C C
Alkynes
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 < ethyneB) ethene < ethane < ethyneC) ethyne < ethene < ethaneD) ethane < ethyne < ethene
H C C
Acidity of Acetyleneand Terminal Alkynes
In general, hydrocarbons are very weak acids
Compound pKa
HF 3.2
H2O 16
NH3 36
45
CH4 60
H2C CH2
Acidity of Hydrocarbons
Acetylene is a weak acid, but not nearlyas weak as alkanes or alkenes.
Compound pKa
HF 3.2
H2O 16
NH3 36
45
CH4 60
H2C CH2
HC CH 26
Acetylene
Question
Which one of the following is the strongest acid?A) waterB) ammoniaC) 1-buteneD) 1-butyne
C H H+ +
H+ +
H+ +
10-60
10-45
10-26
sp3C :
sp2
sp
H
C C
C C H
C C
C C :
:
Electrons in an orbital with more s character are closer to thenucleus and more strongly held.
Carbon: Hybridization and Electronegativity
Question
Which one of the following statements best explains the greater acidity of terminal alkynes (RCºCH) 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
Solution: Use a stronger base. Sodium amideis a stronger base than sodium hydroxide.
NH3NaNH2 + HC CH NaC CH +
–H2N
..: H C CH H
..+ + C CH:
–
stronger acidpKa = 26
weaker acidpKa = 36
Ammonia is a weaker acid than acetylene.The position of equilibrium lies to the right.
H2N
Sodium Acetylide
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
H—C C—H
R—C C—H
R—C C—R
Alkylation of Acetylene and Terminal Alkynes
R XSN2
X–:+C–:H—C C—RH—C +
The alkylating agent is an alkyl halide, andthe reaction is nucleophilic substitution.The nucleophile is sodium acetylide or the sodium salt of a terminal (monosubstituted) alkyne.
Alkylation of Acetylene and Terminal Alkynes
NaNH2
NH3
CH3CH2CH2CH2Br
(70-77%)
HC CH HC CNa
HC C CH2CH2CH2CH3
Example: Alkylation of Acetylene
Question
Which alkyl halide will react faster with the acetylide ion (HCºCNa) in an SN2 reaction?
A) bromopropaneB) 2-bromopropaneC) tert-butyl iodideD) 1-bromo-2-methylbutane
NaNH2, NH3
CH3Br
CH(CH3)2CHCH2C
CNa(CH3)2CHCH2C
(81%)
C—CH3(CH3)2CHCH2C
Example: Alkylation of a Terminal Alkyne
1. NaNH2, NH3
2. CH3CH2Br
(81%)
H—C C—H
1. NaNH2, NH3
2. CH3Br
C—HCH3CH2—C
C—CH3CH3CH2—C
Example: Dialkylation of Acetylene
Effective only with primary alkyl halides
Secondary and tertiary alkyl halides undergo elimination
Limitation
Question
A.
B.
1) NaNH2
2) Pentyl chloride
C.
D.
NaCl + NH3 + ????
What is the product of the following reaction?
Answer
A.
B.
1) NaNH2
2) Pentyl chloride
C.
D.
NaCl + NH3 + ????
What is the product of the following reaction?
B) SEE: Skillbuilder 10.5.
E2 predominates over SN2 when alkyl halide is secondary or tertiary.
C–:H—C
E2
+CH—C —H C C X–:+
Acetylide Ion as a Base
H C
C X
Question
Consider the reaction of each of the following with cyclohexyl bromide. For which one isthe ratio of substitution to elimination highest?A) NaOCH2CH3, ethanol, 60°C
B) NaSCH2CH3, ethanol-water, 25°C
C) NaNH2, NH3, -33°C
D) NaCºCH, NH3, -33°C
Preparation of Alkynesby Elimination Reactions
Geminal dihalide Vicinal dihalide
X
C C
X
H
H
X X
C C
HH
The most frequent applications are in preparation of terminal alkynes.
