of three. There is currently no treatment, but Tay-Sachs disease can be diagnosed in a fetus by assaying the amniotic fluid (amniocentesis) for hexosaminidase A. A blood test can identify Tay-Sachs carriers—people who inherit a defective gene from only one rather than both parents—because they produce only half the normal amount of hexosaminidase A, although they do not exhibit symptoms of the disease.
Looking Closer: Recombinant DNA Technology
More than 3,000 human diseases have been shown to have a genetic component, caused or in some way modulated by the person’s genetic composition. Moreover, in the last decade or so, researchers have succeeded in identifying many of the genes and even mutations that are responsible for specific genetic diseases. Now scientists have found ways of identifying and isolating genes that have specific biological functions and placing those genes in another organism, such as a bacterium, which can be easily grown in culture. With these techniques, known as recombinant DNA technology, the ability to cure many serious genetic diseases appears to be within our grasp.
Isolating the specific gene or genes that cause a particular genetic disease is a monumental task. One reason for the difficulty is the enormous amount of a cell’s DNA, only a minute portion of which contains the gene sequence. Thus, the first task is to obtain smaller pieces of DNA that can be more easily handled. Fortunately, researchers are able to use restriction enzymes (also known as restriction endonucleases), discovered in 1970, which are enzymes that cut DNA at specific, known nucleotide sequences, yielding DNA fragments of shorter length. For example, the restriction enzyme EcoRI recognizes the nucleotide sequence shown here and cuts both DNA strands as indicated:
Once a DNA strand has been fragmented, it must be cloned; that is, multiple identical copies of each DNA fragment are produced to make sure there are sufficient amounts of each to detect and manipulate in the laboratory. Cloning is accomplished by inserting the individual DNA fragments into phages (bacterial viruses) that can enter bacterial cells and be replicated. When a bacterial cell infected by the modified phage is placed in an appropriate culture medium, it forms a colony of cells, all containing copies of the original DNA fragment. This technique is used to produce many bacterial colonies, each containing a different DNA fragment. The result is a DNA library, a collection of bacterial colonies that together contain the entire genome of a particular organism.
The next task is to screen the DNA library to determine which bacterial colony (or colonies) has incorporated the DNA fragment containing the desired gene. A short piece of DNA, known as a hybridization probe, which has a nucleotide sequence complementary to a known sequence in the gene, is synthesized, and a radioactive phosphate group is added to it as a “tag.” You might be wondering how researchers are able to prepare such a probe if the gene has not yet been isolated. One way is to use a segment of the desired gene isolated from another organism. An alternative method depends on knowing all or part of the amino acid sequence of the protein produced by the gene of interest: the amino acid sequence is used to produce an approximate genetic code for the gene, and this nucleotide sequence is then produced synthetically. (The amino acid sequence used is carefully chosen to include, if possible, many amino acids such as methionine and tryptophan, which have only a single codon each.)
After a probe identifies a colony containing the desired gene, the DNA fragment is clipped out, again using restriction enzymes, and spliced into another replicating entity, usually a plasmid. Plasmids are tiny mini-chromosomes found in many bacteria, such as Escherichia coli (E. coli). A recombined plasmid would then be inserted into the host organism (usually the bacterium E. coli), where it would go to work to produce the desired protein.
Proponents of recombinant DNA research are excited about its great potential benefits. An example is the production of human growth hormone, which is used to treat children who fail to grow properly. Formerly, human growth hormone was available only in tiny amounts obtained from cadavers. Now it is readily available through recombinant DNA technology. Another gene that has been cloned is the gene for epidermal growth factor, which stimulates the growth of skin cells and can be used to speed the healing of burns and other skin wounds. Recombinant techniques are also a powerful research tool, providing enormous aid to scientists as they map and sequence genes and determine the functions of different segments of an organism’s DNA.
In addition to advancements in the ongoing treatment of genetic diseases, recombinant DNA technology may actually lead to cures. When appropriate genes are successfully inserted into E. coli, the bacteria can become miniature pharmaceutical factories, producing great quantities of insulin for people with diabetes, clotting factor for people with hemophilia, missing enzymes, hormones, vitamins, antibodies, vaccines, and so on. Recent accomplishments include the production in E. coli of recombinant DNA molecules containing synthetic genes for tissue plasminogen activator, a clot-dissolving enzyme that can rescue heart attack victims, as well as the production of vaccines against hepatitis B (humans) and hoof-and-mouth disease (cattle).
Scientists have used other bacteria besides E. coli in gene-splicing experiments and also yeast and fungi. Plant molecular biologists use a bacterial plasmid to introduce genes for several foreign proteins (including animal proteins) into plants. The bacterium is Agrobacterium tumefaciens, which can cause tumors in many plants, but which can be treated so that its tumor-causing ability is eliminated. One practical application of its plasmids would be to enhance a plant’s nutritional value by transferring into it the gene necessary for the synthesis of an amino acid in which the plant is normally deficient (for example, transferring the gene for methionine synthesis into pinto beans, which normally do not synthesize high levels of methionine).
Restriction enzymes have been isolated from a number of bacteria and are named after the bacterium of origin. EcoRI is a restriction enzyme obtained from the R strain of E. coli. The roman numeral I indicates that it was the first restriction enzyme obtained from this strain of bacteria.
Summary
• The nucleotide sequence in DNA may be modified either spontaneously or from exposure to heat, radiation, or certain chemicals and can lead to mutations.
• Mutagens are the chemical or physical agents that cause mutations.
• Genetic diseases are hereditary diseases that occur because of a mutation in a critical gene.
19.06: Viruses
Learning Objectives
• To explain how viruses reproduce in cells.
Viruses are visible only under an electron microscope. They come in a variety of shapes, ranging from spherical to rod shaped. The fact that they contain either deoxyribonucleic acid (DNA) or ribonucleic acid (RNA)—but never both—allows them to be divided into two major classes: DNA viruses and RNA viruses (Figure \(1\)).
Most RNA viruses use their nucleic acids in much the same way as the DNA viruses, penetrating a host cell and inducing it to replicate the viral RNA and synthesize viral proteins. The new RNA strands and viral proteins are then assembled into new viruses. Some RNA viruses, however, called retroviruses (Figure \(2\)), synthesize DNA in the host cell, in a process that is the reverse of the DNA-to-RNA transcription that normally occurs in cells. The synthesis of DNA from an RNA template is catalyzed by the enzyme reverse transcriptase.
In 1987, azidothymidine (AZT, also known as zidovudine or the brand name Retrovir) became the first drug approved for the treatment of AIDS. It works by binding to reverse transcriptase in place of deoxythymidine triphosphate, after which, because AZT does not have a 3′OH group, further replication is blocked. In the past 10 years, several other drugs have been approved that also act by inhibiting the viral reverse transcriptase.
Raltegravir (Isentress) is a newer anti-AIDS drug that was approved by the FDA in October 2007. This drug inhibits the integrase enzyme that is needed to integrate the HIV DNA into cellular DNA, an essential step in the production of more HIV particles.
A major problem in treating HIV infections is that the virus can become resistant to any of these drugs. One way to combat the problem has been to administer a “cocktail” of drugs, typically a combination of two reverse transcriptase inhibitors along with a protease inhibitor. These treatments can significantly reduce the amount of HIV in an infected person.
Career Focus: Genetics Counselor
A genetics counselor works with individuals and families who have birth defects or genetic disorders or a family history of a disease, such as cancer, with a genetic link. A genetics counselor may work in a variety of health-care settings (such as a hospital) to obtain family medical and reproductive histories; explain how genetic conditions are inherited; explain the causes, diagnosis, and care of these conditions; interpret the results of genetic tests; and aid the individual or family in making decisions regarding genetic diseases or conditions. A certified genetics counselor must obtain a master’s degree from an accredited program. Applicants to these graduate programs usually have an undergraduate degree in biology, psychology, or genetics.
Summary
Viruses are very small infectious agents that contain either DNA or RNA as their genetic material. The human immunodeficiency virus (HIV) causes acquired immunodeficiency syndrome (AIDS). | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/19%3A_Nucleic_Acids/19.05%3A_Mutations_and_Genetic_Diseases.txt |
Concept Review Exercises
1. Identify the three molecules needed to form the nucleotides in each nucleic acid.
1. DNA
2. RNA
2. Classify each compound as a pentose sugar, a purine, or a pyrimidine.
1. adenine
2. guanine
3. deoxyribose
4. thymine
5. ribose
6. cytosine
Answers
1. nitrogenous base (adenine, guanine, cytosine, and thymine), 2-deoxyribose, and H3PO4
2. nitrogenous base (adenine, guanine, cytosine, and uracil), ribose, and H3PO4
1. purine
2. purine
3. pentose sugar
4. pyrimidine
5. pentose sugar
6. pyrimidine
Exercises
1. What is the sugar unit in each nucleic acid?
1. RNA
2. DNA
2. Identify the major nitrogenous bases in each nucleic acid.
1. DNA
2. RNA
3. For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide.
4. For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide.
5. For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine.
6. For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine.
Answers
1. ribose
2. deoxyribose
Concept Review Exercises
1. Name the two kinds of nucleic acids.
2. Which type of nucleic acid stores genetic information in the cell?
1. What are complementary bases?
2. Why is it structurally important that a purine base always pair with a pyrimidine base in the DNA double helix?
Answers
1. deoxyribonucleic acid (DNA) and ribonucleic acid (RNA)
2. DNA
1. the specific base pairings in the DNA double helix in which guanine is paired with cytosine and adenine is paired with thymine
2. The width of the DNA double helix is kept at a constant width, rather than narrowing (if two pyrimidines were across from each other) or widening (if two purines were across from each other).
Exercises
1. For this short RNA segment,
1. identify the 5′ end and the 3′ end of the molecule.
2. circle the atoms that comprise the backbone of the nucleic acid chain.
3. write the nucleotide sequence of this RNA segment.
2. For this short DNA segment,
1. identify the 5′ end and the 3′ end of the molecule.
2. circle the atoms that comprise the backbone of the nucleic acid chain.
3. write the nucleotide sequence of this DNA segment.
3. Which nitrogenous base in DNA pairs with each nitrogenous base?
1. cytosine
2. adenine
3. guanine
4. thymine
4. Which nitrogenous base in RNA pairs with each nitrogenous base?
1. cytosine
2. adenine
3. guanine
4. thymine
5. How many hydrogen bonds can form between the two strands in the short DNA segment shown below?
5′ ATGCGACTA 3′ 3′ TACGCTGAT 5′
6. How many hydrogen bonds can form between the two strands in the short DNA segment shown below?
5′ CGATGAGCC 3′ 3′ GCTACTCGG 5′
Answers
1. c. ACU
1. guanine
2. thymine
3. cytosine
4. adenine
1. 22 (2 between each AT base pair and 3 between each GC base pair)
Concept Review Exercises
1. In DNA replication, a parent DNA molecule produces two daughter molecules. What is the fate of each strand of the parent DNA double helix?
