Saturated hydrocarbons, those of the form CnH2n+2 , and known as alkanes, are relatively unreactive. There are only a few common reactions of saturated hydrocarbons, the two most common being combustion with oxygen andfree-radical halogenation. We will examine these reactions and some special reactions of strained cyclic hydrocarbons. First, however, we will discuss the conformational properties of hydrocarbons.
2.1.1 Common Structural Notations for Alkanes
In Section 1.5, we introduced the common chemical notations used to represent saturated and unsaturated hydrocarbon compounds: the flagged bond notation, sawhorse notation, and the Newman projection. We review them now for n-butane, C4H10• In flagged bond notation, solid lines that connect atoms imply that the atoms lie in the same plane as the page, bold lines (or wedges) indicate atoms that are coming out of the plane of the page, and dashed wedges indicate atoms that are behind the plane of the page. The sawhorse notation is similar to the flagged bond notation except that the n1olecule is drawn such that the viewer looks down on the molecule from above. In the Newman projection the molecule is observed directly down the C2-C3 bond. The large circle represents the back carbon, while the point where the three lines meet in the front represents the front carbon.
Flagged bond notation Sawhorse notation Newman projection
H
H
2.1.2 Conformational Analysis of Alkanes and Cycloalkanes
Using these notations, we tum our attention to the conformational analysis of hydrocarbons as demonstrated for n-butane. Remember that conformational isomers are molecules that differ from one another only by the rotation about a cr (single) bond. For saturated hydrocarbons there are two orientations of crbonds attached to adjacent sp3 hybridized carbons that we will concentrate on. These are the staggered conforn1ation and the eclipsed conformation. In staggered c nformations a cr bond on one carbon bisects the angle formed by two cr bonds on the adjacent carbon. In a~nformation a cr bond on one carbon directly lines up with a cr bond on an adjacent carbon atom.
Sawhorse and Newman Representations For Staggered and Eclipsed Conformations of n-Butane
H
H
H
Two notations for a staggered conformation
H
More Stable Less Crowded
H H
Two notations for an eclipsed conformation
Less Stable More Crowded Electronic Repulsion
The cr bonds should actually directly line up with one another.
This Newman structure is not directly aligned for purposes of clarity.
412 Chapter 2 MCAT Organic Chemistry
The fact that cyclopropane and cyclobutane can undergo these hydrogenation reactions is probably because the hybridization of the carbons is not strictly sp3 • The C-C bonds have considerable p orbital character and are therefore able to undergo some common reactions of carbon-carbon n bonds.
Next we come to the five-membered ring, cyclopentane (C5H10). The pentagonal bond angle is 108° (close to normal tetrahedral of 109°), so we might expect cyclopentane to be a planar structure. If all of the carbons of cyclopentane were in a plane, however, all of the carbon-hydrogen cr bonds on adjacent carbons would eclipse each other. In order to compensate for the eclipsed C-H cr bonds, cyclopentane has one carbon out of the plane of the other carbons and so adopts a puckered conformation. This puckering allows the carbon-hydrogen cr bonds on adjacent carbons to be somewhat staggered, and thus reduces the energy of the compound. This puckered form of cyclopentane is refered to as the "envelope" form.
H
0 H H H
Cyclopentane
H H
The final cycloalkane we will examine is cyclohexane (C6H12). If cyclohexane were planar, it would have bond angles of 120°. This would produce considerable strain on sp3 hybridized carbons whose ideal bond angles are around 109°. Instead, the most stable confor-mation of cyclohexane is a very puckered molecule referred to as the chai:r In the chair conformation, four of the carbons of the ring are in a plane with one carbon above the plane and the remaining carbon below the plane of the four carbons. There are two chair conformations for cyclohexane, and they easily interconvert at room temperature.
H H H H
11 kcal/mol H
H H
H H
Chair representations of cyclohexane
As one chair conformation flips to the other chair conformation, it must pass through several other less stable conformations including some (referred to as half-chair conformations) that reside at energy maxima and one (the boat conformation) at a local energy minimum (but still of much higher energy than the chair conformations). The boat conformation represents a transition state between twist boat conformations. It is important to remember, however, that all of these conformations are much more unstable than the chair conformations and thus do not play an important role in cyclohexane chemistry.
H H
H
H H
Boat conformation H H
Notice that there are two distinct types of hydrogens in the chair forms of cyclohexane. Six of the hydrogens lie on the equator of the ring of carbons. These hydrogens are referred to ?-S equatorial hydrogens. The other six hydrogens lie above or below the ring of carbons, three above and three below; these are called axial hydrogens.
