Alkynes
C
C
Synthesis of Acetylene
Heat coke with lime in an electric to form
calcium carbide.
Then drip water on the calcium carbide.
coke
lime
*
*This reaction was used to produce light
for miners’ lamps and for the stage.
The Structure of Alkynes
A triple bond is composed of a s bond and two p bonds
Question
Arrange ethane, ethene, and ethyne in order of
increasing C-C bond length.
A) ethane < ethene < ethyne
B) ethene < ethane < ethyne
C) ethyne < ethene < ethane
D) ethane < ethyne < ethene
Acidity of Acetylene
and Terminal Alkynes
H
C
C
Acidity of Hydrocarbons
In general, hydrocarbons are
very weak acids
Compound
pKa
HF
3.2
H2 O
16
NH3
36
H2C
45
CH2
CH4
60
Acetylene
Acetylene is a weak acid, but not nearly
as weak as alkanes or alkenes.
HC
CH
Compound
pKa
HF
3.2
H2 O
16
NH3
36
H2C
45
CH2
CH4
26
60
Question
Which one of the following is the strongest acid?
A) water
B) ammonia
C) 1-butene
D) 1-butyne
Carbon: Hybridization and Electronegativity
10-60
C
H
H
C
10-45
C
H+ +
H+ +
C
C
:
sp3
:
sp2
C
10-26
C
C
H
H+ +
C
C :
sp
Electrons in an orbital with more s character are closer to the
nucleus and more strongly held.
Question
Which one of the following statements best explains the
greater acidity of terminal alkynes (RCCH) compared
with monosubstituted alkenes (RCH=CH2)?
A) The sp-hybridized carbons of the alkyne are less
electronegative than the sp2 carbons of the alkene.
B) The two p bonds of the alkyne are better able to
stabilize the negative charge of the anion by
resonance.
C) The sp-hybridized carbons of the alkyne are
more electronegative than the sp2 carbons of
the alkene.
D) The question is incorrect - alkenes are more acidic
than alkynes.
The stronger the acid, the weaker its conjugate base
top 252
Sodium Acetylide
Solution: Use a stronger base. Sodium amide
is a stronger base than sodium hydroxide.
NaNH2 + HC CH
NaC CH + NH3
.. –
H2N : +
H
C
CH
stronger acid
pKa = 26
..
H2N
–
H + :C
weaker acid
pKa = 36
Ammonia is a weaker acid than acetylene.
The position of equilibrium lies to the right.
CH
Question
Which of the following bases is strong enough
to completely deprotonate propyne?
A) NH3
B) CH3OH
C) NaNH2
D) NaOH
Preparation of Various Alkynes
by alkylation reactions with
Acetylide or Terminal Alkynes
Synthesis Using Acetylide Ions:
Formation of C–C Bond
Alkylation of Acetylene and Terminal Alkynes
H—C
C—H
R—C
C—H
R—C
C—R
Alkylation of Acetylene and Terminal Alkynes
H—C
–
C: +
R
X
SN2
H—C
C—R + : X–
The alkylating agent is an alkyl halide, and
the reaction is nucleophilic substitution.
The nucleophile is sodium acetylide or the
sodium salt of a terminal (monosubstituted)
alkyne.
Example: Alkylation of Acetylene
HC
CH
NaNH2
HC
NH3
CNa
CH3CH2CH2CH2Br
HC
C
CH2CH2CH2CH3
(70-77%)
Question
Which alkyl halide will react faster with the
acetylide ion (HCCNa) in an SN2 reaction?
A) bromopropane
B) 2-bromopropane
C) tert-butyl iodide
D) 1-bromo-2-methylbutane
Example: Alkylation of a Terminal Alkyne
(CH3)2CHCH2C
CH
NaNH2, NH3
(CH3)2CHCH2C
CNa
CH3Br
(CH3)2CHCH2C
(81%)
C—CH3
Example: Dialkylation of Acetylene
H—C
C—H
1. NaNH2, NH3
2. CH3CH2Br
CH3CH2—C
C—H
1. NaNH2, NH3
2. CH3Br
CH3CH2—C
C—CH3
(81%)
Limitation
Effective only with primary alkyl halides
Secondary and tertiary alkyl halides
undergo elimination
Question
What is the product of the following reaction?
