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The Relationship between Bond Order and Bond Energy
Triple bonds between like atoms are shorter than double bonds, and because more energy is required to completely break all three bonds than to completely break two, a triple bond is also stronger than a double bond. Similarly, double bonds between like atoms are stronger and shorter than single bonds. Bonds of the same order between different atoms show a wide range of bond energies, however. Table 8.6 lists the average values for some commonly encountered bonds. Although the values shown vary widely, we can observe four trends:
Table 8.6 Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K
Single Bonds Multiple Bonds
H–H 432 C–C 346 N–N ≈167 O–O ≈142 F–F 155 C=C 602
H–C 411 C–Si 318 N–O 201 O–F 190 F–Cl 249 C≡C 835
H–Si 318 C–N 305 N–F 283 O–Cl 218 F–Br 249 C=N 615
H–N 386 C–O 358 N–Cl 313 O–Br 201 F–I 278 C≡N 887
H–P ≈322 C–S 272 N–Br 243 O–I 201 Cl–Cl 240 C=O 749
H–O 459 C–F 485 P–P 201 S–S 226 Cl–Br 216 C≡O 1072
H–S 363 C–Cl 327 S–F 284 Cl–I 208 N=N 418
H–F 565 C–Br 285 S–Cl 255 Br–Br 190 N≡N 942
H–Cl 428 C–I 213 S–Br 218 Br–I 175 N=O 607
H–Br 362 Si–Si 222 I–I 149 O=O 494
H–I 295 Si–O 452 S=O 532
Source: Data from J. E. Huheey, E. A. Keiter, and R. L. Keiter, Inorganic Chemistry, 4th ed. (1993).
1. Bonds between hydrogen and atoms in the same column of the periodic table decrease in strength as we go down the column. Thus an H–F bond is stronger than an H–I bond, H–C is stronger than H–Si, H–N is stronger than H–P, H–O is stronger than H–S, and so forth. The reason for this is that the region of space in which electrons are shared between two atoms becomes proportionally smaller as one of the atoms becomes larger (part (a) in Figure 8.11).
2. Bonds between like atoms usually become weaker as we go down a column (important exceptions are noted later). For example, the C–C single bond is stronger than the Si–Si single bond, which is stronger than the Ge–Ge bond, and so forth. As two bonded atoms become larger, the region between them occupied by bonding electrons becomes proportionally smaller, as illustrated in part (b) in Figure 8.11. Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., N, O, F), which are unusually weak compared with single bonds between their larger congeners. It is likely that the N–N, O–O, and F–F single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms. The trend in bond energies for the halogens is therefore
Cl–Cl > Br–Br > F–F > I–I
Similar effects are also seen for the O–O versus S–S and for N–N versus P–P single bonds.
Note
Bonds between hydrogen and atoms in a given column in the periodic table are weaker down the column; bonds between like atoms usually become weaker down a column.
3. Because elements in periods 3 and 4 rarely form multiple bonds with themselves, their multiple bond energies are not accurately known. Nonetheless, they are presumed to be significantly weaker than multiple bonds between lighter atoms of the same families. Compounds containing an Si=Si double bond, for example, have only recently been prepared, whereas compounds containing C=C double bonds are one of the best-studied and most important classes of organic compounds.
Figure 8.11 The Strength of Covalent Bonds Depends on the Overlap between the Valence Orbitals of the Bonded Atoms. The relative sizes of the region of space in which electrons are shared between (a) a hydrogen atom and lighter (smaller) vs. heavier (larger) atoms in the same periodic group; and (b) two lighter versus two heavier atoms in the same group. Although the absolute amount of shared space increases in both cases on going from a light to a heavy atom, the amount of space relative to the size of the bonded atom decreases; that is, the percentage of total orbital volume decreases with increasing size. Hence the strength of the bond decreases.
4. Multiple bonds between carbon, oxygen, or nitrogen and a period 3 element such as phosphorus or sulfur tend to be unusually strong. In fact, multiple bonds of this type dominate the chemistry of the period 3 elements of groups 15 and 16. Multiple bonds to phosphorus or sulfur occur as a result of d-orbital interactions, as we discussed for the SO42− ion in Section 8.6. In contrast, silicon in group 14 has little tendency to form discrete silicon–oxygen double bonds. Consequently, SiO2 has a three-dimensional network structure in which each silicon atom forms four Si–O single bonds, which makes the physical and chemical properties of SiO2 very different from those of CO2.
Note
Bond strengths increase as bond order increases, while bond distances decrease.
Table: Average bond energies:
Bond
(kJ/mol)
C-F
485
C-Cl
328
C-Br
276
C-I
240
C-C
348
C-N
293
C-O
358
C-F
485
C-C
348
C=C
614
C=C
839
Contributors
• Kim Song (UCD), Donald Le (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/01%3A_Structure_and_Bonding/1.05%3A_Bond_Length_and_Bond_Strength.txt |
Electronegativity
The elements with the highest ionization energies are generally those with the most negative electron affinities, which are located toward the upper right corner of the periodic table. Conversely, the elements with the lowest ionization energies are generally those with the least negative electron affinities and are located in the lower left corner of the periodic table.
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity (represented by the Greek letter chi, χ, pronounced “ky” as in “sky”), defined as the relative ability of an atom to attract electrons to itself in a chemical compound. Elements with high electronegativities tend to acquire electrons in chemical reactions and are found in the upper right corner of the periodic table. Elements with low electronegativities tend to lose electrons in chemical reactions and are found in the lower left corner of the periodic table.
Unlike ionization energy or electron affinity, the electronegativity of an atom is not a simple, fixed property that can be directly measured in a single experiment. In fact, an atom’s electronegativity should depend to some extent on its chemical environment because the properties of an atom are influenced by its neighbors in a chemical compound. Nevertheless, when different methods for measuring the electronegativity of an atom are compared, they all tend to assign similar relative values to a given element. For example, all scales predict that fluorine has the highest electronegativity and cesium the lowest of the stable elements, which suggests that all the methods are measuring the same fundamental property.
Note
Electronegativity is defined as the ability of an atom in a particular molecule to attract electrons to itself. The greater the value, the greater the attractiveness for electrons.
Molecular Dipole Moments
You previously learned how to calculate the dipole moments of simple diatomic molecules. In more complex molecules with polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether there is a net dipole moment. Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO2, a linear molecule (part (a) in Figure 9.2.8). Each C–O bond in CO2 is polar, yet experiments show that the CO2 molecule has no dipole moment. Because the two C–O bond dipoles in CO2 are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO2 molecule has no net dipole moment even though it has a substantial separation of charge. In contrast, the H2O molecule is not linear (part (b) in Figure 9.2.8); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H2O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H2O to hydrogen-bond to other polarized or charged species, including other water molecules.
Figure 9.2.8 How Individual Bond Dipole Moments Are Added Together to Give an Overall Molecular Dipole Moment for Two Triatomic Molecules with Different Structures. (a) In CO2, the C–O bond dipoles are equal in magnitude but oriented in opposite directions (at 180°). Their vector sum is zero, so CO2 therefore has no net dipole. (b) In H2O, the O–H bond dipoles are also equal in magnitude, but they are oriented at 104.5° to each other. Hence the vector sum is not zero, and H2O has a net dipole moment.
Other examples of molecules with polar bonds are shown in Figure 9.2.9. In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl3 is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment.
Figure 9.2.9: Molecules with Polar Bonds. Individual bond dipole moments are indicated in red. Due to their different three-dimensional structures, some molecules with polar bonds have a net dipole moment (HCl, CH2O, NH3, and CHCl3), indicated in blue, whereas others do not because the bond dipole moments cancel (BCl3, CCl4, PF5, and SF6).
Note
Molecules with asymmetrical charge distributions have a net dipole moment.
Example
Which molecule(s) has a net dipole moment?
1. H2S
2. NHF2
3. BF3
Given: three chemical compounds
Asked for: net dipole moment
Strategy:
For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment.
Solution:
1. The total number of electrons around the central atom, S, is eight, which gives four electron pairs. Two of these electron pairs are bonding pairs and two are lone pairs, so the molecular geometry of H2S is bent (Figure 9.2.6). The bond dipoles cannot cancel one another, so the molecule has a net dipole moment.
2. Difluoroamine has a trigonal pyramidal molecular geometry. Because there is one hydrogen and two fluorines, and because of the lone pair of electrons on nitrogen, the molecule is not symmetrical, and the bond dipoles of NHF2 cannot cancel one another. This means that NHF2 has a net dipole moment. We expect polarization from the two fluorine atoms, the most electronegative atoms in the periodic table, to have a greater affect on the net dipole moment than polarization from the lone pair of electrons on nitrogen.
3. The molecular geometry of BF3 is trigonal planar. Because all the B–F bonds are equal and the molecule is highly symmetrical, the dipoles cancel one another in three-dimensional space. Thus BF3 has a net dipole moment of zero:
Exercise
Which molecule(s) has a net dipole moment?
1. CH3Cl
2. SO3
3. XeO3
Answer: CH3Cl; XeO3 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/01%3A_Structure_and_Bonding/1.06%3A_Electronegativity_and_Bond_Polarity.txt |
Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor into the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule.
Introduction
When atoms in a molecule share electrons unequally, they create what is called a dipole moment. This occurs when one atom is more electronegative than another, resulting in that atom pulling more tightly on the shared pair of electrons, or when one atom has a lone pair of electrons and the difference of electronegativity vector points in the same way. One of the most common examples is the water molecule, made up of one oxygen atom and two hydrogen atoms. The differences in electronegativity and lone electrons give oxygen a partial negative charge and each hydrogen a partial positive charge.
Dipole Moment
When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, a dipole is established. The size of a dipole is measured by its dipole moment ((\mu\)). Dipole moment is measured in debye units, which is equal to the distance between the charges multiplied by the charge (1 debye equals 3.34 x 10-30 coulomb-meters). The equation to figure out the dipole moment of a molecule is given below:
μ=qr(1)
where
• μ is the dipole moment,
• q is the magnitude of the charge, and
• r is the distance between the charges.
The dipole moment acts in the direction of the vector quantity. An example of a polar molecule is H2O. Because of the lone pair on oxygen, the structure of H2O is bent, which means it is not symmetric. The vectors do not cancel each other out, making the molecule polar.
Figure 1: Dipole moment of water
The vector points from positive to negative, on both the molecular (net) dipole moment and the individual bond dipoles. The table above shows the electronegativity of some of the common elements. The larger the difference in electronegativity between the two atoms, the more electronegative that bond is. To be considered a polar bond, the difference in electronegativity must be large. The dipole moment points in the direction of the vector quantity of each of the bond electronegativities added together.
Example 1: Water
The water molecule picture from Figure 1 can be used to determine the direction and magnitude of the dipole moment. From the electronegativities of water and hydrogen, the difference is 1.2 for each of the hydrogen-oxygen bonds. Next, because the oxygen is the more electronegative atom, it exerts a greater pull on the shared electrons; it also has two lone pairs of electrons. From this, it can be concluded that the dipole moment points from between the two hydrogen atoms toward the oxygen atom. Using the equation above, the dipole moment is calculated to be 1.85 D by multiplying the distance between the oxygen and hydrogen atoms by the charge difference between them and then finding the components of each that point in the direction of the net dipole moment (remember the angle of the molecule is 104.5˚).
The bond moment of O-H bond =1.5 D, so the net dipole moment =2(1.5)×cos(104.5/2)=1.84 D.
Dipole Moments
It is relatively easy to measure dipole moments. Place substance between charged plates--polar molecules increase the charge stored on plates and the dipole moment can be obtained (has to do with capacitance). Polar molecules align themselves:
1. in an electric field
2. with respect to one another
3. with respect to ions
Nonpolar CCl4 is not deflected; moderately polar acetone deflects slightly; highly polar water deflects strongly.
Figure 2: Polar molecules align themselves in an electric field (left), with respect to one another (middle), and with respect to ions (right)
When proton & electron close together, the dipole moment (degree of polarity) decreases. However, as proton & electron get farther apart, the dipole moment increases. In this case, the dipole moment calculated as:
μ=Qr=(1.60×1019C)(1.00×1010m)=1.60×1029Cm
The debye characterizes size of dipole moment. When a proton & electron 100 pm apart, the dipole moment is 4.80D:
μ=(1.60×1029Cm)1D3.336×1030Cm=4.80D
4.80D is a key reference value and represents a pure charge of +1 & -1 100 pm apart. If the charge separation were greater then the dipole moment increases (linearly):
• When proton & electron are separated by 120 pm,
μ=120100(4.80D)=5.76D
• When proton & electron are separated by 150 pm,
μ=150100(4.80D)=7.20D
• When proton & electron are separated by 200 pm,
μ=200100(4.80D)=9.60D
Polarity and Structure of Molecules
The shape of a molecule and the polarity of its bonds determine the OVERALL POLARITY of that molecule. A molecule that contains polar bonds, might not have any overall polarity, depending upon its shape. The simple definition of whether a complex molecule is polar or not depends upon whether its overall centers of positive and negative charges overlap. If these centers lie at the same point in space, then the molecule has no overall polarity (and is non polar).
Figure 3: Charge distrubtions
If a molecule is completely symmetric, then the dipole moment vectors on each molecule will cancel each other out, making the molecule nonpolar. A molecule can only be polar if the structure of that molecule is not symmetric.
A good example of a nonpolar molecule that contains polar bonds is carbon dioxide. This is a linear molecule and the C=O bonds are, in fact, polar. The central carbon will have a net positive charge, and the two outer oxygens a net negative charge. However, since the molecule is linear, these two bond dipoles cancel each other out (i.e. vector addition of the dipoles equals zero). And the overall molecule has no dipole (μ=0.
Although a polar bond is a prerequisite for a molecule to have a dipole, not all molecules with polar bonds exhibit dipoles
Geometric Considerations
For ABn molecules, where A is the central atom and B are all the same types of atoms, there are certain molecular geometries which are symmetric. Therefore, they will have no dipole even if the bonds are polar. These geometries include linear, trigonal planar, tetrahedral, octahedral and trigonal bipyramid.
Figure 4: Molecular geometries with exact cancelation of polar bonding to generate a non-polar molecule (mu=0)
Example 2: C2Cl4
Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and Cl2C=CCl2 does not have a net dipole moment.
Example 3: CH3Cl
C-Cl, the key polar bond, is 178 pm. Measurement reveals 1.87 D. From this data, % ionic character can be computed. If this bond were 100% ionic (based on proton & electron),
μ=178100(4.80D)=8.54D
Example 4: HCl
Since measurement 1.87 D,
% ionic = (1.7/8.54)x100 = 22%HCl
u = 1.03 D (measured) H-Cl bond length 127 pm
If 100% ionic,
μ=127100(4.80D)=6.09D
ionic = (1.03/6.09)x100 = 17%
Compound
Bond
Length (Å)
Electronegativity
Difference
Dipole
Moment (D)
HF
0.92
1.9
1.82
HCl
1.27
0.9
1.08
HBr
1.41
0.7
0.82
HI
1.61
0.4
0.44
Although the bond length is increasing, the dipole is decreasing as you move down the halogen group. The electronegativity decreases as we move down the group. Thus, the greater influence is the electronegativity of the two atoms (which influences the charge at the ends of the dipole).
1.08: L-DopaA Representative Organic Molecule
An 3D image of L-Dopa can be found at ChemTube3D with many other molecules.
A discussion about the role L-Dopa plays in Parkinson's Disease and its discovery is covered in an article by Oleh Hornykiewicz in the Journal of Parkinson's Disease which can be read at PubMed | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/01%3A_Structure_and_Bonding/1.07%3A_Polarity_of_Molecules.txt |
Resonance Structures
Sometimes, even when formal charges are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Such is the case for ozone (O3), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°.
O3
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn in Section 4.8, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures . The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
Before the development of quantum chemistry it was thought that the double-headed arrow indicates that the actual electronic structure is an average of those shown, or that the molecule oscillates between the two structures. Today we know that the electrons involved in the double bonds occupy an orbital that extends over all three oxygen molecules, combining p orbitals on all three.
Figure 5.3.4 The resonance structure of ozone involves a molecular orbital extending all three oxygen atoms. In ozone, a molecular orbital extending over all three oxygen atoms is formed from three atom centered pz orbitals. Similar molecular orbitals are found in every resonance structure.
We will discuss the formation of these molecular orbitals in the next chapter but it is important to understand that resonance structures are based on molecular orbitals not averages of different bonds between atoms. We describe the electrons in such molecular orbitals as being delocalized, that is they cannot be assigned to a bond between two atoms.
Note the Pattern
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure involves a molecular orbital which is a linear combination of atomic orbitals from each of the atoms.
CO32−
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the Lewis structures describing CO32 has three equivalent representations.
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
As the case for ozone, the actual structure involves the formation of a molecular orbital from pz orbitals centered on each atom and sitting above and below the plane of the CO32 ion.
Example
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (C6H6) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
A Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
B Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
C Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
This combination of p orbitals for benzene can be visualized as a ring with a node in the plane of the carbon atoms.
Exercise
The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO2).
Answer:
Resonance structures are particularly common in oxoanions of the p-block elements, such as sulfate and phosphate, and in aromatic hydrocarbons, such as benzene and naphthalene.
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally. (eg. benzene)
The above resonance structures show that the electrons are delocalized within the molecule and through this process the molecule gains extra stability. Ozone with both of its opposite charges creates a neutral molecule and through resonance it is a stable molecule. The extra electron that created the negative charge on either terminal oxygen can be delocalized by resonance through the terminal oxygens.
Benzene is an extremely stable molecule and it is accounted for its geometry and molecular orbital interaction, but most importantly it's due to its resonance structures. The delocalized electrons in the benzene ring make the molecule very stable and with its characteristics of a nucleophile, it will react with a strong electrophile only and after the first reactivity, the substituted benzene will depend on its resonance to direct the next position for the reaction to add a second substituent.
The next molecule, the Amide, is a very stable molecule that is present in most biological systems, mainly in proteins. By studies of NMR spectroscopy and X-Ray crystallography it is confirmed that the stability of the amide is due to resonance which through molecular orbital interaction creates almost a double bond between the Nitrogen and the carbon.
Example: Multiple Resonance of other Molecules
Molecules with more than one resonance form
Some structural resonance conformations are the major contributor or the dominant forms that the molecule exists. For example, if we look at the above rules for estimating the stability of a molecule, we see that for the third molecule the first and second forms are the major contributors for the overall stability of the molecule. The nitrogen is more electronegative than carbon so, it can handle the negative charge more than carbon. A carbon with a negative charge is the least favorable conformation for the molecule to exist, so the last resonance form contributes very little for the stability of the Ion.
The Hybrid Resonance forms show the different Lewis structures with the electron been delocalized. This is very important for the reactivity of chloro-benzene because in the presence of an electrophile it will react and the formation of another bond will be directed and determine by resonance. The long pair of electrons delocalized in the aromatic substituted ring is where it can potentially form a new bond with an electrophile, as it is shown there are three possible places that reactivity can take place, the first to react will take place at the para position with respect to the chloro substituent and then to either ortho position.
Contributors
• Sharon Wei (UCD), Liza Chu (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/01%3A_Structure_and_Bonding/1.09%3A_Resonance.txt |
Bond lengths and angles
The length of a chemical bond the distance between the centers of the two bonded atoms (the internuclear distance.) Bond lengths have traditionally been expressed in Ångstrom units, but picometers are now preferred (1Å = 10-8 cm = 100 pm.) Bond lengths are typically in the range 1-2 Å or 100-200 pm. Even though the bond is vibrating, equilibrium bond lengths can be determined experimentally to within ±1 pm.
Bond lengths depend mainly on the sizes of the atoms, and secondarily on the bond strengths, the stronger bonds tending to be shorter. Bonds involving hydrogen can be quite short; The shortest bond of all, H–H, is only 74 pm. Multiply-bonded atoms are closer together than singly-bonded ones; this is a major criterion for experimentally determining the multiplicity of a bond. This trend is clearly evident in the above plot which depicts the sequence of carbon-carbon single, double, and triple bonds.
The most common method of measuring bond lengths in solids is by analysis of the diffraction or scattering of X-rays when they pass through the regularly-spaced atoms in the crystal. For gaseous molecules, neutron- or electron-diffraction can also be used.
The complete structure of a molecule requires a specification of the coordinates of each of its atoms in three-dimensional space. This data can then be used by computer programs to construct visualizations of the molecule as discussed above. One such visualization of the water molecule, with bond distances and the HOH bond angle superimposed on a space-filling model, is shown here. (It is taken from an excellent reference source on water). The colors show the results of calculations that depict the way in which electron charge is distributed around the three nuclei.
In most cases the focus of configuration is a carbon atom so the lines specifying bond directions will originate there. As defined in the diagram on the right, a simple straight line represents a bond lying approximately in the surface plane. The two bonds to substituents A in the structure on the left are of this kind. A wedge shaped bond is directed in front of this plane (thick end toward the viewer), as shown by the bond to substituent B; and a hatched bond is directed in back of the plane (away from the viewer), as shown by the bond to substituent D. Some texts and other sources may use a dashed bond in the same manner as we have defined the hatched bond, but this can be confusing because the dashed bond is often used to represent a partial bond (i.e. a covalent bond that is partially formed or partially broken).
Molecular Shape
The following examples make use of this notation, and also illustrate the importance of including non-bonding valence shell electron pairs (colored blue) when viewing such configurations.
Methane Ammonia Water
Bonding configurations are readily predicted by valence-shell electron-pair repulsion theory, commonly referred to as VSEPR in most introductory chemistry texts. This simple model is based on the fact that electrons repel each other, and that it is reasonable to expect that the bonds and non-bonding valence electron pairs associated with a given atom will prefer to be as far apart as possible. The bonding configurations of carbon are easy to remember, since there are only three categories.
Configuration Bonding Partners Bond Angles Example
Tetrahedral 4 109.5º
Trigonal 3 120º
Linear 2 180º
In the three examples shown above, the central atom (carbon) does not have any non-bonding valence electrons; consequently the configuration may be estimated from the number of bonding partners alone. For molecules of water and ammonia, however, the non-bonding electrons must be included in the calculation. In each case there are four regions of electron density associated with the valence shell so that a tetrahedral bond angle is expected. The measured bond angles of these compounds (H2O 104.5º & NH3 107.3º) show that they are closer to being tetrahedral than trigonal or linear. Of course, it is the configuration of atoms (not electrons) that defines the the shape of a molecule, and in this sense ammonia is said to be pyramidal (not tetrahedral). The compound boron trifluoride, BF3, does not have non-bonding valence electrons and the configuration of its atoms is trigonal. Nice treatments of VSEPR theory have been provided by Oxford and Purdue. The best way to study the three-dimensional shapes of molecules is by using molecular models. Many kinds of model kits are available to students and professional chemists.
Two Electron Groups
Our first example is a molecule with two bonded atoms and no lone pairs of electrons, BeH2.
AX2: BeH2
1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is
2. There are two electron groups around the central atom. We see from Figure 9.2.2 that the arrangement that minimizes repulsions places the groups 180° apart.
3. Both groups around the central atom are bonding pairs (BP). Thus BeH2 is designated as AX2.
4. From Figure 9.2.3 we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH2 is linear.
AX2: CO2
1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is
2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH2, the arrangement that minimizes repulsions places the groups 180° apart.
3. Once again, both groups around the central atom are bonding pairs (BP), so CO2 is designated as AX2.
4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO2 is linear (Figure 9.2.3). The structure of CO2 is shown in Figure 9.2.2.1.
Three Electron Groups
AX3: BCl3
1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is
2. There are three electron groups around the central atom. To minimize repulsions, the groups are placed 120° apart (Figure 9.2.2).
3. All electron groups are bonding pairs (BP), so the structure is designated as AX3.
4. From Figure 9.2.3 we see that with three bonding pairs around the central atom, the molecular geometry of BCl3 is trigonal planar, as shown in Figure 9.2.2.1.
AX3: CO32−
1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as
2. The structure of CO32 is a resonance hybrid. It has three identical bonds, each with a bond order of 113. We minimize repulsions by placing the three groups 120° apart (Figure 9.2.2).
3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX3.
4. We see from Figure 9.2.3 that the molecular geometry of CO32 is trigonal planar.
In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.
AX2E: SO2
1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below.
2. There are three electron groups around the central atom, two double bonds and one lone pair. We initially place the groups in a trigonal planar arrangement to minimize repulsions (Figure 9.2.2).
3. There are two bonding pairs and one lone pair, so the structure is designated as AX2E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure 9.2.4). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO2, we have one BP–BP interaction and two LP–BP interactions.
4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent, or V shaped, which can be viewed as a trigonal planar arrangement with a missing vertex (Figures 9.2.2.1 and 9.2.3).
Figure 9.2.4: The Difference in the Space Occupied by a Lone Pair of Electrons and by a Bonding Pair
As with SO2, this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom.
Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH2O (AX3), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).
Four Electron Groups
One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions.
AX4: CH4
1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is
2. There are four electron groups around the central atom. As shown in Figure 9.2.2, repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.
3. All electron groups are bonding pairs, so the structure is designated as AX4.
4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure 9.2.3).
AX3E: NH3
1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure
2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
3. With three bonding pairs and one lone pair, the structure is designated as AX3E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal. In essence, this is a tetrahedron with a vertex missing (Figure 9.2.3). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure 9.2.3 and Figure 9.2.4).
AX2E2: H2O
1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure
2. There are four groups around the central oxygen atom, two bonding pairs and two lone pairs. Repulsions are minimized by directing the bonding pairs and the lone pairs to the corners of a tetrahedron Figure 9.2.2.
3. With two bonding pairs and two lone pairs, the structure is designated as AX2E2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles.
4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent, or V shaped, with an H–O–H angle that is even less than the H–N–H angles in NH3, as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one.. This molecular shape is essentially a tetrahedron with two missing vertices. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/01%3A_Structure_and_Bonding/1.10%3A_Determining_Molecular_Shape.txt |
It is necessary to draw structural formulas for organic compounds because in most cases a molecular formula does not uniquely represent a single compound. Different compounds having the same molecular formula are called isomers, and the prevalence of organic isomers reflects the extraordinary versatility of carbon in forming strong bonds to itself and to other elements. When the group of atoms that make up the molecules of different isomers are bonded together in fundamentally different ways, we refer to such compounds as constitutional isomers. There are seven constitutional isomers of C4H10O, and structural formulas for these are drawn in the following table. These formulas represent all known and possible C4H10O compounds, and display a common structural feature. There are no double or triple bonds and no rings in any of these structures.
Structural Formulas for C4H10O isomers
Kekulé Formula Condensed Formula Shorthand Formula
Simplification of structural formulas may be achieved without any loss of the information they convey. In condensed structural formulas the bonds to each carbon are omitted, but each distinct structural unit (group) is written with subscript numbers designating multiple substituents, including the hydrogens. Shorthand (line) formulas omit the symbols for carbon and hydrogen entirely. Each straight line segment represents a bond, the ends and intersections of the lines are carbon atoms, and the correct number of hydrogens is calculated from the tetravalency of carbon. Non-bonding valence shell electrons are omitted in these formulas.
Developing the ability to visualize a three-dimensional structure from two-dimensional formulas requires practice, and in most cases the aid of molecular models. As noted earlier, many kinds of model kits are available to students and professional chemists, and the beginning student is encouraged to obtain one.
Kekulé Formula
A structural formula displays the atoms of the molecule in the order they are bonded. It also depicts how the atoms are bonded to one another, for example single, double, and triple covalent bond. Covalent bonds are shown using lines. The number of dashes indicate whether the bond is a single, double, or triple covalent bond. Structural formulas are helpful because they explain the properties and structure of the compound which empirical and molecular formulas cannot always represent.
Ex. Kekulé Formula for Ethanol:
Condensed Formula
Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space and make it more convenient and faster to write out. Condensed structural formulas are also helpful when showing that a group of atoms is connected to a single atom in a compound. When this happens, parenthesis are used around the group of atoms to show they are together.
Ex. Condensed Structural Formula for Ethanol: CH3CH2OH (Molecular Formula for Ethanol C2H6O).
Shorthand Formula
Because organic compounds can be complex at times, line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. A carbon atom is present wherever a line intersects another line. Hydrogen atoms are then assumed to complete each of carbon's four bonds. All other atoms that are connected to carbon atoms are written out. Line angle formulas help show structure and order of the atoms in a compound making the advantages and disadvantages similar to structural formulas.
Ex.Shorthand Formula for Ethanol:
Contributors
• Jean Kim (UCD), Kristina Bonnett (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/01%3A_Structure_and_Bonding/1.11%3A_Drawing_Organic_Structures.txt |
Formation of sigma bonds: the H2 molecule
The simplest case to consider is the hydrogen molecule, H2. When we say that the two electrons from each of the hydrogen atoms are shared to form a covalent bond between the two atoms, what we mean in valence bond theory terms is that the two spherical 1s orbitals overlap, allowing the two electrons to form a pair within the two overlapping orbitals.
These two electrons are now attracted to the positive charge of both of the hydrogen nuclei, with the result that they serve as a sort of ‘chemical glue’ holding the two nuclei together.
Bonding in Methane
Now let’s turn to methane, the simplest organic molecule. Recall the valence electron configuration of the central carbon:
This picture, however, is problematic. How does the carbon form four bonds if it has only two half-filled p orbitals available for bonding? A hint comes from the experimental observation that the four C-H bonds in methane are arranged with tetrahedral geometry about the central carbon, and that each bond has the same length and strength. In order to explain this observation, valence bond theory relies on a concept called orbital hybridization. In this picture, the four valence orbitals of the carbon (one 2s and three 2p orbitals) combine mathematically (remember: orbitals are described by equations) to form four equivalent hybrid orbitals, which are named sp3 orbitals because they are formed from mixing one s and three p orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single sp3 orbital.
The sp3 hybrid orbitals, like the p orbitals of which they are partially composed, are oblong in shape, and have two lobes of opposite sign. Unlike the p orbitals, however, the two lobes are of very different size. The larger lobes of the sp3 hybrids are directed towards the four corners of a tetrahedron, meaning that the angle between any two orbitals is 109.5o.
This geometric arrangement makes perfect sense if you consider that it is precisely this angle that allows the four orbitals (and the electrons in them) to be as far apart from each other as possible.This is simply a restatement of the Valence Shell Electron Pair Repulsion (VSEPR) theory that you learned in General Chemistry: electron pairs (in orbitals) will arrange themselves in such a way as to remain as far apart as possible, due to negative-negative electrostatic repulsion.
Each C-H bond in methane, then, can be described as an overlap between a half-filled 1s orbital in a hydrogen atom and the larger lobe of one of the four half-filled sp3 hybrid orbitals in the central carbon. The length of the carbon-hydrogen bonds in methane is 1.09 Å (1.09 x 10-10 m).
While previously we drew a Lewis structure of methane in two dimensions using lines to denote each covalent bond, we can now draw a more accurate structure in three dimensions, showing the tetrahedral bonding geometry. To do this on a two-dimensional page, though, we need to introduce a new drawing convention: the solid / dashed wedge system. In this convention, a solid wedge simply represents a bond that is meant to be pictured emerging from the plane of the page. A dashed wedge represents a bond that is meant to be pictured pointing into, or behind, the plane of the page. Normal lines imply bonds that lie in the plane of the page.
This system takes a little bit of getting used to, but with practice your eye will learn to immediately ‘see’ the third dimension being depicted.
Example
Imagine that you could distinguish between the four hydrogens in a methane molecule, and labeled them Ha through Hd. In the images below, the exact same methane molecule is rotated and flipped in various positions. Draw the missing hydrogen atom labels. (It will be much easier to do this if you make a model.)
Example
Describe, with a picture and with words, the bonding in chloroform, CHCl3.
Solutions
The bonding arrangement here is also tetrahedral: the three N-H bonds of ammonia can be pictured as forming the base of a trigonal pyramid, with the fourth orbital, containing the lone pair, forming the top of the pyramid.
Recall from your study of VSEPR theory in General Chemistry that the lone pair, with its slightly greater repulsive effect, ‘pushes’ the three N-H sbonds away from the top of the pyramid, meaning that the H-N-H bond angles are slightly less than tetrahedral, at 107.3˚ rather than 109.5˚.
VSEPR theory also predicts, accurately, that a water molecule is ‘bent’ at an angle of approximately 104.5˚. It would seem logical, then, to describe the bonding in water as occurring through the overlap of sp3-hybrid orbitals on oxygen with 1sorbitals on the two hydrogen atoms. In this model, the two nonbonding lone pairs on oxygen would be located in sp3 orbitals.
Some experimental evidence, however, suggests that the bonding orbitals on the oxygen are actually unhybridized 2p orbitals rather than sp3 hybrids. Although this would seem to imply that the H-O-H bond angle should be 90˚ (remember that p orbitals are oriented perpendicular to one another), it appears that electrostatic repulsion has the effect of distorting this p-orbital angle to 104.5˚. Both the hybrid orbital and the nonhybrid orbital models present reasonable explanations for the observed bonding arrangement in water, so we will not concern ourselves any further with the distinction.
Example
Draw, in the same style as the figures above, an orbital picture for the bonding in methylamine.
Solution
Formation of $\pi$ bonds - $sp^2$ and $sp$ hybridization
The valence bond theory, along with the hybrid orbital concept, does a very good job of describing double-bonded compounds such as ethene. Three experimentally observable characteristics of the ethene molecule need to be accounted for by a bonding model:
1. Ethene is a planar (flat) molecule.
2. Bond angles in ethene are approximately 120o, and the carbon-carbon bond length is 1.34 Å, significantly shorter than the 1.54 Å single carbon-carbon bond in ethane.
3. There is a significant barrier to rotation about the carbon-carbon double bond.
Clearly, these characteristics are not consistent with an sp3 hybrid bonding picture for the two carbon atoms. Instead, the bonding in ethene is described by a model involving the participation of a different kind of hybrid orbital. Three atomic orbitals on each carbon – the 2s, 2px and 2py orbitals – combine to form three sp2 hybrids, leaving the 2pz orbital unhybridized.
The three sp2 hybrids are arranged with trigonal planar geometry, pointing to the three corners of an equilateral triangle, with angles of 120°between them. The unhybridized 2pz orbital is perpendicular to this plane (in the next several figures, sp2 orbitals and the sigma bonds to which they contribute are represented by lines and wedges; only the 2pz orbitals are shown in the 'space-filling' mode).
The carbon-carbon double bond in ethene consists of one sbond, formed by the overlap of two sp2 orbitals, and a second bond, calleda π (pi) bond, which is formed by the side-by-side overlap of the two unhybridized 2pz orbitals from each carbon.
spacefilling image of bonding in ethene
The pi bond does not have symmetrical symmetry. Because they are the result of side-by-side overlap (rather then end-to-end overlap like a sigma bond), pi bonds are not free to rotate. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2pz orbitals that make up the pi bond. The presence of the pi bond thus ‘locks’ the six atoms of ethene into the same plane. This argument extends to larger alkene groups: in each case, the six atoms of the group form a single plane.
Conversely, sbonds such as the carbon-carbon single bond in ethane (CH3CH3) exhibit free rotation, and can assume many different conformations, or shapes - this is one of the main subjects of Chapter 3.
Example 1.20
Circle the six atoms in the molecule below that are ‘locked’ into the same plane.
Example
What kinds of orbitals are overlapping in bonds a-d indicated below?
Example
What is wrong with the way the following structure is drawn?
Solutions
A similar picture can be drawn for the bonding in carbonyl groups, such as formaldehyde. In this molecule, the carbon is sp2-hybridized, and we will assume that the oxygen atom is also sp2 hybridized. The carbon has three sigma bonds: two are formed by overlap between two of its sp2 orbitals with the 1sorbital from each of the hydrogens, and the third sigma bond is formed by overlap between the remaining carbon sp2 orbital and an sp2 orbital on the oxygen. The two lone pairs on oxygen occupy its other two sp2 orbitals.
The pi bond is formed by side-by-side overlap of the unhybridized 2pz orbitals on the carbon and the oxygen. Just like in alkenes, the 2pz orbitals that form the pi bond are perpendicular to the plane formed by the sigma bonds.
Example
Describe and draw the bonding picture for the imine group shown below. Use the drawing of formaldehyde above as your guide.
Solution | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/01%3A_Structure_and_Bonding/1.12%3A_Hybridization.txt |
Bonding in Ethane
In the ethane molecule, the bonding picture according to valence orbital theory is very similar to that of methane. Both carbons are sp3-hybridized, meaning that both have four bonds arranged with tetrahedral geometry. The carbon-carbon bond, with a bond length of 1.54 Å, is formed by overlap of one sp3 orbital from each of the carbons, while the six carbon-hydrogen bonds are formed from overlaps between the remaining sp3 orbitals on the two carbons and the 1s orbitals of hydrogen atoms. All of these are sigma bonds.
Because they are formed from the end-on-end overlap of two orbitals, sigma bonds are free to rotate. This means, in the case of ethane molecule, that the two methyl (CH3) groups can be pictured as two wheels on a hub, each one able to rotate freely with respect to the other.
In chapter 3 we will learn more about the implications of rotational freedom in sigma bonds, when we discuss the ‘conformation’ of organic molecules.
The sp3 bonding picture is also used to described the bonding in amines, including ammonia, the simplest amine. Just like the carbon atom in methane, the central nitrogen in ammonia is sp3-hybridized. With nitrogen, however, there are five rather than four valence electrons to account for, meaning that three of the four hybrid orbitals are half-filled and available for bonding, while the fourth is fully occupied by a (non-bonding) pair of electrons.
C2H4, also known as ethylene or ethene, is a gaseous material created synthetically through steam cracking. In nature, it is released in trace amounts by plants to signal their fruits to ripen. Ethene consists of two sp2-hybridized carbon atoms, which are sigma bonded to each other and to two hydrogen atoms each. The remaining unhybridized p orbitals on the carbon form a pi bond, which gives ethene its reactivity.
Bonding in Ethene
A key component of using Valence Bond Theory correctly is being able to use the Lewis dot diagram correctly. Ethene has a double bond between the carbons and single bonds between each hydrogen and carbon: each bond is represented by a pair of dots, which represent electrons. Each carbon requires a full octet and each hydrogen requires a pair of electrons. The correct Lewis structure for ethene is shown below:
For more information on how to use Lewis Dot Structures refer to http://chemwiki.ucdavis.edu/Wikitext...wis_Structures.
Valence Shell Electron Pair Repulsion (VSEPR) Theory is used to predict the bond angles and spatial positions of the carbon and hydrogen atoms of ethene and to determine the bond order of the carbon atoms (the number of bonds formed between them). Each carbon atom is of the general arrangement AX3, where A is the central atom surrounded by three other atoms (denoted by X); compounds of this form adopt trigonal planar geometry, forming 120 degree bond angles. In order for the unhybridized p orbitals to successfully overlap, the CH2 must be coplanar: therefore, C2H4 is a planar molecule and each bond angle is about 120 degrees. The diagram below shows the bond lengths and hydrogen-carbon-carbon bond angles of ethene:
According to valence bond theory, two atoms form a covalent bond through the overlap of individual half-filled valence atomic orbitals, each containing one unpaired electron. In ethene, each hydrogen atom has one unpaired electron and each carbon is sp2 hybridized with one electron each sp2 orbital. The fourth electron is in the p orbital that will form the pi bond. The bond order for ethene is simply the number of bonds between each atom: the carbon-carbon bond has a bond order of two, and each carbon-hydrogen bond has a bond order of one. For more information see http://chemwiki.ucdavis.edu/Wikitexts/UCD_Chem_124A%3a_Kauzlarich/ChemWiki_Module_Topics/VSEPR
Bonding in acetylene
Finally, the hybrid orbital concept applies well to triple-bonded groups, such as alkynes and nitriles. Consider, for example, the structure of ethyne (common name acetylene), the simplest alkyne.
This molecule is linear: all four atoms lie in a straight line. The carbon-carbon triple bond is only 1.20Å long. In the hybrid orbital picture of acetylene, both carbons are sp-hybridized. In an sp-hybridized carbon, the 2s orbital combines with the 2px orbital to form two sp hybrid orbitals that are oriented at an angle of 180°with respect to each other (eg. along the x axis). The 2py and 2pz orbitals remain unhybridized, and are oriented perpendicularly along the y and z axes, respectively.
The C-C sigma bond, then, is formed by the overlap of one sp orbital from each of the carbons, while the two C-H sigma bonds are formed by the overlap of the second sp orbital on each carbon with a 1s orbital on a hydrogen. Each carbon atom still has two half-filled 2py and 2pz orbitals, which are perpendicular both to each other and to the line formed by the sigma bonds. These two perpendicular pairs of p orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds).
The hybrid orbital concept nicely explains another experimental observation: single bonds adjacent to double and triple bonds are progressively shorter and stronger than ‘normal’ single bonds, such as the one in a simple alkane. The carbon-carbon bond in ethane (structure A below) results from the overlap of two sp3 orbitals.
In alkene B, however, the carbon-carbon single bond is the result of overlap between an sp2 orbital and an sp3 orbital, while in alkyne C the carbon-carbon single bond is the result of overlap between an sp orbital and an sp3 orbital. These are all single bonds, but the bond in molecule C is shorter and stronger than the one in B, which is in turn shorter and stronger than the one in A.
The explanation here is relatively straightforward. An sp orbital is composed of one s orbital and one p orbital, and thus it has 50% s character and 50% p character. sp2 orbitals, by comparison, have 33% s character and 67% p character, while sp3 orbitals have 25% s character and 75% p character. Because of their spherical shape, 2s orbitals are smaller, and hold electrons closer and ‘tighter’ to the nucleus, compared to 2p orbitals. Consequently, bonds involving sp + sp3 overlap (as in alkyne C) are shorter and stronger than bonds involving sp2 + sp3 overlap (as in alkene B). Bonds involving sp3-sp3overlap (as in alkane A) are the longest and weakest of the group, because of the 75% ‘p’ character of the hybrids. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/01%3A_Structure_and_Bonding/1.13%3A_Ethane_Ethylene_and_Acetylene.txt |
In 1923, chemists Johannes Brønsted and Martin Lowry independently developed definitions of acids and bases based on compounds abilities to either donate or accept protons (H+ ions). Here, acids are defined as being able to donate protons in the form of hydrogen ions; whereas bases are defined as being able to accept protons. This took the Arrhenius definition one step further as water is no longer required to be present in the solution for acid and base reactions to occur.
Brønsted-Lowery Definition
J.N. Brønsted and T.M. Lowry independently developed the theory of proton donors and proton acceptors in acid-base reactions, coincidentally in the same region and during the same year. The Arrhenius theory where acids and bases are defined by whether the molecule contains hydrogen and hydroxide ion is too limiting. The main effect of the Brønsted-Lowry definition is to identify the proton (H+) transfer occurring in the acid-base reaction. This is best illustrated in the following equation:
\[HA+Z \rightleftharpoons A^− + HZ^+\]
Acid Base
Donates hydrogen ions Accepts hydrogen ions.
HCl+ HOH → H3O+ + Cl-
HOH+ NH3 NH4+ + OH-
The determination of a substance as a Brønsted-Lowery acid or base can only be done by observing the reaction. In the case of the HOH it is a base in the first case and an acid in the second case.
To determine whether a substance is an acid or a base, count the hydrogens on each substance before and after the reaction. If the number of hydrogens has decreased that substance is the acid (donates hydrogen ions). If the number of hydrogens has increased that substance is the base (accepts hydrogen ions). These definitions are normally applied to the reactants on the left. If the reaction is viewed in reverse a new acid and base can be identified. The substances on the right side of the equation are called conjugate acid and conjugate base compared to those on the left. Also note that the original acid turns in the conjugate base after the reaction is over.
Acids are Proton Donors and Bases are Proton Acceptors
For a reaction to be in equilibrium a transfer of electrons needs to occur. The acid will give an electron away and the base will receive the electron. Acids and Bases that work together in this fashion are called a conjugate pair made up of conjugate acids and conjugate bases.
\[ HA + Z \rightleftharpoons A^- + HZ^+ \]
A stands for an Acidic compound and Z stands for a Basic compound
• A Donates H to form HZ+.
• Z Accepts H from A which forms HZ+
• A- becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibrium
• HZ+ becomes a conjugate acid of Z and in the reverse reaction it donates a H to A- recreating Z in order to remain in equilibrium
Questions
1. Why is \(HA\) an Acid?
2. Why is \(Z^-\) a Base?
3. How can A- be a base when HA was and Acid?
4. How can HZ+ be an acid when Z used to be a Base?
Now that we understand the concept, let's look at an an example with actual compounds!
\[HCl + H_2O \rightleftharpoons H_3O^+ + Cl^¯\]
• HCL is the acid because it is donating a proton to H2O
• H2O is the base because H2O is accepting a proton from HCL
• H3O+ is the conjugate acid because it is donating an acid to CL turn into it's conjugate acid H2O
• Cl¯ is the conjugate base because it accepts an H from H3O to return to it's conjugate acid HCl
How can H2O be a base? I thought it was neutral?
Answers
1. It has a proton that can be transferred
2. It receives a proton from HA
3. A- is a conjugate base because it is in need of a H in order to remain in equilibrium and return to HA
4. HZ+ is a conjugate acid because it needs to donate or give away its proton in order to return to it's previous state of Z
5. In the Brønsted-Lowry Theory what makes a compound an element or a base is whether or not it donates or accepts protons. If the H2O was in a different problem and was instead donating an H rather than accepting an H it would be an acid!
Conjugate Acid–Base Pairs
In aqueous solutions, acids and bases can be defined in terms of the transfer of a proton from an acid to a base. Thus for every acidic species in an aqueous solution, there exists a species derived from the acid by the loss of a proton. These two species that differ by only a proton constitute a conjugate acid–base pair.
All acid–base reactions contain two conjugate acid–base pairs.
For example, in the reaction of HCl with water, HCl, the parent acid, donates a proton to a water molecule, the parent base, thereby forming Cl-. Thus HCl and Cl- constitute a conjugate acid–base pair. By convention, we always write a conjugate acid–base pair as the acid followed by its conjugate base. In the reverse reaction, the Cl- ion in solution acts as a base to accept a proton from H2O and HCl. Thus H3O+ and H2O constitute a second conjugate acid–base pair. In general, any acid–base reaction must contain two conjugate acid–base pairs, which in this case are HCl/Cl- and H3O+/H2O.
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, \(H_3O^+\) is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid–base pairs: the parent acid and its conjugate base (CH3CO2H/CH3CO2-) and the parent base and its conjugate acid (H3O+/H2O).
In the reaction of ammonia with water to give ammonium ions and hydroxide ions (Figure 16.3), ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are NH4+/NH3 and H2O/OH-.
Some common conjugate acid–base pairs are shown in Figure \(4\).
2.02: Reactions of BrnstedLowry Acids and Bases
The Brønsted-Lowry definition of acidity
We’ll begin our discussion of acid-base chemistry with a couple of essential definitions. The first of these definitions was proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry, and has come to be known as the Brønsted-Lowry definition of acids and bases. An acid, by the Brønsted-Lowry definition, is a species which is able to donate a proton (H+), while a base is a proton acceptor. We have already discussed in the previous chapter one of the most familiar examples of a Brønsted-Lowry acid-base reaction, between hydrochloric acid and hydroxide ion:
In this reaction, a proton is transferred from HCl (the acid, or proton donor) to hydroxide (the base, or proton acceptor). As we learned in the previous chapter, curved arrows depict the movement of electrons in this bond-breaking and bond-forming process.
After a Brønsted-Lowry acid donates a proton, what remains – in this case, a chloride ion – is called the conjugate base. Chloride is thus the conjugate base of hydrochloric acid. Conversely, when a Brønsted-Lowry base accepts a proton it is converted into its conjugate acid form: water is thus the conjugate acid of hydroxide.
We can also talk about conjugate acid/base pairs: the two acid/base pairs involved in our first reaction are hydrochloric acid/chloride and hydroxide/water. In this next acid-base reaction, the two pairs involved are acetate/acetic acid and methyl ammonium/methylamine:
Throughout this text, we will often use the abbreviations HA and :B in order to refer in a general way to acidic and basic reactants:
In order to act as a proton acceptor, a base must have a reactive pair of electrons. In all of the examples we shall see in this chapter, this pair of electrons is a non-bonding lone pair, usually (but not always) on an oxygen, nitrogen, sulfur, or halogen atom. When acetate acts as a base in the reaction shown above, for example, one of its oxygen lone pairs is used to form a new bond to a proton. The same can be said for an amine acting as a base. Clearly, methyl ammonium ion cannot act as a base – it does not have a reactive pair of electrons with which to accept a new bond to a proton.
Later, in chapter 15, we will see several examples where the (relatively) reactive pair of electrons in a $\pi$ bond act in a basic fashion.
In this chapter, we will concentrate on those bases with non-bonding (lone pair) electrons.
Example
Exercise 7.1: Draw structures for the missing conjugate acids or conjugate bases in the reactions below. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/02%3A_Acids_and_Bases/2.01%3A_BrnstedLowry_Acids_and_Bases.txt |
You are no doubt aware that some acids are stronger than others. Sulfuric acid is strong enough to be used as a drain cleaner, as it will rapidly dissolve clogs of hair and other organic material.
Not surprisingly, concentrated sulfuric acid will also cause painful burns if it touches your skin, and permanent damage if it gets in your eyes (there’s a good reason for those safety goggles you wear in chemistry lab!). Acetic acid (vinegar), will also burn your skin and eyes, but is not nearly strong enough to make an effective drain cleaner. Water, which we know can act as a proton donor, is obviously not a very strong acid. Even hydroxide ion could theoretically act as an acid – it has, after all, a proton to donate – but this is not a reaction that we would normally consider to be relevant in anything but the most extreme conditions.
The relative acidity of different compounds or functional groups – in other words, their relative capacity to donate a proton to a common base under identical conditions – is quantified by a number called the dissociation constant, abbreviated Ka. The common base chosen for comparison is water.
We will consider acetic acid as our first example. When a small amount of acetic acid is added to water, a proton-transfer event (acid-base reaction) occurs to some extent.
Notice the phrase ‘to some extent’ – this reaction does not run to completion, with all of the acetic acid converted to acetate, its conjugate base. Rather, a dynamic equilibrium is reached, with proton transfer going in both directions (thus the two-way arrows) and finite concentrations of all four species in play. The nature of this equilibrium situation, as you recall from General Chemistry, is expressed by an equilibrium constant, Keq. The equilibrium constant is actually a ratio of activities (represented by the symbol $a$), but activities are rarely used in courses other than analytical or physical chemistry. To simplify the discussion for general chemistry and organic chemistry courses, the activities of all of the solutes are replaced with molarities, and the activity of the solvent (usually water) is defined as having the value of 1.
In our example, we added a small amount of acetic acid to a large amount of water: water is the solvent for this reaction. Therefore, in the course of the reaction, the concentration of water changes very little, and the water can be treated as a pure solvent, which is always assigned an activity of 1. The acetic acid, acetate ion and hydronium ion are all solutes, and so their activities are approximated with molarities. The acid dissociation constant, or Ka, for acetic acid is therefore defined as:
$K_{eq} = \dfrac{a_{CH_3COO^-}·a_{H_3O^+}}{a_{CH_3COOH}·a_{H_2O}} ≈ \dfrac{[CH_3COO^-][H_3O^+]}{[CH_3COOH][1]}$
Because dividing by 1 does not change the value of the constant, the "1" is usually not written, and Ka is written as:
$K_{eq} = K_{a} = \dfrac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]} = 1.75 \times 10^{-5}$
In more general terms, the dissociation constant for a given acid is expressed as:
$K_a = \dfrac{[A^-][H_3O^+]}{[HA]} \label{First}$
or
$K_a = \dfrac{[A][H_3O^+]}{[HA^+]} \label{Second}$
The first expression applies to a neutral acid such as like HCl or acetic acid, while the second applies to a cationic acid like ammonium (NH4+).
The value of Ka = 1.75 x 10-5 for acetic acid is very small - this means that very little dissociation actually takes place, and there is much more acetic acid in solution at equilibrium than there is acetate ion. Acetic acid is a relatively weak acid, at least when compared to sulfuric acid (Ka = 109) or hydrochloric acid (Ka = 107), both of which undergo essentially complete dissociation in water.
A number like 1.75 x 10- 5 is not very easy either to say or to remember. Chemists have therefore come up with a more convenient term to express relative acidity: the pKa value.
pKa = -log Ka
Doing the math, we find that the pKa of acetic acid is 4.8. The use of pKa values allows us to express the acidity of common compounds and functional groups on a numerical scale of about –10 (very strong acid) to 50 (not acidic at all). Table 7 at the end of the text lists exact or approximate pKa values for different types of protons that you are likely to encounter in your study of organic and biological chemistry. Looking at Table 7, you see that the pKa of carboxylic acids are in the 4-5 range, the pKa of sulfuric acid is –10, and the pKa of water is 14. Alkenes and alkanes, which are not acidic at all, have pKa values above 30. The lower the pKa value, the stronger the acid.
It is important to realize that pKa is not at all the same thing as pH: the former is an inherent property of a compound or functional group, while the latter is the measure of the hydronium ion concentration in a particular aqueous solution:
pH = -log [H3O+]
Any particular acid will always have the same pKa (assuming that we are talking about an aqueous solution at room temperature) but different aqueous solutions of the acid could have different pH values, depending on how much acid is added to how much water.
Our table of pKa values will also allow us to compare the strengths of different bases by comparing the pKavalues of their conjugate acids. The key idea to remember is this: the stronger the conjugate acid, the weaker the conjugate base. Sulfuric acid is the strongest acid on our list with a pKa value of –10, so HSO4- is the weakest conjugate base. You can see that hydroxide ion is a stronger base than ammonia (NH3), because ammonium (NH4+, pKa = 9.2) is a stronger acid than water (pKa = 14.0).
While Table 7 provides the pKa values of only a limited number of compounds, it can be very useful as a starting point for estimating the acidity or basicity of just about any organic molecule. Here is where your familiarity with organic functional groups will come in very handy. What, for example, is the pKaof cyclohexanol? It is not on the table, but as it is an alcohol it is probably somewhere near that of ethanol (pKa = 16). Likewise, we can use Table 7 to predict that para-hydroxyphenyl acetaldehyde, an intermediate compound in the biosynthesis of morphine, has a pKa in the neighborhood of 10, close to that of our reference compound, phenol.
Notice in this example that we need to evaluate the potential acidity at four different locations on the molecule.
Aldehyde and aromatic protons are not at all acidic (pKavalues are above 40 – not on our table). The two protons on the carbon next to the carbonyl are slightly acidic, with pKa values around 19-20 according to the table. The most acidic proton is on the phenol group, so if the compound were to be subjected to a single molar equivalent of strong base, this is the proton that would be donated.
As you continue your study of organic chemistry, it will be a very good idea to commit to memory the approximate pKa ranges of some important functional groups, including water, alcohols, phenols, ammonium, thiols, phosphates, carboxylic acids and carbons next to carbonyl groups (so-called a-carbons). These are the groups that you are most likely to see acting as acids or bases in biological organic reactions.
A word of caution: when using the pKa table, be absolutely sure that you are considering the correct conjugate acid/base pair. If you are asked to say something about the basicity of ammonia (NH3) compared to that of ethoxide ion (CH3CH2O-), for example, the relevant pKa values to consider are 9.2 (the pKa of ammonium ion) and 16 (the pKa of ethanol). From these numbers, you know that ethoxide is the stronger base. Do not make the mistake of using the pKa value of 38: this is the pKa of ammonia acting as an acid, and tells you how basic the NH2- ion is (very basic!)
Example
Exercise 7.2: Using the pKa table, estimate pKa values for the most acidic group on the compounds below, and draw the structure of the conjugate base that results when this group donates a proton.
Solution
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
2.04: Predicting the Outcome of AcidBase Reactions
Using pKa values to predict reaction equilibria
By definition, the pKa value tells us the extent to which an acid will react with water as the base, but by extension, we can also calculate the equilibrium constant for a reaction between any acid-base pair. Mathematically, it can be shown that:
Keq (for the acid base reaction in question) = 10ΔpKa
where ΔpKa = pKa of product acid minus pKa of reactant acid
Consider a reaction between methylamine and acetic acid:
First, we need to identify the acid species on either side of the equation. On the left side, the acid is of course acetic acid, while on the right side the acid is methyl ammonium. The specific pKa values for these acids are not on our very generalized pKa table, but are given in the figure above. Without performing any calculations, you should be able to see that this equilibrium lies far to the right-hand side: acetic acid has a lower pKa, is a stronger acid, and thus it wants to give up its proton more than methyl ammonium does. Doing the math, we see that
Keq = 10ΔpKa = 10(10.6 – 4.8) = 105.8 = 6.3 x 105
So Keq is a very large number (much greater than 1) and the equilibrium lies far to the right-hand side of the equation, just as we had predicted. If you had just wanted to approximate an answer without bothering to look for a calculator, you could have noted that the difference in pKa values is approximately 6, so the equilibrium constant should be somewhere in the order of 106, or one million. Using the pKa table in this way, and making functional group-based pKa approximations for molecules for which we don’t have exact values, we can easily estimate the extent to which a given acid-base reaction will proceed.
Exercise \(1\)
Show the products of the following acid-base reactions, and estimate the value of Keq. Use the pKa table from Section 2.3 and/or from the Reference Tables.
Answer
Add texts here. Do not delete this text first.
Example 7.3
Exercise 7.3 Show the products of the following acid-base reactions, and estimate the value of Keq. Use the pKa table from Section 2.8 and/or from the Reference Tables.
Solution
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/02%3A_Acids_and_Bases/2.03%3A_Acid_Strength_and_%28pK_a%29.txt |
Periodic trends
First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table. We’ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules, such as the side chains of alanine, lysine, and serine.
We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable (weaker) the conjugate base, the stronger the acid. Look at where the negative charge ends up in each conjugate base. In the ethyl anion, the negative charge is borne by carbon, while in the methylamine anion and ethoxide anion the charges are located on a nitrogen and an oxygen, respectively. Remember the periodic trend in electronegativity: it also increases as we move from left to right along a row, meaning that oxygen is the most electronegative of the three, and carbon the least. The more electronegative an atom, the better it is able to bear a negative charge. Thus, the ethoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl anion is the least stable (highest energy, most basic).
We can use the same set of ideas to explain the difference in basicity between water and ammonia.
By looking at the pKavalues for the appropriate conjugate acids, we know that ammonia is more basic than water. Oxygen, as the more electronegative element, holds more tightly to its lone pair than the nitrogen. The nitrogen lone pair, therefore, is more likely to break away and form a new bond to a proton - it is, in other words, more basic. Once again, a more reactive (stronger) conjugate base means a less reactive (weaker) conjugate acid.
When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity. This is best illustrated with the halides: basicity, like electronegativity, increases as we move up the column.
Conversely, acidity in the haloacids increases as we move down the column.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodine ion, the negative charge is spread out over a significantly larger volume:
This illustrates a fundamental concept in organic chemistry that is important enough to put in red:
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one atom.
We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts. For now, the concept is applied only to the influence of atomic radius on anion stability. Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than acetic acid. HI, with a pKa of about -9, is one the strongest acids known.
More importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols. The pKa of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pKa for the hydroxyl on the serine side chain is on the order of 17.
To reiterate: acid strength increases as we move to the right along a row of the periodic table, and as we move down a column.
Example 7.6
Draw the structure of the conjugate base that would form if the compound below were to react with 1 molar equivalent of sodium hydroxide:
Solution
The Resonance Effect
In the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements. Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about the same element acting as the proton donor/acceptor. The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups.
Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are very different. What makes a carboxylic acid so much more acidic than an alcohol? As before, we begin by considering the conjugate bases.
In both species, the negative charge on the conjugate base is held by an oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: a resonance contributor can be drawn in which the negative charge is localized on the second oxygen of the group. The two resonance forms for the conjugate base are equal in energy, according to our ‘rules of resonance’. What this means, you may recall, is that the negative charge on the acetate ion is not located on one oxygen or the other: rather it is shared between the two. Chemists use the term ‘delocalization of charge’ to describe this situation. In the ethoxide ion, by contrast, the negative charge is ‘locked’ on the single oxygen – it has nowhere else to go.
Now is the time to think back to that statement from the previous section that was so important that it got printed in bold font in its own paragraph – in fact, it is so important that we’ll just say it again: "Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one atom." Now, we are seeing this concept in another context, where a charge is being ‘spread out’ (in other words, delocalized) by resonance, rather than simply by the size of the atom involved.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pKa units between ethanol and acetic acid (and remember, pKa is a log expression, so we are talking about a difference of over 1012 between the acidity constants for the two molecules). The acetate ion is that much more stable than the ethoxide ion, all due to the effects of resonance delocalization.
The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but not basic when it is part of an amide group. Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a second resonance contributor in which the nitrogen lone pair is part of a p bond.
While the electron lone pair of an amine nitrogen is ‘stuck’ in one place, the lone pair on an amide nitrogen is delocalized by resonance. Notice that in this case, we are extending our central statement to say that electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too ‘comfortable’ being part of the delocalized pi-bonding system. The lone pair on an amine nitrogen, by contrast, is not part of a delocalized p system, and is very ready to form a bond with any acidic proton that might be nearby.
Often it requires some careful thought to predict the most acidic proton on a molecule. Ascorbic acid, also known as Vitamin C, has a pKa of 4.1.
There are four hydroxyl groups on this molecule – which one is the most acidic? If we consider all four possible conjugate bases, we find that there is only one for which we can delocalized the negative charge over two oxygen atoms.
Example 7.7
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
The inductive effect
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
The presence of the chlorines clearly increases the acidity of the carboxylic acid group, but the argument here does not have to do with resonance delocalization, because no additional resonance contributors can be drawn for the chlorinated molecules. Rather, the explanation for this phenomenon involves something called the inductive effect. A chlorine atom is more electronegative than a hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. In effect, the chlorine atoms are helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect. In this context, the chlorine substituent is called an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects.
The inductive electron-withdrawing effect of the chlorines takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.
Example
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/02%3A_Acids_and_Bases/2.05%3A_Factors_That_Determine_Acid_Strength.txt |
Acidity of Carboxylic Acids
The pKa 's of some typical carboxylic acids are listed in the following table. When we compare these values with those of comparable alcohols, such as ethanol (pKa = 16) and 2-methyl-2-propanol (pKa = 19), it is clear that carboxylic acids are stronger acids by over ten powers of ten! Furthermore, electronegative substituents near the carboxyl group act to increase the acidity.
Compound
pKa
Compound
pKa
HCO2H 3.75 CH3CH2CH2CO2H 4.82
CH3CO2H 4.74 ClCH2CH2CH2CO2H 4.53
FCH2CO2H 2.65 CH3CHClCH2CO2H 4.05
ClCH2CO2H 2.85 CH3CH2CHClCO2H 2.89
BrCH2CO2H 2.90 C6H5CO2H 4.20
ICH2CO2H 3.10 p-O2NC6H4CO2H 3.45
Cl3CCO2H 0.77 p-CH3OC6H4CO2H 4.45
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
2.07: Aspirin
Prostaglandins were first discovered and isolated from human semen in the 1930s by Ulf von Euler of Sweden. Thinking they had come from the prostate gland, he named them prostaglandins. It has since been determined that they exist and are synthesized in virtually every cell of the body. Prostaglandins, are like hormones in that they act as chemical messengers, but do not move to other sites, but work right within the cells where they are synthesized.
Introduction
Prostaglandins are unsaturated carboxylic acids, consisting of of a 20 carbon skeleton that also contains a five member ring. They are biochemically synthesized from the fatty acid, arachidonic acid. See the graphic on the left. The unique shape of the arachidonic acid caused by a series of cis double bonds helps to put it into position to make the five member ring.
Structure of prostaglandin E2 (PGE2)
Prostaglandin Structure
Prostaglandins are unsaturated carboxylic acids, consisting of of a 20 carbon skeleton that also contains a five member ring and are based upon the fatty acid, arachidonic acid. There are a variety of structures one, two, or three double bonds. On the five member ring there may also be double bonds, a ketone, or alcohol groups. A typical structure is on the left graphic.
Functions of Prostaglandins
There are a variety of physiological effects including:
1. Activation of the inflammatory response, production of pain, and fever. When tissues are damaged, white blood cells flood to the site to try to minimize tissue destruction. Prostaglandins are produced as a result.
2. Blood clots form when a blood vessel is damaged. A type of prostaglandin called thromboxane stimulates constriction and clotting of platelets. Conversely, PGI2, is produced to have the opposite effect on the walls of blood vessels where clots should not be forming.
3. Certain prostaglandins are involved with the induction of labor and other reproductive processes. PGE2 causes uterine contractions and has been used to induce labor.
4. Prostaglandins are involved in several other organs such as the gastrointestinal tract (inhibit acid synthesis and increase secretion of protective mucus), increase blood flow in kidneys, and leukotriens promote constriction of bronchi associated with asthma.
Ball-and-stick model of the aspirin molecule, as found in the solid state. Single-crystal X-ray diffraction data from Kim, Y.; Machida, K.; Taga, T.; Osaki, K. (1985). "Structure Redetermination and Packing Analysis of Aspirin Crystal". Chem. Pharm. Bull. 33 (7): 2641-2647. ISSN 1347-5223.