H H
H
H
~ equatorial hydrogen H
H..___
axial hydrogen
There is an energy barrier of about 11 kcal/mol between the two equivalent chair conformations of cyclohexane. At room temperature there is sufficient thermal energy to inter- convert the two chair conformations about 10,000 times per second. Note that when a hydrogen (or any substituent group) is axial in one chair conformation, it becomes equatorial when cyclohexane flips to the other chair conformation. The same is also true for an equatorial hydrogen which flips to an axial position when the chair forms interconvert. This property is demonstrated for deuterocyclohexane:
414
H
H H
H
H
D ~axial
Chapter 2
chair flip
MCAT Organic Chemistry
H H
H
H D
H I
equatorial
These factors become important when examining substituted cyclohexanes. Let's first consider methylcyclohexane. The methyl group can occupy either an equatorial or axial position:
H
~
H r - - c H3
H H
H H
two 1,3-diaxial CH3-H interactions
H
H H
H
H H
no 1,3-diaxial CH3-H interactions
Is one conformation more stable than the other@- It is more favorable for large groups to occupy the equatorial position rather than a crowded axial position. For a methyl group, the equatorial position is more stable by about 1. 7 kcal/mol over the axial position. This is because in the axial position, the me thy 1 group is crowded by the other two hydrogens that are also occupying axial positions on the san1e side of the ring. This is referred to as a 1,3-diaxial interaction. It is more favorable for methyl to be in an equatorial position where it is pointing out away from other atoms.
Unlike cyclopropane and cyclobutane, cyclopentane has a low degree of ring strain, and cyclohexane is strain free. Consequently, these cycloalkanes do not undergo hydrogenation reactions under normal conditions.
0 No reaction
cyclopentane
0 No reaction cyclohexane
Example 2. Identify the more stable isomer in each of the following pairs of substituted cydohexanes:
- (b)
Solution.
(c)
(d)
VS. vs.
Draw chair conformations of each isomer and compare them to see which is more stable. As a good rule of thumb, it is best to first put the bulkier (i.e., the larger) substituent in a roomier equatorial position and decide if it's the more stable of the two chair conformations; it usually is.
(a)
H
H H
two 1,3-diaxial CH3-H interactions
vs. H
This is the more stable isomer.
H H
H no 1 ,3-diaxial CH 3-H
interactions
416
(b)
H
(c)
(d)
H
H
H
two 1 ,3-diaxial interactions
H
H
This is the more stable isomer.
H
H H
no 1 ,3-diaxial CH3-H interactions
H
H H
two 1 ,3-diaxial interactions
Chapter 2
vs. H
vs. H
MCAT Organic Chemistry
This is the more stable isomer.
H
H
H H
H no 1 ,3-diaxial
interactions
H
two 1 ,3-diaxial CH3-H interactions
This is the more stable isomer.
H
no 1,3-diaxial CH3-H interactions
Combustion Reactions of Hydrocarbons
In an efficient combustion reaction, hydrocarbons are completely converted to carbon dioxide and water, thereby liberating the energy stored in the carbon-hydrogen and carbon-carbon bonds. This very exothermic reaction occurs only in the presence of molecular oxygen (02 ) and high temperatures. The stoichiometry for the combustion reaction of hydrocarbons is as follows:
h:X2tmple 3. Write a balanced chemical equation for the combustion of the following compounds:
(a) (b) methylcyclopropane, (c) methane.
Solution.
We will now examine the basic principles of free-radical halogenation focusing on the mechanism of this reaction. Free-radical halogenation is a reaction that proceeds by a multi-step mechanism that includes initiation, propagation, and termination (and is often subject to inhibition by molecular oxygen).
Initiation
A free radical reaction can be initiated by light or heat. In a light-initiated reaction, a photon (E =
hv) collides with a molecule causing homolytic cleavage of a bond (see section 1.5). This results in the formation of two radicals. In a homolytic bond cleavage, one electron goes with each atom of the bond being cleaved. (In a heterolytic bond cleavage, both electrons go with only one of the atoms.)
Homolytic Cleavage: One electron goes with each atom of the bond being broken. This produces two radicals.
,;-....~
R-X R• + X•
418 Chapter 2 MCAT Organic Chemistry
Heterolytic Cleavage: Both electrons of the bond being broken go with one of the atoms.
