Chapter 7: Alkene reactions – conversion to new functional groups
Preparation of alkenes: two common elimination reactions
1. Dehydration of alcohols
Dehydration is a common biochemical reaction in carbohydrate and fatty acid
metabolism and terpene biosynthesis – it’s catalyzed in vivo by specific enzymes.
In the lab, dehydration is an acid-catalyzed elimination reaction. The mechanism
involves formation of a carbocation intermediate (more on eliminations in Ch. 11)
Example: Formation of 2-butene from 2-butanol
H3C
H H2
C C CH3
OH
H2SO4
heat
H3C
C
H
C
H
CH3 + H2O
The regiochemistry of elimination leads to the most substituted
alkene possible = Zaitsev’s rule
2. Dehydrohalogenation of alkyl halides
Alkyl halides also
undergo elimination
to produce alkenes:
What’s the major alkene product of this elimination, based on Zaitsev’s rule?
Heat
Addition Reactions of Alkenes
A. Additions to produce alkyl halides and halohydrins
1. From Ch. 6, Alkene + HX Æ alkyl halide
2. Halogenation: Br2 and Cl2 can add to alkenes just like HBr, HCl and H2O!
Note: Similar reactions do not take place with F2 and I2
H
H
C
Br2
C
H3C
CH3
H3C
H
H
C
C
Br
Br
Product is a
“vicinal"
dibromide
CH3
Where's the electrophile? Br2 makes both electrophile (Br +) and nucleophile (Br -)
Mechanism is a little different: the intermediate is a cyclic cation
H
H
C
H
Br2
C
H
C
C
H
CH 3
H3 C
Br
H3C
H
Br
Br
H
H
C
C
Br
Br
CH 3
Br-
"bromonium ion"
*Rearrangements do not occur with the cyclic cation mechanism!*
Solvent effects:
When planning a synthetic reaction in the lab, solvent should be chosen carefully
For halogenation reactions, solvents such as CH2Cl2 & CCl4 do not interfere
3. Halohydrins: Use of a nucleophilic solvent like H2O alters the outcome of the
reaction, because both the halogen + water react with the alkene
H
H
C
H
C
CH 3
H
Cl2
H2O
H
C
C
H
H
CH3
Cl
O
H
H
H
C
C
Cl
OH
H
a halohydrin
How would the product change if an alcohol was the solvent?
CH3
Stereochemistry of halogenations:
Use of NBS to provide a bromine:
Some key points about this reaction:
1) Br2 furnished by NBS is the electrophile, water is the nucleophile
2) The organic solvent used (DMSO) doesn’t take part in the reaction
3) The benzene double bonds don’t react:
Aromatic C=C do NOT undergo electrophilic addition like alkene C=C!
B. More addition on the "Markovnikov" theme: Alcohol formation rxns
Alkene Æ Alcohol
1. Acid-Catalyzed Hydration:
R
R
C
H 2O
C
R
R
R
H+
R
R
C
C R
H
OH
Mechanism: similar to addition of H - X:
H
H
C
H+
C
H
H
C
C
CH3
H
CH3
H
H
O
H
H
H
C
C CH3
H
H
H
H
O H
H
H
C
C CH3
H
OH
H+
H
Same rules apply as for addition of H - X:
1)
The bond is attracted to an electrophile and a new σ bond forms
2)
The intermediate is the most stable carbocation
3)
The nucleophile reacts with the carbocation to form another new σ bond
4)
Markovnikov's rule is obeyed and the OH group ends up on the most
substituted C.
CH3
Rearrangements
can occur:
H3C
C
H
C
H
CH2
CH3
H3O+
mostly
H3C
H2
C C CH3
OH
Another variation on this theme: alkenes Æ ethers when alcohols are present,
they become the nucleophile and add a -OR group to the molecule:
H
CH3
C
H
C
C 2 H5 OH
H+
CH3
H
H
CH3
C
C
H
H
CH3
O
H
H
CH3
C CH3
C
H
H 3CH 2C
H
O CH 2CH 3
H
H
C
H
C CH3
H
OCH2CH3
H+
an ether
In summary:
Alkenes
Alkenes
Alkenes
Alkenes
+
+
+
+
H-X
HX, H2O
H2O, H+
ROH, H+
Æ
Æ
Æ
Æ
Alkyl halides
Halohydrin
Alcohols
Ethers
2. Oxymercuration-Demercuration: Metal complexes in organic reactions
Some organic reactants are sensitive to harsh reagents (can cause
decomposition!) How do we avoid this in the case of hydration?