2. What is the role of DNA in transcription? What is produced in transcription?
3. Which type of RNA contains the codon? Which type of RNA contains the anticodon?
Answers
1. Each strand of the parent DNA double helix remains associated with the newly synthesized DNA strand.
2. DNA serves as a template for the synthesis of an RNA strand (the product of transcription).
3. codon: mRNA; anticodon: tRNA
Exercises
1. Describe how replication and transcription are similar.
2. Describe how replication and transcription differ.
3. A portion of the coding strand for a given gene has the sequence 5′‑ATGAGCGACTTTGCGGGATTA‑3′.
1. What is the sequence of complementary template strand?
2. What is the sequence of the mRNA that would be produced during transcription from this segment of DNA?
4. A portion of the coding strand for a given gene has the sequence 5′‑ATGGCAATCCTCAAACGCTGT‑3′.
1. What is the sequence of complementary template strand?
2. What is the sequence of the mRNA that would be produced during transcription from this segment of DNA?
Answers
1. Both processes require a template from which a complementary strand is synthesized.
3.
1. 3′‑TACTCGCTGAAACGCCCTAAT‑5′
2. 5′‑AUGAGCGACUUUGCGGGAUUA‑3′
Concept Review Exercises
1. What are the roles of mRNA and tRNA in protein synthesis?
2. What is the initiation codon?
3. What are the termination codons and how are they recognized?
Answers
1. mRNA provides the code that determines the order of amino acids in the protein; tRNA transports the amino acids to the ribosome to incorporate into the growing protein chain.
2. AUG
3. UAA, UAG, and UGA; they are recognized by special proteins called release factors, which signal the end of the translation process.
Exercises
1. Write the anticodon on tRNA that would pair with each mRNA codon.
1. 5′‑UUU‑3′
2. 5′‑CAU‑3′
3. 5′‑AGC‑3′
4. 5′‑CCG‑3′
2. Write the codon on mRNA that would pair with each tRNA anticodon.
1. 5′‑UUG‑3′
2. 5′‑GAA‑3′
3. 5′‑UCC‑3′
4. 5′‑CAC‑3′
3. The peptide hormone oxytocin contains 9 amino acid units. What is the minimum number of nucleotides needed to code for this peptide?
4. Myoglobin, a protein that stores oxygen in muscle cells, has been purified from a number of organisms. The protein from a sperm whale is composed of 153 amino acid units. What is the minimum number of nucleotides that must be present in the mRNA that codes for this protein?
5. Use Figure \(3\) to identify the amino acids carried by each tRNA molecule in Exercise 1.
6. Use Figure \(3\) to identify the amino acids carried by each tRNA molecule in Exercise 2.
7. Use Figure \(3\) to determine the amino acid sequence produced from this mRNA sequence: 5′‑AUGAGCGACUUUGCGGGAUUA‑3′.
8. Use Figure \(3\) to determine the amino acid sequence produced from this mRNA sequence: 5′‑AUGGCAAUCCUCAAACGCUGU‑3′
Answers
1. 3′‑AAA‑5′
2. 3′‑GUA‑5′
3. 3′‑UCG‑5′
4. 3′‑GGC‑5′
1. 27 nucleotides (3 nucleotides/codon)
1. 1a: phenyalanine; 1b: histidine; 1c: serine; 1d: proline
1. met-ser-asp-phe-ala-gly-leu
Concept Review Exercises
1. What effect can UV radiation have on DNA?
2. Is UV radiation an example of a physical mutagen or a chemical mutagen?
1. What causes PKU?
2. How is PKU detected and treated?
Answers
1. It can lead to the formation of a covalent bond between two adjacent thymines on a DNA strand, producing a thymine dimer.
2. physical mutagen
1. the absence of the enzyme phenylalanine hydroxylase
2. PKU is diagnosed by assaying a sample of blood or urine for phenylalanine or one of its metabolites; treatment calls for an individual to be placed on a diet containing little or no phenylalanine.
Exercises
1. A portion of the coding strand of a gene was found to have the sequence 5′‑ATGAGCGACTTTCGCCCATTA‑3′. A mutation occurred in the gene, making the sequence 5′‑ATGAGCGACCTTCGCCCATTA‑3′.
1. Identify the mutation as a substitution, an insertion, or a deletion.
2. What effect would the mutation have on the amino acid sequence of the protein obtained from this mutated gene (use Figure 19.14)?
2. A portion of the coding strand of a gene was found to have the sequence 5′‑ATGGCAATCCTCAAACGCTGT‑3′. A mutation occurred in the gene, making the sequence 5′‑ATGGCAATCCTCAACGCTGT‑3′.
1. Identify the mutation as a substitution, an insertion, or a deletion.
2. What effect would the mutation have on the amino acid sequence of the protein obtained from this mutated gene (use Figure 19.14)?
1. What is a mutagen?
2. Give two examples of mutagens.
3. For each genetic disease, indicate which enzyme is lacking or defective and the characteristic symptoms of the disease.
1. PKU
2. Tay-Sachs disease
Answers
1. substitution
2. Phenylalanine (UUU) would be replaced with leucine (CUU).
1. a chemical or physical agent that can cause a mutation
2. UV radiation and gamma radiation (answers will vary)
Questions
1. Describe the general structure of a virus.
2. How does a DNA virus differ from an RNA virus?
3. Why is HIV known as a retrovirus?
4. Describe how a DNA virus invades and destroys a cell.
1. Describe how an RNA virus invades and destroys a cell.
2. How does this differ from a DNA virus?
5. What HIV enzyme does AZT inhibit?
6. What HIV enzyme does raltegravir inhibit?
Answers
1. A virus consists of a central core of nucleic acid enclosed in a protective shell of proteins. There may be lipid or carbohydrate molecules on the surface.
2. A DNA virus has DNA as its genetic material, while an RNA virus has RNA as its genetic material.
3. In a cell, a retrovirus synthesizes a DNA copy of its RNA genetic material.
4. The DNA virus enters a host cell and induces the cell to replicate the viral DNA and produce viral proteins. These proteins and DNA assemble into new viruses that are released by the host cell, which may die in the process.
5. -
6. reverse transcriptase
7. -
Additional Exercises
1. For this nucleic acid segment,
1. classify this segment as RNA or DNA and justify your choice.
2. determine the sequence of this segment, labeling the 5′ and 3′ ends.
2. For this nucleic acid segment,
1. classify this segment as RNA or DNA and justify your choice.
2. determine the sequence of this segment, labeling the 5′ and 3′ ends.
3. One of the key pieces of information that Watson and Crick used in determining the secondary structure of DNA came from experiments done by E. Chargaff, in which he studied the nucleotide composition of DNA from many different species. Chargaff noted that the molar quantity of A was always approximately equal to the molar quantity of T, and the molar quantity of C was always approximately equal to the molar quantity of G. How were Chargaff’s results explained by the structural model of DNA proposed by Watson and Crick?
4. Suppose Chargaff (see Exercise 3) had used RNA instead of DNA. Would his results have been the same; that is, would the molar quantity of A approximately equal the molar quantity of T? Explain.
5. In the DNA segment
5′‑ATGAGGCATGAGACG‑3′ (coding strand) 3′‑TACTCCGTACTCTGC‑5′ (template strand)
1. What products would be formed from the segment’s replication?
2. Write the mRNA sequence that would be obtained from the segment’s transcription.
3. What is the amino acid sequence of the peptide produced from the mRNA in Exercise 5b?
6. In the DNA segment
5′‑ATGACGGTTTACTAAGCC‑3′ (coding strand) 3′‑TACTGCCAAATGATTCGG‑5′ (template strand)
1. What products would be formed from the segment’s replication?
2. Write the mRNA sequence that would be obtained from the segment’s transcription.
3. What is the amino acid sequence of the peptide produced from the mRNA in Exercise 6b?
7. A hypothetical protein has a molar mass of 23,300 Da. Assume that the average molar mass of an amino acid is 120.
1. How many amino acids are present in this hypothetical protein?
2. What is the minimum number of codons present in the mRNA that codes for this protein?
3. What is the minimum number of nucleotides needed to code for this protein?
8. Bradykinin is a potent peptide hormone composed of nine amino acids that lowers blood pressure.
1. The amino acid sequence for bradykinin is arg-pro-pro-gly-phe-ser-pro-phe-arg. Postulate a base sequence in the mRNA that would direct the synthesis of this hormone. Include an initiation codon and a termination codon.
2. What is the nucleotide sequence of the DNA that codes for this mRNA?
9. A particular DNA coding segment is ACGTTAGCCCCAGCT.
1. Write the sequence of nucleotides in the corresponding mRNA.
2. Determine the amino acid sequence formed from the mRNA in Exercise 9a during translation.
3. What amino acid sequence results from each of the following mutations?
1. replacement of the underlined guanine by adenine
2. insertion of thymine immediately after the underlined guanine
3. deletion of the underlined guanine
10. A particular DNA coding segment is TACGACGTAACAAGC.
1. Write the sequence of nucleotides in the corresponding mRNA.
2. Determine the amino acid sequence formed from the mRNA in Exercise 10a during translation.
3. What amino acid sequence results from each of the following mutations?
1. replacement of the underlined guanine by adenine
2. replacement of the underlined adenine by thymine
11. Two possible point mutations are the substitution of lysine for leucine or the substitution of serine for threonine. Which is likely to be more serious and why?
12. Two possible point mutations are the substitution of valine for leucine or the substitution of glutamic acid for histidine. Which is likely to be more serious and why?
Answers
1.
1. RNA; the sugar is ribose, rather than deoxyribose
2. 5′‑GUA‑3′
1. In the DNA structure, because guanine (G) is always paired with cytosine (C) and adenine (A) is always paired with thymine (T), you would expect to have equal amounts of each.
1.
1. Each strand would be replicated, resulting in two double-stranded segments.
2. 5′‑AUGAGGCAUGAGACG‑3′
3. met-arg-his-glu-thr
1.
1. 194
2. 194
3. 582
1.
1. 5′‑ACGUUAGCCCCAGCU‑3′
2. thr-leu-ala-pro-ala
1. thr-leu-thr-pro-ala
2. thr-leu-val-pro-ser
3. thr-leu-pro-gin
1. substitution of lysine for leucine because you are changing from an amino acid with a nonpolar side chain to one that has a positively charged side chain; both serine and threonine, on the other hand, have polar side chains containing the OH group. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/19%3A_Nucleic_Acids/19.E%3A_Nucleic_Acids_%28Exercises%29.txt |
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
A cell’s hereditary information is encoded in chromosomes in the cell’s nucleus. Each chromosome is composed of proteins and deoxyribonucleic acid (DNA). The chromosomes contain smaller hereditary units called genes, which are relatively short segments of DNA. The hereditary information is expressed or used through the synthesis of ribonucleic acid (RNA). Both nucleic acids—DNA and RNA—are polymers composed of monomers known as nucleotides, which in turn consist of phosphoric acid (H3PO4), a nitrogenous base, and a pentose sugar.
