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The structure of ionic solids is determined by how the cations and anions can pack together. Generally, one of the ions adopts a standard packing structure, like the metal atoms in a metallic solid. The counterions then fit into the holes or interstitial spaces among these ions. It is pretty common for the anions to form a close-packed structure, and for cations to find room in the resulting holes, but sometimes it is the other way around.
Exercise \(1\)
Match each isoelectronic ion with the correct picture of its radius.
Exercise \(2\)
Why might anions more commonly pack into a close-packed structure, rather than the cations?
Answer
Generally, but not always, anions are bigger than cations, so cations can pack efficiently into the holes between the anions.
The holes between the atoms have particular coordination numbers and geometries. There are many possible holes of different shapes where a counterion can find room. However, some of these geometries are very common.
For example, in the center of a simple cube, there is room for an additional atom. This atom is described as sitting in a cubic hole.
The above drawing emphasizes the relationship between the central atom and the atoms that form the corners of the cube. The central atom is in a cubic coordination geometry. Alternatively, we could describe the coordination number of the central atom. Instead of describing the shape formed by the surrounding atoms, we simply count the number of near neighbors. In this case, the coordination number of the central atom is 8. We can think of the central atom as sharing ionic bonding with its eight near neighbors.
An atom in a cubic hole might be viewed more easily if we draw lines between the various atoms at the corners of the cube, however. In that way,we can see more clearly the cubic shape of the cage in which the central atom is sitting.
Another common interstitial space in ionic solids is an octahedral hole. An octahedral hole forms in between two close-packed layers. Because the atoms in the layers are packed more tightly than in a simple cube, this octahedral hole is generally a little smaller than a cubic hole.
A third, common type of interstitial space is a tetrahedral hole. Tetrahedral holes, like octahedral holes, are found between two close-packed layers.
Exercise \(3\)
In the following structures, the anions are represented in red and the cations are represented in blue. For each structure,
1. identify the type of unit cell that is visible
2. identify the type of hole occupied by the counter ion
3. identify the fraction of those holes that are occupied
4. identify the number of anions and cations in the unit cell
5. state the empirical formula (the lowest possible ratio of atoms in the material)
i. The ions are formed from cesium and chlorine
ii. The ions are formed from sodium and chlorine.
iii. The cations are formed from calcium and the anions are formed from fluorine.
iv. The cations are formed from zinc and the anions are formed from sulfur.
v. The cations are formed from zinc and the anions are formed from sulfur.
Additional Information on solid structures:
Liverpool Solid State
Visualization of solid state structures: unit cells, etc.
Davidson College Crystals
Visualization of solid state structures: unit cells, etc.
Oxford University Solid Structures
Visualization of solid state structures: unit cells, etc. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/03%3A_Ionic_Compounds/3.05%3A_Structures_of_Ionic_Solids.txt |
Exercise $1$
Gallium arsenide is used in the electronics industry as a semiconductor. You can start by thinking of gallium arsenide as arsenide ions forming a face-centered cubic unit cell.
a) Draw that face-centered cubic unit cell.
b) Place the gallium ions in half of the tetrahedral holes.
c) Shade one kind of atom in and add a legend to your diagram.
d) Calculate the total number of arsenic atoms in the unit cell.
e) Calculate the total number of gallium atoms in the unit cell.
f) What is the formula of gallium arsenide, based on that ratio?
g) What would be the charge on each ion, based on its location in the periodic table?
h) The melting point of gallium arsenide would be (higher / lower) than CsI.
i) In reality, both ions have much less charge than expected. Why isn’t gallium arsenide fully ionic?
Answer a), b), c
Answer d
There are eight arsenic atoms in the corner; each is shared by eight neighbouring cubes. There are six atoms on the faces; each is shared by two neighbouring cubes.
# atoms = $8 (\frac{1}{8}) + 6 ( \frac{1}{2}) = 1+ 3 = 4$ arsenic atoms
Answer e
Each gallium atom is entirely within the cell. There are 4 gallium atoms.
Answer f
The formula is just the ratio of each atom in the material. Since the 4:4 ratio is the same thing as a 1:1 ratio, the formula is just GaAs.
Answer g
Arsenic is to the right of gallium in the periodic table, so arsenic must be more electronegative; it must be the anion, and gallium the cation.
Arsenic is three atoms from the right edge of the periodic table; as an ion, a 3- charge makes sense. To balance the charge, gallium must have a 3+ charge.
Answer h
Cesium is larger than gallium, and iodine is larger than arsenic, so in terms of internuclear distance, there would be less attraction between the cesium and iodide ions than between the (closer) gallium and arsenide ions. Also, if the charge on gallium and arsenide are 3+ and 3-, respectively, and the charges on cesium and iodide are only 1+ and 1-, respectively, then there should be much greater attraction between the gallium and arsenide ions.
The melting point of gallium arsenide (mp = 1,238 oC) should be higher than that of cesium iodide (mp = 632 oC).
Answer i
The atoms are so close together in the periodic table that it isn't likely that arsenic should completely steal three electrons away from gallium; the difference in electronegativity isn't great enough.
Exercise $2$
Electronic devices rely on an interface between copper, a conductor of electricity, and silicon, a semiconductor. However, copper atoms tend to diffuse through silicon, disrupting its semiconductor properties. A barrier layer is therefore used in between, usually magnesium oxide or titanium nitride. Let's take a look at the structure of these materials.
1. Based on its position in the periodic table, what is the charge on oxide ion in magnesium oxide?
2. Based on its position in the periodic table, what is the charge on magnesium ion in magnesium oxide?
3. Therefore, to balance charge, what is the formula of magnesium oxide?
Magnesium oxide has a rock salt structure. Oxide ions form a face-centered cubic cell, and magnesium ions occupy the octahedral holes.
d) Add the atoms to the figure according to the legend.
e) Calculate how many atoms of each kind there are in the cell.
f) What is the electronic configuration of a Mg atom (you can use noble gas abbreviation)?
g) What is the electronic configuration of a Mg ion in magnesium oxide?
h) Which is bigger: magnesium, or magnesium ion? Why?
i) What is the electronic configuration of an oxide ion in magnesium oxide?
j) Which is bigger: oxide ion, or magnesium ion? Why?
Alternatively, sometimes titanium nitride (TiN) is used as the barrier layer instead of magnesium oxide.
k) Based on its position in the periodic table, what is the charge on nitride ion in titanium nitride?
l) Therefore, what is the charge on titanium in titanium nitride?
Titanium nitride, like magnesum oxide, adopts a rock salt structure. It has a melting point that is a little higher than that of magnesium oxide (2 930 oC vs. 2 852 oC).
m) Explain, in terms of lattice energy, why the melting point of titanium nitride is a little higher than the melting point of magnesium oxide.
In the structure of silicon, imagine a face-centered cubic cell, but with extra silicon atoms that occupy half of the tetrahedral holes.
n) Add the atoms to the figure to show the structure of silicon.
o) What is the coordination number of a silicon in a tetrahedral hole?
Copper usually forms a face-centered cubic cell. However, researchers found that body-centered cubic copper can be formed during laser ablation at room temperature (Wu, F; Narayan, J. Cryst. Growth. Des. 2013, 13, 5018-5024.). When grown on a magnesium oxide surface, this body centered copper forms a more durable layer than the normal FCC form.
p) Add the atoms to the figure to form body-centered cubic copper.
q) What is the electronic configuration of copper (you can use noble gas configuration)?
r) Explain why copper is a good conductor of electricity. Add a picture to support your explanation.
s) Why isn’t silicon as good a conductor of electricity as copper?
Answer a
O2-
Answer b
Mg2+
Answer c
MgO
Answer d
Answer e
O: 8(1/8) corners + 6(1/2) faces = 4 atoms; Mg: 12(1/4) edges + 1 center = 4 atoms; MgO
Answer f
[Ne]3s2
Answer g
[Ne] or [He]2s22p6
Answer h
magnesium; the positive charge on the ion causes the remaining electrons to contract
Answer i
[Ne] or [He]2s22p6
Answer j
oxide; they have the same electron configuration, but magnesium has more protons in its nucleus to attract the electrons.
Answer k
N3-
Answer l
Ti3+
Answer m
The higher charge on the ions in TiN would lead to a higher force of attraction between the ions, and thus a higher lattice energy would be required to remove ions from the lattice. That attraction thus lowers the mobility of the ions, so the compound does not melt as easily. However, the fact that both Ti3+ and N3- are slightly larger than Mg2+ and O2- helps explain why the melting points are not extremely different; Coulomb's Law says that the increased attraction of the higher charges should be partly offset by the greater size of the ions.
Answer n
Answer o
CN = 4
Answer p
Answer q
[Ar]4s13d10 (remember, this is one of those cases in which the 3d level is slightly below the 4s level)
Answer r
Copper's low electronegativity allows it to form an electron sea model of metallic bonding. Copper ions are surrounded by the loosely-held electrons that are allowed to flow through the material.
Answer s
Silicon is too electronegative to form an electron sea structure like a metal. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/03%3A_Ionic_Compounds/3.06%3A_Application_Problems.txt |
Sometimes, it is useful to be able to recognize the name of a compound and know how to translate it into the structure. You might need to go the other direction, too; if you see a formula -- the ratio of atoms in a compound -- you may want to be able to name the compound.
Take a common ionic compound such as table salt or rock salt, which has the formula NaCl. To name this compound systematically, we just name the two ions of which it is composed: sodium and chlorine. We leave the name of the cation exactly as it is in the periodic table, but we change the ending of the anion's name to -ide. We name the cation first and the anion second.
• Name the cation first and the anion second.
• Change the ending of the anion to -ide.
In this case, sodium is the cation and chlorine is the anion, so the name for NaCl is sodium chloride.\
How did we know which one was the cation and which was the anion? For one thing, when we write the formula of a simple ionic compound, we usually writethe elements in the same order: cation before anion. If you didn't know that, or if for some reason the formula did not appear in that order, you should still be able to work it out based on the layout of the periodic table. The key is usually found in the elements that form the first two columns of the periodic table. These are the alkali and alkaline earth metals. They are very commonly found in ionic compounds, and they are reliably found as cations in those compounds.
The alkali metals are always found as 1+ or monovalent cations, such as Na+, K+, or Li+. The reason, of course, is that these metals all have one electron in a shell that has been newly started; that shell is far from the nucleus, so the electron is ot held very tightly and it is lost pretty easily.
A similar thing is true for the alkaline earth metals such as magnesium or calcium. These metals have two electrons in the outermost shell, and there still are not enough protons in that distant nucleus to hold those electrons tightly. As a result, both of these electrons are easily lost, and so magnesium is most often found as Mg2+ and calcium is usually found as Ca2+.
Once you have identified the cation, you can quickly decide that the other atom in the formula must be the anion.
Exercise \(1\)
Provide formulae for the following names.
a) lithium fluoride b) sodium iodide c) potassium bromide d) magnesium chloride
e) calcium oxide f) beryllium sulfide g) sodium oxide h) lithium nitride
Answer a:
LiF
Answer b:
NaI
Answer c:
KBr
Answer d:
MgCl2
Answer e:
CaO
Answer f:
BeS
Answer g:
Na2O
Answer h:
Li3N
A similar line of reasoning works with atoms on the other side of the periodic table. Halogens, such as fluorine and chlorine, are frequently found as halide ions, such as fluoride, F-, and chloride, Cl-. However, the halogens and other elements toward the right hand side of the periodic table display a little more versatility that the alkali and alkaline earth metals. Consequently, if you spot chlorine or oxygen in a structure, you can't always be sure it is acting as a simple anion.
For example, there are host of compounds that are "polyatomic ions", clusters of atoms bound tightly together that have an overall charge. Many of these ions contain oxygen. Although oxygen does exist as a simple oxide anion, O2-, that isn't the case in these examples. Because the oxoanions are very common, they have their own names. That lets us distinguish chloride, Cl-, and oxide, O2-, from chlorate, ClO3-. Some of the most common oxoanions are found in the table below.
Most Common Oxoanions
series lowest # oxygens highest # oxygens
chlorine hypochlorite, ClO- chlorite, ClO2- chlorate, ClO3- perchlorate, ClO4-
sulfur sulfite, SO32- sulfate, SO42- persulfate, SO52-
phosphorus hypophosphite, H2PO2- phosphite, HPO32- phosphate, PO43-
nitrogen nitrite, NO2- nitrate, NO3-
All of these anions have oxygen atoms bound to a central atom of another kind. For example, in chlorate, four oxygen atoms are attached to a central chlorine atom. (The only exception, and it probably isn't important at this point, is persulfate; it has four oxygen atoms attached to a central sulfur atom, and the fifth oxygen hangs from one of the other oxygen atoms.) The names of each of these ions seem a little random until you consider them together. Then, it becomes clear that the names offer clues to the structure. The ending -ate always means more oxygens than -ite, although neither one corresponds to a specific number of oxygen atoms. The prefix per- always means an extra oxygen (beyond what we see in the -ate ion), whereas the prefix hypo- means even fewer oxygens than in the -ite ion.
However, none of those names would signal to us that hydrogen atoms are also attached to the central atom, in addition to the oxygens, but that's what we have in phosphite and hypophosphite.
In addition to these examples, there are a few carbon-based oxoanions that are pretty common: carbonate (CO32-), formate (HCO2-) and acetate (CH3CO2-). Like some of the phosphorus examples, the last two are rather complicated, with hydrogens attached to the central carbons as well as the oxygens seen in all the other cases.
There are also a few polyatomic cations, but they aren't as common as these anions. The most prevalent one in ammonium, NH4+.
Exercise \(2\)
Provide formulae for the following names.
a) sodium nitrate b) potassium perchlorate c) lithium sulfate
d) ammonium phosphate e) magnesium carbonate f) calcium hypochlorite
g) sodium acetate h) beryllium chlorate
Answer a:
NaNO3
Answer b:
KClO4
Answer c:
Li2SO4
Answer d:
(NH4)3PO4
Answer e:
MgCO3
Answer f:
Ca(OCl)2
Answer g:
NaO2CCH3
Answer h:
Be(ClO3)2
Towards the middle of the periodic table, it becomes a lot harder to predict the charges on cations. It's not like in those first few columns, where if something is from the first column, its charge is 1+, and if it is from the second column, its charge is 2+. What if a metal comes from the seventh column in the periodic table? Is it guaranteed to be a 7+ cation? Not exactly, because that is an awful lot of electrons to lose. In the middle portion of the periodic table, the transition metals, charges can vary from one case to another, and the same metal could be found with different charges in different compounds. One of the familiar examples if iron, which can most commonly be found as a ferric ion, Fe3+, or as a ferrous ion, Fe2+ (although other charges are also possible).
To help sort that problem, the charge on these cations is often given in the name. For example, we might refer to iron (III) chloride, FeCl3, vs. iron (II) chloride, FeCl2 (in addition to the common names ferric chloride and ferrous chloride, respectively). The Roman numeral tells you about the positive charge on the cation.
Arabic numeral to Roman numeral Conversion Table
Arabic numeral Roman numeral
1 I
2 II
3 III
4 IV
5 V
6 VI
7 VII
8 VIII
Exercise \(3\)
Provide formulae for the following names.
a) tantalum (V) chloride b) chromium (III) fluoride c) lead (II) bromide
d) mercury (II) chloride e) tungsten (VI) oxide f) molybdenum (IV) sulfide
Answer a:
TaCl5
Answer b:
CrF3
Answer c:
PbBr2
Answer d:
HgCl2
Answer e:
WO3
Answer f:
MoS2
Exercise \(4\)
Provide names for the following formulae.
a) LiCl b) Na2S c) LiO2CCH3 d) Mg(NO2)2 e) Na2(SO3)2
f) PbSO4 g) OsO4 h) Pb(NO3)4
Answer a:
lithium chloride
Answer b:
sodium sulfide
Answer c:
lithium acetate
Answer d:
magnesium nitrite
Answer e:
sodium sulfite
Answer f:
lead (II) sulfate
Answer g:
osmium (VIII) oxide
Answer h:
lead (IV) nitrate | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/03%3A_Ionic_Compounds/3.07%3A_Naming_Ionic_Compounds.txt |
Exercise 3.1.1
1. Ta (tantalum) is lower in the periodic table than V (vanadium)
2. Hg (mercury) is lower in the periodic table than Zn (zinc)
3. Si (silicon) is to the left in the periodic table compared to S (sulfur)
4. W (tungsten) is to the left and lower in the periodic table compared to Cu (copper)
Exercise 3.1.2
1. [He]2s22px22py12pz1
2. [He]2s22px22py22pz2
3. [Ar]4s2
4. [Ar]
5. [Ne]3s22px1
6. [Ne]
7. [He]2s22px12py12pz1
8. [He]2s22px22py22pz2
i) [Ne]2s22px22py22pz1
j) [Ne]2s22px22py22pz2
Exercise 3.1.3
1. 0.076 nm; smaller
2. 0.140 nm; larger
3. 0.167 nm; smaller
4. 0.144; larger
Exercise 3.1.4
1. 0.154; larger
2. 0.071; smaller
3. 0.128; larger
4. 0.133; smaller
Exercise 3.2.1
1. 1:1 K:Cl or KCl
2. 1:3 Fe:Cl or FeCl3
3. 1:6 Mo:Cl or MoCl6
4. 1:4 Zr:Cl or ZrCl4
Exercise 3.2.2
1. Li:O 2:1 or Li2O
2. Fe:O 2:3 or Fe2O3
3. Cr:O 1:3 or CrO3
4. Ti:O 1:2 or TiO2
Exercise 3.2.3
1. Li:N 3:1 or Li3N
2. Ta:N 1:1 or TaN
3. W:N 1:2 or WN2
4. Co:N 3:2 or Co3N2
Exercise 3.2.4
1. Li4WO4 because tungstate anion would be 4-
2. Li4V4O12 because the tetravanadate ion would be 4-
3. Li2Mo4O14 because the tetramolybdate ion would be 2-
4. Li2Cr(OH)6Mo6O18 because the complex polyoxoanion would be 2-
Exercise 3.3.1
1. KCl would have the lowest melting point. It would be easier to melt than LiCl because the smaller Li+ ion would more strongly attract the counterion, owing to the smaller distance separating the opposite charges.
2. NaBr would have a lower melting point than NaF.
3. CaO would have a lower melting point than BeO.
4. KBr would have a lower melting point than LiF.
Exercise 3.3.2
The answer is (a). The sum of the cation and anion radius is longer; therefore the distance betwwen the two is longer, so the force of attraction will be weaker.
1. CaCl2
2. Na2O
3. CaO
Exercise 3.3.4
1. A little lower since K+ is a little larger than Na+ (actual= 770°C)
2. A little higher since F- is a little smaller than Cl-. (actual =930°C)
3. A lot higher since the anion charge is 2x higher. (actual =1275°C (sublimes but doesn’t melt))
4. A whole lot higher since cation and anion charge are 2x higher. (actual = 2000+ °C (decomposes above this temp))
Exercise 3.4.1
1. 25 waters.
2. 2 units (2 anions and 2 cations).
3. Half the water might only dissolve half the salt: 1 unit.
4. Four times the water may dissolve four times the salt: 4 units.
Exercise 3.4.2
1. LiFshould be more soluble.
2. KF should be more soluble (but keep reading).
3. LiF should be more soluble.
Exercise 3.4.3
The MgO contain more highly charged ions (Mg2+ and O2-) than LiF (Li+ and F-) and so it is more difficult to separate the ions from their solid state.
Exercise 3.4.4
The ions interact with the water via electrostatic interactions, too. The same distance factor that allows small ions to attract each other more strongly also allows small ions to interact more strongly with the water.
Exercise 3.4.5
1. The lattice energy of an ionic solid is a measure of the strength of bonds in that ionic compound.
2. Charge of ion (directly related to lattice energy); Radius of each ion (inversely related to lattice energy)
3. Stronger lattice energy results in a stronger bond. The stronger the bond, the more energy required to separate ions.
4. Stronger lattice energy results higher mp or bp.
5. Stronger lattice energy results in less soluble crystal lattice.
1. NaCl NaBr
2. KF CaF2
3. MgO Na2O
4. KF CsCl
5. RbBr CaCl2
Exercise 3.5.2
Generally, but not always, anions are bigger than cations, so cations can pack efficiently into the holes between the anions.
Exercise 3.5.3
i. a) simple cubic; b) cubic hole; c) coordination number = 8; d) all occupied; e) 8 x 1/8 Cl and 1 Cs (or vice versa, depending on how you define a unit cell); f) CsCl
ii. a) face centered cubic; b) octahedral hole; c) coordination number = 6; d) all occupied; e) 6 x 1/2 plus 8 x 1/8 = 4 Cl and 1 plus 12 x 1/4 = 4 Na (or vice versa, depending on how you define a unit cell); f) NaCl
iii. a) face centered cubic; b) tetrahedral hole; c) coordination number = 4; d) all occupied; e) 6 x 1/2 plus 8 x 1/8 = 4 Ca and 8 F ; f) CaF2
Exercise 3.6.1
d) There are eight arsenic atoms in the corner; each is shared by eight neighbouring cubes. There are six atoms on the faces; each is shared by two neighbouring cubes.
$\# \: atoms = 8 (\frac{1}{8}) + 6(\frac{1}{2}) = 1 + 3 = 4$ arsenic atoms
e) Each gallium atom is entirely within the cell. There are 4 gallium atoms.
f) The formula is just the ratio of each atom in the material. Since the 4:4 ratio is the same thing as a 1:1 ratio, the formula is just GaAs.
g) Arsenic is to the right of gallium in the periodic table, so arsenic must be more electronegative; it must be the anion, and gallium the cation.
Arsenic is three atoms from the right edge of the periodic table; as an ion, a 3- charge makes sense. To balance the charge, gallium must have a 3+ charge.
h) Cesium is larger than gallium, and iodine is larger than arsenic, so in terms of internuclear distance, there would be less attraction between the cesium and iodide ions than between the (closer) gallium and arsenide ions. Also, if the charge on gallium and arsenide are 3+ and 3-, respectively, and the charges on cesium and iodide are only 1+ and 1-, respectively, then there should be much greater attraction between the gallium and arsenide ions.
The melting point of gallium arsenide (mp = 1,238 oC) should be higher than that of cesium iodide (mp = 632 oC).
i) The atoms are so close together in the periodic table that it isn't likely that arsenic should completely steal three electrons away from gallium; the difference in electronegativity isn't great enough.
Exercise 3.6.2
a) O2- b) Mg2+ c) MgO
e) O: 8(1/8) corners + 6(1/2) faces = 4 atoms; Mg: 12(1/4) edges + 1 center = 4 atoms; MgO
f) [Ne]3s2 g) [Ne] or [He]2s22p6
h) magnesium; the positive charge on the ion causes the remaining electrons to contract
i) [Ne] or [He]2s22p6
j) oxide; they have the same electron configuration, but magnesium has more protons in its nucleus to attract the electrons.
k) N3- l) Ti3+
m) The higher charge on the ions in TiN would lead to a higher force of attraction between the ions, and thus a higher lattice energy would be required to remove ions from the lattice. That attraction thus lowers the mobility of the ions, so the compound does not melt as easily. However, the fact that both Ti3+ and N3- are slightly larger than Mg2+ and O2- helps explain why the melting points are not extremely different; Coulomb's Law says that the increased attraction of the higher charges should be partly offset by the greater size of the ions.
n)
o) CN = 4
p)
q) [Ar]4s13d10 (remember, this is one of those cases in which the 3d level is slightly below the 4s level)
r) Copper's low electronegativity allows it to form an electron sea model of metallic bonding. Copper ions are surrounded by the loosely-held electrons that are allowed to flow through the material.
s) Silicon is too electronegative to form an electron sea structure like a metal.
Exercise 3.7.1
a) LiF b) NaI c) KBr d) MgCl2 e) CaO f) BeS g) Na2O Li3N
Exercise 3.7.2
a) NaNO3 b) KClO4 c) Li2SO4 d) (NH4)3PO4 e) MgCO3 f) Ca(OCl)2 g) NaO2CCH3 h) Be(ClO3)2
Exercise 3.7.3
a) TaCl5 b) CrF3 c) PbBr2 d) HgCl2 e) WO3 f) MoS2
Exercise 3.7.4
a) lithium chloride b) sodium sulfide c) lithium acetate d) magnesium nitrite e) sodium sulfite
f) lead (II) sulfate g) osmium (VIII) oxide h) lead (IV) nitrate | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/03%3A_Ionic_Compounds/3.08%3A_Ionic_Compounds-_Solutions_for_Selected_Prob.txt |
Molecules aren't everything
To get a good idea of what a molecule is, let's first look at what a molecule isn't.
Not all things are composed of molecules. Many common rocks and minerals are composed mostly of silica. Silica is a non-molecular network solid with the approximate formula SiO2. Its structure keeps going on in all directions. The chunk of silica shown below shows just 14 silicon atoms, but a real chunk of silica could contain 14 thousand or 14 million silicon atoms, or just about any other number.
In contrast, a molecule is made from a particular number of atoms, and so it has a particular size. The atoms in silica are not collected into discrete units of a particular size; instead, they can be arranged in very large arrays of different sizes. A unit of silica could contain a million silicon atoms or a billion of them.
Molecules contain a specific combination of atoms in a specific ratio. That isn't exactly true in silica. There really are about twice as many oxygen atoms as silicon in silica and the atoms are arranged in a particular way, but the Si:O ratio can vary slightly, and there are also small amounts of hydrogen (but so small an amount that we ignore it in the formula).
Copper metal is also a non-molecular solid, with legions of copper atoms bonded together to form one united mass. The copper atoms are arranged in a particular way, and the metal has a very specific formula (Cu), but like silica, a unit of copper could contain a million copper atoms or a billion of them.
Rocks and metals are non-living, inorganic materials. We are often interested in with chemicals used by biological systems. Those compounds are overwhelmingly molecular.
• not all compounds are molecules
• some compounds form larger structures that don't have a well-defined size, although they may have a specific ratio of atoms
• metals form large structures that do not come in regular chunks
Origins and scope of inorganic and organic chemistry
Sometimes you will hear the terms "organic chemistry" and "inorganic chemistry", as in "I've got an inorganic chem lab today" or "I just did really well on my o-chem test". Sometimes high school or first year chemistry is called "inorganic", but that isn't really a good description. What do these terms really mean?
The realization during the European Renaissance that there is an entire chemistry unique to living things gave rise to the study of organic chemistry. Researchers working in medical schools labored to discover the active principles that could be found in herbal remedies and spices, which at that time formed an important part of the treatment of disease. The chemistry of the biological world seemed completely separate from that of the mineral world, and consequently organic chemistry developed as a separate discipline from inorganic chemistry.
• organic chemistry is based on the chemistry of organisms
• the chemistry of organisms overwhelmingly involves carbon compounds
• more generally, organic chemistry is the chemistry of carbon compounds
• inorganic chemistry is the chemistry of the rest of the periodic table
Organic compounds are based on only a few elements from the upper part of the periodic table. The principle component of organic molecules is carbon. Carbon atoms are able to bond with each other to form bigger and bigger structures.
In addition, organic compounds frequently contain hydrogen, oxygen and nitrogen; phosphorus and sulfur are occasionally seen as well. With just these atoms, the major classes of structural biomolecules can be made (lipids, proteins and nucleic acids). Most small, biologically relevant molecules contain only this small group of atoms, including hormones, neurotransmitters and compounds involved in chemical ecology.
Some marine organisms produce halogenated compounds containing bromine, chlorine or iodine. These compounds don't perform an obvious function in these organisms. Many halogenated organic compounds are toxic, and so it is possible that marine organisms use these compounds as a self-defense measure; chemical warfare is not unheard of in nature.
In addition, halogenated compounds are sometimes useful in carrying out organic reactions in the laboratory or in industry.
Occasionally other elements play extremely important roles in biological molecules; for example, vitamin B12 contains a cobalt atom bonded to a carbon, and plays an important role as a regulatory agent. However, by and large, organic and biological compounds contain very few elements: C, H, O, N, occasionally S and P, and sometimes halogens.
Applications of organic chemistry are wide-ranging. An obvious practical area is the pharmaceutical industry, because small organic molecules are frequently biologically active. However, one of the earliest industrial applications of organic chemistry was the dyestuff and paint industry, because many pigments can be made from organic compounds. It has been argued that the development of this industry in Germany in the 1800's was closely connected to the rise of that country as a major industrial power. Today, Germany is still an important center for research and development in chemistry. One of the biggest applications is polymer chemistry, which involves very large molecules or "macromolecules" that are useful as plastics and other materials. Most organic polymers are derived from petroleum, but a growing fraction of them come from agricultural materials such as corn and soybeans.
• pharmaceuticals are an important application of organic chemistry
• pigments, paints and dyes are historically very important in organic chemistry
• one of the biggest applications of organic chemistry is polymer chemistry
Inorganic chemistry is just the rest of chemistry. It is the part that is not principally based on carbon. It turns out that there is a lot of important inorganic chemistry in biology, as well. Frequently, metal ions play crucial roles in There are important applications of inorganic chemistry, too. Many important materials are inorganic, such as ceramics and solid-state materials used as conductors and semiconductors. Many industrial catalytic processes, such as refining petroleum or making detergent, require catalysis by transition metals. Metals themselves are important in society, and often inorganic chemistry is required to extract metals out of natural ores.
• materials science offers important applications of inorganic chemistry
• catalysis offers important applications of inorganic chemistry
• mining and metallurgy often depends on inorganic chemistry
In Living Systems, the Focus is Often on Molecules
Biological structures like DNA, proteins and cells all depend on molecules. A strand of DNA is a very long molecule, and so is a protein. Cells are complicated, organized structures built up from huge numbers of smaller molecules.
There are lots of important small molecules in biology, too. Neurotransmitters like epinephrine are pretty simple molecules, but without them our nervous system couldn't work. Oxygen, containing just two oxygen atoms, is a very simple molecule, and it is certainly essential to living things.
Molecules aren't only found in biology. Sulfur is a molecule that is found naturally in the earth. It contains a group of sulfur atoms connected in a ring, like a daisy chain. Carbon nanotubes and fullerenes are carbon-containing molecules that can be found in meteorites as well as some kinds of steel.
In general, this is a course about structural chemistry. Chemistry is specifically the study of matter and energy: the stuff from which the universe is made. Because many of the readers of this text will be biology majors and pre-medical students, we will be especially careful to pay attention to the stuff living things are made of. Organic chemistry is the study of organic matter, meaning the stuff from which living things are made. Many of the examples in these pages on molecules and their properties are related somehow to living things.
In a molecule, a specific combination of atoms is arranged in a specific way. The atoms are all held in place by bonds that connect them to other atoms. The atoms in a specific compound are always found in the same places from one molecule to the next. This information has been shown using x-ray crystallography and tunneling electron microscopy, two methods that can give us pretty good pictures of where the individual atoms can be found within a compound or an element. This unique arrangement gives the molecule unique physical and biological properties.
While there are a number of ways in which eight carbons, two oxygens, four nitrogens and ten hydrogens can combine to form molecules with the formula C8H10N4O2, one unique arrangement of these atoms forms the structure of caffeine, with the accompanying stimulant effect on your nervous system.
You wouldn't want to ingest just any molecule that happened to have the same formula as caffeine. Other combinations of atoms in the ratio C8H10N4O2 could be ineffective at best or poisonous at worst. The structures may be connected using a different order of atoms; sometimes, we say they have a different connectivity. These other structures are described as "isomers" of caffeine.
• Compounds that have the same formulae but different connectivities are called isomers.
These atoms remain bonded together in the same way throughout the lifetime of the molecule, regardless of whether the molecule is frozen in the solid phase, has melted into the liquid phase, or floats through the air in the gas phase.
When a female bark beetle produces and secretes a sex pheromone in order to attract a mate, the pheromone can be carried on currents of air to male bark beetles, whose antennae can intercept individual pheromone molecules, completely intact, still in the same form as when they were secreted by the female, no matter how far away she is. The male recognizes the structure of the molecule; it can clearly tell if it has just intercepted the proper pheromone or some subtle imposter that also happens to have the formula C10H16O.
Exercise \(1\)
Draw the following molecules. In each case, the hydrogens are all bonded to the non-hydrogen atom.
a) water, H2O b) ammonia, NH3 c) methane, CH4
Answer
Exercise \(2\)
Draw three isomers for each the following formulae. Assume each carbon can make 4 bonds and each hydrogen can make only one bond.
a) C5H12 b) C4H10 c) C6H14
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.01%3A_Why_do_Molecules_Matter.txt |
In order to think about some of the compounds that can be made with this small collection of elements, we will need a way to visualize on paper how these atoms are connected with one another.
At one time, people thought of the atoms in these compounds as charged spheres (such as oxygen anions, O2-, or carbon cations, C4+) that were simply pulled together by their opposing charges, like piles of sticky jellybeans. However, there are some limitations to this way of looking at compounds. We know that in most organic compounds, it is common to find two carbon atoms connected together, which is not likely to happen if all carbons are found as charged cations; in that case, they would simply repel each other.
In the early twentieth century, G.N. Lewis used some simple observations about the periodic table to propose an alternate model of how atoms can be bonded together. This model still influences the way we think about molecules today.
First of all, Lewis noted that if you count up the number of electrons in all the atoms in almost any known molecular compound, such as water (H2O), ammonia (NH3) or carbon dioxide (CO2), you get an even number (10, 10 and 22, respectively, in these examples).
From this observation he eventually proposed the notion that electrons are more stable if found in pairs. Two atoms with odd numbers of electrons may therefore gain stability if they come together to form an even number overall.
• molecules usually contain even numbers of total electrons
For example, hydrogen, with 1 electron, plus fluorine, with 9 electrons, come together to form hydrogen fluoride, with 10 electrons. The new pair of electrons that forms from the unpaired electron in fluorine and the single electron in hydrogen, Lewis thought, is the bond that holds the two atoms together.
To simplify electron counting, one need only consider the valence electrons when thinking about bonding; one could ignore the core electrons, which are simply groups of electrons that already number the same as a noble gas. Thus, we consider fluorine's seven valence electrons instead of all nine; we consider nitrogen's five valence electrons and not all seven.
• valence electrons are the electrons in an outer shell of an atom
• counting only valence electrons simplifies electron counting
Lewis made a couple of additions to this idea based on the layout of the periodic table. People were aware of the remarkable stability of the noble gases, and so it seemed that there was something particularly stable about the number of electrons found in these elements. These are even numbers, of course: 2 for helium, 10 for neon, 18 for argon, 36 for krypton and so on. In terms of valence electrons, neon has 8, argon has 8, krypton has 18 and so on. Lewis suggested that any atom could become more stable if it could share enough electrons from its neighbors to obtain the same electron count as a noble gas.
• when an atom is found in a molecule, it usually has the same number of valence electrons as the noble gas in the same row
• this electron count includes unshared electrons as well as shared electrons
• for electron-counting, shared pairs are counted for both atoms that share the electrons
The fact that organic and biological chemistry, which constitutes a majority of familiar molecules, takes place mostly in the second row of the periodic table allows us to think of the Lewis model as "the octet rule", from the Latin word for eight: second-row elements form bonds so that they can have eight valence shell electrons, like neon.
• most common biological or organic molecules are made from atoms in the second row of the periodic table
• it is common to see atoms with 8 electrons, like neon, in many molecules
• the Lewis "octet" rule says atoms frequently have 8 valence electrons in molecules
• atoms in other rows sometimes have larger numbers of electrons
Lewis structures are drawings that we use to show how the electrons in a molecule are arranged in a stable way. They let us verify that each atom in the molecule has an octet.
Note that a pair of electrons that forms a bond is usually written as a line instead of a pair of dots. This modification of a Lewis structure is sometimes called a Kekule structure. For the most part, Kekule structures will be used in these pages.
Exercise \(1\)
Try drawing Kekule structures for the following diatomic molecules:
a) Cl2 b) H2 c) HCl d) F2
Answer
What happens if a compound is actually an ion? A positive or negative charge tells you something about the structure. If there is a negative charge there must be an extra electron in the structure, so that the proton : electron ratio is no longer balanced. If the charge is two minus, there must be two extra electrons. For the same reason, a positive charge means an electron is missing from the structure.
• a negative charge means extra electron(s)
• a positive charge means missing electron(s)
Exercise \(2\)
Try drawing Kekule structures for the following ions:
a) HO- b) MgCl+ c) HS- d) OCl-
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.02%3A_Lewis_Structures.txt |
The Basics
Let's go over some of the main things you need to consider if you want to draw Lewis structures. Lewis structures will be the principal way you communicate ideas about molecules in this course, so it is best to get some practice with them as soon as possible.
Here's what you need to do:
1. count up the valence electrons in the atoms that will be bonded together.
2. place the atoms beside each other on paper.
3. arrange the given electrons around the atoms so that each atom has an octet.
4. these electrons can be shared between the atoms (these are called bonding pairs) or held by one atom or the other (these are called nonbonding pairs or "lone pairs")
Occasionally, you will need to share more than one pair of electrons between two atoms so that all atoms can obtain an octet. That's what happens in the structure of carbon monoxide, CO.
Carbon has 4 valence electrons and oxygen has 6, for a total of 10 electrons. That's 5 pairs. If carbon and oxygen share a pair of electrons, there will be four pairs left. You could put three pairs on the oxygen and one pair on the carbon. The oxygen has four pairs, including the lone pair and the shared pairs. Carbon only has two pairs. That isn't an octet, and if possible the molecule will avoid that situation.
There could be one bond, and two lone pairs on each atom. That isn't much better. Now neither atom has an octet.
That means we can only have three bonds between the carbon and oxygen -- a total of six electrons. The other four electrons could become lone pairs -- one on carbon and one on oxygen. Now both atoms have an octet.
Exercise \(1\)
Try drawing Lewis structures for the following molecules:
a) O2 b) N2 c) O3
Answer
Consequences of multiple bonding: bond distances
Are there really three bonds between the carbon and oxygen in carbon monoxide? How could we tell?
Multiple bonds have important consequences. A bond is a pair of electrons shared by two atoms. Since both atoms need the pair of electrons to complete an octet, the atoms must stay close to the electron pair and close to each other in order to be stable. Hence, a bond is also an attractive force between atoms. That idea means additional bonds between two atoms leads to additional attraction between the atoms. Two atoms with multiple bonds should be more tightly bound to each other than two atoms with fewer bonds.
• a double bond between two atoms is shorter than a single bond between the same two atoms
• a triple bond between two atoms is shorter than a double bond between the same two atoms
• a quadruple bond between two atoms is shorter than a triple bond between the same two atoms
Experimental data suggests that this is true. Microwave spectroscopy is a technique that can be used to measure bond distances (the distance from one atom to the next) in molecules in the gas phase. The data on carbon monoxide indicates that the C-O distance is about 1.13 A (1 A = 10-10 m). In carbon dioxide, CO2, the distance is 1.16 A. The shorter distance in carbon monoxide suggests the carbon and oxygen are held a little more tightly together than they are in carbon dioxide, in which the Lewis structure shows only a double bond.
The additional bond between a pair of atoms is often called a pi bond (pronounce "pie" bond). The first bond between a pair of atoms is sometimes called a sigma bond. Carbon monoxide has one sigma bond and two pi bonds.
Exercise \(2\)
Draw structures for the following molecules and ions.
a) NO+ b) C22- c) PN
Answer
Exercise \(3\)
NO2 is a brown gas (connectevity O-N-O) that comes from car exhausts. It is partly responsible for smog in large cities. NO2 dimerizes at low temperature to form N2O4, a colorless gas. Dimerization refers to two molecules coming together to form one, bigger structure. Show why this phenomenon happens with NO2, with the help of structures.
Answer
Exercise \(4\)
Do you expect the NO2- anion to dimerize? Explain. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.03%3A_Lewis_Structures_and_Multiple_Bond.txt |
Connectivity
Thinking about how atoms and electrons are arranged in a molecule gets harder when there are more than two atoms. Often you need to decide how the atoms are arranged before writing the Lewis structure.
Let's draw the structure for carbonic acid, H2CO3. First we need to know which atoms are connected together. That will tell us which atoms are sharing electrons.
We could put one atom in the middle of the structure and arrange all the others around it. That may work sometimes, but often it gives the central atom too many bonds. A second-row atom can't have more than eight electrons in its valence shell. As a result, a carbon can't have more than four bonds, because there is no room for more electrons.
We could try to attach all the atoms in a string. Long strings of atoms like this are not very common. There may be several reasons why this sort of arrangement isn't very stable. First, when we consider formal charge a little later, you will see that placing all the atoms in a row may cause unnecessary charge separation. Second, some atoms such as oxygen are not very stable when bonded to other atoms of the same kind. Repulsion between lone pairs may contribute to this instability. Because carbon often has no lone pairs, it is an exception and it can bond to itself.
Note that when we try to arrange the atoms in either way, we would probably avoid putting a hydrogen in the middle. That's because hydrogen's octet is just two electrons, so it can usually only make one bond. Knowing the number of bonds that other atoms usually form can also be helpful.
Valence
Valence is the number of bonds an element usually forms - for example, the valence of carbon is four, nitrogen is three, oxygen is two, fluorine and hydrogen are one. Valence usually corresponds to the number of electrons needed to form an octet. However, there are exceptions: boron would need 5 electrons to form an octet, but since it only has three electrons to share, it can only form three covalent bonds.
Table \(1\) Typical valences (number of bonds formed) for several second-row atoms.
Table \(1\) Typical valences (number of bonds formed) for several second-row atoms.
atom B C N O F
valence 3 4 3 2 1
Keeping valence in mind can help Lewis structures go more easily. In the case of carbonic acid, carbon might go in the middle, since it must form the most bonds to obtain an octet. Also, having two of the oxygens connected to hydrogens as well as to carbon helps them to attain their normal valence as well.
Once the connectivity has been filled in, we have a skeletal structure of the compound. Now we just need to fill in the extra electrons. Six from each oxygen is 18; four from carbon makes 22; one from each hydrogen makes 24 total electrons.
Coming up with a Lewis structure requires a number of steps, but it usually follows a familiar pattern.
Exercise \(1\)
Use the idea of valence to construct Lewis structures of the following compounds:
a) hydroxylamine, H3NO b) ethene, C2H4 c) ethane, C2H6
Answer
Exercise \(2\)
Elements in the third row of the periodic table often behave similarly to elements in the second row. Use the idea of valence to construct Lewis structures of the following compounds:
a) PH3 b) CS2 c) SiH4 d) PCl3
Answer
Sometimes there will be more than one "correct" way to draw a Lewis structure for a given set of atoms. As an example, there are two different structures in nature with the formula C2H6O. These compounds, methyl ether and ethanol, are called isomers, meaning they are formed from the same (iso) things (mer).
To help distinguish between isomers, formulae are sometimes written with structural abbreviations. Instead of writing C2H6O for both these compounds, we would write CH3OCH3 for methyl ether, suggesting the first carbon is connected to three hydrogens and an oxygen, the oxygen is connected to the second carbon, and the second carbon is connected to another three hydrogens.
In a similar way, ethanol would be abbreviated as CH3CH2OH.
Exercise \(3\)
Draw Lewis or Kekule structures for the following compounds:
a) propanol, CH3CH2CH2OH b) acetic acid (or ethanoic acid), CH3CO2H
c) methylamine, CH3NH2 d) butanal, CH3CH2CH2CHO
Answer
Exercise \(4\)
Draw at least two isomers for each of the following formulae:
a) C3H6 b) C4H10 c) C4H8 d) C3H6O
Answer
Sometimes, the decision about connectivity involves additional factors, other than valence. Here is a more detailed set of instructions:
Steps for Guessing Connectivity
IF the connectivity is not provided and you need to draw a Lewis Structure,
A. The first atom listed in the formula is often in the middle of the structure.
B. Most simple structures, especially for inorganic species, have an atom in the middle with several atoms connected to and branching out from this central one (as opposed to a linear chain of atoms). Thus, choose the middle atom to be
1. The atom in the lowest row of the periodic table (most likely to be able to use d orbitals to form more than 2 bonds as needed for “branching” structures)
2. The atom more in the center of the periodic table (group 4, then 3 or 5, etc) -has highest valence (valence=number of bonds formed)
3. The least EN atom is often in the middle.
C. Many simple structures are symmetric. This is not a law of nature but a simple observation!
X-M-X often favored over X-X-M
D. Oxyacids often have H’s on O’s; for example, HClO4 has the connectivity H-O-ClO3. The chlorine is attached to four different oxygen atoms, but one oxygen atom is also attached to a hydrogen.
E. Avoid three and four membered rings--they are not common, especially among inorganic species.
F. Avoid bonds between electronegative atoms (F-F, O-O, O-F, etc) . Such bonds are weak and tend to make molecules unstable, if not non-existent. Obviously the situation cannot be avoided in molecules like O2, OF2, F2, etc.
Exercise \(5\)
Provide structures for the following compounds. Remember to think about common valences.
a) NF3 b) N2H4 c) BF3
d) SiF4 e) Ge2H6 f) SnCl4
g) ClNO h) AlBr3 i) HCN
j) S8 k) P4 l) P4O6
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.04%3A_Lewis_Structures_and_Polyatomic_Mo.txt |
Looking at the structure of a molecule can help us to understand or to predict the behavior of that compound. One of the tools that we will eventually use to understand reactivity is formal charge. That is because reactivity has to do with the reorganization of electrons between atoms. New chemical bonds are formed by sharing electrons. Old chemical bonds are broken when one atom takes the bonding electrons away from another atom.
• chemical reactions occur via attraction and donation of electrons
Formal charge can help us to understand the behavior of carbon monoxide, CO. When exposed to transition metal cations such as the iron in hemoglobin (Fe2+), the carbon is attracted to and binds to the metal. In the case of hemoglobin, because the carbon monoxide binds very strongly to the iron, the CO blocks the position where oxygen would normally be bound and carbon monoxide poisoning results.
• formal charge can help us predict how a molecule behaves
• atoms with positive formal charges often attract electrons
• atoms with negative formal charges often donate electrons
Why does the molecule behave in this way? There are actually a number of reasons. However, the fact that the carbon is attracted to a metal cation begs the question: Is the carbon an anion? Yes, in a sense. In a Lewis structure of the compound, the carbon has a formal negative charge. You will see why below.
Formal charges are an important book-keeping device that we use in Lewis structures. They tell us if one atom is donating extra electrons to another to give it an octet. If an atom needs to donate more electrons than normal in order for everyone to get an octet, it will have a positive formal charge. If an atom donates fewer electrons than normal and everyone still has an octet, it must be getting extra electrons from somewhere else. It will have a negative formal charge.
• formal charge is often present if the atom does not have its usual number of bonds
• valence rules can act as flags to alert you that formal charges are present
To help us think about formal charges, let's look at a few small molecules that all contain carbon-oxygen multiple bonds but that are slightly different from each other. Formaldehyde (CH2O) is a chemical that is used to preserve tissues; you may be familiar with its odour from anatomy lab. Carbon monoxide results from burning fossil fuels; it is also an important industrial chemical used in manufacturing detergents. Carbonate is an anion that is found in many forms. Calcium carbonate is found in limestone and chalk, for instance.
Look at the structure of formaldehyde. Oxygen has a normal valence of two, and it has two bonds in formaldehyde, so there is no formal charge on the oxygen. Carbon has a normal valence of four, and it has four bonds here. There is no formal charge on carbon. There are no formal charges on the hydrogens either.
Carbon monoxide has a structure that is very similar to formaldehyde. It does not have any hydrogens, though. With ten electrons total, the only way to get an octet on both atoms is to make three bonds between carbon and oxygen.
Oxygen has normal valence two, but here it is making three bonds. It is sharing an extra pair of its electrons with carbon to make that third bond. If it is sharing a pair of electrons, we can think of it keeping one for itself and giving the other to carbon. Since it gives one of its electrons to carbon, it has formal charge +1.
Carbon has normal valence four, but here it is only making three bonds, even though it has an octet. How did it get an octet with only three bonds? It got an extra electron from somewhere (the oxygen). It has formal charge -1.
Notice that overall the carbon monoxide molecule is neutral. Oxygen has a plus charge and carbon has a minus charge. These charges cancel to give an overall neutral molecule.
What we are really doing when we assign formal charge is comparing how many electrons the atom brought with it from the periodic table to how many it has now. If the atom brought four electrons of its own and it is now sharing eight, things are even. It brought four to share and got four from its neighbors in an even trade. If it only brought three of its own and is now sharing eight, it got more electrons than it gave, and it will have a negative charge.
To determine formal charge:
• check the number of electrons on the atom in the periodic table
• check the number of electrons entirely owned by the atom in the molecule; this is different than looking for an octet
• "entirely owned" electrons include any electrons in lone pairs, since they belong completely to one atom
• "entirely owned" electrons also include half of the electrons in the bonds to the atom, since it is sharing each of those pairs with other atoms.
Compare the number of entirely owned valence electrons in the periodic table to those entirely owned by the atom in the molecule.
• if the number of entirely owned electrons on the atom in the molecule is higher than in the periodic table, the atom has a negative charge
• if the number of entirely owned electrons on the atom in the molecule is lower than in the periodic table, the atom has a positive charge
• the formal charge is additive: if the atom has two extra electrons in the molecule, it has a two minus charge. If it is two short, it has a two plus charge.
Remember, electron counting to determine an octet counts all of the bonding and nonbonding electrons equally. It is done simply to determine whether the atom has a noble gas configuration right now. Electron counting to determine formal charge is done to keep track of who has given electrons to whom when making the molecule. If, in getting to an octet, atoms have received more electrons than they have given, their electron/proton ratio has changed, and they become charged.
Exercise \(1\)
Draw Lewis or Kekule structures for the following molecules, remembering to include formal charges, if any (and notice that some of these molecules are ions):
a) NO+ b) CN- c) CH3O- d) CH3+ e) HNO3 f) CH3CO2-
Answer
Exercise \(2\)
Given the structures below, assign any missing formal charges.
Exercise \(3\)
Given the structures below, draw in the missing electrons, if any.
Exercise \(4\)
Label all atoms in the following compounds with the correct non-zero formal charge.
Exercise \(5\)
Provide structures for the following oxoanions (anions with oxygens attached to another atom) of chlorine:
a) hypochlorite, ClO- b) chlorite, ClO2- c) chlorate, ClO3- d) perchlorate, ClO4
Exercise \(6\)
Provide structures for the following oxoanions of sulfur:
a) sulfite, SO32- b) sulfate, SO42- c) thiosulfate, S2O32-
d) disulfate, S2O72- e) persulfate or peroxomonosulfate, SO52-
Exercise \(7\)
Provide structures for the following oxoanions of phosphorus:
a) phosphate, PO43- b) phosphite, HPO32- c) hypophosphite, H2PO2-
Note that hydrogen atoms in some phosphorus oxoanions are attached to the phosphorus atom. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.05%3A_Lewis_and_Formal_Charge.txt |
The carbonate anion shown below has one double bond and two single bonds. Normally, the number of bonds between two atoms in the Lewis structure can tell you how closely the two atoms are held. That means there is one C-O bond in the molecule that is shorter than the other two.
However, carbonate ion is an exception to the rule. In fact, carbonate ion is a symmetric, trigonal planar molecule. All three carbon-oxygen bond distances are about 1.28 Angstroms long.
You cannot draw a Lewis structure that would suggest all three bonds are the same length. Instead, you can use resonance structures to understand this fact. Resonance structures are just alternate Lewis structures that take into account different possible arrangements of lone pairs and multiple bonds for a given skeletal structure. In carbonate, there are three ways to arrange these extra electrons.
If you label the oxygen atoms A, B and C, as shown below, then A has a double bond in 1/3 of the structures you could draw, but it would have a single bond in the other 2/3 of the ways in which you could draw the structure. The same is true for the atoms B and C.
In fact, carbonate really has about 1-1/3 bonds between each carbon and oxygen. Furthermore, there is not a full negative charge on any of the oxygens but about two-thirds of a negative charge on each of the three oxygens.
Occasionally we might draw a structure that shows partial bonds between atoms as dashed lines to suggest, in this case, 1-1/3 bonds instead of either one bond or two. We might also write ("delta minus") to denote a partial negative charge.
The limitation of this type of drawing is that it fails to show us exactly how many electrons we are dealing with. Later in the course, it will be very important to keep track of all the electrons in molecules as they undergo chemical reactions. Consequently, we almost always draw out different possible Lewis structures and keep in mind that the structure is something like a combination of all of them.
When we have structures that differ only in the way their electrons are arranged, but have exactly the same connectivity between the atoms, we refer to the set of structures as resonance structures. The double ended arrows between the resonance structures is the symbol we use to say that we have two or more different ways of drawing the same structure, and that we need to take all of the structures into account to understand how the molecule really behaves.
There is a subtlety here. The compound is not one structure or the other, and it is not an equilibrium where the structure changes back and forth rapidly. It is a superposition, in which a single molecule can behave like all three structures at the same time.
Exercise \(1\)
Draw sets of resonance structures for the following compounds.
a) cyanate ion, NCO - (a linear structure) e) azide ion, NNN-
b) formate ion, HCO2- f) diazomethane, CH2NN
c) ozone, O3 (a bent or "boomerang" structure) g) enolate ion, CH2CHO-
d) nitrate ion, NO3-
Answer
Exercise \(2\)
All of the bonds in benzene (below) are the same bond distance (1.39 A). Show why this is true.
Exercise \(3\)
There are several resonance structures for each of the following ions. Show at least three for each.
Exercise \(4\)
Problem IM6.4.
Ions can be greatly stabilized by delocalization. That is, if a charge can be spread out over several atoms by resonance, the charge is more stable. Use this idea to predict the relative stabilities of the following ions (rank them from most stable to least stable).
Exercise \(5\)
Draw resonance structures for the following molecules. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.06%3A_The_Need_for_Resonance_Structures.txt |
Cells contain lots of water. One of the roles of the water is to dissolve different materials. For example, there are many different ionic compounds (salts) in cells. Ions are used to maintain cell potentials and are important in cell signaling and muscle contraction.
How can you tell if a compound is ionic or covalent?
There is not a simple answer to this question. Many bonds are somewhere in between. In a polar covalent bond, a pair of electrons is shared between two atoms in order to fulfill their octets, but the electrons lie closer to one end of the bond than the other. There is more negative charge toward one end of the bond, and that leaves more positive charge at the other end.
Looking at the electronegativity values of different atoms helps us to decide how evenly a pair of electrons in a bond is shared. Electronegativity increases toward the upper right hand corner of the periodic table because of a combination of nuclear charge and shielding factors. Atoms in the upper right hand corner of the periodic table have a greater pull on their shared bonding electrons, while those in the lower left hand corner have a weaker attraction for the electrons in covalent bonds.
In a carbon-oxygen bond, more electrons would be attracted to the oxygen because it is to the right of carbon in its row in the periodic table. Compounds like , dimethyl ether, CH3OCH3, are a little bit polar. Formaldehyde, CH2O, is even more polar. Electrons in pi bonds are held more loosely than electrons in sigma bonds, for reasons involving quantum mechanics. That allows the oxygen to pull the electrons toward it more easily in a multiple bond than in a sigma bond.
Not all polarities are easy to determine by glancing at the periodic table. The direction of the dipole in a boron-hydrogen bond would be difficult to predict without looking up the electronegativity values, since boron is further to the right but hydrogen is higher up. As it turns out, the hydrogen is slightly negative.
A bond is ionic if the electronegativity difference between the atoms is great enough that one atom could pull an electron completely away from the other one. That situation is common in compounds that combine elements from the left-hand edge of the periodic table (sodium, potassium, calcium, etc.) with elements in the extreme upper right hand corner of the periodic table (most commonly oxygen, fluorine, chlorine). Sodium chloride is an ionic compound.
Many bonds can be covalent in one situation and ionic in another. For instance, hydrogen chloride, HCl, is a gas in which the hydrogen and chlorine are covalently bound, but if HCl is bubbled into water, it ionizes completely to give the H+ and Cl- of a hydrochloric acid solution. Even in gaseous HCl, the charge is not distributed evenly. The chlorine is partially negative and the hydrogen is partially positive.
Potassium hydroxide, KOH, contains one bond that is covalent (O-H) and one that is ionic (K-O). Hydrogen is tricky because it is at the top of the periodic table as well as the left side. It is just electropositive enough to form ionic bonds in some cases. It is just electronegative enough to form covalent bonds in other cases.
In KOH, the K-O bond is ionic because the difference in electronegativity between potassium and oxygen is large. The difference in electronegativity between oxygen and hydrogen is not small. An O-H bond can sometimes ionize, but not in all cases.
Sometimes ionization depends on what else is going on within a molecule. Because the K-O bond in potassium hydroxide is ionic, the O-H bond is not very likely to ionize. There is already a negative charge on oxygen. Charge separation costs energy, so it is more difficult to put a second negative charge on the oxygen by ionizing the O-H bond as well. Frequently first ionizations in molecules are much easier than second ionizations.
Exercise \(1\)
Predict the direction of polarity in a bond between the atoms in the following pairs:
a) sulfur-oxygen b) silicon-fluorine
c) hydrogen-sodium d) chlorine-aluminum
Answer
Because it is so common that an element from the extreme left hand of the periodic table is present as a cation, and that elements on the extreme right carry negative charge, we can often assume that a compound containing an example of each will have at least one ionic bond.
Exercise \(2\)
Draw structures of the following compounds. Each one contains at least one anion and cation.
a) KBr b) LiOH c) KNO3 d) MgSO4 e) Na3PO4 f) Na2SO3
g) LiClO4 h) NaClO3 i) KNO2 j) Ca(ClO2)2 k) Ca2SiO4 l) Na3PO3
m) NaOCl n) Mg2SnO4
Answer
Exercise \(3\)
Ammonium ion, NH4+, is a common molecular ion. Draw structures for the following compounds that include this ion.
a) NH4Cl b) (NH4)2CO3 c) (NH4)3PO3 d) NH4CH3CO2 e) NH4HSO4
Answer
Exercise \(4\)
Many anions have names that tell you something about their structure.
Table with Anions names
prefix- or -suffix
common meaning
example name example formula
-ide atom is present as anion chloride Cl -
-ate atom is present as an oxyanion; usually a common form chlorate ClO3-
-ite atom is present as an oxyanion, but with fewer oxygens (or lower "oxidation state") than another common form chlorite ClO2-
per- atom is present as an oxyanion, but with even more oxygens than the "-ate" form perchlorate ClO4-
hypo- atom is present as an oxyanion, but with even fewer oxygens than the "-ite" form hypochlorite ClO-
Using the table as a guide, propose names for the following anions:
a) Br- b) O2- c) F- d) CO32- (common oxyanion) e) NO3- (common oxyanion) f) NO2-
g) S2- h) SO42- (common oxanin) i) SO32- j) SO52- k) C4- l) N3- m) As3-
n) PO43- (common oxyanion) o) PO33- p) I- q) IO3- (common oxyanion) r) IO4-
Answer a
bromide
Answer b
oxide
Answer c
fluoride
Answer d
carbonate
Answer e
nitrate
Answer f
nitrite
Answer g
sulfide
Answer h
sulfate
Answer i
sulfite
Answer j
persulfate
Answer k
carbide
Answer l
nitride
Answer m
arsenide
Answer n
phosphate
Answer o
phosphite
Answer p
iodide
Answer q
iodate
Answer r
periodate | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.07%3A_Which_Bonds_are_Ionic_and_Which_ar.txt |
Line drawings
Lewis structures can tell us a lot about how atoms come together to make molecules. They can also be cumbersome, especially if we are dealing with very large molecules. Drawing a line instead of a pair of dots for bonding pairs of electrons makes it easier to draw structures. There are other abbreviations that are helpful in some situations.
Because organic chemistry is based on the compounds of carbon, we would have to draw the letter C over and over again unless we had a shortcut. In line structures, we drop the label "C" for the carbon atoms. Anytime there is a joint between two bonds (drawn as a vertex in a zig-zag line), the atom attached to that bond is assumed to be a carbon unless written otherwise.
The atom at the end of a zig-zag line would also be a carbon, unless it is explicitly written as another atom.
Taken even further, we will omit the hydrogens from our structures, since the compounds of carbon almost always contain hydrogen as well. Since we know carbon has a valence of four, we always know how many hydrogen atoms are attached to each carbon in order to reach that valence. A carbon with two bonds drawn in must have two hydrogens on it. A carbon with only one bond drawn to it must have three hydrogens.
Note that the hydrogens are not normally omitted if they are attached to heteroatoms (atoms other than carbon, such as oxygen or nitrogen).
You will also note that lone pairs are frequently left out when we use line structures, so you will have to add them back in to think about Lewis structures. Adding the lone pairs back to the heteroatoms in line structures is a good habit to get into, because later in the course we will be very concerned with keeping track of where all the electrons are.
Below is a summary, showing the relationship between Lewis/Kekule structures, line structures and condensed formulae for a few different compounds.
Remember, in phenol, the hydrogen attached to oxygen was labeled in the line structure. Sometimes there are exceptions in line structures, in which atoms that you might not think about labeling usually do get labels. The most common exceptions are shown below.
Exercise \(1\)
Translate the following condensed formulae into line drawings.
a) CH3CH2NHCH2CH3 b) CH3CHFCH2CH2CH2Cl
c) CH2CHOCH2CH3 d) CH3CHClCH2SCH2CH3
Answer
Exercise \(2\)
Translate the following structures into condensed formula.
Exercise \(3\)
Try filling in the missing line drawing, Lewis / Kekule structures or condensed formulae in each line of the table below.
Exercise \(4\)
Look at the following pairs of compounds. Are the compounds constitutional isomers? (Hint: You may need to add in all hydrogens)
4.09: Three Dimensional Drawings
Our purpose in this chapter is to be able to represent the structure of a molecule. When we draw the Lewis structure for methane, it is only a 2-dimensional representation, but really the compound has a three-dimensional shape. The central carbon in methane is tetrahedral, with the four hydrogens at the alternate corners of a cube.
We can draw a tetrahedron by making a vertex between two lines, then attaching a wedged line and a hatched line to the outside of the corner. The black, wedged line is read as "this atom is towards us". The hatched line is read as "this atom is farther away from us". Although the drawing is flat, the hatched and wedged lines is a shorthand notation for 3-dimensionality.
Representing the three-dimensional shape of a molecule can be done on paper, but only if we are very careful about drawing conventions such as wedge-dash drawings.
Carbons that have bonds to four different neighbors are always tetrahedral. The neighbors arranged around a tetrahedral carbon are around 109° apart. Sometimes, for reasons we won't go into here, tetrahedral carbons are called sp3 carbons.
Pentane is a compound that contains four tetrahedral carbons in a row. Each tetrahedral carbon could be shown with wedge - dash projections.
It is also very common to use computer drawing programs to show the shapes of molecules. You can view the three-dimensional structure of pentane in many different ways. Three of the most common ways are shown here. Ball & Stick is an easy way to see where all the atoms are and how they are connected. Wireframe allows you to see connections without having your view obscured by atoms. Space-filling models are meant to most closely show you the shape of the molecule, although the atoms involved get obscured in this view.
Go to Animation IM9.1. A three-dimensional model of pentane.
Carbons double bonded to one neighbor, and consequently bonded to three different neighbors overall, are trigonal planar. The neighbors are all in the same plane and are about 120° apart. Trigonal planar carbons are sometimes described as sp2 carbons.
Propene has two trigonal planar carbons as well as one tetrahedral carbon. Notice that a trigonal planar carbon is flat if viewed from one direction, but not if viewed from the other. Sometimes you will see trigonal carbons drawn differently, if the drawer wants you to look at the molecule from one direction or the other.
See if you can find both kinds of carbon on the model below.
Go to Animation IM9.2. A three-dimensional model of propene.
Carbons that have only two different neighbors are linear; each neighbor is 180° apart from the other. Linear carbons are sometimes called sp carbons.
Propyne has two linear carbons as well as one tetrahedral carbon. See if you can find both kinds of carbon on the model below.
Go to Animation IM9.3. A three-dimensional model of propyne.
Saying a carbon is sp or sp2 is only giving the shape of one atom within a molecule. The molecule itself may have a particular shape which is determined by the geometries of it constituent atoms, but which can't be described simply in words as "tetrahedral" or "linear".
Pentane is roughly a zig-zag shape. In a space-filling model, it resembles a fat caterpillar inching along. Propene is almost boomerang-shaped. Propyne is shaped like a dart; it has a pointed end and a feathered end.
Other molecules have more complicated shapes. The shape of ginkgolide B (a natural product isolated from the medicinally interesting ginkgo biloba tree) is hard to describe. However, it is clear that individual carbons in ginkgolide B are mostly tetrahedral, with just a couple of trigonal planar ones.
Go to Animation IM9.4. A three-dimensional model of ginkgolide B.
Understanding the overall shape of a molecule is particularly important in various aspects of biochemistry. For example, the effect of any drug in the body is strongly influenced by the shape of the drug and how that shape can interact with specific receptors.
Exercise \(1\)
Cis and trans cyclic isomers have two substituents attached to a ring. The cis isomer has both substituents on the same face of the ring. The trans isomer has substituents on opposite faces of the ring. Use wedge-dash projections to draw the shapes of the following molecules:
1. cis-1,2-dichlorocyclopropane, C3H4Cl2 (the numbers mean there is a chlorine on the first carbon and a chlorin on the second carbon)
2. trans-1,2-dichlorocyclopropane, C3H4Cl2 c) cis-1,3-dibromocyclobutane, C4H6Br2 d) trans-1,3-dibromocyclobutane, C4H6Br2
3. cis-1,3-dibromocyclobutane, C4H6Br2
4. trans-1,3-dibromocyclobutane, C4H6Br2
Answer
Exercise \(2\)
Identify the geometry at the indicated carbons in the following molecules: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.08%3A_Line_Drawings.txt |
Previously, we looked at a few common geometries of carbon atoms in molecules. Carbon most often has four neighbouring atoms in the molecule, and in that case it is tetrahedral. However, sometimes there are only two or three neighbouring atoms, and so linear or trigonal planar geometries are also pretty common.
These geometries are found throughout the periodic table for atoms that have 2, 3 or 4 neighbouring atoms within a molecule. However, there are also some other molecular geometries that are pretty common.
Earlier, we looked at the environments of atoms or ions within solid structures and considered the "coordination number" of the atom or ion. Remember, the coordination number is just the number of atoms in close contact with a central atom. We can think about the geometry around a particular atom in a molecule the same way.
Coordination number 4
There are actually two different geometric options that occur with some regularity for coordination number 4. Tetrahedral geometry, which we have seen, maximizes the distances between neighbors. That could be advantageous because it minimizes crowding. In fact, tetrahedral geometry is the most common geometry for tjis coordination number. However, a second geometry is available for a limited number of compounds. In square planar geometry, all four neighbors are found in the same plane and are roughly 90 degrees apart from each other. This geometry is most often observed for certain compounds of palladium and platinum as well as a few other transition metals.
Coordination number 5
In the case of five neighboring atoms, there are two geometries that are roughly equally prevalent. In a trigonal bipyramidal structure, the central atom is surrounded by three neighbors that form a trigonal plane around it. The last two neighbors are above and below this plane. These two neighbors are roughly colinear with the central atom. The trigonal planar positions referred to as "equatorial"; the other two positions are called "axial".
The other common five coordinate geometry is square pyramidal. In this case, the central atom sits above a base of four neighbors; those four positions form a square plane together. A fifth atom occupies the space above the central atom.
Coordination number 6
Almost all examples of atoms with coordination number six can be described as having an octahedral geometry. Four of the neighbors form a square plane around the central atom, with one atom above this plane and one below it. As with the trigonal bipyramid, these positions can be described as equatorial and axial, respectively.
Electronic vs. molecular geometry
There is a striking difference between geometries around transition metals, in the middle of the periodic table, compared to main group p-block elements, in the right hand part of the periodic table. The geometry around a transition metal is simply based on the number of atoms bonded to it. Each surrounding atom takes up some space and so a geometry is adopted to give those atoms roughly equal space around the central atom.
In the p-block, things are generally quite different. It's not just the neighbouring atoms that take up space; the valence electrons on the central atom also have to be accommodated. Thus, in ammonia, NH3, we need to consider not just the three hydrogen atoms that are bonded to the nitrogen, but nitrogen's own lone pair as well, in order to determine the geometry. That's four objects to fit around the nitrogen atom. The normal geometry for four things is tetrahedral. Thus, ammonia takes on a tetrahedral shape. It isn't trigonal planar.
That's sometimes called the overall electronic geometry. It is the geometry of everything, including lone pair electrons. To see the relationship between the atoms -- the nitrogen and its three nieghbouring hydrogens -- just ignore the electron pair for a minute. The nitrogen sits above a trigonal base of three hydrogen atoms. The molecular geometry -- that shape formed by the nitrogen and the three hydrogens -- is pyramidal.
This phenomenon, in which the geometric relationship between atoms is influenced by where the electrons are on a central atom, is pretty common around chalcogens and pnictogens -- that is, oxygen and the other atoms in its column of the periodic table, as well as nitrogen and the other atoms in its group. In those cases, lone pairs appear to force the surrounding atoms a little closer to each other.
Exercise \(1\)
Suppose a compound had a central atom surrounded by six groups, including bonds to other atoms as well as its own lone pairs.
1. What is its overall "electronic" geometry?
2. What would be its molecular geometry if it had one lone pair?
3. What would be its molecular geometry if it had two lone pairs?
Answer a
octahedral
Answer b
Answer c
Exercise \(2\)
Suppose a compound had a central atom surrounded by five groups, including bonds to other atoms as well as its own lone pairs.
1. What is its overall "electronic" geometry?
2. What would be its molecular geometry if it had one lone pair?
3. What would be its molecular geometry if it had two lone pairs?
Answer a
trigonal bipyramidal
Answer b
This time there could be two different answers.
If the lone pair occupies one of the axial positions, it would be pretty close to three other atoms.
If the lone pair occupies one of the equatorial positions, it would be pretty close to only two other atoms. The other equatorial atoms are pretty far away.
The rule is that the lone pair goes in the less crowded position, so this molecule would be see-saw shaped.
Answer c
Again, there are two possible geometries. One of them would be trigonal planar, a pretty common geometry.
However, that places the lone pairs in the more crowded positions. Instead, the molecule is going to be tee shaped.
Exercise \(3\)
Draw structures for the following compounds.
a) H2O b) (CH3)3O+ c) PCl3 d) SF4 e) ClF3 f) PCl5
g) SF6 h) ClF5 i) XeF4
Answer
Exercise \(4\)
What is the molecular geometry of each of the compounds in problem Exercise \(3\) ?
Answer a
bent
Answer b
pyramidal at O, although tetrahedral at C
Answer c
pyramidal
Answer d
see-saw
Answer e
tee
Answer f
trigonal bipyramidal
Answer g
octahedral
Answer h
square pyramidal
Answer i
square planar
Some common geometries are summarized in the table below.
In cases where the molecular geometry is different from the electronic geometry, we sometimes need to make choices about which position a lone pair is more likely to occupy. The general rule is that lone pairs occupy the less crowded site. It is as if the lone pair takes up more space than a bond to an atom. That might seem surprising, because it leaves the atoms in the molecule crowded together. Although these cases are not very common, they are observed in a number of compounds from the p-block. Some examples are summarized in the table below.
Exercise \(5\)
Draw structures for the following compounds.
a) IBr b) ICl3 c) ClF5 d) BrF3 e) SeF4 f) IOF3 g) ArF4
Answer
Exercise \(6\)
Occasionally, atoms can have an even higher number of bonds than we have seen so far. Predict a reasonable geometry for iodine heptafluoride, IF7.
Answer
You may be able to imagine some other possibilites for this number of neighbors, but IF7 adopts a pentagonal bipyramid shape. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.10%3A_Other_Geometries.txt |
There is sometimes controversy in science. Argument can be an important part of how we arrive at a better understanding of things. In the context of molecules and structures, there have been long-running controversies about how to draw certain phosphorus and sulfur compounds. A look at that controversy can tell us some things about how different chemists think about bonding.
The controversy concerns how to draw, and think about, the bonding in a number of "high-valent" sulfur and phosphorus compounds, in which the sulfur or phosphorus atom is bonded to more than the usual number of neighbors.
Sulfate ion, SO42-, is an example of a high-valent sulfur compound. Sulfur, like oxygen, frequently forms two bonds. In sulfate, the sulfur is attached to four different atoms.
We could draw that structure in two ways. A structure that obeys the octet rule would have a single bond to each oxygen. That would satisfy the octet of sulfur. In order to fill in the right number of electrons for this group of atoms, each oxygen will have a negative formal charge. Sulfur will have a charge of plus two.
A different way to draw the structure would include double bonds between sulfur and two of the oxygen atoms. That structure would have less charge separation than the other structure. However, the sulfur would have twelve electrons in its valence shell, not eight. It would exceed its octet.
Can an atom have more than eight electrons in its valence shell? Lots of them can, if they are further down the periodic table. Whether sulfur and phosphorus can is much less clear. Frequently, the fact that these are 3rd row atoms is considered significant. After all, the d electrons begin in the third row. Perhaps sulfur and phosphorus can accommodate extra electrons because they have d orbitals. The d orbitals don't have any electrons in them, yet, so maybe they can accept some from other aroms via bonds.
Many theoretical chemists (including experts in sulfur and phosphorus chemistry) are convinced that this scenario just isn't possible. The d level in these atoms is just too high in energy. Other atoms can't possibly donate electrons to sulfur and phosphorus in this way.
And yet, many chemists (including experts in sulfur and phosphorus chemistry) still draw sulfate with double bonds. Why would they make such a mistake in the face of reason?
Partly, it has to do with experimental evidence. Remember, "experimental" sounds wishy-washy and tentative to non-scientists, but to a chemist the term really means "reality-based". One of the reasons people draw double bonds in many sulfur and phosphorus compounds is that the bonds simply behave like double bonds. That is, they are stronger and shorter than single bonds. We might not think about those bonds in exactly the way we think about double bonds in other situations, but the experimental evidence says that there is some additional attraction between the sulfur and oxygen atoms in these cases.
There is additional evidence that sulfur and phosphorus can "exceed the octet" or form "Lewis-disobedient" structures. Compounds like phosphorus pentafluoride (PF5) clearly require more than four bonds to sulfur. Without five bonds, we would need to draw one of the fluorides as an anion. It's certainly possible that this could be the true structure, but there is evidence from X-ray crystallography that all of the fluorines are attached to phosphorus in PF5, as well as in the related PF6- anion (hexafluorophosphate, a very common molecular ion in X-ray crystallographic studies).
So, sulfur and phosphorus really can form additional bonds. The only true controversy about the octet rule in sulfur and phosphorus chemistry is a little too subtle to explore very deeply at this point: it's whether sulfur can form a double bond in these cases. The issue has to do with ideas from molecular orbital theory, which is a later chapter, and how well d orbitals can form "pi bonds" with the p orbital on oxygen.
Exercise \(1\)
Fill in any missing lone pairs on the following structures.
Exercise \(2\)
Indicate whether each of the structures in Exercise \(1\) (IM11.1) obeys the octet rule. If not, provide a resonance structure that obeys the octet rule.
Answer
Exercise \(3\)
Many oxyphosphorus compounds are commonly drawn in a way that is inconsistent with the octet rule. Draw two structures for each of the following compounds: one that obeys the octet rule but has higher charge separation, and one that has lower charge separation but exceeds the octet on phosphorus. (Note: in a few of the compounds, hydrogen is attached to phosphorus, as suggested in the condensed formula.)
a) H3PO4 or PO(OH)3 b) H3PO3 or HPO(OH)2 c) H3PO2 or H2PO(OH)
d) (CH3)3PO e) (CH3O)PO
Answer
Exercise \(4\)
Draw two structures for each of the following compounds: one that obeys the octet rule but has higher charge separation, and one that has lower charge separation but exceeds the octet on chlorine.
a) hypochlorite, ClO- b) chlorite, ClO2- c) chlorate, ClO3- d) perchlorate, ClO4-
Exercise \(5\)
Draw two structures for each of the following compounds: one that obeys the octet rule but has higher charge separation, and one that has lower charge separation but exceeds the octet on sulfur.
a) sulfite, SO32- b) sulfate, SO42- c) thiosulfate, S2O32-
d) disulfate, S2O72- e) persulfate or peroxomonosulfate, SO52-
Exercise \(6\)
Global warming is a controversial topic in society but there is not a great deal of controversy about it in the scientific community. A summary of evidence can be found at the National Oceanic and Atmospheric Administration's website.
Explain some of the evidence that global warming is occurring and is linked to human activity. Why do you think there is little debate about these ideas among scientists, but lots of debate in the US public as a whole? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.11%3A_Controversial_Lewis_Structures.txt |
Organic compounds are very common in chemistry and biology. Some familiarity with some common types of compounds based on carbon and a few other elements will help you to understand organic chemistry and related fields.
The most common structural piece in organic chemistry is the carbon-carbon bond. Most organic compounds are filled with carbon-carbon bonds. Very often, a network of carbon atoms are single-bonded to each other, and to hydrogen atoms, to build up a complicated structure.
Other pieces, called functional groups, are attached to this framework. Functional groups are specific collections of atoms bonded together in a certain way. These collections of atoms are seen over and over in organic chemistry, and so they are given specific names. This appendix is meant to help you learn to recognize these functional groups. It will also help you learn how to talk about them.
Before looking at functional groups, let's look at simpler compounds that have no functional groups. Alkanes are compounds that contain only carbon and hydrogen, and that contain no double or triple bonds.
Exercise \(1\)
Translate the above structures into Lewis - Kekule structures.
Specific alkanes are given names that say something about their structures. Their names end in "ane"; that means they contain only single bonds, not double or triple bonds. A compound that contains no double or triple bonds is sometimes called "saturated". They are sometimes called "paraffinic" hydrocarbons, which means the same thing as saturated hydrocarbons.
The name also contains a part that tells how many carbons are found connected in the chain. These prefixes are used throughout organic chemistry to name other kinds of compounds, so you need to memorize them.
Prefix and the number of Carbons they have
prefix # carbons
meth 1
eth 2
prop 3
but 4
pent 5
hex 6
hept 7
oct 8
non 9
dec 10
Exercise \(2\)
Provide names for the following alkanes.
The chain of carbons in an alkane is called an alkyl chain. Sometimes there are smaller alkyl chains attached to the main chain of an alkane. The same prefixes can be used to tell how many carbons are in these smaller branches.
Numbers are used to count how far along the chain these branches occur. You need to start on the end of the chain and count the carbons until you get to the place where the branch occurs. If there are two different possible chains, choose the longest chain as the base name. If there are two different directions possible, start at the one that gives the lowest numbers for the branches.
If there are two different branches of the same size on a chain, you need to say so. A different prefix is used to say how many of the same piece are present.
Prefix and the number groups they have
prefix # of groups
di 2
tri 3
tetra 4
penta 5
Exercise \(3\)
Provide names for the following branched alkanes.
"Aliphatic hydrocarbons" specifically refers to branched chains of saturated hydrocarbons. "Alicyclic hydrocarbons" are very similar but they contain rings of carbons. If a chain of carbons wraps around to form a ring, the prefix "cyclo" is used.
All these elements will be used in naming other compounds as well.
Exercise \(4\)
Provide names for the following cyclic alkanes.
Homoatomic functional groups
Any feature other than an alkyl chain is called a "functional group". A functional group is also just a place where reactions take place.
Homoatomic functional groups contain only carbon and hydrogen. They differ from alkyl groups only in having multiple C-C bonding.
The C=C functional group is called an alkene. Simple compounds that contain C=C double bonds are also called alkenes. The term "olefin" also means alkene, and "unsaturated" or "olefinic" hydrocarbons contain double bonds.
In naming alkenes, the suffix "ene" is used instead of "ane". The base name of the alkene describes the length of the chain containing the double bond. A number says how far the double bond is from the nearest end of the chain.
Exercise \(5\)
Provide names for the following alkenes.
The C-C triple bond is called an alkyne. "Acetylenic" hydrocarbons or alkynes are compounds that contain C-C triple bonds. These compounds are also examples of "unsaturated hydrocarbons".
Exercise \(6\)
Provide names for the following hydrocarbons.
Exercise \(7\)
Describe the geometries of the following carbons:
a) any carbon in pentane b) the second carbon in 2-butene c) the first carbon in 1-hexyne
Answer a
tetrahedral
Answer b
trigonal planar
Answer c
linear
Aromatic compounds appear to contain C-C double bonds but they are very different from alkenes. The most common aromatic is benzene. Because benzene is a very common structural piece, you should be familiar with some of the language associated with it.
Benzene rings with two groups attached are common enough that different terms are used to describe their isomers. The terms ortho-, meta- and para- are used only to describe the relationship between two groups around benzene, and not any other compounds. These terms are sometimes abbreviated to o-, m- or p-. Alternatively, two groups attached to a benzene can simply be numbered in order to make it clear where they are. If more than two groups are attached, numbering is used; terms such as "ortho" no longer apply.
