Hanadi SLEIMANCHEM 212 Chapter 1
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I. Lewis Structures
3
• How to write Lewis structures
3
II. VSEPR: The Shape of Molecules
5
III. Bond Polarity & MoleculE Polarity
7
IV. Orbital Theory 8
• Atomic Orbitals 8
• Molecular Orbitals 9
• Hybridization 12
• Summary 14
• Applications 15
V. Acids and Bases 17
• Trends 18
• Lewis Definition of Acids and Bases 20
VI. Exercises 20
VII. Solutions 22
VIII. Suggested Problems from Solomons, 9th edition
26
I. Lewis Structures G.N. Lewis: Theory of Bonding based on the octet rule: atoms of C, N, O, F attain stable configurations when they have 8 electrons in their outer = valence shell. How to write Lewis structures Step 1 Count the number of valence electrons available; do not forget to add 1 e- for each (-) charge, and subtract 1 e- for each (+) charge on the molecule. Examples CF4 : C = 4 valence e- ; F = 7 valence e- ⇒ Total valence electrons = -2 e
-4 e
4 1 7 4 3× + × =
1 2× + × + =NO3
- : N = 5 valence e- ; O = 6 valence e- ; (-) = 1 valence e- ⇒Total valence electrons = 5 1 6 3 Step 2 Connect the bonded atoms by a line, representing 2 shared electrons.
C
F
F
F
F
N
O
O
O Step 3 Count the number of shared electrons, and subtract this from the total number of valence electrons; this gives the number of electrons to be added to each atom to complete the structure. Examples: CF4 = 32 – 8 = 24 NO3
- = 24 – 6 = 18 Step 4 Add electrons in pairs to each atom to complete its octet (add 2 electrons for Hydrogen). This is not always possible for all atoms.
C
F
F F
F
N
O
O
O
In CF4 all atoms have their octet, whereas in NO3
-, the Nitrogen atom has only 6 e-. Step 5 If one or more atoms have fewer than 8 electrons, form double or triple bonds to complete their octet.
N
O
O
O
An electron pair on Oxygen is shared with Nitrogen to form a double bond.
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Step 6 Assigning Formal Charges to atoms; by definition: Formal Charge = # (valence e-) - # (e- assigned to atom) Formal Charge = # (valence e-) – [# (bonds) + #(unpaired e-)] Examples:
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C
F
F F
F
F = 7 - (1+6) = 0C = 4 - (4 + 0) = 0
N
O
O
O
O = 6 - (1+6) = -1
O = 6 - (2+4) = 0
O = 6 - (1+6) = -1
N = 5 - (4+0) = +1 Thus we write:
N
O
O
O
Direct Application: Write the Lewis structures of NH4
+ , CO32- , C2H4 , CH3NO2 , CO2 , C2H5Cl.
Solution:
N
H
H H
H
CO
O
O C C
H
HH
H
C
H
H N
H O
O
CO O C C
H
H
H
H
H
Cl
II. The Shape of Molecules: VSEPR Method The Lewis structure of a molecule gives no information on the shape of this molecule. VSEPR stands for Valence-Shell Electron Pair Repulsion; the VSEPR theory is a model which helps to predict the geometry of molecules; its premise is that electron pairs on an atom will tend to be as far apart from each other as possible. Coordination Number CN = electron pairs on an atom = bonding pairs + lone pairs. Note that double and triple bonds are counted as 1 “bonding pair”. Basic shapes relevant to Organic Molecules:
Coordination Number CN Shape Bond Angles CN = 4 Tetrahedral 109.5 °CN = 3 Trigonal Planar 120 °CN = 2 Linear 180 °
Examples: In Methane CH4 , the central Carbon has CN = 4 (4 bonding pairs and 0 lone pairs); thus the molecule has a Tetrahedral geometry.
C
H
HH
H
109.5°
Dash-Wedge Model Ball-and-Stick Model Space-Filling Model
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In ammonia NH3 , the central nitrogen has CN = 4 electron pairs(3 bonding pairs, 1 lone pair). The basic geometry is tetrahedral, however the lone pair exercises more repulsion than a bond pair; therefore bond angles are less than 109 (they are 107 ). The shape of the molecule, described according to the position of its atoms, is Trigonal Pyramidal.
.5 NH
HH
107°
OH
H
In water H2O, the central oxygen has CN = 4 electron pairs (2 bonding pairs, 2 lone pairs). The basic geometry is Tetrahedral, however the bond angle is less than 109.5 (it is 10 ). 5The shape of the molecule is Bent.
105°
In Boron trifluoride, BF3 , central boron has CN = 3 bonding electron pairs (note that B has only 6 e-). The geometry is Trigonal Planar, with bond angles 12 . 0
B F
F
F 120°
In Carbon dioxide, CO2 , central Carbon has CN = 2 “bonding pairs”. C OOThe geometry is Linear, with bond angles of 18 . 0 180° Direct Application: Determine the shape of NH4
+ , C2H4 , C2H2 , CO32- , NO2
+. Solution:
C C
H
HH
H
C O
N
H
H
H
H
+
Tetrahedral, CN = 4 Trigonal Planar, CN = 3
C C HH
Linear, CN = 2
O
O
2-
Trigonal Planar, CN = 3
N OO+
Linear, CN = 2
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III. Bond Polarity and Molecule Polarity Electronegativity measures the ability of a bonded atom to attract the shared electrons in a covalent bond. In the periodic table, electronegativity increases from left to right, and decreases down a group.
The red arrows indicate an increase in electronegativity. Thus we have the following ranking of electronegativities: Li < Be < B < C < N < O < F and I < Br < Cl < F When 2 atoms of different electronegativities form a covalent bond, the bonding electrons are unequally distributed. The more electronegative atom draws these electrons closer to it, and thus acquires a partial negative charge (-δ); the less electronegative atom acquires a partial positive charge of equal magnitude (+δ): an electrical dipole μ is thus created. This covalent bond is termed polar. A molecule is polar if the resultant of its bond dipoles is non-zero. Molecule polarity depends on: 1. The individual bond polarities 2. The geometry of the molecule 3. The lone pairs (large contribution to dipole moment) Examples: CCl4 is unpolar, μ = 0
C
Cl
ClCl
Cl
Similarly, perfectly trigonal planar (e.g. BF3) and linear (e.g. CO2) molecules are unpolar. Consider NH3 vs. NF3 :
NH
H
Hvs. N
FF
F
The net dipole moment in NH3 is larger than in NF3.
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IV. Orbital Theory Although the Lewis and VSEPR models can be used for many applications, they do not explain what a bond really is. This is because electrons do not just behave as particles localized between nuclei. A large number of experiments show that the electrons also behave like waves. Atomic Orbitals
The wave-particle duality of matter was expressed mathematically by Louis de Broglie as
h hp mv
λ = =
where λ = wavelength of the particle, m = mass of the particle, v = velocity of the particle, and h = Planck’s constant. Note that mv = momentum of the particle = p. Basically, the wave nature of a particle (expressed by λ) is important only when the particle is very small, e.g. the electron. The Heisenberg Uncertainty Principle states that we cannot simultaneously know the position and the momentum of a particle with absolute accuracy. Thus, when the energy (or momentum) of the electron is known with high accuracy, we cannot be certain of the position of this electron; we can only define regions of space where this electron is most likely to be found, the orbitals. The Schrodinger equation is a mathematical expression which describes the motion of an electron, the latter treated as a wave. Its solutions are called wavefunctions (Ψ). Each solution defines a certain region of space where the electron is most likely to be found, called orbital, and each of these orbitals is associated with a certain energy: (Ψ1,E1), (Ψ2,E2), (Ψ3,E3) and so on. An electron in an atom is only allowed these energies (E1, E2, E3 etc.). The energy of an atom is therefore quantized (hence the name quantum mechanics). The square of the wavefunction, Ψ2, represents the probability of finding an electron in a certain region of space, or electron density. The 3D plot of Ψ2 generates the shape of atomic orbitals. The orbitals relevant to Carbon are the 1s, 2s, and 2p orbitals. The 1s and 2s orbitals are spherical in shape. The 2s orbital contains a nodal surface – area where Ψ = 0, and the probability of finding the 2s electron is Ψ2 = 0. A node defines a change in sign of the wavefunction Ψ. There are three 2p orbitals, 2px, 2py, and 2pz. Each is dumbbell-shaped, and possesses a nodal plane perpendicular to its direction (e.g. the 2px node is the yz plane). The sign of the wavefunction is positive in one lobe and negative in the other lobe of the orbital; the energies of these orbitals are E(1s) < E(2s) < E(2p); the three 2p orbitals have the same energy, thus they are called degenerate.
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z
y
x
z
y
x
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1s
+ +
-
+
Node2s
z
y
x
+-
z
y
x
+
-
z
y
x
+-
2py 2pz2px Now that we have the regions of space where the electrons are most likely found in the atom, we need some rules to place the electrons in these orbitals: 1. The Pauli Exclusion Principle: a maximum of 2 electrons can occupy one orbital, and these electrons must have paired spins. 2. The Aufbau Principle: orbitals are filled with electrons in order of increasing energy. 3. Hund’s Rule: when electrons are added to degenerate orbitals, one electron is added to each orbital with their spins unpaired, until each degenerate orbital contains 1 electron. Then a second electron – paired with the first – is added to each orbital. Example: The Carbon electronic configuration is 6C : 1s2 2s2 2p2 (or 1s2 2s2 2px
12py1)
6C
1s
2s
2p
Molecular Orbitals Covalent bonding is very important for Carbon compounds: it occurs when 2 electrons are shared between 2 atoms. In orbital terms, covalent bonding results from the overlap of 2 AO’s, when the 2 atoms approach each other. These AO’s combine to form Molecular Orbitals, regions of space where you are most likely to find the electrons of a molecule. MO’s are generated as linear combinations of AO’s
(LCAO’s), that is, from the addition and subtraction of the AO’s wavefunctions. The number of resulting MO’s is always equal to the number of starting AO’s. Consider H2 : Two 1s orbitals on each atom (HA and HB) combine to form Two molecular orbitals: (1sA + 1sB) and (1sA – 1sB).
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+ ++
1sA
=
1sB
+
Overlap region: Reinforcement of ΨRegion of highest electron density
σ = 1sA + 1sBBonding MO
σ is the + (additive) combination of the 1s AO’s. In the overlap region, the two 1s wavefunctions have the same phase sign, thus the overall Ψ as well as Ψ2 is larger. This means that in the σ MO, the electron density is very high, that is, the electrons are most likely found in the region between the nuclei. The σ electrons are thus attracted to the 2 nuclei, and thus act as the glue which holds the atoms together. σ is a Bonding MO; electrons in this orbital are stable, they have lower energy. A molecular orbital is called σ if it is cylindrically symmetrical around the axis of the nuclei.
+ --
1sA
=
Overlap region: Interference, Node between NucleiΨ= 0
1sBσ∗ = 1sA - 1sBAntibonding MO
++
σ* is the – (subtractive) combination of the 1s AO’s: the overlap region, the two 1s wavefunctions have opposite phase signs, thus the overall Ψ, as well Ψ2 cancel out. In the σ* MO, the electron density is zero, that is, there is a nodal plane in the region between the nuclei. Since the attraction of these electrons to the nuclei is small, the dominant forces are the repulsion between the electrons and between the nuclei. Electron, which reside in the σ* MO tend to pull the atoms apart, and thus cancel the bonding effects of electrons in the σ MO; σ* is an Antibonding Molecular Orbital; electrons in this orbital are unstable, they have higher energy. A node between the nuclei defines an Antibonding MO. The molecular energy diagram for H2 then is:
HA HBH2
1sA 1sB
σ*
σ
The H2 electronic configuration is σ2 (σ*)0
[# (Bonding electrons) - # (Antibonding electrons)]Number of bonds in a molecule = 2
H2 has 2 - 0 12
= single σ bond.
Consider F2 : The electronic configuration of F is 1s2 2s2 2px
2 2py2 2pz
1 There are many possible ways to generate MO’s from two 2pz AO’s: 1. AO’s pointing towards each other: bonding is σ, cylindrically symmetrical around the internuclear axis.
2. The p orbitals are parallel: bonding is π. A π orbital possesses a nodal plane which passes through the axis of the nuclei. There is less overlap in π bonding than in σ bonding, thus a π bond is weaker than a σ bond.
3. The p orbitals are at an angle from each other; there is less overlap than either π or σ. This mode of bonding does not usually occur. 4. The π orbitals are perpendicular to each other: one orbital is thus in the nodal plane of the other; therefore no overlap is possible. Perpendicular orbitals cannot bond to each other.
In the F2 molecule, the bonding mode is thus σ; the configuration of F2 is σ2 (σ*)0 A single bond between 2 atoms is a σ bond.
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Hybridization Consider the molecule CH4 (Methane): The electronic configuration of C is 1s2 2s2 2px
1 2py1; according to
this configuration, Carbon has 2 unpaired electrons, and should form 2 bonds with 2 Hydrogen atoms. In Methane, however, Carbon is bonded to 4 H’s; thus 4 unpaired electrons are needed.
1s
2s
2p 96 kcal/mol+ 96 kcal/molPromote one 2s e- to 2p 1s
2s
2p
Ground StateLowest Energy
Excited State
The 4 bonds that can be formed with the four H’s are: one σ(C2s , H1s) and three σ(C2p , H1s). However, the four bonds of methane are experimentally equivalent. Therefore, the 2s and three 2p orbitals are mixed, or Hybridized together, yielding 4 equivalent hybrid orbitals. One s + Three p = Four sp3 hybrid orbitals. Number of Hybrid AO’s = Number of starting AO’s.
1s
2s
2p
Excited State
No Energy Cost
1s
sp3 hybridized Carbon
sp3
Shape of the sp3 orbitals:
+
s p Hybrid Atomic Orbital
≡+ - + +-- + +
Note that sp3 orbitals have their electron density concentrated in one direction. They are capable of better overlap than either s or p orbitals. Hybrid orbitals are more directional, and therefore capable of stronger bonding than s or p orbitals. The four sp3 orbitals, each containing 1 e-, orient themselves as far apart from each other as possible: along the corners of a tetrahedron.
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C
The Csp3 hybrid orbitals and the H1s orbitals overlap to form σ bonds.
C
H
H
H
H C
H
H
H
H
σ(Csp3, H1s)
The C-H bond energy is 104 kcal/mol, that is, each σ(C-H) lowers the energy by 104 kcal/mol. Consider CH3
+, the Methyl Carbocation: C has 3 electrons (instead of 4) in its valence shell. The three C-H bonds are equivalent; thus 3 hybrid orbitals, and 3 unpaired electrons are needed. Promote one 2s electron to the 2p level, then combine s + px + py, to give three sp2 hybrid orbitals, in the xy plane; this leaves the pz orbital, which is perpendicular to the xy plane.
1s
2s
2p
Promote one 2s e- to 2p 1s
2s
2p
Ground StateLowest Energy
Excited State
Hybridize
s + px + py1s
sp2
2pz
sp2 hybridized Carbon
To be as far apart from each other as possible, the three sp2 orbitals are oriented at the corners of an equilateral triangle; this is a trigonal planar geometry, with 12 angles between the orbitals. Each orbital contains an unpaired electron, and will σ bond to a Hydrogen 1s orbital. The pz orbital is perpendicular to these, and empty in the case of CH3
+. Note that this cation is unstable, since the Carbon does not possess an octet of electrons.
0
xy
pz
H
H
H C H
H
H
+
σ(Csp2, H1s)
C
Consider BeH2: Be has 2 valence electrons. Lewis Structure: H – Be – H; two equivalent Be – H bonds; thus two unpaired electrons in two equivalent hybrid orbitals are needed:
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1s
2s
2p
Promote one 2s e- to 2p 1s
2s
2pHybridize
s + py1s
sp
2px, 2pz
Be Promote 1 e- from 2s to 2p, then combine s + py to give two sp hybrid orbitals, along the y-axis; this leaves the px and pz orbitals, which are perpendicular to the axis of the sp hybrid orbitals (y-axis), and to each other. To be as far apart from each other as possible, the two sp orbitals are oriented in a linear geometry, with 18 angles. Each orbital contains an unpaired electron, and will σ bond to a H1s orbital. The px and pz orbitals do not contain electrons.
0
z
y
x
Be
H HBe
HH
σ(Be sp, H1s)
Summary To describe bonding in a molecule: Step 1 Draw Lewis structure of molecule. Step 2 Find the coordination number of the atom (same as VSEPR). Step 3 If CN = 4, hybridization is sp3, the atom has four sp3 hybrid orbitals, in a tetrahedral geometry, with
angles. 109.5If CN = 3, hybridization is sp2, the atom has three sp2 hybrid orbitals, in a trigonal planar geometry, with
angles, as well as a p orbital, perpendicular to the plane of the sp2 orbital. 120If CN = 2, hybridization is sp, the atom has two sp hybrid orbitals, in a linear geometry, with 18 angles, as well as two p orbitals perpendicular to the sp axis.
0
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Applications Ethane C2H6 Lewis structure:
CH
H
H
C
H
H
HCN = 4 Each Carbon is bonded to 4 atoms, thus CN = 4 and hybridization is sp3. Therefore 4 sp3 hybrid orbitals oriented along the corners of a tetrahedron.
C C
sp3
One of these Csp3 orbitals undergoes σ bonding with another Csp3; the other six sp3 orbitals undergo σ bonding with the H1s orbitals.
C C
HH
H H
HH
σ(Csp3, H1s)
σ(Csp3, Csp3) Since the σ bonding is cylindrically symmetrical around the axis of the nuclei, rotation around this bond will not weaken it or break it. Therefore this rotation around the σ bond requires relatively little energy (less than 10 kcal/mol) and occurs readily at room temperature.
C C
HH
H H
HH
C
H
H
HC
H
HH
Ethene C2H4 Lewis structure:
C C
H
H
H
HCN = 3, sp2
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Thus the orbital picture is:
C C
π
σ
π
Two Csp2 orbitals undergo σ bonding to each other; the other 4 Csp2 orbitals undergo σ bonding to the four H1s orbitals. This leaves the 2p orbitals on the Carbons, which can be oriented parallel to each other, and can thus undergo π bonding.
C C
π(Cp, Cp)
σ(Csp2, Csp2)
H
HH
H
σ(Csp2, H1s)
C CH H
H H
The C = C double bond is made up of 1 σ bond, and 1 π bond. It is stronger than a C – C single bond and more difficult to break. Rotation around C = C: Twisting around C = C would not affect the σ bond, which is cylindrically symmetrical around the axis of nuclei. However, a full rotation around C = C causes the p orbitals to become perpendicular, and thus to lose their overlap. The π bond would thus be broken, and this process would require about 70 kcal/mol. This is a large energy barrier, and rotation around the C = C double bond does not occur at room temperature.
C C
H H
Cl ClC C
H Cl
Cl H
C C
H Cl
Cl H
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Ethyne C2H2
Lewis structure: C CH H
CN = 2, sp
The Carbons are sp-hybridized, they each have two sp hybrid orbitals, 18 apart, as well as two perpendicular p orbitals. Two Csp orbitals will σ bond to each other, and the other two sp orbitals will σ bond to the H1s orbitals. This leaves two p orbitals on each Carbon. These will form two mutually perpendicular π bonds. The C≡C triple bond is thus made up of 1 σ bond and 2 π bonds. The bond energies (strengths) are: E(C≡C) > E(C=C) > E(C-C).
0
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C C
π(Cp, Cp)
σ(Csp2, Csp2)
H
H
π(Cp, Cp)σ(Csp, H1s)
V. Acids and Bases Bronsted Definition of Acids and Bases Acid = Proton (H+) Donor Base = Proton (H+) Acceptor
HSO4- H3O+
Stronger Base Weaker Acid(Conjugate Acid)
H2O
Stronger Acid Weaker Base(Conjugate Base)
+
Equilibrium favors this side
H2SO4 +
The acid-base equilibrium favors the formation of the weaker acid/weaker base pair. An acid HA is strong if: 1. It loses a proton readily: this occurs if the electron density is pulled towards A (δ+H-Aδ-, when δ is large). 2. A- can accommodate the extra electron density after H+ dissociates; it can stabilize the negative charge. A base B- is strong if it readily accepts H+; that is, B- has a large amount of negative charge. The stronger the acid, the weaker its conjugate base; the weaker the acid, the stronger its conjugate base.
