very often, instead of seeing “cis” or “trans” in a name ... · very often, instead of...
TRANSCRIPT
Very often, instead of seeing “cis” or “trans” in a name, you will see an “E” (for trans) or a “Z” (for cis). The E, Z Designation: used for differentiating isomers that are trisubstituted or tetrasubstituted 1. First assign Priorities (High or Low) to Groups connected to each Carbon of C=C.
2. If the High priorities are on the same side of the alkene, it is called a Z isomer (German – “zusammen” – same). If the High priorities are on opposite sides, it is called an E isomer (German – “entgagen” – opposite).
Assigning High/Low Priorities – Cahn-Ingold-Prelog (CIP) Sequence Rules 1. Examine the first atom attached to each C atom of the double bond. Rank according to Atomic Number (higher number = higher priority). Consider 2-bromo-2-butene: Is it E or Z, as shown?
Specifically, this is (2E)-2-bromo-2-butene
2. If the high and low priority cannot be determined by the first atoms that are attached to alkene C’s, continue out branches, considering every three atoms on each round (usually), until a differentiation (assignment) can be made.
orHigh
Low
High
Low
High
Low
Low
High
High
Low
High
Low
High
Low
Low
High
Z E
Br
H3C
H
CH3
Br
H3C
H
CH3
Break down the right side:
And these?
3. Multiple-bonded atoms are equivalent to the same number of single-bonded atoms (i.e. triple bond to C equals three single bonds to C, double bond to O equals two single bonds to O, etc).
E or Z?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Alkene Stability: The more stable the alkene, the less reactive. Not all alkenes are of equal energy. Consider:
H
H3C
CH2CH3
CH2CH2OH
C H, H, C H, H, HH
H3C
CH2CH3
CH2CH2OH C H, H, C H, H, O
H
F
CH2NH2
CH2CH2OH
C H, H, NH
H3C
CH2NH2
C C C, C, CCH3
CH3CH3
CO
CO
O C
C N C NN
N
C
C
would be equivalent to
would be equivalent to
H3C
H
CH2OH
C OH
H3C
H
CH2OH
C OH O C
The 24:76 ratio equates to a specific energy difference of 0.66 kcal/mol (ΔG), favoring the trans alkene. The trans is more stable – it formed more often. Lower energy than the cis alkene. The stability of an alkene is measured by measuring the energy released upon removal of the C=C (a reaction called hydrogenation). More stable alkenes give off LESS energy. Less stable alkenes give off MORE energy. If we remove the double bonds from both cis-2-butene and trans-2-butene, we form the same product, butane, while giving off energy:
The trans alkene gives off less energy than the cis isomer. Trans is more stable than Cis. Note that the energy difference (ΔHº) is 1 kcal/mol. This is very close to that Gibbs Free Energy value (ΔGº) of 0.66 kcal, which we saw earlier, as determined from the 24:76 ratio. 0.66 would round to 1, right? Approximation… Order of Stability for Alkenes:
catalyst
catalyst
catalyst50:50 Mix
24:76
+
24:76
+
24:76
+
H2, Pd/C
Butane+ -28.6 kcal/mol
H2, Pd/C
Butane
+ -27.6 kcal/mol
More alkyl groups equates to more stability in the system. What helps stabilize an alkene (C=C)? 1. Bond strength – There are an increased number of sigma bonds formed by sp2 hybrid orbitals. Sp2-sp3 is stronger and more stable that sp3-sp3 (more s character in sp2 orbitals = better overlap and stronger bonds). More alkyl groups means more sp3-sp2 bonds.