Propagation
This results in the formation of a cation and an anion.
R-X ~
For every halogen radical formed in the initiation step, about 10,000 alkyl halide molecules are formed in the propagation steps of this chain reaction. In the first step of the propagation reactions, a halogen radical collides with an alkane molecule (R-H) causing homolytic cleavage of a C-H bond with formation of a molecule of hydrogen halide (H-X) and an alkyl radical (R • ).
In the next step of the propagation, the alkyl halide product is formed. This is accomplished by the collision of an alkyl radical (R •) with molecular halogen (X-X). This collision results in the homolytic cleavage of the molecular halogen so that a molecule of alkyl halide (R-X) product is formed and a molecule of halogen radical (X •) is generated.
The halogen radical (Xã) then proceeds to collide with another alkane molecule, forming a hydrogen halide (H-X) molecule and an alkyl radical (Rã). The alkyl radical in tum reacts with another molecule of molecular halogen (X-X) forming the alkyl halide product (R-X) and another halogen radical (Xã) which reacts with another alkane (R-H) molecule to continue propagation of the chain reaction.
Termination
The propagation of the chain reaction continues until one of the reactive radicals of the propagation steps combines with another radical. This can be accomplished by the combination of two halogen radicals (Xã) to form a molecule of molecular halogen (X-X), or the combination of two alkyl radicals (Rã) to form a molecule of alkane (R-R), or finally, by the combination of an alkyl radical (R ã)with a halogen radical (Xã) to form an alkyl halide (R-X). The consumption of reactive radicals stops the propagation of the chain reaction.
Inhibition
Finally, as previously stated, the free-radical halogenation reaction is inhibited by molecular oxygen. This occurs when an alkyl radical reacts with a molecule of molecular oxygen to form an alkyl peroxy radical (R-0-0ã ). The reaction slows down because the concentration of reactive radical intermediate is reduced.
Summary of the Mechanism of Free-Radical Halogenation of Alkanes
ll:Jjj'jgJJ'QE A step that yields a net increase in the number of radicals.
,;--.._rt\
(1) X-X hv
or heat
(2) (3)
R-H + ãX
Rã + X-X Then (2), (3), (2), (3), ....
(4) X• + Xã
(5) R • + Rã
(6) R" + Xã
2 X•
Rã + H-X
R-X + X•
X-X R-R
R-X
R - 0 - 0 •
Inhibition by molecular oxygen slows down the reaction by reducing the amount of reactive radical intermediate. This is a reversible reaction.
Stereochemistry of Free-Radical Halogenation
When the halogen radical collides with the alkane, it abstracts a hydrogen fron1 a sp3 hybridized tetrahedral carbon atom in a homolytic cleavage that results in the formation of a molecule of H-X and a carbon radical intermediate. The next important point to consider is the hybridization of the intermediate alkyl carbon radical. As the hydrogen radical is abstracted, the resulting carbon radical rehybridizes to place the single electron in an unhybridized p orbital. The geometry is planar with 120° bond angles, and the lone electron resides above and below the plane of the molecule in an unhybridized p orbital.
0 \\\\CH3
H - C , + HBr
0 CH3
sp3 hybridized carbon sp2 hybridized carbon
420 Chapter 2 MCAT Organic Chemistry
When the alkyl radical then reacts with a molecule of molecular halogen in the next step of the reaction, the carbon-halogen bond can form on either side of the plane defined by the s p2 hybridized atom. This leads to racemization at carbon if its substitution is unsymmetrical, since bond formation can occur to an equal extent from either side of the unhybridized p orbital.
Racemization of an Unsymmetrical Alkyl Radical
2 o alky 1 radical with 3 different groups attached to the carbon radical
Bond formed from top lobe of p orbital
Bond formed from bottom lobe of p orbital
Stability of Alkyl Radicals
Br
~CH3
H C2H5
(S)-2-bromobutane
H fH3
yc2Hs
Br
(R)-2-bromobutane
+ Brã
A 50150 mixture of enantiomers (a racemic mixture) is formed.
+ Brã
Next we examine the relative stability of carbon alkyl radicals. Free radicals are like carbocations in the sense that they have an unfilled p orbital (one electron for a radical vs. zero electrons for a carbocation). Also like carbocations (see section 2.2.2.2), alkyl substituents on carbon increase the relative stability of the radical.