The two types of nitrogenous bases most important in nucleic acids are purines—adenine (A) and guanine (G)—and pyrimidines—cytosine (C), thymine (T), and uracil (U). DNA contains the nitrogenous bases adenine, cytosine, guanine, and thymine, while the bases in RNA are adenine, cytosine, guanine, and uracil. The sugar in the nucleotides of RNA is ribose; the one in DNA is 2-deoxyribose. The sequence of nucleotides in a nucleic acid defines the primary structure of the molecule.
RNA is a single-chain nucleic acid, whereas DNA possesses two nucleic-acid chains intertwined in a secondary structure called a double helix. The sugar-phosphate backbone forms the outside the double helix, with the purine and pyrimidine bases tucked inside. Hydrogen bonding between complementary bases holds the two strands of the double helix together; A always pairs with T and C always pairs with G.
Cell growth requires replication, or reproduction of the cell’s DNA. The double helix unwinds, and hydrogen bonding between complementary bases breaks so that there are two single strands of DNA, and each strand is a template for the synthesis of a new strand. For protein synthesis, three types of RNA are needed: messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA). All are made from a DNA template by a process called transcription. The double helix uncoils, and ribonucleotides base-pair to the deoxyribonucleotides on one DNA strand; however, RNA is produced using uracil rather than thymine. Once the RNA is formed, it dissociates from the template and leaves the nucleus, and the DNA double helix reforms.
Translation is the process in which proteins are synthesized from the information in mRNA. It occurs at structures called ribosomes, which are located outside the nucleus and are composed of rRNA and protein. The 64 possible three-nucleotide combinations of the 4 nucleotides of DNA constitute the genetic code that dictates the sequence in which amino acids are joined to make proteins. Each three-nucleotide sequence on mRNA is a codon. Each kind of tRNA molecule binds a specific amino acid and has a site containing a three-nucleotide sequence called an anticodon.
The general term for any change in the genetic code in an organism’s DNA is mutation. A change in which a single base is substituted, inserted, or deleted is a point mutation. The chemical and/or physical agents that cause mutations are called mutagens. Diseases that occur due to mutations in critical DNA sequences are referred to as genetic diseases.
Viruses are infectious agents composed of a tightly packed central core of nucleic acids enclosed by a protective shell of proteins. Viruses contain either DNA or RNA as their genetic material but not both. Some RNA viruses, called retroviruses, synthesize DNA in the host cell from their RNA genome. The human immunodeficiency virus (HIV) causes acquired immunodeficiency syndrome (AIDS). | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/19%3A_Nucleic_Acids/19.S%3A_Nucleic_Acids_%28Summary%29.txt |
Metabolism is the set of life-sustaining chemical transformations within the cells of living organisms. The three main purposes of metabolism are the conversion of food/fuel to energy to run cellular processes, the conversion of food/fuel to building blocks for proteins, lipids, nucleic acids, and some carbohydrates, and the elimination of nitrogenous wastes. These enzyme-catalyzed reactions allow organisms to grow and reproduce, maintain their structures, and respond to their environments. Metabolism is usually divided into two categories: catabolism, the breaking down of organic matter, for example, by cellular respiration, and anabolism, the building up of components of cells such as proteins and nucleic acids. Usually, breaking down releases energy and building up consumes energy.
• 20.0: Prelude to Energy Metabolism
The insulin receptor is located in the cell membrane and consists of four polypeptide chains: two identical chains called α chains and two identical chains called β chains. The α chains, positioned on the outer surface of the membrane, consist of 735 amino acids each and contain the binding site for insulin. The β chains are integral membrane proteins, each composed of 620 amino acids.
• 20.1: ATP- the Universal Energy Currency
The hydrolysis of ATP releases energy that can be used for cellular processes that require energy.
• 20.2: Stage I of Catabolism
During digestion, carbohydrates are broken down into monosaccharides, proteins are broken down into amino acids, and triglycerides are broken down into glycerol and fatty acids. Most of the digestion reactions occur in the small intestine.
• 20.3: Overview of Stage II of Catabolism
Acetyl-CoA is formed from the breakdown of carbohydrates, lipids, and proteins. It is used in many biochemical pathways.
• 20.4: Stage III of Catabolism
The acetyl group of acetyl-CoA enters the citric acid cycle. For each acetyl-CoA that enters the citric acid cycle, 2 molecules of carbon dioxide, 3 molecules of NADH, 1 molecule of ATP, and 1 molecule of FADH2 are produced. The reduced coenzymes produced by the citric acid cycle are reoxidized by the reactions of the electron transport chain. This series of reactions also produces a pH gradient across the inner mitochondrial membrane that drives the synthesis of ATP from ADP.
• 20.5: Stage II of Carbohydrate Catabolism
The monosaccharide glucose is broken down through a series of enzyme-catalyzed reactions known as glycolysis. For each molecule of glucose that is broken down, two molecules of pyruvate, two molecules of ATP, and two molecules of NADH are produced. In the absence of oxygen, pyruvate is converted to lactate, and NADH is reoxidized to NAD+. In the presence of oxygen, pyruvate is converted to acetyl-CoA and then enters the citric acid cycle. More ATP can be formed from the breakdown of glucose.
• 20.6: Stage II of Lipid Catabolism
Fatty acids, obtained from the breakdown of triglycerides and other lipids, are oxidized through a series of reactions known as β-oxidation. In each round of β-oxidation, 1 molecule of acetyl-CoA, 1 molecule of NADH, and 1 molecule of FADH2 are produced. The acetyl-CoA, NADH, and FADH2 are used in the citric acid cycle, the electron transport chain, and oxidative phosphorylation to produce ATP.
• 20.7: Stage II of Protein Catabolism
Generally the first step in the breakdown of amino acids is the removal of the amino group, usually through a reaction known as transamination. The carbon skeletons of the amino acids undergo further reactions to form compounds that can either be used for the synthesis of glucose or the synthesis of ketone bodies.
• 20.E: Energy Metabolism (Exercises)
Problems and select solutions for the chapter.
• 20.S: Energy Metabolism (Summary)
To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
20: Energy Metabolism
The discovery of the link between insulin and diabetes led to a period of intense research aimed at understanding exactly how insulin works in the body to regulate glucose levels. Hormones in general act by binding to some protein, known as the hormone’s receptor, thus initiating a series of events that lead to a desired outcome. In the early 1970s, the insulin receptor was purified, and researchers began to study what happens after insulin binds to its receptor and how those events are linked to the uptake and metabolism of glucose in cells.
The insulin receptor is located in the cell membrane and consists of four polypeptide chains: two identical chains called α chains and two identical chains called β chains. The α chains, positioned on the outer surface of the membrane, consist of 735 amino acids each and contain the binding site for insulin. The β chains are integral membrane proteins, each composed of 620 amino acids. The binding of insulin to its receptor stimulates the β chains to catalyze the addition of phosphate groups to the specific side chains of tyrosine (referred to as phosphorylation) in the β chains and other cell proteins, leading to the activation of reactions that metabolize glucose. In this chapter we will look at the pathway that breaks down glucose—in response to activation by insulin—for the purpose of providing energy for the cell.
Life requires energy. Animals, for example, require heat energy to maintain body temperature, mechanical energy to move their limbs, and chemical energy to synthesize the compounds needed by their cells. Living cells remain organized and functioning properly only through a continual supply of energy. But only specific forms of energy can be used. Supplying a plant with energy by holding it in a flame will not prolong its life. On the other hand, a green plant is able to absorb radiant energy from the sun, the most abundant source of energy for life on the earth. Plants use this energy first to form glucose and then to make other carbohydrates, as well as lipids and proteins. Unlike plants, animals cannot directly use the sun’s energy to synthesize new compounds. They must eat plants or other animals to get carbohydrates, fats, and proteins and the chemical energy stored in them. Once digested and transported to the cells, the nutrient molecules can be used in either of two ways: as building blocks for making new cell parts or repairing old ones or “burned” for energy.
The thousands of coordinated chemical reactions that keep cells alive are referred to collectively as metabolism. In general, metabolic reactions are divided into two classes: the breaking down of molecules to obtain energy is catabolism, and the building of new molecules needed by living systems is anabolism.
Definition: Metabolite
Any chemical compound that participates in a metabolic reaction is a metabolite.
Most of the energy required by animals is generated from lipids and carbohydrates. These fuels must be oxidized, or “burned,” for the energy to be released. The oxidation process ultimately converts the lipid or carbohydrate to carbon dioxide (CO2) and water (H2O).
Carbohydrate:
$\ce{C6H12O6 + 6O2 → 6CO2 + 6H2O + 670 kcal} \nonumber$
Lipid:
$\ce{C16H32O2 + 23O2 → 16CO2 + 16H2O + 2,385 kcal} \nonumber$
These two equations summarize the biological combustion of a carbohydrate and a lipid by the cell through respiration. Respiration is the collective name for all metabolic processes in which gaseous oxygen is used to oxidize organic matter to carbon dioxide, water, and energy.
Like the combustion of the common fuels we burn in our homes and cars (wood, coal, gasoline), respiration uses oxygen from the air to break down complex organic substances to carbon dioxide and water. But the energy released in the burning of wood is manifested entirely in the form of heat, and excess heat energy is not only useless but also injurious to the living cell. Living organisms instead conserve much of the energy respiration releases by channeling it into a series of stepwise reactions that produce adenosine triphosphate (ATP) or other compounds that ultimately lead to the synthesis of ATP. The remainder of the energy is released as heat and manifested as body temperature. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.00%3A_Prelude_to_Energy_Metabolism.txt |
Learning Objectives
• To describe the importance of ATP as a source of energy in living organisms.
Adenosine triphosphate (ATP), a nucleotide composed of adenine, ribose, and three phosphate groups, is perhaps the most important of the so-called energy-rich compounds in a cell. Its concentration in the cell varies from 0.5 to 2.5 mg/mL of cell fluid.
Energy-rich compounds are substances having particular structural features that lead to a release of energy after hydrolysis. As a result, these compounds are able to supply energy for biochemical processes that require energy. The structural feature important in ATP is the phosphoric acid anhydride, or pyrophosphate, linkage:
The pyrophosphate bond, symbolized by a squiggle (~), is hydrolyzed when ATP is converted to adenosine diphosphate (ADP). In this hydrolysis reaction, the products contain less energy than the reactants; there is a release of energy (> 7 kcal/mol). One reason for the amount of energy released is that hydrolysis relieves the electron-electron repulsions experienced by the negatively charged phosphate groups when they are bonded to each other (Figure 20.1.1).