Note that a benzene group is sometimes called a phenyl group. It is not called a benzyl group, nor a benzoyl group. Those two groups both contain a phenyl ring but they are not exactly the same.
Exercise \(8\)
Provide names for the following aromatics.
Heteroatomic Functional Groups: Carbonyls
Heteroatomic functional groups contain atoms other than carbon and hydrogen.
Probably the most important set of heteroatomic functional groups is the set that contains carbon-oxygen double bonds. The C=O group is called a carbonyl (carbon-EEL). Carbonyl compounds all contain the carbonyl group.
There are two subsets of carbonyl compounds: regular carbonyls and heteroatom-substituted carbonyls. In regular carbonyls, the carbon in the C=O group is attached only to carbon or hydrogen. It is not attached to additional heteroatoms such as nitrogen or oxygens.
There are two kinds of regular carbonyls. In ketones, the carbonyl carbon is attached only to other carbons. In aldehydes, the carbonyl carbon is attached to a hydrogen atom as well as a carbon.
Aldehydes and ketones have specific naming conventions. The suffix for the name of an aldehyde is "-al"; whereas the suffix for the name of a ketone is "-one". In the case of a ketone, the location of the carbonyl along the chain can be described with a number. Otherwise, the names of aldehydes and ketones follow the rules we have seen for other organic compounds.
Exercise \(9\)
Provide names for the following compounds.
Heteroatom-substituted carbonyls all have a heteroatom attached to the carbonyl carbon. A heteroatom is an atom other than hydrogen or carbon, such as nitrogen, oxygen or chlorine. Heteroatom-substituted carbonyls are often called "carboxylic acid derivatives" or sometimes "carboxyloids".
In carboxylic acids, the carbonyl carbon is attached to an OH group. The OH group is often called a hydroxyl group. Fatty acids and amino acids are examples of biological compounds that contain carboxylic acids.
Two biologically important carboxyloids are amides and esters
In esters, the carbonyl carbon is attached to an oxygen, like in a carboxylic acid. However, the oxygen is not attached to a hydrogen. The oxygen is attached to a carbon. The properties of esters differ enough from carboxylic acids that they are given a different name. Glycerides are biological compounds that contain ester groups. Glycerides are found in fats.
In amides, the carbonyl carbon is attached to a nitrogen. Unlike esters and carboxylic acids, it does not matter whether the nitrogen is attached to a hydrogen or to another carbon. The presence of a hydrogen on the nitrogen changes the properties of the amide less dramatically than in esters and carboxylic acids.
Because the amide does change subtly if there is a hydrogen attached to the nitrogen, there is a way to describe the presence ar absence of hydrogens. An amide with two hydrogens on the nitrogen is called a primary amide. The nitrogen is connected to only one (Latin for first: primus) carbon: the carbonyl carbon.
An amide with two carbons connected to the nitrogen is called a secondary amide (Latin for second: secondus). An amide with three carbons connected to the nitrogen is called a tertiary amide (Latin for third: tertius).
Naming simple examples of these compounds, as before, involves a change in the suffix of the name. The suffix for a carboxylic acid is "-oic acid". Esters really have a two-part name. The first part describes the group attached to the enchained oxygen atom of the ester, with the suffix "-yl". The second part describes the portion that contains the carbonyl, with the suffix "-oate". For an amide, the suffix is "amide". However, other groups attached to the nitrogen are usually prefixed with "N-"; this is a little like numbering the position of a group in previous examples, but this time it underscores that the group is attached to a nitrogen.
Exercise \(10\)
Provide names for the following carboxyloids.
A third biologically important carboxylic acid derivative contains a sulfur attached to the carbonyl carbon. Usually the sulfur is attached to another carbon as well. Because sulfur is in the same group in the periodic table as oxygen, this functional group is similar to an ester. It is called a thioester. It is much less common than amides and esters. A biological example of a thioester is acetyl conezyme A, which plays an important role in many biological reactions. In fact, the thioester plays the key role in the reactions of acetyl coenzyme A.
Naming thioesters is just like naming other esters, except that the prefix "thio-" is inserted into the second part of the name.
Two other important carboxylic acid derivatives are not normally seen in biology. They are important because they are used synthetically or industrially to make the other carboxylic acid derivatives.
Acid anhydrides are prepared by heating carboxylic acids at high temperatures until water is lost. Acid anhydrides contain two carbonyl units, connected by an oxygen. Usually those two units are the same as each other. The name for an anhydride indicates the number of carbons in just one of those pieces (since the other one is the same) with the suffix "-oic (space) anhydride".
Acid chlorides have a chlorine atom attached to the carbonyl carbon. The name for an acid chloride contains the suffix "-oyl (space) chloride". Other acid halides (such as acid fluorides) are known, but they are less common than acid chlorides.
Simple Heteroatomic Functional Groups (No Carbonyls)
There are several heteroatomic functional groups that do not contain carbonyls.
Alcohols contain an OH or hydroxyl group, but the hydroxyl is not attached to a carbonyl.
Sometimes alcohols are classed into sub-groups, as "primary", "secondary" or "tertiary". Those words just describe the carbon attached to the OH group. How many carbons are attached, in turn, to that carbon? Are there two, three, or just one? Those classifications have some influence on how reactive the alcohol will be under different conditions.
The suffix in the name of an alcohol is "ol". It does not matter whether the alcohol is described as a primary, secondary or tertiary one They are still named according to the same rule.
Phenols are alcohols in which the hydroxyl group is directly attached to a phenyl group, or benzene ring. Their behavior is different enough from other alcohols that they are sometimes thought of as a separate group. For one thing, they are somewhat acidic. Coming into contact with phenols, whether they are found in pine resin or, in a much more extreme case, poison ivy, can cause severe itching in the skin. Regular alcohols are much less likely to cause that reaction.
Ethers are kind of like alcohols in that they contain oxygen atoms. Ethers contain a C-O-C group, but neither carbon is part of a carbonyl. Generally, to name an ether, we list the two parts attached to the oxygen atom, and follow those two names with "ether".
Amines contain nitrogen atoms connected to three different atoms, but not connected to a carbonyl. If the nitrogen is attached only to one carbon, the functional group is a primary amine. If it is attached to two carbons, it is a secondary amine. If it is attached to three carbons, it is a tertiary amine. This classification system is a little bit different from the one used with alcohols, in that it refers to the number of things attached to the nitrogen itself. In oxygen compounds, the number of things attached to the oxygen determined whether we called it an alcohol or an ether.
Naming amines is a little like naming ethers. We list the groups attached to the nitrogen, followed by "amine". Usually these groups are listed in alphabetical order. Sometimes, as with amides, the idea that a group is attached to the nitrogen is usually reinforced with the prefix "N-".
Thiols and thioethers are the sulfur analogues of alcohols and ethers. Phosphines are the phosphorus analogues of amines. None of these groups are as common as alcohols, ethers and amines. The sulfur-containing compounds are named in a similar way to their oxygen analogues, but with the suffix "-thioether" or "-thiol" used instead of "-ether" or "-ol". Similarly, phosphines have the suffix "-phosphine" instead of "-amine".
Alkyl halides, or haloalkanes, are compounds that contain a halogen that is not attached to a carbonyl. These compounds are not common biologically, but they are important industrially. Usually halogens are treated like alkyl groups in terms of naming compounds that contain them. A halogen is described as a prefix ending in "-o", rather than the prefix ending in "-yl" that is used with alkyl groups. For example, an alkyl bromide is named as a bromoalkane, such as bromoethane or bromohexane.
There are a few other functional groups that contain multiple bonds to nitrogen. An imine is the nitrogen analogue of an aldehyde. An imine contains a carbon-nitrogen double bond. The carbon almost always has a hydrogen attached as well. The word "imine" is just appended to the name of the related aldehyde to describe a specific imine.
A nitrile contains a carbon-nitrogen triple bond. The suffix for a nitrile is just "-nitrile".
A nitro group contains the NO2 unit. "Nitro-" is usually added as a prefix to signal the presence of this group in a compound.
The fun group and their suffixes
fun group suffix
none ane
alkene ene
alkyne yne
ketone one
aldehyde al
carboxylic acid oic acid
ester -yl ...-oate
amide amide
thioester
Exercise \(11\)
Provide names for each of the following compounds.
Exercise \(12\)
Each of the following compounds contains three functional groups. Identify them.
Exercise \(13\)
Identify the circled functional groups.
Exercise \(14\)
Identify the circled functional groups (more advanced level). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.12%3A_Organic_Functional_Groups.txt |
We have seen a number of common "functional groups" from organic chemistry. Functional groups are just collections of atoms that we frequently see, so that it becomes useful to recognize them. Organic compounds, by definition, are carbon-based compounds, usually derived from living things. However, there are a few specific classes of organic molecules that are so common in biology that they are termed "biomolecules". We are going to take a look at them here.
One common group of biomolecules is called "lipids". Lipids can really have lots of different structures. However, the most common type of lipid is an ester. An ester contains an O-C=O unit. Sometimes that group is termed an "ester linkage". We can think of that link as being formed when a carboxylic acid comes together with an alcohol. The carboxylic acid is shown below, in red; the alcohol is in blue.
This carboxylic acid is a particular kind called a fatty acid. It has a long carbon chain. When the alcohol oxygen connects to the carbonyl carbon in the fatty acid, the fatty acid loses its OH group. The alcohol also loses the hydrogen, or proton, from its OH group. That combination of things lost, H + OH, adds up to a water molecule. So every time a fatty acid combines with a alcohol molecule to form an ester linkage, a water molecule is made.
That formation of a water molecule gave rise to an old name for this reaction and others like it. It was called a condensation reaction. That term, "condensation", referred to the water or moisture that was formed during the reaction, like the condensation on the bathroom mirror when you take a shower.
In reality, exactly how these two molecules come together to form one molecule is slightly more complicated than what we are seeing here. At this point, however, this is all you need to know.
The most common kind of lipid in biology is a specific kind of ester called a "glyceride". A glyceride forms between a fatty acid and a glycerol unit. The glycerol unit, in blue, has three hydroxy groups (the OH part). These hydroxy groups are each capable of attaching to the carbonyl carbon in a fatty acid.
When the glycerol oxygen connects to the carbonyl carbon in the fatty acid, the fatty acid loses its OH group, and a proton is lost from the glycerol. So every time a fatty acid combines with a glycerol molecule, a water molecule is made, in addition to the lipid. That's three molecules of water lost, in total, from the original glycerol and fatty acids.
Ester linkages are important partly because they are reversible. Fatty acids can be conveniently stored as glycerides by attaching them to glycerol with ester linkages. However, ester linkages can break open again if water is added. The triglycerides can go back to being glycerol and fatty acids again. Those fatty acids can be used as fuel for the cells. The formation and breaking of ester linkages allows for these processes to be controlled, depending on whether energy is needed right now or should be stored for later.
Incidentally, there are a variety of fatty acids that can form triglycerieds, leading to a diversity of lipids with slightly different properties. There are fatty acids containing carbon chains of various lengths. There are fatty acids with one, two, or three C=C double bonds somewhere in their chains, and others without any. The ones that don't have any double bonds are called saturated fatty acids. Saturated fats are sometimes described by a shorthand notation that tells how long the chain is, followed by a colon, followed by a zero, indicating there are no C=C double bonds.
Unsaturated fats are sometimes described using a similar shorthand notation, but with a number after the colon indicating how many double bonds are present. That's followed by the symbol Δ, the Greek letter delta, with a supercsript number that tells how far the double bond is from the end of the chain.
Sometimes you will hear about omega-3 or omega-6 fatty acids, or some other variation. That's the Greek letter, ω, which is the last letter in the Greek alphabet. It stands for the last carbon in the chain, and the number after it tells you how far the double bond is from the end of the chain.
Exercise \(1\)
Draw the triglycerides that would result in each of the following cases.
Exercise \(2\)
Phospholipids are important in the formation of cell membranes. In a phospholipid, a phosphate replaces one of the alcohols that would otherwise make a triglyceride. Draw the pjospholipid that would result in the following case.
Exercise \(3\)
Draw the following fatty acids, based on the notation given.
1. linolenic acid, 18:3 Δ3,6,9
2. linoleic acid, 18:2 Δ6,9
3. stearidonic acid, 18:4 Δ3,6,9,12
4. myristic acid, 14:0
5. palmitoleic acid, 16:1 Δ7
6. adrenic acid, 22:4 Δ6,9,12,15
7. arachidonic acid, 20:4 Δ6,9,12,15
8. palmitic acid, 16:0
Answer
Carbohydrates are an even more crucial form of fuel storage for the cells. Carbohydrates are carbon-based molecules containing lots of OH groups. Glucose is an example. The drawings below are just a reminder that we usually work with skeletal line structures, like the one on the right, rather than the structures with atom labels, like the one on the left, because things get cluttered pretty quickly in the more explicit drawings. However, if you need to re-draw things with atoms to keep track of where everything is, don't be afraid to do that.
Carbohydrates, like fatty acids, are stored in a form in which they are linked together. When needed, they can be unlinked and the individual units, such as glucose, can be sent into energy-generating reactions.
Carbohydrates (or sugars) are connected by "ether linkages". An ether is just a C-O-C unit. In forming an ether linkage, two carbohydrates could come together, attaching one oxygen on one carbohydrate to a carbon on the next, and releasing a water molecule. It's another condensation reaction, like the ester-forming one. This time, a two-carbohydrate pair is formed, called a "disaccharide". The individual carbohydrates are called "monosaccharide"s.
There are lots of carbons that the oxygen atom on a neighbouring carbohydrate might attach to, but the same one is always used. It's the carbon that already has two oxygens attached to it. The carbon in the right-hand corner of the red glucose molecules is an example. As a result, the particular kind of ether linkage between carbohydrates always looks more like C-O-C-O-C. You can see that pattern if you start from the left-hand corner of the blue glucose in the disaccharide, and move leftward across the gap onto the red glucose.
This sort of double-ether linkage is called a "glycoside linkage" in biochemistry. Don't get it confused with the triglyceride made from a fatty acid plus glycerol. In organic chemistry, this C-O-C-O-C unit is called an "acetal".
These bonds can be used to connect lots of carbohydrates together. Two monosaccharides can be linked to make a disaccharide. A third one would make a trisaccharide. Lots of them connected in a chain would be called a polysaccharide. Other small molecules can be connected in chains, too, although not always by glycoside or acetal linkages. When many small molecules have been connected into one big molecule, we call that molecule a "polymer". That word just means it is made of many things, all put together. A polysaccharide such as cellulose (below) is an example of a biological polymer.
Exercise \(4\)
Draw the disaccharides that would result in each of the following cases. Make sure to use the acetal (glycoside) carbon on one molecule; you can link to the primary alcohol (the -CH2-OH group) on the other.
Amino acids are a third general class of biological molecules. Amino acids are important because they are collected together into polymers called proteins. Proteins function like machines; they carry out most of the tasks needed to keep an organism alive.
Amino acids get their name from the fact that they contain two different organic functional groups: an amine (NH2) and a carboxylic acid (CO2H). In order to make proteins, the amine nitrogen on one amino acid connects to the carbonyl carbon (on the C=O group) of the next. Once again, an OH is lost from one molecule and a proton from the other, to make water.
As in lipids, exactly how these two molecules eventually come together is slightly more complicated than that, but this picture is good enough for now. Two amino acids (two monomers) connect together to make one bigger molecule (a dimer) and a water molecule. That dimer is often called a "dipeptide". The bond that has formed between the nitrogen and the carbonyl is called an "amide" bond in organic chemistry, becasue the funtional group formed is called an amide. In biochemistry, this bond is often called a "peptide linkage". It's what holds the two pieces together in a dipeptide.
As with carbohydrates, amino acids can be linked into long chains. They can make polymers. A polymer of amino acids is called a polypeptide. A protein is just an extremely large polypeptide, in which hundreds of amino acids are connected together in a long chain.
Lipids and carbohydrates are used for storage, among other things, and so it is useful to be able to undo ester linkages and acetal linkages, to get individual fatty acids and carbohydrates again when we need to consume energy. If we get desperate, we could also break down our proteins and polypeptides and get energy from them, but we don't usually use polypeptides for energy storage, because they have so many more valuable roles to play.
In fact, peptide bonds are a little stronger and harder to break than either ester linkages or acetal (glycosidic) linkages. That's because of a resonance contributor to the structure. Notice that oxygen is more electronegative than nitrogen. That oxygen is able to pull the lone pair on the nearby nitrogen closer, forming an N=C π bond that makes the amide stronger and more stable. We still have special enzymes that can break down proteins in our food, but it's not easy.
There are a huge variety of proteins that can result from linking amino acids together in this way. That diversity largely results from the fact that lots of different compounds have the same basic amino acid structure, but with modified groups called "side chains". A few examples are shown below.
There are actually twenty common amino acids in biology. To help organize these compounds, people normally use four different categories that correspond to the properties of the side chains. Two groups of amino acids commonly have charged side chains under biological pH conditions. Acidic side chains give up a proton easily (H+) and become negatively charged as a result; there are two of these, aspartate and glutamate. (Aspartate is just the word for the negatively charged form of aspartic acid; that name is often used because it is shorter than "aspartic acid".) Basic side chains pick up protons easily and are positively charged as a consequence. There are three of them, although only two are usually protonated at neutral pH; the other becomes protonated under slightly more acidic conditions (lower pH).
The polar (or "hydrophilic") amino acids are those that would interact easily with polar water molecules in the cell. You can usually spot them by the presence of strong dipoles in the side chains. Strong dipoles are charge separations between two different atoms that have very different electronegativity values. For example, C-O and O-H bonds are polar. In contrast, the non-polar (or "hydrophobic") amino acids have side chains that lack strong dipoles. For example, C-C and C-H bonds are not very polar, because the electronegativity differences between the atoms are small.
Here are all of the common amino acids, arranged according to category, as they would appear when linked in a peptide chain.
A couple of the amino acid side chains are sort of on the edge in terms of polarity, and some textbooks will place them in a different category. For example, tyrosine has an O-H bond and so is capable of hydrogen bonding with water. That makes it pretty hydrophilic. However, the majority of the side chain contains non-polar C-C and C-H bonds, so sometimes you will see it grouped with hydrophobic side chains.
One more pair of sub-groups you will sometimes see are the aromatic and aliphatic amino acids. Aliphatic ones contain carbon chains, but not other functional groups, whereas aromatics contain "aromatic rings" such as benzene. Aromatics are usually six-membered rings with double bonds alternating all the way around. They are flat groups that can stack on top of each other and sometimes that property can be important.
Exercise \(5\)
Draw the dipeptides that would result in each of the following cases.
If you don't know what the wedged and dashed lines in the drawing mean, don't worry about it. They just represent different orientations in space. You will learn about these representations in a later topic called "stereochemistry".
Exercise \(6\)
Draw the possible dipeptides that would result in each of the following cases.
Exercise \(7\)
Draw the possible tripeptides that would result in each of the following cases.
Possibly the most well-known biomolecules are the nucleic acids, DNA and RNA. These compounds are examples of co-polymers. Like complex carbohydrates, they are composed of individual molecules that have become covalently bonded to each other to form polymer chains. However, DNA and RNA always contain three different types of monomer: bases, sugars, and phosphates.
The sugar is the part that gives rise to the D in DNA or the R in RNA. Ribose, found in RNA, is a five-membered ring with oxygens on each carbon, mostly as OH groups. Deoxyribose, found in DNA, is almost identical, but is missing one OH group.
One of four different bases can attached to this sugar in DNA, and one of four different bases can be attached to this sugar in RNA. Three of those bases can be found in either molecule: adenine, cytosine, and guanine. The fourth is unique to either DNA (thymine) or RNA (uracil).
These five bases (so called because, like the basic amino acid side chains, they can pick up protons) fall into two categories. Pyrimidines (peer-IM-id-eens) each contain a six membered ring. Purines (PURE-eens) contain both a six-membered and a five-membered ring. These categories are important because of "base-pairing"; that's when a strand of DNA can bind to a complementary strand via hydrogen bonding between a base on one strand and another base on another. In base-pairing, a purine always binds to a pyrimidine partner: guanine to cytosine, and adenine to thymine (in DNA) or uracil (in RNA).
The bases are covalently bonded to sugars at the glycoside position - that same place sugars link to each other in carbohydrates. The new bond is called a N-glycoside bond. The new compound is called a nucleoside.
The third type of molecule that comes together to form the nucleic acid co-polymer is a phosphate. It becomes covalently bonded to the primary alcohol group at the other end of the ring from the base. The bond between the phosphate and the sugar is called a phosphate ester bond. The new compound is called a nucleotide.
To form a co-polymer, another phosphate ester bond forms with the sugar of another nucleotide. The phosphate between the two sugars is sometimes called a phosphodiester linkage, meaning that the phosphate is forming two esters at once to bridge the two sugars.
Sometimes, people will refer to the 3'-end or the 5'-end of a nucleic acid strand. Those numbers refer to a system of numbering the carbons in the sugars. If you are talking about the end of the DNA chain that is closest to carbon number 3 of the nearest sugar, you are talking about the 3' end.
Exercise \(8\)
Identify the base in each of the following nucleosides. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.13%3A_Common_Biomolecules.txt |
Line structures will be the most common type of structural drawing you see in these pages. You will also see some others used for specific purposes, and those representations are explained here.
Sometimes even drawing a line structure is too much. Proteins often contain thousands of atoms. Presenting all that information about bonding could make the drawing hard to read, especially in a three-dimensional shape.
A picture of a molecule can be presented in several ways. A ball-and-stick model gives a pretty good idea of where all of the atoms are, and how they are connected. The balls represent the atoms and the sticks represent the bonds. An example of a ball-and-stick picture of a protein, bovine low molecular weight protein tyrosyl phosphate, binding to inorganic phosphate, is shown below.
Go to Animation IM14.1. A three-dimensional model of bovine low molecular weight protein tyrosyl phosphatase, binding to inorganic phosphate.
There are a couple of problems with viewing a protein this way, though. The first problem is that we usually don't know where the hydrogen atoms are, so they are missing from the ball-and-stick structure. Why is that?
It's because protein structure is usually determined experimentally using X-ray crystallography. That means someone has gone to the trouble of isolating a pure sample of protein from an organism. They dissolved the protein very carefully, then coaxed the protein to form crystals. There are different ways to do that, but usually it involves cooling the solution, so that the solvent no longer has enough energy to keep all of the molecules dissolved, or else evaporating a little of the solvent, so that there is no longer enough solvent to keep everything dissolved. This process might take weeks. At that point, they could carefully lift a single crystal out of the solution and mount in on an X-ray diffractometer. When X-rays shine through the crystal, they essentially bounce off the atoms and scatter in different directions. By measuring where the X-rays land, and using some computer software that exploits those imaginary numbers that you learned about in high school but didn't think were useful for anything, the researcher can determine the location of every atom in the molecule (if they get lucky). Hydrogen atoms are just too small to locate with any certainty, however.
• X-ray diffractometry can reveal the location of every atom in a molecule, other than hydrogen.
• X-ray diffractometry requires that the compound is crystalline.
The other problem is that a ball-and-stick model of a protein contains an awful lot of atoms. The structure gets pretty complicated. The example here is a pretty small protein; its molecular weight is only about 35,000 D (D stands for Daltons; a Dalton is just another term for atomic mass unit or amu, but it is used more commonly in biochemistry and macromolecular chemistry). In larger proteins, its difficult to make sense of the pile of atoms we see in a ball-and-stick structure.
Similarly, we could view the protein in a space-filling model. The space-filling model probably gives a better picture of what the molecule would look like in terms of how much room it would take up, or what its overall shape would be. The advantage of this view is that it allows a biochemist to see how other molecules might interact with the protein.
For example, maybe a protein acts as an enzyme; it catalyses reactions that are essential for our survival. Bovine low molecular weight protein tyrosyl phosphatase is an example. It's bovine, which means it is found in cows. It's a phosphatase, which means it removes phosphate groups from other proteins. That is an important regulatory event in biochemistry. Adding a phosphate group or taking a phosphate group off another molecule, including another protein, often acts as an "on/off" switch, so that an organism can control when certain reactions take place. This is a tyrosyl phosphatase, so it specifically removes phosphate that has been added to a tyrosine residue in the substrate protein. By looking at the space-filling model, a biochemist may be able to see where the tyrosine from the substrate can fit into the bovine low molecular weight tyrosyl phosphatase, so that the enzyme can do its job and chop off the phosphate.
The trouble is, most proteins look pretty much like blobs to the untrained eye. To make things a little easier, there are shortcuts that biochemists use when visualising proteins.
One abbreviation used in biochemistry is sometimes called a wireframe structure. In a wireframe structure, none of the atoms are labeled and all of the hydrogens are omitted. That means it is difficult to tell a carbon from a nitrogen. However, biological macromolecules like DNA and proteins are made of a few simple building blocks. A very experienced reader would remember the structures of these building blocks. She would be able to tell what atoms go in which place based on the shapes of the building blocks.
That's still a pretty complicated picture. There are other views that are simplified further.
A protein is made of amino acids strung together in a chain. The chain does not hang out in a long, straight line. It folds up into a coil, and the shape of that coil influences its biological activity. Sometimes a biochemist may be interested in how the chain of amino acids coils up. A backbone structure only shows the atoms that are connected directly in a row from one end of the chain to the next. Any atoms that hang out along the sides of the chain are left off.
Those parts that were left off are pretty important in terms of the function of the protein. They are the parts that make alanine different from lysine, for example. However, if you are trying to see some basic shapes in the protein, they just end up getting in the way in the picture.
The order in which a chain of amino acids is connected together is called the primary structure of a protein. For example, the amino acids glycine, alanine, tyrosone and arginine could be connected in the order gly-ala-tyr-arg, or ala-gly-arg-tyr, or tyr-gly-ala-arg, and so on. These compounds are isomers of each other. Knowing which isomer we have tells us something about the protein.
The way that chain arranges itself is called the secondary structure. Very often, proteins coil up into spirals or helices. This helical structure is similar to that of DNA. Another possible secondary structure is a beta-sheet, which is sort of wavy, a bit like a sheet of corrugated steel. These secondary structures are held in place by hydrogen bonds between different amino acids. (You can see the structure-property relationship chapter to review hydrogen bonding).
In the backbone drawing of bovine low molecular weight tyrosyl phosphatase, you can see where parts of the backbone are coiled up into alpha-helices. These parts are coded in pink in the picture. You can also see some sections that form pleated beta-sheets. Those are the yellow, zig-zag sections next to each other.
Usually, the entire protein does not adopt the same secondary structure. There may be a section that coils into a helix and another section that forms a beta-sheet. The helical domain and the beta-sheet domain then arrange themselves and pack together somehow. The arrangement of domains in the protein is called tertiary structure.
Seeing the tertiary structure in a protein can be difficult even with a backbone structure. Instead, ribbon drawings or cartoon structures are used to convey how beta-sheets and helices are arranged in a protein.
Sometimes a protein is made of more than one chain of amino acids. The arrangement of different coils of amino acids all stuck together is called quaternary structure. Bovine low molecular weight tyrosyl phosphatase is a monomeric structure; it is just made of one protein chain, so we don't see a quaternary structure in this case. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.14%3A_Drawings_for_Large_Biological_Comp.txt |
Exercise $1$
Tantalum nitride (TaN) is a very hard material used in automobile transmissions and other high-performance applications.
1. Given its position in the periodic table, what is the charge on nitride ion?
2. Therefore, what is the charge on tantalum in tantalum nitride?
3. Tantalum nitride most commonly adopts a hexagonal structure. The drawing below shows one unit cell. Place a tantalum atom in each corner of the unit cell.
1. Tantalum nitride is a layered structure. The nitrides lie in the same planes as the tantalum atoms. Within each layer, there is one nitrogen in between each pair of tantalum atoms. Add the nitrogens to each layer.
2. How many tantalums are in a unit cell? Show how you know.
3. How many nitrides are in a unit cell? Show how you know.
4. The resulting formula doesn’t match our expectations. Add atoms to the middle layer to balance things out.
5. What is the coordination geometry of the atoms in the middle layer? Heating tantalum metal under dinitrogen or ammonia leads to formation of hexagonal tantalum nitride. What if other methods could lead to other structures of the same compound? These new structures could also have useful properties.
6. Tantalum nitride also forms a rare cubic phase. In this phase, assume the nitrides form a face-centered cubic array, with tantalum in the octahedral holes. Fill in the atoms.
Wolczanski (Cornell University) reported the synthesis of a molecular tantalum nitride compound (adapted with permission from Banaszak Holl, M. M.; Wolczanski, P. T.; Van Duyne, G. D. J. Am. Chem. Soc. 1990, 112, 7989-7994; Copyright 1994 American Chemical Society).
j) There are two different molecular geometries at tantalum. Add two labels pointing to two tantalum atoms (example on right) describing their different geometries.
k) There are two different molecular geometries at nitrogen. Add two labels pointing to two nitrogen atoms (example on right) describing their different geometries.
l) There are two different tantalum-nitrogen bonds in this compound. Identify which bond length corresponds to which tantalum nitrogen bond(s) by adding two more labels to the drawing.
TaN = 194 nm and TaN = 212 nm
Heating this compound to 800 °C results in formation of the less common cubic form of tantalum nitride.
m) Propose a reason why this compound leads to formation of cubic tantalum nitride instead of hexagonal.
Answer a
N3- would get to a noble gas configuration.
Answer b
Ta3+ would balance the charge in TaN.
Answer c
Answer d
Answer e
#Ta = (\frac{1}{6})(\frac{1}{2})(4)\) for the acute corners and (1/3)(1/2)(4) for the obtuse corners
$= \frac{4}{12} + \frac{4}{6} \nonumber$
$= \frac{4}{12} + \frac{8}{12} \nonumber$
= 1 (note that it's the same outcome as the corners of a cube)
Answer f
$\# = \frac{1}{4}(8)$ for the edges and $frac{1}{2}(2)$ for the faces
$= 2 + 1 \nonumber$
$=3 \nonumber$
Answer g
Need 2 more Ta.
Answer h
Each tantalum has three nitrogens above and three below it. It's almost octahedral, but the top layer of nitrogens is lined up above the bottom layer rather than being twisted 120 degrees to form an octahedron. The geometry is a trigonal prism.
Answer i
Answer j
Answer k
Answer l
The double bonds hold the atoms more closely together than the single bonds.
Answer m
You can imagine the molecules stacking together to make a cubic array of TaN. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.15%3A_Application_Problems.txt |
Exercise 4.7.4:
a) bromide b) oxide c) fluoride d) carbonate e) nitrate f) nitrite
g) sulfide h) sulfate i) sulfite j) persulfate k) carbide l) nitride m) arsenide
n) phosphate o) phosphite p) iodide q) iodate r) periodate
Exercise 4.8.2:
a) (CH3)2CHCH2CH2CN b) CH3CH2CH(OH)CH3
c) (CH3)2CHCOCH2CH3 d) CH3CH2CONHCH2CH3
a) octahedral.
b)
c)
Exercise 4.10.2:
1. trigonal bipyramidal
2. This time there could be two different answers.
If the lone pair occupies one of the axial positions, it would be pretty close to three other atoms.
If the lone pairs occupies one of the equatorial positions, it would be pretty close to only two other atoms. The other equatorial atoms are pretty far away.
The rule is that the lone pair goes in the less crowded position, so this molecule would be see-saw shaped.
c) Again, there are two possible geometries. One of them would be trigonal planar, a pretty common geometry.
Exercise 4.10.4
a) bent b) pyramidal at O, although tetrahedral at C c) pyramidal
d) see-saw e) tee f) trigonal bipyramidal
g) octahedral h) square pyramidal i) square planar
Exercise 4.10.6:
You may be able to imagine some other possibilites for this number of neighbors, but IF7 adopts a pentagonal bipyramid shape.
Exercise 4.12.2:
a) propane b) pentane c) hexane
Exercise 4.12.3:
a) 3-methylhexane b) 2,2-dimethylpentane c) 2,3-dimethylbutane
d) 2,2,3,3-tetramethylpentane e) 3,5-dimethylheptane f) 4-ethyl-3,6-dimethyloctane
Exercise 4.12.4
a) cyclopentane b) cyclohexane c) cyclooctane
d) methylcyclobutane e) 1,1,3-trimethylcyclopentane f) 1,3-dimethylcycloheptane
Exercise 4.12.5:
a) 1-hexene b) 2-methyl-2-pentene c) 1-methylcyclohexene d) 2,4,6-trimethyl-2-heptene
Exercise 4.12.6:
a) cyclopentene b) 1,1-dimethylcyclohexane c) 3-hexyne
d) 4-methylcyclohexene e) 1-hexyne
Exercise 4.12.7:
a) tetrahedral b) trigonal planar c) linear
Exercise 4.12.8:
a) methylbenzene b) propylbenzene c) 1,2-dimethylbenzene or o-dimethylbenzene (also o-xylene)
d) 1,3-dimethylbenzene or m-dimethylbenzene (also m-xylene) e) 1,4-diethylbenzene or p-diethylbenzene
f) 2-ethyl-1,4-dimethylbenzene
Exercise 4.12.9:
a) 2,2-dimethylhexanal b) 2-methylcyclopentanone
c) 3-nonanone d) 2,4-dimethyl-2-hexenal
Exercise 4.12.10:
a) butyl propanoate b) N,N-diethylbutanamide
c) 6-methylheptanoic acid d) 4-pentenoic acid
Exercise 4.12.11:
a) 1-chloro-2-methylcyclohexane b) cyclooctanol c) ethyl cyclopentyl ether
d) N-propylcyclohexylamine e), 5,5-dimethylheptan-2-ol f) 3-bromo-4,4-dimethyloctane
g) dibutylamine i) methyl phenyl ether (or anisole) j) ethane thiol k) diethyl thioether
l) triethylphosphine m) butanenitrile n) nitromethane
Note that sometimes a number is located directly in front of the suffix for the group to which it refers.
Exercise 4.12.12
1. benzene (or aromatic), ketone and ether
2. bromide, amine and aldehyde
3. alcohol, thiol and ester
4. thioether, amide and alkene
5. alkyne, alcohol and carboxylic acid
Exercise 4.13.5:
If you don't know what the wedged and dashed lines in the drawing mean, don't worry about it. They just represent different orientations in space. You will learn about these representations in a later topic called "stereochemistry".
Exercise 4.15.1:
1. N3- would get to a noble gas configuration.
2. Ta3+ would balance the charge in TaN.
c)
d)
e) $\# Ta = (\frac{1}{6})(\frac{1}{2})(4)$ for the acute corners and $(\frac{1}{3})(\frac{1}{2})(4)$ for the obtuse corners = $\frac{4}{12} + \frac{4}{6} = \frac{4}{12} + \frac{8}{12} =1$
(note that it's the same outcome as the corners of a cube)
f) $\# N = \(\frac{1}{4})(8)$ for the edges and $(\frac{1}{2})(2)$ for the faces = $2 + 1 =3$
g) Need 2 more Ta.
h) Each tantalum has three nitrogens above and three below it. It's almost octahedral, but the top layer of nitrogens is lined up above the bottom layer rather than being twisted 120 degrees to form an octahedron. The geometry is a trigonal prism.
i)
j)
k)
l) The double bonds hold the atoms more closely together than the single bonds.
m) You can imagine the molecules stacking together to make a cubic array of TaN. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/04%3A_Introduction_to_Molecules/4.16%3A_Solutions_to_Selected_Problems.txt |
One of the most important properties of molecules in the biological world is three-dimensional shape. Organisms interact with molecules based to a large extent on molecular shape. Many compounds fit together with cell-based receptors (which are themselves molecules of very particular shapes). When this happens, a biological response results: a signal to the brain, the onset of cell division, or other possibilities.
Because shape can influence how two molecules fit together, it can affect physical properties as well as biological ones. For example, if a compound is made of molecules that cannot fit together easily, London dispersion interactions are limited, and so the compound is less likely to be a solid.
Stereochemistry refers to the three-dimensional arrangement of atoms. When two compounds contain the same atoms that are connected in the same order, but have different intrinsic shapes, they are steroisomers.
In biology, stereoisomers result from chiral centers. Because biological molecules are comprised largely of carbon, and because carbon usually adopts a tetrahedral shape, there are often two different ways for four groups to be arranged around a carbon atom.
Sterochemistry are not just found in biology, however. Certain minerals contain right- or left-handed helices, and chiral molecules have been observed by astronomers. It is even possible that some of the chiral molecules found in early biology were influenced by templates from meteorites or other astronomical debris.
5.02: Stereoisomers
Two compounds that have the same formula and the same connectivity do not always have the same shape. There are two reasons why this may happen. In one case, the molecule may be flexible, so that it can twist into different shapes via rotation around individual sigma bonds. This phenomenon is called conformation, and it is covered in a different chapter. The second case occurs when two molecules appear to be connected the same way on paper, but are connected in two different ways in three dimensional space. These two, different molecules are called stereoisomers.
• There may be more than one way to arrange the same groups around the same atom with the same geometry.
• This phenomenon is called stereochemistry.
One simple example of stereoisomers from inorganic chemistry is diammine platinum dichloride, (NH3)2PtCl2. This important compound is sometimes called "platin" for short. As the formula implies, it contains a platinum ion that is coordinated to two ammonia ligands and two chloride ligands (remember, a ligand in inorganic chemistry is an electron donor that is attached to a metal atom, donating a pair of electrons to form a bond).
Platin is an example of a coordination compound. The way the different pieces of coordination compounds bond together is discussed in the chapter of Lewis acids and bases.
Go to Animation SC2.1. A three-dimensional model of cis-platin.
Go to Animation SC2.2. A three-dimensional model of trans-platin.
For reasons arising from molecular orbital interactions, platin has a square planar geometry at the platinum atom. That arrangement results in two possible ways the ligands could be connected. The two sets of like ligands could be connected on the same side of the square or on opposite corners.
These two arrangements result in two different compounds; they are isomers that differ only in three-dimensional space.
• The one with the two ammines beside each other is called cis-platin.
• These two ligands are 90 degrees fom each other.
• The one with the ammines across from each other is trans-platin.
• These two ligands are 180 degrees from each other.
Although these two compounds are very similar, they have slightly different physical properties. Both are yellow compounds that decompose when heated to 270 degrees C, but trans-platin forms pale yellow crystals and is more soluble than cis-platin in water.
• Cis and trans isomers have different physical properties.
Cis-platin has clinical importance in the treatment of ovarian and testicular cancers. The biological mechanism of the drug's action was long suspected to involve binding of the platinum by DNA. Further details were worked out by MIT chemist Steve Lippard and graduate student Amy Rosenzweig in the 1990's. Inside the cell nucleus, the two ammines in cis-platin can be replaced by nitrogen donors from a DNA strand. To donate to the Lewis acidic platinum, the DNA molecule must bend slightly. Normally that bend is detected and repaired by proteins in the cell. However, ovarian and testicular cells happen to contain a protein that is just the right shape to fit around this slightly bent DNA strand. The DNA strand becomes lodged in the protein and can't be displaced, and so it is unable to bind with other proteins used in DNA replication. The cell becomes unable to replicate, and so cancerous growth is stopped.
• Cis and trans isomers have different biological properties.
Exercise \(1\)
Draw the cis and trans isomers of the following compounds:
a) (NH3)2IrCl(CO) b) (H3P)2PtHBr c) (AsH3)2PtH(CO)
Answer
Exercise \(2\)
Only one isomer of (tmeda)PtCl2 is possible [tmeda = (CH3)2NCH2CH2N(CH3)2; both nitrogens connect to the platinum]. Draw this isomer and explain why the other isomer is not possible.
Answer
Still pictures of models obtained using Spartan 14 from Wavefunction, Inc., Irvine, California. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.01%3A_Introduction.txt |
Optical rotation
Exercise $1$
The (+) enantiomer of compound A has an optical rotation of 125o. If a pure sample of compound A has an optical rotation of 100o, what is the composition of the sample?
Answer
Pure = 125o
Optical purity = $\frac{100}{125} = 0.80$
$\% Major = 80 + \frac{20}{2} = 90 \% \nonumber$
$\% Minor = 100-90 = 10 \nonumber$
Exercise $2$
The (+) enantiomer of compound B has an optical rotation of 100o. If a pure sample of compound B has an optical rotation of 95o, what is the composition of the sample?
Answer
Pure = 100o
Optical purity = $\frac{95}{100} = 0.95$
$\% Major = 95 + \frac{5}{2} = 97.5\% \nonumber$
$\% Minor = 100 -97.5 = 2.5 \% \nonumber$
Exercise $3$
The (+) enantiomer of compound C has an optical rotation of 18o. If a pure sample of C contains 30% of the (+) enantiomer and 60% of the (-) enantiomer, what is the optical rotation value?
Answer
Pure = 18o
$\% Major = 60 \% \nonumber$
$\% Minor = 40 \% \nonumber$
Optical purity $= \frac{X}{18} = 20 \%$
Solve for X:
$X = 3.6^{o} \nonumber$
Exercise $4$
The (+) enantiomer of compound D has an optical rotation of 25o. If a pure sample of D contains 80% of the (+) enantiomer and 20% of the (-) enantiomer, what is the optical rotation value?
Answer
Pure = 25o
$\% Major = 80 \% \nonumber$
$\% Minor = 20 \% \nonumber$
Optical purity $= \frac{X}{25} = 80-20 = 60 \%$
Solve for X:
$X = 15^{o} \nonumber$
The polarimetry experiment
Exercise $5$
A pure sample of the naturally-occurring, chiral compound A (0.050 g) is dissolved in water (2.0 mL) and the solution is placed in a 0.5 dm cell. Three polarimetry readings are recorded with the sample: 0.625o, 0.706o, 0.682o. What is [a]?
Answer
$[a] = \frac{observed \: rotation}{(l)(c)} \nonumber$
$c = \frac{0.050 g}{2.0mL} = 0.025 \frac{g}{mL} \nonumber$
Average observed rotation $= \frac{0.625 + 0.706 + 0.682}{3} = 0.671^{o}$
$[a] = \frac{0.671^{o}}{(0.025 \frac{g}{mL})(0.5 dm)} = 53.68^{o} \nonumber$
Exercise $6$
A pure sample of the chiral compound B (0.540 g) is dissolved in ether (2.0 mL) and the solution is placed in a 1.0 dm cell. Three polarimetry readings are recorded with the sample: 1.225o, 1.106o, 1.182o. What is [a]?
Answer
$[a] = \frac{observed \: rotation}{(l)(c)} \nonumber$
$c = \frac{0.540g}{2.0 mL} = 0.27 \frac{g}{mL} \nonumber$
Average observed rotation $= \frac{1.225 + 1.106 + 1.182}{3} = 1.171^{o}$
$[a] = \frac{1.171^{o}}{(0.27 \frac{g}{mL})(1.0 dm)} = 4.34^{o} \nonumber$
Exercise $7$
A pure sample of the (+) enantiomer of compound C shows [a] = 42o. What would be the observed a if a solution of the sample was made by dissolving 0.250 g in 2.0 mL of acetontirile and was then placed in a 0.5 dm cell?
Answer
$[a] = \frac{observed \: rotation}{(l)(c)} \nonumber$
$c = \frac{0.250g}{2.0 mL} = 0.125 \frac{g}{mL} \nonumber$
$42^{o} = \frac{observed \: rotation}{(0.125 \frac{g}{mL})(0.5dm)} \nonumber$
observed rotation = 2.625o
Exercise $8$
A student has prepared compound D in lab. She is sure the compound contains no impurities; a number of physical analyses have confirmed the structure and purity of the compound. A sample of compound D (0.10 g) is dissolved in methanol (2.0 mL) and the solution is placed in a 1.0 dm cell. Three polarimetry readings are recorded with the sample: 0.995o, 0.904o, 0.936o.
1. What is [a]?
2. The optical rotation of D has previously been reported as 25o. What is the optical purity of this sample?
3. What is the enantiomeric excess of this sample?
4. What is the composition of this sample?
5. Why did the previous analyses show that there was only one compound present?
Answer a
$[a] = \frac{observed \: rotation}{(l)(c)} \nonumber$
$c = \frac{0.10g}{2.0 mL} = 0.05 \frac{g}{mL} \nonumber$
Average observed rotation = $frac{0.995 + 0.904 + 0.936}{3} = 0.945^{o}$
$[a] = \frac{0.945^{o}}{(0.05 \frac{g}{mL})(1.0 dm)} = 18.9^{o} \nonumber$
Answer b
% optical purity $= \frac{(100)(18.9)}{25} = 75.6 \%$
Answer c
enantiomeric excess = 75.6%
Answer d
$\frac{100-75.6}{2} = 12.2 \% \nonumber$
12.2% one enantiomer
87.8% other enantiomer
Answer e
Enantiomers differ in how they interact with plane polarized light, but not in other physical analyses.
Exercise $9$
Suppose a student prepared a sample of a compound with known [a] = 40. The student did not realize that her sample contained 10% (by mass) of a non-chiral impurity. After correcting for c and l, what did she measure for [a]?
Answer
$[a] = \frac{observed \: rotation}{(l)(c)} \nonumber$
$40^{o} = \frac{observed \: rotation}{(l)(c)} \nonumber$
$40^{o} = \frac{observed \: rotation}{(l)(1.1c)} \nonumber$
Observed rotation = 44o
diastereomers and physical properties
Exercise $10$
From the following group of molecules, select:
1. a pair that are the same compound.
2. a pair that would have the same physical properties but opposite optical activities.
3. a pair that have different physical properties but the same connectivity.
4. a pair that have different connectivity.
5. a compound that contains chiral centers but has no optical activity.
Exercise $11$
From the following group of molecules, select:
1. a pair that are the same compound.
2. a pair that would have the same physical properties but opposite optical activities.
3. a pair that have different physical properties but the same connectivity.
4. a pair that have different connectivity.
Exercise $12$
From the following group of molecules, select:
1. a pair that are the same compound.
2. a pair that would have the same physical properties but opposite optical activities.
3. a pair that have different physical properties but the same connectivity. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.03%3A__Stereochemistry_-_Extra_Problems..txt |
An atom with four groups attached to it can also adopt a tetrahedral geometry. This geometry often occurs when the central atom is a little smaller. A tetrahedral geometry allows neighbouring groups to get a little farther from each other. The following model of methane shows a pretty simple example. The grey ball is a carbon atom and the four white balls are hydrogen atoms.
Go to Animation SC3.1. A three-dimensional model of methane, CH4.
Unlike square planar compounds, simple tetrahedral compounds do not have the same kind of cis- and trans- isomers. That is, two groups can't be placed on a tetrahedron so that they are opposite each other or beside each other. The relationship between any two groups on a tetrahedron is the same as the relationship between any other two groups on a tetrahedron.
Dichlorodimethylsilane is a compound that can be used to make silicone polymers. Like platin, it has two each of two groups attached to the central atom. However, the central tom is tetrahedral. There is only one way to arrange these four groups.
Go to Animation SC3.1. A three-dimensional model of dichlorodimethylsilane.
However, if four different groups are attached to a tetrahedral atom, the four groups can be arranged in two possible ways. The two compounds that result are mirror images of each other. These two isomers are called enantiomers.
For example, suppose that tetrahedral silicon atom had four different groups attached to it: a hydrogen atom, a chlorine atom, a methyl group (that's a carbon with three hydrogens), and a hydroxy group (that's an oxygen with one hydrogen). There would be two slightly different ways to organize those four groups in space. In the picture below, hydrogen is white, chlorine is green, and oxygen is red, whereas silicon and carbon are both grey.
Go to Animation SC3.2. A three-dimensional model of R-chlorohydroxymethylsilane.
Exercise \(1\)
Turn the model so that the hydrogen attached to the silicon is hidden behind the molecule. In what direction would you travel if you traced from the chlorine to the oxygen to the carbon: clockwise or counter-clockwise?
Answer
Clockwise
This next picture shows a very similar molecule. The same groups are attached to the central silicon atom, but the are arranged in a slightly different way.
Go to Animation SC3.3. A three-dimensional model of S-chlorohydroxymethylsilane.
Exercise \(2\)
Turn the model so that the hydrogen attached to the silicon is hidden behind the molecule. In what direction would you travel if you traced from the chlorine to the oxygen to the carbon: clockwise or counter-clockwise?
Answer
Counter-clockwise
Full disclosure: the molecule we are looking at isn't really a stable compound that you could pour out of a bottle from the chemistry stockroom. That's because one molecule would react with another one just like it, and then another one, and so on to form a long chain called a silicone polymer. That's the kind of material found in things like silly putty and silicone caulking.
Exercise \(3\)
Let's try working without the three-dimensional model. The following ball-and-stick picture shows a silicon attached to a bromine (red), a chlorine (green), a fluorine (yellow) and a hydrogen. If you were looking at the molecule with the hydrogen hiddent behind it, in what direction would you travel if you traced from the bromine to the chlorine to the fluorine: clockwise or counter-clockwise?
Exercise \(4\)
If you were looking at the molecule with the hydrogen hidden behind it, in what direction would you travel if you traced from the bromine to the chlorine to the fluorine: clockwise or counter-clockwise?
There don't have to be four different atoms attached to the central atom to get two stereoisomers. There just have to be four different groups. Even if all four atoms attached to the central one are the same, if they have different things attached to them, then two three-dimensional arrangements are possible. The following compounds illustrate a more subtle example like that. This compound, by the way, is stable and can be isolated and stored; it's a real compound.
Go to Animation SC2.1. A three-dimensional model of the (-) enantiomer
Go to Animation SC2.1. A three-dimensional model of the (+) enantiomer.
• Enantiomers are pairs of compounds with exactly the same connectivity but opposite three-dimensional shapes.
• Enantiomers are not the same as each other; one enantiomer cannot be superimposed on the other.
• Enantiomers are mirror images of each other.
Two compounds with the exact same connectivity, that are mirror images of each other but that are not identical to each other are called enantiomers. The more common definition of an enantiomer is that it is not superimposable on its mirror image. It can be distinguished easily from its mirror image, just as a right hand can easily be identified and distinguished from a left hand.
• Compounds that occur in these pairs are called "chiral".
• "Chiral" comes from the Greek word for "hand".
It can be shown using group theory, the mathematics of symmetry, that an enantiomer may also be defined as a molecule that does not contain a mirror plane, meaning it cannot be divided into two identical and opposite halves.
• Enantiomers contain no mirror planes.
• Enantiomers do not contain two equal and opposite halves.
Unlike cis- and trans-isomers, two enantiomers have the same physical properties. they have the same melting point, the same solubility, and so on. Two compounds that are almost identical, but mirror images of each other, have exactly the same kinds of intermolecular attraction, so it may not be a surprise that their physical properties are identical.
• Enantiomers are another example of a type of stereoisomers.
• Two enantiomers have identical physical properties, except for optical rotation.
Optical rotation involves the interaction of plane-polarized light with a material. If a material is not symmetric, the light that passes through it will be rotated. That means if the waves making up the light are oscillating in one direction as they enter the material, they will have tilted slightly to oscillate in another direction when they emerge from the material. We will look at this phenomenon later.
• Two enantiomers have an equal but opposite rotational effect on plane-polarized light.
• (+) enantiomers rotate light in a clockwise direction.
• (-) enantiomers rotate light in a counterclockwise direction.
For example, in the chiral silicon compound shown above, the (+) enantiomer rotates plane-polarized light in a clockwise direction. It has a "standard optical rotation" of [a] = +12 ( +/-2)o. The (-) enantiomer rotates plane-polarized light in a counterclockwise direction. It has a "standard optical rotation" of [a] = -9.9 ( +/-2)o.
Exercise \(5\)
A certain compound exists in two forms; enantiomer A and enantiomer B. Enantiomer A has a molecular weight of 126 g/mol, a density of 0.995 g/mL, an optical rotation of [a] = 26o, a melting point of 65 oC, a boiling point of 225 oC, and an odour of citrus fruit. What can you say about the corresponding properties of enantiomer B?
Answer
Enantiomer B has a molecular weight of 126 g/mol, a density of 0.995 g/mL, an optical rotation of [a] = -26o, a melting point of 65oC and a boiling point of 225oC.
Exercise \(6\)
Are these molecules chiral?
Draw the mirror image of each compound and determine if it is superimposable.
Exercise \(7\)
Find a mirror plane of symmetry in the following molecules. Some may have no mirror planes and thus are chiral. (Some may have several mirror planes, but it is only necessary to find one in order to determine that the molecule is not chiral.) Hint—there are low barriers to rotation about most single bonds. If any conformer can be drawn in which a molecule has a mirror plane, the molecule will not be chiral.
Exercise \(8\)
Select the picture that illustrates a mirror plane of symmetry in the following compound.
Exercise \(9\)
Select the picture that illustrates a mirror plane of symmetry in the following compound.
Exercise \(10\)
Select the picture that illustrates a mirror plane of symmetry in the following compound.
Exercise \(11\)
Select the picture that illustrates a mirror plane of symmetry in the following compound.
Still pictures of models obtained using Spartan 14 from Wavefunction, Inc., Irvine, California. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.04%3A_Enantiomers.txt |
Stereochemistry is most important to us because of the role it plays in biology. Biological compounds usually belong to a class of compounds called organic compounds, originally meaning compounds that come from organisms. More generally, organic compounds contain carbon. because carbon is usually a tetrahedral atom, there is great potential for enantiomers in organic compounds, including biological ones.
The compound shown below contains a carbon connected to a hydrogen, a bromine, a chlorine and a fluorine. A tetrahedral atom connected to four different things is called a chiral center; it is a place where two different arrangements are possible.
This is not likely to be a naturally-occurring compound, although some marine organisms do make organic compounds containing bromine and chlorine. However, compounds similar to this one are sometimes used as general anaesthetics for surgical purposes. It is used as an example here because the chiral center is easy to see.
• A chiral center is a tetrahedral atom connected to four different groups.
• R and S configurations refer to the three-dimensional relationship of these groups around the chiral center.
• Assignment of R and S configuration follows a well-defined set of arbitrary rules. Anyone can follow these rules and arrive at the same configuration for the same structure.
• About half of S enantiomers rotate light in the (+) direction and about half rotate light in the (-) direction. R and S configurations do not correlate directly with optical rotation values; these are two unrelated systems for describing enantiomers.
Go to Animation SC4.1. A three-dimensional model of (R)-bromochlorofluoromethane.
Go to Animation SC4.2. A three-dimensional model of (S)-bromochlorofluoromethane.
Rules for assigning R & S configuration:
• Find the chiral center(s).
• Compare the four atoms attached to the chiral center. What are their atomic numbers from highest to lowest?
• The highest atomic number has first priority; the second highest has second priority and so on.
• Turn the molecule so that the group with the lowest priority is away from you.
• Look at the three other groups. If they proceed in a clockwise direction from highest to second to third highest priority; the chiral center has the R configuration.
• If they proceed in a counterclockwise direction from highest to second to third highest priority; the chiral center has the S configuration.
Note
If two atoms connected to the chiral center have the same atomic number, you have to use a tie-breaker to decide which one gets higher priority. The tie-breaking rules are described on the next page; basically, you look at the next atoms attached to the ones that are the same.
Exercise \(1\)
Explain the reasons for the assignment of configurations R and S in the models above. Assume red is bromine, bright green is chlorine and pale green is fluorine.
Answer
Priority of groups:
1 Br (red)
2 Cl (bright green)
3 F (pale green)
4 H (white)
In the molecule in Figure \(2\) (SC4.2), with the low-priority hydrogen pointed away, bromine is at the top, chlorine is clockwise from the bromine, and fluorine is clockwise from the chlorine. It therefore has an assigned configuration of R.
In the molecule in Figure \(3\) (SC4.3), with the low-priority hydrogen pointed away, bromine is at the top, chlorine is counterclockwise from the bromine, and fluorine is counterclockwise from the chlorine. Thus, it has an assigned configuration of S.
Exercise \(2\)
Which of the following compounds have enantiomers? If they exist, draw them.
Exercise \(3\)
Circle the stereocenters in the following compounds.
Exercise \(4\)
Assign configuration of the stereocenters (R or S) in the following compounds.
Exercise \(5\)
Assign configurations (R or S) to the chiral centers in these molecules.
Exercise \(6\)
Assign configurations (R or S) to the chiral centers in these molecules. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.05%3A_Simple_Organic_Enantiomers-_R_and_S_configur.txt |
There are a number of important biomolecules that could occur as enantiomers, including amino acids and sugars. In most cases, only one enantiomer occurs (although some fungi, for example, are able to produce mirror-image forms of these compounds). We will look later at some of these biomolecules, but first we will look at a compound that occurs naturally in both enantiomeric forms.
Carvone is a secondary metabolite. That means it is a naturally-occurring compound that is not directly connected to the very basic functions of a cell, such as self-replication or the production of energy. The role of secondary metabolites in nature is often difficult to determine. However, these compounds often play roles in self-defense, acting as deterrents against competitor species in a sort of small-scale chemical warfare scenario. They are also frequently used in communications; this role has been studied most extensively among insects, which use lots of compounds to send information to each other.
Carvone is produced in two enantiomeric forms. One of these forms, called (-)-carvone, is found in mint leaves, and it is a principal contributor to the distinctive odor of mint. The other form, (+)-carvone, is found in caraway seeds. This form has a very different smell, and is typically used to flavour rye bread and other Eastern European foods. The structural connectivity in these two compounds appears to be identical, but they aren't the same in three dimensions, as you will find if you build models of them. Instead, (+)- and (-)-carvone are mirror images of each other, like a right hand and a left hand.
Figure \(2\): Ball-and-stick model of (R)-carvone.
Go to Animation SC5.1. A three-dimensional model of (R)-carvone.
Go to Animation SC5.2. A three-dimensional model of (S)-carvone.
Note that (+)-carvone is the same thing as (S)-carvone. The (+) designation is based on its positive optical rotation value, which is experimentally measured. The (S) designation is determined by the Cahn-Ingold-Prelog rules for designating stereochemistry, which deal with looking at the groups attached to a chiral center and assigning priority based on atomic number. However, carvone's chiral center actually has three carbons attached to it; they all have the same atomic number. We need a new rule to break the tie.
• If two substituent groups have the same atomic number, go one bond further to the next atom.
• If there is a difference among the second tier of atoms, stop.
• The group in which you have encountered a higher atomic number gets the highest priority.
• If there is not a clear difference, proceed one additional bond to the next set of atoms, and so on, until you find a difference.
In carvone, this decision tree works as follows:
• .The chiral center is connected to a H, a C, a C and a C.
• The H is lowest priority.
• One C eventually leads to a C=O. However, at the second bond from the chiral center, this C is connected to a C and two H's.
• A second C is also part of the six-membered ring, but the C=O is farther away in this direction. At the second bond from the chiral center, this C is connected to a C and two H's, just like the first one.
• The third C is part of a little three-carbon group attached to the six-membered ring. At the second bond from the chiral center, it is connected to only one H and has two bonds to another C (this is counted as two bonds to C and one to H).
• Those first two carbon groups are identical so far.
• However, the third group is different; it has an extra bond to C, whereas the others have an extra bond to H. C has a higher atomic number than H, so this group has higher priority.
• The second-highest priority is the branch that reaches the oxygen at the third bond from the chiral center.
How different, exactly, are these two compounds, (+)- and (-)-carvone? Are they completely different isomers, with different physical properties? In most ways, the answer is no. These two compounds have the same appearance (colorless oil), the same boiling point (230 ° C), the same refractive index (1.499) and specific gravity (0.965). However, they have optical rotations that are almost exactly opposite values.
• Two enantiomers have the same physical properties.
• Enantiomers have opposite optical rotations.
Clearly they have different biological properties; since they have slightly different odors, they must fit into slightly different nasal receptors, signaling to the brain whether the person next to you is chewing a stick of gum or a piece of rye bread. This different shape complimentarity is not surprising, just as it isn't surprising that a left hand only fits into a left handed baseball glove and not into a right handed one.
There are other reasons that we might concern ourselves with an understanding of enantiomers, apart from dietary and olfactory preferences. Perhaps the most dramatic example of the importance of enantiomers can be found in the case of thalidomide. Thalidomide was a drug commonly prescribed during the 1950's and 1960's in order to alleviate nausea and other symptoms of morning sickness. In fact, only one enantiomer of thalidomide had any therapeutic effect in this regard. The other enantiomer, apart from being therapeutically useless in this application, was subsequently found to be a teratogen, meaning it produces pronounced birth defects. This was obviously not a good thing to prescribe to pregnant women. Workers in the pharmaceutical industry are now much more aware of these kinds of consequences, although of course not all problems with drugs go undetected even through the extensive clinical trials required in the United States. Since the era of thalidomide, however, a tremendous amount of research in the field of synthetic organic chemistry has been devoted to methods of producing only one enantiomer of a useful compound and not the other. This effort probably represents the single biggest aim of synthetic organic chemistry through the last quarter century.
• Enantiomers may have very different biological properties.
• Obtaining enantiomerically pure compounds is very important in medicine and the pharmaceutical industry.
Exercise \(1\)
a) Draw thalidomide and identify the chiral center with an asterisk.
b) Draw the two possible enantiomeric forms of thalidomide.
Answer
Exercise \(2\)
Draw the two enantiomeric forms of 2-butanol, CH3CH(OH)CH2CH3. Label their configurations.
Answer
Exercise \(3\)
Sometimes, compounds have many chiral centers in them. For the following compounds, identify four chiral centers in each, mark them with asterisks, and identify each center as R or S configuration.
The following is the structure of dysinosin A, a potent thrombin inhibitor that consequently prevents blood clotting.
Ginkgolide B (below) is a secondary metabolite of the ginkgo tree, extracts of which are used in Chinese medicine.
Sanglifehrin A, shown below, is produced by a bacteria that may be found in the soil of coffee plantations in Malawi. It is also a promising candidate for the treatment of organ transplant patients owing to its potent immuno-suppressant activity. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.06%3A_Biological_Small_Molecules.txt |
• An equal mixture of two enantiomers is called a racemic mixture or racemate.
• If two enantiomers rotate plane-polarized light in opposite directions, a racemate will not rotate light at all. The effects of the two enantiomers will cancel out.
e "optical purity" is a comparison of the optical rotation of a pure sample of unknown stereochemistry versus the optical rotation of a sample of pure enantiomer. It is expressed as a percentage. If the sample only rotates plane-polarized light half as much as expected, the optical purity is 50%.
Optical purity also corresponds to "enantiomeric excess". If the unknown sample rotates light 50% as much as a sample of pure enantiomer, it must contain 50% enantiomeric excess; the other 50% is a racemic mixture. In other words, if the sample is 75% of one enantiomer and 25% of the other, 50% of the mixture will simply cancel out in terms of optical activity. The remaining 50% will still exert optical activity, but only half as much as if the sample were 100% of that enantiomer.
These relationships could be expressed in formulae:
$\textrm{Optical purity (op)} = \frac{(\textrm{optical rotation of pure compound})}{(\textrm{optical rotation of pure enantiomer})} \times 100\% \nonumber$
Enantiomeric excess (ee) = optical purity (that is, these numbers are always the same, although they represent different things)
$\% \textrm{major enantiomer} = \textrm{enantiomeric excess} + \frac{100 - \textrm{enantiomeric excess}}{2} = 50 + \frac{\textrm{enantiomeric excess}}{2} \nonumber$
$\% \textrm{minor enantiomer} = 100 - \% \: \textrm{major enantiomer} \nonumber$
Exercise $1$
The (+) enantiomer of compound A has an optical rotation of 75o. If a sample containing only compound A has an optical rotation of 50o, what is the composition of the sample?
Answer
A pure sample of A would have $[\alpha] = 75^{o}$
Optical purity or enatiomeric excess $= \frac{50}{75} = 66 \%$
% major enantiomer $= 66 + \frac{34}{2} = 83 \%$
% minor enantiomer $= 100 - 83 = 17 \%$
Exercise $2$
The (+) enantiomer of compound B has an optical rotation of 50o. If a sample containing only B contains 10% of the (+) enantiomer and 90% of the (-) enantiomer, what is the optical rotation value?
Answer
$\% major = 90 \% \nonumber$
$\% minor = 10 \% \nonumber$
Optical purity or enantiomeric excess $= \frac{X}{-50} = 90 -10 - 80 \%$
Solve for X.
$X = -40 ^{\circ} \nonumber$
Exercise $3$
The (-) enantiomer of compound C has an optical rotation of -60. A sample of compound C is shown by chiral gas chromatography to contain only (-) C and no (+) C, but NMR analysis suggests the sample is about 50% ethyl acetate by weight. Predict the measured optical rotation of a 1 g/mL solution in dichloromethane, measured in a 1 dm cell.
Exercise $4$
Which of the following compounds are optically active?
5.08: The Polarimetry Experiment
In measuring optical rotation, plane-polarized light travels down a long tube containing the sample. If it is a liquid, the sample may be placed in the tube as a pure liquid (it is sometimes called a neat sample). Usually, the sample is dissolved in a solvent and the resulting solution is placed in the tube.
There are important factors affecting the outcome of the experiment.
• Optical rotation depends on the number of molecules encountered by the light during the experiment.
• Two factors can be controlled in the experiment and must be accounted for when comparing an experimental result to a reported value.
• The more concentrated the sample (the more molecules per unit volume), the more molecules will be encountered.
• Concentrated solutions and neat samples will have higher optical rotations than dilute solutions.
• The value of the optical rotation must be corrected for concentration.
• The longer the path of light through a solution of molecules, the more molecules will be encountered by the light, and the greater the optical rotation.
• The value of the optical rotation must be corrected for the length of the cell used to hold the sample.
In summary:
$[\alpha] = \frac{\alpha}{c \times l} \nonumber$
• a is the measured optical rotation.
• c is the sample concentration in grams per deciliter (1 dL = 10 mL).
• That is, c = m / V (m = mass in g, V = volume in dL).
• l is the cell length in decimeters (1 dm = 10 cm = 100 mm)
• The square brackets mean the optical rotation has been corrected for these variables.
Exercise $1$
A pure sample of the naturally-occurring, chiral compound A (0.250 g) is dissolved in acetone (2.0 mL) and the solution is placed in a 0.5 dm cell. Three polarimetry readings are recorded with the sample: 0.775o, 0.806o, 0.682o.
a) What is [a]?
b) What would be the [a] value of the opposite enantiomer?
Answer a:
$[a] = \frac{a}{(c)(l)} \nonumber$
$c = (\frac{0.250g}{2mL})(\frac{10mL}{1 dL}) = 1.25 \frac{g}{dL} \nonumber$
$a = \frac{0.775 ^{o} + 0.806^{o} + 0.682^{o}}{3} = 0.754 ^{o} \nonumber$
$[a] = \frac{a}{(c)(l)} = \frac{0.754^{o}}{(1.25 \frac{g}{dL})(0.5dm)} = + 1.21 ^{o} \nonumber$
Answer b:
-1.21o
Exercise $2$
A pure sample of the (+) enantiomer of compound B shows [a] = 32o. What would be the observed a if a solution of the sample was made by dissolving 0.150 g in 1.0 mL of dichloromethane and was then placed in a 0.5 dm cell?
Answer
$[a] = \frac{a}{(c)(l)} \nonumber$
$[a] = 32^{o} \nonumber$
$c = (\frac{0.150g}{1mL})(\frac{10mL}{1dL}) = 1.5 \frac{g}{dL} \nonumber$
$[a] = \frac{a}{(c)(l)} = 32^{o} = \frac{a}{(1.5 \frac{g}{dL})(0.5 dm} \nonumber$
Solve for a.
$a = + 24^{o} \nonumber$ | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.07%3A_Optical_Rotation.txt |
So far, we have mostly been looking at compounds that contain one chiral center -- that is, one tetrahedral carbon atom with four different groups attached to it. Could there be more than one chiral center? Of course. For example, if carbohydrates have an OH group on every carbon in a chain, we might get a number of chiral centers in one molecule. Thinking in Fischer projections, that means each of the OH groups on the chiral centers could be drawn either on the right or on the left.
Of course, two compounds with multiple chiral centers could still be enantiomers. Maybe we have two carbohydrates, one with all of the OH groups on the right in the Fischer projection, the other with all of them on the left. Those two compounds would still be the exact opposite of each other. They would still be enantiomers.
Suppose instead that only some of the chiral centers were opposite, and others were the same. The compounds would no longer be exact mirror images of each other. The mirror would be flawed. These isomers would be called diastereomers.
We know that enantiomers have the same physical properties. Diastereomers do not. For example, they have different melting points, because the two isomers cannot pack together in exactly the same way, and the intermolecular attractions are different in each case.
• Diastereomers only occur in compounds containing more than one chiral center.
• In a pair of diastereomers, some of the chiral centers are the same in the two molecules, but others are different.
• Diastereomers are stereoisomers that are not mirror images of each other.
Threose is an example of a biological molecule (a carbohydrate) that contains two chiral centers.
Fischer projections are sometimes used in showing carbohydrates and other chain compounds with many chiral centers. Fischer projections are Picasso-esque drawings in which the point of view alternates from one carbon to the next. The chiral centers are easy to compare in Fischer projections because you simply have to decide whether groups are on the same side or opposite sides of the vertical line.
D-threose is chiral and it does have an enantiomer. Its enantiomer is called L-threose. L-threose has exactly the same physical properties as D-threose, except for one. A solution of D-threose rotates a beam of plane-polarized light to the right, whereas a solution of L-threose rotates the plane to the left.
However, D-threose also has other stereoisomers that are NOT its mirror image. These isomers are called diastereomers. These compounds have different physical properties from threose, so they have a completely different common name. That's because common names for compounds were often coined before anyone knew about the structure of the compound.
The relationships between D-threose and its stereoisomers can be seen in wedge-dash projections. The family of aldobutanoses -- that is, four-carbon dugars containing an aldehyde group -- is presented together below.
Alternatively, these relationships can be viewed in Fischer projections.
Alternatively, these relationships can be viewed in Fischer projections.
• In a pair of diastereomers, some chiral centers are the same and some are opposite. The molecule is neither identical to nor the mirror image of its diastereomer.
• D-threose is the enantiomer of L-threose. The two are non-identical, but they are mirror images of each other.
It may help to look at three-dimensional models of these sugars.
Go to Animation SC9.1. A three-dimensional model of D-threose.
Go to Animation SC9.2. A three-dimensional model of L-threosee.
Go to Animation SC9.3. A three-dimensional model of D-erythrose.
Go to Animation SC9.4. A three-dimensional model of L-erythrose.
Exercise \(1\)
Take a look at the three-dimensional model of D-threose using the stick model.
1. If you turn it so that the carbonyl is at the top and the OH and H of the first chiral carbon are coming towards you, is the first OH on the right or on the left?
2. If you turn it so that the carbonyl is at the top and the OH and H of the second chiral carbon are coming towards you, is the first OH on the right or on the left?
Answer a:
left
Answer b:
right
Exercise \(2\)
Take a look at the three-dimensional model of L-threose using the stick model.
1. If you turn it so that the carbonyl is at the top and the OH and H of the first chiral carbon are coming towards you, is the first OH on the right or on the left?
2. If you turn it so that the carbonyl is at the top and the OH and H of the second chiral carbon are coming towards you, is the first OH on the right or on the left?
Answer a:
right
Answer b:
left
Exercise \(3\)
Take a look at the three-dimensional model of D-erythrose using the stick model.
1. If you turn it so that the carbonyl is at the top and the OH and H of the first chiral carbon are coming towards you, is the first OH on the right or on the left?
2. If you turn it so that the carbonyl is at the top and the OH and H of the second chiral carbon are coming towards you, is the first OH on the right or on the left?
Answer a:
right
Answer b:
right
Exercise \(4\)
Take a look at the three-dimensional model of L-erythrose using the stick model.
1. If you turn it so that the carbonyl is at the top and the OH and H of the first chiral carbon are coming towards you, is the first OH on the right or on the left?
2. If you turn it so that the carbonyl is at the top and the OH and H of the second chiral carbon are coming towards you, is the first OH on the right or on the left?
Answer a:
left
Answer b:
left
• When there is one chiral center present, two stereoisomers result.
• When there are two chiral centers present, four stereoisomers result.
• For each additional chiral center, the number of stereoisomers doubles.
Carbohydrates can be classified according to their stereochemistry, forming a sort of carbohydrate family tree. They are considered to originate from the simplest carbohydrate, glyceraldehyde. Addition of a new carbon to the C=O unit of glyceraldehyde produces two possible isomers, because a new steresocenter is formed. From there, additional carbohydrates can be built up, one carbon at a time. In the carbohydrate family tree, glyceraldehyde is great-grandmother.
Exercise \(5\)
What are the absolute configurations of the two chiral centers in D-threose? (You will need to number the carbons on which the chiral centers are found. For example, if an S center is found on the second carbon along the chain and an R center is found on the fourth carbon along the chain, the configuration is 2S,4R. Note that you should number the chain from the highest-priority end, which has a carbon with the most bonds to oxygen.)
Answer
D-threose
2S, 3R
Exercise \(6\)
What are the absolute configurations in L-threose? What do you notice about the relationship between the configurations in D- and L- threose?
Answer
L-threose
2R, 3S
D- and L-threose are enantiomers of one another
Exercise \(7\)
What are the absolute configurations in L-erythrose? What do you notice about the relationships between the configurations in L-erythrose and L- threose?
Answer
L-erythrose
2S, 3S
L-erythrose and L-threose are diastereomers of one another.
Exercise \(8\)
Suppose a compound contained three chiral centers.
1. How many possible stereoisomers would there be?
2. How many different pairs of enantiomers would there be?
3. Suppose you selected a pair of diastereomers from this group. How many possible pairs could you choose from?
Answer a:
(2)3 = 8 possible stereoisomers
RRR; SSS; RRS; SSR; RSS; SRR; SRS; RSR
Answer b:
4 pairs
Answer c:
12 different possible pairs of diastereomers
Exercise \(9\)
Draw one enantiomer and one diastereomer for each of the following compounds.
Exercise \(10\)
Draw one enantiomer and one diastereomer for each of the following compounds. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.09%3A_Carbohydrates_and_Diastereomers.txt |
• In a pair of diastereomers, some chiral centers are the same and some are opposite. The molecule is neither identical to nor the mirror image of its diastereomer.
Threitol is a metabolite closely related to carbohydrates. D-threitol is the enantiomer of L-threitol. The two are non-identical, but they are mirror images of each other. Both compounds can be crystallized, forming needle-like crystals. Both have melting points of 88-89oC. D-threose has an optical rotation [a]D= - 4.0 (c, 7 in H2O, but L-threose has an optical rotation [a]D= + 4.6 (c, 6 in H2O). You might notice that these two numbers are not exactly opposites, but if you have ever tried measuring an optical rotation yourself, you know that this is pretty good.
• Enantiomers have identical physical properties, except that the optical rotation is opposite.
Here is the other anantiomer. Once again, the Fischer projection makes it really easy to see the difference, but navigating back and forth between Fischer and wedge-dash can be tricky. The ball-and-stick model is rotated a little compared to the model of D-threose; see if can confirm that this is really the enantiomer.
Go to Animation SC10.2. A three-dimensional model of L-threitol.
Erythritol: a meso compound
Erythritol is a diastereomer to both L-threose and D-threose. It has one identical chiral center and one opposite one. Erythritol is a solid with a melting point of 121oC and no optical rotation. Erythritol has properties that are different from threitol.
Go to Animation SC10.3. A three-dimensional model of erythritol.
• Diastereomers have different physical properties.
• They do not have the same overall shape.
• They fit together in different ways, and experience different intermolecular attractions.
The fact that erythritol contains chiral centers but has no optical rotation is unusual. Erythritol is chiral but not optically active. It is a special case called a meso compound.
• A meso compound contains two identical, opposite chiral centers.
• Rotation of plane-polarized light is canceled between these two centers, just like in a racemic mixture.
• Meso compounds are not optically active and do not have enantiomers, because the mirror image of a meso compound is itself.
It may be easier to see the mirror symmetry within erythritol if it is in a less-stable conformation. In the following model, the carbon backbone is held in a curled-up, eclipsed conformation rather than the usual zig-zag, staggered conformer. You might notice in the animation that this is really what we are looking at in a Fischer projection: the backbone is always curling away from us, with attached groups projecting towards us on either side. Here, the top half of the molecule is just a reflection of the bottom half.
Go to Animation SC10.4. Another three-dimensional model of erythritol.
Exercise \(1\)
What are the absolute configurations of each chiral carbon in:
1. D-threitol?
2. L-threitol?
3. erythritol?
Answer a:
D-threitol → 2R, 3S
Answer b:
L- threitol → 2S, 3S
Answer c:
erythritol → 2S, 3R or 2R, 3S (a meso compound)
Exercise \(2\)
In problem \(1\)., you should have numbered the carbons corresponding to each chiral center when you were denoting the absolute configuration. How did you know which end of the chain to start counting from?
Answer
It does not matter which end you start counting from on these compounds since they are constituted the same.
Exercise \(3\)
From the following group of molecules, select:
1. a pair that are the same compound.
2. a pair that would have the same physical properties but opposite optical activities.
3. a pair that have different physical properties.
4. a compound that contains chiral centers but has no optical activity.
Exercise \(4\)
What is the isomeric relationship between the following pairs of molecules (diastereomers, enantiomers, constitutional isomers, identical)? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.10%3A_Diastereomers_and_Physical_Properties.txt |
Carbohydrates are complicated molecules. In solution, they slowly change into different isomers. In water, most carbohydrates can change from one form to another quickly; they are described as being in equilibrium with different structures. That means they can change back and forth.
The most prevalent form for most carbohydrates is a ring. One of the oxygens farther along the chain can reach around and bond to the carbon in the C=O at the head of the chain. When that happens, there are two possible orientations of the oxygen at the head of the chain. The two stereoisomers that result are diastereomers. Some of the chiral centers are the same, but one is different.
For example, D-threose can form two different five-membered rings (that is, rings made from a circle of five atoms). In carbohydrates, a five-membered ring with an oxygen in it is called a furanose (another common form, a six-membered ring, is called a pyranose). There are two diastereomers formed. they are sometimes called anomers. That means they differ in 3D space at the anomeric center; the anomeric center is the C=O carbon to which an oxygen binds to form the ring.
These drawings, by the way, are called Haworth projections. They are commonly used in biochemistry to depict the cyclic forms of sugars. They seem to suggest a funny see-saw shape for the carbon atoms, but the idea here is not to convey the shape of the molecule perfectly. Just like in Fischer projections, the main concern is to quickly convey stereochemical relationships. In a Haworth projection, you can easily see whether two groups are cis to each other or trans to each other.
Note that a Haworth projection of an odd-numbered ring is always drawn with the ring oxygen in the back. We are looking at the ring from the edge and slightly above, so the "back" is the upper edge of the polygon. It's just like if you were looking at a swimming pool, and the far corner of the pool appeared in your field of vision above the near corner. For an even-numbered ring, the same idea applies, but the ring oxygen is always drawn on the top edge and to the right.
If you want to get an idea of the true shape of the molecule, you can see it in the ball-and-stick models below. [NOT AVAILABLE]
Go to Animation SC11.1. A three-dimensional model of β-D-threofuranose.
Go to Animation SC11.2. Another three-dimensional model of α-D-threofuranose.
• Carbohydrates are normally found in their cyclic forms, not their chain forms.
• For every ring a carbohydrate can form, there are always two diastereomeric forms called anomers.
The two terms, α and β, that described the two different stereoisomers of the cyclic form of threose refer to the position of the OH group on the newly formed chiral center (the old C=O was trigonal planar, not tetrahedral, so it was not chiral). The terms are defined by looking at the Haworth projection of the molecule. If the new OH is in the upper position, it is termed β (think butterfly); it is is in the lower position, it is termed α (think ant).
Carbohydrates do not always make five-membered rings. Sometimes they make six-membered rings instead. A five-membered ring with an oxygen in it is called a furanose. A six-membered ring with an oxygen in it is called a pyranose. Ribose is a five-carbon sugar that could make either a five-membered or a six-membered ring, although we usually see it in its five-membered ring.
The presence of different diastereomers that can change back and forth complicates things. If you wanted to measure the optical activity of a sugar, you would probably open a bottle of the solid carbohydrate, dissolve some up in water and put the sample in a polarimeter. However, the optical rotation would change over time as the pure solid slowly turned into other isomers. This process is called mutarotation (which means changing optical rotation). For this reason, the time at which a measurement was performed is often reported with optical rotation values of carbohydrates. For example, the optical rotation value for D-threose may be listed as [α]D -12.3 (20 min, c = 4, water).
Carbohydrates are important partly because they are structural units in biology. Carbohydrates are incorporated into DNA and RNA as well as important enzymatic cofactors such as ATP, NADH and Acetyl Coenzyme A. Very frequently, the unique pieces attached to the carbohydrates to make these different molecules are attached via the anomeric carbon.