R_Q ... R
0---..R > R_Q ... R
0---..H > R-~<~ > H_Q ... O'-H H
3° radical 2° radical 1° radical methyl radical
Selectivity
The varying stabilities of alkyl radicals have a profound effect on the selectivity of these reactions. First we will look at the free radical chlorination of 2-methylpropane.
CH3 CH3 CH3
H3C+CH3 H3C+CH2Cl
+ H 3 C~CH 3
hv
H Cl
2-methylpropane 63% 37%
1-chloro-2-methylpropane 2-chloro-2-methylpropane
Upon inspection of 2-methylpropane we see that there are two distinct types of hydrogens, which we shall refer to as A and lB. The nine A hydrogens are 1°, while the lone JB hydrogen is 3°. If the product distribution were determined solely by statistics and the abundance of each type of hydrogen, the products of the reaction would be formed in the ratio 9:1 (= A:lB). However, as you can see, this is clearly not the case; 63/37 is not equal to 9/1.
A
CH3 CH3 CH3
H3C+CH3 H3C+CH2Cl
+ H3C~CH3
A hv
H A Cl
1B
2-methy lpropane 63% 37%
1-chloro-2-methylpropane 2-chloro-2-me thy I propane
This is because not all of the C-H bonds are of equal reactivity. Since there is much more (than we would expect based on statistics) of the product derived from reaction at position B, we can infer that position 1B is more reactive than position A is. In order to calculate the selectivity of the different positions, one needs to factor out the number of reactive sites. This is shown below:
1 . . reactivity
se ecttvtty = - - - ' - - - -
# of sites available to react
selectivity of 3° (reactivity at 3°)/(# of 3° sites) 37/1 37 5.3 selectivity of 1° (reactivity at 1 °)/(# of 1° sites) 63/9 7
422 Chapter 2 MCAT Organic Chemistry
We now consider the corresponding free radical bromination of 2-methylpropane. The bromine radical is less reactive than the corresponding chlorine radical. The lower reactivity of the bromine radical results in a much higher selectivity (3° > 2° > 1 °) in the bromination reaction compared to the corresponding chlorination reaction. The reason for this is that the lower energy bromine radical reaches the transition state for abstraction of a hydrogen from the alkane to produce the intermediate alky 1 radical much later than the higher energy chlorine radical. This higher degree of selectivity has been quantified and is on the order of 1640 : 82 : 1 (see the calcu-lations below).
A
Bromination of alkanes is much more selective than chlorination:
Bromination:
Chlorination:
A CH3
H3C+CH3
H A
B\
hv
1640.0 5.3
>
-reaction rpte- R2CH2 (2°)
CH3 82.0
3.9
H3C+CH2Br
1
CH3 + H3C~CH3 Br
2-methylpropane 1-bromo-2-methylpropane 2-bromo-2-methylpropane
1 -bromo - 2 methylpropane species ratio = _ _ _ _ _ _ ...:...;;;.___;::.____
2 methylpropane
9 0.5% and
9 + 1640
# 1° H reactivity of 1° H
- - X - - _ . . : ; _ _ _ _
# 3° H reactivity of 3° H 1640
9+ 1640 99.5%
9 1 9
- X =
1 1640 1640
The reason for the lower selectivity in the chlorine case is straightforward. The chlorination of an alkane is more exothermic than the corresponding bromination reaction. In the bromine case, only one of the two propagation steps is exothermic (the other is endothermic). For this reason, bromination is slower and more selective than chlorination.
Enthalpies for Radical Halogenation of Methane: CH4 + X2 ~ CH3X + HX X
F Cl Br I
Propagation Steps of Radical Bromination of Methane:
Step
CH4 + Br-~ CH3• + HBr CH3• + Br2 ~ CH3Br + Br-
!ili0 (kcallmol) -102.8
-24.7 -7.3 +12.7
!ili0 (kcal/mol) +18 -25
It is predicted from the enthalpy values in the table above that fluorine should be a very unselective reagent. This is, in fact, experimentally observed.
Example4. Predict the organic product(s) from the following free-radical halogenation reactions. For the brominations, determine the major and minor products.
Cl2 Br2
hv hv
(c)~
hv
(d)~
hv
(e)~
hv
424 Chapter 2 MCAT Organic Chemistry
Solution.
+
cis and trans isomers cis and trans isomers
Cl
cis and trans
major minor cis and trans
Br
(c) ~Cl
+ ~ Cl
(d)
~+ Br ~r
major minor minor minor
(e) Cl
+ +
~ + ~Cl
2.2 Alcohols
Alcohols are a very useful class of chemicals, not only for their physical properties, but also for their diverse reactivity. We begin with the basic structural features and common reactivity of alcohols.