Trends 1. Electronegativity Within a period, acidity increases as electronegativity increases i.e. from left to right. Example:
C
H
H H
H
N HH
H
OH H H F< < <
-H+
Weakest Acid Strongest Acid
C
H
H
H
N
H HOH F
< < <
Strongest Base Weakest Base
2. Polarizability Within a group, acidity increases as size increases i.e. from top to bottom. Example: (Weakest Acid) H-F < H-Cl < H-Br < H-I (Strongest Acid) I- has its valence electrons in large orbitals, and the negative charge is diffuse; it is a Soft, or Polarizable ion. On the other hand, F- has its valence electrons in smaller orbitals and the charge is concentrated; F- is thus a Hard ion. The strongest acid-base interactions are the hard-hard or soft-soft interactions; interactions between soft and hard ions are the weakest. The proton H+ is a hard ion; it interacts better with F- than I-. Therefore, it is not held as tightly by the soft ion I-, and it is easy to dissociate from it. 3. Inductive effects Let us focus on the difference in acidity between the following 2 molecules: CH4 and CHF3. In the latter, the F being more electronegative than C, it pulls electrons towards it (cf. bond polarity). For this reason, a positive partial charge appears on C. Normally, as in CH4, C does not pull electrons of the C – H bond towards it (C and H have comparable electronegativities). On the other hand, in CHF3 there is an Induced Dipole that is due to the neighboring atoms (in our case, the three F’s) and not the atom directly bonded to the H being considered – hence it’s called “induced”.
C
F
FH
F
+δ
C
F
FH
F
+δ
In other words, the F’s are now pulling the C – F bond’s electrons towards them (recall that F is the most electronegative element of the periodic table) such that the C now, in turn, is obliged to pull the C – H bond’s electrons towards it.
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Therefore, a positive partial charge appears on the H in CHF3 which makes the dissociation into H+ + CF3-
easier: CHF3 is stronger an acid than CH4.
C
F
FH
F
+δC
H
>More acidicH H
H Furthermore, if we confront Acetic acid and Ethanol, it becomes obvious that the former is more acidic because of the inductive effects due to the Carbonyl C = O group. Note that in these molecules, it is the H attached to the O that we consider, since it’s the most acidic (O being more electronegative than C).
+δH3C C
O
O H H3C C O H
H
H
>More acidic
Now consider the molecule CH3 – CHF2 which has two types of Hydrogens, as portrayed below. The blue H is more acidic since it is closer to the induction site (inductive effects decrease with distance).
More Acidic
C
F
H
F
C
H
H
H Finally, CH3OH vs. H2O: note that Alkyl group (e.g. Methyl – CH3) release electrons. So the O in Methanol has enough electron density and frees the H (partially) from its electronegative effect. It becomes obvious that Water is more acidic in this couple. Release of e- density
>More acidic
H O HH3C O H 4. Hybridization The more s character in the orbitals of A-, the stronger the acid HA.
Orbital s character sp3 25% sp2 33% sp 50%
Since s orbitals are closer to the nucleus than p orbitals, the more s character in an orbital, the more tightly the electrons are held by the atom A, the stronger the acid HA. Thus, the terminal protons of alkynes are fairly acidic. In fact, acetylene is a stronger acid than NH3.
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C
H
H C
H
< <
Weakest AcidStrongest Acid H
H H
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C CH H C C H
H HHsp sp2 sp3
Stronger Acid < <Weakest Acid
C CH HH2O NH3 Lewis Definition of Acids and Bases (encompasses the Bronsted-Lowry definition) Acid = Electron pair acceptor ; Base = Electron pair donor. Thus, a Lewis acid lacks a full octet of electrons, e.g. BF3 and AlCl3. A Lewis base has a lone pair of electrons, e.g. NH3 and H2O. A Lewis acid reacts with a Lewis base to form an adduct, where a new coordinate covalent bond results from the sharing of the lone pair with the acid.
BF
F
F
N
H
H
H
Lewis Acid Lewis Base
B N
F
F
F
H
H
H
Adduct VI. Exercises Exercise 1 Given the following Lewis structures for CH2N2
H2C N N H2C N N H2C N N (a) Complete with the proper Formal Charges (b) Which structure is the least stable? (c) Which structure is the most stable? (d) What is the Geometry and Hybridization of the central N in each structure? Exercise 2 What are the Hybridization and Geometry of the atoms in each of the following structures? (a) CH3(b) CH3(c) CH2(d)
N
N
CH3
(Nicotine)
Exercise 3 Show the products of the following reactions; if no reaction takes place, write ‘No Reaction’.
(a) HF + HO
(b) HF + I
(c) CH3OH + NaNH2
(d) CH3OH + LiH
(e) CH3OH + H2SO4 Exercise 4 Arrange the following in the order of increasing basicity:
H3C CH2 H C C H2C CH Exercise 5 Show the products of the following reactions; if no reaction takes place, write ‘No Reaction’.
C CH H +(a) NaOH
(b)H3C C
O
OH + CH3ONa
(c)N
H
+N
H
(d)
(e)
(f)
(CH3)3N + BF3
CF3OCH2CH2OCH3 + BCl31 mol 1 mol
CH3CH2CH3 + F3C C
O
O - 21 -
VII. Solutions Exercise 1 (a) Formal Charges are shown in red:
H2C N N
H2C N N H2C N N In detail:
H2C N N H2C N N H2C N N
FC = 4 - (3 + 2) = -1
FC = 5 - (4 + 0) = +1 FC = 5 - (2 + 4) = -1
FC = 4 - (3 + 0) = +1
(b) The following structure is the least stable – C has 6 e- instead of an octet:
H2C N N Violating the Octet rule is the most serious source of instability. (c) The most stable structure is:
H2C N N Note that in the C carries the (-) charge, whereas in it is the N that has a (-) charge. Being less electronegative than N, the C does not like to carry a negative charge.
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(d)
H2C N N H2C N N H2C N N
CN = 2 CN = 2 CN = 3Linear Linear Trigonal Planar sp sp sp2
CH
Exercise 2 (a)
H
H C
H
HH
σ(Csp3, H1s)
CN = 4 sp3 hybridized Tetrahedral
H2C N N H2C N N
(b)
CH
H
H
CN = 3sp2 hybridized Trigonal Planar
C
H
H
H
σ(Csp2, H1s)
Empty p orbital (c)
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CN = 3sp2 hybridized Trigonal Planar
CH
H
σ(Csp2, H1s)
Empty p orbital
C
HH
sp2
Note that the 2 e- were placed in the sp2 orbital, rather than the 2p orbital, because it has lower energy i.e. E(sp2) < E(2p). (d)
N
N
CH3
sp2
sp2
sp2
H H
H
H
H H
H
HH
H
H
sp3
sp3
sp3
sp3
Exercise 3 (a)
H FHO HO H F++ Note that HF is a stronger acid than H2O, F being more electronegative than O. (b) No Reaction because HI is a stronger acid than HF, so the equilibrium lies towards the side of the reactants.
H FI I H F++ (c)
- 24 -
++NH2 NH3 H3C OH CH3O
Note that CH3OH is a stronger acid than NH3 since O is more electronegative than N (so the reaction does occur). Also note that NaNH2 ≡ (Na+, NH2
-); it is important to point out that alkali (Li, Na, K) and alkaline earthes (Ca, Mg, Ba) are salts; the cations behave like spectator ions and do not take part in reactions.
LiA Li A and CaA Ca2+ A2-
(d)
++H H2 H3C OH CH3O
Recall that CH3OH is a stronger acid than H2, since O (in O-H) is more electronegative than H (in H-H). (e)
H3C O H S
O
O
O OHH + S
O
O
O OHH3C OH
H
+
H2SO4 is a very strong acid (by inductive effects), and CH3OH2
+ is weaker. Exercise 4
H3C CH3H2C CH2HC CH > >
Most Acidic
sp sp2 sp3
H3C CH2H2C CHHC C
> >
Most Basic Note that in H-C≡C- the Carbon’s hybridization state is sp, so the base has more s character, and therefore it attracts the e- most.
Exercise 5 (a) No Reaction: H2O is stronger an acid than H-C≡C-H.
C CH H + OH C CH + H2O (b)
H3C C
O
O + CH3OHH3C C
O
O CH3OH Recall that CH3COOH is stronger than CH3OH due to inductive effects. (c)
N
H
+N
H
N
H H
N
+
N(sp2) N(sp3) Stronger Acid Weaker Acid
Choice of the acid reactant: suggested by the (+) charge. The reactant acid has an sp2 hybridized N, so the electrons are more attracted to the N nucleus, which makes it stronger than the product acid. (d) Lewis acid / Lewis base - adduct
N CH3
H3C
H3C
B F
F
F
N
CH3
CH3
H3C B
F
F
F (e) The O next to the CH3 group is stronger as a base than the O next to the CF3 group (by inductive effects).
B Cl
Cl
Cl
O
CH3
F3COH2CH2C B
Cl
Cl
Cl
CF3OCH2CH2OCH3
(f) No Reaction: CF3COOH being a stronger acid than propane (inductive effects).
- 25 -
- 26 -
VIII. The textbook: Solomons, 9th edition Suggested Reading Chapter 1 (except 1.8) Chapter 3 [3.1; 3.7] Suggested Problems Chapter 1 1, 2, 3, 6, 8, 16, 17, 20, 23, 26 Chapter 2 1, 5, 6 Chapter 3 1, 2, 11, 13, 15, 16, 17, 18, 19, 25, 26, 27, 28, 29, 33
1
H H
HHHH
HH
HH H
H
HHH
HH
H H
HH
HH
H
HH
HH
HH
HH
HHH
H
HH
H
HHH
HH
HHH
HH
H
HHH H
H HHH
H H
HHH
H
H
HH
H
H
HH
HH
HH
H
H
HH
HH
HH H H
H HH H H H
H HH H H
HHH
HH
HH
HH
HH
H H
HH
HH
H
H
H
HH
HHH
HH
H
HH
HH
HH
H
HH
H
HH
HHH
HH
HH
HH
HHH
HH
HH
HHH
HH
HHHH
H HH
HH
HH
HH
HH
HH
HH
HHH
H
HH
HH
HH
HH
HHH
H H
HH
HHH
H
HH H
H
HHH
HH
H H
HH
HH
H
HH
HH
HH
HH
HHH
H
HH
H
HHH
HH
HHH
HH
H
HHH H
H HHH
H H
HH
H
H
HH
HH
HH
HH
HH
HH
HHHH
HH
HH
HH
HH
H H
HH
HH
HH
HH
HH
HH
HH
HH
HH
HHHH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
H
H
H
H
H
H HH H
H HH H
H H
H
H
H
H
H
HHH
H
HHHH
H
HH
HH
H
H
H
HH
H H H H H HH H
H HH
H
H
H H
HH
HH
HH
H
H
H
H
H
HH H
H H
H H
H HHH
HH
HH
HH
HH
H
H
H H
H H
H H
H H
H H
HH
H
H
H
HH
H
H
H
H
H
H H
H H
H H
HH
H H
HH
H
H3C
Hanadi SLEIMAN
CHEM 212 Chapter 2
Cl
H
H
2
I. Hydrocarbons
3
• Alkanes 3
• cykloalkanes 3
II. conformational analysis 3
• Non-cyclic alkanes 3
III. cykloalkanes and ring strain 6
• Heats of formation of alkanes 6
• Heats of formation of cykloalkanes 6
• Ring strain 6
• hybridization 7
IV. laboratory sources of alkanes 8
• From alkenes and alkynes 8
• From alkyl halides 8
• Grignard reaction 8
• Corey-house coupling 9
• Wurtz coupling 9
• R-X with Metal and acid 10
V. reactions of alkanes 10
• combustion 10
• halogenation 10
VI. Exercises 19
VII. Solutions 23
VIII. the textbook: Solomons, 9th edition 29
3
I. Hydrocarbons Hydrocarbons are the simplest organic compounds; they contain only C and H.
AlkanesAliphatic Alkenes
HydrocarbonsAlkynes
Aromatic (Benzene compounds)
⎧ ⎧⎪⎪
⎪ ⎨⎨ ⎪
⎩⎪⎪⎩
Alkanes Formula: CnH2n+2
These alkanes form a homologous series: each member differs from the previous by the addition of the same fragment CH2. Nomenclature, Physical Properties: Refer to the lectures. Cycloalkanes Formula: CnH2n Examples: For n = 3, C3H6 Cyclopropane; for n = 4, C4H8 Cyclobutane. Cyclohexane: Refer to the lectures. II. Conformational Analysis Non-cyclic Alkanes Representations of Alkanes:
H
H HH
HH
H
H
H
H
H
H
C C
H
HH
H
H
H
Dash-Wedge Sawhorse Newman Projection Newman Projection: The molecule is viewed along the C-C axis.
4
Ethane: Rotation around the C-C bond (very low energy barrier for rotation around the σ bond). Staggered Conformation Eclipsed Conformation
The Eclipsed conformation is 3 kcal/mol higher in energy than the Staggered conformation, because of the steric repulsion between the C-H electrons. At room temperature, this rotation will occur constantly. To go from Staggered to Eclipsed to Staggered conformations, we have to climb a 3 kcal/mol energy hill, or Torsional Barrier. The ethane molecule will spend most of its time in the Staggered conformation; the Eclipsed conformation has 3 kcal/mol of Torsional Strain.
H
HH
H
HH
HH
H
H
H
H
5
Eclipsed and Staggered ethane are conformational isomers of ethane: isomers that are interconvertible by rotation around single bonds. Consider the rotation around the C2-C3 bond in Butane:
CH3
HH
CH3
HH
CH3H
H
H
H
CH3
+60° +60°
H
CH3H
CH3
HH
CH3CH3
H
H
H
H
+60°
H
HH3C
CH3
HH
CH3H
CH3
H
H
HCH3
HH
CH3
HH
+60°
+60°+60°
Anti Eclipsed Gauche
Fully EclipsedAnti Eclipsed Gauche
The Anti conformation does not have torsional strain, because the H’s are staggered and the methyl groups are far apart. The Gauche conformations are equal in energy; in these, the methyl groups are closer to each other, which gives a steric repulsion (torsional strain) of 0.9 kcal/mol. The Eclipsed conformations are equal in energy; they have two H – CH3 eclipsing interactions (2*1.4 = 2.8 kcal/mol) in addition to one H–H eclipsing interaction (1 kcal/mol); thus Torsional strain = 2.8 kcal/mol + 1 kcal/mol = 3.8 kcal/mol. The Fully Eclipsed conformation has the highest energy, because it possesses the most serious steric interaction, CH3 – CH3 eclipsed (2.5 kcal/mol) in addition to two H–H eclipsed (2*1 = 2 kcal/mol); thus Torsional strain = 2.5 + 2 = 4.5 kcal/mol.
6
III. Cycloalkanes and Ring Strain Heats of formation of alkanes 5 C + 6 H2 ΔHf = -35.1 kcal/mol 6 C + 7 H2 ΔHf = -39.9 kcal/mol In this homologous series, every time a CH2 group is added, ca. -5kcal/mol is added to ΔHf. Heats of formation of cycloalkanes Cyclohexane: ΔHf = -29.5 kcal/mol. For Cyclopentane, the expected ΔHf would be -29.5 – (-5) = -24.5 kcal/mol. However, the experimentally determined ΔHf was -18.5 kcal/mol. Thus cyclopentane is ca. -6 kcal/mol less stable than expected; this energy destabilization is called Ring Strain.
Molecule Expected ΔHf (kcal/mol)
Experimental ΔHf (kcal/mol)
Ring Strain (kcal/mol)
Cyclohexane -29.5 -29.5 0 Cyclopentane -24.5 -18.5 6 Cyclobutane -19.5 +6.8 26
Cyclopropane -14.5 +12.5 27 Ring Strain (a) Angle Strain: difference between optimal and actual C C C− − angle (b) Torsional Strain: Eclipsing and Gauche interactions (c) Steric Strain: Close approach of two large groups
7
Example: Cyclopropane (a) The C atoms are sp3 hybridized, optimal angle is 109.5○; actual angle is close to 60○. Thus angle strain = 109.5 – 60 = 49.5○
C
C C 109.5°Expected
C
C C 60°Experimental
(b) All bonds are eclipsed: Torsional Strain H
H
H
H
H
H
(c) If large groups on cyclopropane – Steric Strain Note that Cyclobutane has less angle strain than Cyclopropane. Hybridization To partially relieve the angle strain, the internal C – C bonds in Cyclopropane will involve C hybrid orbitals with more p character than normal sp3 orbitals. In turn, the C – H bonds will involve C hybrid orbitals with more s character than normal sp3 orbitals (to compensate for the internal bond hybridization change). This is because the optimal C – C – C angle decreases as the p character increases in hybrid orbitals. sp: 50% p; C-C-C angle = 180○
sp2 : 66% p; C-C-C angle = 120○ sp3 : 75% p; C-C-C angle = 109.5○ p (pure p orbitals): angle = 90○
8
IV. Laboratory Sources of Alkanes Form Alkenes and Alkynes A catalyst is necessary for these reactions to occur.
C C
H
H3C CH3
H
H
CH3
H
H
CH3
H
H2
Pd, Pt, or Ni
2-Butene Butane
C C CH3H3C
2-Butyne
2H2
Pd, Pt, or NiH
CH3
H
H
CH3
HButane
From Alkyl Halides Grignard Reaction
C
H
Cl
H
H + Mg C
H
MgCl
H
H
Grignard Reagent This Grignard reagent is an organometallic compound; since Mg (and most other metals) are much less electronegative than C, the Grignard reagent can be thought of as The Grignard reagent RMgX will undergo an acid-base reaction with any acid which is stronger than an alkane R-H. Since R-H is an extremely weak acid, RMgX is a very strong base. Reaction of RMgX with water:
C
H
MgCl
H
HO
H H C
H
H
H
H+ + [HO]- [MgCl]+
Stronger Base Stronger Acid Weaker Acid Weaker Base This reaction is useful for making Deuterated alkane derivatives (D = 2H, H isotope with mass number = 2), by reacting the Grignard reagent with deuterated acids D-A.
C
H
MgCl
H
H
9
Corey-House Coupling of Alkyl Halides with Organocopper compounds This reaction makes R-R’ from R-X and R’-X (where R and R’ are alkyl groups, and X = Cl, Br, I).
R XLi
R Li
Alkyllithium
CuXCuLi
R
R
Lithium dialkyl copper
R' X
R R'
In this sequence, the alkyl halide is reacted with Lithium metal, to give ab alkyllithium reagent. This organometallic compound is a very strong base, and is quite reactive. This compound is reacted with a Copper (I) halide, to give a Lithium dialkyl copper. This organocuprate R2CuLi is coupled to an alkyl halide R’-X to give a new alkane: R-R’. Mechanism of Coupling: The strongly electron-rich Carbon in R2CuLi, with its pair of electrons, attacks the partially positive Carbon in R’X, and substitutes for X.
CuLi
R
R R' X R R'+
δ-
δ+
Note that a primary carbon is bonded to one other C; a secondary carbon is bonded two carbons; a tertiary carbon is bonded to three carbons; a quaternary carbon is bonded to four carbons.
C
H
H Br
H
C
CH3
H Br
H
C
CH3
H3C Br
H
C
CH3
H3C Br
CH3
C
CH3
H3C CH3
CH3Methyl C 1°C 2°C 3°C 4°C
Requirement of the Corey-House reaction: In R’X, R’can be Methyl or 1○ only. In R2CuLi, R can be 1○, 2○, or 3○. Wurtz Coupling This reaction makes R-R from R-X.