• Specific OH groups on carbohydrates are frequently substituted with other groups to make more complex molecules.
One more piece of terminology may be useful to review here. There are terms used in ring structures to describe whether two groups are attached to the same face or on opposite faces. For example, in β-D-threofuranose, the second hydroxy group is cis to the one on the anomeric center, but in α-D-threofuranose, the second hydroxy group is trans to the one on the anomeric center.
• Two groups on the same face of a ring are described as being cis to each other.
• Two groups on the opposite faces of a ring are described as being trans to each other.
• The terms cis and trans can be applied to any rings, not just carbohydrates.
Exercise \(1\)
In the following carbohydrate, glucose, show the cyclic structures that would form if the indicated oxygen bonded to the carbonyl carbon (the C=O in the CHO group at the top). You should move a proton from one oxygen to another so that all the atoms have the usual number of bonds.
Exercise \(2\)
β-deoxyribofuranose (below) is a building block of DNA.
1. What is the relationship between the two hydroxy (OH) groups that are directly attached to the 5-membered ring?
2. In the other possible 5-membered ring form of deoxyribose, what is the relationship between the two hydroxy (OH) groups that are directly attached to the 5-membered ring?
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.11%3A_Carbohydrates_in_Cyclic_Form.txt |
One of the most important classes of biological chiral compounds is the family of amino acids. Amino acids are building blocks that are used to make peptides and proteins. Peptides can be used as hormones, carrying messages throughout the body. Proteins can be used as factories to carry out transformations, such as the formation of a neurotransmitter or the degradation of a toxin. The structure of peptides and proteins depend on the structures of the amino acids that make them. In turn, these shapes determine into what receptors peptide-based hormones can fit, and also what substrates protein-based enzymes can accept.
• Almost all of the naturally-occurring amino acids are chiral. In general, these compounds all have the same L-configuration.
• The peptides and proteins made from amino acids are therefore chiral as well.
• Chirality plays an important role in recognizing substrates and producing other chiral products.
Go to Animation SC12.1. A three-dimensional model of L-leucine.
A few bacteria and fungi are capable of making the opposite enantiomer of the normal amino acids. In those compounds, the amino group and the hydrogen on the chiral carbon have switched positions.
Go to Animation SC12.1. A three-dimensional model of D-alanine.
Amino acids get their name from two "functional groups" or groups of atoms common to all these compounds. One of these groups is a carboxylic acid, the (C=O)OH on each amino acid. Later, we will look at how carboxylic acids can donate protons. The other common group is an amine, the NH2 group connected to a tetrahedral carbon in most of these compounds. Later, we will see how amines can accept protons. In general, amines are bases.
Exercise \(1\)
One of the amino acids is not chiral. Which one?
Answer
Glycine
Exercise \(2\)
One of the amino acids has an enantiomer and also has diastereomers. Which one? Draw both diastereomers.
Answer
Exercise \(3\)
One of the amino acids does not have an NH2 group. Which one?
Answer
Proline
Exercise \(4\)
Which of the amino acids contain more than one carboxylic acid?
Answer
Glutamic acid and aspartic acid
Exercise \(5\)
Which of the amino acids contains more than one amine?
Answer
Arginine, asparagine, glutamine, lysine; also tryptophan contains an aromatic heterocycle, although it is not basic.
Exercise \(6\)
Draw D-alanine.
Answer
Exercise \(7\)
One of the amino acids has a configuration (R or S) that is the opposite of all the others. Which one?
Answer
Cysteine
5.13: Macromolecular Structures- Alpha-Helices
Amino acids are very important building blocks. They form larger structures such as polypeptides, oligopeptides and proteins. The first two cases are just terms for chains of amino acids bonded together; the third refers to extremely long chains, and sometimes several long chains wrapped together.
Peptides are formed via amide bonds between an amino group on one amino acid and a carboxylic acid group on another.
Exercise \(1\)
Sometimes more than one combination of amino acids is possible in a peptide.
1. Show the two possible combinations of alanine and glycine.
2. Would these compounds have identical properties? Why or why not?
Answer
Exercise \(2\)
Peptide names are written by describing the chain of amino acids from the amine end of the peptide to the carboxylic acid end. Draw the peptide Ala-Phe-Lys.
Answer
There are a number of possible "macromolecular structures" that peptides and proteins can adopt. One of them is an alpha helix, in which a long chain of peptides coils up like a phone cord (go to a museum if you don't know what a phone cord looks like).
• Alpha helices are a common structural motif in proteins.
• Alpha helices are chiral; a mirror-image form is possible.
• The shape of the alpha-helix is rooted in the shape of the L-amino acids that compose the protein.
• The determination of helical structures in proteins was a very recent discovery at the time that Rosalind Franklin obtained x-ray crystallography data on a DNA sample. It was during a visit to her lab that Francis Crick and James Watson began to think that DNA also had a helical structure. Franklin's unusual data resembled results for proteins that Watson and Crick had recently been shown by Linus Pauling. Watson and Crick correctly surmised that DNA had a similar structure. Pauling won the Nobel Prize, twice, Watson and Crick won the Nobel Prize for their work on DNA structure, but Rosalind Franklin unfortunately died very young, and the Nobel Prize is not awarded posthumously. Partly as a result, her name is very often not associated with this important work, despite her crucial contribution.
• Alpha helices are also found in DNA.
• The chiral component of DNA is a chiral carbohydrate.
In proteins, the alpha helix is chiral. It almost always adopts a right-handed curl. If you start at one end of the helix, with your right thumb pointing along the chain from the end, the fingers of your right hand will follow the direction of the chain as it curls along the helix. In a left-handed helix, the same thing is true with your left hand. Although the relationship is complicated, this overwhelming preference to adopt a right-handed helix is enforced by the configurations of the chiral centers in the individual amino acids. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.12%3A_Biological_Building_Blocks-_Amino_Acids.txt |
Optical resolution refers to the separation of two different enantiomers from each other. Very rarely, this has been done if the crystalline forms of two enantiomers are visibly different from each other. For example, Louis Pasteur was able to separate the two enantiomers of tartaric acid from each other because the two compounds just happened to crystallize separately. The crystals that formed had multifaceted habits that were visibly mirror images of each other. He painstakingly separated the two enantiomers using a pair of tweezers and a microscope.
Most of the time, we don't get so lucky. Two enantiomers are normally so similar to each other physically that we can't separate them.
However, there can be a pressing need to do so. That's because two enantiomers may have identical physical properties, but they may have very different biological properties. In the cell, they may encounter other chiral materials, including proteins and sugars. Their interactions with those chiral molecules may be strongly influenced by how they fit together. The result may be the difference between a compound relieving a child's athsma or making it worse.
Chiral compounds, of course, are often compared to hands. Two enantiomers are like a right hand and a left hand, mirror images of each other that look exactly the same but are instead completely opposite.
(You don't like hands with three fingers? Start reading the comics page. It's a well-established convention.)
The fact that the interaction of chiral compounds with other chiral compounds is influenced by their stereochemistry is often compared to the fit of your two hands to the same glove. Not surprisingly, one hand fits the glove better than the other one.
One way to separate two enantiomers is through chiral column chromatography. Chiral column chromatography uses something like a hand-glove approach to separate two enantiomers from each other.
In case you aren't familiar with chromatography, we should take a look at it in general terms first. Chromatography describes the separation of compounds through their different physical interactions between two different phases as the compounds pass through a long column or tube.
The column is packed with one material, called the stationary phase, whose job it is to stick to the compounds and keep them from moving through the tube. The stationary phase could be simple alumina or silica (like very finely powdered sand) or it could be a complicated synthetic material protected by a patent.
Another material, called the mobile phase, flows through the column. The job of the mobile phase is to tug the compounds along and get them through the column. The mobile phase could be a gas, allowed to flow through the column from a compressed tank, or a liquid, pushed through with a pump.
A tug-of-war ensues between the mobile phase and the stationary phase. Different compounds will have different affinities for each phase. Compounds that are more strongly attracted to the mobile phase will move more quickly. Compounds that are held more firmly by the stationary phase will move more slowly.
Chiral column chromatography involves the use of one of those patented stationary phases. The stationary phase has a specific kind of chiral molecule (or a chiral moiety, just a piece of a molecule) bonded all over it. The differing interactions between the two enantiomers and the chiral elements of the stationary phase result in the two compounds moving through the column at different rates. The compound that fits together with the chiral stationary phase will slow down and take longer to come through the column.
A second major way of separating two enantiomers is through the use of a chiral resolving agent. A chiral resolving agent is a compound that already has a chiral center of its own. It is used in its pure form: only one enantiomer and not the other, having already been purified somehow (or selectively synthesized). This compound has to be able to react somehow with the target compound. When it does, it will form a third, new compound. The third compound should contain both chiral centers: the one from the target compound and the one from the chiral resolving agent.
That changes everything. The compound contains two chiral centers. One of these is identical among all the molecules in the sample; that's the one from the pure chiral resolving agent. The other one is present in either of two configurations; that's the one from the enantiomeric target compound. That means we now have a mixture of diastereomers.
• Two enantiomers have the same physical properties. They cannot be separated easily by standard laboratory techniques.
• Two diastereomers have different physical properties. they can often be separated via standard laboratory techniques.
• If an additional chiral center can be incorporated into a pair of enantiomers so that they become diastereomers, they can be separated.
The classic example of this method of obtaining one isomer of a compound involves the formation of a diastereomeric salt. The salt has multiple chiral centers, and so diastereomers are possible.
If a racemic mixture of phenylsuccinic acid is mixed with a pure sample of (-)-proline (a naturally available amino acid), a proton transfer (or Brønsted acid-base) reaction occurs. Two protons are transferred from the phenylsuccinic acid to the proline. One proline has a position available for one extra proton, so the two protons end up on two different prolines. Because protons have +1 charge, each proline is cationic. The phenylsuccinic acid gave up two positives, so it is a dianion. Together, these three ions form a salt.
If a racemic mixture of phenylsuccinic acid is used, but pure (-)-proline is added, two possible diastereomers result. One chiral center, in the proline, is always the same. The other chiral center, in the phenylsuccinic acid, can be in two different configurations. As a result, cations and anions may pack differently together in each case, so different melting points and solubilities result. One salt precipitates or forms a solid from the solution, but the other stays dissolved. The two diastereomers can be separated by filtration.
Exercise \(1\)
Suppose you have a pure sample of L-phenylalanine. Write equations for reactions, using structures, that show how you could use it to obtain a sample of (S)-naproxen, an analgesic, from a racemic mixture of naproxen.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.14%3A_Optical_Resolution.txt |
Isomers related by chirality are not the only kind of geometric isomers. One other example is cis-trans isomerism in alkenes; that is, different orientations around a carbon-carbon double bond (and other double bonds, in some cases).
In cis isomers, there is a non-hydrogen group at each end of the alkene, and they are on the same side of the double bond axis.
Figure \(1\): A cis-isomer of an alkene.
Go to Animation SC12.1. A three-dimensional model of cis-2-butene.
In trans isomers, there is a non-hydrogen group at each end of the alkene, and they are on the opposite sides of the double bond axis.
Go to Animation SC12.1. A three-dimensional model of trans-2-butene.
• In cis isomers, the non-hydrogen groups at each end of an alkene are on the same side.
• In trans isomers, the non-hydrogen groups at each end of an alkene are on opposite sides.
This kind of isomerism is different from the shape changes seen in saturated hydrocarbons such as butane; that is, compounds that have only single bonds, and not double bonds, don't have these specific isomers. (However, saturated ring structures can have cis and trans isomers; in that case, the terms describe whether two groups are on the same face of a ring or opposite faces). There is free rotation in butane, because the single bonds that hold the carbons together are not as sensitive to the orientation of the tetrahedral carbons at either end of the bond. In contrast, turning the trigonal planar carbon at one end of a double bond would disrupt the p orbital overlap that forms the pi bond. That would cost a lot of energy.
Cis and trans isomers do play important biological roles. For example, one of the proteins responsible for vision is rhodopsin. Rhodopsin has bound to it an organic compound called retinal (the oxygen in retinal is replaced by a nitrogen in an amino acid side chain in the protein to form active rhodopsin).
One of the bonds in the active form of retinal has a cis configuration. You may know from conformational analysis that there would be some crowding between the groups that are held close together on one side of the double bond. That crowding, sometimes called steric strain, causes an increase in energy.
Go to Animation SC12.1. A three-dimensional model of cis-retinal.
However, if a photon is absorbed by the molecule, an electron can be excited from a ground state level to an excited state level. This electron would then occupy a pi antibonding orbital, weakening the pi bonding in the system. That makes it easier for the molecule to rotate, and it does so.
Once free to rotate, the molecule more easily adopts a more ope n, less folded shape. It becomes the all-trans isomer.
Go to Animation SC12.1. A three-dimensional model of trans-retinal.
The shape change in retinal is important because it is bound to the rhodopsin protein. The protein is folded around the retinal, so that when the retinal portion changes shape, the protein has to undergo a conformational change in order to fit around the retinal. The resulting conformational change of the protein allows a signal to be sent telling the brain that light has been detected at a particular place in the eye.
Exercise \(1\)
Determine whether the following molecules are cis, trans or neither. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.15%3A_Another_Kind_of_Stereochemistry-_Alkene_Isom.txt |
Double bonds can exhibit stereoisomerism if there is more than one way to arrange two groups at opposite ends of a double bond: either on the same side of the bond axis (commonly cis) or opposite sides of the double bond axis (commonly trans). There is a different way of denoting these two stereochemical configurations that is always used if the compound has a more complicated structure. Compounds that have more than one non-hydrogen group at either end of the double bond are not normally referred to as cis or trans isomers, but are instead called E or Z isomers.
• In Z isomers, the highest-priority groups at each end of an alkene are on the same side of the double bond axis. The double bond axis is a straight line drawn through the two atoms that are double bonded to each other.
• Z is from the German zusammen (together).
• In E isomers, the highest-priority groups at each end of an alkene are on opposite sides of the double bond axis.
• E is from the German entgegen (against).
Highest-priority is decided based on atomic number.
If the first atoms attached to one end of the double bond are the same, go one bond further on each chain and compare all the atoms that come next. If one of the chains at this point contains an atom with a higher atomic number than any of the atoms on the other chain, it is given higher priority.
Sometimes, going one bond further leads to the same set of atomic numbers along both branches, but there are more of the higher atomic numbered atom on one branch than the other.
Remember, this is an arbitrary set of rules that have been adopted to name these compounds so that people can refer to one isomer or the other. They may not be the rules that you would have come up with on your own.
Exercise \(1\)
Determine whether the following molecules are E or Z.
5.17: Stereoisomers in Other Geometries- Octahedra
Note
This page shows examples of coordination complexes, in which a number of groups called "ligands" bond to a central metal atom or ion. For more information about these complexes, see the brief introduction here.
Earlier we looked at cis- and trans-isomers in square planar platinum complexes. Other transition metal complexes display cis-trans isomerism. Octahedral complexes can also have two particular ligands adjacent to each other or on opposite sides of the metal atom. For example, the cation [(NH3)4CoCl2]+ has a cis-isomer and a trans-isomer.
Go to Animation SC17.1. A three-dimensional model of cis-(NH3)4CoCl2.
Go to Animation SC17.2. A three-dimensional model of trans-(NH3)4CoCl2.
Exercise \(1\)
Draw cis and trans isomers for the following compounds.
a) [Cr(OH2)4Cl2]+ b) [Co(OH2)4(SCN)2]+ c) [Co(NH3)4(ONO)2]+ d) Pt(NH3)2Cl4 e) [Pt(NH3)4Cl2 ]2+
Answer
Cis and trans isomers are found when there are two ligands whose spatial relationship can be described easily as "beside each other" or "across from each other". What if there are more than two ligands that can adopt different geometric arrangements around a metal?
An example is found in the complex Co(NH3)3Cl3. In that compound, which has an octahedral geometry, the three chlorines might be found all in a row, or they might be found clustered into a triangle. When three ligands are found in a row on an octahedron, the geometry is called "meridional" or simply "mer". When the three cluster together in a triangle, the geometric relationship is called "facial" or simply "fac".
Go to Animation SC17.3. A three-dimensional model of fac-(NH3)3CoCl3.
Go to Animation SC17.4. A three-dimensional model of mer-(NH3)3CoCl3.
The fac designation is derived from the location of the three ligands whose positions are being compared: they are all found along one of the faces of an octahedron.
The mer designation comes from the fact that the three ligands are found along one meridian. A meridian is simply a line from one apex of the compound all the way to the opposite apex. It is like a meridian on the earth, from the north pole to the south pole.
Exercise \(2\)
Draw fac and mer isomers of the following compounds.
1. RhCl3py3 (py is a six-membered ring with five carbons and a nitrogen)
2. Mo(CO)3(PPh3)3
3. Co(NH3)3(NO2)3
Answer
Some of the compounds on this page demonstrate another type of isomerism. It is called "linkage isomerism" and it involves which of the atoms in the ligand is connected to the metal. For example, the atoms at either end of a thiocyanate ion, SCN-, both have lone pairs. One or the other might bind to a metal under different circumstances. The sulfur might be attached to the metal, or else it could be the nitrogen. These two options form two different linkage isomers.
Often, to communicate which linkage isomer we are talking about, the desired isomer is indicated by placing the atom attached to the metal first in the formula of the compound. For example, writing the formula [Co(OH2)4(SCN)2]+ suggests the isomer in which the sulfur is attached to the cobalt. Writing the formula [Co(OH2)4(NCS)2]+ suggests the nitrogen is attached to the cobalt.
Exercise \(3\)
Draw structures for the following linkage isomers.
1. [Co(NH3)4(NO2)2]+ and [Co(NH3)4(ONO)2]+
2. [Co(OH2)4(SCN)2]+ and [Co(OH2)4(NCS)2]+
3. [Cr(OH2)5(CN)]2+ and [Cr(OH2)5(NC)]2+ | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.16%3A_E_and_Z_Alkene_Isomers.txt |
Enantiomers are another kind of isomer that occur in octahedral metal complexes. Like the square planar platinum complexes seen before, these compounds consist of metal ions with other atoms or groups bound to them. More information about the binding of ligands to metals is found in the section on Lewis acids and bases.
These groups that bind to metal ions are called ligands. One example of a type of octahedral compounds that can form enantiomers is bidentate complexes. In bidentate complexes (from the Greek two teeth) or chelating complexes (from the Greek crab), a ligand binds very tightly to the metal because it holds onto the metal via more than one atom. Ethylenediamine is one example of a bidentate ligand.
• Bidentate ligands bind very tightly to a metal because they form two bonds with it, rather than just one.
• In an octahedral complex, the two donor atoms in a bidentate ligand bind cis to each other. They can't reach all the way around the molecule to bind trans to each other.
The spatial relationship between the metal and the two atoms connected to it from the same ligand forms a plane. If more than one bidentate ligand is connected to the metal, the relative orientation of one plane to another creates the possibility of mirror images. A complex containing three bidentate ligands can take on the shape of a left-handed propeller or a right-handed propeller.
If you turned this propellor, in what direction would the air be pushed? Maybe it is an airplane propellor or maybe it is a fan blade. If you turned a lambda fan blade counter-clockwise, it would blow cool air on you. You would have to turn a delta fan clockwise to cool off.
These shapes are alternatively described as a left-handed screw and a right handed screw. If you want to screw a right-handed screw into a floor, you turn the screwdriver clockwise. If you want to screw a right-handed screw into a floor, you turn the screwdriver counterclockwise. Screws are chiral in much the same way that a helix is chiral. The threads on the screw cycle either clockwise or counterclockwise as they run down the length of the screw.
If you can picture turning the "delta" or "lambda" propellor shape so that it screws into the page behind it, which direction would you turn it?
You would turn the lambda screw counter-clockwise to send it into the page behind it. You would have to turn the delta screw the opposite way; turning it clockwise would send it into the page.
Examples of this enantiomeric relationship include cobalt complexes of bidentate ligands.
Go to Animation SC17.3. A three-dimensional model of Δ-Co(en)3.
Go to Animation SC17.3. A three-dimensional model of Λ-Co(en)3.
These relationships are also seen when there are just two bidentate ligands in the compound.
Go to Animation SC17.3. A three-dimensional model of ΛCo(en)2(NH3Cl.
Go to Animation SC17.3. A three-dimensional model of ΔCo(en)2(NH3Cl.
What do we know about enantiomers?
• enantiomers have identical physical properties, except...
• enantiomers have opposite optical rotation.
These kinds of complexes were historically important in demonstrating how small molecules and ions bound to metal cations. By showing that some metal complexes were chiral and displayed optical activity, early 20th century workers such as Alfred Werner were able to rule out some competing ideas about the structures of metal compounds.
Today, we know that metal complexes play important roles in enzymes in biology, and Werner's work on metal complexes laid the groundwork for how we think about these complexes.
In addition, stereochemistry in metal complexes became very important in the late 20th century, especially as pharmaceutical companies looked for catalysts that could aid in the production of one enantiomer of a drug, and not the other, in order to maximize pharmaceutical effectiveness and minimize side effects.
Exercise \(1\)
Identify the relationships between the following pairs of iron(III) oxalate ions.
Exercise \(2\)
Indicate whether each of the following compounds is the Δ or Λ isomer and draw its enantiomer.
Exercise: How many stereoisomers does an octahedral complex have? (contribution from B.J. Johnson)
5.19: Chiral Catalysts for Production of Enantiome
Chiral metal complexes became very important over the last 30 years of the twentieth century, particularly as companies sought to synthesize rare and difficult-to-obtain natural products for use as pharmaceuticals.
One of the landmark developments in this area was the production of chiral metal phosphine complexes. Phosphines are molecules containing phosphorus that bind to metal ions and promote catalytic activity. Among other workers, William Knowles of Monsanto was able to isolate complexes with chiral phosphorus atoms and use them to catalyze reactions producing almost entirely one enantiomer. The resulting synthesis of the anti-Parkinson's drug L-DOPA produces 97% of the desired enantiomer and only 3% of the opposite enantiomer.
Go to Animation SC19.1. A three-dimensional model of Knowles' chiral catalyst.
Chiral catalysts work because metal-catalyzed reactions require the substrate to bind to the metal ion. If there is a chiral center at or very close to the metal ion, very often a substrate is forced to bind to the metal ion in a very specific way. In the case of the Monsanto L-DOPA synthesis, only one face of the C=C double bond can be presented to the metal atom (it won't fit together the other way), and the metal delivers two hydrogen atoms to that face of the double bond and not to the other.
Go to Animation SC19.2. A three-dimensional model of Noyori's chiral catalyst, (R)-configuration.
Go to Animation SC19.3. A three-dimensional model of Noyori's chiral catalyst, (S)-configuration.
Another pharmaceutical produced in highe enantiomeric excess employs a chiral metal complex that contains no chiral center. Ryoji Noyori is a professor at Nagoya University in Japan and also president of RIKEN, a major Japanese research institute involving about 3000 scientists. Among numerous other scientific contributions for which he is responsible, Noyori developed a number of complexes using BINAP ligands, which contain large, planar rings.
This case is roughly analogous to enantiomeric octahedral complexes like Co(en)33+, in which the the planes of the BINAP can twist in either one direction or the other. The rings cannot twist past each other, so when the BINAP complexes to the ruthenium, they get locked into either of two enantiomeric shapes, despite the fact that the compound does not really have a chiral center. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.18%3A_Enantiomers_in_Octahedral_Complexes.txt |
Exercise 5.2.1:
Exercise 5.2.2:
Exercise 5.3.1:
Clockwise.
Exercise 5.3.2:
Counter-clockwise.
Exercise 5.3.3:
Counter-clockwise.
Exercise 5.3.4:
Clockwise.
Exercise 5.3.5:
Enantiomer B has a molecular weight of 126 g/mol, a density of 0.995 g/mL, an optical rotation of [α] = – 26°, a melting point of 65°C and a boiling point of 225°C.
Exercise 5.3.6:
Exercise 5.3.7:
A. The plane of the page is a mirror plane. There is also one perpendicular to the page that reflects one H into the other.
B. The plane of the page contains one P-Cl bond and bisects the other Cl’s.
C. The plane of the page is a mirror plane.
D. Mirror plane contains P-Br bond and bisects the Cl’s.
E. There is no lone pair on the B. Therefore all atoms lie in a mirror plane.
F. No mirror planes--the molecule is therefore chiral.
G. There is a plane perpendicular to the page that contains the Br and Cl and bisects the cyclopropane ring.
H. No mirror planes--the molecule is therefore chiral.
I. There is a plane perpendicular to the page that contains the Br and Cl and bisects the cyclopropane ring.
J. The C-C bond can be rotated by 60 degrees so that there is a plane perpendicular to the C-C bond axis.
K. The C-C bond in one of the chlorine-containing arms can be rotated so that there is a mirror plane that goes through the ethyl group (with no Cl’s) and the P, and one chlorine containing arm is the reflection of the other.
L. Since a double bond is planar, there is a mirror plane that contains all six atoms.
M. There is a mirror plane that contains two C’s and bisects the two Cls.
N. No mirror planes--the molecule is therefore chiral. The rings are not in the same plane due to the CH3 and NH2 groups, which bump into each other. They also prevent rotation around the C-C bond between the rings.
Exercise 5.3.8
Picture (a)
Exercise 5.3.9:
Picture (b)
Exercise 5.3.10:
Picture (d)
Exercise 5.3.11:
Picture (c)
Exercise 5.4.1:
Priority of groups:
1 Br (red)
2 Cl (bright green)
3 F (pale green)
4 H (white)
In the molecule in figure 5.4.2, with the low-priority hydrogen pointed away, bromine is at the top, chlorine is clockwise from the bromine, and fluorine is clockwise from the chlorine. It therefore has an assigned configuration of R.
In the molecule in figure 5.4.3, with the low-priority hydrogen pointed away, bromine is at the top, chlorine is counterclockwise from the bromine, and fluorine is counterclockwise from the chlorine. Thus, it has an assigned configuration of S.
Exercise 5.4.2
Exercise 5.4.3:
Exercise 5.4.4:
Exercise 5.4.5:
Exercise 5.4.6:
Exercise 5.5.1:
Exercise 5.5.2:
Exercise 5.5.3:
Exercise 5.6.1:
A pure sample of A would have $[\alpha] = 75^{\circ}$
Optical purity or enatiomeric excess = $\frac{50}{75} = 66 \%$
% major enantiomer = $66 + \frac{34}{2} = 83 \%$
% minor enantiomer = $100 - 83 = 17 \%$
Exercise 5.6.2
$\% \: major \: = 90 \% \nonumber$
$\% \: major \: = 10 \%$
Optical purity or enantiomeric excess = $\frac{X}{-50} = 90-10 = 80 \%$
Solve for X. $X = -40^{\circ}$
Exercise 5.6.4:
Exercise 5.7.1:
a) $[\alpha] = \frac{\alpha}{(c)(l)}$
$c = (\frac{0.250g}{2mL})(\frac{10mL}{1dL}) = 1.25 \frac{g}{dL}$
$\alpha = \frac{(0.775 ^{\circ} + 0.806 ^{\circ} + 0.682 ^{\circ})}{3} = 0.754 ^ {\circ}$
$[\alpha] = \frac{\alpha}{(c)(l)} = \frac{(0.754 ^{\circ})}{(1.25 \frac{g}{dL})(0.5dm)} = + 1.21 ^{\circ}$
b)
– 1.21°
Exercise 5.7.2:
$[\alpha] = \frac{\alpha}{(c)(l)} \nonumber$
$[\alpha] = 32 \circ$
$c = (\frac{0.150g}{1mL})(\frac{10mL}{1dL}) = 1.5 \frac{g}{dL}$
c = (0.150g/ 1 mL)(10 mL/1 dL) = 1.5g/dL
$[\alpha] = \frac{\alpha}{(c)(l)} = 32 ^{\circ} = \frac{\alpha}{(1.5 \frac{g}{dL})(0.5dm)}$
Solve for α. $\alpha = + 24^{\circ}$
Exercise 5.8.1:
D-glyceraldehyde is R
L-glyceraldehyde is S
Exercise 5.8.2:
Exercise 5.8.3:
Exercise 5.9.1:
a) left b) right
Exercise 5.9.2:
a) right b) left
Exercise 5.9.1:
a) right b) right
Exercise 5.9.2:
a) left b) left
Exercise 5.9.5:
D-threose
2S, 3R
Exercise 5.9.6:
L-threose
2R, 3S
D- and L-threose are enantiomers of one another
Exercise 5.9.7:
L-erythrose
2S, 3S
L-erythrose and L-threose are diastereomers of one another.
Exercise 5.9.8:
a) (2)3 = 8 possible stereoisomers
RRR; SSS; RRS; SSR; RSS; SRR; SRS; RSR
b) 4 pairs
c) 12 different possible pairs of diastereomers
Exercise 5.9.9:
Exercise 5.9.10:
Exercise 5.10.1:
a) D-threitol → 2R, 3S
b) L- threitol → 2S, 3S
c) erythritol → 2S, 3R or 2R, 3S (a meso compound)
Exercise 5.10.2:
It does not matter which end you start counting from on these compounds since they are constituted the same.
Exercise 5.10.3:
Exercise 5.10.4:
Exercise 5.11.1:
Exercise 5.11.2:
Exercise 5.12.1:
Glycine
Exercise 5.12.2:
Exercise 5.12.3:
Proline
Exercise 5.12.4:
Glutamic acid and aspartic acid
Exercise 5.12.5:
Arginine, asparagine, glutamine, lysine; also tryptophan contains an aromatic heterocycle, although it is not basic.
Exercise 5.12.6:
Exercise 5.12.7:
Cysteine
Exercise 5.13.1:
Exercise 5.13.2:
Exercise 5.14.1:
Exercise 5.15.1:
A. Cis
B. Trans
C. Trans
D. Neither (There is free rotation around the C-C single bond.)
E. Neither (C-C triple bonds have substituents at 180 degrees to each other (linear).
F. cis
Exercise 5.16.1:
A. Z
B. E
C. E
D. Z
E. E
F. E
G. Z
H. E
Exercise 5.17.1:
Exercise 5.17.2:
Exercise 5.18.1:
a) Enantiomers
b) Enantiomers
c) Identical
d) Identical
e) Enantiomers
f) Identical
g) Enantiomers
Exercise 5.18.2:
a) Δ b) Λ c) Δ d) Λ e) Δ f) Δ g) Λ h) Λ
Exercise 5.20.1:
Pure = 125°
Optical purity = $\frac{100}{125} = 0.80$
$\% \: Major \: = 80 + \frac{20}{2} = 90\% \nonumber$
$\% \: Minor \: 100-90 = 10 \% \nonumber$
Exercise 5.20.2
Pure = 100°
Optical purity = $\frac{95}{100} = 0.95$
$\% \: Major \: = 95 + \frac{5}{2} = 97.5 \%$
$\% \: Minor \: 100-97.5=2.5 \% \nonumber$
Exercise 5.20.3
$Pure \: = 18^{\circ} \nonumber$
$\% \: Major \: = 60 \% \nonumber$
$\% \: Minor \: = 40 \%$
Optical purity = $\frac{X}{18} = 20 \%$
Solve for X:
$X = 3.6^{\circ} \nonumber$
Exercise 5.20.4:
$Pure \: = 25^{\circ} \nonumber$
$\% Major \: = 80 \% \nonumber$
$\% Minor \: = 20 \%$
Optical purity = $\frac{X}{25} = 80-20= 60 \%$
Solve for X:
$X = 15^{\circ} \nonumber$
Exercise 5.20.5
$[\alpha] = \frac{observed \: rotation}{(l)(c)} \nonumber$
$c = \frac{0.050g}{2.0mL} = 0.025 \frac{g}{mL}$
Average observed rotation = $\frac{(0.625 + 0.706 + 0.682)}{3} = 0.671 ^{\circ}$
$[\alpha] = \frac{0.671 ^{\circ}}{(0.025 \frac{g}{mL})(0.5 dm)} = 53.68 ^{\circ} \nonumber$
Exercise 5.20.6:
$[\alpha] = \frac{observed \: rotation}{(l)(c)} \nonumber$
$c = \frac{0.540g}{2.0mL} = 0.27 \frac{g}{mL}$
Average observed rotation = $\frac{(1.225 + 1.106 + 1.182)}{3} = 1.171 ^{\circ}$
$[\alpha] = \frac{1.171 ^{\circ}}{(0.27 \frac{g}{mL})(1.0 dm)} = 4.34 ^{\circ} \nonumber$
Exercise 5.20.7:
$[\alpha] = \frac{observed \: rotation}{(l)(c)} \nonumber$
$c = \frac{0.250g}{2.0mL} = 0.125 \frac{g}{mL} \nonumber$
$42 ^{\circ} = \frac{observed \: rotation}{(0.125 \frac{g}{mL})(0.5dm)}$
observed rotation = 2.625°
Exercise 5.20.8
a) $[\alpha] = \frac{observed \: rotation}{(l)(c)}$
$c = \frac{0.10g}{2.0mL} = 0.05 \frac{g}{mL} \nonumber$
$c = \frac{0.10g}{2.0mL} = 0.05 \frac{g}{mL}$
Average observed rotation = $\frac{(0.995 + 0.904 + 0.936)}{3} = 0.945 \circ$
$[\alpha] = \frac{0.945 ^{\circ}}{(0.05 \frac{g}{mL})(1.0 dm)} = 18.9 \circ$
b) % optical purity = $\frac{(100)(18.9)}{25}= 75.6 \%$
c) enantiomeric excess = 75.6%
d) $\frac{(100-75.6)}{2} = 12.2 \%$
12.2 % one enantiomer
87.8% other enantiomer
e) Enantiomers differ in how they interact with plane polarized light, but not in other physical analyses.
Exercise 5.20.9
$[\alpha] = \frac{observed \: rotation}{(l)(c)} \nonumber$
$40 \circ = \frac{observed \: rotation}{(l)(c)} \nonumber$
$40 \circ = \frac{observed \: rotation}{(l)(1.1c)}$
Observed rotation = 44°
Exercise 5.20.10
Exercise 5.20.11:
Exercise 5.20.12: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.20%3A_Solutions_to_Selected_Problems.txt |
Carbohydrates are one of the major classes of biological molecules. Of course, they play an important role in energy storage and transport. They are also crucial parts of important structures such as DNA.
Carbohydrates, in general, are polyhydroxylated aldehydes and ketones. That means their structure can be thought of as having a carbonyl, C=O, for one of its carbons. All of the other carbons in the structure are attached to a hydroxyl group, OH. It's a little more complicated than that because carbohydrates are sort of shape-shifting molecules: they undergo very simple reactions that can slightly change their structure, and they can change back again.
The simplest carbohydrate is glyceraldehyde. It is a three-carbon chain with a carbonyl on one end and hydroxy groups on the other two atoms. The second carbon in the chain is chiral.
The naturally-occurring carbohydrates are chiral.
Go to Animation SC8.1. A three-dimensional model of D-glyceraldehyde.
Go to Animation SC8.2. A three-dimensional model of L-glyceraldehyde.
Exercise \(1\)
What is the absolute configuration of L-glyceraldehyde? What about D-glyceraldehyde?
Answer
D-glyceraldehyde is R
L-glyceraldehyde is S
Biological molecules are frequently homochiral. That means that we have one enantiomer of glyceraldehyde in most organisms, but not the other. The one we have is D-glyceraldehyde. There are always a few organisms that have figured out over the eons how to make the other enantiomer of a naturally-occuring chiral molecule, but almost everything in nature shares our preference for D-glyceraldehyde.
Not only that, but all of the other carbohydrates share that same chiral center. You can think of the other carbohydrates as being based on glyceraldehyde. To make a bigger carbohydrate, additional carbons must be added into the glyceraldehyde chain. They are added to the carbonyl end, leaving the chiral center of glyceraldehyde unchanged. Just as the second-last carbon (really, the middle carbon) in glyceraldehyde is chiral and has a specific configuration, so do all of the others, and they all have the same configuration at that carbon.
• The "last" chiral center in other carbohydrates all have the same configuration as the chiral center in glyceraldehyde.
Most of the time, if you look in a biology or biochemistry textbook, you will see carbohydrates drawn in very particular ways. Now, carbohydrates can always take a couple of different forms (remember, they are shape-shifters). They usually have a ring form and a chain form, although very small carbohydrates only have a chain form. The chain form is usually drawn in a cross shape or tree shape, with a long vertical line crossed by horizontal lines. These drawings are called Fischer projections. If there is a chiral center in a Fischer projection, then it matters whether groups on the horizontal line are drawn on the right or on the left.
• Fischer projections are drawings that are sometimes used to show stereochemistry in a different way.
• The groups to left and right of the Fischer projection are coming towards you.
• The groups on top and bottom are going away from you.
• The carbon chain is drawn vertically with attachments at the sides.
• Normally, the chain is oriented so that the "highest priority functional group" on the chain is at or near the top of the chain; that usually means the group with the most bonds to oxygen. In glyceraldehyde, the CH=O group is drawn at the top rather than at the bottom.
The big advantage of Fischer projections is that you can easily tell at a glance the relationship between two molecules, and the relationship between specific chiral centers along the chain. Are the molecule the same? Are they mirror images? Are the chiral centers the same? Are they opposite?
On the next page, we will see some slightly bigger carbohydrates.
Exercise \(2\)
Draw the following molecules as Fischer projections.
Exercise \(3\)
Convert the following Fischer projections into wedge-dash drawings. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/05%3A_Stereochemistry/5.21%3A_Biological_Building_Blocks-_Carbohydrates.txt |
When we smell something, the information travels to us via molecules, and almost always these are organic molecules. These molecules can be detected by a variety of organs, including noses in dogs and antennae in crickets, but no matter what organ is sensing the smell, one of the crucial factors in determining how an organism reacts to a compound is the shape of the molecule. The sense of smell depends on thousands of different receptors in the organ, working conceptually on a lock-and-key basis: a molecule with a given shape can fit into a given receptor, and when it does, a signal is sent telling the nervous system that the organism has encountered that particular type of molecule, and the organism reacts appropriately.
What gives a molecule its shape? Given a structural formula, could you determine the shape of the corresponding molecule in three dimensions? Could you predict its biological activity, including not only its smell, but also a host of other behaviors linked to the shape of molecular messengers, such as anti-cancer activity or narcotic properties? These questions are at the cutting edge of biological chemistry. Although they are best answered through computer modeling, we can develop some of the qualitative ideas used in these models.
Computer modeling employs "basis sets", small sets of information that the computer could apply to any molecule in order to predict its properties. In order to understand computational conformational analysis, in this chapter we will develop a very simple basis set as an example.
6.02: Simple Molecules
Ethene (CH2=CH2) contains two trigonal planar carbon atoms. Each carbon is bonded to two hydrogens as well as the other carbon.
This structure is illustrated in the Jmol animation linked below. These animations are completely manipulable with a mouse. (They are grabbable and movable. Use your mouse.)
Go to Animation CA2.1. A three-dimensional model of ethene.
From molecular orbital considerations, the two trigonal carbons must be coplanar with each other in order to maximize carbon-carbon pi bonding. As a result, the entire molecule must be planar.
Ethane (CH3-CH3) is very similar to ethene, except that each carbon is tetrahedral and is bonded to three hydrogen atoms. There is no need to arrange the two carbon atoms with a certain orientation to each other because the sigma bond holding the two carbons together is symmetric about the C-C axis. As a result, there are a number of ways the two ends of the molecules could be oriented.
Go to Animation CA2.2. A three-dimensional model of ethane in staggered conformer.
Go to Animation CA2.3. A three-dimensional model of ethane in eclipsed conformation.
Working with models often reflects reality quite well. While holding a model of ethane, you could grab one CH3 group and twist it while holding the other one still, and as a result you could put the molecule into different poses. The ethane model is flexible via rotation around the C-C bond.
Ethane can adopt different shapes, because one end of the molecule can rotate with respect to the other. These different shapes are seen most dramatically while viewing ethane along the carbon-carbon bond. From this vantage point, rotation about the C-C bond results in a changing shape that alternates between a six-pointed star and a three-pointed one. In the three-pointed star, some of the hydrogens are hidden or eclipsed behind the corresponding hydrogens on the carbon that is closest to the viewer; in the six pointed star they have come out into view again.
• A drawing that shows a molecule viewed along the axis of a rotating bond is called a Newman Projection.
• The different shapes obtained via bond rotation are called conformations.
• The angle between a specific C-H bond on one carbon and a specific C-H bond on another carbon, when viewed in a Newman projection, is called a dihedral angle.
• The conformation in which the Newman projection shows all six hydrogens of ethane in plain view is a staggered conformation.
• The conformation in which three of the hydrogens are hidden behind the others is an eclipsed conformation.
• In the eclipsed conformation, the dihedral angle could be 0, 120, 240 or 360 degrees; in a staggered conformer, the dihedral angle could be 60, 180, 300 degrees.
Many conformations of ethane are possible, but not all of these conformations are equally likely. The most probable conformation is a staggered one.
• Staggered conformations are lower in energy than eclipsed ones.
• The term "conformer" is used for a stable conformation, one that has a practical chance of occurring.
The energy barrier to rotation about the C-C bond in an ethane molecule can be measured using microwave spectroscopy. This technique is similar to infrared spectroscopy but the molecular motion connected to microwaves is bond rotation rather than vibration. Experimentally, the barrier to rotation is about 3 kcal/mol. (Keep in mind that, in chemistry, the term "experimental" gives a finding more weight, not less, because it reflects reality.) This barrier is small enough that at room temperature an ethane molecule can freely rotate, but high enough that an ethane molecule would not be found sitting in the eclipsed state.
• The factor that destabilizes eclipsed conformations is called torsional strain.
• Torsional strain in ethane apparently arises from a molecular orbital interaction between a populated C-H σ bond and the empty C-H σ* orbital on the neighbouring carbon.
• When the two C-H bonds are in the same plane but pointing in opposite directions, the orbitals have proper symmetry to interact. This is called an antiperiplanar orientation, from anti, meaning "opposite", and periplanar, meaning "in the same plane".
• This orientation is only possible when the neighbouring bonds are staggered.
• Eclipsed conformations lack this interaction and are higher in energy by about 3 kcal/mol (about 1 kcal/mol per C-H σ - σ* interaction.
Looking at a space filling model of ethane, it seems that the overall shape difference between staggered and eclipsed ethane is pretty small. The important point is that most organic structures are not static, but flexible. Also, torsional strain is found in all carbon-carbon bonds, so we will keep using the experimental result from ethane as we build a "basis set" for understanding conformation in larger molecules.
Still pictures of models obtained using Spartan 14 from Wavefunction, Inc., Irvine, California. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.01%3A_Introduction_to_Conformation.txt |
In the case of ethane, conformational changes are very subtle, but in others they are more obvious. Butane (CH3CH2CH2CH3) has four tetrahedral carbons and three carbon-carbon bonds connecting them together. Let's number the carbons along the chain C1, C2, C3 and C4. Rotating around the C-C bonds connected to the terminal carbons -- C1-C2 and C3-C4 -- only subtle changes in shape would be apparent. However, rotating about C2-C3 produces some pretty obvious shape changes.
Pay attention to where the two methyl groups are with respect to each other. If we call C1-C2-C3-C4 the dihedral angle, then at 0 degrees the molecule is in an eclipsed conformation, apparent when looking at the Newman projection. Looked at from other vantage points, the molecule is curled up into the shape of a letter C.
Go to Animation CA3.1. A three-dimensional model of butane in an eclipsed conformation.
The "least stable eclipsed" conformation is described that way because there are two different eclipsed conformations; we will see the other one soon. Of the two eclipsed conformations, this one is less stable than the other one. In an eclipsed conformation, bonds coming from two neighbouring atoms are lined up with each other. The dihedral angle between these bonds -- the C1-C2 bond and the C3-C4 bond in the case of butane's least stable eclipsed conformation -- is 0 °. That situation introduces torsional strain. As a result, neither one of these eclipsed conformations is very stable.
At 60 degrees, the molecule is no longer eclipsed, and just as in ethane the energy is a little bit lower, but the overall shape when viewed from the side is still a sort of twisted C. This shape is one of two possible staggered conformations, in which there is no torsional strain.
Go to Animation CA3.2. A three-dimensional model of butane in a gauche conformation.
At 120 degrees the molecule is eclipsed again , but from the side it has now twisted almost into a shape like the letter Z. It is clearly different from the other eclipsed conformation. This one is a little more stable than the other eclipsed conformation, for reasons that we will look into shortly. It is called the "more stable eclipsed conformation".
Go to Animation CA3.3. A three-dimensional model of butane in an eclipsed conformation.
At 180 degrees, the molecule is staggered again and has settled into a regular, zig-zag, letter Z shape. There is no torsional strain, so this is one of the more stable shapes that the molecule can adopt. However, it is not the same shape as the other staggered conformer (the one we called "gauche"). This one is called the "anti" conformer.
Go to Animation CA3.4. A three-dimensional model of butane in anti conformer.
These conformations of butane are really pretty different. Which shape would the molecule prefer? From what we learned about ethane, we could probably rule out the eclipsed conformations. Each of those would have 3 kcal/mol of torsional strain.
However, there is another factor that destabilizes the initial conformation at 0 degrees, in favor of the Z-shaped one with the methyl groups 180 degrees apart from each other. This factor is called "sterics" and it refers to the idea that molecules, or parts of molecules, take up space, and so two parts of the butane can't occupy the same place at the same time. Put more simply, sterics refers to crowdedness. When the two methyl groups in butane are too close together, they are too crowded, and they are at higher energy. When they get farther apart, crowding subsides and the energy in the molecule goes down.
• Torsional strain forces bonds on neighbouring carbons to be staggered.
• Steric strain forces groups away from each other to relieve crowding.
• Torsion and sterics both contribute to strain energy.
• Strain can be released by getting the molecule in a lower energy conformation.
For butane, that means getting those two methyl groups away from each other and keeping the bonds staggered.
There is some additional jargon that is used to describe these butane conformations:
• In an anti conformer, the largest groups are 1800 from each other.
• In a gauche conformer, the bonds are staggered but the largest groups are 600 from each other.
Because of different steric interactions, the most stable conformer is the anti conformer. The second-most stable conformer is the gauche conformer. The other shapes we have looked at, the two eclipsed conformations, are not very stable at all. At room temperature, a bottle of butane gas has enough energy that the individual molecules can allow their bonds to spin around, sampling different shapes. However, at any given time, most of the molecules will be found in the anti conformer, because each of those molecules will spend most of its time in the most stable conformation. Some of the molecules will be found in the gauche conformer, because that is the second-most stable conformation. None of the molecules will be found in the eclipsed conformations, because those are points of maximum energy. The molecules have enough energy to rock back and forth a little bit, and sometimes they will not rotate all the way through an eclipsed conformation, but they won't stop there. Instead, they will keep rotating until they get to another staggered conformer.
• A butane molecule spends most of its time in the anti confrmer.
• A butane molecule spends some of its time in the gauche conformer.
• A butane molecule spends almost none of its time in eclipsed conformations.
Note that we have two very similar words: conformation and conformer. They are related, but not the same thing.
• A conformation is any shape that a molecule can adopt.
• A conformer is a conformation that is stable enough so that a molecule can adopt for significant amount of time.
Exercise \(1\)
Draw Newman projections of the following compounds in a gauche and an anti conformation.
1. 1,2-dichloroethane (\(\ce{ClCH2CH2Cl}\))
2. 1,2-butanediol (\(\ce{CH3CH2CH(OH)CH2OH}\))
3. 2-chloroethylamine (\(\ce{ClCH2CH2NH2}\))
Answer
Still pictures of models obtained using Spartan 14 from Wavefunction, Inc., Irvine, California. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.03%3A_Conformation_of_Butane.txt |
Using information on the rotational barrier of butane, which is about 4.5 kcal/mol, we can add more information to our basis set. We will also use information from other sources to build a more complete picture.
• In ethane, there is no steric strain, because the hydrogens are too far apart to crowd each other.
• In propane (CH3CH2CH3), the hydrogens on the first and third carbon are close to each other, but hardly cause any crowding.
• In butane's anti conformer, there is no steric strain, because the methyl groups are far apart.
• In butane's least stable eclipsed conformer, with the methyls eclipsed, there is about 1.5 kcal/mol of steric strain. The 1.5 kcal sterics plus 3 kcal torsion equals the rotational barrier.
• In butane's gauche conformer, there is no torsion, but the amount of sterics appears to be only a little less than when the methyls are eclipsed.
One simple method of assessing steric strain, without memorizing a big table of information, is use an atom counting method. Count from one hydrogen, to the next atom, and so on to the other hydrogen that seems likely to interact with the first. In ethane, this distance is four atoms (H-C-C-H). In propane it is five. In butane it is six. The potential steric interaction in butane is accordingly called a six atom interaction. The basis set used in this method is given in the table below.
Table \(1\). Atom-Counting Basis Set for Estimating Strain Energy
Type of strain Strain energy (kcal/mol)
torsion (three pairs of adjacent bonds eclipsed) 3
4-atom sterics 0
5-atom sterics 0.1
6-atom sterics, gauche 1
6-atom sterics, eclipsed 1.5
7-atom sterics, gauche 3
Note that a methyl (CH3) group can always rotate, so multiple hydrogens on one methyl group could potentially interact with multiple hydrogens on another, but not at one time. We only need to take into account the full interactions that could happen simultaneously.
Exercise \(1\)
Calculate the torsional strain in the following cases.
Answer
Exercise \(2\)
Find all 6 atom interactions in the following pictures.
Answer
Exercise \(3\)
Draw Newman Projections every 60° for the following molecules. Assess the strain energy for each Newman Projection found by rotating about the C2-C3 bond. Use the basis set in Table CA1. Plot strain energy vs. dihedral angle and label
1. 2-methylpropane (CH3CH(CH3)CH3)
2. 2,3-dimethylbutane (CH3CH(CH3)CH(CH3)CH3)
3. pentane (CH3CH2CH2CH2CH3)
Answer a:
Answer b:
Answer c:
6.05: Molecular Modeling
The shapes of molecules can be predicted by tallying up estimates of steric and torsional interactions. This task is referred to as a molecular mechanics calculation. The basis set used may vary from the one given here, but the idea is the same: a limited set of information is applied to molecules, no matter how large or complex, in order to measure the relative energy of different conformations. Typically a calculation is followed by a change in a dihedral angle somewhere, the calculation is repeated and so on, so that eventually the most stable conformation is found.
A calculation like this could take a very long time. Even on a computer, there is usually a need to speed things up. A common short cut for employs the following approach. If a change in dihedral angle results in a decrease in energy, the calculation continues with a new change in that dihedral angle. Eventually, if changes lead to increased energy, the program assumes the lowest energy has already been passed and it backtracks to that point again. It may then try another dihedral using the same method. In this way the computer does not waste time on calculations involving high energy eclipsed conformations or ones with steric problems. As a result, a molecular mechanics calculation can usually estimate the lowest energy conformer in a few seconds.
The limitation of that approach can be seen in a generic plot of energy vs. conformational change in which there is more than one low-energy conformer. How does the calculation determine that it has found the absolute lowest energy conformer, and not just a dip in the energy surface?
For example, in a plot of butane's relative energy vs its dihedral angle around C2-C3. The gauche conformation occupies a "local energy minimum", meaning that the energy is lower here than when the dihedral angle is either greater or smaller, because in either direction the molecule would arrive at an eclipsed conformation with higher energy. In the approach described above, typically the default approach in most calculations, the computer could get stuck at the gauche conformation, because a change in either direction results in an increase in energy. However, if the computer would just push past those energy maxima then it would arrive at the anti conformation, which is the "global energy minimum", with the lowest energy of all.
• Beware of local minima when looking for the most stable conformer on a computer.
• Usually, you can tell whether the computer has the right answer based on your understanding of sterics.
Complete conformational analysis sometimes requires some method of "dihedral angle driving". In this kind of approach, the software user can tell the computer to calculate energies a given number of times around a given dihedral angle; for example, maybe a calculation would be done every sixty degrees for the dihedral around the C2-C3 bond in butane, or maybe every thirty degrees. That way, the computer is sure to find the global minimum for a particular dihedral angle.
More advanced methods may be required to arrive at correct low-energy structures for more complicated molecules. For example, hydrogen bonding is clearly a major factor controlling the structure of molecules such as DNA and proteins. We have looked only at steric and torsional strains because they are ubiquitous in molecular mechanics, but different basis sets, and different types of calculations, could take into account additional interactions such as hydrogen bonding or other electronic phenomena which would help correctly establish the shapes of more complex molecules. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.04%3A_The_Basis_Set.txt |
Cyclic systems are a little different from open-chain systems. In an open chain, any bond can be rotated 360 degrees, going through many different conformations. That complete rotation isn't possible in a cyclic system, because the parts that you would be trying to twist away from each other would still be connected together. Cyclic systems have fewer "degrees of freedom" than aliphatic systems; they have "restricted rotation".
Because of the restricted rotation of cyclic systems, most of them have much more well-defined shapes than their aliphatic counterparts. Let's take a look at the basic shapes of some common rings.
• Many biologically important compounds are built around structures containing rings, so it's important that we become familiar with them.
• In nature, three- to six-membered rings are frequently encountered, so we'll focus on those.
A three membered ring has no rotational freedom whatsoever. A plane is defined by three points, so the three carbon atoms in cyclopropane are all constrained to lie in the same plane.
Furthermore, if you look at a model you will find that the neighboring C-H bonds (C-C bonds, too) are all held in eclipsed conformations.
Go to Animation CA6.1. A three-dimensional model of cyclopropane.
Cyclopropane is always at maximum torsional strain. This strain can be illustrated in a line drawing of cyclopropane as shown from the side. In this oblique view, the dark lines mean that those sides of the ring are closer to you.
However, the ring isn't big enough to introduce any steric strain, which does not become a factor until we reach six membered rings. Until that point, rings are not flexible enough for two atoms to reach around and bump into each other.
The really big problem with cyclopropane is that the C-C-C bond angles are all too small.
• All the carbon atoms in cyclopropane appear to be tetrahedral.
• These bond angles ought to be 109 degrees.
• The angles in an equilateral triangle are actually 60 degrees, about half as large as the optimum angle.
• This factor introduces a huge amount of strain in the molecule, called ring strain.
Cyclobutane is a four membered ring. In two dimensions, it is a square, with 90 degree angles at each corner.
However, in three dimensions, cyclobutane is flexible enough to buckle into a "butterfly" shape, relieving torsional strain a little bit. When it does that, the bond angles get a little worse, going from 90 degrees to 88 degrees.
Go to Animation CA6.2. A three-dimensional model of cyclobutane.
In a line drawing, this butterfly shape is usually shown from the side, with the near edges drawn using darker lines.
• With bond angles of 88 rather than 109 degrees, cyclobutane has a lot of ring strain, but less than in cyclopropane.
• Torsional strain is still present, but the neighbouring bonds are not exactly eclipsed in the butterfly.
• Cyclobutane is still not large enough that the molecule can reach around to cause crowding. Steric strain is very low.
• Cyclobutanes are a little more stable than cyclopropanes and are also a little more common in nature.
Cyclopentanes are even more stable than cyclobutanes, and they are the second-most common paraffinic ring in nature, after cyclohexanes. In two dimensions, a cyclopentane appears to be a regular pentagon.
In three dimensions, there is enough freedom of rotation to allow a slight twist out of this planar shape.
Go to Animation CA6.3. A three-dimensional model of cyclopentane
In a line drawing, this three-dimensional shape is drawn from an oblique view, just like cyclobutane.
• The ideal angle in a regular pentagon is about 107 degrees, very close to a tetrahedral bond angle.
• Cyclopentane distorts only very slightly into an "envelope" shape in which one corner of the pentagon is lifted up above the plane of the other four, and as a result, ring strain is entirely removed.
• The envelope removes torsional strain along the sides and flap of the envelope. However, the neighbouring carbons are eclipsed along the "bottom" of the envelope, away from the flap. There is still some torsional strain in cyclopentane.
• Again, there is no steric strain in this system.
Still pictures of models obtained using Spartan 14 from Wavefunction, Inc., Irvine, California. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.06%3A_Cyclic_System.txt |
Finally, cyclohexane is the most common paraffinic ring of all, partly because it is possible to form a conformation of cyclohexane having very little strain energy. However, it has enough flexibility that things become more complicated than in the other rings we have looked at. If cyclohexane were flat, the C-C-C bond angles would be 120 degrees, like the corners of a regular hexagon. That's too big for a tetrahedral carbon, so the cyclohexane uses its flexibility to adopt a shape in which this strain energy is relieved.
There are many shapes that a cyclohexane could adopt. We will look only at two of them and use our basis set to think about which one is more stable and more likely. One of these shapes is called a "boat" conformation. In this conformation, the opposite corners of the cyclohexane ring are pulled up above the plane of the other four carbons, forming the prow and stern of a boat.
Go to Animation CA7.1. A three-dimensional model of cyclohexane, in a strained conformation
The bond angles in the boat are 109 degrees. The boat has no ring strain.
There are torsional strains present along the two sides of the boat, although not at the prow and the stern.
The boat conformation suffers from steric strains. By pulling the carbons at the prow and stern up above the plane of the ring, the attached hydrogens get very close to each other.
In our basis set, this is a six atom interaction. We count the number of atoms to assess the potential steric interaction, we start at one of these crowded hydrogens, go through the first to fourth carbon, and then finish at another crowded hydrogen. That's a total of six atoms including the two atoms that appear to be leaning towards each other, so this is a significant steric interaction. If we then look along one of the sides of the boat in a Newman projection, we find that the stern and prow of the boat actually eclipse each other, so this six atom interaction is pretty strong, about 1.5 kcal/mol.
Add to that the 6 kcal/mol of torsional strain, from three pairs of eclipsed bonds on one side of the boat and three pairs on the other, and there is a total of 7.5 kcal/mol strain energy in the boat conformer of cyclohexane.
In reality, the boat conformation would not occur if there were another conformation significantly lower in energy.
• The boat conformation is an unstable conformation because of both torsional and steric strain.
A second shape is called the chair conformation. In a hand-held model, this shape can be obtained by grasping the prow of the boat and twisting it down below the boat, while keeping the stern in its same position above the boat.
Go to Animation CA7.2. A three-dimensional model of cyclohexane in its stable conformer
A thorough investigation of your model will show that there are no longer any torsional strains in this conformation.
There aren't any steric strains, either, because although you can identify pairs of hydrogens that are six atoms apart from each other, these pairs always seem to be pointing well away from each other.
In fact, if the model is viewed from above, so that it looks like a hexagon, six of the hydrogen atoms point directly outward away from the ring; they can't get much further apart than that. These atoms are all called equatorial hydrogens.
The only atoms that get close to each other at all are the three hydrogens above the ring, which are all roughly in the same area of space, and the three below the ring. These atoms are called axial hydrogens.
The worst steric strain in the chair is the 5-atom interaction between axial hydrogens. In our basis set, this interaction costs 0 kcal/mol. The chair is more stable than the boat by 7.5 kcal/mol.
This conformation of cyclohexane is the preferred conformer.
• Cyclohexane is always found in the chair conformer. This conformer has no torsional or steric strain.
The chair can be shown in a line drawing as shown below. Notice that three hydrogens point up from the chair and three point down. The other hydrogens are pointing outward from the center.
Still pictures of models obtained using Spartan 14 from Wavefunction, Inc., Irvine, California.
6.08: Diamond Lattice Drawings
It will be helpful for you to be able to draw cyclohexane in the chair form because it is such a common structural element in natural compounds. Let's look at the way we would most easily draw cyclohexane: it's a six membered ring, a hexagon. Notice that the opposite sides of this ring are parallel. This feature is the key to drawing a chair.
Start by drawing two parallel lines of equal length, one slightly higher than the other and one slightly offset to the side. Most people find it easiest to make this first pair of lines horizontal.
A second pair of parallel lines will connect to these two at an angle, but at opposite ends of the original lines, and going in opposite directions: one upward, one downward.
At one end of the structure, we are beginning to draw the uplifted back of the chair, and at the other end is the downward footrest. Next, bridge the gaps at the two ends of the drawing with another pair of parallel lines.
That's the carbon skeleton of cyclohexane, but for careful conformational analysis we will also need to consider the positions of the hydrogens. Remember that there are three axial hydrogens above the ring and three below. Look at the ring you have drawn: three of the vertices are pointing downward and three are pointing upward. These vertices are pointing the way to the axial hydrogens. Extend these vertices with six parallel lines, three pointing up and three pointing down, from alternating corners.
You have just formed the axial C-H bonds.
Now, the useful thing to know about the drawing is that lines can only be drawn in four different directions. The remaining lines to the equatorial hydrogens will have to be chosen from among these directions. We already have lines in three directions on each carbon; we just need to add a fourth to each. To decide which direction is missing, skip one bond in either direction along the ring and you will find a parallel line.
Finally, remember that the equatorial hydrogens point outward, away from the ring, so keep these lines pointing away from the center of the chair.
This type of drawing is called a diamond lattice drawing, because the all-carbon structure of diamond can be constructed this way by extending the axial and equatorial lines into additional chairs. Diamond lattice drawings can also be used to depict staggered conformations of aliphatic hydrocarbons, but only if you are very careful. Taking a few minutes to practice these drawings now may be helpful before you proceed to the next section.
Exercise \(1\)
Using the cyclohexane chair as a template, draw diamond lattice representations of the following compounds in staggered conformations. If you have to, draw a whole chair and erase parts until you have only what you need.
1. ethane
2. propane
3. butane in a gauche conformation
4. butane in an anti conformation
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.07%3A_Cyclohexane.txt |
Most rings found in nature are part of a larger structure, so understanding the shape of the ring is just the beginning of the question of conformation. How will substituents that are attached to the ring affect the shape? Let's look at substituted cyclohexanes, because they are very common and because their flexibility allows them to adopt very different shapes in order to accommodate attached groups.
Remember that in the chair form of cyclohexane, which has already been established as the most stable conformation by several kcal/mol, there are hydrogens in two distinct positions:
• axial positions
• equatorial positions.
Replacing one of these hydrogens with a methyl group would result in the methyl group sitting in either an axial or equatorial position. Let's assume it's in an axial position, which is above or below the ring rather than out to the sides.
• We know that there is no torsional strain in the chair, and we'll assume that the methyl group can be oriented such that it has no torsional strain either.
• We do need to look for steric strains. You will find that the methyl hydrogens are close to the two axial hydrogens on the same face of the ring. This is a six atom interaction in each case.
• We will assume that the methyl group is rotating freely such that only one of its hydrogens is close enough to these axial neighbors to cause steric strain, but there will be strain with both of those axial hydrogens.
• Sighting along the CH2--C(H)Me bond on the edge of the ring, you can see that the interaction between the axial methyl and an axial hydrogen resembles the gauche interaction in butane. These two steric interactions will cause an increase in strain energy of about 2 kcal/mol.
Go to Animation CA9.1. A three-dimensional model of methylcyclohexane, in an axial conformation
This chair conformation is undoubtedly somewhat stable; there is much less strain in this compound, for instance, than there was in the boat conformation of cyclohexane. However, it's possible that there is an even more stable conformation, and if so we should be able to find that conformation by twisting around some of the bonds in the molecule.
In cyclohexanes, there are usually two different chair conformations that can be accessed via a "ring flip". To see this, grasp a model of methylcyclohexane and identify two corners at opposite ends of the hexagon (the first and fourth carbons in the ring). One of these corners will be pointing down; push it up. The other corner will be pointing up; push it down. In doing so, all the other corners of the ring will change positions from down to up and from up to down as well, so that you end up with a new chair.
If you carry out a ring flip on methylcyclohexane, you will find that the methyl group moves between an axial position and an equatorial one depending on which form of the chair conformation it begins in. The actual compound can do this pretty easily at room temperature; although these two chairs occupy energy minima, the energy barrier between them is pretty low. It's possible that one of these conformations is a global minimum, however.
Take a look at the new conformation of methylcyclohexane, with the methyl group in an equatorial position.
• There need be no torsional strains.
• The methyl group is not very close to any neighbouring hydrogens. It is 5 atoms away from a couple of the other equatorial hydrogens, but 5 atom interactions are not large enough to worry about.
• As a result, there is no significant strain energy in this conformation. This is the most stable conformer of methylcyclohexane.
Go to Animation CA9.2. A three-dimensional model of methylcyclohexane, in an equatorial conformation
It is usually the case that when substituents are in equatorial positions, the conformation is more stable, for the same reasons that we saw with methylcyclohexane: it is pointing outward, away from the rest of the ring, rather than sharing the space above the ring with other atoms. However, it may not always be possible for substituents to be in that favored position.
Exercise \(1\)
Practice drawing these monosubstituted cyclohexane structures in the chair forms.
a. Draw the compound on a diamond lattice.
b. Draw the other chair conformation.
c. Find all strain.
d. Determine which conformer is more stable.
Answer
Exercise \(2\)
How will other substituents behave in axial vs. equatorial positions? Use a statistical assumption to estimate how much steric strain will be caused by a methyl group vs. an amino, a hydroxyl or a fluoro substituent. Apply your assumption to estimate the energy difference between axial and equatorial conformers of the following compounds:
a) cyclohexanol (c-C6H11OH) b) cyclohexylamine (c-C6H11NH2)
c) fluorocyclohexane (c-C6H11F)
Answer
Exercise \(3\)
Label the substituents as axial or equatorial and as up or down.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.09%3A_Substituted_Cyclohexanes.txt |
Take a look at 1,2-dimethylcyclohexane. The two methyl (CH3) groups in this compound are on adjacent carbons.
There are two ways these methyl groups could be attached to the ring: they can either be on the same face, as they are in the cis isomer; or they can be on opposite faces, as they are in the trans isomer.
The difference between these two compounds is pretty clear from a regular hexagon view of the molecule, with wedges and dashes denoting the spatial relationship between the methyl groups. Wedges show things coming towards us, on the near face of the ring, and dashes show things away from us, on the far face of the ring.
The diamond lattice drawing can be a little confusing, in this case.
• Remember that alternating corners of the ring point either up or down, and that each has an attached axial and equatorial position.
• The two methyl groups in cis-1,2-dimethylcyclohexane must occupy the axial position on one carbon and the equatorial position on the next.
That's because on a carbon that occupies a lower corner of the ring, the upper position that a methyl can occupy is equatorial, but for the next carbon, set in an upper corner of the chair, the upper position is an axial one. Viewed from the edge of the chair, two cis methyl groups would both be attached to either the upper face or the lower face of the chair, and so one would be axial and the other equatorial.
Go to Animation CA10.1. A three-dimensional model of cis-dimethylcyclohexane.
In trans-1,2-dimethylcyclohexane, the two methyl groups would occupy opposite faces of the ring. The easiest way to view this relationship is in a conformation in which the two methyl groups are in axial positions on adjacent carbons, one pointing up and the other down.
This view can also be seen in a model.
Go to Animation CA10.2. A three-dimensional model of trans-dimethylcyclohexane, in a diaxial conformation.
However, a ring flip converts the same molecule into a different conformer, in which both these groups are equatorial. The methyl groups no longer appear to be trans to each other, but they must be, because we haven't broken any bonds and this is still the same molecule we were looking at in the diaxial conformation.
Go to Animation CA10.3. A three-dimensional model of trans-dimethylcyclohexane, in a diequatorial conformation.
In fact, given the two positions available at each ring carbon, one of the methyls is always in the uppermost position, whether it is in an axial or equatorial orientation, and the other methyl is always in the lowermost position.
Exercise \(1\)
Convert the following chair structures to the wedge-dash projection.
Answer
Exercise \(2\)
Determine the stereochemical relationship between these pairs of compounds.
Answer
Exercise \(3\)
3. For the following the disubstituted cyclohexane rings:
a. Circle all stereocenters and label them with the correct R/S designation.
b. Draw the two chair conformers on the diamond lattice. (Be careful not to draw the enantiomer).
Answer
Exercise \(4\)
Perform an analysis of steric interactions in the chair conformation of trans-1,2-dimethylcyclohexane. After a ring flip, perform this analysis again. Compare the energy of the two conformers.
Answer
It's tempting when working with models to take a group off and snap it back on somewhere else in order to think about different conformations, but that isn't a good idea. Taking a group off the model requires breaking a bond, and that isn't what happens when molecules convert between conformers. Instead, molecules simply twist into new shapes without breaking any bonds. If you snap pieces on and off your model while trying to evaluate conformations, you may end up with a different isomer altogether; for instance, it would be easy to accidentally convert cis-1,2-dimethylcyclohexane into trans-1.2-dimethylcyclohexane.
Exercise \(5\)
Perform an analysis of steric interactions in the two possible chair conformations of the following compounds.
a) cis-1,3-dimethylcyclohexane b) trans-1,3-dimethylcyclohexane
c) cis-1,4-dimethylcyclohexane d) trans-1,4-dimethylcyclohexane
Answer
Still pictures of models obtained using Spartan 14 from Wavefunction, Inc., Irvine, California. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.10%3A_Disubstituted_Cyclohexanes.txt |
We have looked at some different six-membered rings, cyclohexanes, and have seen how substituents attached to these rings might influence the shape of the overall molecule. Cyclohexanes are particularly important because of they are quite common in nature. There are other variations of cyclic systems that are worth looking at, including different cyclohexane derivatives and other rings.
Bicyclics and Other Polycyclics
Decalins are examples of fused ring systems. In fused ring systems, two rings share an edge. Decalins contain ten carbons total, but the ten carbons are divided into two rings, each of which is six carbons around.
There are two possible isomers of decalin: cis and trans. The easiest way to distinguish the two isomers is to look for the two hydrogens on the shared edge between the two rings. Those two hydrogens are found on the same face of the cis decalin.
Go to Animation CA11.1. A three-dimensional model of cis-decalin.
The two hydrogens along the junction are seen on the opposite faces in trans decalin.
Go to Animation CA11.2. A three-dimensional model of trans-decalin.
Exercise \(1\)
Draw cis-decalin and trans-decalin using a wedge/dash picture.
Answer
Exercise \(2\)
Calculate steric strains in cis-decalin and trans-decalin.
Answer
Fused ring systems like decalin are very common. In fact, similar ring systems are found in steroids, which are an important class of lipids. Steroids generally have similar structures that include a series of fused rings as seen in the parent compound, cholesterol.
The similarity is striking between estradiol, a female sex hormone, and testosterone, a male sex hormone.
Go to Animation CA11.3. A three-dimensional model of estradiol.
Go to Animation CA11.4. A three-dimensional model of testosterone.
Exercise \(3\)
Consider a steroid structure such as cholesterol.
1. Based on the relationships along the fused edges between the ring, does the structure more closely resemble that of cis-decalin or that of trans-decalin?
2. The overall shape of the structure could probably best be described as (choose one: wide and wavy / compact and curled up / boxy like a cage).
Answer a:
The substituents are always trans along the junctions between each pair of rings. The steroids resemble a series of trans-decalin structures.
Answer b:
The overall structure would be more wide and wavy like a trans-decalin, rather than curled or boxy like a cis-decalin.
Polycyclic systems do not have to be fused along one edge like decalin. They can be connected in a variety of ways. Norbornane is an example of a bicyclic system in which the two rings actually share more than two carbons.
Norbornane can be used to illustrate the systematic way to describe fused ring systems. The suffix describes the total number of carbons in the ring system. The prefix describes how many rings are fused together. The middle part, in brackets, indicates how many atoms form the "bridges" between the "bridgehead" atoms, where the system splits off into two different rings. One number is given to describe each bridge. In norbornane, there are two 2-carbon bridges and a single 1-carbon bridge.
Exercise \(4\)
Write a systematic name for decalin.
Answer
Bicyclo[2.2.0]decane
Exercise \(5\)
Provide systematic names for the following structures.
Answer a:
Bicyclo[2.1.1]hexane
Answer b:
Bicyclo[3.2.1]octane
Answer c:
Bicyclo[2.1.0]pentane (more commonly called "housane")
Answer d:
Bicyclo[2.2.2]octane
Answer e:
cis-Bicyclo[3.3.0]octane
Answer f:
cis-Bicyclo[1.1.0]butane
Answer g:
Bicyclo[1.1.1]pentane
Answer h:
Bicyclo[4.3.3]dodecane
Exercise \(6\)
How many different rings can you trace in the structure of adamantane?
Answer
Although we could sketch out many rings using adamantane, just three rings are needed to include all the carbon atoms in the structure. Thus, adamantane is considered a tricyclic system. The systematic nomenclature of tricyclic systems gets a little more complicated, so we won't worry about that.
Bicyclic systems are common in nature. For example, the tropane alkaloids all contain a bicyclo[3.2.1] ring system. Alkaloids are a group of natural products that contain basic nitrogen atoms. The tropane alkaloids contain this nitrogen atom in the 1-atom bridge. Tropinone is a simple example.
Common examples of natural products containing the tropinone structure include atropine and cocaine. Atropine, derived from deadly nightshade or belladonna, is a heart medication; it is used to increase heart rate if the rate gets too low. Atropine can also be fatal if used incorrectly, causing a wide range of physiological effects; in some dosages, it can even decrease the heart rate. Cocaine, of course, is a narcotic and a controlled substance.
These compounds have a special place in the history of organic and medicinal chemistry. During World War I there was a shortage of atropine, prompting Sir Robert Robinson to develop an efficient synthesis of tropinone as an inexpensive source of synthetic atropine.
Larger Rings
Larger rings of carbon atoms are possible, although they are not as common as five- and six-membered rings. Ring formation is thought to be easiest from a six-carbon chain, because the two ends of the chain can easily wrap around to connect together yet they are not so distant that such an encounter becomes unlikely.
Intermediate-sized rings, such as cyclodecane, become distorted to avoid steric interactions. Internal steric interactions are an added complication in the formation of some rings.
Exercise \(7\)
Show the steric interactions within cyclodecane.
Answer
If cyclodecane adopted a regular diamond lattice conformation, there would be a whopping 8 atom interaction in the middle of the ring. That interaction isn't even included in our basis set. It would cost at least 6-7 kcal/mol. As a result, the cyclodecane adopts a twisted structure to avoid this interaction.
At some point, in very large rings, steric problems are not as severe. The hydrogens on the inside of the ring may be far enough away from each other to avoid interactions. Cyclooctadecane is an example of such a ring. Although interior 6-atom interactions are possible, the ring is flexible enough to twist in such a way as to avoid these problems.
Exercise \(8\)
Look at the model for progesterone and make a line drawing of it.
Go to Animation CA11.5. A three-dimensional model of progesterone.
Answer
Exercise \(9\)
Look at the model for androstanediol and make a line drawing of it.
Go to Animation CA11.6. A three-dimensional model of androstanediol.
Answer
Still pictures of models obtained using Spartan 14 from Wavefunction, Inc., Irvine, California. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.11%3A_Other_Rings.txt |
Rings often contain atoms other than carbon. In many cases, the same basic shape of the carbocyclic compound is still adopted. For example, replacing a carbon of cyclohexane with an oxygen, as in tetrahydropyran, preserves the same basic chair shape.
Go to Animation CA12.1. A three-dimensional model of tetrahydropyran.
The tetrahydropyran structure is common in nature, showing up most frequently in the form of sugars such as glucose. Glucose has a number of different forms that exist in equilibrium. The predominant form in aqueous solution is β-D-glucopyranose.
Go to Animation CA12.2. A three-dimensional model of β-D-glucopyranose.
Exercise \(1\)
Draw β-D-glucopyranose in a chair. Label each substituent axial or equatorial. Perform a ring flip and compare steric strain in the two conformers.
Answer
One of the other isomers in equilibrium with β-D-glucopyranose in solution is α-D-glucopyranose. The two isomers are almost identical to each other, except one hydroxyl group is pointing a different direction.
Go to Animation CA12.3. A three-dimensional model of α-D-glucopyranose.
Exercise \(2\)
Compare steric strain in the two glocopyranose isomers. Which isomer should be the major one?
Answer
We can isolate that core tetrahydropyran structure of the sugar again, bringing with it the one hydroxyl group that differs between the two glucose isomers. In one of the glucose isomers, the hydroxyl group is in an axial position, as it is in this axial conformation of 1-hydroxytetrahydropyran.
Go to Animation CA12.4. A three-dimensional model of 1-hydroxytetrahydropyran.
In the other glucose isomer, the hydroxy group is in an equatorial position, as it is in this equatorial conformation of 1-hydroxytetrahydropyran.
Go to Animation CA12.5. A three-dimensional model of 1-hydroxytetrahydropyran.
Exercise \(3\)
An unusual effect is seen in cases in which an oxygen in a ring is attached to a carbon that also holds an OH group (see drawings above). This O-C-OH group is called a hemiacetal. In a six membered ring, the hemiacetal could be found in two different isomeric positions. In sugars, these are called alpha and beta positions.
a) Based on steric considerations, which of these two isomers should be more stable?
b) These two isomers exist in equilibrium with each other. Which isomer should be the major species present?
c) In solvents of modest polarity, such as dichloromethane, a factor other than sterics actually reverses this equilibrium; the sterically disfavored compound becomes more stable. What factor do you think is responsible for this phenomenon?
d) In solvents of high polarity, such as water, the factor in question (c) becomes less important, and sterics dominate the equilibrium. Why would a highly polar environment diminish the effect discussed in part (c)?
Answer a:
The β-D-glucose isomer should be the more stable isomer. The β-D-glucose isomer places the C1 hydroxyl group in the equatorial position.
Answer b:
The β-D-glucose isomer should be the more abundant isomer.
Answer c:
This is due to something called the anomeric effect. In solvents of modest polarity, such as dicholoromethane, the α-D-glucose isomer is not as polar as the β-D-glucose isomer. In the α-D-glucose isomer the dipoles of the ring oxygen and the C1 hydroxyl group opposing each other (therefore the overall effect is the molecule is less polar). In addition, the α-D-glucose isomer is stabilized by hyper conjugation of the ring oxygen and C1. For more information see http://en.Wikipedia.org/wiki/Anomeric_effect
Answer d:
A more polar environment would promote having more of the β-D-glucose isomer around. In the β-D-glucose isomer, the dipoles of the ring oxygen and the C1 hydroxyl group align each other (therefore the overall effect is the molecule is more polar).
Inflexible Rings
Many rings do not undergo conformational changes because they are not very flexible. Frequently these rings contain a number of double bonds; remember that rotation is not allowed around a double bond because the bond would break.
For example, cyclohexane is a six-membered carbon ring that has a chair structure. It is flexible and can undergo conformational changes. Benzene is also a six-membered carbon ring, but it does not have a chair structure. It is flat and rigid. It cannot undergo conformational changes, because none of its C-C bonds can rotate.
• Benzene does not undergo conformational changes because of delocalized bonding around the ring.
• This aromatic ring is completely flat, maximizing interaction between the p bonding orbitals.
Go to Animation CA12.6. A three-dimensional model of benzene.
Other rings also display aromatic delocalization in which heteroatoms also participate. In some cases, the heteroatom simply participates in a regular double bond. In others, the heteroatom donates a lone pair into the π system.
Furan is an example in which a lone pair from the oxygen is delocalized in order to reach aromatic stability.
Go to Animation CA12.7. A three-dimensional model of furan.
Pyridine is an example in which the ring has aromatic delocalization, but the lone pair on the nitrogen does not participate. Because the nitrogen already has a p bond, the lone pair is orthogonal to that bond, and lies in the plane of the ring.
Go to Animation CA12.8. A three-dimensional model of pyridine.
Still pictures of models obtained using Spartan 14 from Wavefunction, Inc., Irvine, California.
6.13: More Practice with Conformation
Exercise \(1\)
Hydrogen bonding can have a strong influence on molecular shape. For example, it is a key factor in determining the shape of DNA. In DNA, "base pairs" in adjacent strands of DNA hydrogen bond to each other. Show how adenine and thymine, shown in the following drawing, hydrogen bond to each other.
Answer
Exercise \(2\)
Show how cytosine and guanine, shown in the following drawing, hydrogen bond to each other.
Answer
Exercise \(3\)
Show why thymine is not well-suited to hydrogen bond to cytosine and guanine.
Answer
Exercise \(4\)
Eric Kool at Stanford University has used "designer nucleotides" to study DNA polymerase and DNA repair mechanisms. He has been particularly interested in how sterics and hydrogen bonding affect these functions. The following set of nuleotides were designed to mimic a natural nucleotide. Perhaps surprisingly, these nucleotides are acceptable substrates as far as DNA polymerase is concerned, and they can be incorporated into the "correct" position in DNA.
1. Which natural nucleotide do these designer molecules resemble most closely in shape?
2. Show how the designer nucleotide, dF, would fit together with the correct base pair partner for that natural nucleotide.
3. Compare and contrast these nucleotides with the natural one, in terms of their ability to hydrogen bond with the correct "base pair partner".