2. Increased electron density in the area of the alkene = more stable double bond system. There are actually empty spaces around an alkene. If you can squeeze more electron density into those empty spaces, you strengthen the double bond. Electron density comes from alkyl group electron clouds. The more alkyl groups, the more electron density around the alkene, the stronger and more stable the alkene! So – Where’s the empty space? Antibonding orbitals - those high-energy orbitals that electrons avoid… A Pi bond: Together: Antibonding Orbitals:
Hyperconjugation – “movement of electron density thru space from a filled sigma bonding molecular orbital of an alkyl group into the empty antibonding pi* molecular orbital of the C=C”.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Electrophilic Addition Reactions – Addition of E+ to C=C
R
R
R
R>
R
R
R
H>
H
R
R
H>
R
H
R
H
R
R
H
H
R
H
H
H>~~ > CH2CH2
π*
C C
H H
H
Consider what must occur: Step One:
Pi bond is nucleophilic (e- rich). Proton (H+) of HBr is electrophilic (e- poor). Pi electrons reach out and form a new sigma bond to the proton, from one of the carbon atoms of the C=C, releasing the Br-. “Carbocation intermediate” forms. Step Two:
The Br- then donates a lone pair of electrons to the positively charged carbon atom of the intermediate, forming a new C-Br bond. Energy Diagram:
The Intermediate: The Carbocation
• Three atoms attached to carbon (no lone pair)
H-Br HBr-
H Br
H-Br Br-Br
H-Br
sp2H3C
H3C H
H H3C
H3C H
H
H
sp2 sp3
Br-
H3C
H3C H
H
H
H3C
H3C H
H
HBr
sp3
Ener
gy
reaction coordinate
R
I
P
• Sp2 hybridized C atom planar, 120º bond angle • Empty perpendicular p orbital (Lewis Acid, E+)
Order of Stability:
What stabilizes a carbocation? Carbocations are carboCATIONS because they are missing an electron so stabilization results when electron density is added to the system. Alkyl groups are quite good at sharing electron density. 1. Hyperconjugation: movement of electron density thru space from a filled sigma bonding molecular orbital of an alkyl group to the empty p orbital of the carbocation.
Surrounding an electron-deficient carbon atom with electron density stabilizes the positive charge. 2. Inductive Effect: movement of electron density thru bonds Electrons in the alkyl groups C-C sigma bonds shift towards the carbon with the positive charge, delocalizing the positive charge, spreading it out over multiple atoms.
Alkyl groups can shift e- density better than solitary H atoms (which are not polarizable – their electron clouds are not big enough to “shift”). The electrophilic addition reaction of HX:
> >RR
R HR
RH
R
H > CH3
3º 2º 1º
>>>>
R
R
HH
H
R
R R
HBr
BrH
Note that in each of the prior cases, only one product forms. Look at that first reaction - the product that forms is very specific:
These types of reactions are called regiospecific reactions: “reactions that could potentially produce two or more products but actually only favor formation of one specific product”. Markovnikov’s Rule: In an electrophilic addition reaction to an alkene, the reaction will always proceed to produce the most stable carbocation intermediate. What does this mean? The carbocation always forms on the more substituted end of the alkene. More substituted end = position of more substituted carbocation = place of more stable carbocation. The more stable carbocation has the lowest activation barrier and thus the fastest reaction. As a result, the H attaches to the end with the fewer alkyl groups and the X attaches to the end with more alkyl groups. Ex.
If the alkene is equally substituted and symmetrical, there is no “more stable carbocation” because they would form exactly the same on either end, so “Markovnikov” does not actually apply:
If the alkene is equally substituted and asymmetrical, a mixture of two different products occurs, from the two different but equally stable carbocations. Getting two
HCl
HCl
KI, H3PO4
I
H
H-Br
Brnot
Br
H-Cl ClH
HBr
Br
H
H
Brand
products from a reaction is generally considered a bad reaction – the products would have to be separated before either could be used in another reaction.
Draw some products:
Last section: How do we know that carbocations really exist? Experimental info tells us so… In the 1930’s, F.C. Whitmore did the addition of HCl to 3-methyl-1-butene. And had some surprising results:
The first product makes sense but the second product was unexpected. Where did it come from? You need to look at the mechanism for the process.