R ClNa
R R
[R-Na]
10
The alkylsodium reagent RNa, which is the initial product of this reaction, is so reactive that it couples immediately with the starting R-X to give R-R. This reaction gives a maximum of 50% yield, and is thus of limited use. Reaction of an Alkyl Halide with Metal and Acid
R X + ZnH+
R H + Zn2+ + X -
V. Reactions of Alkanes Combustion (Oxidation)
CH4 + 2 O2Flame
CO2 + 2 H2O + Heat
Exothermic reaction between alkanes and oxygen, initiated by a flame; heat is usually the desired product of this reaction. Halogenation
+ + +C
H
H
H
H + Cl2UV light, 25°C
Or Dark, 450°CC
H
Cl
H
H C
Cl
Cl
H
H C
Cl
Cl
Cl
H C
Cl
Cl
Cl
Cl HCl+
Chloromethane Dichloromethane (Methylene Chloride)
Trichloromethane (Chloroform)
Tetrachloromethane(Carbon Tetrachloride)
(1) This is a substitution reaction: Cl is substituted for H. (2) This is a vigorous reaction: it generates heat, and it is fast. (3) Light or heat is required: without them, no reaction. (4) To get only CH3Cl, excess CH4 is needed. (5) This reaction is slowed down, or inhibited by oxygen O2. (6) Order of reactivity for the halogens: F2 >> Cl2 > Br2 (>>I2). Fluorination is too vigorous to be practical, iodination is very slow. Chlorination and bromination are the only practical halogenations. Reactivity: F2 is more reactive with CH4 than Br2 means that the reaction of CH4 with F2 is faster than the reaction of CH4 with Br2. Mechanism: Detailed, Step-by-Step description of a chemical reaction. A mechanism for this reaction which is consistent with the above observations follows.
+Cl Clhν(Light)
Or Δ (Heat)Cl Cl
Free Radical
Step 1: Initiation
11
(1) This step is a bond homolysis: the bond is broken and the shared electrons are divided equally between atoms. Bond heterolysis is unsymmetrical bond-breaking. (2) Energy is required to break this bond: ΔH = 58 kcal/mol; this is the Cl-Cl bond dissociation energy. (3) This step generates a free radical: a chemical species with an odd electron. Most free radicals are unstable and very reactive. Potential Energy Diagram of this step:
Chain Propagation Steps
Cl C
H
H
HH+ H atom abstraction
Cl H + C
H
H
H
Methyl Free Radical
Step 2
ClC
H
H
H + Cl atom abstraction
+ Cl
Step 3
Cl C
H
H
H
Cl
Chloromethane In step 2, the Cl free radical abstracts a hydrogen atom from methane to form a methyl free radical. In step 3, the methyl free radical abstracts a chlorine atom from Cl2 to form chloromethane, and regenerate a chlorine free radical. This Cl radical undergoes step 2 again, then 3, etc., until all the methane is consumed.
12
Each Cl free radical can undergo thousands of chain propagation steps; thus, only a small concentration of these free radicals is needed. Chain Termination Steps
+Cl Cl
+Cl CH3
CH3H3C +
Cl Cl
Cl CH3
H3C CH3 During the chain propagation, the concentration of free radicals is too low for these to react together. As soon as they are generated, they react with the starting materials (CH4, CH3Cl, CH2Cl2, CHCl3). When these starting materials are consumed, the chain termination steps can occur. Overall (Step 2 + Step 3): CH4 + Cl2 CH3Cl + HCl Inhibition by Oxygen The oxygen molecule, which acts like a diradical, reacts with 3CHi to give a Peroxy free radical. This latter species is less reactive than alkyl free radicals, and each time this occurs, one chain is inhibited. The reaction slows down.
H3C + O O
Diradical
H3C O O
Peroxy Free Radical If CH4 is not in excess, then:
Cl C
H
H
ClH+ H atom abstraction
Cl H + C
H
H
Cl
ClC
H
H
Cl + Cl atom abstraction
+ ClCl C
Cl
H
H
Cl
Dichloromethane The chlorine free radical abstracts a hydrogen atom from CH3Cl, and this chain yields CH2Cl2. Similarly, reaction with CH2Cl2 yields CHCl3, and reaction with CHCl3 yields CCl4.
13
Heat of Reaction – Thermodynamics H CH3 + Cl Cl H3C Cl + H Cl
Break C – H Break Cl – Cl Make C – Cl Make H – Cl + 104 kcal/mol + 58 kcal/mol - 84 kcal/mol - 103 kcal/mol
These values are the homolytic bond dissociation energies. Overall Enthalpy of reaction: ΔH = +104 + 58 - 84 -103 = - 25 kcal/mol is a thermodynamic value. Since ΔH (reaction) < 0 then we are dealing with an Exothermic reaction. Thermodynamics of a reaction A → B: Only depends on the reactant A and the product B. They do not depend the mechanism of the reaction. Energy of Activation – Kinetics Kinetics of a Reaction A → B: Describe how fast the reaction goes, the rate of the reaction. Depend on the mechanism of the reaction. Consider Step 2 Cl + CH3H Cl H + CH3 For this step to occur: (1) The Cl free radical and CH3-H must collide (2) The orientation of the molecules during the collision must be the correct one
Cl C
H
H H
H (3) The collision must be effective: the Cl free radical and CH4 must have enough energy to interact.
Their energy must be at least equal to the energy of activation.
14
As the reactants are brought together, repulsion between their electronic clouds occurs; the potential energy of the system thus rises, until it reaches a maximum, called the transition state. The transition state is approximately midway between the reactant and the product; the Cl-H bond is partly formed, and the CH3-H bond is partly broken. The odd electron density is partly on the Cl and partly on C.
Cl H C
H
H
H
δ δ
TS2 The energy of activation Ea is the difference in energy between the reactant and the transition state. It is a kinetic term – whereas ΔH is a thermodynamic term. Rate of Reaction = Collision factor Probability factor Energy factor× × Collision factor Frequency of collisions – depends on the concentration, the temperature Probability factor Fraction of collisions with the proper orientation – depends on the geometry of
the molecule Energy factor Fraction of collisions with an energy greater than, or equal to the energy of
activation Ea - increases as the temperature increases.
Note: Ea itself does not depend on temperature or concentration (only a catalyst can reduce its value).
If we know the thermodynamics of a reaction, we do not necessarily know its kinetics (and vice versa). For example, there are many reactions which are thermodynamically favored (ΔH < 0), but which do not occur because they are very slow (Ea very high).
Potential Energy Diagram:
15
Ea2= activation energy for reaction 2; TS2:
Cl H C
H
H
H
δ δ
TS2
Ea3 = activation energy for reaction 3; TS3:
ClClC
H
H
H
TS3
δδ
Ea2 >> Ea3, step 2 is slower than step 3, it is the rate-determining step (RDS).
A transition state is the top of a hill, in the energy diagram, a state between reactant and product (both at the bottom of a hill).
An intermediate is at the bottom of a hill, but has high energy, and thus has a short lifetime, is unstable.
Reactivity and Selectivity The monochlorination of butane yields 2-chlorobutane (72%) and 1-chlorobutane (28%). The monobromination yields 2-bromobutane (98%) and 1-bromobutane (2%). Br2 is more selective, and less reactive than Cl2.
C
H
C C C
H
H
H
H
H
H H
H
H
1°H2°H
Cl2hν
C
H
C C C
H
H
H
H
H
H H
H
Cl + C
H
C C C
H
H
H
H
H
Cl H
H
H
C
H
C C C
H
H
H
H
H
H H
H
Br + C
H
C C C
H
H
H
H
H
Br H
H
HC
H
C C C
H
H
H
H
H
H H
H
H
1°H2°H
Br2
hν
1-Chlorobutane 28%
2-Chlorobutane 72%1°Cl 2°Cl
1-Bromobutane 2%
2-Bromobutane 98%
1°Br
2°Br
16
The secondary halide is the major product in both cases, it is formed faster than the primary halide. This means that the Cl (or Br) free radical abstracts the 2o hydrogen of butane faster than the 1o hydrogen.
Why the difference in rate between primary and secondary H?
Probability factor – Orientation of collision; there are 6 primary hydrogens, and 4 secondary hydrogens in
butane; 6:4 or 3:2 in favor of 1-chlorobutane.
Collision factor – The same two molecules are colliding to form 1-, or 2-chlorobutane; no difference.
Energy factor – To evaluate it, we define the Hammond postulate.
The Hammond Postulate
The transition state structure more closely resembles the state which is closest to it in energy. That is, in reactions which form a high-energy intermediate (such as a free radical), the transition state looks like this intermediate, or is a late TS. The more stable the intermediate, the more stable the transition state, the faster this intermediate is formed.
17
Thermodynamic Stability of Free Radicals: o o o33 >2 >1 > CH ⇒i
Rate of formation of Free Radicals o o o33 >2 >1 > CH i
Rate of abstraction of H: 3o>2o>1o
Only in the case of intermediates (like free radicals) can you predict the rate of formation from the thermodynamic stability.
Why is Bri more selective?
Bri is less reactive than Cli .Since Cli is more reactive, it is higher in energy, and according to the Hammond postulate, the transition state is later for Bri than Cli ; thus, it looks more like the alkyl free radical in the bromination reaction, and is more affected by the stability of the free radical formed. Thus, Bri is more selective than Cli .
R H +HCl R H Cl+ TS [R- -H- - -Cl] Earlier
R H + Br R H Br+ TS [R- - -H- - Br] Later
δ δ'
δ'' δ'''
δ > δ''
For chlorination, the rates of abstraction are : 1oH:2oH:3oH = 1: 3.8 : 5
For bromination, the rates of abstraction are : 1oH:2oH:3oH = 1: 82 : 1600 (more selective, less reactive) o o
o o
Rate of Formation of 1-chlorobutane #(1 H) Reactivity of 1 H 6 1×Rate of Formation of 2-chlorobutane #(2 H) Reactivity of 2 H 4 3.8Rate of Formation of 1-bromobutaneRate of Formation of 2-bromobutan
28%72%
= = × =
o o
o o
#(1 H) Reactivity of 1 H 6 1×e #(2 H) Reactivity of 2 H 4 82
2%98%
= = × =
ΔH(kcal/mol)
CH3
CH2-CH3
CH3-CH-CH3CH3C CH3
CH3
CH4 CH3-CH3 CH3-CH2-CH3CH3C CH3
CH3
H
Methyl Free Radical
1° 2° 3°
10498
9592
18
Do Free Radicals Rearrange?
C
CH3
H3C CH2
H H
BrH3C C
CH3
CH2
H
H3C CH
CH3
CH2
H
δ δH3C C
CH3
CH2
H
? Br2 H3C C
CH3
CH2
H Br
?
t-butyl free radical isobutyl free radical Isobutylbromide
To find out whether free radicals undergo rearrangement, we can use a deuterium labelled alkyl halide (H. C. Brown).
C
CH3
H3C CH2
D H
H3C C
CH3
CH2
H
?
H3C C
CH3
CH2
Cl HCl
Abstract D
Abstract H
D Cl +Cl2
H3C C
CH3
CH2 H3C C
CH3
CH2
D Cl
H Cl +Cl2
D
Rearrangement
Rearrangement
?
Cl2
Cl2
H3C C
CH3
CH2
H Cl
H3C C
CH3
CH2
Cl D
isobutyl chloride
t-butyl chloride
t-butyl chloride
isobutyl chloride
If free radicals do not undergo rearrangement, then each time a deuterium is abstracted and a DCl is formed, t-butyl chloride should form; each time a hydrogen is abstracted and an HCl is formed, isobutyl chloride should form.
Experimentally, it was found that: DCl/HCl = t-butyl chloride/isobutyl chloride; this is evidence that alkyl free radicals do not undergo rearrangement.
19
VI. Exercises Exercise 1 Give the products of the following reactions; if no reaction takes place, write “No Reaction”.
C
CH3
H3C CF2H
CH3
+ CH3MgBr
CuLi
H3C
H3C
+ CH3CH2Br
Lithium dimethyl cuprate
Ethyl bromide
Lithium disec-butyl cuprate + Isobutyl bromide
CuLiCCH
H3C
H3C
2
H
H
+ (CH3)2CHI
Br
Na
CuLi
2
+Br
(a)
(b)
(c)
(d)
(e)
(f)
Exercise 2 (a) Write the structure of Bicyclo[3.3.0]octane. (b) Write the structure of 2-t-butylbicyclo[4.4.0]decane. (c) Give the IUPAC name of
CH3
20
Exercise 3 Give the structure of the most stable conformation of the product of the following reaction:
C
CH3
CH3
CH3Br2
hν
Exercise 4 Express Quantitatively the difference in stability between the following 2 structures:
H3C CH3vs.
CH3
H3C
(I) (II) Exercise 5 (a) Give the Newman projection (along C2-C3) of 2,3-dibromobutane with zero dipole moment. (b) Write the most stable form of 1,4-dimethylcyclohexane. (c) Write the most stable form of trans-3-methyl-1-isopropylcyclohexane. (d) Give the Newman projection (along C2-C3) of the most stable conformation of 2-Methylbutane. Exercise 6 1. What is the simplest alkane which possesses 1○, 2○, and 3○ Carbon atoms ? (A) 2-Methylpropane (B) 2-Methylbutane (C) 2-Methylpentane (D) 3-Methylpentane (E) 2,2-Dimethylbutane 2. In the most stable conformation of cis-1,4-Dimethylcyclohexane, the methyl groups are: (A) One axial, one equatorial (B) Both axial (C) Both equatorial (D) Alternating between both axial and both equatorial (E) None of the above 3. Which of the following can be reduced with Zn/H+ to form 2-Methylpentane ? (A) 1-Bromo-2-methylpentane (B) 2-Bromo-3-methylpentane (C) 3-Bromo-3-methylpentane (D) 1-Bromo-2,2-dimethylpentane (E) 1-Bromo-3,3-dimethylpentane
21
4. A molecule with IU = 3 (Index of Unsaturation) absorbs 1 mol of H2 – per mole of this molecule – upon catalytic hydrogenation. Which of these is it ?
A B C D E
5. Which sequence is best to prepare CH3–C ≡ C–D ?
(A)
(B)
(C)
(D)
(E) None of the above
H3C C C H1. NaH
2. D2O
H3C C C H1. NaOH
2. D2O
H3C C C H
H3C C C H
1. CH3ONa
2. D2O
DOH
Exercise 7 (a) Draw the structures of the products expected in the photochemical chlorination of Methylcyclohexane. Assume that only monochlorination takes place. (b) Calculate the percent composition of the mixture, given that the rates of attack of Cli at 1○, 2○, 3○ Hydrogens are 1 : 3.9 : 5.2. Exercise 8 Consider 2 compounds A and B with the molecular formula C5H10. (A) reacts with Cl2 in the light to give one substance C5H9Cl. (B) reacts under the same conditions to give six different C5H9Cl isomers. Write the structures of A and B, given that neither reacts with H2/Ni.
22
Exercise 9 Show a Synthesis of the following compounds from the indicated starting materials and any other needed reagents:
CH2CH2CH3
from
Br
(a)
(b) C
CH3
CH3 CH2
H
CH3 from CH3 CH2 C CH3
Br
H
C
CH3
CH3 CH2
D
CH3 from CH3CH2CH3(c)
Exercise 10 Carry out a conformational analysis of 2,3-Dimethylpentane along the C2-C3 bond.
23
VII. Solutions Exercise 1
C
CH3
H3C C
CH3
(a)
F
F
H
Most acidic H, by inductive effects
CH3 MgBr C
CH3
H3C C
CH3
F
F
MgBr+ + CH4
Note that the equilibrium lies towards the side of the products, CH4 being a weaker acid than (CH3)3CHF2.
CuLi
H3C
H3C
+(b)H3C C Br
H
H
H3C C CH3
H
H
Note that if we had a 2○ or 3○ alkyl halide instead of the Methyl here, the reaction would not take place. Also note that (CH3)2CuLi is equivalent to CH3
-.
(c)
CuLi
CH
CH3
CH2CH3
CH
CH3
CH2CH3
CH2CH
CH3
CH3
Br CH2CH
CH3
CH3
CH
CH3
CH2 CH3
Note that the C directly bonded to the Br is 1○. (d) No Reaction: the alkyl halide is 2○.
Br
Na(e)
This is the Wurtz coupling.
CuLi +Br
(f)
24
Exercise 2 (a) Bicyclo[3.3.0]octane
(b) t-Butyl-bicyclo[4.4.0]decane
(c) 2-Methyl-bicyclo[4.3.0]nonane
CH3 Exercise 3 Note the formation of the 3○ alkyl halide (work out the mechanism, keeping in mind that the most stable intermediate is formed i.e. the 3○ free radical). Also note that the t-butyl is equatorial since it’s bigger than Br.
C
CH3
CH3
CH3Br2
hν
1°
2°3°
Br
CC
CH3
CH3
CH3
Br
CH3
H3CCH3
Exercise 4 In conformation (II) there is one axial methyl, thus two 1,3-diaxial interactions. Therefore (I) is more stable than (II) by 2*0.9 = 1.8 kcal/mol.
H3C CH3vs.
CH3
H3C
(I) (II)
H
H
25
Exercise 5 (a) All Anti.
H
Br
CH3
H
Br
H3C Br
H CH3
Br
HH3C
(b) Both equatorial; note that the methyl groups are in trans.
CH3H3C
H
H (c) Isopropyl being bigger than Methyl, it should be the one placed equatorially.
CH3
H
H
CHH3C
H3C (d)
H
CH3
H
CH3
CH3
HCH3
H HCH3
CH3H
Exercise 6 1. (B) 2-Methylbutane
CH3 CH CH2 CH3
CH3
2°
1°3°
2. (A) One axial, one equatorial
CH3
H
H
H3C
3. (A) 1-Bromo-2-methylpentane
Br
CH3
Zn/H+
CH3
+ Zn2+ + Br-
26
4. The molecule has one double bond only – because 1 mole of this substance absorbs 1 mol of H2. Since IU = 3, this leaves only one possibility: one double bond + 2 rings.
(C)
5. ANSWER: (A)
(A)
(B)
(C)
(D)
H3C C C H Na H H3C C C + H2
Weaker acidStronger Acid
H3C C CO
D D H3C C C D + DO
Weaker acidStronger acid
H3C C C H Na OHStronger Acid
H3C C C + H2OWeaker acid
H3C C C H Na CH3O H3C C C + CH3OHWeaker acid Stronger Acid
H3C C C H H3C C C +Weaker acid Stronger Acid
O
D HO
D
H
H
No Reaction
No Reaction
No Reaction
Exercise 7 (a)
CH3
Cl2hν
CH3
Cl
H3C Cl CH3
Cl
CH3
Cl
CH2Cl
+ + + +
cis + trans
cis + trans
cis + trans
A B C D E
27
(b)
12 5 2 2 13 13236 391 0.16539 236
A B C D E
E E E E E
E E
+ + + + =
+ + + + =
= ⇒ = ≈
Thus we have in the mixture: 11% A; 6.5% B; 33% C; 33% D; 16.5% E. Exercise 8
For C5H10 we have IU = Index of Unsaturation = 10 10 2 12
− += site. It is mentioned that neither A nor B
reacts with H2/Ni, therefore this site of unstauration is not a double bond, it is a ring. Now that we know C5H10 is a cycloalkane, we have 4 possibilities:
(E) (F) (G) (H) We are told that A + Cl2 gives one product, and B + Cl2 gives 6 products. The blue arrows indicate the different possible sites of chlorination:
2
2
(2)
(E) (F) (G) (H) By monochlorination, E would yield 1 product; F would yield 6 products; trans-G would yield 3 products; cis-G would yield 4 products ; and H would yield 2 products.
CH3
B, 1°, 3H
A, 3°, 1H
C, 2°, 4HC
D, 2°, 4H
E, 2°, 2H
D
1 5.2 2 22 3.9 3 33 1 5 52 3.9 13 13
4 3.9 2 22 3.9
= × = ⇒ =
= × = ⇒ =
= = × = ⇒ = =
A A EEB B EEC D C D EE E
28
It becomes clear that A ≡ E and B ≡ F.
A B
A
B
Cl2
hνCl
Cl2
hν
Cl CH2Cl Cl
Cl
++ +
cis + trans cis + trans Exercise 9
Br
(a)
(b) H3CH2C C CH3
Br
H
CH3CH2CH3(c)
Li
Li
CuI CuLi CH3CH2CH2Br
CH2CH2CH3
1. Li
2. CuICuLiCH
H3C
CH2
H3C
2
CH3Br CH3CH
H3C
CH2
H3C
Br2
hνCH3CHCH3
Br
1. Li
2. CuI3. CH3CH2Br
CH3CHCH3
CH2CH3
Br2
hν CH3CCH3
CH2CH3
Br
1. Mg
2. D2OCH3CCH3
CH2CH3
D Note that in (c), Retrosynthetic analysis was applied:
C
CH3
H3C CH2
D
CH3
R R'
R2CuLi R'X+1°2°
29
Exercise 10
CH3
CH3H
C2H5
HCH3 CH3
H H
CH3
H
CH3H3C
HCH3
CH3H
H
CH3
HH3C
H
H
CH3H
CH3
+60° +60°
+60°
+60°
+60°
+60°
C2H5
C2H5
C2H5
C2H5
C2H5
CH3
CH3
CH3
CH3
CH3
Most Stable
Least Stable VIII. The Textbook Suggested Reading from Solomons 9th Edition: Chapter 4 (except 4.18C and 4.20B) Chapter 10 [10.1-10.8] Chapter 2 – Physical Properties of Alkanes – 2.14
Suggested Problems from Solomons: Chapter 4: 3; 5; 6; 10; 13; 14; 16; 20; 29; 31; 33; 34; 39; 41(a; c; f).