4. Compare and contrast these nucleotides with the natural one, in terms of their ability to fit together with the correct "base pair partner" in a DNA molecule.
Answer a:
Thymine 5'-triphosphate
Answer b:
Answer c:
dF is only able to hydrogen bond with adinine in one position while thymine can hydrogen at two positions with thymine.
Answer d:
dF is similar in size and electronic nature to thymine. The fluorine groups on dF are somewhat smaller than oxygen on thymine, but present a comprable polar group that is able to act as a hydrogen bond acceptor. dF lacks the hydrogen bond donating ability as seen in thymine.
Exercise \(5\)
Another one of Kool's "designer nucleotides" is shown below.
a) Which natural nucleotide would xC interact with most strongly? Show the interaction.
b) Compare and contrast xC with its natural analog, in terms of their ability to hydrogen bond with the correct "base pair partner".
c) Compare and contrast xC with its natural analog, in terms of their ability to fit together with the correct "base pair partner" in a DNA molecule.
Answer a:
Answer b:
xC is similar to cytosine. Both molecules present hydrogen bond donors and acceptors in the same positions.
Answer c:
While xC is similar to cytosine in terms of placement of hydrogen bond donors and acceptors, the additional ring creates steric problems in that the hydrogen bond donors and acceptors are placed too close to the correct "base pair partner". | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.12%3A_Rings_with_Heteroatoms.txt |
Exercise \(1\)
Researchers at Bristol-Myers-Squibb developed a new antidepressant (compound 2, below). The drug binds to GluN2B receptors, shutting down a central nervous system signal that has been linked to depression. Unfortunately, heart problems appeared as a side effect of the medication; the drug also binds to hERG receptors, interfering with potassium ion channels that are important for heart function.
The team designed a new set of compounds (including 5, 10, 11, above. Adapted with permission from Lawrence R. Marcin, Jayakumar Warrier, Srinivasan Thangathirupathy, et al, ACS Med. Chem. Lett. 2018, 9, 472−477. Copyright 2018 American Chemical Society. Used with permission.).
They wanted to make sure the new compounds would still fit tightly in the GluN2B receptor, which is about 9-11 Ångstroms long.
1. On compound 5, circle bonds that run roughly parallel to the length of the molecule (i.e. left-right; don’t circle any that are more up-down).
2. Count up the number of these in a row. If each one is about 1 Å long, estimate the length of the molecule.
3. The functional groups are also important because they control interactions with the receptor. On compound 10, circle and name four of the functional groups. Place an asterisk beside any functional groups found in both compounds 2 and 10.
4. What other factor are the researchers clearly looking at between compounds 5, 10, 11? Why would the researchers be interested in that factor?
5. State the relationships between compounds 5, 10, 11.
6. Find three chiral centers in compound 10. Draw enough of the structure around each chiral center to make it obvious which one you are considering. Rank the priority of the substituents. Assign the configuration (R or S) to each center.
7. Researchers added an extra ring in the new compounds. Circle the new ring in compound 10 that has no equivalent in compound 2.
8. They incorporated this ring because were interested in adding “conformational rigidity”. To illustrate this idea, draw a Newman projection looking down
i) the middle bond of butane ii) any bond of cyclobutane
Explain why a ring introduces “conformational rigidity” or “conformational restrictions”.
9. Why were the researchers interested in adding conformational rigidity?
10. Only one of the rings in compound 5 can adopt chair conformations (the others are flat). Draw the two possible chairs. Asses steric strains in each conformer. Indicate which conformer is more stable.
Compound 5 was the most promising candidate; it required the lowest dose for GluN2B but the highest dose to bind hERG.
Answer a:
Answer b:
about 9 Å
Answer c:
Answer d:
Stereochemistry. Because biological receptors are generally proteins containing unique, and chiral, binding sites, they hope to find a stereoisomer that fits the GluN2B receptor, but not the hERG receptor.
Answer e:
These compounds are diastereomers of each other.
Answer f:
Answer g:
Answer h:
The ring introduces conformational rigidity by preventing the bond from undergoing complete rotation; it is tied back by the connection on the other side of the ring.
Answer i:
They were trying to decrease the number of possible conformations, or shapes, that the compound could adopt. That way, it would be less likely to bind in both the GluN2B receptor and the hERG receptor.
Answer j:
The substituents around the flexible six-membered ring could adopt a number of conformations. We will choose ways that minimize additional steric interactions so that we can focus on the most basic interactions.
Note that amines do not usually have fixed stereochemistry; the fourth substituent on an amine nitrogen is a lone pair, which does not occupy a fixed position in space. Unlike a carbon, the nitrogen can readily switch from one stereochemistry to another.
The one on the left, with more equatorial substituents and thus fewer steric interactions, is the more stable one.
Exercise \(2\): Tramadol
Tramadol (below) is an opioid analgesic. It is used to treat chronic pain. It is administered as a racemic mixture; each compound targets a different pain receptor.
1. Draw the enantiomer and any diastereomer(s) of the molecule shown. Label the configurations at the chiral centers.
2. If the original molecule is part of the therapeutic tramadol mixture, which other compound is also administered?
3. The original compound contains a ring that could adopt different chair conformers. Draw both conformers and assess steric strain in each.
4. Which conformer is the more stable one?
5. The most stable conformer is not necessarily the medicinally active one. How would dosage need to be adjusted if the least stable conformer were actually the active form?
Answer a:
Answer b:
The original and its enantiomer; the first two above, from the left.
Answer c:
Answer d:
The one on the left, with both groups equatorial and less steric strain, is more stable.
Answer e:
If the one on the left is active, a lower dose can be used because it will spend more time in the active form and be able to bond more receptors. If the less stable one, on the right, is the active form, a higher dose will be needed because only a small fraction of the molecules will be ready to bind the receptor at any given time. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.14%3A_Application_Problems.txt |
Exercise 6.3.1:
Exercise 6.4.1:
Exercise 6.4.2:
Exercise 6.4.3a:
Exercise 6.4.3b:
Exercise 6.4.3c:
Exercise 6.8.1:
Exercise 6.9.1
Exercise 6.9.2:
Exercise 6.9.3:
Exercise 6.10.1:
Exercise 6.10.2:
Exercise 6.10.3:
Exercise 6.10.4:
Exercise 6.10.5:
Exercise 6.11.1:
Exercise 6.11.2:
Exercise 6.11.3:
1. The substituents are always trans along the junctions between each pair of rings. The steroids resemble a series of trans-decalin structures.
2. The overall structure would be more wide and wavy like a trans-decalin, rather than curled or boxy like a cis-decalin.
Exercise 6.11.4:
Bicyclo[2.2.0]decane
Exercise 6.11.5:
a) Bicyclo[2.1.1]hexane b) Bicyclo[3.2.1]octane c) Bicyclo[2.1.0]pentane (more commonly called "housane")
d) Bicyclo[2.2.2]octane e) cis-Bicyclo[3.3.0]octane
f) cis-Bicyclo[1.1.0]butane g) Bicyclo[1.1.1]pentane h) Bicyclo[4.3.3]dodecane
Exercise 6.11.6:
Although we could sketch out many rings using adamantane, just three rings are needed to include all the carbon atoms in the structure. Thus, adamantane is considered a tricyclic system. The systematic nomenclature of tricyclic systems gets a little more complicated, so we won't worry about that.
Exercise 6.11.7:
If cyclodecane adopted a regular diamond lattice conformation, there would be a whopping 8 atom interaction in the middle of the ring. That interaction isn't even included in our basis set. It would cost at least 6-7 kcal/mol. As a result, the cyclodecane adopts a twisted structure to avoid this interaction.
Exercise 6.11.8:
Exercise 6.11.9:
Exercise 6.12.1:
Exercise 6.12.2:
Exercise 6.12.3:
a) The β-D-glucose isomer should be the more stable isomer. The β-D-glucose isomer places the C1 hydroxyl group in the equatorial position.
b) The β-D-glucose isomer should be the more abundant isomer.
c) This is due to something called the anomeric effect. In solvents of modest polarity, such as dicholoromethane, the α-D-glucose isomer is not as polar as the β-D-glucose isomer. In the α-D-glucose isomer the dipoles of the ring oxygen and the C1 hydroxyl group opposing each other (therefore the overall effect is the molecule is less polar). In addition, the α-D-glucose isomer is stabilized by hyper conjugation of the ring oxygen and C1. For more information see http://en.Wikipedia.org/wiki/Anomeric_effect
d) A more polar environment would promote having more of the β-D-glucose isomer around. In the β-D-glucose isomer, the dipoles of the ring oxygen and the C1 hydroxyl group align each other (therefore the overall effect is the molecule is more polar).
Exercise 6.13.1:
Exercise 6.13.2:
Exercise 6.13.3:
Exercise 6.13.4:
Exercise 6.13.5:
Exercise 6.14.1:
a)
b) about 9 Å
c)
d) Stereochemistry. Because biological receptors are generally proteins containing unique, and chiral, binding sites, they hope to find a stereoisomer that fits the GluN2B receptor, but not the hERG receptor.
e) These compounds are diastereomers of each other.
f)
g)
h)
The ring introduces conformational rigidity by preventing the bond from undergoing complete rotation; it is tied back by the connection on the other side of the ring.
i) They were trying to decrease the number of possible conformations, or shapes, that the compound could adopt. That way, it would be less likely to bind in both the GluN2B receptor and the hERG receptor.
j) The substituents around the flexible six-membered ring could adopt a number of conformations. We will choose ways that minimize additional steric interactions so that we can focus on the most basic interactions.
Note that amines do not usually have fixed stereochemistry; the fourth substituent on an amine nitrogen is a lone pair, which does not occupy a fixed position in space. Unlike a carbon, the nitrogen can readily switch from one stereochemistry to another.
The one on the left, with more equatorial substituents and thus fewer steric interactions, is the more stable one.
Exercise 6.14.2:
b) The original and its enantiomer; the first two above, from the left.
c)
d) The one on the left, with both groups equatorial and less steric strain, is more stable.
e) If the one on the left is active, a lower dose can be used because it will spend more time in the active form and be able to bond more receptors. If the less stable one, on the right, is the active form, a higher dose will be needed because only a small fraction of the molecules will be ready to bind the receptor at any given time. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.15%3A_Solutions_to_Selected_Problems.txt |
There is something very special about water, H2O. For starters, it is essential for life. A human being can live for only three days without fresh water. Indeed, the cells in our bodies are made mostly of water.
For most organisms, it is important that cellular water remains in liquid form. Frostbite or death results if cells freeze.
Fortunately, we live on a remarkable planet. The median temperature on the surface of the earth is roughly 4 degrees celsius. That's a pretty good temperature for liquid water. Water is a liquid between 0 degrees celsius and 100 degrees celsius. If the earth were just a little colder -- if it were a few million miles farther from the sun, like Mars -- then it would be too cold for liquid water. Almost all the water on earth would be frozen. If the earth were a few million miles closer to the sun, like Venus, then the earth would be a little too hot for liquid water. Most of the water on earth would be water vapor in the atmosphere.
One of the reasons water is so special is the broad range of temperatures over which it stays in the liquid state. Molecules of similar size and complexity remain liquid over much narrower temperature ranges. For example, methane, CH4, melts at -182 oC and boils at -164 oC. It is liquid only over a range of only 18 degrees celsius. On a summer day in Minnesota, the afternoon high may be 95 oF (35 oC) and the overnight low could drop to 60 oF (15 oC). That's a range of 20 oC. Methane goes all the way from a solid to a gas over a similar temperature span. If there were some methane-based organism subjected to a similar temperature change on a distant moon of Saturn, it could start its day as a popsicle, thaw out for a few productive hours of grazing in the morning, and float away like a balloon by lunchtime.
Not only is there a huge difference in the temperature range available to liquid water and liquid methane, but these two similar compounds freeze at very different temperatures. Other common small molecules like oxygen, O2, nitrogen, N2, carbon monoxide, CO, and hydrogen sulfide, H2S, are gases at 25 C at atmospheric pressure. That is a comfortable room temperature. Water freezes at 0 oC: a cold day in Chicago. Methane freezes at -182 oC: a cold day on Saturn.
Exercise \(1\)
Explain why water other small molecules like methane and dinitrogen are gases at room temperature, whereas water is a liquid.
Answer
One difference between water and these other molecules is that water is polar: there is a significant electronegativity difference between the oxygen and the hydrogen. The charges in one water molecule may be interacting with charges in other water molecules.
7.02: Changes of State
By now you are probably well aware that all the matter around you exists in one of three states: it is a solid, like the earth; a liquid, like the water; or a gas, like the air. In some places on earth, however, materials are not restricted to one particular state, but can change from one to another depending on the weather. On a winter day Yellowstone Park, you might walk past a hot spring and see frozen ice and snow, liquid water, and clouds of steam all at once.
There are several different ways in which you could describe these three phases of matter. Solids have a definite shape, regardless of the container in which they are placed. A marble is still a marble, whether it is rolling around in a bowl or a flowerpot. Liquids take up the shapes of their containers; a liter of water could get round like a fishbowl or square like a brownie pan. Gases are harder to see and more difficult to imagine. They expand to take up the spaces in which they are placed. A gas will even escape from a container that does not have a tight lid on it, much like the carbon dioxide bubbles rise out of a can of soda once it has been opened.
Another way of thinking about these states of matter has to do with the speed of the molecules in the matter, and how much energy those molecules contain. If you don't feel like you have much energy, you probably lay around a lot near the bottom of your container or dorm room, like a lump of coal on the bottom of a coal bin. If you have more energy, you might move around throughout your room, walking from place to place on the floor, but largely staying within its walls. If you have a great deal of energy, you may begin to dance or leap into the air or escape through the doorway to the outside.
Matter is a little bit like that, too. How much energy a group of molecules has -- and consequently how quickly they are moving on average -- strongly influences whether that material is in the solid, liquid or vapor phase. Solids tend to be composed of very slowly moving molecules while gases generally consist of speedier ones.
Exercise \(1\)
Draw a picture of a group of methane (CH4) molecules in the solid state, liquid state and phase.
Answer
Remember, each molecule has the same form whether it is in the gas phase, liquid phase or solid phase. The only thing that changes is the amount of energy the molecules have. As a result, the speed of the molecules and the ability of the molecules to attract each other also changes.
7.03: Kinetic-molecular Theory
Temperature is an indicator of how much energy is available in the surroundings. The higher the temperature, the more energy there is available, and the faster molecules can go. Different kinds of matter require different amounts of energy to get moving. That's why some compounds are gases at room temperature while others are solids.
There are many reasons for these differences. One factor is the mass of a molecule. If you hit a tennis ball with a tennis racket, you can be sure that it will sail through the air, but the same thing might not be true if you hit a bowling ball with a tennis racket. By putting an equal amount of energy into a tennis ball and a bowling ball, you would make the tennis ball move much faster than the bowling ball. Lighter objects need less energy to reach high speeds.
For a molecular example, let's look at halogens. All the elemental halogens are simple diatomic molecules. Their chemical properties have many similarities, but they have very different boiling points and melting points. Iodine, with the greatest mass, is a solid at room temperature, while fluorine has the lowest mass and is a gas at pretty much all the temperatures you are likely to encounter on Earth.
Name & formula MW, g/mol mp, ° C bp, ° C
fluorine, F2 38.0 -219.6 -188.4
chlorine, Cl2 70.9 -100.98 -34.6
bromine, Br2 159.8 -7.2 58.8
iodine, I2 253.8 113.5
184.3
Clearly the mass of a molecules influences its state. Does the mass of a molecule alone allow us to predict its behavior? No. A much more important factor involves another difference between gases, liquids and solids: attraction between molecules. The molecules in solids are not just sitting still; they are sitting still together. Getting a crystal of iodine to evaporate requires coaxing molecules away from each other, and overcoming this attraction between them.
There are many different types of intermolecular attractions. The halogens in the table above cling to each other via the weakest type: London dispersion interactions. We will start by looking at those interactions, and then move to stronger kinds, such as dipole interactions and hydrogen bonding.
Exercise \(1\)
Based on the data in the table above, what would you predict for the melting point and boiling point of nitrogen, N2, and oxygen, O2?
Answer
As far as we can tell from these data, the melting point of nitrogen and oxygen would be somewhere around -220oC, whereas the boiling point would be around -180oC. We would expect these numbers to be similar to fluorine, which has a similar mass. In fact, the melting and boiling points of oxygen are about -219oC and -183oC, respectively. The melting and boiling points of nitrogen are about -210oC and -196oC, respectively. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.01%3A_The_Properties_of_Water.txt |
All molecules can experience London interactions or dispersion forces (sometimes called van der Waals interactions, although many people use this term to describe any kind of intermolecular attraction). These interactions are temporary electrostatic attractions caused by the random movement of electrons around a molecule. If at one moment more of the electrons are found on one side of the molecule than the other, then the molecule will develop a dipole.
A dipole is a separation of charge. If electrons slosh over to one side of the molecule, that end of the molecule becomes negatively charged, because electrons are negative. The other end of the molecule, however, becomes positively charged, because the negatively charged electrons that should be balancing the positive charge of the nucleus have drifted to the other end of the molecule. We symbolize this charge separation with a dipole arrow. One end of the arrow looks like a plus sign; that's the more positive end. The pointy end of the arrow represents the negative end.
This dipole is transient, meaning it does not last very long. However, in its short lifetime it could actually induce another dipole in a neighboring molecule. The electrons gathering on one side of one molecule could repel the electrons on the near side of the next molecule. This situation would leave the more negative end of one molecule next to the more positive end of its neighbor, and a weak attraction would result.
In most molecules, and especially relatively non-polar ones, these kinds of interactions last for tiny fractions of seconds. That means that if molecules rely entirely on London interactions to stick together, they probably don't stick together very well.
However, there are a couple of factors that can make these small interactions add up until they are pretty significant. First, molecules with many, loosely-held electrons would experience more London interactions than molecules with fewer, more tightly-held electrons. That's because the more electrons there are, and the more freely they are moving, the more likely they are to create that initial transient dipole. This factor is pretty significant when comparing the halogens, since fluorine has nine tightly-held electrons while iodine has fifty three, and most of them are so far from the nucleus that they are hard to keep track of.
In addition, this type of attraction does not work very well between two molecules that are far away from each other. Because it is such a weak attraction, it works best when the two molecules are very close to each other. The more contact there is between the surfaces of the two molecules, the more likely it is that one molecule can induce a dipole in its neighbor, and the stronger the attraction will be.
It's all a matter of surface area. The more surface contact between two molecules or two materials, the more likely they are to interact with each other. In fact, this sort of phenomenon can lead to some surprising results if taken to the extreme. Many researchers believe that the ability of geckos to walk on walls and ceilings, even if the surface is very smooth, is entirely due to van der Waals forces. Millions of tiny fibers (or setae) cover the bottoms of the lizard's feet, and each of these fibers is covered in hundreds of tinier pads (or spatulae). These structures ensure that there is enough surface contact for many, many weak interactions to add up to one strong grip. Some companies are even trying to use this idea to make very strong but removable adhesives based on this gecko model.
Now consider the difference between the attractive forces that would be found in two different shapes of molecules. In one case, the molecules are small and round, like ping-pong balls. In the other case, the molecules are long, and linear, like ribbons of spaghetti or lasagna.
You can arrange ping pong balls in any way you like (without squashing them flat), and they will never be able to do more than barely touch their neighbors.
Lasagna noodles stack very easily on top of each other, and have a good deal of contact with the other lasagna noodles.
If London interactions rely on temporary dipoles that are induced only if molecules can get very close to each other; it would be best if the two molecules could almost touch each other.
This factor results from the physical law that electric forces fall off very quickly with increasing distance. The farther apart the molecules, the less strongly they can have an effect on each other.
So, the long, ribbon-like molecules, having more surface area in contact with their neighbors, are more likely to induce dipoles in their neighbors, and are more likely to stick together. The spherical molecules, with less surface area in contact, will attract each other more weakly.
What sort of molecules are shaped like ping pong balls and spaghetti? Two very similar molecules, both with the formula C5H12, can serve as examples.
Normal pentane, or n-pentane, is a molecule in which all five carbons are arranged in a string, with the hydrogens completing the valence of each carbon. Neopentane, or 2,2-dimethylpropane, is a molecule in which one central carbon is bonded to each of the four other carbons, and each of those four carbons has three hydrogens on it. On paper, one might imagine pentane is spaghetti shaped, but neopentane looks more like a cross. You need to look at a space-filling model to see that neopentane is very much like a ball.
It turns out that pentane has a boiling point of 35 oC, while neopentane, with the exact same molecular formula, has a boiling point of -9 oC. That means that, at comfortable room temperature, pentane is a liquid, while neopentane is a gas even on a pretty cold day. The pentane molecules stick together more because there is more contact between them; the neopentane molecules don't, and they each go their own way, forming a gas.
Pentane and neopentane are isomers. They have the same formula but different structures. Isomers is from the Greek, meaning "same things", referring to their identical content. However, isomers often have very different physical and biological properties.
Exercise \(1\)
Compare London forces in the following pairs of molecules, and predict the relative melting points of the two compounds.
a) CH3(CH2)6CH3 and CH3(CH2)100CH3
b) CH3(CH2)6CH3 and CH3(CH2)3C(CH3)3
c) CH3Br and CH3Cl
Answer a:
A chain that is 8 carbons long would have much lower London interactions than a chain that is 102 atoms long. The greater the contact between molecules, the greater the chance for weak attractions due to small charges arising from random electron movement. The molecule with the longer chain would have the higher melting point, because more energy would have to be added in order to overcome those attractions and get the molecules moving.
Answer b:
The first chain is more linear whereas the second chain is more branched. The more linear molecules will pack together more easily, allowing greater surface contact for increased London interactions. The straighter chain would have a higher melting point.
Answer c:
Bromomethane has a lot more electrons than chloromethane. Based on London interactions alone, we would expect bromomethane to have a higher melting point. In fact, the melting point of bromomethane is about -94oC, compared to about -97oC for chloromethane; there must be another factor that compensates for the difference in numbers of electrons and makes these values so close.
Exercise \(2\)
Predict the relative boiling points of 3,4-dibromohexane and 3,3-dimethylhexane.
Answer
Because of the large numbers of electrons in bromine, we would expect a higher boiling point in 3,4-dibromohexane.
Exercise \(3\)
What would you predict about the relative melting points of dibenzyl ether and bis(cyclohexylmethyl) ether. (Note: "bis" is often used instead of "di" as a prefix for more complicated groups.)
Answer
Remember benzyl refers to the group C6H5CH2 (below right); cyclohexylmethyl suggests the superficially similar C6H12CH2 (below left).
The benzyl group contains trigonal planar carbons, whereas the cyclohexylmethyl group contains only tetrahedral carbons. The benzyl groups, being flatter, could pack more easily together and interact more strongly. We would expect dibenzyl ether to have a higher melting point. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.04%3A_London_Interactions.txt |
When two different kinds of atoms are connected to each other, the electrons between them are typically not shared evenly. That's because in most cases, two different atoms would have two different electronegativity values. One atom would be more electronegative than the other. It would have a stronger attraction for the electrons in the shared bond, and the electrons would be a little bit more attracted to that atom than the other.
Remember, electronegativity is a periodic trend. As we move to the right in the periodic table, more and more protons are added to the nucleus, and so atoms on the right attract electrons more strongly that do the atoms on the left. Also, as we move from top to bottom of the periodic table, the valence shells of the atoms become larger; that means valence electrons are getting farther from the nucleus, so they are less tightly held. Those two factors, distance and amount of charge, control the strength of interaction between charged particles.
The periodic table above shows Pauling electronegativity values. There are different ways to calculate electronegativity, but the Pauling scale is very commonly used. Calculating the Pauling scale depends on being able to observe compounds, so the noble gases are not usually given a value in this scale, because they normally don't form any compounds.
For example, when a carbon atom is bonded to a fluorine atom, there is an electronegativity difference between the two atoms. There is a difference in electronegativity values of almost 1.5 units between these two atoms. Fluorine is more electronegative than carbon because it has more protons in the nucleus than carbon. The electrons are pulled closer to the fluorine than to the carbon. A carbon-fluorine bond is polarized towards the fluorine. A molecule like fluoromethane, CH3F, has a permanent dipole.
Note that there are also dipoles in C-H bonds, but they are so much smaller than the ones in the C-F bond that they do not matter. The overall dipole has a buildup of negative charge on the fluorine.
You can imagine that molecules with permanent dipoles would interact with each other much more strongly than molecules that rely on temporary dipoles in order to stick together. Molecules with no natural dipole will stick only loosely together, but molecules with permanent dipoles will stick to each other easily.
Ethane, C2H6 (sometimes written CH3CH3, suggesting two carbons are each connected to three hydrogens, and also to each other), and formaldehyde, CH2O, have different formulae but the same molecular weight. Based on weight alone, it would take about the same amount of energy to move an ethane molecule and a molecule of formaldehyde.
The two molecules also have somewhat similar shapes, unlike neopentane and pentane. At room temperature, ethane and formaldehyde are both gases. Nevertheless the two compounds have very different boiling points; formaldehyde becomes liquid around -20 oC, which would be a very cold winter day in, say, Chicago. Ethane does not become liquid unless it is cooled to around -90 oC, a cold winter day on Neptune, at which point formaldehyde is just about ready to freeze solid.
The difference between these two molecules must be due to the oxygen atom in formaldehyde. Oxygen, the second most electronegative element in the periodic table, can form some very polar bonds. The difference between carbon and oxygen is 0.89 units, almost a full electronegativity unit. The permanent dipole that results between the oxygen and carbon makes the formaldehyde molecules much stickier than the ethane molecules, which depend on fleeting London interactions if they are going to hold on to each other.
In contrast, the difference in electronegativity between carbon and hydrogen is actually pretty small (about 0.35 units, based on the Pauling scale). The tiny dipole in the C-H bond is so much smaller than the one in the C=O bond that it really isn't a factor here.
Taking that story a little further, hydrocarbons (molecules made up of hydrogen and carbon) are the most common example of non-polar compounds, because their bonds are either completely nonpolar C-C bonds or else C-H bonds with very low polarity. There is a tiny dipole in a C-H bond, but it isn't enough to make the molecules attract each other very well. Furthermore, there tend to be a lot of C-H bonds in a hydrocarbon pointing in all sorts of directions; an approaching molecule would be unlikely to encounter either a fully positive end or a fully negative end of the molecule. In the end, all those little dipoles in all those C-H bonds end up cancelling each other out.
There are other variations on dipole interactions that are pretty common. For example, you could imagine that a dipolar molecule would interact pretty strongly with an ion. We will take a look at that situation shortly when we think about how mixtures of different compounds interact with each other. However, perhaps the most important variation in organic chemistry is hydrogen bonding.
Exercise \(1\)
Comment on dipole moments in the following compounds and predict their relative boiling points.
a) 1-chlorobutane vs. pentane
b) triethylamine vs. 3-ethylpentane
c) diethyl ether vs. pentane
Answer
These compounds all contains lots of non-polar C-C and C-H bonds, which would not have appreciable dipoles. However, in each pair, there is a compound that contains mroe polar bonds as well: C-Cl, C-N and C-O.
Sometimes a molecule that contains polar bonds is not really polar overall. Carbon dioxide is a good example. Polar molecules strongly attract each other because the more negative end of one molecule is attracted to the more positive end of the next molecule via electrostatics. Opposites attract. In carbon dioxide, however, both ends of the molecule are more negative. How can they attract each other?
There's more to it. Dipoles are "vector quantities". That means it isn't just the amount of charge separation that matters; it's also the direction of separation. In carbon dioxide, a linear molecule, there are two polar bonds. Those polar bonds point directly away from each other. The dipole of one bond balances out the dipole of the other bond. Overall, the molecule does not have a dipole.
Water, on the other hand, is a different case. Like carbon dioxide, water has two polar bonds. However, the bonds in water do not lie in a straight line like the ones in carbon dioxide. They are at an angle to each other instead. That means their dipoles are not quite pointing in opposite directions. As a result, water has an overall dipole moment.
At the extreme, a polar bond can get to the point that one atom is pulling the electrons so strongly toward itself that it pulls them completely away from the other atom. The first atom becomes an anion, and the second atom becomes a cation. At this point, we have an ionic bond, rather than a covalent one. How different must the electronegativity values be in order to have an ionic bond? The usual rule of thumb is a difference of about 1.6 electronegativity units. So, for example, all of the blue atoms in the periodic table above would form ionic bonds with all of the red atoms. The green atoms would form ionic bonds with most of the dark red atoms, but not with the light red ones.
Exercise \(2\)
The dipole moment of a molecule (essentially the size of its dipole) can be calculated from the dielectric constant of a material, which can be measured experimentally on a bulk quantity of the compound. The dipole moment of chloromethane, 12.9 D, is higher than that of dichloromethane, 9.08 D, even though dichloromethane has a greater number of polar bonds than chloromethane. Explain why.
Answer
Dipoles are vectors. If we have two polar bonds, they will add together using vector addition. Any vector can be shown as the sum of two other vectors. Below, the red arrow points down and to the left. How far down and how far to the left? We can show how far using the blue arrows. The blue arrow down and the blue arrow to the left add up to give the red arrow that goes down and to the left.
In dichloromethane, there are two polar bonds. Both bonds are polarized towards the more electronegative chlorine. As drawn below, one bond has a dipole straight to the right. The other bond has a dipole that goes down and to the left. If the molecule has an overall dipole, with a positive end of the whole molecule and a negative end of the whole molecule, which end would be positive and which would be negative?
It seems reasonable that the positive end would be somewhere over on the hydrogen side and the negative end would be somewhere over on the chlorine side. The overall dipole, as opposed to the individual bond dipoles, would point somewhere in the direction of the blue, dashed arrow. But how big would it be? Surely it's bigger than either of the two individual dipoles, because they are adding together, right?
Let's take a very qualitative look. The first red arrow is just to the right. We'll break the second arrow into a smaller arrow down and an even smaller arrow to the left. The sum of the left/right arrows is a very small arrow to the right.
The sum of the up/down arrows is just a small arrow down. Overall, the sum is a green arrow going down and to the right. The result doesn't really seem any bigger than an individual bond dipole.
Exercise \(3\)
Dichloromethane nonetheless has a higher boiling point (39 °C) than chloromethane ( -24 °C). Explain why.
Answer
Despite the relatively similar dipole moments, dichloromethane's boiling point is much higher than chloromethane's. The difference probably lies in the much greater mass of dichloromethane.
Exercise \(4\)
Comment on dipole moments in the following compounds and predict their relative boiling points.
1. butane and ethyl methyl ether
2. ethyl methyl ether and 2-propanone
3. trimethylamine and propanenitrile
4. 2-propanone and 2-chloropropane
Answer
In each case, the one on the right has a larger dipole and a higher melting point.
Exercise \(5\)
1. Draw the Lewis structure for each molecule with the correct electronic geometry.
2. Draw in dipole moments.
3. Decide if the molecule is polar. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.05%3A_Dipole_Interactions.txt |
Methanol, CH4O (or CH3OH, a CH3 group attached to an OH), is another example of a molecule with a similar molecular weight to ethane. Like formaldehyde, methanol freezes somewhere around-90 oC, but it does not become a gas until it is heated to 65 oC. Methanol is certainly similar to formaldehyde in some ways. It contains oxygen and is very polar. The huge difference in their boiling points is due to the very strong hydrogen bonds in methanol.
Hydrogen bonding occurs when there is a significant amount of positive charge building up on a hydrogen atom. That happens because the hydrogen is attached to an atom that is much more electronegative than the hydrogen. As a result of that positive charge, a lone pair on another molecule strongly interacts with the hydrogen.
Hydrogen bonds occur only when a hydrogen is attached to one of the few most electronegative elements in the periodic table: fluorine, oxygen or nitrogen. It is these strong hydrogen bonds that are responsible for the relatively high boiling point of methanol; there is so much positive charge on the hydrogen of the OH group that it can essentially form a real bond with the lone pair on another methanol molecule.
Water is an even better example of a hydrogen bonding compound; in fact, water molecules are so tightly bound to each other that this very small molecule freezes at 0 oC and does not boil until 100 oC (way, way hotter than an August day in Las Vegas). Water molecules are much less able to move around freely than are ethane molecules, even though ethane molecules weigh almost twice as much.
For practical purposes, hydrogen bonding usually involves compounds that contain N-H and O-H bonds. That's because oxygen and nitrogen are the second- and third-most electronegative elements in the periodic table, respectively. There are many, many examples of compounds like that. It also occurs in HF, of course, because fluorine is usually considered the most electronegative element, but there is really only one example of a compound with that bond, so you are much less likely to have to worry about that example. Slightly weaker hydrogen bonds have also been shown to occur involving other species, such as S-H bonds, but the most significant ones involve N-H and O-H bonds.
Exercise \(1\)
Which of the following molecules are capable of strong hydrogen bonding?
a) ethanal, CH3CHO b) ethylamine, CH3CH2NH2 c) acetic or ethanoic acid, CH3CO2H d) dimethyl peroxide, CH3OOCH3
Answer
Only ethylamine and ethanoic acid have the N-H or O-H bonds that are polar enough for hydrogen bonding.
Exercise \(2\)
Predict the relative order of melting points in the following amines:
hexylamine, CH3(CH2)4CH2NH2; triethylamine, (CH3CH2)3N; dipropylamine, (CH3CH2CH2)N
Answer
hexylamine > dipropylamine > triethylamine
The one with the greatest potential for hydrogen bonding will have the highest melting point.
Exercise \(3\)
Identify the strongest IMF in each case.
Answer
Biological Molecules and IMFs
Because compounds containing oxygen and nitrogen are so common in biology, hydrogen bonding is a fairly common feature in organic and biological chemistry.
Carbohydrates are one of the major classes of biomolecules. Carbohydrates are compounds that contain lots of OH groups, so they have lots of potential to form hydrogen bonds.
Proteins and peptides contain amide groups. Amides contain N-H bonds; hydrogen bonds involving amides are crucial influences in the structure of proteins.
Large biological molecules like proteins consists of 1000s of atoms, which interact through "intramolecular" ion-ion, London dispersion, dipole-dipole, and hydrogen bonding to create a unique 3D structure. Other molecules, both small and large, can bind to biological macromolecule using the same intermolecular interactions. An example of a protein, bovine low molecular weight protein tyrosyl phosphate, binding to inorganic phosphate, is shown below.
The displays above show three different ways to render the LMW-PTPase. You will recognize the spacefill and wireframe models, but note their complexity compared to the smaller molecules you have studied thus far. The cartoon molecules removes much of the detail by stripping away atoms and bonds, and replacing them with lines, ribbons (representing alpha helices) and block arrows (representing beta structure). [Not Available]
Inorganic phosphate, an inhibitor of this enzyme, binds to the enzyme through the intramolecular forces mentioned above. It is rendered in spacefill with the protein in cartoon and wireframe modes.
DNA may be the most well-known example of a compound that displays hydrogen bonding. In DNA, complementary "base pairs" hydrogen bond to each other to held two strands of DNA together.
Weak, "Secondary" Hydrogen Bonds
Hydrogen bonding is really a special case of dipole forces. It occurs under very particular circumstances: in the presence of O-H or N-H bonds, which provide both a very positive hydrogen and a lone pair, the two features necessary for a hydrogen bonding interaction.
In general, we think of a hydrogen bond as the electrostatic interaction between an electron pair on one atom and a hydrogen atom that has built up a partial positive charge. There are some examples where that might be happening, but that we would rather keep separate from the general idea of a hydrogen bond. For example, a lone pair on an oxygen atom in formaldehyde might be attracted to a hydrogen on another formaldehyde, which would be at the more positive end of the molecule. There is evidence from x-ray crystallography that such weak "hydrogen bonds" occur in the solid state in molecules containing aldehyde groups, similar to the one in formaldehyde.
However, these kinds of interactions are really very weak. They shouldn't be thought of in the same category as regular hydrogen bonds. For that reason, we will give them a separate term, "secondary" hydrogen bond. This type of interaction is fairly esoteric and you are not that likely to see it. We're looking at it here really to underscore the idea that there is something very special about the hydrogen bonds resulting from compounds that contain O-H and N-H bonds.
Later, you will see that whenever there is a case in which there is an opportunity for regular hydrogen bonding, the properties of the molecule are dramatically affected. That isn't true otherwise. Secondary hydrogen bonds may be responsible for some very subtle effects in the solid state, when molecules are already at pretty low energy and are not moving around much. However, they hardly change the properties of a molecule beyond what we would expect based on simple dipole-dipole interactions.
7.07: Ionic Attractions
So far, we have seen that all molecules can potentially have weak attractions that make them cling to each other. Because the electrons in an atom or molecule are not stationary, but are always in motion, there is always a chance that a molecule will become lop-sided temporarily. Its electrons can lean a little to one side, leading to a small buildup of negative charge on that side. The molecule would be slightly low on electron density on the other side, leading to a small buildup of positive charge there. This temporary dipole can induce temporary dipoles in other molecules nearby. That slightly positive side can tug on the electrons on a molecule nearby. And so on; a chain of these events can cascade through a group of molecules. We get an induced dipole-induced dipole interaction.
We have also seen that some molecules have permanent dipoles. If two atoms are bonded together and they have a sufficiently large difference in electronegativity, then there is a significant charge separation between the two atoms, and a permanent dipole results. The difference between these permanent dipoles and the ones we saw in London interactions is that the London interactions rise and fall in an instant. The London interactions typically don't last long enough to hold two molecules together very tightly, unless there are an awful lot of these interactions. Because a fixed dipole is permanent, there is little opportunity for two molecules with a dipole-dipole interaction to squirm away from each other.
Hydrogen bonds are a particularly strong example of dipole-dipole interactions because of the significant electronegativity difference between oxygen and hydrogen. The interaction is also strengthened by the availability of a lone pair on oxygen to facilitate the interaction with the partial positive charge on the hydrogen. Furthermore, they are really important when you live on such a watery planet.
So, these interactions get progressively stronger based on the amount of charge separation in the bond. The logical conclusion to that progression is a bond that is completely ionized. What if one end of the bond has a full negative charge and the other end has a full positive charge? Obviously, ions should have even more electrostatic attraction for each other than mere dipoles.
Now, many treatments of intermolecular attractions don't even consider ionic attractions; they've got that covered under "ionic bonding". Nevertheless, it's useful to look at ionic attractions in the present context, as well, because they help to complete the picture of how two molecules might interact with each other.
To take a biological example, we might consider the strength of interaction of two amino acid side chains. A serine and histidine would have significant interactions with each other, because both contain side chains capable of hydrogen bonding. A serine would have a stronger interaction with the side chain of a positively charged lysine, however. To go one step further, a positively charged lysine would have an even stronger attraction to a negatively charged aspartate. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.06%3A_Hydrogen_Bonding.txt |
Sometimes there is more than one way to connect a given group of atoms into a molecular structure. Given the formula C2H6O, two different isomers are possible: methyl ether and ethanol. They have the same formula but different structures. The word, "isomer", is from the Greek, meaning "same things", referring to their identical content. However, isomers often have very different physical and biological properties.
These two compounds have different physical properties, as you can see in the table below.
compound melting pt, ° C boiling pt, ° C
dimethyl ether -138 -25
ethanol -117 78
Remember, typical room temperature is about 25 °C. That means that at room temperature, dimethyl ether is a gas, because it is above its boiling point. Ethanol, on the other hand, is a liquid at room temperature. Ethanol at that temperature is well above its freezing point but has not yet reached its boiling point.
Why is that? What is the origin for the differences in physical properties between these two, similar compounds, one of which is a gas at room temperature while the other is a liquid? Both contain carbon-hydrogen and carbon-oxygen bonds. The ethanol also contains a carbon-carbon bond and an oxygen-hydrogen bond.
Certain bonds, or groups of bonds, confer specific behaviors on the compounds in which they are found. For instance, when an OH group is found in an organic compound, it can make the compound moderately acidic. It also makes the compound a little more likely to be a solid or liquid rather than a gas. Why?
The dominant intermolecular attraction between ethanol molecules is hydrogen bonding. What we mean is, although two ethanol molecules may be attracted by London dispersion forces and even by dipoles, both of those factors pale in comparison to the attraction that arises from hydrogen bonding. Hydrogen bonding is the single most important factor holding two ethanol molecules together.
In dimethyl ether, on the other hand, hydrogen bonding is not a factor. In this molecule, the major intermolecular attraction involves regular dipole moments. Because hydrogen bonds are typically much stronger attractions than ordinary dipole moments, a group of ethanol molecules is much harder to separate from each other than a group of dimethyl ether molecules. The ethanol has a much higher boiling point.
Exercise \(1\)
Compounds don't really have to be isomers in order to compare them in this way. They just have to be similar enough that we can reasonably compare them.
For example, propan-2-one or acetone, CH3(CO)CH3, has a boiling point of 56-57 °C whereas 2-propanol, CH3CH(OH)CH3, has a boiling point near 83 °C. Explain the difference.
Answer
The hydrogen bonding between 2-propanol molecules is a stronger interaction and is more difficult to overcome than the dipole-dipole interactions between 2-propanone molecules. Thus, 2-propanone molecules can more easily escape into the vapour phase.
Different collections of bonds that are commonly found in organic compounds are referred to as functional groups. Functional groups influence the physical properties of the compounds in which they are found. They can also influence the biological activity of a compound; that is, they can help determine whether a compound might be active as a specific type of drug, a hormone, or other types of compounds that regularly interact with organisms in nature.
In the case we looked at above, we were comparing an alcohol to an ether. In an alcohol, there is an OH group attached to a carbon. In an ether, there is an O attached to two carbons. (There are exceptions to these definitions: if an OH group is attached to a carbonyl, a C=O group, it is not an alcohol but a carboxylic acid. If an oxygen is attached to a carbonyl and to another carbon, it is not an ether but an ester.) In general, we can say that, because of the difference in intermolecular attractions, an alcohol has a higher boiling point than a comparably-sized ether.
• Properties of materials are closely related to their structure
• Two isomeric compounds, composed of exactly the same elements in exactly the same ratio, may have very different properties because of the different ways that the atoms are connected within the compounds.
Another example of isomers is found by looking at the formula C3H9N. This time there are three different structures that can occur with the given formula. All three isomers contain the "amine" functional group. An amine is just a nitrogen connected to a mixture of carbons and hydrogens. Again, the three isomers have different properties, and these differences are not just restricted to melting points and boiling points. What factor do you think is responsible for these differences.
compound melting pt, ° C boiling pt, ° C density, g/mL
trimethylamine -117 3 0.6356
N-methylethylamine not reported 37 0.688
n-propylamine -83 48 0.7173
The differences in melting points and boiling points are quite straightforward. The first example possesses no potential for hydrogen bonding. The second one does because it contains an N-H bond and the third even more so, because it contains two N-H bonds rather than one.
Unlike alcohols and ethers, whether or not an amine contains an N-H does not change the name of the functional group. It is still called an amine. That may be an unfortunate artifact of history, because clearly an amine that can hydrogen bond is a little different from an amine that cannot, in just the same way that an alcohol is different from an ether. To compensate for that lack of distinction between amines, an additional adjective is used to describe them. An amine in which the nitrogen is attached to only one carbon and two hydrogens is called a "primary amine" (primary means one, as in one carbon attachment). An amine in which the nitrogen is attached to two carbons and one hydrogens is called a "secondary amine" (secondary means attachment to two carbons). An amine in which the nitrogen is attached to three carbons and no hydrogens is called a "tertiary amine" (attachment to three carbons).
Isomers have a variety of properties that are different from the properties of other members of their group. Not all of these properties can be easily explained by looking at different structures, but some can. The density of a material is the ratio of its weight to its volume (in more "scientific" language: mass/volume). A material with a higher density than another can be thought of as having the same weight, but shrunken to a smaller volume. Alternatively, it could be thought of as having the same volume, but packed full of more mass.
Note that the isomers in the table above have different densities. The densities in the table refer to liquid densities. Despite their similarities, the three compounds in the liquid state pack together a little differently. That, too, is a result of structural differences.
Exercise \(2\)
Explain the differences between the densities in the table above. The values in the table all refer to the materials in their liquid states.
Answer
There is hydrogen bonding between molecules of N-methylethylamine, but not between molecules of triethylamine, which are held together by relatively weak dipoles. Trimethylamine molecules are not held together as tightly as the N-methylethylamine molecules, which therefore pack more densely. On the other hand, n-propylamine contains two N-H bond, compared to only one in N-methylethylamine. Consequently, n-propylamine has a greater propensity to form hydrogen bonds, and it packs together even more densley than the N-methylethylamine.
Exercise \(3\)
Although this information is not noted in the table above, density values are usually listed along with the temperature at which the value was measured. That's because density is actually a function of temperature. Most materials become more dense as they get colder and less dense as they get warmer. Explain why density varies with temperature.
Answer
Temperature is a measure of how much energy is available in the environment. As temperature increases, molecules obtain more energy and they transfer that energy into a variety of molecular motions. As they move around more, they are less able to remain densely packed. The density decreases as the temperature goes up. Conversely, things tend to shrink as they cool, because molecules slow down and begin to pack more tightly. They become more dense.
There are other properties that can be compared and explained based on the structures of compounds. Vapour pressure is closely related to boiling point. It refers to the pressure exerted by molecules escaping from a liquid into the vapour phase. The more easily molecules can get away from each other in the liquid and escape into the vapour phase above the liquid, the higher the vapour pressure. The higher the vapour pressure, the more easily the liquid reaches the boiling point. In a simple sense, we can think of strong attractions between molecules holding them in the liquid phase and lowering their vapour pressure.
Viscosity refers to the resistance to flow in a liquid. The higher the viscosity of a liquid, the more slowly it flows. Maple syrup is more viscous than water, but honey is more viscous than maple syrup. In a simplified way, we can think of molecules clinging to their neighbors as they flow past each other, hindering flow and increasing viscosity. Thus, the intermolecular attractions between molecules in a liquid affect its viscosity.
It's a little like a group of friends rushing down a crowded street in Manhattan. If they all hold hands, they don't make as rapid progress as they would individually. Individually, each member of the group is able to step aside, dart forward and get past the oncoming pedestrians. Clinging together, they all have to turn, come to a halt and wait for people to let them past.
Surface tension is a measure of how strongly a liquid resists external forces. You might think of it as how tightly the molecules in a material cling to each other when given the chance to cling to something else instead.
The most familiar picture of surface tension is a drop of water. Water forms drops because the molecules in the water cling to each other more strongly than they do to the molecules in the air. Those are mostly nitrogen molecules, N2, with a significant fraction of oxygen molecules, O2. The molecules in the air are quite non-polar, because they are made of the same kinds of atoms: two nitrogens, for instance. There is no electronegativity difference that would impart a dipole to the molecule. That's in stark contrast to a highly polar water molecule; it's hard to find molecules more polar than that. If a water molecule were to interact with a molecule of dinitrogen, it would be trading in some strong hydrogen bonds with other molecules. That wouldn't be worth it.
Exercise \(4\)
Explain the differences in the vapour pressures of the following compounds (listed in hectoPascals, hPa).
1. acetic acid, CH3CO2H (15.3) vs. propan-2-one, CH3(CO)CH3 (240)
2. ethyl acetate, CH3CO2CH2CH3 (97) vs. diethyl ether, CH3CH2OCH2CH3 (587)
3. pentane, CH3CH2CH2CH2CH3, (573) vs. heptane, CH3CH2CH2CH2CH2CH2CH3 (48)
Answer a:
Acetic acid posesses hydrogen bonding capability, whereas 2-propanone has the ability to form only dipole-dipole interactions. Thus, 2-propanone molecules are not held in the liquid phase as strongly as acetic acid, and the former compound has a higher vapour pressure.
Answer b:
The C=O bond of ethyl acetate has a substantial dipole, but the dipoles in ether don't add up to very much; they are smaller to begin with and they partially cancel out via vectorial addition. Ether molecules can escape the liquid more easily, resulting in a higher vapour pressure.
Answer c:
Both molecules are limited to relatively weak London dispersion interactions. Because heptane forms a longer chain than pentane, there will be a slightly greater interaction between heptane molecules than between pentane molecules. Pentane thus has the higher vapour pressure.
Exercise \(5\)
Explain the differences in the viscosities of the following compounds (listed in centiPoise, cP, or milliPascal-seconds, mPa.s).
1. diethyl ether, CH3CH2OCH2CH3 (0.22) vs. 2-butanone, CH3(CO)CH2CH3 (0.41)
2. decane, CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 (0.70) vs hexane, CH3CH2CH2CH2CH2CH3 (0.26)
3. diethyl ether, CH3CH2OCH2CH3 (0.22) vs. 1-butanol, CH3CH2CH2CH2OH (2.54)
Answer a:
The C=O bond 2-butanone has a substantial dipole, but the dipoles in ether don't add up to very much; they are smaller to begin with and they partially cancel out via vectorial addition. The 2-butanone molecules cling together more strongly, resulting in a higher viscosity.
Answer b:
Both molecules are limited to relatively weak London dispersion interactions. Because decane forms a longer chain than hexane, there will be a slightly greater interaction between decane molecules than between hexane molecules. Decane thus has the higher viscosity. Later, if you study macromolecules, you will see that in very, very long chains, "entanglement" becomes a factor as well. Just as the name implies, really long chains get tangled up together and have a difficult time moving past each other; that's why materials like cooking oil and motor oil are so viscous.
Answer c:
The 1-butanol can hydrogen bond together, but the ether only has weak dipole-dipole interactions. The 1-butanol is therefore more viscous.
Exercise \(6\)
Explain the differences in the surface tensions of the following compounds (listed in dynes/cm).
1. 1-butanol, CH3CH2CH2CH2OH (25) vs diethyl ether, CH3CH2OCH2CH3 (16)
2. 1-butanol, CH3CH2CH2CH2OH (25) vs 1,3-butanediol, HOCH2CH2CH(OH)CH3 (47)
3. formamide, H(CO)NH2 (57) vs. nitromethane, CH3NO2 (36)
Answer a:
The 1-butanol can hydrogen bond together, but the ether only has weak dipole-dipole interactions. The 1-butanol therefore has greater surface tension.
Answer b:
The 1-butanol can hydrogen bond together, but the 1,3-butanediol has two OH groups and can form even more hydrogen bonds than 1-butanol can. The 1,3-butanediol therefore has greater surface tension.
Answer c:
The formamide has N-H bonds and so it is capable of hydrogen bonding. Despite having a pretty darned big dipole, the nitromethane can't really hydrogen bond together. The formamide therefore has greater surface tension.
You don't have to have data in front of you to compare two compounds' properties. At least to some extent, you may be able to look at two similar structures and make an educated guess about which one will melt or evaporate more readily than the other, provided they have different functional groups that you can compare. Knowing something about structure has explanatory value; you can explain differences based on structure. It also has predictive value: you can predict differences based on structure.
Exercise \(7\)
In the following pairs of isomers, which one would be expected to have the highest boiling point?
1. 2-propenol, CH2CHCH2OH vs propanal, CH3CH2CHO
2. ethyl pentanoate, CH3CH2O2CCH2CH2CH2CH3 vs heptanoic acid, CH3CH2CH2CH2CH2CH2CO2H
3. 1-hexen-3-one, CH2CHCOCH2CH2CH3 vs diallyl ether, (CH2CHCH2)2O
Answer a:
2-propenol. It has hydrogen bonding capability but propanal does not.
Answer b:
heptanoic acid. It has hydrogen bonding capability but ethyl pentanoate does not.
Answer c:
1-hexen-3-one. Its dipole is greater than that in the ether, because it comes from a C=O bond rather than C-O bonds.
Exercise \(8\)
In the following pairs of isomers, which one would be expected to have the highest viscosity?
1. 2-butanone, CH3COCH2CH3 vs. methyl allyl ether, CH3OCH2CHCH2
2. pentane, CH3CH2CH2CH2CH3 vs octane, CH3CH2CH2CH2CH2CH2CH2CH3
3. dimethoxyethane, CH3OCH2CH2OCH3 vs 1,3-butanediol, HOCH2CH2CH(OH)CH3
Answer a:
2-butanone. Its dipole is greater than that in the ether, because it comes from a C=O bond rather than C-O bonds.
Answer b:
octane. It has greater London dispersion interactions that pentane because of the greater surface contact area in octane.
Answer c:
1,3-butanediol. It has hydrogen bonding but the DME does not.
Exercise \(9\)
In each of the following questions, you have a choice between A and B. Choose which option in which stronger attractions are present.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.08%3A_Comparing_Properties_of_Iso.txt |
So far, we've been thinking about how one molecule could interact with another molecule of the same compound to form a liquid or a solid. What about two different compounds? It turns out that whether two different compounds interact with each other depends on the same kinds of interactions we have looked at so far.
Two liquids that do not really mix well together, such as oil and water, are described as immiscible. Two liquids that appear to mix completely together are said to be miscible. Water and ethanol are one example of a pair of miscible liquids, because you can take any amount of ethanol and mix it with any amount of water and you will always end up with a clear, colorless liquid just like the ones you started with.
Exercise \(1\)
What kind of intermolecular attraction allows water and ethanol to mix together?
Answer
Hydrogen bonding.
Exercise \(2\)
The following pairs of liquids are miscible. Identify the predominant type of intermolecular attraction between each pair.
1. 2-propanone and ethyl ethanoate (ethyl acetate)
2. pentane and octane
3. dichloromethane and 2-butanone
4. methanol and ethanoic acid
Answer a:
dipole interactions
Answer b:
London interactions
Answer c:
dipole interactions
Answer d:
hydrogen bonding
Exercise \(3\)
Explain, in terms of intermolecular forces, why water and octane are not miscible.
Answer
Water molecules interact with each other mainly through hydrogen bonding, whereas octane molecules interact with each other via London interactions. The main problem here is the strong hydrogen bonding between the water molecules. All molecules, in principle, could interact with each other via London interactions. However, if octane molecules were introduced among the water molecules, they would take up space. Some of the water molecules would not be able to get close enough to each other to hydrogen bond anymore. The loss of that very stabilizing interaction would be too costly.
Exercise \(4\)
Explain, in terms of intermolecular forces, why ethanenitrile and hexane are not miscible.
Answer
This problem is similar to the previous one, but in this case the attraction between the strong dipoles of the nitrile groups would be too much to overcome.
Sometimes, the attractions between molecules are a little more complicated. Two molecules may have different interactions on their own, but when placed together still manage to interact with each other. For example, dichloromethane and hexane mix together pretty well.
How can that be? Dichloromethane has dipole interactions. Hexane has nothing more than London interactions. Ethanenitrile and hexane didn't mix for that very reason. But the problem was not whether those two molecules could interact with each other; they could. The problem was that ethanenitrile would not give up its dipole-dipole interactions for the small amount it could gain by mixing with hexane.
In this case, the dipoles between dichloromethane are much smaller; they aren't held back from the hexane molecules as strongly. On the other hand, the interaction between the the hexane and dichloromethane is actually amplified a little bit. Whereas hexane molecules rely solely on weak, transient London interactions to cling to each other, dichloromethane has a permanent dipole. That permanent dipole is able to enhance the transient dipole in hexane.
This is called a dipole / induced dipole interaction. The slight tug dichloromethane exerts on hexane's electrons actually helps the two different molecules interact more strongly.
Exercise \(5\)
Below are several solvents and some possible solutes. For each pair,
i. Draw a solvent molecule interacting with several other (identical) solvent molecules. Label the strongest intermolecular force holding them together.
ii. Draw a solute molecule interacting with several other (identical) solute molecules. Label the strongest intermolecular force holding them together.
iii. Draw one solute molecule interacting with several solvent molecules. Label the strongest intermolecular force holding them together. Predict whether the solvent will dissolve significant amounts of the solute.
A. Solvent = Ethylene glycol (HOCH2CH2OH); Solute = NH3
B. Solvent = Pentane (CH3(CH2)2CH3); Solute = triethylamine, [(CH3CH2)3N]
C. Solvent = CH2Cl2; Solute = NaCl
Answer
Exercise \(6\)
Describe the intermolecular interaction between each of the following pairs.
a) 2-propanone; benzene
b) 2-propanol; dichloromethane
c) ethanol; ethanoic acid
d) octanol; methylbenzene
Answer a:
dipole / induced dipole
Answer b:
dipole / dipole
Answer c:
hydrogen bonding
Answer d:
dipole / induce dipole
Solubility is conceptually similar to miscibility. If you can mix a solid with a liquid and the solid particle gets smaller and smaller until it disappears, it must have dissolved in the liquid. At that point it may look like you are left only with the original liquid, but really the individual molecules of the solid are just distributed throughout the liquid instead of all sitting together on the bottom. You can see this when blue kool-aid powder dissolves in water to give blue kool-aid. The blue substance has not disappeared; it has become intimately mixed with the water. Just as you cannot see an individual "chunk" of water, you cannot see an individual "chunk" of the original solid anymore -- it has separated into individual units (molecules) that are too small to be seen. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.09%3A_Miscibility.txt |
In order for two substances to be miscible, or in order for one substance to be soluble in another, the two kinds of molecules must interact with each other. Not only that, but there must be strong enough attraction of one type of molecule for the other so that they are willing to move away from their identical siblings. If the attractive forces between the unlike molecules are strong enough to overcome those forces between both sets of like molecules, dissolution results.
Table salt dissolves in water because the very polar water molecules attract both the positively charged sodium ions and the negatively charged chloride ions. This interaction is called an ion / dipole interaction.
In order to distribute the sodium cations and chloride ions throughout the solution, some of the water molecules may need to back away from each other to make room for the ions. Some of the strong hydrogen bonds may be lost, but the energy needed to overcome those interactions is compensated by the attraction between the dipole of the water and the sodium or chloride ions.
Other salts dissolve in water, too, but some of them dissolve more easily than others. You could easily dissolve about 360 g of table salt in a liter of water, but the solubility of calcium carbonate is only about 0.01 grams per liter. That's partly due to the fact that the ions in sodium chloride, Na+ and Cl-, have lower charges than the ions in calcium carbonate, Ca2+ and CO32-. The higher the charges on the ions, the stronger their electrostatic attraction for each other, and the harder it is for the water to pull them apart. It's a good thing, too; if the calcium carbonate in marble were to dissolve too easily, think of how many buildings and statues would dissolve in the rain.
Exercise \(1\)
Rank the following salts from most easily to least easily dissolved in water: MgSO4, LiCl, AlPO3
Answer
The predicted order, from most soluble to least, would be LiCl > MgSO4 > AlPO4 because the ions increase in charge from 1+/- to 2+/- to 3+/-
The factors that control the solubility of ions can be complex, though. For example, the charges on the ions also affect the strength of the interaction with the water molecules. Highly charged ions interact more strongly with each other, but they also interact more strongly with water molecules.
There are other solvents that can dissolve salts via ion / dipole interactions. Typically they would have strong dipoles. However, not many solvents have dipoles as strong as that of water.
Exercise \(2\)
Which of the following solvents would be most capable of dissolving some LiCl? Rank from greatest solubility to least solubility.
triethylamine, dimethylsulfoxide, dichloromethane, acetonitrile, pyridine.
Answer
The predicted order, from most soluble to least, would be
dimethylsulfoxide > acetonitrile > pyridine > dichloromethane > triethylamine.
Of course, it isn't just salts that can dissolve in solvents. Other solids can, too, if they can interact with solvent molecules strongly enough. Sugar also dissolves pretty well in water. Why do you think that is so?
Sugars, of course, can hydrogen bond easily with water, so they are usually pretty soluble.
7.11: Hydrogen Bond Acceptors
Formaldehyde is another example of a compound that dissolves well in water, and in fact the most common way to obtain formaldehyde is as an aqueous solution. You may have encountered such a solution in an anatomy or general biology lab, because formaldehyde solutions are used as preservatives for biological specimens. Of course, water and formaldehyde are both polar molecules, so it is easy to imagine their dipoles interacting together. However, formaldehyde does not have very strong hydrogen bonds like water does, so at first glance it seems as if formaldehyde might not interact with water molecules as strongly as water molecules interact with each other.
Actually, the interaction between these molecules may be stronger than it first seems. Although formaldehyde does not engage in strong hydrogen bonds by itself, in the presence of water or another protic compound -- one that contains a very positive hydrogen, such as H2O, HF or NH3 -- strong hydrogen bonds do form.
This occurs because formaldehyde has an oxygen atom with lone pairs and so it can act as a hydrogen bond acceptor. That means it can engage in hydrogen bonding with something that does have positive hydrogens that can interact with its lone pairs. The water in this case is acting as the hydrogen bond donor for formaldehyde. Hydrogen bond acceptors are often important in biological systems, where nearly everything takes place in the presence of water.
Exercise \(1\)
Which of the following compounds are capable of hydrogen bonding? Which are hydrogen bond acceptors? Which are neither?
a) triethylamine
b) pentanal
c) ethylamine
d) diethyl ether
e) butanamide
f) hexene
Answer a:
H-bond acceptor
Answer b:
H-bond acceptor
Answer c:
Fully H-bonding
Answer d:
H-bond acceptor
Answer e:
Fully H-bonding
Answer f:
neither | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.10%3A_Solubility.txt |
Why does not oil dissolve in water? Oil is composed, in part, of long chains of carbon atoms with hydrogens attached. These chains aren't very polar. It shouldn't be too hard to pull them apart, because they are held together only by London interactions. The chains really aren't long enough to create the strong London interactions that would prevent oil from mixing with water.
On the other hand, it is pretty difficult to pull water molecules away from each other, and the oil does not have the means to do so; it just isn't polar enough. If the water molecules don't move away from each other, there will be no room between them for the individual oil molecules to become dissolved. These two substances will not mix together very well.
Consequently, if placed in the same vessel, they will remain separate and form two different layers. The more dense layer (the water) will sink to the bottom while the lighter, less dense one (the oil) will float to the top.
The same situation is true for a number of other, non-polar organic compounds, such as benzene and toluene. These liquids are too non-polar to dissolve very well in water. Consequently if you mix benzene and water, the two liquids will form two separate layers. Benzene has a specific gravity or density of 0.874 g/mL, whereas the density of water is 1.0 g/mL. As a result, benzene would float on the top, while water would sink to the bttom. Mixing the two layers up as hard as you can may produce temporary mixing (the mixture would form a sort of cloudy, sparkly mess called schlieren), but once left alone the benzene and water would separate out again.
There's one more useful example we should look at. Suppose we have a molecule that is very polar on one end, but non-polar at the other. Soap, for instance, is an ionic compound, but while the cation is usually just a sodium ion, the anion is more complicated. This molecular anion most often contains a very polar "carboxylate group", composed of a carbon with two attached oxygens. It also contains a very long carbon chain, just like in the oil. So one part of the molecule should dissolve well in water, while the other does not. There is a trade-off, and a balance will be struck that determines exactly how soluble the soap is in the water. Interestingly, when placed in water, these soap molecules will arrange themselves in groups so that the polar ends face outward, towards the water, while the nonpolar ends are tucked on the inside. Think of the "circle the wagons" scene in a classic western movie.
There are two reasons why this phenomenon is useful. Micelle formation, as the behavior is called, allows nonpolar substances, such as dirt and oils, to be dissolved in the polar water. The dirt can interact perfectly well with the nonpolar soap tails, and so it will end up in the middle of the micelles, and something that could not be dissolved in plain water turns out to be perfectly soluble in soapy water. But at another level, micelle formation is a very good model for some of the phenomena of cell and molecular biology. For instance, cell membranes are composed of molecules that are somewhat similar to soap molecules. These molecules form groupings similar to large micelles, but with an additional layer of molecules on the inside of the circle, with their polar ends pointing inward. That leaves the nonpolar ends of both layers sandwiched together, out of the water. Proteins also have polar and non-polar regions, and getting the non-polar regions away from the water leads the protein to adopt a specific shape that, in turn, determines the behavior of the protein.
Exercise \(1\)
Answer a:
Answer b:
Tail is nonpolar and hydrophobic. Head is polar and hydrophilic.
Answer c:
induced dipole - induced dipole
ion-induced dipole
dipole-induced dipole
ion-dipole
Answer d:
No. The dipole-induced dipole is too weak to overcome hydrogen bonding between waters.
Answer e:
Ion-dipole between water and the hydrophilic head of the soap.
Answer f:
outside
Answer f:
Water is attracted to the hydrophilic heads to form this strong ion-dipole interaction whereas the tails cannot overcome the strong hydrogen bonding of water-water interactions.
Answer g:
Hydrophobic dirt and grease will prefer to interact with the hydrophobic tails (inside micelles).
Answer h:
The grease will be encapsulated in the micelle and washed away in water. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.12%3A_Heterogenous_Mixtures.txt |
Intermolecular attractions contribute a crucial part to the behavior of everyday materials around us. In biology, attractions involving biomolecules are essential to their function. To get an idea of the role played by intermolecular attractions, such as hydrogen bonding, in biology, let's look at a common type of biomolecule: proteins.
Remember that proteins are polymers made up of lots of amino acids joined together ny peptide bonds. An explicit, atom-labeled structure is shown below. In addition, there is an abbreviated, skeletal structure below that. This line drawing does not label the carbon atoms explicitly; it does not even show where hydrogen atoms are if they are connected to carbons. You are just supposed to know that they are there. The trade-off is a much cleaner, simpler picture.
In the line structure, we can easily see the zig-zag backbone of the protein chain. Periodically along that chain we can see an amide bond (N-C=O). The amide bonds have been formed by joining the amino end of one amino acid to the carboxylic end of another.
This amino acid chain isn't very long; we probably wouldn't even call it a protein, because proteins are so much bigger than that. We would just call it a polypeptide or an oligopeptide -- the latter term means there are a bunch of amino acids in this chain, but not even enough to consider it a true polymer. The particular oligopeptide we are looking at is a poly(glycine) molecule. All of the amino acids are the same monomer: glycine.
Oligopeptides, although not really proteins, are important in their own way. They are often used as signaling molecules because their small size allows them to travel relatively easily throughout an organism. Most proteins and oligopeptides are composed of different amino acids. Just to illustrate this idea, an example is shown below; it is a seven amino acid chain.
In order to help describe their structure, proteins and peptides are usually numbered, starting from the amino end of the chain (notice "oligopeptide" and "polypeptide" are a bit of a mouthful, so they often get shortened to the more general term "peptide"). Often, if a researcher wants to discuss a particular amino acid along the chain, they will list both its number and its amino acid identity. The amino acids differ from each other by their "side chain" or "residue"; that's the part hanging from the main backbone.If the water molecules don't move away from each other. For example, serine's side chain is a CH2OH group; methionine's is a CH2CH2SCH3 group. So, when you're reading the latest article in that issue of Biochemistry you've been waiting for all week, and they mention that Ser237 plays a crucial role in catalysing the formation of an important neurotransmitter, they are talking about a particular serine residue attached at the 237th position of the protein, counting from the nitrogen end.
Those protein structures can get a little bit complicated, simply because the molecules are so large. However, the basic idea of the structure is prettty simple. The same zig zag backbone runs throughout the entire molecule, and the residues are evenly spaced along the backbone. Here, we can represent those different residues as colored dots. The structure starts to seem a little like a series of different balloons tied to a clothesline.
That isn't to see that the structures of those side chains are not important. They are very important, and we will get back to them a little later. Sometimes, though, you have to decide what you need to focus on right now, and it might be that we want to look at the backbone without the distraction of all of those side chain structures.
In fact, even the backbone might be too much detail. Maybe we just want to keep in mind that we are dealing with a polymer, a long chain of amino acids bound together. We might fall back to an even simpler drawing of colored beads on a string to signify that chain.
Now let's just stick with the simple beads on a string picture for a minute to think about protein structure. That simple order of amino acids -- which one is at the amino end of the chain, who comes after that, and who comes after that, all the way to the carboxylate end of the chain -- is called the primary structure of the protein. The primary structure is just what amino acids are found in the protein, and in what order.
The secondary structure is a second level of detail about the protein. It begins to describe the three-dimensional shape of the structure. That's where intermolecular attractions come in to the picture. A protein, being a very long chain, can usually reach back and interact with itself. Really, we should call these intramolecular attractions, since they are within the same molecule. The interactions could involve different side chains of the protein, but frequently they also involve the backbone. Remember, the backbone has lots of carbonyl oxygen capable of donating a lone pair, and lots of amide N-H bonds capable of donating a proton, so hydrogen bonding along the backbone is very important. The cartoon below tries to suggest a chain that has looped back alongside itself, and is held in place by some sort of interactions (the dashed lines).
That sort of motif, in which a protein can turn and run parallel to itself, and can be held in place by an interaction (specifically hydrogen bonds), is found in beta-sheets. Beta-sheets are very common in proteins.
Unfortunately, the cartoon does not really convey any details about the intramolecular attractions that hold it in place. To see that, we should take a look at a more detailed example. Here we have that polyglycine oligomer again, and we are using it to illustrate hydrogen bonding between two chains to form a beta-sheet. In this case, we have drawn the two peptides as two separate molecules. However, as in the cartoon, we could also have one long peptide that turns back to run alongside itself.
Note that, in the example above, the two chains are running in the same direction (the amino ends are on the left and the carboxylate ends are on the right). That's an example of what's called a parallel beta-sheet. You can also have an anti-parallel beta sheet, in which the two chains are running in opposite directions. That is illustrated in the following diagram, using our simplified balloons-on-a-clothesline drawing.
The hydrogen bonds are spaced slightly differently in these two cases. Once again, these might be two different peptide chains interacting with each other, but it is very common for one chain to undergo a turn and reach back to hydrogen bond with itself. That would put the two sections of chain running antiparallel to each other. This idea is illustrated in the simple cartoon below.
The earlier case, of a parallel beta-sheet, is also possible in a single chain; it simply involves a much longer turn, so that the chain loops around to run in the same direction as the original section. This case is illustrated in the cartoon below.
Beta sheets, of both kinds, are very common in proteins. So are alpha-helices, in which a peptide chain coils around like a spiral staircase. It is held in that shape, once again, by hydrogen bonding along the backbone of the protein.
Beta-sheets and alpha-helices are two common examples of secondary structure. They are units of structural organization that frequently show up in protein structures. Look at the following example; it's a picture of a real crystal structure result for a protein. Researchers took this protein (in this case, it's called a GTP Binding Protein) and carefully got it into a crystalline form. Then they shone X-rays through the crystal. By analysing how the X-rays were scattered as they passed through the crystal, they could go back and determine where all of the atoms were. For that, they could get this "space-filling" picture, showing the positions of all the atoms.
That picture is a little complicated, but each round ball in that mass represents a carbon, oxygen, nitrogen, or sulfur (hydrogen atoms are usually too small to spot using X-rays). The colors were added by the researchers to keep track of parts of the structure they were interested in, and, you'll see in a minute, the secondary structures.
If we go to a simpler model, we will ignore individual atoms and just trace out that backbone chain of the protein. Things will suddenly look much less complicated.
Now, if you take a look at the pink part on the left side, you can almost see how the backbone forms a coil, like a spiral staircase. That's an alpha helix. And those yellow zig-zags in the middle appear to be tracking each other. As one chain zigs up, so does its neighbour. When one chain zigs down, its neighbour follows. That's because they are connected. They are forming a beta sheet.
So the researchers simplify further, replacing part of the backbone with a pink helix. They replace other parts of the backbone with yellow ribbons; where the yellow ribbons run alongside each other, they are forming beta sheets. The arrows at the ends of the ribbons point towards the carboxylate end of the chain.
So a protein can contain both alpha helices and beta sheets. It can also contain less organized loops and turns. That higher level of structure -- how are the beta sheets and alpha helices and other things organized in space -- is called the tertiary structure. In a very simple cartoon, the protein might start with a length of chain, go into a beta sheet, run through another open length of chain or a loop, and finish with an alpha helix. The overall shape it forms is the tertiary structure.
There is one more level of structure in a protein, although not all proteins have this feature. It is called the quaternary structure. A quaternary structure is really a cluster of more than one protein, stuck together. Below, the cartoon shows a dimer. Two identical tertiary structures are held together by intermolecular attractions. QUaternary structure is an example of what's called a "supramolecular assembly". That's when two or more molecules come together and hold on tightly, forming a bigger structure that's bigger than a molecule.
Notice that there are gaps within that cartoon structure. Those gaps can be very important. Usually a protein packs very tightly and does not leave many spaces; when a space is there, it is often for a specific reason. A gap in a protein can hold another molecule. The other molecule might be another protein that arrives and "docks" temporarily while some important task is performed. It might also be a small molecules that needs to bind with the protein to accomplish some goal.
These binding sites are crucial for your survival as an organism. At the same time, they are also exploited by physicians when they prescribe a medication. The drug that the doctor gives you, more often than not, binds in a specific place in a specific protein, tweaking its activity a little. Usually it gets the protein to slow down and take a break, but in some cases it may pep the protein up a little bit and make it work harder. Of course, whatever the small molecule it, and whatever it's doing there, it binds into that "active site" through intermolecular attractions.
That's it for protein structures. Let's just take a minute, though, to get to know the amino acids that make up proteins. The amino acids are usually divided into four different categories, partly based on their potential for intermolecular attraction. Here are the nonpolar amino acids. They are sometimes called the hydrophobic amino acids, because they can't hydrogen bond with things like water. They are shown here connected into a peptide chain, so that you can see what they would look like in a protein.
In contrast, the following group shows the members of the polar amino acids. They are much better at hydrogen bonding and other polar interaction, such as ion-dipole interactions.
There are a few polar amino acids that were not included in that last group because they are special. These are the basic amino acids. There are just three of them.
Finally, there are two more special polar ones. These are the acidic amino acids.
Those last two groups are special because each one actually has two possible structures. They will vary depending on their environment. The basic amino acids are each able to pick up an extra proton (an H+ ion). That's very important; they help the protein carry out lots of important jobs by moving protons around.
The acidic amino acids, on the other hand, can each give away a proton. They are also needed to carry out different tasks by the protein.
Not only that, but the acidic amino acids usually have a negative charge on their side chains, because usually they are in the state in which they have lost a proton. The basic amino acids often have positive charge on their side chains, for similar reasons: they are usually in the state with an extra proton. That means that these two sets of amino acids are capable of additional intermolecular attractions, both within the protein structure and with other molecules that may come along and bind to the protein.
Exercise \(1\)
Intermolecular attractions play a crucial role in other biomolecules, such as DNA. A DNA "molecule" is really a supramolecular assembly of two separate molecules. Show how the following DNA base pairs would hold onto each other.
Answer
Note that the last example is actually found in RNA, not DNA. RNA is found as a single strand, rather than a pair of strands. It hydrogen bonds to itself in a particular shape, not unlike the way a protein adopts a specific shape.
Exercise \(2\)
Intermolecular attractions can be important in other biomolecules, such as cellulose. Show how the following cellulose chains could hold onto each other.
Answer
Exercise \(3\)
Cellulose is an industrially important commodity: we use it to make clothing and money (usually from cotton) as well as paper (usually from wood), for example. The pulp & paper industry expends a great deal of energy processing wood in order to make paper, with a major goal being the removal of lignin (which makes up about 15-30% of most woods, compared to about 30-50% cellulose content). The following drawing shows two chains of cellulose and a chain of lignin. Why would paper companies want to remove the lignin?
Answer
Although lignin is capable of some hydrogen bonding, it cannot make nearly as many hydrogen bonds as cellulose can. Paper containing lignin would be much weaker, and more likely to tear or to fall apart when it gets wet, because the cellulose molecules would not be able to bind each other as tightly with lignin mixed in. In fact, brown paper bags from the grocery store are made from kraft, which has not had lignin removed; they are much weaker than most paper, especially if they get wet.
Exercise \(4\)
Vegetable oils, such as canola oil, are high in "unsaturated" fatty acids, in which the fatty acid chains in the triglyderides contain C=C double bonds. For example, canola oil contains about 60% oleic acid, an 18-carbon chain with a double bond in the middle. Animal fats, such as lard, contain mostly saturated fatty acids (with no C=C double bonds). Lard contains about 30% stearic acid; it's the same chain length as oleic acid but without the double bond. Why is canola an oil, whereas lard is a solid?
Answer
The dominant intermolecular attraction here is just London dispersion (or induced dipole only). London dispersion is very weak, so it depends strongly on lots of contact area between molecules in order to build up appreciable interaction. The stearic acid-containing triglycerides will be able to pack much more closely together than the more jumbled oleic acid-containing ones, so they will have much more surface contact and much stronger attractions. They will hold together into a solid.
Exercise \(5\)
Identify the following pictures as illustrations of either alpha helices or beta sheets.
Answer a:
alpha helix
Answer b:
beta sheet
Answer c:
alpha helix
Answer d:
beta sheet
Answer e:
beta sheet
Answer f:
alpha helix
Answer g:
alpha helix
Answer h:
beta sheet
Answer i:
alpha helix
Answer j:
alpha helix
X-ray crystal structures for space-filling, backbone, and cartoon illustrations: Filchtinski, D.; Sharabi, O.; Ruppel, A.; Vetter, I.R.; Herrmann, C.; Shifman, J.M. What makes Ras an efficient molecular switch: a computational, biophysical, and structural study of Ras-GDP interactions with mutants of Raf. m>J. Mol. Biol. 2010 399: 422-435. Images obtained via RCSB Protein Data Bank (3KUD). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.13%3A_Intermolecular_Attractions_.txt |
Exercise \(1\)
The cleaner, Goof Off (the MIRACLE remover)®, is advertised to remove grease, oil and wax from surfaces. Goof Off is composed of a mixture of xylene and ethyl benzene. The structures of these compounds are shown below:
Grease, oil and wax are hydrocarbon compounds. Why does Goof Off® work better than water to remove these substances? To support your answer, use IMF tables to determine:
a) the strongest IMF between water and water;
b) Goof Off® and Goof Off;
c) Goof Off® and grease; and
d) strongest IMF between grease and water.
Why does Goof-off work better to dissolve grease and oil than if you just use water?
Answer a
Answer b
Answer c
Answer d
The key to why goof-off works is that it engages in the same London dispersion interactions with the grease as the grease does itself. Water is good at dissolving many things, but not grease. That's mostly because the water would have to give up some of the strong hydrogen bonds with other water molecules in order to fit grease molecules among the water molecules.
Exercise \(2\)
Often, a pint of gas-line antifreeze (commonly called "HEET", after a popular brand) is added to the gas tanks of cars, especially in the coldest days of winter, to prevent gas-line freeze-up. (Small amounts of water in gasoline can freeze when it is very cold. This blocks the fuel line and stops the engine.)
a) Assume that gasoline is pure octane (C8H18). What is the strongest IMF force between octane molecules in gasoline?
b) One type of gas-line antifreeze contains almost 100% isopropyl alcohol, shown below. What is the strongest IMF between two isopropyl alcohol molecules?
c) Isopropyl alcohol is completely soluble in both water and gasoline. Why? Consider the IMF involved when isopropyl alcohol is:
i) interacting with water
ii) Interacting with octane
d) When HEET is added to a tank of gas it tends to cause water to dissolve (or at least stay suspended) in gasoline. Show in a picture and explain how this happens, on the molecular level.
e) All gasoline in Minnesota is required to contain 10% ethanol (CH3CH2OH). Given this formulation, would a Minnesotan need to use HEET? Explain.
Answer a
Answer b
Answer c
Answer e
No, the ethanol should do the same thing that the HEET does.
Exercise \(3\)
The laboratory of Samuel Stupp at Northwestern designs molecules with potential biomedical applications. The compound below was made to promote bone regeneration at the cell surface. Bone is mostly composed of calcium phosphate, Ca3(PO4)2 (Adapted with permission from Liam C. Palmer, Christina J. Newcomb, Stuart R. Kaltz, Erik D. Spoerke and Samuel I. Stupp, Chem. Rev. 2008, 108, 4754-4783. Copyright 2018 American Chemical Society.).
a) This molecule is a synthetic amphiphile; it has a polar end and a non-polar end. Put a rectangle around the polar part of this molecule. Circle the nonpolar part of the molecule.
b) The amphiphile inserts into the cell membrane. A cartoon of a portion of a cell membrane is shown below. Add a cartoon of the synthetic amphiphile to the membrane. What is the IMF through which it interacts with the membrane?
c) Based on its position in the periodic table, what is the charge on a calcium ion?
d) Based on the overall neutral charge of calcium phosphate, Ca3(PO4)2, what is the charge on phosphate?
e) What is the advantage of bone being composed of calcium and phosphate rather than more common ions like sodium and chloride?
f) Show how this amphiphile would promote bone growth.
Answer a:
Answer b:
Answer c:
Ca2+ (two atoms from the left edge).
Answer d:
PO43- because 3 x (2+) + 2 x (3-) = 0; neutral charge overall.
Answer e:
These high charges (2+/3-) bind much more tightly than the low charges (1+/1-) in sodium chloride.
Answer f:
Exercise \(4\)
Let's make some paper.