H-Cl
Cl
H
+Cl
H
KI, H3PO4
H-Br
H-Cl
H-Br
H-Cland
Cl
Cl
H-Cl
Cl
Cl
Cl
Cl
And how do you go from the first carbocation to the second carbocation? It’s called a hydride shift. That’s a hydrogen with TWO electrons and a negative charge as a result. Hydride Shift:
Other things can also shift… Another example would be a methide shift:
A methide shift is in the following:
What shifted in these?
Chapter 8: Alkenes – Reactions and Synthesis Synthesis is the construction of a molecule through the use of two or more consecutive reactions, whether to increase the carbon skeleton or change the functional groups. Preparations of Alkenes: Elimination reactions a. Dehydrohalogenation – occurs with an alkyl halide by the addition of a strong base Strong Bases: KOtBu
HH H
H
H H
H-Cl andCl
Cl
H-BrBr
H-Cl
Cl
Br
KOtBu or
O CCH3
CH3
CH3
K
Mechanism (revisit in more detail in Ch. 11)
Draw the Products: Ex.
Ex.
Ex:
b. Dehydration – occurs with alcohols in the presence of a mineral acid Common acids – H2SO4, H3PO4
Mechanism (revisit in more detail again in Ch. 11):
Draw the Products: Ex.
H
X!
" Baseα
β
BrO C
CH3
CH3
CH3
H
KOtBuI
ClKOtBu
BrKOtBu
heatH2SO4
OH
or
H+
OH OH2
H H2O
Ex.
Reactions of Alkenes: Recall “Regiospecific”: a reaction that potentially could produce two or more products but favors the formation of one. 1. Addition of HX (X = Cl, Br) – Markovnikov
Don’t forget KI, H3PO4! 2. Acid-Catalyzed Hydration – Addition of H, OH Markovnikov
Draw the Products: Ex.
Ex.
Ex.
OH
H2SO4
heat
heatH2SO4
OH
HX
H2O, H2SO4OH
H
H2O, H3PO4
H2O, H2SO4
H2O, H3PO4
Mechanism: Similar first step as in addition of HX… Remember? Pi bond reaches out to grab H+…
Reaction occurs in presence of mineral acid, higher temps can be required. Rather brutal conditions. Carbocation intermediates can rearrange. Thus this reaction is really only suitable for very simple alkenes. On paper though, everything works… 3. Oxymercuration-Demercuration – Addn of H, OH Markovnikov
Ex.
Ex.
Ex.
Step 2 is the conversion of mercury to a hydrogen atom (commonly referred to as a reduction reaction or “demercuration”). The mechanism will not be discussed.
H-O-HH+
H
O
H
H H
H
OH
H-O-H
1. Hg(OAc)2, H2O2. NaBH4
H
OH
Hg(OAc)2 = Mercury (II) diacetate OHg
O
O O
1. Hg(OAc)2, H2O2. NaBH4
1. Hg(OAc)2, H2O2. NaBH4
1. Hg(OAc)2, H2O2. NaBH4
The regiochemistry of this reaction is determined in Step 1, which is the only step we will look at mechanistically. This step occurs via a three-membered mercurinium ion:
Why “Markovnikov”? First, remember than any positive charge, full or partial, is more stable on a more substituted carbon. This makes one C-Hg bond weaker and easier to break. Whichever end is more substituted will result in a δ+ charge that is more stable and faster to form and react.
For clarification: As the nucleophile approaches the mercurinium ion, the partial charge on the right carbon is 3º (Option A), the most stabilized, lowest energy partial charge, with the lowest activation barrier to reach the transition state:
The partial charge on the left carbon is 2º (Option B), whose partial charge will be less stable, higher in energy and a higher energy transition state, with higher activation barrier:
H CH3
OAcHg
H CH3
Hg Hg
H CH3O
HH
Hg
H CH3O
H
OAcAcO
OAc
H-O-H
AcO
OAcAcO
HgAcO
H CH3
Hg
H CH3
Nuc
δ+Hg
H CH3
Nuc
Option AOAc OAc
The more substituted side would form a more stable partial charge.