A milder reagent for hydration:
Mercuric acetate in THF & water
A two-step synthetic procedure:
H 3C
C
H
CH2
1) Hg(OAc)2, H2O, THF
H
C
H3C
2) NaBH4, OH-
CH3
OH
THF (tetrahydrofuran) is a cyclic ether; a moderately polar solvent
O
“OAc” = CH3COONo rearrangements occur with this process due to cyclic cation intermediate:
OAc
H
H
C
C
Hg(OAc)2 H
H
H
C
C
H3C
H
H3C
Hg
O
-OAc
H3C
H
H
HgOAc
C
C
OH
H
H
+ HOAc
H
Demercuration step: NaBH4 provides a hydride ion, HH
H3C
HgOAc
C
C
OH
H
H
NaBH4
NaOH
H3C
H
C
CH3
+
Hg + -OAc
OH
A similar reaction, alkoxymercuration: Using ROH instead of H2O produces ethers
Advantages of this procedure over hydration:
1) No rearrangements can occur
2) Conditions for reaction are less harsh
How would you prepare each of these alcohols?
3. 1o alcohols by Hydroboration-Oxidation (“Anti-Markovinikov orientation”)
1) BH3
CH3 – CH = CH2
CH3CH2CH2OH
2) OH-, H2O2, H2O
What’s new about this reaction:
1.
2.
The electrophile is a Lewis acid, BH3
The nucleophile is a hydride (H-) from the borane…watch this:
CH3 – HC = CH2
CH3 – CH2 – CH2 – BH2
alkylborane
H – BH2
3.
4.
5.
It’s a concerted reaction (one step, no intermediate) and pericyclic
Cyclic transition state means no rearrangements
Product has “Anti-Markovnikov orientation” but…it only APPEARS to violate
Markovnikov’s rule…the nucleophile (H-) still ends up on the most substituted C,
because this gives the more stable transition state:
Figure from p. 153 here
What happens next:
The alkylborane reacts with 2 more moles of alkene to produce a trialkylborane.
Steric hindrance of alkyl groups promotes further addition to less-substituted C:
CH3-CH=CH2
CH3CH2CH2BH2
CH3-CH=CH2
(CH3CH2CH2)2BH
(CH3CH2CH2)3B
Trialkylborane
The trialkylborane reacts with peroxide and hydroxide ion to release 3 moles of alcohol:
(CH3CH2CH2)3B
H2O2, OH-, H2O
3 CH3CH2CH2OH + BO3-
Summary: Hydroboration is an effective route to 1o alcohols
C. Alkanes by Catalytic Hydrogenation (the "Crisco" reaction)
Suppose you want to prepare alkanes from alkenes? An example from real life:
Vegetable oil
Semi-solid “shortening”
Addition of H2 is catalyzed by Pt or Pd on charcoal, PtO2, or Ni
(heterogeneous catalysts):
R
R
C
C
R
R
Pt/C
R
H2
R
R
C
C R
H
H
The H - H bond is particularly strong; requires a catalyst to help bond break.
Addition occurs on one side of the molecule (syn addition)…see OWL tutorial
H2 adsorbs on the catalyst surface and the reaction occurs there.
D. Cyclopropanation by using Carbenes: carbon adds to the C=C in a
stereospecific way
Carbenes are an electron-deficient, sp2-hybridized species with formula (R)2C:
These are reactive species generated “in situ” by deprotonation of chloroform:
CHCl3
+
KOH
-:CCl3
-:CCl2
+
Cl-
Or generated from similar reagents: Simmons-Smith reaction
Zn/Cu
:CH2
CH2I2
Ether
As electrophiles, carbenes can react with C = C to form a cyclopropane ring:
Cl
H2C = CH2 +
Cl
:CCl2
Stereospecific: the original
arrangement of groups around
C = C bond is retained.
H
H
C
H3C
C
CH3
H
CH2I2
Zn/Cu/ether
H3C
H
C
C
C
H2
CH3
E.
Oxidation of C = C bonds to produce oxygenated functional groups
Oxidation: Reaction resulting in an increased number of bonds from carbon to
oxygen, and a decrease in bonds to hydrogen
Increasingly oxidized functional groups
H
R-CH2CH2-R
H
C
C
CH3
H3 C
alkanes
R1
alkenes
alkynes
C
H
O
O
OH
R2
R1
alcohols
C
R2
ketones
aldehydes
R1
C
OH
acids
esters
Oxidation of C-C pi bonds is a versatile way to introduce new functional groups to
molecules containing the alkene group.