Paper is made from cellulose fibers that stick together to form a sheet. Cellulose, in turn, is a polymer made from individual glucose molecules bonded together in a chain.
a) Circle an individual glucose unit in the cellulose.
b) Name and draw the IMF that holds two cellulose chains together.
One of the problems in making paper is that the most important source of cellulose is generally wood pulp. Wood pulp differs from other plant materials in that it contains lignin.
The lignin is pretty important in the plants -- it provides structural rigidity or stiffness that allows trees to grow a lot taller than other plants -- but it isn't useful in paper. It is left in brown paper bags (kraft paper), but those tear more easily than paper that has had lignin removed.
c) Show why lignin would weaken the interaction between cellulose molecules.
One of the problems with paper is that it can absorb water and swell; it can even dissolve if there is enough water.
d) Show why paper absorbs water so easily.
One way to combat moisture absorption is to add sizing to the paper. Sizing can be mixed in with the cellulose mash before it is rolled out into sheets ("internal sizing"), or it can be sprayed onto the paper sheet after it is rolled out ("surface sizing"). Rosin, also obtained from trees, is a common source of sizing; one of its major components is abietic acid.
e) In order to modify the properties of the paper, the rosin will have to stick to the paper. Explain how it does that.
f) How does rosin make the paper more water-resistant?
A very old source of sizing was rabbitskin glue; gelatin is a related material that is still used as sizing today (in addition to being the basis for a jiggly dessert). These materials are mostly proteins, rich in glycine, proline, alanine, phenylalanine, leucine and valine, among other amino acids.
g) How would a protein stick to the paper?
h) What kind of amino acid residues would make the paper water-resistant?
Raw paper is quite transparent, so opaquifiers are usually added. They make a more opaque material so that you can write on both sides. Titanium dioxide is a common opaquifier.
i) In order to get to a noble gas configuration, what would be the charge on a Ti ion? An O ion?
j) Therefore, what is the formula of titanium oxide (the ratio of the ions in the compound)?
k) Titanium oxide does not dissolve very well in water. Why not?
l) Titanium oxide does stick in the fabric of the paper. Name the IMF.
m) In the picture, the O is dark and the Ti light (picture credit: Ben Mills; public domain). Calculate the number of each atom in the unit cell.
n) Describe the type of unit cell formed by the Ti atoms alone.
We will need to try out the paper by writing on it. In a ballpoint pen, the ink (such as eosin red, used to grade your compositions in English class) is suspended or dissolved in a light oil such as benzyl alcohol.
o) How does eosin dissolve in benzyl alcohol? Show and name one kind of IMF on the drawing.
p) Why will the eosin stick to the paper?
Answer a:
Answer b:
Answer c:
Answer d:
Answer e:
The carboxylic acid group provides sites for hydrogen bonding, including both a hydrogen bond donor in the OH group and an additional hydrogen bond acceptor in the C=O group.
Answer f:
It introduces a "hydrophobic" part in which the major intermolecular force with water would be a dipole-induced dipole interaction. If the water molecules keep to themselves, they will be able to maximize the amount of hydrogen bonding they can do with each other. The water is better off not mixing in with the paper.
Answer g:
All proteins have polar backbones formed by their amide bonds or peptide linkages (-NH-C=O). These portions can hydrogen bond with the cellulose in the paper. In addition, there may be polar side chains such as serine or threonine, for example, containing OH groups that can hydrogen bond with the paper.
Answer h:
Hydrophobic residues would add water resistance, for the same reason as the hydrophobic part of the abietic acid in rosin.
Answer i:
Ti4+ would have the same electronic configuration as argon. O2- would have the same electronic configuration as neon.
Answer j:
The formula is TiO2, so that the positive charges (4+) and the negative charges (2 x 2- = 4-) are balanced.
Answer k:
The 4+ and 2- charges are pretty large. Highly charged ions are difficult to dissolve, because the ion-dipole attraction of the ion with the water cannot compete with the very strong ionic attractions between the ions.
Answer l:
Ion-dipole. You can imagine the partially negative oxygen atoms of cellulose interacting with the titanium cations and the partially positive hydrogens of the OH groups in the cellulose interacting with the oxide anions.
Answer m:
Ti: one full atom in the center and eight in the corners, so 1 + 8x(1/8) = 2 Ti atoms.
O: two full atoms (towards the left and right ends of the picture) plus four half atoms (on the upper and lower faces of the cell), so 2 + 4x(1/2) = 4 O atoms.
ratio: 1 Ti : 2 O or TiO2 in formula notation.
Answer n:
The Ti are packed in a cell that resembles a body-centered cubic cell. However, the cell is not really cubic, because it is stretched out a little. Sometimes we call that shape "tetragonal".
Answer o:
The benzyl alcohol could interact with the eosin in a number of ways, because eosin is a fairly large, complicated molecule (at least at this stage in your studies). We could imagine ion-dipole interactions, in which the partially positive hydrogens on the OH groups in the benzyl alcohol interact with the anionic oxygens of the eosin, or the partially negative oxygens in the benzyl alcohol interact with the positive calcium ion.
However, the polar part of benzyl alcohol is really the smaller part of the molecule. The bigger part of the molecules is the benzene, a six-membered carbon ring with three double bonds. The carbons at the ends of those double bonds are trigonal planar, so because the benzene is built up from planar atoms, it is completely flat. C-C bonds are not polar, and C-H bonds are hardly polar at all, so the dominant IMF in this part of the molecule is London dispersion or induced dipole-induced dipole interaction. That's a pretty weak interaction generally, but it can get a little stronger between large molecules, or between flat ones, because two flat molecules can have more surface area in contact with each other. That means the benzenes in benzyl alcohol and the benzenes in eosin have appreciable London forces that allow them to interact with each other.
Answer p:
With the cellulose, covered in OH groups, the polar interactions outlined above will become more prominent.
Exercise \(5\)
Let's make a Post-It Note©. Post-It Notes are basically sticky pieces of paper. If we are going to make one, we will need some paper. Once you have the paper, we will need an adhesive to make it sticky.
Researchers in Teresa Reineke’s lab at University of Minnesota have developed new adhesives based on sugars, a renewable resource (Nasiri, M.; Saxon, D. L.; Reineke, T. M. Macromolecules 2018, 51, 2456-2465. Copyright 2018, American Chemical Society. Used with permission.). The adhesives are long polymer chains containing about 800 X groups in a row, with between 12 and 90 Y groups at each end.
They tried two different kinds of Y groups: AAI and GATA.
a) What kind of functional group is OAc?
b) How many chiral centers are found in the AAI group? In the GATA group?
A good glue has a balance between adhesion – sticking to something else – and cohesion – sticking together. If there are lots of these chains in the adhesive (like strands of spaghetti), they can stick to each other via intermolecular forces.
c) Show and name the IMFs that would occur between (i) two alkyl chains on group X; (ii) the ester groups on GATA.
d) What IMF leads to good adhesion of the polymer to paper?
In a static shear test, a 1-pound weight is stuck on a vertical surface with the adhesive, and the time it takes to fall off is measured. The AAI version outperformed the GATA version (6000 vs. 800 minutes). However, the lab made a slight alteration to the GATA and the adhesion time doubled (to 1600 minutes).
Thinking only of cohesion, why did the strength of the material improve?
e) Show and name the IMF on a drawing of the modified GATA. Compare its strength to the IMF it replaces in the original GATA.
The lab also performed a peel adhesion test. They measured the force required to peel the “post-it note” from a surface. Both GATA and AAI versions out-performed Scotch© Tape, duct tape, and genuine Post-It© Notes.
Answer a:
OAc is an ester.
Answer b:
AAI has three chiral centres; GATA has five.
Answer c:
The chains in the adhesive would stick to each other via intermolecular attractions. Dipole-dipole attractions would be the strongest. However, induced dipole-induced dipole or London dispersion forces would actually be significant because there would be hundreds of such interactions along the chains.
Answer d:
The most significant interaction with the paper would be hydrogen bonding. The cellulose in the paper would provide hydrogen bond donors (the OH groups) and the side groups in the adhesive chains would provide hydrogen bond acceptors (all those oxygens). However, if the paper contains a lot of sizing, induced dipole-induced dipole or London dispersion forces may form a significant part of the interaction. As with cohesion, this would not be a significant factor except that there are so many alkyl chains involved, so these tiny attractions would add up.
Answer e: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.14%3A_Application_Problems.txt |
Exercise 7.1.1:
One difference between water and these other molecules is that water is polar: there is a significant electronegativity difference between the oxygen and the hydrogen. The charges in one water molecule may be interacting with charges in other water molecules.
Exercise 7.2.1:
Exercise 7.3.1:
As far as we can tell from these data, the melting point of nitrogen and oxygen would be somewhere around -220°C, whereas the boiling point would be around -180°C. We would expect these numbers to be similar to fluorine, which has a similar mass. In fact, the melting and boiling points of oxygen are about -219°C and -183°C, respectively. The melting and boiling points of nitrogen are about -210°C and -196°C, respectively.
Exercise 7.4.1:
1. A chain that is 8 carbons long would have much lower London interactions than a chain that is 102 atoms long. The greater the contact between molecules, the greater the chance for weak attractions due to small charges arising from random electron movement. The molecule with the longer chain would have the higher melting point, because more energy would have to be added in order to overcome those attractions and get the molecules moving.
2. The first chain is more linear whereas the second chain is more branched. The more linear molecules will pack together more easily, allowing greater surface contact for increased London interactions. The straighter chain would have a higher melting point.
3. Bromomethane has a lot more electrons than chloromethane. Based on London interactions alone, we would expect bromomethane to have a higher melting point. In fact, the melting point of bromomethane is about -94°C, compared to about -97°C for chloromethane; there must be another factor that compensates for the difference in numbers of electrons and makes these values so close.
Exercise 7.4.2:
Because of the large numbers of electrons in bromine, we would expect a higher boiling point in 3,4-dibromohexane.
Exercise 7.4.3:
Remember benzyl refers to the group C6H5CH2 (below right); cyclohexylmethyl suggests the superficially similar C6H12CH2 (below left).
The benzyl group contains trigonal planar carbons, whereas the cyclohexylmethyl group contains only tetrahedral carbons. The benzyl groups,being flatter, could pack more easily together and interact more strongly. We would expect dibenzyl ether to have a higher melting point.
Exercise 7.5.1:
These compounds all contains lots of non-polar C-C and C-H bonds, which would not have appreciable dipoles. However, in each pair, there is a compound that contains more polar bonds as well: C-Cl, C-N and C-O.
Exercise 7.5.2:
Dipoles are vectors. If we have two polar bonds, they will add together using vector addition. Any vector can be shown as the sum of two other vectors. Below, the red arrow points down and to the left. How far down and how far to the left? We can show how far using the blue arrows. The blue arrow down and the blue arrow to the left add up to give the red arrow that goes down and to the left.
In dichloromethane, there are two polar bonds. Both bonds are polarized towards the more electronegative chlorine. As drawn below, one bond has a dipole straight to the right. The other bond has a dipole that goes down and to the left. If the molecule has an overall dipole, with a positive end of the whole molecule and a negative end of the whole molecule, which end would be positive and which would be negative?
It seems reasonable that the positive end would be somewhere over on the hydrogen side and the negative end would be somewhere over on the chlorine side. The overall dipole, as opposed to the individual bond dipoles, would point somewhere in the direction of the blue, dashed arrow. But how big would it be? Surely it's bigger than either of the two individual dipoles, because they are adding together, right?
Let's take a very qualitative look. The first red arrow is just to the right. We'll break the second arrow into a smaller arrow down and an even smaller arrow to the left. The sum of the left/right arrows is a very small arrow to the right.
The sum of the up/down arrows is just a small arrow down. Overall, the sum is a green arrow going down and to the right. The result does not really seem any bigger than an individual bond dipole.
Exercise 7.5.3:
Despite the relatively similar dipole moments, dichloromethane's boiling point is much higher than chloromethane's. The difference probably lies in the much greater mass of dichloromethane.
Exercise 7.5.4:
In each case, the one on the right has a larger dipole and a higher melting point.
Exercise 7.6.1:
Only ethylamine and ethanoic acid have the N-H or O-H bonds that are polar enough for hydrogen bonding.
Exercise 7.6.2:
hexylamine > dipropylamine > triethylamine
The one with the greatest potential for hydrogen bonding will have the highest melting point.
Exercise 7.6.3:
Exercise 7.8.1:
The hydrogen bonding between 2-propanol molecules is a stronger interaction and is more difficult to overcome than the dipole-dipole interactions between 2-propanone molecules. Thus, 2-propanone molecules can more easily escape into the vapour phase.
Exercise 7.8.2:
There is hydrogen bonding between molecules of N-methylethylamine, but not between molecules of triethylamine, which are held together by relatively weak dipoles. Trimethylamine molecules are not held together as tightly as the N-methylethylamine molecules, which therefore pack more densely. On the other hand, n-propylamine contains two N-H bond, compared to only one in N-methylethylamine. Consequently, n-propylamine has a greater propensity to form hydrogen bonds, and it packs together even more densley than the N-methylethylamine..
Exercise 7.8.3:
Temperature is a measure of how much energy is available in the environment. As temperature increases, molecules obtain more energy and they transfer that energy into a variety of molecular motions. As they move around more, they are less able to remain densely packed. The density decreases as the temperature goes up. Conversely, things tend to shrink as they cool, because molecules slow down and begin to pack more tightly. They become more dense.
Exercise 7.8.4:
1. Acetic acid posesses hydrogen bonding capability, whereas 2-propanone has the ability to form only dipole-dipole interactions. Thus, 2-propanone molecules are not held in the liquid phase as strongly as acetic acid, and the former compound has a higher vapour pressure.
2. The C=O bond of ethyl acetate has a substantial dipole, but the dipoles in ether don't add up to very much; they are smaller to begin with and they partially cancel out via vectorial addition. Ether molecules can escape the liquid more easily, resulting in a higher vapour pressure.
3. Both molecules are limited to relatively weak London dispersion interactions. Because heptane forms a longer chain than pentane, there will be a slightly greater interaction between heptane molecules than between pentane molecules. Pentane thus has the higher vapour pressure.
Exercise 7.8.5:
1. The C=O bond 2-butanone has a substantial dipole, but the dipoles in ether don't add up to very much; they are smaller to begin with and they partially cancel out via vectorial addition. The 2-butanone molecules cling together more strongly, resulting in a higher viscosity.
2. Both molecules are limited to relatively weak London dispersion interactions. Because decane forms a longer chain than hexane, there will be a slightly greater interaction between decane molecules than between hexane molecules. Decane thus has the higher viscosity. Later, if you study macromolecules, you will see that in very, very long chains, "entanglement" becomes a factor as well. Just as the name implies, really long chains get tangled up together and have a difficult time moving past each other; that's why materials like cooking oil and motor oil are so viscous.
3. The 1-butanol can hydrogen bond together, but the ether only has weak dipole-dipole interactions. The 1-butanol is therefore more viscous.
Exercise 7.8.6:
1. The 1-butanol can hydrogen bond together, but the ether only has weak dipole-dipole interactions. The 1-butanol therefore has greater surface tension.
2. The 1-butanol can hydrogen bond together, but the 1,3-butanediol has two OH groups and can form even more hydrogen bonds than 1-butanol can. The 1,3-butanediol therefore has greater surface tension.
3. The formamide has N-H bonds and so it is capable of hydrogen bonding. Despite having a pretty darned big dipole, the nitromethane can't really hydrogen bond together. The formamide therefore has greater surface tension.
Exercise 7.8.7:
1. 2-propenol. It has hydrogen bonding capability but propanal does not.
2. heptanoic acid. It has hydrogen bonding capability but ethyl pentanoate does not.
3. 1-hexen-3-one. Its dipole is greater than that in the ether, because it comes from a C=O bond rather than C-O bonds.
Exercise 7.8.8:
1. 2-butanone. Its dipole is greater than that in the ether, because it comes from a C=O bond rather than C-O bonds.
2. octane. It has greater London dispersion interactions that pentane because of the greater surface contact area in octane.
3. 1,3-butanediol. It has hydrogen bonding but the DME does not.
Exercise 7.8.9:
Exercise 6.9.1:
Hydrogen bonding.
Exercise 6.9.2:
a) dipole interactions b) London interactions c) dipole interactions d) hydrogen bonding
Exercise 6.9.3:
Water molecules interact with each other mainly through hydrogen bonding, whereas octane molecules interact with each other via London interactions. The main problem here is the strong hydrogen bonding between the water molecules. All molecules, in principle, could interact with each other via London interactions. However, if octane molecules were introduced among the water molecules, they would take up space. Some of the water molecules would not be able to get close enough to each other to hydrogen bond anymore. The loss of that very stabilizing interaction would be too costly.
Exercise 6.9.4:
This problem is similar to the previous one, but in this case the attraction between the strong dipoles of the nitrile groups would be too much to overcome.
Exercise 6.9.5:
Exercise 7.9.6:
a) dipole / induced dipole b) dipole / dipole c) hydrogen bonding d) dipole / induced dipole
Exercise 7.10.1
The predicted order, from most soluble to least, would be LiCl > MgSO4 > AlPO4 because the ions increase in charge from 1+/- to 2+/- to 3+/-
Exercise 7.10.2:
The predicted order, from most soluble to least, would be dimethylsulfoxide > acetonitrile > pyridine > dichloromethane > triethylamine.
Exercise 7.11.1:
a) H-bond acceptor b) H-bond acceptor c) fully H-bonding d) H-bond acceptor e) fully H-bonding f) neither
Exercise 7.12.1:
Exercise 7.13.1:
Note that the last example is actually found in RNA, not DNA. RNA is found as a single strand, rather than a pair of strands. It hydrogen bonds to itself in a particular shape, not unlike the way a protein adopts a specific shape.
Exercise 7.13.2:
Exercise 7.13.3:
Although lignin is capable of some hydrogen bonding, it cannot make nearly as many hydrogen bonds as cellulose can. Paper containing lignin would be much weaker, and more likely to tear or to fall apart when it gets wet, because the cellulose molecules would not be able to bind each other as tightly with lignin mixed in. In fact, brown paper bags from the grocery store are made from kraft, which has not had lignin removed; they are much weaker than most paper, especially if they get wet.
Exercise 7.13.4:
The dominant intermolecular attraction here is just London dispersion (or induced dipole only). London dispersion is very weak, so it depends strongly on lots of contact area between molecules in order to build up appreciable interaction. The stearic acid-containing triglycerides will be able to pack much more closely together than the more jumbled oleic acid-containing ones, so they will have much more surface contact and much stronger attractions. they will hold together into a solid.
Exercise 7.13.5:
a) alpha helix b) beta sheet c) alpha helix d) beta sheet e) beta sheet
f) alpha helix g) alpha helix h) beta sheet i) alpha helix j) alpha helix
Exercise 7.14.1:
a)
water water
ionic no no
H-bond acceptor yes yes
H-bond donor yes yes
dipole yes yes
induced dipole yes yes
Water hydrogen bonds with water.
b)
goof goof
ionic no no
H-bond acceptor no no
H-bond donor no no
dipole no no
induced dipole yes yes
Goof-off forms induced dipole/induced dipole interactions (London dispersion forces) with itself.
c)
goof grease
ionic no no
H-bond acceptor no no
H-bond donor no no
dipole no no
induced dipole yes yes
Goof-off forms induced dipole/induced dipole interactions (London dispersion forces) with grease.
d)
grease water
ionic no no
H-bond acceptor no yes
H-bond donor no yes
dipole no yes
induced dipole yes yes
Water forms dipole/induced dipole interactions with grease.
The key to why goof-off works is that it engages in the same London dispersion interactions with the grease as the grease does itself. Water is good at dissolvng many things, but not grease. That's mostly because the water would have to give up some of the strong hydrogen bonds with other water molecules in order to fit grease molecules among the water molecules.
Exercise 7.14.2:
a)
octane octane
ionic no no
H-bond acceptor no no
H-bond donor no no
dipole no no
induced dipole yes yes
That's going to be induced dipole-induced dipole interactions in octane.
b)
isopropanol isopropanol
ionic no no
H-bond acceptor yes yes
H-bond donor yes yes
dipole yes yes
induced dipole yes yes
Hydrogen bonding in isopropanol.
c)
i)
isopropanol water
ionic no no
H-bond acceptor yes yes
H-bond donor yes yes
dipole yes yes
induced dipole yes yes
Hydrogen bonding between isopropanol and water.
ii)
octane isopropanol
ionic no no
H-bond acceptor no yes
H-bond donor no yes
dipole no yes
induced dipole yes yes
Dipole-induced dipole interactions between octane and isopropanol.
d)
e) No, the ethanol should do the same thing that the HEET does.
Exercise 7.14.3:
a)
b) The amphiphile inserts into the cell membrane. Draw a cartoon of a portion of a cell membrane. Add a cartoon of the synthetic amphiphile to the membrane. What is the IMF through which it interacts with the membrane?
c) Ca2+ (two atoms from the left edge).
d) PO43- because $3 \times (2^{+}) + 2 \times (3^{-}) = 0$; neutral charge overall.
e) These high charges (2+/3-) bind much more tightly than the low charges (1+/1-) in sodium chloride.
f)
Exercise 7.14.4:
a)
b)
c)
d)
e) The carboxylic acid group provides sites for hydrogen bonding, including both a hydrogen bond donor in the OH group and an additional hydrogen bond acceptor in the C=O group.
f) It introduces a "hydrophobic" part in which the major intermolecular force with water would be a dipole-induced dipole interaction. If the water molecules keep to themselves, they will be able to maximize the amount of hydrogen bonding they can do with each other. The water is better off not mixing in with the paper.
g) All proteins have polar backbones formed by their amide bonds or peptide linkages (-NH-C=O). These portions can hydrogen bond with the cellulose in the paper. In addition, there may be polar side chains such as serine or threonine, for example, containing OH groups that can hydrogen bond with the paper.
h) Hydrophobic residues would add water resistance, for the same reason as the hydrophobic part of the abietic acid in rosin.
i) Ti4+ would have the same electronic configuration as argon. O2- would have the same electronic configuration as neon.
j) The formula is TiO2, so that the positive charges (4+) and the negative charges (2 x 2- = 4-) are balanced.
k) The 4+ and 2- charges are pretty large. Highly charged ions are difficult to dissolve, because the ion-dipole attraction of the ion with the water cannot compete with the very strong ionic attractions between the ions.
l) Ion-dipole. You can imagine the partially negative oxygen atoms of cellulose interacting with the titanium cations and the partially positive hydrogens of the OH groups in the cellulose interacting with the oxide anions.
m) Ti: one full atom in the center and eight in the corners, so 1 + 8x(1/8) = 2 Ti atoms.
O: two full atoms (towards the left and right ends of the picture) plus four half atoms (on the upper and lower faces of the cell), so 2 + 4x(1/2) = 4 O atoms.
ratio: 1 Ti : 2 O or TiO2 in formula notation.
n) The Ti are packed in a cell that resembles a body-centered cubic cell. However, the cell is not really cubic, because it is stretched out a little. Sometimes we call that shape "tetragonal".
o) The benzyl alcohol could interact with the eosin in a number of ways, because eosin is a fairly large, complicated molecule (at least at this stage in your studies). We could imagine ion-dipole interactions, in which the partially positive hydrogens on the OH groups in the benzyl alcohol interact with the anionic oxygens of the eosin, or the partially negative oxygens in the benzyl alcohol interact with the positive calcium ion.
However, the polar part of benzyl alcohol is really the smaller part of the molecule. The bigger part of the molecules is the benzene, a six-membered carbon ring with three double bonds. The carbons at the ends of those double bonds are trigonal planar, so because the benzene is built up from planar atoms, it is completely flat. C-C bonds are not polar, and C-H bonds are hardly polar at all, so the dominant IMF in this part of the molecule is London dispersion or induced dipole-induced dipole interaction. That's a pretty weak interaction generally, but it can get a little stronger between large molecules, or between flat ones, because two flat molecules can have more surface area in contact with each other. That means the benzenes in benzyl alcohol and the benzenes in eosin have appreciable London forces that allow them to interact with each other.
p) With the cellulose, covered in OH groups, the polar interactions outlined above will become more prominent.
Exercise 7.14.5:
1. OAc is an ester.
2. AAI has three chiral centers; GATA has five.
3. The chains in the adhesive would stick to each other via intermolecular attractions. Dipole-dipole attractions would be the strongest. However, induced dipole-induced dipole or London dispersion forces would actually be significant because there would be hundreds of such interactions along the chains.
4. The most significant interaction with the paper would be hydrogen bonding. The cellulose in the paper would provide hydrogen bond donors (the OH groups) and the side groups in the adhesive chains would provide hydrogen bond acceptors (all those oxygens). However, if the paper contains a lot of sizing, induced dipole-induced dipole or London dispersion forces may form a significant part of the interaction. As with cohesion, this would not be a significant factor except that there are so many alkyl chains involved, so these tiny attractions would add up. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/07%3A_Structure-Property_Relationships/7.15%3A_Solutions_to_Selected_Probl.txt |
Fatty Acids: Single Chain Amphiphiles
Consider stearic acid, a type of lipid known as a fatty acid. It has an 18 C atom chain with a carboxylic acid group (RCO2H) at one end.
The carboxyl end (shown in red) containing the two oxygens is polar, but the rest of the molecule is completely nonpolar. We can draw a "cartoon" model of this molecule as a circle representing the carboxyl polar end or "head group" with a single connecting line representing the long, nonpolar "tail".
A molecule that contains a distinct polar region and a distinct single chain nonpolar region is called a single chain amphiphile.
How would this fatty acid interact with water? A small number of these molecules might be soluble in water (remember even insoluble salts dissociate to a small degree to form some ions). The nonpolar tail is not stable in water, but the polar head is. Some of the molecules migrate to the surface of the water, with the nonpolar tails protruding into the air and away from water, forming a monolayer on the surface. Others will self-aggregate through IMFs to form a spherical structure in which the nonpolar tails are sequestered from water and the polar heads are facing the water. This structure is called a micelle. In the figure below, the polar head groups are shown in blue.
Which IMFs are involved? First consider attractive IMFs. The buried nonpolar chains can interact and be stabilized through London forces. Buried inside of the micelle, the nonpolar chains are sequestered from water. This view fits our simple axiom of "like-dissolves like". The polar head groups can be stabilized by ion-dipole interactions between charged head groups and water. Likewise H-bonds between water and the head group stabilize the exposed head groups in water. Repulsive forces are also involved. Head groups can repel each other through steric factors, or ion-ion repulsion from like-charged head groups. The attractive forces must be greater than the repulsive forces, leading to micelle formation.
Detergents are molecules, like fatty acids, that have distinct and separate polar and nonpolar regions. They form micelles in water. Grease from clothes or foods, normally not soluble in water, can "dive" into the middle of the micelle and be carried off by this structure. This effectively "solubilizes" the nonpolar molecule. Notice that there is no water inside the micelle as you can see in the figure and Jmol computer model above.
Phospholipids: Double Chain Amphipiles
Now consider another lipid, a phospholipid, the example shown below. X represents a polar group containing a charged phosphate.
Note that it has a polar end (shown in red), but in contrast to stearic acid, it has two long nonpolar tails shown in black. We can draw a "cartoon" model of this as a circle representing the polar end or "head group" with two connecting lines representing the two long nonpolar "tails". It is a double chain amphiphile. Phospholipids are the major component of biological membranes. Again, in the figure below, the polar head groups are shown in blue.
How would this phospholipid interact with water? Again, as with single chain detergents, there are several possible ways. A small number of these molecules might be soluble in water. Nonpolar tails are unstable in water, while the polar head are stable in water. As in the case of single chain detergents, some of the molecules migrate to the surface of the water, with the nonpolar tails sticking out into air, away from water, to form a monolayer on the top of the water. Others will self- aggregate through IMFs to form a bilayer or membrane instead of a micelle, which only contains single chain amphiphiles. The polar head groups in a double chain amphiphile can get closer together than in a micelle since there are extra stabilizing interactions among the side chains. The result is the formation of a membrane bilayer which affords less space for each head group than in a micelle. Imagine the bilayer membrane curving around and eventually meeting itself. A structure like this would look like a small biological cell. In contrast to a micelle, the interior of this multimolecular lipid aggregate, called a liposome or vesicle, is filled with water which can interact through IMFs with the head groups of the inner leaflet of this membrane. The head groups of the outer leaflet of the membrane interact through IMFs with the bulk water.
Liposomes are useful since they are simple models of actual biological membranes. In addition, they can be used therapeutically. Toxic drugs, like those used in chemotherapy, can be incorporated into the aqueous volume inside a liposome where they can be targeted to specific tumor cells.
Liposomes and micelles seem to be complicated structures. However their formation and structure can be predicted from the simple solubility properties of these molecules and through an understanding of IMFs.
Ever wonder why a solution of sodium hydroxide (a base) feels slippery on your skin? The hydroxide base cleaves the bond between fatty acids connected to more complex skin phospholipids. The free fatty acids then aggregate spontaneously to form micelles which act like detergents which as you know feel slippery as well.
Triacylglycerides
We've seen the structures of a single fatty acid and of a phospholipid which contains two fatty acids linked to glycerol. Another very abundant lipid contains three fatty acids connected to glycerol to form a triacylglyeride, as shown below.
This molecule overall is so nonpolar that it self-associates to form lipid drops which are stored predominantly in fat or adipose tissue in the body. Instead of having a structural role, it functions mainly as a source of energy.
Exercise \(1\)
We've discussed how lipids can form aggregates in aqueous solution. Now let's reverse the scenario. Draw a diagram showing what might happen if lots of stearic acid molecules were added to the solvent hexane, C6H14.
Exercise \(2\)
An important property of a micelle is the surface area/head group (AS/HG). This is equal to the total surface area of the micelle divided by the number of single chain amphiphiles in the micelle, Atot/#amphiphiles.
What kind of repulsive forces exist between head groups? What happens to AS/HG if the total charge on the head group increases? Decreases? How would IMFs between the nonpolar chains affect micelle stability?
Exercise \(3\)
Imagine now that you squish the spherical micelle into a bilayer. Think of this as flattening a tennis ball. What would happen to the (AS/HG)? (Hint: Think of pencils sticking out radially from the tennis ball, with the erasers representing head groups. What happens to the eraser spacing as it the tennis ball distorts to a bilayer?. How does this explain why charge single chain amphiphile don't form bilayers?
Exercise \(4\)
Now try a different thought experiment. Add more and more of the same single chain amphiphile to a spherical micelle. The shape must deviate from a sphere. Why? Draw a cartoon diagram to illustrate your point. In this case the lipid aggregate can grow if it adopts a cylindrical shape. Compare the (AS/HG) for the spherical, cylindrical and bilayer shapes. Which of these shapes are more likely to form
Exercise \(5\)
Now lets consider a bilayer made of double chain amphiphiles (shown below). Compare the AS/HG for a bilayer to a scenario where the double chain amphiphile forms a micelle? Now draw a cartoon diagram to explain the observation that double chain amphiphiles don't form micelles (note the bilayer shown).
Exercise \(6\)
Now explain how the double chain amphiphiles can tolerate the close packing of head groups found in the bilayer while single chain amphiphiles usually can't.
Exercise \(7\)
There are many other different types of lipids. An example, cholesterol, is shown below. This molecule has many biological functions. It is often found in lipid bilayers. Draw a cartoon showing how cholesterol might fit into a bilayer. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/08%3A_Introduction_to_Biomolecules/8.01%3A_Lipids.txt |
Proteins are large biological molecules that have molecular weights ranging from the thousand to the millions. Humans have about 24,000 different proteins which catalyze chemical reactions, recognize foreign molecules and pathogens, allow cellular and organism movement, and regulate cell response, including cell division and death.
Proteins are polymers consisting of monomers called amino acids. There are twenty different naturally occurring amino acids that differ in one of the four groups connected to a central carbon atom. In an amino acid, the central (alpha) carbon has an amine group (RNH2, RNH3+), a carboxylic acid group (RCOOH, RCOO-), a hydrogen (H), and one of twenty different R groups (also called side chains) attached to it.
The R groups are classified as generally nonpolar, polar charged, or polar uncharged. The smallest amino acid is glycine (Gly) which has a hydrogen atom as its R group. All of the other 19 naturally-occurring amino acids have one stereocenter (at the carbon containing the amine and carboxyl groups) and can exist as two possible enantiomers; only the L-enantiomer occurs in proteins. All amino acids in proteins have the absolute configuration shown above. With the exception of the amino acid cysteine (Cys) with a -CH2SH for an R group (which happens to have an R stereocenter), all of the remaining amino acids found in proteins have an S stereocenter.
Amino acids form polymers when an amino group of an amino acid is covalently attached to the carbonyl carbon (C=O) of the carboxyl group of the next amino acid. The resulting link between the amino acids is an amide bond which biochemists call a peptide bond. In this reaction, water is released (condensation). In a reverse reaction, the peptide bond can be cleaved by water (hydrolysis). When two amino acids link together to form an amide link, the resulting structure is called a dipeptide. Likewise, we can have tripeptides, tetrapeptides, and other polypeptides. At some point, when the structure is long enough, it is called a protein. There are many different ways to represent the structure of a polypeptide or protein. Each shows differing amounts of information. A heptapeptide, Aspartic Acid-Lysine-Glutamine-Histidine-Cysteine-Arginine-Phenylalanine is shown below. Each amino acid is denoted by a three letter code (Asp-Lys-Gln-His-Cys-Arg-Phe).
Notice that the protein chain has a beginning (an N-terminus with a amino group) and an end (a C-terminus with a carboxyl group). Also note that every atom in the backbone has a slight charge arising from the presence of the electronegative atoms O and N. Hence the backbone is polar. The R groups on each amino acid in the peptide are also called side chains.
The actual linear sequence of amino acids that make up a protein is called its primary (1o) structure. Both the sequence of a amino acids and total chain length differ from one protein to another. Just for an octapeptide, there are over 25 billion different possible arrangements of amino acids. Hence the diversity of possible proteins is enormous.
Most proteins do not form an elongated structure as implied by the extended structures shown above. Rather they collapse on themselves to form compact, mostly globular (roughly spherical) structures. They do so as groups local and distant on the chain attract each other through IMFs which are now exerted within the large protein and not between different proteins. What kinds of IMFs are involved?
To simplify the process, lets consider first just the polar backbone without the side chains. The main chain can clearly form hydrogen bonds with itself and to water. If the hydrogen bonds are between the amide H (a hydrogen bond "donor" and a carbonyl O (a hydrogen bond "acceptor") a fixed number of amino acids distant from the amide H, a regular, repetitive secondary (2o) structures called a helix can form. One especially prevalent helix, the alpha helix, forms within a short stretch of amino acids when the amide H of an amino acid (given the number ith) in the backbone forms a hydrogen bond to the carbonyl C four amino acids in the protein sequence (ith + 4). There are 3.6 amino acids/turn of the alpha helix.
Alpha Helix (dotted yellow lines represent hydrogen bonds)
Note that all the side chains (R groups) are pointing away from the helix axis. Evident from the space-filling model, there is no opening in the helix as you look down the axis because the actual atoms are densely packed. The trace of an alpha helix in a protein is usually represented by a red or purple curly ribbon.
Beta strands/sheets, another type of secondary structure, also occur when H bonds form between adjacent short stretches of amino acids in which the backbone of the short stretches are running either in the same N-C direction (parallel beta strands) or in opposite directions (antiparallel beta strands). The hydrogen bonds in secondary structures are all among main chain atoms in the backbone, not among side chains. Beta structure is usually represented in "cartoon" form by yellow flat ribbons with an arrow showing the direction of the protein backbone from the N-terminus to C terminus direction.
Antiparallel Beta Strands (yellow lines represent hydrogen bonds)
Parallel Beta Strands (yellow lines represent hydrogen bonds)
The side chains in the beta sheet are perpendicular to the plane of the sheet, extending out from the plane on alternating sides. Parallel sheets characteristically distribute nonpolar (or hydrophobic) side chains on both side of the sheet, while antiparallel sheets are usually arranged with all the hydrophobic residues on one side. This requires an alternation of hydrophilic and hydrophobic side chains in the primary sequence. Antiparallel sheets are found in silk with the sheets running parallel to the silk fibers. The following repeat is found in the primary sequence: (Ser-Gly-Ala-Gly)n), with Gly pointing out from one face and Ser or Ala from the other.
Protein folding is determined by much more than the formation of hydrogen bonds between backbone donors and acceptors. We must consider the effects of the 20 different R groups (side chains) which complicate the folding process. A protein ultimately folds in space to form a unique 3D shape, which usually contains some alpha helices and beta sheets. The overall 3 D structure is called the tertiary (3o) structure of the protein. The 3D structure of a protein determines the function of the protein. Protein shape and surface charge characteristics determine which molecules, both small and large, bind to the protein.
Here are some models of proteins showing secondary and tertiary structures.
The structure of proteins is much more complicated than micelles and bilayers. To a first approximation, a protein consists of a polar main chain/backbone from which amino acid side chains of varying polarity and charge hang. These side chains can be polar uncharged, polar charged, or nonpolar. In general the nonpolar side chains are more stable buried in the center of the protein, where they are surrounded by other nonpolar side chains and are oriented away from polar water. Compare this to the structure of a micelle. Given the greater complexity of protein primary and tertiary structure, however, not all nonpolar side groups can be buried. Some are on the surface exposed to solvent. Likewise, polar and charged polar side chains like to be on the surface exposed to water, but some will find themselves buried. If they are, they will be surrounded by polar side chains or interact with buried hydrogen bond donors and acceptors on the backbone that stabilize the buried polar group. Here are some findings about proteins derived from the known 3D structure of thousands of different proteins:
• On average, about 50% of the amino acids in a protein are in secondary structure with an average of about 27% alpha helix and 23% beta structure.
• The side chain location varies with polarity. 83% of nonpolar side chains (such as Val, Leu, Ile, Met, and Phe) are in the interior in the folded protein.
• Charged polar side chains are almost equally partitioned between being buried or exposed on the surface.
• Uncharged polar groups such as Asn, Gln, Ser, Thr, and Tyr are mostly (63%) buried, and not on the surface.
• Globular (spherical) proteins are quite compact, with water excluded. The packing density (Vvdw/Vtot) is about 0.74, which is like the NaCl crystal and equals the closest packing density of 0.74. This compares to organic liquids, whose density is about 0.6-0.7.
The packing around a buried nonpolar side chain of the amino acid phenylalanine (Phe) is shown in the Jmol below. It shows the structure of a small protein (protein tyrosine phosphatase) and the amino acids groups surrounding the buried Phe.
Exercise \(1\)
Secondary structure in proteins involves
1. hydrogen bonds between side chain atoms and backbone atoms
2. hydrogen bonds between side chain atoms and other side chain atoms
3. London dispersion forces among side chain atoms
4. hydrogen bond between backbone atoms and other backbone atoms.
Exercise \(2\)
The structure of the amino acid Ser is shown below:
Which groups are involved in forming covalent bonds to other amino acids in a protein?
1. -CH2OH and -CO2H
2. -CH2OH and -NH3 +
3. -CO2H and -NH3 +
4. -H and -CH2OH
Exercise \(3\)
The structure of the amino acid Ser is shown below:
If Ser was in an alpha helix, which group would most likely be projecting away from the helix axis?
1. -CH2OH
2. -NH3 +
3. -CO2H +
4. -H
Exercise \(4\)
If Ser were part of a protein, which of the following would most likely NOT describe the environment around the side chain?
1. it could be buried in the protein surrounded by nonpolar amino acid side chains
2. it could be on the surface surrounded by water
3. it would buried adjacent to a glutamine side chain
4. it is polar so it could not be buried in the protein
Exercise \(5\)
Molecules of a given protein have the following trait(s):
1. a defined amino acid sequence
2. an invariant molecular weight
3. a fixed amino acid composition
4. all of the above
Exercise \(6\)
______ between tightly packed amino acid side chains in the interior of a protein are a major contribution to the stability of the native state.
1. Dipole-dipole interactions
2. Ion - ion interactions
3. Covalent bonds
4. London dispersion forces
Exercise \(7\)
If the following section of a polypeptide is folded into an alpha helix, to which amino acid is the carbonyl group of Ala H-bonded?
N term Ala-Leu-Ser-Asp-Glu-Val C term
1. Leu
2. Ser
3. Asp
4. Glu
5. Val
Exercise \(8\)
Which form of a 10 amino acid peptide would most likely have the shortest length?
1. the peptide is in solution and not part of a proteins structure
2. the peptide is part of a protein and in a antiparallel beta sheet
3. the peptide is part of a protein and in a parallel beta sheet
4. the peptide is part of a protein and in an alpha helix | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/08%3A_Introduction_to_Biomolecules/8.02%3A_Proteins.txt |
The nucleic acids RNA and DNA are involved in the storage and expression of genetic information in a cell. Both are polymers of monomeric nucleotides. DNA exists in the cell as double-stranded helices while RNA typically is a single-stranded molecule which can fold in 3D space to form complex secondary (double-stranded helices) and tertiary structures in a fashion similar to proteins. The complex 3D structures formed by RNA allow it to perform functions other than simple genetic information storage, such as catalysis. Hence most scientists believe that RNA preceded both DNA and proteins in evolution as it can both store genetic information and catalyze chemical reactions.
DNA
DNA is a polymer, consisting of monomers call deoxynucleotides. The monomer contains a simple sugar (deoxyribose, shown in black below), a phosphate group (in red), and a cyclic organic R group (in blue) that is analogous to the side chain of an amino acid.
Only four bases are used in DNA (in contrast to the 20 different side chains in proteins) which we will abbreviate, for simplicity, as A, G, C and T. They are bases since they contain amine groups that can accept protons. The polymer consists of a sugar - phosphate - sugar - phosphate backbone, with one base attached to each sugar molecule. As with proteins, the DNA backbone is polar but also charged. It is a polyanion. The bases, analogous to the side chains of amino acids, are predominately polar. Given the charged nature of the backbone, you might expect that DNA does not fold to a compact globular (spherical) shape, even if positively charged cations like Mg bind to and stabilize the charge on the polymer. Instead, DNA exists usually as a double-stranded (ds) structure with the sugar-phosphate backbones of the two different strands running in opposite directions (5'-3' and the other 3'-5'). The strands are held together by hydrogen bonds between bases on complementary strands. Hence like proteins, DNA has secondary structure; but in this case, rather than being within the backbone, hydrogen bonds are between the "side chain" bases on opposing strands. It is actually a misnomer to call dsDNA a molecule, since it really consists of two different, complementary strands held together by hydrogen bonds. A structure of ds-DNA showing the opposite polarity of the strands is included below.
In double-stranded DNA, the guanine (G) base on one strand can form three H-bonds with a cytosine (C) base on another strand (this is called a GC base pair). The thymine (T) base on one strand can form two H-bonds with an adenine (A) base on the other strand (this is called an AT base pair). Double-stranded DNA has a regular geometric structure with a fixed distance between the two backbones. This requires the base pairs to consists of one base with a two-ring (bicyclic) structure (these bases are called purines) and one with a single ring structure (these bases are called pyrimidines). Hence a G and A or a T and C are not possible base pair partners.
Double-stranded DNA varies in length (number of sugar-phosphate units connected), base composition (how many of each base set) and sequence (the order of the bases in the backbone). The following links provide interactive Jmol models of dsDNA made by Angel Herráez, Univ. de Alcalá (Spain) and Eric Martz.
Chromosomes consist of one dsDNA with many different bound proteins. The human genome has about 3 billion base pairs of DNA. Therefore, on average, each single chromosome of a pair has about 150 million base pairs and lots of proteins bound to it. dsDNA is a highly charged molecule, and can be viewed, to a first approximation, as a long rod-like molecule with a large negative charge. It is a polyanion. This very large molecule must somehow be packed into a small nucleus of a tiny cell. In complex (eukaryotic) cells, this packing problem is solved by coiling DNA around a core complex of four different pairs (eight proteins total) of histone proteins (H2A, H2B, H3, and H4) which have net positive charges. The histone core complex with dsDNA wound around approximately 2.5 times is called the nucleosome.
DNA can adopt two other types of double-helical forms. The one discovered by Watson and Crick and found in most textbooks is called B-DNA. Depending on the actual DNA sequence and the hydration state of the DNA, it can be coaxed to form two other types of double-stranded helices, Z and A DNA. The A form is much more open then the B form.
The 3.2 billion base pairs of DNA in humans contain about 24,000 short stretches (genes) that encode different proteins. These genes are interspersed among DNA that helps determining if the gene is decoded into RNA molecules (see below) and ultimately into proteins. For a particular gene to be activated (or "turned on"), specific proteins must bind to the region of a particular gene. How can binding proteins find specific binding targets among the vast number of base pairs that to a first approximation have a repetitive sugar-phosphate-base repeat? The Jmol below shows how specificity can be achieved. When DNA winds into a double helix through base-pairs between AT and GC, hydrogen bond donors (amide H's) and acceptors (O's) on the bases that are not used in intrastrand base pairing are still available in the major and minor grove of the ds-DNA helix (see Jmol below). Unique base pair sequences will display unique patterns of H bond donors and acceptors in the major grove. These donors/acceptors can be recognized by specific DNA binding proteins which can lead to gene activation upon binding.
RNA
The monomers that make up RNA are identical to those for DNA with two differences. The 2'C of the ribose ring has an OH instead of H (as is found in deoxyribose). In addition, the thymine (T) base found in DNA is replaced with uracil (U). The structure of uridine-5-monophosphate is shown below.
Based on analogy with DNA, you might expect RNA to adopt a double-stranded helical conformation. It can do this in short stretches, but instead, even in the presence of positively charged divalent cations, it is found predominantly in a single-stranded form. This single strand folds to form regions of secondary structure (double-stranded helices) imbedded in a complex tertiary structures (not unlike a protein).
The addition of the 2'OH on the ribose ring or RNA alters the pucker of the ribose ring and destabilizes ds-RNA in comparison to ss-RNA through steric interference. The five-membered ribose or deoxyribose ring can adopt an envelope-like structure in which one ring atom, representing the tip of the envelope, is above, or endo, to the other four coplanar atoms. In deoxyribose, the 2' ring carbon atom prefers the endo position; in ribose, the 3' C is endo (see Jmol model below). When double-stranded RNA forms, it is actually very similar in structure to the more open (and less abundantly found) A-DNA form.
Here are some examples of complex tertiary structures formed from single-stranded RNA which folds to form a compact 3D structure with some double-stranded A-type helices stabilized by intramolecular H bonds between G/C and A/U base pairs. .
Exercise \(1\)
The figure below shows an actual base pair in a dsDNA.
• identify the base pair and the deoxyribose sugars
• place a negative charge on any atom with a negative charge
• label atoms other than those between the bases that could form H bonds with other DNA binding molecules
Exercise \(2\)
Do the same for the following base pair
• identify the base pair and the deoxyribose sugars
• place a negative charge on any atom with a negative charge
• label atoms other than those between the bases that could form H bonds with other DNA binding molecules
Exercise \(3\)
The backbone of dsDNA is highlighted below. Place a circle on every atoms with a negative charge.
Exercise \(4\)
The structure of a fully protonated peptide, poly-Lys, composed just of the amino acid lysine, is shown below.
Representing this molecule in cartoon form as an elongated ellipse, show how it might interact with on the DNA helix
Exercise \(5\)
Based on your answer to the above question, what general feature might be present in DNA binding proteins? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/08%3A_Introduction_to_Biomolecules/8.03%3A_Nucleic_Acids.txt |
Carbohydrates (CHOs) are among the most complex of biological molecules. If you have ever "counted" your carbs, you know that one biological function of CHOs is to store and provide energy (upon oxidation) to the body for required functions. Instead of concentrating on how CHOs are used for energy production, we will focus predominantly on their structures, which allows them to elicit their main function to provide binding interactions with other biomolecules, either in solution or on cell membranes. Binding, promoted through IMFs, initiates and modulates all biological interactions and processes. First, we will consider the structure of simple and more complex carbohydrates.
Monosaccharides
The simplest carbohydrates are monosaccharides, small organic molecules that contain more than one OH (alcohol) group and a single aldehyde (RCOH) or ketone (RCOR). Hence the simplest monosaccharides, glyceraldehyde and dihydroxyacetone, contain three carbons. In contrast to the alkane hydrocarbons which have the generic formula CnH2n+2, simple monosaccharides have the generic formula Cn(H2O)n which corresponds to their designation as carbo(hydrates).
Glyceraldehyde has one stereocenter. The naturally occurring isomer shown above is R-glyceraldehyde. Instead of using the R/S systems to denote the absolute configuration of specific stereocenters in sugars (as well as amino acids), biochemists often use the D/L system. The D designation on sugars refers to any monosaccharide whose last stereocenter (or in the case of glyceraldehyde, the only stereocenter) has the same absolute configuration as R-glyceraldehyde. All naturally occurring amino acids have the L configuration as they have the same absolute configuration when compared to L-glyceraldehyde.
Monosaccharides can exist in solution in an equilibrium mixture of a straight chain and cyclic form. An example is shown below for the common sugar D-glucose (Glc). The R/S designation for each stereocenter is included to make it easier to see the orientation of each OH group as the straight chain form cyclizes. In the process of forming the 6-membered ring, a lone pair on the slightly negative OH on C5 forms a bond to the slightly positive C1. This process requires rotation around the various C-C bonds to allow the OH on C5 to approach C1. Since the carbon can only have 4 bonds, bond formation to C1 from the OH on C5 causes the C=O (an aldehyde) to become C-OH (an alcohol). The chair, wedge/dash and "Haworth" projections of the resulting ring are shown. Note that the OH on C1 can point up (beta form, remembered by the phrase butterflies UP) or down (alpha form, remembered by ants DOWN). In the beta form, all ring substituents are in the equatorial position (in the chair form) and in alternating up and down forms in the wedge structure, making the beta form the most stable of all possible 6-membered cyclic forms of 6C sugars.
Since there are 4 stereocenters in a 6C sugar aldehyde, there are 24 or 16 possible stereoisomers. Most 6C sugars are D sugars which are defined to have the same R,S designation as D-glyceraldehyde at the last stereocenter in the sugar. Common 6C sugars are D-mannose (Man) and D-galactose (Gal). D-glucose, D-mannose, and D-galactose, all of which have the R configuration at C5, differ in the orientation of the OH groups in the ring structures at C2, C3, and C4. Hence they are diastereomers. The most common 6C sugar containing a ketone (RCOR) at C2 is fructose.
Since sugars are polyhydroxy aldehydes or ketones, can the OH's on C2, C3, and C4 also form a bond to C1 (as illustrated for the C5 OH above) and form rings. They can but the most common form is the 6-membered ring form shown above for D-glucose. How many atoms would be in the rings if the other OH's were involved in ring formation? Why is the 6-membered ring most abundant in nature?
Another type of structural representation, the Fisher projection, is often used in biology texts. In this form, the structure is written in a vertical direction with vertical and horizontal lines (not wedges or dashes). It is simple to translate a Fisher project for simple sugars into a Haworth projection, then to a wedge/dash ring structure, and then to a chair form. An example of how to convert a Fisher projection to a Haworth for D-glucose is shown below. If the OH group points to the right in the Fisher projection, it points down in the Haworth projections. If the OH points to the left, it points up in the Haworth projection.
Chemical Derivatives of Monosaccharides
Many derivatives of monosaccharides are found in nature. These include:
• oxidized forms in which the aldehyde and/or alcohol functional groups are converted to carboxylic acids
• phosphorylated forms in which phosphate is added by ATP to form phosphoester derivatives
• amine derivatives such as glucosamine or galactosamine
• acetylated amine derivatives such as N-Acetyl-GlcNAc (GlcNAc) or GalNAc
• more complex condensation products of sugar derivatives with lactate (CH3CHOHCO2-) and pyruvate, (CH3COCO2- ) to form muramic acid and neuraminic acids, (also called sialic acids), respectively.
Di- and Polysaccharides
As with proteins, which are polymers of amino acid monomers, covalent bonds can form between different sugar monomers to form disaccharides, trisaccharides, and ultimately polysaccharides, as shown below for two disaccharides, lactose and sucrose..
Longer polysaccharides can be quite simple in structure, but given the diversity of monosaccharides that each have multiple OH groups available to connect two monosaccharides in a polymer, the complexity of polysaccharide structure can be overwhelming to study and understand. Compare this to a protein, a polymer of amino acids that, with the exception of the side chains, each have only two functional groups involved in the linkages between the amino acids.
Homopolysaccharides
The simplest polysaccharides consist of only one monosaccharide repeating unit. Three examples of homopolysaccharides, glycogen/starch, cellulose, and chitin are shown below. The monomer in glycogen/starch and cellulose is glucose. The differences between them is the linkage between the glucose monomers. In starch and glycogen, which are energy storage polysaccharides, the linkage is alpha 1-4. In cellulose, the most abundant biomolecule, the linkage is beta 1-4. The beta linkage ensures that all bulky groups on the glucose chairs are in the more stable, equatorial position. This is yet another example of the fact that structure can account for properties and function of molecules. A small covalent change to glucose, the substitution of a CH3CONH- for a OH group on C2 of glucose gives the monosaccharide N-acetylglucosamine. This polymerizes, like cellulose, with a beta 1-4 linkage, to form a new polysaccharide, chitin, with markedly different properties than cellulose. Chitin, is the main component of the exoskeleton of arthropods.
Starch consist of 20-30 % amylose, an unbranched alpha 1-4 polymer of D-glucose, and 70-80% amylopectin, which consists of alpha 1-4 links with alpha 1-6 links every 25-30 glucose residues. Glycogen is similar to amylopectin and also has a protein, glycogenin, at the core of the glycogen particle.
Heteropolysaccharides with disaccharide repeat
Polysaccharides can also be made of a repeating disaccharide unit, instead of repeating monosaccharides in the homopolysaccharides shown above. One example is a glycosoaminoglycan called heparin (an anticoagulant, shown in the figure below) that is composed repeating units of the disaccharide, D-glucuronate-2-sulfate (alpha 1,4) GlcNSulfo-6-sulfate. This disaccharides consists of two glucoses that have been chemically modified. Note the charges on this repeat unit. Another example is the disaccharide repeat in the peptidoglycan (contains both CHO and amino acids) cell walls of bacteria (see figure below). The repeat again consists of chemically modified glucose units. Compare the repeat with the main component of the cytoskeleton, chitin. Nature has chosen variants of this repeat to create a rigid and protective exoskeleton and cell wall for arthropods and bacteria.
Here is a figure that provides a better view of the overall cell wall of Gram-positive bacteria.
Carbohydrates in biomolecule recognition
Carbohydrates are covalently attached to many different biomolecules, including lipids, to form glycolipids, and proteins, to form glycoproteins. Glycoproteins and glycolipids are often found in biological membranes, to which they are anchored by nonpolar interactions. A special kind of glycoprotein, a proteoglycan, actually has more carbohydrate mass than protein. What is the function of these carbohydrates? Two are apparent. First, glycosylation of proteins helps protect the protein from degradation by enzyme catalysts within the body. However, a second main function arises from the facilitation of biomolecule recognition. Covalently attached carbohydrates that "decorate" the surface of glycoproteins or glycolipids provide new binding sites for interactions with other biomolecules. Hence glycosylation allows for cell:cell, cell:protein, or protein:protein interactions. Unfortunately, bacteria and viruses often recognize glycosylated molecules on cell membranes, allowing for their import into the cell.
Here are some "cartoon" examples of carbohydrates covalently linked to the amino acid asparagine (Asn) on a glycoprotein.
Here are some examples of biomolecular interactions promoted by IMFs involving carbohydrates.
Influenza Virus binding to Cell Surface Glycoproteins with Neu5Ac - A protein on the surface of influenza virus, hemagluttinin, binds to sialic acid (Sia), which is covalently attached to many cell membrane glycoproteins on host cells. The sialic acid is usually connected through an alpha (2,3) or alpha (2,6) link to galactose on N-linked glycoproteins. The subtypes found in avian (and equine) influenza isolates bind preferentially to Sia (alpha 2,3) Gal which predominates in avian GI tract where viruses replicate. Human virus of H1, H2, and H3 subtype (cause of the 1918, 1957, and 1968 pandemics) recognize Sia (alpha 2,6) Gal, the major form in human respiratory tract. The swine influenza HA bind to Sia (alpha 2,6) Gal and some Sia (alpha 2,3)both of which found in swine.
Leukocyte: Cell Wall binding - During inflammation, circulating leukocytes (a type of white blood cell) tether and roll on the walls of blood vessels where they become active. E-, L- and P-selectin proteins are the primary proteins responsible for the tethering and rolling of these leukocytes. P-selectin binds, in part, to a tetrasaccharide, sialyl-Lewisx (SLEX) on the cell surface.. The interaction between P-selectin and the cell mediates the initial binding/rolling of the leukocyte on the vessel wall.
Exercise \(1\)
Draw a Haworth and wedge/dash ring structure for the five carbon sugar D-ribose, whose straight chain form is shown below. D-ribose is extremely important in the biological world since its 5-membered ring form is found in RNA. A variant, deoxyribose, in which the OH on C2 is replaced with an H atom, is part of the main chain of DNA.
Exercise \(2\)
Humans utilize two kinds of energy storage molecules: glycogen and triacylglycerides. (Review the structure of triacyglycerides from Chapter 8.1). Given intermolecular forces between these different types of molecules and water, predict which one might be biologically advantageous from a weight/mass perspective for an organism who must support this storage mass.
Exercise \(3\)
Using geometric figures to represent monomers, draw cartoon figures of the following
a. a straight chain homopolymer
b. a highly branched chain homopolymer
c. a linear heteropolymer with a repeat unit consisting of two different monomers
d. a highly branch polymer containing 4 different monomers
Exercise \(4\)
Nature uses all of these motifs in the biological world of carbohydrates. There must be advantages of these varying structure. Offer a possible reason for the following types of polysaccharide structures being adopted by nature.
a. straight chain and elongated polymer of glucose which can form fibrils in cellulose
b. a highly branched polymer of glycogen which must be broken down by enzymes to form glucose in times of urgent energy need
c. a highly branched polymer of diverse monomers and chemically-modified monomers attached to proteins.
Exercise \(5\)
The repeating disaccharide motif of heparin (discussed above) is shown below once more.
a. Would you expect this molecule to collapse into a compact spherical structure or have a more open and extended structure? Explain.
b. Heparin is used widely as an anticoagulant. It affects the properties of thrombin, an enzyme that causes the blood to clot. Thrombin is a globular (spherical) protein. For heparin to have a direct effect on thrombin, it must interact with it. What structural features of thrombin might be involved in this interaction? Draw a cartoon model showing the interaction. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/08%3A_Introduction_to_Biomolecules/8.04%3A_Carbohydrates.txt |
Exercise \(1\)
The laboratory of Samuel Stupp at Northwestern designs molecules with potential biomedical applications. The molecule below helps recruit growth factor to the cell (Christina J. Newcomb, Shantanu Sur, Sungsoo S. Lee, Jeong Min Yu, Yan Zhou, Malcolm L. Snead, and Samuel I. Stupp, Nano. Lett. 2016, 16, 3042-3050. Copyright 2016 American Chemical Society).
1. This molecule is a synthetic amphiphile. Put a rectangle around the polar part of this molecule. Circle the nonpolar part of the molecule.
2. The amphiphile inserts into the cell membrane. Draw a cartoon of a portion of a cell membrane. Add a cartoon of the synthetic amphiphile to the membrane. What is the IMF through which it interacts with the membrane?
3. How would the amphiphile attract / bind growth factor and other proteins?
4. This effect might be amplified if several of the amphiphiles bind to the membrane and cluster together. Use an abbreviated drawing (not the whole molecule) to show how two synthetic amphiphiles would interact with each other.
5. This drawing is reminiscent of a specific structural type seen in proteins. Which one?
6. The first amphiphile is not very good at interacting with others because it tends to spiral into a corkscrew shape. The amphiphile below is better at interacting with others because it keeps a flatter shape. Propose a reason why.
Exercise \(2\)
Here is a second type of amphiphile from the Stupp lab. This molecule was designed to bind a DNA strand (Adapted with permission from Mustafo Guler, Jonathan K. Pokorski, Daniel H. Appella, and Samuel I. Stupp, Bioconjugate Chem.. 2005, 16, 501-503. Copyright 2018 American Chemical Society).
1. Circle the part of the molecule that would bind to a DNA base.
2. The base in the amphiphile binds to another DNA base. Draw the three other options.
3. Which DNA base would the amphiphile bind best? Show how with a drawing.
4. The melting point of DNA increases by 2°C for every AT pair; 4°C for every CG pair. Assuming all of the bases in the amphiphile bind to DNA, predict the melting point of the adduct.
5. DNA actually binds more tightly to this amphiphile than to another DNA strand. Propose a reason why.
Exercise \(3\)
Problem IB5.3.
Liposomes are artificially prepared vesicles made of lipid bilayers. A drawing of a liposome is shown below along with its structural relative, a micelle.
1. Compare and contrast a liposome with a micelle.
2. What type of molecules would be contained in the center of a liposome?
3. What type of molecules would be contained in the center of a micelle?
1. Liposomes are used for drug delivery due to their unique properties. A liposome encapsulates a drug that cannot easily pass through the lipid bilayer.
2. A liposome can carry both hydrophobic molecules and/or hydrophilic molecules:
4. [ Hydrophobic / Hydrophilic ] chemicals can be dissolved into the membrane. (Choose one.)
5. [ Hydrophobic / Hydrophilic ] chemicals can be dissolved into the liposome's center. (Choose one.)
Another interesting property of liposomes is their ability to target cancer. Tumors do not have tight junctions between cells and are leaky. Liposomes of certain sizes, typically less than 200 nm, can rapidly enter tumor sites from the blood, but are kept out of healthy tissue by the endothelial wall. Anti-cancer drugs such as Doxorubicin and Camptothecin (shown below) are currently being marketed in liposome delivery systems.
f) Identify each of the circled functional groups.
g) To encapsulate these drugs in liposomes, you will need to predict whether they are water soluble or not. Choose the best attribute for each compound.
Doxorubicin [water-soluble / insoluble in water]
Camptothecin [water-soluble / insoluble in water]
To deliver the molecules to sites of action, the lipid bilayer can fuse with other bilayers such as a cell membrane, thus delivering the liposome contents.
Since cholesterol can improve the stability of lipid bilayers, some researchers designed and synthesized a category of phospholipids that incorporate cholesterol covalently linked to the glycerol backbone of phosphatidylcholine (shown below).
h) On the structure of the phosphatidylcholine, put a circle around the polar head, a rectangle around the cholesterol group, and a dashed ellipse around the fatty acid group.
i) Draw a picture of the lipid bilayer formed with this phospholipid (use the cartoon on the right) vs a regular phospholipid with cholesterol added in.
Data on animal models suggest that liposomes formed with the sterol modified phospholipid are very effective for delivery of anti-cancer agents.
j) Suggest at least two advantages to the sterol modified liposomes. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/08%3A_Introduction_to_Biomolecules/8.05%3A_Biomolecule_Application_Problem.txt |
You have probably studied the cell many times, either in high school, or in college biology classes. There are many websites available that review both prokaryotic (bacterial), archeal and eukaryotic cells (see links at bottom). This tutorial is designed specifically from the viewpoint of chemistry. It explores four classes of biomolecules (lipids, proteins, nucleic acids and carbohydrates) and describes in a simplified pictorial manner where they are found, made, and degraded in the cell (i.e. their history). It focuses on eukaryotic cells, which in contrast to simpler prokaryotic cells, have internal organelles surrounded by membranes that compartmentalize chemical reactions. First a general overview of the cell is presented. Subsequent chapters will concentrate on specific biomolecules.
Let's think of a cell as a chemical factory which designs, imports, synthesizes, uses, exports and degrades a variety of chemicals (in the case of cell, these include lipids, proteins, nucleic acids and carbohydrates). It also must determine or sense the amount of raw and finished chemicals it has available and respond to its own and external needs by ramping up or shutting off production. The figure of the cell below and in other linked sites based on it was made available with the kind permission of Liliana Torres.
http://torresbioclan.pbworks.com/w/page/22377234/Spikefish%20About%20Cells
Use with permission from Liliana Torres . Also at www.animalport.com/animal-cells.html
Design - The design for a cell mostly resides in the blueprint for the cell, the genetic code, which is comprised of the DNA in the cell nucleus and a small amount in the mitochondria. Of course, the DNA blueprint must be read out (transcribed) by protein enzymes which themselves were encoded by the DNA. The genetic code has the master plan that determines the sequence of all cellular proteins, which then catalyze almost all other activities in the cell, including catalysis, motility, architectural structure, etc. In contrast to DNA, RNA and protein polymers, the length and sequence of a polysaccharide polymer is not driven by a template but rather by the enzymes that catalyzes the polymerization.
Import/Export: Many of the chemical constituents of the cell arise not from direct synthesis but from import of both small and large molecules. The imported molecules must pass through the cell membrane and in some cases through additional membranes if they need to reside inside membrane-bound organelles. Molecules can move into the cell by passive diffusion across the membrane but usually their movement is "facilitated" by a membrane receptor. Molecules can also move up a concentration gradient in a process called "active transport". Given the amphiphilic nature of the bilayer (polar head group exterior, nonpolar interior), you would expect that polar molecule like glucose would have difficulty in moving across the membrane. In fact membrane transport proteins are involved in the movement of both nonpolar and polar molecules.
- transporters, carrier proteins and permeases: These membrane proteins move specific ligand molecules across a membrane, typically down a concentration gradient. Computer simulations of the facilitated diffusion of lactose across the membrane is shown in the following link. Animation of lactose diffusion through the LacY receptor
(The link above and immediately below are from the Theoretical and Computational Biophysics group at the Beckman Institute, University of Illinois at Urbana-Champaign. These molecular dynamic simulations were made with VMD/NAMD/BioCoRE/JMV/other software support developed by the Group with NIH support.)
- ion channels - These membrane proteins allow the flow of ions across membranes. Some are permanently open (nongated) while others are gated open or closed depending on the presence of ligands that bind the protein channel and the local environment of the protein in the membrane. Flow of ions through the channel proceeds in a thermodynamically favored direction, which depends on their concentration and voltage gradients across the membrane.
- pores: Some membranes (nuclear, mitochondria) assemble proteins (such as porins) to form large, but regulated pores. Porins are found in mitochondrial membranes while nucleoporins are found in the nuclear membrane. Small molecules can generally pass through these membrane pores while large one are selected based on their tendency to form transient intermolecular attractive forces with the pore proteins. The following link shows the diffusion of water through aquaporin. animation of water diffusion through the aquaporin channel,
- endocytosis: Very large particles (for example, LDL or Low Density Lipoproteins or viruses) can enter a cell through a process called endocytosis. Initially the LDL or virus binds to a receptor on the surface of the cell. This triggers a series of events which leads to the invagination of the cell membrane at that point. This eventually pinches off to form an endosomal vesicle which is surrounded by a protein called clathrin. "Early" endosomes can pick up new proteins and other constituents as well as shed them as they move and mature through the cell. During this maturation process, protein pumps in the endosome lead to a decrease in the endosomal pH which can lead to conformation changes in protein structure and shedding of proteins. Eventually the "late" endosome reaches and fuses with the lysosome, an internal organelle which contains degradative enzymes. Undegraded components like viral nucleic acids or cholesterol, are delivered to the cell. This transport can also go in the reverse direction and recycle receptors to the cell membrane. Likewise vesicles pinched off from the Golgi complex can fuse with endosomes, with some components surviving the process to reenter the Golgi.
Synthesize/Degrade: Cells have to both synthesize and degrade small molecules as well as larger polymeric protein, carbohydrates and nucleic acids. The anabolic (synthetic) and catabolic (degradative) pathways are often compartmentalized in time and space within a cell. For example fatty acid synthesis is carried out in the cytoplasm but fatty acid oxidation is carried out in the mitochondria. Proteins are synthesized in the cytoplasm or completed in the endoplasmic reticulum (for membrane and exported proteins) while they are degraded in the lysosome or more importantly in a large multimolecular structure in the cell called the proteasome.
Let's consider some key characteristics of a cell before we get into the details.
Cells and their internal compartments have regulated concentrations of ions and hydronium ion.
As expected the pH of the cytosol (the aqueous substance surrounding all the organelles within the cell) varies from about 7.0-7.4, depending on the metabolic state of the cell. Some organelles have proton transporters that can significantly alter the pH inside an organelle. For example the pH inside the lysosome, a degradative organelle, is about 4.8. The collapse of the pH gradient across the inner mitochondrial membrane is sufficient to drive the thermodynamically unfavored synthesis of ATP.
Compared to the extracellular fluid, the concentration of potassium ion is higher inside the cell, while concentrations of sodium, chloride and calcium ions are higher on the outside of the cell (see table below). These concentration gradients are maintained by ion transporters and channels and require energy expenditure ultimately in the form of ATP hydrolysis. Changes in these concentrations are integral to the signaling system used by the cell to sense and respond to changes in its external and internal environments.
Water is found in bulk form as well as bound to macromolecules like proteins and polysaccharides. These waters would be expected to have different properties. The table below show approximate ion concentrations in the cell.
Ion Inside (mM) Outside (mM)
Na+ 140 5
K+ 12 140
Cl- 4 15
Ca2+ 1 uM 2
Cells have an internal framework that provides architectural and internal structural support.
The "cytoskeletal" architecture of a (with molecular "cables"- and "girder-like" structures) is not dissimilar from a factory.
The internal framework or cytoskeleton of a cell, is composed of microfilaments, intermediate filaments, and microtubules. These are comprised of monomeric proteins which self assemble to form the internal architecture. Parts of the cytoskeleton can be seen in the photo above (taken from Wikipedia).
Microfilaments of actin monomers (which are stained with a red/orange fluorophore) and microtubules which offer more structural support made of tubulin monomers (stained green) along with the blue-stained nucleus, are shown in the image. Organelles are supported and organized by the cytoskeleton (primarily microtubules). Even the cell membrane is supported underneath the inner leaflet by actin (stained orange) and spectrin microfilaments. Motor proteins like myosin (that moves along actin microfilaments) and dynein and kinesin (that move along tubulin microtubules) carry cargo (vesicles, organelles) in a directional fashion. The cell is not a disorganized collection of molecules and organelles. Rather is a highly organized for optimal chemical production, use and degradation.
Cell have a variety of shapes. Some circulating immune cells must slip through the cells that line capillary walls to migrate to sites of infection. The same process occurs when tumor cells metastasize and escape to other sites in the body. In order to do so, the cell must drastically change shape, a response that requires dissociation of the cytoskeleton polymers into monomers which are available later for repolymerization.
The cell is an amazingly crowded place.
In chemistry labs, we typically work with dilute solutions of solute molecules in a solvent. You have probably heard that the body is comprised of 68% water, but the water concentration is obviously dependent on the cellular environment. Solute molecules like protein and carbohydrates are densely packed. Cells are so crowded that the space between larger molecules like protein is less than the size of protein. Studies have shown that the stability of a protein is increased in such condition, which would help keep the protein in the folded, native state. Another consequence of high concentrations is that it would promote the binding or self aggregation of like molecules as well as dislike ones which from an equilibrium perspective would not occur in dilute solutions. Hence the study of biomolecules in dilute solutions in the lab may not reveal the actual complexities of interactions and activities of the same molecule in vivo. Recently investigators have added a neutral copolymer of sucrose and epichlorhydrine to cells in vitro. These particles induced organization of extracellular molecules secreted by cell, forming an organized extracellular "matrix" which induced the organization of the microfilaments on the inside of the cell as well as inducing changes in cell activity.1 In vitro enzyme activity of a key enzyme in glycolysis dramatically increases under crowded conditions.2 Another result of crowding may be the spatial and temporal association of key enzymes in specific metabolic pathways, allowing easier diffusion of substrates and products within the colocalized enzymes.
The computer simulation used 50 different types of the most abundant macromolecules of the E. coli cytoplasm and 1008 individual molecules. Rendering of the cytoplasm model at the end of a dynamics simulation. RNA is shown as green and yellow. This figure was prepared with VMD
Cell components undergo phase transitions to form substructures within the cell.
A perplexing question is how do substructures form within a cell. This includes not only the biogenesis of organelles like mitochondria but also smaller particle such as polysaacharide granules, lipid droplets, protein/RNA particles (including the ribosome) as well as the nucleolus of the cell nucleus. It might be easiest to consider this problem using two examples from the lipid world, lipid droplets and membrane rafts. You are very familiar with phase transitions that occur when a sparing soluble nonpolar liquid is added to water. At a high enough concentration, the solubility of the nonpolar liquid is exceeded and a phase transition occur as evidenced by the appearance of two separate liquid phases. The same process occurs when triglycerides coalesce into lipid droplets with proteins associated on their outside. Another example occurs within a cell membrane when lipids with saturated alkyl chains self associate with membrane cholesterol (which contains a rigid planar ring system) to form a lipid raft characterized by greater packing efficiency, rigidity and thickness that other parts of the membrane. These lipid rafts often recruit protein involved in signaling processes within the cell membranes. This process of phase separation is also called liquid/liquid demixing as two "liquid-like" substances separate.
It appears that proteins that interact with RNA are composed of less diverse amino acid and have more flexible ("more liquid-like) structures allowing their preferential interaction with RNA to form large RNA-protein particles (like the ribosome and other RNA processing structures) in a fashion that mimics liquid/liquid demixing. All of these interactions are just manifestations of the various intermolecular forces you studied in earlier chemistry classes.
You have studied lipid, protein, nucleic acid and carbohydrate structure in introductory biology and/or chemistry classes.
A list of web sites and applets on eukaryotic cell structure:
• Cell Tutorial from UCLA
• Plant and animal cell simulation (requires Java)
• Download Apple Store Applet for IPad, Itouch, IPhone: 3D Cell
References:
1. Zeiger AS, Loe FC, Li R, Raghunath M, Van Vliet KJ (2012) Macromolecular Crowding Directs Extracellular Matrix Organization and Mesenchymal Stem Cell Behavior. PLoS ONE 7(5): e37904. doi:10.1371/journal.pone.0037904 .
2. Apratim Dhara, Antonios Samiotakisb, Simon Ebbinghausa, Lea Nienhausa, Dirar Homouzb, Martin Gruebelea, and Margaret S. Cheungb. Structure, function, and folding of phosphoglycerate kinase are strongly perturbed by macromolecular crowding. Proc. Natl Acad Sci. www.pnas.org/cgi/doi/10.1073/pnas.1006760107
3. Anthony A. Hyman and Kai Simons. Beyond Oil and Water--Phase Transitions in Cells. Science 337, 1047 (2012); DOI: 10.1126/science.1223728
4. McGuffee SR, Elcock AH (2010) Diffusion, Crowding & Protein Stability in a Dynamic Molecular Model of the Bacterial Cytoplasm. PLoS Comput Biol 6(3): e1000694. doi:10.1371/journal.pcbi.1000694
9.02: The Cell- Lipids
French Translation √ by Anna Chekovsky
http://torresbioclan.pbworks.com/w/page/22377234/Spikefish%20About%20Cells
Use with permission from Liliana Torres . Also at www.animalport.com/animal-cells.html
Lipids are small biological molecules which are soluble in organic solvents, such as chloroform/methanol, and are sparingly soluble in aqueous solutions. The simple classification of lipids belies the complexity of possible lipid structures as over 1000 different lipids are found in eukaryotic cells. This complexity has led to the development of a comprehensive classification system for lipids. In this system, lipids are given a very detailed as well as all-encompassing definition: "hydrophobic or amphipathic small molecules that may originate entirely or in part by carbanion-based condensations of thioesters (fatty acyl, glycerolipids, glycerophospholipids, sphingolipds, saccharolipds and polyketides) and/or by carbocation-based condensations of isoprene units (prenol lipids and sterol lipids)."
The following site will show you the "life" history of lipids in cells.
The sites linked above are written and maintained by Henry V. Jakubowski, Ph.D., College of Saint Benedict / Saint John's University. These materials are available for educational use.
Send corrections to [email protected]
9.03: The Cell - Proteins
Henry Jakubowski
http://torresbioclan.pbworks.com/w/page/22377234/Spikefish%20About%20Cells
Use with permission from Liliana Torres . Also at www.animalport.com/animal-cells.html
Proteins are polymers of bifunctional alpha-amino acids each containing a different R group at the alpha carbon. Encoded by the DNA genome, proteins are involved in all cellular activities including recognition, catalysis, movement, and signalling.
The following site will show you the "life" history of proteins in cells.
The sites linked above are written and maintained by Henry V. Jakubowski, Ph.D., College of Saint Benedict / Saint John's University. These materials are available for educational use.
Send corrections to [email protected]
9.04: The Cell- Nucleic Acids
Henry Jakubowski
http://torresbioclan.pbworks.com/w/page/22377234/Spikefish%20About%20Cells
Use with permission from Liliana Torres . Also at www.animalport.com/animal-cells.html
Nucleic acids are polymers of deoxyribonucleotides (DNA) or ribonucleotides (RNA). DNA contains the genetic code that when transcribed into RNA followed by translation of the RNA for around 22,000 protein. Only about 2% of the 3.2 billion base pairs of DNA encodes protein. The function of 80% of the remaining genome has recently been found to regulate the transcription of DNA, mostly through various types of RNA transcribed from the DNA.
The following site will show you the "life" history of nucleic acids in cells.
The sites linked above are written and maintained by Henry V. Jakubowski, Ph.D., College of Saint Benedict / Saint John's University. These materials are available for educational use.
Send corrections to [email protected]
9.05: The Cell- Carbohydrates
Figure \(1\)
http://torresbioclan.pbworks.com/w/page/22377234/Spikefish%20About%20Cells
Use with permission from Liliana Torres . Also at www.animalport.com/animal-cells.html
Carbohydrates are A.
The following site will show you the "life" history of nucleic acids in cells. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/09%3A_Cell_Tutorial/9.01%3A_The_Cell-_An_Overview.txt |
• 10.1: A Network Solid - Diamond
Some materials don't fall into any of the categories that we have seen so far. They are not metals, so they can't be described as lattices of identical ions surrounded by delocalized electrons. They are not ionic solids, so they can't be thought of as arrays of one type of ion with the counterions packed into the interstitial holes to balance the charge. They are not molecules, so we wouldn't draw them as discrete, self-contained collections of connected atoms. Diamond, for example, is a networ
• 10.2: Other Forms of Carbon
Diamond is just one form in which elemental carbon can be found. Other forms include graphite, lonsdaleite and fullerenes.
• 10.3: Silicates and Silica
About 90% of the earth's crust is composed of silicate minerals. Silicates are compounds composed of silicon and oxygen; these compounds have negative charges on them. That means there are positive counterions found together with these anions. The variety of possible cations, ranging from sodium and potassium to copper and iron, is one of the factors that leads to a dazzling array of silicate minerals.
• 10.4: Aluminosilicates
Aluminosilicates are compounds containing oxides of both silicon and aluminum. These compounds are quite common in the earth's crust, so it is worth taking a look at a few examples. Two of the most prevalent types are phyllosilicates, with extended layered structures, and tectosilicates, with structures that extend through a three dimensional network of covalent bonds.
• 10.5: Solutions to Selected Problems
10: Network Solids
Some materials don't fall into any of the categories that we have seen so far. They are not metals, so they can't be described as lattices of identical ions surrounded by delocalized electrons. They are not ionic solids, so they can't be thought of as arrays of one type of ion with the counterions packed into the interstitial holes to balance the charge. They are not molecules, so we wouldn't draw them as discrete, self-contained collections of connected atoms.
Diamond, for example, is a network solid. Diamond is an allotrope of carbon - it is one of several forms of elemental carbon found in nature. It looks something like this, on an atomic scale. The lines are bonds between the carbon atoms. Each carbon atom makes four bonds, one to each of four different neighbors.
Diamond is composed entirely of carbon atoms, but carbon is too electronegative to allow its electrons to completely delocalize into an electron sea, like metallic elements do. It forms a crystalline, solid structure, but it does not dissolve even a tiny bit in water like ionic compounds would, because it has no ions to form ion-dipole interactions with the water molecules. It forms covalent bonds with its neighboring atoms, sharing these electrons rather than exchanging them, but it forms an extended solid rather than individual units.
• Network solids are like molecules because they have covalent bonds connecting their atoms.
• Network solids are unlike molecules because they don't have a limited or specific size; their structures can extend "seemingly forever" on the atomic scale.
What do most people know about diamonds? They are very expensive, of course. They are very shiny. They are very, very hard. In fact, there is a scale used by minerologists to describe the hardness of materials called "Mohs scale of hardness". The scale was developed by German geologist Friedrich Mohs in the early 1800's. It simply places ten different minerals in order from softest to hardest and assigns each of them a number. Diamond is a 10, meaning it is the hardest material on the scale. Diamond is, in fact, the hardest naturally-ocurring substance in the world.