Energy is released because the products (ADP and phosphate ion) have less energy than the reactants [ATP and water (H2O)].
The general equation for ATP hydrolysis is as follows:
$ATP + H_2O → ADP + P_i + 7.4\; kcal/mol \nonumber$
If the hydrolysis of ATP releases energy, its synthesis (from ADP) requires energy. In the cell, ATP is produced by those processes that supply energy to the organism (absorption of radiant energy from the sun in green plants and breakdown of food in animals), and it is hydrolyzed by those processes that require energy (the syntheses of carbohydrates, lipids, proteins; the transmission of nerve impulses; muscle contractions). In fact, ATP is the principal medium of energy exchange in biological systems. Many scientists call it the energy currency of cells.
$P_i$ is the symbol for the inorganic phosphate anions $H_2PO_4^−$ and $HPO_4^{2−}$.
ATP is not the only high-energy compound needed for metabolism. Several others are listed in Table $1$. Notice, however, that the energy released when ATP is hydrolyzed is approximately midway between those of the high-energy and the low-energy phosphate compounds. This means that the hydrolysis of ATP can provide energy for the phosphorylation of the compounds below it in the table. For example, the hydrolysis of ATP provides sufficient energy for the phosphorylation of glucose to form glucose 1-phosphate. By the same token, the hydrolysis of compounds, such as creatine phosphate, that appear above ATP in the table can provide the energy needed to resynthesize ATP from ADP.
Table $1$: Energy Released by Hydrolysis of Some Phosphate Compounds
Type Example Energy Released (kcal/mol)
acyl phosphate
1,3-bisphosphoglycerate (BPG) −11.8
acetyl phosphate −11.3
guanidine phosphates
creatine phosphate −10.3
arginine phosphate −9.1
pyrophosphates
PPi* → 2Pi −7.8
ATP → AMP + PPi −7.7
ATP → ADP + Pi −7.5
ADP → AMP + Pi −7.5
sugar phosphates
glucose 1-phosphate −5.0
fructose 6-phosphate −3.8
AMP → adenosine + Pi −3.4
glucose 6-phosphate −3.3
glycerol 3-phosphate −2.2
*PPi is the pyrophosphate ion.
Summary
The hydrolysis of ATP releases energy that can be used for cellular processes that require energy. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.01%3A_ATP-_the_Universal_Energy_Currency.txt |
Learning Objectives
• To describe how carbohydrates, fats, and proteins are broken down during digestion.
We have said that animals obtain chemical energy from the food—carbohydrates, fats, and proteins—they eat through reactions defined collectively as catabolism. We can think of catabolism as occurring in three stages (Figure \(1\)). In stage I, carbohydrates, fats, and proteins are broken down into their individual monomer units: carbohydrates into simple sugars, fats into fatty acids and glycerol, and proteins into amino acids. One part of stage I of catabolism is the breakdown of food molecules by hydrolysis reactions into the individual monomer units—which occurs in the mouth, stomach, and small intestine—and is referred to as digestion.
In stage II, these monomer units (or building blocks) are further broken down through different reaction pathways, one of which produces ATP, to form a common end product that can then be used in stage III to produce even more ATP. In this chapter, we will look at each stage of catabolism—as an overview and in detail.
The conversion of food into cellular energy (as ATP) occurs in three stages.
Digestion of Carbohydrates
Carbohydrate digestion begins in the mouth (Figure \(2\)) where salivary α-amylase attacks the α-glycosidic linkages in starch, the main carbohydrate ingested by humans. Cleavage of the glycosidic linkages produces a mixture of dextrins, maltose, and glucose. The α-amylase mixed into the food remains active as the food passes through the esophagus, but it is rapidly inactivated in the acidic environment of the stomach.
The primary site of carbohydrate digestion is the small intestine. The secretion of α-amylase in the small intestine converts any remaining starch molecules, as well as the dextrins, to maltose. Maltose is then cleaved into two glucose molecules by maltase. Disaccharides such as sucrose and lactose are not digested until they reach the small intestine, where they are acted on by sucrase and lactase, respectively. The major products of the complete hydrolysis of disaccharides and polysaccharides are three monosaccharide units: glucose, fructose, and galactose. These are absorbed through the wall of the small intestine into the bloodstream.
Digestion of Proteins
Protein digestion begins in the stomach (Figure \(3\)), where the action of gastric juice hydrolyzes about 10% of the peptide bonds. Gastric juice is a mixture of water (more than 99%), inorganic ions, hydrochloric acid, and various enzymes and other proteins.
The pain of a gastric ulcer is at least partially due to irritation of the ulcerated tissue by acidic gastric juice.
The hydrochloric acid (HCl) in gastric juice is secreted by glands in the stomach lining. The pH of freshly secreted gastric juice is about 1.0, but the contents of the stomach may raise the pH to between 1.5 and 2.5. HCl helps to denature food proteins; that is, it unfolds the protein molecules to expose their chains to more efficient enzyme action. The principal digestive component of gastric juice is pepsinogen, an inactive enzyme produced in cells located in the stomach wall. When food enters the stomach after a period of fasting, pepsinogen is converted to its active form—pepsin—in a series of steps initiated by the drop in pH. Pepsin catalyzes the hydrolysis of peptide linkages within protein molecules. It has a fairly broad specificity but acts preferentially on linkages involving the aromatic amino acids tryptophan, tyrosine, and phenylalanine, as well as methionine and leucine.
Protein digestion is completed in the small intestine. Pancreatic juice, carried from the pancreas via the pancreatic duct, contains inactive enzymes such as trypsinogen and chymotrypsinogen. They are activated in the small intestine as follows (Figure \(4\)): The intestinal mucosal cells secrete the proteolytic enzyme enteropeptidase, which converts trypsinogen to trypsin; trypsin then activates chymotrypsinogen to chymotrypsin (and also completes the activation of trypsinogen). Both of these active enzymes catalyze the hydrolysis of peptide bonds in protein chains. Chymotrypsin preferentially attacks peptide bonds involving the carboxyl groups of the aromatic amino acids (phenylalanine, tryptophan, and tyrosine). Trypsin attacks peptide bonds involving the carboxyl groups of the basic amino acids (lysine and arginine). Pancreatic juice also contains procarboxypeptidase, which is cleaved by trypsin to carboxypeptidase. The latter is an enzyme that catalyzes the hydrolysis of peptide linkages at the free carboxyl end of the peptide chain, resulting in the stepwise liberation of free amino acids from the carboxyl end of the polypeptide.
Aminopeptidases in the intestinal juice remove amino acids from the N-terminal end of peptides and proteins possessing a free amino group. Figure \(5\) illustrates the specificity of these protein-digesting enzymes. The amino acids that are released by protein digestion are absorbed across the intestinal wall into the circulatory system, where they can be used for protein synthesis.
This diagram illustrates where in a peptide the different peptidases we have discussed would catalyze hydrolysis the peptide bonds.
Digestion of Lipids
Lipid digestion begins in the upper portion of the small intestine (Figure \(6\)). A hormone secreted in this region stimulates the gallbladder to discharge bile into the duodenum. The principal constituents of bile are the bile salts, which emulsify large, water-insoluble lipid droplets, disrupting some of the hydrophobic interactions holding the lipid molecules together and suspending the resulting smaller globules (micelles) in the aqueous digestive medium. These changes greatly increase the surface area of the lipid particles, allowing for more intimate contact with the lipases and thus rapid digestion of the fats. Another hormone promotes the secretion of pancreatic juice, which contains these enzymes.
The lipases in pancreatic juice catalyze the digestion of triglycerides first to diglycerides and then to 2‑monoglycerides and fatty acids:
The monoglycerides and fatty acids cross the intestinal lining into the bloodstream, where they are resynthesized into triglycerides and transported as lipoprotein complexes known as chylomicrons. Phospholipids and cholesteryl esters undergo similar hydrolysis in the small intestine, and their component molecules are also absorbed through the intestinal lining.
The further metabolism of monosaccharides, fatty acids, and amino acids released in stage I of catabolism occurs in stages II and III of catabolism.
Summary
During digestion, carbohydrates are broken down into monosaccharides, proteins are broken down into amino acids, and triglycerides are broken down into glycerol and fatty acids. Most of the digestion reactions occur in the small intestine. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.02%3A_Stage_I_of_Catabolism.txt |
Learning Objectives
• To describe the role of acetyl-CoA in metabolism.
A metabolic pathway is a series of biochemical reactions by which an organism converts a given reactant to a specific end product. There are specific metabolic pathways—which are different for carbohydrates, triglycerides, and proteins—that break down the products of stage I of catabolism (monosaccharides, fatty acids, and amino acids) to produce a common end product, acetyl-coenzyme A (acetyl-CoA) in stage II of catabolism.
Acetyl-CoA is shown in Figure \(1\). The acetyl unit, derived (as we will see) from the breakdown of carbohydrates, lipids, and proteins, is attached to coenzyme A, making the acetyl unit more reactive. Acetyl-CoA is used in a myriad of biochemical pathways. For example, it may be used as the starting material for the biosynthesis of lipids (such as triglycerides, phospholipids, or cholesterol and other steroids). Most importantly for energy generation, it may enter the citric acid cycle and be oxidized to produce energy, if energy is needed and oxygen is available. The various fates or uses of acetyl-CoA are summarized in Figure \(1\).
Glycolysis
Glycolysis is the catabolic process in which glucose is converted into pyruvate via ten enzymatic steps. There are three regulatory steps, each of which is highly regulated that are separated into two phases:
1. the "priming phase" because it requires an input of energy in the form of 2 ATPs per glucose molecule and
2. the "pay off phase" because energy is released in the form of 4 ATPs, 2 per glyceraldehyde molecule.
The end result of Glycolysis is two new pyruvate molecules which can then be fed into the Citric Acid cycle (also known as the Kreb's Cycle) if oxygen is present, or can be reduced to lactate or ethanol in the absence of of oxygen using a process known as fermentation.
Glycolysis occurs within almost all living cells and is the primary source of Acetyl-CoA, which is the molecule responsible for the majority of energy output under aerobic conditions. The structures of Glycolysis intermediates can be found in Figure \(3\).
Phase 1: The "Priming Step"
The first phase of Glycolysis requires an input of energy in the form of ATP (adenosine triphosphate).
1. alpha-D-Glucose is phosphorolated at the 6 carbon by ATP via the enzyme Hexokinase (Class: Transferase) to yield alpha-D-Glucose-6-phosphate (G-6-P). This is a regulatory step which is negatively regulated by the presence of glucose-6-phosphate.