Chapter 10: 2; 5; 8; 12.
1
Hanadi SLEIMAN
CHEM 212 Chapter 3
N
O
O
N
H
OO
2
I. Definitions
3
II. Physical Properties of Enantiomers
4
• The polarimeter 5
III. The fischer projection 6
• The (R,S) convention 7
iV. Diastereomers 8
V. Isomers - Summary 10
VI. reactions which forM a chiral center 11
VII. resolution of a racemic mixture
12
viii. Biochemical examples 12
ix. exercises 13
x. solutions 15
xi. the textbook: Solomons, 9th edition 17
3
C
CH3
C2H5H
H C
C2H5
H3C
H
H
I. Definitions: A molecule which is not superimposable on its mirror image is a chiral molecule. These two isomers of 2-iodobutane are non-superimposable mirror images of each other: they are enantiomers. Each of these enantiomers is a chiral molecule. They cannot be converted into each other by rotation around bonds. To go from one to the other, you would have to break two bonds of the central atom and exchange two groups; this process requires a large amount of energy; thus enantiomers cannot be converted into each other under normal conditions. The chirality of these molecules is due to the presence of a carbon with four different groups around it: Methyl, Ethyl, Iodide and Hydrogen. This carbon is called a chiral carbon (also called: stereocenter, asymmetric carbon, stereogenic center). Consider the following molecule: These two mirror images are superimposable, and the molecule is not chiral, or achiral; note that none of the carbons of this molecule are chiral. When a molecule has a chiral center, it usually is chiral; when a molecule is chiral, it usually has a chiral center. However, there are exceptions to this.
C
CH3
C2H5H
I C
C2H5
H3C
H
I
Chiral Center Chiral Molecule⇔Usually, but Not Always
4
II. Physical Properties of Enantiomers Enantiomers have identical physical properties: exactly the same melting points, boiling points, solubilities, etc. They are different only in the way that they interact with plane-polarized light. Light is an electromagnetic wave phenomenon, its electric and magnetic field waves are perpendicular to the direction of light propagation:
Molecules interact with the electric field of light; the electric field wave propagates in an infinite number of planes. When light passes through a polarizer, the beam which emerges propagates in only one plane; this is plane-polarized light. A chiral molecule rotates the plane of polarized light by a certain angle α; it is called an optically active substance. The enantiomer of the above molecule rotates the plane of polarized light by – α.
5
Measurement of α : The Polarimeter A monochromatic light source (usually the sodium D line) is passed through a polarizer, and the plane-polarized light passes through a sample of the optically active substance; the observer can measure the rotation of the emerging beam. Specific Rotation: In the polarimeter tube, each chiral molecule rotates the plane of light by a certain angle β; the rotation α is the sum of all these β’s; thus α depends on concentration and tube length.
We define specific rotation as: [ ]Concentration × Length of tubeD
αα =
The specific rotation only depends on the substance measured:
CH
I
C2H5
H3C
CH
I
C2H5
CH3
[α] = + 15.9° (+) - 2 - Iodobutanedextrorotatory (d)
[α] = - 15.9° (-) - 2 - Iodobutanelevorotatory (l)
There is no known relation between the three-dimensional structure and the sign/magnitude of the specific rotation.
6
A mixture of enantiomers of equal amounts is a racemic mixture; it has an [α] = 0, and is optically inactive.
Similarly, a sample of an achiral substance has equal amounts of the molecule and its mirror image (which are identical), and it will thus have zero optical rotation; all achiral substances are optically inactive. A mixture of enantiomers of unequal amounts, or an optically impure substance has a net optical rotation, corresponding to the excess of one enantiomer over the other. Optical Purity = Enantiomeric Excess (ee)
mixture
pure
[ ] = 100 = % (one enantiomer) - % (the other enantiomer) [ ]
ee αα
×
Thus if a mixture of 2-iodobutanes has 1[ ] 7.95 [ ] 50%2 pure eeα α= + = ⇒ = i.e. we have in the
mixture 75% of one enantiomer (the d form) and 25% of the other. III. Representation and the Fischer Projection
H
C2H5 CH3
I
Fischer Projection
≡ ≡C
H
IC2H5
H3C
C
H
C2H5 CH3
I
Fischer projection: - For chiral carbons only - Horizontal bonds point toward you, vertical bonds are away from you - You can only rotate a Fischer projection in the plane of the paper by 180o - Do not remove the Fischer projection mentally from the paper, do not rotate by any angle other than 180o
7
Nomenclature: R, S convention (Cahn, Ingold and Prelog) 1. Number the four groups on the chiral carbon. (a) Give the lowest number to the atom bonded to the chiral center with the highest atomic number. (b) If two atoms are isotopes, give the lowest number to the atom of highest mass number (e.g., D has a lower number than H). (c) If two atoms are identical, move outwards to the next atom, and compare; do not compare the entire groups attached to the chiral carbon. 2. Look away from group number 4; go from1 2 3→ → , if the motion is clockwise, then the 3D arrangement, or configuration is R; if the motion is counterclockwise, then the configuration is S.
H
C CH3
I
H3C
H
H 1
2 3
1
2 3
CW R4
One exchange of two groups on a Fischer projection (or any other 3D-structure) gives the enantiomer; thus two exchanges give back the initial chiral center.
R
H
C2H5 CH3
I
CH3
C2H5 H
I
1 exchange
CH3
H C2H5
I
S R
1 exchange
We can apply this to the case when group 4 is on a horizontal position; it is more convenient to place it on a vertical position, because then the viewer is looking at the Fischer projection away from group 4; do this by the double exchange method:
≡1
2
3
4
1
2
3
CW
1 exchange
R
1 exchange
CH3
I H
C2H5
H
I CH3
C2H5
H
H3C I
C2H5 Note: Count C = C as
C C
C C
8
IV. Diastereomers Consider the following molecule: 2-Chloro-3-iodobutane:
H3C C
Cl
C
I
CH3
H H
∗ ∗
It possesses two chiral centers (marked with a *); the number of stereoisomers of a molecule possessing n chiral centers is 2n. There are four possibilities (RR, SS, RS, SR).
CH3
H Cl
CH3
H I
I (2S, 3R)CH3
Cl H
CH3
I H
II (2R, 3S)
CH3
H Cl
CH3
I H
III (2S, 3S)
CH3
Cl H
CH3
H I
IV (2R, 3R)
R
S
R
R
I and II are Enantiomers. III and IV are Enantiomers. I and III are stereoisomers which are not mirror images; they are called Diastereomers (in two diastereomers, one part of the molecule is the same, while the other part is the mirror image of the first). Diastereomers have different physical properties.
9
Consider the following molecule:
H3C C
CH3
CH3
C
Br
C
Br
C
CH3
CH3
CH3H H
∗ ∗
The four possible stereoisomers are:
R
SR
S
S R
t-Bu
Br H
t-Bu
H Br
I (3R, 4R)t-Bu
H Br
t-Bu
Br H
II (3S, 4S)
t-Bu
H Br
t-Bu
H Br
III (3S, 4R)
t-Bu
Br H
t-Bu
Br H
IV (3R, 4S)
t-Bu
H Br
t-Bu
H BrR S
Rotate 180° S
R
III (3S, 4R) I, II are Enantiomers. I, III are Diastereomers. II, III are Diastereomers. III and IV are Superimposable: III is a Meso compound: a compound which contains chiral centers and is superimposable on its mirror image; meso compounds contain a mirror plane. When a molecule contains a mirror plane, it is automatically superimposable on its mirror image, and thus achiral. (Note: compound I, for example, is called: (3R, 4R)-3,4-Dibromo-2,2,5,5-tetramethylhexane).
10
Consider the following dichlorocyclohexanes:
Cl
Cl
1,1Achiral
(no chiral centers)
56
4
1
23
ClCl
cis-1,3Achiral
Mirror plane passes through C2 and C5
23
4 5
1
6
Cl
Cl
Mirror plane passesthrough C1 and C4
Achiralcis-1,4
12
3
45 6
ClCl
Mirror plane passesthrough C1 and C4
trans-1,4 Achiral
ClCl
∗
∗
trans-1,2 Chiral
No mirror plane
Cl
Cl
trans-1,3 Chiral
No mirror plane
∗
∗
ClCl
cis-1,2
Cl
Cl
Cl
Cl
I II Superimposable on I
Achiral because I and II are mirror images but are also conformational isomers i.e. interconvertible at room temperature
V. Isomers Constitutional Isomers are different in the way the atoms are connected to each other. Stereoisomers are only different in the three-dimensional arrangement of their atoms.
EnantiomersConfigurational Isome
StereoisomConforma
rsDias
ti
teer
re
onal Is
omer
omes
rs
s⎧⎨⎩
⎧⎪⎨⎪⎩
11
Conformational Isomers: Can be interconverted by rotation around single bonds – occurs at room temperature. Configurational Isomers: One has to break bonds to interconvert them – does not occur at room temperature. VI. Reactions Which Form a Chiral Center Optically inactive starting materials give optically inactive products. Thus if in a reaction, a chiral center is formed, the two possible enantiomers are formed in equal amounts (a racemic mixture). Example:
H3C CH2 CH2 CH3Br2
hνCH3 C CH2 CH3
Br
H
∗
Optically Inactive Racemic Mixture
If we start with a chiral molecule, and we subject it to free radical bromination:
H3C C
C2H5
C3H7
H
∗ HBr
H abstractione.g. R
There is equal probability of attacking the upper or lower lobe of the p orbital. Thus, the 2 enantiomers are formed in equal amounts i.e. racemic mixture.
CC3H7
C2H5
CH3
Br Br
C
Br
C3H7
C2H5
CH3
CC3H7
C2H5
CH3
Br Br
C
Br
C3H7C2H5
CH3
Pathway 1
Pathway 2
50% Probability
50% Probability
R
S
Both enantiomers are formed because the intermediate free radical is planar; this reaction is not enantioselective (enantioselective reactions are reactions where one enantiomer is formed in a higher amount than the other).
12
CH3
O
CH CH2
CH3
∗
CarvoneS- Caraway FlavorR- Spearmint
CH3
CH CH2
CH3
∗
LimoneneS- Lemon FlavorR- Orange Flavor
N
O
O
N
H
OO
∗
ThalidomideR- Sedative, HarmlessS- Causes fetal limb deformation
∗
H2C
HO
CHHOH2C
HN C CH3
CH3
CH3
OH
AlbuterolSold as racemate, VentolinR- BronchodilatorS- Causes Bronchial Hyperactivity
∗C
O H
NHCH3
CF3
FluoxetineSold by Lilly as racemate, Prozac R- Anti-DepressorS- Inactive
H2CHC CH3
CH3
IbuprofenSold as racemate, Advil S- Anti-inflammatory, AnalgesicR-Inactive
H3C H
C H
O
VII. Resolution of a Racemic Mixture Since enantiomers have the same physical properties, they cannot be separated by physical methods; however, they can be converted to diastereomers, which have different physical properties; these diastereomers can be separated, and then each pure diastereomer can be converted back to the pure enantiomer. For example, a racemic mixture of an acidic substance HA can be reacted with an optically pure base B:
H AR
S H A
+ BR[BH] [A]
[BH] [A]
R
R
R
S
SeparateH AR
S H A
[BH] [A]RR
[BH] [A]R S
H+
H+
VIII. Examples of Biologically Active Chiral Molecules
13
IX. Exercises Exercise 1 In the following pairs of compounds, one is chiral and the other is achiral. Identify each.
CHCH2 CH2 OHOH
Cl
CHCH2 CH2 OHCl
OH
(a)
(b)
ClH
Cl
and
and
Exercise 2 Specifiy the configuration as R or S.
OHH
∗
CO2−
H3N H
CH2CH2CO2− Na
NH
H2C
HO
C
H
NH3
CO2−
∗
(a)
(b)
(c)
Exercise 3 Match these relationships to the following pairs of compounds: Identical - Constitutional Isomers – Enantiomers – Diastereomers
14
C
H CH3
HO CH2BrC
H3C H
Br CH2OH
C
H CH3
H3CH2C Br
C
H Br
H3CH2C CH3
CH2OH
H OH
OH
H H
CH2OH
HO H
OH
H H
Cl
H
H
H3C
Cl
H
H3C
H
C C
H3C
H C
H
CH3
OH
H
C C
H3C
H C
H
CH3
H
OH
CH2OHCH2OH
H H
HO
HO
CH2OH
HO HCH3
OHH
H
CH2OH
OH
H
CH3
HO
H
H3C
BrBr
COOH
H
CH3
Br H
COOH
H Br
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
and
and
and
and
and
and
and
and
15
X. Solutions Exercise 1
CHCH2 CH2 OHOH
Cl
CCH2 CH2 OHCl
OH
(a) and
H
∗
Chiral Molecule Achiral Molecule In the Achiral molecule, central C has two –CH2OH groups, so it is not a chiral center. Note that if a molecule has one chiral center, then it is chiral.
(b)
ClH
Cland
Achiral Molecule
∗
Chiral Molecule There are no chiral centers in the Achiral molecule displayed above. Exercise 2
OHH
CO2−
H3N H
CH2CH2CO2−
NH
H2C
HO
C
H
NH3
CO2−
(a)
(b)
(c)
1
23
4
R
1
2
3
4
1
2
3
4
Double exchange
1
23
4
S
Double exchange1
2
3
4
S
16
Exercise 3 (a) Constitutional isomers (follow the colors)
andCH3 CH CH2
OH
Br CH3 CH CH2
Br
OH
(b) Enantiomers (since the H is towards us, don’t forget to perform a double exchange)
C
H CH3
H3CH2C Br
C
H Br
H3CH2C CH3
and
1
23
4
12
34
S R (c) Identical (there are no chiral centers in this molecule)
CH2OH
H OH
CH2OH
CH2OH
HO H
CH2OH
and
(d) Diastereomers Note that (S,S) and (R,S) are diastereomers: one part of the molecule is the same, the other is the mirror image of the second part. Also note that these two molecules are geometric isomers as well (trans and cis).
1
2
3
4
S R
Cl
H
H
H3C
Cl
H
H3C
H
and
1
3
4
1
3
4
2
21
3
4
2
SS
trans cis (e) Enantiomers
R1
3
4
2
S
C C
H3C
H C
H
CH3
OH
H
C C
H3C
H C
H
CH3
H
OH
and
17
(f) Constitutional isomers (g) Enantiomers
RS
1
3
4
2
H
CH2OH
OH
H
CH3
HO
and
1
3
4
2
HO
CH2OH
H
OH
CH3
H
1
3
4
2
1
3
4
2
R S (h) Identical
R
1
3
4
2
R
H
CH3
BrBr
COOH
H
CH3
Br H
COOH
H Brand
1
3
4
2
1
3
4
2
1
3
4
2
R
R
XI. The Textbook: Solomons, 9th Edition Assigned Reading: Chapter 5 Assigned Exercises: 4; 5; 11; 12; 13; 15; 26; 27; 33; 35; 38
- 1 -
C C
H
X
HO
+ HO H + X
Hanadi SLEIMAN CHEM 212 Chapter 4
- 2 -
I. Structure of alkyl halides
3
II. Preparation
3
III. Reactions 4
• SN1 / SN2 4
• Basicity vs. Nucleophilicity 4
• Leaving group 5
• mechanism 5
• e1 / e2 9
• summary 11 IV. Exercises 12
V. Solutions 14
VI. The Textbook: solomons, 9th edition 15
- 3 -
I. Structure of alkyl halides Alkyl halide: R-X, where R is the alkyl group, and X is the functional group. Note that X = F, Cl, Br, or I. Example: 3-Chloro-2,6-dimethyloctane
Cl
Vinylic Halide:
X
II. Preparation 1. From Alcohols – most common method: R OH HX R X
(X = Cl, Br, I)
R OHPX3 R X
2. From Alkenes and Alkynes:
HX C C
H X
C C2 HX C C
H
H
X
X 3. From Alkanes (only certain alkanes, where halogenation yields one product):
Br2
hν
Br
4. From Alkyl Halides (used to prepare alkyl iodides)
R IR X (X = Cl, Br)+ I− + X−
- 4 -
III. Reactions Nucleophilic aliphatic substitution
C
H
H Cl
H
I + C
H
H I
H
+ Cl
Nucleophile
Electrophile substrate
Leaving group
Acid-Base reaction:
HO H Cl HO H + Cl
Base Acid+
General equation for a nucleophilic aliphatic substitution:
+ XR X + Y R Y
R Y Y R OH Alcohol OH
Hydroxide
R OR' Ether R'O
Alkoxide
R C C R' Alkyne R' C C
Alkynyl anion
R R' Alkane R2CuLi Lithium dialkyl copper R I Alkyl iodide I
Iodide
R CN Nitrile CN Cyanide Basicity vs. Nucelophilicity
Basicity (Bronsted-Lowry) Nucleophilicity 1. Equilibrium property The stronger the acid, the farther the equilibrium is to the side of the products.
1. Rate property The stronger the nucleophile, the faster the reaction.
2. Interaction with H-X 2. Interaction with C-X 3. No concern with size 3. Cannot be sterically bulky e.g. Lithium diisopropyl amide: strong base, poor nucleophile
N
CH
CH
H3C
CH3
H3C
CH3
Li
- 5 -
Leaving group The weaker the base, the better it is as a leaving group. e.g. p-toluenesulfonyl, or tosyl group: weak base, good leaving group.
H3C S
O
O
O
Mechanism Two possibilities, compared below: Mechanism A
+ XR X+Y R Y Mechanism B
+ XR X
Y
R
+ R R Y
Step 1
Step 2 Mechanism A: SN2 Mechanism B: SN1 Mechanism
+ XR X+Y R Y
+ XR X
Y
R
+ R R Y
Step 1
Step 2 Type of Mechanism
The R-X bond breaks and the R-Y bond forms simultaneously: Concerted mechanism.
The R-X bond breaks first, then the R-Y bond forms: Stepwise mechanism.
Kinetics Rate [ ][ ]k RX Y −= Where k = rate constant; property of the reaction, depends on temperature. Kinetics are first-order in [RX], first-order in [Y-], overall kinetics are second-order.
Step 1 is the rate-determining step; the rate is not affected by step 2. Rate '[ ]k RX= Kinetics are first-order in [RX], zeroth-order in [Y-], overall kinetics are first-order.
- 6 -
Pote
ntia
l Ene
rgy
Progress of Reaction
RX + Y-
TS
RY + X-
Pote
ntia
l Ene
rgy
Progress of Reaction
RXRY
Ea1
Ea2
R+ X-
Ea1 >> Ea2
Molecularity (mechanistic property)
Number of species present in the rate-determining step: 2. Bimolecular: SN2 (Substitution Nucleophilic Bimolecular)
Number of species present in the rate-determining step: 1. Unimolecular: SN1 (Substitution Nucleophilic Unimolecular)
Stereochem-istry
SN2 occurs with backside attack; results in inversion of configuration at the chiral carbon.
C
Br
R2R1
R3
OCH3
Backsideattack C R2
R3
R1
Br
OCH3δ−
δ−
C
OCH3
R2
R3R1
(Inversion of configuration)
Step 1 forms an sp2, flat carbocation R+.
C R2
R3
R1
C R2
R3
R1
C
Br
R2R1
R3
H3CO
H
a
b
C
O
R2R1
R3
H
CH3C
O
R2
R3R1
H
CH3
+
From aFrontside attack
Retention of configuration
From bBackside attack
Inversion of configuration
- H+
C
OCH3
R2R1
R3
C
OCH3
R2
R3R1
+
Enantiomers Equal probability of attack from either side; complete racemization, loss of optical activity.