Table \(1\): Mohs scale of hardness.
Mineral Hardness
talc 1
gypsum 2
calcite 3
fluorite 4
apatite 5
orthoclase 6
quartz 7
topaz 8
corundum 9
diamond 10
Mohs scale is qualitative, not quantitative; each mineral in the scale is harder than the one before it. Mohs scale has been widely used by field geologists because of its simplicity. If you pick up a mineral and can scratch it with that diamond you keep in your toolbox, but the corundum does not leave a mark, then the hardness of the new sample is around 9.
Exercise \(1\)
Rank the following minerals using Mohs scale.
1. graphite, which scratches talc but is itself scratched by gypsum
2. stishovite, which can scratch corundum but not diamond; diamond scratches stishovite
3. opal, which scratches orthoclase and fluorite but not quartz or topaz
4. obsidian, which can be scratched by quartz and orthoclase but not apatite; obsidian does not scratch apatite, either
5. ruby, which is scratched by diamond; however, neither corundum nor ruby can scratch each other.
6. emerald, which can be scratched by ruby and topaz but scratches quartz
Answer a:
graphite is between 1 & 2
Answer b:
stishovite is between 9 & 10
Answer c:
opal is between 6 & 7
Answer d:
obsidian is about 5
Answer e:
ruby is about 9
Answer f:
emerald is between 7 & 8
Exercise \(2\)
"Absolute hardness" can be measured in a number of ways. For example, a sample can be scored by a diamond needle using carefully controlled pressure, and the width or depth of the score can be measured to get a quantitative comparison of the hardness of different materials.
Plot the hardness of each material according to Mohs scale on the x axis and the hardness determined by a sclerometer test to see the relationship between Mohs scale and absolute hardness.
material
hardness
talc 1
gypsum 3
calcite 9
fluorite 21
apatite 48
orthoclase 72
quartz 100
topaz 200
corundum 400
diamond 1600
What makes diamond so hard? Let's compare it to a few of the softer minerals in the scale. Gypsum is a crystalline hydrate of calcium sulfate: CaSO4.2H2O. Calcite is a mineral form of calcium carbonate: CaCO3. Fluorite is composed of calcium fluoride: CaF2.
These materials are all ionic solids. Gypsum contains calcium cations and sulfate anions, as well as bound water. Calcite contains calcium ions and carbonate ions. Fluorite contains calcium ions and fluoride ions. Surely the ionic bonds that hold these ions together are very strong. Why can they be deformed and scratched?
An ionic bond is not directional. It does not matter whether a sulfate ion is above or below a calcium ion, or to the left or the right of it. The attraction between the ions is still the same.
If we imagine a shearing force on a crystal of calcium sulfate, meaning that we are pushing against only one layer of the crystalline material, then we might see that layer slide in response to the force. As the ions in that layer slide, they begin to lose their attraction to the ions in the layer beneath them, but then they become attracted to new ions that are sliding towards. When we are finished, we still have an ionic solid in which ions are atracted to counterions around them, but the partners have changed.
We can't do that in diamond because the carbon atoms are all covalently bonded to each other. The electrons in those bonds appear to be localized in particular regions of space around the carbon atoms, making a framework of tetrahedra. That strong framework holds each carbon atom in place with four bonds, so that it is difficult to displace any one atom or layer of atoms.
Silicon: an analog of carbon
Silicon is found below carbon in the periodic table. Consequently, it has some chemical similarities with its neighbor. For example, pure silicon adopts a structure that is very similar to that of diamond.
Elemental silicon, Si, is indispensable for modern electronics because, in the presence of trace impurities, it has semiconducting properties. However, elemental silicon really does not exist in nature. It must go through chemical reduction from silica, SiO2, in an electric arc furnace:
\[\ce{SiO2 + 2C -> Si + 2CO}\]
The resulting silicon is painstakingly purified by a number of chemical processes. For example, it can be converted to SiCl4, which is then purified repeatedly via distillation to remove trace impurities, then turned back into silicon by reduction with zinc:
\[\ce{SiCl4 + 2Zn -> Si + 2 ZnCl2}\]
One of the final steps is recrystallization of the silicon from its molten phase. A rod with a small "seed" of pure silicon is dipped into the molten silicon and slowly drawn upwards. As the molten silicon cools against the cooler surface of the rod, it recrystallizes. The rod is pulled slowly upward, and pure silicon continues to grow on the cooler, crystalline silicon being pulled from the melt, forming a long crystal.
Exercise \(3\)
Draw the structure of elemental silicon.
Exercise \(4\)
Silicone is not the same thing as silicon. Silicone is a class of polymer. It exists as long, flexible chains in which the same group of atoms is repeated over and over. There are lots of kinds of silicone, but one example of a silicone formula is ((CH3)2SiO)n. The subscript n means the unit in parentheses is repeated over and over in a chain.
1. Draw a section of silicone several units long.
2. Explain how the structures of silicon and silicone might give them very different physical properties. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/10%3A_Network_Solids/10.01%3A_A_Network_Solid_-_Diamond.txt |
Diamond is just one form in which elemental carbon can be found. Other forms include graphite, lonsdaleite and fullerenes.
Graphite
Graphite is a mineral that, like diamond, is composed of carbon. However, the carbon atoms in graphite are arranged in a different way than they are in diamond. There are significant consequences for the relative properties of diamond and graphite.
Graphite is not nearly so hard as diamond. It has a hardness of about 1.5 on the Mohs scale, compared to diamond's 10 rating. That difference is related to the fact that graphite is composed of separate sheets, called graphene sheets, whereas diamond is firmly bonded in three dimensions. A force exerted on graphite can easily displace one layer away from its neighbors.
However, graphite isn't equally hard in all directions. If a force is exerted parallel to the graphene sheets, one sheet can be moved with respect to the next one. That isn't true perpendicular to the sheets. On the picture, the sheets could slide on the x and y directions but not in the z direction.
That general idea is true in other materials. Even diamond is not equally hard on all surfaces. Minerals often have typical shapes because they are cleaved more easily along specific planes, but are harder to break along others.
Other properties also depend on the plane of the crystal in network solids. For example, graphite is also a conductor of electricity along its graphene planes, but not perpendicular to them.
Lonsdaleite
Lonsdaleite is named in honour of Kathleen Lonsdale, the Irish crystallographer who first demonstrated the planar structure of benzene via X-ray diffraction. It is a very rare form of carbon generated during the impact of meteors that contain graphite. In the high energy of impact, layers of graphite fuse together to form a structure that at first glance looks a lot like diamond. A rough sketch of the lonsdaleite structure is depicted below.
Lonsdaleite formed in nature is much softer than diamond, probably because the chaotic nature of its origin leaves lots of graphite-like domains embedded in the lonsdaleite. Instead of having bonds regularly spaced throughout the material, there are probably flawed portions in which part of one layer is still unconnected from the next. It has a hardness of about 8 on the Mohs scale.
Fullerenes
The fullerenes are a form of elemental carbon that really does not fall completely under the category of network solids. They are a little like network solids but they do form discrete units, and so they are really molecules. A fullerene is a little bit like a small sheet of graphite that has been curved into a ball. The shapes of the spherical ones resemble the Geodesic Domes designed by architect Buckminster Fuller in the 1960's; that's where they get their name. Spherical fullerenes are also called buckyballs.
Fullerenes were first discovered in the 1990's by Harold Kroto. There was an ensuing rush to develop new applications for the materials. For example, people have proposed using buckyballs as lubricants (like little ball bearings, they would roll between two surfaces). Much of the interest in this area, however, has focussed on potential electronic properties of these materials.
Carbon nanotubes are conceptually similar to fullerenes. Instead of being spherical, they are tube-shaped. They can also be though of as graphite sheets that have been curved into a new shape.
Coal and Other Forms of Carbon
Many other forms of carbon that are commonly found on earth are not pure carbon. We often think of coal as being equivalent to carbon, a notion that is reinforced by the common knowledge that diamonds are a result of immense geological pressures and high temperatures acting on coal seams over long periods of time. Coal actually comes in many different grades containing varying amounts of carbon.
Lignite or "brown coal" can contain as little as 60% carbon. The rest of it is composed of varying amounts of hydrogen, oxygen, nitrogen and sulfur -- all of which are common elements found in living or "organic" things. The most valuable grade of coal, anthracite, contains over 90% carbon. There are varying grades in between.
Why does coal contain all of those elements found in living things? Coal forms from the compression of organic material over geological time scales. Most of the atoms present in that organic material are still present in the coal.
Other non-living but organic materials are important ecologically and agriculturally. As plant and animal matter decays, it turns eventually into "humus" (hyoo-muss) or the organic material of the soil. Humus comes from those portions of organic material that are not easily broken down. Humus contains a significant amount of carbon, as well hydrogen and oxygen. Different kinds of soils contain different levels of humus, which confers varying properties on the soil such as moisture- and nutrient-retention.
Coal and humus are not exactly network solids. What fullerenes have in common with network solids is a regular, crystalline pattern of covalent bonds extended across the solid, but they are limited in size. Coal and humus demonstrate an opposite case: they do not have a limited size and their bonded networks can extend seemingly endlessly, but there is nothing regular or repeating about their structure. Instead, these materials are "amorphous" extended solids.
• Crystalline materials display regularly repeating patterns in three dimensions
• Amorphous materials do not display regular patterns in any direction; they are arranged seemingly at random | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/10%3A_Network_Solids/10.02%3A_Other_Forms_of_Carbon.txt |
About 90% of the earth's crust is composed of silicate minerals. Silicates are compounds composed of silicon and oxygen; these compounds have negative charges on them. That means there are positive counterions found together with these anions. The variety of possible cations, ranging from sodium and potassium to copper and iron, is one of the factors that leads to a dazzling array of silicate minerals.
In addition to having a wide range of possible cations, silicate anions themselves have a breadth of available structures. These structures range from individual silicate anions (SiO44-) to three dimensional networks of tetrahedra.
Silicates are not necessarily network solids, but we will see that their structures range from straightforward ionic solids at one extreme to something that looks more and more like an extended network at the other. By looking at these materials we can learn a little bit more about the related network solids.
Nesosilicates
Nesosilicates (meaning "island" silicates) are individual silicate anions arranged in crystalline ionic solids with their counterions. Those individual ions mean that nesosilicates are not network solids at all. They represent the extreme, other end of the silicate spectrum of structures.
The nesosilicate anion forms a tetrahedral shape. The silicon atom is at the center and the oxygen atoms are at the four corners. That tetrahedral shape is often illustrated in structural drawings rather than drawing the atoms explicitly. Instead of showing the SiO44- anion with labelled atoms on the left, the tetrahedron on the right is shown to represent it.
Silicates can have lots of different cations. Frequently, several different cations may be found in one material. For example, olivine is a pretty common nesosilicate with magnesium and iron cations (Mg2+ and Fe2+). If the olivine is of very high quality, it can be a gemstone, called peridot.
There are two limiting forms of olivine, called "endmembers". At one end is forsterite, Mg2SiO4. At the other end is fayalite, Fe2SiO4. In between is every possible combination of magnesium and iron. There could be equal amounts of magnesium and iron, or there could be just a few magnesium ions and many, many iron ions in a sample.
• Minerals frequently occur in families, in which one, similar ion can be substituted for another.
• Frequently, both ions can be present in varying amounts, but the properties of the material (such as color) change with changing ratios of these ions.
• The minerals that have just one of the possible ions but not the other are called the endmembers of the family.
Another common nesosilicate is garnet. Garnets are used as gems as well as abrasives -- they are sometimes used in sandpaper. There are many kinds of garnet, but a common one is almandine, the red-brown color of which we usually think of as "garnet". Garnets always contain two different cations. One of them is a M2+ ion and the other is a M3+ ion (M here just stands for "metal"). These two counterions are always present in a 3:2 ratio in garnet.
Exercise $1$
Show that forsterite and fayalite are overall neutral.
Answer a:
forsterite: Mg2SiO4: If each nesosilicate unit has a charge of 4- because of the formal charges on the four oxygens, then the two magnesiums must combine to balance that charge with a 4+ charge. Each magnesium must have a 2+ charge of its own. That seems reasonable, since magnesium is two atoms from the left edge of the periodic table; it would have a 2+ charge in a noble gas configuration.
Answer b:
fayalite: Fe2SiO4: In order for this to work, each iron would have to have a 2+ charge. Transition metal charges are harder to predict than alkali and alkaline earth metals because they are farther from the edge of the periodic table, but it turns out that the two most common charges on an iron ion are 2+ and 3+, so this charge seems reasonable.
Exercise $2$
Suppose an olivine has 100 nesosilicate anions for every 25 magnesium ions.
1. How many iron ions (assume Fe3+) would there be in the neutral olivine?
2. What would be the values of x and y in MgxFeySiO4?
Answer a:
If each nesosilicate has a charge of 4-, then the total negative charge is 100 x 4- = 400-. If there are 25 magnesium ions, each with a 2+ charge, the total positive charge is 25 x 2+ = 50+. That leaves a net negative charge of 400- + 50+ = 350-. The irons would balance that charge with an equal positive charge. The number of iron ions would be 350+/3+ = 150.
Answer b:
If we divide these ratios by 100 to arrive at the one nesosilicate in the formula, we get Mg0.25Fe1.5SiO4. It doesn't mean there is a quarter of a magnesium ion anywhere; this formula is just the overall ratio of atoms.
Exercise $3$
Almandine is a nesosilicate. It contains aluminum (which is almost always Al3+ in nature) and iron (in this case Fe2+) in a 2:3 ratio. What is its formula?
Answer
So far we have Al2Fe3. The positive charge = 2 x 3+ + 3 x 2+ = 12+. That charge must be balanced by the nesosilicate anions, which are each 4-. So the number of silicates is 12-/4- = 3. The formula is Al2Fe3(SiO4)3.
Sorosilicates
Sorosilicates ("sister" silicates) are silicate anion dimers. The two silicate units share an oxygen atom. In the polyhedral depiction, the two tetrahedra share a corner. Sorosilicates clearly aren't network solids, either. They just contain slightly bigger anions than nesosilicates.
Sorosilicates are not nearly as common as nesosilicates. Where they exist, sorosilicates are found in the presence of other anions, as well as cations. In some minerals, they are even found in combination with nesosilicates. For example, epidote is a more common form of sorosilicate-containing mineral. It could be a number of colors but is usually green. It contains nesosilicate, sorosilicate, oxide and hydroxide anions, as well as calcium, iron and aluminum cations.
The sorosilicate anion can actually adopt different shapes in different cases. It contains two units that can twist relative to each other. The different shapes that result are called "conformations". A different conformation is shown below. In the picture on the left, we are looking at it from the same angle as the orosilicate shown above. In the picture on the right we are looking at it from a different angle to emphasize that the two tetrahedra are pointing in opposite directions. We are looking at the top of one tetrahedron but the bottom of the other one.
Exercise $4$
Epidote contains just one of each of its four different anions in its formula.
1. Write the formula, with appropriate charge, for each of its anions. Note that the oxygen atoms in the hydroxide (OH) and oxide (O) ions each have an octet.
2. Epidote also contains two Ca2+ ions in its formula, but varying amounts of Fe3+ and Al3+. How much iron and aluminum, total, does it have in its formula?
Answer a:
The anions are nesosilicate (4-) + sorosilicate (6-) + oxide (2-) + hydroxide (1-). The total charge is 13-.
Answer b:
Calcium is an alkaline earth metal, two atoms from the left edge of the periodic table. The two calciums would partially balance that charge with 2 x 2+ = 4+ charge; the net charge is now 13- + 4+ = 9-.
If each Al or Fe has charge 3+, then the number needed to balance the charge is 9+/3+ = 3 Fe/Al ions.
Exercise $5$
Use your three-dimensional drawing tools (lines, wedges and dashes) to show the two different conformations of sorosilicate via Lewis structures.
Neither nesosilicates not sorosilicates are network solids. They are really just families of ionic solids. However, by looking at the relationship between the nesosilicate ion and the sorosilicate ion, we can see why many silicates do start to look more like network solids. If one oxygen atoms can be shared between two silicate units, maybe more can, too.
Inosilicates
If two silicate units can share an oxygen, more can, too. In fact, long chain silicates can form. To do so, two of the oxygens in a silicate unit have to be shared with two other silicate units. The same is true in the next unit, and so on, forming a long chain. Materials containing these long chain silicates are called inosilicates ("sinew" or "string" silicates). The polyhedral drawing of a section of inosilicate ion is shown below.
Inosilicates are not exactly network solids because they do not form three-dimensional networks. However, at least in one direction, they are covalently bonded and can extend indefinitely. There could be hundreds of silicate units in a row. In a sense, inosilicates could also be described as polymers. They contain individual units that keep repeating all along the chain. In the polyhedral drawing, the repeating units are simply tetrahedra.
The inosilicate chain drawn above does illustrate an important consideration that it has in common with network solids. All network solids are different on the inside than they are on the outside. We can illustrate that difference in the inosilicate chain because it is a relatively simple structure. In the inosilicate chain, each tetrahedron along the chain looks just like the next, but the ones on the end are different. They are only attached to one neighbor, not two. Compared to the other repeating units, these ones will have an extra oxygen with a negative charge.
Extending that idea to network solids, in which the covalent bonds extend in three directions, we will encounter a similar problem as we move out from the center of the solid and we reach the surface. The structure at the surface will be different from the structure in the interior of the solid, because we will run out atoms to keep connecting to.
• Network solids have different structures at the surface of the material than they do in the interior of the material.
• The surface of the material must have slightly different groups present because the network has come to an end.
One of the most important inosilicates is wollastonite, which is used in a number of industries. Wollastonite is a bright white mineral in which the counterion for the inosilicate is calcium ion, Ca2+. It is added to ceramic tiles, paints and plastics, partly because it increases their durability and strength. It is sometimes used to make brake pads for cars; it provides friction when pressed against the rotor, slowing down the wheels on your car, but because of its durability it does not wear out too quickly.
Exercise $6$
Convert the polyhedral drawing of inosilicate ion into a Lewis structure. You can stop after five repeating units.
Exercise $7$
Looking at the Lewis structure for inosilicate ion, how many calcium ions would accompany each repeating unit in Wollastonite?
Exercise $8$
In your Lewis structure for inosilicate ion, circle the repeat units. Propose a formula for Wollastonite.
Exercise $9$
Suggest a reason why Wollastonite might confer strength on ceramics and other materials.
Cyclosilicates
If silicate units can form chains, it shouldn't be a surprise that they can form other shapes, too. For example, the silicate units can wrap around to form rings. They come in a veriety of different sizes. These types of silicates are called cyclosilicates ("wheel" silicates).
On of the most familar cyclosilicate mineral groups is the beryl family. The most prominent member of the family is emerald. It contains the cyclic anion shown below.
The polyhedral form is a gorgeous star.
All beryls have the same cyclosilicate anion. They even have the same counterions, with the general formula Be3Al2(Si6O18). However, beryls come in different colors because of small amounts of cation impurities. In emerald, a tiny amount of chromium ion provides the green color. In the closely related aquamarine, a tiny amount of Fe2+ results in a pale blue; however, when Fe3+ is present, the mineral is yellow, instead.
Exercise $10$
The California state gem, benitoite, contains a cyclosilicate anion with just three silicon atoms in it.
1. Draw the Lewis structure of this cyclosilicate.
2. Draw the polyhedral depiction of this cyclosilicate.
3. This anion is accompanied by a barium and a titanium ion. Given their places in the periodic table, what is the likely charge on each cation?
Clearly we have turned back from the edge of network solids again. These anions may be large, but they certainly don't extend on and on in any direction. However, previously, we looked at single unit silicates and extended them to doubles, then linked them together to make longer chains. What would happen if we could link some cyclosilicates together?
Instead of getting a one-dimensional chain, we would get a two-dimensional network.
Phyllosilicates
The name "phyllosilicates" means, roughly, "leaf"-silicates. These anions are sheets or leaves in which silicates link together to make two-dimensional networks. The structure shown below would extend on and on through the top, bottom and sides of the page (or the screen).
The most familiar examples of phyllosilicates are the mica family. If you have ever seen mica, with its thin sheets, you can easily imagine how it would be composed of two-dimensional networks like the one above. One of the most interesting applications of mica was its use as "Muscovy glass" in the old days in Russia. Its thin, transparent layers made very good windows.
Micas are a slightly more complicated example of phyllosilicates because they are also aluminosilicates. That is, they contain aluminum as well as silicon within their extended oxide network.
Another familiar example is chrysotile, which is one of many minerals that we think of as "asbestos". It is commonly called "white asbestos" and it accounts for the vast majority of asbestos mining in the world. Chrysotile has the formula Mg3(Si2O5)(OH)4.
Exercise $11$
Consider the formula for chrysotile.
1. What is the usual charge on a magnesium ion, given its place in the periodic table?
2. What is the charge on hydroxide ion?
3. The phyllosilicate ion is written as Si2O5, representing the overall ratio of those two atoms in the structure, but of course it does not come in units of that size. What is the total negative charge per Si2O5 unit?
4. What does that translate into in terms of negative charge per individual silicate unit (i.e. one individual silicate unit per silicon atom)? Explain whether that makes sense in terms of the polyhedral structure for a phyllosilicate shown above.
For decades, chrysotile was prized for its naturally insulating and fire-resistant qualities. It was commonly used in homes and other buildings until we realized that it produces a carcinogenic dust. Typically, the danger from this dust only arises from repeated exposure during the mining of asbestos and manufacture of asbestos-containing products. However, it could also be a long-term problem because of dust produced during building renovation projects.
Other, less common forms of asbestos include "blue asbestos" or crocidolite, with formula Na2FeII3FeIII2Si8O22(OH)2, and grey-to-brown amosite, (Mg,Fe)7Si8O22(OH)2.
Note that there are some complications in the way these formulae are written. There are two different kinds of iron ion in crocidolite. One is an Fe2+ ion and the other is an Fe3+ ion. Often, in structures and formulae, the charges on metal ions are denoted by the "oxidation state", a number written using Roman numerals. Hence, FeII means Fe2+ in this context and FeIII means Fe3+.
The formula for amosite is actually suggesting something that we have seen before. In amosite, iron and magnesium ions can easily replace each other in the structure. In this case, the ions are Mg2+ and Fe2+. The (Mg,Fe) just means we need some combination of these ions for a total of seven ions.
However, crocidolite and amosite are actually examples of inosilicates rather than phyllosilicates. They are a little bit different from the inosilicates that we saw before, however. Rather than containing single polysilicate chains, these materials contain double polysilicate ribbons.
In the picture, the inosilicate ribbon extends through the top and bottom of the page (or the screen) but not to the sides.
If you have ever seen asbestos, which is a fibrous mineral (a little like a clump of long hair), you may be able to imagine a connection between these structures and the physical appearance of the mineral.
Exercise $12$
Provide formulae for the endmembers of the amosite family.
Tectosilicates ("building" silicates) form three-dimensional frameworks.
In sorosilicates, individual silicate unit share oxygens with their neighbors via one of the corners of their tetrahedra. In inosilicates, individual silicate unit share oxygens with their neighbors via two of the corners of their tetrahedra. In phyllosilicates, individual silicate unit share oxygens with their neighbors via three of the corners of their tetrahedra.
In tectosilicates, individual silicate unit share oxygens with their neighbors via four of the corners of their tetrahedra.
These structures are pretty complicated and generally include aluminum as well as silicon oxides. However, an idealized drawing of such a network is shown below.
The idea of the drawing is that the silicate structure extends in two dimensions like the phyllosilicates, but also in a third direction, bonding to additional tetrahedra above and below the plane of the drawing.
Silica
Silica is truly a covalent network solid. Its formula is SiO2. Unlike silicates, which are anions, it is a neutral compound, so it does not need counterions to balance a charge. Like the tectosilicates, it forms a three-dimensional network, with one individual silicate unit sharing an an oxygen with its neghbours via all four corners of the tetrahedron.
Although it is a three-dimensional extended solid, it is useful to look for sub-structures within the network, just as it is useful to identify unit cells of ionic solids. In silica, a pattern arises by following "chains" of silica tetrahedra through the network. As we follow along from one tetrahedron to the next, we find a spiral pattern in which each tetrahedron is twisted relative to the one before it. Eventually, every fourth (or in some cases, seventh) tetrahedron echoes the original and the pattern repeats itself.
This spiral shape is reminiscent of DNA. Like DNA, the chains within the three-dimensional structure form helices. They could form right-handed helices or left handed helices. As a result, silica is a "chiral" material and it is "optically active", topics that are covered in stereochemistry.
As in other network solids, silica must have a different structure on its surface than in the interior. On the surface, there are no more individual silicate units with which to form links.
Instead, dangling oxygen atoms may form covalent bonds to hydrogen, so that the surface is covered with OH groups. They may also remain as siloxide ions just as they do in silicates; in that case, they might have counterions such as sodium or potassium.
However, compared to the bulk interior of the solid, the surface contributes very little to the overall mass of the material. As a result, these extra hydrogens and sodiums are not included in the formula of silica. In reality, the true formula of a sample of silica might be something like this (just a fictional example to emphasize the idea):
$\ce{Si1,022,000,000,000,000,000,000O2,044,000,000,000,000,000,000H36Na42} \nonumber$
but nobody would ever write that. The material is one millionth of a percent hydrogen, one millionth of a percent sodium, about 66.7% oxygen and approximately 33.3% silicon. It's pretty darned close to SiO2.
Silica is a very common material. Its most familiar pure form is quartz. In addition to being found as large crystals, microcrystalline particles of quartz is found in a wide rage of other minerals. It's also the major component of sand.
Exercise $13$
Zircon is a nesosilicate containing Zr4+ ion. What is its formula?
Exercise $14$
Propose a formula for a simple, single-chain inosite containing only sodium counterions.
Exercise $15$
Jadeite is beautiful green mineral; high quality stones are one of the forms of the gemstone, jade. Jadeite is a simple, single-chain inosite containing equal numbers of Na+ and Al3+ counterions. Propose a formula for this mineral. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/10%3A_Network_Solids/10.03%3A_Silicates_and_Silica.txt |
Aluminosilicates are compounds containing oxides of both silicon and aluminum. These compounds are quite common in the earth's crust, so it is worth taking a look at a few examples. Two of the most prevalent types are phyllosilicates, with extended layered structures, and tectosilicates, with structures that extend through a three dimensional network of covalent bonds.
Phyllosilicates
The clay minerals are a class of layered aluminosilicates formed by weathering of other silicate minerals. You are probably familiar with clay in some form or other. It is much finer-grained than sand, which is another product of weathering that is commonly found in the soil or exposed at beaches. Unlike sand,clay can sometimes be formed into shapes. We will take a look at two important examples.
The first example, kaolinite, is the major component of kaolin, a clay that is heavily used in making "china" and other ceramics. It is usually white but small amounts of iron can make it pink, yellow or orange. Kaolin contains hexagonally arrayed layers of silicate tetrahedra, as we have seen in other phyllosilicates. The difference is that this layer sits on top of a layer of aluminate octahedra, with some of the oxygens shared between the two layers.
In the picture below, the lines are meant to reinforce the idea of the geometries involved: tetrahedra are outlined in blue and octahedra are outlined in green. The bonds between silicon and oxygen are not shown, nor are the bonds between aluminum and oxygen. We have shown the tetrahedra on top and the octahedra below but there really is no top or bottom to the layer, so it can just as easily be drawn the other way around.
Kaolinite is an example of what is sometimes called a 1:1 clay mineral or a T-O clay mineral. These designations convey the idea that there is one tetrahedral layer and one octahedral layer bonded together by shared oxygen atoms.
If you take a look at the structure, you will see that some of the oxygen atoms are bonded to hydrogens; they are hydroxides. These hydroxides play an important role in the long-range structure of kaolin.
In the picture below, we have simplified the structural drawing to a cartoon. We have also expanded the view to see cross sections of two phyllosilicate sheets and to see a little bit further down each sheet.
In the drawing, the basal layer of hydroxides (that's the layer of OH groups on the bottom face of a sheet) interacts with the basal layer of oxides on the next sheet (that's the layer of oxides on the top of the sheet). These sheets are held together via hydrogen bonding. Of course, there are more sheets than that. You could easily picture a third layer above the ones that are shown, with its basal hydroxides hydrogen bonding with the oxygens of the tetrahedral layer on top of the drawing.
The other feature that is apparent in this drawing is that the long-range structure of the layer varies a little bit. Some of the octahedra are missing. Only 2/3 of the possible sites in the octahedral are actually occupied, although they are not necessarily spaced out evenly like in the drawing. There are other clay minerals that don't have these absences, although we aren't going to look at any here.
One of the unique features in kaolinite is an electrically neutral layer. If we were to consider all of the oxygens to be O2-, with the cations Si4+, Al3+ or H+ based on the positions of these atoms in the perioidc table, then the positive and negative charges would exactly balance each other. That contrasts with what we will see in the other cases that we will look at.
Exercise \(1\)
Kaolinite has a general formula Al2Si2O5(OH)4. Show that kaolinite is electronically neutral overall.
Montmorillonite is an example of another type of clay mineral. It is important as a soil additive in gardens because it helps with moisture retention. It is also used in the petroleum industry when drilling for oil; a slurry of montmorillonite in water helps keep the drill bit cool.
The structure, shown below, has a lot in common with kaolinite.
However, there are some key differences. Monmorillonite is a 2:1 clay mineral or a T-O-T clay mineral. Its octahedral layer is sandwiched between two tetrahedral layers. Also, there are fewer hydroxide groups in monmorillonite than in kaolinite; both basal layers are oxides, with hydroxides being found only in the interior of the layer. Finally, montmorillonite does not form electrically neutral layers. The layer is overall anionic, requiring counterions to balance out the charge. These ions are usually Na+, but also include Ca2+.
A more extended view of monmorillonite once again reveals the absence of a third of the aluminate octahedra, as well as some other, subtle differences.
The most obvious difference is the presence of water molecules between the layers. That difference is a key feature of montmorillonite's properties. However, the long-term structure also reveals occasional substitutions, in which a magnesium has replaced an aluminum in an octahedral site, or an aluminum has replaced a silicon in a tetrahedral site.
The moisture-retaining ability of montmorillonite is related to the cushion of water molecules between the layers of the material. More water can always be allowed in.
Exercise \(2\)
Predict the effect of the following substitutions on the charge of a montmorillonite sheet.
1. A silicon is replaced by an aluminum.
2. An aluminum is replaced by a magnesium.
Exercise \(3\)
Potters and ceramicists typically wouldn't use montmorillonite clay in their work. Explain why.
Micas provide additional examples of phllosilicates. The most common form of mica is muscovite, which at one time was used in Russian windows. It has a very distinctive appearance, forming blocks that can be peeled apart into thin, solid sheets of transparent material.
Muscovite is not a clay mineral. It does not form from weathering of other rocks. However, it clearly has some structural similarities to clay minerals, especially to montmorillonite.
Like montmorillonite, muscovite sheets contain octahedral layers sandwiched between tetrahedral layers. Also, there are some substitutions of aluminum for silicon in the tetrahedral layer.
A striking difference between montmorillonite and muscovite is the absence of water molecules between layers in the mica. Instead, the space between the sheets is occupied only by potassium ions.
Exercise \(4\)
What interaction holds the sheets of muscovite together?
Exercise \(5\)
How is the structure of muscovite related to the physical appearance of the mineral?
Tectosilicates
Once again, tectosilicate structures get very complicated and we will only look at them briefly.
One of the most common groups of aluminosilicates is the feldspars. They are thought to make up about 50% of the earth's crust. Feldspars, like silica, have extended three-dimensional structures. However, like silica, we can see some patterns emerging by following individual chains of tetrahedra through the structure. In silica, we saw spirals or helices; in feldspars, we see zig-zags instead.
Feldspars are significantly different from silica because of frequent substitution of aluminum for silicon and because their structures contain counterions. These counterions are frequently potassium, sodium or calcium.
Exercise \(6\)
The feldspar endmembers are potassium feldspar, albite and anorthite. Potassium feldspar is the potassium endmember of the group, KAlSi3O8. Determine the formulae for:
1. albite, the sodium endmember.
2. anorthite, the calcium endmember.
Feldspars are the most common class of tectosilicates, but one of the most interesting classes is the zeolites. Zeolites are tremendously unique and useful materials. They are characterized by a great capacity for storing gases and small molecules. In fact, the name zeolite translates from the Greek to mean "boiling stones", because they may give off steam when heated. Zeolites are also used in catalysis in the petroleum industry, helping to convert petroleum distillates into more useful forms.
The reason zeolites are able to store small molecules is directly tied to their very porous, cage-like structure. In order to emphasize that feature, zeolites are frequently drawn as octahedral line drawings. The following sketch shows the structure of a faujasite, a naturally-occurring zeolite that is also made synthetically for industrial purposes.
The structure is truncated; it would extend further through the attachment of similar units in different directions. However, this view lets us see some key features. A large cavity is clearly seen in the middle of the structure. In addition, there are smaller cavities inside the hexagonal "globes" on the corners, as well as tunnels through the hexagonal "drums" connecting the globes. The hexagons themselves are really pore-like openings into the interior of the zeolite.
The connection between this polyhedral structure and the atomic makeup of faujasite is illustrated below. There, the atoms around one of the hexagonal pores have been filled in.
Each vertex or corner of the structure represents a silicon. However, aluminum can substitute in for silicon to varying degrees. along each line in the polyhedral drawing there is an oxygen atom. Thus, the hexagonal pores in the drawing are really rings formed by twelve atoms, not six: six silicons (or aluminums) and six oxygens. Also illustrated in this picture is the anionic nature of the zeolite framework, necessitating the presence of counterions such as sodium.
Zeolites are useful because molecules can be stored inside of them. Also, molecules can be made to react with each other while trapped inside a zeolite at high temperature. Researchers are very interested in making new architectures of zeolites, and in modifing them to contain catalytic sites that will carry out important reactions.
The structures of zeolites can have a kind of architectural beauty. However, there is one zeolite that has been used in art since ancient times. Ultramarine is a deep blue pigment obtained by grinding lapis lazuli. The characteristic color comes from deep blue thiozonide anions (S3-) trapped within the zeolite structure.
10.05: Solutions to Selected Problems
a) graphite is between 1 & 2 b) stishovite is between 9 & 10 c) opal is between 6 & 7
d) obsidian is about 5 e) ruby is about 9 f) emerald is between 7 & 8
1. forsterite: Mg2SiO4: If each nesosilicate unit has a charge of 4- because of the formal charges on the four oxygens, then the two magnesiums must combine to balance that charge with a 4+ charge. Each magnesium must have a 2+ charge of its own. That seems reasonable, since magnesium is two atoms from the left edge of the periodic table; it would have a 2+ charge in a noble gas configuration.
2. fayalite: Fe2SiO4: In order for this to work, each iron would have to have a 2+ charge. Transition metal charges are harder to predict than alkali and alkaline earth metals because they are farther from the edge of the periodic table, but it turns out that the two most common charges on an iron ion are 2+ and 3+, so this charge seems reasonable.
1. If each nesosilicate has a charge of 4-, then the total negative charge is 100 x 4- = 400-. If there are 25 magnesium ions, each with a 2+ charge, the total positive charge is 25 x 2+ = 50+. That leaves a net negative charge of 400- + 50+ = 350-. The irons would balance that charge with an equal positive charge. The number of iron ions would be 350+/3+ = 150.
2. If we divide these ratios by 100 to arrive at the one nesosilicate in the formula, we get Mg0.25Fe1.5SiO4. It does not mean there is a quarter of a magnesium ion anywhere; this formula is just the overall ratio of atoms.
So far we have Al2Fe3. The positive charge = 2 x 3+ + 3 x 2+ = 12+. That charge must be balanced by the nesosilicate anions, which are each 4-. So the number of silicates is 12-/4- = 3. The formula is Al2Fe3(SiO4)3.
The anions are nesosilicate (4-) + sorosilicate (6-) + oxide (2-) + hydroxide (1-). The total charge is 13-.
Calcium is an alkaline earth metal, two atoms from the left edge of the periodic table. The two calciums would partially balance that charge with 2 x 2+ = 4+ charge; the net charge is now 13- + 4+ = 9-.
If each Al or Fe has charge 3+, then the number needed to balance the charge is 9+/3+ = 3 Fe/Al ions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/10%3A_Network_Solids/10.04%3A_Aluminosilicates.txt |
Transition metals are found in the middle of the periodic table. In addition to being found in the metallic state, they also form a range of compounds with different properties. Many of these compounds are ionic or network solids, but there are also some molecular compounds, in which different atoms are arranged around a metal ion. These compounds are called transition metal complexes or coordination complexes. They are often brightly-colored compounds and they sometimes play very useful roles as catalysts or even as pharmaceuticals.
Because of their relatively low electronegativity, transition metals are frequently found as positively-charged ions, or cations. These metal ions are not found by themselves, instead, they attract other ions or molecules to themselves. These species bind to the metal ions, forming coordination complexes.
Hexaamminecobalt(III) chloride, [Co(NH3)6]Cl3, is an example of a coordination complex. It is a yellow compound. The "complex" part refers to the fact that the compound has a bunch of different pieces. There is a cationic part, which itself is a moderately complicated structure, plus three chloride anions.
Potassium hexachloroplatinate, K2[PtCl6], is another good example. It is another bright yellow compound. This time, the anion is the more complex part, and there are two potassium ions as well.
The formulae for coordination complexes always give you hints about the structures. The stuff inside the square brackets always makes up one of the ions. In that part, there are a number of things attached to the transition metal. Those things attached to the transition metal are called ligands; we'll take a closer look at them later. The part outside the square brackets tells you what the counterions are; those are there to balance out the charge of the ion inside the square brackets.
Very often, the counterions are individual atomic ions, like chloride anions (Cl-) or potassium cations (K+). So, when you see the K2 within the formula, the potassium atoms are not connected together; they are two separate potassium ions: 2 K+. Likewise, the Cl3 at the end of a formula does not really mean a group of three chlorine atoms clustered together; they are three separate chloride ions, 3 x Cl-.
Sometimes, polyatomic ions that act as counterions to these complexes; this is especially common in the case of anions. Most often, they are oxoanions, in which an atom has some number of oxygens attached to it. A couple of the most common examples are nitrate (NO3-) and sulfate (SO42-); these ions have been known for hundreds of years. Tetrafluoroborate (BF4-) and hexafluorophosphate (PF6-) are a couple of twentieth-century anions.
A number of common ions are listed in the table below. Most of them have a +1 or -1 charge. Sulfate is the only common example listed with a 2- charge.
Note that some of these structures use a charge-minimised Lewis structure. In third-row elements, these structures may contain additional bonds to the central atom, going past the octet, in order to lower the number of + and - charges. Sulfate and perchlorate can also be drawn as octet-obedient structures, in which the sulfur and chlorine have true octets, and charge separation occurs. However, hexafluorophosphate has no octet-obedient Lewis structure; regardless of how you draw it, the phosphorus has six bonds.
Exercise \(1\)
Draw the octet-obedient structures for (a) sulfate and (b) perchlorate.
Answer a
Answer b
Exercise \(2\)
Indicate the individual ions in the following complexes.
1. [Co(NH3)4Cl2]Cl
2. [Co(NH3)5Cl]Cl2
3. [Co(NH3)5NO2]Cl2
4. [Co(NH3)5OH2]Cl3
5. K2[PtCl4]
6. Na2[Co(SCN)4]
7. [Pt(NH3)2(OH2)2](NO3)2
8. [Co(NH3)4(OH2)2]2(SO4)3
Answer a
Answer b
Answer c
Answer d
Answer e
Answer f
Answer g
Answer h
We haven't drawn proper Lewis structures for these coordination complexes so far. For the hexachloroplatinate complex, the Lewis structure is shown below:
The electron accounting looks like this:
atom electrons
Pt 10 e-
Cl 6 x 7 e-
charge add 2 e-
total 54 e-
Two electrons for each bond takes up 12 electrons. That leaves 42 more. If each chlorine gets three lone pairs, that uses up another 36 electrons. There are six left, and they could be left on the platinum.
That is a lot of electrons. But remember where platinum sits in the periodic table: it's a transition metal. How many electrons does the next noble gas have? Eighteen: that's radon. This structure works out perfectly in terms of reaching a noble gas configuration.
For another example, let's take a look at the cobalt complex.
This time, the electron counting looks like:
atom electrons
Co 9 e-
N 6 x 5 e-
H 18 x 1 e-
charge lose 3 e-
total 54 e-
A dozen of those electrons get used for the six Co-N bonds, leaving 42 more. We need another 36 for the eighteen N-H bonds; that leaves six electrons. We can just put those on the cobalt, like last time.
Each nitrogen ends up with its octet (eight electrons in four bonds). Each hydrogen has its "octet" (two electrons in a bond, corresponding to helium's noble gas configuration. Cobalt also gets its "octet" (eighteen electrons corresponding to krypton's noble gas configuration).
These are still really big, unwieldy numbers. We need a simplification. We won't really count up all of the electrons in coordination complexes because they tend to be assembled from pre-existing parts (the "ligands") that already have their octets. The ligands are just donating their electrons to the metal in the center. Instead, we focus on that metal, and see how many electrons it has once all of the ligands have been attached.
Also, notice that in the Lewis structure of the cobalt complex, we neglected the formal charges. That's actually common practice with transition metal complexes. The reason for that is simply that the structure is getting pretty crowded with all the lone pairs and formal charges. Normally, in order to simplify an already complicated structure, the lone pairs on the transition metal and the formal charges are not shown.
On the next page, we will practice electron counting some more. We will also take a look at some common ligands: the pieces that are directly attached to the transition metal ions.
See a more in-depth discussion of coordination complexes in a later course. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/11%3A_Transition_Metal_Complexes/11.01%3A_Introduction.txt |
Transition metal complexes or coordination complexes are molecules that contain groups arranged around a central metal ion. In a way, these are like "lego-molecules", easily assembled from smaller parts, and sometimes they are easily transformed into new molecules by switching out old parts for new ones. That rapid assembly and disassembly is part of what makes these compounds very useful in both industrial and biological catalysis.
What are the building blocks that go into a coordination complex? The key, central part is usually a transition metal ion (although cations from other parts of the periodic table are also seen in some cases). The following table shows some common examples of ions formed by each of the transition metals.
Table \(1\): Some common transition metal ions.
Sc3+ Ti4+
Ti3+
V5+
V4+
Cr3+
Cr6+
Cr2+
Mn2+
Mn4+
Mn7+
Fe3+
Fe2+
Co2+
Co3+
Ni2+ Cu2+
Cu+
Zn2+
Y3+ Zr4+ Nb5+
Nb3+
Mo6+
Mo4+
Mo5+
(Tc7+)*
(Tc4+)*
Ru3+
Ru4+
Ru2+
Rh3+
Rh+
Pd2+
Pd4+
Ag+
Ag2+
Cd2+
La3+ Hf4+ Ta5+ W6+
W4+
W5+
Re4+
Re6+
Re7+
Os4+
Os3+
Ir4+
Ir3+
Pt2+
Pt4+
Au3+
Au+
Hg2+
Hg1+
For each transition metal, the most common form of ion is listed first. For example, iron is often found in compounds as Fe3+. Other common ions are also shown below that; iron is seen pretty frequently as Fe2+. Other charges are possible; iron has been reported with charges all the way from Fe0 to Fe6+, but these cases are less common.
Also, it's worth noting that the top row of transition metals is generally much more common than the next two rows. That does not mean that the others are less important; gold (Au), silver (Ag) and platinum (Pt) are certainly important economically. However, you may be more likely to encounter examples of complexes from the first row. That's especially true in biological chemistry, because organisms have evolved to make use of those metal ions that are most readily available to them.
Exercise \(1\)
1. The metals on the left hand side of the table tend to have relatively high charges compared to the metals on the right. What do ions such as Sc3+, Zr4+, and Ta5+ have in common that would explain this trend?
2. The metals on the right side of the table have relatively low charges. In nature, metals such as copper (Cu), silver (Ag), and gold (Au) are frequently found as native metals (e.g. Cu0, with no charge at all) rather than as compounds. Explain this preference for low charges in this part of the transition metals.
3. The metals in the middle, on the other hand, have very wide ranges of charges (that's why the table lists three common charges for those, although even more exist). Why?
Answer a:
All of these ions have the electronic configuration of the preceding noble gas, Ar. These metals have just started filling the next energy levels; these electrons are relatively high in energy, with relatively few protons attracting them (compared to later elements in the row). The valence electrons are relatively easy to use, although not the core electrons.
Answer b:
These metals have relatively high numbers of protons compared to the elements to their left in the same row. They have relatively high electronegativity, so they do not lose many electrons.
Answer c:
These metals in the middle display a mixture of behaviours: they are electronegative enough to stabilize relatively low charges, but far enough to the left that they can become fully oxidised to reach a noble gas configuration.
The charge on the metal ion is sometimes called the oxidation state. This term refers to the fact that metals that are exposed to the elements sometimes become positively charged, forming compounds such as metal oxides. The oxygen from the air provides the oxide or hydroxide ion to counter the charge when the metal atom loses electrons and becomes a metal cation. Familiar examples include the oxidation of aluminum metal to form silvery-white aluminum oxide; you may have seen aluminum screen doors that are actually covered in a coating of aluminum oxide. The Statue of Liberty was originally covered in copper metal, but quickly became coated in money-green copper oxide.
Sometimes, within a complex, the charge or oxidation state of the metal is denoted using Roman numerals. For example, Co3+ is sometimes writted Co(III); Mn2+ might be written Mn(II).
So, if you need to brush up on Roman numerals:
1 = I; 2 = II; 3 = III; 4 = IV; 5 = V; 6 = VI; 7 = VII; 8 = VIII; 9 = IX
After that, we won't need to worry about it. Sometimes osmium is found as Os(VIII), for example, but we don't see a 9+ charge very often.
The metal ion is the first building block. The second building block is the ligand. The ligands are the pieces that are arranged around the metal ion. On the last page, we saw chloride anions (Cl-) and ammonia (NH3) act as ligands in different transition metal complexes. It is easy to imagine how a positively charged metal ion would attract a negatively charge anion such as chloride. As it happens, neutral ligands are just as common. The main requirement for sticking to a metal ion is a non-bonding pair of electrons, or a lone pair (at least, that's the case at this stage of your education).
The following table illustrates a variety of ligands for transition metal complexes.
Exercise \(2\)
Calculate the overall charge on the following complexes, if any.
a) [Fe(II)(OH2)6] b) [Cr(III)(NH3)5Cl] c) [(py)4Mn(II)(SCN)2]
d) [Fe(II)(CN)6] e) [Co(III)(CN)5CO] f) [Fe(II)(CN)5NH3]
Answer a:
H2O = 0 charge; Fe(II) = 2+ charge; total = 2+
Answer b:
H3N = 0 charge; Cl = 1- charge; Cr(III) = 3+ charge; total = 2+
Answer c:
py = 0 charge; SCN = 1- charge; Mn(II) = 2+ charge; total = 0
Answer d:
CN = 1- charge; Fe(II) = 2+ charge; total = 4-
Answer e:
CN = 1- charge; CO = 0 charge; Co(III) = 3+ charge; total = 2-
Answer f:
CN = 1- charge; NH3 = 0 charge; Fe(II) = 2+ charge; total = 3-
A coordination complex has a central metal ion with a number of ligands arranged around it. The last piece is the counterions, which would balance out the charge of the transition metal complex ion. We saw examples of common counterions on the introduction page.
Exercise \(3\)
Calculate the charge or oxidation state on the metal ion in each of the following complexes.
a) [Cr(OH2)6](NO3)3 b) K3[FeF6] c) [Cr(SCN)(NH3)5]SO4
d) K4[Mn(CN)6] e) [Au(NCCH3)2]ClO4 f) [(Ph3P)Ag(CN)]
Answer a:
Overall charge = 3+; ligands charge = 0; charge on Cr = 3+
Answer b:
Overall charge = 3-; ligands charge = 6-; charge on Fe = 3+
Answer c:
Overall charge = 2+; ligands charge = 1-; charge on Cr = 3+
Answer d:
Overall charge = 4-; ligands charge = 6-; charge on Mn = 2+
Answer e:
Overall charge = 1+; ligands charge = 0; charge on Au = 1+
Answer f:
Overall charge = 0; ligands charge = 1-; charge on Ag = 1+
On the next page, we will see how some ligands can bind to a metal more than once. That helps them hold on more tightly. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/11%3A_Transition_Metal_Complexes/11.02%3A_Building_Blocks-_Metal_Ions_and_.txt |
Transition metals bind to ligands in order to edge closer to electronic saturation. In other words, the electrons donated by the ligands to form the bonds to the metal help that metal get to the next noble gas configuration. It does not always get there, and sometimes it overshoots, but that's a rough guide.
Consider the coordination complex below. [Co(NH3)6]Cl3 contains a cobalt complex ion and three chloride counterions; the counterions just there to balance the charge. We could think of that complex as being assembled from a cobalt ion and six ammonia molecules. The six ammonia molecules each lend a lone pair to help the cobalt towards its octet. If we go backwards, and disassemble the complex into its components, we see that each of those ammonia molecules was neutral. There is no formal charge on any of those nitrogens. If they are all neutral, but the complex has a positive charge, then where did the positive charge come from?
The answer has to be the cobalt ion. The cobalt has a 3+ charge (or "oxidation state"). Knowing that, we can figure out how many electrons cobalt has in its valence shell within the complex.
In the periodic table, cobalt has nine valence electrons. However, if the overall complex has a charge of 3+, then the cobalt also had a charge of 3+. It has lost three electrons, so it only has six left. Once it forms the complex, each ammonia donates a pair of electrons; that's twelve total. The valence shell around cobalt includes its own d electrons plus that twelve donated by the ligands, for a total of eighteen.
We can summarise in a table:
metal valence e- (or d e-) 9 e-
charge on complex 3+
charge on ligands 0
charge on metal 3+
metal ion d e- 6 e-
e- donated by ligands 6 x 2 = 12 e-
total 18 e-
[Co(NH3)5Cl]Cl2 is a very similar example. This time, when we dissect the complex, we find a chlorine that would have a formal charge of 1- (we always assume the ligands all had octets before they joined the metal). The overall complex has a 2- charge; taking into account that there is a charge of -1 on the chloride within that complex, then the cobalt ion must have a charge of 3+ in order to arrive at that overall 2+ charge.
Again,we can use a table to count the electrons on cobalt in the complex.
metal valence e- (or d e-) 9 e-
charge on complex 2+
charge on ligands 1-
charge on metal 3+
metal ion d e- 6 e-
e- donated by ligands 12 e-
total 18 e-
Exercise \(1\)
Confirm the electron count on the metal is 18 electrons in each of the following complexes.
a) Pd(PPh3)4 b) Cr(CO)6 c) [Cu(NCCH3)4]ClO4
d) Fe(CO)5 e) K4[Fe(CN)6]
Answer a:
Pd valence e- 10 e-
charge on complex 0
charge on ligands 0
charge on Pd 0
revised Pd e- 10 e-
e- donated by ligands 4 x 2 = 8 e-
total 18 e-
Answer b:
Cr valence e- 6 e-
charge on complex 0
charge on ligands 0
charge on Cr 0
revised Cr e- 6 e-
e- donated by ligands 6 x 2 = 12 e-
total 18 e-
Answer c:
Cu valence e- 11 e-
charge on complex +1
charge on ligands 0
charge on Cu +1
revised Cu e- 10 e-
e- donated by ligands 4 x 2 = 8 e-
total 18 e-
Answer d:
Fe valence e- 8 e-
charge on complex 0
charge on ligands 0
charge on Fe 0
revised Fe e- 8 e-
e- donated by ligands 5 x 2 = 10 e-
total 18 e-
Answer e:
Fe valence e- 8 e-
charge on complex -4
charge on ligands -6
charge on Fe +2
revised Fe e- 6 e-
e- donated by ligands 6 x 2 = 12 e-
total 18 e-
Exercise \(2\)
Determine the electron count on the metal in each of the following complexes.
a) RhClH2(PPh3)2 b) [Ni(OH2)4](ClO4)2 c) [Cu(NH3)4](BF4)2 d) K2[PtCl6]
Answer a:
Answer b:
Answer c:
Answer d:
On the next page, we will see how some ligands can bind to a metal more than once. That helps them hold on more tightly. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/11%3A_Transition_Metal_Complexes/11.03%3A_Electron_Counting.txt |
Each ligand that binds to a transition metal donates a pair of its electrons to the metal, sharing those electrons in a metal-ligand bond. Counting the total number of electrons in the metal's valence shell is fairly easy when you know that each ligand contributes two electrons.
But what if a ligand could contribute more than two electrons? Take a look at the following table. Each of the ligands shown here can bind to a metal twice. The ligand forms two bonds to the metal, donating two pairs of electrons. It might seem obvious, but forming two bonds to the metal means the ligand binds more tightly to the metal. Remember the lego-like nature of transition metal complexes: ligands can come and go, but these ligands are less likely to go; they hold on.
These ligands are called bidentate ligands. That means, literally, that they have two teeth. That does not sound like much, but the ligands we have seen previously are described as monodentate; they have only one tooth. Bidentate ligands can bite into the metal, and hold onto it, more strongly than monodentate ones.
Another term used for these kinds of ligands is derived from the Greek chele, for claw. These ligands grab onto the metal like the claw of a lobster or crab; they chelate. The "chelate effect" is the tendency of these ligands to bind firmly to a metal, whereas monodentate ligands might come off more easily.
Now, just because a ligand could be bidentate does not mean that it always binds that way. That's often true with the acetate ligand, for example, because the four-membered ring that forms when it binds through both oxygens is a little too strained. Consequently, there are examples binding through both oxygen atoms, and there are also examples of binding through only one. Sometimes, acetate uses one oxygen to bind to one metal atom and the other oxygen to bind to a second metal atom, forming a bridge.
Usually, if a ligand is capable of chelation, assume it binds that way. However, there may be cases in which you are asked specifically to draw it binding in one way or another.
Exercise \(1\)
Determine the denticity of each ligand in the following complexes.
a) [Cu(en)2(OH)2] b) [RuCl2(dmpe)(bpy)] c) [Ni(acac)2(OH2)2] d) K2[Cu(ox)2(OH2)2]
e) [FeH2(dmpe)2] f) [RuH(OAc)(PPh3)3] g) [Co(en)2Cl2]BF4 h) [Ru(bpy)2(HOCH2CH3)2](ClO4)2
Answer a:
en = bidentate; OH = monodentate
Answer b:
dmpe = bidentate; bpy = bidentate; Cl = monodentate
Answer c:
acac = bidentate; H2O = monodentate
Answer d:
ox = bidentate; H2O = monodentate
Answer e:
dmpe = bidentate; H = monodentate
Answer f:
H = monodentate; PPh3 = monodentate; OAc = bidentate. In this case, the acetate gets the complex to 18 electrons by binding twice.
Answer g:
en = bidentate; Cl = monodentate
Answer h:
bpy = bidentate; HOCH2CH3 = monodentate
Exercise \(2\)
Determine the charges (or oxidation states) on the metals in the following complexes:
a) K[Cr(ox)2(OH2)2] b) K[Mn(acac)3] c) [Cr(en)2Cl2]PF6 d) [Co(en)2(OH)Cl]ClO4
e) Na3[Mn(ox)3] f) K3[Cr(ox)3] g) Na[Au(bpy)(CN)2]
Answer a:
Overall charge = 1-; ligands charge = 4-; charge on Cr = 3+
Answer b:
Overall charge = 1-; ligands charge = 3-; charge on Mn = 2+
Answer c:
Overall charge = 1+; ligands charge = 2-; charge on Cr = 3+
Answer d:
Overall charge = 1+; ligands charge = 2-; charge on Co = 3+
Answer e:
Overall charge = 3-; ligands charge = 6-; charge on Mn = 3+
Answer f:
Overall charge = 3-; ligands charge = 6-; charge on Cr = 3+
Answer g:
Overall charge = 1-; ligands charge = 2-; charge on Au = 1+
Exercise \(3\)
Determine the electron count on the metal in each of the complexes from the previous problem.
Answer a:
Cr = d6; Cr3+ = d3; ox = 2 x 4e- = 8 e-; water = 2 x 2e- = 4 e-; total = 15 e-
Answer b:
Mn = d7; Mn2+ = d5; acac = 3 x 4e- = 12 e-; total = 17 e-
Answer c:
Cr = d6; Cr3+ = d3; en = 2 x 4e- = 8 e-; Cl = 2 x 2e- = 4 e-; total = 15 e-
Answer d:
Co = d9; Co3+ = d6; en = 2 x 4e- = 8 e-; OH = 2e-; Cl = 2 e-; total = 15 e-
Answer e:
Mn = d7; Mn3+ = d4; ox = 3 x 4e- = 12 e-; total = 16 e-
Answer f:
Cr = d6; Cr3+ = d3; ox = 3 x 4e- = 12 e-; total = 15 e-
Answer g:
Au = s1d10; Au1+ = d10; bpy = 2 x 2e- = 4 e-; cyanide = 2 x 2e- = 4 e-; total = 18 e-
See a more in-depth discussion of coordination complexes in a later course.
11.05: Isomers
Sometimes, two compounds can have the same molecular formula but different structures. In transition metal complexes, that can happen if the same ligands are arranged around the metal in different ways.
For example, the complex ion [CoCl2(NH3)4]+ contains two chlorine ligands that can be placed in two different ways around the cobalt. The cobalt in this complex has octahedral geometry. Every ligand in the complex is 90 degrees away from four other ligands, and 180 degrees away from a fifth. If there are two chloride ligands, then they might be found 90 degreed away from each other, or they might be found 180 degrees away from each other. The former case exists, and it is called the cis-isomer; the latter case also exists, and is called the trans-isomer (cis means same, as in same side, as opposed to trans, which means opposite).
In this case, the two compounds have different physical properties, most readily seen in the two different colors of the complexes: one is purple and the other is green. The exact reasons for the colors are subtle; you're not expected to know why one is green and the other is purple at this stage.
Exercise \(1\)
Draw structures for the following compounds:
1. trans-Fe(py)4Cl2
2. trans-Mn(OH2)4F2
3. cis-(bpy)V(acac)Cl2
4. cis-Mo(CO)4(PMe3)2
A group of three ligands can also adopt two different geometries in an octahedron. The three ligands can be found clustered together on one side of the complex. They occupy one face of the octahedron, in what is called the fac-geometry. Alternatively, they could lie all in a row. Sitting along a meridian of the octahedron, they adopt a mer-geometry.
Exercise \(2\)
Draw structures of the following complexes:
1. fac-W(CO)3(NCCH3)3
2. mer-Mo(CO)3(PMe3)3
3. fac-Mo(CO)3(PF3)3
4. mer-Mo(CO)3(PCl3)3
5. fac-Na[Mn(CO)3(CN)3]
Answer a:
Answer b:
Answer c:
Answer d:
Answer e:
See a more in-depth discussion of coordination complexes in a later course. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/11%3A_Transition_Metal_Complexes/11.04%3A_Chelation.txt |
It's sometimes useful to be able to be able to read the name of a compound and understand what that means. The names of coordination compounds tend to be pretty long because of the number of pieces involved, but they usually follow a trend.
• Coordination compounds often contain a "complex" ion, with ligands attached to the metal ion, as well as a counterion, to balance charge. If there is a cation and an anion, the cation is named first.
[Ni(OH2)6]Cl2, hexaaquanickel(II) chloride (nickel cation before chloride anion)
In the formula, you can always tell one ion from the other because the complex ion (metal plus ligands) is always enclosed in square brackets. Anything outside of brackets is the counterion or counterions.
• Ligands are listed in front of the metal.
[Cu(NH3)6]Br2, hexaamminecopper(II) bromide (ammine ligand before copper metal)
Note that anionic ligands generally end in "o" in the name of a coordination compound. Chloride becomes chloro. Acetate becomes acetato. Neutral ligands keep their regular names, but the two common molecules H2O and NH3 are called "aqua" and "ammine", respectively.
If two different ligands are found in the complex, they are listed alphabetically.
[Co(OH2)5Cl](NO3)2, pentaaquachlorocobalt(III) nitrate (aqua before chloro)
• If there is more than one ligand of the same kind, then the number of ligands is indicated as follows:
ligands prefix
2 di
3 tri
4 tetra
5 penta
6 hexa
In the examples above, there were six waters, "hexaaqua", six ammines, "hexaammine", and five waters, "pentaaqua".
Sometimes ligand names are complicated. For example, they may already contain prefixes denoting a number, such as ethylenediamine, NH2CH2CH2NH2. The di means that there are two amino groups in the ligand. We avoid saying "di" to describe something that already contains a "di". Instead, for more complicated ligands like this, we use the following system:
ligands prefix
2 bis
3 tris
4 tetrakis
5 pentakis
6 hexakis
For example, [(Ph3P)4Pd] is tetrakis(triphenylphosphine)palladium(0).
If you need to stress that only one ligand of a certain type is present, you could use the numbering adjective "mono". For example, if you are comparing a compound with two CO ligands to another that has only one, you might call them "the dicarbonyl compound" and "the monocarbonyl compound", respectively.
• The metal is named following the ligands, and its charge or oxidation state is listed using Roman numerals.
[Co(en)2Cl2]ClO4, dichlorobis(ethylenediamine)cobalt(III) perchlorate.
However, if the complex ion containing the metal is an anion, the metal is given the suffix "-ate", sometimes replacing its normal ending.
K2[PtCl6], potassium hexachloroplatinate(IV)
If the complex ion is an anion and the element's symbol does not match its name in English, the Latin name of the metal is usually used rather than the English one. For example, an iron-containing anion would contain the term "ferrate" instead of "iron"; a copper-containing one would be called a "cuprate", one with gold would be called an "aurate" and so on.
• There is no need to describe the number of counterions, because that information can be worked out given the charge on the metal and the charges on the ligands.
Na2[CuF4], sodium tetrachlorocuprate(II) (no need to say there are two sodium ions).
Exercise \(1\)
Provide names for the following complexes:
a) K[Cr(ox)2(OH2)2] b) [Co(NH3)5Br](NO3)2 c) [Cr(en)2Cl2]PF6 d) [Co(bpy)2(OH)Cl]ClO4
e) [TiCl3(OH2)3] f) K3[Fe(CN)6] g) Na[Au(bpy)(CN)2]
Answer a:
potassium diaquabis(oxalato)chromate(III)
Answer b:
pentaamminebromocobalt(III) nitrate
Answer c:
dichlorobis(ethylenediamine)chromium(III) hexafluorophosphate
Answer d:
bis(bipyridine)chlorohydroxocobalt(III) perchlorate
Answer e:
triaquatrichlorotitanium(III)
Answer f:
potassium hexacyanoferrate(III)
Answer g:
sodium bipyridinedicyanoaurate(I)
Bipyridine is sometimes called "bipyridyl".
Exercise \(2\)
Draw the following structures:
1. dicarbonylbis(1,2-dimethylphosphino)ethaneruthenium (0)
2. potassium bipyridyltetracyanoferrate(II)
3. pentaamminechlorochromium(III) perchlorate
4. tetraacetonitrilecopper(I) tetrafluoroborate
5. sodium ethylenediaminebis(oxalato)cobalt(III)
6. chlorotris(triphenylphosphine)rhodium(I)
Answer a:
Answer b:
Answer c:
Answer d:
Answer e:
Answer f:
See a more in-depth discussion of coordination complexes in a later course.
11.07: Geometry in Transition Metal Com
If you have studied Lewis structures, then you know that the number of lone pairs and atoms around a specific atoms tells you the geometry at that specific atom. Something very similar is true in coordination complexes, but it's slightly simpler. In transition metal complexes, the lone pairs on the metal do not contribute to the metal complex's overall geometry. All that matters is the number of ligands attached to the metal.
• In coordination complexes, the number of ligands determines the geometry.
As a result, there is a pretty straightforward correlation between coordination number (the number of ligands) and geometry.
# ligands geometry
2 linear
3 trigonal planar
4 tetrahedral; sometimes square planar
5 trigonal bipyramidal; sometimes square pyramidal
6 octahedral
There are two exceptions to that straightforward rule. Coordination number four and coordination number five both have two possible geometries. How do we know which will happen?
For coordination number five, we really can't make an easy prediction. In terms of the angles between ligands, these two possible geometries are really pretty close to each other, so the distinction is subtle. If you call everything that's five-coordinate a trigonal bipyramidal geometry, you will be in the right ballpark.
Coordination number four is a different story; we do have some predictive ability here. However, there are two different factors that come into play, so there are still some subtle cases.
The first factor is sterics or crowding. In a tetrahedron, the four ligands are 109 degrees away from each other. In square planar geometry, the four ligands are only 90 degrees away from each other. Since a tetrahedral geometry is less crowded, sterics makes this geometry more favorable and common.
To form a square planar geometry, there has to be some factor that offsets the steric preference for tetrahedral geometry. Electronic destabilization can occur in tetrahedral geometry. The same goes for square planar geometry; however, destabilization occurs to a lesser extent when there are eight electrons on the metal in this geometry. As a result, square planar geometry is most common when the metal has eight electrons.
See a more in-depth discussion of coordination complexes in a later course. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/11%3A_Transition_Metal_Complexes/11.06%3A_Naming_Transition_Metal_Complexe.txt |
Exercise 11.1.1:
Exercise 11.1.2:
Exercise 11.2.1:
1. All of these ions have the electronic configuration of the preceding noble gas, Ar. These metals have just started filling the next energy levels; these electrons are relatively high in energy, with relatively few protons attracting them (compared to later elements in the row). The valence electrons are relatively easy to use, although not the core electrons.
2. These metals have relatively high numbers of protons compared to the elements to their left in the same row. They have relatively high electronegativity, so they do not lose many electrons.
3. These metals in the middle display a mixture of behaviors: they are electronegative enough to stabilize relatively low charges, but far enough to the left that they can become fully oxidised to reach a noble gas configuration.
Exercise 11.2.2:
1. H2O = 0 charge; Fe(II) = 2+ charge; total = 2+
2. H3N = 0 charge; Cl = 1- charge; Cr(III) = 3+ charge; total = 2+
3. py = 0 charge; SCN = 1- charge; Mn(II) = 2+ charge; total = 0
4. CN = 1- charge; Fe(II) = 2+ charge; total = 4-
5. CN = 1- charge; CO = 0 charge; Co(III) = 3+ charge; total = 2-
6. CN = 1- charge; NH3 = 0 charge; Fe(II) = 2+ charge; total = 3-
Exercise 11.2.3:
1. Overall charge = 3+; ligands charge = 0; charge on Cr = 3+
2. Overall charge = 3-; ligands charge = 6-; charge on Fe = 3+
3. Overall charge = 2+; ligands charge = 1-; charge on Cr = 3+
4. Overall charge = 4-; ligands charge = 6-; charge on Mn = 2+
5. Overall charge = 1+; ligands charge = 0; charge on Au = 1+
6. Overall charge = 0; ligands charge = 1-; charge on Ag = 1+
Exercise 11.3.1:
a)
Pd valence e- 10 e-
charge on complex 0
charge on ligands 0
charge on Pd 0
revised Pd e- 10 e-
e- donated by ligands 4 x 2 = 8 e-
total 18 e-
b)
Cr valence e- 6 e-
charge on complex 0
charge on ligands 0
charge on Cr 0
revised Cr e- 6 e-
e- donated by ligands 6 x 2 = 12 e-
total 18 e-
c)
Cu valence e- 11 e-
charge on complex +1
charge on ligands 0
charge on Cu +1
revised Cu e- 10 e-
e- donated by ligands 4 x 2 = 8 e-
total 18 e-
d)
Fe valence e- 8 e-
charge on complex 0
charge on ligands 0
charge on Fe 0
revised Fe e- 8 e-
e- donated by ligands 5 x 2 = 10 e-
total 18 e-
e)
Fe valence e- 8 e-
charge on complex -4
charge on ligands -6
charge on Fe +2
revised Fe e- 6 e-
e- donated by ligands 6 x 2 = 12 e-
total 18 e-
Exercise 11.3.2:
a)
Rh valence e- 9 e-
charge on complex 0
charge on ligands -3
charge on Rh +3
revised Rh e- 6 e-
e- donated by ligands 5 x 2 = 10 e-
total 16 e-
b)
Ni valence e- 10 e-
charge on complex 2+
charge on ligands 0
charge on Ni 2+
revised Ni e- 8 e-
e- donated by ligands 4 x 2 = 8 e-
total 16 e-
c)
Cu valence e- 11 e-
charge on complex 2+
charge on ligands 0
charge on Cu 2+
revised Cu e- 9 e-
e- donated by ligands 4 x 2 = 8 e-
total 17 e-
d)
Pt valence e- 10 e-
charge on complex 2-
charge on ligands 6-
charge on Pt 4+
revised Pt e- 6 e-
e- donated by ligands 6 x 2 = 12 e-
total 16 e-
Exercise 11.4.1:
1. en = bidentate; OH = monodentate
2. dmpe = bidentate; bpy = bidentate; Cl = monodentate
3. acac = bidentate; H2O = monodentate
4. ox = bidentate; H20 = monodentate
5. dmpe = bidentate; H = monodentate
6. H = monodentate; PPh3 = monodentate; OAc = bidentate. In this case, the acetate gets the complex to 18 electrons by binding twice.
7. en = bidentate; Cl = monodentate
8. bpy = bidentate; HOCH2CH3 = monodentate
Exercise 11.4.2:
1. Overall charge = 1-; ligands charge = 4-; charge on Cr = 3+
2. Overall charge = 1-; ligands charge = 3-; charge on Mn = 2+
3. Overall charge = 1+; ligands charge = 2-; charge on Cr = 3+
4. Overall charge = 1+; ligands charge = 2-; charge on Co = 3+
5. Overall charge = 3-; ligands charge = 6-; charge on Mn = 3+
6. Overall charge = 3-; ligands charge = 6-; charge on Cr = 3+
7. Overall charge = 1-; ligands charge = 2-; charge on Au = 1+
Exercise 11.4.3:
1. Cr = d6; Cr3+ = d3; ox = 2 x 4e- = 8 e-; water = 2 x 2e- = 4 e-; total = 15 e-
2. Mn = d7; Mn2+ = d5; acac = 3 x 4e- = 12 e-; total = 17 e-
3. Cr = d6; Cr3+ = d3; en = 2 x 4e- = 8 e-; Cl = 2 x 2e- = 4 e-; total = 15 e-
4. Co = d9; Co3+ = d6; en = 2 x 4e- = 8 e-; OH = 2e-; Cl = 2 e-; total = 15 e-
5. Mn = d7; Mn3+ = d4; ox = 3 x 4e- = 12 e-; total = 16 e-
6. Cr = d6; Cr3+ = d3; ox = 3 x 4e- = 12 e-; total = 15 e-
7. Au = s1d10; Au1+ = d10; bpy = 2 x 2e- = 4 e-; cyanide = 2 x 2e- = 4 e-; total = 18 e-
Exercise 11.5.1:
Exercise 11.5.2:
Exercise 11.6.1:
a) potassium diaquabis(oxalato)chromate(III) b) pentaamminebromocobalt(III) nitrate c) dichlorobis(ethylenediamine)chromium(III) hexafluorophosphate
d) bis(bipyridine)chlorohydroxocobalt(III) perchlorate e) triaquatrichlorotitanium(III) f) potassium hexacyanoferrate(III)
g) sodium bipyridinedicyanoaurate(I)
Bipyridine is sometimes called "bipyridyl".
Exercise 11.6.2: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/11%3A_Transition_Metal_Complexes/11.08%3A_Solutions_for_Selected_Problems.txt |
Macromolecules are very large molecules. Their molecular weights can range from the thousands to the millions. They can have very different shapes, although the most common structure involves a long chain. Typically they are constructed from small, repeating units linked together to form this long chain.
These macromolecules are also called "polymers". Polymers are made of many small molecules linked together. These molecules are linked or "enchained" by covalent bonds. The small molecules that come together to form polymers are called "monomers". Sometimes small numbers of monomers can come together to form medium-sized molecules; there might be two repeating units, forming a dimer, or three repeating units, forming a trimer, and so on. These mini-polymers are often called oligomers in general.
The most common, naturally-occurring polymer in the world is cellulose. Cellulose is found in plants, where it plays an important structural role, lending strength to cell walls. We use cellulose fibers in a variety of applications. Cellulose from wood is the major component of paper and cardboard. Cellulose is also obtained from cotton fibers and is used to make fabric, usually for clothing. Paper currency is made of cellulose, too; usually the source is cotton, as well as some amount of linen.
Cellulose is made of a long chain of sugar units strung together in a chain. In the picture, the dashed lines indicate that the chain is continuing on. The individual sugar units are all D-glucose. These polyhydroxylated rings are all enchained in exactly the same way, forming a repeating pattern from one end of the molecule to the other.
It is very common to depict polymers with this repeating unit shown in parentheses. A subscript number after the parentheses indicates how many times this unit is repeated. If the number isn't known, this variable is indicated with the letter n.
The polymer is formed by enchaining a number of glucose monomers together. The particular form of glucose is α-D-glucopyranose. At the ends of the polymer chain, there are two hydroxyl end groups.
When drawn this way, the etching of the polymer hardly differs from how we would depict the monomer.
The world's second-most common, naturally-occurring polymer is lignin. Like cellulose, lignin is found in the cell walls of plants, where it provides strength and structural support. However, lignin is generally associated with woody plants, and it is much stronger, less flexible and more durable than cellulose. In fact,because of this durability, the removal of lignin from wood pulp poses a significant obstacle during the paper manufacturing process.
Lignin has a more complicated structure than cellulose. It is also formed from monomers, but the structures of those monomers are less regular than in cellulose. Furthermore, they are put together in more complicated ways, rather than all being joined together in an identical fashion.
Like lignin, proteins are also formed from multiple types of monomers. However, these monomers are all enchained in exactly the same way. Proteins have a perfectly repeating "backbone", although the "pendant" groups attached to this backbone vary from one unit to another.
In the picture, this backbone is shown in red. We can simplify the drawing by denoting all of the black, variable pendant groups as "R".
As with cellulose, the picture of the polymer, when drawn this way, looks very much like a picture of the monomer. In either case, there is an acid-base equilibrium involving the amine group and carboxylic acid group in the monomer, or what are termed the N-terminus and C-terminus in the polymer.
Commercial polymers are used in almost all aspects of daily life. Like the common biopolymers, they are constructed of small monomers assembled into long chains. Many common polymers are familiar to us from the recycling codes on the bottom of bottles and other packaging.
Polymer Recycling Code Example Uses
polyethylene terephthalate (PETE) 1 drink bottles
high density polyethylene (HDPE) 2 bottles for food and household cleaners
polyvinyl chloride (PVC) 3 plumbing, siding, window frames, shampoo bottles
low density polyethylene (LDPE) 4 carpet, fabric, squeezable bottles
polypropylene (PP) 5 bottle caps, straws, medicine bottles
polystyrene (PS) 6 CDs, disposable cups & plates
everything else (PLA, PC, ABS, etc.) 7 food packaging, electronics casing, signs
Drawings of these polymers (see problems below) illustrate that most of them have a long "backbone" with "pendant groups" attached periodically along that chain. Mcuh of the properties of commercial properties arise from that long chain structure. The different properties of these structures stem from the differences in those pandant groups, as well as alteartions in the backbone structure from one polymer to another.
Exercise \(1\)
Revise the following drawings to show the repeating unit in parentheses.
Exercise \(2\)
Expand the following polymer structures to show at least three repeating units.
Exercise \(3\)
Find an application for each of the polymers in the previous question.
Exercise \(4\)
HDPE and LDPE have very similar structures. Like several other common polymers, such as PVC, PP and PS, they are olefin polymers in which the monomer makes a two-carbon repeating unit that appears in the backbone.
1. Looking at the structures of both polymers, provide a structural picture in which the repeating unit is shown in parentheses.
2. LDPE has a slightly lower density than HDPE. Explain why.
Answer a
Answer b
Chains of HDPE are able to pack more tightly together than LDPE, in which the branches prevent tight packing of the backbone. That difference makes LDPE less dense than HDPE. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/12%3A_Macromolecules_and_Supramolecular_Assemblies/12.01%3A_What_are_Macro.txt |
In their liquid state, polymers seem thicker than regular liquids. They are more like honey than tea. They are difficult to stir, vexing to pour, and impossible to draw up into a pipette. With tea, it's much easier to do all of these things. We describe this property of "thickness" as "viscosity".
• Those tea-like liquids have low viscosity.
• Those honey-like liquids have high viscosity.
One way to think about motion through liquids is to imagine the liquid divided into layers. Of course it isn't, really. The reason we are thinking of it this way is to imagine what will happen when an object passes through the liquid, parallel to these layers. Maybe we're looking at a bullet being fired from a gun into a test chamber on CSI.
In a low-viscosity liquid, as the bullet moves through the liquid, it pushes a couple of layers out of the way. It does not have much effect on the surounding layers.
As the bullet moves through a high-viscosity liquid, it still displaces the layers in front of them, but those layers drags the surrounding layers along, too. Those surrounding layers, in turn, drag along the layers next to them.
What is the difference between these two scenarios? On the molecular level, it has to do with the interaction between the molecules in one layer and the next. If the molecules are completely isolated, they don't interact with each other at all. The bullet just pushes the molecules ahead of it forward without disturbing the ones on either side.
A macromolecule is different from regular molecules because a force on one part of the molecule will be transmitted very far away. We can imagine a macromolecule spanning many layers, so that when the bullet hits one end of the molecule, the other end of the molecule transmits this force to its neighbors.
Viscosity is not a phenomenon limited to macromolecules. In fact, liquids don't really exhibit the no-viscosity case outlined above, because the molecules do interact with each other. However, different liquids do exhibit different viscosities, and macromolecules are on the high-viscosity end of the spectrum.
In reality, macromolecules are not like long sticks tumbling over each other. They are floppy and can change shape because of bond rotations; they can undergo "conformational change". As a result, macromolecules become "entangled" with each other.
Entanglement is a very important feature in polymer behavior. It influences the viscosity of a polyer in its melted state, which affects how easily the polymer can be molded into different shapes for different applications. Not only that, but it strongly influences the physical strength of a plastic in its regular, room-temperature state. The more entangled the polymer chains, the stronger the material.
• Entanglement increases viscosity of a polymer in the melt phase.
• Entanglement increases the strength of a polymer in the "solid" phase.
Exercise \(1\)
Using water as a comparison point, explain the relative values of viscosity (η) of the following liquids, recorded near ambient temperature, 25 °C). Water: η =1.00 mPa s. (e.g. this one is a little lower than water because...)
1. ethylene glycol or 1,2-hydroxyethane: η = 16.1 mPa s.
2. olive oil: η = 84 mPa s.
3. methanol: η = 0.58 mPa s.
4. 2-propanone: η = 0.34 mPa s.
5. hexane: η = 0.294 mPa s.
Answer a:
Water can in principle form four hydrogen bonds: it can donate two and accept two. Ethylene glycol has two OH groups, so it can donate two H bonds and accept four. That enhanced clinginess is responsible for its higher viscosity.
Answer b:
Olive oil does not hydrogen bond; it isn't even very polar. However, it is a triglyceride and its fatty acid chains are about 80% oleic acid and linoleic acid, which are both 18-carbon chains. So the average olive oil molecule contains three 18 carbon branches, leading to a modest level of entanglement. Its viscosity is much higher than water's.
Answer c:
Methanol has only one OH bond, so it has a little less hydrogen bonding than water, but not enough of anything else to form other interactions. ts viscosity is a little lower than water's.
Answer d:
2-propanone has no hydrogen bonding at all, although it does have a dipole moment to hold different molecules together. Its viscosity is a little lower than methanol's.
Answer e:
Hexane has no dipole and no hydrogen bonding. It is really too short to become entangled. Molecules only attract each other through weak London dispersion interactions. Its viscosity is a little lower than 2-propanone's.
Exercise \(2\)
Explain why the viscosity of a melted sample of polyethylene, (CH2CH2)n, depends on the molecular weight of the polymers in the sample.
Answer
The greater the number of repeating units, the higher the molecular weight. Also, the greater the number of repeating units, the greater the entanglement and the higher the viscosity.
A curious example of a very high-viscosity liquid is illustrated by the pitch-drop experiment currently being performed at the University of Queensland, Australia. In this experiment, bitumen was poured into a glass funnel and the rate at which the bitumen drips through the funnel has been monitored subsequently. The experiment was set up in 1927. Ten drops have fallen so far. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/12%3A_Macromolecules_and_Supramolecular_Assemblies/12.02%3A_Viscosity.txt |
Note
comprehension of this section may be aided by an understanding of conformational analysis.