1) NaNH2
NaCl + NH3 + ????
2) Pentyl chloride
A.
C.
B.
D.
Answer
What is the product of the following reaction?
1) NaNH2
NaCl + NH3 + ????
2) Pentyl chloride
A.
C.
B.
D.
B)
SEE: Skillbuilder 10.5.
Acetylide Ion as a Base
E2 predominates over SN2 when alkyl
halide is secondary or tertiary.
H—C
–
C:
H
C
C
X
E2
H—C
C —H +
C
C
+
: X–
Question
Consider the reaction of each of the following
with cyclohexyl bromide. For which one is
the ratio of substitution to elimination highest?
A) NaOCH2CH3, ethanol, 60°C
B) NaSCH2CH3, ethanol-water, 25°C
C) NaNH2, NH3, -33°C
D) NaCCH, NH3, -33°C
Preparation of Alkynes
by Elimination Reactions
Preparation of Alkynes
by "Double Dehydrohalogenation"
H
X
H
H
C
C
C
C
H
X
X
X
Geminal dihalide
Vicinal dihalide
The most frequent applications are in preparation
of terminal alkynes.
Geminal dihalide  Alkyne
(CH3)3CCH2—CHCl2
1. 3NaNH2, NH3
2. H2O
(CH3)3CC
CH
(56-60%)
Geminal dihalide  Alkyne
(CH3)3CCH2—CHCl2
NaNH2, NH3 (slow)
(CH3)3CCH
CHCl
NaNH2, NH3 (slow)
(CH3)3CC
CH
H2O
(CH3)3CC
CNa
NaNH2, NH3 (fast)
Question
In addition to NaNH2, what other base can be
used to convert 1,1-dichlorobutane into
1-butyne?
A) NaOCH3
B) NaOH
C) NaOCH2CH3
D) KOC(CH3)3
Vicinal dihalide  Alkyne
CH3(CH2)7CH—CH2Br
Br
1. 3NaNH2, NH3
2. H2O
CH3(CH2)7C
(54%)
CH
Question
Which of the following compounds yield 1heptyne on being treated with three moles of
sodium amide (in liquid ammonia as the
solvent) followed by adding water to the
reaction mixture?
A) 1,1,2,2-tetrachloroheptane
B) 1-bromo-2-chloroheptane
C) 1,1,2-trichloropentane
D) all of the above
Reactions of Alkynes
Reactions of Alkynes
Acidity
Hydrogenation
Metal-Ammonia Reduction
Addition of Hydrogen Halides
Hydration
Addition of Halogens
Ozonolysis
Hydrogenation of Alkynes
Atomic Force Microscopy of Acetylene
Lawrence Berkeley Laboratory (LBL)
H
C
C
H
Imaging: acetylene on Pd(111) at 28 K
Molecular Image
Tip cruising altitude ~700 pm
Δz = 20 pm
H
C
C
H
Why don’t we see the Pd atoms?
Because the tip needs to be very close to
image the Pd atoms and would knock
the molecule away
Surface atomic profile
Tip cruising altitude
~500 pm
Δz = 2 pm
TIP
pz
H
+
O
p orbital
Calculated image
(Philippe Sautet)
If the tip was made as big as an airplane, it would be
flying at 1 cm from the surface and waving up an down
by 1 micrometer
The STM image is a map of the pi-orbital
of distorted acetylene
M. Salmeron (LBL)
1 cm
(± 1 μm)
Excitation of frustrated rotational modes in
acetylene molecules on Pd(111) at T = 30 K
Tip
e-
((( ) (
)))
M. Salmeron (LBL)
Hydrogenation of Alkynes
RC
CR'
+
2H2
cat
RCH2CH2R'
catalyst = Pt, Pd, Ni, or Rh
alkene is an intermediate
Partial Hydrogenation
RC
CR'
H2
cat
RCH
CHR'
H2
cat
RCH2CH2R'
Alkenes could be used to prepare alkenes if a
catalyst were available that is active enough to
catalyze the hydrogenation of alkynes, but not
active enough for the hydrogenation of alkenes.