- 7 -
Reactivity of substrates
SN2 is a concerted reaction: it depends on the access of the nucleophile to the electrophilic carbon; the more sterically hindered this carbon, the slower the SN2 reaction.
3CH X 1 halide 2 halide 3 halide> >Most reactive by SN2
3 halides do not react via the SN2 mechanism.
• SN1: step 1 is rate determining. • Step 1 involves formation of an intermediate carbocation R+. • The more stable this carbocation, the faster step 1 is (by the Hammond Postulate), the faster the overall SN1 is. • Since alkyl groups release electrons, the order of stability of carbocations is: +
33 2 1 CH> > > Therefore, Most reactive by SN1
33 R-X 2 R-X 1 R-X CH X> >
1 halides (and CH3X) do not react via the SN1 mechanism.
Rearrange-ment
Concerted mechanism, no intermediate formed; no rearrangement.
• Step 1 forms an intermediate carbocation, which can rearrange by a 1,2-alkyl shift, or a 1,2-hydride shift. • Rearrangement occurs to form a more stable carbocation: 1 2 ; 1 3 ; 2 3 → → →
Dependence on nucleophile
SN2 depends on the nucleophile’s strength and concentration. What makes a nucleophile stronger? 1. If two nucleophiles have the same attacking atom, basicity and nucleophilicity follow the same trend. e.g. CF3O- vs. CH3O- The latter (CH3O-) is a stronger base, and thus a better nucleophile. 2. If two nucleophiles have attacking atoms in the same period, basicity and nucleophilicity follow the same trend. e.g. CH3O- vs. CH3NH- (the latter is a stronger base and thus a better nucleophile). 3. If two nucleophiles have attacking atoms in the same group, nucleophilicity depends on the solvent • In polar, protic solvents, nucleophilicity follows the opposite trend to basicity: e.g. in solvents like H2O, CH3OH we have:
F- > Cl- > Br- > I-Strongest base
SN1 does not depend on the nucleophile. It can take place with weaker nucleophiles; often, the solvent acts as nucleophile (solvolysis reactions).
- 8 -
Strongest nucleophile
F- < Cl- < Br- < I-
This is because protic solvents can H-bond better to the hard ion F-, than the soft ion I-. Thus when F- attacks an electrophile, it is surrounded by a large, tightly bound solvent sphere; it is less nucleophilic than the relatively ‘naked’ I-. • In polar, aprotic solvents, nucleophilicity follows the same trend as basicity. e.g. in solvents like THF, DMF, DMSO:
F- > Cl- > Br- > I-
Strongest nucleophile
Strongest base
This is because no H-bonding interactions can solvate the anions; thus, F- and I- are both relatively ‘naked’; their nucleophilicity is related to their electron richness, or basicity. An important consequence to this is that anionic nucleophiles are more nucleophilic in aprotic solvents than protic solvents. • Neutral nucleophiles: regardless of solvent, nucleophilicity follows the opposite trend to basicity. e.g. (CH3)3P is more nucleophilic and less basic than (CH3)3N.
Leaving groups
Work with good leaving groups. The weakest bases are the best leaving groups: -OTs > I- > Br- > Cl- > F-. In R-OH, -OH is a very poor leaving group; it can become a better leaving group by protonation to +
2R-OH ; the leaving group is then the neutral H2O molecule. ROH + HX → RX + H2O
Polar solvents
In general, polar solvents slow down SN2. • For SN2, in general, the transition state has more dispersal of charge than the reactant; thus the transition state is less polar than the reactant. • Polar solvents thus stabilize the reactant more than the transition state. • SN2 is slower in polar solvents (but see exceptions).
In general, polar solvents accelerate SN1. • For SN1, in general, the transition state has more separation of unlike charges than the reactant; thus, the transition state is more polar than the reactant. • Polar solvents thus stabilize the transition state more than the reactant. • SN1 is faster in polar solvents (but see exceptions).
- 9 -
SN2 vs. SN1 - Summary
SN2 SN1
• R-X: R = Methyl, 1 , 2 • Strong nucleophile • Aprotic solvents
• R-X: R = 3 • Weak nucleophile • Polar, protic solvents
Elimination reactions generate alkenes
C C
H
X
αβ
Alkyl halide
Base
(X = Cl, Br, I) Dehydrohalogenation reaction: loss of H-X. Two possible mechanisms, compared below. Mechanism A: Concerted
C C
H
X
HO
+ HO H + X
Mechanism B: Stepwise
Step 1
Step 2
C C
H
X
C C
H
+ X
C C
H
+ OH
H2O
H
H
- 10 -
Mechanism A: E2 Mechanism B: E1 Mechanism
C C
H
X
HO
+ HO H + X
Step 1
Step 2
C C
H
X
C C
H
+ X
C C
H
+ OH
H2O
H
H
SlowRDS
Type of Mechanism
Concerted Stepwise
Kinetics Overall, kinetics are second-order. Rate [Alkyl halide][Base]k=
Overall, kinetics are first-order. Rate [Alkyl halide]k= First step is rate-determining.
Molecularity
Bimolecular: E2 Unimolecular: E1
H-isotope effect
• C-H easier to break than C-D • C-H + B → C + H-B ( Hk ) • C-D + B → C + D-B ( Dk ) • If the C-H bond is broken in the RDS,
then 7H
D
kk
≈ primary isotope effect.
• If C-H bond is not broken in the RDS,
then 1.H
D
kk
≈
• In the E2 mechanism, C-H bond
breaking occurs in the the RDS ⇒ 7H
D
kk
≈ .
In the E1 mechanism, the C-H bond is not broken in the RDS (it breaks in the second
step i.e. after the RDS) therefore 1.H
D
kk
≈
Rearrange-ment
E2 concerted mechanism; does not form an intermediate carbocation; no rearrangement.
E1 stepwise mechanism, forms a carbocation; rearrangement does occur.
Regiochem-istry
Transition state of E2 mechanism looks like the alkene product; thus this reaction results in the formation of the more stable alkene as the major product. Zaitsev’s rule: the more substituted the alkene, the more stable it is. Most stable R2C = CR2 > R2C = CHR > trans-RCH = CHR > cis-RCH = CHR > CH2 = CR2 > CH2 = CHR > CH2 = CH2.
Stability of the cation formed determines the rate of E1:
Formed fastest by E1
Most stable
> >3 2 1
Nature of base
E2 does depend on base; a stronger base or a higher concentration of base accelerates the reaction.
E1 does not depend on the strength or concentration of the base, it is not involved in the RDS.
Leaving group
The better the leaving group, the faster the Elimination reaction: ROTs > RI > RBr > RCl > RF
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Stereochem-istry
Elimination of H and X is anti: gives only one alkene stereoisomer.
C2H5
Cl
CH3
CH3
H
C2H5
HO
H3C
C2H5
C2H5
CH3
cis-alkene
The trans-alkene is only a minor product.
Forms carbocation: planar, gives both alkene stereoisomers (cis + trans).
Solvent polarity
E2 transition state:
C C
H
X
HO
C C
H
X
HOδ−
δ−
+ HO H + X
TS is less polar than reactant; polar solvents slow down E2 (in general).
E1, transition state of the first step (RDS):
C C
H
Xδ−
C C
H
X
C C
H
+ X
δ+
TS is more polar than reactant; polar solvents increase the rate of E1 (in general).
SN2, SN1, E2 and E1 When the base or nucleophile is strong (e.g. negatively charged) the reaction is SN2 or E2 (Unimolecular mechanisms SN1 and E1 cannot compete). Elimination reactions (E2 and E1) require higher temperatures than substitution (SN2 and SN1). SN2 vs. E2
SN2 E2 • Less branching: 1 , 2 alkyl halides • More branching at both α and β carbons • Strong nucleophile • Strong base • Hβ − not necessary • Requires Hβ − • Can occur readily at room temperature • Favored by heat
SN1 vs. E1: usually mixtures are obtained with one mechanism slightly predominant.
SN1 E1 • β -carbon less crowded for nucleophile to attack carbocation
• β -carbon can be more crowded
• Can occur at room temperature • Favored by heat
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IV. Exercises Exercise 1 Write the products of the following reactions, and decide which is faster: 1. Couple 1
O S
O
O
CF3O S
O
O
CH3
CH3O CH3O
(a) (b)
2. Couple 2
(a) (b)NCH3OH N
CH3OH
Exercise 2 Which of the following nucleophilic substitutions is not likely to occur?
Cl
Br
Br
OH
Cl
I−
I−
I−
CH3O−
OH−
+
+
+
+
+
(a)
(b)
(c)
(d)
(e) Exercise 3 Which of the following alkyl halides undergoes solvolysis most rapidly, in a mixture of methanol and water?
(a)
(b)
(c)
(d)(e)
Cl
Br
I
F
Same rate
- 13 -
Exercise 4 Which alkyl halide is most reactive in SN1? Write the structure of the final product if the nucleophile used is methanol.
(a)
(b)
(c)
(d)
(e)Br Br
Br
BrBr
Exercise 5 What is the product of the following reaction:
CH3
BrCH3OH
Exercise 6 In the reaction of butyl bromide with sodium hydroxide, what would be the effect of doubling the concentration of both reactants on the rate of the reaction? (a) New rate = Old rate (b) New rate = 2 x Old rate (c) New rate = 3 x Old rate (d) New rate = 4 x Old rate (e) New rate = ¼ x Old rate
- 14 -
V. Solutions Exercise 1 1. ANSWER: (a) is faster since the leaving group is better (i.e. weaker base – due to inductive effects).
O S
O
O
CF3
O S
O
O
CH3
CH3O
CH3O
(a)
(b)
SN2
SN2
OCH3
+ O S
O
O
CF3
OCH3
+ O S
O
O
CH3
Weaker Base Better leaving group
2° C
2. ANSWER: (a) is faster; (b) does not even occur since the leaving group is a stronger base than OH-.
+(a)
(b)
NCH3OH
NCH3OH
O N
H
CH3
-H+ OCH3SN2
SN2
1° C-H+ O
CH3+ N
Weaker Base Better leaving group
No reaction
Exercise 2 ANSWER: (D) OH- is a bad leaving group. Exercise 3 ANSWER: (D) I- is the best leaving group among the 4 proposed, since it is the weakest base. Note that this is an SN1 reaction; the final product is portrayed below:
OCH3
- 15 -
Exercise 4 ANSWER: (D) The stability of the carbocation intermediate obtained in the RDS should be considered – a tertiary carbocation is the most stable – thus the corresponding reaction is the fastest. Final product is displayed below:
OCH3
Exercise 5
CH3OH
-H+SN1
2° C
CH3
Br CH3OH
CH3
1,2-Hydride shift
CH3H3C OCH3
A 1,2-hydride shift (and not 1,2-methyl shift) occurs because it leads to the most stable carbocation in this case: H3C
1,2-Hydride shift
CH3H
H3C
1,2-Methyl shift
H
H
CH3
H
3° C+
2° C+
Exercise 6 ANSWER: (D)
Br + OHOH + Br
× 2 × 2 The rate of an SN2 reaction is represented by: [ ][ ]Rate k RX Y −= . In our case: [R - Br][OH ] = 2[R - Br] 2[OH ] 4 − −= ⇒ × × = ×Old Rate k New Rate k Old Rate VI. The textbook: Solomons, 9th edition Suggested reading Chapter 6 + Chapter 7 [7.1; 7.8] Suggested problems Chapter 6: 1, 4-9, 13-17, 21, 25, 32, 33
- 1 -
Hanadi SLEIMAN CHEM 212 Chapter 5
- 2 -
I. Structure and Nomenclature of Alkenes
3
II. Reactions of Alkenes
5
• Hydrogenation 5
• Electrophilic Addition Reactions 6
• Free Radical Reactions 13
• Oxidations 14
• Cycloadditions 15
III. Structure and Nomenclature of Alkynes 21
IV. Preparation of Alkynes 21
V. Reactions of Alkynes 23
• Hydrogenation 23
• Additions of HX and X2 23
• Hydration (Keto-Enol Tautomerism) 24
• Ozonolysis 25
VI. Exercises 25
VII. Solutions 29
VIII. the textbook: Solomons, 9th edition 33
- 3 -
I. Structure and Nomenclature of alkenes Molecular formula: CnH2n
n Molecular formula IUPAC name Common name 2 C2H4 ethene Ethylene 3 C3H6 propene Propylene 4 C4H8 butene butylene
Structure Ethene:
C C
H
H H
Hsp2
C C
π(Cp, Cp)
σ(Csp2, Csp2)
H
HH
H
σ(Csp2, H1s)
π bond strength: ~ 68 kcal/mol σ bond strength: ~ 95 kcal/mol
Bond lengths: ( o
101A 10 m−= )
C C
1.341.53 A°A°
C C
π bonds are weaker than σ bonds; the 2 electrons that make a π bond can be given to Lewis acids. Alkenes are Lewis bases, they react with electrophiles. Vinyl chloride: C-Cl: σ (Csp2, Cl); more s character than a C-Cl for an alkyl halide (R-Cl); the two electrons in the vinylic bond are held closer to the C nucleus, and the vinylic bond is shorter and stronger than an alkyl C-Cl bond. This bond is more difficult to break, and vinyl halides do not normally engage in substitution or elimination reactions.
C C
σ (stronger)
π (weaker)
C C
H
H Cl
H
- 4 -
Rotation around a C=C double bond
To go from I to II, we need to rotate around the C = C double bond. Are these stereoisomers interconvertible under normal conditions? Rotation around a C=C double bond does not occur under normal conditions, because it involves breaking a π bond. These stereoisomers are not interconvertible under normal conditions; they are called geometric isomers (or diastereomers: one part of the molecule is the same, the other is the mirror image of the first). Nomenclature
CH2 CH CH2 CH3
1-Butene
CH2
1
CH2
CH3
CH3
4
CH3
3-Methyl-1-butene
CH2 CH
CH2 CH CH2
Vinyl group
Allyl group
CH3
1-Methylcyclopentene
1
2
3Cl
3-Chlorocycloheptene
C C
H
H3C H
CH3
trans-2-Butene
C C
H3C
H H
CH3
cis-2-Butene
C C
Cl
H F
Br1 1
C C
H
Cl F
Br
1
1
For this carbon, Cl has a higher atomic numberFor this carbon, Br has a higher atomic number
Z-alkene
E-alkene
The 2 groups having the higher atomic number - for each carbon separately - are on the same side of the C = C
⇑
C C
Cl
H Cl
H
C C
Cl
H H
Cl
I II
- 5 -
II. Reactions of alkenes 1. Hydrogenation of Alkenes
C CH2
Pd / Pt / NiC C
H H A catalyst lowers the energy of activation of a reaction by taking part in the mechanism:
Platinum (palladium or nickel) is a heterogeneous catalyst; it is insoluble in the reaction medium, and the alkene hydrogenation occurs on the metal surface. The process of surface attraction is called adsorption. The alkene and hydrogen are adsorbed on the catalyst surface, and the p bond of the alkene, as well as the s bond of the hydrogen are weakened; the hydrogenation can thus occur easily on the metal surface. The two hydrogens are transferred from the same side of the alkene double bond, or in a syn fashion. This reaction is thus diastereoselective (not enantioselective, because the hydrogens can attack the two sides of the alkene double bond at equal rates).
D
H
H3CD
H
C2H5
C2H5
H
DH3C
H
D
C C
H3C
D D
C2H5
H2
Pd+
RS SR (Not RR + SS) Note that this reaction is diastereoselective, not enantioselective.
- 6 -
Heats of Hydrogenation In hydrogenation, HΔ is negative because you are making two C-H σ bonds and breaking one C-C π bond and one H-H σ bond. Therefore, a potential energy diagram where isomeric alkenes give the same alkane looks like this:
CH3CH2CH2CH3
CH2 CH CH2CH3
C C
H
H3C CH3
H
C C
H
H3C H
CH3
CH3CH CHCH3cis-
trans- CH3CH CHCH3
30.1 |ΔH| in kcal/mol 28.4 27.4
For these double bond isomers, the lower the heat of hydrogenation (in absolute value), the more stable the alkene. We can see how Saytzeff’s rule can be derived: CH2=CHCH2CH3 < CH3CH=CHCH3 and trans-2-butene is more stable than cis-2-butene (for steric reasons): cis-CH3CH=CHCH3 < trans- CH3CH=CHCH3 2. Electrophilic Addition Reactions Alkenes are Lewis bases, they can attack a variety of electrophiles. Addition of HX
C C + HX C C
H X The addition of H-X across the alkene double bond occurs with the following mechanism:
C C H+RDS
C C
H
C C
H
+XFast Step
C C
HX
HX H + X
- 7 -
When the alkene is unsymmetrical, there are two possibilities for attack of H+:
CH2 H+CH3CH
CH3 CH CH2
H
CH2 CH CH3
H
a
b
2° C
1° C
X
X
CH3 CH CH2
HX
CH2 CH CH3
HX
Major
Minor
Since the transition state for this reaction looks like the carbocation formed (Hammond’s postulate), the formation of a secondary carbocation is favored; thus the hydrogen adds to the carbon holding the most hydrogens, in order to produce the most stable carbocation. Markovnikov’s rule (determines the regiochemistry of addition of HX): In the addition of H-X to an alkene, the hydrogen atom adds to the carbon atom of the double bond that has the greatest number of hydrogen atoms. This reaction is not stereoselective, because it produces a flat carbocation, which can be attacked by X- from either side; thus it results in equal syn and anti addition of H+ and X-. HX: can be HF < HCl <HBr < HI (order of reactivity). HX can also be H2SO4: this reaction forms an alkyl hydrogen sulfate, which can be hydrolyzed to an alcohol:
C C C C
OSO3HH
Conc. H2SO4 H2OC C
OHH
alkyl hydrogen sulfate alcohol Hydration of an alkene: (when HX is H2SO4 , H2O); this reaction forms alcohols, and its reverse is the dehydration of an alcohol (or E1 reaction). The principle of microscopic reversibility states that this reaction and its reverse should occur via the same mechanism.
C C
OH H
HC C + C C
H2OH
C C
HO
H H
- H
To drive the reaction from the alkene to the alcohol, we use dilute H2SO4 at room temperature.
- 8 -
To drive the reverse reaction, we heat the alcohol in the presence of H2SO4 and we distill the alkene out; (the alkene should have a significantly lower boiling point than the alcohol, since the alcohol’s intermolecular forces are hydrogen bonds, which are much stronger than the Van der Waals forces holding the alkene molecules together). The alkene hydration reaction has Markovnikov regiochemistry, and is not stereoselective. Addition of Aδ+-Bδ-:
CH3CH CH2 A Bδ−δ+
CH3 CH CH2
A
B CH3 CH CH2
A
+
B
Aδ+ adds to the alkene carbon that has the highest number of hydrogens; for example: + -
I - Clδ δ
Hydroboration of Alkenes Here the Lewis acid is BH3 (borane); this molecule is not stable as the monomer, and either diborane B2H6 (dimer of borane) or BH3.THF (borane, tetrahydrofuran reagent) are good sources of BH3. Hydroboration of an alkene, followed by treatment with H2O2 under basic conditions, yields an alcohol; thus, H and OH are added across the alkene double bond:
C C
R2
R1 H
R3
B2H6
Or BH3 . THF
H2O2
NaOHR1
H
R2
H
OH
R3
+ enantiomer
Mechanism:
H3C C
CH3
CH4
BH
R
R
C C
CH3 H
HH3C
H BR2δ−
δ+C C
CH3 H
H
BR2
H3C
H
H2O2
OH−C C
CH3 H
H
OH
H3C
H
syn retention
B goes to the least crowded C; δ+ is preferable on alkene C with most alkyl groups. This reaction results in an anti-Markovnikov regiochemistry of addition of H and OH; that is H adds to the alkene carbon with most alkyl groups; the stereochemistry of addition of H and OH is syn. When BH3 is added to an alkene, three alkene molecules insert in the three B-H bonds to form a trialkylborane; this then undergoes oxidation to three alcohol molecules:
H2C CH2BH33 B
CH2
CH2H2C
CH2
H
CH2H
H2CH
H2O2
OH−CH2OH CH2 H3
- 9 -
Addition of Halogens (X2 = Cl2 or Br2)
C C C C
Br
Br
vicinal dihalide (1,2-dihalide)
Br2
(brown)
(colorless) The bromination reaction can be used as a color test to determine the presence of a C=C double (or triple) bond; bromine is a brown liquid, and its brown color rapidly disappears upon contact with the alkene (or alkyne). Mechanism: Temporary dipole in Br2
C C
Br
C C +
Br Brδ−δ+
Br
Bromonium ion The bromonium ion is a more stable form of a carbocation (which does not form in thi reaction), where every atom has an octet.