Even though it appears to be a solid, a plastic bottle that you might throw into your recycling bin has some things in common with liquids. We think of the molecules in a solid as mostly static; maybe they rock back and forth a little bit, but they mostly stay put. The long chains in a polymer are constantly in motion, however.
These chains slide over each other like snakes in a process called "reptation" or "creep". This flowing motion of the chains is responsible for the flexibility of many plastics, which can often be bent easily even when they are solids.
When we exert a force upon a polypropylene bottle, it changes shape. Rather than being locked together, the polypropylene chains give way and move past each other. When we let go, the chains may just slide back to their original position, or close to it, returning the bottle to its original shape.
The chains slide back because they were originally in a stable conformation. Rapidly deforming the plastic twists the chains into less favorable positions. Left alone, they twist back into positions that are energetically more favorable.
Of course, it is always possible that the plastic is deformed a little too far and does not bounce back to its original shape. In that case, the chains have become stretched out in such a way that they have become stuck. They may have become tangled in new ways. New obstacles have appeared that make it difficult for them to slide back again.
Sometimes, a force is exerted on a plastic material over a long period of time, causing the chains to slowly slide away from their original positions. The chains adopt a new equilibrium distribution. When the force is removed, the plastic has adopted a new shape. A formerly straight bar might become bowed under a weight, for example.
Exercise \(1\)
In the drawing below, we are looking down a C-C bond in a polymer chain (left). We can see hydrogens attached to the carbon in front and the carbon behind, although the carbons are not labelled. We can see the where the polymer chain is attached, labelled "CH2P". A cross-section of the neighbouring chain is also shown (in black).
Show, with the help of drawings, how a 360° rotation about that bond will cause the polymer chain on the left to creep over the one on the right.
One of the most commonly-used diagnostic tests for a polymer is the measurement of the glass transition temperature (Tg). The glass transition temperature is related to the ease of chain flow in the material.
The glass transition represents a phase change. As a polymeric material is cooled, it changes from a more flexible, rubbery state to a more rigid, glassy state. This transition occurs because of the change in volume with temperature. Generally, solids shrink as they are cooled. At some point, the volume of a polymer decreases to the point at which the chains can no longer move past each other; there just isn't enough room.
Answer
You can think about a length of chain flopping over a neighbouring strand like a grappling hook, then pulling itself along as the bond continues to rotate, vaulting over the other chain.
The glass transition temperatures of a number of common polymers are listed below. These data indicate whether a given polymer will exist in a rubbery state or a glassy state at a particular temperature.
Name Abbreviation Approx. Tg, °C
low-density polyethylene LDPE -120
polyethylene oxide PEO -50
polypropylene PP -10
nylon-6 - 45
polylactide PLA 65 (L isomer)
poly(ethylene terephthalate) PETE 70
polyvinylchloride PVC or V 90
polystyrene PS or S 100
poly(methyl methacrylate) PMMA 110
polycarbonate PC 145
polynorbonene - 215
kevlar - 240
These values are only approximate. The glass transition temperature actually depends on a number of factors, including the average molecular weight of the polymer and how the data is measured.
Exercise \(2\)
Explain why glass transition temperature would depend on molecular weight.
Exercise \(3\)
Glass transition temperature is influenced by several factors, including the following ones.
1. sterics (crowdedness, or physical interference between groups as they move through space)
2. polarity
3. backbone rigidity (ability for bond rotation along backbone)
Answer a:
The greater the number of repeating units, the greater the entanglement. Entanglement acts as an impediment of the movement of chains, increasing the temperature at which chain motion becomes restricted (in other words, increasing the minimum volume needed to get the chains moving again). For example, polypropylene has a higher Tg than polyethylene, and that of polystyrene is even higher (although additional factors influencethe Tg of PS).
Answer b:
Increased polarity increases the attraction between chains and can even lead to formation of physical crosslinks between chains. This increased interaction hinders chains from moving freely and raises the glass transition temperature. For example, polylactide, poly(vinyl chloride) and poly(methyl methacrylate) all have much higher values of Tg than polypropylene.
Answer c:
If the backbone is rigid, reptation is hindered because the chain can't adopt different conformations. Chain flow is restricted more easily and the Tg is higher. Kevlar and polynorbornene both have stiff backbones because of the aromatics and aliphatic rings, respectively, that form part of the backbone. Both polymers have much higher values of Tg than polystyrene, in which the aromatics hang from a flexible chain.
In each case, explain how the factor would influence the glass transition temperature, and choose some examples from the table to support your claim.
Exercise \(4\)
Polymers also undergo other phase transitions. For example, the melting point of polystyrene is around 240 °C. Explain, with the help of drawings, the difference between the glass transition temperature and the melting temperature.
Exercise \(5\)
Indicate whether the following materials would be found in a rubbery state or a glassy state at a comfortable room temperature, about 25°C.
a) LDPE b) PS c) PVC d) PP
Answer a:
rubbery
Answer b:
glassy
Answer c:
glassy
Answer d:
rubbery
Exercise \(6\)
Indicate whether the following materials would be found in a rubbery state or a glassy state at the boiling point of water, about 100°C.
a) PEO b) PETE c) PMMA d) PC
Exercise \(7\)
Why do you think PVC is used in household plumbing, rather than other materials such as PLA or PEO? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/12%3A_Macromolecules_and_Supramolecular_Assemblies/12.03%3A_Glass_Transiti.txt |
The basic structure of a polymer is a chain. The chain is formed by the backbone of the polymer. It may be decorated with pendant groups, such as methyls in polypropylene, chlorines in polyvinyl chloride or phenyls in polystyrene, but the basic structure is like a long string.
Sometimes, the architecture of a polymer can be more complex than that. For example, low-density polyethylene has branches coming off the main chain. Consequently, it is described as a branched polymer.
In some cases, branching is so extensive that the polymer does not resemble a chain at all. Some of these highly-branched polymers are called dendrimers. Dendrimers are tree-like structures that branch out in all directions from a central point. Viewed from a distance, a dendrimer would have an overall shape that is more like a fuzzy ball than like a coiled chain.
If the branches on one branched chain connect directly to other chains, then the polymer is said to be crosslinked. Crosslinked polymers can have extra strength and rigidity because they are less flexible than polymers in which the chains are able to move independently. However, just the right amount of crosslinking can make a polymer more elastic, making it bounce back to its original form after it is flexed.
Crosslinking can actually be of two different types: "chemical" or "physical". Chemical crosslinks are permanent covalent bonds between the chains. Physical crosslinks are more temporary; they are formed by strong intermolecular attractions, such as hydrogen bonds, between chains.
Exercise \(1\)
Choose an architecture that would work well for each of the following applications, and explain your choice.
1. a thread for weaving into fabric
2. a rubber band
3. a "drug delivery device" that would slowly release trapped pharmaceuticals into the body
4. a plastic film to protect a surface
Answer a:
A regular chain structure could be wound into a thread easily. There would be enough entanglement to held the threed together.
Answer b:
A cross-linked structure would help a rubber band snap back to its original shape. However, too much cross-linking would make it too difficult to stretch the rubber band.
Answer c:
A dendrimer has lots of nooks and crannies that might be used to sponge up pharmaceuticals and slowly release them later.
Answer d:
A branched or cross-linked structure might form a good two-dimensional network to provide a protective coating.
Copolymers
The structure of polymers can vary in how their chains are arranged. Additional structural complexity arises if the polymer is a copolymer. Regular polymers are made from one kind of monomer. Copolymers are formed from two different monomers.
If there are two different monomers, they may be arranged in different ways along the chain. Maybe the two monomers simply alternate, one after the other: that's called an alternating copolymer. Maybe they are distributed at random: that's a random or statistical copolymer. Maybe they are clustered in two separate groups: that's called a block copolymer.
These copolymers are different from polymer blends. Polymer blends are just mixtures of two different kinds of polymers. Both polymer blends and copolymers are important. Either one might allow different properties of each polymer, such as strength and flexibility, to be brought together in one material.
There are different kinds of block copolymers, depending on how the blocks of each monomer are arranged in the chain. The simplest kind of chain would contain one long block of the first monomer followed by one long block of the other. That's called a diblock (or AB diblock) copolymer . If two blocks of the same polymer are found on the ends, with a different one in the middle, that would be called an ABA triblock copolymer. If there were three different blocks from three different monomers, we would have an ABC triblock copolymer instead. There are lots of possible variations.
Sometimes, the blocks in a block copolymer are not found along the chain. For example, in some cases, the entire backbone may be composed of one monomer, with pendant chains attached to the backbone. A material with this type of architecture is called a graft copolymer, or sometimes a comb or a brush copolymer. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/12%3A_Macromolecules_and_Supramolecular_Assemblies/12.04%3A_Polymer_Archit.txt |
Morphology is the study of shapes. In the context of polymeric materials, morphology describes the distribution of phases within the material.
In general, polymers tend to exist as amorphous solids. Unlike metals or ionic solids, their components are not found in ordered layers or rows; those long chains don't line up nicely, but they get all tangled up instead. Whether in the glassy state or the rubbery state, the polymer chains aren't organized.
Oligomers -- smaller molecules made of just a few monomers -- crystallize more easily because their chains are too short to really get tangled up. If they are cooled down from the liquid phase, they can get arranged into orderly rows and form crystals.
Well, in reality, short sections of polymer chains can act like oligomers. If a polymer is cooled down from the melt, several parallel lengths of chain might actually settle into a crystalline pattern together. However, at some point one or more of the chains will get twisted up, bringing this ordered section to a halt as it trails off into an amorphous mess.
Frequently, polymers contain small "microcrystalline domains" surrounded by "amorphous domains." Most of the chains are tangled up, but little sections of them are crystalline. This type of phenomenon is called "microphase separation"; the material is not actually in the same phase all the way through.
Microphase separation is particularly important in copolymers. Sometimes, two different blocks don't completely mix together in the solid state, especially when doing so would disrupt strong interactions between the chains within one of the blocks. Maybe one of the blocks is actually more crystalline than the other. In that case, the different blocks might microphase separate into crystalline and amorphous domains.
That kind of behavior can be very useful. The amorphous domains are very flexible. Those chains can be stretched and twisted without much problem. The crystalline domains, on the other hand, are very strong. They are like anchor points within the structure. Together, these two structures combine to make something that can be stretched, but only so far before it will snap back. They could make an elastic material, or an elastomer.
In many cases, the phase separation isn't between a completely crystalline and amorphous domains, but between a glassy domain and a rubbery domain. In that case, it's the same concept: parts of the material flow easily, whereas other parts do not. The latter segments hold the material firmly in place and help snap it back to its original shape after stretching.
Exercise \(1\)
The following ABA triblock copolymers can be used to make chewing gum. In each case, identify the part that would form the more amorphous microphase and the part that would form the more crystalline microphase, and explain your reasoning.
Answer
The polar vinyl acetate block of the first polymer contains carbonyl groups that would attract each other by dipole-dipole interactions. It is easy to imagine these segments forming physical crosslinks. The hydrocarbon chain in the middle block would prvide flexibility.
Although nonpolar, the benzene groups in styrene are flat and can stack together with relatively strong London dispersion forces. The higher Tg of polystyrene would make these blocks much stiffer than the more flexible polyisoprene block in the middle. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/12%3A_Macromolecules_and_Supramolecular_Assemblies/12.05%3A_Morphology_and.txt |
Macromolecules are very large molecules. Their molecular weights can range from the thousands to the millions. Typically they are constructed from small, repeating units linked together in some way.
How many repeating units are there in a chain? Because of the way synthetic polymers are made, we may be able to predict how long the chain is by the conditions we choose when we make it. Polymers are frequently made through the action of an initiator, which gets the monomer units to start attaching to each other. Without the initiator, the monomers just sit there. (There are exceptions; nylon 6-6 and other condensation polymers are sometimes made just by mixing two different kinds of monomers together, without an initiator.)
If each initiator in the process starts its own chain growing, then the ratio of monomers to intiators that we start with may tell us the chain length we will end up with.
We can confirm that outcome experimentally by measuring the ratio of monomer to initiator in the polymer that is produced. That evidence is sometimes obtained by integration of a 1H NMR spectrum. The "degree of polymerization" is the measured number of repeating units in the chain. In the case below, we started with 10 monomers per initiator and ended up with DP = 10.
Things don't always work out that way. Maybe for some reason the chain stops growing unexpectedly (an event that is often called "chain death"). Maybe we stopped the reaction a little too early, before all the monomers had time to react. Maybe there is a reverse reaction, in which a polymer can break down into monomers again, and we end up at some balance between polymer formation and polymer breakdown (an "equilibrium").
In that case, the degree of polymerization might not reflect the original monomer to initiator ratio. It might also depend upon the percent conversion -- that is, how much of the original monomers were converted into polymer. If we start with 10 monomers per initiator, but only 70% of the monomers are enchained in the reaction, then the degree of polymerization will be 7, not 10.
If we are dealing with a block copolymer, we may be interested in more than just the total number of monomers in the chain. We may instead be interested in the ratio of one monomer to the other (for example, A:B:A = 6:6:6 in the example below).
Note that there are a couple of different ways of making block copolymers. We can simply start growing the chain from one end, as shown above. Alternatively, we might start with a difunctional initiator. That's an initiator that starts gorwing the chain in two directions at once. The result is called a "telechelic polymer". This method of growth can be more efficient in some cases.
Exercise \(1\)
Explain why a telechelic polymer might grow more quickly than a regular one.
Answer
The telechelic polymer has two growing ends per chain, so it grows twice as fast as a normal chain.
Exercise \(2\)
If a polymerization is performed using a 500:1 ratio of initiators to monomers and the reaction reaches 85% conversion, what is the degree of polymerization?
Answer
If this chain reached 100% conversion, then the degree of polymerization would be 500. At 85% conversion, DP = 425. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/12%3A_Macromolecules_and_Supramolecular_Assemblies/12.06%3A_Polymer_Synthe.txt |
Macromolecules are very large molecules. Their molecular weights can range from the thousands to the millions. Typically they are constructed from small, repeating units linked together in some way.
Describing the molecular weight of a polymer is not as straightforward as it is in a small molecule. The fact that each macromolecule is built by joining together a large number of monomers complicates things. Exactly how many monomers are there in one polymer molecule? Twenty? Twenty-five? Thirty? Each of those possibilities would have a very different molecular weight.
This is generally the case in macromolecules, whether we are dealing with commercial or biological samples. Polymers are grown, often through the action of some "initiator" that picks up a monomer and starts knitting together a chain. What if one of the initiators starts a little later than the others? What if one of the chains grows a little faster than the others? What if something goes wrong and one chain stops growing?
As a result of a number of factors, polymers generally display a distribution of molecular weights representing a distribution of chain lengths. There is no exact molecular weight. Instead, an average must be determined.
Exercise \(1\)
Explain the general relationship between degree of polymerization and molecular weight of the chain.
Answer
As the number of repeat units increases, the molecular weight increases.
Exercise \(2\)
Calculate the molecular weight of each of the following polymer chains. For simplicity, assume the end groups are just hydrogen atoms in each case.
1. polystyrene (PS) with DP = 1,000
2. polylactide (PLA) with DP = 200
3. poly(ethylene oxide) (PEO) with DP = 50
4. polypropylene (PP) with DP = 100,000
Answer a:
The molecular weight is the sum of the weights of all the repeat units plus the weight of the end groups.
MW = 1,000 units x 104.15 Da/unit + 2 Da = 104,152 Da
Answer b:
MW = 200 units x 72.06 Da/unit + 2 Da = 14,412 Da (based on 200 simple repeating units, but this is a complicated case. In common practice, PLA is made from a dimeric monomer, with each monomer forming two repeat units in the chain, so a degree of polymerization of 200 really corresponds to 400 repeat units in the chain, which would have twice the molecular weight.)
Answer c:
MW = 50 units x 44.05 Da/unit + 2 Da = 2,202 Da
Answer d:
MW = 100,000 units x 42.08 Da/unit + 2 Da = 4,208,002 Da
Note that, in this case, the end groups make no appreciable difference in the weight because they are so small.
There are many different ways of estimating the average molecular weight of a polymer. These methods fall into two categories, giving rise to two different estimates of the average molecular weight.
In some techniques, the total weight of a sample is measured followed by a determination of the number of molecules present. The molecular weight is just the total weight divided by the number of molecules. This quantity is called the number average molecular weight, or Mn
Exercise \(3\)
1. A sample weighing 3.205 g is determined to contain 5.50 x 10-4 moles. What is Mn?
2. A sample weighing 1.315 g is determined to contain 1.25 x 10-3 moles. What is Mn?
3. A sample weighing 0.225 g is determined to contain 3.20 x 10-7 moles. What is Mn?
Answer a:
MW = weight (in g) / moles
5827 Da
Answer b:
1052 Da
Answer c:
703,000 Da or 703 kDa
A classic example of this type of measurement is a freezing point depression. A pure liquid freezes at a specific temperature, but impurities in the liquid tend to interfere with crystal formation, lowering the freezing point. The change in freezing point depends on the number of molecules present in the impurity. In this approach, a measured amount of polymer would be dissolved in a specific quantity of toluene and the freezing point would be measured very carefully. The polymer in this context is acting as an impurity. The change in freezing point would indicate the number of molecules (or moles) of the impurity. The average molecular weight could then be calculated.
Other techniques rely on measurements that depend on the relative size of the macromolecules, rather than the number of them. The quantity determined this way is called the weight average molecular weight, or Mw.
A classic example of this type of measurement is a light-scatterring experiment. A beam of laser light is shone into a solution of polymer. Some of the light is scattered when it encounters polymer molecules, and the interference patterns in the scattered light can be decoded to give information about the particle size.
Typically, techniques that depend on particle size are more heavily influenced by the larger particles, and so Mw > Mn. The closer these numbers are to each other, the more uniform the partciles must be. The more different they are, the broader the distribution present. Consequently, the ratio Mw/Mn, historically called the polydispersity index (PDI), is an important indicator of how uniform the molecules are in size. "Polydisperse" as an adjective means that there is a wide distribution of sizes, as opposed to "monodisperse", meaning everything is the same size.
IUPAC has recently decided that the term "polydispersity index" should be abandoned in favor of the more concise term, dispersity (Đ).
Exercise \(4\)
Determine the polydispersity in each of the following cases.
1. Mn = 2,105; Mw = 2,408
2. Mn = 102,365; Mw = 124,710
3. Mn = 56,300; Mw = 87,530
4. Mn = 8,105; Mw = 14,220
5. Mn = 1,170,000; Mw = 1,899,000
Answer a:
Đ = Mw / Mn
1.14
Answer b:
1.22
Answer c:
1.56
Answer d:
1.76
Answer e:
1.62
These days, two of the most common techniques for estimating molecular weight are 1H NMR spectroscopy and size exclusion chromatography (although other techniques such as light scattering are still used). We'll take a closer look at NMR analysis in a minute. First we'll start with a simplified view of size exclusion chromatography (SEC) or, as it is sometimes called, gel permeation chromatography (GPC).
In SEC, a sample is sent through a column containing a porous gel. Generally, smaller molecules that wander into the pores within the gel take longer to get through the column. Larger molecules that can't fit into the pores emerge sooner. The effluent from the column is monitored by a technique such as UV-Vis spectroscopy or refractive index. The response from this detector changes over time, with a stronger response indicating more molecules are being detected as they come out of the column. The output from the instrument shows a distribution, not unlike a bell curve, indicating the distribution of molecular weights by comparison with a standard sample.
The drawing above shows you what a chromatogram or trace would look like for two different samples, one of higher molecular weight and one of lower molecular weight. The retention time in a GPC trace does not directly indicate the molecular weight of the polymer. A calibration has to be run. In a calibration, polymers of known molecular weight are run through the GPC and their chromatograms are obtained. The GPC trace of an unknown polymer, run on the same instrument under the same conditions, is then compared to the standard ones. Software is used to estimate the molecular weight and the polydispersity.
Exercise \(5\)
Suppose the following drawing shows calibration traces for the chromatogram shown above (D stands for Daltons, another name for the units "g/mol"). Estimate the molecular weight of
1. the low molecular weight unknown.
2. the high molecular weight unknown.
Answer a:
3,000 D
Answer b:
5,000 D
Exercise \(6\)
SEC, which is based on the size of a coiled polymer molecule, can sometimes give misleading estimates of the molecular weight. Explain why with the help of the following diagram.
Answer
The longer the chain in polymer A, the bigger the sphere it coils into. There isn't much problem comparing samples of polymer A to each other, provided we have some way of calibrating sphere size to chain length.
However, polymer B appears to be more tightly coiled than A. A relatively small sphere could pack a lot more chain. We would have to be cautious about comparing the sphere size of B to that of A and drawing any conclusions about chain length or molecular weight.
NMR spectroscopy is a method that uses magnetic fields and radio waves to detect different kinds of atoms in a sample. 1H NMR detects the hydrogen atoms in a sample. It is conceptually very similar to MRI used in hospitals and clinics. In the experiment, the sample is placed in a very strong magnetic field (somewhere on the order of a hundred thousand times greater than the earth's magnetic field on the planet's surface). Radio waves are sent through the sample. Most of them pass straight through, but some of them are absorbed by the hydrogen atoms.
The result of the NMR experiment is an NMR spectrum. A spectrum is really just a graph. On the x axis, it shows all the possible frequencies of the radio waves that were sent through the sample. On the y axis, it shows the absorption level: in other words, how much of the radio waves at each frequency that were absorbed by the sample. Did a particular frequency of radio waves just pass right through the sample? The absorbance at that frequency would be zero. If a particular frequency is absorbed, there will be a "peak" on the graph (a high point) at that frequency.
Each peak in the NMR spectrum corresponds to a hydrogen in a different environment. Maybe the hydrogen is attached to an oxygen atom, or a carbon atom, and so on. In the drawing shown above, the spectrum indicates three different locations for hydrogen atoms in the molecule. Note that "frequency" in NMR is called "chemical shift", for historical reasons. The units of chemical shift are parts per million, or ppm, for reasons that have to do with the mechanics of the experiment.
But how does that help us with molecular weight? We're getting there.
1H NMR spectroscopy also allows us to measure how many hydrogens we have of each type, at least in a relative sense. It could tell us we have equal amounts of hydrogens in these three locations, or maybe twice as many in one location as there are in the other two. The relationship is pretty simple: the more hydrogens there are in a particular location, the bigger the peak associated with them. That's because if one hydrogen atom absorbs one radio frequency photon, two hydrogen atoms would absorb two radio frequency photons. That's twice as much. The peak should be twice as big for two hydrogens as it would be for one hydrogen.
What do we mean by "bigger", though? It turns out that the height of the peak does not really tell us anything. What matters is the area of the peak. That's because even for two seemingly identical protons in a sample, there are lots of factors that mess with the radio frequency absorbed, ever so slightly. So, if a specific hydrogen in one molecule absorbs a radio photon at 1.7005 ppm, a hydrogen in the exact same position in a second molecule might absorb at 1.7007 ppm. There are lots of complicated reasons for that. The result, however, is that NMR peaks get smeared out a little bit, so that it isn't the height that matters, but the area, which is the sum of a whole lot of very similar radio frequencies absorbed by those identical hydrogens in different molecules.
So we need to measure the areas of those peaks. Fortunately, NMR spectrometers are equipped with software that can do this. Classically, the instrument would print a red line across the spectrum. The red line would hop up every time it passed over a peak. The amount of each hop corresponded to the relative area of the peak.
Notice that we don't even bother to label the y axis anymore. It does not tell us anything, anyway.
The drawing below emphasizes what we are measuring with the integral line. It is the vertical distance that the integral line hops up as it passes a peak. That distance is called the "integral". In this case, the integrals of the three peaks are all about equal.
Now it's time to see how 1H NMR spectroscopy can reveal molecular weight through a measurement called "end group analysis".
Consider the polymer poly(ethylene glycol), or PEG. PEG is commonly used in medicine. By itself it can be used as a laxative, depending on its molecular weight. It can also be mixed with pharmaceuticals; it helps control side effects and spreads the medicinal effect over a longer time, so you don't have to take the medicine as often.
Frequently, there is an OH group at both ends of the chain. In the middle, the polymer is made of CH2CH2O repeating units.
PEG can be thought of as a polymer of ethylene glycol (or ethane-1,2-diol). The structure of ethylene glycol is provided below. The structure of ethylene glycol shows two different kinds of hydrogens. There are the hydrogens along the chain, attached to carbons, and there are hydrogens at the ends of the chain, attached to oxygens. The 1H NMR spectrum of ethylene would show a different peak for each of those types of hydrogens. There would be two peaks.
The spectrum also shows an integration line. We have measured the integral for each peak. It looks like a 2:1 ratio, because the vertical distance in one case is 24 mm and in the other case it is only 12 mm. Even if you don't know anything else about NMR spectroscopy, you can see that the peak around 4 ppm must correspond to the hydrogens on the carbons, whereas the peak around 1 ppm must correspond to the hydrogens on the oxygens. That's because there are four hydrogens on the carbons and only two on the oxygens. The ratio is 2:1.
The structure of PEG is pretty similar to that of ethylene glycol, except that it has a bunch of units strung in a row. The NMR spectrum of PEG is pretty similar to that of ethylene glycol, too.
In end group analysis, we compare the integral of a hydrogen found at the end of the chain to the integral of one found along the chain, in the repeating unit. In PEG, the end group is the OH group at either end. The repeat unit corresponds to the hydrogens on the carbons.
In the case above, there are two hydrogen in the end groups (one at each end). The four repeat units have four hydrogens each, for a total of sixteen. The integration ratio should be 16:2, or 8:1. The measured ratio is 48:6, or 8:1.
Contrast that result with the case of ethylene glycol, in which there isbasically just one repeat unit. The ratio in the oligomer (short polymer) is four times greater than the ratio in the monomer. That's because there are four times as many repeat units in the oligomer. We sometimes call the number of repeat units in the chain the "degree of polymerization".
Exercise \(7\)
Expand the above structure of the 4-unit PEG oligomer to show the whole chain, rather than just the repeat unit in parentheses.
Exercise \(8\)
What would be the integration ratio in the following oligomer?
Answer
16:1
To estimate chain length in NMR, a peak must be found from one of the repeating units along the chain as well as a peak from an end group, found only at the end of the chain. The integration of the repeating unit to the end group indicates the average number of repeating units per polymer. This technique essentially provides Mn.
Exercise \(9\)
What is the molecular weight of the 4-unit PEG oligomer, the NMR spectrum of which was shown above?
Answer
It's OH + 4 x (CH2CH2O) + H. That's 17 + (4 x 44) + 1 = 18 + 176 = 194 D.
In the next example, you are going to see a peculiar fact of life. The truth is, OH groups often don't show up well in 1H NMR spectroscopy (for complicated reasons). That can complicate things. To get around that problem, people often make polymers in such a way that there is a different end group, other than an OH, so that the integration ratio can be measured more easily.
Exercise \(10\)
Use the integration ratio in the following spectrum to estimate the degree of polymerization.
Answer
The ratio of c to b is 145:18 = 8:1. However, the number of hydrogens in the repeat unit c is 4; the number of hydrogens in the end group b is 2. That ratio is 2:1. That means the comparison we need is (145/4)/(18:2) = 4:1. The degree of polymerization is 4. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/12%3A_Macromolecules_and_Supramolecular_Assemblies/12.07%3A_Molecular_Weig.txt |
Supramolecular assemblies are a different type of large structure, related to macromolecules. In a supermolecular assembly, parts of the structure are held together by very strong interactions, but not necessarily by covalent bonds. For example, a polymer in which chains are held together by physical crosslinks such as hydrogen bonds may be considered to form a supermolecular assembly.
DNA is an example of a supramolecular assembly. Each strand of DNA is an alternating sugar-phosphate copolymer. Along this backbone, there are pendant "base" units, adenine, cytosine, guanine or thymine.
Two DNA strands form a helical supramolecular assembly through hydrogen bonding interactions that form between the bases. Thymine can hydrogen bond preferentially with adenine.
Cytosine hydrogen bonds preferentially with guanine.
There are a number of different ways in which synthetic supramolecular assemblies can be formed. These kinds of structures have been getting increased attention in recent years as researchers try to develop new materials with useful properties. There are a few common methods through which the assemblies are formed. We will look at hydrogen bonding, host-guest chemistry and coordination chemistry.
Hydrogen Bonding
Hydrogen bonding is a common way to build supramoleucular assemblies. Bert Meijer at Eindhoven Technical University in the Netherlands has developed a number of systems that use this approach. For example, graft copolymers have been assembled in his lab by hanging pendant chains onto a polymer backbone via hydrogen bonds.
This approach can be pictured in cartoon, below, in which the hydrogen bonding elements are boxes and the lines are flexible chains or tails.
Exercise \(1\)
The diffusion coefficients cited in the picture above are related to how easily compounds can move through a solution. The higher the diffusion coefficient, the more easily the compound moves through solution.
1. How is the diffusion coefficient related to viscosity?
2. Explain the difference between the diffusion coefficients of the two compounds above.
Answer a:
Viscosity is related to the amount of drag in the fluid. The more drag, the slower solutes will be able to move, so the diffusion coefficient is inversely related to viscosity (i.e. higher diffusion coefficientshould correlate with lower viscosity).
Answer b:
The regular chain polymer diffuses more easily than the graft copolymer / supramolecular assembly because the pendant chains in the latter lead to increased entanglement.
Host-Guest Complexes
Another approach to supramolecular assemblies is to exploit host-guest interactions to bring different parts of the assembly together. Host-guest complexes act a little bit like enzymes and their substrates in biology, or like materials bound inside zeolites. The idea is that one molecule has a cavity or opening big enough for another molecule to fit inside. It's a little like a basketball fitting into a peach basket.
There might be a physical attraction (an intermolecular force) between a host and a guest, but there does not have to be. Julius Rebek at the Scripps Institute has developed host complexes that will hold normally aloof guests such as N2. Part of the reason guests can become comfortably ensconced without intermolecular attractions involves the fact that the molecule inside would have to take a very particular path in order to escape.
A number of host compounds are shown below. Each of them is designed to have an enclosed space of some kind. In some cases it's no more than a ring, but that is enough to keep the guest from escaping on those sides.
Exercise \(2\)
The following host-guest system was designed in the lab of Taichi Ikeda at the National Institute for Materials Science in Japan. In the polymer notation, "x" and "n-x" are just numbers of repeating units. When solutions of the two polymers are mixed together, an elastic gel is formed.
1. Draw a cartoon of each of the polymers and of the host-guest linked assembly.
2. What makes this host and guest especially attractive for each other?
3. Describe the structure of this assembly, in terms of polymer architectures seen previously.
4. Why is the gel elastic?
5. This gel is "self-healing": it can be cut with a knife, but the pieces can then be stuck back together again. Explain how this re-attachment occurs.
Answer b:
The lone pairs on the oxygens in the host would be attracted to the positively charged nitrogens in the guest. The N-H groups in the guest would even act as hydrogen bond donors to the oxygens.
Answer c:
This supramolecular assembly forms a cross-linked architecture.
Answer d:
The gel is elastic because, although it contains flexible chains and can be distorted up to a point, at some point the crosslinks pull it back to its original position.
Answer e:
When the gel is cut, probably some guests are pulled away from their hosts along the surface of the cut. WHen the edges are placed back together, the guests can find new hosts and form the cross-linked structure again.
Coordination Compounds
If you don't know about coordination compounds, you might want to start with Lewis acid-base complexes.
A number of such assemblies have been reported from the lab of Stephen Craig at Duke University. For example, mixing a solution of certain palladium compounds into a solution of poly(vinylpyridine) (PVP) produces a thickening of the liquid into a honey-like consistency.
As a cartoon, this system can be represented as shown below. The palladium complexes form physical crosslinks between the pendant pyridine units on the polymer chains.
Exercise \(3\)
This is a challenge problem for students of Chem 251.
Craig's lab reported the following kinetic data for the poly(vinylpyridine) / MLn system in DMSO (dimethylsulfoxide) solvent. Remember, MLn just means "coordination complex".
1. The rate constant kd refers to the dissociation of a PVP/MLn linkage. Show a likely mechanism for that reaction.
2. Explain why the reaction is faster with the methyl-substituted ligand than the ethyl-substituted one.
3. Is Kb the equilibrium constant for formation of the PVP/MLn linkage or for dissociation of the linkage?
4. Explain the large difference in values of Kb between the palladium complexes and the platinum complexes.
Answer a:
The palladium in the complex is Pd(II), so it is d8 and the structure of the crosslink is square planar at palladium. Ligand exchange likely occurs through an associative process. That means that before the PVP dissociates, a new ligand would bind; this ligand is probably a solvent molecule, DMSO.
Answer b:
The increased crowding in the ethyl-substituted ligand, compared to the methyl, would hinder the approach of DMSO during substitution, slowing down the reaction.
Answer c:
Kb is defined as ka/kd, the rate constant for the association step over the rate constant for the dissociation step. It is the binding constant for PVP-Pd formation.
Answer d:
Palladium is a second row transition metal whereas platinum is a third row transition metal. Third row metals have greater ligand field splitting energies than second row metals and are generally less labile. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/12%3A_Macromolecules_and_Supramolecular_Assemblies/12.08%3A_Supramolecular.txt |
Exercise 12.1.1:
Exercise 12.1.2:
Exercise 12.1.4:
b) Chains of HDPE are able to pack more tightly together than LDPE, in which the branches prevent tight packing of the backbone. That difference makes LDPE less dense than HDPE.
Exercise 12.2.1:
1. Water can in principle form four hydrogen bonds: it can donate two and accept two. Ethylene glycol has two OH groups, so it can donate two H bonds and accept four. That enhanced clinginess is responsible for its higher viscosity.
2. Olive oil does not hydrogen bond; it isn't even very polar. However, it is a triglyceride and its fatty acid chains are about 80% oleic acid and linoleic acid, which are both 18-carbon chains. So the average olive oil molecule contains three 18 carbon branches, leading to a modest level of entanglement. Its viscosit is much higher than water's.
3. Methanol has only one OH bond, so it has a little less hydrogen bonding than water, but not enough of anything else to form other interactions. Its viscosity is a little lower than water's.
4. 2-propanone has no hydrogen bonding at all, although it does have a dipole moment to hold different molecules together. Its viscosity is a little lower than methanol's.
5. Hexane has no dipole and no hydrogen bonding. It is really too short to become entangled. Molecules only attract each other through weak London dispersion interactions. Its viscosity is a little lower than 2-propanone's.
Exercise 12.2.2:
The greater the number of repeating units, the higher the molecular weight. Also, the greater the number of repeating units, the greater the entanglement and the higher the viscosity.
Exercise 12.3.1:
You can think about a length of chain flopping over a neighbouring strand like a grappling hook, then pulling itself along as the bond continues to rotate, vaulting over the other chain.
Exercise 12.3.2:
1. The greater the number of repeating units, the greater the entanglement. Entanglement acts as an impediment of the movement of chains, increasing the temperature at which chain motion becomes restricted (in other words, increasing the minimum volume needed to get the chains moving again). For example, polypropylene has a higher Tg than polyethylene, and that of polystyrene is even higher (although additional factors influencethe Tg of PS).
2. Increased polarity increases the attraction between chains and can even lead to formation of physical crosslinks between chains. This increased interaction hinders chains from moving freely and raises the glass transition temperature. For example, polylactide, poly(vinyl chloride) and poly(methyl methacrylate all have much higher values of Tg than polypropylene.
3. If the backbone is rigid, reptation is hindered because the chain can't adopt different conformations. Chain flow is restricted more easily and the Tg is higher. Kevlar and polynorbornene both have stiff backbones because of the aromatics and aliphatic rings, respectively, that form part of the backbone. Both polymers have much higher values of Tg than polystyrene, in which the aromatics hang from a flexible chain.
Exercise 12.3.5:
a) rubbery b) glassy c) glassy d) rubbery
Exercise 12.4.1:
There may be different architectures that lend themselves to the same applications.
1. A regular chain structure could be wound into a thread easily. There would be enough entanglement to held the threed together.
2. Across-linked structure would help a rubber band snap back to its original shape. However, too much cross-linking would make it too difficult to stretch the rubber band.
3. A dendrimer has lots of nooks and crannies that might be used to sponge up pharmaceuticals and slowly release them later.
4. A branched or cross-linked structure might form a good two-dimensional network to provide a protective coating.
Exercise 12.5.1:
The polar vinyl acetate block of the first polymer contains carbonyl groups that would attract each other by dipole-dipole interactions. It is easy to imagine these segments forming physical crosslinks. The hydrocarbon chain in the middle block would prvide flexibility.
Although nonpolar, the benzene groups in styrene are flat and can stack together with relatively strong London dispersion forces. The higher Tg of polystyrene would make these blocks much stiffer than the more flexible polyisoprene block in the middle.
Exercise 12.6.1:
The telechelic polymer has two growing ends per chain, so it grows twice as fast as a normal chain.
Exercise 12.6.2:
If this chain reached 100% conversion, then the degree of polymerization would be 500. At 85% conversion, DP = 425.
Exercise 12.6.3:
As the number of repeat units increases, the molecular weight increases.
Exercise 12.7.2:
The molecular weight is the sum of the weights of all the repeat units plus the weight of the end groups.
1. $MW = 1000 units \times 104.15 \frac{Da}{unit} + 2 Da = 104152 Da$
2. $MW = 200 units \times 72.06 \frac{Da}{unit} + 2 Da = 14412 Da$ (based on 200 simple repeating units, but this is a complicated case. In common practice, PLA is made from a dimeric monomer, with each monomer forming two repeat units in the chain, so a degree of polymerization of 200 really corresponds to 400 repeat units in the chain, which would have twice the molecular weight.)
3. $MW = 50 units \times 44.05 \frac{Da}{unit} + 2Da = 2202 Da$
4. $MW = 100000 units \times 42.08 \frac{Da}{unit} + 2 Da = 4208002 Da$
Note that, in this case, the end groups make no appreciable difference in the weight because they are so small.
Exercise 12.7.3:
MW = weight (in g) / moles
1. 5827 Da
2. 1052 Da
3. 703,000 Da or 703 kDa
Exercise 12.7.4:
Đ = Mw / Mn
1. 1.14
2. 1.22
3. 1.56
4. 1.76
5. 1.62
Exercise 12.7.5:
1. 3,000 D
2. 5,000 D
Exercise 12.7.6:
The longer the chain in polymer A, the bigger the sphere it coils into. There isn't much problem comparing samples of polymer A to each other, provided we have some way of calibrating sphere size to chain length.
However, polymer B appears to be more tightly coiled than A. A relatively small sphere could pack a lot more chain. We would have to be cautious about comparing the sphere size of B to that of A and drawing any conclusions about chain length or molecular weight.
Exercise 12.7.7:
Exercise 12.7.8:
16:1.
Exercise 12.7.9:
It's $OH + 4 \times (CH_{2}CH_{2}O) + H.$ That's $17 + (4 \times 44) + 1 = 18 + 176=194D$
Exercise 12.7.10:
The ratio of c to b is 145:18 = 8:1. However, the number of hydrogens in the repeat unit c is 4; the number of hydrogens in the end group b is 2. That ratio is 2:1. That means the comparison we need is (145/4)/(18:2) = 4:1. The degree of polymerization is 4.
Exercise 12.8.1:
a) Viscosity is related to the amount of drag in the fluid. The more drag, the slower solutes will be able to move, so the diffusion coefficient is inversely related to viscosity (i.e. higher diffusion coefficientshould correlate with lower viscosity).
b) The regular chain polymer diffuses more easily than the graft copolymer / supramolecular assembly because the pendant chains in the latter lead to increased entanglement.
Exercise 12.8.2:
b) The lone pairs on the oxygens in the host would be attracted to the positively charged nitrogens in the guest. The N-H groups in the guest would even act as hydrogen bond donors to the oxygens.
c) This supramolecular assembly forms a cross-linked architecture.
d) The gel is elastic because, although it contains flexible chains and can be distorted up to a point, at some point the crosslinks pull it back to its original position.
e) When the gel is cut, probably some guests are pulled away from their hosts along the surface of the cut. WHen the edges are placed back together, the guests can find new hosts and form the cross-linked structure again.
Exercise 12.8.3:
1. The palladium in the complex is Pd(II), so it is d8 and the structure of the crosslink is square planar at palladium. Ligand exchange likely occurs through an associative process. That means that before the PVP dissociates, a new ligand would bind; this ligand is probably a solvent molecule, DMSO.
2. The increased crowding in the ethyl-substituted ligand, compared to the methyl, would hinder the approach of DMSO during substitution, slowing down the reaction.
3. Kb is defined as ka/kd, the rate constant for the association step over the rate constant for the dissociation step. It is the binding constant for PVP-Pd formation.
4. Palladium is a second row transition metal whereas platinum is a third row transition metal. Third row metals have greater ligand field splitting energies than second row metals and are generally less labile. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/12%3A_Macromolecules_and_Supramolecular_Assemblies/12.09%3A_Solutions_for_.txt |
The nature of the chemical bond
At an elementary level, we can think about atoms being held together by simple electrostatic attraction. It is a fundamental principal that opposite charges attract. A positively charged ion and a negatively charged ion are held together by this force of attraction.
• This idea works well for ionic compounds, such as sodium chloride.
• It does not work well for compounds in which similar atoms are connected, such as ethane, C2H6.
Maybe the carbons are negatively charged and the hydrogens are positively charged, but what is holding the two carbons together if they have like charges?
A similar problem is encountered in diatomic molecules such as H2. Is one of these hydrogen atoms negative while the other hydrogen atom is positive?
In the early twentieth century, G. N. Lewis noticed a trend in the characteristics of compounds that he called "the rule of two". If you were to count up the total number of electrons in any stable compound, you would always come up with an even number -- that is, some number that is divisible by two. Perhaps, Lewis reasoned, this predominance of even numbers arises because electrons need to be in pairs.
What does an element do if it has an odd number of electrons? One solution is to steal an electron from another element, or to allow one to be stolen away; these arrangements lead to ionic bonds. However, those elements not adept at stealing electrons may have a problem; they may need to share them instead. In order to share electrons, elements will have to form close associations with each other. They will become bonded together.
• An "ionic bond" is an electrostatic interaction between an anion and a cation.
• A "covalent bond" is one pair of electrons shared between two atoms.
Lewis took this idea further. If you count up the valence shell electrons around each of the atoms in stable compounds, not only are there even numbers, but there are almost always the same number of electrons as there are in one of the noble gases: He, Ne, Ar, Kr (2, 8, 8, or 18). This observation is sometimes called the Lewis octet rule because so many common atoms that form stable compounds obtain 8 electrons in their outermost shell as a result. Neon is the nearest noble gas to carbon, oxygen and nitrogen, and all of these atoms adopt 8-electron configurations in stable compounds.
Lewis structures illustrate how atoms can maintain these numbers of electrons by sharing with other atoms. These simple structural drawings are used to convey most of our ideas about molecular chemistry. However, additional information can often be found through quantum mechanics and a molecular orbital approach to bonding.
13.02: Wave Behavior and Bonding in the H
In the early 20th Century, our understanding of light and matter was progressing very quickly, but some things still didn't make sense to people. Things really started to click when a number of physicists started thinking about things in a different way. One of the revolutionary things they did was to treat seriously, with mathematical equations, the wave nature of light and matter.
In the case of light, the idea actually dates back to the mid-17th Century. Christiaan Huygens, a Dutch clockmaker and astronomer, believed that light traveled in waves because he could see evidence of interference patterns in light, much like the way ripples on a pond can interfere with each other to produce new patterns. Other scientists, such as Newton, really thought light was made of small particles. In a sense, both were right, and we now think of particles of light, or photons, as having some of the properties of waves.
What do we know about waves? Waves on the surface of a lake or an ocean are good examples. You probably took a detailed look at them around 6th grade, although you may have forgotten some of the details. So, let's take another look.
Maybe you have stood beside a wave tank at a science museum and seen something that looks like this. The surface of the water undulates up and down. There are crests where the water mounds up in hills and troughs where the water slips away into valleys. Somewhere in between, there is an average water level, where the surface would settle if things were calm and there were no waves at all.
We can also think of waves as having null points or nodes; those are the positions halfway in between the crest and the trough.
Max Planck and Albert Einstein each took a closer look at that idea in the early 1900's when they were trying to determine the relationships between electrons in materials and their interaction with electromagnetic radiation. Electromagnetic radiation includes things like ultraviolet and visible light.
If we are standing at the seaside, one of the first things we notice about the waves is how tall they are. Maybe it's a really calm day and the waves are almost flat. Maybe it's a really rough day and the waves are really tall. The feature of waves that you are noticing is called the amplitude. That's the height of the wave from the average water level to the crest, or, conversely, the depth down to the trough. In the macroscopic world, the amplitude of the waves is really important. You need to know whether it's OK to stand and watch the waves, or whether it's getting too rough and you should seek higher ground. If you aren't careful, a really high amplitude wave could come crashing down on you.
Waves on the ocean can have high amplitude or low amplitude, and there is a physical consequence. Sound waves can also have high amplitude (those are the loud ones) or low amplitude (those are the quiet ones). Of course, if the amplitude were really, really low, we would almost have no wave at all (when that person is whispering a secret but you can't even hear them).
In the nanoscale world, the world of photons and electrons, the amplitude of a wave is a much more subtle thing. It isn't our first consideration. What's much more important is the wavelength. The wavelength is just the distance from one crest to the next, or from one trough to the next.
The wavelength is really important because, according to Planck and Einstein's analysis of experimental data, it's what determines the energy of an individual photon, or the energy of an individual electron. The shorter the wavelength of a photon or electron, the higher its energy. The longer the wavelength of a photon or electron, the lower its energy. This relationship is an example of inverse proportionality: when one gets bigger, the other gets smaller.
There is a complementary attribute of waves, and that is frequency. We can think of frequency as how often a wave crests. Alternatively, if we picture an electromagnetic wave passing a stationary object, the frequency would describe how often a crest passes the object. (All light travels at the same speed, about 3 x 108 m s-1, so we don't have to worry about that being a factor.)
Frequency is inversely proportional to wavelength: the longer the wavelength, the lower the frequency. Consequently, frequency is directly proportional to the energy of the photon or electron. The higher the frequency, the higher the energy.
Exercise \(1\)
Between each pair, which wave has greater energy?
Answer
In each case, the wave with the higher frequency and shorter wavelength would have the higher energy. a) i; b) ii; c, ii; e) i.
Let's return to Christiaan Huygens' original observation of interference patterns in light. What does that mean?
Suppose two waves come together. We'll think about two different cases. In the first case, the two waves are in-phase with each other. That means that they both crest at the same time and their troughs are always aligned, too. What happens then? You can imagine the crests will just build on each other, getting taller, and the troughs will fall into each other, getting deeper. The amplitude of the waves, together, increases. This is constructive interference.
Now suppose that the two waves are out-of-phase. When one crests, the other bottoms out. In that case, the crest of one wave falls into the trough of the other. Everything levels out. This is destructive interference.
Constructive interference and destructive interference also occurs with waves on the water. That leads to lots of ripple patterns if you toss two stones into a still pond. Photons can also produce those kinds of patterns, and so can electrons.
Lewis suggests that two atoms with unpaired electrons may share their electrons to form a bond. In a Lewis structure, we can draw those two electrons between the symbols for the two atoms to show that a bond has formed. Quantum mechanics offers a quantitative mathematical view of molecules, and is considered to give an accurate picture of the behavior of atoms and molecules. Now, instead of putting two atoms beside each other and letting two dots get together to form a bond, like in a Lewis structure, we'll think about bonding in quantum mechanical terms.
• Two hydrogen atoms are placed next to each other.
• The two atoms are surrounded by the waves formed by their electrons.
• If the two atoms are close together, these two waves overlap in between the two atoms.
What happens when waves overlap? Interference patterns develop. Think of a flat, calm pond, into which you and a friend throw two small stones, a few feet apart. The resulting ripples move out in circles from the two points where the stones fell in, and when the two sets of waves meet each other, new patterns result. The circular sets of waves combine to form new waves with new shapes.
When waves meet, there are two basic combinations that result. These combinations depend on whether the two waves are in-phase or out-of-phase with each other.
• If the waves are in-phase, so that the peak of one wave lands on top of the peak of another, and the trough of one wave coincides with the trough on the other, constructive interference occurs. The waves add up, and get bigger where they overlap.
• If the waves are exactly out-of-phase, the peak of one falls into the trough of another, and where they overlap the two waves are canceled out. This is destructive interference.
We can apply these ideas to the electrons in atoms that come together to make a bond. In Lewis structures, individual electrons on two different atoms are paired up in a bond between the two atoms. The electrons have to be combined to make a bond. If electrons have wave properties, then there will be interference patterns that result when these atoms come together.
Before we start, note that the waves that we will be dealing with look a little different from the standing waves we looked at before. These waves will be three-dimensional and will occupy different regions of space. Instead of drawing phase information as peaks and troughs, we are going to use a common convention of shading to signify whether we are looking at the positive part (in terms of phase) or the negative part of a wave. For example, we might show the crests in blue and the troughs in red, or we might show one in black and the other in white. In a standing wave, the phases would look like this:
In a hydrogen atom, the electron is found in a spherical area around the nucleus. We won't be able to see two distinct phases - a crest and a trough - in this wave at one time. However, we could look at two different hydrogen atoms and illustrate whether their electrons are in-phase or out-of-phase with each other.
When those two electrons come together and they are in-phase with each other, constructive interference occurs. They won't overlap completely; these are atoms and they take up space. However, they will be able to sit pretty close to each other, sort of smushing their electrons together. That constructive interference will happen in the middles, between the atoms.
When the two electrons come together and they are out-of-phase, destructive interference occurs. Again, that will happen in the middle, between the atoms.
So, in the pair of hydrogen atoms, two waves can combine to produce two possible outcomes.
• In one, the height of the combined wave is highest between the two atoms, where they are overlapping and in-phase. The amplitude of the wave is greater here.
• In the other case, the combined wave is composed of two halves that are out-of-phase, and there is no wave at all between the two atoms, where they have canceled out; the amplitude of this wave is zero between the two nuclei.
Those two possible combinations produce two different results. We keep track of these outcomes using a molecular orbital interaction diagram. The diagram says that the two orbitals can combine in two different ways: in-phase or out-of-phase. The diagram also tells us that the in-phase combination results in a decrease in energy when the two atoms come together. The out-of-phase combination results in an increase in energy when the atoms come together.
Why does the in-phase combination go down in energy, whereas the out-of-phase combination goes up?
Consider the idea that these waves each have a wavelength, and that this wavelength is related to kinetic energy. For the 1s electron on a hydrogen atom, we can really only visualize half the wave; this half-wave is basically the size of a hydrogen atom, so the total wavelength would be two hydrogen atoms long (for lack of a better yardstick).
For the out-of-phase combination, we can visualize the entire wave, both its "peak" on one side and the "trough" on the other. The two hydrogen atoms are not merely sitting beside each other, they are actually overlapping, so that the distance from one end of the wave to the next is actually somewhat less than two hydrogen atoms. Let's say it's one and a half hydrogen atoms long. This wave has a shorter wavelength than the waves on either individual hydrogen atom, and so it is at higher energy.
For the in-phase case, again, the two hydrogen atoms are overlapping. But this time we can only visualize half the wave; so the total wavelength would be longer than two hydrogen atoms. Maybe it's three hydrogen atoms long. This wave has a longer wavelength than the electron in an individual hydrogen atom, so it is at lower energy.
• The combination of atomic orbitals results in new, molecular orbitals.
• The number of atomic orbitals combined = the number of molecular orbitals produced.
• In-phase (low energy) and out-of-phase (high energy) combinations are possible.
And now for the weird part:
• According to quantum mechanics, the in-phase and out-of-phase combinations occur simultaneously.
Where would the two electrons actually go? The aufbau process used for atomic electron configurations holds true here. Two electrons can be found in the same space provided they are spin-paired; one would have spin "up" and the other one spin "down". Both electrons could be found in the lower energy orbital. That means both hydrogen electrons go down in energy when H2 forms; the formation of a hydrogen molecule from two hydrogen atoms is thus exothermic.
• The occupied orbital is called a bonding orbital; it is responsible for the bond energy in H2.
• Bond energy refers to the amount of energy given off when a bond forms.
What would happen if each hydrogen atom brought two electrons with it? Would two hydride ions form a bond?
• There would be four electrons in an H22- molecule.
• The first two electrons would occupy the lower, bonding orbital.
• The next two electrons would occupy the upper orbital
• Overall, there would be no net loss of energy, so a bond would not form.
• The out-of-phase combination is called an antibonding orbital, because its increase in energy offsets the decrease in energy that occurs when electrons go into a bonding orbital.
• Placing electrons in an antibonding orbital weakens or breaks a bond.
What does quantum mechanics tell us about bonding that we can't learn from Lewis structures? We learn that electrons can sometimes lower their kinetic energy by forming bonds. If that is the case, the bond will form. On the other hand, if kinetic energy cannot be lowered, the bond won't form at all.
• Covalent bonds form because electrons can adopt longer wavelengths.
• Electrons with longer wavelengths have lower kinetic energy.
Exercise \(2\)
Add a few words to explain the ideas conveyed in these drawings.
Answer
When two s orbitals combine out-of-phase, destructive interference occurs.
There is a node between the atoms.
The energy of the electrons increases.
When two s orbitals combine in-phase, constructive interference occurs.
There is no node between the atoms; the electrons are found between the atoms.
The energy of the electrons decreases. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.01%3A_Introductions.txt |
In the case of the hydrogen molecule, we took two atomic orbitals and combined them to form two molecular orbitals. These new molecular orbitals had different wavelengths than the two atomic orbitals: one had a longer wavelength and was a little lower in energy, while the other had a shorter wavelength and was a little higher in energy. If we take into account the energy of the two original atomic waves, and compare them to the total energy of the two new molecular waves, there is no change overall.
We started with two atomic orbitals, and by combining them we produced two molecular orbitals. Both of these ideas are useful in considering the formation of more complex molecules from individual atoms.
• The average energy of the orbitals has remained almost constant.
• Also, the number of waves has remained constant.
Of course, from the point of view of the two real electrons, some remarkable changes have occurred. Both of these electrons have adopted a longer wavelength and a lower energy and that has made all the difference. There is an occupied molecular orbital and an unoccupied molecular orbital; only the occupied orbital makes a real energetic contribution to the overall stability of the molecule. The unoccupied orbital is completely imaginary.
A bonding picture of He2 would look exactly the same, because it would also involve the overlap of 1s electrons on one atom with 1s electrons on the other atom. There would be a different electronic energy, however. That difference would affect the prospects of helium-helium bond formation.
• The electrons have lower kinetic energy in the bond than they had before bonding.
• Electronic energy has decreased. A stable bond has formed.
Exercise \(1\)
Construct molecular orbital diagrams for the following diatomic species and discuss the likelihood of bond formation in each case.
1. He2.
2. Li2.
3. Be2.
Answer
13.04: Sigma Bonding with p Orbitals
Other diatomic molecules in the upper right corner of the periodic table can be constructed in a similar way. Look at dinitrogen, \(\ce{N2}\). We can think about how dinitrogen would form if two nitrogen atoms were placed close enough together to share electrons. Nitrogen has more electrons than hydrogen, so this interaction is more complicated.
In our qualitative examination of bonding in main group diatomics, we will take the approach used in Lewis structures and just look at the valence electrons. A quantitative molecular orbital calculation with a computer would not take this shortcut, but would include all of the electrons in the atoms that are bonding together.
Nitrogen has five valence electrons, and these electrons are found in the 2s and 2p levels. There are three possible atomic orbitals in the 2p level where some of these electrons could be found: px, py and pz. We need to look at the interaction between the s and px, py and pz orbitals on one nitrogen atom with the s and px, py and pz orbitals on the other nitrogen. That process could be extremely complicated, but:
• Orbital interactions are governed by symmetry.
Orbitals interact most easily with other orbitals that have the same element of symmetry. For now, we can simplify and say that orbitals on one atom only interact with the same type of orbitals on the other atom.
• s orbitals interact with s orbitals. We can already see how that will work out in dinitrogen, because that is what happened in dihydrogen.
• px orbitals interact with px orbitals.
• py orbitals interact with py orbitals.
• pz orbitals interact with pz orbitals.
Another complication here is that the s and p orbitals do not start out at the same energy level. When the orbitals mix, one combination goes up in energy and one goes down. Does the s antibonding combination go higher in energy than the combinations from p orbitals? Do the p bonding combinations go lower in energy than the combinations from s orbitals? We will simplify and assume that the s and p levels remain completely separate from each other. This is not always true, but the situation varies depending on what atoms we are dealing with.
• The combination of one s orbital with another is just like in hydrogen. Two old orbitals will combine and rearrange to produce two new orbitals.
• There is a bonding combination in which the orbitals are in phase. The new orbital produced has longer wavelength than the original orbital. It is lower in energy.
• There is an antibonding combination in which the orbitals are out of phase. The new orbital produced has shorter wavelength than the original orbital. It is higher in energy.
In considering the interaction of two p orbitals, we have to keep in mind that p orbitals are directional. A p orbital lies along a particular axis: x, y or z. The three p orbitals on nitrogen are all mutually perpendicular or orthogonal to each other. That situation is in contrast to s orbitals, which are spherical and thus look the same from any direction.
We first need to define one axis as lying along the N-N bond. It does not really matter which one. We arbitrarily say the N-N bond lies along the z axis. The pz orbitals have a different spatial relationship to each other compared to the py and px. The pz orbitals lie along the bond axis, whereas the py and px are orthogonal to it.
As the nitrogen atoms are brought together, one lobe on one pz orbital overlaps strongly with one lobe on the other pz orbital. The other lobes point away from each other and do not interact in any obvious way.
As with the s orbital, the pz orbitals can be in phase or out of phase. The in-phase combination results in constructive interference. (By "in-phase" here we mean the lobes that overlap are in-phase; for that to happen the two p orbitals are actually completely out of phase with each other mathematically, so that one orbital is the mirror image of the other.) This combination is at longer wavelength than the original orbital. It is a lower energy combination.
The out-of-phase combination (meaning in this case that the overlapping lobes are out of phase) results in destructive interference. This combination is at shorter wavelength than the original orbital. It is a higher energy combination.
As a result, we have two different combinations stemming from two different p orbitals coming together in two different ways. We get a low-energy, in-phase, bonding combination and a high-energy, out-of-phase, antibonding combination.
What about those other p orbitals, the ones that do not lie along the bond axis? We'll take a look at that problem on the next page.
Exercise \(1\)
Draw an MO cartoon of a sigma bonding orbital formed by the overlap of two p orbitals between two oxygen atoms. Label the positions of the oxygen nuclei with the symbol "O". Label the O-O bond axis.
Answer
Exercise \(2\)
Consider the following reaction.
• Draw an MO mixing diagram for the reaction above.
• Draw the orbital from the base that is likely to donate its electrons.
• Draw the orbital from the acid that is likely to accept electrons.
• Complete the MO mixing diagram of these two orbitals:
• Label the electron donating orbital
• Label the electron accepting orbital
• Populate the MO mixing diagram with electrons
• Draw a cartoon showing the mixing of these orbitals.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.03%3A_Molecular_Orbitals-_Lessons_from_D.txt |
Earlier, we saw that p orbitals that lie along the same axis can interact to form bonds.
Parallel, but not collinear, p orbitals can also interact with each other. They would approach each other side by side, above and below the bond axis between the two atoms. They can be close enough to each other to overlap, although they do not overlap as strongly as orbitals lying along the bond axis. They can make an in-phase combination, as shown below.
They could also make an out-of-phase combination, as shown below.
• parallel p orbitals can overlap to produce bonding and antibonding combinations.
• the resulting orbitals contain nodes along the bond axis.
• the electron density is found above and below the bond axis.
• this is called a p (pi) bond.
The illustration above is for one set of p orbitals that are orthogonal to the bond axis. The second picture shows the result of the constructive (or destructive) interference. A similar picture could be shown for the other set of p orbitals.
In a main group diatomic species like dinitrogen, one p orbital lying along the bond axis can engage in s bonding. The two p orbitals orthogonal to the bond axis can engage in p bonding. There will be both bonding and antibonding combinations.
Just as the sigma-bonding orbitals display progressively shorter wavelengths along the bonding axis as they go to higher energy, so do the pi bonding orbitals. In other words, there are more nodes in the higher-energy orbitals than in the lower-energy ones.
An important consequence of the spatial distribution or "shape" of a p orbital is that it is not symmetric with respect to the bond axis. A s orbital is not affected when the atom at one end of the bond is rotated with respect to the other. A p orbital is affected by rotation. If one atom turns with respect to the other, the p orbital would have to stretch to maintain the connection. The orbitals would not be able to overlap, so the connection between the atoms would be lost.
Exercise \(1\)
When the 1s orbitals of two hydrogen atoms combine to form a hydrogen molecule, which molecular orbitals are formed?
Draw a picture.
Answer
When 2 atomic orbitals are combined, 2 molecular orbitals are formed: one in-phase bonding orbital and one out-of-phase antibonding orbital.
Exercise \(2\)
In-phase combinations of atomic orbitals give ______________ orbitals.
Draw a picture
Answer
AIn-phase combinations of atomic orbitals give bonding orbitals.
Exercise \(3\)
Out-of-phase combinations of atomic orbitals give ______________ orbitals.
Draw a picture.
Answer
Out-of-phase combinations of atomic orbitals give antibonding orbitals.
Exercise \(4\)
The combinations of ______________ atomic orbitals leads to σ orbitals.
Draw pictures.
Answer
The combinations of s + s OR s + p OR p + p OR s + d OR p + d atomic orbitals can lead to σ orbitals.
Exercise \(5\)
The combinations of ______________ atomic orbitals leads to π orbitals.
Draw pictures.
Answer
The combinations of side by side p + p or p + d atomic orbitals leads to π orbitals.
Exercise \(6\)
Which molecular orbital is typically the highest in energy?
a. p
b. σ
c. π*
d. π
e. σ*
Answer
e) σ*
Exercise \(7\)
Why can 1s orbital not interact with a 2s orbital?
Hint: Why is a Li2O bond stronger than a K2O bond?
Answer
Li+ and O2- are more similar in size than K+ and O2-, so the bond between Li+ and O2- is stronger.
The energy difference between the 1s orbitals and 2s orbitals is too large, so they cannot interact. In order for orbitals to interact, the orbitals need to have the same symmetry, be in the same plane, and be similar in energy.
Exercise \(8\)
Add a few words to explain the ideas conveyed in these drawings.
Answer
When two parallel p orbitals combine out-of-phase, destructive intereference occurs.
There is a node between the atoms.
The energy of the electrons increases.
When two parallel p orbitals combine in-phase, constructive interference occurs.
There is no node between the atoms; the electrons are found above and below the axis connecting the atoms.
The energy of the electrons decreases. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.05%3A_Pi_Bonding_with_p_Orbitals.txt |
So far, we have looked at the ways in which pairs of atomic orbitals could combine to form molecular orbitals -- to form bonds. Just as we think of there being a progression of atomic orbitals from lowest energy to highest (1s, 2s, 2p, 3s...), we can organize these molecular orbitals by order of their energy.
To a great extent, the order of molecular orbitals in energy can be considered to follow from the order of the atomic orbitals from which they are constructed. There are some departures from that rule, sometimes, but that's the simplest place to start. So, in a molecule, the lowest-energy molecular orbitals would be the ones formed from the lowest-energy atomic orbitals, the 1s orbitals.
What we see here is a molecular orbital interaction diagram. The middle of the diagram is just the molecular orbital energy diagram. It is analogous to the atomic orbital energy diagram (which goes 1s, 2s, 2p, 3s...). The order of energy so far is σ1s, σ1s*. The sides of the diagram just refer back to where those molecular orbitals came from, with dotted lines to guide you from one place to another. Altogether, the picture says that the 1s orbital on one atom and the 1s orbital on the other atom can combine in two different ways, producing the lower-energy, bonding σ1s and the higher-energy, antibonding σ1s*.
Note that we have not added any electrons to that molecular orbital energy diagram yet, but when we do, we will just fill them in from the bottom up, just like we would an atomic orbital energy diagram.
The next lowest set of atomic orbitals is the 2s level. These spherical orbitals would combine very much like 1s orbitals, and we would get a similar diagram, only at a slightly higher energy level.
Most of the time, we aren't going to see both the σ1s and the σ2s displayed in the diagram. That's because if there are any 2s electrons, then those 1s electrons are really core electrons, not valence. They are buried a little deeper in the atom, and they don't play a very important role in bonding. Ignoring the core electrons is pretty common; if you recall, in atomic electron configurations we might write [He]2s22p4 instead of 1s22s22p4 for oxygen; we were ignoring the core. When we drew Lewis structures, we gave oxygen six electrons, rather than eight; we were ignoring the core.
In the context of MO, suppose we do have 2s electrons. That must mean that each atom has two 1s electrons of its own, for a total of four. When those four electrons are filled into the MO diagram from the bottom up, they will occupy both the bonding σ1s and the antibonding σ1s*. The effect of both those combinations being occupied is to cancel out the bonding; those two pairs of electrons remain non-bonding. So we can ignore them and we aren't really missing anything.
The 2s orbitals aren't the only ones in the second shell. There are also 2p orbitals. Remember, there are a couple of very different ways in which p orbitals can combine with each other, depending upon which axis they lie. If they do not lie parallel to each other -- that is, if they are perpendicular to each other, such as a px and a py -- then they cannot interact with each other at all. The pz on one atom could interact with the pz on the other atom, however, because they are parallel to each other.
Usually, we define the z axis as lying along the line between the two atoms we are looking at. Two pz orbitals would lie along that axis, each with a lobe extending into the space between the atoms, and each with another lobe extending away, in the other direction
The resulting combinations are called σ because they do lie along the axis between the atoms (that's exactly what σ means, in terms of bonding). There is a σ combination, if the overlapping lobes are in phase with each other, and σ* combination, if those lobes are out of phase with each other. Because these new orbitals arise from the atomic 2p orbitals, we call them σ2p and σ2p*.
There are also those p orbitals that do not lie along the bond axis, or the axis between the two atoms. The px orbitals are perpendicular to the pz orbitals we just looked at, and therefore perpendicular to the axis between the bonds. However, they are still parallel to each other, and they can still form combinations. These two orbitals would form an in-phase combination and an out-of-phase combination.
Note that the energetic separation between these two combinations is a little smaller that the gap between the σ2p and σ2p* levels. The difference is related to the degree of overlap between the atomic orbitals. The on-axis orbitals project strongly into the same space; they overlap a lot, and they interact strongly. The off-axis orbitals brush against each other, interacting less strongly, and resulting in smaller energetic changes. The gap between the π2p orbital and π2p* orbital is therefore much smaller than the one between the σ2p and σ2p* orbitals.
There are actually two of those off-axis p orbitals. In addition to the px set, we would have a py set. If the px set is in the plane of the screen, the py set has one orbital sticking out in front and one hidden behind. Nevertheless, the combinations between the two py orbitals are exactly the same as what we saw between the two px orbitals. They are just rotated into a perpendicular plane with respect to the px combinations.
We can put all of those 2p-based orbitals together in one diagram. It's starting to get a little more crowded, but this diagram is just a combination of the pieces we have already seen. Note that the px, py, and pz atomic orbitals all start out at the same energy (we have stacked them here so that you can still see the correlation between the atomic and molecular orbitals). That means that the π2p & π2p* orbitals will be "nested" between the σ2p & σ2p* orbitals.
Finally, keeping in mind that the 2p orbitals are higher in energy than the 2s orbitals, we can combine those pictures into one diagram. Again, we have seen these individual pieces before; we are just assembling them now.
While we are at it, we can can add in the electrons. How? It's just the total number of valence electrons. For an example, we have used N2. Each nitrogen has five valence electrons, for a total of ten, so we have just filled in ten electrons, starting at the bottom of the molecular orbital energy level diagram. If this were another molecule, such as F2 or O2, we would construct the overall diagram in a similar way, but just use a different number of electrons.
The orbital picture we have described above is really just a potential picture of the electronic structure of dinitrogen (and any other main group or p-block diatomic). We won't get a real picture of dinitrogen's structure until we populate these potential levels with electrons.
• Only the energy levels with electrons have an effect on the energy (and behavior) of the molecule.
In other words, the energy of the electrons determines the behavior of the molecule. The other energy levels are only possibilities that remain unfulfilled.
Think about the picture of dinitrogen.
• Each nitrogen has five valence electrons.
• There are a total of ten electrons.
• Two each go into the s s bonding and s s* antibonding levels. Remember, we kept these separate from the p set as a simplification.
• Two each go into the s p bonding and each of the p bonding levels.
The remaining orbitals (s p* antibonding and each of the p * antibonding levels) are unoccupied. These are imaginary levels that do not play a role in determining the energy of dinitrogen. In a real molecular orbital calculation, the electrons in these levels would contribute to the overall energy of the molecule.
We get additional information from this picture. For example, we can see the bond order in dinitrogen.
• Bond order is just the number of bonds between a pair of atoms.
• The bond order is one of several factors that influences the strength of the covalent bond.
• The higher the bond order, the more electrons are shared between the atoms, and the stronger the bond.
In dinitrogen, the s s bonding s s* antibonding levels cancel each other out. One pair is lower in energy than it was in the atom, but the other is higher. There is no net lowering of energy. These electrons do not contribute to a nitrogen-nitrogen bond. These are non-bonding electron pairs.
The six electrons in the s p bonding and the p bonding levels, however, represent a decrease in energy from the energy levels in the free nitrogen atoms. These three, low-energy pairs of electrons indicate three bonds between the nitrogen atoms.
Remember, we have made some short-cuts in this picture, and a real molecular orbital calculation could give slightly different results. Nevertheless, it would still reveal a bond order of three as well as two non-bonding electron pairs.
In addition, sometimes molecular orbital pictures are shown in different ways. A molecular orbital interaction diagram shows how atomic or molecular orbitals combine together to make new orbitals. Sometimes, we may be interested in only the molecular orbital energy levels themselves, and not where they came from. A molecular orbital energy level diagram just shows the energy levels in the molecule. Frequently, but not always, energy level diagrams are shown without any pictures of the orbitals, in order to focus attention on the energy levels, which in a fundamental way are the most important part of the picture. Furthermore, because only the occupied energy levels actually contribute to the energy of the molecule, sometimes the higher-energy, unoccupied orbitals are left out of the picture.
Very often the results of molecular orbital calculations reinforce what we would predict from Lewis structures. If you draw a Lewis structure of dinitrogen, you will also predict a triple nitrogen-nitrogen bond. The chief advantage of molecular orbital theory is that it allows quantitative prediction of energy when we do a real calculation on a computer. In addition, it is important to realize that there is no real reason for the octet rule unless we consider quantum mechanics. Lewis structures are founded on an empirical observation that electrons form pairs and octets, without attempting to explain why. Molecular orbital theory takes some fundamental relationships from physics and applies them to very complicated molecules with very good success. Just by knowing the number of electrons in the molecule, and by knowing approximately where the nuclei are located in the structure, molecular orbital calculations give very useful information about energy. In addition, in more complicated cases than N2, these calculations can even correct our first guess about molecular geometry and where the bonds are located.
Exercise $1$
A Molecular Orbital Diagram for a diatomic molecule (two atoms) always has the same basic pattern.
• Draw a picture of the levels.
• Label each level with σ, σ*, π, π*
Answer
Exercise $2$
A Molecular Orbital Diagram for a diatomic molecule (two atoms) varies in the number of electrons. How do you populate the electrons?
Answer
• Count the valence electrons on the molecule. That's the number of valence electrons on each atom, adjusted for any charge on the molecule. (eg C22- has 10 valence electrons: 4 from each carbon -- that's 8 -- and two more for the 2- charge).
• Fill electrons into the lowest energy orbitals first.
• Pair electrons after all orbitals at the same energy level have one electron.
Exercise $3$
Construct a qualitative molecular orbital diagram for chlorine, Cl2. Compare the bond order to that seen in the Lewis structure (remember that an electron in an antibonding orbital cancels the stabilization due to bonding of an electron in a bonding orbital).
Answer
Exercise $4$
1. Construct a qualitative molecular orbital diagram for oxygen, O2.
2. Compare the bond order to that seen in the Lewis structure.
3. How else does this MO picture of oxygen compare to the Lewis structure? What do the two structures tell you about electron pairing?
4. Compounds that have all of their electrons paired are referred to as diamagnetic. Those with unpaired electrons are referred to as paramagnetic. Paramagnetic materials are attracted by a magnetic field, but diamagnetic things are not. How would you expect molecular oxygen to behave?
Answer
Exercise $5$
1. Construct a qualitative molecular orbital diagram for peroxide anion, O22-.
2. Compare the bond order to that seen in the Lewis structure.
3. How else does this MO picture of oxygen compare to the Lewis structure? What do the two structures tell you about electron pairing?
4. Based on molecular orbital pictures, how easily do you think dioxygen could be reduced to peroxide (through the addition of two electrons)?
Answer
Exercise $6$
Construct a qualitative molecular orbital diagram for diboron, B2. Do you think boron-boron bonds could form easily, based on this picture?
Answer
Exercise $7$
1. Construct a qualitative molecular orbital diagram for dicarbon, C2.
2. Compare the bond order to that seen in the Lewis structure.
3. How else does this MO picture of oxygen compare to the Lewis structure? What do the two structures tell you about electron pairing?
Answer
Exercise $8$
1. Construct a qualitative molecular orbital diagram for acetylide anion, C22-.
2. Compare the bond order to that seen in the Lewis structure.
3. How else does this MO picture of oxygen compare to the Lewis structure? What do the two structures tell you about electron pairing?
4. Based on molecular orbital pictures, how easily do you think dicarbon could be reduced to acetylide (through the addition of two electrons)?
Answer
Exercise $9$
Make drawings and notes to summarize the effect of populating antibonding orbitals.
Answer
Exercise $10$
Researchers at Johns Hopkins recently reported the formation of Na4Al2 in a pulsed arc discharge (they put a lot of electric current through a sample of sodium and aluminum; Xinxing Zhang, Ivan A. Popov, Katie A. Lundell, Haopeng Wang, Chaonan Mu, Wei Wang, Hansgeorg Schnöckel, Alexander I. Boldyrev, Kit H. Bowen, Angewandte Chemie International Edition, 2018, 57(43), 14060-14064. Copyright 2018, John Wiley & Sons. Used with permission.).
1. The compound is ionic. Explain which atoms form the cations, based on periodic trends.
2. Therefore, what atoms form the anion?
3. The anion is one molecule. What is the charge on this molecule?
4. Show how to calculate the total valence electrons in this molecular anion.
5. Draw a Lewis structure for this molecular anion.
6. Construct a diatomic molecular orbital energy level diagram for this molecule. Label the energy levels (sigma, pi, etc.) and add in the correct number of electrons.
7. Show how to calculate the bond order in the molecule.
Answer
a) Na, because Na has a lower ionization potential (and a lower electronegativity) than Al.
b) Al
c) 4-, because there are four Na+
d) total e- = 2 x 3 e- (per Al) + 4 e- (for the negative charge) = 10 e-
g) $\textrm{bond order} = \frac{( \# bonding \: e^{-} - \# antibonding \: 3^{-})}{2} = \frac{8-2}{2}= 3$ | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.06%3A_Assembling_the_Complete_Diagram_an.txt |
The molecular orbital picture of dioxygen differs from the Lewis picture. Both models predict an oxygen-oxygen double bond, but one model suggests unpaired electrons whereas the other indicates an electron-paired system. Often, there is experimental evidence available to check the reliability of predictions about structure. These data include measurements of bond lengths and bond strengths as well as magnetic properties.
Bond dissociation energy data tells us how difficult it is to separate one atom from another in a molecule. Bond order is one of the factors that influences bond strength. Thus, measuring a bond dissociation energy is one way to confirm that dioxygen really does contain an oxygen-oxygen double bond.
First, we need something to compare it to. Peroxides (such as hydrogen peroxide, \(\ce{H2O2}\), or sodium peroxide, \(\ce{Na2O2}\)) probably contain oxygen-oxygen single bonds, according to their Lewis structures. These bonds are relatively weak, costing about 35 kcal/mol to break. In contrast, the bond in dioxygen costs about 70 kcal/mol to break. Its bond is about twice as strong; it is a double bond.
• Bond dissociation energies can be used to determine how many bonds there are between two atoms.
Bond dissociation energies can be complicated to measure. They require a comparison of energy changes in numerous chemical reactions so that the energy change resulting from cleavage of a specific bond can be inferred. In contrast, infrared absorption frequencies are easy to measure. They simply require shining infrared light through a sample and measuring what frequencies of the light are absorbed by the material. (A related technique, Raman spectroscopy, gives similar information by measuring subtle changes in the frequency of laser light that is scattered off a sample). The frequencies absorbed depend on what bonds are present in the material. These frequencies vary according to two basic factors: the weights of the atoms at the ends of the bond and the strength of the bond between them. The stronger the bond, the higher the absorption frequency.
Peroxides absorb infrared light at around 800 cm-1 (this unusual frequency unit is usually pronounced "wavenumbers"). Dioxygen absorbs infrared light around 1300 cm-1. Since the atoms at the ends of the bond in both peroxide and dioxygen are oxygens, we can be sure that this difference in frequency is not due to a difference in mass. It is due to a difference in bond strength. The bond in dioxygen is much stronger than the O-O bond in peroxide, because the former is a double bond and the latter is a single bond.
• Vibrational spectroscopies (IR and Raman spectroscopy) can give information about the bond order between two atoms.
A third measure of bond order is found in bond length measurements. The more strongly bound two atoms are, the closer they are together. An O=O bond should be shorter than an O-O bond. Bond lengths can be measured by microwave spectroscopy (usually for gas-phase molecules), in which frequencies absorbed depend on the distance between the molecules. Alternatively, bond lengths can be measured by x-ray crystallography. X-rays can be diffracted through crystals of solid materials. The interference pattern that is produced can be mathematically decoded to produce a three-dimensional map of where all the atoms are in the material. The distances between these atoms can be measured very accurately.
The O-O bond in peroxides are about 1.49 Angstroms long (an Angstrom is \(10^{-10}\, m\); this unit is often used for bond lengths because it is a convenient size for this task. Covalent bonds are generally one to three Angstroms long). The O-O bond in dioxygen is about 1.21 A long. The O-O bond in dioxygen is shorter and stronger than in a peroxide.
• Bond length data provides insight into the bond order.
In addition to bond order, there is the question of electron pairing in dioxygen. The Lewis structure suggests electrons are paired in dioxygen. The molecular orbital picture suggests two unpaired electrons.
Compounds with paired electrons are referred to as diamagnetic. Those with unpaired electrons are called paramagnetic. Paramagnetic substances interact strongly with magnetic fields.
It turns out that oxygen does interact with a magnetic fields. A sample of liquid-phase oxygen can be held between the poles of a magnet. Oxygen has unpaired electrons. This finding is consistent with molecular orbital theory, but not with simple Lewis structures. Thus, MO theory tells us something that the Lewis picture cannot.
• Magnetic information, and measurements of magnetism, give us experimental evidence of spin states. We can tell if electrons are paired, unpaired, and how many unpaired spins there are.
A final important source of experimental data is photoelectron spectroscopy. Photoelectron spectroscopy gives information about the electron energy levels in an atom or compound. In this technique, gas-phase molecules are subjected to high-energy electromagnetic radiation, such as ultraviolet light or X-rays. Electrons are ejected from various energy levels in the molecule, and the binding energies of electrons in those levels is determined. Thus, photoelectron spectroscopy provides verification for exactly the sort of information that quantitative molecular orbital calculations are designed to deliver.
• Photoelectron spectroscopy tells how much energy is needed to remove electrons from various energy levels in a molecule. This technique gives us an accurate experimental picture of the energy levels that we predict with molecular orbital calculations.
Exercise \(1\)
In the previous section, you were asked to draw MO diagrams for some molecules. Determine whether the molecules in problems MO 6.3-6.8 are paramagnetic or diamagnetic.
Answer
From MO6: 3. Diamagnetic (no unpaired electrons); 4. Paramagnetic; 5. Diamagnetic; 6. Diamagnetic; 7. Paramagnetic; 8. Diamagnetic
Exercise \(2\)
Use a MO diagram to determine which species in the following pairs will have the longer bond. Give an explanation for your choice.
1. N2 or N2+
2. N2 or N2-
Answer a:
N2+. From the MO diagrams, N2+ has one less bonding electron. Thus, the bond order will be lower and the bond will be longer than in N2.
Answer b:
N2-. From the MO diagrams, N2- has one more antibonding electron. Thus the bond order will be lower and the bond will be longer than N2.
Exercise \(3\)
Use a MO diagram to determine which species in the following pairs will have the stronger bond. Give an explanation for your choice.
1. O2 or O2+
2. O2 or O2-
Answer a:
O2+ . From the MO diagram, O2+ has one less antibonding electron. Thus the bond order will be higher and the bond will be stronger than in O2.
