Alkynes II
Reading: Wade chapter 9, sections 9-8 – 9-10
Study Problems: 9-32, 9-33, 9-34, 9-37, 9-38, 9-39
Key Concepts and Skills:
•
Propose effective single- and multistep syntheses of alkynes
•
Predict the products of additions, oxidations, reductions, and cleavages of
alkynes, including regiochemistry and stereochemistry.
Lecture Topics:
I.
Reduction of Alkynes
a. Hydrogenation
Treatment of an alkyne with hydrogen over palladium results in complete reduction to the
alkane, via syn addition of two equivalents of hydrogen across the triple bond:
H
2 H–H
H3C
H
H3C
CH3
Pd(0)
CH3
H
H
b. Lindlar reduction.
Treatment of an alkyne with a poisoned palladium catalyst known as Lindlar’s catalyst
(Pd(0), BaSO4, quinoline) leads to formation of cis alkenes though addition of one
equivalent of H2 across the triple bond. The cis alkene formed cannot undergo further
addition because of its steric size relative to the alkyne; the poisoned palladium catalyst
will not accept the bulky alkene as a ligand
broken: H-H
104 kcal/mol
H2, Pd(0)
H 3C
H3C
CH3
ΔH=-42 kcal/mol
CH3
BaSO4,
quinoline
broken:
alkyne pi bond:54kcal/mol
H
H
cis-alkene
2 (100kcal/mol)
2 C-H bonds formed
c. Alkali-Metal/ ammonia reduction
Alkynes are reduced to trans alkenes by treatment with lithium or sodium in ammonia.
The alkali-metal ammonia complex is a deep blue solution of “solvated” electrons which
adde to the alkyne to give a radical anion and thence a vinyl radical after protonation.
Note that the vinyl radical prefers the trans orientation of substituents at low
temperatures, and thus trans stereochemistry is usually observed in the product alkenes.
Na + NH3 ==>
(H3N)e- +
solvated
electrons
Na+
H3C
eH3C
CH3
CH3
radical anion
H H H
N
fast
H3C
H3C
eCH3
H
CH3
H
slow
vinyl radical
(prefers trans)
vinyl anion
H H H
N
H3C
H
CH3
H
trans alkene
II.
Addition of X2 and HX to alkynes
a. Addition of halogens, X2
Addition of one equivalent of Br2 or Cl2 to an alkyne results in cis-trans mixtures of dihal
alkenes; a vinyl cation intermediate is postulated as giving rise to this result. Addition of
two equivalents of halide leads to a practical preparation of tetrahalides:
Br2, 1 mole
H 3C
H3C
CH3
CH3
H 3C
Br
+
Br
Br
Br
CH3
Br
2 Br2
CH3
H 3C
Br
Br
H 3C
sp
vinyl cation intermediate
CH3
Br
Br
Cl
H3CH2C
H
Cl2, 2 mole
Cl
H3CH2C
H
Cl
Cl
b. Addition of hydrogen halides, H–X
These additions follow Markovnikov’s rule, with the halide ending up on the more
substituted carbon atom. A vinyl cation intermediate is involved. Internal alkynes give
cis/trans mixtures of vinyl halides. Addition of a second mole of H-X yields a dibromo
alkane. Terminal alkynes give one vinyl halide and one dihalo alkane; internal alkynes
can give four vinyl halides and two dihaloalkanes.
H–Br
H3CH2C
H3CH2C
H
Br
H
Markovnikov product
H
1mole
Br-
HBr (second mole)
H
H3CH2C
Br
H
H3CH2C
more substituted vinyl
cation produced
H3CH2C
CH3
Br
1 mole
HBr
CH3
H3CH2C
Br
H3CH2C
CH3
Br
H
H
Br
H
Br
H
+
Br
H
H3CH2C
CH3
second mole
HBr
Br
H
H3CH2C
CH3
H
CH3
CH3
+
+
H
H3CH2C
Br
Br
CH3
H3CH2C
c. Anti-Markovnikov addition of H-X
Radical-promoted addition of HBr to alkynes results in formation of anti-Markovnikov
vinyl bromides; a vinyl radical intermediate is postulated:
H3CH2C
H
HBr
H
ROOR,
heat
H
H3CH2C
H
+
H3CH2C
H
Br
Br
Br
H–Br
sp2
H
H
H3CH2C
H
H3CH2C
H3CH2C
Br
Br
Br
p
Br–H
sp2
At elevated temperatures, the vinyl radical equilibrates so that a mixture of cis and trans
vinyl bromides are obtained.