C C
Br
C C
Br
carbocationC has 6 e−
Does not formEvery atom has its octet
The bromonium ion undergoes an SN2 reaction with the nucleophile Br-, resulting in a backside attack. The bromination reaction thus results in the addition of two Br’s in an anti fashion; this reaction is thus diastereoselective.
C C
Br
Br
SN2C C
Br
Br
- 10 -
H3C
DD
C2H5
Cl2D
DH3C
C2H5
Cl
ClC2H5
Cl
D
D
Cl
H3C
+
RR SS (RR + SS not RS + SR) Halohydrin formation The reaction of X2 in H2O (X2 = Cl2 or Br2) results in addition of X and OH across the double bond; the product is a halohydrin:
H3C
C2H5
CH3
H
Cl2C2H5
CH3
H3C
H
OH
Cl
(+ enantiomer)H2O
The OH adds to the alkene carbon with the most alkyl groups, the X adds to the alkene carbon with the most hydrogens; the OH and X add in an anti fashion. Mechanism:
C CR2
H
R3
Cl
R1
C C +
Cl Clδ−δ+
R2
R1
H
R3
Clδ+
The chloronium ion formed is unsymmetrical; the chlorine is closer to the carbon whose positive charge needs the most stabilization; it is thius closer to the carbon atom with the most hydrogens. The other carbon thus holds a partial positive charge and the nucleophile H2O attacks this carbon.
C CR2
H
R3
Cl
R1
δ+
O
H
H
SN2C C
R2
H
R3
Cl
R1
OH
H
- H+C C
R2
H
R3
Cl
R1
HO
This SN2 backside attack leads to an anti stereochemistry of OH, X addition. This reaction is thus diastereoselective.
- 11 -
Epoxides Halohydrins are useful precursors to epoxides, or oxacyclopropanes:
C C
OH
Br
NaOHC C
O
Br
Intramolecular SN2 C C
O
Epoxides, in turn, are very useful starting materials in organic chemistry. Because of ring strain, nucleophiles can readily attack the epoxide carbons, resulting in ring-opened products. For example, the reaction of NaOH with an epoxide leads to a vicinal diol:
C C
O
HONaOH C C
O
HO
H2O C C
OH
HO From the starting alkene, this series of reactions (alkene → halohydrin → epoxide → vic-diol) has resulted in the anti addition of two –OH groups across the double bond. A more direct method to prepare epoxides from alkenes involves treatment of the alkene with a peroxyacid (RCO3H); this results in oxygen transfer from the peroxy acid to the alkene, in a syn-fashion, to form an epoxide directly:
C CR C
O
O OC C
O
e.g.
MCPBA Peracetic acid
H
R C
O
O O H C
O
O O H
Cl
H3C C
O
O O H
Peroxyacid
The following compound is a naturally occuring epoxide: dispalure is the sex attractant pheromone of the female gypsy moth. There has been increasing interest in using insect pheromones to control insect populations. These are not as toxic as pesticides, and insects cannot develop immunity to their own pheromones.
OH H
Disparlure
- 12 -
Cationic Polymerization of Alkenes When an alkene in high concentration is placed in the presence of an acid, a polymer is formed:
H3C CH CH2H
High conc.CH3 CH
CH2 H
CH3CHCH2CH3 CH
CH3
CH2 CH
CH3
CH3 CH
CH3
CH2 CH
CH3
CH3CHCH2CH3 CH
CH3
CH2 CH
CH3
CH2 CH
CH3
CH2 CH
CH3 n
Polymer(Polypropylene)
Common Polymers:
Name Formula Examples of Uses polyethylene
(branched, low density) -(CH2-CH2)n- water, shampoo bottles
plastic bags cling wrap
polyethylene (linear, high density)
-(CH2-CH2)n- plastic trays, baby bottles toys, radio/TV cabinets
polystyrene CH2 CHn
plastic drinking cups CD cases
styrofoam cups computer cases/ hairdryers
polyvinyl chloride -(CH2-CHCl)n- pipes, linoleum
vinyl music LP’s polytetrafluoroethylene
(Teflon) -(CF2-CF2)n- non-stick coating for frying
pans
polymethyl methacrylate CH2 C
nCOCH3
O
CH3
Plexiglas Latex paints
polyvinylidene chloride -(CH2-CCl2)n- Saran wrap polyvinylidene fluoride -(CH2-CF2)n- piezoelectric material for
acoustic tweeters polyvinylpyrrolidone CH2 CH
N On
hairspray
nylon (polyamide) e.g.,
NH CH2 NH C CH2 CO O
6 6n
fibers for stockings, toothbrushes, clothing, etc.
polycarbonates O C
OO C C
n eyeglasses
CD’s
- 13 -
The monomer cyanoacrylate shown below polymerizes upon contact with the moisture in air. This monomer is sold under the name “crazy glue” and strongly binds two surfaces by forming this polymer spontaneously.
CH2 C
C
n
C
CN
O
OCH3
H2O
CN
OCH3
O
3. Free Radical Additions Addition of HBr/peroxides
CH2 CH CH3
HBr
ROORCH2 CH CH3
Br H The addition of HBr in the presence of peroxides ROOR (or light) produces the alkyl bromide in an anti-Markonikov regiochemistry. Mechanism:
ROOR 2 RO
RO H Br+ +OR H BrInitiation Steps
Br CH2 CH CH3 CH2 CH CH3
Br
(2° free radical)
CH2 CH CH3
Br
H Br CH2 CH CH3
Br H
+ BrPropagation Steps
Not Br CH2CHCH3 CH3 CH CH2 (1° free radical)
Br
Termination Steps ...
∗
∗
This is a free radical reaction, where Br. adds first to the alkene, to produce the most stable alkyl free radical (3o>2o>1o). Thus the net result is that Br adds to the alkene carbon with the highest number of hydrogens, and the hydrogen adds to the carbon with the most alkyl groups, i.e. in an anti-Markovnikov sense.
- 14 -
Free Radical Polymerization Polymers are formed from alkenes when a high concentration of the alkene is heated in the presence of a peroxide:
CH2 CH CH3ROOR
CH2 CH
CH3 n
Polymer(Polypropylene)
(high concentration) Δ
Mechanism:
CH2 CH CH3
ROOR 2 RO
RO CH2 CH CH3 CH2 CH
CH3
RO CH2 CH
CH3
RO CH2 CH
CH3
CH2 CH
CH3 n 4. Alkene Oxidations Calculation of formal oxidation states for carbon • When C is atached to more electronegative atom (e.g. N, O, S, Cl, etc.); add +1 for every bond • When C is attached to less electronegative atom (e.g., H, B, metals like Li, Mg...); add -1 for each bond • When C is attached to C; zero contribution to the formal oxidation state • Add these numbers to obtain the formal oxidation state of each carbon in molecule. • For example:
C C
H
H H
CH3-1
-1 -1-2
-1
0
C C
H
H
OH
CH3
H
OH
-1
-1
-1
+1 +1-1
00
1 e− lost, Oxidation
1 e− lost, Oxidation Hydroxylation- Formation of 1,2-Diols
C CKMnO4
Cold, BasicC C
OH OH
syn- addition
KMnO4 is used as a cold, basic solution. Upon addition to the alkene, the purple color of this solution disappears, and a brown solid (MnO2) is formed. This is another color test for a double (or triple) bond. The product of this reaction is a vicinal diol, and it results from a syn addition of two –OH groups to the alkene. This is due to the formation of an intermediate which is converted to the syn- diol. Recall that we learned a method to form an anti- 1,2- diol from an alkene.
- 15 -
Ozonolysis When an alkene is reacted with ozone gas, followed by zinc/H2O treatment, the alkene double bond breaks, and aldehydes and ketones are formed:
C C
H3C
H CH3
CH3
O3 Zn
H2OC O
H3C
H
O C
CH3
CH3
+
"Breakage"
Aldehyde Ketone Mechanism:
C C
Zn
H2OC O O C+
O
OO
C C
O O
O
C C
O O
O
Ozonide
This reaction has been extensively used to break down unknown alkenes to the aldehydes and ketones, identify these and deduce the structure of the alkene. Nowadays, the chemist possesses a large variety of instrumental analysis methods to identify unknown molecules. Vigorous oxidation: formation of carboxylic acids KMnO4 can also serve as a vigorous oxidizing agent: when heated with alkenes, the alkene double bond is broken, resulting in carboxylic acids and ketones:
C C
H3C
H CH3
CH3
C O
H3C
HO
O C
CH3
CH3
+
Carboxylic Acid Ketone
KMnO4Δ
5. Cycloadditions Formation of Cyclopropanes Structure of Carbenes
Methylene; the carbon is neutral, but has six electrons; this is another unstable intermediate in organic chemistry (in addition to carbocations and free radicals). The carbon in methylene is sp2 hybridized; thus, one possible structure of carbenes is the Singlet carbene:
CH H
sp2
Methylene
- 16 -
C
H
HC
H
H
Here, the two electrons are paired in the lower energy sp2 orbital; this is a singlet carbene, and it will behave like an electrophile (it has an empty p orbital which accepts two electrons). Triplet carbene:
C
H
HCH
H
Here, one electron is in the sp2 orbital, and the other is in the higher energy p orbital; the advantage of this is that the two electrons are not paired. This is a triplet carbene, and it will behave like a diradical (it will accept one electron at a time from the alkene). When a singlet carbene adds to an alkene, the addition is concerted and syn; thus, a cis-alkene will give a cis-cyclopropane (trans-alkene → trans-cyclopropane).
C C
R
H H
R
C
H H
C
CH2
CRR
H H
When a triplet carbene adds to an alkene, the addition is stepwise, forms free radicals which can rotate around the C-C single bonds; this results in both syn- and anti- addition; thus, a cis- alkene will yield both cis- and trans-cyclopropanes.
C C
R
H H
R
C
H
H
C
C
C
RR
H H
H
H
C
C
C
RR
H H
H
H
Rotation around C-C
C
C
C
RR
H H
H H
C
C
C
HR
H R
H
H
C
C
C
HR
H R
H H
- 17 -
Sources of Carbenes (a) Diazomethane:
C C +hν
C C
CH2
H2C N N
Diazomethane
hνCH2 + N2
CH2N2
Upon photolysis, diazomethane is converted to methylene, which adds to alkenes to form cyclopropanes; the addition is usually (but not always) syn-. (b) Simmons-Smith Reagent: When treated with a Zn(Cu) couple (zinc particles activated with copper), diiodomethane is converted to the Simmons-Smith reagent:
CH2I2Zn(Cu) CH2I ZnI
Simmons-Smith Reagent This reagent behaves like a carbene; it adds in a concerted fashion to alkenes to produce cyclopropanes:
CH2
I ZnI
C C
R
H H
R
C
CH2
CRR
H H+ ZnI2
The Simmons-Smith reagent is usually generated in the presence of the alkene with which it will react; this is a more useful method than the diazomethane reaction, because it is a syn-addition, and it is safer (diazomethane is explosive under certain conditions).
CH2I2
Zn(Cu)
H3C
H
CH3
H
(c) α-Elimination: Recall from previous chapters β-Elimination:
C C
H
Br
O
E2β-elimination
C C
- 18 -
Here is α-Elimination:
C ClH
Cl
Cl
O Acid-Base reaction C Cl
Cl
Cl
- Cl C
ClCl
Note that H and Cl are eliminated from the same C. Usually, dihalocarbenes (:CX2) are in the singlet state; they thus add to alkenes in a concerted fashion to form dihalocyclopropanes; a cis-alkene yields a cis cyclopropane (and trans-alkene → trans-cyclopropane).
C
C
C
HCH3
H CH3
Cl Cl
+ CHCl3 + tBuOK
Diels-Alder Reaction
+
Diene Dienophile
1
2
3
4
5
6
1
2
3
4
5
61
2
3
4
5
6
This reaction involves heating together a 1,3-diene, and an alkene (called dienophile), to form a cyclohexene. It is a concerted reaction, involving the breaking of the C1-C2, C3-C4, and C5-C6 π bonds, and the simultaneous formation of new σ bonds between C1-C6 and C4-C5, as well as a new π bond between C2-C3. a. For this reaction to occur under normal conditions, the diene must be electron-rich (most unsubstituted or alkyl substituted dienes are) and the alkene (dienophile) must be electron-poor (must have electron-withdrawing groups attached to it, e.g.
CH2 CH C
O
R CH2 CH C N
e− withdrawing groups
- 19 -
+200°C vs. +
30°CH3C
H3C
H
O O
H
b. The diene must be in the s-cis conformation:
1
5
6
1
2
3
4
5
62
34
100°C+ H
OO
H
s-trans s-cis The s-trans-conformation would give an unstable trans-cyclohexene transition state:
Locked in s-cis Locked in s-trans Thus, dienes locked in the s-cis conformation are excellent substrates for the Diels-Alder reaction, and dienes locked in the s-trans conformation do not react. c. The reaction is concerted and is thus stereoselective:
+2
34
1
5
6
O
O
H
H
cis-alkene
O
O
H
H
O
H
OH cis
d. Addition of the dienophile to the alkene results in the endo- product as the major product. This is because the electron poor C=O and C≡N π clouds like to stack underneath the electron rich π-cloud of the diene:
- 20 -
ENDO
EXO
This phenomenon is called π-stacking and leads to the positioning of the C=O of the dienophile on the same side of the new C2-C3 π bond. The π-π interaction is also found between the parallel bases in double helical DNA.
π-stacking interaction
- 21 -
III. Structure and Nomenclature of alkynes Molecular Formula CnH2n-2 C2H2 ethyne (common name: acetylene) C3H4 propyne Structure:
C C
π(Cp, Cp)
σ(Csp2, Csp2)
H
H
π(Cp, Cp)σ(Csp, H1s)
C CH H
sp
The C-H bond in acetylene is σ (Csp, Hs); the C orbital possesses a large amount of s character, and the electrons in this bond are closer to the carbon atom; the acetylene proton is acidic.
C CH Hδ+
Example of nomenclature:
C1
C2
CH2
3
CH2
4
CH3
5C
6CH3
7
CH3
CH3
6,6-Dimethyl-3-heptyne
IV. Preparation of Alkynes From Vicinal Dihalides - Elimination In a vicinal dihalide, the first E2 reaction (using KOH as base) gives a vinylic halide. Because a vinylic bond is difficult to break, a stronger base is needed for the second elimination. NaNH2 is used (it can be used for both eliminations).
C C
H
Br
H
Br
KOHC C
H
Br
KOHNo Reaction
NaNH2 C C
C C
H
Br
H
Br
2 NaNH2C C
vinyl halidevicinal dihalide
- 22 -
A geminal dihalide (where the two halogens are on the same carbon) can also be used for this reaction.
C C
H
Br
Br
H
2 NaNH2C C
geminal dihalide From shorter alkynes This hydrogen is acidic → H-C≡C-R; Terminal alkyne (has H-C≡C). Relative acidities: H2O > ROH > H-C≡C-H > NH3 > CH2=CH2 > CH3-CH3 To generate the conjugate base of this terminal alkyne:
NaNH2C CR H C CR Na + NH3
Weaker acidStronger acid
C CR H C CR Na + H2O
Weaker acid Stronger acidNaOH
This conjugate base R-C≡C- can react with an alkyl halide R’-X in an SN2 mechanism to give a longer alkyne, R-C≡C-R’; note that the alkyl halide can be primary (or secondary, in some cases), but tertiary alkyl halides result in elimination (E2).
C CRCH3 CH2 Br
1°, SN2C CR CH2 CH3
C CR CH2 C CH33°, E2
CH3
Br
H
C CR H C
CH3
CH3
+ CH2 (Cannot be used to generate R - C ≡ C - R')
- 23 -
V. Reaction of alkynes 1. Hydrogenation of Alkynes When using H2/Pt, it is difficult to stop an alkyne from undergoing hydrogenation all the way to an alkane:
C C
H
H
H
H
C CH2
PtC C
H H
H2
Pt
To hydrogenate an alkyne to an alkane, two methods may be used: a. syn-Hydrogenation: Use a catalyst which has been modified by depositing palladium on CaCO3 and treatment with quinoline (Lindlar’s catalyst). Alkynes can bind to this catalyst and undergo hydrogenation; however alkenes cannot bind to this catalyst. Thus, hydrogenation stops at the alkene.
C CH2
Pd / CaCO3C C
H HQuinoline
cis-alkene
This method results in the syn-addition of hydrogen across the triple bond, and a cis-alkene is obtained. b. anti-Hydrogenation: This occurs when alkynes are treated with Lithium or Sodium metal in liquid ammonia or ethylamine (at –78oC). This reaction results in anti-addition of hydrogen across the triple bond to yield trans-alkenes.
C CNa
liq. NH3C C
H
H
trans-alkene
2. Addition of HX and X2 Alkynes add 2 molar equivalents of HX (HCl, HBr, HI) to produce geminal dihalides; the additions occur in a Markovnikov fashion:
C CR H C CR
X H
HHX C CR
X H
H
X H
HX
When alkynes react with HBr in the presence of peroxides ROOR, the geminal dihalide is obtained in an anti-Markovnikov fashion:
- 24 -
C CR H C CR
H Br
H
H Br
2 HBr
ROOR
Alkynes can add two molar equivalents of X2 (Cl2, Br2) to give the tetrahalide:
C C C C
X X
X X
2 X2
With the addition of one molar equivalent of X2, the trans-dihalide is obtained.