Answer b:
O2. From the MO diagram, O2 has one less antibonding electron. Thus the bond order will be higher and the bond will be stronger than in O2-. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.07%3A_Experimental_Evidence_for_Molecula.txt |
At first, we approximated and said that only orbitals of the same kind can interact with each other: s orbitals with s orbitals, px orbitals with px orbitals, and so on. One of the qualitative differences between a real MO calculation and our scrap-paper approach is that mixing does occur between different kinds of orbitals. However, restrictions still apply.
Orbital interactions are governed by symmetry. Without taking an entire course in Group Theory (the branch of mathematics that deals with symmetry), we can learn a little bit about symmetry as we need it.
One of the most basic aspects of molecular symmetry applies to diatomic molecules. That element is symmetry about the bond axis. If a molecule is spun around its bond axis -- as if the two atoms are the two wheels on an axle of a wagon or a car -- does the orbital change its orientation?
• An s orbital is so symmetric that it looks the same from any direction. No matter how you spin it, it looks the same. In a diatomic molecule, an s orbital is symmetric with respect to rotation about the bond axis.
• What about p orbitals? Suppose the bond axis is the z axis. Rotation about the z axis produces no change in orientation of a pz orbital. The orbital spins but does not move through space.
• Rotation about the z axis does produce a visible change in orientation of a px orbital. The orbital spins like a propeller around the axis.
Rotation about the z axis does produce a visible change in orientation of a py orbital. The orbital spins like a propeller around the axis.
Our picture of main group diatomics changes slightly with this news. The p bonding orbitals remain unchanged. However, the s bonding orbitals share axial symmetry. That means that the s bonding orbitals, far from keeping completely separate from each other, actually mix together in some cases. That means there can be an orbital that is composed of an s and pz orbital from one atom as well as an s and pz orbital from the other. The only restriction is that, because four atomic orbitals are allowed to mix, four new molecular orbitals must be produced.
• As a result, in diatomics, an s orbital shares an element of symmetry with one p orbital, but not with the others.
• The rules of mathematics and quantum mechanics allow s orbitals in diatomics to interact with one p orbital with which it shares symmetry, but not with the other two.
Another way of thinking about this mixing effect is to take the molecular orbital picture we already made for dinitrogen and then say that the sigma bonding orbitals can interact with each other. That means they can push each other up or down in energy, mixing together to create new combinations with new energies. The extent of this mixing varies across the periodic table.
For main group diatomics like \(\ce{N2}\), the effect is to lower the energy of the lower sigma orbitals (related to the atomic s levels) and raise the energy of the higher sigma orbitals (related to the atomic p levels). The result is that the second-highest sigma orbital is shifted slightly above the energy of the pi levels, rather than being slightly below them. This does not change the overall conclusions very much, but it is a favorable interaction because the lower-lying sigma levels, which are occupied by electrons, move to lower energy. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.08%3A_Symmetry_and_Mixing.txt |
Two different atoms from the main group often bond together to make diatomic molecules. In most cases, the molecular orbital picture does not change very much from what we have already seen.
From a Lewis point of view, some very simple compounds include the hydrogen halides. For example, HF contains a hydrogen bonded to a fluorine.
A molecular orbital picture of HF could involve an s orbital on hydrogen interacting with an s orbital on fluorine. It may be that the three p orbitals on fluorine stay out of the way, do not interact with the hydrogen, and remain unchanged in the molecule. In that case, we are dealing with a picture of two s orbitals combining, as in dihydrogen.
The only difference is in the starting energy levels of the atomic electrons. The electrons in fluorine are at lower energy than the electron in hydrogen. Thus, the molecular orbital picture is skewed, with one side of the diagram at lower energy than the other.
• The energetic differences of the two atoms are shown in the molecular orbital interaction diagram.
There is a physical interpretation for this lopsidedness. Notice that the bonding orbital is closer in energy to fluorine than it is to hydrogen. Often, energetic similarity correlates with structural similarity. We think of these bonding electrons as being physically closer to the fluorine as well. In other words, the electrons in this bond are not equally shared, but are held more closely by fluorine.
• More energetic similarity to one atom implies electrons are found closer to that atom.
This idea is consistent with our ideas about bonding between dissimilar atoms. We would think of the H-F bond as a polar, covalent one.
The idea that fluorine's electrons are at lower energy may be confusing. A 1s electron in fluorine is certainly at lower energy than a 2s or 2p electron in fluorine. However, a 1s electron in hydrogen is not necessarily at a lower energy than a 2s or 2p electron in fluorine. All the electrons in fluorine are at lower energy than the electron in hydrogen because fluorine has a much higher nuclear charge, with relatively little shielding from its own 1s electrons. Fluorine has a high Zeff, as evinced by its high electron affinity (high attraction for more electrons) and high ionization potential (large amount of energy needed to strip an electron away).
Exercise \(1\)
Construct molecular orbital interaction diagrams for the following diatomic molecules. In each case, what is the bond order? Is the compound diamagnetic or paramagnetic?
1. carbon monoxide, CO.
2. nitrous oxide, NO.
3. hypochlorite ion, ClO-.
Answer a
Answer b
Answer c
13.10: Another Complication in HF- Orbita
Earlier, we talked about the fact that axially symmetric orbitals can mix together to produce new combinations. In HF, we looked at the overlap of the 2s orbital on fluorine with the 1s orbital on hydrogen. However, one of the 2p orbitals can lie along the bond axis and share axial symmetry with the hydrogen 1s orbital.
We will call this fluorine orbital the 2pz orbital. We could just as well label it px or py, but many people refer to the p orbital along the bond axis as the pz. Note that the 2p orbitals that are perpendicular to the bond axis can't interact with the hydrogen 1s orbital. Because the bond axis crosses the p orbital at a node, where the p orbital changes phase, the hydrogen atom can't form either an in-phase or out-of-phase combination with this p orbital. It would actually be in-phase with one part of the orbital and out-of-phase with the other. Instead, these orbitals simply do not interact.
• The hydrogen 1s can combine with the fluorine 2s and the fluorine 2pz.
• Now there are three orbitals that may combine. They will produce three new orbitals.
• We should consider these three orbitals together. What different combinations can they make?
• The orbitals could all be in-phase -- that is, the 1s of hydrogen is in-phase with the 2s of fluorine, and also in-phase with the part of the 2pz that overlaps with it.
• This will be a low-energy, bonding combination.
• The orbitals could be completely out-of-phase. The 1s of hydrogen is out-of-phase with the 2s of fluorine, and also out-of-phase with the part of the 2pz that overlaps with it.
• This will be a high-energy, antibonding combination.
The third combination is a little more subtle. The other possibility is that the 1s orbital of hydrogen is in-phase with one orbital of fluorine, but out-of-phase with the other. However, that means it could be in-phase with the 2s but out-of-phase with the 2pz, or vice versa. That means there are two more combinations, for a total of four. However, since we are using only three atomic orbitals, we can produce only three molecular orbitals. These two possible combinations must somehow be mathematically reduced to one.
Once again, mixing that occurs based on symmetry has some quantitative effects on the energies of the electrons, but in this case the qualitative effect is a subtle change in the ordering of energy levels.
• A third combination will be partially bonding and partially antibonding.
• This combination will be somewhat like a non-bonding orbital (although its exact energy may be slightly stabilized or destabilized with respect to the original atoms). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.09%3A_When_Different_Atoms_Bond_Together.txt |
What if a molecule had three or more atoms in it? We can no longer draw a simple molecular orbital interaction diagram, in which we show on atom on each side and the molecular results in the middle, because we don't have two atoms coming together to share electrons.
Instead, we look at slightly simplified MO diagrams in which we keep track of the result all those atomic orbitals coming together to make new orbitals. At a very simple level, the qualitative result is always the same and can be roughly predicted based on the Lewis structure.
• There is a bonding MO for every bond you see in the Lewis structure. These levels are low on the diagram.
• Every bonding MO has a corresponding antibonding MO. These levels are high on the diagram.
• Lone pairs in Lewis structure correspond to non-bonding orbitals. These levels are intermediate on the diagram.
• The diagram is populated by the same number of electrons that you see in the Lewis structure.
Exercise \(1\)
a) This is the MO diagram for ethane.
Draw a Lewis structure of ethane.
Populate the electrons into the MO diagram.
b) This is the MO diagram for propane.
Draw a Lewis structure of propane.
Populate the electrons into the MO diagram.
Add in the missing MO cartoons for the different levels.
Label the HOMO and LUMO.
c) This is the MO diagram for butane.
Draw a Lewis structure of butane.
Populate the electrons into the MO diagram.
What happpens to the MO diagram as the alkane chain length increases?
Exercise \(2\)
a) This is the MO diagram for ethene.
Draw a Lewis structure of ethene.
Populate the electrons into the MO diagram.
b) This is the MO diagram for propene.
Draw a Lewis structure of propene.
Populate the electrons into the MO diagram.
Add in the missing MO cartoons for the different levels.
Label the HOMO and LUMO.
c) This is the MO diagram for 1-butene.
Draw a Lewis structure of 1-butene.
Populate the electrons into the MO diagram.
What happpens to the MO diagram as the alkane chain length increases?
d) This is the MO diagram for 2-butene.
Draw a Lewis structure of 2-butene.
Populate the electrons into the MO diagram.
Does the location of the pi bond change the MO diagram?
e) This is the MO diagram for 1-butyne.
Draw a Lewis structure of 1-butyne.
Populate the electrons into the MO diagram.
How does an alkyne change the MO diagram?
Methane is a moderately small molecule. Its Lewis structure is pretty easy to draw, with a carbon in the middle bonded to four hydrogens. Trying to find the interactions of the hydrogen 1s orbitals with the orbitals on carbon could be challenging. However, molecular orbital calculations generally start with a good guess at the geometry of the structure.
In quantitative calculations, that starting geometry is often modified after the energy of the molecule has been calculated. A long series of iterative calculations are done, changing bond lengths or bond angles each time, until the geometry with the lowest possible energy has been found. If this geometry has the lowest energy, then it is probably the shape that the molecule would really adopt in nature. In fact, it is assumed in quantum chemistry that if a lower energy structure can be found, it is always a more accurate picture of the real molecule (assuming no mistakes have been made in the calculation).
Since the carbon in methane has four groups attached to it, we can be pretty sure that the carbon is tetrahedral.
We will assume the four hydrogens are arranged at the corners of a tetrahedron around the central carbon. Overlap with the carbon 2s orbital is easy to see, because the s orbital can overlap with neighbors in all directions. Of course, there could be an in-phase combination and an out-of-phase combination.
The complicated part arises when we want to introduce the carbon electrons and have them interact with the hydrogens. There is no universal orientation of p orbitals. Which way should they point?
In principle, it does not matter too much what direction we choose for the p orbitals. We should eventually end up with the same conclusions in the end. However, things may get extremely complicated and mistakes are likely to occur if we don't choose an orientation that makes the interactions simple to see. Since it does not matter which way the p orbitals are pointing, we should choose a direction that makes the calculation easiest.
• In a tetrahedron, the four corners point to alternating corners of a cube.
• The hydrogen atoms are in alternating corners of an imaginary cube around the carbon atom.
• The simplest way to align the p orbitals is to have them pointing at the faces of the cube.
Orienting the p orbitals this way simplifies things a lot. It means each lobe of each p orbital is directed at a space (on the face of a cube) in between two hydrogen atoms (on the opposite corners of the face). The other lobe of the same p orbital is directed at a space (on the opposite face of the cube) in between the other two hydrogen atoms (on the opposite corners of that face). The same is true for each p orbital, but they point at different faces and they interact with different combinations of the same four hydrogens.
As a result of this choice, there are really only a few interactions to consider in methane.
• The carbon 2s orbital can be either in phase or out of phase with the entire group of hydrogen orbitals. That makes two combinations.
• The carbon 2p orbitals can be either in phase or out of phase with their adjacent hydrogen atoms. With three p orbitals, that makes six more combinations.
• Eight atomic orbitals combine to make eight molecular orbital combinations.
Note that here the hydrogen atoms are often treated as being in phase with each other, or at least pairs of them are. This assumption simplifies the calculation and cuts down on the number of non-allowed combinations (i.e. combinations that produce the total number of orbitals over eight).
Organizing these combinations in terms of energy:
• The carbon 2s + H in-phase combination is entirely bonding. It is very low in energy.
• The carbon 2s - H out-of-phase combination is entirely antibonding. It is very high in energy.
• The carbon 2p + H in-phase combinations are mostly bonding. However, there is a node through the middle of the molecule owing to the node in the p orbital. As a result, the two hydrogens adjacent to one lobe are out of phase with the two hydrogens on the other. This combination is medium-low in energy.
• The carbon 2p - H out-of-phase combinations is highly antibonding. There is a node through the middle of the molecule owing to the node in the p orbital. There are also nodes between the p orbitals and the hydrogen s orbitals. This orbital has the most nodes through the structure and therefore corresponds to the shortest wavelength and highest energy.
In general, this idea that the geometry around an atom determines what atomic orbitals are needed to form direct (sigma) bonds with its neighbors is closely tied to hybridization. That's the idea that s and p orbitals in the same shell will need to work together to bond to neighbouring atoms. In molecules, we sometimes blur the distinction between the ones that are needed to form this set of bonds, and refer to them instead as "hybrids". Hybridization is closely tied to the geometry of the central atom, so much so that the terms, geometry and hybridization, are occasionally used interchangeably.
• linear geometry corresponds to sp hybridization
• trigonal planar geometry correxponds to sp2 hybridization
• tetrahedral geometry corresponds to sp3 hybridization
Exercise \(3\)
Construct a molecular orbital picture for borohydride anion, BH4-.
a) Use the geometry for BH4- that you would predict from its Lewis structure.
b) Select an orientation for the p orbitals on boron.
c) What interaction are possible between the set of hydrogens and the boron 2s orbital?
d) What interactions are possible between the set of hydrogens and one boron 2p orbital?
e) What interactions are possible between the set of hydrogens and the second boron 2p orbital?
f) What interactions are possible between the set of hydrogens and the third boron 2p orbital?
g) Can all the hydrogen orbitals interact in each case? If not, why not?
h) Order the resulting combinations in energy, show the resulting energy levels and populate them with the correct number of electrons.
Answer
Exercise \(4\)
Construct a molecular orbital picture for borane, BH3.
a) Use the geometry for BH3 that you would predict from its Lewis structure.
b) Select an orientation for the p orbitals on boron.
c) What interaction are possible between the set of hydrogens and the boron 2s orbital?
d) What interactions are possible between the set of hydrogens and one boron 2p orbital?
e) What interactions are possible between the set of hydrogens and the second boron 2p orbital?
f) What interactions are possible between the set of hydrogens and the third boron 2p orbital?
g) Can all the hydrogen orbitals interact in each case? If not, why not?
h) Order the resulting combinations in energy, show the resulting energy levels and populate them with the correct number of electrons.
Answer
Exercise \(5\)
Based upon your results of the previous two questions about borane and borohydride, comment on the energetics involved in the addition of a hydride ion (H-) to borane to produce borohydride. For this question, just assume the hydride electrons start out at a similar energy to the 2p electrons of boron.
Exercise \(6\)
Construct a molecular orbital picture for carbene, CH2, in two different ways.
a) In one approach, assume the H-C-H forms a 180 degree angle (linear geometry). Only one of the p orbitals lies along this line.
b) In a second approach, assume the H-C-H form a 120 degree angle (angular geometry). These three atoms define a plane. Two of the p orbitals can lie in the plane of these three atoms.
c) Based on your molecular orbital energy level diagrams, which of these two possibilities do you think is most favorable? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.11%3A_Geometry_and_Orbital_Contribution_.txt |
Very often, in qualitative discussions of molecular orbital theory, approximations are made to greatly simplify the view of the molecule. Information is lost whenever we approximate, so this is not as good as a more careful approach. However, there are cases in which we want to focus on one aspect of the molecule and don't want a complicated picture to distract from that focus.
For example, a common approximation of the structure of methane involves making two realizations:
• This structure requires that the carbon form bonds with hydrogens arranged in three dimensions around it. An s orbital could do this easily, because it is symmetric in all directions and can overlap with neighbors in any direction.
• In order to bond with hydrogens in three dimensions, all three p orbitals will be required, since together their axes define three dimensional space.
In an approximate approach, we don't worry about the specific interactions between specific pairs of orbitals. We simply say that the carbon 2s and 2p orbitals will all act together to bond with the four neighbouring hydrogens.
• This structure requires that the carbon form bonds with hydrogens arranged in three dimensions around it. An s orbital could do this easily, because it is symmetric in all directions and can overlap with neighbors in any direction.
• In order to bond with hydrogens in three dimensions, all three p orbitals will be required, since together their axes define three dimensional space.
• The result is four bonding orbitals and four antibonding orbitals.
• We don't worry about the relative energies of the four bonding orbitals; we assume they are all similar.
This result is not consistent with photoelectron spectroscopy. Our more rigorous approach to methane is more consistent with experimental data. However, this shortcut, which is sometimes referred to as a "hybridization" approach, is often used to discuss more complicated molecules.
• A tetrahedral atom can use an s orbital and three p orbitals to bond with its neighbors.
• When using a hybridization model, a tetrahedral atom. It is sometimes referred to by the label "sp3 hybrid".
Exercise \(1\)
Suppose you wanted to use a hybridization approach to show a rough molecular orbital diagram for borane, BH3.
a) What is the expected geometry at boron in borane?
b) Could the s orbital bond with the hydrogens?
c) Given the geometry of boron, could all of the p orbitals bond with the hydrogens?
d) Are any orbitals not properly oriented to bond with the hydrogens? If so, they would be non-bonding and would not go up or down in energy upon formation of the molecule.
e) Show the resulting molecular orbital energy level diagram for borane.
f) Tetrahedral atoms are referred to as sp3. What hybrid label would you give your boron with this geometry?
Answer
Exercise \(2\)
Suppose you wanted to use a hybridization approach to show a rough molecular orbital diagram for beryllium hydride, BeH2.
a) What is the expected geometry at beryllium?
b) Could the s orbital bond with the hydrogens?
c) Given the geometry of beryllium, could all of the p orbitals bond with the hydrogens?
d) Are any orbitals not properly oriented to bond with the hydrogens? If so, they would be non-bonding and would not go up or down in energy upon formation of the molecule.
e) Show the resulting molecular orbital energy level diagram for beryllium hydride.
f) Tetrahedral atoms are referred to as sp3. What hybrid label would you give your beryllium with this geometry?
Answer
Exercise \(3\)
Indicate hybridization of each carbon atom in the following molecules. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.12%3A_Approximations_in_More_Complicated.txt |
Sometimes, compounds are complicated enough that it can be easier to think about the bonding in one portion of the molecule at a time. Ethane, for example, does not contain one central atom like methane does. It contains two atoms that we would think of as being tetrahedral. That makes the potential orbital interactions slightly harder to think about.
That does not mean this calculation poses any problem for a computational chemist. A quantitative molecular orbital approach on a computer would begin with a spreadsheet of data to submit for calculations. This data would include a list of all the atoms in the structure and the approximate position of each atom in three-dimensional space. These positions could be thought of in Cartesian coordinates (x,y,z), polar coordinates (r,q,f) or internal coordinates (a slightly complicated system relating the relative positions of the atoms with respect to each other: atom B is x distance from atom A; atom C is y distance from atom B and A-B-C forms an angle of z degrees, and so on). The computer does not need to know where the bonds are or where the electrons are; it will figure that out from its calculations. Ethane would prove to be very simple to calculate with a computer.
On paper, we might need to break ethane down into easier pieces. The obvious thing to do is break ethane into two smaller parts and then bring those parts together. Ethane could easily be thought of as two CH3 units interacting together.
The geometry of a CH3 unit may be debatable. Since we expect the carbon to be tetrahedral in ethane, it is probably easiest to think of CH3 as pyramidal, in terms of the relationship of the four atoms. The three hydrogens form a triangle at the base of the pyramid, with the carbon at the apex of the pyramid.
Now we have two methyl groups. We can bring them both together and allow the previously non-bonding orbitals to interact with each other.
• If we take a hybrid approach to the methyl group, we can use the 2s orbital and all three 2p orbitals to bond with the hydrogens.
• There are only three hydrogens, so we only need three of the sp3 orbital set to bond with the hydrogens.
• There will be three bonding combinations using these orbitals and three antibonding combinations.
• These interactions describe the C-H bonds.
• The last of the sp3 set will be left non-bonding. We don't care which orbital this is because we are taking a shortcut.
• The remaining orbital from the sp3 set on one methyl interacts with the remaining orbital from the sp3 set on the other methyl.
• There is a bonding and an antibonding combination.
• This interaction describes the C-C bond.
• The C-C bond is probably not at the same energy level as the C-H bonds.
• We know from experiment that C-C bonds are weaker than C-H bonds, so we will place it higher in energy than the C-H bonds.
This approach does not agree well with the exact energy levels of the C-H bonds as determined by both quantitative molecular orbital theory and photoelectron spectroscopy. However, it is a method that will allow us to tackle bigger problems without worrying about some of the details.
Exercise \(1\)
Use the partial structure approach, with a hybridization model, to construct an approximate molecular orbital interaction diagram of ethene, connected CH2-CH2 (two H on the first carbon, two H on the second, the two carbons connected together).
1. Build this model up from two methylene groups, CH2 and CH2. What will be the geometry of each carbon in ethylene?
2. Given that geometry, choose the appropriate set of carbon s & p orbitals to bond to the neighbors. Build the methylene molecular orbital energy level diagram.
3. Bring the two methylenes together. Let an unused orbital from your hybrid set interact with the same kind of orbital from the other methylene.
4. When finished, are there other unused atomic orbitals on each methylene? If so, could they form an additional bond?
Answer a
Answer b
Answer c
Answer d
Exercise \(2\)
Use the partial structure approach, with a hybridization model, to construct an approximate molecular orbital interaction diagram of ethyne, connected CH-CH (one H on the first carbon, one H on the second, the two carbons connected together).
1. Build this model up from two methyne groups, CH and CH. What will be the geometry of each carbon in ethyne?
2. Given that geometry, choose the appropriate set of carbon s & p orbitals to bond to the neighbors. Build the methyne molecular orbital energy level diagram.
3. Bring the two methynes together. Let an unused orbital from your hybrid set interact with the same kind of orbital from the other methyne.
4. When finished, are there other unused atomic orbitals from each methyne? If so, could they form an additional bond?
Answer a
Answer b
Answer c
Answer d
Exercise \(3\)
Use the partial structure approach, with a hybridization model, to construct an approximate molecular orbital interaction diagram of formaldehyde, connected CH2-O (two H on the carbon, carbon connected to oxygen).
1. Build this model up from a methylene group, CH2. What will be the geometry of each carbon in ethylene? What set of carbon s & p orbitals will bond to the neighbors? Build the methylene molecular orbital energy level diagram.
2. Bring the methylene with an oxygen. Let an unused orbital from your hybrid set on methylene interact with any orbital you think will overlap with it on the oxygen
3. When finished, are there other unused atomic orbitals on the methylene? If so, could they form an additional bond with any atomic orbital on oxygen?
Answer a
Answer b
Answer c | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.13%3A_Building_a_Molecule_from_Pieces.txt |
Ozone is a fairly simple molecule, with only three atoms. However, in order to focus on one aspect of ozone's structure, we will use a hybrid approximation in order to simplify the picture.
The Lewis structure of ozone is somewhat unsatisfactory. That's because the true structure of ozone can't be drawn easily using Lewis conventions. Ozone is an angular structure in which both oxygen-oxygen bonds are about 1.278 Angstroms long.
However, the Lewis structure of ozone does not reflect that reality. In the Lewis structure, one pair of oxygens is double-bonded and the other is single-bonded. If this were true, there would be two different bond lengths in ozone. One bond would be about 1.49 Angstroms long, like the O-O bond in peroxide. The other bond would be about 1.208 Angstroms long, like the O=O bond in dioxygen. The bond in ozone looks pretty close to a double bond, does not it? It is just a little longer, however.
In Lewis structures, we fix this discrepancy by drawing two resonance structures for ozone. In one structure, the double bond is between one pair of oxygens. In the other structure, the double bond is between the other pair. The resonance structures imply that the real structure is somewhere in between the two that are shown. Roughly speaking, there should be one-and-a-half bonds between the neighbouring oxygens. Nevertheless, Lewis structures have trouble illustrating the nature of the double bond in ozone, which seems to be both there and not there at the same time.
Exercise \(1\)
Allyl cation, CH2=CHCH2+, is another conjugated system. Use resonance structures to show how its double bond is delocalized.
Answer
These two resonance structures follow the Lewis rules, but both are necessary to illustrate the delocalize electrons
Exercise \(2\)
Use resonance structures to show that the negative charge in a formate ion (HCO2-, C is in the middle and attached to the three other atoms) is spread out (delocalized) over more than one oxygen atom.
Answer
These two resonance structures follow the Lewis rules, but both are necessary to illustrate the delocalize electrons
We could get another look at bonding in ozone using a molecular orbital approach. We are basically concerned with one question: what is the nature of the double bond?
We already know about double bonds. We have seen them in compounds like nitrogen. A second bond is generally made through a pi bonding interaction. Therefore, we are only going to worry about the orbitals that will form pi bonds.
We have three orbitals to combine. Therefore, there will be three combinations.
• Ozone is made of three atoms at an angle to each other. They are, by definition, a plane.
• A planar system can be described in a simple hybridization model using the s orbital and two of the p orbitals on each oxygen.
• We will assume some combination of these orbitals interact within the plane to form the first bonds between the oxygens. We won't worry about the details.
• Each oxygen on ozone has a p orbital that was left out of this sp2 set.
• These leftover p orbitals could interact with each other to form a pi bond.
We have three orbitals to combine. Therefore, there will be three combinations.
• In one combination, all three orbitals are in phase. This combination will have a node through the plane of the molecule (because they are p orbitals) but none cutting through the molecule crosswise. This is a low energy, highly bonding combination.
• In another combination, all three orbitals are out of phase. This phase will have a node through the plane of the molecule (because they are p orbitals) and two more nodes cutting through the molecule crosswise. This is a high energy, highly antibonding combination.
• In a third combination, the middle p orbital can be thought of as out of phase with one neighbour but in phase with the other. This phase will have a node through the plane of the molecule (because they are p orbitals) and one more nodes cutting through the molecule crosswise. Because the bonding and antibonding interactions within this orbital cancel out, this is nonbonding combination.
• More correctly, this combination is usually drawn as a p orbital on each end of the molecule, out of phase with each other. The middle p orbital might as well sit out because overall it isn't doing anything. This depiction stil has one node cutting through the molecule crosswise, and is energetically equivalent to the other way we drew it. However, the first way we drew it is disqualified by symmetry rules (it is too lopsided).
Populating these orbitals, and getting an exact energy, is not possible given the huge approximations we have made. However, in focusing on the pi bonding, we see something that we can't see in Lewis terms. There really is a pi bond that stretches the entire length of the ozone molecule. This is the lowest energy combination, with a wavelength steretching over twice the length of the molecule.
• Pi bonding in ozone is delocalized over all three oxygens.
• Delocalization is highly stabilizing.
• Delocalization allows electrons to achieve longer wavelength and lower energy
Because it is low in energy, the extended pi bond is pretty certain to be populated by electrons, and it will make some contribution to the structure of ozone. How much impact it has would depend on the population of the other combinations, which we can't predict without a more careful approach.
Exercise \(3\)
There is delocalization in the following species.
In each case:
1. add any missing lone pairs.
2. show delocalization using resonance structures.
3. show delocalization using drawings of the orbitals used in pi bonding.
Answer a
Answer b
Answer c
Answer d
Exercise \(4\)
Construct a Huckel MO diagram for each of the cases in Problem MO14.3.
Answer a
Answer b
Answer c
Answer d
Exercise \(5\)
In each of the following cases, there may or may not be conjugation involving lone pairs and pi bonds. Show why or why not, using drawings of the orbitals involved.
Answer a
Answer b
Answer c
Answer d
Exercise \(6\)
The molecule acetamide is shown in problem MO14.1. Explain the following structural features.
a. In acetamide, the C-N and C-O bond lengths are 1.334 and 1.260 angstroms, respectively. For comparison, some typical bond lengths are C-N (1.47 A); C=N (1.38 A); C-O (1.43 A), C=O (1.20 A).
b. The two C atoms, plus the O, the N and the two hydrogens on the N lie in a plane.
c. The barrier to rotation about the C-N bond is approximately 11 kcal/mol, while the barrier to rotation about the C-N bond in CH3NH2 is about 2.4 kcal/mol.
d. The two hydrogens on the N are not in identical chemical environments.
Answer
All of the answers depend on an understanding of the contributions of two resonance structures to the overall picture of acetaminde, or alternatively, that actetamide forms a conjugated pi system with four electrons delocalized over the O, C and N.
Answer a
Contribution of the second resonance structure introduces some double bond character to the C-N bond and some single bond character to the C-O bond. Thus, both of these bonds are intermediate in length between single and double bonds.
Answer b
Since the C, N and O atoms are sp2 hybridized, the C-N pi bond can only form if the remaining p orbitals on these atoms align. This places the atoms participating in the sp2 sigma bonds in the same plane.
Answer c
Because of the partial double bond character there is a larger barrier to rotation than is typically found in molecules with only single bonds.
Answer d
Because of the partial double bond character and the restricted rotation, the two H’s are not identical. (One is nearer the O and one is nearer the CH3 and the restricted rotation prevents their interconversion. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.14%3A_Delocalization.txt |
Bonding in ozone can be explained with conjugation. In conjugation, several p orbitals can interact with each other to create a delocalized pi bond.
Conjugation can also occur in polyunsaturated alkenes or olefins. Alkenes are compounds that contain carbon-carbon double bonds. A carbon-carbon double bond is sometimes described as unsaturated. This term distinguishes alkenes from alkanes or paraffins, which have only carbon-carbon single bonds. Alkanes can be made from alkenes through the addition of hydrogen gas in the presence of a transition metal catalyst. Whereas unsaturated alkenes can take up additional hydrogens to add to double-bonded carbons, alkanes are already full: they are saturated with hydrogen and can accept no more.
The conversion of unsaturated carbon chains into saturated ones is very common. A familiar example concerns cooking fats and oils. Many vegetable oils contain unsaturated fatty acids, which contain long chains of carbon atoms having at least one carbon-carbon double bond. These double bonds produce a kink in the carbon chain, making these compounds oils at room temperature. When dairy products were rationed during World War II, vegetable oils were converted into saturated fats through catalytic hydrogenation. Saturated fats have a regularly repeating, zig-zag shape in their carbon chains and they are solids at room temperature. The product of this conversion was called oleo, or later, margarine. Margarine remains a common substitute for butter today.
Exercise \(1\)
Why does a kink in the carbon chain make a compound an (liquid) oil, when otherwise it would be a (solid) fat or wax?
Answer
• Polyunsaturated alkenes, or polyenes, are compounds that have several carbon-carbon double bonds.
• The double bonds can be spaced out along the carbon chain, with several carbon-carbon single bonds separating them. In this case they are called isolated polyenes.
• Two double bonds could be attached to the same carbon, so that there are two double bonds in a row, but this situation is rare. This compound would be called a cumulene.
• Two double bonds could be separated by one single bond. Alternatively, a number of double bonds could be separated by single bonds, so that a carbon chain had alternating double and single bonds. This compound is a conjugated polyene.
Conjugated polyenes have a few unusual properties.
• First of all, they are slightly more difficult to hydrogenate than are other polyenes. Their double bonds somehow resist taking up hydrogen.
• Second, these compounds have "high rotational barriers" around their carbon-carbon single bonds.
Normally, single bonds between atoms are not difficult to rotate, and the atoms on one end of a bond can spin with respect to the other end. Double bonds, on the other hand, do not rotate easily, because the p orbitals forming the pi bond need to remain parallel to each other in order to bond. The single bonds in conjugated polyenes do not rotate freely; they behave a little more like double bonds.
• Finally, conjugated polyenes show a pronounced "red shift" in their ultraviolet-visible spectra.
When alkenes absorb ultraviolet light, an electron is promoted to a higher energy level. In alkenes, one of the electronic transitions that can occur involves the promotion of an electron from a p to a p * level. The frequency necessary to induce this electronic transition is usually in the ultraviolet range. However, increasing the number of conjugated double bonds moves this transition to lower and lower frequency, as shown in the table below. Some molecules, such as carotene, have many conjugated double bonds, and absrob visible light. Carotene absorbs blue and green light, reflecting red and orange. It is one of the pigments responsible for the orange and red color of leaves in the fall.
Compound
Absorption wavelength (nm)
CH2=CH2
180
CH2=CH-CH=CH2
210
CH2=CH-CH=CH-CH=CH2
240
All of these properties of polyenes can be explained using quantitative molecular orbital calculations. We can also get an idea of where some of the results come from by taking a very approximate qualitative MO approach.
Start with the first entry in the table above.
• Ethene, CH2=CH2, contains two trigonal carbons.
• As an approximation, we will say that these two planar carbons each use a set of sp2 orbitals to bond to the hydrogens and each other.
• That leaves a leftover p orbital on each carbon to form a pi bond.
We aren't going to bother drawing a complete molecular orbital energy level diagram for ethene. We will only draw a Huckel diagram, which shows only the portion of the molecule involved in pi bonding.
• In the Huckel diagram, there is a bonding combination and an antibonding combination.
• In the absence of a complete calculation, we will guess from the Lewis structure and say that there is one pair of electrons to populate this set of orbital.
• They will go into the lower, bonding orbital.
• These electrons are delocalized over two carbons, with a half wavelength stretching over both carbons. A full wavelength would stretch to four carbons long.
For comparison, we can look at the next entry in the table. It is butadiene, CH2=CH-CH=CH2.
• As an approximation, we will say that these four planar carbons each use a set of sp2 orbitals to bond to the hydrogens and each other.
• That leaves a leftover p orbital on each carbon to form a pi bond.
• In the Huckel diagram, there is a fully bonding combination (all in phase with their neighbors) and a fully antibonding combination (all out of phase with their neighbors, with three nodes).
There must be two other combinations, because we started with four p orbitals.
• The two pairs of orbitals on each end could be in phase, but out of phase in the middle. This combination contains one node.
• The opposite could be true: The two in the middle could be in phase, but out of phase with both ends. This final combination contains two nodes.
Notice that the number of nodes increases one by one, from the all-bonding combination to the all-antibonding.
• We will guess from the Lewis structure and say that there are two pairs of electrons to populate this set of orbitals.
• They will go into the lower two orbitals.
• The lowest energy electrons are delocalized over four carbons. A full wavelength would stretch to eight carbons long.
This pair of low-energy electrons must be at least partly responsible for the stability of polyenes. These electrons are much lower in energy than they would be in two separate double bonds, which would be similar in energy to the double bonds in ethene.
In addition, this orbital explains the difficulty of rotating around the single bonds in butadiene. In a sense, there are no single bonds in butadiene. There are only double bonds.
The spectroscopy results require more attention to subtleties. Note that the next pair of electrons goes into an orbital that contains a node. In a quantitative MO calculation, the presence of this antibonding interaction in an otherwise bonding orbital raises its energy slightly above the energy of the pi bond in ethene. Also, the lowest lying empty orbital has a bonding interaction, lowering its energy slightly below that of the purely antibonding orbital in ethene. The energy gap between the p to a p * levels is slightly smaller in butadiene, requiring a lower-energy photon to promote an electron.
Exercise \(2\)
Problem MO15.2.
Construct a Huckel MO diagram for 1,3,5-hexatriene, CH2=CH-CH=CH-CH=CH2.
Answer
Exercise \(3\)
Problem MO15.3.
Another possible UV-visible absorption leads to n to p * transition in oxygen-containing compounds; the n refers to a nonbonding electron in a lone pair on oxygen. Construct Huckel MO diagrams for CH3-CH2-CHO (the last C is bonded to an H and also to an oxygen, as well as the next carbon) and CH2=CH-CHO. Include the oxygen nonbonding energy levels in your diagrams. Compare the n to p * energy gaps in these two compounds. How do you expect their absorption wavelengths will compare?
Answer
Polyenes, and other compounds in which orbitals are conjugated, form very predictable Huckel MO patterns. If there is no conjugation, then we just have one orbital, all by itself. If two orbitals are conjugated, then there are two combinations.
Exercise \(4\)
Problem MO15.4.
Match each of the structures below to the correct Huckel MO diagram. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.15%3A_Polyenes.txt |
Aromatics are a special class of delocalized compounds. Benzene is the common example of an aromatic compound.
Go to Animation CA12.6. A three-dimensional model of benzene.
Lots of compounds in nature contain "benzene units" within larger structures. It is so common that it has gotten a great deal of attention over the years.
These benzene units are sometimes called "phenyls" or, more generally, "arenes".
Benzene is a good example of a delocalized structure. It is something that could be studied using a Huckel MO approach. One of the pieces of experimental evidence that benzene has delocalized bonding comes from crystal structures. The first X-ray crystal structure of benzene was performed in 1929 by Kathleen Lonsdale. This experiment allowed Lonsdale to measure the distances between the carbons in benzene. A C-C bond in benzene is 1.40 Angstroms long. By comparison, a typical C-C bond is 1.47 Angstroms long, and a typical C=C bond is 1.35 Angstroms long. The C-C bonds in benzene appear to be somewhere between a double and a single bond, and they are all the same. They are not alternating, double-single-double as the Lewis structure suggests.
• Benzene has delocalized pi bonding.
• Despite what the Lewis structure says, all of benzene's C-C bonds are the same length.
We can try to show that benzene's bonds are all the same by using two resonance structures. When we do that, we are saying that the real molecule behaves a little like one Lewis structure and a little like the other.
Exercise \(1\)
Draw two Lewis structures that together illustrate the delocalization in benzene.
Answer
In some ways, aromatics behave differently than other compounds with C=C bonds. For example, there are reactions that alkenes undergo pretty easily, but benzene does not. Bromine, Br2 adds easily across alkene double bonds. Ethene, or CH2=CH2, would add bromine to form dibromoethane, BrCH2-CH2Br. As a result, the red-brown color of bromine disappears instantly if it is added to an alkene, because the bromine is rapidly consumed.
Alkenes also react with hydrogen, H2. Hydrogen can add across double bonds to form "saturated hydrocarbons" (hydrocarbons that don't have any double bonds). Hexene, CH2=CHCH2CH2CH2CH3, could react with hydrogen to form hexane, CH3CH2CH2CH2CH2CH3. This reaction is a lot more sluggish than bromination, though. Bromination often happens instantly. Hydrogenating something takes a lot of work. In general, some kind of catalyst, like powdered palladium metal, must be added before anything happens, and the alkene must be placed under an atmosphere of hydrogen for a long time.
Benzene can be hydrogenated (to form cyclohexane, C6H12), but it is even more difficult than usual. For a catalyst, palladium really won't do. This requires platinum or platinum oxide, which are a little more reactive. Also, high temperatures may be needed as well as high pressures of hydrogen (often several times greater than atmospheric pressure).
Maybe benzene can be brominated if the right catalyst is added. Ferric chloride works, but it turns out a completely different reaction takes place. C6H6 turns into C6H5Br. All the double bonds are still there. The bromine atom just replaces a hydrogen atom.
• Benzene does not undergo the same reactions as other compounds with C=C bonds.
• In some ways, aromatics are more stable than expected.
Historically, much has been made about benzene being a ring. When first isolated by Michael Faraday in the 1820's, people didn't have much idea about its structure, and what it might look like was debated for decades. There is a story that August Kekule reported about nodding off in an armchair and dreaming of a snake biting its tail; he says he woke up and had the idea that benzene was composed of a circle of carbon atoms (confirmed by Lonsdale about 65 years later). Maybe the special ring structure of benzene lends it extra stability.
• Benzene is delocalized and cyclic.
So maybe benzene's stability comes with having conjugation and being cyclic. But maybe that isn't enough, either. There are other compounds that are both delocalized and cyclic, but that do not display benzene's properties. For example, cyclopentadiene has long C-C bonds and short C=C bonds, and it behaves like an alkene. It is not aromatic. Why not? Maybe the difference is that cyclopentadiene does not have alternating pi bonds all around the ring. It does not have a conjugated p orbital on each carbon. There is a tetrahedral CH2 unit in between two of the double bonds. That tetrahedral carbon, or sp3 carbon, does not have a free p orbital it can use to make double bonds.
Go to Animation MO16.2. A three-dimensional model of cyclopentadiene.
• Benzene is fully conjugated, all the way around the ring.
Not only is benzene cyclic, and not only does it display conjugation, but the conjugation goes around and around and around the ring in an endless pi bond. That must make it different from other compounds with C=C bonds. However, cyclooctatetraene is cyclic and fully conjugated, with four double bonds and four single bonds in an eight membered ring. Cyclooctatetraene behaves exactly like other alkenes. It gets hydrogenated and it gets brominated and it loses its double bonds.
Go to Animation MO16.3. A three-dimensional model of cyclooctatetraene.
Cyclooctatetraene is a basket. You could hold fruit in it.
• Benzene is flat.
Not only is benzene cyclic and fully conjugated all the way around the ring, but it's flat. Maybe that's important. After all, if it's flat, the p orbitals on adjacent carbons could line up right beside each other. They would be perfectly parallel, and they would have really good overlap. To be aromatic, a molecule needs all three of these characteristics.
A very similar molecule should fit these criteria. Cyclobutadiene, C4H4 , would be cyclic, conjugated and flat. It should be aromatic and extra-stable, just like benzene
Go to Animation MO16.4. A three-dimensional model of cyclobutadiene.
The trouble is, cyclobutadiene is incredibly hard to make. People spent years trying. When they finally succeeded in making it (or compounds that contained a cyclobutadiene unit), they could just catch a fleeting glimpse of it using spectroscopic methods. Then it was gone. It reacted as soon as it came into being.
• Cyclobutadiene is even more reactive than most alkenes (other compounds with C=C bonds).
The spectroscopic techniques people used to detect cyclobutadiene seemed to give some additional information about it. Cyclobutadiene is not square; it's rectangular.
• Cyclobutadiene does not behave as if it has delocalized pi bonding. It does not seem to be conjugated.
What's the difference? It's a simple question of numbers.
• Cyclobutadiene has an even number of pi bonds. It is not aromatic. It is called "anti-aromatic".
• Benzene has an odd number of pi bonds. It is aromatic.
A Huckel MO treatment of benzene and cyclobutadiene does reveal some differences. Before looking at that, we should see the way people often draw the p orbitals combinations on rings. Because people often have trouble drawing rings edge-on, they look at the ring from the top. If you look at the ring from the top, the p orbitals look like circles, because you can only see the lobe that is closest to you. That's much easier to draw.
Looking at benzene from above, we get the following molecular orbitals (below).
Note that:
• There are six p orbitals involved in pi bonding, so six MO combinations result.
• There is only one way to get all orbitals in-phase (either they are in phase or they are not).
• This combination has the most bonding interactions between neighbouring p orbitals. It produces a new MO that is all bonding.
There is one, lowest-energy molecular orbital that has no nodes through the edges of the ring. The electrons in this orbital have a much more extended wavelength than they would in either a p orbital or a regular pi bond.
• There is only one way to get all orbitals out-of-phase (either they are out-of-phase or they are not).
• This combination has six nodes. It is entirely antibonding.
Because there can only be even nodes in a ring (try drawing one in which p orbitals are out of phase in one spot but nowhere else), there can also be combinations with two nodes or four nodes. There are always alternate ways to arrive at these pictures, and there are always two of each of these combinations.
• The combination with two nodes is still mostly bonding. It has more in-phase interactions between neighbors than out-of-phase interactions.
• The combination with four nodes is mostly antibonding. It has more out-of-phase interactions between neighbors than in-phase interactions.
Because there are only six pi electrpons in benzene (count them in the Lewis structure), all of the pi electrons are in bonding orbitals. They are delocalized.
In cyclobutadiene, the orbital combinations work in a similar way. There is an all in phase combination and an all out of phase combinations. In addition, there are two ways of getting a combination that is equally in phase and out of phase.
Note that, like in benzene, when there are multiple ways to represent a combination, one molecular orbital is usually represented to show both bonding and antibonding interactions. The other usually shows the net result, either just bonding or antibonding or, in the case of cyclobutadiene, no direct interaction between adjacent atoms because it is a non-bonding orbital. These two different representations are part of how two orbitals at the same energy level are ensured to be orthogonal to each other, just like two p orbitals on the same atom could be along the x axis or along the y axis, but not both along the same axis. They need to get out of each other's way.
If delocalized, all the pi electrons in benzene are low in energy; they are bonding. However, in cyclobutadiene, only two of the pi electrons are actually bonding; the other two are non-bonding. What's more, the orbital diagram suggests two of the electrons are unpaired, a situation that usually makes organic molecules very reactive. It has been suggested that cyclobutadiene distorts into a rectangle to avoid this sitiuation.
The energy level diagrams for cyclic, fully conjugated systems can be drawn very quickly using the Frost circle method. To use the method, draw a circle. Draw the shape of the cyclic compound inside this circle. One corner of the shape must be at the very bottom of the circle. You are going to add levels around the edge of the circle, one for each place where a corner touches the circle.
If you look at enough such systems, a pattern emerges. Huckel's rule is used to determine whether a system is aromatic or "anti-aromatic." To apply the rule, the molecule must fit the following criteria:
• it is cyclic
• it is fully conjugated all the way around the ring
• it is flat
After that, whether it is aromatic or anti-aromatic depends on the following situations:
• Aromatic systems have an odd number of electron pairs participating in the pi system
• Anti-aromatic systems have an even number of electron pairs participating in the pi system
The pi system refers to delocalized bonds formed from p orbitals that are perpendicular to the plane of the ring. In a Lewis structure, these may be obvious pi bonds, or they may be lone pairs or even "empty" p orbitals.
Exercise \(2\)
Indicate which of the following compounds would be aromatic, anti-aromatic, or neither. Just assume they are all flat.
The rules of aromaticity do not apply only to regular hydrocarbons. Sometimes, anionic or cationic hydrocarbons may have aromatic properties. As with benzene, the aromatic species have regular geometries all around, even though the Lewis structure suggests some of the carbon-carbon bonds are double and some are single. The cyclopentadienyl anion, C5H5-, is a common example. It is frequently observed as a ligand in transition metal complexes.
Figure \(5\): Ball-and-stick model of cyclopentadienyl anion.
Go to Animation MO16.5. A three-dimensional model of cyclopentadienyl anion.
Exercise \(3\)
Indicate which of the following cations would be aromatic, anti-aromatic, or neither. Just assume they are all flat.
Exercise \(4\)
Indicate whether the following anions would be aromatic, anti-aromatic, or neither. Just assume they are all flat.
Exercise \(5\)
Sometimes, hydrocarbons can be deprotonated (H+ removed by a very, very strong base) if the resulting anion is particularly stable. In the following cases, draw the anion that would result by removing a proton from a tetrahedral carbon, leaving behind a lone pair. Predict whether the anion would be especially stable.
Exercise \(6\)
Use the Frost circle method to generate the pi orbital MO diagram for the following species. For E, you’ll need to be creative in how to apply the Frost circle idea. (How many energy levels should be in the diagram?)
Exercise \(7\)
A recent research report examined the influence of hydrogen bonding on aromaticity. (Adapted from Judy I. Wu, James E. Jackson, P. v. R. Schleyer, J. Am. Chem. Soc. 2014, 136, 13526-13529. Copyright 2014, American Chemical Society. Used with permission.)
a) Draw a resonance structure of tropone in which there is charge separation (a cation and an anion).
b) Which of the two structures shows aromatic character: the original one, or the one you drew?
Answer b
In tropone, the resonance structure on the right shows more aromatic character, because it clearly shows a fully-conjugated ring with an odd number of electron pairs in the pi system. The carbonyl in the picture on the left makes that conjugation less obvious: does its carbonyl pi bond contribute to the conjugated system of the ring? If it does, that would mean four pairs of electrons in the pi system, and that would be an even number, so it would be anti-aromatic.
c) Hydroxytropolone was found to have more aromatic character than tropone. Draw a resonance structure to help explain why. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.16%3A_Delocalization_in_Aromatics.txt |
Many important biomolecules behave as aromatics, even though they don't contain benzene. Histidine is a very common amino acid with a cyclic side chain. The bases in nucleic acids, DNA and RNA, are also aromatic. Closer examination shows that these compounds follow some variation of Huckel's rules for aromaticity. The variations involve the inclusion of lone pairs in the pi bonding system.
Pyrrole is a five-membered ring with a nitrogen in it. It contains two C=C bonds. Pyrrole is aromatic. It usually reacts in a way similar to benzene, rather than like alkenes. But how does it fit the rules for aromaticity?
Pyrrole is cyclic. It is flat. Is it conjugated? It does have alternating double bond-single bond-double-single-lone pair. We have seen before that lone pairs can be conjugated with pi bonds. So, pyrrole is conjugated all the way around the ring. Counting the lone pair as a pair of pi electrons gives it an odd number.
• a lone pair can act like a pair of pi electrons if the molecule will become aromatic as a result.
Does a lone pair always have to act as a pi pair? Pyridine is a six membered ring, exactly like benzene, except that one CH unit is replaced by a nitrogen. The nitrogen has a lone pair. Pyridine already has three pi bonds. It is flat and fully conjugated. Does the lone pair add in to the system and make it anti-aromatic?
It turns out it can't. There is already a pi bond on that nitrogen, so it would be difficult for the lone pair to occupy the same space, above and below the ring. Instead, the lone pair remains orthogonal to the pi system. It is in the plane of the ring, rather than above and below it. It completely avoids interacting with the pi system.
• a lone pair can't act like a pair of pi electrons if there is already another pair of pi electrons on the same atom.
Exercise \(1\)
Create a Hückel MO diagram for each of the following 5-membered rings.
• Add electrons.
• Label the HOMO and the LUMO.
• Draw pictures of the molecular orbitals.
Indicate the number of nodes for each orbital energy level.
• Label the energy levels (π, σ, n, π*, σ*)
Answer
All of the compounds are aromatic and have the same Hückel MO diagram.
Exercise \(2\)
Create a Hückel MO diagram for each of the following 6-membered rings.
• Add electrons.
• Label the HOMO and the LUMO.
• Draw pictures of the molecular orbitals.
• Indicate the number of nodes for each orbital energy level.
• Label the energy levels (π, σ, n, π*, σ*)
13.18: Frontier Orbitals
We said before that the energy of the electrons determine the behavior of the molecule. The populated energy levels contribute to the energy and behavior of the molecule. The other energy levels are only possibilities. The only case in which these orbitals are considered is when we think about adding additional electrons to the structure. That can happen during a reduction reaction, for instance. In that case, the next energy level may be a reasonable place to put another electron.
This description may be an approximation. Once another electron is added to the molecule, repulsive and attractive forces have been changed within the molecule, so the energies of all the electrons would have to adjust. Nevertheless, the LUMO is a good first guess at where an additional electron would go and what its energy would be.
Conversely, if a molecule is going to donate an electron (or an electron pair) in a reaction, frequently the electrons come from the highest occupied molecular orbital (HOMO). These electrons are often the most reactive, both because of their high energy and their accessible location, usually farther from the nucleus or on the edge of the molecule.
HOMO and LUMO orbitals are together called frontier orbitals, because they are at the edge of the real, occupied orbitals and the imaginary, unoccupied ones. Frontier orbitals are often considered when trying to understand reactions.
• The lowest unoccupied molecular orbital (LUMO) may describe an additional electron added to a molecule.
• An additional electron added to the molecule could have this energy and could be located in the area of space described by this orbital. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.17%3A_Heteroaromatics.txt |
Exercise 13.2.1:
When two s orbitals combine out-of-phase, destructive intereference occurs.
There is a node between the atoms.
The energy of the electrons increases.
When two s orbitals combine in-phase, constructive interference occurs.
There is no node between the atoms; the electrons are found between the atoms.
The energy of the electrons decreases.
Exercise 13.3.1:
Exercise 13.4.1:
Exercise 13.4.2:
Exercise 13.5.1:
When 2 atomic orbitals are combined, 2 molecular orbitals are formed: one in-phase bonding orbital and one out-of-phase antibonding orbital.
Exercise 13.5.2:
In-phase combinations of atomic orbitals give bonding orbitals.
Exercise 13.5.3:
Out-of-phase combinations of atomic orbitals give antibonding orbitals.
Exercise 13.5.4:
The combinations of s + s OR s + p OR p + p OR s + d OR p + d atomic orbitals can lead to σ orbitals.
Exercise 3.5.5:
The combinations of side by side p + p or p + d atomic orbitals leads to π orbitals.
e) σ*
Exercise 13.5.7:
Li+ and O2- are more similar in size than K+ and O2-, so the bond between Li+ and O2- is stronger.
The energy difference between the 1s orbitals and 2s orbitals is too large, so they cannot interact. In order for orbitals to interact, the orbitals need to have the same symmetry, be in the same plane, and be similar in energy.
Exercise 13.5.8:
When two parallel p orbitals combine out-of-phase, destructive intereference occurs.
There is a node between the atoms.
The energy of the electrons increases.
When two parallel p orbitals combine in-phase, constructive interference occurs.
There is no node between the atoms; the electrons are found above and below the axis connecting the atoms.
The energy of the electrons decreases.
Exercise 13.6.1:
Exercise 13.6.2:
• Count the valence electrons on the molecule. That's the number of valence electrons on each atom, adjusted for any charge on the molecule. (eg C22- has 10 valence electrons: 4 from each carbon -- that's 8 -- and two more for the 2- charge).
• Fill electrons into the lowest energy orbitals first.
• Pair electrons after all orbitals at the same energy level have one electron.
Exercise 13.6.3:
Exercise 13.6.4:
Exercise 13.6.5:
Exercise 13.6.6:
Exercise 13.6.7:
Exercise 13.6.8:
Exercise 13.6.9:
Exercise 13.6.10
1. Na, because Na has a lower ionization potential (and a lower electronegativity) than Al.
2. Al
3. 4-, because there are four Na+
4. total e- = 2 x 3 e- (per Al) + 4 e- (for the negative charge) = 10 e-
g) $\textrm{bond order} = \frac{\textrm{# bonding e- - # antibonding e-}}{2} = \frac{8-2}{2}=3$
Exercise 13.7.1:
From MO6: 3. Diamagnetic (no unpaired electrons); 4. Paramagnetic; 5. Diamagnetic; 6. Diamagnetic; 7. Paramagnetic; 8. Diamagnetic
Exercise 13.7.2:
a. N2+. From the MO diagrams, N2+ has one less bonding electron. Thus, the bond order will be lower and the bond will be longer than in N2.
b. N2-. From the MO diagrams, N2- has one more antibonding electron. Thus the bond order will be lower and the bond will be longer than N2.
Exercise 13.7.3:
a. O2+ . From the MO diagram, O2+ has one less antibonding electron. Thus the bond order will be higher and the bond will be stronger than in O2.
b. O2. From the MO diagram, O2 has one less antibonding electron. Thus the bond order will be higher and the bond will be stronger than in O2-.
Exercise 13.9.1:
a)
b)
c)
Exercise 13.11.1:
a)
b)
c)
Exercise 13.11.2:
a)
b)
c)
d)
e)
Exercise 13.11.3:
a)
b)
c)
d)
e)
f)
g) In each case all four H1s orbitals interact with each boron orbital.
h) The constructive overlap of the four H1s orbitals and the B2s orbital results in the lowest energy combination as it contains no nodes. (LOWEST ENERGY)
The constructive overlap of the four H1s orbitals with the three B2p orbitals results in bonding orbitals containing a single node. (MEDIUM-LOW ENERGY)
The destructive overlap of the four H1s orbitals with the B2s orbital results in an anti-bonding orbital. (HIGH ENERGY)
The destructive overlap of the four H1s orbitals with the three B2p orbitals results in anti-bonding orbitals. (HIGHEST ENERGY)
Exercise 13.11.4:
a)
b)
c)
d)
e)
f)
g) In some of the cases the one or more Hs may be in a nodal plane
h) The constructive overlap of the H1s orbitals and the B2s orbital results in the lowest energy combination. (LOWEST ENERGY)
The constructive overlap of the three H1s orbitals with the B2p orbital results in a bonding orbital (MEDIUM-LOW ENERGY)
The constructive overlap of the two H1s orbitals with the B2p orbital results in a bonding orbital. (MEDIUM-LOW ENERGY)
The B2p orbital that is perpendicular to the plane of hydrogen atoms results in a non-bonding orbital. (MEDIUM ENERGY)
The destructive overlap of the two H1s orbitals with the B2p orbital results in an anti-bonding orbital (HIGH ENERGY)
The destructive overlap of the three H1s orbitals with the B2p orbital results in an anti-bonding orbital (HIGH ENERGY)
The destructive overlap of the three H1s orbitals with the B2p orbital results in an anti-bonding orbital (HIGHEST ENERGY)
Exercises 13.12.1 & 13.12.2:
Exercise 13.12.3:
Exercises 13.13.1-13.13.3:
Exercises 13.14.1-13.14.5:
Exercise 13.14.6:
All of the answers depend on an understanding of the contributions of two resonance structures to the overall picture of acetaminde, or alternatively, that actetamide forms a conjugated pi system with four electrons delocalized over the O, C and N.
a. Contribution of the second resonance structure introduces some double bond character to the C-N bond and some single bond character to the C-O bond. Thus, both of these bonds are intermediate in length between single and double bonds.
b. Since the C, N and O atoms are sp2 hybridized, the C-N pi bond can only form if the remaining p orbitals on these atoms align. This places the atoms participating in the sp2 sigma bonds in the same plane.
c. Because of the partial double bond character there is a larger barrier to rotation than is typically found in molecules with only single bonds.
d. Because of the partial double bond character and the restricted rotation, the two H’s are not identical. (One is nearer the O and one is nearer the CH3 and the restricted rotation prevents their interconversion.
Exercises 13.15.1-13.15.3:
Exercise 13.15.4:
Exercise 13.16.1:
Because the two resonance structures shows double bonds in two different places, the implication is that all of the bonds in benzene have some double bond and some single bond character. You can think of them all as being about 1.5 bonds.
Exercise 13.16.2:
a) non-aromatic b) non-aromatic c) aromatic d) aromatic e) anti-aromatic f) anti-aromatic
Exercise 13.16.3:
a) aromatic b) non-aromatic c) anti-aromatic d) aromatic
Exercise 13.16.4:
a) aromatic b) anti-aromatic c) aromatic d) anti-aromatic
Exercise 13.16.5:
Exercise 13.16.6:
Exercise 13.16.7:
In tropone, the resonance structure on the right shows more aromatic character, because it clearly shows a fully-conjugated ring with an odd number of electron pairs in the pi system. The carbonyl in the picture on the left makes that conjugation less obvious: does its carbonyl pi bond contribute to the conjugated system of the ring? If it does, that would mean four pairs of electrons in the pi system, and that would be an even number, so it would be anti-aromatic.
The hydroxy group in hydroxytropolone would stabilize the compound in the structure shown on the right, via an ion-dipole interaction with the anionic oxygen. That makes the right-hand, explicitly aromatic structure the dominant one.
Exercise 13.17.1:
All of the compounds are aromatic and have the same Hückel MO diagram.
Exercise 13.17.2:
All of the compounds are aromatic and have the same Hückel MO diagram. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/13%3A_Molecular_Orbital_Theory/13.19%3A_Solutions_to_Selected_Problems.txt |
Physical changes result in the transition of a molecular material from one form into another without any change in the composition of the material. A liquid compound can be cooled until it freezes or heated until it evaporates, but the atoms that make up the material are still connected together in the same way.
Chemical reactions result in a change in the composition of a material. Atoms become associated in different ways. Changes in bonding occur. Because bonding involves some kind of shared distribution of electrons between atoms, a chemical reaction involves some change in how electrons are arranged in the material.
• The standard way to think about a chemical reaction is to consider the movement of electrons in the reaction.
Reactions often involve many changes, so that one material is transformed into others via numerous redistributions of electrons. These individual steps within an overall reaction are sometimes called elementary reactions. One of the most common ways to analyze an elementary reaction is to understand in a very basic way where the electrons are moving from and where they are moving towards.
• An atom or molecule that supplies a pair of electrons to form a new bond is an electron donor. An electron donor is often called a Lewis base.
• An atom or molecule that accepts a pair of electrons to form a new bond is an electron acceptor. An electron acceptor is often called a Lewis acid.
One of the most common Lewis acids, or electron acceptors, is a proton. It is so common that people often use the term "acid" just to describe compounds or solutions that supply protons.
A Lewis base is anything with a lone pair. For example, water is a Lewis base because the oxygen atom has lone pairs. If we take a source of protons, such as hydrogen chloride (HCl, a colorless gas) and bubbler it into a flask of water, a reaction will result. The water will donate electrons to the proton in the HCl, producing hydrochloric acid, or aqueous hydrogen chloride: HCl(aq).
Note that a hydrogen atom can only have one bond, so when the water donates its electrons to the hydrogen, the bond between the hydrogen and the chlorine has to be broken. The electrons in the bond move to the chlorine, which is more electronegative than the hydrogen.
Also, notice that the hydrogen atom did not even need a full positive charge to attract the water. A partial positive charge will still attract electrons.
Acids don't have to be protons or protons donors. Very often, metal ions are able to attract electrons, because metal ions are positively charged. Ca2+, Sc2+, Ti4+ are just a few common examples.
Figure \(4\): A calcium ion can attract electrons from a donor atom.
These ideas will be explored more fully in this chapter. They are some of the most important ideas in chemistry, forming the basis for most of what we know about the relationships between structure, properties, and reactivity.
14.02: Lewis Bases
What makes a molecule (or an atom or ion) a Lewis base? It must have a pair of electrons available to share with another atom to form a bond. The most readily available electrons are those that are not already in bonds. Bonding electrons are low in energy. Non-bonding electrons are higher in energy and may be stabilized when they are delocalized in a new bond.
• Lewis bases usually have non-bonding electrons or lone pairs.
Ammonia, NH3, has a lone pair and is a Lewis base. It can donate to compounds that will accept electrons.
Lewis bases may be anionic or neutral. The basic requirement is that they have a pair of electrons to donate. Examples of Lewis bases include halide ions such as bromide or chloride. To the right of the halides in the periodic table are Noble gases such as neon. Noble gases do have lone pairs, but are stable enough that they do not usually react. They are not very good Lewis bases. To the left of the halides, however, are other examples in oxygen and nitrogen compounds. Water also has lone pairs and is a common Lewis base, and so is hydroxide ion, HO-.
• Halides, water, ammonia and hydroxide ion are examples of Lewis bases.
One column further to the left in the periodic table from nitrogen is carbon. Carbon does not normally have a lone pair. For example, methane, CH4, has all of its valence electrons in bonding pairs. These bonding pairs are too stable to donate under normal conditions. Methane is not a Lewis base.
Even further to the left is boron. A simple boron compound is borane, BH3. Borane has no lone pairs; all its valence electrons are in bonds. Boron is not a good Lewis base.
Exercise \(1\)
What structural feature is required in order to be a Lewis base?
Answer
Exercise \(2\)
Which of the following compounds appear to be Lewis bases?
a) SiH4 b) AlH3 c) PH3 d) SH2 e) -SH
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.01%3A_General_Acidity_and_Basicity.txt |
Borane is unusual because it is a compound without an octet. The central boron atom has only six valence electrons. It needs one more pair of electrons to obtain an octet. The boron is a Lewis acid.
• Lewis acids are often short of a complete octet.
In the main group of the periodic table, atoms in the Group IIIA column (including boron and aluminum) have three valence electrons to share in order to make bonds. Sharing these electrons with three electrons from neighbors would make three bonds, and provide six electrons, not eight, in the valence shell. Another pair of electrons must be accepted from a donor to achieve an octet.
• Boron, aluminum and indium compounds are often Lewis acids.
The eight-electron rule does not hold throughout the periodic table. In order to obtain noble gas configurations, some atoms may need eighteen electrons in their valence shell. For example, transition metals such as titanium often follow an eighteen-electron rule. Titanium has four valence electrons and can form four bonds in compounds such as titanium tetrakis(isopropoxide), below, or titanium tetrachloride, TiCl4. However, the titanium atom in that compound has only eight valence electrons, not eighteen. It can easily accept electrons from donors.
• Transition metals such as titanium, iron and nickel may have up to eighteen electrons and can frequently accept electron pairs from Lewis bases. Transition metals are often Lewis acids.
• Lanthanides such as cerium and samarium could conceivably have up to thirty-two electrons in their valence shells! They never do. However, they are usually strong Lewis acids.
Positive ions are often Lewis acids because they have an electrostatic attraction for electron donors. Examples include alkali and alkaline earth metals in the group IA and IIA columns. K+, Mg2+ and Ca2+ are sometimes seen as Lewis acidic sites in biology, for example. These ions are very stable forms of these elements because of their low electron ionization potentials. However, their positive charges do attract electron donors.
In a similar way, "early" transition metals -- those that are close to the left hand side of the periodic table, especially in groups IIIB, IVB and VB -- have low ionization potentials and have high positive charges or oxidation states. For example, Sc3+, Zr4+ and V5+ are common forms of some early transition metals, and they are strong Lewis acids.
• Many cations such as Ca2+ or Sc3+ are good Lewis acids. Their positive charges attract electrons.
Sodium, potassium, calcium and magnesium are common ions in biology. Sodium, potassium and calcium have well-known roles in cell signaling. The build-up of these ions on one side or another of a cell membrane results in a charge separation across the membrane, called a membrane potential. However, the Lewis acidity of these ions are also important in biology. Because they attract electron donors, these ions are generally found in biological situations with water molecules stuck to them; remember that the oxygen atom in water is Lewis basic. That means that the proteins that form ion channels in cell membranes must somehow cope with the water molecules that are all around when they are transporting potassium ions or sodium ions across the membranes. A potassium ion covered in water molecules is a much larger object than a potassium ion by itself. The pore that opens up between the proteins must be big enough to allow the entire water-potassium complex through, or else there must be a way to separate the water molecules from the potassium ion. At the same time, the ion channel may exploit the Lewis acidity of the potassium ion; for example, the channel proteins may contain oxygen atoms situated in such a way that they draw the ion into the channel. Exactly how ion channels work is a very active topic of current research at the interface of chemistry and biology.
There are other biological situations in which the Lewis acidity of these ions plays a role. For example, potassium ions bind to nucleic acids in DNA to form the telomeres at the ends of chromosomes. This binding involves the donation of lone pairs on the nucleic acids (guanidines in this case) to a potassium ion. Another common biological structure in which donor atoms bind to a Lewis acid is chlorophyll, in which a porphyrin ring binds to a central magnesium ion. Chlorophyll is involved in the absorption of sunlight and the initiation of electron transport to photosystem I and II, which are the engines of photosynthesis. One of the roles of the magnesium ion in this structure is to affect the color of the chlorophyll and hence the wavelengths of light absorbed; when other metal ions are bound by the porphyrin, different colors are absorbed.
A number of biological molecules that contain metal ions perform tasks that require Lewis acidity. Many enzymes that contain metal ions, sometimes called metalloproteins, bind a substrate and carry out a transformation on it. The substrate is usually able to donate a pair of electrons to the metal ion and stick to it while a reaction occurs. For example, vanadium bromoperoxidase is a protein found in some kinds of marine algae. It can bind dissolved oxygen molecules as well as bromide ions from the surrounding saltwater. A subsequent series of reactions allows the bromine atom to be incorporated into large organic molecules. Halogens in organic molecules are extremely rare in nature, but it is perhaps not surprising that organisms surrounded by salt water have managed to make them. The resulting brominated compounds seem to be used as chemical defenses, making the algae less likely to get eaten.
A number of other metal ions are common in biology. Manganese is prominent in photosystem I and II, iron is found in hemoglobin, and cobalt is in vitamin B12. Hemoglobin may be familiar because of its role in transporting oxygen in the blood.
Atoms in the lower part of the periodic table often have variable oxidation states. They can form different numbers of bonds. Although carbon generally forms four bonds, forming compounds like carbon tetrachloride, CCl4, tin can sometimes form two bonds as well as four. Tin forms stannous chloride, SnCl2, as well as stannic chloride, SnCl4. It is possible that stannous chloride, with sixteen valence electrons, could accept another pair of electrons.
It may be surprising that stannic chloride is also a Lewis acid. The tin atom in stannic chloride has a full octet (eighteen valence electrons), but it still attracts electron donors. Given the difference in electronegativity between tin and chlorine, however, it seems reasonable to think of these compounds in terms of very polar bonds.
• Atoms with an octet can still be Lewis acidic if they bear a large, partial positive charge.
Carbon atoms can be Lewis acidic even though carbon does not generally form ionic bonds. For example, formyl chloride, H-(C=O)-Cl, contains a carbon atom that is bonded to a chlorine and also double bonded to an oxygen atom. Although these bonds are certainly covalent, bond polarity places a large, partial positive charge on the carbon. Formyl chloride easily attracts a pair of electrons from a Lewis base.
Exercise \(1\)
What structural feature is required to be a Lewis acid?
Answer
Exercise \(2\)
Which of the following compounds appear to be Lewis acids? Explain your reasoning in each case.
a) CeCl3 b) BF3 c) CH4 d) CH2O e) N2
Answer
Exercise \(3\)
The Smith research group at University of Notre Dame has been working with molecules that can be used for imaging, sensing and related health care applications.
1. Circle the ring that contains several oxygens.
2. The part of the molecule you have circled would bind (choose one):
anions cations
c) Explain how these ions would be bound.
d) Put a box around the portion of the molecule capable of acting as a hydrogen bond donor.
e) The part of the molecule you have circled would bind (choose one):
anions cations
f) The Smith lab has shown that these molecules bind chloride ions. Draw a chloride ion in the correct position in the molecule.
g) In the presence of potassium ions, the molecule can bind both potassium and chloride. Add both ions in the correct positions in the molecule.
Measurement of the equilibrium of bound vs. unbound chloride ions showed that chloride binding was stronger when potassium ions were also bound.
h) Explain why chloride binding is stronger when a potassium ion is also bound to the sensor molecule.
Answer
a)
b) cations
c) possibly ion-dipole forces; alternatively, the oxygen atoms could act as Lewis bases, donating lone pairs to a cation.
d)
e) anions
f)
g)
h) When the chloride ion binds to the sensor molecule it gives the overall complex a negative charge and addition of the potassium ion cancels out this charge | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.03%3A_Lewis_Acids.txt |
What happens when a Lewis base donates a pair of electrons to a Lewis acid? The arrow formulism we have been using to illustrate the behavior of Lewis acids and Lewis bases is meant to show the direction of electron movement from the donor to the acceptor. However, given that a bond can be thought of as a pair of electrons that are shared between two atoms (in this case, between the donor and the acceptor), these arrows also show where bonds are forming.
The electrons donated from a Lewis base to a Lewis acid form a new bond. A new, larger compound is formed from the smaller Lewis acid and Lewis base. This compound is called a Lewis acid-base complex.
A simple example of Lewis acid-base complexation involves ammonia and boron trifluoride. The nitrogen atom has a lone pair and is an electron donor. The boron has no octet and is an electron acceptor. The two compounds can form a Lewis acid-base complex or a coordination complex together.
When the nitrogen donates a pair of electrons to share with the boron, the bond that forms is sometimes called a coordinate bond. Another term for this kind of bond is a dative bond. A coordinate or dative bond is any covalent bond that arose because one atom brought a pair of its electrons and donated them with another.
There is another piece of terminology you should get used to here. Sometimes, the electron donor is called a nucleophile and the electron acceptor is called an electrophile. Ammonia is a nucleophile and boron trifluoride is an electrophile.
• Because Lewis bases are attracted to electron-deficient atoms, and because positive charge is generally associated with the nucleus of an atom, Lewis bases are sometimes refered to as "nucleophiles". Nucleophile means nucleus-loving.
• Because Lewis acids attract electron pairs, Lewis acids are sometimes called "electrophiles". Electrophile meanse electron-loving.
Lewis acid-base complexes frequently have very different properties from the separate compounds from which they were formed. For example, titanium tetrachloride is a yellow liquid at room temperature. It is so Lewis acidic that it reacts with moisture in the air, undergoing a reaction that generates HCl gas in the form of white smoke. Tetrahydrofuran (or THF), a mild Lewis base, is a colorless liquid. When THF and TiCl4 are combined, a Lewis acid-base complex is formed, TiCl4(THF)2. TiCl4(THF)2 is a yellow solid at room temperature. Although it still reacts with the air, it does so very slowly, and shows no visible change when exposed to the air for several minutes.
Exercise \(1\)
Use curved arrow notation to show the electron movement for the following reactions.
Answer
Exercise \(2\)
Show, using arrow notation, the two-step reaction between THF and titanium tetrachloride to form the Lewis acid-base complex, TiCl4(THF)2. Also show the structures of the intermediate and final complexes.
Answer
Exercise \(3\)
A similar Lewis acid-base complex is formed between THF and borane, BH3.
a) Which compound is the Lewis acid? Which one is the Lewis base?
b) Which atom in the Lewis acid is the acidic site? Why?
c) Which atom in the Lewis base is the basic site? Why?
d) How many donors would be needed to satisfy the acidic site?
e) Show, using arrow notation, the reaction to form a Lewis acid-base complex.
f) Borane is highly pyrophoric; it reacts violently with air, bursting into flames. Show, using arrow notation, what might be happening when borane contacts the air.
g) Borane-THF complex is much less pyrophoric than borane. Why do you suppose that is so?
Answer
Exercise \(4\)
When a neutral Lewis acid combines with an anionic Lewis base, the product is called a complex ion. The same is true if a cationic Lewis acid combines with a neutral Lewis base.
Show the formation of the following polyatomic anions from the Lewis acid-base pairs that were combined in each case.
a) BF4- b) PF6- c) AlCl4- d) AlH4- e) Ag(NH3)2
Exercise \(5\)
Consider the reaction below.
• Draw an MO mixing diagram for the reaction above, using the following steps:
o Draw the orbital from the base that is likely to donate its electrons.
o Draw the orbital from the acid that is likely to accept electrons.
o Complete the MO mixing diagram of these two orbitals:
• Label the electron donating orbital
• Label the electron accepting orbital
• Populate the MO mixing diagram with electrons
o Draw a cartoon showing the mixing of these orbitals.
Answer
Exercise \(6\)
a) A typical Lewis acid/base pair is shown below.
• Label the two compounds below as either Lewis acid or Lewis base.
• Draw curved arrows to show movement of electrons to form a bond.
• Draw the product of the reaction.
b) The Lewis pair below reacts to form an adduct, but much more slowly than the one above.
• Suggest two reasons why the Lewis acid/base adduct forms so slowly.
Answer
a)
b)
• The benzene rings take up more space than a fluorine atom; they may get in the way, making it harder for the water molecule to approach.
Also, the aromatic rings may be able to form a conjugated system with the empty p orbital; if that p orbital is partially filled, the boron atom becomes less Lewis acidic.
• A larger group than fluorine may cause more steric hindrance. Replacing the two fluorines closest to the boron on each arene (aromatic ring) with a CH3 or CF3 group would slow down the formation of an adduct. Alternatively, a less electronegative group than fluorine would also make the boron seem less positive; a CH3 or OCH3 are two possibilities.
• An even more electron-withdrawing group than fluorine would make the Lewis acid more reactive. Examples include nitro (NO2) and carbonyl groups (such as CH3C=0); these groups are resonance-withdrawing. Alternatively, a smaller group such a hydrogen would lower steric resistance, but would also lead to lower electrophilicity at boron, owing to the lower electronegativity of hydrogen compared to fluorine.
Chemical reactions involve bond-making and bond-breaking events, as well as the movement of electrons. When we think about chemical reactions, we often think about where the electrons are coming from, and where they are going. In a Lewis structure picture, we most often think of the electrons as coming from a lone pair -- a non-bonding pair of electrons on one particular atom. We picture the electrons becoming attracted toward an atom that lacks electrons, maybe because it does not have a filled valence shell, or maybe because it has some amount of positive charge.
In addition to a Lewis picture, it's often useful to think about reactions in terms of molecular orbital interactions. That kind of consideration is especially useful in computational chemistry where, through the use of the right software, we can calculate energy changes that occur over the course of a reaction. It's also helpful to develop some conceptual understanding of these approaches qualitatively. This qualitative approach to molecular orbital interactions is routinely used by chemists because of the insight it can give into reactions.
One common way of thinking about reactions in this way is through the concept of frontier orbitals. This idea says that if one species is going to donate electrons to another in order to form a new bond, then the donated electrons are most likely going to come from the highest occupied energy level. In this level, called the highest occupied molecular orbital (HOMO), the electrons are further from the nucleus and therefore less tightly held by the protons in the nucleus. The electrons would be donated, in turn, to the lowest empty energy level on the other species, called the lowest unoccupied molecular orbital (LUMO).
Consider an example of such an interaction, between a hydroxide ion and a proton. A hydroxide ion, HO-, is a Lewis base. The oxygen atom has three lone pairs, any of which might be donated to a Lewis acid. A proton, H+, is a Lewis acid. For a hydrogen atom, in the very first little row of the periodic table, the "octet rule" is two electrons, so a proton would be able to accept a pair of electrons from another atom and form a covalent bond.
The atomic orbital diagram for a proton is very simple. Hydrogen has only a 1s orbital, and in H+ that energy level is empty. This orbital corresponds to the LUMO for a proton.
The molecular orbital diagram for hydroxide ion is not much more complicated. This molecule is diatomic; it comes from the combination of an oxygen atom with a hydrogem atom, with the addition of an extra electron to provide the negative charge of the ion. In the diagram below, the hydrogen atom interacts with one of the p orbitals on the oxygen, but it does not matter exactly which oxygen orbital we use.
In most cases, we could come up with the MO diagram in another way. If we take the shortcut of working out an approximate MO diagram of a molecule based on its Lewis structure, and we know that hydroxide ion has one O-H bond and three lone pairs on oxygen, then we know there should be a bonding orbital at low energy, an antibonding orbital at high energy, and three non-bonding orbitals in the middle.
In a Lewis acid-base interaction, a pair of electrons would be donated from the non-bonding level on hydroxide (the HOMO) to the empty 1s orbital on the proton (the LUMO).
What we have here is an interaction between two orbitals. A pair of electrons in one orbital is being shared with another orbital. We already know that an interaction between two orbitals results in two new orbitals. One of the new orbitals, resulting from constructive interference, is lower in energy than either of the original orbitals. The other new orbital results from destructive interference and is higher in energy than either of the original orbitals.
In the end, the two electrons being donated slide down in energy to become an O-H bond. The combination that rises in energy does not really matter because there are no electrons at that level, anyway. Overall, the net energy of the proton and the hydroxide ion has decreased as the pair came together to form a water molecule.
Note that the MO diagram for the resulting water molecule resempbles what we would expect from its Lewis structure. There are two low-energy O-H bonding pairs and two correspondingly high-energy antibonding orbitals. There are also two intermediate-level nonbonding pairs corresponding to the two lone pairs we see in the Lewis structure.
We can use the same approach to look at Lewis acid-base interaction in bigger molecules. The MO diagrams are a little busier, but the ideas are the same. For example, we could imagine a fluoride ion donating electrons to a molecule of borane, BH3. The fluoride is a Lewis base because it has lone pairs. The borane is a Lewis acid because the boron atom lacks an octet; it has only six valence electrons in its structure.
In this case, the borane contains three B-H bonds, so there will be three B-H bonding pairs and three empty B-H antibonding levels. There would be an empty orbital as well, corresponding to an empty p orbital on the boron. That empty p orbital is the lowest unoccupied molecular orbital (LUMO). A fluoride ion would have four lone pairs, and we would probably imagine a pair of electrons from one of its p orbitals as the highest occupied molecular orbital (HOMO). The interaction therefore involves donation from one of these orbitals on fluoride to the empty p orbital on borane.
Once again, this interaction would result in a new molecule with a new molecular orbital diagram. The only appreciable changes would involve the two orbitals that interact with each other, the HOMO and LUMO. The diagram for BH3F- ion is really a superposition of the two diagrams before, except that the HOMO and LUMO have formed a new bonding and antibonding orbital for the new B-F bond.
Exercise \(7\)
For the following reactions, show (i) the HOMO-LUMO interaction and (ii) the molecular orbital interaction diagram for formation of the new molecule.
1. A fluoride ion donates to a boron trifluoride molecule (BF3), forming a tetrafluoroborate ion (BF4-).
2. An ammonia molecule (NH3) donates to a borane molecule (BH3), forming the adduct BH3NH3.
Answer
Answer a.(i
Answer a.(ii
Answer b.(i
Answer b.(ii | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.04%3A_Lewis_Acid-Base_Complexes_and_Molecular.txt |
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