Preparation of Alkynesby "Double Dehydrohalogenation"
(CH3)3CCH2—CHCl2
1. 3NaNH2, NH3
2. H2O
(56-60%)
(CH3)3CC CH
Geminal dihalide Alkyne
NaNH2, NH3
H2O
(CH3)3CCH2—CHCl2
(CH3)3CCH CHCl(slow)
NaNH2, NH3(CH3)3CC CH
(slow)
NaNH2, NH3(CH3)3CC CNa
(fast)
Geminal dihalide Alkyne
Question
In addition to NaNH2, what other base can be used to convert 1,1-dichlorobutane into1-butyne?A) NaOCH3
B) NaOHC) NaOCH2CH3
D) KOC(CH3)3
CH3(CH2)7CH—CH2Br
Br
1. 3NaNH2, NH3
2. H2O
(54%)
CH3(CH2)7C CH
Vicinal dihalide Alkyne
Question
Which of the following compounds yield 1-heptyne 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-tetrachloroheptaneB) 1-bromo-2-chloroheptaneC) 1,1,2-trichloropentaneD) all of the above
Reactions of Alkynes
Acidity Hydrogenation Metal-Ammonia Reduction Addition of Hydrogen HalidesHydrationAddition of HalogensOzonolysis
Reactions of Alkynes
Hydrogenation of Alkynes
Atomic Force Microscopy of Acetylene Lawrence Berkeley Laboratory (LBL)
C
C
H
H
CC
H
H
Calculated image (Philippe Sautet)
p orbital
pz
TIP
HO
+
Imaging: acetylene on Pd(111) at 28 K
Molecular Image Tip cruising altitude ~700 pmΔz = 20 pm
Surface atomic profile
Tip cruising altitude ~500 pm
Δz = 2 pm
1 cm(± 1 μm)
The STM image is a map of the pi-orbital of distorted acetylene
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
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
M. Salmeron (LBL)
Excitation of frustrated rotational modes in acetylene molecules on Pd(111) at T = 30 K
Tip
e-
((( ) ( )))
M. Salmeron (LBL)
RCH2CH2R'cat
catalyst = Pt, Pd, Ni, or Rh
alkene is an intermediate
RC CR' + 2H2
Hydrogenation of Alkynes
RCH2CH2R'
Alkenes could be used to prepare alkenes if acatalyst were available that is active enough to catalyze the hydrogenation of alkynes, but notactive enough for the hydrogenation of alkenes.
catH2
RC CR' catH2
RCH CHR'
Partial Hydrogenation
There is a catalyst that will catalyze the hydrogenationof alkynes to alkenes, but not that of alkenes to alkanes.
It is called the Lindlar catalyst and consists ofpalladium supported on CaCO3, which has been
poisoned with lead acetate and quinoline.
syn-Hydrogenation occurs; cis alkenes are formed.
RCH2CH2R'catH2
RC CR' catH2
RCH CHR'
Lindlar Palladium
+ H2
Lindlar Pd
CH3(CH2)3 (CH2)3CH3
H H(87%)
CH3(CH2)3C C(CH2)3CH3
CC
Example
Alkynes trans-Alkenes
Metal-Ammonia Reductionof Alkynes
RCH2CH2R'
Another way to convert alkynes to alkenes isby reduction with sodium (or lithium or potassium)in ammonia.
trans-Alkenes are formed.
RC CR' RCH CHR'
Partial Reduction
CH3CH2
CH2CH3H
H
(82%)
CH3CH2C CCH2CH3
CC
Na, NH3
Example
Question
How would you accomplish the following conversion?
A) NaNH2
B) H2, Lindlar Pd
C) Na, NH3
D) either B or C
four steps
(1) electron transfer
(2) proton transfer
(3) electron transfer
(4) proton transfer
Metal (Li, Na, K) is reducing agent; H2 is not involved; proton comes from NH3
Mechanism
Question
Select the most effective way to synthesize cis-2-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 PdD) 1) NaNH2 2) CH3CH2Br 3) Na,NH3
Question
Which reagent would accomplish the transformation of 3-hexyne into trans-3-hexene?A) H2/Ni
B) H2, Lindlar Pd
C) Na, NH3
D) NaNH2, NH3
Suggest an efficient syntheses of (E)- and (Z)-2-heptene from propyne and any necessary organic or inorganic reagents.