2. alpha-D-Glucose-6-phosphate is then converted into D-Fructose-6-phosphate (F-6-P) by Phosphoglucoisomerase (Class: Isomerase)
3. D-Fructose-6-phosphate is once again phosphorolated this time at the 1 carbon position by ATP via the enzyme Phosphofructokinase (Class: Transferase) to yield D-Fructose-1,6-bisphosphate (FBP). This is the committed step of glycolysis because of its large \(\Delta G\) value.
4. D-Fructose-1,6-bisphosphate is then cleaved into two, three carbon molecules; Dihydroxyacetone phosphate (DHAP) and D-Glyceraldehyde-3-phosphate (G-3-P) by the enzyme Fructose bisphosphate aldolase (Class: Lyase)
5. Because the next portion of Glycolysis requires the molecule D-Glyceraldehyde-3-phosphate to continue Dihydroxyacetone phosphate is converted into D-Glyceraldehyde-3-phosphate by the enzyme Triose phosphate isomerase (Class: Isomerase)
Phase 2: The "Pay Off Step"
The second phase of Glycolysis where 4 molecules of ATP are produced per molecule of glucose. Enzymes appear in red:
1. D-Glyceraldehyde-3-phosphate is phosphorolated at the 1 carbon by the enzyme Glyceraldehyde-3-phosphate dehodrogenase to yield the high energy molecule 1,3-Bisphosphoglycerate (BPG)
2. ADP is then phosphorolated at the expense of 1,3-Bisphosphoglycerate by the enzyme Phosphoglycerate kinase (Class: Transferase) to yield ATP and 3-Phosphoglycerate (3-PG)
3. 3-Phosphoglycerate is then converted into 2-Phosphoglycerate by Phosphoglycerate mutase in preparation to yield another high energy molecule
4. 2-Phosphoglycerate is then converted to phosphoenolpyruvate (PEP) by Enolase. H2O, potassium, and magnesium are all released as a result.
5. ADP is once again phosphorolated, this time at the expense of PEP by the enzyme pyruvate kinase to yield another molecule of ATP and and pyruvate. This step is regulated by the energy in the cell. The higher the energy of the cell the more inhibited pyruvate kinase becomes. Indicators of high energy levels within the cell are high concentrations of ATP, Acetyl-CoA, Alanine, and cAMP.
Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the "Pay Off" phase occurs twice per molecule of glucose.
Beta-Oxidation
The best source of energy for eukaryotic organisms are fats. Glucose offers a ratio 6.3 moles of ATP per carbon while saturated fatty acids offer 8.1 ATP per carbon. Also the complete oxidation of fats yields enormous amounts of water for those organisms that do not have adequate access to drinkable water. Camels and killer whales are good example of this, they obtain their water requirements from the complete oxidation of fats.
There are four distinct stages in the oxidation of fatty acids. Fatty acid degradation takes place within the mitochondria and requires the help of several different enzymes. In order for fatty acids to enter the mitochondria the assistance of two carrier proteins is required, Carnitine acyltransferase I and II. It is also interesting to note the similarities between the four steps of beta-oxidation and the later four steps of the TCA cycle.
Entry into Beta-oxidation
Most fats stored in eukaryotic organisms are stored as triglycerides as seen below. In order to enter into beta-oxidation bonds must be broken usually with the use of a Lipase. The end result of these broken bonds are a glycerol molecule and three fatty acids in the case of triglycerides. Other lipids are capable of being degraded as well.
Activation Step
• Once the triglycerides are broken down into glycerol and fatty acids they must be activated before they can enter into the mitochondria and proceed on with beta-oxidation. This is done by Acyl-CoA synthetase to yield fatty acyl-CoA.
• After the fatty acid has been acylated it is now ready to enter into the mitochondria.
• There are two carrier proteins (Carnitine acyltransferase I and II), one located on the outer membrane and one on the inner membrane of the mitochondria. Both are required for entry of the Acyl-CoA into the mitochondria.
• Once inside the mitochondria the fatty acyl-CoA can enter into beta-oxidation.
Oxidation Step
A fatty acyl-CoA is oxidized by Acyl-CoA dehydrogenase to yield a trans alkene. This is done with the aid of an [FAD] prosthetic group.
Hydration Step
The trans alkene is then hydrated with the help of Enoyl-CoA hydratase
Oxidation Step
The alcohol of the hydroxyacly-CoA is then oxidized by NAD+ to a carbonyl with the help of Hydroxyacyl-CoA dehydrogenase. NAD+ is used to oxidize the alcohol rather then [FAD] because NAD+ is capable of the alcohol while [FAD] is not.
Cleavage
Finally acetyl-CoA is cleaved off with the help of Thiolase to yield an Acyl-CoA that is two carbons shorter than before. The cleaved acetyl-CoA can then enter into the TCA and ETC because it is already within the mitochondria.
Summary
Acetyl-CoA is formed from the breakdown of carbohydrates, lipids, and proteins. It is used in many biochemical pathways. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.03%3A_Overview_of_Stage_II_of_Catabolism.txt |
Learning Objectives
• Describe the reactions of the citric acid cycle.
• Describe the function of the citric acid cycle and identify the products produced.
• Describe the role of the electron transport chain in energy metabolism.
• Describe the role of oxidative phosphorylation in energy metabolism.
The acetyl group enters a cyclic sequence of reactions known collectively as the citric acid cycle (or Krebs cycle or tricarboxylic acid [TCA] cycle). The cyclical design of this complex series of reactions, which bring about the oxidation of the acetyl group of acetyl-CoA to carbon dioxide and water, was first proposed by Hans Krebs in 1937. (He was awarded the 1953 Nobel Prize in Physiology or Medicine.) Acetyl-CoA’s entrance into the citric acid cycle is the beginning of stage III of catabolism. The citric acid cycle produces adenosine triphosphate (ATP), reduced nicotinamide adenine dinucleotide (NADH), reduced flavin adenine dinucleotide (FADH2), and metabolic intermediates for the synthesis of needed compounds.
Steps of the Citric Acid Cycle
At first glance, the citric acid cycle appears rather complex (Figure $1$). All the reactions, however, are familiar types in organic chemistry: hydration, oxidation, decarboxylation, and hydrolysis. Each reaction of the citric acid cycle is numbered, and in Figure $1$, the two acetyl carbon atoms are highlighted in red. Each intermediate in the cycle is a carboxylic acid, existing as an anion at physiological pH. All the reactions occur within the mitochondria, which are small organelles within the cells of plants and animals.
1. In the first step, acetyl-CoA enters the citric acid cycle, and the acetyl group is transferred onto oxaloacetate, yielding citrate. Note that this step releases coenzyme A. The reaction is catalyzed by citrate synthase.
2. In the next step, aconitase catalyzes the isomerization of citrate to isocitrate. In this reaction, a tertiary alcohol, which cannot be oxidized, is converted to a secondary alcohol, which can be oxidized in the next step.
3. Isocitrate then undergoes a reaction known as oxidative decarboxylation because the alcohol is oxidized and the molecule is shortened by one carbon atom with the release of carbon dioxide (decarboxylation). The reaction is catalyzed by isocitrate dehydrogenase, and the product of the reaction is α-ketoglutarate. An important reaction linked to this is the reduction of the coenzyme nicotinamide adenine dinucleotide (NAD+) to NADH. The NADH is ultimately reoxidized, and the energy released is used in the synthesis of ATP, as we shall see.
4. The fourth step is another oxidative decarboxylation. This time α-ketoglutarate is converted to succinyl-CoA, and another molecule of NAD+ is reduced to NADH. The α-ketoglutarate dehydrogenase complex catalyzes this reaction. This is the only irreversible reaction in the citric acid cycle. As such, it prevents the cycle from operating in the reverse direction, in which acetyl-CoA would be synthesized from carbon dioxide.
So far, in the first four steps, two carbon atoms have entered the cycle as an acetyl group, and two carbon atoms have been released as molecules of carbon dioxide. The remaining reactions of the citric acid cycle use the four carbon atoms of the succinyl group to resynthesize a molecule of oxaloacetate, which is the compound needed to combine with an incoming acetyl group and begin another round of the cycle.
In the fifth reaction, the energy released by the hydrolysis of the high-energy thioester bond of succinyl-CoA is used to form guanosine triphosphate (GTP) from guanosine diphosphate (GDP) and inorganic phosphate in a reaction catalyzed by succinyl-CoA synthetase. This step is the only reaction in the citric acid cycle that directly forms a high-energy phosphate compound. GTP can readily transfer its terminal phosphate group to adenosine diphosphate (ADP) to generate ATP in the presence of nucleoside diphosphokinase.
Succinate dehydrogenase then catalyzes the removal of two hydrogen atoms from succinate, forming fumarate. This oxidation-reduction reaction uses flavin adenine dinucleotide (FAD), rather than NAD+, as the oxidizing agent. Succinate dehydrogenase is the only enzyme of the citric acid cycle located within the inner mitochondrial membrane. We will see soon the importance of this.
In the following step, a molecule of water is added to the double bond of fumarate to form L-malate in a reaction catalyzed by fumarase.
One revolution of the cycle is completed with the oxidation of L-malate to oxaloacetate, brought about by malate dehydrogenase. This is the third oxidation-reduction reaction that uses NAD+ as the oxidizing agent. Oxaloacetate can accept an acetyl group from acetyl-CoA, allowing the cycle to begin again.
Video: "The Citric Acid Cycle: An Overview". In the matrix of the mitochondrion, the Citric Acid Cycle uses Acetyl CoA molecules to produce energy through eight chemical reactions. This animation provides an overview of the pathway and its products. NDSU VCell Production's animation; for more information please see Vcell, NDSU, Animations(opens in new window) [vcell.ndsu.edu].
Cellular Respiration
Respiration can be defined as the process by which cells oxidize organic molecules in the presence of gaseous oxygen to produce carbon dioxide, water, and energy in the form of ATP. We have seen that two carbon atoms enter the citric acid cycle from acetyl-CoA (step 1), and two different carbon atoms exit the cycle as carbon dioxide (steps 3 and 4). Yet nowhere in our discussion of the citric acid cycle have we indicated how oxygen is used. Recall, however, that in the four oxidation-reduction steps occurring in the citric acid cycle, the coenzyme NAD+ or FAD is reduced to NADH or FADH2, respectively. Oxygen is needed to reoxidize these coenzymes. Recall, too, that very little ATP is obtained directly from the citric acid cycle. Instead, oxygen participation and significant ATP production occur subsequent to the citric acid cycle, in two pathways that are closely linked: electron transport and oxidative phosphorylation.