1. Hydroxylation of alkenes: diol preparation
Alkenes can be oxidized by transition metal oxides with a high metal oxidation state
H
H
C
H3C
cold KMnO4, OH-, H2O
C
CH3
1. OsO4
2. H2O, NaHSO 3
H3C
H
H
C
C
OH
OH
CH3
„ Stereospecific syn addition occurs to produce vicinal diols
„ The positive charge on the metal attracts electrons and sets a pericyclic
reaction in motion; π electrons form σ bonds
„ As the organic functional group gets oxidized, the inorganic reagent gets
reduced (by products: MnO2 or OsO3)
„ KMnO4 is cheaper but harsher (can completely oxidize C=C, see next
page)
„ OsO4 is expensive, highly toxic
2. Oxidative Cleavage of alkenes: produces carbonyl compounds by
breaking both σ and π bonds
The 1,2-diols produced by oxidation of alkenes can be further oxidized to carbonyl
compounds by a second pericyclic reaction:
H3C
H
H
C
C
OH
OH
H
H
HIO 4
CH3
C
O
O +
C
+ HIO3
CH3
H3C
„ Products may be aldehydes or ketones depending on structure at diol carbons
„ Significant reaction because C - C bonds are not broken easily
A similar reaction occurs when KMnO4 is used under acid conditions or heat:
R1
R3
C
C
C
H
R2
OH
R1
KMnO 4
H+ or heat
O +
O
C
R3
R2
„ A disubstituted C= (two R groups attached) becomes a ketone carbon
„ A monosubstituted C= (one H attached) becomes fully oxidized to a carboxylic acid
(no aldehydes are produced under these conditions)
„ A =CH2 from a terminal alkene becomes CO2 instead of a carboxylic acid:
H3C
C
H3C
O
H
C
C
H
H3C
+
O C O
CH2
Oxidation of cyclic alkenes opens up the ring resulting in a diacid or a diketone:
KMnO4, H3O +
HOOC
COOH
KMnO4, H3O+
O
O
3. Ozonolysis: A milder, more efficient, "greener" (?) way to do oxidative
cleavage!
Problems with transition metal-mediated oxidations:
„ Reagents form toxic metal compounds as by-products (hazardous waste!)
„ Conditions may be rather harsh, require heat, acids, bases
„ Side reactions: KMnO4 will oxidize any OH or C=O groups in the molecule too
The solution: Ozone!
Cycloaddition of ozone to C=C produces a molozonide which rearranges to an ozonide.
Cleavage under oxidizing or reducing conditions to yield different products as shown:
R2
R1
R2
R1
O
C
O
C
O
R3
H
O3
C
C
R3
R1
O
O
O
H
O
R2
C
O
Z n,
H 3O
C
O
+
C
R2
R1
O
R3
O
+
H
C
R3
O
2O
2
H
O
C
R1
H
C
+
R2
R3
OH
„ A C bearing a single R group yields an aldehyde under reducing cond.
„ Terminal alkenes form formaldehyde or CO2 as the second product
„ Tetra-substituted alkenes form only ketones under any conditions
Ozonolysis is a useful way to determine the structure of an unknown alkene:
„ React the unknown with ozone under controlled conditions
„ Determine the identity of the oxidative cleavage products (the simpler the
molecule, the easier it is to determine structure!)
„ Figure out how the pieces would fit together
“Road map” problems!
Goal: Piece together information from reactions to figure out structures of
unknown compounds. You are given some key pieces of information to help you
figure out what is happening to the molecule in each step.
Example: Compound A has the formula C10H16. On catalytic hydrogenation over palladium (H2,
Pd) it reacts with only one molar equivalent of H2. Compound A also undergoes reaction with
ozone (O3), followed by zinc treatment (Zn, H3O+) to yield a symmetrical diketone, B which has
formula (C10H16O2). Propose plausible structures for A and B.
What reagents would be best to carry out these reactions?
(f)
KMnO4
H3O+
?
H2O, H+
or
1.Hg(OAc)2
2.NaBH4
1.BH3/THF
2.H2O2, OH-
H2/catalyst
Cl2 or Br2, H2O
Cl2 or Br2, CH2Cl2
1.OsO4,
2.NaHSO3, H2O
CH 3
CH2I2,
Zn(Cu), ether
HBr or HCl
Alkene + HX
Æ
Alkene + X2 (Br2, Cl2)
Alkene + X2, H2O
Æ
Alkene + H+, H2O Æ
KMnO4, H3O+
or 1. O3
2. Zn, H3O+ or H2O2
Alkyl halide
Æ
Alkyl dihalide
Halohydrin
Alcohol (Markovnikov)
Alkene + Hg(OAc)2, NaBH4 Æ Alcohol (Markovnikov)
Alkene + BH3/THF, H2O2, OH- Æ Alcohol (Non-Markovnikov)
Alkene + H2/ Pd, Pt or Ni catalyst Æ Alkane
Alkene + CH2I2, Zn/Cu Æ Cyclopropyl alkane
Alkene + OsO4, H2O, NaHSO3 Æ diol + HIO4 Æ aldehydes & ketones
Alkene + KMnO4, H3O+, heat Æ carboxylic acids & ketones
Alkene + O3, Zn, H3O+ Æ aldehydes & ketones