Effects of Aspirin and other Pain Killers
When you see that prostaglandins induce inflammation, pain, and fever, what comes to mind but aspirin. Aspirin blocks an enzyme called cyclooxygenase, COX-1 and COX-2, which is involved with the ring closure and addition of oxygen to arachidonic acid converting to prostaglandins. The acetyl group on aspirin is hydrolzed and then bonded to the alcohol group of serine as an ester. This has the effect of blocking the channel in the enzyme and arachidonic can not enter the active site of the enzyme. By inhibiting or blocking this enzyme, the synthesis of prostaglandins is blocked, which in turn relives some of the effects of pain and fever. Aspirin is also thought to inhibit the prostaglandin synthesis involved with unwanted blood clotting in coronary heart disease. At the same time an injury while taking aspirin may cause more extensive bleeding.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook
2.08: Lewis Acids and Bases
According to the Lewis theory, an acid is an electron pair acceptor, and a base is an electron pair donor. Lewis bases are also Brønsted bases; however, many Lewis acids, such as BF3, AlCl3 and Mg2+, are not Brønsted acids. The product of a Lewis acid-base reaction, is a neutral, dipolar or charged complex, which may be a stable covalent molecule. As shown at the top of the following drawing, coordinate covalent bonding of a phosphorous Lewis base to a boron Lewis acid creates a complex in which the formal charge of boron is negative and that of phosphorous is positive. In this complex, boron acquires a neon valence shell configuration and phosphorous an argon configuration. If the substituents (R) on these atoms are not large, the complex will be favored at equilibrium. However, steric hindrance of bulky substituents may prohibit complex formation. The resulting mixture of non-bonded Lewis acid/base pairs has been termed "frustrated", and exhibits unusual chemical behavior.
Two examples of Lewis acid-base equilibria that play a role in chemical reactions are shown in equations 1 & 2 below.
In the first example, an electron deficient aluminum atom bonds to a covalent chlorine atom by sharing one of its non-bonding valence electron pairs, and thus achieves an argon-like valence shell octet. Because this sharing is unilateral (chlorine contributes both electrons), both the aluminum and the chlorine have formal charges, as shown. If the carbon chlorine bond in this complex breaks with both the bonding electrons remaining with the more electronegative atom (chlorine), the carbon assumes a positive charge. We refer to such carbon species as carbocations. Carbocations are also Lewis acids, as the reverse reaction demonstrates.
Many carbocations (but not all) may also function as Brønsted acids. Equation 3 illustrates this dual behavior; the Lewis acidic site is colored red and three of the nine acidic hydrogen atoms are colored orange. In its Brønsted acid role the carbocation donates a proton to the base (hydroxide anion), and is converted to a stable neutral molecule having a carbon-carbon double bond.
The interaction between a magnesium cation (Mg+2) and a carbonyl oxygen is a common example of a Lewis acid-base reaction. The carbonyl oxygen (the Lewis base) donates a pair of electrons to the magnesium cation (the Lewis acid).
As we will see we begin the study of reactions involving carbonyl groups, this interaction has the very important effect of increasing the polarity of the carbon-oxygen double bond.
The Brønsted-Lowry equivalent of the reaction above is simply protonation of the carbonyl group. This, too, has the effect of increasing the polarity of the carbonyl double bond.
While it is important to be familiar with the Lewis definition, the focus throughout the remainder of this chapter will be on acid-base reactions of the Brønsted-Lowry type, where an actual proton transfer event takes place. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/02%3A_Acids_and_Bases/2.06%3A_Common_Acids_and_Bases.txt |
Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity when treated with certain reagents. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups. In the following table the atoms of each functional group are colored red and the characteristic IUPAC nomenclature suffix that denotes some (but not all) functional groups is also colored.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
3.02: An Overview of Functional Groups
Functional Group Tables
Exclusively Carbon Functional Groups
Group Formula
Class Name Specific Example IUPAC Name Common Name
Alkene H2C=CH2 Ethene Ethylene
Alkyne HC≡CH Ethyne Acetylene
Arene C6H6 Benzene Benzene
Functional Groups with Single Bonds to Heteroatoms
Group Formula
Class Name
Specific Example
IUPAC Name
Common Name
Halide H3C-I Iodomethane Methyl iodide
Alcohol CH3CH2OH Ethanol Ethyl alcohol
Ether CH3CH2OCH2CH3 Diethyl ether Ether
Amine H3C-NH2 Aminomethane Methylamine
Nitro Compound H3C-NO2 Nitromethane
Thiol H3C-SH Methanethiol Methyl mercaptan
Sulfide H3C-S-CH3 Dimethyl sulfid
Functional Groups with Multiple Bonds to Heteroatoms
Group Formula
Class Name
Specific Example
IUPAC Name
Common Name
Nitrile H3C-CN Ethanenitrile Acetonitrile
Aldehyde H3CCHO Ethanal Acetaldehyde
Ketone H3CCOCH3 Propanone Acetone
Carboxylic Acid H3CCO2H Ethanoic Acid Acetic acid
Ester H3CCO2CH2CH3 Ethyl ethanoate Ethyl acetate
Acid Halide H3CCOCl Ethanoyl chloride Acetyl chloride
Amide H3CCON(CH3)2 N,N-Dimethylethanamide N,N-Dimethylacetamide
Acid Anhydride (H3CCO)2O Ethanoic anhydride Acetic anhydride
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
3.03: Intermolecular Forces
Introduction
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.
Note
The properties of liquids are intermediate between those of gases and solids but are more similar to solids.
Intermolecular forces determine bulk properties such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid.
Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures. (For more information on the behavior of real gases and deviations from the ideal gas law,.)
In this section, we explicitly consider three kinds of intermolecular interactions:There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. The first two are often described collectively as van der Waals forces.
Dipole–Dipole Interactions
Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in part (a) in Figure 11.3. These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (part (c) in Figure 11.3). Hence dipole–dipole interactions, such as those in part (b) in Figure 11.3, are attractive intermolecular interactions, whereas those in part (d) in Figure 11.3 are repulsive intermolecular interactions. Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure 11.4. On average, however, the attractive interactions dominate.
Figure 11.3 Attractive and Repulsive Dipole–Dipole Interactions
(a and b) Molecular orientations in which the positive end of one dipole (δ+) is near the negative end of another (δ) (and vice versa) produce attractive interactions. (c and d) Molecular orientations that juxtapose the positive or negative ends of the dipoles on adjacent molecules produce repulsive interactions.
Figure 11.4 Both Attractive and Repulsive Dipole–Dipole Interactions Occur in a Liquid Sample with Many Molecules
Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/r, where r is the distance between the ions. Doubling the distance (r → 2r) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/r6, so doubling the distance between the dipoles decreases the strength of the interaction by 26, or 64-fold. Thus a substance such as HCl, which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure, whereas NaCl, which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table 11.2. Using what we learned about predicting relative bond polarities from the electronegativities of the bonded atoms, we can make educated guesses about the relative boiling points of similar molecules.
Table 11.2 Relationships between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Molar Mass
Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K)
C3H6 (cyclopropane) 42 0 240
CH3OCH3 (dimethyl ether) 46 1.30 248
CH3CN (acetonitrile) 41 3.9 355
Note
The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.
Example 1
Arrange ethyl methyl ether (CH3OCH2CH3), 2-methylpropane [isobutane, (CH3)2CHCH3], and acetone (CH3COCH3) in order of increasing boiling points. Their structures are as follows:
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.
Solution:
The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds. The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point. Ethyl methyl ether has a structure similar to H2O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point. Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point. Thus we predict the following order of boiling points: 2-methylpropane < ethyl methyl ether < acetone. This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.
Exercise 1
Arrange carbon tetrafluoride (CF4), ethyl methyl sulfide (CH3SC2H5), dimethyl sulfoxide [(CH3)2S=O], and 2-methylbutane [isopentane, (CH3)2CHCH2CH3] in order of decreasing boiling points.
Answer: dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)
London Dispersion Forces
Thus far we have considered only interactions between polar molecules, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table 11.3).
What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances.
Table 11.3 Normal Melting and Boiling Points of Some Elements and Nonpolar Compounds
Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C)
Ar 40 −189.4 −185.9
Xe 131 −111.8 −108.1
N2 28 −210 −195.8
O2 32 −218.8 −183.0
F2 38 −219.7 −188.1
I2 254 113.7 184.4
CH4 16 −182.5 −161.5
Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure 11.5, the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/r6. Doubling the distance therefore decreases the attractive energy by 26, or 64-fold.
Figure 11.5 Instantaneous Dipole Moments. The formation of an instantaneous dipole moment on one He atom (a) or an H2 molecule (b) results in the formation of an induced dipole on an adjacent atom or molecule.
Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H2 molecules in part (b) in Figure 11.5, tends to become more pronounced as atomic and molecular masses increase (Table 11.3). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones.
Note
For similar substances, London dispersion forces get stronger with increasing molecular size.
The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure 11.6). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure 11.6 shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C5H12. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C).
Figure 11.6 Mass and Surface Area Affect the Strength of London Dispersion Forces. (a) In this series of four simple alkanes, larger molecules have stronger London forces between them than smaller molecules and consequently higher boiling points. (b) Linear n-pentane molecules have a larger surface area and stronger intermolecular forces than spherical neopentane molecules. As a result, neopentane is a gas at room temperature, whereas n-pentane is a volatile liquid.
All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.
Example 2
Arrange n-butane, propane, 2-methylpropane [isobutene, (CH3)2CHCH3], and n-pentane in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Determine the intermolecular forces in the compounds and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n-butane (−0.5°C) < n-pentane (36.1°C).
Exercise 2
Arrange GeH4, SiCl4, SiH4, CH4, and GeCl4 in order of decreasing boiling points.
Answer: GeCl4 (87°C) > SiCl4 (57.6°C) > GeH4 (−88.5°C) > SiH4 (−111.8°C) > CH4 (−161°C)
Hydrogen Bonds
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure 11.7. Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H2Te and H2Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.
Figure 11.7 The Effects of Hydrogen Bonding on Boiling Points. These plots of the boiling points of the covalent hydrides of the elements of groups 14–17 show that the boiling points of the lightest members of each series for which hydrogen bonding is possible (HF, NH3, and H2O) are anomalously high for compounds with such low molecular masses.
Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure 11.8. A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water rather than sinks.
Figure 11.8 The Hydrogen-Bonded Structure of Ice
Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cagelike structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.
Note
Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.
Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze.
Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH3. Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.
Example 3
Considering CH3OH, C2H6, Xe, and (CH3)3N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Given: compounds
Asked for: formation of hydrogen bonds and structure
Strategy:
1. Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.
2. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.
Solution:
A Of the species listed, xenon (Xe), ethane (C2H6), and trimethylamine [(CH3)3N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.
B The one compound that can act as a hydrogen bond donor, methanol (CH3OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows:
Exercise 3
Considering CH3CO2H, (CH3)3N, NH3, and CH3F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Answer: CH3CO2H and NH3;
Example 4
Arrange C60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N2O in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N2O have very similar molar masses (40 and 44 g/mol, respectively), but N2O is polar while Ar is not. Consequently, N2O should have a higher boiling point. A C60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N2O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses: He (−269°C) < Ar (−185.7°C) < N2O (−88.5°C) < C60 (>280°C) < NaCl (1465°C).
Exercise 4
Arrange 2,4-dimethylheptane, Ne, CS2, Cl2, and KBr in order of decreasing boiling points.
Answer: KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS2 (46.6°C) > Cl2 (−34.6°C) > Ne (−246°C)
Summary
Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid or the liquid and gas phases are due to changes in intermolecular interactions but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds. Dipole–dipole interactions arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/r6, where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules. Like dipole–dipole interactions, their energy falls off as 1/r6. Larger atoms tend to be more polarizable than smaller ones because their outer electrons are less tightly bound and are therefore more easily perturbed. Hydrogen bonds are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the hydrogen bond donor) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor). Because of strong OHhydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cagelike structure that is less dense than liquid water.
Key Takeaway
• Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds.
Conceptual Problems
1. What is the main difference between intramolecular interactions and intermolecular interactions? Which is typically stronger? How are changes of state affected by these different kinds of interactions?
2. Describe the three major kinds of intermolecular interactions discussed in this chapter and their major features. The hydrogen bond is actually an example of one of the other two types of interaction. Identify the kind of interaction that includes hydrogen bonds and explain why hydrogen bonds fall into this category.
3. Which are stronger—dipole–dipole interactions or London dispersion forces? Which are likely to be more important in a molecule with heavy atoms? Explain your answers.
4. Explain why hydrogen bonds are unusually strong compared to other dipole–dipole interactions. How does the strength of hydrogen bonds compare with the strength of covalent bonds?
5. Liquid water is essential for life as we know it, but based on its molecular mass, water should be a gas under standard conditions. Why is water a liquid rather than a gas under standard conditions?
6. Describe the effect of polarity, molecular mass, and hydrogen bonding on the melting point and boiling point of a substance.
7. Why are intermolecular interactions more important for liquids and solids than for gases? Under what conditions must these interactions be considered for gases?
8. Using acetic acid as an example, illustrate both attractive and repulsive intermolecular interactions. How does the boiling point of a substance depend on the magnitude of the repulsive intermolecular interactions?
9. In group 17, elemental fluorine and chlorine are gases, whereas bromine is a liquid and iodine is a solid. Why?
10. The boiling points of the anhydrous hydrogen halides are as follows: HF, 19°C; HCl, −85°C; HBr, −67°C; and HI, −34°C. Explain any trends in the data, as well as any deviations from that trend.
11. Identify the most important intermolecular interaction in each of the following.
1. SO2
2. HF
3. CO2
4. CCl4
5. CH2Cl2
12. Identify the most important intermolecular interaction in each of the following.
1. LiF
2. I2
3. ICl
4. NH3
5. NH2Cl
13. Would you expect London dispersion forces to be more important for Xe or Ne? Why? (The atomic radius of Ne is 38 pm, whereas that of Xe is 108 pm.)
14. Arrange Kr, Cl2, H2, N2, Ne, and O2 in order of increasing polarizability. Explain your reasoning.
15. Both water and methanol have anomalously high boiling points due to hydrogen bonding, but the boiling point of water is greater than that of methanol despite its lower molecular mass. Why? Draw the structures of these two compounds, including any lone pairs, and indicate potential hydrogen bonds.
16. The structures of ethanol, ethylene glycol, and glycerin are as follows:
Arrange these compounds in order of increasing boiling point. Explain your rationale.
17. Do you expect the boiling point of H2S to be higher or lower than that of H2O? Justify your answer.
18. Ammonia (NH3), methylamine (CH3NH2), and ethylamine (CH3CH2NH2) are gases at room temperature, while propylamine (CH3CH2CH2NH2) is a liquid at room temperature. Explain these observations.
19. Why is it not advisable to freeze a sealed glass bottle that is completely filled with water? Use both macroscopic and microscopic models to explain your answer. Is a similar consideration required for a bottle containing pure ethanol? Why or why not?
20. Which compound in the following pairs will have the higher boiling point? Explain your reasoning.
1. NH3 or PH3
2. ethylene glycol (HOCH2CH2OH) or ethanol
3. 2,2-dimethylpropanol [CH3C(CH3)2CH2OH] or n-butanol (CH3CH2CH2CH2OH)
21. Some recipes call for vigorous boiling, while others call for gentle simmering. What is the difference in the temperature of the cooking liquid between boiling and simmering? What is the difference in energy input?
22. Use the melting of a metal such as lead to explain the process of melting in terms of what is happening at the molecular level. As a piece of lead melts, the temperature of the metal remains constant, even though energy is being added continuously. Why?
23. How does the O–H distance in a hydrogen bond in liquid water compare with the O–H distance in the covalent O–H bond in the H2O molecule? What effect does this have on the structure and density of ice?
1. Explain why the hydrogen bonds in liquid HF are stronger than the corresponding intermolecular HI interactions in liquid HI.
2. In which substance are the individual hydrogen bonds stronger: HF or H2O? Explain your reasoning.
3. For which substance will hydrogen bonding have the greater effect on the boiling point: HF or H2O? Explain your reasoning.
Answers
1. Water is a liquid under standard conditions because of its unique ability to form four strong hydrogen bonds per molecule.
2. As the atomic mass of the halogens increases, so does the number of electrons and the average distance of those electrons from the nucleus. Larger atoms with more electrons are more easily polarized than smaller atoms, and the increase in polarizability with atomic number increases the strength of London dispersion forces. These intermolecular interactions are strong enough to favor the condensed states for bromine and iodine under normal conditions of temperature and pressure.
1. The V-shaped SO2 molecule has a large dipole moment due to the polar S=O bonds, so dipole–dipole interactions will be most important.
2. The H–F bond is highly polar, and the fluorine atom has three lone pairs of electrons to act as hydrogen bond acceptors; hydrogen bonding will be most important.
3. Although the C=O bonds are polar, this linear molecule has no net dipole moment; hence, London dispersion forces are most important.
4. This is a symmetrical molecule that has no net dipole moment, and the Cl atoms are relatively polarizable; thus, London dispersion forces will dominate.
5. This molecule has a small dipole moment, as well as polarizable Cl atoms. In such a case, dipole–dipole interactions and London dispersion forces are often comparable in magnitude.
3. Water has two polar O–H bonds with H atoms that can act as hydrogen bond donors, plus two lone pairs of electrons that can act as hydrogen bond acceptors, giving a net of four hydrogen bonds per H2O molecule. Although methanol also has two lone pairs of electrons on oxygen that can act as hydrogen bond acceptors, it only has one O–H bond with an H atom that can act as a hydrogen bond donor. Consequently, methanol can only form two hydrogen bonds per molecule on average, versus four for water. Hydrogen bonding therefore has a much greater effect on the boiling point of water.
4. Vigorous boiling causes more water molecule to escape into the vapor phase, but does not affect the temperature of the liquid. Vigorous boiling requires a higher energy input than does gentle simmering.
Anonymous | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/03%3A_Introduction_to_Organic_Molecules_and_Functional_Groups/3.01%3A_Functional_Groups.txt |
The molecule is the smallest observable group of uniquely bonded atoms that represent the composition, configuration and characteristics of a pure compound. Our chief focus up to this point has been to discover and describe the ways in which atoms bond together to form molecules. Since all observable samples of compounds and mixtures contain a very large number of molecules (ca.!020), we must also concern ourselves with interactions between molecules, as well as with their individual structures. Indeed, many of the physical characteristics of compounds that are used to identify them (e.g. boiling points, melting points and solubilities) are due to intermolecular interactions.
All atoms and molecules have a weak attraction for one another, known as van der Waals attraction. This attractive force has its origin in the electrostatic attraction of the electrons of one molecule or atom for the nuclei of another. If there were no van der Waals forces, all matter would exist in a gaseous state, and life as we know it would not be possible. It should be noted that there are also smaller repulsive forces between molecules that increase rapidly at very small intermolecular distances.
Boiling and Melting Points
For general purposes it is useful to consider temperature to be a measure of the kinetic energy of all the atoms and molecules in a given system. As temperature is increased, there is a corresponding increase in the vigor of translational and rotation motions of all molecules, as well as the vibrations of atoms and groups of atoms within molecules. Experience shows that many compounds exist normally as liquids and solids; and that even low-density gases, such as hydrogen and helium, can be liquified at sufficiently low temperature and high pressure. A clear conclusion to be drawn from this fact is that intermolecular attractive forces vary considerably, and that the boiling point of a compound is a measure of the strength of these forces. Thus, in order to break the intermolecular attractions that hold the molecules of a compound in the condensed liquid state, it is necessary to increase their kinetic energy by raising the sample temperature to the characteristic boiling point of the compound.
The following table illustrates some of the factors that influence the strength of intermolecular attractions. The formula of each entry is followed by its formula weight in parentheses and the boiling point in degrees Celsius. First there is molecular size. Large molecules have more electrons and nuclei that create van der Waals attractive forces, so their compounds usually have higher boiling points than similar compounds made up of smaller molecules. It is very important to apply this rule only to like compounds. The examples given in the first two rows are similar in that the molecules or atoms are spherical in shape and do not have permanent dipoles. Molecular shape is also important, as the second group of compounds illustrate. The upper row consists of roughly spherical molecules, whereas the isomers in the lower row have cylindrical or linear shaped molecules. The attractive forces between the latter group are generally greater. Finally, permanent molecular dipoles generated by polar covalent bonds result in even greater attractive forces between molecules, provided they have the mobility to line up in appropriate orientations. The last entries in the table compare non-polar hydrocarbons with equal-sized compounds having polar bonds to oxygen and nitrogen. Halogens also form polar bonds to carbon, but they also increase the molecular mass, making it difficult to distinguish among these factors.
Boiling Points (ºC) of Selected Elements and Compounds
Increasing Size
Atomic Ar (40) -186 Kr (83) -153 Xe (131) -109
Molecular CH4 (16) -161 (CH3)4C (72) 9.5 (CH3)4Si (88) 27 CCl4 (154) 77
Molecular Shape
Spherical: (CH3)4C (72) 9.5 (CH3)2CCl2 (113) 69 (CH3)3CC(CH3)3 (114) 106
Linear: CH3(CH2)3CH3 (72) 36 Cl(CH2)3Cl (113) 121 CH3(CH2)6CH3 (114) 126
Molecular Polarity
Non-polar: H2C=CH2 (28) -104 F2 (38) -188 CH3C≡CCH3 (54) -32 CF4 (88) -130
Polar: H2C=O (30) -21 CH3CH=O (44) 20 (CH3)3N (59) 3.5 (CH3)2C=O (58) 56
HC≡N (27) 26 CH3C≡N (41) 82 (CH2)3O (58) 50 CH3NO2 (61) 101
The melting points of crystalline solids cannot be categorized in as simple a fashion as boiling points. The distance between molecules in a crystal lattice is small and regular, with intermolecular forces serving to constrain the motion of the molecules more severely than in the liquid state. Molecular size is important, but shape is also critical, since individual molecules need to fit together cooperatively for the attractive lattice forces to be large. Spherically shaped molecules generally have relatively high melting points, which in some cases approach the boiling point. This reflects the fact that spheres can pack together more closely than other shapes. This structure or shape sensitivity is one of the reasons that melting points are widely used to identify specific compounds. The data in the following table serves to illustrate these points.
Compound Formula Boiling Point Melting Point
pentane CH3(CH2)3CH3 36ºC –130ºC
hexane CH3(CH2)4CH3 69ºC –95ºC
heptane CH3(CH2)5CH3 98ºC –91ºC
octane CH3(CH2)6CH3 126ºC –57ºC
nonane CH3(CH2)7CH3 151ºC –54ºC
decane CH3(CH2)8CH3 174ºC –30ºC
tetramethylbutane (CH3)3C-C(CH3)3 106ºC +100ºC
Notice that the boiling points of the unbranched alkanes (pentane through decane) increase rather smoothly with molecular weight, but the melting points of the even-carbon chains increase more than those of the odd-carbon chains. Even-membered chains pack together in a uniform fashion more compactly than do odd-membered chains. The last compound, an isomer of octane, is nearly spherical and has an exceptionally high melting point (only 6º below the boiling point).
Hydrogen Bonding
The most powerful intermolecular force influencing neutral (uncharged) molecules is the hydrogen bond. If we compare the boiling points of methane (CH4) -161ºC, ammonia (NH3) -33ºC, water (H2O) 100ºC and hydrogen fluoride (HF) 19ºC, we see a greater variation for these similar sized molecules than expected from the data presented above for polar compounds. This is shown graphically in the following chart. Most of the simple hydrides of group IV, V, VI & VII elements display the expected rise in boiling point with molecular mass, but the hydrides of the most electronegative elements (nitrogen, oxygen and fluorine) have abnormally high boiling points for their mass.
The exceptionally strong dipole-dipole attractions that cause this behavior are called the hydrogen bond. Hydrogen forms polar covalent bonds to more electronegative atoms such as oxygen, and because a hydrogen atom is quite small, the positive end of the bond dipole (the hydrogen) can approach neighboring nucleophilic or basic sites more closely than can other polar bonds. Coulombic forces are inversely proportional to the sixth power of the distance between dipoles, making these interactions relatively strong, although they are still weak (ca. 4 to 5 kcal per mole) compared with most covalent bonds. The unique properties of water are largely due to the strong hydrogen bonding that occurs between its molecules. In the following diagram the hydrogen bonds are depicted as magenta dashed lines.
The molecule providing a polar hydrogen for a hydrogen bond is called a donor. The molecule that provides the electron rich site to which the hydrogen is attracted is called an acceptor. Water and alcohols may serve as both donors and acceptors, whereas ethers, aldehydes, ketones and esters can function only as acceptors. Similarly, primary and secondary amines are both donors and acceptors, but tertiary amines function only as acceptors. Once you are able to recognize compounds that can exhibit intermolecular hydrogen bonding, the relatively high boiling points they exhibit become understandable. The data in the following table serve to illustrate this point.
Compound Formula Mol. Wt. Boiling Point Melting Point
dimethyl ether CH3OCH3 46 –24ºC –138ºC
ethanol CH3CH2OH 46 78ºC –130ºC
propanol CH3(CH2)2OH 60 98ºC –127ºC
diethyl ether (CH3CH2)2O 74 34ºC –116ºC
propyl amine CH3(CH2)2NH2 59 48ºC –83ºC
methylaminoethane CH3CH2NHCH3 59 37ºC
trimethylamine (CH3)3N 59 3ºC –117ºC
ethylene glycol HOCH2CH2OH 62 197ºC –13ºC
acetic acid CH3CO2H 60 118ºC 17ºC
ethylene diamine H2NCH2CH2NH2 60 118ºC 8.5ºC
Alcohols boil considerably higher than comparably sized ethers (first two entries), and isomeric 1º, 2º & 3º-amines, respectively, show decreasing boiling points, with the two hydrogen bonding isomers being substantially higher boiling than the 3º-amine (entries 5 to 7). Also, O–H---O hydrogen bonds are clearly stronger than N–H---N hydrogen bonds, as we see by comparing propanol with the amines. As expected, the presence of two hydrogen bonding functions in a compound raises the boiling point even further. Acetic acid (the ninth entry) is an interesting case. A dimeric species, shown on the right, held together by two hydrogen bonds is a major component of the liquid state. If this is an accurate representation of the composition of this compound then we would expect its boiling point to be equivalent to that of a C4H8O4 compound (formula weight = 120). A suitable approximation of such a compound is found in tetramethoxymethane, (CH3O)4C, which is actually a bit larger (formula weight = 136) and has a boiling point of 114ºC. Thus, the dimeric hydrogen bonded structure appears to be a good representation of acetic acid in the condensed state.
A related principle is worth noting at this point. Although the hydrogen bond is relatively weak (ca. 4 to 5 kcal per mole), when several such bonds exist the resulting structure can be quite robust. The hydrogen bonds between cellulose fibers confer great strength to wood and related materials.
Solubility in Water
Water has been referred to as the "universal solvent", and its widespread distribution on this planet and essential role in life make it the benchmark for discussions of solubility. Water dissolves many ionic salts thanks to its high dielectric constant and ability to solvate ions. The former reduces the attraction between oppositely charged ions and the latter stabilizes the ions by binding to them and delocalizing charge density. Many organic compounds, especially alkanes and other hydrocarbons, are nearly insoluble in water. Organic compounds that are water soluble, such as most of those listed in the above table, generally have hydrogen bond acceptor and donor groups. The least soluble of the listed compounds is diethyl ether, which can serve only as a hydrogen bond acceptor and is 75% hydrocarbon in nature. Even so, diethyl ether is about two hundred times more soluble in water than is pentane.
The chief characteristic of water that influences these solubilities is the extensive hydrogen bonded association of its molecules with each other. This hydrogen bonded network is stabilized by the sum of all the hydrogen bond energies, and if nonpolar molecules such as hexane were inserted into the network they would destroy local structure without contributing any hydrogen bonds of their own. Of course, hexane molecules experience significant van der Waals attraction to neighboring molecules, but these attractive forces are much weaker than the hydrogen bond. Consequently, when hexane or other nonpolar compounds are mixed with water, the strong association forces of the water network exclude the nonpolar molecules, which must then exist in a separate phase. This is shown in the following illustration, and since hexane is less dense than water, the hexane phase floats on the water phase.
It is important to remember this tendency of water to exclude nonpolar molecules and groups, since it is a factor in the structure and behavior of many complex molecular systems. A common nomenclature used to describe molecules and regions within molecules is hydrophilic for polar, hydrogen bonding moieties and hydrophobic for nonpolar species. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/03%3A_Introduction_to_Organic_Molecules_and_Functional_Groups/3.04%3A_Physical_Properties.txt |
The essential dietary substances called vitamins are commonly classified as "water soluble" or "fat soluble". Water soluble vitamins, such as vitamin C, are rapidly eliminated from the body and their dietary levels need to be relatively high. The recommended daily allotment (RDA) of vitamin C is 100 mg, and amounts as large as 2 to 3 g are taken by many people without adverse effects. The lipid soluble vitamins, shown in the diagram below, are not as easily eliminated and may accumulate to toxic levels if consumed in large quantity. The RDA for these vitamins are:
• Vitamin A 800 μg ( upper limit ca. 3000 μg)
• Vitamin D 5 to 10 μg ( upper limit ca. 2000 μg)
• Vitamin E 15 mg ( upper limit ca. 1 g)
• Vitamin K 110 μg ( upper limit not specified)
From this data it is clear that vitamins A and D, while essential to good health in proper amounts, can be very toxic. Vitamin D, for example, is used as a rat poison, and in equal weight is more than 100 times as poisonous as sodium cyanide.
From the structures shown here, it should be clear that these compounds have more than a solubility connection with lipids. Vitamins A is a terpene, and vitamins E and K have long terpene chains attached to an aromatic moiety. The structure of vitamin D can be described as a steroid in which ring B is cut open and the remaining three rings remain unchanged. The precursors of vitamins A and D have been identified as the tetraterpene beta-carotene and the steroid ergosterol, respectively.
3.06: Application of Solubility- Soap
Carboxylic acids and salts having alkyl chains longer than eight carbons exhibit unusual behavior in water due to the presence of both hydrophilic (CO2) and hydrophobic (alkyl) regions in the same molecule. Such molecules are termed amphiphilic (Gk. amphi = both) or amphipathic. Fatty acids made up of ten or more carbon atoms are nearly insoluble in water, and because of their lower density, float on the surface when mixed with water. Unlike paraffin or other alkanes, which tend to puddle on the waters surface, these fatty acids spread evenly over an extended water surface, eventually forming a monomolecular layer in which the polar carboxyl groups are hydrogen bonded at the water interface, and the hydrocarbon chains are aligned together away from the water. This behavior is illustrated in the diagram on the right. Substances that accumulate at water surfaces and change the surface properties are called surfactants.
Alkali metal salts of fatty acids are more soluble in water than the acids themselves, and the amphiphilic character of these substances also make them strong surfactants. The most common examples of such compounds are soaps and detergents, four of which are shown below. Note that each of these molecules has a nonpolar hydrocarbon chain, the "tail", and a polar (often ionic) "head group". The use of such compounds as cleaning agents is facilitated by their surfactant character, which lowers the surface tension of water, allowing it to penetrate and wet a variety of materials.
Very small amounts of these surfactants dissolve in water to give a random dispersion of solute molecules. However, when the concentration is increased an interesting change occurs. The surfactant molecules reversibly assemble into polymolecular aggregates called micelles. By gathering the hydrophobic chains together in the center of the micelle, disruption of the hydrogen bonded structure of liquid water is minimized, and the polar head groups extend into the surrounding water where they participate in hydrogen bonding. These micelles are often spherical in shape, but may also assume cylindrical and branched forms, as illustrated on the right. Here the polar head group is designated by a blue circle, and the nonpolar tail is a zig-zag black line.
The oldest amphiphilic cleaning agent known to humans is soap. Soap is manufactured by the base-catalyzed hydrolysis (saponification) of animal fat (see below). Before sodium hydroxide was commercially available, a boiling solution of potassium carbonate leached from wood ashes was used. Soft potassium soaps were then converted to the harder sodium soaps by washing with salt solution. The importance of soap to human civilization is documented by history, but some problems associated with its use have been recognized. One of these is caused by the weak acidity (pKa ca. 4.9) of the fatty acids. Solutions of alkali metal soaps are slightly alkaline (pH 8 to 9) due to hydrolysis. If the pH of a soap solution is lowered by acidic contaminants, insoluble fatty acids precipitate and form a scum. A second problem is caused by the presence of calcium and magnesium salts in the water supply (hard water). These divalent cations cause aggregation of the micelles, which then deposit as a dirty scum.
These problems have been alleviated by the development of synthetic amphiphiles called detergents (or syndets). By using a much stronger acid for the polar head group, water solutions of the amphiphile are less sensitive to pH changes. Also the sulfonate functions used for virtually all anionic detergents confer greater solubility on micelles incorporating the alkaline earth cations found in hard water. Variations on the amphiphile theme have led to the development of other classes, such as the cationic and nonionic detergents shown above. Cationic detergents often exhibit germicidal properties, and their ability to change surface pH has made them useful as fabric softeners and hair conditioners. These versatile chemical "tools" have dramatically transformed the household and personal care cleaning product markets over the past fifty years.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/03%3A_Introduction_to_Organic_Molecules_and_Functional_Groups/3.05%3A_Application_-_Vitamins.txt |
Phospholipids
Phospholipids are the main constituents of cell membranes. They resemble the triglycerides in being ester or amide derivatives of glycerol or sphingosine with fatty acids and phosphoric acid. The phosphate moiety of the resulting phosphatidic acid is further esterified with ethanolamine, choline or serine in the phospholipid itself. The following diagram shows the structures of some of these components. Clicking on the diagram will change it to display structures for two representative phospholipids. Note that the fatty acid components (R & R') may be saturated or unsaturated.
To see a model of a phospholipid Click Here.
As ionic amphiphiles, phospholipids aggregate or self-assemble when mixed with water, but in a different manner than the soaps and detergents. Because of the two pendant alkyl chains present in phospholipids and the unusual mixed charges in their head groups, micelle formation is unfavorable relative to a bilayer structure. If a phospholipid is smeared over a small hole in a thin piece of plastic immersed in water, a stable planar bilayer of phospholipid molecules is created at the hole. As shown in the following diagram, the polar head groups on the faces of the bilayer contact water, and the hydrophobic alkyl chains form a nonpolar interior. The phospholipid molecules can move about in their half the bilayer, but there is a significant energy barrier preventing migration to the other side of the bilayer. To see an enlarged segment of a phospholipid bilayer Click Here.
This bilayer membrane structure is also found in aggregate structures called liposomes. Liposomes are microscopic vesicles consisting of an aqueous core enclosed in one or more phospholipid layers. They are formed when phospholipids are vigorously mixed with water. Unlike micelles, liposomes have both aqueous interiors and exteriors.
This bilayer membrane structure is also found in aggregate structures called liposomes. Liposomes are microscopic vesicles consisting of an aqueous core enclosed in one or more phospholipid layers. They are formed when phospholipids are vigorously mixed with water. Unlike micelles, liposomes have both aqueous interiors and exteriors.
A cell may be considered a very complex liposome. The bilayer membrane that separates the interior of a cell from the surrounding fluids is largely composed of phospholipids, but it incorporates many other components, such as cholesterol, that contribute to its structural integrity. Protein channels that permit the transport of various kinds of chemical species in and out of the cell are also important components of cell membranes.
The interior of a cell contains a variety of structures (organelles) that conduct chemical operations vital to the cells existence. Molecules bonded to the surfaces of cells serve to identify specific cells and facilitate interaction with external chemical entities. The sphingomyelins are also membrane lipids. They are the major component of the myelin sheath surrounding nerve fibers. Multiple Sclerosis is a devastating disease in which the myelin sheath is lost, causing eventual paralysis.
Ionophores
Because the health of cells depends on maintaining the proper levels of cations in intracellular fluids, any change that affects the normal flux of metal ions across cell membranes could well cause an organism to die. Molecules that facilitate the transport of metal ions across membranes are generally called ionophores (ion plus phore from the Greek phorein, meaning “to carry”). Many ionophores are potent antibiotics that can kill or inhibit the growth of bacteria. An example is valinomycin, a cyclic molecule with a central cavity lined with oxygen atoms (part (a) in Figure 21.14 "Valinomycin Is an Antibiotic That Functions Like an Ionophore") that is similar to the cavity of a crown ether (part (a) in Figure 13.7 "Crown Ethers and Cryptands"). Like a crown ether, valinomycin is highly selective: its affinity for K+ is about 1000 times greater than that for Na+. By increasing the flux of K+ ions into cells, valinomycin disrupts the normal K+ gradient across a cell membrane, thereby killing the cell (part (b) in Figure 21.14 "Valinomycin Is an Antibiotic That Functions Like an Ionophore").
Figure 21.14 Valinomycin Is an Antibiotic That Functions Like an Ionophore
(a) This model of the structure of the K+–valinomycin complex, determined by x-ray diffraction, shows how the valinomycin molecule wraps itself around the K+ ion, shielding it from the environment, in a manner reminiscent of a crown ether complex. (For more information on the crown ethers, see Chapter 13 "Solutions", Section 13.2 "Solubility and Molecular Structure".) (b) Valinomycin kills bacteria by facilitating the transport of K+ ions across the cell membrane, thereby disrupting the normal distribution of ions in the bacterium. At the surface of the membrane, valinomycin binds a K+ ion. Because the hydrophobic exterior of the valinomycin molecule forms a “doughnut” that shields the positive charge of the metal ion, the K+–valinomycin complex is highly soluble in the nonpolar interior of the membrane. After the K+–valinomycin complex diffuses across the membrane to the interior of the cell, the K+ ion is released, and the valinomycin is free to diffuse back to the other side of the membrane to bind another K+ ion. Valinomycin thereby destroys the normal K+ gradient across the membrane, killing the cell.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/03%3A_Introduction_to_Organic_Molecules_and_Functional_Groups/3.07%3A_Application-_The_Cell_Membrane.txt |
Organic chemistry encompasses a very large number of compounds ( many millions ), and our previous discussion and illustrations have focused on their structural characteristics. Now that we can recognize these actors ( compounds ), we turn to the roles they are inclined to play in the scientific drama staged by the multitude of chemical reactions that define organic chemistry.
We begin by defining some basic terms that will be used frequently as this subject is elaborated.
Chemical Reaction: A transformation resulting in a change of composition, constitution and/or configuration of a compound ( referred to as the reactant or substrate ).
Reactant or Substrate: The organic compound undergoing change in a chemical reaction. Other compounds may also be involved, and common reactive partners ( reagents ) may be identified. The reactant is often ( but not always ) the larger and more complex molecule in the reacting system. Most ( or all ) of the reactant molecule is normally incorporated as part of the product molecule.
Reagent: A common partner of the reactant in many chemical reactions. It may be organic or inorganic; small or large; gas, liquid or solid. The portion of a reagent that ends up being incorporated in the product may range from all to very little or none.
Product(s) The final form taken by the major reactant(s) of a reaction.
Reaction Conditions The environmental conditions, such as temperature, pressure, catalysts & solvent, under which a reaction progresses optimally. Catalysts are substances that accelerate the rate ( velocity ) of a chemical reaction without themselves being consumed or appearing as part of the reaction product. Catalysts do not change equilibria positions.
If you scan any organic textbook you will encounter what appears to be a very large, often intimidating, number of reactions. These are the "tools" of a chemist, and to use these tools effectively, we must organize them in a sensible manner and look for patterns of reactivity that permit us make plausible predictions. Most of these reactions occur at special sites of reactivity known as functional groups, and these constitute one organizational scheme that helps us catalog and remember reactions.
Ultimately, the best way to achieve proficiency in organic chemistry is to understand how reactions take place, and to recognize the various factors that influence their course.
First, we identify four broad classes of reactions based solely on the structural change occurring in the reactant molecules. This classification does not require knowledge or speculation concerning reaction paths or mechanisms. The four main reaction classes are additions, eliminations, substitutions, and rearrangements.
Rearrangement
In an addition reaction the number of σ-bonds in the substrate molecule increases, usually at the expense of one or more π-bonds. The reverse is true of elimination reactions, i.e.the number of σ-bonds in the substrate decreases, and new π-bonds are often formed. Substitution reactions, as the name implies, are characterized by replacement of an atom or group (Y) by another atom or group (Z). Aside from these groups, the number of bonds does not change. A rearrangement reaction generates an isomer, and again the number of bonds normally does not change.
The examples illustrated above involve simple alkyl and alkene systems, but these reaction types are general for most functional groups, including those incorporating carbon-oxygen double bonds and carbon-nitrogen double and triple bonds. Some common reactions may actually be a combination of reaction types. The reaction of an ester with ammonia to give an amide, as shown below, appears to be a substitution reaction ( Y = CH3O & Z = NH2 ); however, it is actually two reactions, an addition followed by an elimination.
The addition of water to a nitrile does not seem to fit any of the above reaction types, but it is simply a slow addition reaction followed by a rapid rearrangement, as shown in the following equation. Rapid rearrangements of this kind are called tautomerizations.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/03%3A_Introduction_to_Organic_Molecules_and_Functional_Groups/3.08%3A_Functional_Groups_and_Reactivity.txt |
Glucose
Glucose is initially synthesized by chlorophyll in plants using carbon dioxide from the air and sunlight as an energy source. Glucose is further converted to starch for storage.
Prostaglandins
Prostaglandins are chemical messengers synthesized in the cells in which their physiological activity is expressed. They are unsaturated fatty acids containing 20 carbon atoms and are synthesized from arachidonic acid—a polyunsaturated fatty acid—when needed by a particular cell. They are called prostaglandins because they were originally isolated from semen found in the prostate gland. It is now known that they are synthesized in nearly all mammalian tissues and affect almost all organs in the body. The five major classes of prostaglandins are designated as PGA, PGB, PGE, PGF, and PGI. Subscripts are attached at the end of these abbreviations to denote the number of double bonds outside the five-carbon ring in a given prostaglandin.
The prostaglandins are among the most potent biological substances known. Slight structural differences give them highly distinct biological effects; however, all prostaglandins exhibit some ability to induce smooth muscle contraction, lower blood pressure, and contribute to the inflammatory response. Aspirin and other nonsteroidal anti-inflammatory agents, such as ibuprofen, obstruct the synthesis of prostaglandins by inhibiting cyclooxygenase, the enzyme needed for the initial step in the conversion of arachidonic acid to prostaglandins.
Their wide range of physiological activity has led to the synthesis of hundreds of prostaglandins and their analogs. Derivatives of PGE2 are now used in the United States to induce labor. Other prostaglandins have been employed clinically to lower or increase blood pressure, inhibit stomach secretions, relieve nasal congestion, relieve asthma, and prevent the formation of blood clots, which are associated with heart attacks and strokes.
Waxes
Waxes are esters formed from long-chain fatty acids and long-chain alcohols. Most natural waxes are mixtures of such esters. Plant waxes on the surfaces of leaves, stems, flowers, and fruits protect the plant from dehydration and invasion by harmful microorganisms. Carnauba wax, used extensively in floor waxes, automobile waxes, and furniture polish, is largely myricyl cerotate, obtained from the leaves of certain Brazilian palm trees. Animals also produce waxes that serve as protective coatings, keeping the surfaces of feathers, skin, and hair pliable and water repellent. In fact, if the waxy coating on the feathers of a water bird is dissolved as a result of the bird swimming in an oil slick, the feathers become wet and heavy, and the bird, unable to maintain its buoyancy, drowns.
DNA
The secondary structure of DNA is actually very similar to the secondary structure of proteins. The protein single alpha helix structure held together by hydrogen bonds was discovered with the aid of X-ray diffraction studies. The X-ray diffraction patterns for DNA show somewhat similar patterns.
Introduction
In addition, chemical studies by E. Chargaff indicate several important clues about the structure of DNA. In the DNA of all organisms:
1. The concentration of adenine equals that of thymine.
2. The concentration of guanine equals that of cytosine.
Chargaff's findings clearly indicate that some type of heterocyclic amine base pairing exists in the DNA structure. X-ray diffraction data shows that a repeating helical pattern occurs every 34 Angstrom units with 10 subunits per turn. Each subunit occupies 3.4 Angstrom units which is the same amount of space occupied by a single nucleotide unit. Using Chargaff's information and the X-ray data in conjunction with building actual molecular models, Watson and Crick developed the double helix as a model for DNA.
The double helix in DNA consists of two right-handed polynucleotide chains that are coiled about the same axis. The heterocyclic amine bases project inward toward the center so that the base of one strand interacts or pairs with a base of the other strand. According to the chemical and X-ray data and model building exercises, only specific heterocyclic amine bases may be paired.
Base Pairing Principle
The Base Pairing Principle is that adenine pairs with thymine (A - T) and guanine pairs with cytosine (G - C)
The base pairing is called complementary because there are specific geometry requirements in the formation of hydrogen bonds between the heterocylic amines. Heterocyclic amine base pairing is an application of the hydrogen bonding principle. In the structures for the complementary base pairs given in the graphic on the left, notice that the thymine - adenine pair interacts through two hydrogen bonds represented as (T=A) and that the cytosine-guanine pair interacts through three hydrogen bonds represented as (C=G).
Although other base pairing-hydrogen bonding combinations may be possible, they are not utilized because the bond distances do not correspond to those given by the base pairs already cited. The diameter of the helix is 20 Angstroms.
DNA Double Helix
The double-stranded helical model for DNA is shown in the graphic on the left. The easiest way to visualize DNA is as an immensely long rope ladder, twisted into a cork-screw shape. The sides of the ladder are alternating sequences of deoxyribose and phosphate (backbone) while the rungs of the ladder (bases) are made in two parts with each part firmly attached to the side of the ladder. The parts in the rung are heterocyclic amines held in position by hydrogen bonding. Although most DNA exists as open ended double helices, some bacterial DNA has been found as a cyclic helix. Occasionally, DNA has also been found as a single strand.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/03%3A_Introduction_to_Organic_Molecules_and_Functional_Groups/3.09%3A_Biomolecules.txt |
Alkanes are organic compounds that consist entirely of single-bonded carbon and hydrogen atoms and lack any other functional groups. Alkanes have the general formula \(C_nH_{2n+2}\) and can be subdivided into the following three groups: the linear straight-chain alkanes, branched alkanes, and cycloalkanes. Alkanes are also saturated hydrocarbons.
Cycloalkanes are cyclic hydrocarbons, meaning that the carbons of the molecule are arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds). There are also polycyclic alkanes, which are molecules that contain two or more cycloalkanes that are joined, forming multiple rings.
Molecular Formulas
Alkanes are the simplest family of hydrocarbons - compounds containing carbon and hydrogen only. Alkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds. The first six alkanes are as follows:
methane CH4
ethane C2H6
propane C3H8
butane C4H10
pentane C5H12
hexane C6H14
You can work out the formula of any of the alkanes using the general formula CnH2n+2
Isomerism
Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. For example, both of the following are the same molecule. They are not isomers. Both are butane.
All of the alkanes containing four or more carbon atoms show structural isomerism, meaning that there are two or more different structural formulae that you can draw for each molecular formula.
Example 1: Butane or MethylPropane
C4H10 could be either of these two different molecules:
These are different molecules named butane (left) and 2-methylpropane (right).
There are also endless other possible ways that this molecule could twist itself. There is completely free rotation around all the carbon-carbon single bonds. If you had a model of a molecule in front of you, you would have to take it to pieces and rebuild it if you wanted to make an isomer of that molecule. If you can make an apparently different molecule just by rotating single bonds, it's not different - it's still the same molecule. In structural isomerism, the atoms are arranged in a completely different order. This is easier to see with specific examples. What follows looks at some of the ways that structural isomers can arise.
Constitutional isomers arise because of the possibility of branching in carbon chains. For example, there are two isomers of butane, C4H10. In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched.
Be careful not to draw "false" isomers which are just twisted versions of the original molecule. For example, this structure is just the straight chain version of butane rotated about the central carbon-carbon bond.
You could easily see this with a model. This is the example we've already used at the top of this page.
Example 2: Constitutional Isomers in Pentane
Pentane, C5H12, has three chain isomers. If you think you can find any others, they are simply twisted versions of the ones below. If in doubt make some models.
Classification of Carbon and Hydrogen Atoms
Carbons have a special terminology to describe how many other carbons they are attached to.
• Primary carbons (1o) attached to one other C atom
• Secondary carbons (2o) are attached to two other C’s
• Tertiary carbons (3o) are attached to theree other C’s
• Quaternary carbons (4o) are attached to four C's
For example, each of the three types of carbons are found in the 2,2 -dimethyl, 4-methylpentane molecule
Hydrogen atoms are also classified in this manner. A hydrogen atom attached to a primary carbon atom is called a primary hydrogen; thus, isobutane, has nine primary hydrogens and one tertiary hydrogen.
• Primary hydrogens (1o) are attached to carbons bonded to one other C atom
• Secondary hydrogens (2o) are attached to carbons bonded to two other C’s
• Tertiary hydrogens (3o) are attached to carbons bonded to theree other C’s
Each of the three types of carbons are found in the 2,2 -dimethyl, 4-methylpentane molecule
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.01%3A_AlkanesAn_Introduction.txt |
Cycloalkanes are cyclic hydrocarbons, meaning that the carbons of the molecule are arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds). There are also polycyclic alkanes, which are molecules that contain two or more cycloalkanes that are joined, forming multiple rings.
Introduction
Many organic compounds found in nature or created in a laboratory contain rings of carbon atoms with distinguishing chemical properties; these compounds are known as cycloalkanes. Cycloalkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds, but in cycloalkanes, the carbon atoms are joined in a ring. The smallest cycloalkane is cyclopropane.
If you count the carbons and hydrogens, you will see that they no longer fit the general formula \(C_nH_{2n+2}\). By joining the carbon atoms in a ring,two hydrogen atoms have been lost. The general formula for a cycloalkane is \(C_nH_{2n}\). Cyclic compounds are not all flat molecules. All of the cycloalkanes, from cyclopentane upwards, exist as "puckered rings". Cyclohexane, for example, has a ring structure that looks like this:
Figure 2: This is known as the "chair" form of cyclohexane from its shape, which vaguely resembles a chair. Note: The cyclohexane molecule is constantly changing, with the atom on the left, which is currently pointing down, flipping up, and the atom on the right flipping down. During this process, another (slightly less stable) form of cyclohexane is formed known as the "boat" form. In this arrangement, both of these atoms are either pointing up or down at the same time
In addition to being saturated cyclic hydrocarbons, cycloalkanes may have multiple substituents or functional groups that further determine their unique chemical properties. The most common and useful cycloalkanes in organic chemistry are cyclopentane and cyclohexane, although other cycloalkanes varying in the number of carbons can be synthesized. Understanding cycloalkanes and their properties are crucial in that many of the biological processes that occur in most living things have cycloalkane-like structures.
Glucose (6 carbon sugar) Ribose (5 carbon sugar)
Cholesterol (polycyclic)
Although polycyclic compounds are important, they are highly complex and typically have common names accepted by IUPAC. However, the common names do not generally follow the basic IUPAC nomenclature rules. The general formula of the cycloalkanes is \(C_nH_{2n}\) where \(n\) is the number of carbons. The naming of cycloalkanes follows a simple set of rules that are built upon the same basic steps in naming alkanes. Cyclic hydrocarbons have the prefix "cyclo-".
Contents
For simplicity, cycloalkane molecules can be drawn in the form of skeletal structures in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens.
same as same as
Cycloalkane Molecular Formula Basic Structure
Cyclopropane C3H6
Cyclobutane C4H8
Cyclopentane C5H10
Cyclohexane C6H12
Cycloheptane C7H14
Cyclooctane C8H16
Cyclononane C9H18
Cyclodecane C10H20
4.03: An Introduction to Nomenclature
The increasingly large number of organic compounds identified with each passing day, together with the fact that many of these compounds are isomers of other compounds, requires that a systematic nomenclature system be developed. Just as each distinct compound has a unique molecular structure which can be designated by a structural formula, each compound must be given a characteristic and unique name.
Introduction
As organic chemistry grew and developed, many compounds were given trivial names, which are now commonly used and recognized. Some examples are:
Name Methane Butane Acetone Toluene Acetylene Ethyl Alcohol
Formula CH4 C4H10 CH3COCH3 CH3C6H5 C2H2 C2H5OH
Such common names often have their origin in the history of the science and the natural sources of specific compounds, but the relationship of these names to each other is arbitrary, and no rational or systematic principles underly their assignments.
The IUPAC Systematic Approach to Nomenclature
A rational nomenclature system should do at least two things. First, it should indicate how the carbon atoms of a given compound are bonded together in a characteristic lattice of chains and rings. Second, it should identify and locate any functional groups present in the compound. Since hydrogen is such a common component of organic compounds, its amount and locations can be assumed from the tetravalency of carbon, and need not be specified in most cases.
The IUPAC nomenclature system is a set of logical rules devised and used by organic chemists to circumvent problems caused by arbitrary nomenclature. Knowing these rules and given a structural formula, one should be able to write a unique name for every distinct compound. Likewise, given a IUPAC name, one should be able to write a structural formula. In general, an IUPAC name will have three essential features:
• A root or base indicating a major chain or ring of carbon atoms found in the molecular structure.
• A suffix or other element(s) designating functional groups that may be present in the compound.
• Names of substituent groups, other than hydrogen, that complete the molecular structure.
As an introduction to the IUPAC nomenclature system, we shall first consider compounds that have no specific functional groups. Such compounds are composed only of carbon and hydrogen atoms bonded together by sigma bonds (all carbons are sp3 hybridized).
An excellent presentation of organic nomenclature is provided on a Nomenclature Page. created by Dave Woodcock. A full presentation of the IUPAC Rules is also available. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.02%3A_Cycloalkanes.txt |
Alkanes
Hydrocarbons having no double or triple bond functional groups are classified as alkanes or cycloalkanes, depending on whether the carbon atoms of the molecule are arranged only in chains or also in rings. Although these hydrocarbons have no functional groups, they constitute the framework on which functional groups are located in other classes of compounds, and provide an ideal starting point for studying and naming organic compounds. The alkanes and cycloalkanes are also members of a larger class of compounds referred to as aliphatic. Simply put, aliphatic compounds are compounds that do not incorporate any aromatic rings in their molecular structure.
The following table lists the IUPAC names assigned to simple continuous-chain alkanes from C-1 to C-10. A common "ane" suffix identifies these compounds as alkanes. Longer chain alkanes are well known, and their names may be found in many reference and text books. The names methane through decane should be memorized, since they constitute the root of many IUPAC names. Fortunately, common numerical prefixes are used in naming chains of five or more carbon atoms.
Table: Simple Unbranched Alkanes
Name Molecular
Formula
Structural
Formula
Isomers Name Molecular
Formula
Structural
Formula
Isomers
methane CH4 CH4 1 hexane C6H14 CH3(CH2)4CH3 5
ethane C2H6 CH3CH3 1 heptane C7H16 CH3(CH2)5CH3 9
propane C3H8 CH3CH2CH3 1 octane C8H18 CH3(CH2)6CH3 18
butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
pentane C5H12 CH3(CH2)3CH3 3 decane C10H22 CH3(CH2)8CH3 75
Some important behavior trends and terminologies
1. The formulas and structures of these alkanes increase uniformly by a CH2 increment.
2. A uniform variation of this kind in a series of compounds is called homologous.
3. These formulas all fit the CnH2n+2 rule. This is also the highest possible H/C ratio for a stable hydrocarbon.
4. Since the H/C ratio in these compounds is at a maximum, we call them saturated (with hydrogen).
Beginning with butane (C4H10), and becoming more numerous with larger alkanes, we note the existence of alkane isomers. For example, there are five C6H14 isomers, shown below as abbreviated line formulas (A through E):
Although these distinct compounds all have the same molecular formula, only one (A) can be called hexane. How then are we to name the others?
The IUPAC system requires first that we have names for simple unbranched chains, as noted above, and second that we have names for simple alkyl groups that may be attached to the chains. Examples of some common alkyl groups are given in the following table. Note that the "ane" suffix is replaced by "yl" in naming groups. The symbol R is used to designate a generic (unspecified) alkyl group.
Group CH3 C2H5 CH3CH2CH2 (CH3)2CH– CH3CH2CH2CH2 (CH3)2CHCH2 CH3CH2CH(CH3)– (CH3)3C– R–
Name Methyl Ethyl Propyl Isopropyl Butyl Isobutyl sec-Butyl tert-Butyl Alkyl
IUPAC Rules for Alkane Nomenclature
1. Find and name the longest continuous carbon chain.
2. Identify and name groups attached to this chain.
3. Number the chain consecutively, starting at the end nearest a substituent group.
4. Designate the location of each substituent group by an appropriate number and name.
5. Assemble the name, listing groups in alphabetical order.
The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing.
Halogen substituents are easily accommodated, using the names: fluoro (F-), chloro (Cl-), bromo (Br-) and iodo (I-).
Example 1: Halogen Substitution
For example, (CH3)2CHCH2CH2Br would be named 1-bromo-3-methylbutane. If the halogen is bonded to a simple alkyl group an alternative "alkyl halide" name may be used. Thus, C2H5Cl may be named chloroethane (no locator number is needed for a two carbon chain) or ethyl chloride.
For the above isomers of hexane the IUPAC names are: B 2-methylpentane C 3-methylpentane D 2,2-dimethylbutane E 2,3-dimethylbutane
Alkyl Groups
Alkanes can be described by the general formula CnH2n+2. An alkyl group is formed by removing one hydrogen from the alkane chain and is described by the formula CnH2n+1. The removal of this hydrogen results in a stem change from -ane to -yl. Take a look at the following examples.
The same concept can be applied to any of the straight chain alkane names provided in the table above.
Name Molecular Formula Condensed Structural Formula
Methane CH4 CH4
Ethane C2H6 CH3CH3
Propane C3H8 CH3CH2CH3
Butane C4H10 CH3(CH2)2CH3
Pentane C5H12 CH3(CH2)3CH3
Hexane C6H14 CH3(CH2)4CH3
Heptane C7H16 CH3(CH2)5CH3
Octane C8H18 CH3(CH2)6CH3
Nonane C9H20 CH3(CH2)7CH3
Decane C10H22 CH3(CH2)8CH3
Undecane C11H24 CH3(CH2)9CH3
Dodecane C12H26 CH3(CH2)10CH3
Tridecane C13H28 CH3(CH2)11CH3
Tetradecane C14H30 CH3(CH2)12CH3
Pentadecane C15H32 CH3(CH2)13CH3
Hexadecane C16H34 CH3(CH2)14CH3
Heptadecane C17H36 CH3(CH2)15CH3
Octadecane C18H38 CH3(CH2)16CH3
Nonadecane C19H40 CH3(CH2)17CH3
Eicosane C20H42 CH3(CH2)18CH3
Three Principles of Naming
1. Choose the longest, most substituted carbon chain containing a functional group.
2. A carbon bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number.
3. Take the alphabetical order into consideration; that is, after applying the first two rules given above, make sure that your substitutes and/or functional groups are written in alphabetical order.
Example 1
What is the name of the follow molecule?
SOLUTION
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example does not contain any functional groups, so we only need to be concerned with choosing the longest, most substituted carbon chain. The longest carbon chain has been highlighted in red and consists of eight carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. Because this example does not contain any functional groups, we only need to be concerned with the two substitutes present, that is, the two methyl groups. If we begin numbering the chain from the left, the methyls would be assigned the numbers 4 and 7, respectively. If we begin numbering the chain from the right, the methyls would be assigned the numbers 2 and 5. Therefore, to satisfy the second rule, numbering begins on the right side of the carbon chain as shown below. This gives the methyl groups the lowest possible numbering.
Rule 3: In this example, there is no need to utilize the third rule. Because the two substitutes are identical, neither takes alphabetical precedence with respect to numbering the carbons. This concept will become clearer in the following examples.
Example 2
What is the name of the follow molecule?
SOLUTION
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. In this example, numbering the chain from the left or the right would satisfy this rule. If we number the chain from the left, bromine and chlorine would be assigned the second and sixth carbon positions, respectively. If we number the chain from the right, chlorine would be assigned the second position and bromine would be assigned the sixth position. In other words, whether we choose to number from the left or right, the functional groups occupy the second and sixth positions in the chain. To select the correct numbering scheme, we need to utilize the third rule.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Example 3
What is the name of the follow molecule?
SOLUTION
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine, and one substitute, the methyl group. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. After taking functional groups into consideration, any substitutes present must have the lowest possible carbon number. This particular example illustrates the point of difference principle. If we number the chain from the left, bromine, the methyl group and chlorine would occupy the second, fifth and sixth positions, respectively. This concept is illustrated in the second drawing below. If we number the chain from the right, chlorine, the methyl group and bromine would occupy the second, third and sixth positions, respectively, which is illustrated in the first drawing below. The position of the methyl, therefore, becomes a point of difference. In the first drawing, the methyl occupies the third position. In the second drawing, the methyl occupies the fifth position. To satisfy the second rule, we want to choose the numbering scheme that provides the lowest possible numbering of this substitute. Therefore, the first of the two carbon chains shown below is correct.
Therefore, the first numbering scheme is the appropriate one to use.
Once you have determined the correct numbering of the carbons, it is often useful to make a list, including the functional groups, substitutes, and the name of the parent chain.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Parent chain: heptane 2-Chloro 3-Methyl 6-Bromo
• Anonymous | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.04%3A_Naming_Alkanes.txt |
Cycloalkanes are cyclic hydrocarbons, meaning that the carbons of the molecule are arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds). There are also polycyclic alkanes, which are molecules that contain two or more cycloalkanes that are joined, forming multiple rings.
Introduction
Many organic compounds found in nature or created in a laboratory contain rings of carbon atoms with distinguishing chemical properties; these compounds are known as cycloalkanes. Cycloalkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds, but in cycloalkanes, the carbon atoms are joined in a ring. The smallest cycloalkane is cyclopropane.
If you count the carbons and hydrogens, you will see that they no longer fit the general formula \(C_nH_{2n+2}\). By joining the carbon atoms in a ring,two hydrogen atoms have been lost. The general formula for a cycloalkane is \(C_nH_{2n}\). Cyclic compounds are not all flat molecules. All of the cycloalkanes, from cyclopentane upwards, exist as "puckered rings". Cyclohexane, for example, has a ring structure that looks like this:
Figure 2: This is known as the "chair" form of cyclohexane from its shape, which vaguely resembles a chair. Note: The cyclohexane molecule is constantly changing, with the atom on the left, which is currently pointing down, flipping up, and the atom on the right flipping down. During this process, another (slightly less stable) form of cyclohexane is formed known as the "boat" form. In this arrangement, both of these atoms are either pointing up or down at the same time
In addition to being saturated cyclic hydrocarbons, cycloalkanes may have multiple substituents or functional groups that further determine their unique chemical properties. The most common and useful cycloalkanes in organic chemistry are cyclopentane and cyclohexane, although other cycloalkanes varying in the number of carbons can be synthesized. Understanding cycloalkanes and their properties are crucial in that many of the biological processes that occur in most living things have cycloalkane-like structures.
Glucose (6 carbon sugar) Ribose (5 carbon sugar)
Cholesterol (polycyclic)
Although polycyclic compounds are important, they are highly complex and typically have common names accepted by IUPAC. However, the common names do not generally follow the basic IUPAC nomenclature rules. The general formula of the cycloalkanes is \(C_nH_{2n}\) where \(n\) is the number of carbons. The naming of cycloalkanes follows a simple set of rules that are built upon the same basic steps in naming alkanes. Cyclic hydrocarbons have the prefix "cyclo-".
Contents
For simplicity, cycloalkane molecules can be drawn in the form of skeletal structures in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens.
same as same as
Cycloalkane Molecular Formula Basic Structure
Cyclopropane C3H6
Cyclobutane C4H8
Cyclopentane C5H10
Cyclohexane C6H12
Cycloheptane C7H14
Cyclooctane C8H16
Cyclononane C9H18
Cyclodecane C10H20
IUPAC Rules for Nomenclature
1. Determine the cycloalkane to use as the parent chain. The parent chain is the one with the highest number of carbon atoms. If there are two cycloalkanes, use the cycloalkane with the higher number of carbons as the parent chain.
2. If there is an alkyl straight chain that has a greater number of carbons than the cycloalkane, then the alkyl chain must be used as the primary parent chain. Cycloalkane acting as a substituent to an alkyl chain has an ending "-yl" and, therefore, must be named as a cycloalkyl.
Cycloalkane Cycloalkyl
cyclopropane cyclopropyl
cyclobutane cyclobutyl
cyclopentane cyclopentyl
cyclohexane cyclohexyl
cycloheptane cycloheptyl
cyclooctane cyclooctyl
cyclononane cyclononanyl
cyclodecane cyclodecanyl
Example
The longest straight chain contains 10 carbons, compared with cyclopropane, which only contains 3 carbons. Because cyclopropane is a substituent, it would be named a cyclopropyl-substituted alkane.
3) Determine any functional groups or other alkyl groups.
4) Number the carbons of the cycloalkane so that the carbons with functional groups or alkyl groups have the lowest possible number. A carbon with multiple substituents should have a lower number than a carbon with only one substituent or functional group. One way to make sure that the lowest number possible is assigned is to number the carbons so that when the numbers corresponding to the substituents are added, their sum is the lowest possible.
(1+3=4) NOT (1+5=6)
5) When naming the cycloalkane, the substituents and functional groups must be placed in alphabetical order.
(ex: 2-bromo-1-chloro-3-methylcyclopentane)
6) Indicate the carbon number with the functional group with the highest priority according to alphabetical order. A dash"-" must be placed between the numbers and the name of the substituent. After the carbon number and the dash, the name of the substituent can follow. When there is only one substituent on the parent chain, indicating the number of the carbon atoms with the substituent is not necessary.
(ex: 1-chlorocyclohexane or cholorocyclohexane is acceptable)
7) If there is more than one of the same functional group on one carbon, write the number of the carbon two, three, or four times, depending on how many of the same functional group is present on that carbon. The numbers must be separated by commas, and the name of the functional group that follows must be separated by a dash. When there are two of the same functional group, the name must have the prefix "di". When there are three of the same functional group, the name must have the prefix "tri". When there are four of the same functional group, the name must have the prefix "tetra". However, these prefixes cannot be used when determining the alphabetical priorities.