All the enzymes and coenzymes for the citric acid cycle, the reoxidation of NADH and FADH2, and the production of ATP are located in the mitochondria, which are small, oval organelles with double membranes, often referred to as the “power plants” of the cell (Figure $2$). A cell may contain 100–5,000 mitochondria, depending on its function, and the mitochondria can reproduce themselves if the energy requirements of the cell increase.
Cellular respiration occurs in the mitochondria
Figure $2$ shows the mitochondrion’s two membranes: outer and inner. The inner membrane is extensively folded into a series of internal ridges called cristae. Thus there are two compartments in mitochondria: the intermembrane space, which lies between the membranes, and the matrix, which lies inside the inner membrane. The outer membrane is permeable, whereas the inner membrane is impermeable to most molecules and ions, although water, oxygen, and carbon dioxide can freely penetrate both membranes. The matrix contains all the enzymes of the citric acid cycle with the exception of succinate dehydrogenase, which is embedded in the inner membrane. The enzymes that are needed for the reoxidation of NADH and FADH2 and ATP production are also located in the inner membrane. They are arranged in specific positions so that they function in a manner analogous to a bucket brigade. This highly organized sequence of oxidation-reduction enzymes is known as the electron transport chain (or respiratory chain).
Electron Transport
Figure $3$ illustrates the organization of the electron transport chain. The components of the chain are organized into four complexes designated I, II, III, and IV. Each complex contains several enzymes, other proteins, and metal ions. The metal ions can be reduced and then oxidized repeatedly as electrons are passed from one component to the next. Recall that a compound is reduced when it gains electrons or hydrogen atoms and is oxidized when it loses electrons or hydrogen atoms.
Electrons can enter the electron transport chain through either complex I or II. We will look first at electrons entering at complex I. These electrons come from NADH, which is formed in three reactions of the citric acid cycle. Let’s use step 8 as an example, the reaction in which L-malate is oxidized to oxaloacetate and NAD+ is reduced to NADH. This reaction can be divided into two half reactions:
• Oxidation half-reaction:
• Reduction half-reaction:
In the oxidation half-reaction, two hydrogen (H+) ions and two electrons are removed from the substrate. In the reduction half-reaction, the NAD+ molecule accepts both of those electrons and one of the H+ ions. The other H+ ion is transported from the matrix, across the inner mitochondrial membrane, and into the intermembrane space. The NADH diffuses through the matrix and is bound by complex I of the electron transport chain. In the complex, the coenzyme flavin mononucleotide (FMN) accepts both electrons from NADH. By passing the electrons along, NADH is oxidized back to NAD+ and FMN is reduced to FMNH2 (reduced form of flavin mononucleotide). Again, the reaction can be illustrated by dividing it into its respective half-reactions.
• Oxidation half-reaction:
• Reduction half-reaction:
Complex I contains several proteins that have iron-sulfur (Fe·S) centers. The electrons that reduced FMN to FMNH2 are now transferred to these proteins. The iron ions in the Fe·S centers are in the Fe(III) form at first, but by accepting an electron, each ion is reduced to the Fe(II) form. Because each Fe·S center can transfer only one electron, two centers are needed to accept the two electrons that will regenerate FMN.
• Oxidation half-reaction:
$\ce{FMNH_2 -> FMN + 2H+ + 2e-} \nonumber$
• Reduction half-reaction:
$\ce{2Fe(III) \cdot S + 2e- -> 2Fe(II) \cdot S} \nonumber$
Electrons from FADH2, formed in step 6 of the citric acid cycle, enter the electron transport chain through complex II. Succinate dehydrogenase, the enzyme in the citric acid cycle that catalyzes the formation of FADH2 from FAD is part of complex II. The electrons from FADH2 are then transferred to an Fe·S protein.
• Oxidation half-reaction:
$\ce{FADH_2 -> FAD + 2H+ + 2e-} \nonumber$
• Reduction half-reaction:
$\ce{2Fe(III) \cdot S + 2e- → 2Fe(II) \cdot S} \nonumber$
Electrons from complexes I and II are then transferred from the $\ce{Fe \cdot S}$ protein to coenzyme Q (CoQ), a mobile electron carrier that acts as the electron shuttle between complexes I or II and complex III.
Coenzyme Q is also called ubiquinone because it is ubiquitous in living systems.
• Oxidation half-reaction:
$\ce{2Fe(II) \cdot S -> 2Fe(III) \cdot S + 2e-} \nonumber$
• Reduction half-reaction:
Complexes III and IV include several iron-containing proteins known as cytochromes. The iron in these enzymes is located in substructures known as iron porphyrins (Figure $4$). Like the Fe·S centers, the characteristic feature of the cytochromes is the ability of their iron atoms to exist as either Fe(II) or Fe(III). Thus, each cytochrome in its oxidized form—Fe(III)—can accept one electron and be reduced to the Fe(II) form. This change in oxidation state is reversible, so the reduced form can donate its electron to the next cytochrome, and so on. Complex III contains cytochromes b and c, as well as Fe·S proteins, with cytochrome c acting as the electron shuttle between complex III and IV. Complex IV contains cytochromes a and a3 in an enzyme known as cytochrome oxidase. This enzyme has the ability to transfer electrons to molecular oxygen, the last electron acceptor in the chain of electron transport reactions. In this final step, water (H2O) is formed.
• Oxidation half-reaction:
$\ce{4Cyt\, a_3–Fe(II) -> 4Cyt\, a_3–Fe(III) + 4e-} \nonumber$
• Reduction half-reaction:
$\ce{O2 + 4H+ + 4e- -> 2H2O} \nonumber$
Video: Cellular Respiration (Electron Transport Chain). Cellular respiration occurs in the mitochondria and provides both animals and plants with the energy needed to power other cellular processes. This section covers the electron transport chain.NDSU Virtual Cell Animations Project animation; ror more information please see Vcell, NDSU, Animations(opens in new window) [vcell.ndsu.edu]
Oxidative Phosphorylation
Each intermediate compound in the electron transport chain is reduced by the addition of one or two electrons in one reaction and then subsequently restored to its original form by delivering the electron(s) to the next compound along the chain. The successive electron transfers result in energy production. But how is this energy used for the synthesis of ATP? The process that links ATP synthesis to the operation of the electron transport chain is referred to as oxidative phosphorylation.
Electron transport is tightly coupled to oxidative phosphorylation. The coenzymes NADH and FADH2 are oxidized by the respiratory chain only if ADP is simultaneously phosphorylated to ATP. The currently accepted model explaining how these two processes are linked is known as the chemiosmotic hypothesis, which was proposed by Peter Mitchell, resulting in Mitchell being awarded the 1978 Nobel Prize in Chemistry.
Looking again at Figure $3$, we see that as electrons are being transferred through the electron transport chain, hydrogen (H+) ions are being transported across the inner mitochondrial membrane from the matrix to the intermembrane space. The concentration of H+ is already higher in the intermembrane space than in the matrix, so energy is required to transport the additional H+ there. This energy comes from the electron transfer reactions in the electron transport chain. But how does the extreme difference in H+ concentration then lead to ATP synthesis? The buildup of H+ ions in the intermembrane space results in an H+ ion gradient that is a large energy source, like water behind a dam (because, given the opportunity, the protons will flow out of the intermembrane space and into the less concentrated matrix). Current research indicates that the flow of H+ down this concentration gradient through a fifth enzyme complex, known as ATP synthase, leads to a change in the structure of the synthase, causing the synthesis and release of ATP.
In cells that are using energy, the turnover of ATP is very high, so these cells contain high levels of ADP. They must therefore consume large quantities of oxygen continuously, so as to have the energy necessary to phosphorylate ADP to form ATP. Consider, for example, that resting skeletal muscles use about 30% of a resting adult’s oxygen consumption, but when the same muscles are working strenuously, they account for almost 90% of the total oxygen consumption of the organism.
Experiment has shown that 2.5–3 ATP molecules are formed for every molecule of NADH oxidized in the electron transport chain, and 1.5–2 ATP molecules are formed for every molecule of FADH2 oxidized. Table $1$ summarizes the theoretical maximum yield of ATP produced by the complete oxidation of 1 mol of acetyl-CoA through the sequential action of the citric acid cycle, the electron transport chain, and oxidative phosphorylation.
Table $1$: Maximum Yield of ATP from the Complete Oxidation of 1 Mol of Acetyl-CoA
Reaction Comments Yield of ATP (moles)
Isocitrate → α-ketoglutarate + CO2 produces 1 mol NADH
α-ketoglutarate → succinyl-CoA + CO2 produces 1 mol NADH
Succinyl-CoA → succinate produces 1 mol GTP +1
Succinate → fumarate produces 1 mol FADH2
Malate → oxaloacetate produces 1 mol NADH
1 FADH2 from the citric acid cycle yields 2 mol ATP +2
3 NADH from the citric acid cycle yields 3 mol ATP/NADH +9
Net yield of ATP: +12
Key Takeaways
• The acetyl group of acetyl-CoA enters the citric acid cycle. For each acetyl-CoA that enters the citric acid cycle, 2 molecules of carbon dioxide, 3 molecules of NADH, 1 molecule of ATP, and 1 molecule of FADH2 are produced.
• The reduced coenzymes (NADH and FADH2) produced by the citric acid cycle are reoxidized by the reactions of the electron transport chain. This series of reactions also produces a pH gradient across the inner mitochondrial membrane.
• The pH gradient produced by the electron transport chain drives the synthesis of ATP from ADP. For each NADH reoxidized, 2.5–3 molecules of ATP are produced; for each FADH2 reoxidized, 1.5–2 molecules of ATP are produced. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.04%3A_Stage_III_of_Catabolism.txt |
Learning Objectives
• Describe the function of glycolysis and identify its major products.
• Describe how the presence or absence of oxygen determines what happens to the pyruvate and the NADH that are produced in glycolysis.
• Determine the amount of ATP produced by the oxidation of glucose in the presence and absence of oxygen.
In stage II of catabolism, the metabolic pathway known as glycolysis converts glucose into two molecules of pyruvate (a three-carbon compound with three carbon atoms) with the corresponding production of adenosine triphosphate (ATP). The individual reactions in glycolysis were determined during the first part of the 20th century. It was the first metabolic pathway to be elucidated, in part because the participating enzymes are found in soluble form in the cell and are readily isolated and purified. The pathway is structured so that the product of one enzyme-catalyzed reaction becomes the substrate of the next. The transfer of intermediates from one enzyme to the next occurs by diffusion.
Steps in Glycolysis
The 10 reactions of glycolysis, summarized in Figures $1$ and $2$, can be divided into two phases. In the first 5 reactions—phase I—glucose is broken down into two molecules of glyceraldehyde 3-phosphate. In the last five reactions—phase II—each glyceraldehyde 3-phosphate is converted into pyruvate, and ATP is generated. Notice that all the intermediates in glycolysis are phosphorylated and contain either six or three carbon atoms.