Lindlar Palladium
RC
CR'
H2
cat
RCH
CHR'
H2
cat
RCH2CH2R'
There is a catalyst that will catalyze the hydrogenation
of alkynes to alkenes, but not that of alkenes to alkanes.
It is called the Lindlar catalyst and consists of
palladium supported on CaCO3, which has been
poisoned with lead acetate and quinoline.
syn-Hydrogenation occurs; cis alkenes are formed.
Example
CH3(CH2)3C
C(CH2)3CH3 + H2
Lindlar Pd
CH3(CH2)3
(CH2)3CH3
C
C
H
H
(87%)
Metal-Ammonia Reduction
of Alkynes
Alkynes  trans-Alkenes
Partial Reduction
RC
CR'
RCH
CHR'
RCH2CH2R'
Another way to convert alkynes to alkenes is
by reduction with sodium (or lithium or potassium)
in ammonia.
trans-Alkenes are formed.
Example
CH3CH2C
CCH2CH3
Na, NH3
CH3CH2
H
C
C
CH2CH3
H
(82%)
Question
How would you accomplish the following
conversion?
A)
B)
C)
D)
NaNH2
H2, Lindlar Pd
Na, NH3
either B or C
Mechanism
Metal (Li, Na, K) is reducing agent;
H2 is not involved; proton comes from NH3
four steps
(1) electron transfer
(2) proton transfer
(3) electron transfer
(4) proton transfer
Question
Select the most effective way to synthesize cis2-pentene from 1-propyne.
A) 1) NaNH2 2) CH3CH2Br 3) H2, Pd
B) 1) NaNH2 2) CH3Br 3) H2, Lindlar Pd
C) 1) NaNH2 2) CH3CH2I 3) H2, Lindlar Pd
D) 1) NaNH2 2) CH3CH2Br 3) Na,NH3
Question
Which reagent would accomplish the
transformation of 3-hexyne into trans-3hexene?
A) H2/Ni
B) H2, Lindlar Pd
C) Na, NH3
D) NaNH2, NH3
Problem
Suggest an efficient syntheses of (E)- and (Z)-2heptene from propyne and any necessary organic
or inorganic reagents.
Problem
Strategy
Problem
Strategy
Problem
Synthesis
1. NaNH2
2. CH3CH2CH2CH2Br
H2, Lindlar Pd
Na, NH3
Question
Which would be the best sequence of reactions to use in
order to prepare cis-3-nonene
from 1-butyne?
A) 1. NaNH2 in NH3; 2. 1-bromopentane; 3. H2,
Lindlar Pd
B) 1. NaNH2 in NH3; 2. 1-bromopentane; 3. Na,
NH3
C) 1. H2, Lindlar Pd; 2. NaNH2 in NH3; 3. 1bromopentane
D) 1. Na, NH3; 2. NaNH2 in NH3; 3. 1bromopentane
Addition of Hydrogen Halides
to Alkynes
Follows Markovnikov's Rule
CH3(CH2)3C
HBr
CH
CH3(CH2)3C
CH2
Br
(60%)
Alkynes are slightly less reactive than alkenes
Two Molar Equivalents of Hydrogen Halide
CH3CH2C
CCH2CH3
2 HF
CH3CH2
H
F
C
C
H
F
(76%)
CH2CH3
Free-radical Addition of HBr
CH3(CH2)3C
CH
HBr
peroxides
CH3(CH2)3CH
(79%)
regioselectivity opposite to Markovnikov's rule
CHBr
Hydration of Alkynes
expected reaction:
RC
CR'
+ H2O
H+
RCH
CR'
OH
enol
observed reaction:
RC
CR'
+ H2O
H+
RCH2CR'
O
ketone
Enols
RCH
CR'
OH
enol
RCH2CR'
O
ketone
enols are regioisomers of ketones, and exist
in equilibrium with them
keto-enol equilibration is rapid in acidic media
ketones are more stable than enols and
predominate at equilibrium
Mechanism of conversion of enol to ketone
..