There must always be commas between the numbers and the dashes that are between the numbers and the names.
Example 2
(2-bromo-1,1-dimethylcyclohexane)
Notice that "f" of fluoro alphabetically precedes the "m" of methyl. Although "di" alphabetically precedes "f", it is not used in determining the alphabetical order.
Example 3
(2-fluoro-1,1,-dimethylcyclohexane NOT 1,1-dimethyl-2-fluorocyclohexane)
8) If the substituents of the cycloalkane are related by the cis or trans configuration, then indicate the configuration by placing "cis-" or "trans-" in front of the name of the structure.
Blue=Carbon Yellow=Hydrogen Green=Chlorine
Notice that chlorine and the methyl group are both pointed in the same direction on the axis of the molecule; therefore, they are cis.
cis-1-chloro-2-methylcyclopentane
9) After all the functional groups and substituents have been mentioned with their corresponding numbers, the name of the cycloalkane can follow.
Reactivity
Cycloalkanes are very similar to the alkanes in reactivity, except for the very small ones, especially cyclopropane. Cyclopropane is significantly more reactive than what is expected because of the bond angles in the ring. Normally, when carbon forms four single bonds, the bond angles are approximately 109.5°. In cyclopropane, the bond angles are 60°.
With the electron pairs this close together, there is a significant amount of repulsion between the bonding pairs joining the carbon atoms, making the bonds easier to break.
Alcohol Substituents on Cycloalkanes
Alcohol (-OH) substituents take the highest priority for carbon atom numbering in IUPAC nomenclature. The carbon atom with the alcohol substituent must be labeled as 1. Molecules containing an alcohol group have an ending "-ol", indicating the presence of an alcohol group. If there are two alcohol groups, the molecule will have a "di-" prefix before "-ol" (diol). If there are three alcohol groups, the molecule will have a "tri-" prefix before "-ol" (triol), etc.
Example 4
The alcohol substituent is given the lowest number even though the two methyl groups are on the same carbon atom and labeling 1 on that carbon atom would give the lowest possible numbers. Numbering the location of the alcohol substituent is unnecessary because the ending "-ol" indicates the presence of one alcohol group on carbon atom number 1.
2,2-dimethylcyclohexanol NOT 1,1-dimethyl-cyclohexane-2-ol
Example 5
3-bromo-2-methylcyclopentanol NOT 1-bromo-2-methyl-cyclopentane-2-ol
Example 5
Blue=Carbon Yellow=Hydrogen Red=Oxygen
trans-cyclohexane-1,2-diol
Other Substituents on Cycloalkanes
There are many other functional groups like alcohol, which are later covered in an organic chemistry course, and they determine the ending name of a molecule. The naming of these functional groups will be explained in depth later as their chemical properties are explained.
Name Name ending
alkene -ene
alkyne -yne
alcohol -ol
ether -ether
nitrile -nitrile
amine -amine
aldehyde -al
ketone -one
carboxylic acid -oic acid
ester -oate
amide -amide
Although alkynes determine the name ending of a molecule, alkyne as a substituent on a cycloalkane is not possible because alkynes are planar and would require that the carbon that is part of the ring form 5 bonds, giving the carbon atom a negative charge.
However, a cycloalkane with a triple bond-containing substituent is possible if the triple bond is not directly attached to the ring.
Example
ethynylcyclooctane
Example
1-propylcyclohexane
Summary
1. Determine the parent chain: the parent chain contains the most carbon atoms.
2. Number the substituents of the chain so that the sum of the numbers is the lowest possible.
3. Name the substituents and place them in alphabetical order.
4. If stereochemistry of the compound is shown, indicate the orientation as part of the nomenclature.
5. Cyclic hydrocarbons have the prefix "cyclo-" and have an "-alkane" ending unless there is an alcohol substituent present. When an alcohol substituent is present, the molecule has an "-ol" ending.
Glossary
• alcohol: An oxygen and hydrogenOH hydroxyl group that is bonded to a substituted alkyl group.
• alkyl: A structure that is formed when a hydrogen atom is removed from an alkane.
• cyclic: Chemical compounds arranged in the form of a ring or a closed chain form.
• cycloalkanes: Cyclic saturated hydrocarbons with a general formula of CnH(2n). Cycloalkanes are alkanes with carbon atoms attached in the form of a closed ring.
• functional groups: An atom or groups of atoms that substitute for a hydrogen atom in an organic compound, giving the compound unique chemical properties and determining its reactivity.
• hydrocarbon: A chemical compound containing only carbon and hydrogen atoms.
• saturated: All of the atoms that make up a compound are single bonded to the other atoms, with no double or triple bonds.
• skeletal structure: A simplified structure in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens.
Problems
Name the following structures. (Note: The structures are complex for practice purposes and may not be found in nature.)
1) 2) 3) 4) 5) 6)
7)
Draw the following structures.
8) 1,1-dibromo-5-fluoro-3-butyl-7-methylcyclooctane 9) trans-1-bromo-2-chlorocyclopentane
10) 1,1-dibromo-2,3-dichloro-4-propylcyclobutane 11) 2-methyl-1-ethyl-1,3-dipropylcyclopentane 12) cycloheptane-1,3,5-triol
Name the following structures.
Blue=Carbon Yellow=Hydrogen Red=Oxygen Green=Chlorine
13) 14) 15) 16) 17)
18) 19)
Answers to Practice Problems
1) cyclodecane 2) chlorocyclopentane or 1-chlorocyclopentane 3) trans-1-chloro-2-methylcycloheptane
4) 3-cyclopropyl-6-methyldecane 5) cyclopentylcyclodecane or 1-cyclopentylcyclodecane 6) 1,3-dibromo-1-chloro-2-fluorocycloheptane
7) 1-cyclobutyl-4-isopropylcyclohexane
8) 9) 10) 11) 12)
13) cyclohexane 14) cyclohexanol 15) chlorocyclohexane 16) cyclopentylcyclohexane 17) 1-chloro-3-methylcyclobutane
18) 2,3-dimethylcyclohexanol 19) cis-1-propyl-2-methylcyclopentane
Contributors
• Pwint Zin
• Jim Clark (ChemGuide) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.05%3A_Naming_Cycloalkanes.txt |
Using Common Names with Branched Alkanes
Certain branched alkanes have common names that are still widely used today. These common names make use of prefixes, such as iso-, sec-, tert-, and neo-. The prefix iso-, which stands for isomer, is commonly given to 2-methyl alkanes. In other words, if there is methyl group located on the second carbon of a carbon chain, we can use the prefix iso-. The prefix will be placed in front of the alkane name that indicates the total number of carbons. Examples:
• isopentane which is the same as 2-methylbutane
• isobutane which is the same as 2-methylpropane
To assign the prefixes sec-, which stands for secondary, and tert-, for tertiary, it is important that we first learn how to classify carbon molecules. If a carbon is attached to only one other carbon, it is called a primary carbon. If a carbon is attached to two other carbons, it is called a seconday carbon. A tertiary carbon is attached to three other carbons and last, a quaternary carbon is attached to four other carbons. Examples:
• 4-sec-butylheptane (30g)
• 4-tert-butyl-5-isopropylhexane (30d); if using this example, may want to move sec/tert after iso disc
The prefix neo- refers to a substituent whose second-to-last carbon of the chain is trisubstituted (has three methyl groups attached to it). A neo-pentyl has five carbons total. Examples:
• neopentane
• neoheptane
4.07: Fossil Fuels
Petroleum
The petroleum that is pumped out of the ground at locations around the world is a complex mixture of several thousand organic compounds, including straight-chain alkanes, cycloalkanes, alkenes, and aromatic hydrocarbons with four to several hundred carbon atoms. The identities and relative abundances of the components vary depending on the source. So Texas crude oil is somewhat different from Saudi Arabian crude oil. In fact, the analysis of petroleum from different deposits can produce a “fingerprint” of each, which is useful in tracking down the sources of spilled crude oil. For example, Texas crude oil is “sweet,” meaning that it contains a small amount of sulfur-containing molecules, whereas Saudi Arabian crude oil is “sour,” meaning that it contains a relatively large amount of sulfur-containing molecules.
Gasoline
Petroleum is converted to useful products such as gasoline in three steps: distillation, cracking, and reforming. Recall from Chapter 1 "Introduction to Chemistry" that distillation separates compounds on the basis of their relative volatility, which is usually inversely proportional to their boiling points. Part (a) in Figure 2.23 "The Distillation of Petroleum" shows a cutaway drawing of a column used in the petroleum industry for separating the components of crude oil. The petroleum is heated to approximately 400°C (750°F), at which temperature it has become a mixture of liquid and vapor. This mixture, called the feedstock, is introduced into the refining tower. The most volatile components (those with the lowest boiling points) condense at the top of the column where it is cooler, while the less volatile components condense nearer the bottom. Some materials are so nonvolatile that they collect at the bottom without evaporating at all. Thus the composition of the liquid condensing at each level is different. These different fractions, each of which usually consists of a mixture of compounds with similar numbers of carbon atoms, are drawn off separately. Part (b) in Figure 2.23 "The Distillation of Petroleum" shows the typical fractions collected at refineries, the number of carbon atoms they contain, their boiling points, and their ultimate uses. These products range from gases used in natural and bottled gas to liquids used in fuels and lubricants to gummy solids used as tar on roads and roofs.
Figure 2.23 The Distillation of Petroleum
(a) This is a diagram of a distillation column used for separating petroleum fractions. (b) Petroleum fractions condense at different temperatures, depending on the number of carbon atoms in the molecules, and are drawn off from the column. The most volatile components (those with the lowest boiling points) condense at the top of the column, and the least volatile (those with the highest boiling points) condense at the bottom.
The economics of petroleum refining are complex. For example, the market demand for kerosene and lubricants is much lower than the demand for gasoline, yet all three fractions are obtained from the distillation column in comparable amounts. Furthermore, most gasolines and jet fuels are blends with very carefully controlled compositions that cannot vary as their original feedstocks did. To make petroleum refining more profitable, the less volatile, lower-value fractions must be converted to more volatile, higher-value mixtures that have carefully controlled formulas. The first process used to accomplish this transformation is cracking, in which the larger and heavier hydrocarbons in the kerosene and higher-boiling-point fractions are heated to temperatures as high as 900°C. High-temperature reactions cause the carbon–carbon bonds to break, which converts the compounds to lighter molecules similar to those in the gasoline fraction. Thus in cracking, a straight-chain alkane with a number of carbon atoms corresponding to the kerosene fraction is converted to a mixture of hydrocarbons with a number of carbon atoms corresponding to the lighter gasoline fraction. The second process used to increase the amount of valuable products is called reforming; it is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons. Using metals such as platinum brings about the necessary chemical reactions. The mixtures of products obtained from cracking and reforming are separated by fractional distillation.
Octane Ratings
The quality of a fuel is indicated by its octane rating, which is a measure of its ability to burn in a combustion engine without knocking or pinging. Knocking and pinging signal premature combustion (Figure 2.24 "The Burning of Gasoline in an Internal Combustion Engine"), which can be caused either by an engine malfunction or by a fuel that burns too fast. In either case, the gasoline-air mixture detonates at the wrong point in the engine cycle, which reduces the power output and can damage valves, pistons, bearings, and other engine components. The various gasoline formulations are designed to provide the mix of hydrocarbons least likely to cause knocking or pinging in a given type of engine performing at a particular level.
Figure 2.24 The Burning of Gasoline in an Internal Combustion Engine
(a) Normally, fuel is ignited by the spark plug, and combustion spreads uniformly outward. (b) Gasoline with an octane rating that is too low for the engine can ignite prematurely, resulting in uneven burning that causes knocking and pinging.
The octane scale was established in 1927 using a standard test engine and two pure compounds: n-heptane and isooctane (2,2,4-trimethylpentane). n-Heptane, which causes a great deal of knocking on combustion, was assigned an octane rating of 0, whereas isooctane, a very smooth-burning fuel, was assigned an octane rating of 100. Chemists assign octane ratings to different blends of gasoline by burning a sample of each in a test engine and comparing the observed knocking with the amount of knocking caused by specific mixtures of n-heptane and isooctane. For example, the octane rating of a blend of 89% isooctane and 11% n-heptane is simply the average of the octane ratings of the components weighted by the relative amounts of each in the blend. Converting percentages to decimals, we obtain the octane rating of the mixture:
0.89(100)+0.11(0)=89
A gasoline that performs at the same level as a blend of 89% isooctane and 11% n-heptane is assigned an octane rating of 89; this represents an intermediate grade of gasoline. Regular gasoline typically has an octane rating of 87; premium has a rating of 93 or higher.
As shown in Figure 2.25 "The Octane Ratings of Some Hydrocarbons and Common Additives", many compounds that are now available have octane ratings greater than 100, which means they are better fuels than pure isooctane. In addition, antiknock agents, also called octane enhancers, have been developed. One of the most widely used for many years was tetraethyllead [(C2H5)4Pb], which at approximately 3 g/gal gives a 10–15-point increase in octane rating. Since 1975, however, lead compounds have been phased out as gasoline additives because they are highly toxic. Other enhancers, such as methyl t-butyl ether (MTBE), have been developed to take their place. They combine a high octane rating with minimal corrosion to engine and fuel system parts. Unfortunately, when gasoline containing MTBE leaks from underground storage tanks, the result has been contamination of the groundwater in some locations, resulting in limitations or outright bans on the use of MTBE in certain areas. As a result, the use of alternative octane enhancers such as ethanol, which can be obtained from renewable resources such as corn, sugar cane, and, eventually, corn stalks and grasses, is increasing.
Figure 2.25 The Octane Ratings of Some Hydrocarbons and Common Additives | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.06%3A_Common_Names.txt |
Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless.
Boiling Points
The boiling points shown are for the "straight chain" isomers of which there is more than one. The first four alkanes are gases at room temperature, and solids do not begin to appear until about \(C_{17}H_{36}\), but this is imprecise because different isomers typically have different melting and boiling points. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers!
Cycloalkanes have boiling points that are approximately 20 K higher than the corresponding straight chain alkane.
There is not a significant electronegativity difference between carbon and hydrogen, thus, there is not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be Van der Waals dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size.
Where you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules.
Example
For example, the boiling points of the three isomers of \(C_5H_{12}\) are:
• pentane: 309.2 K
• 2-methylbutane: 301.0 K
• 2,2-dimethylpropane: 282.6 K
The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"!
Solubility
Alkanes (both alkanes and cycloalkanes) are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds.
Solubility in Water
When a molecular substance dissolves in water, the following must occur:
• break the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces.
• break the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds.
Breaking either of these attractions requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is relatively negligible; this is not true of the hydrogen bonds in water.
As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water. The alkane does not dissolve.
Note: This is a simplification because entropic effects are important when things dissolve.
Solubility in organic solvents
In most organic solvents, the primary forces of attraction between the solvent molecules are Van der Waals - either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility.
Contributors
Jim Clark (Chemguide.co.uk)
4.09: Conformations of Butane
Now let us consider butane, a slightly larger molecule. There are now three rotating carbon-carbon bonds to consider, but we will focus on the middle bond between C2 and C3. Below are two representations of butane in a conformation which puts the two CH3 groups (C1 and C4) in the eclipsed position.
This is the highest energy conformation for butane, due to what is called ‘van der Waals repulsion’, or ‘steric repulsion’, between the two rather bulky methyl groups.
What is van der Waals repulsion? Didn’t we just learn in Chapter 2 that the van der Waals force between two nonpolar groups is an attractive force? Consider this: you probably like to be near your friends, but no matter how close you are you probably don’t want to share a one-room apartment with five of them. When the two methyl groups are brought too close together, the overall resulting noncovalent interaction is repulsive rather than attractive. The result is that their respective electron densities repel one another.
If we rotate the front, (blue) carbon by 60°clockwise, the butane molecule is now in a staggered conformation.
This is more specifically referred to as the ‘gauche’ conformation of butane. Notice that although they are staggered, the two methyl groups are not as far apart as they could possibly be. There is still significant steric repulsion between the two bulky groups.
A further rotation of 60°gives us a second eclipsed conformation (B) in which both methyl groups are lined up with hydrogen atoms.
Due to steric repulsion between methyl and hydrogen substituents, this eclipsed conformation B is higher in energy than the gauche conformation. However, because there is no methyl-to-methyl eclipsing, it is lower in energy than eclipsed conformation A.
One more 60 rotation produces the ‘anti’ conformation, where the two methyl groups are positioned opposite each other and steric repulsion is minimized.
This is the lowest energy conformation for butane.
The diagram below summarizes the relative energies for the various eclipsed, staggered, and gauche conformations.
At room temperature, butane is most likely to be in the lowest-energy anti conformation at any given moment in time, although the energy barrier between the anti and eclipsed conformations is not high enough to prevent constant rotation except at very low temperatures. For this reason (and also simply for ease of drawing), it is conventional to draw straight-chain alkanes in a zigzag form, which implies anti conformation at all carbon-carbon bonds.
Template:ExampleStart
Exercise 3.1: Draw Newman projections of the eclipsed and staggered conformations of propane.
Exercise 3.2: Draw a Newman projection, looking down the C2-C3 bond, of 1-butene in the conformation shown below.
Solutions
Template:ExampleEnd
The following diagram illustrates the change in potential energy that occurs with rotation about the C2–C3 bond.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.08%3A_Physical_Properties_of_Alkanes.txt |
Cycloalkanes have one or more rings of carbon atoms. The simplest examples of this class consist of a single, unsubstituted carbon ring, and these form a homologous series similar to the unbranched alkanes. The IUPAC names of the first five members of this series are given in the following table. The last (yellow shaded) column gives the general formula for a cycloalkane of any size. If a simple unbranched alkane is converted to a cycloalkane two hydrogen atoms, one from each end of the chain, must be lost. Hence the general formula for a cycloalkane composed of n carbons is CnH2n. Although a cycloalkane has two fewer hydrogens than the equivalent alkane, each carbon is bonded to four other atoms so such compounds are still considered to be saturated with hydrogen.
Table 4.11.1: Examples of Simple Cycloalkanes
Name Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cycloalkane
Molecular
Formula
C3H6 C4H8 C5H10 C6H12 C7H14 CnH2n
Structural
Formula
(CH2)n
Line
Formula
The Baeyer Theory on the Strain in Cycloalkane Rings
Many of the properties of cyclopropane and its derivatives are similar to the properties of alkenes. In 1890, the famous German organic chemist, A. Baeyer, suggested that cyclopropane and cyclobutane derivatives are different from cyclopentane and cyclohexane, because their C—C—C angles cannot have the tetrahedral value of 109.5°. At the same time, Baeyer hypothesized that the difficulties encountered in synthesizing cycloalkane rings from C7 upward was the result of the angle strain that would be expected if the large rings were regular planar polygons (see Table 12-3). Baeyer also believed that cyclohexane had a planar structure like that shown in Figure 12-2, which would mean that the bond angles would have to deviate 10.5° from the tetrahedral value. However, in 1895, the then unknown chemist H. Sachse suggested that cyclohexane exists in the strain-free chair and boat forms discussed in Section 12-3. This suggestion was not accepted at the time because it led to the prediction of several possible isomers for compounds such as chlorocyclohexane (cf. Exercise 12-4). The idea that such isomers might act as a single substance, as the result of rapid equilibration, seemed like a needless complication, and it was not until 1918 that E. Mohr proposed a definitive way to distinguish between the Baeyer and Sachse cyclohexanes. As will be discussed in Section 12-9, the result, now known as the Sachse-Mohr theory, was complete confirmation of the idea of nonplanar large rings.
Table 12-3: Strain in Cycloalkane Rings and Heats of Combustion of Cycloalkanes
Compound n Angle Strain at each CH2 Heat of Combustion ΔHo (kcal/mol) Heat of Combustion ΔHo per CH2/N (kcal/mol) Total Strain (kcal/mol)
ethene 2 109.5 337.2 168.6 22.4
cyclopropane 3 49.5 499.9 166.6 27.7
cyclobutane 4 19.5 655.9 164.0 26.3
cyclopentane 5 1.5 793.4 158.7 6.5
cyclohexane 6 10.5 944.8 157.5 0.4
cycloheptane 7 19.1 1108.1 158.4 6.3
cyclooctane 8 25.5 1268.9 158.6 9.7
cyclononane 9 30.5 1429.5 158.8 12.9
cyclodecane 10 34.5 1586.1 158.6 12.1
cyclopentadecane 15 46.5 2362.5 157.5 1.5
open chain alkane 157.4 -
One of the most interesting developments in ‘the stereochemistry of organic compounds in recent years has been the demonstration that transcyclooctene (but not the cis isomer) can be resolved into stable chiral isomers (enantiomers, Section 5-IB). In general, a trans-cycloalkene would not be expected to be resolvable because of the possibility for formation of achiral conformations with a plane of symmetry. Any conformation with all of the carbons in a plane is such an achiral conformation (Figure 12-20a). However, when the chain connecting the ends of the double bond is short, as in trans-cyclooctene, steric hindrance and steric strain prevent easy.
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.10%3A_An_Introduction_to_Cycloalkanes.txt |
Although the customary line drawings of simple cycloalkanes are geometrical polygons, the actual shape of these compounds in most cases is very different.
Cyclopropane is necessarily planar (flat), with the carbon atoms at the corners of an equilateral triangle. The 60º bond angles are much smaller than the optimum 109.5º angles of a normal tetrahedral carbon atom, and the resulting angle strain dramatically influences the chemical behavior of this cycloalkane. Cyclopropane also suffers substantial eclipsing strain, since all the carbon-carbon bonds are fully eclipsed. Cyclobutane reduces some bond-eclipsing strain by folding (the out-of-plane dihedral angle is about 25º), but the total eclipsing and angle strain remains high. Cyclopentane has very little angle strain (the angles of a pentagon are 108º), but its eclipsing strain would be large (about 10 kcal/mol) if it remained planar. Consequently, the five-membered ring adopts non-planar puckered conformations whenever possible.
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring).
Conformations of Cyclohexane
A planar structure for cyclohexane is clearly improbable. The bond angles would necessarily be 120º, 10.5º larger than the ideal tetrahedral angle. Also, every carbon-carbon bond in such a structure would be eclipsed. The resulting angle and eclipsing strains would severely destabilize this structure. If two carbon atoms on opposite sides of the six-membered ring are lifted out of the plane of the ring, much of the angle strain can be eliminated.
This boat structure still has two eclipsed bonds and severe steric crowding of two hydrogen atoms on the "bow" and "stern" of the boat. This steric crowding is often called steric hindrance. By twisting the boat conformation, the steric hindrance can be partially relieved, but the twist-boat conformer still retains some of the strains that characterize the boat conformer. Finally, by lifting one carbon above the ring plane and the other below the plane, a relatively strain-free 'chair' conformer is formed. This is the predominant structure adopted by molecules of cyclohexane.
Investigations concerning the conformations of cyclohexane were initiated by H. Sachse (1890) and E. Mohr (1918), but it was not until 1950 that a full treatment of the manifold consequences of interconverting chair conformers and the different orientations of pendent bonds was elucidated by D. H. R. Barton (Nobel Prize 1969 together with O. Hassel). The following discussion presents some of the essential features of this conformational analysis.
On careful examination of a chair conformation of cyclohexane, we find that the twelve hydrogens are not structurally equivalent. Six of them are located about the periphery of the carbon ring, and are termed equatorial. The other six are oriented above and below the approximate plane of the ring (three in each location), and are termed axial because they are aligned parallel to the symmetry axis of the ring.
In the figure above, the equatorial hydrogens are colored blue, and the axial hydrogens are in bold. Since there are two equivalent chair conformations of cyclohexane in rapid equilibrium, all twelve hydrogens have 50% equatorial and 50% axial character. The figure below illustrates how to convert a molecular model of cyclohexane between two different chair conformations - this is something that you should practice with models. Notice that a 'ring flip' causes equatorial hydrogens to become axial, and vice-versa.
How to draw stereo bonds ("up" and "down" bonds)
There are various ways to show these orientations. The solid (dark) "up wedge" I used is certainly common. Some people use an analogous "down wedge", which is light, to indicate a down bond; unfortunately, there is no agreement as to which way the wedge should point, and you are left relying on the lightness of the wedge to know it is "down". The "down bond" avoids this wedge ambiguity, and just uses some kind of light line. The down bond I used (e.g., in Figure 5B) is a dashed line; IUPAC encourages a series of parallel lines, something like . What I did is a variation of what is recommended by IUPAC: http://www.chem.qmul.ac.uk/iupac/stereo/intro.html.
In ISIS/Draw, the "up wedge" and "down bond" that I used, along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the tool button for ordinary C-C bonds.
In Symyx Draw, the "up wedge" and "down bond", along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the "Chain" tool button.
ChemSketch provides up and down wedges, but not the simple up and down bonds discussed above. The wedges are available from the second toolbar across the top. For an expanded discussion of using these wedges, see the section of my ChemSketch Guide on Stereochemistry: Wedge bonds.
As always, the information provided on these pages in intended to help you get started. Each program has more options for drawing bonds than discussed here. When you feel the need, look around!
How to draw chairs
Most of the structures shown on this page were drawn with the free program ISIS/Draw. I have posted a guide to help you get started with ISIS/Draw. ISIS/Draw provides a simple cyclohexane (6-ring) hexagon template on the toolbar across the top. It provides templates for various 6-ring chair structures from the Templates menu; choose Rings. There are templates for simple chairs, without substituents (e.g., Fig 1B), and for chairs showing all the substituents (e.g., Fig 2B). In either case, you can add, delete, or change things as you wish. Various kinds of stereo bonds (wedges and bars) are available by clicking the left-side tool button that is just below the regular C-C single bond button. It may have a wedge shown on it, but this will vary depending on how it has been used. To choose a type of stereo bond, click on the button and hold the mouse click; a new menu will appear to the right of the button.
The free drawing program Symyx Draw, the successor to ISIS/Draw, provides similar templates and tools. A basic chair structure is provided on the default template bar that is shown. More options are available by choosing the Rings template. See my page Symyx Draw for a general guide for getting started with this program.
The free drawing program ChemSketch provides similar templates and tools. To find the special templates for chairs, go to the Templates menu, choose Template Window, and then choose "Rings" from the drop-down menu near upper left. See my page ChemSketch for a general guide for getting started with this program.
If you want to draw chair structures by hand (and if you are going on in organic chemistry, you should)... Be careful. The precise zigs and zags, and the angles of substituents are all important. Your textbook may offer you some hints for how to draw chairs. A short item in the Journal of Chemical Education offers a nice trick, showing how the chair can be thought of as consisting of an M and a W. The article is V Dragojlovic, A method for drawing the cyclohexane ring and its substituents. J Chem Educ 78:923, 7/01. (I thank M Farooq Wahab, Chemistry, Univ Karachi, for suggesting that this article be noted here.)
Aside from drawing the basic chair, the key points in adding substituents are:
• Axial groups alternate up and down, and are shown "vertical".
• Equatorial groups are approximately horizontal, but actually somewhat distorted from that, so that the angle from the axial group is a bit more than a right angle -- reflecting the common 109 degree bond angle.
• As cautioned before, it is usually easier to draw and see what is happening at the four corners of the chair than at the two middle positions. Try to use the corners as much as possible.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.11%3A_Cyclohexane.txt |
Because axial bonds are parallel to each other, substituents larger than hydrogen generally suffer greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which the larger substituents assume equatorial orientation.
When the methyl group in the structure above occupies an axial position it suffers steric crowding by the two axial hydrogens located on the same side of the ring.
The conformation in which the methyl group is equatorial is more stable, and thus the equilibrium lies in this direction.
The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane may be determined by the conformational equilibrium of the compound. The corresponding equilibrium constant is related to the energy difference between the conformers, and collecting such data allows us to evaluate the relative tendency of substituents to exist in an equatorial or axial location.A table of these free energy values (sometimes referred to as A values) may be examined by clicking here.
Looking at the energy values in this table, it is clear that the apparent "size" of a substituent (in terms of its preference for equatorial over axial orientation) is influenced by its width and bond length to cyclohexane, as evidenced by the fact that an axial vinyl group is less hindered than ethyl, and iodine slightly less than chlorine.
We noted earlier that cycloalkanes having two or more substituents on different ring carbon atoms exist as a pair (sometimes more) of configurational stereoisomers. Now we must examine the way in which favorable ring conformations influence the properties of the configurational isomers. Remember, configurational stereoisomers are stable and do not easily interconvert, whereas, conformational isomers normally interconvert rapidly. In examining possible structures for substituted cyclohexanes, it is useful to follow two principles:
(i) Chair conformations are generally more stable than other possibilities.
(ii) Substituents on chair conformers prefer to occupy equatorial positions due to the increased steric hindrance of axial locations.
A Selection of AG° Values for the Change from Axial to Equatorial Orientation of Substituents for Monosubstituted Cyclohexanes
Substituent \(-\Delta{G}^o\) kcal/mol Substituent \(-\Delta{G}^o\) kcal/mol
\(\ce{CH_3\bond{-}}\) 1.7 \(\ce{O_2N\bond{-}}\) 1.1
\(\ce{CH_2H_5\bond{-}}\) 1.8 \(\ce{N#C\bond{-}}\) 0.2
\(\ce{(CH_3)_2CH\bond{-}}\) 2.2 \(\ce{CH_3O\bond{-}}\) 0.5
\(\ce{(CH_3)_3C\bond{-}}\) \(\geq 5.0\) \(\ce{HO_2C\bond{-}}\) 0.7
\(\ce{F\bond{-}}\) 0.3 \(\ce{H_2C=CH\bond{-}}\) 1.3
\(\ce{Cl\bond{-}}\) 0.5 \(\ce{C_6H_5\bond{-}}\) 3.0
\(\ce{Br\bond{-}}\) 0.5
\(\ce{I\bond{-}}\) 0.5
Chlorocyclohexane
This is an example of the next level of complexity, a mono-substituted cycloalkane. See Fig 3.
Figure 3
So what is new here? Not much, with the hexagon formula, Fig 3A. That type of formula shows the basic "connectivity" of the atoms -- who is connected to whom. This chemical has one Cl on the ring, and it does not matter where we show it. There is now only one H on that C, but since we are not showing H explicitly here, that is not an issue in drawing the structure. (It is an issue when you look at it and want to count H.)
With the chair formula (Fig 3B), which shows information not only about connectivity but also about conformation, there is important new information here. In a chair, there are two "types" of substituents: those pointing up or down, and called axial, and those pointing "outward", and called equatorial. I have shown the chlorine atom in an equatorial position. Why? Two reasons: it is what we would predict, and it is what is found. Why do we predict that the Cl is equatorial? Because it is bigger than H, and there is more room in the equatorial positions.
Helpful Hints...
If possible, examine a physical model of cyclohexane and chlorocyclohexane, so that you can see the axial and equatorial positions. Common ball and stick models are fine for this. It should be easy to see that the three axial H on one side can get very near each other.
If you do not have access to physical models, examining computer models can also be useful. See my RasMol page for a program and source of structures for this.
When putting substituents on chair structures, I encourage you to use the four corner positions of the chair as much as possible. It is easier to see the axial and equatorial relationship at the corners.
In Fig 3B I have shown the H atom that is on the same carbon as the Cl atom. This is perhaps not necessary, since the correct number of H atoms is understood, by counting bonds on C. But showing the H explicitly at key C atoms helps to make the structure clearer. This may be particularly important with hand-drawn structures. I often see structures where I am not sure whether a particular atom is shown axial or equatorial. But if both atoms at the position (the H as well as the Cl) are shown, then hopefully it becomes clearer which is which. I also encourage students who are not sure of their art work to annotate their drawing. Say what you mean. That allows me to distinguish whether you are unsure which direction things point or simply unsure how to draw them.)
Again, a reminder... For notes on how to draw chairs (by hand or using a drawing program), see the section E.2. Note: How to draw chairs.
Dichlorocyclohexanes: an introduction
The next level of complexity is a di-substituted cycloalkane, "dichlorocyclohexane". The first question we must ask is which C the two chlorine substituents are on. For now, I want to discuss 1,3- dichlorocyclohexane. This introduces another issue: are the two Cl on the same side of the ring, or on opposite sides? We call these "cis" (same side) and "trans" (opposite sides). We focus on one of these, cis-1,3-dichlorocyclohexane. And for now, we will just look at hexagon structural formulas, leaving the question of conformation for later. Let's go through this one step at a time.
1,3-Dichlorocyclohexane
Our first attempt to draw 1,3-dichlorocyclohexane might look something like Fig 4.
Figure 4
The structure in Fig 4 is indeed a dichlorocyclohexane. It is even a 1,3-dichlorocyclohexane. However, this structure provides no information about the orientation of the two Cl atoms relative to the plane of the ring. To show a specific isomer -- cis or trans -- we must somehow show how the two Cl atoms are oriented relative to the plane of the ring.
cis-1,3-Dichlorocyclohexane
Fig 5 shows two common ways to show how the substituents are oriented relative to the plane of the ring. The compound shown here is cis-1,3-dichlorocyclohexane.
Figure 5
The basic idea in both of these is that we can imagine the ring to be planar, and then show the groups above or below the plane of the ring. How we show this is different in the two parts of Fig 5. In Fig 5A, we look at the ring "edge-on". The thick line for the bottom bond is intended to convey the edge-on view (or side view); this is sometimes omitted, especially with hand-drawn structures, but be careful then that the meaning is clear. Once we understand that we are now looking at the ring edge-on, it is clear that the two Cl atoms are both above the ring, hence cis. In Fig 5B, we view the ring "face-on" (or top view), and use special bond symbols -- "stereo bonds" -- to convey up and down: the heavy wedge -- an "up wedge" -- points upward, toward you, and the dashed bond -- a "down bond" -- points downward, away from you. Again, both Cl are "up", hence cis.
Notes...
For some notes on how to draw the stereo bonds, see the section E.1. Note: How to draw stereo bonds ("up" and "down" bonds).
In discussing Fig 5, I started by saying that we imagine the ring to be planar. Emphasize that cycloalkane rings are not really planar (except for cyclopropane rings). As so often, the structural formula represents the general layout of the atoms, but not the actual molecular geometry.
Those with the Ouellette book can see examples of these two ways of showing up/down on p 81 (top) and p 80 (middle). Most organic chemistry books will show you this.
The conformation of cis-1,3-dichlorocyclohexane
Fig 6, at the right, is one way to show this. Fig 6 shows features of the compound that we have already noted -- plus one more. Let's go through these features, emphasizing what is new.
Figure 6
The structure shows cis-1,3-dichlorocyclohexane: a 6-ring; 2 Cl atoms, at positions 1 and 3; and cis, with both Cl on the same side of -- above -- the H that is on the same C.
I showed the 2 Cl atoms at corner positions, and I showed the H at the key positions explicitly. These points follow from some of the Helpful Hints discussed earlier. It is not required that you do these things, but they can make things easier for you -- and for anyone reading your structures.
Now, what is new here? The conformation. We start with the notion that the conformation of cyclohexane derivatives is based on the "chair". At each position, one substituent is axial (loosely, perpendicular to the ring), and one is equatorial (loosely, in the plane of the ring). There is more room in the equatorial positions (not easily seen with these simple drawings, but ordinary ball and stick models do help with this point). Thus we try to put the larger substituents in the equatorial positions. In this case, we put the Cl equatorial and the H axial at each position 1 and position 3.
We are now done with this compound, cis-1,3-dichlorocyclohexane. However, we have missed one very important concern -- because it is not an issue in this case. So let's look at another compound.
cis-1,2-Dichlorocyclohexane
cis-1,2-Dichlorocyclohexane is like the previous compound, except that the two chloro groups are now at 1,2, rather than at 1,3.
Fig 7 shows a simple structural formula for 1,2-dichlorocyclohexane, without orientation. This is analogous to Fig 4 above for the 1,3 isomer.
Figure 7
Figure 8
Fig 8 shows two ways to show the cis orientation in cis-1,2-dichlorocyclohexane. This Fig is analogous to Fig 5 above for the 1,3 isomer.
Figures 7 and 8 above introduce no new ideas or complications. These two figures should be straightforward.
So, what is the preferred conformation of cis-1,2-dichlorocyclohexane? This requires careful consideration; an important lesson from this exercise is to realize that we cannot propose a good conformation based simply on what we have learned so far.
The two guidelines we have so far for conformation of 6-rings are:
• The carbon ring is in a chair.
• Larger substituents are in equatorial positions.
Let's explore the difficulty here by looking at some things people might naively draw.
Figure 9
Fig 9 shows an attempt to draw a chair conformation of cis-1,2-dichlorocyclohexane. It satisfies both of the guidelines listed above. But it is wrong.
Why is Fig 9 wrong? It is the wrong chemical. The structure shown in Fig 9 is trans, not cis. Look carefully at the 1 and 2 positions. At one of them, the H is above the Cl; at the other, the Cl is above the H. Trans. Wrong chemical. The structure shown in Fig 9 is not the requested chemical.
Figure 10
Fig 10 attempts to fix the problem with Fig 9. But it is also wrong.
Why is Fig 10 wrong? After all, it seems to address the criteria presented. It contains both of the larger atoms (Cl) equatorial, and they are cis as desired. However, in Fig 10, the two axial groups on carbons # 1 and 2 (the two H that are shown) are both pointing up. This is impossible. In a valid chair, the axial groups alternate up/down as one goes around the ring; see Figure 2B, above. This follows from the tetrahedral bonding of C. Adjacent axial groups, as relevant here, must point in opposite directions; that condition is violated here. That is, Fig 10 is not a valid chair.
Those who find the above point new or surprising should check their textbook. If possible, look at models of cyclohexane and simple derivatives such as the one here. Figure 2B, above, is correct. In my experience, many students have not yet noticed this feature of chair conformations.
If you think you have an alternative that is better (or even satisfactory), please show it to me. I suspect it will turn out to be equivalent to Fig 9 or Fig 10, above. But if you think it is good, let's discuss it.
So now what? We have a contradiction. And that really is the most important point here. It is important to realize that we cannot draw a conformation for cis-1,2-dichlorocyclohexane which easily fits the criteria we have used so far: a chair, with large groups equatorial. So what do we do? Clearly, the conformation of this compound must, in some way, involve more issues than what we have considered so far.
Instructors and books will vary in how much further explanation they want to give on this matter. Therefore, how you proceed from here must take into account the preferences in your course. Here is one way to proceed.
One simple way to proceed is to re-examine the two criteria we have been using, and then state a generality about what to do in the event of a conflict. That generality is: use the chair, and then fit the groups as best you can. That is, try to put as many of the larger groups equatorial as you can, but realize that you may not get them all equatorial.
Figure 11
Fig 11 shows a plausible chair conformation of cis-1,2-dichlorocyclohexane.
What have we accomplished here? First, this is the correct compound. Convince yourself that this really is cis-1,2-dichlorocyclohexane. In particular, it is cis because at each substituted position the Cl is "above" the H; that is, both Cl are on the same side of the ring.
Second, this is a proper chair. Adjacent axial groups point in opposite directions. Only two of them are shown here: the axial Cl on the upper right C is "up", and the axial H on the lower right C is "down". Thus, these aspects satisfy the proper way to orient things, in general, on a cyclohexane chair.
How good a conformation is the one shown in Fig 11? Actually, it is quite good, in this case. The actual conformation has been measured, and it follows the basic ideas shown here.
Is this the end of the story? No, but it is about enough for now. The main purpose here was to show how one must carefully look at conformation, within the constraints of the specific isomer one is trying to draw. Some compounds cannot be easily drawn within the common "rules". cis-1,2-dichlorocyclohexane is one such example. In this case, we kept the basic chair conformation, but put one larger group in the less favored axial orientation. Measurements on many such chemicals have shown that the energetic penalty of moving a cyclohexane ring much away from the basic chair conformation is quite large -- certainly larger than the energetic penalty of putting one "somewhat large" group in an axial position. Of course, with larger groups or more groups, this might not hold.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.12%3A_Substituted_Cycloalkanes.txt |
Oxidation and Reduction of Alkanes
You are undoubtedly already familiar with the general idea of oxidation and reduction: you learned in general chemistry that when a compound or atom is oxidized it loses electrons, and when it is reduced it gains electrons. You also know that oxidation and reduction reactions occur in pairs: if one species is oxidized, another must be reduced at the same time - thus the term 'redox reaction'.
Most of the redox reactions you have seen previously in general chemistry probably involved the flow of electrons from one metal to another, such as the reaction between copper ion in solution and metallic zinc:
Cu+2(aq) + Zn(s) → Cu(s) + Zn+2(aq)
In organic chemistry, redox reactions look a little different. Electrons in an organic redox reaction often are transferred in the form of a hydride ion - a proton and two electrons. Because they occur in conjunction with the transfer of a proton, these are commonly referred to as hydrogenation and dehydrogenation reactions: a hydride plus a proton adds up to a hydrogen (H2) molecule. Be careful - do not confuse the terms hydrogenation and dehydrogenation with hydration and dehydration - the latter refer to the gain and loss of a water molecule (and are not redox reactions), while the former refer to the gain and loss of a hydrogen molecule.
When a carbon atom in an organic compound loses a bond to hydrogen and gains a new bond to a heteroatom (or to another carbon), we say the compound has been dehydrogenated, or oxidized. A very common biochemical example is the oxidation of an alcohol to a ketone or aldehyde:
When a carbon atom loses a bond to hydrogen and gains a bond to a heteroatom (or to another carbon atom), it is considered to be an oxidative process because hydrogen, of all the elements, is the least electronegative. Thus, in the process of dehydrogenation the carbon atom undergoes an overall loss of electron density - and loss of electrons is oxidation.
Conversely, when a carbon atom in an organic compound gains a bond to hydrogen and loses a bond to a heteroatom (or to another carbon atom), we say that the compound has been hydrogenated, or reduced. The hydrogenation of a ketone to an alcohol, for example, is overall the reverse of the alcohol dehydrogenation shown above. Illustrated below is another common possibility, the hydrogenation (reduction) of an alkene to an alkane.
Hydrogenation results in higher electron density on a carbon atom(s), and thus we consider process to be one of reduction of the organic molecule.
Notice that neither hydrogenation nor dehydrogenation involves the gain or loss of an oxygen atom. Reactions which do involve gain or loss of one or more oxygen atoms are usually referred to as 'oxygenase' and 'reductase' reactions, and are the subject of section 16.10 and section 17.3.
For the most part, when talking about redox reactions in organic chemistry we are dealing with a small set of very recognizable functional group transformations. It is therefore very worthwhile to become familiar with the idea of 'oxidation states' as applied to organic functional groups. By comparing the relative number of bonds to hydrogen atoms, we can order the familiar functional groups according to oxidation state. We'll take a series of single carbon compounds as an example. Methane, with four carbon-hydrogen bonds, is highly reduced. Next in the series is methanol (one less carbon-hydrogen bond, one more carbon-oxygen bond), followed by formaldehyde, formate, and finally carbon dioxide at the highly oxidized end of the group.
This pattern holds true for the relevant functional groups on organic molecules with two or more carbon atoms:
Alkanes are highly reduced, while alcohols - as well as alkenes, ethers, amines, sulfides, and phosphate esters - are one step up on the oxidation scale, followed by aldehydes/ketones/imines and epoxides, and finally by carboxylic acid derivatives (carbon dioxide, at the top of the oxidation list, is specific to the single carbon series).
Notice that in the series of two-carbon compounds above, ethanol and ethene are considered to be in the same oxidation state. You know already that alcohols and alkenes are interconverted by way of addition or elimination of water (section 14.1). When an alcohol is dehydrated to form an alkene, one of the two carbons loses a C-H bond and gains a C-C bond, and thus is oxidized. However, the other carbon loses a C-O bond and gains a C-C bond, and thus is considered to be reduced. Overall, therefore, there is no change to the oxidation state of the molecule.
You should learn to recognize when a reaction involves a change in oxidation state in an organic reactant . Looking at the following transformation, for example, you should be able to quickly recognize that it is an oxidation: an alcohol functional group is converted to a ketone, which is one step up on the oxidation ladder.
Likewise, this next reaction involves the transformation of a carboxylic acid derivative (a thioester) first to an aldehyde, then to an alcohol: this is a double reduction, as the substrate loses two bonds to heteroatoms and gains two bonds to hydrogens.
An acyl transfer reaction (for example the conversion of an acyl phosphate to an amide) is not considered to be a redox reaction - the oxidation state of the organic molecule is does not change as substrate is converted to product, because a bond to one heteroatom (oxygen) has simply been traded for a bond to another heteroatom (nitrogen).
It is important to be able to recognize when an organic molecule is being oxidized or reduced, because this information tells you to look for the participation of a corresponding redox agent that is being reduced or oxidized- remember, oxidation and reduction always occur in tandem! We will soon learn in detail about the most important biochemical and laboratory redox agents.
Template:ExampleStart
Exercise 16.1: is an aldol condensation a redox reaction? Explain.
Exercise 16.2: Is the reaction catalyzed by squalene epoxidase a redox reaction? How about squalene cyclase? Explain.
Template:ExampleEnd
he combustion of carbon compounds, especially hydrocarbons, has been the most important source of heat energy for human civilizations throughout recorded history. The practical importance of this reaction cannot be denied, but the massive and uncontrolled chemical changes that take place in combustion make it difficult to deduce mechanistic paths. Using the combustion of propane as an example, we see from the following equation that every covalent bond in the reactants has been broken and an entirely new set of covalent bonds have formed in the products. No other common reaction involves such a profound and pervasive change, and the mechanism of combustion is so complex that chemists are just beginning to explore and understand some of its elementary features.
$\ce{CH3 - CH2 - CH3} + 5 \ce{O2} \rightarrow 3 \ce{CO2} + 4\ce{H2O} + \text{heat}$
Two points concerning this reaction are important:
1. Since all the covalent bonds in the reactant molecules are broken, the quantity of heat evolved in this reaction is related to the strength of these bonds (and, of course, the strength of the bonds formed in the products). Precise heats of combustion measurements can provide useful information about the structure of molecules.
2. The stoichiometry of the reactants is important. If insufficient oxygen is supplied some of the products will consist of the less oxidized carbon monoxide $\ce{CO}$ gas.
$\ce{CH3 - CH2 - CH3} + 4 \ce{O2} \rightarrow \ce{CO2} + 2 \ce{CO} + 4\ce{H2O} + \text{heat}$
Heat of Combustion
From the previous discussion, we might expect isomers to have identical heats of combustion. However, a few simple measurements will disabuse this belief. Thus, the heat of combustion of pentane is –782 kcal/mole, but that of its 2,2-dimethylpropane (neopentane) isomer is –777 kcal/mole. Differences such as this reflect subtle structural variations, including the greater bond energy of 1º-C–H versus 2º-C–H bonds and steric crowding of neighboring groups. In small-ring cyclic compounds ring strain can be a major contributor to thermodynamic stability and chemical reactivity. The following table lists heat of combustion data for some simple cycloalkanes and compares these with the increase per CH2 unit for long chain alkanes.
Cycloalkane
(CH2)n
CH2 Units
n
ΔH25º
kcal/mole
ΔH25º
per CH2 Unit
Ring Strain
kcal/mole
Cyclopropane n = 3 468.7 156.2 27.6
Cyclobutane n = 4 614.3 153.6 26.4
Cyclopentane n = 5 741.5 148.3 6.5
Cyclohexane n = 6 882.1 147.0 0.0
Cycloheptane n = 7 1035.4 147.9 6.3
Cyclooctane n = 8 1186.0 148.2 9.6
Cyclononane n = 9 1335.0 148.3 11.7
Cyclodecane n = 10 1481 148.1 11.0
CH3(CH2)mCH3 m = large 147.0 0.0
The chief source of ring strain in smaller rings is angle strain and eclipsing strain. As noted elsewhere, cyclopropane and cyclobutane have large contributions of both strains, with angle strain being especially severe. Changes in chemical reactivity as a consequence of angle strain are dramatic in the case of cyclopropane, and are also evident for cyclobutane. Some examples are shown in the following diagram. The cyclopropane reactions are additions, many of which are initiated by electrophilic attack. The pyrolytic conversion of β-pinene to myrcene probably takes place by an initial rupture of the 1:6 bond, giving an allylic 3º-diradical, followed immediately by breaking of the 5:7 bond.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.13%3A_Oxidation_of_Alkanes.txt |
Waxes
Waxes are esters of fatty acids with long chain monohydric alcohols (one hydroxyl group). Natural waxes are often mixtures of such esters, and may also contain hydrocarbons. The formulas for three well known waxes are given below, with the carboxylic acid moiety colored red and the alcohol colored blue.
spermaceti
beeswax
carnuba wax
CH3(CH2)14CO2-(CH2)15CH3
CH3(CH2)24CO2-(CH2)29CH3
CH3(CH2)30CO2-(CH2)33CH3
Waxes are widely distributed in nature. The leaves and fruits of many plants have waxy coatings, which may protect them from dehydration and small predators. The feathers of birds and the fur of some animals have similar coatings which serve as a water repellent. Carnuba wax is valued for its toughness and water resistance.
Prostaglandins Thromboxanes & Leukotrienes
The members of this group of structurally related natural hormones have an extraordinary range of biological effects. They can lower gastric secretions, stimulate uterine contractions, lower blood pressure, influence blood clotting and induce asthma-like allergic responses. Because their genesis in body tissues is tied to the metabolism of the essential fatty acid arachadonic acid (5,8,11,14-eicosatetraenoic acid) they are classified as eicosanoids. Many properties of the common drug aspirin result from its effect on the cascade of reactions associated with these hormones.
The metabolic pathways by which arachidonic acid is converted to the various eicosanoids are complex and will not be discussed here. A rough outline of some of the transformations that take place is provided below. It is helpful to view arachadonic acid in the coiled conformation shown in the shaded box.
Leukotriene A is a precursor to other leukotriene derivatives by epoxide opening reactions. The prostaglandins are given systematic names that reflect their structure. The initially formed peroxide PGH2 is a common intermediate to other prostaglandins, as well as thromboxanes such as TXA2. To see a model of prostaglandin PGE2 Click Here.
Steroids
The important class of lipids called steroids are actually metabolic derivatives of terpenes, but they are customarily treated as a separate group. Steroids may be recognized by their tetracyclic skeleton, consisting of three fused six-membered and one five-membered ring, as shown in the diagram to the right. The four rings are designated A, B, C & D as noted, and the peculiar numbering of the ring carbon atoms (shown in red) is the result of an earlier misassignment of the structure. The substituents designated by R are often alkyl groups, but may also have functionality. The R group at the A:B ring fusion is most commonly methyl or hydrogen, that at the C:D fusion is usually methyl. The substituent at C-17 varies considerably, and is usually larger than methyl if it is not a functional group. The most common locations of functional groups are C-3, C-4, C-7, C-11, C-12 & C-17. Ring A is sometimes aromatic. Since a number of tetracyclic triterpenes also have this tetracyclic structure, it cannot be considered a unique identifier.
Steroids are widely distributed in animals, where they are associated with a number of physiological processes. Examples of some important steroids are shown in the following diagram. Different kinds of steroids will be displayed by clicking the "Toggle Structures" button under the diagram. Norethindrone is a synthetic steroid, all the other examples occur naturally. A common strategy in pharmaceutical chemistry is to take a natural compound, having certain desired biological properties together with undesired side effects, and to modify its structure to enhance the desired characteristics and diminish the undesired. This is sometimes accomplished by trial and error.
The generic steroid structure drawn above has seven chiral stereocenters (carbons 5, 8, 9, 10, 13, 14 & 17), which means that it may have as many as 128 stereoisomers. With the exception of C-5, natural steroids generally have a single common configuration. This is shown in the last of the toggled displays, along with the preferred conformations of the rings.
Chemical studies of the steroids were very important to our present understanding of the configurations and conformations of six-membered rings. Substituent groups at different sites on the tetracyclic skeleton will have axial or equatorial orientations that are fixed because of the rigid structure of the trans-fused rings. This fixed orientation influences chemical reactivity, largely due to the greater steric hindrance of axial groups versus their equatorial isomers. Thus an equatorial hydroxyl group is esterified more rapidly than its axial isomer.
To see a model of the steroid cholesterol Click Here | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.14%3A_LipidsPart_1.txt |
Conformational isomerism involves rotation about sigma bonds, and does not involve any differences in the connectivity or geometry of bonding. Two or more structures that are categorized as conformational isomers, or conformers, are really just two of the exact same molecule that differ only in terms of the angle about one or more sigma bonds.
Ethane Conformations
Although there are seven sigma bonds in the ethane molecule, rotation about the six carbon-hydrogen bonds does not result in any change in the shape of the molecule because the hydrogen atoms are essentially spherical. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations.
In order to better visualize these different conformations, it is convenient to use a drawing convention called the Newman projection. In a Newman projection, we look lengthwise down a specific bond of interest – in this case, the carbon-carbon bond in ethane. We depict the ‘front’ atom as a dot, and the ‘back’ atom as a larger circle.
The six carbon-hydrogen bonds are shown as solid lines protruding from the two carbons at 120°angles, which is what the actual tetrahedral geometry looks like when viewed from this perspective and flattened into two dimensions.
The lowest energy conformation of ethane, shown in the figure above, is called the ‘staggered’ conformation, in which all of the C-H bonds on the front carbon are positioned at dihedral angles of 60°relative to the C-H bonds on the back carbon. In this conformation, the distance between the bonds (and the electrons in them) is maximized.
If we now rotate the front CH3 group 60°clockwise, the molecule is in the highest energy ‘eclipsed' conformation, where the hydrogens on the front carbon are as close as possible to the hydrogens on the back carbon.
This is the highest energy conformation because of unfavorable interactions between the electrons in the front and back C-H bonds. The energy of the eclipsed conformation is approximately 3 kcal/mol higher than that of the staggered conformation.
Another 60°rotation returns the molecule to a second eclipsed conformation. This process can be continued all around the 360°circle, with three possible eclipsed conformations and three staggered conformations, in addition to an infinite number of variations in between.
The carbon-carbon bond is not completely free to rotate – there is indeed a small, 3 kcal/mol barrier to rotation that must be overcome for the bond to rotate from one staggered conformation to another. This rotational barrier is not high enough to prevent constant rotation except at extremely cold temperatures. However, at any given moment the molecule is more likely to be in a staggered conformation - one of the rotational ‘energy valleys’ - than in any other state.
The potential energy associated with the various conformations of ethane varies with the dihedral angle of the bonds, as shown below.
Although the conformers of ethane are in rapid equilibrium with each other, the 3 kcal/mol energy difference leads to a substantial preponderance of staggered conformers (> 99.9%) at any given time.
The animation below illustrates the relationship between ethane's potential energy and its dihedral angle
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/04%3A_Alkanes/4.15%3A_Conformations_of_Acyclic_AlkanesEthane.txt |
The polysaccharides are the most abundant carbohydrates in nature and serve a variety of functions, such as energy storage or as components of plant cell walls. Polysaccharides are very large polymers composed of tens to thousands of monosaccharides joined together by glycosidic linkages. The three most abundant polysaccharides are starch, glycogen, and cellulose. These three are referred to as homopolymers because each yields only one type of monosaccharide (glucose) after complete hydrolysis. Heteropolymers may contain sugar acids, amino sugars, or noncarbohydrate substances in addition to monosaccharides. Heteropolymers are common in nature (gums, pectins, and other substances) but will not be discussed further in this textbook. The polysaccharides are nonreducing carbohydrates, are not sweet tasting, and do not undergo mutarotation.
Starch
Starch is the most important source of carbohydrates in the human diet and accounts for more than 50% of our carbohydrate intake. It occurs in plants in the form of granules, and these are particularly abundant in seeds (especially the cereal grains) and tubers, where they serve as a storage form of carbohydrates. The breakdown of starch to glucose nourishes the plant during periods of reduced photosynthetic activity. We often think of potatoes as a “starchy” food, yet other plants contain a much greater percentage of starch (potatoes 15%, wheat 55%, corn 65%, and rice 75%). Commercial starch is a white powder.
Starch is a mixture of two polymers: amylose and amylopectin. Natural starches consist of about 10%–30% amylase and 70%–90% amylopectin. Amylose is a linear polysaccharide composed entirely of D-glucose units joined by the α-1,4-glycosidic linkages we saw in maltose (part (a) of Figure 5.1.1). Experimental evidence indicates that amylose is not a straight chain of glucose units but instead is coiled like a spring, with six glucose monomers per turn (part (b) of Figure 5.1.1). When coiled in this fashion, amylose has just enough room in its core to accommodate an iodine molecule. The characteristic blue-violet color that appears when starch is treated with iodine is due to the formation of the amylose-iodine complex. This color test is sensitive enough to detect even minute amounts of starch in solution.
Figure 5.1.1: Amylose. (a) Amylose is a linear chain of α-D-glucose units joined together by α-1,4-glycosidic bonds. (b) Because of hydrogen bonding, amylose acquires a spiral structure that contains six glucose units per turn.
Amylopectin is a branched-chain polysaccharide composed of glucose units linked primarily by α-1,4-glycosidic bonds but with occasional α-1,6-glycosidic bonds, which are responsible for the branching. A molecule of amylopectin may contain many thousands of glucose units with branch points occurring about every 25–30 units (Figure 5.1.2). The helical structure of amylopectin is disrupted by the branching of the chain, so instead of the deep blue-violet color amylose gives with iodine, amylopectin produces a less intense reddish brown.
Figure 5.1.2: Representation of the Branching in Amylopectin and Glycogen. Both amylopectin and glycogen contain branch points that are linked through α-1,6-linkages. These branch points occur more often in glycogen.
Dextrins are glucose polysaccharides of intermediate size. The shine and stiffness imparted to clothing by starch are due to the presence of dextrins formed when clothing is ironed. Because of their characteristic stickiness with wetting, dextrins are used as adhesives on stamps, envelopes, and labels; as binders to hold pills and tablets together; and as pastes. Dextrins are more easily digested than starch and are therefore used extensively in the commercial preparation of infant foods.
The complete hydrolysis of starch yields, in successive stages, glucose:
starch → dextrins → maltose → glucose
In the human body, several enzymes known collectively as amylases degrade starch sequentially into usable glucose units.
Glycogen
Glycogen is the energy reserve carbohydrate of animals. Practically all mammalian cells contain some stored carbohydrates in the form of glycogen, but it is especially abundant in the liver (4%–8% by weight of tissue) and in skeletal muscle cells (0.5%–1.0%). Like starch in plants, glycogen is found as granules in liver and muscle cells. When fasting, animals draw on these glycogen reserves during the first day without food to obtain the glucose needed to maintain metabolic balance.
Note
About 70% of the total glycogen in the body is stored in muscle cells. Although the percentage of glycogen (by weight) is higher in the liver, the much greater mass of skeletal muscle stores a greater total amount of glycogen.
Glycogen is structurally quite similar to amylopectin, although glycogen is more highly branched (8–12 glucose units between branches) and the branches are shorter. When treated with iodine, glycogen gives a reddish brown color. Glycogen can be broken down into its D-glucose subunits by acid hydrolysis or by the same enzymes that catalyze the breakdown of starch. In animals, the enzyme phosphorylase catalyzes the breakdown of glycogen to phosphate esters of glucose.
Cellulose
Cellulose, a fibrous carbohydrate found in all plants, is the structural component of plant cell walls. Because the earth is covered with vegetation, cellulose is the most abundant of all carbohydrates, accounting for over 50% of all the carbon found in the vegetable kingdom. Cotton fibrils and filter paper are almost entirely cellulose (about 95%), wood is about 50% cellulose, and the dry weight of leaves is about 10%–20% cellulose. The largest use of cellulose is in the manufacture of paper and paper products. Although the use of noncellulose synthetic fibers is increasing, rayon (made from cellulose) and cotton still account for over 70% of textile production.
Like amylose, cellulose is a linear polymer of glucose. It differs, however, in that the glucose units are joined by β-1,4-glycosidic linkages, producing a more extended structure than amylose (part (a) of Figure 5.1.3). This extreme linearity allows a great deal of hydrogen bonding between OH groups on adjacent chains, causing them to pack closely into fibers (part (b) of Figure 5.1.3). As a result, cellulose exhibits little interaction with water or any other solvent. Cotton and wood, for example, are completely insoluble in water and have considerable mechanical strength. Because cellulose does not have a helical structure, it does not bind to iodine to form a colored product.
Figure 5.1.3: Cellulose. (a) There is extensive hydrogen bonding in the structure of cellulose. (b) In this electron micrograph of the cell wall of an alga, the wall consists of successive layers of cellulose fibers in parallel arrangement.
Cellulose yields D-glucose after complete acid hydrolysis, yet humans are unable to metabolize cellulose as a source of glucose. Our digestive juices lack enzymes that can hydrolyze the β-glycosidic linkages found in cellulose, so although we can eat potatoes, we cannot eat grass. However, certain microorganisms can digest cellulose because they make the enzyme cellulase, which catalyzes the hydrolysis of cellulose. The presence of these microorganisms in the digestive tracts of herbivorous animals (such as cows, horses, and sheep) allows these animals to degrade the cellulose from plant material into glucose for energy. Termites also contain cellulase-secreting microorganisms and thus can subsist on a wood diet. This example once again demonstrates the extreme stereospecificity of biochemical processes. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.01%3A_Starch_and_Cellulose.txt |
Isomers are different compounds that have the same molecular formula. When the group of atoms that make up the molecules of different isomers are bonded together in fundamentally different ways, we refer to such compounds as constitutional isomers. For example, in the case of the C4H8hydrocarbons, most of the isomers are constitutional. Shorthand structures for four of these isomers are shown below with their IUPAC names.
Note that the twelve atoms that make up these isomers are bonded in very different ways. As is true for all constitutional isomers, each different compound has a different IUPAC name. Furthermore, the molecular formula provides information about some of the structural features that must be present in the isomers. Since the formula C4H8 has two fewer hydrogens than the four-carbon alkane butane (C4H10), all the isomers having this composition must incorporate either a ring or a double bond. A fifth possible isomer of formula C4H8 is CH3CH=CHCH3. This would be named 2-butene according to the IUPAC rules; however, a close inspection of this molecule indicates it has two possible structures. These isomers may be isolated as distinct compounds, having characteristic and different properties. They are shown here with the designations cis and trans.
The bonding patterns of the atoms in these two isomers are essentially equivalent, the only difference being the relative configuration of the two methyl groups and the two associated hydrogen atoms about the double bond. In the cis isomer the methyl groups are on the same side; whereas they are on opposite sides in the trans isomer. Isomers that differ only in the spatial orientation of their component atoms are called stereoisomers. Stereoisomers always require that an additional nomenclature prefix be added to the IUPAC name in order to indicate their spatial orientation.
Stereoisomers are also observed in certain disubstituted (and higher substituted) cyclic compounds. Unlike the relatively flat molecules of alkenes, substituted cycloalkanes must be viewed as three-dimensional configurations in order to appreciate the spatial orientations of the substituents. By agreement, chemists use heavy, wedge-shaped bonds to indicate a substituent located above the average plane of the ring, and a hatched line for bonds to atoms or groups located below the ring. As in the case of the 2-butene stereoisomers, disubstituted cycloalkane stereoisomers may be designated by nomenclature prefixes such as cis and trans. The stereoisomeric 1,2-dibromocyclopentanes below are an example.
In general, if any two sp3 carbons in a ring have two different substituent groups (not counting other ring atoms) stereoisomerism is possible. This is similar to the substitution pattern that gives rise to stereoisomers in alkenes; indeed, one might view a double bond as a two-membered ring. Four other examples of this kind of stereoisomerism in cyclic compounds are shown below.
If more than two ring carbons have different substituents (not counting other ring atoms) the stereochemical notation distinguishing the various isomers becomes more complex. However, we can always state the relationship of any two substituents using cis or trans. For example, in the trisubstitutued cyclohexane below, we can say that the methyl group is cis to the ethyl group, and trans to the chlorine. We can also say that the ethyl group is trans to the chlorine. We cannot, however, designate the entire molecule as a cis or trans isomer.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.02%3A_The_Two_Major_Classes_of_Isomers.txt |
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existance of these molecules are determined by concept known as chirality. The word "chiral" was derived from the Greek word for hand, because our hands display a good example of chirality since they are non-superimposable mirror images of each other.
Introduction
The opposite of chiral is achiral. Achiral objects are superimposable with their mirror images. For example, two pieces of paper are achiral. In contrast, chiral molecules, like our hands, are non superimposable mirror images of each other.
Try to line up your left hand perfectly with your right hand, so that the palms are both facing in the same directions. Spend about a minute doing this. Do you see that they cannot line up exactly? The same thing applies to some molecules
A Chiral molecule has a mirror image that cannot line up with it perfectly- the mirror images are non superimposable. The mirror images are called enantiomers.
But why are chiral molecules so interesting? A chiral molecule and its enantiomer have the same chemical and physical properties(boiling point, melting point,polarity, density etc...). It turns out that many of our biological molecules such as our DNA, amino acids and sugars, are chiral molecules.
It is pretty interesting that our hands seem to serve the same purpose but most people are only able to use one of their hands to write. Similarily this is true with chiral biological molecules and interactions. Just like your left hand will not fit properly in your right glove, one of the enantiomers of a molecule may not work the same way in your body.
This must mean that enantiomers have properties that make them unique to their mirror images. One of these properties is that they cannot have a plane of symmetry or an internal mirror plane. So, a chiral molecule cannot be divided in two mirror image halves. Another property of chiral molecules is optical activity.