C C
X
X
C C X2
3. Hydration of Alkynes- Keto-Enol Tautomerism Alkynes undergo a hydration reaction with H2O in the presence of acid (H2SO4) and mercuric ions (HgSO4). However, instead of giving vinyl alcohols (enols), a ketone is obtained:
C C
H R
OH
C CH R
H
H2OH2SO4 / HgSO4
C C
H
H
H
O
R
Enol(Unstable)
Ketone
Keto-Enol Tautomerism:
C C
OH
Enol
C C
H O
Keto Mechanism:
C C
OH
Enol
C C
H
O
KetoH
C C
H
O H- H
- 25 -
Tautomers: constitutional isomers which exist in rapid equilibrium with one another. Thus, hydration of an alkyne leads to a ketone:
C CH3C HH2O
H2SO4 / HgSO4H3C C
O
CH3
And hydroboration/oxidation of an alkyne leads to an aldehyde:
C CH3C H CH2 C
O
H1. B2H6
2. H2O2 / HOCH3
4. Ozonolysis Ozonolysis of alkynes, followed by treatment with H2O (or acetic acid) gives carboxylic acids:
C CH3C CH2CH31. O3
2. H2OC O O C+
HO
H3C CH2CH3
OH
"Breakage"
VI. Exercises Exercise 1 Synthesize the following molecules from the indicated starting material and any needed reagent: (a) 2-propanol from 1-propanol (b) 1,2-dibromopropane from 2-bromopropane (c) 1-bromo-2-propanol from 2-propanol (d) t-butyl iodide from isobutyl iodide (e) 1,2-epoxypropane from 2-propanol (f) trans-2-chlorocyclohexanol from cyclohexyl chloride (g) cyclopentyl iodide from cyclopentane Exercise 2 Suggest a reasonable mechanism for the following reaction:
Br2
CH3OHOCH3
Br
+ enantiomer
- 26 -
Exercise 3 Give the products of the following reactions:
CH3
1. B2H6
2. H2O2 / HO
CH3Cl2H2O
1. O3
2. Zn, H2O
CH3
CH3
H3C CH3
H2
Pd
H2SO4
H2O
H2C C C CH2
CH3 CH3
+ O
O
O
C C
H
H3C C2H5
H
KMnO4
Cold, Basic
CH3
C2H5
O + C
O
C C C
O
OCH3H3CO
C C
H3C
H CH3
H
CH2I2
Zn(Cu)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
- 27 -
Exercise 4 Suggest reagents suitable for carrying out each step of the following synthesis:
a
Br
Br
b
Br
Br
O
cO
O
O
O
O
d
Exercise 5 Show how you could prepare the following compounds from cyclopentene:
C O
H Br
Br(a) (b)
Exercise 6 Synthesize the following compounds:
OO O
from
H
H3C
H
CH3
Cl Cl
from C CH3C CH3
CH3
H OH
CH3
HO H
from C CH3C CH3
H
O
O
from
Br
Br
Br
fromBr
(a)
(b)
(c)
(d)
(e)
- 28 -
Exercise 7 Give the major product of the following reactions:
Br2
C CH H1. NaNH2
2. CH3I1. NaNH22. CH3I
H2
Pd / CaCO3 / Quinoline
CH2I2
Zn(Cu)
Br Cl
CH2
1. O32. Zn / H2O
1. O32. Zn / H2O
(excess)
Cl2H2O
NaOH
Br
Br
NaNH2
(excess)H2O
H2SO4 / HgSO4
+
O
HΔ
CH3
H3C
KMnO4Cold, Basic
(a)
(b)
(c)
(d)
(f)
(g)
(h)
(i)
(e)
Exercise 8 Give the structures of A, B, and C:
AC10H18O
H2SO4
ΔB
C10H16
+ CC10H16
B1. O3
2. Zn / H2O2 Oand
- 29 -
VII. Solutions Exercise 1
OH H2SO4
Δ
H2SO4
H2O
OH
(Markovnikov)
Br
KOHΔ
Br2
Br
Br
OH
H2SO4
Δ
Br2
H2O
OH
Br
IKOH
Δ
HI
I
(Markovnikov)
OH
H2SO4
Δ
Br2
H2O
OH
Br
NaOHO
Cl
KOHΔ
Cl2H2O
OH
Cl
Br2
hν
BrKOH
ΔHI
I
II2
hν
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Exercise 2
Br Brδ+ δ−
BrCH3OH
SN2
O
Br
H
CH3
- H
O
Br
CH3
+ enantiomer + enantiomer+ enantiomer(when Br attacks from below)
- 30 -
Exercise 3 CH3
1. B2H6
2. H2O2 / HO
CH3Cl2H2O
1. O3
2. Zn, H2O
CH3
CH3
H3C CH3
H2
Pd
CH2 CH CH
CH3
CH3
CH2
C
C
CH2
H3C
H3C
+ O
O
O
C C
H
H3C C2H5
H
KMnO4
Cold, Basic
C2H5
H OH
CH3
H OH
O+
C O
C
C
C O
OCH3
OCH3
C C
H3C
H CH3
H
CH2I2
Zn(Cu)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
CH3
H
OH
Hsyn-additionanti-Markovnikov
+ enantiomer
CH3
Cl
OH
H
+ enantiomeranti-addition-OH on most substituted C
OO
CH3
CH3
H3C CH3
H
H
syn-additionFrom opposite side of CH3
HCH3 CH C
CH3
CH3
H
RearrangementCH3 CH C
CH3
CH3
H
H2OCH3 CH2 C
CH3
CH3
OH
H3C
H3C
O
O
O
OH
HC2H5
OH
HH3C
C2H5
OHH
CH3
OHH
C2H5
HHO
CH3
HHO
O
O
O
OCH3
OCH3
H
H3C
H
CH3
H
H
≡≡ ≡
- 31 -
Exercise 4
Br
Br
Br
Br
O
OO
O
O
O
Cl
C
Br
Br
Br2
O
O OH
Br
Br
O OLDA (excess)
Δ(2 E2 reactions)
O
O
O
(a)
(b)
(c)
(d) (Diels-Alder)
N
Li
LDA
Exercise 5
C O
H
(a)
(b)
+
O
H
Br2
Br
Br
2 KOHΔ
O
H
ΔC O
H
CHBr3
K OBr
Br
α-elimination C
Br
H Br
Br
O C
Br
Br
Br
C
BrBr
- 32 -
Exercise 6
O
O
O
H
H3C
H
CH3
Cl Cl
C CH3C CH3
CH3
H OH
CH3
HO H
C CH3C CH3
H
O
O
Br
Br
Br
Br
(a)
(b)
(c)
(d)
(e)
+ O
O
O
H2
Pd / CaCO3 / QuinolineCHCl3
K O
H2
Pd / CaCO3 / QuinolineC C
H3C CH3
H H
CH3CO3H 1. NaOH2. H2O
HHO
CH3
OHH
CH3 ≡
1. Li2. CuI3. CH3I
CH3
Br2
hν
CH3
Br KOH
Δ
CH3
1. O3
2. Zn / H2O
+ C CH
HC CH NaNH2
C CH2 HBrROOR
Exercise 7
Br2
C CH H1. NaNH22. CH3I
1. NaNH22. CH3I
H2
Pd / CaCO3 Quinoline
CH2I2
Zn(Cu)
Br Cl
CH2
1. O32. Zn / H2O
(a)
(b)
(c)
(d)
H3C
Br
H
CH3
Br
H
+ enantiomer
C CH CH3 C CH3C CH3
H3C
H H
CH3
Br
Cl
O
H2C O+
- 33 -
1. O3
2. Zn / H2O(excess)
Cl2H2O
NaOH
Br
Br
NaNH2
(excess)
H2OH2SO4 / HgSO4
+
O
HΔ
CH3
H3C
KMnO4Cold, Basic
(f)
(g)
(h)
(i)
(e)
O O
O O
H
H
H
H
CH3
Cl
OH
CH3
O
C CH3C H H3C C CH3
O
O
H
CH3H3C
H
OH
OH
H
Exercise 8
A
H2SO4
Δ+
CB
1. O3
2. Zn / H2O2 O
OH
B
VIII. the textbook: Solomons, 9th edition Suggested Reading: Chapter 7 + Chapter 8 + Chapter 13 section 11 + Chapter 16 section 5B Suggested Problems: Chapter 7: 3, 5, 7, 10, 15, 16, 19, 21, 24, 26, 33, 40, 43. Chapter 8: 2, 3, 8, 11-13, 21, 24, 26, 32, 35, 39.
- 1 -
Hanadi SLEIMAN CHEM 212 Chapter 6
- 2 -
I. benzene and its properties
3
• Proposed structures for benzene
3
• Additional facts about benzene 4
II. resonance 6
Iii. aromaticity 14
iV. Nomenclature of benzene compounds 16
V. reactions of benzene 17
• halogenation 18
• nitration 19
• sulfonation and desulfonation 20
• friedel-crafts 21
VI. reactivity and orientation 24
• alkyl-substituted benzene 24
• aniline 25
• nitrobenzene 27
• halobenzenes 27
• summary 29
VIi. Exercises 30
VIiI. Solutions 31
ix. the textbook: Solomons, 9th edition 34
- 3 -
I. Benzene and its Properties Isolated by M. Faraday in 1825, benzene was extensively investigated for its unusual properties:
1. Its structural formula is C6H6 (it has C6H14 – C6H6 = 8H 42
= sites of unsaturation; thus benzene has 4
double bonds/rings). 2. It is unusually inert. 3. It does not behave like an alkene: it does not decolorize Br2, and does not react with cold KMnO4. 4. It undergoes substitution with Br2/FeBr3 (Lewis acid). C6H6 + Br2/FeBr3 → C6H5Br (one isomer) C6H5Br + Br2/FeBr3 → C6H5Br2 (three isomers: 1,2-, 1,3- and 1,4-) Proposed Structures for Benzene Kekule (1865):
But it has double bonds (should react with Br2 and KMnO4); it also has two 1,2-dibromo isomers (across the single, and the double bonds).
Br
Br Br
Br
Kekule proposed that benzene was a rapidly equilibrating mixture of isomers:
So that on average, all benzene bonds looked equivalent; this may explain the fact that only one 1,2-dibromobenzene can be detected. Armstrong (1866):
- 4 -
Dewar (1867):
But it has double bonds (should react with KMnO4 and Br2), and has two monobromo isomers; Dewar’s benzene has been synthesized and shown to convert to benzene upon heating. Ladenburg (1869): was also synthesized and shown to convert to benzene when heated.
Additional facts about benzene X-Ray structure:
1. Flat molecule 2. All C-C-C angles 120o (all carbons sp2)
3. Equivalent C-C bonds: C-C bond length 1.4 A ; so it falls between C-C single bonds (1.54 A ) and C=C
double bonds (1.33 A ). Heat of Hydrogenation
H2, Pd
2 H2, Pd
3 H2, Pd
ΔH = - 30 kcal/mol
Predicted ΔH ~ - 57 kcal/molExperimental ΔH = - 55 kcal/mol (close)
Predicted ΔH ~ - 85.8 kcal/molExperimental ΔH = - 49.8 kcal/mol
- 5 -
cyclohexatriene (predicted)
30 55 85.8 49.8
36 kcal/mol:resonance stabilization of benzene
|ΔH| (kcal/mol)
Orbitals of Benzene Instead of forming three localized π-bonds, the six parallel p orbitals of benzene share their six electrons in delocalized π-bonding; this delocalization of electrons is called resonance and is a stabilizing factor.
Resonance, or electron delocalization can only occur with electrons in parallel p-orbitals (π electrons). Thus, benzene cannot be represented by this structure (which has localized π-bonds):
Nor this structure:
Benzene’s structure is a combination, or hybrid, of these two structures:
Where each bond is 1 σ, and ½ π- bond (for a total of three π-bonds).
- 6 -
II. Resonance 1. Resonance occur when a molecule can be represented by two or more Lewis structures, with the same position of atoms, but a different distribution of electrons; these Lewis structures are called resonance contributing structures (or resonance structures); a double headed arrow is used to represent resonance:
C C HH
H
H
O
C CH
H O
H
H
tautomers, not resonance structures
2. The resonance structures are not real; the real structure of the molecule is a combination, or hybrid, of these resonance structures:
⇒ Benzene is
3. Only delocalize : π-bond lone pair + charge (atom with 6 electrons) odd electron (free radical) These are called resonance elements; they can only be delocalized if separated by one bond: CH2 CH CH2 CH2 CH CH2 CH2 CH CH2 CH CH2Yes NoYes
Why?
CH2 CH CH CH CH3 CH2 CH CH CH CH3 (delocalization)
CH2 CH CH2 CH CH2 CH2 CH C CH CH2
H
H
(localized π bonds)
Resonance requires a continuous set of parallel p-orbitals.
- 7 -
4. Resonance Energy: Stabilization upon delocalization of electrons. a. The higher the number of resonance structures, the more stable the hybrid (the higher the resonance energy). b. The closer the resonance structures are in energy, the more stable the resonance hybrid. 5. All resonance structures have the same number of paired and unpaired electrons. Examples: 1. The Allyl Carbocation:
CH2 CH CH2 CH2 CH CH2CH2 CH CH2
2 π e-
CH2 CH CH2
δ+δ+
Allyl carbocation
The allyl carbocation is resonance stabilized (2 π-electrons can be delocalized over the three parallel p orbitals); its + charge is equally divided between two atoms; this stabilization is significant enough that a primary allylic carbocation is more stable than a secondary carbocation; this means that allyl halides and allyl alcohols readily undergo SN1 reactions: 3° carbocation > 1° allylic > 2° >> 1° Another consequence of this resonance stabilization is the allyl rearrangement: once the allylic carbocation is generated, the nucleophile can attack either of the positively charged carbons, resulting in two products:
CH2 CH CH CH3
Cl
2°, allylic
CH3OH
SN1CH2 CH CH CH3 CH2 CH CH CH3
CH2 CH CH CH3
δ+ δ+
CH3OH- H+ - H+
CH2 CH CH CH3
OCH3
CH2 CH CH CH3
OCH3
a b
The benzylic carbocation is also resonance stabilized:
CH2 CH2 CH2 CH2 CH2
- 8 -
δ+
CH2Hybrid δ+
δ+
A primary benzylic carbocation is as stable as a primary allylic carbocation: 3° carbocation > > 2° >> 1° carbocation1° allyl
1° benzyl 2. The Heteroatom-Substituted Cation
CH
H
O CH3
I
δ+ δ+
C
H
H
O CH3
II(especially stable,
every atom has its octet)
C
H
H
O CH3
2π e-
C
H
H
O CH3
Hybrid
X = heteroatom (atom other than C or H, e.g. O, N, S, halogen). This carbocation is resonance stabilized (2 π-electrons delocalized over two p-orbitals); resonance structure II is especially stable because every atom has an octet; it stabilizes the overall hybrid, and this carbocation looks more like II. But, in fact, there are two opposing effects for this cation: • Heteroatoms like N, O, etc. are more electronegative than carbon; they withdraw electrons by inductive effects, and thus destabilize a carbocation:
C
H
H
OCH3
• Heteroatoms usually have a lone pair, and thus donate electrons to the carbocation by resonance; they can thus stabilize the carbocation by resonance effects.
CH
H
O CH3 C
H
H
O CH3
If X = N, O, P, S; resonance effects win out; they stabilize carbocations when directly attached to them. If X = F, Cl, Br, I; inductive effects win out; they destabilize carbocations when directly attached to them.
- 9 -
HCCH2 O CH3
H+
HIHCCH2 O CH3
H
I-
favored carbocation over
HCCH2 O CH3
H
HCCH2 O CH3
H I
Note: In order to delocalize the lone pair electrons with the carbocation p orbital, we had to place the lone pair electrons in a p orbital parallel to the carbocation p orbital; this means that we had to change the hybridization of the oxygen from sp3 to (sp2,p) in order to create a p orbital for this lone pair; this is the only violation of the CN (coordination number rule): thus, a lone pair which can be delocalized into a π-system is placed in a p-orbital (by changing the hybridization from sp3 to sp2,p).
C
H
H
OCH3
2π e-
H2C OCH3
p sp3
H2C OCH3
p
sp2
3. The Allyl Carbanion
CH2 CH CH2 CH2 CH CH2CH2 CH CH2
4 π e-
CH2 CH CH2
δ-δ-
Hybrid The allyl carbanion is also resonance stabilized (4 π-electrons can be delocalized over the three p-orbitals); its – charge is delocalized over two carbons; this means that an allyl carbanion is a weaker base than an alkyl carbanion (less likely to give its electrons).
CH2 CH CH2 weaker base than CH3CH2CH2
4. The Carboxylate anion Similarly, the carboxylate anion (conjugate base of a carboxylic acid) is resonance stabilized; the – charge is distributed equally over the two oxygens in the resonance hybrid:
H3C C
O
O
H3C C
O
O
H3C C
O
O
δ−
δ−
- 10 -
This also means that the carboxylate anion is a weaker base than an alkoxide anion, and thus a carboxylic acid is a stronger acid than an alcohol:
H3C C
O
OH
+ H2O H3O+ H3C C
O
O
+Stronger acid Weaker base
CH3CH2OH CH3CH2O+ H2O H3O+ +Weaker acid Stronger base
The overall order of acidity is: RCOOH > H2O > ROH > H-C≡C-H > NH3 > Alkenes > Alkanes 5. Conjugated Dienes CH2 CH CH CH2 Two double bonds separated by one bond are conjugated; they can delocalize their electrons; 1,3-butadiene, for example, has three resonance contributing structures:
CH2 CH CH CH2 CH2 CH CH CH2
CH2 CH CH CH2 CH2 CH CH CH2
I II
I III I is more stable than II or III; thus 1,3-butadiene will have a small contribution of structures II and III; essentially, this gives a small amount of π-character to the C2-C3 bond of conjugated dienes; thus, this introduces a barrier to rotation around the C2-C3 bond (recall that there is no rotation around a π-bond under normal conditions). CH2 CH CH CH21 2 3 4
CH2 CH CH CH2
4 π e−
Conjugated dienes are resonance stabilized, and are thus more stable than isolated dienes; cumulated dienes (or 1,2-dienes) possess a significant amount of strain, and are the least stable dienes:
CH2 CH CH CH CH3 CH2 CH CH2 CH CH2 CH2 C CH CH2 CH3
Conjugated Isolated Cumulated
More stable> >
- 11 -
Conjugated dienes are thus preferentially formed in elimination reactions:
Br
Br
2 KOH
Δ
Major Minor
ΔCH2 CH CH2 CH3
Br
KOHCH CH CH2 CH3 CH2 CH CH CH3
Major Minor
Another consequence of resonance stabilization of conjugated dienes, is their modified reactivity with electrophiles. Consider the reaction of 1,3-butadiene with HCl (H+, Cl-): H+ can either attack the outer carbon, to form a secondary, and allylic carbocation:
CH2 CH CH CH2
H
aCH2 CH CH CH2
H 2°, allyl
Or H+ can attack the inner carbon and form a primary (non-allylic) carbocation:
CH2 CH CH CH2
H
bCH2 CH CH CH2
1°H
Pathway a is preferred; this produces an allylic carbocation, with two possible resonance contributing structures, I and II; thus the + charge on the allyl carbocation is distributed over two carbons; this means that the nucleophile Cl- can attack either of these carbons, resulting in two products: the 1,2-addition and the 1,4-addition product:
CH2 CH CH CH2
H
CH2 CH CH CH2
H
I II
CH2 CH CH CH2
H
δ+δ+
Cla b
CH2 CH CH CH2
H Cl
CH2 CH CH CH2
H Cl
a 1,2-product
b 1,4-product
- 12 -
Which of these products is the major product? Experimentally, the 1,2-product is the major product at low temperature, and the 1,4-product is the major product at higher temperature:
HCl
HCl
-80°C
+40°C
H
Cl
H
Cl
+
H
Cl
H
Cl
+
Kinetic control
Thermodynamic control
80 % 20%
80 % 20%How can we explain this? First, we would predict that the 1,4-product is more stable than the 1,2-product, because its double bond has more alkyl substituents than the 1,2-product. Second, we would predict that the 1,2-product should be formed faster: this is because the carbocation intermediate looks more like the resonance structure I (secondary carbocation) than II (primary carbocation); this means that there is more + charge on the secondary carbon in the carbocation; thus the nucleophile attacks the carbocation at this secondary carbon faster than at the primary carbocation:
CH2 CH CH CH2
H
CH2 CH CH CH2
H
I II
CH2 CH CH CH2
H
δ'+δ+
Clmore stable less stable
slowerfaster
δ+ > δ'+
• The low temperature reaction occurs under kinetic control: the major product is the product which is formed faster (1,2-product). • The higher temperature reaction occurs under thermodynamic control: the major product is the product which is most stable (1,4-product). This is what the potential energy diagram looks like:
At low temperature, the carbocation intermediate can go through the lowest Ea pathway: I → II → III; thus the 1,2-product is formed (it is called the kinetic product).
- 13 -
At higher temperature, molecules have more energy: they can go through: I → II → III; III → II → I; (which means the reaction is reversible) then also I → II’ → III’; but once at the 1,4-product (III’), it is more difficult to go through III’ → II’ → I’. Thus the 1,4-product is the major product, under thermodynamic control, i.e. when the reaction is reversible (it is called the thermodynamic product). Other examples of 1,2- and 1,4-additions:
HBrROOR
Low Temperature
Br
H
Br
H
HBrROOR
High Temperature
Br2
Low Temperature
Br
Br
BrBr2
High Temperature Br
6. The Allyl Free Radical The allyl free radical is resonance stabilized (3 π-electrons can be delocalized over the three parallel p orbitals); its odd electron density is equally divided between two atoms; this stabilization is significant enough that a primary allylic free radical is more stable than a tertiary free radical; similarly, the benzylic free radical is resonance stabilized, and as stable as the allyl free radical:
CH2 CH CH2 CH2 CH CH2 CH2 CH CH2
3 π e-
CH2 CH CH2
δ δ
This means that these free radicals are preferentially formed in free radical substitution reactions: 1° allyl, 1° benzyl > 3° > 2° > 1° free radical
CH2CH2CH3Br2
hνCH CH2CH3
Br
There are two methods to carry out allylic halogenations: → Use low concentrations of Cl2 (or Br2) at high temperature. → Use N-bromosuccinimide, as a source of bromine in low concentration. These two methods can give exclusive allylic chlorination or bromination:
CH2 CH CH2 CH3Low conc. Cl2
500°CCH2 CH CH CH3
Cl
CH2 CH CH2 CH3N-Bromosuccinimide (NBS)
Δ
N
Br
O O
CH2 CH CH CH3
Br
- 14 -
III. Aromaticity Normal delocalization of electrons gives 1,3-butadiene, for example, ca. 3 kcal/mol stabilization. The amount of resonance stabilization of benzene (36 kcal/mol) is unusually high, and indicates more unique stabilizing properties. Benzene is, in fact, a member of a class of unusually stable molecules, called aromatic compounds; they all possess the following features: 1. They are cyclic. 2. They are planar and fully conjugated; this means that every member of the cycle possesses a p orbital parallel to the others (continuous π-system). 3. They possess 4 2n + π-electrons ; where n = integer = 0,1,2,3, etc.; that is they can have 2, or 6, or 10, or 14, etc. π-electrons. The combination of these features imparts extra stability on the molecule, which is termed aromaticity, and the molecule is called aromatic (this phenomenon is very similar to the octet for atoms, and how the closed outer shell of noble gases gives stability to these atoms). Examples: 1.