2.2.1 Structure, Hydrogen Bonding, and Acidity in Alcohols and Phenols 2.2.1.1 Hydrogen Bonding in Alcohols
In order to examine the effect of hydrogen bonding in alcohols, let us examine two molecules that are isomers of one another, n-butanol and diethyl ether. Both have the same molecular formula (C4H100), yet there is a dramatic difference in their boiling points (ll7°C for n-butanol vs. 34.6°C for diethyl ether). This difference arises from the ability of n-butanol to form intermolecular hydrogen bonds, while diethyl ether cannot. Alcohols form intermolecular hydrogen bonds because they have hydroxyl ( -OH) groups. This results from a strong dipole in which the hydroxyl group's proton acquires a substantial partial positive charge (8+) and the oxygen acquires a substantial partial negative charge (8-). The partial positive hydrogen can interact electrostatically with a non-bonding pair of electrons on a nearby oxygen, resulting in a hydrogen bond. On the other hand, diethyl ether has an oxygen atom with non-bonding electrons but all hydrogen atoms are bound to carbons. Since carbon and hydrogen have similar electronegativity values, the bond is not very polarized, and these hydrogens cannot participate in hydrogen bonding. It is important to remember that a hydrogen bond is not a covalent bond;
in this case it is an intermolecular interaction, but it can also be an intramolecular interaction.
Intermolecular hydrogen bonding between molecules of n-butanol.
molecular weight= 74 b.p. = ll7°C
Intermolecular hydrogen bonding is not possible between molecules of diethyl ether
molecular weight = 7 4 b.p. = 34.6°C
426 Chapter 2 MCA T Organic Chemistry
Example 5. For each of the following pairs, predict which molecule has the higher
OoH or OBr
(b) 0 or ()OH
(c) ~OH 0
or ~O_...CH3
0
(d) OH
~ OH or ;; OH
or o-CH3
Solution.
(a) (b)
()OH (c) ~OH
OoH 0
(d) OH (e)
o-Br
~ (primarily because of its greater mass)
OH
The hydrogen bonding pattern in phenols provides insight into intermolecular vs.
intramolecular hydrogen bonding. Let's consider the two isomers, para-nitrophenol (also called 4-nitrophenol) and ortho-nitrophenol (also called 2-nitrophenol). (For more on the nomenclature of aromatic compounds, see section 3 .2.2.) First, examine the hydrogen bonding pattern in para-
nitrophenol. Notice that hydrogen bonding can occur with both the nitro and the hydroxyl groups in this molecule and that the bonding is exclusively intermolecular. That is, all hydrogen bonding takes place between individual molecules of para-nitrophenol. These hydrogen bonding interactions hold molecules of para-nitrophenol together and increase their boiling and melting points. Now, examine the hydrogen bonding pattern in ortho-nitrophenol. Notice that for this molecule, the nitro group and the hydroxyl group are in close proximity so that intramolecular hydrogen bonding can occur between the hydrogen of the hydroxyl group and a lone pair of electrons on the nitro group on the same molecule. These intramolecular hydrogen bonding interactions decrease the an1ount of intermolecular hydrogen bonding interactions that can occur between n1olecules thereby decreasing the melting and boiling points of ortho-nitrophenol , relative to para-nitrophenol.
Intermolecular: o+ •• 8
H ãoã
I ãão// '\\ (±) / ••
para-nitrophenol o-:o: IIIIIIH -0 'I ~ N
Q Nđ 0+ 8~ - \)ã .. ~~e
. ~ ... .. ããa'\\ đ ,.o .•
ão .. .O.mwH-< 0 . . .. .'1 ~ N ~
e <>~~-~~ "oãã
\-~~\.;t (,.'." \ ••
~YJ~~'' ~0~
. ~\v;~
'---ãã \;
2.2.1.2 Acidity of Alcohols
Intramolecular:
ortho-nitrophenol
The acidity of a compound is determined by the ease with which it can lose a proton (H+).
Alcohols are a relatively acidic functional group for the same reason that they engage in hydrogen bonding; the large difference in electronegativity between oxygen and hydrogen. If an alcohol like methanol is deprotonated, an alkoxide ion is formed. The alkoxide is a relatively stable species since the negative charge that results from this reaction is located on the very electronegative oxygen atom.
H3C~H ~ H3C~:8 + H@
Methanol Methoxide Proton ion