Organic compounds, molecules created around a chain of carbon atom (more commonly known as carbon backbone), play an essential role in the chemistry of life. These molecules derive their importance from the energy they carry, mainly in a form of potential energy between atomic molecules. Since such potential force can be widely affected due to changes in atomic placement, it is important to understand the concept of an isomer, a molecule sharing same atomic make up as another but differing in structural arrangements. This article will be devoted to a specific isomers called stereoisomers and its property of chirality (Figure 1).
Figure 1. Two enantiomers of a tetrahedral complex.
The concepts of steroisomerism and chirality command great deal of importance in modern organic chemistry, as these ideas helps to understand the physical and theoretical reasons behind the formation and structures of numerous organic molecules, the main reason behind the energy embedded in these essential chemicals. In contrast to more well-known constitutional isomerism, which develops isotopic compounds simply by different atomic connectivity, stereoisomerism generally maintains equal atomic connections and orders of building blocks as well as having same numbers of atoms and types of elements.
What, then, makes stereoisomers so unique? To answer this question, the learner must be able to think and imagine in not just two-dimensional images, but also three-dimensional space. This is due to the fact that stereoisomers are isomers because their atoms are different from others in terms of spatial arrangement.
Spatial Arrangement
First and foremost, one must understand the concept of spatial arrangement in order to understand stereoisomerism and chirality. Spatial arrangement of atoms concern how different atomic particles and molecules are situated about in the space around the organic compound, namely its carbon chain. In this sense, spatial arrangement of an organic molecule are different another if an atom is shifted in any three-dimensional direction by even one degree. This opens up a very broad possibility of different molecules, each with their unique placement of atoms in three-dimensional space .
Stereoisomers
Stereoisomers are, as mentioned above, contain different types of isomers within itself, each with distinct characteristics that further separate each other as different chemical entities having different properties. Type called entaniomer are the previously-mentioned mirror-image stereoisomers, and will be explained in detail in this article. Another type, diastereomer, has different properties and will be introduced afterwards.
Enantiomers
This type of stereoisomer is the essential mirror-image, non-superimposable type of stereoisomer introduced in the beginning of the article. Figure 3 provides a perfect example; note that the gray plane in the middle demotes the mirror plane.
Figure 3.
Note that even if one were to flip over the left molecule over to the right, the atomic spatial arrangement will not be equal. This is equivalent to the left hand - right hand relationship, and is aptly referred to as 'handedness' in molecules. This can be somewhat counter-intuitive, so this article recommends the reader try the 'hand' example. Place both palm facing up, and hands next to each other. Now flip either side over to the other. One hand should be showing the back of the hand, while the other one is showing the palm. They are not same and non-superimposable.
This is where the concept of chirality comes in as one of the most essential and defining idea of stereoisomerism.
Chirality
Chirality essentially means 'mirror-image, non-superimposable molecules', and to say that a molecule is chiral is to say that its mirror image (it must have one) is not the same as it self. Whether a molecule is chiral or achiral depends upon a certain set of overlapping conditions. Figure 1 shows an example of two molecules, chiral and achiral, respectively. Notice the distinct characteristic of the achiral molecule: it possesses two atoms of same element. In theory and reality, if one were to create a plane that runs through the other two atoms, they will be able to create what is known as bisecting plane: The images on either side of the plan is the same as the other (Figure 4).
Figure 4.
In this case, the molecule is considered 'achiral'. In other words, to distinguish chiral molecule from an achiral molecule, one must search for the existence of the bisecting plane in a molecule. All chiral molecules are deprive of bisecting plane, whether simple or complex.
As a universal rule, no molecule with different surrounding atoms are achiral. Chirality is a simple but essential idea to support the concept of stereoisomerism, being used to explain one type of its kind. The chemical properties of the chiral molecule differs from its mirror image, and in this lies the significance of chilarity in relation to modern organic chemistry.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.03%3A_Looking_Glass_ChemistryChiral_and_Achiral_Molecules.txt |
A consideration of the chirality of molecular configurations explains the curious stereoisomerism observed for lactic acid, carvone and a multitude of other organic compounds. Tetravalent carbons have a tetrahedral configuration. If all four substituent groups are the same, as in methane or tetrachloromethane, the configuration is that of a highly symmetric "regular tetrahedron". A regular tetrahedron several planes of symmetry and is achiral.
A carbon atom that is bonded to four different atoms or groups loses all symmetry, and is often referred to as an asymmetric carbon. The configuration of such a molecular unit is chiral, and the structure may exist in either a right-handed configuration or a left-handed configuration (one the mirror image of the other). This type of configurational stereoisomerism is termed enantiomorphism, and the non-identical, mirror-image pair of stereoisomers that result are called enantiomers. In the general figure below, A and B are nonsuperposable mirror images of one another, and thus are a pair of enantiomers.
The structural formulas of lactic acid and carvone are drawn on the right with the asymmetric carbon colored red. Consequently, we find that these compounds exist as pairs of enantiomers. The presence of a single asymmetrically substituted carbon atom in a molecule is sufficient to render the whole configuration chiral, and modern terminology refers to such groupings as chiral centers. Most of the chiral centers we shall discuss are asymmetric carbon atoms, but it should be recognized that other tetrahedral or pyramidal atoms may become chiral centers if appropriately substituted. When more than one chiral center is present in a molecular structure, care must be taken to analyze their relationship before concluding that a specific molecular configuration is chiral or achiral. This aspect of stereoisomerism will be treated later.
A useful first step in examining structural formulas to determine whether stereoisomers may exist is to identify all stereogenic elements. A stereogenic element is a center, axis or plane that is a focus of stereoisomerism, such that an interchange of two groups attached to this feature leads to a stereoisomer. Stereogenic elements may be chiral or achiral. An asymmetric carbon is often a chiral stereogenic center, since interchanging any two substituent groups converts one enantiomer to the other. Alkenes having two different groups on each double bond carbon constitute an achiral stereogenic element, since interchanging substituents at one of the carbons changes the cis/trans configuration of the double bond.
Some of the structures in the figure above are chiral and some are achiral. First, try to identify all chiral stereogenic centers. Formulas having no chiral centers are necessarily achiral. Formulas having one chiral center are always chiral; and if two or more chiral centers are present in a given structure it is likely to be chiral, but in special cases, to be discussed later, may be achiral.
Structures F and G are achiral. The former has a plane of symmetry passing through the chlorine atom and bisecting the opposite carbon-carbon bond. The similar structure of compound E does not have such a symmetry plane, and the carbon bonded to the chlorine is a chiral center (the two ring segments connecting this carbon are not identical). Structure G is essentially flat. All the carbons except that of the methyl group are sp2 hybridized, and therefore trigonal-planar in configuration. Compounds C, D & H have more than one chiral center, and are also chiral. Remember, all chiral structures may exist as a pair of enantiomers. Other configurational stereoisomers are possible if more than one stereogenic center is present in a structure.
In the 1960’s, a drug called thalidomide was widely prescribed in the Western Europe to alleviate morning sickness in pregnant women.
Thalidomide had previously been used in other countries as an antidepressant, and was believed to be safe and effective for both purposes. The drug was not approved for use in the U.S.A. It was not long, however, before doctors realized that something had gone horribly wrong: many babies born to women who had taken thalidomide during pregnancy suffered from severe birth defects.
Researchers later realized the that problem lay in the fact that thalidomide was being provided as a mixture of two different isomeric forms.
One of the isomers is an effective medication, the other caused the side effects. Both isomeric forms have the same molecular formula and the same atom-to-atom connectivity, so they are not constitutional isomers. Where they differ is in the arrangement in three-dimensional space about one tetrahedral, sp3-hybridized carbon. These two forms of thalidomide are stereoisomers.
Note that the carbon in question has four different substituents (two of these just happen to be connected by a ring structure). Tetrahedral carbons with four different substituent groups are called stereocenters.
Example 1
Exercise 3.5: Locate all of the carbon stereocenters in the molecules below.
Solution
Looking at the structures of what we are referring to as the two isomers of thalidomide, you may not be entirely convinced that they are actually two different molecules. In order to convince ourselves that they are indeed different, let’s create a generalized picture of a tetrahedral carbon stereocenter, with the four substituents designated R1-R4. The two stereoisomers of our simplified model look like this:
If you look carefully at the figure above, you will notice that molecule A and molecule B are mirror images of each other (the line labeled 's' represents a mirror plane). Furthermore, they are not superimposable: if we pick up molecule A, flip it around, and place it next to molecule B, we see that the two structures cannot be superimposed on each other. They are different molecules!
If you make models of the two stereoisomers of thalidomide and do the same thing, you will see that they too are mirror images, and cannot be superimposed (it well help to look at a color version of the figure below).
Thalidomide is a chiral molecule. Something is considered to be chiral if it cannot be superimposed on its own mirror image – in other words, if it is asymmetric (lacking in symmetry). The term ‘chiral’ is derived from the Greek word for ‘handedness’ – ie. right-handedness or left-handedness. Your hands are chiral: your right hand is a mirror image of your left hand, but if you place one hand on top of the other, both palms down, you see that they are not superimposable.
A pair of stereoisomers that are non-superimposable mirror images of one another are considered to have a specific type of stereoisomeric relationship – they are a pair of enantiomers. Thalidomide exists as a pair of enantiomers. On the macro level, your left and right hands are also a pair of enantiomers.
Here are some more examples of chiral molecules that exist as pairs of enantiomers. In each of these examples, there is a single stereocenter, indicated with an arrow. (Many molecules have more than one stereocenter, but we will get to that that a little later!)
Here are some examples of molecules that are achiral (not chiral). Notice that none of these molecules has a stereocenter.
It is difficult to illustrate on the two dimensional page, but you will see if you build models of these achiral molecules that, in each case, there is at least one plane of symmetry, where one side of the plane is the mirror image of the other. Chirality is tied conceptually to the idea of asymmetry, and any molecule that has a plane of symmetry cannot be chiral. When looking for a plane of symmetry, however, we must consider all possible conformations that a molecule could adopt. Even a very simple molecule like ethane, for example, is asymmetric in many of its countless potential conformations – but it has obvious symmetry in both the eclipsed and staggered conformations, and for this reason it is achiral.
Looking for planes of symmetry in a molecule is useful, but often difficult in practice. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions (see section 3.7B), the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral. Carbon stereocenters are also referred to quite frequently as chiral carbons.
When evaluating a molecule for chirality, it is important to recognize that the question of whether or not the dashed/solid wedge drawing convention is used is irrelevant. Chiral molecules are sometimes drawn without using wedges (although obviously this means that stereochemical information is being omitted). Conversely, wedges may be used on carbons that are not stereocenters – look, for example, at the drawings of glycine and citrate in the figure above. Just because you see dashed and solid wedges in a structure, do not automatically assume that you are looking at a stereocenter.
Other elements in addition to carbon can be stereocenters. The phosphorus center of phosphate ion and organic phosphate esters, for example, is tetrahedral, and thus is potentially a stereocenter.
We will see in chapter 10 how researchers, in order to investigate the stereochemistry of reactions at the phosphate center, incorporated sulfur and/or 17O and 18O isotopes of oxygen (the ‘normal’ isotope is 16O) to create chiral phosphate groups. Phosphate triesters are chiral if the three substituent groups are different.
Asymmetric quaternary ammonium groups are also chiral. Amines, however, are not chiral, because they rapidly invert, or turn ‘inside out’, at room temperature.
Example 2
Exercise 3.6: Label the molecules below as chiral or achiral, and locate all stereocenters.
Solution
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The thalidomide that was used in the 1960s to treat depression and morning sickness was sold as a 50:50 mixture of both the R and the S enantiomer – this is referred to as a racemic mixture. A 'squiggly' bond in a chemical structure indicates a racemic mixture - thus racemic (R/S) thalidomide would be drawn as:
The problem with racemic thalidomide, as we know, was that only the R enantiomer was an effective medicine, while the S enantiomer caused mutations in the developing fetus. How does such a seemingly trivial structural variation lead to such a dramatic (and in this case, tragic) difference in biological activity? Virtually all drugs work by interacting in some way with important proteins in our cells: they may bind to pain receptor proteins to block the transmission of pain signals, for instance, or clog up the active site of an enzyme that is involved in the synthesis of cholesterol. Proteins are chiral molecules, and are very sensitive to stereochemistry: just as a right-handed glove won't fit on your left hand, a protein that is able to bind tightly to (R)-thalidomide may not bind well at all to (S)-thalidomide (it will help to view a color version of the figure below).
Instead, it seems that (S)-thalidomide interacts somehow with a protein involved in the development of a growing fetus, eventually causing the observed birth defects.
The over-the-counter painkiller ibuprofen is currently sold as a racemic mixture, but only the S enantiomer is effective.
Fortunately, the R enantiomer does not produce any dangerous side effects, although its presence does seem to increase the amount of time that it takes for (S)-ibuprofen to take effect.
You can, with the assistance your instructor, directly experience the biological importance of stereoisomerism. Carvone is a chiral, plant-derived molecule that smells like spearmint in the R form and caraway (a spice) in the S form.
The two enantiomers interact differently with smell receptor proteins in your nose, generating the transmission of different chemical signals to the olfactory center of your brain.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
5.06: Labeling Stereogenic Centers with R or S
To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The method for this is formally known as R/S nomenclature.
Introduction
The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and, as such, is also often called the Cahn-Ingold-Prelog rules. In addition to the Cahn-Ingold system, there are two ways of experimentally determining the absolute configuration of an enantiomer:
1. X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer.
2. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction.
However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of optical rotation, although different for the two enantiomers of a chiral molecule,at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes.
Stereocenters are labeled R or S
The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S.
Consider the first picture: a curved arrow is drawn from the highest priority (1) substituent to the lowest priority (4) substituent. If the arrow points in a counterclockwise direction (left when leaving the 12 o' clock position), the configuration at stereocenter is considered S ("Sinister" → Latin= "left"). If, however, the arrow points clockwise,(Right when leaving the 12 o' clock position) then the stereocenter is labeled R ("Rectus" → Latin= "right").
The R or S is then added as a prefix, in parenthesis, to the name of the enantiomer of interest.
Example 1
(R)-2-Bromobutane
(S)-2,3- Dihydroxypropanal
Sequence rules to assign priorities to substituents
Before applying the R and S nomenclature to a stereocenter, the substituents must be prioritized according to the following rules:
Rule 1
First, examine at the atoms directly attached to the stereocenter of the compound. A substituent with a higher atomic number takes precedence over a substituent with a lower atomic number. Hydrogen is the lowest possible priority substituent, because it has the lowest atomic number.
1. When dealing with isotopes, the atom with the higher atomic mass receives higher priority.
2. When visualizing the molecule, the lowest priority substituent should always point away from the viewer (a dashed line indicates this). To understand how this works or looks, imagine that a clock and a pole. Attach the pole to the back of the clock, so that when when looking at the face of the clock the pole points away from the viewer in the same way the lowest priority substituent should point away.
3. Then, draw an arrow from the highest priority atom to the 2nd highest priority atom to the 3rd highest priority atom. Because the 4th highest priority atom is placed in the back, the arrow should appear like it is going across the face of a clock. If it is going clockwise, then it is an R-enantiomer; If it is going counterclockwise, it is an S-enantiomer.
When looking at a problem with wedges and dashes, if the lowest priority atom is not on the dashed line pointing away, the molecule must be rotated.
Remember that
• Wedges indicate coming towards the viewer.
• Dashes indicate pointing away from the viewer.
Rule 2
If there are two substituents with equal rank, proceed along the two substituent chains until there is a point of difference. First, determine which of the chains has the first connection to an atom with the highest priority (the highest atomic number). That chain has the higher priority.
If the chains are similar, proceed down the chain, until a point of difference.
For example: an ethyl substituent takes priority over a methyl substituent. At the connectivity of the stereocenter, both have a carbon atom, which are equal in rank. Going down the chains, a methyl has only has hydrogen atoms attached to it, whereas the ethyl has another carbon atom. The carbon atom on the ethyl is the first point of difference and has a higher atomic number than hydrogen; therefore the ethyl takes priority over the methyl.
Rule 3
If a chain is connected to the same kind of atom twice or three times, check to see if the atom it is connected to has a greater atomic number than any of the atoms that the competing chain is connected to.
• If none of the atoms connected to the competing chain(s) at the same point has a greater atomic number: the chain bonded to the same atom multiple times has the greater priority
• If however, one of the atoms connected to the competing chain has a higher atomic number: that chain has the higher priority.
Example 2
A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below:
However:
Remember that being double or triple bonded to an atom means that the atom is connected to the same atom twice. In such a case, follow the same method as above.
Caution!!
Keep in mind that priority is determined by the first point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant.
When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.
After all your substituents have been prioritized in the correct manner, you can now name/label the molecule R or S.
1. Put the lowest priority substituent in the back (dashed line).
2. Proceed from 1 to 2 to 3. (it is helpful to draw or imagine an arcing arrow that goes from 1--> 2-->3)
3. Determine if the direction from 1 to 2 to 3 clockwise or counterclockwise.
i) If it is clockwise it is R.
ii) if it is counterclockwise it is S.
USE YOUR MODELING KIT: Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S).
IF YOU DO NOT HAVE A MODELING KIT: remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models.
If you have a modeling kit use it to help you solve the following practice problems.
Problems
Are the following R or S?
Solutions
1. S: I > Br > F > H. The lowest priority substituent, H, is already going towards the back. It turns left going from I to Br to F, so it's a S.
2. R: Br > Cl > CH3 > H. You have to switch the H and Br in order to place the H, the lowest priority, in the back. Then, going from Br to Cl, CH3 is turning to the right, giving you a R.
3. Neither R or S: This molecule is achiral. Only chiral molecules can be named R or S.
4. R: OH > CN > CH2NH2 > H. The H, the lowest priority, has to be switched to the back. Then, going from OH to CN to CH2NH2, you are turning right, giving you a R.
5. (5) S: \(\ce{-COOH}\) > \(\ce{-CH_2OH}\) > \(\ce{C#CH}\) > \(\ce{H}\). Then, going from \(\ce{-COOH}\) to \(\ce{-CH_2OH}\) to \(\ce{-C#CH}\) you are turning left, giving you a S configuration.
Contributors
• Ekta Patel (UCD), Ifemayowa Aworanti (University of Maryland Baltimore County) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.05%3A_Stereogenic_Centers_in_Cyclic_Compounds.txt |
Diastereomers are stereoisomers that are not related as object and mirror image and are not enantiomers. Unlike enatiomers which are mirror images of each other and non-sumperimposable, diastereomers are not mirror images of each other and non-superimposable. Diastereomers can have different physical properties and reactivity. They have different melting points and boiling points and different densities. They have two or more stereocenters.
Introduction
It is easy to mistake between diasteromers and enantiomers. For example, we have four steroisomers of 3-bromo-2-butanol. The four possible combination are SS, RR, SR and RS (Figure 1). One of the molecule is the enantiomer of its mirror image molecule and diasteromer of each of the other two molecule (SS is enantiomer of RR and diasteromer of RS and SR). SS's mirror image is RR and they are not superimposable, so they are enantiomers. RS and SR are not mirror image of SS and are not superimposable to each other, so they are diasteromers.
Figure 1
Diastereomers vs. Enantiomers
Tartaric acid, C4H6O6, is an organic compound that can be found in grape, bananas, and in wine. The structures of tartaric acid itself is really interesting. Naturally, it is in the form of (R,R) stereocenters. Artificially, it can be in the meso form (R,S), which is achiral. R,R tartaric acid is enantiomer to is mirror image which is S,S tartaric acid and diasteromers to meso-tartaric acid (figure 2).
(R,R) and (S,S) tartaric acid have similar physical properties and reactivity. However, meso-tartaric acid have different physical properties and reactivity. For example, melting point of (R,R) & (S,S) tartaric is about 170 degree Celsius, and melting point of meso-tartaric acid is about 145 degree Celsius.
Figure 2
We turn our attention next to molecules which have more than one stereocenter. We will start with a common four-carbon sugar called D-erythrose.
A note on sugar nomenclature: biochemists use a special system to refer to the stereochemistry of sugar molecules, employing names of historical origin in addition to the designators 'D' and 'L'. You will learn about this system if you take a biochemistry class. We will use the D/L designations here to refer to different sugars, but we won't worry about learning the system.
As you can see, D-erythrose is a chiral molecule: C2 and C3 are stereocenters, both of which have the R configuration. In addition, you should make a model to convince yourself that it is impossible to find a plane of symmetry through the molecule, regardless of the conformation. Does D-erythrose have an enantiomer? Of course it does – if it is a chiral molecule, it must. The enantiomer of erythrose is its mirror image, and is named L-erythrose (once again, you should use models to convince yourself that these mirror images of erythrose are not superimposable).
Notice that both chiral centers in L-erythrose both have the S configuration. In a pair of enantiomers, all of the chiral centers are of the opposite configuration.
What happens if we draw a stereoisomer of erythrose in which the configuration is S at C2 and R at C3? This stereoisomer, which is a sugar called D-threose, is not a mirror image of erythrose. D-threose is a diastereomer of both D-erythrose and L-erythrose.
The definition of diastereomers is simple: if two molecules are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers, then they are diastereomers by default. In practical terms, this means that at least one - but not all - of the chiral centers are opposite in a pair of diastereomers. By definition, two molecules that are diastereomers are not mirror images of each other.
L-threose, the enantiomer of D-threose, has the R configuration at C2 and the S configuration at C3. L-threose is a diastereomer of both erythrose enantiomers.
In general, a structure with n stereocenters will have 2n different stereoisomers. (We are not considering, for the time being, the stereochemistry of double bonds – that will come later). For example, let's consider the glucose molecule in its open-chain form (recall that many sugar molecules can exist in either an open-chain or a cyclic form). There are two enantiomers of glucose, called D-glucose and L-glucose. The D-enantiomer is the common sugar that our bodies use for energy. It has n = 4 stereocenters, so therefore there are 2n = 24 = 16 possible stereoisomers (including D-glucose itself).
In L-glucose, all of the stereocenters are inverted relative to D-glucose. That leaves 14 diastereomers of D-glucose: these are molecules in which at least one, but not all, of the stereocenters are inverted relative to D-glucose. One of these 14 diastereomers, a sugar called D-galactose, is shown above: in D-galactose, one of four stereocenters is inverted relative to D-glucose. Diastereomers which differ in only one stereocenter (out of two or more) are called epimers. D-glucose and D-galactose can therefore be refered to as epimers as well as diastereomers.
Examples
Exercise 3.10: Draw the structure of L-galactose, the enantiomer of D-galactose.
Exercise 3.11: Draw the structure of two more diastereomers of D-glucose. One should be an epimer.
Solutions
Erythronolide B, a precursor to the 'macrocyclic' antibiotic erythromycin, has 10 stereocenters. It’s enantiomer is that molecule in which all 10 stereocenters are inverted.
In total, there are 210 = 1024 stereoisomers in the erythronolide B family: 1022 of these are diastereomers of the structure above, one is the enantiomer of the structure above, and the last is the structure above.
We know that enantiomers have identical physical properties and equal but opposite degrees of specific rotation. Diastereomers, in theory at least, have different physical properties – we stipulate ‘in theory’ because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to separate them. In addition, the specific rotations of diastereomers are unrelated – they could be the same sign or opposite signs, and similar in magnitude or very dissimilar.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.07%3A_Diastereomers.txt |
A meso compound is an achiral compound that has chiral centers. It is superimposed on its mirror image and is optically inactive although it contains two or more stereocenters.
Introduction
In general, a meso compound should contain two or more identical substituted stereocenters. Also, it has an internal symmetry plane that divides the compound in half. These two halves reflect each other by the internal mirror. The stereochemistry of stereocenters should "cancel out". What it means here is that when we have an internal plane that splits the compound into two symmetrical sides, the stereochemistry of both left and right side should be opposite to each other, and therefore, result in optically inactive. Cyclic compounds may also be meso.
Identification
If A is a meso compound, it should have two or more stereocenters, an internal plane, and the stereochemistry should be R and S.
1. Look for an internal plane, or internal mirror, that lies in between the compound.
2. The stereochemistry (e.g. R or S) is very crucial in determining whether it is a meso compound or not. As mentioned above, a meso compound is optically inactive, so their stereochemistry should cancel out. For instance, R cancels S out in a meso compound with two stereocenters.
trans-1,2-dichloro-1,2-ethanediol
(meso)-2,3-dibromobutane
Tips: An interesting thing about single bonds or sp3-orbitals is that we can rotate the substituted groups that attached to a stereocenter around to recognize the internal plane. As the molecule is rotated, its stereochemistry does not change. For example:
Another case is when we rotate the whole molecule by 180 degree. Both molecules below are still meso.
Remember the internal plane here is depicted on two dimensions. However, in reality, it is three dimensions, so be aware of it when we identify the internal mirror.
Example 1
1 has a plane of symmetry (the horizonatal plane going through the red broken line) and, therefore, is achiral; 1 has chiral centers. Thus, 1 is a meso compound.
Example 2
This molecules has a plane of symmetry (the vertical plane going through the red broken line perpendicular to the plane of the ring) and, therefore, is achiral, but has has two chiral centers. Thus, its is a meso compound.
Other Examples of meso compounds
Meso compounds can exist in many different forms such as pentane, butane, heptane, and even cyclobutane. They do not necessarily have to be two stereocenters, but can have more.
Optical Activity Analysis
When the optical activity of a meso compound is attempted to be determined with a polarimeter, the indicator will not show (+) or (-). It simply means there is no certain direction of rotation of the polarized light, neither levorotatory (-) and dexorotatory (+).
Problems
Beside meso, there are also other types of molecules: enantiomer, diastereomer, and identical. Determine if the following molecules are meso.
Answer key: A C, D, E are meso compounds.
• Duy Dang
5.09: R and S Assignments in Compounds with Two or More Stereogenic Centers
The Chinese shrub Ma Huang (Ephedra vulgaris) contains two physiologically active compounds ephedrine and pseudoephedrine. Both compounds are stereoisomers of 2-methylamino-1-phenyl-1-propanol, and both are optically active, one being levorotatory and the other dextrorotatory. Since the properties of these compounds (see below) are significantly different, they cannot be enantiomers. How, then, are we to classify these isomers and others like them?
Ephedrine: m.p. 35 - 40 º C, [α]D = –41º, moderate water solubility [this isomer may be referred to as (–)-ephedrine]
Pseudoephedrine: m.p. 119 º C, [α]D = +52º, relatively insoluble in water [this isomer may be referred to as (+)-pseudoephedrine]
Since these two compounds are optically active, each must have an enantiomer. Although these missing stereoisomers were not present in the natural source, they have been prepared synthetically and have the expected identical physical properties and opposite-sign specific rotations with those listed above. The structural formula of 2-methylamino-1-phenylpropanol has two stereogenic carbons (#1 & #2). Each may assume an R or S configuration, so there are four stereoisomeric combinations possible. These are shown in the following illustration, together with the assignments that have been made on the basis of chemical interconversions
We turn our attention next to molecules which have more than one stereocenter. We will start with a common four-carbon sugar called D-erythrose.
A note on sugar nomenclature: biochemists use a special system to refer to the stereochemistry of sugar molecules, employing names of historical origin in addition to the designators 'D' and 'L'. You will learn about this system if you take a biochemistry class. We will use the D/L designations here to refer to different sugars, but we won't worry about learning the system.
As you can see, D-erythrose is a chiral molecule: C2 and C3 are stereocenters, both of which have the R configuration. In addition, you should make a model to convince yourself that it is impossible to find a plane of symmetry through the molecule, regardless of the conformation. Does D-erythrose have an enantiomer? Of course it does – if it is a chiral molecule, it must. The enantiomer of erythrose is its mirror image, and is named L-erythrose (once again, you should use models to convince yourself that these mirror images of erythrose are not superimposable).
Notice that both chiral centers in L-erythrose both have the S configuration. In a pair of enantiomers, all of the chiral centers are of the opposite configuration.
What happens if we draw a stereoisomer of erythrose in which the configuration is S at C2 and R at C3? This stereoisomer, which is a sugar called D-threose, is not a mirror image of erythrose. D-threose is a diastereomer of both D-erythrose and L-erythrose.
The definition of diastereomers is simple: if two molecules are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers, then they are diastereomers by default. In practical terms, this means that at least one - but not all - of the chiral centers are opposite in a pair of diastereomers. By definition, two molecules that are diastereomers are not mirror images of each other.
L-threose, the enantiomer of D-threose, has the R configuration at C2 and the S configuration at C3. L-threose is a diastereomer of both erythrose enantiomers.
In general, a structure with n stereocenters will have 2n different stereoisomers. (We are not considering, for the time being, the stereochemistry of double bonds – that will come later). For example, let's consider the glucose molecule in its open-chain form (recall that many sugar molecules can exist in either an open-chain or a cyclic form). There are two enantiomers of glucose, called D-glucose and L-glucose. The D-enantiomer is the common sugar that our bodies use for energy. It has n = 4 stereocenters, so therefore there are 2n = 24 = 16 possible stereoisomers (including D-glucose itself).
In L-glucose, all of the stereocenters are inverted relative to D-glucose. That leaves 14 diastereomers of D-glucose: these are molecules in which at least one, but not all, of the stereocenters are inverted relative to D-glucose. One of these 14 diastereomers, a sugar called D-galactose, is shown above: in D-galactose, one of four stereocenters is inverted relative to D-glucose. Diastereomers which differ in only one stereocenter (out of two or more) are called epimers. D-glucose and D-galactose can therefore be refered to as epimers as well as diastereomers.
Examples
Exercise 3.10: Draw the structure of L-galactose, the enantiomer of D-galactose.
Exercise 3.11: Draw the structure of two more diastereomers of D-glucose. One should be an epimer.
Solutions
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.08%3A_Meso_Compounds.txt |
Configurational Stereoisomers of Cycloalkanes
Stereoisomers are also observed in certain disubstituted (and higher substituted) cyclic compounds. Unlike the relatively flat molecules of alkenes, substituted cycloalkanes must be viewed as three-dimensional configurations in order to appreciate the spatial orientations of the substituents. By agreement, chemists use heavy, wedge-shaped bonds to indicate a substituent located above the average plane of the ring (note that cycloalkanes larger than three carbons are not planar), and a hatched line for bonds to atoms or groups located below the ring. As in the case of the 2-butene stereoisomers, disubstituted cycloalkane stereoisomers may be designated by nomenclature prefixes such as cis and trans. The stereoisomeric 1,2-dibromocyclopentanes shown to the right are an example.
Many of the stereochemical concepts we have covered are illustrated nicely by looking at disubstituted cyclohexanes. We will now examine several dichlorocyclohexanes.
The 1,1-dichloro isomer has no centers of chirality. The 1,2- and 1,3-dichlorocyclohexanes each have two centers of chirality, bearing the same set of substituents. The cis & trans-1,4-dichlorocyclohexanes do not have any chiral centers, since the two ring groups on the substituted carbons are identical.
There are three configurational isomers of 1,2-dichlorocyclohexane and three configurational isomers of 1,3-dichlorocyclohexane. These are shown below:
1,2-substitution
1,3-substitution
All the 1,2-dichloro isomers are constitutional isomers of the 1,3-dichloro isomers. In each category (1,2- & 1,3-), the (R,R)-trans isomer and the (S,S)-trans isomer are enantiomers. The cis isomer is a diastereomer of the trans isomers.
Finally, all of these isomers may exist as a mixture of two (or more) conformational isomers.The chair conformer of the cis 1,2-dichloro isomer appears to be chiral. However, It exists as a 50:50 mixture of chair conformations, which interconvert so rapidly they cannot be resolved (ie. separated). In fact, a boat conformation can be constructued in which there exists a plane of symmetry, therefore the cis isomer is, over, achiral. Since the cis isomer has two centers of chirality (asymmetric carbons) but is achiral, it is a meso-compound. The corresponding trans isomers also exist as rapidly interconverting chiral conformations, but they are chiral. The diequatorial conformer predominates in each case, the (R,R) conformations being mirror images of the (S,S) conformations. All these conformations are diastereomeric with the cis conformations.
The diequatorial chair conformer of the cis 1,3-dichloro isomer is achiral. It is the major component of a fast equilibrium with the diaxial conformer, which is also achiral. This isomer is also a meso compound. The corresponding trans isomers also undergo a rapid conformational interconversion. For these isomers, however, this interconversion produces an identical conformer, so each enantiomer (R,R) and (S,S) has predominately a single chiral conformation. These enantiomeric conformations, both of which are chiral, are diastereomeric with the cis (meso) isomer.
Cyclic compounds can also be meso. One of many such examples is cis-1,2-dihydroxycyclohexane. Note, however, that if the hydroxyl groups are trans to each other, the molecule is chiral.
Fortunately for overworked organic chemistry students, the meso compound is a very special case, and it is difficult to find many naturally occurring examples. They do, however, seem to like showing up in organic chemistry exam questions.
Example
Exercise: Which of the molecules shown below are meso? Which are chiral? Which are achiral, but not meso?
Exercise: Draw the structure of another dimethylcyclopentane isomer that is meso (do not use structures from the previous problem).
Solutions
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.10%3A_Disubstituted_Cycloalkanes.txt |
Conformational Isomers
The C–C single bonds in ethane, propane, and other alkanes are formed by the overlap of an sp3 hybrid orbital on one carbon atom with an sp3 hybrid orbital on another carbon atom, forming a σ bond. Each sp3 hybrid orbital is cylindrically symmetrical (all cross-sections are circles), resulting in a carbon–carbon single bond that is also cylindrically symmetrical about the C–C axis. Because rotation about the carbon–carbon single bond can occur without changing the overlap of the sp3 hybrid orbitals, there is no significant electronic energy barrier to rotation. Consequently, many different arrangements of the atoms are possible, each corresponding to different degrees of rotation. Differences in three-dimensional structure resulting from rotation about a σ bond are called differences in conformation, and each different arrangement is called a conformational isomer (or conformer).
Structural Isomers
Unlike conformational isomers, which do not differ in connectivity, structural isomers differ in connectivity, as illustrated here for 1-propanol and 2-propanol. Although these two alcohols have the same molecular formula (C3H8O), the position of the –OH group differs, which leads to differences in their physical and chemical properties.
In the conversion of one structural isomer to another, at least one bond must be broken and reformed at a different position in the molecule. Consider, for example, the following five structures represented by the formula C5H12:
Of these structures, (a) and (d) represent the same compound, as do (b) and (c). No bonds have been broken and reformed; the molecules are simply rotated about a 180° vertical axis. Only three—n-pentane (a) and (d), 2-methylbutane (b) and (c), and 2,2-dimethylpropane (e)—are structural isomers. Because no bonds are broken in going from (a) to (d) or from (b) to (c), these alternative representations are not structural isomers. The three structural isomers—either (a) or (d), either (b) or (c), and (e)—have distinct physical and chemical properties.
Stereoisomers
Molecules with the same connectivity but different arrangements of the atoms in space are called stereoisomers. There are two types of stereoisomers: geometric and optical. Geometric isomers differ in the relative position(s) of substituents in a rigid molecule. Simple rotation about a C–C σ bond in an alkene, for example, cannot occur because of the presence of the π bond. The substituents are therefore rigidly locked into a particular spatial arrangement (part (a) in Figure 2.16. Thus a carbon–carbon multiple bond, or in some cases a ring, prevents one geometric isomer from being readily converted to the other. The members of an isomeric pair are identified as either cis or trans, and interconversion between the two forms requires breaking and reforming one or more bonds. Because their structural difference causes them to have different physical and chemical properties, cis and trans isomers are actually two distinct chemical compounds.
Note
Stereoisomers have the same connectivity but different arrangements of atoms in space.
Optical isomers are molecules whose structures are mirror images but cannot be superimposed on one another in any orientation. Optical isomers have identical physical properties, although their chemical properties may differ in asymmetric environments. Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands, feet, and ears. As shown in part (a) in Figure 5.11.1., your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image, as shown by the superimposed flasks in part (b) in Figure 5.11.1..
Figure 5.11.1: Chiral and Achiral Objects. (a) Objects that are nonsuperimposable mirror images of each other are chiral, such as the left and the right hand. (b) The unmarked flask is achiral because it can be superimposed on its mirror image.
Most chiral organic molecules have at least one carbon atom that is bonded to four different groups, as occurs in the bromochlorofluoromethane molecule shown in part (a) in Figure 5.11.2. This carbon, often designated by an asterisk in structural drawings, is called a chiral center or asymmetric carbon atom. If the bromine atom is replaced by another chlorine (part (b) in Figure 5.11.2), the molecule and its mirror image can now be superimposed by simple rotation. Thus the carbon is no longer a chiral center. Asymmetric carbon atoms are found in many naturally occurring molecules, such as lactic acid, which is present in milk and muscles, and nicotine, a component of tobacco. A molecule and its nonsuperimposable mirror image are called enantiomers (from the Greek enantiou, meaning “opposite”).
Figure 5.11.2: Comparison of Chiral and Achiral Molecules. (a) Bromochlorofluoromethane is a chiral molecule whose stereocenter is designated with an asterisk. Rotation of its mirror image does not generate the original structure. To superimpose the mirror images, bonds must be broken and reformed. (b) In contrast, dichlorofluoromethane and its mirror image can be rotated so they are superimposable.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• Dan Chong | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.11%3A_IsomersA_Summary.txt |
Identifying and distinguishing enantiomers is inherently difficult, since their physical and chemical properties are largely identical. Fortunately, a nearly two hundred year old discovery by the French physicist Jean-Baptiste Biot has made this task much easier. This discovery disclosed that the right- and left-handed enantiomers of a chiral compound perturb plane-polarized light in opposite ways. This perturbation is unique to chiral molecules, and has been termed optical activity.
Polarimetry
Plane-polarized light is created by passing ordinary light through a polarizing device, which may be as simple as a lens taken from polarizing sun-glasses. Such devices transmit selectively only that component of a light beam having electrical and magnetic field vectors oscillating in a single plane. The plane of polarization can be determined by an instrument called a polarimeter, shown in the diagram below.
Monochromatic (single wavelength) light, is polarized by a fixed polarizer next to the light source. A sample cell holder is located in line with the light beam, followed by a movable polarizer (the analyzer) and an eyepiece through which the light intensity can be observed. In modern instruments an electronic light detector takes the place of the human eye. In the absence of a sample, the light intensity at the detector is at a maximum when the second (movable) polarizer is set parallel to the first polarizer (α = 0º). If the analyzer is turned 90º to the plane of initial polarization, all the light will be blocked from reaching the detector.
Chemists use polarimeters to investigate the influence of compounds (in the sample cell) on plane polarized light. Samples composed only of achiral molecules (e.g. water or hexane), have no effect on the polarized light beam. However, if a single enantiomer is examined (all sample molecules being right-handed, or all being left-handed), the plane of polarization is rotated in either a clockwise (positive) or counter-clockwise (negative) direction, and the analyzer must be turned an appropriate matching angle, α, if full light intensity is to reach the detector. In the above illustration, the sample has rotated the polarization plane clockwise by +90º, and the analyzer has been turned this amount to permit maximum light transmission.
The observed rotations (α) of enantiomers are opposite in direction. One enantiomer will rotate polarized light in a clockwise direction, termed dextrorotatory or (+), and its mirror-image partner in a counter-clockwise manner, termed levorotatory or (–). The prefixes dextro and levo come from the Latin dexter, meaning right, and laevus, for left, and are abbreviated d and l respectively. If equal quantities of each enantiomer are examined , using the same sample cell, then the magnitude of the rotations will be the same, with one being positive and the other negative. To be absolutely certain whether an observed rotation is positive or negative it is often necessary to make a second measurement using a different amount or concentration of the sample. In the above illustration, for example, α might be –90º or +270º rather than +90º. If the sample concentration is reduced by 10%, then the positive rotation would change to +81º (or +243º) while the negative rotation would change to –81º, and the correct α would be identified unambiguously.
Since it is not always possible to obtain or use samples of exactly the same size, the observed rotation is usually corrected to compensate for variations in sample quantity and cell length. Thus it is common practice to convert the observed rotation, α, to a specific rotation, [α], by the following formula:
Specific Rotation = where l = cell length in dm, c = concentration in g/ml
D is the 589 nm light from a sodium lamp
Compounds that rotate the plane of polarized light are termed optically active. Each enantiomer of a stereoisomeric pair is optically active and has an equal but opposite-in-sign specific rotation. Specific rotations are useful in that they are experimentally determined constants that characterize and identify pure enantiomers. For example, the lactic acid and carvone enantiomers discussed earlier have the following specific rotations.
Carvone from caraway: [α]D = +62.5º this isomer may be referred to as (+)-carvone or d-carvone
Carvone from spearmint: [α]D = –62.5º this isomer may be referred to as (–)-carvone or l-carvone
Lactic acid from muscle tissue: [α]D = +2.5º this isomer may be referred to as (+)-lactic acid or d-lactic acid
Lactic acid from sour milk: [α]D = –2.5º this isomer may be referred to as (–)-lactic acid or l-lactic acid
A 50:50 mixture of enantiomers has no observable optical activity. Such mixtures are called racemates or racemic modifications, and are designated (±). When chiral compounds are created from achiral compounds, the products are racemic unless a single enantiomer of a chiral co-reactant or catalyst is involved in the reaction. The addition of HBr to either cis- or trans-2-butene is an example of racemic product formation (the chiral center is colored red in the following equation).
CH3CH=CHCH3 + HBr (±) CH3CH2CHBrCH3
Chiral organic compounds isolated from living organisms are usually optically active, indicating that one of the enantiomers predominates (often it is the only isomer present). This is a result of the action of chiral catalysts we call enzymes, and reflects the inherently chiral nature of life itself. Chiral synthetic compounds, on the other hand, are commonly racemates, unless they have been prepared from enantiomerically pure starting materials.
There are two ways in which the condition of a chiral substance may be changed:
1. A racemate may be separated into its component enantiomers. This process is called resolution.
2. A pure enantiomer may be transformed into its racemate. This process is called racemization.
Enantiomeric Excess
The "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%.
Because R and S enantiomers have equal but opposite optical activity, it naturally follows that a 50:50 racemic mixture of two enantiomers will have no observable optical activity. If we know the specific rotation for a chiral molecule, however, we can easily calculate the ratio of enantiomers present in a mixture of two enantiomers, based on its measured optical activity. When a mixture contains more of one enantiomer than the other, chemists often use the concept of enantiomeric excess (ee) to quantify the difference. Enantiomeric excess can be expressed as:
For example, a mixture containing 60% R enantiomer (and 40% S enantiomer) has a 20% enantiomeric excess of R: ((60-50) x 100) / 50 = 20 %.
Expressed in terms of optical rotation (use absolute values):
Example
The specific rotation of (S)-carvone is (+)61°, measured 'neat' (pure liquid sample, no solvent). The optical rotation of a neat sample of a mixture of R and S carvone is measured at (-)23°. Which enantiomer is in excess, and what is its ee? What are the percentages of (R)- and (S)-carvone in the sample?
Chiral molecules are often labeled according to the sign of their specific rotation, as in (S)-(+)-carvone and (R)-(-)-carvone, or (±)-carvone for the racemic mixture. However, there is no relationship whatsoever between a molecule's R/S designation and the sign of its specific rotation. Without performing a polarimetry experiment or looking in the literature, we would have no idea that (-)-carvone has the R configuration and (+)-carvone has the S configuration.
Separation of Chiral Compounds
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used in this technique are known as '
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.12%3A_Physical_Properties_of_Stereoisomers.txt |
Some Chiral Organic Molecules
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.
Figure SC5.1. The two naturally-occurring enantiomers of carvone.
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.
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.
Figure SC5.2. Comparing atoms step-by-step to assign configuration.
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 (colourless 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.
Figure SC5.3. Thalidomide.
Problem SC5.1.
1. Draw thalidomide and identify the chiral center with an asterisk.
2. Draw the two possible enantiomeric forms of thalidomide.
Problem SC5.2.
Draw the two enantiomeric forms of 2-butanol, CH3CH(OH)CH2CH3. Label their configurations.
Problem SC5.3.
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/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/05%3A_Stereochemistry/5.13%3A_Chemical_Properties_of_Enantiomers.txt |
Organic chemistry encompasses a very large number of compounds ( many millions ), and our previous discussion and illustrations have focused on their structural characteristics. Now that we can recognize these actors ( compounds ), we turn to the roles they are inclined to play in the scientific drama staged by the multitude of chemical reactions that define organic chemistry.
We begin by defining some basic terms that will be used frequently as this subject is elaborated.
Chemical Reaction: A transformation resulting in a change of composition, constitution and/or configuration of a compound ( referred to as the reactant or substrate ).
Reactant or Substrate: The organic compound undergoing change in a chemical reaction. Other compounds may also be involved, and common reactive partners ( reagents ) may be identified. The reactant is often ( but not always ) the larger and more complex molecule in the reacting system. Most ( or all ) of the reactant molecule is normally incorporated as part of the product molecule.
Reagent: A common partner of the reactant in many chemical reactions. It may be organic or inorganic; small or large; gas, liquid or solid. The portion of a reagent that ends up being incorporated in the product may range from all to very little or none.
Product(s) The final form taken by the major reactant(s) of a reaction.
Reaction Conditions The environmental conditions, such as temperature, pressure, catalysts & solvent, under which a reaction progresses optimally. Catalysts are substances that accelerate the rate ( velocity ) of a chemical reaction without themselves being consumed or appearing as part of the reaction product. Catalysts do not change equilibria positions.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
6.02: Kinds of Organic Reactions
If you scan any organic textbook you will encounter what appears to be a very large, often intimidating, number of reactions. These are the "tools" of a chemist, and to use these tools effectively, we must organize them in a sensible manner and look for patterns of reactivity that permit us make plausible predictions. Most of these reactions occur at special sites of reactivity known as functional groups, and these constitute one organizational scheme that helps us catalog and remember reactions.
Ultimately, the best way to achieve proficiency in organic chemistry is to understand how reactions take place, and to recognize the various factors that influence their course.
First, we identify four broad classes of reactions based solely on the structural change occurring in the reactant molecules. This classification does not require knowledge or speculation concerning reaction paths or mechanisms. The four main reaction classes are additions, eliminations, substitutions, and rearrangements.
Rearrangement
In an addition reaction the number of σ-bonds in the substrate molecule increases, usually at the expense of one or more π-bonds. The reverse is true of elimination reactions, i.e.the number of σ-bonds in the substrate decreases, and new π-bonds are often formed. Substitution reactions, as the name implies, are characterized by replacement of an atom or group (Y) by another atom or group (Z). Aside from these groups, the number of bonds does not change. A rearrangement reaction generates an isomer, and again the number of bonds normally does not change.
The examples illustrated above involve simple alkyl and alkene systems, but these reaction types are general for most functional groups, including those incorporating carbon-oxygen double bonds and carbon-nitrogen double and triple bonds. Some common reactions may actually be a combination of reaction types. The reaction of an ester with ammonia to give an amide, as shown below, appears to be a substitution reaction ( Y = CH3O & Z = NH2 ); however, it is actually two reactions, an addition followed by an elimination.
The addition of water to a nitrile does not seem to fit any of the above reaction types, but it is simply a slow addition reaction followed by a rapid rearrangement, as shown in the following equation. Rapid rearrangements of this kind are called tautomerizations. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.01%3A_Writing_Equations_for_Organic_Reactions.txt |
The Arrow Notation in Mechanisms
Since chemical reactions involve the breaking and making of bonds, a consideration of the movement of bonding ( and non-bonding ) valence shell electrons is essential to this understanding. It is now common practice to show the movement of electrons with curved arrows, and a sequence of equations depicting the consequences of such electron shifts is termed a mechanism. In general, two kinds of curved arrows are used in drawing mechanisms:
A full head on the arrow indicates the movement or shift of an electron pair:
A partial head (fishhook) on the arrow indicates the shift of a single electron:
The use of these symbols in bond-breaking and bond-making reactions is illustrated below. If a covalent single bond is broken so that one electron of the shared pair remains with each fragment, as in the first example, this bond-breaking is called homolysis. If the bond breaks with both electrons of the shared pair remaining with one fragment, as in the second and third examples, this is called heterolysis.
Bond-Breaking Bond-Making
Other Arrow Symbols
Chemists also use arrow symbols for other purposes, and it is essential to use them correctly.
The Reaction Arrow
The Equilibrium Arrow
The Resonance Arrow
The following equations illustrate the proper use of these symbols:
Reactive Intermediates
The products of bond breaking, shown above, are not stable in the usual sense, and cannot be isolated for prolonged study. Such species are referred to as reactive intermediates, and are believed to be transient intermediates in many reactions. The general structures and names of four such intermediates are given below.
A pair of widely used terms, related to the Lewis acid-base notation, should also be introduced here.
Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile.
Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile (or Lewis acid).
Using these definitions, it is clear that carbocations ( called carbonium ions in the older literature ) are electrophiles and carbanions are nucleophiles. Carbenes have only a valence shell sextet of electrons and are therefore electron deficient. In this sense they are electrophiles, but the non-bonding electron pair also gives carbenes nucleophilic character. As a rule, the electrophilic character dominates carbene reactivity. Carbon radicals have only seven valence electrons, and may be considered electron deficient; however, they do not in general bond to nucleophilic electron pairs, so their chemistry exhibits unique differences from that of conventional electrophiles. Radical intermediates are often called free radicals.
The importance of electrophile / nucleophile terminology comes from the fact that many organic reactions involve at some stage the bonding of a nucleophile to an electrophile, a process that generally leads to a stable intermediate or product. Reactions of this kind are sometimes called ionic reactions, since ionic reactants or products are often involved. Some common examples of ionic reactions and their mechanisms may be examined by Clicking Here
The shapes ideally assumed by these intermediates becomes important when considering the stereochemistry of reactions in which they play a role. A simple tetravalent compound like methane, CH4, has a tetrahedral configuration. Carbocations have only three bonds to the charge bearing carbon, so it adopts a planar trigonal configuration. Carbanions are pyramidal in shape ( tetrahedral if the electron pair is viewed as a substituent ), but these species invert rapidly at room temperature, passing through a higher energy planar form in which the electron pair occupies a p-orbital. Radicals are intermediate in configuration, the energy difference between pyramidal and planar forms being very small. Since three points determine a plane, the shape of carbenes must be planar; however, the valence electron distribution varies. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.03%3A_Bond_Breaking_and_Bond_Making.txt |
The homolytic bond dissociation energy is the amount of energy needed to break apart one mole of covalently bonded gases into a pair of radicals. The SI units used to describe bond energy are kiloJoules per mole of bonds (kJ/Mol). It indicates how strongly the atoms are bonded to each other.
Introduction
Breaking a covalent bond between two partners, A-B, can occur either heterolytically, where the shared pair of electron goes with one partner or another
$A-B \rightarrow A^+ + B:^-$
or
$A-B \rightarrow A:^- + B^+ $
or homolytically, where one electron stays with each partner.
$A-B \rightarrow A^• + B^•$
The products of homolytic cleavage are radicals and the energy that is required to break the bond homolytically is called the Bond Dissociation Energy (BDE) and is a measure of the strength of the bond.
Calculation of the BDE
The BDE for a molecule A-B is calculated as the difference in the enthalpies of formation of the products and reactants for homolysis
$BDE = \Delta_fH(A^•) + \Delta_fH(B^•) - \Delta_fH(A-B)$
Officially, the IUPAC definition of bond dissociation energy refers to the energy change that occurs at 0 K, and the symbol is $D_o$. However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation enthalpy, which generally refers to the enthalpy change at room temperature (298K). Although there are technically differences between BDEs at 0 K and 298 K, those difference are not large and generally do not affect interpretations of chemical processes.
Bond Breakage/Formation
Bond dissociation energy (or enthalpy) is a state function and consequently does not depend on the path by which it occurs. Therefore, the specific mechanism in how a bond breaks or is formed does not affect the BDE. Bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using Hess's Law can be used to estimate reaction enthalpies.
Example $1$: Chlorination of Methane
Consider the chlorination of methane
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$
the overall reaction thermochemistry can be calculated exactly by combining the BDEs for the bonds broken and bonds formed
CH4 → CH3• + H• BDE(CH3-H)
Cl2 → 2Cl• BDE(Cl2)\]
H• + Cl• → HCl -BDE(HCl)
CH3• + Cl• → CH3Cl -BDE(CH3-Cl)
---------------------------------------------------
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$
$\Delta H = BDE(R-H) + BDE(Cl_2) - BDE(HCl) - BDE(CH_3-Cl)$
Because reaction enthalpy is a state function, it does not matter what reactions are combined to make up the overall process using Hess's Law. However, BDEs are convenient to use because they are readily available.
Alternatively, BDEs can be used to assess individual steps of a mechanism. For example, an important step in free radical chlorination of alkanes is the abstraction of hydrogen from the alkane to form a free radical.
RH + Cl• → R• + HCl
The energy change for this step is equal to the difference in the BDEs in RH and HCl
$\Delta H = BDE(R-H) - BDE(HCl)$
This relationship shows that the hydrogen abstraction step is more favorable when BDE(R-H) is smaller. The difference in energies accounts for the selectivity in the halogenation of hydrocarbons with different types of C-H bonds.
Representative C-H BDEs in Organic Molecules
R-H Do, kJ/mol D298, kJ/mol R-H Do, kJ/mol D298, kJ/mol
CH3-H 432.7±0.1 439.3±0.4 H2C=CH-H 456.7±2.7 463.2±2.9
CH3CH2-H 423.0±1.7 C6H5-H 465.8±1.9 472.4±2.5
(CH3)2CH-H 412.5±1.7 HCCH 551.2±0.1 557.8±0.3
(CH3)3C-H 403.8±1.7
H2C=CHCH2-H 371.5±1.7
HC(O)-H 368.6±0.8 C6H5CH2-H 375.3±2.5
CH3C(O)-H 374.0±1.2
Trends in C-H BDEs
It is important to remember that C-H BDEs refer to the energy it takes to break the bond, and is the difference in energy between the reactants and the products. Therefore, it is not appropriate to interpret BDEs solely in terms of the "stability of the radical products" as is often done.
Analysis of the BDEs shown in the table above shows that there are some systematic trends:
1. BDEs vary with hybridization: Bonds with sp3 hybridized carbons are weakest and bonds with sp hybridized carbons are much stronger. The vinyl and phenyl C-H bonds are similar, reflecting their sp2 hybridization. The correlation with hybridization can be viewed as a reflection of the C-H bond lengths. Longer bonds formed with sp3 orbitals are consequently weaker. Shorter bonds formed with orbitals that have more s-character are similarly stronger.
2. C-H BDEs vary with substitution: Among sp3 hybridized systems, methane has the strongest C-H bond. C-H bonds on primary carbons are stronger than those on secondary carbons, which are stronger than those on tertiary carbons.
Interpretation of C-H BDEs for sp3 Hybridized Carbons
The interpretation of the BDEs in saturated molecules has been subject of recent controversy. As indicated above, the variation in BDEs with substitution has traditionally been interpreted as reflecting the stabilities of the alkyl radicals, with the assessment that more highly substituted radicals are more stable, as with carbocations. Although this is a popular explanation, it fails to account fo the fact the bonds to groups other than H do not show the same types of variation.
R BDE(R-CH3) BDE(R-Cl) BDE(R-Br) BDE(R-OH)
CH3- 377.0±0.4 350.2±0.4 301.7±1.3 385.3±0.4
CH3CH2-
372.4±1.7
354.8±2.1 302.9±2.5 393.3±1.7
(CH3)2CH- 370.7±1.7 356.5±2.1 309.2±2.9 399.6±1.7
(CH3)3C- 366.1±1.7 355.2±2.9 303.8±2.5 400.8±1.7
Therefore, although C-CH3 bonds get weaker with more substitution, the effect is not nearly as large as that observed with C-H bonds. The strengths of C-Cl and C-Br bonds are not affected by substitution, despite the fact that the same radicals are formed as when breaking C-H bonds, and the C-OH bonds in alcohols actually increase with more substitution.
Gronert has proposed that the variation in BDEs is alternately explained as resulting from destabilization of the reactants due to steric repulsion of the substituents, which is released in the nearly planar radicals.1 Considering that BDEs reflect the relative energies of reactants and products, either explanation can account for the trend in BDEs.
Another factor that needs to be considered is the electronegativity. The Pauling definition of electronegativity says that the bond dissociation energy between unequal partners is going to be dependent on the difference in electrongativities, according to the expression
$D_o(A-B) = \dfrac{D_o(A-A) + D_o(B-B)}{2} + (X_A - X_B)^2$
where $X_A$ and $X_B$ are the electronegativities and the bond energies are in eV. Therefore, the variation in BDEs can be interpreted as reflecting variation in the electronegativities of the different types of alkyl fragments.
There is likely some merit in all three interpretations. Since Gronert's original publication of his alternate explanation, there have been many desperate attempts to defend the radical stability explanation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.04%3A_Bond_Dissociation_Energy.txt |
Equilibrium Constant
For the hypothetical chemical reaction:
$aA + bB \rightleftharpoons cC + dD \tag{6.5.1}$
the equilibrium constant is defined as:
$K_C = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{6.5.2}$
The notation [A] signifies the molar concentration of species A. An alternative expression for the equilibrium constant involves partial pressures:
$K_P = \dfrac{P_C^c P_D^d}{P_A^aP_B^b} \tag{6.5.3}$
Note that the expression for the equilibrium constant includes only solutes and gases; pure solids and liquids do not appear in the expression. For example, the equilibrium expression for the reaction
$CaH_2 (s) + 2H_2O (g) \rightleftharpoons Ca(OH)_2 (s) + 2H_2 (g) \tag{6.5.4}$
is the following:
$K_C = \dfrac{[H_2]^2}{[H_2O]^2} \tag{6.5.5}$
Observe that the gas-phase species H2O and H2 appear in the expression but the solids CaH2 and Ca(OH)2 do not appear.
The equilibrium constant is most readily determined by allowing a reaction to reach equilibrium, measuring the concentrations of the various solution-phase or gas-phase reactants and products, and substituting these values into the Law of Mass Action.
Gibbs Energy
The interaction between enthalpy and entropy changes in chemical reactions is best observed by studying their influence on the equilibrium constants of reversible reactions. To this end a new thermodynamic function called Free Energy (or Gibbs Free Energy), symbol ΔG, is defined as shown in the first equation below. Two things should be apparent from this equation. First, in cases where the entropy change is small, ΔG ≅ ΔH. Second, the importance of ΔS in determining ΔG increases with increasing temperature.
$ΔGº = ΔHº – TΔSº \tag{6.5.6}$
where $T$ is the absolute temperature measured in kelvin.
The free energy function provides improved insight into the thermodynamic driving forces that influence reactions. A negative ΔGº is characteristic of an exergonic reaction, one which is thermodynamically favorable and often spontaneous, as is the melting of ice at 1 ºC. Likewise a positive ΔGº is characteristic of an endergonic reaction, one which requires an input of energy from the surroundings.
Example 6.5.1: Decomposition of Cyclobutane
For an example of the relationship of free energy to enthalpy consider the decomposition of cyclobutane to ethene, shown in the following equation. The standard state for all the compounds is gaseous.
This reaction is endothermic, but the increase in number of molecules from one (reactants) to two (products) results in a large positive ΔSº.
At 25 ºC (298 ºK), ΔGº = 19 kcal/mol – 298(43.6) cal/mole = 19 – 13 kcal/mole = +6 kcal/mole. Thus, the entropy change opposes the enthalpy change, but is not sufficient to change the sign of the resulting free energy change, which is endergonic. Indeed, cyclobutane is perfectly stable when kept at room temperature.
Because the entropy contribution increases with temperature, this energetically unfavorable transformation can be made favorable by raising the temperature. At 200 ºC (473 ºK), ΔGº = 19 kcal/mol – 473(43.6) cal/mole = 19 – 20.6 kcal/mole = –1.6 kcal/mole. This is now an exergonic reaction, and the thermal cracking of cyclobutane to ethene is known to occur at higher temperatures.
$ΔGº = –RT \ln K = –2.303RT \log K$
with
• R = 1.987 cal/ ºK mole
• T = temperature in ºK
• K = equilibrium constant
A second equation, shown above, is important because it demonstrates the fundamental relationship of ΔGº to the equilibrium constant, K. Because of the negative logarithmic relationship between these variables, a negative ΔGº generates a K>1, whereas a positive ΔGº generates a K<1. When ΔGº = 0, K = 1. Furthermore, small changes in ΔGº produce large changes in K. A change of 1.4 kcal/mole in ΔGº changes K by approximately a factor of 10. This interrelationship may be explored with the calculator on the right. Entering free energies outside the range -8 to 8 kcal/mole or equilibrium constants outside the range 10-6 to 900,000 will trigger an alert, indicating the large imbalance such numbers imply.
Substituted Cyclohexanes
A Values
Substituents on a cyclohexane prefer to be in the equatorial position. When a substituent is in the axial position, there are two gauche butane interactions more than when a substituent is in the equatorial position. We quantify the energy difference between the axial and equatorial conformations as the A-value, which is equivalent to the negative of the ∆G°, for the equilibrium shown below. Therefore the A-value, or -∆G°, is the preference for the substituent to sit in the equatorial position.
Recall that the equilibrium constant is related to the change in Gibbs Energies for the reaction:
$\Delta{G^o} = -RT\ln {K_{eq}}$
The balance between reactants and products in a reaction will be determined by the free energy difference between the two sides of the reaction. The greater the free energy difference, the more the reaction will favor one side or the other (Table 6.5.1).
Table 6.5.1: Below is a table of A-values for some common substituents.
Substituent ∆G° (kcal/mol) A-value
-F
-0.28-0.24 0.24-0.28
-Cl -0.53 0.53
-Br -0.48 0.48
-I -0.47 0.47
-CH3 (-Me) -1.8 1.8
-CH2CH3 (-Et) -1.8 1.8
-CH(CH3)2 (-i-Pr) -2.1 2.1
-C(CH3)3 (-t-Bu) <-4.5 >4.5
-CHCH2 -1.7 1.7
-CCH -0.5 0.5
-CN -0.25-0.15 0.15-0.25
-C6H5 (-Ph) -2.9 2.9
Polysubstituted Cyclohexanes
1,4-disubstitution
The A-values of the substituents are roughly additive in either the cis- or trans-diastereomers.
1,3-disubstitution
A-values are only additive in the trans-diastereomer:
When there are cis-substituents on the chair, there is a new interaction in the di-axial conformation:
In the above example, each methyl group has one 1,3-diaxial interaction with a hydrogen. The methyl groups also interact with each other. This new diaxial interaction is extremely unfavorable based on their steric interaction (see: double-gauche pentane conformation). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.05%3A_Thermodynamics.txt |
Enthalpy
Thermodynamics is the study of the relationship between heat (or energy) and work. Enthalpy is a central factor in thermodynamics. It is the heat content of a system. The heat that passes into or out of the system during a reaction is the enthalpy change. Whether the enthalpy of the system increases (i.e. when energy is added) or decreases (because energy is given off) is a crucial factor that determines whether a reaction can happen.
Sometimes, we call the energy of the molecules undergoing change the "internal enthalpy". Sometimes, we call it the "enthalpy of the system." These two phrases refer to the same thing. Similarly, the energy of the molecules that do not take part in the reaction is called the "external enthalpy" or the "enthalpy of the surroundings".
Roughly speaking, the energy changes that we looked at in the introduction to thermodynamics were changes in enthalpy. We will see in the next section that there is another energetic factor, entropy, that we also need to consider in reactions. For now, we will just look at enthalpy.
• Enthalpy is the heat content of a system.
• The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.
• A reaction is favored if the enthalpy of the system decreases over the reaction.
That last statement is a lot like the description of energetics on the previous page. If a system undergoes a reaction and gives off energy, its own energy content decreases. It has less energy left over if it gave some away. Why does the energy of a set of molecules change when a reaction occurs? To answer that, we need to think about what happens in a chemical reaction.
In a reaction, there is a change in chemical bonding. Some of the bonds in the reactants are broken, and new bonds are made to form the products. It costs energy to break bonds, but energy is released when new bonds are made.
Whether a reaction is able to go forward may depend on the balance between these bond-making and bond-breaking steps.
• A reaction is exothermic if more energy is released by formation of new bonds than is consumed by breaking old bonds.
• A reaction is exothermic if weaker bonds are traded for stronger ones.
• A reaction is endothermic if bond-breaking costs more energy than what is provided in bond-making.
Bond energies (the amount of energy that must be added in order to break a bond) are an important factor in determining whether a reaction will occur. Bond strengths are not always easy to predict, because the strength of a bond depends on a number of factors. However, lots of people have done lots of work measuring bond strengths, and they have collected the information in tables, so if you need to know how strong a bond is, you can just look up the information you need.
Bond Bond Energy (kcal/mol) Bond Bond Energy (kcal/mol)
H-H 104 O-H 111
C-C 83 C-H 99
O=O 119 N-H 93
N=N 226 C=O 180
For example, suppose you wanted to know whether the combustion of methane were an exothermic or endothermic reaction. I am going to guess that it's exothermic, because this reaction (and others like it) is used to provide heat for lots of homes by burning natural gas in furnaces.
The "combustion" of methane means that it is burned in air, so that it reacts with oxygen. The products of burning hydrocarbons are mostly carbon dioxide and water. The carbon atom in methane (CH4) gets incorporated into a carbon dioxide molecule. The hydrogen atoms get incorporated into water molecules. There are four hydrogen atoms in methane, so that's enough to make two molecules of H2O.
• Four C-H bonds must be broken in the combustion of methane.
• Four new O-H bonds are made when the hydrogens from methane are added into new water molecules.
• Two new C=O bonds are made when the carbon from methane is added into a CO2 molecule.