6 π e-
Benzene is cyclic, fully conjugated (every atom has a p orbital parallel to the others), and has 6 π-electrons (three π-bonds); it is thus aromatic (36 kcal/mol extra stability). 2.
H Hsp3
Cyclopentadiene is cyclic, but not fully conjugated; it is not aromatic.
6 π e-
H The cyclopentadienyl anion is cyclic and fully conjugated (remember, we can place the lone pair in a p orbital); it is thus aromatic, and extra stable; this means that cyclopentadiene is quite acidic (since its conjugate base is stable).
- 15 -
3.
Cyclooctatetraene is cyclic, fully conjugated, but has 8 π-electrons; actually when a molecule is cyclic, fully conjugated, and has 4n π-electrons, it is antiaromatic, and unstable (this electron count gives an electronic configuration with two unpaired electrons, i.e. a diradical, and an unstable molecule). In fact, cyclooctatetraene is so unstable as an antiaromatic molecule that it prefers to pucker up as a tub-shaped molecule; this way, its p-orbitals are no longer parallel, and it is not antiaromatic:
4.
6 π e-
N
H
N
Pyrrole is cyclic, fully conjugated (lone pair on N in a p orbital) and has 6 π-electrons; it is aromatic. 5.
NNsp2
6 π e-
In pyridine, the N lone pair is in an sp2 orbital perpendicular to the π-system; thus it is not part of the system; pyridine then is cyclic, fully conjugated and has 6 π-electrons: it is aromatic. This means, in practice, that if you count one pair of electrons, or π-bond on an atom as part of the π-system, you cannot count any other lone pairs on the same atom as part of this π-system; for example, in Furan, you can only count one lone pair on oxygen as part of the π-system, and the other lone pair resides in an orbital perpendicular to this π-system:
6 π e-
O O
- 16 -
IV. Nomenclature of Benzene Compounds
Ph- φ- Phenyl group CH2 Benzyl group
Monosubstituted Benzenes
Cl NO2 CH3 OH NH2 OCH3
Chlorobenzene Nitrobenzene Toluene Phenol Aniline Anisole
C C SO3H
O
H
O
OH
Benzaldehyde Benzoic acid Benzene sulfonic acid
Disubstituted Benzenes
Cl Cl Cl
Cl
Cl
Clortho meta para
o-Dichlorobenzene m-Dichlorobenzene p-Dichlorobenzene
- 17 -
Polysubstituted Benzenes
CH3 NH2 NH2
NO2
NO2
O2N H3C CH3
NO2
NO2
2,4,6-Trinitrotoluene 2-Methyl-5-nitroaniline 2-Methyl-3-nitroaniline (sweet taste) (tasteless)
12
3
4
5
6
V. Reactions of Benzene: Electrophilic Aromatic Substitution Alkenes carry out electrophilic addition:
C C E Yδ+ δ-
+ C C
E
Y+ Y C C
EY Benzene, which has an electron-rich π-system, should react similarly with electrophiles E-Y, to give a carbocation; this carbocation is resonance stabilized, and is a hybrid of three resonance contributing structures:
E Yδ+ δ-
H
E
H
E
H
E
H
E
Not aromatic However, this carbocation is no longer aromatic (has an sp3 center in the cycle); now if the nucleophile Y- attacks this carbocation in an electrophilic addition (like alkenes), aromaticity is completely lost in the product. This does not occur with benzene:
YH
E
H
EH Y
Aromaticity lost
Does not occurNo reaction
- 18 -
On the other hand, if Y- acts like a base and removes the H+ from the β-hydrogen (like the second step of an E1 reaction), then aromaticity is restored.
H YY +H
EE
The net result of this is substitution of the electrophile for the hydrogen; this is electrophilic aromatic substitution. Its overall mechanism is thus:
E Yδ+ δ-
H
E
H
E
H
E
RDSSlow
+ YStep 1
Step 2 H YY +
H
EE
Fast
The first step, which forms the carbocation is the slow, or rate-determining step. Because C-H bond
breaking does not occur in the rate-determining step, 1H
D
kk
≈ for this reaction.
Because benzene is more inert than alkenes, an activated electrophile is needed (e.g. Br2 does not react with benzene, but Br2/FeBr3 does). Halogenation of Benzene (chlorination or bromination)
Cl2AlCl3
Lewis acid
Cl
+ HCl Br2
FeBr3Lewis acid
Br
+ HBr
Mechanism Step 1: The activated electrophile is generated
Br Br FeBr3 Br Br FeBr3
Step 2
Br Br H
Br
H
Br
H
Br
+FeBr3 Fe
Br
Br Br
Br
- 19 -
Step 3
Fe
Br
Br Br
Br
H Br+H
BrBr
+ FeBr3catalyst
Nitration of Benzene
HNO3
NO2
H2SO4
Nitrobenzene Mechanism Step 1: Generation of the activated electrophile NO2
+(nitronium ion).
NOH
O
O
+ H2SO4 NOH
O
OH
+ HSO4- H2O + N
O
ONitronium ion
Step 2
N
O
O
H
N
H
NO2
H
NO2
O
O
Step 3
+H
NO2
NO2
H2O H3O+
We can also obtain aniline by reduction of nitrobenzene:
NO2 NH2
Nitrobenzene Aniline
SnHCl
- 20 -
Sulfonation and Desulfonation This is a reversible reaction; to sulfonate benzene, we use concentrated sulfuric acid at 30oC; a better reagent is fuming sulfuric acid (SO3/H2SO4):
conc. H2SO4 , 30°CSO3 / H2SO4Or
SO3H
Benzene sulfonic acid
To desulfonate benzenesulfonic acid to benzene, we use dilute acid at high temperature (100-175oC):
SO3H
dilute H2SO4
Δ
H
Mechanism Step 1: The activated electrophile SO3 is generated:
S
O
O
OHHO + H2SO4 S
O
O
OHOH
H
+ HSO4− S
O
O O H+ H2O S
O
O O+ H3O+
Sulfur trioxide Step 2
H
SO3
H
SO3
H
SO3
S
O
O O
Step 3
+H
SO3
SO3
H2O H3O+
Step 4
+
SO3
H2SO4
SO3H
+ HSO4−
- 21 -
Friedel-Crafts Reactions 1. Friedel-Crafts Alkylation
+ R ClAlCl3
R
+ H Cl
Mechanism Step 1: The activated electrophile, R+ (carbocation) is generated:
R Cl AlCl3 R Cl AlCl3 R + Al
Cl
Cl Cl
Clcarbocation
Step 2
H
R
H
R
H
R
R
Step 3
Al
Cl
Cl Cl
Cl
H Cl+H
RR
+ AlCl3
There are other methods to generate a carbocation (by treating alcohols with acid / by treating alkenes with acid); these can also be used in a Friedel-Crafts Alkylation:
C
CH3
H3C
CH3
OH
H2SO4
CCH3
CH3
CH3
C
CH3
H2C
CH3
HF
CCH3
CH3
CH3
C
CH3
H3C
CH3
OH H2SO4+ C
CH3
H3C
CH3
C
CH3
H2C
CH3
HF+ C
CH3
H2C
CH3H
- 22 -
Problems with Friedel-Crafts Alkylation 1. Alkyl carbocations undergo rearrangement; therefore, for example, n-propylbenzene cannot be synthesized from benzene by a Friedel-Crafts alkylation:
?
CH3CH2CH2ClAlCl3 CH3CH2CH2 CH3CHCH3
CH3CH2CH2Cl
AlCl3
CHCH3
CH3
⇒
2. Alkylation of a benzene ring makes this ring more electron-rich, because alkyl groups are electron-releasing; thus this ring is more susceptible to attack by electrophile than benzene itself; thus overalkylation is a common problem of Friedel-Crafts alkylation reactions:
AlCl3
RCl
R
AlCl3
RCl
R
R
overalkylation
more reactive
more reactive 3. Friedel-Crafts reactions cannot place a vinyl, or a phenyl group on a benzene ring. This is because the vinylic (sp2) C-Cl bond in vinyl halides and phenyl halides is difficult to break:
AlCl3
⇒
CH2 CH Cl No reaction
AlCl3 No reaction
AlCl3
AlCl3
CH2 CH Cl
Cl φ-Cl
CH CH2 Not obtained
Not obtained
2. Friedel-Crafts Acylation
+ R C
O
Cl
C
O
R
+ HCl
Acyl chloride
AlCl3
- 23 -
Mechanism Step 1: The activated electrophile RCO+ (acylium ion) is generated:
R C
O
Cl + AlCl4-AlCl3 AlCl3R C
O
Cl R C O R C O6 e-
more stableevery atom has octet
Acylium ion
Step 2
R
C
O
H
C
H
C
H
C
O
R
O O
R R
Step 3
Al
Cl
Cl Cl
Cl
H Cl+H
CC
+ AlCl3
O
RO
R
The Friedel-Crafts acylation reaction does not have the problems associated with alkylation: a. Acylium ions do not undergo rearrangement. b. Acylation of benzene gives a products which react less readily with electrophiles (than benzene); thus overacylation does not occur.
C
O
R
less reactive than benzene The acylation product can be converted to an alkylation product by reduction, using one of the following two methods:
C
O
RZn(Hg)
HCl
CH2R
Clemmensen Reduction
C
O
R
NH2NH2KOH, Δ
CH2R
Wolff-Kischner Reduction
- 24 -
Problems with both Friedel-Crafts Alkylation and Acylation 1. These reactions do not occur with strongly electron-withdrawing groups on benzene (e.g. CF3, NO2, CN, SO3H, COR). 2. They do not occur with amino substituents (NH2, NHR, NR2); the basic electron pair on nitrogen ties up the Lewis acid AlCl3. VI. Reactivity and Orientation in Electrophilic Aromatic Substitution In electrophilic aromatic substitution, step 1 is the rate-determining step. Since step 1 forms a carbocation, the more stable the carbocation, the faster this step (Hammond postulate) and the faster the overall reaction. It is thus the stability of the carbocation which determines the rate of electrophilic aromatic substitution. Alkyl-Substituted Benzene Consider the electrophilic aromatic substitution of toluene. Since the methyl group is electron-releasing, toluene is more electron-rich, and thus more reactive than benzene towards electrophiles. Thus, alkyl groups are activators of benzene (for electrophilic substitution). What is the orientation of attack of the electrophile (o-, p- or m-)? a. ortho attack
Br Br H
Br
H
Br
H
Br
+FeBr3 Fe
Br
Br Br
Br
Fe
Br
Br Br
Br
H Br+H
BrBr
+ FeBr3
CH3 CH3 CH3 CH3
CH3CH3
b. para attack
Br Br FeBr3
CH3 CH3 CH3CH3
BrH BrH BrH
I'
- 25 -
Fe
Br
Br Br
Br
H Br+ + FeBr3
CH3
BrH
CH3
Br
c. meta attack
Br Br FeBr3
CH3 CH3 CH3CH3
Br
H
Br
H
Br
H
FeBr4-
CH3
Br
The ortho and para attack lead to carbocation resonance structure (I and I’) where the + charge is on the carbon holding the alkyl group, and is thus stabilized (tertiary carbocation). No such structure can be obtained from the meta attack. Thus, the ortho and para attack occur faster than the meta attack (which does not occur). Usually (but not always), the para attack is favored over the ortho attack because of steric effects. Thus, alkyl groups are activators and ortho-, para-directors. Aniline Recall that when N can manifest inductive effects (electron-withdrawing) and resonance effects (electron-releasing), resonance effects win out. Thus NH2 (as well as O, S, P) is an activator towards electrophilic aromatic substitution. What about the orientation of attack? a. ortho attack
Br Br H
Br
H
Br
H
Br
FeBr3
NH2 NH2 NH2 NH2
H
Br
NH2
II - especially stableevery atom has octet
FeBr4-
Br
NH2
- 26 -
b. para attack
NH2
II' - especially stableevery atom has octet
FeBr4-
NH2
Br Br FeBr3
NH2 NH2 NH2NH2
BrH BrH BrH
H Br
Br
c. meta attack
Br Br FeBr3
NH2 NH2 NH2NH2
Br
H
Br
H
Br
H
FeBr4-
NH2
Br
In the ortho and the para attack, especially stable resonance structures of the carbocation (II and II’) are obtained. In these structures, every atom has an octet of electrons, and thus the overall carbocation is extremely stabilized. No structure like II or II’ can be obtained from the meta- attack. In fact, these structures (II and II’) are much more stable than the structures (I and I’) obtained from the electrophilic substitution of toluene. This means that: • NH2 is a strong activator; it activates benzene so well towards electrophiles, that no Lewis acid catalyst is usually needed, and that often, trisubstituted benzenes are obtained:
NH2
Br2
H2O
Br
Br
NH2
Br • NH2 is an ortho-, para-director.
- 27 -
Nitrobenzene The nitro group is strongly electron withdrawing (the nitrogen is positively charged, has no lone pairs, and is bonded to two oxygens); thus nitrobenzene is less reactive than benzene towards electrophiles. The nitro group is thus a deactivator toward electrophilic aromatic substitution. What about orientation? a. ortho attack
Br Br H
Br
H
Br
H
Br
FeBr3
NO2 NO2 NO2 NO2
FeBr4-
Br
NO2
III
b. para-attack
FeBr4-
NO2
Br Br FeBr3
NO2 NO2 NO2NO2
BrH BrH BrH Br
III' c. meta attack
Br Br FeBr3
NO2 NO2 NO2NO2
Br
H
Br
H
Br
H
FeBr4-
NO2
Br
While all three carbocations from the o-, p- or m-attacks are destabilized by the nitro group, the effect is more serious in the ortho- and para- attacks, where structures III and III’, which are especially unstable (because the + charge is on the carbon holding the nitro group) are resonance structures of the carbocation. Thus, the meta attack is favored in this case. The NO2 group is a deactivator, and meta director. Halobenzenes Recall that when halogens can manifest inductive effects (electron-withdrawing) and resonance effects (electron-releasing), inductive effects win out. Thus F, Cl, Br, I are deactivators towards electrophilic aromatic substitution. What about orientation?
- 28 -
a. ortho attack
Br Br H
Br
H
Br
H
Br
FeBr3
ClCl Cl Cl
H
Br
Cl
IV - stableevery atom has octet
FeBr4-
Br
Cl
b. para attack
Cl
IV' - stableevery atom has octet
FeBr4-
Cl
Br Br FeBr3
ClCl ClCl
BrH BrH BrH
H Br
Br
c. meta attack
Br Br FeBr3
Cl Cl ClCl
Br
H
Br
H
Br
H
FeBr4-
Cl
Br
Even though a halobenzene is less reactive than benzene towards electrophiles, the ortho and para attacks are favored over the meta- attack, because they lead to the formation of resonance structures IV and IV’, which result in some stabilization of the carbocation. Thus, halogens are deactivators, and ortho-, para- directors.
- 29 -
Summary Substituents, in decreasing order of activation (towards electrophilic aromatic substitution): 1. ortho, para-directors • Strong activators: NH2, NR2, NHR, OH, SH • Moderate activators: -NH-C(O)-CH3, -OCH3, -O-C(O)-R • Weak activators: alkyl, phenyl • Moderate deactivators: F, Cl, Br, I 2. meta-directors • Strong deactivators: NO2, +NR3, -C(O)X (X = alkyl, halogen, OH, OR), -S(O)2X, CN, CX3 (X=halogen) If two of these groups are on a benzene ring, the more activating, or less deactivating group wins out in its directing ability:
CH3
NO2
1
21
1 2
2 Cl2AlCl3
CH3
NO2
ClMajor
HN
NO2
C CH3
O1
2
1
HNO3
H2SO4
HN
NO2
C CH3
O
NO2
(1 wins over 2)
C
Br
O
OH1
2
2
2
(2 wins over 1)
Cl2AlCl3
+
COOH
Cl
Br
Cl
COOH
Br
- 30 -
VII. Exercises Exercise 1 Which anion is more stable?
CH CH3 CH
O
NOO
CH2O
NO2
or or
CH3 CH CH2 CH CH2 CH3 CH CH CH CH3or
(a) (b)
(c) Exercise 2 Classify the following molecules as aromatic, non-aromatic, or antiaromatic:
O
N
S(A) (B) (C) (D) (E) (F) (G)
Exercise 3 Rank the following in terms of their SN1 reactivity:
NO2
CH2Cl CH2Cl
OCH3
CH2Cl
OCH3
CH2Cl
(A) (B) (C) (D) Exercise 4 Write resonance contributing structures for the following:
H3C N N H2C C
O
H H2C N NH2 S C N NH
(a) (b) (c) (d) (e)
- 31 -
Exercise 5 Synthesize:
O
O
O
from
CH3
from(a) (b)
Exercise 6 Draw the products of the following sequence:
Br2hν
KOHΔ
NBSΔ
H2Pd NaNH2
HC CH 1. B2H6
2. H2O2 , OH-
VIII. Solutions Exercise 1
CH
O
NOO
CH2O
NO2
CH3 CH CH2 CH CH2
CH3 CH CH CH CH3
(a)
(b)
(c)
Source of instability
2 benzene rings ⇒ more conjugated ⇒ more stable
this negative charge is not conjugated
O
NOO
O
NOO
O
NOO
O
NOO
O
NOO
more stable
conjugated - the negative charge is stabilized by resonance ⇒ more stable
the negative charge is isolated
- 32 -
Exercise 2 (A) Aromatic (B) Antiaromatic (C) Antiaromatic (D) Antiaromatic (E) Aromatic (count only one lone pair on oxygen) (F) Antiaromatic (G) Aromatic (N: don’t count lone pair; S: count only one lone pair)
N
Scount
count count
Exercise 3
NO2
CH2 CH2
OCH3
CH2
OCH3
CH2
OCH3
CH2
OCH3
CH2
OCH3
CH2
especially stabeevery atom has octet
OCH3
CH2
OCH3
CH2
OCH3
CH2
OCH3
CH2
(A) (B) (C)
(D)
least stable
The positive charge skips the carbon holding the -OCH3, thus the resonance effects of OCH3 cannot be felt.Can only feel electronegativity of -O ⇒ C destabilized by -OCH3.
C > B > D > A
most reactive for SN1 (most stable carbocation)
- 33 -
Exercise 4
H3C N N
H2C C
O
H
S C N
N
H3C NH
Nmore stable - every atom has octet
H2C C
O
H
more stable - O holds the negative charge
H2C N NH2 H2C N NH2
more stable
S C N
H
N
H
N
H
N
H
N
H
N
H
(a)
(b)
(c)
(d)
(e)
Exercise 5
O
O
O
1
2
3
4
5
6 O
O
O
(a) +
Br2hν
2 KOHΔ
Br
KOHΔ
Br2
Br
BrO
O O
O
O
O
CH3
NBSΔ
Br
Li
Li
CuI
CuI
2
CH3I(b)
- 34 -
Exercise 6
Br2hν
KOHΔ
NBSΔ
H2
Pd
NaNH2
HC CH 1. B2H6
2. H2O2 , OH-
Br
Br Br
Br C CH CH2 C
O
HHC Ci.e.
IX. The Textbook: Solomons 9th edition Suggested Reading Chapter 13 (excluding 13.7C, 13.9) + Chapter 14 (excluding 14.7B, 14.11) + Chapter 15. Suggested Problems Chapter 13: 3, 4, 14, 19, 24, 25, 26, 33, 40 Chapter 14: 1, 20, 22 Chapter 15: 5, 7, 26-29, 31, 35, 36, 49, 51