III.
Hydration of alkynes
a. Oxymercuration/Markovnikov Hydration
Reaction of terminal alkynes with mercuric sulfate in the presence of aqueous acid results
in the production of a vinyl alcohol (enol) intermediate that rapidly tautomerizes to a
more stable carbonyl compound known as a ketone. The driving force for this
isomerization is the formation of the more stable C=O (164kcal/mol) from the C=C (144
kcal/mol). Internal alkynes give regioisomeric ketone mixtures.
HgSO4
H3CH2C
CH3
H3CH2C
H
O
H2O, H2SO4
Hg2+
H3CH2C
Hg+
Hg+
H3CH2C
H+
H
H3CH2C
H
HO
H
H
HO
vinyl alcohol
(enol)
H2O
H3CH2C
H3CH2C
H
H
H
H
O
H
O
H
enol-keto tautomerism
Acid-catalyzed Keto-enol tautomerism
H+
H3CH2C
H3CH2C
H
H3CH2C
H
H
H
H
H
O
H
H
O
H
H
O
H2O
HgSO4
H3CH2C
CH3
H2
C
H3CH2C
H2O, H2SO4
+
CH3
O
H2
C
CH3
H3CH2C
O
b.
Anti-Markovnikov Hydration: Hydroboration of Alkynes
Hydroboration of terminal alkynes is usually performed with a bulky borane to prevent
addition of two molecules of borane across the triple bond. Disiamyl borane is usually
used for this purpose. A vinyl alcohol (enol) intermediate is again formed, but because of
the preference for addition of boron to the terminal carbon of alkynes, aldehydes are the
carbonyl products obtained.
H
B
H3CH2C
O
H
H
H
aldehyde
H3CH2C
disiamylborane
H
H2O2, HO-OH
B
H
H3CH2C
O
H
H2O2
H
vinyl alcohol (enol)
H
H3CH2C
H
O H
H
H
O
H
O
H
H3CH2C
H3CH2C
H
H
enolate
aldehyde
c. Permanganate Oxidation/Oxidative cleavage of alkynes
Treatment of alkynes with dilute permanganate under neutral conditions results in
oxidation to 1,2-diketones. Under heating and basic conditions, oxidative cleavage to
carboxylic acids takes place; terminal alkynes liberate a mole of CO2.
H
KMnO4
H3CH2C
CH3
O
OH
OH
O
H3CH2C
CH3
neutral, dilute
H
-2H2O
O
CH3
H3CH2C
O
O
KMnO4
H3CH2C
CH3
CH3
H3CH2C
KOH,
heat
HO
O
H3CH2C
CH3
O
HCl, H2O
OH
H3CH2C
O
1. KMnO4
H3CH2C
K+ -O
O
CH3
+
O-
K+
O
+
H
KOH,
heat
2. HCl, H2O
O
H3CH2C
OH
CO2
d. Ozonolysis of alkynes
When subjected to ozonolysis, alkynes give the same products as obtained from
permanganate cleavage. The products of an ozonolytic or permanganate cleavage give
important clues as to the structure of the parent alkyne.
O3, H2O
H3CH2C
HO
O
CH3
n eutral
conditions
CH3
+
H3CH2C
OH
O
What is the structure of the unknown alkyne?
O3, H2O
unknown alkyne
HOOC–(CH2)4–COOH + 2 CH3COOH
Additional Problems for practice:
1. Design a synthesis of the following racemic cyclopropane starting from acetylene,
organic compounds containing three or fewer carbons, and any inorganic reagents
necessary. Note with a star each step that creates a chiral product, explain why
each starred step affords a racemic mixture.
CH3
H3C
2. Outline a preparation of racemic disparlure from acetylene and any other
compound containing not more than five carbon atoms. Note with a star each step
that creates a chiral product, explain why each starred step affords a racemic
mixture
O
H3C(H2C)9
(CH2)4CH(CH3)2
3. Explain how you could accomplish the following isomerization in a three-step
sequence:
H
H
H
H
4. Propose a structure for the unknown compound given the following data:
COOH
KMnO4, conc
unknown compound
KOH
COOH
HOOC
O
+ 1 mole
H 3C
no CO2 evident
OH