The other piece of the puzzle is the oxygen source for the reaction. Oxygen is present in the atmosphere mostly as O2. Because we need two oxygen atoms in the CO2 molecule and two more oxygen atoms for the two water molecules, we need a total of four oxygen atoms for the reaction, which could be provided by two O2 molecules.
• Two O=O bonds must be broken to provide the oxygen atoms for the products.
Altogether, that's four C-H and two O=O bonds broken, plus two C=O and four O-H bonds made. That's 4 x 99 kcal/mol for the C-H bonds and 2 x 119 kcal/mol for the O=O bonds, a total of 634 kJ/mol added. The reaction releases 2 x 180 kcal/mol for the C=O bonds and 4 x 111 kcla/mol for the OH bonds, totaling 804 kcal/mol. Overall, there is 170 kcal/mol more released than is consumed.
That means the reaction is exothermic, so it produces heat. It's probably a good way to heat your home.
Entropy
Observations of natural processes led a surprising number of chemists of the late 19th century (including Berthelot and Thomsen) to conclude that all spontaneous reactions must be exothermic since:
• Objects roll downhill spontaneously (i.e., energy is "lost" from the system)
• Objects do not roll uphill spontaneously (i.e., energy does not suddenly appear from nowhere)
If this were true all we would need to predict whether a reaction is spontaneous is the change of enthalpy. If $ΔH$ were negative, the process should be able to occur by itself. If ΔH were positive, the reaction could not occur by itself.
Indeed, almost all exothermic reactions are spontaneous at standard thermodynamic conditions (1 atmosphere pressure) and $25^oC$. However a number of common processes which are both endothermic and spontaneous are known. The most obvious are simple phase changes, like ice melting at room temperature. Also, many solids dissolve in water and simultaneously absorb heat.
So the energy is now dispersed among the molecules of liquid water which have access to all kinds of molecular motion states that were not available in the solid. At the same time, the ordered structure of the solid ice has given way to a much less organized flowing liquid:
But the actual change has occurred in the energy dispersal. This subtle property that matter possesses in terms of the way energy is dispersed in it is known as entropy. Entropy is sometimes erroneously referred to as "randomness" or even "disorder" but these descriptions do not fit the state of energy as well as they seem to describe some of the often obvious results.
Just as reactions which form stronger bonds tend to occur spontaneously, energy is constantly being dispersed or "spread out" in any process which either happens on its own or which we make happen.
It is the Second Law of Thermodynamics which gives us the criterion we are seeking to decide whether a reaction will be spontaneous or not (well, almost...):
In a spontaneous process the entropy of the universe increases.
The "universe" is a pretty big place. Recall the definitions we used when we introduced chemical thermodynamics. The system is that part of the universe on which we focus our attention--generally chemicals. The surroundings are everything else. Taken together they constitute the universe.
So the Second Law could be written this way for a spontaneous process:
$ΔS_{sys} + ΔS_{surr} > 0$
Entropy trends and physical properties
(values in )
1. Entropy increases with mass
• $F_{2 (g)}$ = 203 J/mol·K
• $Cl_{2 (g)}$ = 224 J/mol·K
• $Br_{2\;(g)}$ = 245 J/mol·K
• $I_{2\; (g)}$ = 261 J/mol·K
2. Entropy increases with melting, vaporization or sublimation
• $I_{2(s)}$ = 117 J/mol·K vs. $I_{2(ℓ)}$ = 261 J/mol·K and
• $H_2O_{(ℓ)}$ = 70 J/mol·K vs. $H_2O_{(g)}$ = 189 J/mol·K
3. Entropy increases when solids or liquids dissolve in water
• CH3OH(ℓ) = 127 J/mol·K vs. $CH_3OH_{(aq)}$ = 132 J/mol·K and
• NaCl(s) = 72J/mol·K vs. Na+(aq) + Cl-(aq) = 115 J/mol·K
4. Entropy decreases when a gas is dissolved in water
• $HCl_{(g)}$ = 187 vs. $H^+_{(aq)}$ + $Cl^-_{(aq)}$ = 55
5. Entropy is lower in hard, brittle substances than in malleable solids like metals
• Diamond (C) = 2.4J/mol·K vs. Pb = 65 J/mol·K
6. Entropy increases with chemical complexity
• $NaCl$ = 72 J/mol·K vs. $MgCl_2$ = 90 J/mol·K vs. $AlCl_3$ = 167 J/mol·K
Of course, the main issue here is how entropy changes during a process. This can be determined by calculation from standard entropy values ($S^o$) in the same way that enthalpy changes are calculated:
$\sum{S^o_{products}} -\sum{S^o_{reactants}} = ΔS^o_{rxn}$ | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.06%3A_Enthalpy_and_Entropy.txt |
You may recall from general chemistry that it is often convenient to describe chemical reactions with energy diagrams. In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products. The energy diagram for a typical one-step reaction might look like this:
Despite its apparent simplicity, this energy diagram conveys some very important ideas about the thermodynamics and kinetics of the reaction. Recall that when we talk about the thermodynamics of a reaction, we are concerned with the difference in energy between reactants and products, and whether a reaction is ‘downhill’ (exergonic, energy releasing) or ‘uphill (endergonic, energy absorbing). When we talk about kinetics, on the other hand, we are concerned with the rate of the reaction, regardless of whether it is uphill or downhill thermodynamically.
First, let’s review what this energy diagram tells us about the thermodynamics of the reaction illustrated by the energy diagram above. The energy level of the products is lower than that of the reactants. This tells us that the change in standard Gibbs Free Energy for the reaction (Δrnx) is negative. In other words, the reaction is exergonic, or ‘downhill’. Recall that the Δrnx term encapsulates both Δrnx, the change in enthalpy (heat) and Δrnx , the change in entropy (disorder):
$ΔG˚ = ΔH˚- TΔS˚$
where T is the absolute temperature in Kelvin. For chemical processes where the entropy change is small (~0), the enthalpy change is essentially the same as the change in Gibbs Free Energy. Energy diagrams for these processes will often plot the enthalpy (H) instead of Free Energy for simplicity.
The standard Gibbs Free Energy change for a reaction can be related to the reaction's equilibrium constant ($K_{eq}\_) by a simple equation: $ΔG˚ = -RT \ln K_{eq}$ where: • Keq = [product] / [reactant] at equilibrium • R = 8.314 J×K-1×mol-1 or 1.987 cal× K-1×mol-1 • T = temperature in Kelvin (K) If you do the math, you see that a negative value for Δrnx (an exergonic reaction) corresponds - as it should by intuition - to Keq being greater than 1, an equilibrium constant which favors product formation. In a hypothetical endergonic (energy-absorbing) reaction the products would have a higher energy than reactants and thus Δrnx would be positive and Keq would be less than 1, favoring reactants. Now, let's move to kinetics. Look again at the energy diagram for exergonic reaction: although it is ‘downhill’ overall, it isn’t a straight downhill run. First, an ‘energy barrier’ must be overcome to get to the product side. The height of this energy barrier, you may recall, is called the ‘activation energy’ (ΔG). The activation energy is what determines the kinetics of a reaction: the higher the energy hill, the slower the reaction. At the very top of the energy barrier, the reaction is at its transition state (TS), which is the point at which the bonds are in the process of breaking and forming. The transition state is an ‘activated complex’: a transient and dynamic state that, unlike more stable species, does not have any definable lifetime. It may help to imagine a transition state as being analogous to the exact moment that a baseball is struck by a bat. Transition states are drawn with dotted lines representing bonds that are in the process of breaking or forming, and the drawing is often enclosed by brackets. Here is a picture of a likely transition state for a substitution reaction between hydroxide and chloromethane: $CH_3Cl + HO^- \rightarrow CH_3OH + Cl^-$ This reaction involves a collision between two molecules: for this reason, we say that it has second order kinetics. The rate expression for this type of reaction is: rate = k[reactant 1][reactant 2] . . . which tells us that the rate of the reaction depends on the rate constant k as well as on the concentration of both reactants. The rate constant can be determined experimentally by measuring the rate of the reaction with different starting reactant concentrations. The rate constant depends on the activation energy, of course, but also on temperature: a higher temperature means a higher k and a faster reaction, all else being equal. This should make intuitive sense: when there is more heat energy in the system, more of the reactant molecules are able to get over the energy barrier. Here is one more interesting and useful expression. Consider a simple reaction where the reactants are A and B, and the product is AB (this is referred to as a condensation reaction, because two molecules are coming together, or condensing). If we know the rate constant k for the forward reaction and the rate constant kreverse for the reverse reaction (where AB splits apart into A and B), we can simply take the quotient to find our equilibrium constant \(K_{eq}$:
This too should make some intuitive sense; if the forward rate constant is higher than the reverse rate constant, equilibrium should lie towards products.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Further Reading
MasterOrganicChemistry
Equilibria
Websites
Reversible and irreversible reactions
6.08: Energy Diagram for a Two-Step Reaction Mechanism
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches:
This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry.
We saw that SN2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of SN1 reactions? In the model SN1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp2-hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an SN1 reaction with water as the incoming nucleophile.
Exercise
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
The SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an SN1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.07%3A_Energy_Diagrams.txt |
Activation Energy
Since exothermic reactions are energetically (thermodynamically) favored, a careless thinker might conclude that all such reactions will proceed spontaneously to their products. Were this true, no life would exist on Earth, because the numerous carbon compounds that are present in and essential to all living organisms would spontaneously combust in the presence of oxygen to give carbon dioxide-a more stable carbon compound. The combustion of methane (eq.1), for example, does not occur spontaneously, but requires an initiating energy in the form of a spark or flame. The flaw in this careless reasoning is that we have focused only on the initial (reactant) and final (product) states of reactions. To understand why some reactions occur readily (almost spontaneously), whereas other reactions are slow, even to the point of being unobservable, we need to consider the intermediate stages of reactions.
Exothermic
Single Step Reaction
Endothermic
Single Step Reaction
Exothermic
Two Step Reaction
Every reaction in which bonds are broken will have a high energy transition state that must be reached before products can form. In order for the reactants to reach this transition state, energy must be supplied and reactant molecules must orient themselves in a suitable fashion. The energy needed to raise the reactants to the transition state energy level is called the activation energy, ΔE. An example of a single-step exothermic reaction profile is shown on the left above, and a similar single-step profile for an endothermic reaction is in the center. The activation energy is drawn in red in each case, and the overall energy change (ΔE) is in green.
The profile becomes more complex when a multi-step reaction path is described. An example of a two-step reaction proceeding by way of a high energy intermediate is shown on the right above. Here there are two transition states, each with its own activation energy. The overall activation energy is the difference in energy between the reactant state and the highest energy transition state. We see now why the rate of a reaction may not correlate with its overall energy change. In the exothermic diagram on the left, a significant activation energy must be provided to initiate the reaction. Since the reaction is strongly exothermic, it will probably generate enough heat to keep going as long as reactants remain. The endothermic reaction in the center has a similar activation energy, but this will have to be supplied continuously for the reaction to proceed to completion.
What is the source of the activation energy that enables a chemical reaction to occur? Often it is heat, as noted above in reference to the flame or spark that initiates methane combustion. At room temperature, indeed at any temperature above absolute zero, the molecules of a compound have a total energy that is a combination of translational (kinetic) energy, internal vibrational and rotational energies, as well as electronic and nuclear energies. The temperature of a system is a measure of the average kinetic energy of all the atoms and molecules present in the system. As shown in the following diagram, the average kinetic energy increases and the distribution of energies broadens as the temperature is raised from T1 to T2. Portions of this thermal or kinetic energy provide the activation energy for many reactions, the concentration of suitably activated reactant molecules increasing with temperature, e.g. orange area for T1 and yellow plus orange for T2. (Note that the area under a curve or a part of a curve is proportional to the number of molecules represented.)
Distribution of Molecular Kinetic Energy
at Two Different Temperatures, T1 & T2
Reaction Rates and Kinetics
Chemical reactivity is the focus of chemistry, and the study of reaction rates provides essential information about this subject. Some reactions proceed so rapidly they seem to be instantaneous, whereas other reactions are so slow they are nearly unobservable. Most of the reactions described in this text take place in from 0.2 to 12 hours at 25 ºC. Temperature is important, since fast reactions may be slowed or stopped by cooling, and slow reactions are accelerated by heating. When a reaction occurs between two reactant species, it proceeds faster at higher concentrations of the reactants. These facts lead to the following general analysis of reaction rates.
Reaction Rate =
Number of Collisions
between Reactant Molecules
per Unit of Time
•
Fraction of Collisions
with Sufficient Energy
to pass the Transition State
•
an Orientational
or Probability
Factor
Since reacting molecules must collide to interact, and the necessary activation energy must come from the kinetic energy of the colliding molecules, the first two factors are obvious. The third (probability) factor incorporates the orientational requirements of the reaction. For example, the addition of bromine to a double bond at the end of a six-carbon chain (1-hexene) could only occur if the colliding molecules came together in a way that allowed the bromine molecule to interact with the pi-electrons of the double bond.
The collision frequency of reactant molecules will be proportional to their concentration in the reaction system. This aspect of a reaction rate may be incorporated in a rate equation, which may take several forms depending on the number of reactants. Three general examples are presented in the following table.
Reaction Type
Rate Equation
Reaction Order
A ——> B
Reaction Rate = k•[A]
First Order Reaction (no collision needed)
A + B ——> C + D
Reaction Rate = k•[A]*[B]
Second Order Reaction
A + A ——> D
Reaction Rate = k•[A]2
Second Order Reaction
These rate equations take the form Reaction Rate = k[X] n[Y] m, where the proportionality constant k reflects the unique characteristics of a specific reaction, and is called the rate constant. The concentrations of reactants X and Y are [X] and [Y] respectively, and n & m are exponential numbers used to fit the rate equation to the experimental data. The sum n + m is termed the kinetic order of a reaction. The first example is a simple first order process. The next two examples are second order reactions, since n + m = 2. The kinetic order of a reaction is usually used to determine its molecularity.
In writing a rate equation we have disconnected the collision frequency term from the activation energy and probability factors defined above, which are necessarily incorporated in the rate constant k. This is demonstrated by the following equation.
The complex parameter A incorporates the probability factor. Because of the exponential relationship of k and the activation energy small changes in ΔE will cause relatively large changes in reaction rate. An increase in temperature clearly acts to increase k, but of greater importance is the increase in average molecular kinetic energy such an increase produces. This was illustrated in a previous diagram, increase in temperature from T1 to T2 producing a larger proportion of reactant molecules having energies equal or greater than the activation energy (designated by the red line.
6.10: Catalysts
We come now to the subject of catalysis. Our hypothetical bowl of sugar (from section 6.2) is still stubbornly refusing to turn into carbon dioxide and water, even though by doing so it would reach a much more stable energy state. There are, in fact, two ways that we could speed up the process so as to avoid waiting several millennia for the reaction to reach completion. We could supply enough energy, in the form of heat from a flame, to push some of the sugar molecules over the high energy hill. Heat would be released from the resulting exothermic reaction, and this energy would push more molecules over their energy hills, and so on - the sugar would literally burn up.
A second way to make the reaction go faster is to employ a catalyst. You probably already know that a catalyst is an agent that causes a chemical reaction to go faster by lowering its activation energy.
How might you catalyze the conversion of sugar to carbon dioxide and water? It’s not too hard – just eat the sugar, and let your digestive enzymes go to work catalyzing the many biochemical reactions involved in breaking it down. Enzymes are proteins, and are very effective catalysts. ‘Very effective’ in this context means very specific, and very fast. Most enzymes are very selective with respect to reactant molecules: they have evolved over millions of years to catalyze their specific reactions. An enzyme that attaches a phosphate group to glucose, for example, will not do anything at all to fructose (the details of these reactions are discussed in section 10.2B).
Glucose kinase is able to find and recognize glucose out of all of the other molecules floating around in the 'chemical soup' of a cell. A different enzyme, fructokinase, specifically catalyzes the phosphorylation of fructose.
We have already learned (section 3.9) that enzymes are very specific in terms of the stereochemistry of the reactions that they catalyze . Enzymes are also highly regiospecific, acting at only one specific part of a molecule. Notice that in the glucose kinase reaction above only one of the alcohol groups is phosphorylated.
Finally, enzymes are capable of truly amazing rate acceleration. Typical enzymes will speed up a reaction by anywhere from a million to a billion times, and the most efficient enzyme currently known to scientists is believed to accelerate its reaction by a factor of about 1017 (see Chemical and Engineering News, March 13, 2000, p. 42 for an interesting discussion about this enzyme, orotidine monophosphate decarboxylase).
We will now begin an exploration of some of the basic ideas about how enzymes accomplish these amazing feats of catalysis, and these ideas will be revisited often throughout the rest of the text as we consider various examples of enzyme-catalyzed organic reactions. But in order to begin to understand how enzymes work, we will first need to learn (or review, as the case may be) a little bit about protein structure.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.09%3A_Kinetics.txt |
The active site
A critical element in the three-dimensional structure of any enzyme is the presence of an ‘active site’, which is a pocket, usually located in the interior of the protein, that serves as a docking point for the enzyme’s substrate(s) (‘substrate’ is the term that biochemists use for a reactant molecule in an enzyme-catalyzed reaction). It is inside the active site pocket that enzymatic catalysis occurs. Shown below is an image of the glycolytic enzyme fructose-1,6-bisphosphate aldolase, with its substrate bound inside the active site pocket.
When the substrate binds to the active site, a large number of noncovalent interactions form with the amino acid residues that line the active site. The shape of the active site, and the enzyme-substrate interactions that form as a result of substrate binding, are specific to the substrate-enzyme pair: the active site has evolved to 'fit' one particular substrate and to catalyze one particular reaction. Other molecules do not fit in this active site nearly so well as fructose 1,6-bisphosphate.
Here are two close-up views of the same active site pocket, showing some of the specific hydrogen-bonding interactions between the substrate and active site amino acids. The first image below is a three-dimensional rendering directly from the crystal structure data. The substrate is shown in 'space-filling' style, while the active site amino acids are shown in the 'ball and stick' style. Hydrogens are not shown. The color scheme is grey for carbon, red for oxygen, blue for nitrogen, and orange for phosphorus.
Below is a two-dimensional picture of the substrate (colored red) surrounded by hydrogen-bonding active site amino acids. Notice that both main chain and side chain groups contribute to hydrogen bonding: in this figure, main chain H-bonding groups are colored blue, and side chain H-bonding groups are colored green.
Looking at the last three images should give you some appreciation for the specific manner in which a substrate fits inside its active site.
Transition state stabilization
One of the most important ways that an enzyme catalyzes any given reaction is through entropy reduction: by bringing order to a disordered situation (remember that entropy is a component of Gibbs Free Energy, and thus a component of the activation energy). Let’s turn again to our previous example (from section 6.1C) of a biochemical nucleophilic substitution reaction, the methylation of adenosine in DNA. The reaction is shown below with non-reactive sections of the molecules depicted by variously shaped 'bubbles' for the sake of simplicity.
In order for this reaction to occur, the two substrates (reactants) must come into contact in precisely the right way. If they are both floating around free in solution, the likelihood of this occurring is very small – the entropy of the system is simply too high. In other words, this reaction takes place very slowly without the help of a catalyst.
Here’s where the enzyme’s active site pocket comes into play. It is lined with various functional groups from the amino acid main and side chains, and has a very specific three-dimensional architecture that has evolved to bind to both of the substrates. If the SAM molecule, for example, diffuses into the active site, it can replace its (favorable) interactions with the surrounding water molecules with (even more favorable) new interactions with the functional groups lining the active site.
In a sense, SAM is moving from one solvent (water) to another 'solvent' (the active site), where many new energetically favorable interactions are possible. Remember: these new contacts between SAM and the active site groups are highly specific to SAM and SAM alone – no other molecule can ‘fit’ so well in this precise active site environment, and thus no other molecule will be likely to give up its contacts to water and bind to the active site.
The second substrate also has a specific spot reserved in the active site. (Because in this case the second substrate is a small segment of a long DNA molecule, the DNA-binding region of the active site is more of a 'groove' than a 'pocket').
So now we have both substrates bound in the active site. But they are not just bound in any random orientation – they are specifically positioned relative to one another so that the nucleophilic nitrogen is held very close to the electrophilic carbon, with a free path of attack. What used to be a very disordered situation – two reactants diffusing freely in solution – is now a very highly ordered situation, with everything set up for the reaction to proceed. This is what is meant by entropy reduction: the entropic component of the energy barrier has been lowered.
Looking a bit deeper, though, it is not really the noncovalent interaction between enzyme and substrate that are responsible for catalysis. Remember: all catalysts, enzymes included, accelerate reactions by lowering the energy of the transition state. With this in mind, it should make sense that the primary job of an enzyme is to maximize favorable interactions with the transition state, not with the starting substrates. This does not imply that enzyme-substrate interactions are not strong, rather that enzyme-TS interactions are far stronger, often by several orders of magnitude. Think about it this way: if an enzyme were to bind to (and stabilize) its substrate(s) more tightly than it bound to (and stabilized) the transition state, it would actually slow down the reaction, because it would be increasing the energy difference between starting state and transition state. The enzyme has evolved to maximize favorable noncovalent interactions to the transition state: in our example, this is the state in which the nucleophilic nitrogen is already beginning to attack the electrophilic carbon, and the carbon-sulfur bond has already begun to break.
In many enzymatic reactions, certain active site amino acid residues contribute to catalysis by increasing the reactivity of the substrates. Often, the catalytic role is that of acid and/or base. In our DNA methylation example, the nucleophilic nitrogen is deprotonated by a nearby aspartate side chain as it begins its nucleophilic attack on the methyl group of SAM. We will study nucleophilicity in greater detail in chapter 8, but it should make intuitive sense that deprotonating the amine increases the electron density of the nitrogen, making it more nucleophilic. Notice also in the figure below that the main chain carbonyl of an active site proline forms a hydrogen bond with the amine, which also has the effect of increasing the nitrogen's electron density and thus its nucleophilicity (Nucleic Acids Res. 2000, 28, 3950).
How does our picture of enzyme catalysis apply to multi-step reaction mechanisms? Although the two-step nucleophilic substitution reaction between tert-butyl chloride and hydroxide (section 6.1C) is not a biologically relevant process, let’s pretend just for the sake of illustration that there is a hypothetical enzyme that catalyzes this reaction.
The same basic principles apply here: the enzyme binds best to the transition state. But therein lies the problem: there are two transition states! To which TS does the enzyme maximize its contacts?
Recall that the first step – the loss of the chloride leaving group to form the carbocation intermediate – is the slower, rate-limiting step. It is this step that our hypothetical enzyme needs to accelerate if it wants to accelerate the overall reaction, and it is thus the energy of TS1 that needs to be lowered.
By Hammond’s postulate, we also know that the intermediate I is a close approximation of TS1. So the enzyme, by stabilizing the intermediate, will also stabilize TS1 (as well as TS2) and thereby accelerate the reaction.
If you read scientific papers about enzyme mechanisms, you will often see researchers discussing how an enzyme stabilizes a reaction intermediate. By virtue of Hammond's postulate, they are, at the same time, talking about how the enzyme lowers the energy of the transition state.
An additional note: although we have in this section been referring to SAM as a 'substrate' of the DNA methylation reaction, it is also often referred to as a coenzyme, or cofactor. These terms are used to describe small (relative to protein and DNA) biological organic molecules that bind specifically in the active site of an enzyme and help the enzyme to do its job. In the case of SAM, the job is methyl group donation. In addition to SAM, we will see many other examples of coenzymes in the coming chapters, a number of which - like ATP (adenosine triphosphate), coenzyme A, thiamine, and flavin - you have probably heard of before. The full structures of some common coenzymes are shown in table 6 in the tables section.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.11%3A_Enzymes.txt |
If you scan any organic textbook you will encounter what appears to be a very large, often intimidating, number of reactions. These are the "tools" of a chemist, and to use these tools effectively, we must organize them in a sensible manner and look for patterns of reactivity that permit us make plausible predictions. Most of these reactions occur at special sites of reactivity known as functional groups, and these constitute one organizational scheme that helps us catalog and remember reactions.
Ultimately, the best way to achieve proficiency in organic chemistry is to understand how reactions take place, and to recognize the various factors that influence their course.
First, we identify four broad classes of reactions based solely on the structural change occurring in the reactant molecules. This classification does not require knowledge or speculation concerning reaction paths or mechanisms. The four main reaction classes are additions, eliminations, substitutions, and rearrangements.
Rearrangement
In an addition reaction the number of σ-bonds in the substrate molecule increases, usually at the expense of one or more π-bonds. The reverse is true of elimination reactions, i.e.the number of σ-bonds in the substrate decreases, and new π-bonds are often formed. Substitution reactions, as the name implies, are characterized by replacement of an atom or group (Y) by another atom or group (Z). Aside from these groups, the number of bonds does not change. A rearrangement reaction generates an isomer, and again the number of bonds normally does not change.
The examples illustrated above involve simple alkyl and alkene systems, but these reaction types are general for most functional groups, including those incorporating carbon-oxygen double bonds and carbon-nitrogen double and triple bonds. Some common reactions may actually be a combination of reaction types. The reaction of an ester with ammonia to give an amide, as shown below, appears to be a substitution reaction ( Y = CH3O & Z = NH2 ); however, it is actually two reactions, an addition followed by an elimination.
The addition of water to a nitrile does not seem to fit any of the above reaction types, but it is simply a slow addition reaction followed by a rapid rearrangement, as shown in the following equation. Rapid rearrangements of this kind are called tautomerizations.
6.13: Thermodynamics
Equilibrium Constant
For the hypothetical chemical reaction:
$aA + bB \rightleftharpoons cC + dD \tag{6.5.1}$
the equilibrium constant is defined as:
$K_C = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{6.5.2}$
The notation [A] signifies the molar concentration of species A. An alternative expression for the equilibrium constant involves partial pressures:
$K_P = \dfrac{P_C^c P_D^d}{P_A^aP_B^b} \tag{6.5.3}$
Note that the expression for the equilibrium constant includes only solutes and gases; pure solids and liquids do not appear in the expression. For example, the equilibrium expression for the reaction
$CaH_2 (s) + 2H_2O (g) \rightleftharpoons Ca(OH)_2 (s) + 2H_2 (g) \tag{6.5.4}$
is the following:
$K_C = \dfrac{[H_2]^2}{[H_2O]^2} \tag{6.5.5}$
Observe that the gas-phase species H2O and H2 appear in the expression but the solids CaH2 and Ca(OH)2 do not appear.
The equilibrium constant is most readily determined by allowing a reaction to reach equilibrium, measuring the concentrations of the various solution-phase or gas-phase reactants and products, and substituting these values into the Law of Mass Action.
Gibbs Energy
The interaction between enthalpy and entropy changes in chemical reactions is best observed by studying their influence on the equilibrium constants of reversible reactions. To this end a new thermodynamic function called Free Energy (or Gibbs Free Energy), symbol ΔG, is defined as shown in the first equation below. Two things should be apparent from this equation. First, in cases where the entropy change is small, ΔG ≅ ΔH. Second, the importance of ΔS in determining ΔG increases with increasing temperature.
$ΔGº = ΔHº – TΔSº \tag{6.5.6}$
where $T$ is the absolute temperature measured in kelvin.
The free energy function provides improved insight into the thermodynamic driving forces that influence reactions. A negative ΔGº is characteristic of an exergonic reaction, one which is thermodynamically favorable and often spontaneous, as is the melting of ice at 1 ºC. Likewise a positive ΔGº is characteristic of an endergonic reaction, one which requires an input of energy from the surroundings.
Example 6.5.1: Decomposition of Cyclobutane
For an example of the relationship of free energy to enthalpy consider the decomposition of cyclobutane to ethene, shown in the following equation. The standard state for all the compounds is gaseous.
This reaction is endothermic, but the increase in number of molecules from one (reactants) to two (products) results in a large positive ΔSº.
At 25 ºC (298 ºK), ΔGº = 19 kcal/mol – 298(43.6) cal/mole = 19 – 13 kcal/mole = +6 kcal/mole. Thus, the entropy change opposes the enthalpy change, but is not sufficient to change the sign of the resulting free energy change, which is endergonic. Indeed, cyclobutane is perfectly stable when kept at room temperature.
Because the entropy contribution increases with temperature, this energetically unfavorable transformation can be made favorable by raising the temperature. At 200 ºC (473 ºK), ΔGº = 19 kcal/mol – 473(43.6) cal/mole = 19 – 20.6 kcal/mole = –1.6 kcal/mole. This is now an exergonic reaction, and the thermal cracking of cyclobutane to ethene is known to occur at higher temperatures.
$ΔGº = –RT \ln K = –2.303RT \log K$
with
• R = 1.987 cal/ ºK mole
• T = temperature in ºK
• K = equilibrium constant
A second equation, shown above, is important because it demonstrates the fundamental relationship of ΔGº to the equilibrium constant, K. Because of the negative logarithmic relationship between these variables, a negative ΔGº generates a K>1, whereas a positive ΔGº generates a K<1. When ΔGº = 0, K = 1. Furthermore, small changes in ΔGº produce large changes in K. A change of 1.4 kcal/mole in ΔGº changes K by approximately a factor of 10. This interrelationship may be explored with the calculator on the right. Entering free energies outside the range -8 to 8 kcal/mole or equilibrium constants outside the range 10-6 to 900,000 will trigger an alert, indicating the large imbalance such numbers imply.
Substituted Cyclohexanes
A Values
Substituents on a cyclohexane prefer to be in the equatorial position. When a substituent is in the axial position, there are two gauche butane interactions more than when a substituent is in the equatorial position. We quantify the energy difference between the axial and equatorial conformations as the A-value, which is equivalent to the negative of the ∆G°, for the equilibrium shown below. Therefore the A-value, or -∆G°, is the preference for the substituent to sit in the equatorial position.
Recall that the equilibrium constant is related to the change in Gibbs Energies for the reaction:
$\Delta{G^o} = -RT\ln {K_{eq}}$
The balance between reactants and products in a reaction will be determined by the free energy difference between the two sides of the reaction. The greater the free energy difference, the more the reaction will favor one side or the other (Table 6.5.1).
Table 6.5.1: Below is a table of A-values for some common substituents.
Substituent ∆G° (kcal/mol) A-value
-F
-0.28-0.24 0.24-0.28
-Cl -0.53 0.53
-Br -0.48 0.48
-I -0.47 0.47
-CH3 (-Me) -1.8 1.8
-CH2CH3 (-Et) -1.8 1.8
-CH(CH3)2 (-i-Pr) -2.1 2.1
-C(CH3)3 (-t-Bu) <-4.5 >4.5
-CHCH2 -1.7 1.7
-CCH -0.5 0.5
-CN -0.25-0.15 0.15-0.25
-C6H5 (-Ph) -2.9 2.9
Polysubstituted Cyclohexanes
1,4-disubstitution
The A-values of the substituents are roughly additive in either the cis- or trans-diastereomers.
1,3-disubstitution
A-values are only additive in the trans-diastereomer:
When there are cis-substituents on the chair, there is a new interaction in the di-axial conformation:
In the above example, each methyl group has one 1,3-diaxial interaction with a hydrogen. The methyl groups also interact with each other. This new diaxial interaction is extremely unfavorable based on their steric interaction (see: double-gauche pentane conformation). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.12%3A_Kinds_of_Organic_Reactions.txt |
You may recall from general chemistry that it is often convenient to describe chemical reactions with energy diagrams. In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products. The energy diagram for a typical one-step reaction might look like this:
Despite its apparent simplicity, this energy diagram conveys some very important ideas about the thermodynamics and kinetics of the reaction. Recall that when we talk about the thermodynamics of a reaction, we are concerned with the difference in energy between reactants and products, and whether a reaction is ‘downhill’ (exergonic, energy releasing) or ‘uphill (endergonic, energy absorbing). When we talk about kinetics, on the other hand, we are concerned with the rate of the reaction, regardless of whether it is uphill or downhill thermodynamically.
First, let’s review what this energy diagram tells us about the thermodynamics of the reaction illustrated by the energy diagram above. The energy level of the products is lower than that of the reactants. This tells us that the change in standard Gibbs Free Energy for the reaction (Δrnx) is negative. In other words, the reaction is exergonic, or ‘downhill’. Recall that the Δrnx term encapsulates both Δrnx, the change in enthalpy (heat) and Δrnx , the change in entropy (disorder):
$ΔG˚ = ΔH˚- TΔS˚$
where T is the absolute temperature in Kelvin. For chemical processes where the entropy change is small (~0), the enthalpy change is essentially the same as the change in Gibbs Free Energy. Energy diagrams for these processes will often plot the enthalpy (H) instead of Free Energy for simplicity.
The standard Gibbs Free Energy change for a reaction can be related to the reaction's equilibrium constant ($K_{eq}\_) by a simple equation: $ΔG˚ = -RT \ln K_{eq}$ where: • Keq = [product] / [reactant] at equilibrium • R = 8.314 J×K-1×mol-1 or 1.987 cal× K-1×mol-1 • T = temperature in Kelvin (K) If you do the math, you see that a negative value for Δrnx (an exergonic reaction) corresponds - as it should by intuition - to Keq being greater than 1, an equilibrium constant which favors product formation. In a hypothetical endergonic (energy-absorbing) reaction the products would have a higher energy than reactants and thus Δrnx would be positive and Keq would be less than 1, favoring reactants. Now, let's move to kinetics. Look again at the energy diagram for exergonic reaction: although it is ‘downhill’ overall, it isn’t a straight downhill run. First, an ‘energy barrier’ must be overcome to get to the product side. The height of this energy barrier, you may recall, is called the ‘activation energy’ (ΔG). The activation energy is what determines the kinetics of a reaction: the higher the energy hill, the slower the reaction. At the very top of the energy barrier, the reaction is at its transition state (TS), which is the point at which the bonds are in the process of breaking and forming. The transition state is an ‘activated complex’: a transient and dynamic state that, unlike more stable species, does not have any definable lifetime. It may help to imagine a transition state as being analogous to the exact moment that a baseball is struck by a bat. Transition states are drawn with dotted lines representing bonds that are in the process of breaking or forming, and the drawing is often enclosed by brackets. Here is a picture of a likely transition state for a substitution reaction between hydroxide and chloromethane: $CH_3Cl + HO^- \rightarrow CH_3OH + Cl^-$ This reaction involves a collision between two molecules: for this reason, we say that it has second order kinetics. The rate expression for this type of reaction is: rate = k[reactant 1][reactant 2] . . . which tells us that the rate of the reaction depends on the rate constant k as well as on the concentration of both reactants. The rate constant can be determined experimentally by measuring the rate of the reaction with different starting reactant concentrations. The rate constant depends on the activation energy, of course, but also on temperature: a higher temperature means a higher k and a faster reaction, all else being equal. This should make intuitive sense: when there is more heat energy in the system, more of the reactant molecules are able to get over the energy barrier. Here is one more interesting and useful expression. Consider a simple reaction where the reactants are A and B, and the product is AB (this is referred to as a condensation reaction, because two molecules are coming together, or condensing). If we know the rate constant k for the forward reaction and the rate constant kreverse for the reverse reaction (where AB splits apart into A and B), we can simply take the quotient to find our equilibrium constant \(K_{eq}$:
This too should make some intuitive sense; if the forward rate constant is higher than the reverse rate constant, equilibrium should lie towards products.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Further Reading
MasterOrganicChemistry
Equilibria
Websites
Reversible and irreversible reactions
6.15: Energy Diagram for a Two-Step Reaction Mechanism
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches:
This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry.
We saw that SN2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of SN1 reactions? In the model SN1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp2-hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an SN1 reaction with water as the incoming nucleophile.
Exercise
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
The SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an SN1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.14%3A_Energy_Diagrams.txt |
Activation Energy
Since exothermic reactions are energetically (thermodynamically) favored, a careless thinker might conclude that all such reactions will proceed spontaneously to their products. Were this true, no life would exist on Earth, because the numerous carbon compounds that are present in and essential to all living organisms would spontaneously combust in the presence of oxygen to give carbon dioxide-a more stable carbon compound. The combustion of methane (eq.1), for example, does not occur spontaneously, but requires an initiating energy in the form of a spark or flame. The flaw in this careless reasoning is that we have focused only on the initial (reactant) and final (product) states of reactions. To understand why some reactions occur readily (almost spontaneously), whereas other reactions are slow, even to the point of being unobservable, we need to consider the intermediate stages of reactions.
Exothermic
Single Step Reaction
Endothermic
Single Step Reaction
Exothermic
Two Step Reaction
Every reaction in which bonds are broken will have a high energy transition state that must be reached before products can form. In order for the reactants to reach this transition state, energy must be supplied and reactant molecules must orient themselves in a suitable fashion. The energy needed to raise the reactants to the transition state energy level is called the activation energy, ΔE. An example of a single-step exothermic reaction profile is shown on the left above, and a similar single-step profile for an endothermic reaction is in the center. The activation energy is drawn in red in each case, and the overall energy change (ΔE) is in green.
The profile becomes more complex when a multi-step reaction path is described. An example of a two-step reaction proceeding by way of a high energy intermediate is shown on the right above. Here there are two transition states, each with its own activation energy. The overall activation energy is the difference in energy between the reactant state and the highest energy transition state. We see now why the rate of a reaction may not correlate with its overall energy change. In the exothermic diagram on the left, a significant activation energy must be provided to initiate the reaction. Since the reaction is strongly exothermic, it will probably generate enough heat to keep going as long as reactants remain. The endothermic reaction in the center has a similar activation energy, but this will have to be supplied continuously for the reaction to proceed to completion.
What is the source of the activation energy that enables a chemical reaction to occur? Often it is heat, as noted above in reference to the flame or spark that initiates methane combustion. At room temperature, indeed at any temperature above absolute zero, the molecules of a compound have a total energy that is a combination of translational (kinetic) energy, internal vibrational and rotational energies, as well as electronic and nuclear energies. The temperature of a system is a measure of the average kinetic energy of all the atoms and molecules present in the system. As shown in the following diagram, the average kinetic energy increases and the distribution of energies broadens as the temperature is raised from T1 to T2. Portions of this thermal or kinetic energy provide the activation energy for many reactions, the concentration of suitably activated reactant molecules increasing with temperature, e.g. orange area for T1 and yellow plus orange for T2. (Note that the area under a curve or a part of a curve is proportional to the number of molecules represented.)
Distribution of Molecular Kinetic Energy
at Two Different Temperatures, T1 & T2
Reaction Rates and Kinetics
Chemical reactivity is the focus of chemistry, and the study of reaction rates provides essential information about this subject. Some reactions proceed so rapidly they seem to be instantaneous, whereas other reactions are so slow they are nearly unobservable. Most of the reactions described in this text take place in from 0.2 to 12 hours at 25 ºC. Temperature is important, since fast reactions may be slowed or stopped by cooling, and slow reactions are accelerated by heating. When a reaction occurs between two reactant species, it proceeds faster at higher concentrations of the reactants. These facts lead to the following general analysis of reaction rates.
Reaction Rate =
Number of Collisions
between Reactant Molecules
per Unit of Time
•
Fraction of Collisions
with Sufficient Energy
to pass the Transition State
•
an Orientational
or Probability
Factor
Since reacting molecules must collide to interact, and the necessary activation energy must come from the kinetic energy of the colliding molecules, the first two factors are obvious. The third (probability) factor incorporates the orientational requirements of the reaction. For example, the addition of bromine to a double bond at the end of a six-carbon chain (1-hexene) could only occur if the colliding molecules came together in a way that allowed the bromine molecule to interact with the pi-electrons of the double bond.
The collision frequency of reactant molecules will be proportional to their concentration in the reaction system. This aspect of a reaction rate may be incorporated in a rate equation, which may take several forms depending on the number of reactants. Three general examples are presented in the following table.
Reaction Type
Rate Equation
Reaction Order
A ——> B
Reaction Rate = k•[A]
First Order Reaction (no collision needed)
A + B ——> C + D
Reaction Rate = k•[A]*[B]
Second Order Reaction
A + A ——> D
Reaction Rate = k•[A]2
Second Order Reaction
These rate equations take the form Reaction Rate = k[X] n[Y] m, where the proportionality constant k reflects the unique characteristics of a specific reaction, and is called the rate constant. The concentrations of reactants X and Y are [X] and [Y] respectively, and n & m are exponential numbers used to fit the rate equation to the experimental data. The sum n + m is termed the kinetic order of a reaction. The first example is a simple first order process. The next two examples are second order reactions, since n + m = 2. The kinetic order of a reaction is usually used to determine its molecularity.
In writing a rate equation we have disconnected the collision frequency term from the activation energy and probability factors defined above, which are necessarily incorporated in the rate constant k. This is demonstrated by the following equation.
The complex parameter A incorporates the probability factor. Because of the exponential relationship of k and the activation energy small changes in ΔE will cause relatively large changes in reaction rate. An increase in temperature clearly acts to increase k, but of greater importance is the increase in average molecular kinetic energy such an increase produces. This was illustrated in a previous diagram, increase in temperature from T1 to T2 producing a larger proportion of reactant molecules having energies equal or greater than the activation energy (designated by the red line. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/06%3A_Understanding_Organic_Reactions/6.16%3A_Kinetics.txt |
Alkyl halides are also known as haloalkanes. This page explains what they are and discusses their physical properties. alkyl halides are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). We will only look at compounds containing one halogen atom. For example:
alkyl halides fall into different classes depending on how the halogen atom is positioned on the chain of carbon atoms. There are some chemical differences between the various types.
Primary alkyl halides
In a primary (1°) halogenoalkane, the carbon which carries the halogen atom is only attached to one other alkyl group.Some examples of primary alkyl halides include:
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH2 group holding the halogen. There is an exception to this: CH3Br and the other methyl halides are often counted as primary alkyl halides even though there are no alkyl groups attached to the carbon with the halogen on it.
Secondary alkyl halides
In a secondary (2°) halogenoalkane, the carbon with the halogen attached is joined directly to two other alkyl groups, which may be the same or different. Examples:
Tertiary alkyl halides
In a tertiary (3°) halogenoalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Contributors
Jim Clark (Chemguide.co.uk)
7.02: Nomenclature
e Learning Objective is to name halogenated hydrocarbons given formulas and write formulas for these compounds given names.
Many organic compounds are closely related to the alkanes. As we noted in Section 12.7, alkanes react with halogens to produce halogenated hydrocarbons, the simplest of which have a single halogen atom substituted for a hydrogen atom of the alkane. Even more closely related are the cycloalkanes, compounds in which the carbon atoms are joined in a ring, or cyclic fashion.
The reactions of alkanes with halogens produce halogenated hydrocarbons, compounds in which one or more hydrogen atoms of a hydrocarbon have been replaced by halogen atoms:
The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane). The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide. The IUPAC system uses the name of the parent alkane with a prefix indicating the halogen substituents, preceded by number indicating the substituent’s location. The prefixes are fluoro-, chloro-, bromo-, and iodo-. Thus CH3CH2Cl has the common name ethyl chloride and the IUPAC name chloroethane. Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names.
Example 3
Give the common and IUPAC names for each compound.
1. CH3CH2CH2Br
2. (CH3)2CHCl
SOLUTION
1. The alkyl group (CH3CH2CH2–) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane.
2. The alkyl group [(CH3)2CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro-) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane.
Skill-Building Exercise
Give common and IUPAC names for each compound.
1. CH3CH2I
2. CH3CH2CH2CH2F
Example 4
Give the IUPAC name for each compound.
SOLUTION
1. The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
2. The parent alkane is hexane. Methyl (CH3) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.
Skill-Building Exercise
Give the IUPAC name for each compound.
A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH4) can react with chlorine (Cl2), replacing one, two, three, or all four hydrogen atoms with Cl atoms. Several halogenated products derived from methane and ethane (CH3CH3) are listed in Table 12.6 "Some Halogenated Hydrocarbons", along with some of their uses.
Table 12.6 Some Halogenated Hydrocarbons
Formula Common Name IUPAC Name Some Important Uses
Derived from CH4
CH3Cl methyl chloride chloromethane refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber
CH2Cl2 methylene chloride dichloromethane laboratory and industrial solvent
CHCl3 chloroform trichloromethane industrial solvent
CCl4 carbon tetrachloride tetrachloromethane dry-cleaning solvent and fire extinguishers (but no longer recommended for use)
CBrF3 halon-1301 bromotrifluoromethane fire extinguisher systems
CCl3F chlorofluorocarbon-11 (CFC-11) trichlorofluoromethane foaming plastics
CCl2F2 chlorofluorocarbon-12 (CFC-12) dichlorodifluoromethane refrigerant
Derived from CH3CH3
CH3CH2Cl ethyl chloride chloroethane local anesthetic
ClCH2CH2Cl ethylene dichloride 1,2-dichloroethane solvent for rubber
CCl3CH3 methylchloroform 1,1,1-trichloroethane solvent for cleaning computer chips and molds for shaping plastic | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.01%3A_Introduction_to_Alkyl_Halides.txt |
Physical properties of alkyl halides
Boiling Points
The chart shows the boiling points of some simple alkyl halides.
Notice that three of these have boiling points below room temperature (taken as being about 20°C). These will be gases at room temperature. All the others you are likely to come across are liquids. Remember:
• the only methyl halide which is a liquid is iodomethane;
• chloroethane is a gas.
The patterns in boiling point reflect the patterns in intermolecular attractions.
van der Waals dispersion forces
These attractions get stronger as the molecules get longer and have more electrons. That increases the sizes of the temporary dipoles that are set up. This is why the boiling points increase as the number of carbon atoms in the chains increases. Look at the chart for a particular type of halide (a chloride, for example). Dispersion forces get stronger as you go from 1 to 2 to 3 carbons in the chain. It takes more energy to overcome them, and so the boiling points rise.
The increase in boiling point as you go from a chloride to a bromide to an iodide (for a given number of carbon atoms) is also because of the increase in number of electrons leading to larger dispersion forces. There are lots more electrons in, for example, iodomethane than there are in chloromethane - count them!
van der Waals dipole-dipole attractions
The carbon-halogen bonds (apart from the carbon-iodine bond) are polar, because the electron pair is pulled closer to the halogen atom than the carbon. This is because (apart from iodine) the halogens are more electronegative than carbon. The electronegativity values are:
C 2.5 F 4.0
Cl 3.0
Br 2.8
I 2.5
This means that in addition to the dispersion forces there will be forces due to the attractions between the permanent dipoles (except in the iodide case). The size of those dipole-dipole attractions will fall as the bonds get less polar (as you go from chloride to bromide to iodide, for example). Nevertheless, the boiling points rise! This shows that the effect of the permanent dipole-dipole attractions is much less important than that of the temporary dipoles which cause the dispersion forces. The large increase in number of electrons by the time you get to the iodide completely outweighs the loss of any permanent dipoles in the molecules.
Example 1: Boiling Points of Some Isomers
The examples show that the boiling points fall as the isomers go from a primary to a secondary to a tertiary halogenoalkane. This is a simple result of the fall in the effectiveness of the dispersion forces.
The temporary dipoles are greatest for the longest molecule. The attractions are also stronger if the molecules can lie closely together. The tertiary halogenoalkane is very short and fat, and won't have much close contact with its neighbours.
Solubility
Solubility in water
The alkyl halides are at best only slightly soluble in water. For a halogenoalkane to dissolve in water you have to break attractions between the halogenoalkane molecules (van der Waals dispersion and dipole-dipole interactions) and break the hydrogen bonds between water molecules. Both of these cost energy.
Energy is released when new attractions are set up between the halogenoalkane and the water molecules. These will only be dispersion forces and dipole-dipole interactions. These aren't as strong as the original hydrogen bonds in the water, and so not as much energy is released as was used to separate the water molecules. The energetics of the change are sufficiently "unprofitable" that very little dissolves.
Solubility in organic solvents
alkyl halides tend to dissolve in organic solvents because the new intermolecular attractions have much the same strength as the ones being broken in the separate halogenoalkane and solvent.
Chemical Reactivity
The pattern in strengths of the four carbon-halogen bonds are:
Notice that bond strength falls as you go from C-F to C-I, and notice how much stronger the carbon-fluorine bond is than the rest. To react with the alkyl halides, the carbon-halogen bond has got to be broken. Because that gets easier as you go from fluoride to chloride to bromide to iodide, the compounds get more reactive in that order. Iodoalkanes are the most reactive and fluoroalkanes are the least. In fact, fluoroalkanes are so unreactive that we shall pretty well ignore them completely from now on in this section! | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.03%3A_Physical_Properties.txt |
Halogen containing organic compounds are relatively rare in terrestrial plants and animals. The thyroid hormones T3 and T4 are exceptions; as is fluoroacetate, the toxic agent in the South African shrub Dichapetalum cymosum, known as "gifblaar". However, the halogen rich environment of the ocean has produced many interesting natural products incorporating large amounts of halogen. Some examples are shown below.
The ocean is the largest known source for atmospheric methyl bromide and methyl iodide. Furthermore, the ocean is also estimated to supply 10-20% of atmospheric methyl chloride, with other significant contributions coming from biomass burning, salt marshes and wood-rotting fungi. Many subsequent chemical and biological processes produce poly-halogenated methanes.
Synthetic organic halogen compounds are readily available by direct halogenation of hydrocarbons and by addition reactions to alkenes and alkynes. Many of these have proven useful as intermediates in traditional synthetic processes. Some halogen compounds, shown in the box. have been used as pesticides, but their persistence in the environment, once applied, has led to restrictions, including banning, of their use in developed countries. Because DDT is a cheap and effective mosquito control agent, underdeveloped countries in Africa and Latin America have experienced a dramatic increase in malaria deaths following its removal, and arguments are made for returning it to limited use. 2,4,5-T and 2,4-D are common herbicides that are sold by most garden stores. Other organic halogen compounds that have been implicated in environmental damage include the polychloro- and polybromo-biphenyls (PCBs and PBBs), used as heat transfer fluids and fire retardants; and freons (e.g. CCl2F2 and other chlorofluorocarbons) used as refrigeration gases and fire extinguishing agents.
7.05: The Polar CarbonHalogen Bond
Halogens and the Character of the Carbon-Halogen Bond
With respect to electronegativity, halogens are more electronegative than carbons. This results in a carbon-halogen bond that is polarized. As shown in the image below, carbon atom has a partial positive charge, while the halogen has a partial negative charge.
The following image shows the relationship between the halogens and electronegativity. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
The influence of bond polarity
Of the four halogens, fluorine is the most electronegative and iodine the least. That means that the electron pair in the carbon-fluorine bond will be dragged most towards the halogen end. Looking at the methyl halides as simple examples:
The electronegativities of carbon and iodine are equal and so there will be no separation of charge on the bond.
One of the important set of reactions of alkyl halides involves replacing the halogen by something else - substitution reactions. These reactions involve either:
• the carbon-halogen bond breaking to give positive and negative ions. The ion with the positively charged carbon atom then reacts with something either fully or slightly negatively charged.
• something either fully or negatively charged attracted to the slightly positive carbon atom and pushing off the halogen atom.
You might have thought that either of these would be more effective in the case of the carbon-fluorine bond with the quite large amounts of positive and negative charge already present. But that's not so - quite the opposite is true! The thing that governs the reactivity is the strength of the bonds which have to be broken. If is difficult to break a carbon-fluorine bond, but easy to break a carbon-iodine one. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.04%3A_Interesting_Alkyl_Halides.txt |
In many ways, the proton transfer process in a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon.
In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) attacks an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base.
In the next few sections, we are going to be discussing some general aspects of nucleophilic substitution reactions, and in doing so it will simplify things greatly if we can use some abbreviations and generalizations before we dive into real examples.
Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. In a similar fashion, we will call the leaving group 'X'. We will see as we study actual reactions that leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. We will also see some examples of nucleophiles that are negatively charged and some that are neutral. Therefore, in this general picture we will not include a charge designation on the 'X' or 'Nu' species. In the same way, we will see later that nucleophiles and leaving groups are sometimes protonated and sometimes not, so for now, for the sake of simplicity, we will not include protons on 'Nu' or 'X'. We will generalize the three other groups bonded on the electrophilic central carbon as R1, R2, and R3: these symbols could represent hydrogens as well as alkyl groups. Finally, in order to keep figures from becoming too crowded, we will use in most cases the line structure convention in which the central, electrophilic carbon is not drawn out as a 'C'.
Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction:
The functional group of alkyl halides is a carbon-halogen bond, the common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing on the right. Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is covalent bond strength. The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, alkyl fluorides and fluorocarbons in general are chemically and thermodynamically quite stable, and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative stability of the corresponding halide anions, which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pKa = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI.
In order to understand why some combinations of alkyl halides and nucleophiles give a substitution reaction, whereas other combinations give elimination, and still others give no observable reaction, we must investigate systematically the way in which changes in reaction variables perturb the course of the reaction. The following general equation summarizes the factors that will be important in such an investigation.
One conclusion, relating the structure of the R-group to possible products, should be immediately obvious. If R- has no beta-hydrogens an elimination reaction is not possible, unless a structural rearrangement occurs first. The first four halides shown on the left below do not give elimination reactions on treatment with base, because they have no β-hydrogens. The two halides on the right do not normally undergo such reactions because the potential elimination products have highly strained double or triple bonds.
It is also worth noting that sp2 hybridized C–X compounds, such as the three on the right, do not normally undergo nucleophilic substitution reactions, unless other functional groups perturb the double bond(s).
Using the general reaction shown above as our reference, we can identify the following variables and observables.
Variables
R change α-carbon from 1º to 2º to 3º
if the α-carbon is a chiral center, set as (R) or (S)
X change from Cl to Br to I (F is relatively unreactive)
Nu: change from anion to neutral; change basicity; change polarizability
Solvent polar vs. non-polar; protic vs. non-protic
Observables
Products substitution, elimination, no reaction.
Stereospecificity if the α-carbon is a chiral center what happens to its configuration?
Reaction Rate measure as a function of reactant concentration.
When several reaction variables may be changed, it is important to isolate the effects of each during the course of study. In other words: only one variable should be changed at a time, the others being held as constant as possible. For example, we can examine the effect of changing the halogen substituent from Cl to Br to I, using ethyl as a common R–group, cyanide anion as a common nucleophile, and ethanol as a common solvent. We would find a common substitution product, C2H5–CN, in all cases, but the speed or rate of the reaction would increase in the order: Cl < Br < I. This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.06%3A_General_Features_of_Nucleophilic_Substitution.txt |
Our general discussion of nucleophilic substitution reactions, we have until now been designating the leaving group simply as “X". As you may imagine, however, the nature of the leaving group is an important consideration: if the C-X bond does not break, the new bond between the nucleophile and electrophilic carbon cannot form, regardless of whether the substitution is SN1 or SN2. In this module, we are focusing on substitution reactions in which the leaving group is a halogen ion, although many reactions are known, both in the laboratory and in biochemical processes, in which the leaving group is something other than a halogen.
In order to understand the nature of the leaving group, it is important to first discuss factors that help determine whether a species will be a strong base or weak base. If you remember from general chemistry, a Lewis base is defined as a species that donates a pair of electrons to form a covalent bond. The factors that will determine whether a species wants to share its electrons or not include electronegativity, size, and resonance.
As Electronegativity Increases, Basicity Decreases: In general, if we move from the left of the periodic table to the right of the periodic table as shown in the diagram below, electronegativity increases. As electronegativity increases, basicity will decrease, meaning a species will be less likely to act as base; that is, the species will be less likely to share its electrons.
As Size Increases, Basicity Decreases: In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons.
Resonance Decreases Basicity:The third factor to consider in determining whether or not a species will be a strong or weak base is resonance. As you may remember from general chemistry, the formation of a resonance stabilized structure results in a species that is less willing to share its electrons. Since strong bases, by definition, want to share their electrons, resonance stabilized structures are weak bases.
Now that we understand how electronegativity, size, and resonance affect basicity, we can combine these concepts with the fact that weak bases make the best leaving groups. Think about why this might be true. In order for a leaving group to leave, it must be able to accept electrons. A strong bases wants to donate electrons; therefore, the leaving group must be a weak base. We will now revisit electronegativity, size, and resonance, moving our focus to the leaving group, as well providing actual examples.
Note
As the Electronegativity of the group Increases, The propensity of the Leaving Group to Leave Increases
As mentioned previously, if we move from left to right on the periodic table, electronegativity increases. With an increase in electronegativity, basisity decreases, and the ability of the leaving group to leave increases. This is because an increase in electronegativity results in a species that wants to hold onto its electrons rather than donate them. The following diagram illustrates this concept, showing -CH3 to be the worst leaving group and F- to be the best leaving group. This particular example should only be used to facilitate your understanding of this concept. In real reaction mechanisms, these groups are not good leaving groups at all. For example, fluoride is such a poor leaving group that SN2 reactions of fluoroalkanes are rarely observed.
As Size Increases, The Ability of the Leaving Group to Leave Increases:Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms.
Example 7.7.1
In each pair (A and B) below, which electrophile would be expected to react more rapidly in an SN2 reaction with the thiol group of cysteine as the common nucleophile?
Solution (8.13)
Contributor
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.07%3A_The_Leaving_Group.txt |
What is a nucleophile?
Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates. More specifically in laboratory reactions, halide and azide (N3-) anions are commonly seen acting as nucleophiles.
Of course, carbons can also be nucleophiles - otherwise how could new carbon-carbon bonds be formed in the synthesis of large organic molecules like DNA or fatty acids? Enolate ions (section 7.5) are the most common carbon nucleophiles in biochemical reactions, while the cyanide ion (CN-) is just one example of a carbon nucleophile commonly used in the laboratory. Reactions with carbon nucleophiles will be dealt with in chapters 13 and 14, however - in this chapter and the next, we will concentrate on non-carbon nucleophiles.
When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Protonation states and nucleophilicity
The protonation state of a nucleophilic atom has a very large effect on its nucleophilicity. This is an idea that makes intuitive sense: a hydroxide ion is much more nucleophilic (and basic) than a water molecule, because the negatively charged oxygen on the hydroxide ion carries greater electron density than the oxygen atom of a neutral water molecule. In practical terms, this means that a hydroxide nucleophile will react in an SN2 reaction with methyl bromide much faster ( about 10,000 times faster) than a water nucleophile.
Periodic trends and solvent effects in nucleophilicity
There are predictable periodic trends in nucleophilicity. Moving horizontally across the second row of the table, the trend in nucleophilicity parallels the trend in basicity:
The reasoning behind the horizontal nucleophilicity trend is the same as the reasoning behind the basicity trend: more electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond.
This horizontal trends also tells us that amines are more nucleophilic than alcohols, although both groups commonly act as nucleophiles in both laboratory and biochemical reactions.
Recall that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger area, which tends to increase stability (and thus reduce basicity).
The vertical periodic trend for nucleophilicity is somewhat more complicated that that for basicity: depending on the solvent that the reaction is taking place in, the nucleophilicity trend can go in either direction. Let's take the simple example of the SN2 reaction below:
. . .where Nu- is one of the halide ions: fluoride, chloride, bromide, or iodide, and the leaving group I* is a radioactive isotope of iodine (which allows us to distinguish the leaving group from the nucleophile in that case where both are iodide). If this reaction is occurring in a protic solvent (that is, a solvent that has a hydrogen bonded to an oxygen or nitrogen - water, methanol and ethanol are the most important examples), then the reaction will go fastest when iodide is the nucleophile, and slowest when fluoride is the nucleophile, reflecting the relative strength of the nucleophile.
Relative nucleophilicity in a protic solvent
This of course, is opposite that of the vertical periodic trend for basicity, where iodide is the least basic. What is going on here? Shouldn't the stronger base, with its more reactive unbonded valence electrons, also be the stronger nucleophile?
As mentioned above, it all has to do with the solvent. Remember, we are talking now about the reaction running in a protic solvent like ethanol. Protic solvent molecules form very strong ion-dipole interactions with the negatively-charged nucleophile, essentially creating a 'solvent cage' around the nucleophile:
In order for the nucleophile to attack the electrophile, it must break free, at least in part, from its solvent cage. The lone pair electrons on the larger, less basic iodide ion interact less tightly with the protons on the protic solvent molecules - thus the iodide nucleophile is better able to break free from its solvent cage compared the smaller, more basic fluoride ion, whose lone pair electrons are bound more tightly to the protons of the cage.
The picture changes if we switch to a polar aprotic solvent, such as acetone, in which there is a molecular dipole but no hydrogens bound to oxygen or nitrogen. Now, fluoride is the best nucleophile, and iodide the weakest.
Relative nucleophilicity in a polar aprotic solvent
The reason for the reversal is that, with an aprotic solvent, the ion-dipole interactions between solvent and nucleophile are much weaker: the positive end of the solvent's dipole is hidden in the interior of the molecule, and thus it is shielded from the negative charge of the nucleophile.
A weaker solvent-nucleophile interaction means a weaker solvent cage for the nucleophile to break through, so the solvent effect is much less important, and the more basic fluoride ion is also the better nucleophile.
Why not use a completely nonpolar solvent, such as hexane, for this reaction, so that the solvent cage is eliminated completely? The answer to this is simple - the nucleophile needs to be in solution in order to react at an appreciable rate with the electrophile, and a solvent such as hexane will not solvate an a charged (or highly polar) nucleophile at all. That is why chemists use polar aprotic solvents for nucleophilic substitution reactions in the laboratory: they are polar enough to solvate the nucleophile, but not so polar as to lock it away in an impenetrable solvent cage. In addition to acetone, three other commonly used polar aprotic solvents are acetonitrile, dimethylformamide (DMF), and dimethyl sulfoxide (DMSO).
In biological chemistry, where the solvent is protic (water), the most important implication of the periodic trends in nucleophilicity is that thiols are more powerful nucleophiles than alcohols. The thiol group in a cysteine amino acid, for example, is a powerful nucleophile and often acts as a nucleophile in enzymatic reactions, and of course negatively-charged thiolates (RS-) are even more nucleophilic. This is not to say that the hydroxyl groups on serine, threonine, and tyrosine do not also act as nucleophiles - they do.
Resonance effects on nucleophilicity
Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity. If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance.
The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group.
Steric effects on nucleophilicity
Steric hindrance is an important consideration when evaluating nucleophility. For example, tert-butanol is less potent as a nucleophile than methanol. This is because the comparatively bulky methyl groups on the tertiary alcohol effectively block the route of attack by the nucleophilic oxygen, slowing the reaction down considerably (imagine trying to walk through a narrow doorway while carrying three large suitcases!).
It is not surprising that it is more common to observe serines acting as nucleophiles in enzymatic reactions compared to threonines - the former is a primary alcohol, while the latter is a secondary alcohol.
Example 7.8.1
Which is the better nucleophile - a cysteine side chain or a methionine side chain? Explain.
Example 7.8.2
In each of the following pairs of molecules/ions, which is the better nucleophile in a reaction with CH3Br in acetone solvent? Explain your choice.
1. phenolate ion (deprotonated phenol) or benzoate ion (deprotonated benzoic acid)
2. water and hydronium ion
3. trimethylamine and triethylamine
4. chloride anion and iodide anion
5. CH3NH- and CH3CH2NH
Contributors
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
7.10: Two Mechanisms for Nucleophilic Substitution
The SN2 Mechanism
As described in the previous section, a majority of the reactions thus far described appear to proceed by a common single-step mechanism. This mechanism is referred to as the SN2 mechanism, where S stands for Substitution, N stands for Nucleophilic and 2 stands for bimolecular. Other features of the SN2 mechanism are inversion at the alpha-carbon, increased reactivity with increasing nucleophilicity of the nucleophilic reagent and steric hindrance to rear-side bonding, especially in tertiary and neopentyl halides. Although reaction 3 exhibits second order kinetics, it is an elimination reaction and must therefore proceed by a very different mechanism, which will be described later.
The SN1 Mechanism
Reaction 7, shown at the end of the previous section, is clearly different from the other cases we have examined. It not only shows first order kinetics, but the chiral 3º-alkyl bromide reactant undergoes substitution by the modest nucleophile water with extensive racemization. In all of these features this reaction fails to meet the characteristics of the SN2 mechanism. A similar example is found in the hydrolysis of tert-butyl chloride, shown below. Note that the initial substitution product in this reaction is actually a hydronium ion, which rapidly transfers a proton to the chloride anion. This second acid-base proton transfer is often omitted in writing the overall equation, as in the case of reaction 7 above.
(CH3)3C-Cl + H2O ——> (CH3)3C-OH2(+) + Cl(–) ——> (CH3)3C-OH + HCl | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.08%3A_The_Nucleophile.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
This is called an 'SN2' mechanism. In the term SN2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way.
The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the SN2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
We will be contrasting about two types of nucleophilic substitution reactions. One type is referred to as unimolecular nucleophilic substitution (SN1), whereby the rate determining step is unimolecular and bimolecular nucleophilic substitution (SN2), whereby the rate determining step is bimolecular. We will begin our discussion with SN2 reactions, and discuss SN1 reactions elsewhere.
Biomolecular Nucleophilic Substitution Reactions and Kinetics
In the term SN2, the S stands for substitution, the N stands for nucleophilic, and the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane.
If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate.
If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate.
The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics:
Bimolecular Nucleophilic Substitution Reactions Are Concerted
Bimolecular nucleophilic substitution (SN2) reactions are concerted, meaning they are a one step process. This means that the process whereby the nucleophile attacks and the leaving group leaves is simultaneous. Hence, the bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electophilic carbon and the halogen.
The potential energy diagram for an SN2 reaction is shown below. Upon nucleophilic attack, a single transition state is formed. A transition state, unlike a reaction intermediate, is a very short-lived species that cannot be isolated or directly observed. Again, this is a single-step, concerted process with the occurrence of a single transition state.