Problem
Problem Strategy
Problem Strategy
1. NaNH2
2. CH3CH2CH2CH2Br
Na, NH3H2, Lindlar Pd
Problem Synthesis
Question
Which would be the best sequence of reactions to use in order to prepare cis-3-nonenefrom 1-butyne?A) 1. NaNH2 in NH3; 2. 1-bromopentane; 3. H2,
Lindlar PdB) 1. NaNH2 in NH3; 2. 1-bromopentane; 3. Na,
NH3
C) 1. H2, Lindlar Pd; 2. NaNH2 in NH3; 3. 1-bromopentaneD) 1. Na, NH3; 2. NaNH2 in NH3; 3. 1-bromopentane
Addition of Hydrogen Halidesto Alkynes
HBr
Br
(60%)
Alkynes are slightly less reactive than alkenes
CH3(CH2)3C CH CH3(CH2)3C CH2
Follows Markovnikov's Rule
(76%)
CH3CH2C CCH2CH3
2 HF
F
F
C C
H
H
CH3CH2 CH2CH3
Two Molar Equivalents of Hydrogen Halide
HBr
(79%)
regioselectivity opposite to Markovnikov's rule
CH3(CH2)3C CH CH3(CH2)3CH CHBrperoxides
Free-radical Addition of HBr
expected reaction:
enolobserved reaction:
RCH2CR'
O
H+
RC CR' H2O+
H+
RC CR' H2O+
OH
RCH CR'
ketone
Hydration of Alkynes
enols are regioisomers of ketones, and exist in equilibrium with themketo-enol equilibration is rapid in acidic mediaketones are more stable than enols andpredominate at equilibrium
enol
OH
RCH CR' RCH2CR'
O
ketone
Enols
O H
C C
H+O
H
H
:
..:
Mechanism of conversion of enol to ketone
O H
C C
H+O
H
H
:
..:
Mechanism of conversion of enol to ketone
O H
C CH+
O
H
H
:
..:
:
Mechanism of conversion of enol to ketone
O H
C C
H
H
O: :
H+
..:
Mechanism of conversion of enol to ketone
Carbocation is stabilized by electron delocalization (resonance).
H O
C CH
..H+
Key Carbocation Intermediate
O
C CH+
..:
O H
C C
H
H
O: :
H+
..:
Mechanism of conversion of enol to ketone
Useful for symmetrical starting alkynes
to produce a single product.
Unsymmetrical starting alkynes that are not terminal produce a mixture of ketones…non-regioselectively.
H2O, H2SO4
HgSO4
CH3(CH2)5CCH3
(91%)via
Markovnikov's rule followed in formation of enol, Useful with terminal alkynes.
CH3(CH2)5C CH2
OH
CH3(CH2)5C CH
O
Regioselectivity
Aldehyde vs. Ketone
Question
What is the product of the acid catalyzed hydration of 1-hexyne?
A) B)
C) D)
Question
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
I) Which reactions give ketones? II) Which reactions give aldehydes?
Addition of Halogens to Alkynes
+ 2 Cl2
Cl
Cl
(63%)
CCl2CH CH3HC CCH3
Example
Br2
CH3CH2
CH2CH3Br
Br
(90%)
CH3CH2C CCH2CH3 C C
Addition is anti
+ NBS Br-C CCH3
NBS Example
NBS = N-bromosuccinimide
HC CCH3
AgNO3
O
O
NBr
gives two carboxylic acids by cleavage of triple bond
Ozonolysis of Alkynes
1. O3
2. H2O
CH3(CH2)3C CH
+CH3(CH2)3COH
(51%)
O
HOCOH
O
Example
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)
AlkynesSynthesis & Functions
Can you identify and name the function?
Example
Question
What is the structure of Compound Y in the following synthetic sequence?
A) pentaneB) cis-2-penteneC) trans-2-penteneD) 2-pentyne
Answer
What is the structure of Compound Y in the following synthetic sequence?
A) pentaneB) cis-2-penteneC) trans-2-penteneD) 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.
+ NBS Br-C CCH3
PolyyneCu (I) Coupling Reactions
HC CCH3
AgNO3