More stable intermediate = Markovnikov “Stereospecific”: a reaction that potentially could produce more than one 3-dimensional arrangement but favors the formation of a specific one. 4. Hydroboration-Oxidation – Addition of H, OH “Non-Markovnikov” SYN Addition (same side!)
Draw the Products: Ex.
Ex.
Ex.
BH3 – Borane – 6 electrons on B (Group III element)
• Hybridization of Boron is Sp2 • Boron has an empty perpendicular p orbital – GREAT Lewis Acid and Electrophile
Borane has three “hydrides” (H:-) attached and can react up to three times with three alkenes.
δ+Hg
H CH3
Nuc
Hg
H CH3
Nuc
Option BOAc OAc
1. BH3
2. NaOH, H2O2
H
OH
1. BH3
2. NaOH, H2O2
1. BH3
2. NaOH, H2O2
1. BH3
2. NaOH, H2O2
Step 2 is the conversion of B to O (as in OH). The mechanism for this step will not be one you are required to know. As this does not affect either the regiochemistry or stereochemistry, we will focus on Step 1 only. Step 1 Mechanism: “Concerted, 4-membered cyclic transition state” (no intermediate formation – no chance for the alkene to rotate!)
Why “Syn”? Because of this required four-membered, cyclic transition state, B and H must add to same face resulting in SYN stereochemistry. Note – in Step 2, the B is converted into an OH group so the H and OH are on the same face. Why “Non-Markovnikov”? B-H bond can line up in 2 possible ways with the alkene pi bond: Option A: Placement of the boron under the less substituted side:
Option B: Place of the boron under the more substituted side:
Which works best and why? The favored transition state is found in Option A and occurs for two reasons: a. Sterics: More unstable steric interactions occur when B is located under extra alkyl group on more substituted side (as seen in Option B)
b. Electronics: as the electron density shifts from alkene towards B of BH3, a δ+ charge develops. The more highly substituted partial charge is more stable, will have a more stable transition state, lower activation barrier and thus forms faster.
H BH2 BH2H BH2H
H CH3 H CH3H2B H
H CH3 H CH3H BH2
H CH3H2B H
H CH3H BH2
vs
5. Cyclopropanation Reactions – SYN additions always (cannot form a three-membered ring on both faces at the same time!)
Carbene Reactions: :CR2 Neutral Only 6 e- (e- poor – no octet) Sp2 hybridized, empty p orbital Both a Lewis Acid and a Lewis Base Reagents: a. CH2N2 Diazomethane
The reactive species forms with the loss of N2 gas:
The carbene then reacts with the alkene:
Draw the products: Ex.
Ex.
H CH3 H CH3HB H BHH
vs3º 2º
d+
d- d
-
d+
+
CH2N2
C N NH
HC
H
H+ N2
CH
HC
H
H
CH2N2
b. CHCl3, KOH
Draw the product: Ex.
Ex.
c. Carbenoid Reaction: Simmons-Smith Reaction (“:CR2”)
Draw the product: Ex.
Ex.
CH2N2
OHCCl
ClCl H
CHCl3, KOH Cl
Cl
CHCl3, KOH
CHCl3, KOH
CH2I2, Zn(Cu) H
H
CH(CH3)I2, Zn(Cu) CH3
H
Ex.
6. Epoxide Formation
mCPBA = meta-chloroperbenzoic acid
Ex.
mCPBA is an electrophilic oxidizing agent (e- poor) and reacts with the most electron-rich alkene present. Ex.
More alkyl groups on an alkene make the alkene more electron rich!
CH(CH2CH3)I2, Zn(Cu) CH2CH3
H
C(CH3)2I2, Zn(Cu) CH3
CH3
mCPBAO O+
OOHCl
O
mCPBA
1 equiv. mCPBA