Sterrically Hindered Substrates Will Reduce the SN2 Reaction Rate
Now that we have discussed the effects that the leaving group, nucleophile, and solvent have on biomolecular nucleophilic substitution (SN2) reactions, it's time to turn our attention to how the substrate affects the reaction. Although the substrate, in the case of nucleophilic substitution of haloalkanes, is considered to be the entire molecule circled below, we will be paying particular attention to the alkyl portion of the substrate. In other words, we are most interested in the electrophilic center that bears the leaving group.
In the section Kinetics of Nucleophilic Substitution Reactions, we learned that the SN2 transition state is very crowded. Recall that there are a total of five groups around the electrophilic center, the nucleophile, the leaving group, and three substituents.
If each of the three substituents in this transition state were small hydrogen atoms, as illustrated in the first example below, there would be little steric repulsion between the incoming nucleophile and the electrophilic center, thereby increasing the ease at which the nucleophilic substitution reaction can occur. Remember, for the SN2 reaction to occur, the nucleophile must be able to attack the electrophilic center, resulting in the expulsion of the leaving group. If one of the hydrogens, however, were replaced with an R group, such as a methyl or ethyl group, there would be an increase in steric repulsion with the incoming nucleophile. If two of the hydrogens were replaced by R groups, there would be an even greater increase in steric repulsion with the incoming nucleophile.
How does steric hindrance affect the rate at which an SN2 reaction will occur? As each hydrogen is replaced by an R group, the rate of reaction is significantly diminished. This is because the addition of one or two R groups shields the backside of the electrophilic carbon, impeding nucleophilic attack.
The diagram below illustrates this concept, showing that electrophilic carbons attached to three hydrogen atoms results in faster nucleophilic substitution reactions, in comparison to primary and secondary haloalkanes, which result in nucleophilic substitution reactions that occur at slower or much slower rates, respectively. Notice that a tertiary haloalkane, that which has three R groups attached, does not undergo nucleophilic substitution reactions at all. The addition of a third R group to this molecule creates a carbon that is entirely blocked.
Substitutes on Neighboring Carbons Slow Nucleophilic Substitution Reactions
Previously we learned that adding R groups to the electrophilic carbon results in nucleophilic substitution reactions that occur at a slower rate. What if R groups are added to neighboring carbons? It turns out that the addition of substitutes on neighboring carbons will slow nucleophilic substitution reactions as well.
In the example below, 2-methyl-1-bromopropane differs from 1-bromopropane in that it has a methyl group attached to the carbon that neighbors the electrophilic carbon. The addition of this methyl group results in a significant decrease in the rate of a nucleophilic substitution reaction.
If R groups were added to carbons farther away from the electrophilic carbon, we would still see a decrease in the reaction rate. However, branching at carbons farther away from the electrophilic carbon would have a much smaller effect.
Frontside vs. Backside Attacks
A biomolecular nucleophilic substitution (SN2) reaction is a type of nucleophilic substitution whereby a lone pair of electrons on a nucleophile attacks an electron deficient electrophilic center and bonds to it, resulting in the expulsion of a leaving group. It is possible for the nucleophile to attack the electrophilic center in two ways.
• Frontside Attack: In a frontside attack, the nucleophile attacks the electrophilic center on the same side as the leaving group. When a frontside attack occurs, the stereochemistry of the product remains the same; that is, we have retention of configuration.
• Backside Attack: In a backside attack, the nucleophile attacks the electrophilic center on the side that is opposite to the leaving group. When a backside attack occurs, the stereochemistry of the product does not stay the same. There is inversion of configuration.
The following diagram illustrates these two types of nucleophilic attacks, where the frontside attack results in retention of configuration; that is, the product has the same configuration as the substrate. The backside attack results in inversion of configuration, where the product's configuration is opposite that of the substrate.
Experimental Observation: All SN2 Reactions Proceed With Nucleophilic Backside Attacks
Experimental observation shows that all SN2 reactions proceed with inversion of configuration; that is, the nucleophile will always attack from the backside in all SN2 reactions. To think about why this might be true, remember that the nucleophile has a lone pair of electrons to be shared with the electrophilic center, and the leaving group is going to take a lone pair of electrons with it upon leaving. Because like charges repel each other, the nucleophile will always proceed by a backside displacement mechanism.
SN2 Reactions Are Stereospecific
The SN2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. For example, if the substrate is an R enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the R enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the S enantiomer.
Conversely, if the substrate is an S enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the S enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the R enantiomer.
In conclusion, SN2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept also applies to substrates that are cis and substrates that are trans. If the cis configuration is the substrate, the resulting product will be trans. Conversely, if the trans configuration is the substrate, the resulting product will be cis.
Contributors
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.11%3A_The_%28S_N2%29_Mechanism.txt |
SAM methyltransferases
Some of the most important examples of SN2 reactions in biochemistry are those catalyzed by S-adenosyl methionine (SAM) – dependent methyltransferase enzymes. We have already seen, in chapter 6 and again in chapter 8, how a methyl group is transferred in an SN2 reaction from SAM to the amine group on the nucleotide base adenosine:
(Nucleic Acids Res. 2000, 28, 3950).
Another SAM-dependent methylation reaction is catalyzed by an enzyme called catechol-O-methyltransferase. The substrate here is epinephrine, also known as adrenaline.
Notice that in this example, the attacking nucleophile is an alcohol rather than an amine (that’s why the enzyme is called an O-methyltransferase). In both cases, though, a basic amino acid side chain is positioned in the active site in just the right place to deprotonate the nucleophilic group as it attacks, increasing its nucleophilicity. The electrophile in both reactions is a methyl carbon, so there is little steric hindrance to slow down the nucleophilic attack. The methyl carbon is electrophilic because it is bonded to a positively-charged sulfur, which is a powerful electron withdrawing group. The positive charge on the sulfur also makes it an excellent leaving group, as the resulting product will be a neutral and very stable sulfide. All in all, in both reactions we have a reasonably good nucleophile, an electron-poor, unhindered electrophile, and an excellent leaving group.
Because the electrophilic carbon in these reactions is a methyl carbon, a stepwise SN1-like mechanism is extremely unlikely: a methyl carbocation is very high in energy and thus is not a reasonable intermediate to propose. We can confidently predict that this reaction is SN2. Does this SN2 reaction occur, as expected, with inversion of stereochemistry? Of course, the electrophilic methyl carbon in these reactions is achiral, so inversion is not apparent. To demonstrate inversion, the following experiment has been carried out with catechol-O-methyltransferase:
Here, the methyl group of SAM was made to be chiral by incorporating hydrogen isotopes tritium (3H, T) and deuterium (2H, D). The researchers determined that the reaction occurred with inversion of configuration, as expected for an SN2 displacement (J. Biol. Chem. 1980, 255, 9124).
Example
Exercise 9.1: SAM is formed by a nucleophilic substitution reaction between methionine and adenosine triphosphate (ATP). Propose a mechanism for this reaction.
Solution
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9.1B: Synthetic parallel – the Williamson ether synthesis
Synthetic organic chemists often use reactions in the laboratory that are conceptually very similar to the SAM-dependent methylation reactions described above. In what is referred to as "O-methylation", an alcohol is first deprotonated by a strong base, typically sodium hydride (this is essentially an irreversible deprotonation, as the hydrogen gas produced can easily be removed from the reaction vessel).
The alkoxide ion, a powerful nucleophile, is then allowed to attack the electrophilic carbon in iodomethane, displacing iodide in an SN2 reaction. Recall (section 7.3A) that iodide ion is the least basic - and thus the best leaving group - among the halogens commonly used in the synthetic lab. CH3I is one of the most commonly used lab reagents for methyl transfer reactions, and is the lab equivalent of SAM. In the synthesis of a modified analog of the signaling molecule myo-inisitol triphosphate, an alcohol group was methylated using sodium hydride and iodomethane (Carbohydrate Research 2004, 339, 51):
Methylation of amines (N-methylation) by iodomethane is also possible. A recent study concerning the optimization of peptide synthesis methods involved the following reaction (J. Org. Chem. 2000 65, 2309):
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.12%3A_Application-_Useful_%28S_N2%29_Reactions.txt |
The SN1 mechanism
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches:
This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry.
We saw that SN2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of SN1 reactions? In the model SN1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp2-hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an SN1 reaction with water as the incoming nucleophile.
Exercise
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
The SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an SN1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding.
Exercise
Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is SN2, and reaction B is SN1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
Influence of the solvent in an SN1 reaction
Since the hydrogen atom in a polar protic solvent is highly positively charged, it can interact with the anionic nucleophile which would negatively affect an SN2, but it does not affect an SN1 reaction because the nucleophile is not a part of the rate-determining step. Polar protic solvents actually speed up the rate of the unimolecular substitution reaction because the large dipole moment of the solvent helps to stabilize the transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.The SN1 reaction of allyl bromide in methanol is an example of what we would call methanolysis, while if water is the solvent the reaction would be called hydrolysis:
The polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the solvolysis (see table below). The dielectric constant of a solvent roughly provides a measure of the solvent's polarity. A dielectric constant below 15 is usually considered non-polar. Basically, the dielectric constant can be thought of as the solvent's ability to reduce the internal charge of the solvent. So for our purposes, the higher the dielectric constant the more polar the substance and in the case of SN1 reactions, the faster the rate.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.13%3A_The_%28S_N1%29_Mechanism.txt |
Stability of carbocation intermediates
We know that the rate-limiting step of an SN1 reaction is the first step - formation of the this carbocation intermediate. The rate of this step – and therefore, the rate of the overall substitution reaction – depends on the activation energy for the process in which the bond between the carbon and the leaving group breaks and a carbocation forms. According to Hammond’s postulate (section 6.2B), the more stable the carbocation intermediate is, the faster this first bond-breaking step will occur. In other words, the likelihood of a nucleophilic substitution reaction proceeding by a dissociative (SN1) mechanism depends to a large degree on the stability of the carbocation intermediate that forms.
The critical question now becomes, what stabilizes a carbocation?
So if it takes an electron withdrawing group to stabilize a negative charge, what will stabilize a positive charge? An electron donating group!
A positively charged species such as a carbocation is very electron-poor, and thus anything which donates electron density to the center of electron poverty will help to stabilize it. Conversely, a carbocation will be destabilized by an electron withdrawing group.
Alkyl groups – methyl, ethyl, and the like – are weak electron donating groups, and thus stabilize nearby carbocations. What this means is that, in general, more substituted carbocations are more stable: a tert-butyl carbocation, for example, is more stable than an isopropyl carbocation. Primary carbocations are highly unstable and not often observed as reaction intermediates; methyl carbocations are even less stable.
Alkyl groups are electron donating and carbocation-stabilizing because the electrons around the neighboring carbons are drawn towards the nearby positive charge, thus slightly reducing the electron poverty of the positively-charged carbon.
It is not accurate to say, however, that carbocations with higher substitution are always more stable than those with less substitution. Just as electron-donating groups can stabilize a carbocation, electron-withdrawing groups act to destabilize carbocations. Carbonyl groups are electron-withdrawing by inductive effects, due to the polarity of the C=O double bond. It is possible to demonstrate in the laboratory (see section 16.1D) that carbocation A below is more stable than carbocation B, even though A is a primary carbocation and B is secondary.
The difference in stability can be explained by considering the electron-withdrawing inductive effect of the ester carbonyl. Recall that inductive effects - whether electron-withdrawing or donating - are relayed through covalent bonds and that the strength of the effect decreases rapidly as the number of intermediary bonds increases. In other words, the effect decreases with distance. In species B the positive charge is closer to the carbonyl group, thus the destabilizing electron-withdrawing effect is stronger than it is in species A.
In the next chapter we will see how the carbocation-destabilizing effect of electron-withdrawing fluorine substituents can be used in experiments designed to address the question of whether a biochemical nucleophilic substitution reaction is SN1 or SN2.
Stabilization of a carbocation can also occur through resonance effects, and as we have already discussed in the acid-base chapter, resonance effects as a rule are more powerful than inductive effects. Consider the simple case of a benzylic carbocation:
This carbocation is comparatively stable. In this case, electron donation is a resonance effect. Three additional resonance structures can be drawn for this carbocation in which the positive charge is located on one of three aromatic carbons. The positive charge is not isolated on the benzylic carbon, rather it is delocalized around the aromatic structure: this delocalization of charge results in significant stabilization. As a result, benzylic and allylic carbocations (where the positively charged carbon is conjugated to one or more non-aromatic double bonds) are significantly more stable than even tertiary alkyl carbocations.
Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in section 7.4). Consider the two pairs of carbocation species below:
In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually destabilize the positive charge, because they are electron withdrawing by induction.
Finally, vinylic carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any reaction.
Example 8.10
Exercise 8.10: In which of the structures below is the carbocation expected to be more stable? Explain.
For the most part, carbocations are very high-energy, transient intermediate species in organic reactions. However, there are some unusual examples of very stable carbocations that take the form of organic salts. Crystal violet is the common name for the chloride salt of the carbocation whose structure is shown below. Notice the structural possibilities for extensive resonance delocalization of the positive charge, and the presence of three electron-donating amine groups.
Example 8.11
Draw a resonance structure of the crystal violet cation in which the positive charge is delocalized to one of the nitrogen atoms.
Solution
When considering the possibility that a nucleophilic substitution reaction proceeds via an SN1 pathway, it is critical to evaluate the stability of the hypothetical carbocation intermediate. If this intermediate is not sufficiently stable, an SN1 mechanism must be considered unlikely, and the reaction probably proceeds by an SN2 mechanism. In the next chapter we will see several examples of biologically important SN1 reactions in which the positively charged intermediate is stabilized by inductive and resonance effects inherent in its own molecular structure.
Example 8.12
State which carbocation in each pair below is more stable, or if they are expected to be approximately equal. Explain your reasoning.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.14%3A_Carbocation_Stability.txt |
Now, back to transition states. Chemists are often very interested in trying to learn about what the transition state for a given reaction looks like, but addressing this question requires an indirect approach because the transition state itself cannot be observed. In order to gain some insight into what a particular transition state looks like, chemists often invoke the Hammond postulate, which states that a transition state resembles the structure of the nearest stable species. For an exergonic reaction, therefore, the transition state resembles the reactants more than it does the products.
If we consider a hypothetical exergonic reaction between compounds A and B to form AB, the distance between A and B would be relatively large at the transition state, resembling the starting state where A and B are two isolated species. In the hypothetical endergonic reaction between C and D to form CD, however, the bond formation process would be much further along at the TS point, resembling the product.
The Hammond Postulate is a very simplistic idea, which relies on an assumption that potential energy surfaces are parabolic. Although such an assumption is not rigorously true, it is fairly reliable and allows chemists to make energetic arguments about transition states by employing arguments about the stability of a related species. Since the formation of a reactive intermediate is very reliably endergonic, arguments about the stability of reactive intermediates can serve as proxy arguments about transition state stability.
The Hammond Postulate and the SN1 Reaction
the Hammond postulate suggests that the activation energy of the rate-determining first step will be inversely proportional to the stability of the carbocation intermediate. The stability of carbocations was discussed earlier, and a qualitative relationship is given below:
Carbocation
Stability
CH3(+) < CH3CH2(+) < (CH3)2CH(+) CH2=CH-CH2(+) < C6H5CH2(+) (CH3)3C(+)
Consequently, we expect that 3º-alkyl halides will be more reactive than their 2º and 1º-counterparts in reactions that follow an SN1 mechanism. This is opposite to the reactivity order observed for the SN2 mechanism. Allylic and benzylic halides are exceptionally reactive by either mechanism.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
7.16: Application- (S N1) Reactions Nitrosamines and Cancer
Secondary amines and nitrous acid
This time there isn't any gas produced. Instead, you get a yellow oil called a nitrosamine. These compounds are powerful carcinogens - avoid them! For example:
Contributors
Jim Clark (Chemguide.co.uk)
7.17: When Is the Mechanism (S N1) or (S N2)
Predicting SN1 vs. SN2 mechanisms; competition between nucleophilic substitution and elimination reactions
When considering whether a nucleophilic substitution is likely to occur via an SN1 or SN2 mechanism, we really need to consider three factors:
1. The electrophile: when the leaving group is attached to a methyl group or a primary carbon, an SN2 mechanism is favored (here the electrophile is unhindered by surrounded groups, and any carbocation intermediate would be high-energy and thus unlikely). When the leaving group is attached to a tertiary, allylic, or benzylic carbon, a carbocation intermediate will be relatively stable and thus an SN1 mechanism is favored.
2. The nucleophile: powerful nucleophiles, especially those with negative charges, favor the SN2 mechanism. Weaker nucleophiles such as water or alcohols favor the SN1 mechanism.
3. The solvent: Polar aprotic solvents favor the SN2 mechanism by enhancing the reactivity of the nucleophile. Polar protic solvents favor the SN1 mechanism by stabilizing the carbocation intermediate. SN1 reactions are frequently solvolysis reactions.
For example, the reaction below has a tertiary alkyl bromide as the electrophile, a weak nucleophile, and a polar protic solvent (we’ll assume that methanol is the solvent). Thus we’d confidently predict an SN1 reaction mechanism. Because substitution occurs at a chiral carbon, we can also predict that the reaction will proceed with racemization.
In the reaction below, on the other hand, the electrophile is a secondary alkyl bromide – with these, both SN1 and SN2 mechanisms are possible, depending on the nucleophile and the solvent. In this example, the nucleophile (a thiolate anion) is strong, and a polar protic solvent is used – so the SN2 mechanism is heavily favored. The reaction is expected to proceed with inversion of configuration.
Example 7.17.1
Determine whether each substitution reaction shown below is likely to proceed by an SN1 or SN2 mechanism.
Solution
Contributors
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.15%3A_The_Hammond_Postulate.txt |
Nucleophilic substitution in the aryl halides
We'll look in some detail at the structure of chlorobenzene. Bromobenzene and iodobenzene are just the same. The simplest way to draw the structure of chlorobenzene is:
To understand chlorobenzene properly, you need to dig a bit deeper than this. There is an interaction between the delocalised electrons in the benzene ring and one of the lone pairs on the chlorine atom. This overlaps with the delocalised ring electron system . . .
to giving a structure rather like this:
This delocalisation is by no means complete, but it does have a significant effect on the properties of both the carbon-chlorine bond and the polarity of the molecule. The delocalisation introduces some extra bonding between the carbon and the chlorine, making the bond stronger. This has a major effect on the reactions of compounds like chlorobenzene.
There is also some movement of electrons away from the chlorine towards the ring. Chlorine is quite electronegative and usually draws electrons in the carbon-chlorine bond towards itself. In this case, this is offset to some extent by the movement of electrons back towards the ring in the delocalisation. The molecule is less polar than you would otherwise have expected.
Simple aryl halides like chlorobenzene are very resistant to nucleophilic substitution. It is possible to replace the chlorine by -OH, but only under very severe industrial conditions - for example at 200°C and 200 atmospheres. In the lab, these reactions do not happen. There are two reasons for this - depending on which of the above mechanisms you are talking about.
1. The extra strength of the carbon-halogen bond in aryl halides: The carbon-chlorine bond in chlorobenzene is stronger than you might expect. There is an interaction between one of the lone pairs on the chlorine atom and the delocalized ring electrons, and this strengthens the bond. Both of the mechanisms above involve breaking the carbon-halogen bond at some stage. The more difficult it is to break, the slower the reaction will be.
2. Repulsion by the ring electrons: This will only apply if the hydroxide ion attacked the chlorobenzene by a mechanism like the first one described above. In that mechanism, the hydroxide ion attacks the slightly positive carbon that the halogen atom is attached to.If the halogen atom is attached to a benzene ring, the incoming hydroxide ion is going to be faced with the delocalized ring electrons above and below that carbon atom. The negative hydroxide ion will simply be repelled. That particular mechanism is simply a non-starter!
Contributors
Jim Clark (Chemguide.co.uk)
7.19: Organic Synthesis
Wöhler synthesis of Urea in 1828 heralded the birth of modern chemistry. The Art of synthesis is as old as Organic chemistry itself. Natural product chemistry is firmly rooted in the science of degrading a molecule to known smaller molecules using known chemical reactions and conforming the assigned structure by chemical synthesis from small, well known molecules using well established synthetic chemistry techniques. Once this art of synthesizing a molecule was mastered, chemists attempted to modify bioactive molecules in an attempt to develop new drugs and also to unravel the mystery of biomolecular interactions. Until the middle of the 20th Century, organic chemists approached the task of synthesis of molecules as independent tailor made projects, guided mainly by chemical intuition and a sound knowledge of chemical reactions. During this period, a strong foundation was laid for the development of mechanistic principles of organic reactions, new reactions and reagents. More than a century of such intensive studies on the chemistry of carbohydrates, alkaloids, terpenes and steroids laid the foundation for the development of logical approaches for the synthesis of molecules.
The job of a synthetic chemist is akin to that of an architect (or civil engineer). While the architect could actually see the building he is constructing, a molecular architect called Chemist is handicapped by the fact that the molecule he is synthesizing is too small to be seen even through the most powerful microscope developed to date. With such a limitation, how does he ‘see’ the developing structure? For this purpose, a chemist makes use of spectroscopic tools. How does he cut, tailor and glue the components on a molecule that he cannot see? For this purpose chemists have developed molecular level tools called Reagents and Reactions. How does he clean the debris and produce pure molecules? This feat is achieved by crystallization, distillation and extensive use of Chromatography techniques. A mastery over several such techniques enables the molecular architect (popularly known as organic chemist) to achieve the challenging task of synthesizing the mirade molecular structures encountered in Natural Products Chemistry, Drug Chemistry and modern Molecular Materials. In this task, he is further guided by several ‘thumb rules’ that chemists have evolved over the past two centuries. The discussions on the topics Name Reactions, Reagents for synthesis, Spectroscopy and Chromatography are beyond the scope of this write-up. Let us begin with a brief look at some of the important ‘Rules’ in organic chemistry that guide us in planning organic synthesis. We would then discuss Protection and Deprotection of some important functional groups. We could then move on to the Logic of planning Organic Synthesis. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/07%3A_Alkyl_Halides_and_Nucleophilic_Substitution/7.18%3A_Vinyl_Halides_and_Aryl_Halides.txt |
So far in this chapter, we have seen several examples of carbanion-intermediate (E1cb) beta-elimination reactions, in which the first step was proton abstraction at a carbon positioned ato an electron-withdrawing carbonyl or imine. Elimination reactions are also possible at positions that are isolated from carbonyls or any other electron-withdrawing groups. This type of elimination can be described by two model mechanisms: it can occur in a single concerted step (proton abstraction at Cα occurring at the same time as Cβ-X bond cleavage), or in two steps (Cβ-X bond cleavage occurring first to form a carbocation intermediate, which is then 'quenched' by proton abstraction at the alpha-carbon).
These mechanisms, termed E2 and E1, respectively, are important in laboratory organic chemistry, but are less common in biological chemistry. As explained below, which mechanism actually occurs in a laboratory reaction will depend on the identity of the R groups (ie., whether the alkyl halide is primary, secondary, tertiary, etc.) as well as on the characteristics of the base.
E1 and E2 reactions in the laboratory
E2 elimination reactions in the laboratory are carried out with relatively strong bases, such as alkoxides (deprotonated alcohols). 2-bromopropane will react with ethoxide, for example, to give propene.
Propene is not the only product of this reaction, however - the ethoxide will also to some extent act as a nucleophile in an SN2 reaction.
Chemists carrying out nonenzymatic nucleophilic substitution or elimination reactions always have to be aware of the competition between the two mechanisms, because bases can also be nucleophiles, and vice-versa. However, a chemist can tip the scales in one direction or another by carefully choosing reagents. Primary carbon electrophiles like 1-bromopropane, for example, are much more likely to undergo substitution (by the SN2 mechanism) than elimination (by the E2 mechanism) – this is because the electrophilic carbon is unhindered and a good target for a nucleophile.
SN1 and E1 mechanisms are unlikely with such compounds because of the relative instability of primary carbocations.
The nature of the electron-rich species is also critical. Acetate, for example, is a weak base but a reasonably good nucleophile, and will react with 2-bromopropane mainly as a nucleophile.
In order to direct the reaction towards elimination, a strong hindered base such as tert-butoxide can be used. The bulkiness of tert-butoxide makes it difficult for the oxygen to reach the carbon (in other words, to act as a nucleophile). It is more likely to pluck off a proton, which is much more accessible than the electrophilic carbon).
E1 reactions occur by the same kinds of carbocation-favoring conditions that have already been described for SN1 reactions (section 9.4): a secondary or tertiary substrate, a protic solvent, and a relatively weak base/nucleophile. In fact, E1 and SN1 reactions generally occur simultaneously, giving a mixture of substitution and elimination products after formation of a common carbocation intermediate. When tert-butyl chloride is stirred in a mixture of ethanol and water, for example, a mixture of SN1 products (tert-butyl alcohol and tert-butyl ethyl ether) and E1 product (2-methylpropene) results.
Template:ExampleStart
Exercise 14.4: A straightforward functional group conversion that is often carried out in the undergraduate organic lab is the phosphoric acid-catalyzed dehydration of cyclohexanol to form cyclohexene. No solvent is necessary in this reaction - pure liquid cyclohexanol is simply stirred together with a few drops of concentrated phosphoric acid. In order to drive the equilibrium of this reversible reaction towards the desired product, cyclohexene is distilled out of the reaction mixture as it forms (the boiling point of cyclohexene is 83 oC, significantly lower than that of anything else in the reaction solution). Any cyclohexyl phosphate that might form from the competing SN1 reaction remains in the flask, and is eventually converted to cyclohexene over time. Draw a mechanism for the cyclohexene synthesis reaction described above. Also, draw a mechanism showing how the undesired cyclohexyl phosphate could form.
Template:ExampleEnd
Next, let’s put aside the issue of competition between nucleophilic substitution and elimination, and focus on the regioselectivity of elimination reactions. In many cases an elimination reaction can result in more than one constitutional isomer or stereoisomer. The elimination products of 2-chloropentane provide a good example:
This reaction is both regiospecific and stereospecific. In general, more substituted alkenes are more stable, and as a result, the product mixture will contain less 1-butene than 2-butene (this is the regiochemical aspect of the outcome, and is often referred to as Zaitsev's rule). In addition, we already know that trans (E) alkenes are generally more stable than cis (Z) alkenes (section 3.7C), so we can predict that more of the E product will form compared to the Z product.
Why might a reaction undergo elimination rather than substitution? The most important reason concerns the nature of the nucleophile. The more basic the nucleophile, the more likely it will induce elimination.
• Basic nucleophiles lead to elimination.
Very strong bases include carbon and nitrogen anions and semi-anions. Examples include butyllithium and sodium amide. Very strong bases are highly likely to engage in elimination, rather than substitution.
Strong bases include non-stabilized oxygen anions. Examples include sodium hydroxide as well as alkoxides such as potassium tert-butoxide or sodium ethoxide. Strong bases favour elimination, too. Nevertheless, they can sometimes undergo either elimination or substitution, depending on other factors (see below).
Weak bases include cyanide, stabilized oxygen anions such as carboxylates and aryloxides, fluoride ion and neutral amines. Weak bases are much more likely to undergo substitution than elimination.
Very weak bases include heavy halides such as chloride, bromide or iodide, as well as phosphorus and sulfur nucleophiles. Very weak bases undergo elimination only rarely.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/08%3A_Alkyl_Halides_and_Elimination_Reactions/8.01%3A_General_Features_of_Elimination.txt |
Ethene is the formal IUPAC name for H2C=CH2, but it also goes by a common name: Ethylene. The name Ethylene is used because it is like an ethyl group ($CH_2CH_3$) but there is a double bond between the two carbon atoms in it. Ethene has the formula $C_2H_4$ and is the simplest alkene because it has the fewest carbons (two) necessary for a carbon-carbon double bond.
Introduction
Bonding in carbon is covalent, containing either sigma or $\pi$ bonds. Carbon can make single, double, or triple bonds. The number of bonds it makes determines the structure. With four single bonds, carbon has a tetrahedral structure, while with one double bond it's structure is trigonal planar, and with a triple bond it has a linear structure.
A solitary carbon atom has four electrons, two in the 2s orbital, and one in each of the 2$p_x$ and 2$p_y$ orbitals, leaving the $2p_z$ orbital empty. A single carbon atom can make up to four bonds, but by looking at its electron configuration this would not be possible because there are only two electrons available to bond with. The other two are in a lone pair state, making them much less reactive to another electron that is by itself. Well it is, in order to make the four bonds, the carbon atom promotes one of the 2s electrons into the empty $2p_z$ orbital, leaving the carbon with four unpaired electrons allowing it to now form four bonds. The electron is not promoted spontaneously. It becomes promoted when a photon of light with the correct wavelength hits the carbon atom. When this photon hits the carbon atom it gives the atom enough energy to promote one of the lone pair electrons to the $2p_z$ orbital.
Sigma and Pi Bonds
All the bonds in Ethene are covalent, meaning that they are all formed by two adjacent atoms sharing their valence electrons. As opposed to ionic bonds which hold atoms together through the attraction of two ions of opposite charges.
Sigma bonds are created when there is overlap of similar orbitals, orbitals that are aligned along the inter-nuclear axis. Common sigma bonds are $s+s$, $p_z+p_z$ and $s+p_z$, $z$ is the axis of the bond on the xyz-plane of the atom.
$\pi$ bonds are created when there is adequate overlap of similar, adjacent $p$ orbitals, such as $p_x$+$p_x$ and $p_y$+$p_y$. Each p orbital has two lobes, one usually indicated by a + and the other indicated by a - (sometimes one may be shaded while the other is not). This + and - (shaded, not shaded) are only meant to indicate the opposite phase $\phi$ the wave functions, they do not indicate any type of electrical charge. For a $\pi$ bond to form both lobes of the $p$ orbital must overlap, + with + and - with -. When a + lobe overlaps with a - lobe this creates an anti-bonding orbital interaction which is much higher in energy, and therefore not a desirable interaction.
Usually there can be no $\pi$ bonds between two atoms without having at least one sigma bond present first. But there are special cases such as dicarbon ($C_2$) where the central bond is a $\pi$ bond not a sigma bond, but in cases like these the two atoms want to have as much orbital overlap as possible so the bond lengths between the atoms are smaller than what is normally expected.
The $\pi$ bond in ethene is weak compared to the sigma bond between the two carbons. This weakness makes the $\pi$ bond and the overall molecule a site of comparatively high chemical reactivity to an array of different substances. This is due to the high electron density in the $\pi$ bond, and because it is a weak bond with high electron density the $\pi$ bond will easily break in order to form two separate sigma bonds. Sites such as these are referred to as functional groups or functionalities. These groups have characteristic properties and they control the reactivity of the molecule as a whole. How these functional groups and other reactants form various products are an important concept in organic chemistry.
Orbital Bonding in Ethene
Ethene is made up of four 1s1 Hydrogen atoms and two 2s2 2$p_x$1 2$p_y$2 carbon atoms. These carbon atoms already have four electrons, but they each want to get four more so that they have a full eight in the valence shell. Having eight valence electrons around carbon gives the atom itself the same electron configuration as neon, a noble gas. Carbon wants to have the same configuration as Neon because when it has eight valence electrons carbon is at its most stable, lowest energy state, it has all of the electrons that it wants, so it is no longer reactive.
Structure of Ethene
Ethene is not a very complicated molecule. It contains two carbon atoms that are double bonded to each other, with each of these atoms also bonded to two Hydrogen atoms.
This forms a total of three bonds to each carbon atom, giving them an $sp^2$ hybridization. Since the carbon atom is forming three sigma bonds instead of the four that it can, it only needs to hybridize three of its outer orbitals, instead of four. It does this by using the $2s$ electron and two of the $2p$ electrons, leaving the other unchanged. This new orbital is called an $sp^2$ hybrid because that's exactly what it is, it is made from one s orbital and two p orbitals.
When atoms are an $sp^2$ hybrid they have a trigonal planar structure. These structures are very similar to a 'peace' sign, there is a central atom with three atoms around it, all on one plane. Trigonal planar molecules have an ideal bond angle of 120° on each side.
The H-C-H bond angle is 117°, which is very close to the ideal 120° of a carbon with $sp^2$ hybridization. The other two angles (H-C=C) are both 121.5°.
Rigidity in Ethene
There is rigidity in the Ethene molecule due to the double-bonded carbons. In Ethane there are two carbons that share a single bond, this allows the two Methyl groups to rotate with respect to each other. These different conformations result in higher and lower energy forms of Ethane. In Ethene there is no free rotation about the carbon-carbon sigma bond. There is no rotation because there is also a $\pi$ bond along with the sigma bond between the two carbons. A $\pi$ bond is only formed when there is adequate overlap between both top and bottom p-orbitals. In order for there to be free rotation the p-orbitals would have to go through a phase where they are 90° from each other, which would break the $\pi$ bond because there would be no overlap. Since the $\pi$ bond is essential to the structure of Ethene it must not break, so there can be not free rotation about the carbon-carbon sigma bond.
Configurational Stereoisomers of Alkenes
The carbon-carbon double bond is formed between two sp2 hybridized carbons, and consists of two occupied molecular orbitals, a sigma orbital and a pi orbital. Rotation of the end groups of a double bond relative to each other destroys the p-orbital overlap that creates the pi orbital or bond. Because the pi bond has a bond energy of roughly 60 kcal/mole, this resistance to rotation stabilizes the planar configuration of this functional group. As a result, certain disubstituted alkenes may exist as a pair of configurational stereoisomers, often designated cis and trans. The essential requirement for this stereoisomerism is that each carbon of the double bond must have two different substituent groups (one may be hydrogen). This is illustrated by the following general formulas. In the first example, the left-hand double bond carbon has two identical substituents (A) so stereoisomerism about the double bond is not possible (reversing substituents on the right-hand carbon gives the same configuration). In the next two examples, each double bond carbon atom has two different substituent groups and stereoisomerism exists, regardless of whether the two substituents on one carbon are the same as those on the other.
Some examples of this configurational stereoisomerism (sometimes called geometric isomerism) are shown below. Note that cycloalkenes smaller than eight carbons cannot exist in a stable trans configuration due to ring strain. A similar restriction holds against cycloalkynes smaller than ten carbons. Since alkynes are linear, there is no stereoisomerism associated with the carbon-carbon triple bond.
Geometric (cis-trans) isomerism
The carbon-carbon double bond doesn't allow any rotation about it. That means that it is possible to have the CH3 groups on either end of the molecule locked either on one side of the molecule or opposite each other.
These are called cis-but-2-ene (where the groups are on the same side) or trans-but-2-ene (where they are on opposite sides).
Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below.
Stability of Alkenes
The stability of alkene can be determined by measuring the amount of energy associated with the hydrogenation of the molecule. Since the double bond is breaking in this reaction, the energy released in hydrogenation is proportional to the energy in the double bond of the molecule. This is a useful tool because heats of hydrogenation can be measured very accurately. The ?H° is usually around -30 kcal/mol for alkenes. Stability is simply a measure of energy. Lower energy molecules are more stable than higher energy molecules. More substituted alkenes are more stable than less substituted ones due to hyperconjugation. They have a lower heat of hydrogenation. The following illustrates stability of alkenes with various substituents:
In disubstituted alkenes, trans isomers are more stable than cis isomers due to steric hindrance. Also, internal alkenes are more stable than terminal ones. See the following isomers of butene:
Trans-2-butene is the most stable because it has the lowest heat of hydrogenation.
Note: In cycloalkenes smaller than cyclooctene, the cis isomers are more stable than the trans as a result of ring strain. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/08%3A_Alkyl_Halides_and_Elimination_Reactions/8.02%3A_AlkenesThe_Products_of_Elimination_Reactions.txt |
So far in this chapter, we have seen several examples of carbanion-intermediate (E1cb) beta-elimination reactions, in which the first step was proton abstraction at a carbon positioned ato an electron-withdrawing carbonyl or imine. Elimination reactions are also possible at positions that are isolated from carbonyls or any other electron-withdrawing groups. This type of elimination can be described by two model mechanisms: it can occur in a single concerted step (proton abstraction at Cα occurring at the same time as Cβ-X bond cleavage), or in two steps (Cβ-X bond cleavage occurring first to form a carbocation intermediate, which is then 'quenched' by proton abstraction at the alpha-carbon).
These mechanisms, termed E2 and E1, respectively, are important in laboratory organic chemistry, but are less common in biological chemistry. As explained below, which mechanism actually occurs in a laboratory reaction will depend on the identity of the R groups (ie., whether the alkyl halide is primary, secondary, tertiary, etc.) as well as on the characteristics of the base.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
8.04: The (E 2) Mechanism
E2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold. Unlike E1 reactions, E2 reactions remove two subsituents with the addition of a strong base, resulting in an alkene.
Introduction
E2 reactions are typically seen with secondary and tertiary alkyl halides, but a hindered base is necessary with a primary halide. The mechanism by which it occurs is a single step concerted reaction with one transition state. The rate at which this mechanism occurs is second order kinetics, and depends on both the base and alkyl halide. A good leaving group is required because it is involved in the rate determining step. The leaving groups must be coplanar in order to form a pi bond; carbons go from sp3 to sp2 hybridization states.
To get a clearer picture of the interplay of these factors involved in a a reaction between a nucleophile/base and an alkyl halide, consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In one pathway, a methanethiolate nucleophile substitutes for bromine in an SN2 reaction. In the other (bottom) pathway, methoxide ion acts as a base (rather than as a nucleophile) in an elimination reaction. As we will soon see, the mechanism of this reaction is single-step, and is referred to as the E2 mechanism.
General Reaction
Below is a mechanistic diagram of an elimination reaction by the E2 pathway:
.
In this reaction Ba represents the base and Br representents a leaving group, typically a halogen. There is one transition state that shows the concerted reaction for the base attracting the hydrogen and the halogen taking the electrons from the bond. The product be both eclipse and staggered depending on the transition states. Eclipsed products have a synperiplanar transition states, while staggered products have antiperiplanar transition states. Staggered conformation is usually the major product because of its lower energy confirmation.
An E2 reaction has certain requirements to proceed:
• A strong base is necessary especially necessary for primary alkyl halides. Secondary and tertirary primary halides will procede with E2 in the precesence of a base (OH-, RO-, R2N-)
• Both leaving groups should be on the same plane, this allows the double bond to form in the reaction. In the reaction above you can see both leaving groups are in the plane of the carbons.
• Follows Zaitsev's rule, the most substituted alkene is usually the major product.
• Hoffman Rule, if a strically hindered base will result in the least substituted product.
The Leaving Group Effect in E2 Reactions
As Size Increases, The Ability of the Leaving Group to Leave Increases: Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
8.05: The Zaitsev Rule
Zaitsev's Rule can be used to predict the regiochemistry of elimination reactions.
Introduction
Zaitsev’s or Saytzev’s (anglicized spelling) rule is an empirical rule used to predict regioselectivity of 1,2-elimination reactions occurring via the E1 or E2 mechanisms. It states that in a regioselective E1 or E2 reaction the major product is the more stable alkene, (i.e., the alkene with the more highly substituted double bond). For example:
If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below.
By using the strongly basic hydroxide nucleophile, we direct these reactions toward elimination. In both cases there are two different sets of beta-hydrogens available to the elimination reaction (these are colored red and magenta and the alpha carbon is blue). If the rate of each possible elimination was the same, we might expect the amounts of the isomeric elimination products to reflect the number of hydrogens that could participate in that reaction. For example, since there are three 1º-hydrogens (red) and two 2º-hydrogens (magenta) on beta-carbons in 2-bromobutane, statistics would suggest a 3:2 ratio of 1-butene and 2-butene in the products. This is not observed, and the latter predominates by 4:1. This departure from statistical expectation is even more pronounced in the second example, where there are six 1º-beta-hydrogens compared with one 3º-hydrogen. These results point to a strong regioselectivity favoring the more highly substituted product double bond, an empirical statement generally called the Zaitsev Rule. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/08%3A_Alkyl_Halides_and_Elimination_Reactions/8.03%3A_The_Mechanisms_of_Elimination.txt |
Unimolecular Elimination (E1) is a reaction in which the removal of an HX substituent results in the formation of a double bond. It is similar to a unimolecular nucleophilic substitution reaction (SN1) in various ways. One being the formation of a carbocation intermediate. Also, the only rate determining (slow) step is the dissociation of the leaving group to form a carbocation, hence the name unimolecular. Thus, since these two reactions behave similarly, they compete against each other. Many times, both these reactions will occur simultaneously to form different products from a single reaction. However, one can be favored over another through thermodynamic control. Although Elimination entails two types of reactions, E1 and E2, we will focus mainly on E1 reactions with some reference to E2.
General Reaction
An E1 reaction involves the deprotonation of a hydrogen nearby (usually one carbon away, or the beta position) the carbocation resulting in the formation of an alkene product. In order to accomplish this, a Lewis base is required. For a simplified model, we’ll take B to be a Lewis base, and LG to be a halogen leaving group.
As can be seen above, the preliminary step is the leaving group (LG) leaving on its own. Because it takes the electrons in the bond along with it, the carbon that was attached to it loses its electron, making it a carbocation. Once it becomes a carbocation, a Lewis Base (\(B^-\)) deprotonates the intermediate carbocation at the beta position, which then donates its electrons to the neighboring C-C bond, forming a double bond. Unlike E2 reactions, which require the proton to be anti to the leaving group, E1 reactions only require a neighboring hydrogen. This is due to the fact that the leaving group has already left the molecule. The final product is an alkene along with the HB byproduct.
Reactivity
Due to the fact that E1 reactions create a carbocation intermediate, rules present in \(S_N1\) reactions still apply.
As expected, tertiary carbocations are favored over secondary, primary and methyl’s. This is due to the phenomena of hyperconjugation, which essentially allows a nearby C-C or C-H bond to interact with the p orbital of the carbon to bring the electrons down to a lower energy state. Thus, this has a stabilizing effect on the molecule as a whole. In general, primary and methyl carbocations do not proceed through the E1 pathway for this reason, unless there is a means of carbocation rearrangement to move the positive charge to a nearby carbon. Secondary and Tertiary carbons form more stable carbocations, thus this formation occurs quite rapidly.
Secondary carbocations can be subject to the E2 reaction pathway, but this generally occurs in the presence of a good / strong base. Adding a weak base to the reaction disfavors E2, essentially pushing towards the E1 pathway. In many instances, solvolysis occurs rather than using a base to deprotonate. This means heat is added to the solution, and the solvent itself deprotonates a hydrogen. The medium can effect the pathway of the reaction as well. Polar protic solvents may be used to hinder nucleophiles, thus disfavoring E2 / Sn2 from occurring.
Acid catalyzed dehydration of secondary / tertiary alcohols
We’ll take a look at a mechanism involving solvolysis during an E1 reaction of Propanol in Sulfuric Acid.
• Step 1: The OH group on the pentanol is hydrated by H2SO4. This allows the OH to become an H2O, which is a better leaving group.
• Step 2: Once the OH has been hydrated, the H2O molecule leaves, taking its electrons with it. This creates a carbocation intermediate on the attached carbon.
• Step 3: Another H2O molecule comes in to deprotonate the beta carbon, which then donates its electrons to the neighboring C-C bond. The carbons are rehybridized from sp3 to sp2, and thus a pi bond is formed between them.
Mechanism for Alkyl Halides
This mechanism is a common application of E1 reactions in the synthesis of an alkene.
Once again, we see the basic 2 steps of the E1 mechanism.
1. The leaving group leaves along with its electrons to form a carbocation intermediate.
2. A base deprotonates a beta carbon to form a pi bond.
In this case we see a mixture of products rather than one discrete one. This is the case because the carbocation has two nearby carbons that are capable of being deprotonated, but that only one forms a major product (more stable).
How are Regiochemistry & Stereochemistry involved?
In terms of regiochemistry, Zaitsev's rule states that although more than one product can be formed during alkene synthesis, the more substituted alkene is the major product. This infers that the hydrogen on the most substituted carbon is the most probable to be deprotonated, thus allowing for the most substituted alkene to be formed.
Unlike E2 reactions, E1 is not stereospecific. Thus, a hydrogen is not required to be anti-periplanar to the leaving group.
In this mechanism, we can see two possible pathways for the reaction. One in which the methyl on the right is deprotonated, and another in which the CH2 on the left is deprotonated. Either one leads to a plausible resultant product, however, only one forms a major product. As stated by Zaitsev's rule, deprotonation of the most substituted carbon results in the most substituted alkene. This then becomes the most stable product due to hyperconjugation, and is also more common than the minor product.
Contributors
• Satish Balasubramanian | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/08%3A_Alkyl_Halides_and_Elimination_Reactions/8.06%3A_The_%28E_1%29_Mechanism.txt |
The E1 mechanism is nearly identical to the SN1 mechanism, differing only in the course of reaction taken by the carbocation intermediate. As shown by the following equations, a carbocation bearing beta-hydrogens may function either as a Lewis acid (electrophile), as it does in the SN1 reaction, or a Brønsted acid, as in the E1 reaction.
Thus, hydrolysis of tert-butyl chloride in a mixed solvent of water and acetonitrile gives a mixture of 2-methyl-2-propanol (60%) and 2-methylpropene (40%) at a rate independent of the water concentration. The alcohol is the product of an SN1 reaction and the alkene is the product of the E1 reaction. The characteristics of these two reaction mechanisms are similar, as expected. They both show first order kinetics; neither is much influenced by a change in the nucleophile/base; and both are relatively non-stereospecific.
(CH3)3CCl + H2O ——> [ (CH3)3C(+) ] + Cl(–) + H2O ——> (CH3)3COH + (CH3)2C=CH2 + HCl + H2O
To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:
1. The cation may bond to a nucleophile to give a substitution product.
2. The cation may transfer a beta-proton to a base, giving an alkene product.
3. The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.
Since the SN1 and E1 reactions proceed via the same carbocation intermediate, the product ratios are difficult to control and both substitution and elimination usually take place.
Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, the most important being the structure of the alkyl group and the nature of the nucleophilic reactant. The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both SN2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).
8.08: Stereochemistry of the (E 2) Reaction
Stereochemistry of the E2 Reaction
E2 elimination reactions of certain isomeric cycloalkyl halides show unusual rates and regioselectivity that are not explained by the principles thus far discussed. For example, trans-2-methyl-1-chlorocyclohexane reacts with alcoholic KOH at a much slower rate than does its cis-isomer. Furthermore, the product from elimination of the trans-isomer is 3-methylcyclohexene (not predicted by the Zaitsev rule), whereas the cis-isomer gives the predicted 1-methylcyclohexene as the chief product. These differences are described by the first two equations in the following diagram.
Unlike open chain structures, cyclic compounds generally restrict the spatial orientation of ring substituents to relatively few arrangements. Consequently, reactions conducted on such substrates often provide us with information about the preferred orientation of reactant species in the transition state. Stereoisomers are particularly suitable in this respect, so the results shown here contain important information about the E2 transition state.
The most sensible interpretation of the elimination reactions of 2- and 4-substituted halocyclohexanes is that this reaction prefers an anti orientation of the halogen and the beta-hydrogen which is attacked by the base. These anti orientations are colored in red in the above equations. The compounds used here all have six-membered rings, so the anti orientation of groups requires that they assume a diaxial conformation. The observed differences in rate are the result of a steric preference for equatorial orientation of large substituents, which reduces the effective concentration of conformers having an axial halogen. In the case of the 1-bromo-4-tert-butylcyclohexane isomers, the tert-butyl group is so large that it will always assume an equatorial orientation, leaving the bromine to be axial in the cis-isomer and equatorial in the trans. Because of symmetry, the two axial beta-hydrogens in the cis-isomer react equally with base, resulting in rapid elimination to the same alkene (actually a racemic mixture). This reflects the fixed anti orientation of these hydrogens to the chlorine atom. To assume a conformation having an axial bromine the trans-isomer must tolerate serious crowding distortions. Such conformers are therefore present in extremely low concentration, and the rate of elimination is very slow. Indeed, substitution by hydroxide anion predominates.
A similar analysis of the 1-chloro-2-methylcyclohexane isomers explains both the rate and regioselectivity differences. Both the chlorine and methyl groups may assume an equatorial orientation in a chair conformation of the trans-isomer, as shown in the top equation. The axial chlorine needed for the E2 elimination is present only in the less stable alternative chair conformer, but this structure has only one axial beta-hydrogen (colored red), and the resulting elimination gives 3-methylcyclohexene. In the cis-isomer the smaller chlorine atom assumes an axial position in the more stable chair conformation, and here there are two axial beta hydrogens. The more stable 1-methylcyclohexene is therefore the predominant product, and the overall rate of elimination is relatively fast.
An orbital drawing of the anti-transition state is shown on the right. Note that the base attacks the alkyl halide from the side opposite the halogen, just as in the SN2 mechanism. In this drawing the α and β carbon atoms are undergoing a rehybridization from sp3 to sp2 and the developing π-bond is drawn as dashed light blue lines. The symbol R represents an alkyl group or hydrogen. Since both the base and the alkyl halide are present in this transition state, the reaction is bimolecular and should exhibit second order kinetics. We should note in passing that a syn-transition state would also provide good orbital overlap for elimination, and in some cases where an anti-orientation is prohibited by structural constraints syn-elimination has been observed.
Instead, in an E2 reaction, stereochemistry of the double bond -- that is, whether the E or Z isomer results -- is dictated by the stereochemistry of the starting material, if it is diastereomeric. In other words, if the carbon with the hydrogen and the carbon with the halogen are both chiral, then one diastereomer will lead to one product, and the other diastereomer will lead to the other product.
The following reactions of potassium ethoxide with dibromostilbene (1,2-dibromo-1,2-diphenylethane) both occurred via an E2 mechanism. Two different diastereomers were used. Two different stereoisomers (E vs. Z) resulted. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/08%3A_Alkyl_Halides_and_Elimination_Reactions/8.07%3A_%28S_N1%29_and_%28E_1%29_Reactions.txt |
Alkynes can be a useful functional group to synthesize due to some of their antibacterial, antiparasitic, and antifungal properties. One simple method for alkyne synthesis is by double elimination from a dihaloalkane.
Introduction
One case in which elimination can occur is when a haloalkane is put in contact with a nucleophile. The table below is used to determine which situations will result in elimination and the formation of a π bond.
Elimination: Haloalkane-Nucleophile Reaction
Type of Haloalkane Weak Base, Poor Nucleophile Weak Base, Good Nucleophile Strong, Unhindered Base Strong, Hindered Base
Primary
Unhindered E2
Branched E2 E2
Secondary E1 E2 E2
Tertiary E1 E1 E2 E2
* Empty Box means no elimination or Pi bond forms
To synthesize alkynes from dihaloalkanes we use dehydrohalogenation. The majority of these reactions take place using alkoxide bases (other strong bases can also be used) with high temperatures. This combination results in the majority of the product being from the E2 mechanism.
E2 Mechanism
Recall that the E2 mechanism is a concerted reaction (occurs in 1 step). However, in this 1 step there are 3 different changes in the molecule. This is the reaction between 2-Bromo-2-methylpropane and Sodium Hyrdoxide.
This is a brief review of the E2 reaction. For further information on why the reaction proceeds as it does visit the E2 reaction page. Now, if we apply this concept using 2 halides on vicinal or geminal carbons, the E2 reaction will take place twice resulting in the formation of 2 Pi bonds and thus an Alkyne.
Dihaloalkane Elimination
This is a general picture of the reaction taking place without any of the mechanisms shown.
or
* With a terminal haloalkane the equation above is modified in that 3 equivalents of base will be used instead of 2.
Lets look at the mechanism of a reaction between 2,3-Dibromopentane with sodium amide in liquid ammonia.
• Liquid ammonia is not part of the reaction, but is used as a solvent
• Notice the intermediate of the alkyne synthesis. It is stereospecifically in its anti form. Because the second proton and halogen are pulled off the molecule this is unimportant to the synthesis of alkynes. For more information on this see the page on preparation of alkenes from haloalkanes.
Preparation of Alkynes from Alkenes
Lastly, we will briefly look at how to prepare alkynes from alkenes. This is a simple process using first halogenation of the alkene bond to form the dihaloalkane, and next, using the double elimination process to protonate the alkane and from the 2 Pi bonds.
This first process is gone over in much greater detail in the page on halogenation of an alkene. In general, chlorine or bromine is used with an inert halogenated solvent like chloromethane to create a vicinal dihalide from an alkene. The vicinal dihalide formed is the reactant needed to produce the alkyene using double elimination, as covered previously on this page.
In The Lab
Due to the strong base and high temperatures needed for this reaction to take place, the triple bond may change positions. An example of this is when reactants that should form a terminal alkyne, form a 2-Alkyne instead. The use of NaNH2 in liquid NH3 is used in order to prevent this from happening due to its lower reacting temperature. Even so, most chemists will prefer to use nucleophilic substitution instead of elimination when trying to form a terminal alkyne.
Questions
Question 1: Why would we need 3 bases for every terminal dihaloalkane instead of 2 in order to form an alkyne?
Question 2: What are the major products of the following reactions:
a.) 1,2-Dibromopentane with sodium amide in liquid ammonia
b.) 1-Pentene first with Br2 and chloromethane, followed by sodium ethoxide (Na+ -O-CH2CH3)
Question 3: What would be good starting molecules for the synthesis of the following molecules:
Question 4: Use a 6 carbon diene to synthesize a 6 carbon molecule with 2 terminal alkynes.
Answers
Answer 1: Remember that hydrogen atoms on terminal alkynes make the alkyne acidic. One of the base molecules will pull off the terminal hydrogen instead of one of the halides like we want.
Answer 2:
a.) 1-Pentyne
b.) 1-Pentyne
Answer 3:
Answer 4: Bromine or chlorine can be used with different inert solvents for the halogenation. This can be done using many different bases. Liquid ammonia is used as a solvent and needs to be followed by an aqueous work-up. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/08%3A_Alkyl_Halides_and_Elimination_Reactions/8.10%3A_%28E_2%29_Reactions_and_Alkyne_Synthesis.txt |
Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, the most important being the structure of the alkyl group and the nature of the nucleophilic reactant.In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable. Strong nucleophile favor substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor elimination.
The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both SN2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).
The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. Note that halogens bonded to sp2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents.
Nucleophile
Anionic Nucleophiles
( Weak Bases: I, Br, SCN, N3,
CH3CO2 , RS, CN etc. )
pKa's from -9 to 10 (left to right)
Anionic Nucleophiles
( Strong Bases: HO, RO )
pKa's > 15
Neutral Nucleophiles
( H2O, ROH, RSH, R3N )
pKa's ranging from -2 to 11
Alkyl Group
Primary
RCH2
Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g.
ClCH2CH2Cl + KOH ——> CH2=CHCl
SN2 substitution. (N ≈ S >>O)
Secondary
R2CH–
SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O)
In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly.
Tertiary
R3C–
E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed.
Allyl
H2C=CHCH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Benzyl
C6H5CH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/08%3A_Alkyl_Halides_and_Elimination_Reactions/8.11%3A_When_Is_the_Reaction_%28S_N1%29_%28S_N2%29_%28E_1%29_or_%28E_2%29.txt |
The E2 elimination of 3º-alcohols under relatively non-acidic conditions may be accomplished by treatment with phosphorous oxychloride (POCl3) in pyridine. This procedure is also effective with hindered 2º-alcohols, but for unhindered and 1º-alcohols an SN2 chloride ion substitution of the chlorophosphate intermediate competes with elimination. Examples of these and related reactions are given in the following figure. The first equation shows the dehydration of a 3º-alcohol. The predominance of the non-Zaitsev product (less substituted double bond) is presumed due to steric hindrance of the methylene group hydrogen atoms, which interferes with the approach of base at that site. The second example shows two elimination procedures applied to the same 2º-alcohol. The first uses the single step POCl3 method, which works well in this case because SN2 substitution is retarded by steric hindrance. The second method is another example in which an intermediate sulfonate ester confers halogen-like reactivity on an alcohol. In every case the anionic leaving group is the conjugate base of a strong acid.
9.02: Conversion of Alcohols to Alkyl Halides with HX
When alcohols react with a hydrogen halide, a substitution takes place producing an alkyl halide and water:
Scope of Reaction
• The order of reactivity of alcohols is 3° > 2° > 1° methyl.
• The order of reactivity of the hydrogen halides is HI > HBr > HCl (HF is generally unreactive).
The reaction is acid catalyzed. Alcohols react with the strongly acidic hydrogen halides HCl, HBr, and HI, but they do not react with nonacidic NaCl, NaBr, or NaI. Primary and secondary alcohols can be converted to alkyl chlorides and bromides by allowing them to react with a mixture of a sodium halide and sulfuric acid:
Mechanisms of the Reactions of Alcohols with HX
Secondary, tertiary, allylic, and benzylic alcohols appear to react by a mechanism that involves the formation of a carbocation, in an \(S_N1\) reaction with the protonated alcohol acting as the substrate.
The \(S_N1\) mechanism is illustrated by the reaction tert-butyl alcohol and aqueous hydrochloric acid (\(H_3O^+\), \(Cl^-\) ). The first two steps in this \(S_n1\) substitution mechanism are protonation of the alcohol to form an oxonium ion. Although the oxonium ion is formed by protonation of the alcohol, it can also be viewed as a Lewis acid-base complex between the cation (\(R^+\)) and \(H_2O\). Protonation of the alcohol converts a poor leaving group (OH-) to a good leaving group (\)H_2O\_), which makes the dissociation step of the \(S_N1\) mechanism more favorable.
In step 3, the carbocation reacts with a nucleophile (a halide ion) to complete the substitution.
When we convert an alcohol to an alkyl halide, we carry out the reaction in the presence of acid and in the presence of halide ions, and not at elevated temperature. Halide ions are good nucleophiles (they are much stronger nucleophiles than water), and since halide ions are present in high concentration, most of the carbocations react with an electron pair of a halide ion to form a more stable species, the alkyl halide product. The overall result is an \(S_n1\) reaction.
Primary Alcohols
Not all acid-catalyzed conversions of alcohols to alkyl halides proceed through the formation of carbocations. Primary alcohols and methanol react to form alkyl halides under acidic conditions by an SN2 mechanism.
In these reactions the function of the acid is to produce a protonated alcohol. The halide ion then displaces a molecule of water (a good leaving group) from carbon; this produces an alkyl halide:
Again, acid is required. Although halide ions (particularly iodide and bromide ions) are strong nucleophiles, they are not strong enough to carry out substitution reactions with alcohols themselves. Direct displacement of the hydroxyl group does not occur because the leaving group would have to be a strongly basic hydroxide ion:
We can see now why the reactions of alcohols with hydrogen halides are acid-promoted.
Carbocation rearrangements are extremely common in organic chemistry reactions are are defined as the movement of a carbocation from an unstable state to a more stable state through the use of various structural reorganizational "shifts" within the molecule. Once the carbocation has shifted over to a different carbon, we can say that there is a structural isomer of the initial molecule. However, this phenomenon is not as simple as it sounds.
Carbocation Rearrangement in the SN1 Reaction
Whenever alcohols are subject to transformation into various carbocations, the carbocations are subject to a phenomenon known as carbocation rearrangement. A carbocation, in brief, holds the positive charge in the molecule that is attached to three other groups and bears a sextet rather than an octet. However, we do see carbocation rearrangements in reactions that do not contain alcohol as well. Those, on the other hand, require more difficult explanations than the two listed below. There are two types of rearrangements: hydride shift and alkyl shift. These rearrangements usualy occur in many types of carbocations. Once rearranged, the molecules can also undergo further unimolecular substitution (SN1) or unimolecular elimination (E1). Though, most of the time we see either a simple or complex mixture of products. We can expect two products before undergoing carbocation rearrangement, but once undergoing this phenomenon, we see the major product.
Hydride Shift
Whenever a nucleophile attacks some molecules, we typically see two products. However, in most cases, we normally see both a major product and a minor product. The major product is typically the rearranged product that is more substituted (aka more stable). The minor product, in contract, is typically the normal product that is less substituted (aka less stable).
The reaction: We see that the formed carbocations can undergo rearrangements called hydride shift. This means that the two electron hydrogen from the unimolecular substitution moves over to the neighboring carbon. We see the phenomenon of hydride shift typically with the reaction of an alcohol and hydrogen halides, which include HBr, HCl, and HI. HF is typically not used because of its instability and its fast reactivity rate. Below is an example of a reaction between an alcohol and hydrogen chloride:
GREEN (Cl) = nucleophile BLUE (OH) = leaving group ORANGE (H) = hydride shift proton RED(H) = remaining proton
The alcohol portion (-OH) has been substituted with the nucleophilic Cl atom. However, it is not a direct substitution of the OH atom as seen in SN2 reactions. In this SN1 reaction, we see that the leaving group, -OH, forms a carbocation on Carbon #3 after receiving a proton from the nucleophile to produce an alkyloxonium ion. Before the Cl atom attacks, the hydrogen atom attached to the Carbon atom directly adjacent to the original Carbon (preferably the more stable Carbon), Carbon #2, can undergo hydride shift. The hydrogen and the carbocation formally switch positions. The Cl atom can now attack the carbocation, in which it forms the more stable structure because of hyperconjugation. The carbocation, in this case, is most stable because it attaches to the tertiary carbon (being attached to 3 different carbons). However, we can still see small amounts of the minor, unstable product. The mechanism for hydride shift occurs in multiple steps that includes various intermediates and transition states. Below is the mechanism for the given reaction above: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.01%3A_Dehydration_Using_%28POCl_3%29_and_Pyridine.txt |
The most common methods for converting 1º- and 2º-alcohols to the corresponding chloro and bromo alkanes (i.e. replacement of the hydroxyl group) are treatments with thionyl chloride and phosphorus tribromide, respectively. These reagents are generally preferred over the use of concentrated HX due to the harsh acidity of these hydrohalic acids and the carbocation rearrangements associated with their use.
Synthetic organic chemists, when they want to convert an alcohol into a better leaving group, have several methods to choose from. One common strategy is to convert the alcohol into an alkyl chloride or bromide, using thionyl chloride or phosphorus tribromide:
Drawbacks to using \(PBr_3\) and \(SOCl_2\)
Despite their general usefulness, phosphorous tribromide and thionyl chloride have shortcomings. Hindered 1º- and 2º-alcohols react sluggishly with the former, and may form rearrangement products, as noted in the following equation.
Below, an abbreviated mechanism for the reaction is displayed. The initially formed trialkylphosphite ester may be isolated if the HBr byproduct is scavenged by base. In the presence of HBr a series of acid-base and SN2 reactions take place, along with the transient formation of carbocation intermediates. Rearrangement (pink arrows) of the carbocations leads to isomeric products.
Reaction of thionyl chloride with chiral 2º-alcohols has been observed to proceed with either inversion or retention. In the presence of a base such as pyridine, the intermediate chlorosulfite ester reacts to form an "pyridinium" salt, which undergoes a relatively clean SN2 reaction to the inverted chloride. In ether and similar solvents the chlorosulfite reacts with retention of configuration, presumably by way of a tight or intimate ion pair. This is classified as an SNi reaction (nucleophilic substitution internal). The carbocation partner in the ion pair may also rearrange. These reactions are illustrated by the following equations. An alternative explanation for the retention of configuration, involving an initial solvent molecule displacement of the chlorosulfite group (as SO2 and chloride anion), followed by chloride ion displacement of the solvent moiety, has been suggested. In this case, two inversions lead to retention.
Example 1: Conversion of Alcohols to Alkyl Chlorides
There’s one important thing to note here: see the stereochemistry? It’s been inverted.*(white lie alert – see below) That’s an important difference between \(SOCl_2\) and TsCl, which leaves the stereochemistry alone. We’ll get to the root cause of that in a moment, but in the meantime, can you think of a mechanism which results in inversion of configuration at carbon?
Formation of Alkyl Chlorides
Since the reaction proceeds through a backside attack (\(S_N2\)), there is inversion of configuration at the carbon
The mechanism for formation of acid chlorides from carboxylic acids is similar. The conversion of caboxylic acids to acid chlorides is similar, but proceeds through a [1,2]-addition of chloride ion to the carbonyl carbon followed by [1,2]-elimination to give the acid chloride, \(SO_2\) and \(HCl\)
Formation of Alkyl Bromides
The PBr3 reaction is thought to involve two successive SN2-like steps:
Notice that these reactions result in inversion of stereochemistry in the resulting alkyl halide.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
9.04: TosylateAnother Good Leaving Group
Alternatively, we can transform an alcohol group into sulfonic ester using para-toluene sulfonyl chloride (Ts-Cl) or methanesulfonyl chloride (Ms-Cl), creating what is termed an organic tosylate or mesylate:
Again, you’ll have a chance to work a mechanism for tosylate and mesylate formation in the chapter 12 problems. Notice, though, that unlike the halogenation reactions above, conversion of an alcohol to a tosylate or mesylate proceeds with retention of configuration at the electrophilic carbon.
Chlorides, bromides, and tosylate / mesylate groups are excellent leaving groups in nucleophilic substitution reactions, due to resonance delocalization of the developing negative charge on the leaving oxygen.
The laboratory synthesis of isopentenyl diphosphate - the 'building block' molecule used by nature for the construction of isoprenoid molecules such as cholesterol and b-carotene - was accomplished by first converting the alcohol into an organic tosylate (step 1), then displacing the tosylate group with an inorganic pyrophosphate nucleophile (step 2) (J. Org. Chem 1986, 51, 4768).
Example
Exercise 8.14: Predict the structures of A and B in the following reaction:
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
9.05: Reaction of Ethers with Strong Acid
The most common reaction of ethers is cleavage of the C–O bond by strong acids. This may occur by SN1 or E1 mechanisms for 3º-alkyl groups or by an SN2 mechanism for 1º-alkyl groups. Some examples are shown in the following diagram. The conjugate acid of the ether is an intermediate in all these reactions, just as conjugate acids were intermediates in certain alcohol reactions.