• When glucose enters a cell, it is immediately phosphorylated to form glucose 6-phosphate, in the first reaction of phase I. The phosphate donor in this reaction is ATP, and the enzyme—which requires magnesium ions for its activity—is hexokinase. In this reaction, ATP is being used rather than being synthesized. The presence of such a reaction in a catabolic pathway that is supposed to generate energy may surprise you. However, in addition to activating the glucose molecule, this initial reaction is essentially irreversible, an added benefit that keeps the overall process moving in the right direction. Furthermore, the addition of the negatively charged phosphate group prevents the intermediates formed in glycolysis from diffusing through the cell membrane, as neutral molecules such as glucose can do.
• In the next reaction, phosphoglucose isomerase catalyzes the isomerization of glucose 6-phosphate to fructose 6-phosphate. This reaction is important because it creates a primary alcohol, which can be readily phosphorylated.
• The subsequent phosphorylation of fructose 6-phosphate to form fructose 1,6-bisphosphate is catalyzed by phosphofructokinase, which requires magnesium ions for activity. ATP is again the phosphate donor.
• Fructose 1,6-bisphosphate is enzymatically cleaved by aldolase to form two triose phosphates: dihydroxyacetone phosphate and glyceraldehyde 3-phosphate.
• Isomerization of dihydroxyacetone phosphate into a second molecule of glyceraldehyde 3-phosphate is the final step in phase I. The enzyme catalyzing this reaction is triose phosphate isomerase.
When a molecule contains two phosphate groups on different carbon atoms, the convention is to use the prefix bis. When the two phosphate groups are bonded to each other on the same carbon atom (for example, adenosine diphosphate [ADP]), the prefix is di.
In steps 4 and 5, aldolase and triose phosphate isomerase effectively convert one molecule of fructose 1,6-bisphosphate into two molecules of glyceraldehyde 3-phosphate. Thus, phase I of glycolysis requires energy in the form of two molecules of ATP and releases none of the energy stored in glucose.
In the initial step of phase II (Figure $2$), glyceraldehyde 3-phosphate is both oxidized and phosphorylated in a reaction catalyzed by glyceraldehyde-3-phosphate dehydrogenase, an enzyme that requires nicotinamide adenine dinucleotide (NAD+) as the oxidizing agent and inorganic phosphate as the phosphate donor. In the reaction, NAD+ is reduced to reduced nicotinamide adenine dinucleotide (NADH), and 1,3-bisphosphoglycerate (BPG) is formed.
• BPG has a high-energy phosphate bond (Table $1$) joining a phosphate group to C1. This phosphate group is now transferred directly to a molecule of ADP, thus forming ATP and 3-phosphoglycerate. The enzyme that catalyzes the reaction is phosphoglycerate kinase, which, like all other kinases, requires magnesium ions to function. This is the first reaction to produce ATP in the pathway. Because the ATP is formed by a direct transfer of a phosphate group from a metabolite to ADP—that is, from one substrate to another—the process is referred to as substrate-level phosphorylation, to distinguish it from the oxidative phosphorylation discussed in Section 20.4.
• In the next reaction, the phosphate group on 3-phosphoglycerate is transferred from the OH group of C3 to the OH group of C2, forming 2-phosphoglycerate in a reaction catalyzed by phosphoglyceromutase.
• A dehydration reaction, catalyzed by enolase, forms phosphoenolpyruvate (PEP), another compound possessing a high-energy phosphate group.
• The final step is irreversible and is the second reaction in which substrate-level phosphorylation occurs. The phosphate group of PEP is transferred to ADP, with one molecule of ATP being produced per molecule of PEP. The reaction is catalyzed by pyruvate kinase, which requires both magnesium and potassium ions to be active.
Table $1$: Maximum Yield of ATP from the Complete Oxidation of 1 Mol of Glucose
Reaction Comments Yield of ATP (moles)
glucose → glucose 6-phosphate consumes 1 mol ATP −1
fructose 6-phosphate → fructose 1,6-bisphosphate consumes 1 mol ATP −1
glyceraldehyde 3-phosphate → BPG produces 2 mol of cytoplasmic NADH
BPG → 3-phosphoglycerate produces 2 mol ATP +2
phosphoenolpyruvate → pyruvate produces 2 mol ATP +2
pyruvate → acetyl-CoA + CO2 produces 2 mol NADH
isocitrate → α-ketoglutarate + CO2 produces 2 mol NADH
α-ketoglutarate → succinyl-CoA + CO2 produces 2 mol NADH
succinyl-CoA → succinate produces 2 mol GTP +2
succinate → fumarate produces 2 mol FADH2
malate → oxaloacetate produces 2 mol NADH
2 cytoplasmic NADH from glycolysis yields 2–3 mol ATP per NADH (depending on tissue) +4 to +6
2 NADH from the oxidation of pyruvate yields 3 mol ATP per NADH +6
2 FADH2 from the citric acid cycle yields 2 ATP per FADH2 +4
3 NADH from the citric acid cycle yields 3 ATP per NADH +18
Net yield of ATP: +36 to +38
In phase II, two molecules of glyceraldehyde 3-phosphate are converted to two molecules of pyruvate, along with the production of four molecules of ATP and two molecules of NADH.
To Your Health: Diabetes
Although medical science has made significant progress against diabetes , it continues to be a major health threat. Some of the serious complications of diabetes are as follows:
• It is the leading cause of lower limb amputations in the United States.
• It is the leading cause of blindness in adults over age 20.
• It is the leading cause of kidney failure.
• It increases the risk of having a heart attack or stroke by two to four times.
Because a person with diabetes is unable to use glucose properly, excessive quantities accumulate in the blood and the urine. Other characteristic symptoms are constant hunger, weight loss, extreme thirst, and frequent urination because the kidneys excrete large amounts of water in an attempt to remove excess sugar from the blood.
There are two types of diabetes. In immune-mediated diabetes, insufficient amounts of insulin are produced. This type of diabetes develops early in life and is also known as Type 1 diabetes, as well as insulin-dependent or juvenile-onset diabetes. Symptoms are rapidly reversed by the administration of insulin, and Type 1 diabetics can lead active lives provided they receive insulin as needed. Because insulin is a protein that is readily digested in the small intestine, it cannot be taken orally and must be injected at least once a day.
In Type 1 diabetes, insulin-producing cells of the pancreas are destroyed by the body’s immune system. Researchers are still trying to find out why. Meanwhile, they have developed a simple blood test capable of predicting who will develop Type 1 diabetes several years before the disease becomes apparent. The blood test reveals the presence of antibodies that destroy the body’s insulin-producing cells.
Type 2 diabetes, also known as noninsulin-dependent or adult-onset diabetes, is by far the more common, representing about 95% of diagnosed diabetic cases. (This translates to about 16 million Americans.) Type 2 diabetics usually produce sufficient amounts of insulin, but either the insulin-producing cells in the pancreas do not release enough of it, or it is not used properly because of defective insulin receptors or a lack of insulin receptors on the target cells. In many of these people, the disease can be controlled with a combination of diet and exercise alone. For some people who are overweight, losing weight is sufficient to bring their blood sugar level into the normal range, after which medication is not required if they exercise regularly and eat wisely.
Those who require medication may use oral antidiabetic drugs that stimulate the islet cells to secrete insulin. First-generation antidiabetic drugs stimulated the release of insulin. Newer second-generation drugs, such as glyburide, do as well, but they also increase the sensitivity of cell receptors to insulin. Some individuals with Type 2 diabetes do not produce enough insulin and thus do not respond to these oral medications; they must use insulin. In both Type 1 and Type 2 diabetes, the blood sugar level must be carefully monitored and adjustments made in diet or medication to keep the level as normal as possible (70–120 mg/dL).
Metabolism of Pyruvate
The presence or absence of oxygen determines the fates of the pyruvate and the NADH produced in glycolysis. When plenty of oxygen is available, pyruvate is completely oxidized to carbon dioxide, with the release of much greater amounts of ATP through the combined actions of the citric acid cycle, the electron transport chain, and oxidative phosphorylation. However, in the absence of oxygen (that is, under anaerobic conditions), the fate of pyruvate is different in different organisms. In vertebrates, pyruvate is converted to lactate, while other organisms, such as yeast, convert pyruvate to ethanol and carbon dioxide. These possible fates of pyruvate are summarized in Figure $2$. The conversion to lactate or ethanol under anaerobic conditions allows for the reoxidation of NADH to NAD+ in the absence of oxygen.
ATP Yield from Glycolysis
The net energy yield from anaerobic glucose metabolism can readily be calculated in moles of ATP. In the initial phosphorylation of glucose (step 1), 1 mol of ATP is expended, along with another in the phosphorylation of fructose 6-phosphate (step 3). In step 7, 2 mol of BPG (recall that 2 mol of 1,3-BPG are formed for each mole of glucose) are converted to 2 mol of 3-phosphoglycerate, and 2 mol of ATP are produced. In step 10, 2 mol of pyruvate and 2 mol of ATP are formed per mole of glucose.
For every mole of glucose degraded, 2 mol of ATP are initially consumed and 4 mol of ATP are ultimately produced. The net production of ATP is thus 2 mol for each mole of glucose converted to lactate or ethanol. If 7.4 kcal of energy is conserved per mole of ATP produced, and the total amount of energy that can theoretically be obtained from the complete oxidation of 1 mol of glucose is 670 kcal (as stated in the chapter introduction), the energy conserved in the anaerobic catabolism of glucose to two molecules of lactate (or ethanol) is as follows:
$\mathrm{\dfrac{2\times 7.4\: kcal}{670\: kcal}\times100=2.2\%} \nonumber$
Thus anaerobic cells extract only a very small fraction of the total energy of the glucose molecule.
Contrast this result with the amount of energy obtained when glucose is completely oxidized to carbon dioxide and water through glycolysis, the citric acid cycle, the electron transport chain, and oxidative phosphorylation as summarized in Table $1$. Note the indication in the table that a variable amount of ATP is synthesized, depending on the tissue, from the NADH formed in the cytoplasm during glycolysis. This is because NADH is not transported into the inner mitochondrial membrane where the enzymes for the electron transport chain are located. Instead, brain and muscle cells use a transport mechanism that passes electrons from the cytoplasmic NADH through the membrane to flavin adenine dinucleotide (FAD) molecules inside the mitochondria, forming reduced flavin adenine dinucleotide (FADH2), which then feeds the electrons into the electron transport chain. This route lowers the yield of ATP to 1.5–2 molecules of ATP, rather than the usual 2.5–3 molecules. A more efficient transport system is found in liver, heart, and kidney cells where the formation of one cytoplasmic NADH molecule results in the formation of one mitochondrial NADH molecule, which leads to the formation of 2.5–3 molecules of ATP.The total amount of energy conserved in the aerobic catabolism of glucose in the liver is as follows:
$\mathrm{\dfrac{38\times7.4\: kcal}{670\: kcal}\times100=42\%} \nonumber$
Conservation of 42% of the total energy released compares favorably with the efficiency of any machine. In comparison, automobiles are only about 20%–25% efficient in using the energy released by the combustion of gasoline.