:O
H
+
:O
H
C
H
C
H
Mechanism of conversion of enol to ketone
..
:O
H
+
:O
H
C
H
C
H
Mechanism of conversion of enol to ketone
..
:O
H
: O:
H
H
C
C
+
H
Mechanism of conversion of enol to ketone
H
..
:O
H
C
C
+
H
: O:
H
Key Carbocation Intermediate
Carbocation is stabilized by
electron delocalization (resonance).
..
:O
H
C
C
+
..
+O
H
H
C
C
H
Mechanism of conversion of enol to ketone
H
..
:O
H
C
C
+
H
: O:
H
Useful for symmetrical starting alkynes
to produce a single product.
Unsymmetrical starting alkynes that are
not terminal produce a mixture of
ketones…non-regioselectively.
Regioselectivity
Markovnikov's rule followed in formation of enol,
Useful with terminal alkynes.
O
H2O, H2SO4
CH3(CH2)5C CH
CH3(CH2)5CCH3
HgSO4
(91%)
via
OH
CH3(CH2)5C
CH2
Aldehyde vs. Ketone
Question
What is the product of the acid catalyzed
hydration of 1-hexyne?
A)
B)
C)
D)
Question
I) Which reactions give ketones? II) Which reactions give aldehydes?
A. I = C, D, E, F; II = A, B
B. I = B, D, E, F; II = A, C
C. I = B, C, D, F; II = A, E
D. I = A, D, E, F; II = B, C
E. I = A, C, D, F; II = B, E
Addition of Halogens to Alkynes
Example
Cl
HC
CCH3 + 2 Cl2
Cl2CH
C
CH3
Cl
(63%)
Addition is anti
Br
CH3CH2
CH3CH2C
CCH2CH3
Br2
C
C
CH2CH3
Br
(90%)
NBS Example
AgNO3
HC
CCH3 + NBS
Br-C
CCH3
O
NBS = N-bromosuccinimide
NBr
O
Ozonolysis of Alkynes
gives two carboxylic acids by cleavage
of triple bond
Example
CH3(CH2)3C
CH
1. O3
2. H2O
O
O
CH3(CH2)3COH
(51%)
+
HOCOH
Question
What product is formed when 2-butyne is
subjected to ozonolysis?
A)
B)
C)
D)
Answer
What product is formed when 2-butyne is
subjected to ozonolysis?
A)
B)
C)
D)
Alkynes
Synthesis & Functions
Can you identify and name the function?
Example
Question
What is the structure of Compound Y in the
following synthetic sequence?
A)
B)
C)
D)
pentane
cis-2-pentene
trans-2-pentene
2-pentyne
Answer
What is the structure of Compound Y in the
following synthetic sequence?
A)
B)
C)
D)
pentane
cis-2-pentene
trans-2-pentene
2-pentyne
Natural Products
Polyynes
Compound 1 is isolated from the root and bark of
mistletoe, Paramacrolobium caeruleum
(Loranthaceae). The stems and leaves of members of
this family have been used for the treatment of cancer
in Indonesia.
Compound 2, thiarubrine B, has been isolated from
Giant Ragweed, Ambrosia trifida. Native cultures in
Canada and Africa use plants with similar natural
products to treat skin infections and intestinal
parasites.
Compound 3, oplopandiolacetate, is found in the bark
and roots of Devil's club, Oplopanax horridus. It is
used medicinally by native Americans to treat a
variety of ailments. Other polyynes found in plants
include oenanthotoxin, cicutoxin, and falcarinol .
Compound 4, dihydromatricaria acid, is found in the soldier
beetle, Cantharidae, who are related to the Lampyridae or firefly
family, but unable to produce light. They provide biological
control of a number of insect pests including grasshoppers,
aphids, caterpillars and other soft bodied insects.
Histrionicotoxin is isolated from the skin of frogs in the
Dendrobatidae family. It comes from insects in their
diet and is used by indigenous South American tribes
as a poison on arrows.
Polyyne
Cu (I) Coupling Reactions
AgNO3
HC
CCH3 + NBS
Br-C
CCH3