The first two reactions proceed by a sequence of SN2 steps in which the iodide or bromide anion displaces an alcohol in the first step, and then converts the conjugate acid of that alcohol to an alkyl halide in the second. Since SN2 reactions are favored at least hindered sites, the methyl group in example #1 is cleaved first. The 2º-alkyl group in example #3 is probably cleaved by an SN2 mechanism, but the SN1 alternative cannot be ruled out. The phenol formed in this reaction does not react further, since SN2, SN1 and E1 reactions do not take place on aromatic rings. The last example shows the cleavage of a 3º-alkyl group by a strong acid. Acids having poorly nucleophilic conjugate bases are often chosen for this purpose so that E1 products are favored. The reaction shown here (#4) is the reverse of the tert-butyl ether preparation described earlier. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.03%3A_Conversion_of_Alcohols_to_Alkyl_Halides_with_%28SOCl_2%29_and_%28PBr_3%29.txt |
Epoxide ring-opening reactions - SN1 vs. SN2, regioselectivity, and stereoselectivity
The nonenzymatic ring-opening reactions of epoxides provide a nice overview of many of the concepts we have seen already in this chapter. Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions. If the epoxide is asymmetric, the structure of the product will vary according to which mechanism dominates. When an asymmetric epoxide undergoes solvolysis in basic methanol, ring-opening occurs by an SN2 mechanism, and the less substituted carbon is the site of nucleophilic attack, leading to what we will refer to as product B:
Conversely, when solvolysis occurs in acidic methanol, the reaction occurs by a mechanism with substantial SN1 character, and the more substituted carbon is the site of attack. As a result, product A predominates.
These are both good examples of regioselective reactions. In a regioselective reaction, two (or more) different constitutional isomers are possible as products, but one is formed preferentially (or sometimes exclusively).
Let us examine the basic, SN2 case first. The leaving group is an alkoxide anion, because there is no acid available to protonate the oxygen prior to ring opening. An alkoxide is a poor leaving group, and thus the ring is unlikely to open without a 'push' from the nucleophile.
The nucleophile itself is potent: a deprotonated, negatively charged methoxide ion. When a nucleophilic substitution reaction involves a poor leaving group and a powerful nucleophile, it is very likely to proceed by an SN2 mechanism.
What about the electrophile? There are two electrophilic carbons in the epoxide, but the best target for the nucleophile in an SN2 reaction is the carbon that is least hindered. This accounts for the observed regiochemical outcome. Like in other SN2 reactions, nucleophilic attack takes place from the backside, resulting in inversion at the electrophilic carbon.
Probably the best way to depict the acid-catalyzed epoxide ring-opening reaction is as a hybrid, or cross, between an SN2 and SN1 mechanism. First, the oxygen is protonated, creating a good leaving group (step 1 below) . Then the carbon-oxygen bond begins to break (step 2) and positive charge begins to build up on the more substituted carbon (recall the discussion from section 8.4B about carbocation stability).
Unlike in an SN1 reaction, the nucleophile attacks the electrophilic carbon (step 3) before a complete carbocation intermediate has a chance to form.
Attack takes place preferentially from the backside (like in an SN2 reaction) because the carbon-oxygen bond is still to some degree in place, and the oxygen blocks attack from the front side. Notice, however, how the regiochemical outcome is different from the base-catalyzed reaction: in the acid-catalyzed process, the nucleophile attacks the more substituted carbon because it is this carbon that holds a greater degree of positive charge.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
9.07: Application- Epoxides Leukotrienes and Asthma
The members of this group of structurally related natural hormones have an extraordinary range of biological effects. They can lower gastric secretions, stimulate uterine contractions, lower blood pressure, influence blood clotting and induce asthma-like allergic responses. Because their genesis in body tissues is tied to the metabolism of the essential fatty acid arachadonic acid (5,8,11,14-eicosatetraenoic acid) they are classified as eicosanoids. Many properties of the common drug aspirin result from its effect on the cascade of reactions associated with these hormones.
The metabolic pathways by which arachidonic acid is converted to the various eicosanoids are complex and will not be discussed here. A rough outline of some of the transformations that take place is provided below. It is helpful to view arachadonic acid in the coiled conformation shown in the shaded box.
Leukotriene A is a precursor to other leukotriene derivatives by epoxide opening reactions. The prostaglandins are given systematic names that reflect their structure. The initially formed peroxide \(PGH_2\) is a common intermediate to other prostaglandins, as well as thromboxanes such as \(TXA_2\).
9.08: Application- Benzoapyrene Epoxides and Cancer
To Your Health: Polycyclic Aromatic Hydrocarbons and Cancer
The intense heating required for distilling coal tar results in the formation of PAHs. For many years, it has been known that workers in coal-tar refineries are susceptible to a type of skin cancer known as tar cancer. Investigations have shown that a number of PAHs are carcinogens. One of the most active carcinogenic compounds, benzopyrene, occurs in coal tar and has also been isolated from cigarette smoke, automobile exhaust gases, and charcoal-broiled steaks. It is estimated that more than 1,000 t of benzopyrene are emitted into the air over the United States each year. Only a few milligrams of benzopyrene per kilogram of body weight are required to induce cancer in experimental animals. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.06%3A_Reactions_of_Epoxides.txt |
Alcohols
We have already seen the simplest possible example of an alcohol functional group in methanol. In the alcohol functional group, a carbon is single-bonded to an OH group (this OH group, by itself, is referred to as a hydroxyl). If the central carbon in an alcohol is bonded to only one other carbon, we call the group a primary alcohol. In secondary alcohols and tertiary alcohols, the central carbon is bonded to two and three carbons, respectively. Methanol, of course, is in class by itself in this respect.
Primary alcohols
In a primary (1°) alcohol, the carbon atom that carries the -OH group is only attached to one alkyl group. Some examples of primary alcohols are shown below:
Notice that the complexity of the attached alkyl group is irrelevant. In each case there is only one linkage to an alkyl group from the CH2 group holding the -OH group. There is an exception to this. Methanol, CH3OH, is counted as a primary alcohol even though there are no alkyl groups attached to the the -OH carbon atom.
Secondary alcohols
In a secondary (2°) alcohol, the carbon atom with the -OH group attached is joined directly to two alkyl groups, which may be the same or different. Examples include the following:
Tertiary alcohols
In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of the same or different groups. Examples of tertiary alcohols are given below:
'
Ethers
With the general formula ROR′, an ether may be considered a derivative of water in which both hydrogen atoms are replaced by alkyl or aryl groups. It may also be considered a derivative of an alcohol (ROH) in which the hydrogen atom of the OH group is been replaced by a second alkyl or aryl group:
$\mathrm{HOH\underset{H\: atoms}{\xrightarrow{replace\: both}}ROR'\underset{of\: OH\: group}{\xleftarrow{replace\: H\: atom}}ROH}$
Simple ethers have simple common names, formed from the names of the groups attached to oxygen atom, followed by the generic name ether. For example, CH3–O–CH2CH2CH3 is methyl propyl ether. If both groups are the same, the group name should be preceded by the prefix di-, as in dimethyl ether (CH3–O–CH3) and diethyl ether CH3CH2–O–CH2CH3.
Epoxides
An epoxide is a cyclic ether with three ring atoms. These rings approximately define an equilateral triangle, which makes it highly strained. The strained ring makes epoxides more reactive than other ethers. Simple epoxides are named from the parent compound ethylene oxide or oxirane, such as in chloromethyloxirane. As a functional group, epoxides feature the epoxy prefix, such as in the compound 1,2-epoxycycloheptane, which can also be called cycloheptene epoxide, or simply cycloheptene oxide.
A generic epoxide.
The chemical structure of the epoxide glycidol, a common chemical intermediate
A polymer formed by reacting epoxide units is called a polyepoxide or an epoxy. Epoxy resins are used as adhesives and structural materials. Polymerization of an epoxide gives a polyether, for example ethylene oxide polymerizes to give polyethylene glycol, also known as polyethylene oxide.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
9.10: Structure and Bonding
Structure of Alcohols
The structure of an alcohol resembles that of water. With both alcohol and water, the bond angles reflect the effect of electron repulsion and increasing steric bulk of the substituents on the central oxygen atom. The electronegativity of oxygen contributes to the unsymmetrical distribution of charge, creating a partial positive charge on hydrogen and a partial negative charge on oxygen. This uneven distribution of electron density in the O-H bond creates a dipole. In addition, because of the high electronegativity of oxygen relative to that of carbon, the O-H bond is shorter (1.41 Å for C-O vs. 1.51 Å for C-C) and stronger (ΔHoO-H = 104 kcal mol-1 for C-O vs. ΔHoC-H = 98 kcal mol-1 for C-C).
Water Methanol
Structure of Epoxides
Epoxides (also known as oxiranes) are three-membered ring structures in which one of the vertices is an oxygen and the other two are carbons.
The carbons in an epoxide group are very reactive electrophiles, due in large part to the fact that substantial ring strain is relieved when the ring opens upon nucleophilic attack.
Structure of Ethers
Ethers are a class of organic compounds that contain an oxygen between two alkyl groups. They have the formula R-O-R', with R's being the alkyl groups. these compounds are used in dye, perfumes, oils, waxes and industrial use. Ethers are named as alkoxyalkanes.
An aliphatic ether is an ether in the molecule of which there are no aryl groups on the ether group.
eg:
An ether molecule may contain aryl groups, nevertheless, be an aliphatic ether.
eg:
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• Jaspreet Dheri | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.09%3A_Introduction.txt |
Naming Alcohols
1. Find the longest chain containing the hydroxy group (OH). If there is a chain with more carbons than the one containing the OH group it will be named as a subsitutent.
2. Place the OH on the lowest possible number for the chain. With the exception of carbonyl groups such as ketones and aldehydes, the alcohol or hydroxy groups have first priority for naming.
3. When naming a cyclic structure, the -OH is assumed to be on the first carbon unless the carbonyl group is present, in which case the later will get priority at the first carbon.
4. When multiple -OH groups are on the cyclic structure, number the carbons on which the -OH groups reside.
5. Remove the final e from the parent alkane chain and add -ol. When multiple alcohols are present use di, tri, et.c before the ol, after the parent name. ex. 2,3-hexandiol. If a carbonyl group is present, the -OH group is named with the prefix "hydroxy," with the carbonyl group attached to the parent chain name so that it ends with -al or -one.
Examples
Ethane: CH3CH3 ----->Ethanol: (the alcohol found in beer, wine and other consumed sprits)
Secondary alcohol: 2-propanol
Other functional groups on an alcohol: 3-bromo-2-pentanol
Cyclic alcohol (two -OH groups): cyclohexan-1,4-diol
Other functional group on the cyclic structure: 3-hexeneol (the alkene is in bold and indicated by numbering the carbon closest to the alcohol)
A complex alcohol:4-ethyl-3hexanol (the parent chain is in red and the substituent is in blue)
In the IUPAC system of nomenclature, functional groups are normally designated in one of two ways. The presence of the function may be indicated by a characteristic suffix and a location number. This is common for the carbon-carbon double and triple bonds which have the respective suffixes -ene and -yne. Halogens, on the other hand, do not have a suffix and are named as substituents, for example: (CH3)2C=CHCHClCH3 is 4-chloro-2-methyl-2-pentene.
Alcohols are usually named by the first procedure and are designated by an -ol suffix, as in ethanol, CH3CH2OH (note that a locator number is unnecessary on a two-carbon chain). On longer chains the location of the hydroxyl group determines chain numbering. For example: (CH3)2C=CHCH(OH)CH3 is 4-methyl-3-penten-2-ol. Other examples of IUPAC nomenclature are shown below, together with the common names often used for some of the simpler compounds. For the mono-functional alcohols, this common system consists of naming the alkyl group followed by the word alcohol. Alcohols may also be classified as primary, , secondary, , and tertiary, , in the same manner as alkyl halides. This terminology refers to alkyl substitution of the carbon atom bearing the hydroxyl group (colored blue in the illustration).
Many functional groups have a characteristic suffix designator, and only one such suffix (other than "-ene" and "-yne") may be used in a name. When the hydroxyl functional group is present together with a function of higher nomenclature priority, it must be cited and located by the prefix hydroxy and an appropriate number. For example, lactic acid has the IUPAC name 2-hydroxypropanoic acid.
Naming Ethers
Ethers are compounds having two alkyl or aryl groups bonded to an oxygen atom, as in the formula R1–O–R2. The ether functional group does not have a characteristic IUPAC nomenclature suffix, so it is necessary to designate it as a substituent. To do so the common alkoxy substituents are given names derived from their alkyl component (below):
Alkyl Group Name Alkoxy Group Name
CH3 Methyl CH3O– Methoxy
CH3CH2 Ethyl CH3CH2O– Ethoxy
(CH3)2CH– Isopropyl (CH3)2CHO– Isopropoxy
(CH3)3C– tert-Butyl (CH3)3CO– tert-Butoxy
C6H5 Phenyl C6H5O– Phenoxy
The smaller, shorter alkyl group becomes the alkoxy substituent. The larger, longer alkyl group side becomes the alkane base name. Each alkyl group on each side of the oxygen is numbered separately. The numbering priority is given to the carbon closest to the oxgen. The alkoxy side (shorter side) has an "-oxy" ending with its corresponding alkyl group. For example, CH3CH2CH2CH2CH2-O-CH2CH2CH3 is 1-propoxypentane. If there is cis or trans stereochemistry, the same rule still applies.
Example
Examples are: CH3CH2OCH2CH3, diethyl ether (sometimes referred to as ether), and CH3OCH2CH2OCH3, ethylene glycol dimethyl ether (glyme).
Common names
Simple ethers are given common names in which the alkyl groups bonded to the oxygen are named in alphabetical order followed by the word "ether". The top left example shows the common name in blue under the IUPAC name. Many simple ethers are symmetrical, in that the two alkyl substituents are the same. These are named as "dialkyl ethers".
Heterocycles
In cyclic ethers (heterocycles), one or more carbons are replaced with oxygen. Often, it's called heteroatoms, when carbon is replaced by an oxygen or any atom other than carbon or hydrogen. In this case, the stem is called the oxacycloalkane, where the prefix "oxa-" is an indicator of the replacement of the carbon by an oxygen in the ring. These compounds are numbered starting at the oxygen and continues around the ring. For example,
If a substituent is an alcohol, the alcohol has higher priority. However, if a substituent is a halide, ether has higher priority. If there is both an alcohol group and a halide, alcohol has higher priority. The numbering begins with the end that is closest to the higher priority substituent. There are ethers that are contain multiple ether groups that are called cyclic polyethers or crown ethers. These are also named using the IUPAC system.
hyl sulphide. Sulphides are chemically more reactive than ethers, reflecting the greater nucleophilicity of sulfur relative to oxygen.
Problems
Name the following ethers:
(Answers to problems above: 1. diethyl ether; 2. 2-ethoxy-2-methyl-1-propane; 3. cis-1-ethoxy-2-methoxycyclopentane; 4. 1-ethoxy-1-methylcyclohexane; 5. oxacyclopropane; 6. 2,2-Dimethyloxacyclopropane)
Common names of some ethers
anisole (try naming anisole by the other two conventions. J )
oxirane
1,2-epoxyethane, ethylene oxide, dimethylene oxide, oxacyclopropane,
furan (this compound is aromatic)
tetrahydrofuran
oxacyclopentane, 1,4-epoxybutane, tetramethylene oxide,
dioxane
1,4-dioxacyclohexane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.11%3A_Nomenclature.txt |
Physical properties of alcohols
Boiling Points
The chart below shows the boiling points of the following simple primary alcohols with up to 4 carbon atoms:
These boiling points are compared with those of the equivalent alkanes (methane to butane) with the same number of carbon atoms.
Notice that:
• The boiling point of an alcohol is always significantly higher than that of the analogous alkane.
• The boiling points of the alcohols increase as the number of carbon atoms increases.
The patterns in boiling point reflect the patterns in intermolecular attractions.
Hydrogen bonding
Hydrogen bonding occurs between molecules in which a hydrogen atom is attached to a strongly electronegative element: fluorine, oxygen or nitrogen. In the case of alcohols, hydrogen bonds occur between the partially-positive hydrogen atoms and lone pairs on oxygen atoms of other molecules.
The hydrogen atoms are slightly positive because the bonding electrons are pulled toward the very electronegative oxygen atoms. In alkanes, the only intermolecular forces are van der Waals dispersion forces. Hydrogen bonds are much stronger than these, and therefore it takes more energy to separate alcohol molecules than it does to separate alkane molecules. This the main reason for higher boiling points in alcohols.
The effect of van der Waals forces
• Boiling points of the alcohols: Hydrogen bonding is not the only intermolecular force alcohols experience. There are also van der Waals dispersion forces and dipole-dipole interactions. The hydrogen bonding and dipole-dipole interactions are much the same for all alcohols, but dispersion forces increase as the alcohols get bigger. These attractions get stronger as the molecules get longer and have more electrons. This increases the sizes of the temporary dipoles formed. This is why the boiling points increase as the number of carbon atoms in the chains increases. It takes more energy to overcome the dispersion forces, and thus the boiling points rise.
• Comparison between alkanes and alcohols: Even without any hydrogen bonding or dipole-dipole interactions, the boiling point of the alcohol would be higher than the corresponding alkane with the same number of carbon atoms.
Compare ethane and ethanol:
Ethanol is a longer molecule, and the oxygen atom brings with it an extra 8 electrons. Both of these increase the size of the van der Waals dispersion forces, and subsequently the boiling point. A more accurate measurement of the effect of the hydrogen bonding on boiling point would be a comparison of ethanol with propane rather than ethane. The lengths of the two molecules are more similar, and the number of electrons is exactly the same.
Solubility of alcohols in water
Small alcohols are completely soluble in water; mixing the two in any proportion generates a single solution. However, solubility decreases as the length of the hydrocarbon chain in the alcohol increases. At four carbon atoms and beyond, the decrease in solubility is noticeable; a two-layered substance may appear in a test tube when the two are mixed.
Consider ethanol as a typical small alcohol. In both pure water and pure ethanol the main intermolecular attractions are hydrogen bonds.
In order to mix the two, the hydrogen bonds between water molecules and the hydrogen bonds between ethanol molecules must be broken. Energy is required for both of these processes. However, when the molecules are mixed, new hydrogen bonds are formed between water molecules and ethanol molecules.
The energy released when these new hydrogen bonds form approximately compensates for the energy needed to break the original interactions. In addition, there is an increase in the disorder of the system, an increase in entropy. This is another factor in deciding whether chemical processes occur. Consider a hypothetical situation involving 5-carbon alcohol molecules.
The hydrocarbon chains are forced between water molecules, breaking hydrogen bonds between those water molecules. The -OH ends of the alcohol molecules can form new hydrogen bonds with water molecules, but the hydrocarbon "tail" does not form hydrogen bonds. This means that many of the original hydrogen bonds being broken are never replaced by new ones.
In place of those original hydrogen bonds are merely van der Waals dispersion forces between the water and the hydrocarbon "tails." These attractions are much weaker, and unable to furnish enough energy to compensate for the broken hydrogen bonds. Even allowing for the increase in disorder, the process becomes less feasible. As the length of the alcohol increases, this situation becomes more pronounced, and thus the solubility decreases.
Comparisons of Alcohols and Ethers
Ether molecules have no hydrogen atom on the oxygen atom (that is, no OH group). Therefore there is no intermolecular hydrogen bonding between ether molecules, and ethers therefore have quite low boiling points for a given molar mass. Indeed, ethers have boiling points about the same as those of alkanes of comparable molar mass and much lower than those of the corresponding alcohols (Table 14.4 "Comparison of Boiling Points of Alkanes, Alcohols, and Ethers").
Table 14.4 Comparison of Boiling Points of Alkanes, Alcohols, and Ethers
Condensed Structural Formula Name Molar Mass Boiling Point (°C) Intermolecular Hydrogen Bonding in Pure Liquid?
CH3CH2CH3 propane 44 –42 no
CH3OCH3 dimethyl ether 46 –25 no
CH3CH2OH ethyl alcohol 46 78 yes
CH3CH2CH2CH2CH3 pentane 72 36 no
CH3CH2OCH2CH3 diethyl ether 74 35 no
CH3CH2CH2CH2OH butyl alcohol 74 117 yes
Ether molecules do have an oxygen atom, however, and engage in hydrogen bonding with water molecules. Consequently, an ether has about the same solubility in water as the alcohol that is isomeric with it. For example, dimethyl ether and ethanol (both having the molecular formula C2H6O) are completely soluble in water, whereas diethyl ether and 1-butanol (both C4H10O) are barely soluble in water (8 g/100 mL of water). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.12%3A_Physical_Properties.txt |
Alcohols Produced by Fermentation
In addition to its preparation from ethylene, ethanol is made by the fermentation of sugars or starch from various sources (potatoes, corn, wheat, rice, etc.). Fermentation is catalyzed by enzymes found in yeast and proceeds by an elaborate multistep mechanism. We can represent the overall process as follows:
Methanol is quite poisonous to humans. Ingestion of as little as 15 mL of methanol can cause blindness, and 30 mL (1 oz) can cause death. However, the usual fatal dose is 100 to 150 mL. The main reason for methanol’s toxicity is that we have liver enzymes that catalyze its oxidation to formaldehyde, the simplest member of the aldehyde family:
Formaldehyde reacts rapidly with the components of cells, coagulating proteins in much the same way that cooking coagulates an egg. This property of formaldehyde accounts for much of the toxicity of methanol.
Note
Organic and biochemical equations are frequently written showing only the organic reactants and products. In this way, we focus attention on the organic starting material and product, rather than on balancing complicated equations.
Ethanol is oxidized in the liver to acetaldehyde:
The acetaldehyde is in turn oxidized to acetic acid (HC2H3O2), a normal constituent of cells, which is then oxidized to carbon dioxide and water. Even so, ethanol is potentially toxic to humans. The rapid ingestion of 1 pt (about 500 mL) of pure ethanol would kill most people, and acute ethanol poisoning kills several hundred people each year—often those engaged in some sort of drinking contest. Ethanol freely crosses into the brain, where it depresses the respiratory control center, resulting in failure of the respiratory muscles in the lungs and hence suffocation. Ethanol is believed to act on nerve cell membranes, causing a diminution in speech, thought, cognition, and judgment.
Rubbing alcohol is usually a 70% aqueous solution of isopropyl alcohol. It has a high vapor pressure, and its rapid evaporation from the skin produces a cooling effect. It is toxic when ingested but, compared to methanol, is less readily absorbed through the skin.
Ethylene Glycol
Polyhydric alcohols in which the hydroxyl groups are situated on different carbons are relatively stable, and, as we might expect for substances with multiple polar groups, they have high boiling points and considerable water solubility, but low solubility in nonpolar solvents:
1,2-Diols are prepared from alkenes by oxidation with reagents such as osmium tetroxide, potassium permanganate, or hydrogen peroxide. However, ethylene glycol is made on a commercial scale from oxacyclopropane, which in turn is made by air oxidation of ethene at high temperatures over a silver oxide catalyst.
Ethylene glycol has important commercial uses. It is an excellent permanent antifreeze for automotive cooling systems because it is miscible with water in all proportions and a 50% solution freezes at —34° (—29°F). It also is used as a solvent and as an intermediate in the production of polymers (polyesters) and other products.
Ethylene oxide
The most important and simplest epoxide is ethylene oxide which is prepared on an industrial scale by catalytic oxidation of ethylene by air.
Ethers as General Anesthetics
A general anesthetic acts on the brain to produce unconsciousness and a general insensitivity to feeling or pain. Diethyl ether (CH3CH2OCH2CH3) was the first general anesthetic to be used.
William Morton, a Boston dentist, introduced diethyl ether into surgical practice in 1846. This painting shows an operation in Boston in 1846 in which diethyl ether was used as an anesthetic. Inhalation of ether vapor produces unconsciousness by depressing the activity of the central nervous system. Source: Painting of William Morton by Ernest Board, from http://commons.wikimedia.org/wiki/Fi...Ether_1846.jpg.
Diethyl ether is relatively safe because there is a fairly wide gap between the dose that produces an effective level of anesthesia and the lethal dose. However, because it is highly flammable and has the added disadvantage of causing nausea, it has been replaced by newer inhalant anesthetics, including the fluorine-containing compounds halothane, enflurane, and isoflurane. Unfortunately, the safety of these compounds for operating room personnel has been questioned. For example, female operating room workers exposed to halothane suffer a higher rate of miscarriages than women in the general population.
These three modern, inhalant, halogen-containing, anesthetic compounds are less flammable than diethyl ether.
Crown Ethers
It is possible to dissolve ionic compounds in organic solvents using crown ethers. Cyclic polyether with four or more oxygen atoms separated by two or three carbon atoms. All crown ethers have a central cavity that can accommodate a metal ion coordinated to the ring of oxygen atoms., cyclic compounds with the general formula (OCH2CH2)n. Crown ethers are named using both the total number of atoms in the ring and the number of oxygen atoms. Thus 18-crown-6 is an 18-membered ring with six oxygen atoms (part (a) in Figure 12.2.4 ). The cavity in the center of the crown ether molecule is lined with oxygen atoms and is large enough to be occupied by a cation, such as K+. The cation is stabilized by interacting with lone pairs of electrons on the surrounding oxygen atoms. Thus crown ethers solvate cations inside a hydrophilic cavity, whereas the outer shell, consisting of C–H bonds, is hydrophobic. Crown ethers are useful for dissolving ionic substances such as KMnO4 in organic solvents such as isopropanol [(CH3)2CHOH] (Figure 12.2.4). The availability of crown ethers with cavities of different sizes allows specific cations to be solvated with a high degree of selectivity.
Figure 12.2.4: Crown Ethers and Cryptands (a) The potassium complex of the crown ether 18-crown-6. Note how the cation is nestled within the central cavity of the molecule and interacts with lone pairs of electrons on the oxygen atoms. (b) The potassium complex of 2,2,2-cryptand, showing how the cation is almost hidden by the cryptand. Cryptands solvate cations via lone pairs of electrons on both oxygen and nitrogen atoms.
Figure 12.2.5. Effect of a Crown Ether on the Solubility of KMnO4 in Benzene. Normally which is intensely purple, is completely insoluble in benzene which has a relatively low dielectric constant. In the presence of a small amount of crown ether, KMnO4 dissolves in benzene as shown by the reddish purple color caused by the permanganate ions in solution.
Cryptands (from the Greek kryptós, meaning “hidden”) are compounds that can completely surround a cation with lone pairs of electrons on oxygen and nitrogen atoms (part (b) in Figure 12.2.4). The number in the name of the cryptand is the number of oxygen atoms in each strand of the molecule. Like crown ethers, cryptands can be used to prepare solutions of ionic compounds in solvents that are otherwise too nonpolar to dissolve them.
Figure 12.2.3 Ion–Dipole Interactions in the Solvation of Li+ Ions by Acetone, a Polar Solvent | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.13%3A_Interesting_Alcohols_Ethers_and_Epoxides.txt |
Alcohols are prepared by SN2 reaction
Alkyl halides can be converted to alcohols by using SN2 reactions with OH- as a nucleophile. Substrates that undergo substitution by SN1 reaction can be converted to alcohols using water as the nucleophile (and it can even be the solvent). Recall that SN1 reactions are promoted in polar, protic solvents.
Ethers are prepared by SN2 reaction
Williamson Ether Reactions involve an alkoxide that reacts with a primary haloalkane or a sulfonate ester. Alkoxides consist of the conjugate base of an alcohol and are comprised of an R group bonded to an oxygen atom. They are written as RO, where R is the organic substituent.
Ethers can be synthesized in standard SN2 conditions by coupling an alkoxide with a haloalkane/sulfonate ester. The alcohol that supplies the electron rich alkoxide can be used as the solvent, as well as dimethyl sulfoxide (DMSO) or hexamethylphosphoric triamide (HMPA).
Epoxides are prepared by an SN2
Sn2 reactions are characterized by the inversion of stereochemistry at the site of the leaving group. Williamson Ether synthesis is not an exception to this rule and the reaction is set in motion by the backside attack of the nucleophile. This requires that the nucleophile and the electrophile are in anti-configuration.
9.15: General FeaturesReactions of Alcohols Ethers and Epoxides
Alcohols as a leaving group
Despite this promising background evidence, alcohols do not undergo the same SN2 reactions commonly observed with alkyl halides. For example, the rapid SN2 reaction of 1-bromobutane with sodium cyanide, shown below, has no parallel when 1-butanol is treated with sodium cyanide. In fact, ethyl alcohol is often used as a solvent for alkyl halide substitution reactions such as this.
CH3CH2CH2CH2–Br + Na(+) CN(–) CH3CH2CH2CH2–CN + Na(+) Br(–)
CH3CH2CH2CH2–OH + Na(+) CN(–) No Reaction
The key factor here is the stability of the leaving anion (bromide vs. hydroxide). HBr is a much stronger acid than water (by more than 18 orders of magnitude), and this difference is reflected in reactions that generate their respective conjugate bases. The weaker base, bromide, is more stable, and its release in a substitution or elimination reaction is much more favorable than that of hydroxide ion, a stronger and less stable base.
A clear step toward improving the reactivity of alcohols in SN2 reactions would be to modify the –OH functional group in a way that improves its stability as a leaving anion. One such modification is to conduct the substitution reaction in strong acid, converting –OH to –OH2(+). Because the hydronium ion (H3O(+)) is a much stronger acid than water, its conjugate base (H2O) is a better leaving group than hydroxide ion. The only problem with this strategy is that many nucleophiles, including cyanide, are deactivated by protonation in strong acid, effectively removing the nucleophilic co-reactant needed for the substitution. The strong acids HCl, HBr and HI are not subject to this difficulty because their conjugate bases are good nucleophiles and are even weaker bases than alcohols. The following equations illustrate some substitution reactions of alcohols that may be effected by these acids. As with alkyl halides, the nucleophilic substitution of 1º-alcohols proceeds by an SN2 mechanism, whereas 3º-alcohols react by an SN1 mechanism. Reactions of 2º-alcohols may occur by both mechanisms and often produce some rearranged products. The numbers in parentheses next to the mineral acid formulas represent the weight percentage of a concentrated aqueous solution, the form in which these acids are normally used.
CH3CH2CH2CH2–OH + HBr (48%) CH3CH2CH2CH2–OH2(+) Br(–) CH3CH2CH2CH2Br + H2O SN2
(CH3)3C–OH + HCl (37%) (CH3)3C–OH2(+) Cl(–) (CH3)3C(+) Cl(–) + H2O (CH3)3C–Cl + H2O SN1
Although these reactions are sometimes referred to as "acid-catalyzed," this is not strictly correct. In the overall transformation a strong HX acid is converted to water, a very weak acid, so at least a stoichiometric quantity of HX is required for a complete conversion of alcohol to alkyl halide. The necessity of using equivalent quantities of very strong acids in this reaction limits its usefulness to simple alcohols of the type shown above. Alcohols with acid sensitive groups do not, of course, tolerate such treatment. Nevertheless, the idea of modifying the -OH functional group to improve its stability as a leaving anion can be pursued in other directions. The following diagram shows some modifications that have proven effective. In each case the hydroxyl group is converted to an ester of a strong acid. The first two examples show the sulfonate esters described earlier. The third and fourth examples show the formation of a phosphite ester (X represents remaining bromines or additional alcohol substituents) and a chlorosulfite ester respectively. All of these leaving groups (colored blue) have conjugate acids that are much stronger than water (by 13 to 16 powers of ten) so the leaving anion is correspondingly more stable than hydroxide ion. The mesylate and tosylate compounds are particularly useful in that they may be used in substitution reactions with a wide variety of nucleophiles. The intermediates produced in reactions of alcohols with phosphorus tribromide and thionyl chloride (last two examples) are seldom isolated, and these reactions continue on to alkyl bromide and chloride products.
Epoxides as a "Leaving Group"
Epoxides (oxiranes) are three-membered cyclic ethers that are easily prepared from alkenes by reaction with peracids. Because of the large angle strain in this small ring, epoxides undergo acid and base-catalyzed C–O bond cleavage more easily than do larger ring ethers. Among the following examples, the first is unexceptional except for the fact that it occurs under milder conditions and more rapidly than other ether cleavages. The second and third examples clearly show the exceptional reactivity of epoxides, since unstrained ethers present in the same reactant or as solvent do not react. The aqueous acid used to work up the third reaction, following the Grignard reagent cleavage of the ethylene oxide, simply neutralizes the magnesium salt of the alcohol product.
The carbons in an epoxide group are very reactive electrophiles, due in large part to the fact that substantial ring strain is relieved when the ring opens upon nucleophilic attack.
Examples
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.14%3A_Preparation_of_Alcohols_Ethers_and_Epoxides.txt |
One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo E1 or E2 mechanisms to lose water and form a double bond.
Introduction
The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
The required range of reaction temperature decreases with increasing substitution of the hydroxy-containing carbon:
• 1° alcohols: 170° - 180°C
• 2° alcohols: 100°– 140 °C
• 3° alcohols: 25°– 80°C
If the reaction is not sufficiently heated, the alcohols do not dehydrate to form alkenes, but react with one another to form ethers (e.g., the Williamson Ether Synthesis).
Alcohol as a Base
Alcohols are amphoteric; they can act both as acid or base. The lone pair of electrons on oxygen atom makes the –OH group weakly basic. Oxygen can donate two electrons to an electron-deficient proton. Thus, in the presence of a strong acid, R—OH acts as a base and protonates into the very acidic alkyloxonium ion +OH2 (The pKa value of a tertiary protonated alcohol can go as low as -3.8). This basic characteristic of alcohol is essential for its dehydration reaction with an acid to form alkenes.
Mechanism for the Dehydration of Alcohol into Alkene
Different types of alcohols may dehydrate through a slightly different mechanism pathway. However, the general idea behind each dehydration reaction is that the –OH group in the alcohol donates two electrons to H+ from the acid reagent, forming an alkyloxonium ion. This ion acts as a very good leaving group which leaves to form a carbocation. The deprotonated acid (the nucleophile) then attacks the hydrogen adjacent to the carbocation and form a double bond.
Primary alcohols undergo bimolecular elimination (E2 mechanism) while secondary and tertiary alcohols undergo unimolecular elimination (E1 mechanism). The relative reactivity of alcohols in dehydration reaction is ranked as the following
Methanol < primary < secondary < tertiary
Primary alcohol dehydrates through the E2 mechanism
Oxygen donates two electrons to a proton from sulfuric acid H2SO4, forming an alkyloxonium ion. Then the nucleophile HSO4 back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.
Secondary and tertiary alcohols dehydrate through the E1 mechanism
Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, in this case the ion leaves first and forms a carbocation as the reaction intermediate. The water molecule (which is a stronger base than the HSO4- ion) then abstracts a proton from an adjacent carbon, forming a double bond. Notice in the mechanism below that the aleke formed depends on which proton is abstracted: the red arrows show formation of the more substituted 2-butene, while the blue arrows show formation of the less substituted 1-butene. Recall the general rule that more substituted alkenes are more stable than less substituted alkenes, and trans alkenes are more stable than cis alkenes. Thereore, the trans diastereomer of the 2-butene product is most abundant.
Dehydration reaction of secondary alcohol
The dehydration mechanism for a tertiary alcohol is analogous to that shown above for a secondary alcohol.
When more than one alkene product are possible, the favored product is usually the thermodynamically most stable alkene. More-substituted alkenes are favored over less-substituted ones; and trans-substituted alkenes are preferred compared to cis-substituted ones.
1. Since the C=C bond is not free to rotate, cis-substituted alkenes are less stable than trans-subsituted alkenes because of steric hindrance (spatial interfererence) between two bulky substituents on the same side of the double bond (as seen in the cis product in the above figure). Trans-substituted alkenes reduce this effect of spatial interference by separating the two bulky substituents on each side of the double bond (for further explanation on the rigidity of C=C bond, see Structure and Bonding in Ethene- The pi Bond).
2. Heats of hydrogenation of differently-substituted alkene isomers are lowest for more-substituted alkenes, suggesting that they are more stable than less-substituted alkenes and thus are the major products in an elimination reaction. This is partly because in more --substituted alkenes, the p orbitals of the pi bond are stabilized by neighboring alkyl substituents, a phenomenon similar to hyperconjugation (see also Catalytic Hydrogenation of Alkenes: Relative Stabilities of Double Bond).
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.16%3A_Dehydration_of_Alcohols_to_Alkenes.txt |
Hydride and Alkyl Shifts
Since the dehydration reaction of alcohol has a carbocation intermediate, hydride or alkyl shifts can occur which relocates the carbocation to a more stable position. The dehydrated products therefore are a mixture of alkenes, with and without carbocation rearrangement. Tertiary cation is more stable than secondary cation, which in turn is more stable than primary cation due to a phenomenon known as hyperconjugation, where the interaction between the filled orbitals of neighboring carbons and the singly occupied p orbital in the carbocation stabilizes the positive charge in carbocation.
• In hydride shifts, a secondary or tertiary hydrogen from a carbon next to the original carbocation takes both of its electrons to the cation site, swapping place with the carbocation and renders it a more stable secondary or tertiary cation.
Similarly, when there is no hydride available for hydride shifting, an alkyl group can take its bonding electrons and swap place with an adjacent cation, a process known as alkyl shift.
Consider the major product of the addition of HCl to 3,3-dimethyl-1-butene:
This may not be the result that you would have predicted, based on what we have learned so far about electrophilic additions! Most likely, you would at first predict that the sole product of the reaction would be the first one shown above, 3-chloro-2,2-dimethylbutane. Why does the second product (2-chloro-2,3-dimethylbutane) form also?
The answer lies in the observation that formation of carbocations is sometimes accompanied by a structural rearrangement. Such rearrangements take place by a shift of a neighboring alkyl group or hydrogen, and are favored when the rearranged carbocation is more stable than the initial carbocation. As you can see in the mechanism below, this holds true in the case of the 3,3-dimethyl-1-butene addition, and explains why the 2-chloro product is actaully the major product: more of it forms than the minor 3-chloro product.
Protonation of the alkene leads to a secondary carbocation, but a methyl shift to a lower-energy tertiary carbocation occurs faster than nucleophilic attack by chlorine, so that when chlorine does attack it does so at carbon #3, rather than carbon #2.
In most examples of carbocation rearrangements that you are likely to encounter, the shifting species is a hydride or methyl group. However, pretty much any alkyl group is capable of shifting. Sometimes, the entire side of a ring will shift over in a ring-expanding rearrangement. Consider the following electrophilic addition of HBr to an alkene: right off the bat, it is very hard to see what is going on here.
Taking into account the possibility of rearrangement, however, we can see how the observed product forms.
The first 1,2-alkyl shift is driven by the expansion of a five-membered ring to a six-membered ring, which has slightly less ring strain. A hydride shift then converts a secondary carbocation to a tertiary carbocation, which is the electrophile ultimately attacked by the bromide nucleophile.
Another factor that may induce rearrangement of carbocation intermediates is strain. The addition of HCl to α-pinene, the major hydrocarbon component of turpentine, gives the rearranged product, bornyl chloride, in high yield. This rearrangement converts a 3º-carbocation to a 2º-carbocation, a transformation that is normally unfavorable. However, the rearrangement also expands a strained four-membered ring to a much less-strained five-membered ring, and this relief of strain provides a driving force for the rearrangement.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/09%3A_Alcohols_Ethers_and_Epoxides/9.17%3A_Carbocation_Rearrangements.txt |
This is an introductory page about alkenes such as ethene, propene and the rest. It deals with their formulae and isomerism, their physical properties, and an introduction to their chemical reactivity.
What are alkenes?
Alkenes are a family of hydrocarbons (compounds containing carbon and hydrogen only) containing a carbon-carbon double bond. The first two are:
ethene C2H4
propene C3H6
You can work out the formula of any of them using: CnH2n The table is limited to the first two, because after that there are isomers which affect the names.
Isomerism in the alkenes
Structural isomerism
All the alkenes with 4 or more carbon atoms in them show structural isomerism. This means that there are two or more different structural formulae that you can draw for each molecular formula.
For example, with C4H8, it isn't too difficult to come up with these three structural isomers:
There is, however, another isomer. But-2-ene also exhibits geometric isomerism.
Geometric (cis-trans) isomerism
The carbon-carbon double bond doesn't allow any rotation about it. That means that it is possible to have the CH3 groups on either end of the molecule locked either on one side of the molecule or opposite each other.
These are called cis-but-2-ene (where the groups are on the same side) or trans-but-2-ene (where they are on opposite sides).
Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below..
Chemical Reactivity
Bonding in the alkenes
We just need to look at ethene, because what is true of C=C in ethene will be equally true of C=C in more complicated alkenes. Ethene is often modeled like this:
The double bond between the carbon atoms is, of course, two pairs of shared electrons. What the diagram doesn't show is that the two pairs aren't the same as each other.
One of the pairs of electrons is held on the line between the two carbon nuclei as you would expect, but the other is held in a molecular orbital above and below the plane of the molecule. A molecular orbital is a region of space within the molecule where there is a high probability of finding a particular pair of electrons.
In this diagram, the line between the two carbon atoms represents a normal bond - the pair of shared electrons lies in a molecular orbital on the line between the two nuclei where you would expect them to be. This sort of bond is called a sigma bond.
The other pair of electrons is found somewhere in the shaded part above and below the plane of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move around anywhere in this shaded region and can move freely from one half to the other.
The pi electrons are not as fully under the control of the carbon nuclei as the electrons in the sigma bond and, because they lie exposed above and below the rest of the molecule, they are relatively open to attack by other things.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.01%3A_Introduction.txt |
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of $\pi$ bonds and/or cyclic rings.
Saturated vs. Unsaturated Molecules
In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated.
CH3CH2CH3 1-methyoxypentane
Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s).
CH3CH=CHCH3 3-chloro-5-octyne
Calculating Degrees of Unsaturation (DoU)
Degree of Unsaturation (DoU) is also known as Double Bond Equivalent. If the molecular formula is given, plug in the numbers into this formula:
$DoU= \dfrac{2C+2+N-X-H}{2}$
• $C$ is the number of carbons
• $N$ is the number of nitrogens
• $X$ is the number of halogens (F, Cl, Br, I)
• $H$ is the number of hydrogens
As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6.
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
DoU
Possible combinations of rings/ bonds
# of rings
# of double bonds
# of triple bonds
1
1
0
0
0
1
0
2
2
0
0
0
2
0
0
0
1
1
1
0
3 3 0 0
2 1 0
1 2 0
0 1 1
0 3 0
1 0 1
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example: Benzene
What is the Degree of Unsaturation for Benzene?
Solution
The molecular formula for benzene is C6H6. Thus,
DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DoB of 4 and have the same molecular formula as benzene.
Problems
1. Are the following molecules saturated or unsaturated:
1. (b.) (c.) (d.) C10H6N4
2. Using the molecules from 1., give the degrees of unsaturation for each.
3. Calculate the degrees of unsaturation for the following molecular formulas:
1. (a.) C9H20 (b.) C7H8 (c.) C5H7Cl (d.) C9H9NO4
4. Using the molecular formulas from 3, are the molecules unsaturated or saturated.
5. Using the molecular formulas from 3, if the molecules are unsaturated, how many rings/double bonds/triple bonds are predicted?
Answers
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3. Use the formula to solve
(a.) 0
(b.) 4
(c.) 2
(d.) 6
4.
(a.) saturated
(b.) unsaturated
(c.) unsaturated
(d.) unsaturated
5.
(a.) 0 (Remember-a saturated molecule only contains single bonds)
(b.) The molecule can contain any of these combinations (i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
Contributors
• Kim Quach (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.02%3A_Calculating_Degrees_of_Unsaturation.txt |
Alkenes contain carbon-carbon double bonds and are unsaturated hydrocarbons with the molecular formula is CnH2n. This is also the same molecular formula as cycloalkanes. Alkenes are named by dropping the -ane ending of the parent and adding -ene.
Introduction
The parent structure is the longest chain containing both carbon atoms of the double bond. The two carbon atoms of a double bond and the four atoms attached to them lie in a plane, with bond angles of approximately 120° A double bond consists of one sigma bond formed by overlap of sp2 hybrid orbitals and one pi bond formed by overlap of parallel 2 p orbitals
The Basic Rules
For straight chain alkenes, it is the same basic rules as nomenclature of alkanes except change the suffix to "-ene."
i. Find the Longest Carbon Chain that Contains the Carbon Carbon double bond. If you have two ties for longest Carbon chain, and both chains contain a Carbon Carbon double bond, then identify the most substituted chain.
ii. Give the lowest possible number to the Carbon Carbon double bond.
1. Do not need to number cycloalkenes because it is understood that the double bond is in the one position.
2. Alkenes that have the same molecular formula but the location of the doble bonds are different means they are constitutional isomers.
3. Functional Groups with higher priority:
iii. Add substituents and their position to the alkene as prefixes. Of course remember to give the lowest numbers possible. And remember to name them in alphabetical order when writting them.
iv. Next is identifying stereoisomers. when there are only two non hydrogen attachments to the alkene then use cis and trans to name the molecule.
In this diagram this is a cis conformation. It has both the substituents going upward. This molecule would be called (cis) 5-chloro-3-heptene.)
Trans would look like this
v. On the other hand if there are 3 or 4 non-hydrogen different atoms attached to the alkene then use the E, Z system.
E (entgegen) means the higher priority groups are opposite one another relative to the double bond.
Z (zusammen) means the higher priority groups are on the same side relative to the double bond.
(You could think of Z as Zame Zide to help memorize it.)
In this example it is E-4-chloro-3-heptene. It is E because the Chlorine and the CH2CH3 are the two higher priorities and they are on opposite sides.
vi. A hydroxyl group gets precedence over th double bond. Therefore alkenes containing alchol groups are called alkenols. And the prefix becomes --enol. And this means that now the alcohol gets lowest priority over the alkene.
vii. Lastly remember that alkene substituents are called alkenyl. Suffix --enyl.
Here is a chart containing the systemic name for the first twenty straight chain alkenes.
Name Molecular formula
Ethene C2H4
Propene C3H6
Butene C4H8
Pentene C5H10
Hexene C6H12
Heptene C7H14
Octene C8H16
Nonene C9H18
Decene C10H20
Undecene C11H22
Dodecene C12H24
Tridecene C13H26
Tetradecene C14H28
Pentadecene C15H30
Hexadecene C16H32
Heptadecene C17H34
Octadecene C18H36
Nonadecene C19H38
Eicosene C20H40
Did you notice how there is no methene? Because it is impossible for a Carbon to have a double bond with nothing.
Geometric Isomers
Double bonds can exist as geometric isomers and these isomers are designated by using either the cis / trans designation or the modern E / Z designation.
cis Isomers
.The two largest groups are on the same side of the double bond.
trans Isomers
...The two largest groups are on opposite sides of the double bond.
E/Z nomenclature
E = entgegan ("trans") Z = zusamen ("cis")
Priority of groups is based on the atomic mass of attached atoms (not the size of the group). An atom attached by a multiple bond is counted once for each bond.
fluorine atom > isopropyl group > n-hexyl group
deuterium atom > hydrogen atom
-CH2-CH=CH2 > -CH2CH2CH3
Example 1
Try to name the following compounds using both conventions...
Common names
Remove the -ane suffix and add -ylene. There are a couple of unique ones like ethenyl's common name is vinyl and 2-propenyl's common name is allyl. That you should know are...
• vinyl substituent H2C=CH-
• allyl substituent H2C=CH-CH2-
• allene molecule H2C=C=CH2
• isoprene
Endocyclic Alkenes
Endocyclic double bonds have both carbons in the ring and exocyclic double bonds have only one carbon as part of the ring.
Cyclopentene is an example of an endocyclic double bond.
Methylenecylopentane is an example of an exocyclic double bond.
Name the following compounds...
1-methylcyclobutene. The methyl group places the double bond. It is correct to also name this compound as 1-methylcyclobut-1-ene.
1-ethenylcyclohexene, the methyl group places the double bond. It is correct to also name this compound as 1-ethenylcyclohex-1-ene. A common name would be 1-vinylcyclohexene.
Try to draw structures for the following compounds...
• 2-vinyl-1,3-cyclohexadiene
Examples
Both these compounds have double bonds, making them alkenes. In example (1) the longest chain consists of six carbons, so the root name of this compound will be hexene. Three methyl substituents (colored red) are present. Numbering the six-carbon chain begins at the end nearest the double bond (the left end), so the methyl groups are located on carbons 2 & 5. The IUPAC name is therefore: 2,5,5-trimethyl-2-hexene.
In example (2) the longest chain incorporating both carbon atoms of the double bond has a length of five. There is a seven-carbon chain, but it contains only one of the double bond carbon atoms. Consequently, the root name of this compound will be pentene. There is a propyl substituent on the inside double bond carbon atom (#2), so the IUPAC name is: 2-propyl-1-pentene.
The double bond in example (3) is located in the center of a six-carbon chain. The double bond would therefore have a locator number of 3 regardless of the end chosen to begin numbering. The right hand end is selected because it gives the lowest first-substituent number (2 for the methyl as compared with 3 for the ethyl if numbering were started from the left). The IUPAC name is assigned as shown.
Example (4) is a diene (two double bonds). Both double bonds must be contained in the longest chain, which is therefore five- rather than six-carbons in length. The second and fourth carbons of this 1,4-pentadiene are both substituted, so the numbering begins at the end nearest the alphabetically first-cited substituent (the ethyl group).
These examples include rings of carbon atoms as well as some carbon-carbon triple bonds. Example (6) is best named as an alkyne bearing a cyclobutyl substituent. Example (7) is simply a ten-membered ring containing both a double and a triple bond. The double bond is cited first in the IUPAC name, so numbering begins with those two carbons in the direction that gives the triple bond carbons the lowest locator numbers. Because of the linear geometry of a triple bond, a-ten membered ring is the smallest ring in which this functional group is easily accommodated. Example (8) is a cyclooctatriene (three double bonds in an eight-membered ring). The numbering must begin with one of the end carbons of the conjugated diene moiety (adjacent double bonds), because in this way the double bond carbon atoms are assigned the smallest possible locator numbers (1, 2, 3, 4, 6 & 7). Of the two ways in which this can be done, we choose the one that gives the vinyl substituent the lower number.
Problems
Try to name the following compounds...
1-pentene or pent-1-ene
2-ethyl-1-hexene or 2-ethylhex-1-ene
Try to draw structures for the following compounds...
• 2-pentene
CH3–CH=CH–CH2–CH3
• 3-heptene
CH3–CH2–CH=CH–CH2–CH2–CH3
b. Give the double bond the lowest possible numbers regardless of substituent placement.
• Try to name the following compound...
• Try to draw a structure for the following compound...
4-methyl-2-pentene J
Name the following structures:
v. Draw (Z)-5-Chloro-3-ethly-4-hexen-2-ol.
Answers
I. trans-8-ethyl-3-undecene
II. E-5-bromo-4-chloro-7,7-dimethyl-4-undecene
III. Z-1,2-difluoro-cyclohexene
IV. 4-ethenylcyclohexanol.
V.
Contributors
• S. Devarajan (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.03%3A_Nomenclature.txt |
Physical state
Ethene, Propene, and Butene exists as colorless gases. Members of the 5 or more carbons such as Pentene, Hexene, and Heptene are liquid, and members of the 15 carbons or more are solids.
Density
Alkenes are lighter than water, therefore, are insoluble in water. Alkenes are only soluble in nonpolar solvent.
Solubility
Alkenes are virtually insoluble in water, but dissolve in organic solvents. The reasons for this are exactly the same as for the alkanes.
Boiling Points
The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids.
Boiling points of alkenes depends on more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules.
Compound Boiling points (oC)
Ethene -104
Propene -47
Trans-2-Butene 0.9
Cis-2-butene 3.7
Trans 1,2-dichlorobutene 155
Cis 1,2-dichlorobutene 152
1-Pentene 30
Trans-2-Pentene 36
Cis-2-Pentene 37
1-Heptene 115
3-Octene 122
3-Nonene 147
5-Decene 170
In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.
Melting Points
Melting points of alkenes depends on the packaging of the molecules. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers.
Compound Melting Points (0C)
Ethene -169
Propene -185
Butene -138
1-Pentene -165
Trans-2-Pentene -135
Cis-2-Pentene -180
1-Heptene -119
3-Octene -101.9
3-Nonene -81.4
5-Decene -66.3
Polarity
Chemical structure and fuctional groups can affect the polarity of alkenes compounds. The sp2 carbon is much more electron-withdrawing than the sp3 hybridize orbitals, therefore, creates a weak dipole along the substituent weak alkenly carbon bond. The two individual dipoles together form a net molecular dipole. In trans-subsituted alkenes, the dipole cancel each other out. In cis-subsituted alkenes there is a net dipole, therefore contributing to higher boiling in cis-isomers than trans-isomers.
• Trung Nguyen
10.05: Interesting Alkenes
Ethene
Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking.
There is not any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:
Or, showing more clearly what happens to the various atoms and bonds:
This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. You will remember that during the polymeriation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene. The reaction is done at high pressures in the presence of a trace of oxygen as an initiator.
Beta-Carotene
The long chain of alternating double bonds (conjugated) is responsible for the orange color of beta-carotene. The conjugated chain in carotenoids means that they absorb in the visible region - green/blue part of the spectrum. So β-carotene appears orange, because the red/yellow colors are reflected back to us.
Vitamin A
Vitamin A has several functions in the body. The most well known is its role in vision - hence carrots "make you able to see in the dark". The retinol is oxidized to its aldehyde, retinal, which complexes with a molecule in the eye called opsin. When a photon of light hits the complex, the retinal changes from the 11-cis form to the all-trans form, initiating a chain of events which results in the transmission of an impulse up the optic nerve. A more detailed explanation is in Photochemical Events.
Other roles of vitamin A are much less well understood. It is known to be involved in the synthesis of certain glycoproteins, and that deficiency leads to abnormal bone development, disorders of the reproductive system, xerophthalmia (a drying condition of the cornea of the eye) and ultimately death.
Vitamin A is required for healthy skin and mucus membranes, and for night vision. Its absence from diet leads to a loss in weight and failure of growth in young animals, to the eye diseases; xerophthalmia, and night blindness, and to a general susceptibility to infections. It is thought to help prevent the development of cancer. Good sources of carotene, such as green vegetables are good potential sources of vitamin A. Vitamin A is also synthetically manufactured by extraction from fish-liver oil and by synthesis from beta-ionone. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.04%3A_Physical_Properties.txt |
Fatty acids are merely carboxylic acids with long hydrocarbon chains. The hydrocarbon chain length may vary from 10-30 carbons (most usual is 12-18). The non-polar hydrocarbon alkane chain is an important counter balance to the polar acid functional group. In acids with only a few carbons, the acid functional group dominates and gives the whole molecule a polar character. However, in fatty acids, the non-polar hydrocarbon chain gives the molecule a non-polar character.
Introduction
The most common fatty acids are listed. Note that there are two groups of fatty acids--saturated and unsaturated. Recall that the term unsaturated refers to the presence of one or more double bonds between carbons as in alkenes. A saturated fatty acid has all bonding positions between carbons occupied by hydrogens.
The melting points for the saturated fatty acids follow the boiling point principle observed previously. Melting point principle: as the molecular weight increases, the melting point increases. This observed in the series lauric (C12), palmitic (C16), stearic (C18). Room temperature is 25oC, Lauric acid which melts at 44o is still a solid, while arachidonic acid has long since melted at -50o, so it is a liquid at room temperature.
Table 1: Common Fatty Acids
Melting Points of Saturated vs. Unsaturated Fatty Acids
Note that as a group, the unsaturated fatty acids have lower melting points than the saturated fatty acids. The reason for this phenomenon can be found by a careful consideration of molecular geometries. The tetrahedral bond angles on carbon results in a molecular geometry for saturated fatty acids that is relatively linear although with zigzags.
Stearic acid
Oleic aci
This molecular structure allows many fatty acid molecules to be rather closely "stacked" together. As a result, close intermolecular interactions result in relatively high melting points.
On the other hand, the introduction of one or more double bonds in the hydrocarbon chain in unsaturated fatty acids results in one or more "bends" in the molecule. The geometry of the double bond is almost always a cis configuration in natural fatty acids. and these molecules do not "stack" very well. The intermolecular interactions are much weaker than saturated molecules. As a result, the melting points are much lower for unsaturated fatty acids.
Percent Fatty Acid Present in Triglycerides
Fat or Oil
Saturated
Unsaturated
Palmitic Stearic Oleic Linoleic Other
Animal Origin
Butter 29 9 27 4 31
Lard 30 18 41 6 5
Beef 32 25 38 3 2
Vegatable Origin
Corn oil 10 4 34 48 4
Soybean 7 3 25 56 9
Peanut 7 5 60 21 7
Olive 6 4 83 7 -
Saturated vs. Unsaturated Fatty Acids in Fats and Oils
The triesters of fatty acids with glycerol (1,2,3-trihydroxypropane) compose the class of lipids known as fats and oils. These triglycerides (or triacylglycerols) are found in both plants and animals, and compose one of the major food groups of our diet. Triglycerides that are solid or semisolid at room temperature are classified as fats, and occur predominantly in animals. Those triglycerides that are liquid are called oils and originate chiefly in plants, although triglycerides from fish are also largely oils. Some examples of the composition of triglycerides from various sources are given in the following table.
Saturated Acids (%)
Unsaturated Acids (%)
Source
C10
& less
C12
lauric
C14
myristic
C16
palmitic
C18
stearic
C18
oleic
C18
linoleic
C18
unsaturated
Animal Fats
butter
15
2
11
30
9
27
4
1
lard
-
-
1
27
15
48
6
2
human fat
-
1
3
25
8
46
10
3
herring oil
-
-
7
12
1
2
20
52
Plant Oils
coconut
-
50
18
8
2
6
1
-
corn
-
-
1
10
3
50
34
-
olive
-
-
-
7
2
85
5
-
palm
-
-
2
41
5
43
7
-
peanut
-
-
-
8
3
56
26
7
safflower
-
-
-
3
3
19
76
-
As might be expected from the properties of the fatty acids, fats have a predominance of saturated fatty acids, and oils are composed largely of unsaturated acids. Thus, the melting points of triglycerides reflect their composition, as shown by the following examples. Natural mixed triglycerides have somewhat lower melting points, the melting point of lard being near 30 º C, whereas olive oil melts near -6 º C. Since fats are valued over oils by some Northern European and North American populations, vegetable oils are extensively converted to solid triglycerides (e.g. Crisco) by partial hydrogenation of their unsaturated components. Some of the remaining double bonds are isomerized (to trans) in this operation. These saturated and trans-fatty acid glycerides in the diet have been linked to long-term health issues such as atherosclerosis.
Triglycerides having three identical acyl chains, such as tristearin and triolein (above), are called "simple", while those composed of different acyl chains are called "mixed". If the acyl chains at the end hydroxyl groups (1 & 3) of glycerol are different, the center carbon becomes a chiral center and enantiomeric configurations must be recognized.
The hydrogenation of vegetable oils to produce semisolid products has had unintended consequences. Although the hydrogenation imparts desirable features such as spreadability, texture, "mouth feel," and increased shelf life to naturally liquid vegetable oils, it introduces some serious health problems. These occur when the cis-double bonds in the fatty acid chains are not completely saturated in the hydrogenation process. The catalysts used to effect the addition of hydrogen isomerize the remaining double bonds to their trans configuration. These unnatural trans-fats appear to to be associated with increased heart disease, cancer, diabetes and obesity, as well as immune response and reproductive problems. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.06%3A_LipidsPart_2.txt |
E2 Reaction
E2 reactions are typically seen with secondary and tertiary alkyl halides, but a hindered base is necessary with a primary halide. The mechanism by which it occurs is a single step concerted reaction with one transition state. The rate at which this mechanism occurs is second order kinetics, and depends on both the base and alkyl halide. A good leaving group is required because it is involved in the rate determining step. The leaving groups must be coplanar in order to form a pi bond; carbons go from sp3 to sp2 hybridization states.
To get a clearer picture of the interplay of these factors involved in a a reaction between a nucleophile/base and an alkyl halide, consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In one pathway, a methanethiolate nucleophile substitutes for bromine in an SN2 reaction. In the other (bottom) pathway, methoxide ion acts as a base (rather than as a nucleophile) in an elimination reaction. As we will soon see, the mechanism of this reaction is single-step, and is referred to as the E2 mechanism.
.
E1 Reaction
An E1 reaction involves the deprotonation of a hydrogen nearby (usually one carbon away, or the beta position) the carbocation resulting in the formation of an alkene product. In order to accomplish this, a Lewis base is required. For a simplified model, we’ll take B to be a Lewis base, and LG to be a halogen leaving group.
As can be seen above, the preliminary step is the leaving group (LG) leaving on its own. Because it takes the electrons in the bond along with it, the carbon that was attached to it loses its electron, making it a carbocation. Once it becomes a carbocation, a Lewis Base (B) deprotonates the intermediate carbocation at the beta position, which then donates its electrons to the neighboring C-C bond, forming a double bond. Unlike E2 reactions, which require the proton to be anti to the leaving group, E1 reactions only require a neighboring hydrogen. This is due to the fact that the leaving group has already left the molecule. The final product is an alkene along with the HB byproduct.
Dehydration
One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo E1 or E2 mechanisms to lose water and form a double bond.
The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
Primary alcohol dehydrates through the E2 mechanism
Oxygen donates two electrons to a proton from sulfuric acid H2SO4, forming an alkyloxonium ion. Then the nucleophile HSO4 back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.
Secondary and tertiary alcohols dehydrate through the E1 mechanism
Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, in this case the ion leaves first and forms a carbocation as the reaction intermediate. The water molecule (which is a stronger base than the HSO4- ion) then abstracts a proton from an adjacent carbon, forming a double bond. Notice in the mechanism below that the aleke formed depends on which proton is abstracted: the red arrows show formation of the more substituted 2-butene, while the blue arrows show formation of the less substituted 1-butene. Recall the general rule that more substituted alkenes are more stable than less substituted alkenes, and trans alkenes are more stable than cis alkenes. Thereore, the trans diastereomer of the 2-butene product is most abundant.
Zaitsev's Rule
Zaitsev’s or Saytzev’s (anglicized spelling) rule is an empirical rule used to predict regioselectivity of 1,2-elimination reactions occurring via the E1 or E2 mechanisms. It states that in a regioselective E1 or E2 reaction the major product is the more stable alkene, (i.e., the alkene with the more highly substituted double bond). For example:
If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
10.08: Introduction to Addition Reactions
Alkenes are found throughout nature. They form the basis of many natural products, such as terpenes, which play a variety of roles in the lives of plants and insects. The C=C bonds of alkenes are very different from the C=O bonds that are also common in nature. The C=C bonds of alkenes are electron-rich and nucleophilic, in contrast to the electron-poor C=O bonds of carbohydrates, fatty acids and proteins. That difference plays a role in how terpenes form in nature.
Alkenes, or olefins, are also a major product of the petroleum industry. Reactions of alkenes form the basis for a significant porion of our manufacturing economy. Commonly used plastics such as polyethylene, polypropylene and polystyrene are all formed through the reactions of alkenes. These materials continue to find use in our society because of their valuable properties, such as high strength, flexibility and low weight.
Alkenes undergo addition reactions like carbonyls do. Often, they add a proton to one end of the double bond and another group to the other end. These reactions happen in slightly different ways, however.
Alkenes are reactive because they have a high-lying pair of π-bonding electrons. These electrons are loosely held, being high in energy compared to σ-bonds. The fact that they are not located between the carbon nuclei, but are found above and below the plane of the double bond, also makes these electrons more accessible.
Alkenes can donate their electrons to strong electrophiles other than protons, too. Sometimes their reactivity pattern is a little different than the simple addition across the double bond, but that straightforward pattern is what we will focus on in this chapter. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.07%3A_Preparation_of_Alkenes.txt |
his page looks at the reaction of the carbon-carbon double bond in alkenes such as ethene with hydrogen halides such as hydrogen chloride and hydrogen bromide.
Symmetrical alkenes (like ethene or but-2-ene) are dealt with first. These are alkenes where identical groups are attached to each end of the carbon-carbon double bond.
Addition to symmetrical alkenes
What happens?
All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other.
For example, with ethene and hydrogen chloride, you get chloroethane:
With but-2-ene you get 2-chlorobutane:
What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product.
The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space.
That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.
Mechanism
The addition of hydrogen halides is one of the easiest electrophilic addition reactions because it uses the simplest electrophile: the proton. Hydrogen halides provide both a electrophile (proton) and a nucleophile (halide). First, the electrophile will attack the double bond and take up a set of π electrons, attaching it to the molecule (1). This is basically the reverse of the last step in the E1 reaction (deprotonation step). The resulting molecule will have a single carbon- carbon bond with a positive charge on one of them (carbocation). The next step is when the nucleophile (halide) bonds to the carbocation, producing a new molecule with both the original hydrogen and halide attached to the organic reactant (2). The second step will only occur if a good nucleophile is used.
Mechanism of Electrophilic Addition of Hydrogen Halide to Ethene
Mechanism of Electrophilic Addition of Hydrogen Halide to Propene
All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals)s.
Reaction rates
Variation of rates when you change the halogen
Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions.
When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.
Variation of rates when you change the alkene
This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be.
Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond.
For example:
There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.
Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this.
Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes.
The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.
The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:
The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.
Addition to unsymmetrical alkenes
What happens?
In terms of reaction conditions and the factors affecting the rates of the reaction, there is no difference whatsoever between these alkenes and the symmetrical ones described above. The problem comes with the orientation of the addition - in other words, which way around the hydrogen and the halogen add across the double bond.