As indicated earlier, the 58% of released energy that is not conserved enters the surroundings (that is, the cell) as heat that helps to maintain body temperature. If we are exercising strenuously and our metabolism speeds up to provide the energy needed for muscle contraction, more heat is produced. We begin to perspire to dissipate some of that heat. As the perspiration evaporates, the excess heat is carried away from the body by the departing water vapor.
Summary
• The monosaccharide glucose is broken down through a series of enzyme-catalyzed reactions known as glycolysis.
• For each molecule of glucose that is broken down, two molecules of pyruvate, two molecules of ATP, and two molecules of NADH are produced.
• In the absence of oxygen, pyruvate is converted to lactate, and NADH is reoxidized to NAD+. In the presence of oxygen, pyruvate is converted to acetyl-CoA and then enters the citric acid cycle.
• More ATP can be formed from the breakdown of glucose when oxygen is present. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.05%3A_Stage_II_of_Carbohydrate_Catabolism.txt |
Learning Objectives
• To describe the reactions needed to completely oxidize a fatty acid to carbon dioxide and water.
Like glucose, the fatty acids released in the digestion of triglycerides and other lipids are broken down in a series of sequential reactions accompanied by the gradual release of usable energy. Some of these reactions are oxidative and require nicotinamide adenine dinucleotide (NAD+) and flavin adenine dinucleotide (FAD). The enzymes that participate in fatty acid catabolism are located in the mitochondria, along with the enzymes of the citric acid cycle, the electron transport chain, and oxidative phosphorylation. This localization of enzymes in the mitochondria is of the utmost importance because it facilitates efficient utilization of energy stored in fatty acids and other molecules.
Fatty acid oxidation is initiated on the outer mitochondrial membrane. There the fatty acids, which like carbohydrates are relatively inert, must first be activated by conversion to an energy-rich fatty acid derivative of coenzyme A called fatty acyl-coenzyme A (CoA). The activation is catalyzed by acyl-CoA synthetase. For each molecule of fatty acid activated, one molecule of coenzyme A and one molecule of adenosine triphosphate (ATP) are used, equaling a net utilization of the two high-energy bonds in one ATP molecule (which is therefore converted to adenosine monophosphate [AMP] rather than adenosine diphosphate [ADP]):
The fatty acyl-CoA diffuses to the inner mitochondrial membrane, where it combines with a carrier molecule known as carnitine in a reaction catalyzed by carnitine acyltransferase. The acyl-carnitine derivative is transported into the mitochondrial matrix and converted back to the fatty acyl-CoA.
Steps in the β-Oxidation of Fatty Acids
Further oxidation of the fatty acyl-CoA occurs in the mitochondrial matrix via a sequence of four reactions known collectively as β-oxidation because the β-carbon undergoes successive oxidations in the progressive removal of two carbon atoms from the carboxyl end of the fatty acyl-CoA (Figure $1$).
The first step in the catabolism of fatty acids is the formation of an alkene in an oxidation reaction catalyzed by acyl-CoA dehydrogenase. In this reaction, the coenzyme FAD accepts two hydrogen atoms from the acyl-CoA, one from the α-carbon and one from the β-carbon, forming reduced flavin adenine dinucleotide (FADH2).
Next, the trans-alkene is hydrated to form a secondary alcohol in a reaction catalyzed by enoyl-CoA hydratase. The enzyme forms only the L-isomer.
The secondary alcohol is then oxidized to a ketone by β-hydroxyacyl-CoA dehydrogenase, with NAD+ acting as the oxidizing agent. The reoxidation of each molecule of NADH to NAD+ by the electron transport chain furnishes 2.5–3 molecules of ATP.
The final reaction is cleavage of the β-ketoacyl-CoA by a molecule of coenzyme A. The products are acetyl-CoA and a fatty acyl-CoA that has been shortened by two carbon atoms. The reaction is catalyzed by thiolase.
The shortened fatty acyl-CoA is then degraded by repetitions of these four steps, each time releasing a molecule of acetyl-CoA. The overall equation for the β-oxidation of palmitoyl-CoA (16 carbon atoms) is as follows:
Because each shortened fatty acyl-CoA cycles back to the beginning of the pathway, β-oxidation is sometimes referred to as the fatty acid spiral.
The fate of the acetyl-CoA obtained from fatty acid oxidation depends on the needs of an organism. It may enter the citric acid cycle and be oxidized to produce energy, it may be used for the formation of water-soluble derivatives known as ketone bodies, or it may serve as the starting material for the synthesis of fatty acids. For more information about the citric acid cycle, see Section 20.4.
Looking Closer: Ketone Bodies
In the liver, most of the acetyl-CoA obtained from fatty acid oxidation is oxidized by the citric acid cycle. However, some of the acetyl-CoA is used to synthesize a group of compounds known as ketone bodies: acetoacetate, β-hydroxybutyrate, and acetone. Two acetyl-CoA molecules combine, in a reversal of the final step of β-oxidation, to produce acetoacetyl-CoA. The acetoacetyl-CoA reacts with another molecule of acetyl-CoA and water to form β-hydroxy-β-methylglutaryl-CoA, which is then cleaved to acetoacetate and acetyl-CoA. Most of the acetoacetate is reduced to β-hydroxybutyrate, while a small amount is decarboxylated to carbon dioxide and acetone.
The acetoacetate and β-hydroxybutyrate synthesized by the liver are released into the blood for use as a metabolic fuel (to be converted back to acetyl-CoA) by other tissues, particularly the kidney and the heart. Thus, during prolonged starvation, ketone bodies provide about 70% of the energy requirements of the brain. Under normal conditions, the kidneys excrete about 20 mg of ketone bodies each day, and the blood levels are maintained at about 1 mg of ketone bodies per 100 mL of blood.
In starvation, diabetes mellitus, and certain other physiological conditions in which cells do not receive sufficient amounts of carbohydrate, the rate of fatty acid oxidation increases to provide energy. This leads to an increase in the concentration of acetyl-CoA. The increased acetyl-CoA cannot be oxidized by the citric acid cycle because of a decrease in the concentration of oxaloacetate, which is diverted to glucose synthesis. In response, the rate of ketone body formation in the liver increases further, to a level much higher than can be used by other tissues. The excess ketone bodies accumulate in the blood and the urine, a condition referred to as ketosis. When the acetone in the blood reaches the lungs, its volatility causes it to be expelled in the breath. The sweet smell of acetone, a characteristic of ketosis, is frequently noticed on the breath of severely diabetic patients.
Because two of the three kinds of ketone bodies are weak acids, their presence in the blood in excessive amounts overwhelms the blood buffers and causes a marked decrease in blood pH (to 6.9 from a normal value of 7.4). This decrease in pH leads to a serious condition known as acidosis. One of the effects of acidosis is a decrease in the ability of hemoglobin to transport oxygen in the blood. In moderate to severe acidosis, breathing becomes labored and very painful. The body also loses fluids and becomes dehydrated as the kidneys attempt to get rid of the acids by eliminating large quantities of water. The lowered oxygen supply and dehydration lead to depression; even mild acidosis leads to lethargy, loss of appetite, and a generally run-down feeling. Untreated patients may go into a coma. At that point, prompt treatment is necessary if the person’s life is to be saved.
ATP Yield from Fatty Acid Oxidation
The amount of ATP obtained from fatty acid oxidation depends on the size of the fatty acid being oxidized. For our purposes here. we’ll study palmitic acid, a saturated fatty acid with 16 carbon atoms, as a typical fatty acid in the human diet. Calculating its energy yield provides a model for determining the ATP yield of all other fatty acids.
The breakdown by an organism of 1 mol of palmitic acid requires 1 mol of ATP (for activation) and forms 8 mol of acetyl-CoA. Recall from Table 20.4.1 that each mole of acetyl-CoA metabolized by the citric acid cycle yields 10 mol of ATP. The complete degradation of 1 mol of palmitic acid requires the β-oxidation reactions to be repeated seven times. Thus, 7 mol of NADH and 7 mol of FADH2 are produced. Reoxidation of these compounds through respiration yields 2.5–3 and 1.5–2 mol of ATP, respectively. The energy calculations can be summarized as follows:
ATP Yield from Fatty Acid Oxidation
1 mol of ATP is split to AMP and 2Pi −2 ATP
8 mol of acetyl-CoA formed (8 × 12) 96 ATP
7 mol of FADH2 formed (7 × 2) 14 ATP
7 mol of NADH formed (7 × 3) 21 ATP
Total 129 ATP
The number of times β-oxidation is repeated for a fatty acid containing n carbon atoms is n/2 – 1 because the final turn yields two acetyl-CoA molecules.
The combustion of 1 mol of palmitic acid releases a considerable amount of energy:
$C_{16}H_{32}O_2 + 23O_2 → 16CO_2 + 16H_2O + 2,340\; kcal \nonumber$
The percentage of this energy that is conserved by the cell in the form of ATP is as follows:
$\mathrm{\dfrac{energy\: conserved}{total\: energy\: available}\times100=\dfrac{(129\: ATP)(7.4\: kcal/ATP)}{2,340\: kcal}\times100=41\%} \nonumber$
The efficiency of fatty acid metabolism is comparable to that of carbohydrate metabolism, which we calculated to be 42%. For more information about the efficiency of fatty acid metabolism, see II of Carbohydrate Catabolism" data-cke-saved-href="/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20:_Energy_Metabolism/20.05:_Stage_II_of_Carbohydrate_Catabolism" href="/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20:_Energy_Metabolism/20.05:_Stage_II_of_Carbohydrate_Catabolism" data-quail-id="91">Section 20.5.
The oxidation of fatty acids produces large quantities of water. This water, which sustains migratory birds and animals (such as the camel) for long periods of time.
Summary
• Fatty acids, obtained from the breakdown of triglycerides and other lipids, are oxidized through a series of reactions known as β-oxidation.
• In each round of β-oxidation, 1 molecule of acetyl-CoA, 1 molecule of NADH, and 1 molecule of FADH2 are produced.
• The acetyl-CoA, NADH, and FADH2 are used in the citric acid cycle, the electron transport chain, and oxidative phosphorylation to produce ATP. | textbooks/chem/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/20%3A_Energy_Metabolism/20.06%3A_Stage_II_of_Lipid_Catabolism.txt |
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