Orientation of addition
If HCl adds to an unsymmetrical alkene like propene, there are two possible ways it could add. However, in practice, there is only one major product.
This is in line with Markovnikov's Rule which says:
When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already.
In this case, the hydrogen becomes attached to the CH2 group, because the CH2 group has more hydrogens than the CH group.
Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH3 group are totally irrelevant. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.09%3A_HydrohalogenationElectrophilic_Addition_of_HX.txt |
very important regarding electrophilic addition reactions is that if the starting alkene is asymmetrical, there are two possible courses that could be followed, depending on which of the two alkene carbons forms the new sigma bond in the first step.
Of course, the two reaction courses involve two different carbocation intermediates, which may have different energy levels. Two different products are possible, and in general the product which predominates will be the one that is derived from the lower-energy carbocation intermediate.
This important regiochemical principle is nicely illustrated by a simple electrophilic addition that is commonly carried out in the organic laboratory: the conversion of an alkene to an alkyl bromide by electrophilic addition of HBr to the double bond. Let's look at a hypothetical addition of HBr to 2-methyl-2-butene, pictured below. Two different regiochemical outcomes are possible:
The initial protonation step could follow two different pathways, resulting in two different carbocation intermediates: pathway 'a' gives a tertiary carbocation intermediate (Ia), while pathway 'b' gives a secondary carbocation intermediate (Ib) We know already that the tertiary carbocation is more stable (in other words, lower in energy). According to the Hammond postulate, this implies that the activation energy for pathway a is lower than in pathway b, meaning in turn that Ia forms faster.
Because the protonation step is the rate determining step for the reaction, the tertiary alkyl bromide A will form much faster than the secondary alkyl halide B, and thus A will be the predominant product observed in this reaction. This is a good example of a non-enzymatic organic reaction that is highly regiospecific.
In the example above, the difference in carbocation stability can be accounted for by the electron-donating effects of the extra methyl group on one side of the double bond. It is generally observed that, in electrophilic addition of acids (including water) to asymmetrical alkenes, the more substituted carbon is the one that ends up bonded to the heteroatom of the acid, while the less substituted carbon is protonated.
This rule of thumb is known as Markovnikov's rule, after the Russian chemist Vladimir Markovnikov who proposed it in 1869.
While it is useful in many cases, Markovikov's rule does not apply to all possible electrophilic additions. It is more accurate to use the more general principle that has already been stated above:
When an asymmetrical alkene undergoes electrophilic addition, the product that predominates is the one that results from the more stable of the two possible carbocation intermediates.
How is this different from Markovnikov's original rule? Consider the following hypothetical reaction, which is similar to the HBr addition shown above except that the six methyl hydrogens on the left side of the double bond have been replaced by highly electron-withdrawing fluorines.
Now when HBr is added, it is the less substituted carbocation that forms faster in the rate-determining protonation step, because in this intermediate the carbon bearing the positive charge is located further away from the electron-withdrawing, cation-destabilizing fluorines. As a result, the predominant product is the secondary rather than the tertiary bromoalkane. This would be referred to as an 'anti-Markovnikov' addition product, because it 'breaks' Markovnikov's rule.
Example
Predict the product of the following reaction:
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
10.11: Stereochemistry of Electrophilic Addition of HX
As illustrated in the drawing below, the pi-bond fixes the carbon-carbon double bond in a planar configuration, and does not permit free rotation about the double bond itself. We see then that addition reactions to this function might occur in three different ways, depending on the relative orientation of the atoms or groups that add to the carbons of the double bond: (i) they may bond from the same side, (ii) they may bond from opposite sides, or (iii) they may bond randomly from both sides. The first two possibilities are examples of stereoselectivity, the first being termed syn-addition, and the second anti-addition. Since initial electrophilic attack on the double bond may occur equally well from either side, it is in the second step (or stage) of the reaction (bonding of the nucleophile) that stereoselectivity may be imposed.
If the two-step mechanism described above is correct, and if the carbocation intermediate is sufficiently long-lived to freely-rotate about the sigma-bond component of the original double bond, we would expect to find random or non-stereoselective addition in the products. On the other hand, if the intermediate is short-lived and factors such as steric hindrance or neighboring group interactions favor one side in the second step, then stereoselectivity in product formation is likely. The following table summarizes the results obtained from many studies, the formula HX refers to all the strong Brønsted acids. The interesting differences in stereoselectivity noted here provide further insight into the mechanisms of these addition reactions.
Reagent H–X X2 HO–X RS–Cl Hg(OAc)2 BH3
Stereoselectivity mixed anti anti anti anti syn | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.10%3A_Markovnikovs_Rule.txt |
Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a carbocation is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane.
What Is Electrophilic Hydration?
Electrophilic hydration is the reverse dehydration of alcohols and has practical application in making alcohols for fuels and reagents for other reactions. The basic reaction under certain temperatures (given below) is the following:
The phrase "electrophilic" literally means "electron loving" (whereas "nucleophilic" means "nucleus loving"). Electrophilic hydrogen is essentially a proton: a hydrogen atom stripped of its electrons. Electrophilic hydrogen is commonly used to help break double bonds or restore catalysts (see SN2 for more details).
How Does Electrophilic Hydration Work?
Temperatures for Types of Alcohol Synthesis
Heat is used to catalyze electrophilic hydration; because the reaction is in equilibrium with the dehydration of an alcohol, which requires higher temperatures to form an alkene, lower temperatures are required to form an alcohol. The exact temperatures used are highly variable and depend on the product being formed.
• Primary Alcohol: Less than 170ºC
• Secondary Alcohol: Less than 100ºC
• Tertiary Alcohol: Less than 25ºC
But...Why Does Electrophilic Hydration Work?
• An alkene placed in an aqueous non-nucleophilic strong acid immediately "reaches out" with its double bond and attacks one of the acid's hydrogen atoms (meanwhile, the bond between oxygen and hydrogen performs heterolytic cleavage toward the oxygen—in other words, both electrons from the oxygen/hydrogen single bond move onto the oxygen atom).
• A carbocation is formed on the original alkene (now alkane) in the more-substituted position, where the oxygen end of water attacks with its 4 non-bonded valence electrons (oxygen has 6 total valence electrons because it is found in Group 6 on the periodic table and the second row down: two electrons in a 2s-orbital and four in 2p-orbitals. Oxygen donates one valence electron to each bond it forms, leaving four 4 non-bonded valence electrons).
• After the blue oxygen atom forms its third bond with the more-substituted carbon, it develops a positive charge (3 bonds and 2 valence electrons give the blue oxygen atom a formal charge of +1).
• The bond between the green hydrogen and the blue oxygen undergoes heterolytic cleavage, and both the electrons from the bond move onto the blue oxygen. The now negatively-charged strong acid picks up the green electrophilic hydrogen.
• Now that the reaction is complete, the non-nucleophilic strong acid is regenerated as a catalyst and an alcohol forms on the most substituted carbon of the current alkane. At lower temperatures, more alcohol product can be formed.
What is Regiochemistry and How Does It Apply?
Regiochemistry deals with where the substituent bonds on the product. Zaitsev's and Markovnikov's rules address regiochemistry, but Zaitsev's rule applies when synthesizing an alkene while Markovnikov's rule describes where the substituent bonds onto the product. In the case of electrophilic hydration, Markovnikov's rule is the only rule that directly applies. See the following for an in-depth explanation of regiochemistry Markovnikov explanation: Radical Additions--Anti-Markovnikov Product Formation
In the mechanism for a 3º alcohol shown above, the red H is added to the least-substituted carbon connected to the nucleophilic double bonds (it has less carbons attached to it). This means that the carbocation forms on the 3º carbon, causing it to be highly stabilized by hyperconjugationelectrons in nearby sigma (single) bonds help fill the empty p-orbital of the carbocation, which lessens the positive charge. More substitution on a carbon means more sigma bonds are available to "help out" (by using overlap) with the positive charge, which creates greater carbocation stability. In other words, carbocations form on the most substituted carbon connected to the double bond. Carbocations are also stabilized by resonance, but resonance is not a large factor in this case because any carbon-carbon double bonds are used to initiate the reaction, and other double bonded molecules can cause a completely different reaction.
If the carbocation does originally form on the less substituted part of the alkene, carbocation rearrangements occur to form more substituted products:
• Hydride shifts: a hydrogen atom bonded to a carbon atom next to the carbocation leaves that carbon to bond with the carbocation (after the hydrogen has taken both electrons from the single bond, it is known as a hydride). This changes the once neighboring carbon to a carbocation, and the former carbocation becomes a neighboring carbon atom.
• Alkyl shifts: if no hydrogen atoms are available for a hydride shift, an entire methyl group performs the same shift.
The nucleophile attacks the positive charge formed on the most substituted carbon connected to the double bond, because the nucleophile is seeking that positive charge. In the mechanism for a 3º alcohol shown above, water is the nucleophile. When the green H is removed from the water molecule, the alcohol attached to the most substituted carbon. Hence, electrophilic hydration follows Markovnikov's rule.
What is Stereochemistry and How Does It Apply?
Stereochemistry deals with how the substituent bonds on the product directionally. Dashes and wedges denote stereochemistry by showing whether the molecule or atom is going into or out of the plane of the board. Whenever the bond is a simple single straight line, the molecule that is bonded is equally likely to be found going into the plane of the board as it is out of the plane of the board. This indicates that the product is a racemic mixture.
Electrophilic hydration adopts a stereochemistry wherein the substituent is equally likely to bond pointing into the plane of the board as it is pointing out of the plane of the board. The 3º alcohol product could look like either of the following products:
Note: Whenever a straight line is used along with dashes and wedges on the same molecule, it could be denoting that the straight line bond is in the same plane as the board. Practice with a molecular model kit and attempting the practice problems at the end can help eliminate any ambiguity.
Is this a Reversible Synthesis?
Electrophilic hydration is reversible because an alkene in water is in equilibrium with the alcohol product. To sway the equilibrium one way or another, the temperature or the concentration of the non-nucleophilic strong acid can be changed. For example:
• Less sulfuric or phosphoric acid and an excess of water help synthesize more alcohol product.
• Lower temperatures help synthesize more alcohol product.
Is There a Better Way to Add Water to Synthesize an Alcohol From an Alkene?
A more efficient pathway does exist: see Oxymercuration - Demercuration: A Special Electrophilic Addition. Oxymercuration does not allow for rearrangements, but it does require the use of mercury, which is highly toxic. Detractions for using electrophilic hydration to make alcohols include:
• Allowing for carbocation rearrangements
• Poor yields due to the reactants and products being in equilibrium
• Allowing for product mixtures (such as an (R)-enantiomer and an (S)-enantiomer)
• Using sulfuric or phosphoric acid
Problems
Predict the product of each reaction.
1)
2) How does the cyclopropane group affect the reaction?
3) (Hint: What is different about this problem?)
4) (Hint: Consider stereochemistry.)
5) Indicate any shifts as well as the major product:
Answers to Practice Problems
1) This is a basic electrophilic hydration.
2) The answer is additional side products, but the major product formed is still the same (the product shown). Depending on the temperatures used, the cyclopropane may open up into a straight chain, which makes it unlikely that the major product will form (after the reaction, it is unlikely that the 3º carbon will remain as such).
3) A hydride shift actually occurs from the top of the 1-methylcyclopentane to where the carbocation had formed.
4) This reaction will have poor yields due to a very unstable intermediate. For a brief moment, carbocations can form on the two center carbons, which are more stable than the outer two carbons. The carbocations have an sp2 hybridization, and when the water is added on, the carbons change their hybridization to sp3. This makes the methyl and alcohol groups equally likely to be found going into or out of the plane of the paper- the product is racemic.
5) In the first picture shown below, an alkyl shift occurs but a hydride shift (which occurs faster) is possible. Why doesn't a hydride shift occur? The answer is because the alkyl shift leads to a more stable product. There is a noticeable amount of side product that forms where the two methyl groups are, but the major product shown below is still the most significant due to the hyperconjugation that occurs by being in between the two cyclohexanes.
Contributors
• Lance Peery (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.12%3A_HydrationElectrophilic_Addition_of_Water.txt |
Halogens can act as electrophiles to attack a double bond in alkene. Double bond represents a region of electron density and therefore functions as a nucleophile. How is it possible for a halogen to obtain positive charge to be an electrophile?
Introduction
As halogen molecule, for example Br2, approaches a double bond of the alkene, electrons in the double bond repel electrons in bromine molecule causing polarization of the halogen bond. This creates a dipolar moment in the halogen molecule bond. Heterolytic bond cleavage occurs and one of the halogens obtains positive charge and reacts as an electrophile. The reaction of the addition is not regioselective but stereoselective.Stereochemistry of this addition can be explained by the mechanism of the reaction.In the first step electrophilic halogen with a positive charge approaches the double carbon bond and 2 p orbitals of the halogen, bond with two carbon atoms and create a cyclic ion with a halogen as the intermediate step. In the second step, halogen with the negative charge attacks any of the two carbons in the cyclic ion from the back side of the cycle as in the SN2 reaction. Therefore stereochemistry of the product is vicinial dihalides through anti addition.
$\ce{R_2C=CR_2 + X_2 \rightarrow R_2CX-CR_2X}$
Halogens that are commonly used in this type of the reaction are: $Br$ and $Cl$. In thermodynamical terms $I$ is too slow for this reaction because of the size of its atom, and $F$ is too vigorous and explosive.
Solvents that are used for this type of electrophilic halogenation are inert (e.g., CCl4) can be used in this reaction.
Because halogen with negative charge can attack any carbon from the opposite side of the cycle it creates a mixture of steric products.Optically inactive starting material produce optically inactive achiral products (meso) or a racemic mixture.
Electrophilic addition mechanism consists of two steps.
Before constructing the mechanism let us summarize conditions for this reaction. We will use Br2 in our example for halogenation of ethylene.
Nucleophile Double bond in alkene
Electrophile Br2, Cl2
Regiochemistry not relevant
Stereochemistry
ANTI
Step 1: In the first step of the addition the Br-Br bond polarizes, heterolytic cleavage occurs and Br with the positive charge forms a intermediate cycle with the double bond.
Step 2: In the second step, bromide anion attacks any carbon of the bridged bromonium ion from the back side of the cycle. Cycle opens up and two halogens are in the position anti.
Summary
Hallogens can act as electrophiles due to polarizability of their covalent bond.Addition of halogens is stereospecific and produces vicinial dihalides with anti addition.Cis starting material will give mixture of enantiomers and trans produces a meso compound.
Problems
1.What is the mechanism of adding Cl2 to the cyclohexene?
2.A reaction of Br2 molecule in an inert solvent with alkene follows?
a) syn addition
b) anti addition
c) Morkovnikov rule
3)
4)
Key:
1.
2. b
3.enantiomer
4.
10.14: Stereochemistry of Halogenation
The halogens chlorine and bromine add rapidly to a wide variety of alkenes without inducing the kinds of structural rearrangements (carbocation shifts) noted for strong acids - this is because a discreet carbocation intermediate does not form in these reactions. The stereoselectivity of halogen additions is strongly anti, as shown in many of the following examples.
We can account both for the high stereoselectivity and the lack of rearrangement in these reactions by proposing a stabilizing interaction between the developing carbocation center and the electron rich halogen atom on the adjacent carbon. This interaction delocalizes the positive charge on the intermediate and blocks halide ion attack from the syn-location.
The stabilization provided by the halogen-carbocation bonding makes rearrangement unlikely, and in a few cases three-membered cyclic halonium cations have been isolated and identified as true intermediates. A resonance description of such a bromonium ion intermediate is shown below. The positive charge is delocalized over all the atoms of the ring, but should be concentrated at the more substituted carbon (where positive charge is more stable), and this is the site to which the nucleophile will bond.
10.15: Halohydrin Formation
The proton is not the only electrophilic species that initiates addition reactions to the double bond of alkenes. Lewis acids like the halogens, boron hydrides and certain transition metal ions are able to bond to the alkene pi-electrons, and the resulting complexes rearrange or are attacked by nucleophiles to give addition products.The electrophilic character of the halogens is well known. Chlorine (Cl2) and bromine(Br2) react selectively with the double bond of alkenes, and these reactions are what we will focus on. Fluorine adds uncontrollably with alkenes,and the addition of iodine is unfavorable, so these are not useful preparative methods.
The addition of chlorine and bromine to alkenes, as shown below, proceeds by an initial electrophilic attack on the pi-electrons of the double bond. Dihalo-compounds in which the halogens are bound to adjacent carbons are called vicinal, from the Latin vicinalis, meaning neighboring.
R2C=CR2 + X2 ——> R2CX-CR2X
Other halogen-containing reagents which add to double bonds include hypohalous acids, HOX, and sulfenyl chlorides, RSCl. These reagents are unsymmetrical, so their addition to unsymmetrical double bonds may in principle take place in two ways. In practice, these addition reactions are regioselective, with one of the two possible constitutionally isomeric products being favored. The electrophilic moiety in both of these reagents is the halogen.
(CH3)2C=CH2 + HOBr ——> (CH3)2COH-CH2Br
(CH3)2C=CH2 + C6H5SCl ——> (CH3)2CCl-CH2SC6H5
The regioselectivity of the above reactions may be explained by the same mechanism we used to rationalize the Markovnikov rule. Thus, bonding of an electrophilic species to the double bond of an alkene should result in preferential formation of the more stable (more highly substituted) carbocation, and this intermediate should then combine rapidly with a nucleophilic species to produce the addition product.
To apply this mechanism we need to determine the electrophilic moiety in each of the reagents. By using electronegativity differences we can dissect common addition reagents into electrophilic and nucleophilic moieties, as shown on the right. In the case of hypochlorous and hypobromous acids (HOX), these weak Brønsted acids (pKa's ca. 8) do not react as proton donors; and since oxygen is more electronegative than chlorine or bromine, the electrophile will be a halide cation. The nucleophilic species that bonds to the intermediate carbocation is then hydroxide ion, or more likely water (the usual solvent for these reagents), and the products are called halohydrins. Sulfenyl chlorides add in the opposite manner because the electrophile is a sulfur cation, RS(+), whereas the nucleophilic moiety is chloride anion (chlorine is more electronegative than sulfur).
Below are some examples illustrating the addition of various electrophilic halogen reagents to alkene groups. Notice the specific regiochemistry of the products, as explained above. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.13%3A_HalogenationAddition_of_Halogen.txt |
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from BH3 or BHR2) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene bouble bond. Furthermore, the borane acts as a lewis acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The Hydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific (syn addition), meaning that the hydroboration takes place on the same face of the double bond, this leads cis stereochemistry.
Introduction
Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes. An Additional feature of this reaction is that it occurs without rearrangement.
The Borane Complex
First off it is very imporatnt to understand little bit about the structure and the properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula B2H6 (diborane). Additionally, the dimer B2H6ignites spontaneously in air. Borane is commercially available in ether and tetrahydrofuran (THF), in these solutions the borane can exist as a lewis acid-base complex, which allows boron to have an electron octet.BH3B2H6
The Mechanism
Step #1
• Part #1: Hydroboration of the alkene. In this first step the addittion of the borane to the alkene is initiated and prceeds as a concerted reaction because bond breaking and bond formation occurs at the same time. This part consists of the vacant 2p orbital of the boron electrophile pairing with the electron pair of the ? bondof the nucleophile.
Transition state
* Note that a carbocation is not formed. Therefore, no rearrangement takes place.
• Part #2: The Anti Markovnikov addition of Boron. The boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon. Both, the boron and the hydrogen add simultaneously on the same face of the double bond (syn addition).
Oxidation of the Trialkylborane by Hydrogen Peroxide
Step #2
• Part #1: the first part of this mechanism deals with the donation of a pair of electrons from the hydrogen peroxide ion. the hydrogen peroxide is the nucleophile in this reaction because it is the electron donor to the newly formed trialkylborane that resulted from hydroboration.
• Part 2: In this second part of the mechanism, a rearrangement of an R group with its pair of bonding electrons to an adjacent oxygen results in the removal of a hydroxide ion.
Two more of these reactions with hydroperoxide will occur in order give a trialkylborate
• Part 3: This is the final part of the Oxidation process. In this part the trialkylborate reacts with aqueous NaOH to give the alcohol and sodium borate.
If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided here that will take you to other pages and media that are very helpful as well.
Stereochemistry of hydroboration
The hydroboration reaction is among the few simple addition reactions that proceed cleanly in a syn fashion. As noted above, this is a single-step reaction. Since the bonding of the double bond carbons to boron and hydrogen is concerted, it follows that the geometry of this addition must be syn. Furthermore, rearrangements are unlikely inasmuch as a discrete carbocation intermediate is never formed. These features are illustrated for the hydroboration of α-pinene.
Since the hydroboration procedure is most commonly used to hydrate alkenes in an anti-Markovnikov fashion, we also need to know the stereoselectivity of the second oxidation reaction, which substitutes a hydroxyl group for the boron atom. Independent study has shown this reaction takes place with retention of configuration so the overall addition of water is also syn.
The hydroboration of α-pinene also provides a nice example of steric hindrance control in a chemical reaction. In the less complex alkenes used in earlier examples the plane of the double bond was often a plane of symmetry, and addition reagents could approach with equal ease from either side. In this case, one of the methyl groups bonded to C-6 (colored blue in the equation) covers one face of the double bond, blocking any approach from that side. All reagents that add to this double bond must therefore approach from the side opposite this methyl.
Problems
What are the products of these following reactions?
#1.
#2.
#3.
Draw the structural formulas for the alcohols that result from hydroboration-oxidation of the alkenes shown.
#4.
#5. (E)-3-methyl-2-pentene
If you need clarification or a reminder on the nomenclature of alkenes refer to the link below on naming the alkenes.
#1.
#2.
#3.
#4.
#5. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.16%3A_HydroborationOxidation.txt |
E2 Reaction
E2 reactions are typically seen with secondary and tertiary alkyl halides, but a hindered base is necessary with a primary halide. The mechanism by which it occurs is a single step concerted reaction with one transition state. The rate at which this mechanism occurs is second order kinetics, and depends on both the base and alkyl halide. A good leaving group is required because it is involved in the rate determining step. The leaving groups must be coplanar in order to form a pi bond; carbons go from sp3 to sp2 hybridization states.
To get a clearer picture of the interplay of these factors involved in a a reaction between a nucleophile/base and an alkyl halide, consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In one pathway, a methanethiolate nucleophile substitutes for bromine in an SN2 reaction. In the other (bottom) pathway, methoxide ion acts as a base (rather than as a nucleophile) in an elimination reaction. As we will soon see, the mechanism of this reaction is single-step, and is referred to as the E2 mechanism.
.
E1 Reaction
An E1 reaction involves the deprotonation of a hydrogen nearby (usually one carbon away, or the beta position) the carbocation resulting in the formation of an alkene product. In order to accomplish this, a Lewis base is required. For a simplified model, we’ll take B to be a Lewis base, and LG to be a halogen leaving group.
As can be seen above, the preliminary step is the leaving group (LG) leaving on its own. Because it takes the electrons in the bond along with it, the carbon that was attached to it loses its electron, making it a carbocation. Once it becomes a carbocation, a Lewis Base (B) deprotonates the intermediate carbocation at the beta position, which then donates its electrons to the neighboring C-C bond, forming a double bond. Unlike E2 reactions, which require the proton to be anti to the leaving group, E1 reactions only require a neighboring hydrogen. This is due to the fact that the leaving group has already left the molecule. The final product is an alkene along with the HB byproduct.
Dehydration
One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo E1 or E2 mechanisms to lose water and form a double bond. The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
Primary alcohol dehydrates through the E2 mechanism
Oxygen donates two electrons to a proton from sulfuric acid H2SO4, forming an alkyloxonium ion. Then the nucleophile HSO4 back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.
Secondary and tertiary alcohols dehydrate through the E1 mechanism
Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, in this case the ion leaves first and forms a carbocation as the reaction intermediate. The water molecule (which is a stronger base than the HSO4- ion) then abstracts a proton from an adjacent carbon, forming a double bond. Notice in the mechanism below that the aleke formed depends on which proton is abstracted: the red arrows show formation of the more substituted 2-butene, while the blue arrows show formation of the less substituted 1-butene. Recall the general rule that more substituted alkenes are more stable than less substituted alkenes, and trans alkenes are more stable than cis alkenes. Thereore, the trans diastereomer of the 2-butene product is most abundant.
Zaitsev's Rule
Zaitsev’s or Saytzev’s (anglicized spelling) rule is an empirical rule used to predict regioselectivity of 1,2-elimination reactions occurring via the E1 or E2 mechanisms. It states that in a regioselective E1 or E2 reaction the major product is the more stable alkene, (i.e., the alkene with the more highly substituted double bond). For example:
If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.17%3A_Keeping_Track_of_Reactions.txt |
Disconnection of bonds
Having chosen the TARGET molecule for synthesis, the next exercise is to draw out synthetic plans that would summarize all reasonable routes for its synthesis. During the past few decades, chemists have been working on a process called RETROSYNTHESIS. Retrosynthesis could be described as a logical Disconnection at strategic bonds in such a way that the process would progressively lead to easily available starting material(s) through several synthetic plans. Each plan thus evolved, describes a ‘ROUTE’ based on a retrosynthesis. Each disconnection leads to a simplified structure. The logic of such disconnections forms the basis for the retroanalysis of a given target molecule. Natural products have provided chemists with a large variety of structures, having complex functionalities and stereochemistry. This area has provided several challenging targets for development of these concepts. The underlining principle in devising logical approaches for synthetic routes is very much akin to the following simple problem. Let us have a look of the following big block, which is made by assembling several small blocks (Fig 1.4.2.1). You could easily see that the large block could be broken down in different ways and then reassembled to give the same original block.
Fig 1.4.2.1
Now let us try and extend the same approach for the synthesis of a simple molecule. Let us look into three possible ‘disconnections’ for a cyclohexane ring as shown in Fig 1.4.2.2.
Fig 1.4.2.2
In the above analysis we have attempted to develop three ways of disconnecting the six membered ring. Have we thus created three pathways for the synthesis of cyclohexane ring? Do such disconnections make chemical sense? The background of an organic chemist should enable him to read the process as a chemical reaction in the reverse (or ‘retro-‘) direction. The dots in the above structures could represent a carbonium ion, a carbanion, a free radical or a more complex reaction (such as a pericyclic reaction or a rearrangement). Applying such chemical thinking could open up several plausible reactions. Let us look into path b, which resulted from cleavage of one sigma bond. An anionic cyclisation route alone exposes several candidates as suitable intermediates for the formation of this linkage. The above analysis describes only three paths out of the large number of alternate cleavage routes that are available. An extended analysis shown below indicates more such possibilities (Fig 1.4.2.3). Each such intermediate could be subjected to further disconnection process and the process continued until we reach a reasonably small, easily available starting materials. Thus, a complete ‘SYNTHETIC TREE’ could be constructed that would summarize all possible routes for the given target molecule.
Fig 1.4.2.3
1.4.3 Efficiency of a route
A route is said to be efficient when the ‘overall yield’ of the total process is the best amongst all routes investigated. This would depend not only on the number of steps involved in the synthesis, but also on the type of strategy followed. The strategy could involve a ‘linear syntheses’ involving only consequential steps or a ‘convergent syntheses’ involving fewer consequential steps. Fig 1.4.3.1 shown below depicts a few patterns that could be recognized in such synthetic trees. When each disconnection process leads to only one feasible intermediate and the process proceeds in this fashion
Fig 1.4.3.1
all the way to one set of starting materials (SM), the process is called a Linear Synthesis. On the other hand, when an intermediate could be disconnected in two or more ways leading to different intermediates, branching occurs in the plan. The processes could be continued all the way to SMs. In such routes different branches of the synthetic pathways converge towards an intermediate. Such schemes are called Convergent Syntheses.
The flow charts shown below (Fig 1.4.3.2) depicts a hypothetical 5-step synthesis by the above two strategies. Assuming a very good yield (90%) at each step (this is rarely seen in real projects), a linier synthesis gives 59% overall yield, whereas a convergent synthesis gives 73% overall yield for the same number of steps..
Fig 1.4.3.2
1.4.4 Problem of substituents and stereoisomers
The situation becomes more complex when you consider the possibility of unwanted isomers generated at different steps of the synthesis. The overall yield drops down considerably for the synthesis of the right isomer. Reactions that yield single isomers (Diastereospecific reactions) in good yields are therefore preferred. Some reactions like the Diels Alder Reaction generate several stereopoints (points at which stereoisomers are generated) simultaneously in one step in a highly predictable manner. Such reactions are highly valued in planning synthetic strategies because several desirable structural features are introduced in one step. Where one pure enantiomer is the target, the situation is again complex. A pure compound in the final step could still have 50% unwanted enantiomer, thus leading to a drastic drop in the efficiency of the route. In such cases, it is desirable to separate the optical isomers as early in the route as possible, along the synthetic route. This is the main merit of the Chiron Approach, in which the right starting material is chosen from an easily available, cheap ‘chiral pool’. We would discuss this aspect after we have understood the logic of planning syntheses. Given these parameters, you could now decide on the most efficient route for any given target.
Molecules of interest are often more complex than the plain cyclohexane ring discussed above. They may have substituents and functional groups at specified points and even specific stereochemical points. Construction of a synthetic tree should ideally accommodate all these parameters to give efficient routes. Let us look into a slightly more complex example shown in Fig 1.4.4.1 . The ketone 1.4.4.1A is required as an intermediate in a synthesis. Unlike the plain cyclohexane discussed above, the substitution pattern and the keto- group in this molecule impose some restrictions on disconnection processes.
Fig 1.4.4.1
Cleavage a: This route implies attack of an anion of methylisopropylketone on a bromo-component. Cleavage b: This route implies simple regiospecific methylation of a larger ketone that bears all remaining structural elements. Cleavage c: This route implies three different possibilities. Route C-1 envisages an acylonium unit, which could come from an acid halide or an ester. Route C-2 implies an umpolung reaction at the acyl unit. Route C-3 suggests an oxidation of a secondary alcohol, which could be obtained through a Grignard-type reaction. Cleavage d: This implies a Micheal addition.
Each of these routes could be further developed backwards to complete the synthetic tree. These are just a few plausible routes to illustrate an important point that the details on the structure would restrict the possible cleavages to some strategic points. Notable contributions towards planning organic syntheses came from E.J. Corey’s school. These developments have been compiles by Corey in a book by the title LOgIC OF CHEMICAL SYNTHESIS. These and several related presentations on this topic should be taken as guidelines. They are devised after analyzing most of the known approaches published in the literature and identifying a pattern in the logic. They need not restrict the scope for new possibilities. Some of the important strategies are outlined below.
1.4.5 Preliminary scan
When a synthetic chemist looks at the given Target, he should first ponder on some preliminary steps to simplify the problem on hand. Is the molecule polymeric? See whether the whole molecule could be split into monomeric units, which could be coupled by a known reaction. This is easily seen in the case of peptides, nucleotides and organic polymers. This could also be true to other natural products. In molecules like C-Toxiferin 1 (1.4.5.1A) (Fig 1.4.5.1), the point of dimerisation is obvious. In several other cases, a deeper insight is required to identify the monomeric units, as is the case with Usnic acid (1.4.5.1B). In the case of the macrolide antibiotic Nonactin (1.4.5.1C), this strategy reduces the possibilities to the synthesis of a monomeric unit (1.4.5.1D). The overall structure has S4 symmetry and is achiral even though assembled from chiral precursors. Both (+)-nonactic acid and (−)-nonactic acid (1.4.5.1D) are needed to construct the macrocycle and they are joined head-to-tail in an alternating (+)-(−)-(+)-(−) pattern. (see J. Am. Chem. Soc., 131, 17155 (2009) and references cited therein).
Fig 1.4.5.1
Is a part of the structure already solved? Critical study of the literature may often reveal that the same molecule or a closely related one has been solved. R.B. Woodward synthesized (1.4.5.2C) as a key intermediate in an elegant synthesis of Reserpine (1.4.5.2A). The same intermediate compound (1.4.5.2C) became the key starting compound for Velluz et.al., in the synthesis of Deserpidine (1.4.5.2B) (Fig 1.4.5.2).
Fig 1.4.5.2
Such strategies reduce the time taken for the synthesis of new drug candidates. These strategies are often used in natural product chemistry and drug chemistry. Once the preliminary scan is complete, the target molecule could be disconnected at Strategic Bonds.
1.4.6 Strategic Bonds, Retrons and Transforms
STRATEGIC BONDS are the bonds that are cleaved to arrive at suitable Starting Materials (SM) or SYNTHONS. For the purpose of bond disconnection, Corey has suggested that the structure could be classified according to the sub-structures generated by known chemical reactions. He called the sub-structures RETRONS and the chemical transformations that generate these Retrons were called TRANSFORMS. A short list of Transforms and Retrons are given below (TABLE 1.4.6.1). Note that when Transforms generate Retrons, the product may have new STEREOPOINTS (stereochemical details) generated that may need critical appraisal.
Fig 1.4.6.1
The structure of the target could be such that the Retron and the corresponding Transforms could be easily visualized and directly applied. In some cases, the Transforms or the Retrons may not be obvious. In several syntheses, transformations do not simplify the molecule, but they facilitate the process of synthesis. For example, a keto- group could be generated through modification of a -CH-NO2 unit through a Nef reaction. This generates a new set of Retron / transforms pair. A few such transforms are listed below, along with the nomenclature suggested by Corey (Fig 1.4.6.2).
Fig 1.4.6.2
A Rearrangement Reaction could be a powerful method for generating suitable new sub-structures. In the following example, a suitable Pinacol Retron, needed for the rearrangement is obtained through an acyloin transform (Fig 1.4.6.3). Such rearrangement Retrons are often not obvious to inexperienced eyes.
Fig 1.4.6.3
Some transforms may be necessary to protect (acetals for ketones), modify (reduction of a ketone to alcohol to avoid an Aldol condensation during a Claisen condensation) or transpose a structural element such as a stereopoint (e.g. SN2 inversion, epimerization etc.,) or shifting a functional group. Such transforms do not simplify the given structural unit. At times, activation at specific points on the structure may be introduced to bring about a C-C bond formation and later the extra group may be removed. For example, consider the following retrosynthesis in which an extra ester group has been introduced to facilitate a Dieckmann Retron. In complex targets, combinations of such strategies could prove to be a very productive strategy in planning retrosynthesis. Witness the chemical modification strategy shown below for an efficient stereospecific synthesis of a trisubstituted olefin (Fig 1.4.6.4)
Fig 1.4.6.4
Fig 1.4.6.4 Examples for FGA / FGR strategies for complex targets
Amongst the molecular architectures, the bridged-rings pose a complex challenge in Structure-Based disconnection procedures. Corey has suggested guidelines for efficient disconnections of strategic bonds.
A bond cleavage for retrosynthesis should lead to simplified structures, preferably bearing five- or six-membered rings. The medium and large rings are difficult to synthesize stereospecifically. Amongst the common rings, a six-membered ring is easily approached and manipulated to large and small rings. Simultaneous cleavage of two bonds, suggesting cycloaddition – retrons are often more efficient. Some cleavages of strategic bonds are shown in Fig 1.4.6.5, suggesting good and poor cleavage strategies based on this approach. However, these guidelines are not restrictive.
Fig 1.4.6.5
Fig 1.4.6.5: Some cleavages at strategic bonds on bridged-ring systems.
Identifying Retron – Transform sets in a given target molecule is therefore a critical component in retrosynthesis. Such an approach could often generate several synthetic routes. The merit of this approach is that starting materials do not prejudice this logic. Retrosyntheses thus developed could throws open several routes that need further critical scrutiny on the basis of known facts.
Identification of Retrons / Transforms sets provided the prerequisite for computer assisted programs designed for generating retrosynthetic routes. A list of Retrons and the corresponding transforms were interlinked and the data was stored in the computer. All known reactions were thus analyzed for their Retron / Transform characteristics and documented. The appropriate literature citations were also documented and linked. Based on these inputs, computer programs were designed to generate retrosynthetic routes for any given structure. Several such programs are now available in the market to help chemists generate synthetic strategies. Given any structure, these programs generate several routes. Once the scientist identifies the specific routes of interest for further analysis, the program generates detailed synthetic steps, reagents required and the appropriate citations. In spite of such powerful artificial intelligence, the intelligence and intuitive genius of a chemist is still capable of generating a new strategy, not yet programmed. Again, human intelligence is still a critical input for the analysis the routes generated using a computer. Based on the experience of the chemists’ team, their projected aim of the project and facilities available, the routes are further screened.
Contributors
• Prof. R Balaji Rao (Department of Chemistry, Banaras Hindu University, Varanasi) as part of Information and Communication Technology | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/10%3A_Alkenes/10.18%3A_Alkenes_in_Organic_Synthesis.txt |
The simplest alkyne—a hydrocarbon with carbon-to-carbon triple bond—has the molecular formula C2H2 and is known by its common name—acetylene. Its structure is H–C≡C–H.
Terminal Alkyne: Internal Alkyne:
3-chloro-1-propyne 4,4-dichloro-2-pentyne
Bonding and Hybridization
Bond Name Location Overlap
Bond 1 s (? bond) bond Formed between 2 sp orbitals of carbon and hydrogen atoms End-on overlap
Bond 2 S (? bond) bond Formed between the 2 sp orbital of 2 unsaturated Carbon atoms. End-on overlap
Bond 3 p-bonds (? bonds) Formed between the 2 p-orbitals among the carbon atoms Side-on overlap
Orbital Name
Location
Orbital 1 sp hybrid orbitals Formed in the linear structure model of carbon atom
Orbital 2 p-orbitals Formed on each carbon
11.02: Nomenclature
Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of \(C_nH_{2n-2}\). They are unsaturated hydrocarbons. Like alkenes have the suffix –ene, alkynes use the ending –yne; this suffix is used when there is only one alkyne in the molecule.
Introduction
Here are the molecular formulas and names of the first ten carbon straight chain alkynes.
Name
Molecular Formula
Ethyne
C2H2
Propyne
C3H4
1-Butyne
C4H6
1-Pentyne
C5H8
1-Hexyne
C6H10
1-Heptyne
C7H12
1-Octyne
C8H14
1-Nonyne
C9H16
1-Decyne
C10H18
The more commonly used name for ethyne is acetylene, which used industrially.
Naming Alkynes
Like previously mentioned, the IUPAC rules are used for the naming of alkynes.
Rule 1
Find the longest carbon chain that includes both carbons of the triple bond.
Rule 2
Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes.
For example:
4-chloro-6-diiodo-7-methyl-2-nonyne
Rule 3
After numbering the longest chain with the lowest number assigned to the alkyne, label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order.
For example:
1-triiodo-4-dimethyl-2-nonyne
If there is an alcohol present in the molecule, number the longest chain starting at the end closest to it, and follow the same rules. However, the suffix would be –ynol, because the alcohol group takes priority over the triple bond.
5- methyl-7-octyn-3-ol
When there are two triple bonds in the molecule, find the longest carbon chain including both the triple bonds. Number the longest chain starting at the end closest to the triple bond that appears first. The suffix that would be used to name this molecule would be –diyne.
For example:
4-methyl-1,5-octadiyne
Rule 4
Substituents containing a triple bond are called alkynyl.
For example:
1-chloro-1-ethynyl-4-bromocyclohexane
Here is a table with a few of the alkynyl substituents:
Name
Molecule
Ethynyl
-C?CH
2- Propynyl
-CH2C?CH
2-Butynyl
-CH3C?CH2CH3
Rule 5
A molecule that contains both double and triple bonds is called an alkenyne. The chain can be numbered starting with the end closest to the functional group that appears first. For example:
6-ethyl-3-methyl-1,4-nonenyne
Reference
1. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. 5th Edition. New York: W. H. Freeman & Company, 2007.
Problems
Name or draw out the following molecules:
1. 4,4-dimethyl-2-pentyne
2. 4-Penten-1-yne
3. 1-ethyl-3-dimethylnonyne
4.
Contributors
• A. Sheth and S. Sujit (UCD)
11.03: Physical Properties
The characteristic of the triple bond helps to explain the properties and bonding in the alkynes.
Importance of Triple Bonds
Hybridization due to triple bonds allows the uniqueness of alkyne structure. This triple bond contributes to the nonpolar bonding strength, linear, and the acidity of alkynes. Physical Properties include nonpolar due to slight solubility in polar solvents and insoluble in water. This solubility in water and polar solvents is a characteristic feature to alkenes as well. Alkynes dissolve in organic solvents.
Boiling Points
Compared to alkanes and alkenes, alkynes have a slightly higher boiling point. Ethane has a boiling point of -88.6 ?C, while Ethene is -103.7 ?C and Ethyne has a higher boiling point of -84.0 ?C.
Alkynes are High In Energy
Alkynes are involved in a high release of energy because of repulsion of electrons. The content of energy involved in the alkyne molecule contributes to this high amount of energy. The pi-bonds however, do not encompass a great amount of energy even though the concentration is small within the molecule. The combustion of Ethyne is a major contributor from CO2, water, and the ethyne molecule
??H = -311 kcal/mol
To help understand the relative stabilities of alkyne isomers, heats of hydrogenation must be used. Hydrogenation of the least energy, results in the release of the internal alkyne. With the result of the production of butane, the stability of internal versus terminal alkynes has significant relative stability due to hyperconjugation.
Problems
1. What is the carbon-carbon, carbon-hydrogen bond length for alkyne? Is it shorter or longer than alkane and alkene?
2. Which is the most acidic and most stable, alkane, alkene, or alkyne? And depends on what?
3. How many pi bonds and sigma bonds are involved in the structure of ethyne?
4. Why is the carbon-hydrogen bond so short?
5. What is the alkyne triple bond characterizes by? How is this contribute to the weakness of the pi bonds?
6. How is heat of hydrogenation effects the stability of the alkyne?
Contributors
• Bao Kha Nguyen, Garrett M. Chin
11.04: Interesting Alkynes
The important class of lipids called steroids are actually metabolic derivatives of terpenes, but they are customarily treated as a separate group. Steroids may be recognized by their tetracyclic skeleton, consisting of three fused six-membered and one five-membered ring.
Steroids are widely distributed in animals, where they are associated with a number of physiological processes. Examples of some important steroids are shown in the following diagram. Norethindrone is a synthetic steroid, all the other examples occur naturally. A common strategy in pharmaceutical chemistry is to take a natural compound, having certain desired biological properties together with undesired side effects, and to modify its structure to enhance the desired characteristics and diminish the undesired. This is sometimes accomplished by trial and error. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/11%3A_Alkynes/11.01%3A_Introduction.txt |
lkynes can be a useful functional group to synthesize due to some of their antibacterial, antiparasitic, and antifungal properties. One simple method for alkyne synthesis is by double elimination from a dihaloalkane.
Introduction
One case in which elimination can occur is when a haloalkane is put in contact with a nucleophile. The table below is used to determine which situations will result in elimination and the formation of a π bond.
Elimination: Haloalkane-Nucleophile Reaction
Type of Haloalkane Weak Base, Poor Nucleophile Weak Base, Good Nucleophile Strong, Unhindered Base Strong, Hindered Base
Primary
Unhindered E2
Branched E2 E2
Secondary E1 E2 E2
Tertiary E1 E1 E2 E2
* Empty Box means no elimination or Pi bond forms
To synthesize alkynes from dihaloalkanes we use dehydrohalogenation. The majority of these reactions take place using alkoxide bases (other strong bases can also be used) with high temperatures. This combination results in the majority of the product being from the E2 mechanism.
E2 Mechanism
Recall that the E2 mechanism is a concerted reaction (occurs in 1 step). However, in this 1 step there are 3 different changes in the molecule. This is the reaction between 2-Bromo-2-methylpropane and Sodium Hyrdoxide.
This is a brief review of the E2 reaction. For further information on why the reaction proceeds as it does visit the E2 reaction page. Now, if we apply this concept using 2 halides on vicinal or geminal carbons, the E2 reaction will take place twice resulting in the formation of 2 Pi bonds and thus an Alkyne.
Dihaloalkane Elimination
This is a general picture of the reaction taking place without any of the mechanisms shown.
or
* With a terminal haloalkane the equation above is modified in that 3 equivalents of base will be used instead of 2.
Lets look at the mechanism of a reaction between 2,3-Dibromopentane with sodium amide in liquid ammonia.
• Liquid ammonia is not part of the reaction, but is used as a solvent
• Notice the intermediate of the alkyne synthesis. It is stereospecifically in its anti form. Because the second proton and halogen are pulled off the molecule this is unimportant to the synthesis of alkynes. For more information on this see the page on preparation of alkenes from haloalkanes.
Preparation of Alkynes from Alkenes
Lastly, we will briefly look at how to prepare alkynes from alkenes. This is a simple process using first halogenation of the alkene bond to form the dihaloalkane, and next, using the double elimination process to protonate the alkane and from the 2 Pi bonds.
This first process is gone over in much greater detail in the page on halogenation of an alkene. In general, chlorine or bromine is used with an inert halogenated solvent like chloromethane to create a vicinal dihalide from an alkene. The vicinal dihalide formed is the reactant needed to produce the alkyene using double elimination, as covered previously on this page.
In The Lab
Due to the strong base and high temperatures needed for this reaction to take place, the triple bond may change positions. An example of this is when reactants that should form a terminal alkyne, form a 2-Alkyne instead. The use of NaNH2 in liquid NH3 is used in order to prevent this from happening due to its lower reacting temperature. Even so, most chemists will prefer to use nucleophilic substitution instead of elimination when trying to form a terminal alkyne.
Questions
Question 1: Why would we need 3 bases for every terminal dihaloalkane instead of 2 in order to form an alkyne?
Question 2: What are the major products of the following reactions:
a.) 1,2-Dibromopentane with sodium amide in liquid ammonia
b.) 1-Pentene first with Br2 and chloromethane, followed by sodium ethoxide (Na+ -O-CH2CH3)
Question 3: What would be good starting molecules for the synthesis of the following molecules:
Question 4: Use a 6 carbon diene to synthesize a 6 carbon molecule with 2 terminal alkynes.
Answers
Answer 1: Remember that hydrogen atoms on terminal alkynes make the alkyne acidic. One of the base molecules will pull off the terminal hydrogen instead of one of the halides like we want.
Answer 2:
a.) 1-Pentyne
b.) 1-Pentyne
Answer 3:
Answer 4: Bromine or chlorine can be used with different inert solvents for the halogenation. This can be done using many different bases. Liquid ammonia is used as a solvent and needs to be followed by an aqueous work-up. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/11%3A_Alkynes/11.05%3A_Preparation_of_Alkynes.txt |
Alkanes are undoubtedly the weakest Brønsted acids commonly encountered in organic chemistry. It is difficult to measure such weak acids, but estimates put the pKa of ethane at about 48. Hybridizing the carbon so as to increase the s-character of the C-H increases the acidity, with the greatest change occurring for the sp-C-H groups found in terminal alkynes. Thus, the pKa of ethene is estimated at 44, and the pKa of ethyne (acetylene) is found to be 25, making it 1023 times stronger an acid than ethane. This increase in acidity permits the isolation of insoluble silver and copper salts of such compounds.
RC≡C-H + Ag(NH3)2(+) (in NH4OH) → RC≡C-Ag (insoluble) + NH3 + NH4(+)
Despite the dramatic increase in acidity of terminal alkynes relative to other hydrocarbons, they are still very weak acids, especially when compared with water, which is roughly a billion times more acidic. If we wish to prepare nucleophilic salts of terminal alkynes for use in synthesis, it will therefore be necessary to use a much stronger base than hydroxide (or ethoxide) anion. Such a base is sodium amide (NaNH2), discussed above, and its reactions with terminal alkynes may be conducted in liquid ammonia or ether as solvents. The products of this acid-base reaction are ammonia and a sodium acetylide salt. Because the acetylide anion is a powerful nucleophile it may displace halide ions from 1º-alkyl halides to give a more highly substituted alkyne as a product (SN2 reaction). This synthesis application is described in the following equations. The first two equations show how acetylene can be converted to propyne; the last two equations present a synthesis of 2-pentyne from propyne.
H-C≡C-H + NaNH2 (in ammonia or ether) → H-C≡C-Na (sodium acetylide) + NH3
H-C≡C-Na + CH3-I → H-C≡C-CH3 + NaI
CH3-C≡C-H + NaNH2 (in ammonia or ether) → CH3-C≡C-Na (sodium propynylide) + NH3
CH3-C≡C-Na + C2H5-BrCH3-C≡C-C2H5 + NaBr
Because RC≡C:(–) Na(+) is a very strong base (roughly a billion times stronger than NaOH), its use as a nucleophile in SN2 reactions is limited to 1º-alkyl halides; 2º and 3º-alkyl halides undergo elimination by an E2 mechanism.
The enhanced acidity of terminal alkynes relative to alkanes also leads to metal exchange reactions when these compounds are treated with organolithium or Grignard reagents. This exchange, shown below in equation 1, can be interpreted as an acid-base reaction which, as expected, proceeds in the direction of the weaker acid and the weaker base. This factor clearly limits the usefulness of Grignard or lithium reagents when a terminal triple bond is present, as in equation 2.
1) RC≡C-H + C2H5MgBr (in ether) → RC≡C-MgBr + C2H6
2) HC≡C-CH2CH2Br + Mg (in ether) → [ HC≡C-CH2CH2MgBr] → BrMgC≡C-CH2CH2H
The acidity of terminal alkynes also plays a role in product determination when vicinal (or geminal) dihalides undergo base induced bis-elimination reactions. The following example illustrates eliminations of this kind starting from 1,2-dibromopentane, prepared from 1-pentene by addition of bromine. The initial elimination presumably forms 1-bromo-1-pentene, since base attack at the more acidic and less hindered 1 º-carbon should be favored. The second elimination then produces 1-pentyne. If the very strong base sodium amide is used, the terminal alkyne is trapped as its sodium salt, from which it may be released by mild acid treatment. However, if the weaker base KOH is used for the elimination, the terminal alkyne salt is not formed, or is formed reversibly, and the initially generated 1-pentyne rearranges to the more stable 2-pentyne via an allene intermediate.
In the case of non-terminal alkynes, sodium and potassium amide, and related strong bases from 1 º-amines, are able to abstract protons from carbon atoms adjacent to the triple bond. The resulting allenic carbanions undergo rapid proton transfer equilibria, leading to the relatively stable terminal alkyne conjugate base. This isomerization may be used to prepare longer chain 1-alkynes, as shown in the following conversion of 3-heptyne to 1-heptyne. The R and R' substituents on the allenic intermediate range from propyl to hydrogen, as the proton transfers proceed.
Conjugate base anions of terminal alkynes (acetylide anions) are nucleophiles, and can do both nucleophilic substitution and nucleophilic addition reactions. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/11%3A_Alkynes/11.06%3A_Introduction_to_Alkyne_Reactions.txt |
Reaction 1: Addition of Hydrogen Halide to an Alkyne
Summary: Reactivity order of hydrogen halides: HI > HB r> HCl > HF.
Follows Markovnikov’s rule:
• Hydrogen adds to the carbon with the greatest number of hydrogens, the halogen adds to the carbon with fewest hydrogens.
• Protination occurs on the more stable carbocation. With the addition of HX, haloalkenes form.
• With the addition of excess HX, you get anti addition forming a geminal dihaloalkane.
Addition of a HX to an Internal Alkyne
As described in Figure 1, the $\pi$ electrons will attack the hydrogen of the HBr and because this is a symmetric molecule it does not matter which carbon it adds to, but in an asymmetric molecule the hydrogen will covalently bond to the carbon with the most hydrogens. Once the hydrogen is covalently bonded to one of the carbons, you will get a carbocation intermediate (not shown, but will look the same as depicted in Figure 1) on the other carbon. Again, this is a symmetric molecule and if it were asymmetric, which carbon would have the positive charge?
The final step is the addition of the Bromine, which is a good nucleophile because it has electrons to donate or share. Bromine, therefore attacks the carbocation intermediate placing it on the highly substituted carbon. As a result, you get 2-bromobutene from your 2-butyne reactant, as shown below.
Now, what if you have excess HBr?
Addition due to excess HX present ? yields a geminal dihaloalkane
Here, the electrophilic addition proceeds with the same steps used to achieve the product in Addition of a HX to an Internal Alkyne. The $\pi$ electrons attacked the hydrogen, adding it to the carbon on the left (shown in blue). Why was hydrogen added to the carbon on left and the one on the right bonded to the Bromine?
Now, you will have your carbocation intermediate, which is followed by the attack of the Bromine to the carbon on the right resulting in a haloalkane product.
Addition of HX to Terminal Alkyne
• Here is an addition of HBr to an asymmetric molecule.
• First, try to make sense of how the reactant went to product and then take a look at the mechanism.
The $\pi$ electrons are attacking the hydrogen, depicted by the electron pushing arrows and the Bromine gains a negative charge. The carbocation intermediate forms a positive charge on the left carbon after the hydrogen was added to the carbon with the most hydrogen substituents.
The Bromine, which has a negative charge, attacks the positively charged carbocation forming the final product with the nucleophile on the more substituted carbon.
Addition due to excess HBr present
Most Hydrogen halide reactions with alkynes occur in a Markovnikov-manner in which the halide attaches to the most substituted carbon since it is the most positively polarized. A more substituted carbon has more bonds attached to 1) carbons or 2) electron-donating groups such as Fluorine and other halides. However, there are two specific reactions among alkynes where anti-Markovnikov reactions take place: the radical addition of HBr and Hydroboration Oxidation reactions. For alkynes, an anti-Markovnikov addition takes place on a terminal alkyne, an alkyne on the end of a chain.
HBr Addition With Radical Yields 1-bromoalkene
The Br of the Hydrogen Bromide (H-Br) attaches to the less substituted 1-carbon of the terminal alkyne shown below in an anti-Markovnikov manner while the Hydrogen proton attaches to the second carbon. As mentioned above, the first carbon is the less substituted carbon since it has fewer bonds attached to carbons and other substituents. The H-Br reagent must also be reacted with heat or some other radicial initiator such as a peroxide in order for this reaction to proceed in this manner. This presence of the radical or heat leads to the anti-Markovnikov addition since it produces the most stable reaction. For more on Anti-Markovnikov additions:Radical Additions--Anti-Markovnikov Product Formation
The product of a terminal alkyne that is reacted with a peroxide (or light) and H-Br is a 1-bromoalkene.
Regioselectivity: The Bromine can attach in a syn or anti manner which means the resulting alkene can be both cis and trans. Syn addition is when both Hydrogens attach to the same face or side of the double bond (i.e. cis) while the anti addition is when they attach on opposite sides of the bond (trans).
Contributors
• Prof. Hilton Weiss (Bard College)
• Aleksandra Milman
• Ali Alvandi | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/11%3A_Alkynes/11.07%3A_Addition_of_Hydrogen_Halides.txt |
Reaction: Halogenation of Alkynes
Summary:
• Stereoslectivity: anti addition
• Reaction proceeds via cyclic halonium ion
Addition of Br2
• The addition of Br2 to an alkyne is analogous to adding Br2 to an alkene.
• Once Br2 approaches the nucleophilic alkyne, it becomes polarized.
• The $\pi$ electrons, from the triple bond, can now attack the polarized bromine forming a C-Br bond and displacing the bromine ion.
• Now, you will get an intermediate electrophilic carbocation, which will immediately react with the bromine ion giving you the dibromo product.
First, you see the polarized Br2 being attacked by the $\pi$ electrons. Once you form the C-Br bond, the other bromine is released as a bromine ion. The intermediate here is a bromonium ion, which is electrophilic and reacts with the bromine ion giving you the dibromo product.
11.09: Addition of Water
Reaction: Hydration of Alkynes
As with alkenes,hydration (addition of water) to alkynes requires a strong acid, usually sulfuric acid, and is facilitated by mercuric sulfate. However, unlike the additions to double bonds which give alcohol products, addition of water to alkynes gives ketone products ( except for acetylene which yields acetaldehyde ). The explanation for this deviation lies in enol-keto tautomerization, illustrated by the following equation. The initial product from the addition of water to an alkyne is an enol (a compound having a hydroxyl substituent attached to a double-bond), and this immediately rearranges to the more stable keto tautomer.
Tautomers are defined as rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers ( acetone, for example, is 99.999% keto tautomer ). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples. The three examples shown below illustrate these reactions for different substitutions of the triple-bond. The tautomerization step is indicated by a red arrow. For terminal alkynes the addition of water follows the Markovnikov rule, as in the second example below, and the final product ia a methyl ketone ( except for acetylene, shown in the first example ). For internal alkynes ( the triple-bond is within a longer chain ) the addition of water is not regioselective. If the triple-bond is not symmetrically located ( i.e. if R & R' in the third equation are not the same ) two isomeric ketones will be formed.
HC≡CH + H2O + HgSO4 & H2SO4 ——> [ H2C=CHOH ] ——> H3C-CH=O
RC≡CH + H2O + HgSO4 & H2SO4 ——> [ RC(OH)=CH2 ] ——> RC(=O)CH3
RC≡CR' + H2O + HgSO4 & H2SO4 ——> [ RHC=C(OH)R' + RC(OH)=CHR' ] ——> RCH2-C(=O)R' + RC(=O)-CH2R'
With the addition of water, alkynes can be hydrated to form enols that spontaneously tautomerize to ketones. Reaction is catalyzed by mercury ions. Follows Markovnikov’s Rule: Terminal alkynes give methyl ketones
• The first step is an acid/base reaction where the π electrons of the triple bond acts as a Lewis base and attacks the proton therefore protinating the carbon with the most hydrogen substituents.
• The second step is the attack of the nucleophilic water molecule on the electrophilic carbocation, which creates an oxonium ion.
• Next you deprotonate by a base, generating an alcohol called an enol, which then tautomerizes into a ketone.
• Tautomerism is a simultaneous proton and double bond shift, which goes from the enol form to the keto isomer form as shown above in Figure 7.
Now let's look at some Hydration Reactions.
Hydration of Terminal Alkyne produces methyl ketones
Just as described in Figure 7 the π electrons will attack a proton, forming a carbocation, which then gets attacked by the nucleophilic water molecules. After deprotination, we generate an enol, which then tautomerizes into the ketone form shown.
Hydration of Alkyne
As you can see here, the π electrons of the triple bond are attacking the proton, which forms a covalent bond on the carbon with the most hydrogen substituents. Once the hydrogen is bound you have a carbocation, which gets attacked by the water molecule. Now you have a positive charge on the oxygen which results in a base coming in and deprotinating the molecule. Once deprotinated, you have an enol, which then gets tautomerized.
Tautomerism is shown here when the proton gets attacked by the double bond π electrons forming a covalent bond between the carbon and the hydrogen on the less substituted carbon. Electrons from the Oxygen end up moving to the carbon, forming a double bond with carbon and giving itself a positive charge, which then gets attacked by the base. The base deprotinates the Oxygen resulting in the more stable final product at equilibrium, which is a ketone. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/11%3A_Alkynes/11.08%3A_Addition_of_Halogen.txt |
Hydroboration Reactions
Diborane reacts readily with alkynes, but the formation of substituted alkene products leaves open the possibility of a second addition reaction. A clever technique for avoiding this event takes advantage of the fact that alkynes do not generally suffer from steric hindrance near the triple-bond (the configuration of this functional group is linear). Consequently, large or bulky electrophilic reagents add easily to the triple-bond, but the resulting alkene is necessarily more crowded or sterically hindered and resists further additions. The bulky hydroboration reagent needed for this strategy is prepared by reaction of diborane with 2-methyl-2-butene, a highly branched alkene. Because of the alkyl branching, only two alkenes add to a BH3 moiety (steric hindrance again), leaving one B-H covalent bond available for reaction with an alkyne, as shown below. The resulting dialkyl borane is called disiamylborane, a contraction of di-secondary-isoamylborane (amyl is an old name for pentyl).
2 (CH3)2C=CHCH3 + BH3 in ether ——> [ (CH3)2CH-CH(CH3) ]2B-H disiamylborane
An important application of disiamylborane is its addition reaction to terminal alkynes. As with alkenes, the B-H reagent group adds in an apparently anti-Markovnikov manner, due to the fact that the boron is the electrophile, not the hydrogen. Further addition to the resulting boron-substituted alkene does not occur, and the usual oxidative removal of boron by alkaline hydrogen peroxide gives an enol which rapidly rearranges to the aldehyde tautomer. Thus, by the proper choice of reagents, terminal alkynes may be converted either to methyl ketones (mercuric ion catalyzed hydration) or aldehydes (hydroboration followed by oxidation).
RC≡CH + (C5H11)2B-H ——> [ RCH=CH-B(C5H11)2 ] + H2O2 & NaOH ——> [ RCH=CH-OH ] ——> RCH2-CH=O
Hydroboration of internal alkynes is not a particularly useful procedure because a mixture of products will often be obtained, unless the triple-bond is symmetrically substituted. Mercuric ion catalyzed hydration gives similar results.
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from BH3 or BHR2) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene bouble bond. Furthermore, the borane acts as a lewis acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The Hydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific (syn addition), meaning that the hydroboration takes place on the same face of the double bond, this leads cis stereochemistry.
The Mechanism
Step #1
• Part #1: Hydroboration of the alkene. In this first step the addittion of the borane to the alkene is initiated and prceeds as a concerted reaction because bond breaking and bond formation occurs at the same time. This part consists of the vacant 2p orbital of the boron electrophile pairing with the electron pair of the ? bondof the nucleophile.
Transition state
* Note that a carbocation is not formed. Therefore, no rearrangement takes place.
• Part #2: The Anti Markovnikov addition of Boron. The boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon. Both, the boron and the hydrogen add simultaneously on the same face of the double bond (syn addition).
Oxidation of the Trialkylborane by Hydrogen Peroxide
Step #2
• Part #1: the first part of this mechanism deals with the donation of a pair of electrons from the hydrogen peroxide ion. the hydrogen peroxide is the nucleophile in this reaction because it is the electron donor to the newly formed trialkylborane that resulted from hydroboration.
• Part 2: In this second part of the mechanism, a rearrangement of an R group with its pair of bonding electrons to an adjacent oxygen results in the removal of a hydroxide ion.
Two more of these reactions with hydroperoxide will occur in order give a trialkylborate
• Part 3: This is the final part of the Oxidation process. In this part the trialkylborate reacts with aqueous NaOH to give the alcohol and sodium borate.
If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided here that will take you to other pages and media that are very helpful as well.
Stereospecific Hydroboration Oxidation of Alkynes
Step 1: Hydroboration of terminal alkynes reacts in an anti-Markovnikov fashion in which the Boron attacks the less substituted carbon which is the least hindered. It is a stereospecific reaction where syn addition is observed as the hydroboration occurs on the same side of the alkyne and results in cis stereochemistry. However, a bulky borane reagent needs to be used to stop at the alkenyl-borane stage. Otherwise, a second hydroboration will occur. In this example, diisoamyl borane or HBsia2, a fairly large and sterically-hindered borane reagent, is used. Both of the alkyne's pi bonds will undergo hydroboration if BH3 (borane) is used by itself.
Step 2: Oxidation is the next step that occurs. The resulting alkylborane is oxidized to a vinyl alcohol due to the reaction with a hydroxide in a basic solution such as aqueous sodium hydroxide. A vinyl alcohol is an alcohol that has both an alkene and an -OH group. After the vinyl alcohol is formed, tautomerization takes place. Tautomerism is the interconversion of isomeric compounds due to the migration of a proton. The alcohol, which in this case is a terminal enol, spontaneously and rapidly rearranges due to tautomerism to become an aldehyde since the aldehyde form is much more stable than the enol.
For additional information on hydroboration oxidation: Hydroboration Oxidation
11.11: Reaction of Acetylide Anions
Acidity of Terminal Alkynes: Formation of Acetylide Anions
Terminal alkynes are much more acidic than most other hydrocarbons. Removal of the proton leads to the formation of an acetylide anion, RC=C:-. The origin of the enhanced acidity can be attributed to the stability of the acetylide anion, which has the unpaired electrons in an sp hybridized orbital. The stability results from occupying an orbital with a high degree of s-orbital character.
There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The enhanced acidity with greater s-character occurs despite the fact that the homolytic C-H BDE is larger.
Compound Conjugate Base Hybridization "s Character" pKa C-H BDE (kJ/mol)
CH3CH3 CH3CH2- sp3 25% 50 410
CH2CH2 CH2CH- sp2 33% 44 473
HCCH HCC- sp 50% 25 523
Consequently, acetylide anions can be readily formed by deprotonation using a sufficiently strong base. Amide anion (NH2-), in the form of NaNH2 is commonly used for the formation of acetylide anions.
Nucleophilic Substitution Reactions of Acetylides
Acetylide anions are strong bases and strong nucleophiles. Therefore, they are able to displace halides and other leaving groups in substitution reactions. The product is a substituted alkyne.
Because the ion is a very strong base, the substitution reaction is most efficient with methyl or primary halides without substitution near the reaction center,
Secondary, tertiary or even bulky primary substrates will give elimination by the E2 mechanism.
Nucleophilic Addition of Acetylides to Carbonyls
Acetylide anions will add to aldehydes and ketones to form alkoxides, which, upon protonation, give propargyl alcohols.
With aldehydes and non-symmetric ketones, in the absence of chiral catalyst, the product will be a racemic mixture of the two enantiomers.
Problems
1. The pKa of ammonia is 35. Estimate the equilibrium constant for the deprotonation of pent-1-yne by amide, as shown above.
Answers
1. Assuming the pKa of pent-1-yne is about 25, then the difference in pKas is 10. Since pentyne is more acidic, the formation of the acetylide will be favored at equilibrium, so the equilibrium constant for the reaction is about 1010 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/11%3A_Alkynes/11.10%3A_HydroborationOxidation.txt |
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