Chapter 8 - ALKYNES This group of compounds contain a carbon-carbon triple bond. The simplest of these is acetylene (a common name). C C Industrial preparation
H C C H
steam 1200˚C
2CH4
Acetylene
H C C H + 3H 2
Structure:
C
H
sigma bond H
C
H
pi bond
s-sp-C-sp-sp-C-sp-s
Acetylene is a linear molecule. The carbons are sp hyridized. Bond between two carbons is a sigma bond from sp-sp overlap. Bonds to hydrogen are sigma bonds from sp-s overlap. There are two pi bonds that are mutually perpendicular to each other and the H-C-C-H bond axis. This is the triple bond.
H
sigma bond pi bond The carbons are sp hydridized. The triple bond consists of one sigma bond formed from the overlap of an sp orbital from each carbon. The remaining p orbitals on each carbon align to overlap to form two pi bonds which are mutually perpendicular to the C-C bond axis and each other. sigma bond
The molecule is linear as the sp orbitals on each carbon are 180˚ apart. The carbon-carbon bond length is 120pm. Bond strength of the triple bond is about 835 kJ/mol (200kcal/mol) --- therefore, shortest and strongest of the CC bonds. If we examine homolytic cleavage of single sigma bond, and the pi bonds of the double and triple C-C bonds we see that the bond bonds are more easily broken than the sigma bond (require less energy to cleave).
1
C
C
C
C
376 kJ/mol
(90kcal/mol)
C
C
C
C
611-376 = 235 kJ/mol (56)
C
C
C
C
835-611 = 224 kJ/mol (4)
NOMEMCLATURE Use the parent name for the number of carbons in the longest chain that contains the triple bond. Drop the -ANE ending from the alkane name and add - YNE. The triple bond gets the lowest number since the alkyne is the PARENT COMPOUND. Diynes and triynes as in alkenes. If both a double and triple bond in compound, called and en-yne. Numbering starts near the lowest multiple bond regardless of whether double ot triple. If a choice, the double bond gets the lower number. 1
2
3
4
5
6
7
8
9
C CH CH2CH 2CH CH3
CH3CH 2C
CH3
CH3
But 1
2 3
4
5
6
7
8
9
HC CCH2 CHCH2CH2CH CHCH3 CH3 To have a cyclic alkyne must have a fairly large ring. WHY?
2
Substituent groups containing a triple bond are possible, just as with alkyl and alkenyl groups.
Do Problems 8.1 & 8.2, p 248 for practice
PREPARATION OF ALKYNES Elimination of Dihalides (Mechanism in Chapter 11) Remember CH3CH 2
Br CH CH2CH 3
KOH or NaNH2
CH3CH
CHCH2CH3
Alkyl halide treated with base yields alkene. Dihalides with two equivalents of base give: Br Br CH3 CH CH CH3
2KOH ethanol
Dihalide
CH3C
CCH3
Alkene
This is a two-fold elimination. First one HX is eliminated and then the secodn HX is eliminated. Goes through a vinylic halide intermediate. H R Br
R vinylic halide
So.... 3
Br
Br CH CH
CH CH
2NaNH2/ether Diphenylacetylene
An alkene can be converted into an alkyne in a two step SYNTHESIS.
REACTIONS OF ALKYNES: 1. Addition of HX and X2 (electrophilic reagents) Pi (π) electrons are available for reaction just as in alkenes. a. Addition of HX
CH3CH 2C CH
HBr HOAc
Br CH3CH 2C CH2
Br CH3CH 2C CH3 Br
HBr HOAc
Follows Markovnikov's rule for regiochemistry. The reaction is step-wise and can be stopped after one equivalent of HX is added. The stereochemistry is usually trans but not always. CH3CH 2C CCH2CH3
CH3CH 2
1 HCl
H C
Cl (Try naming the product by E-Z nomenclature of alkenes). Remember - in addition of HX to an alkyne • if a terminal alkyne, get one product (Markovnikov's rule) • if a symmetrical internal alkyne, one product • if an unsymmetrical alkyne, get a mixture
4
C CH2CH 3
CH3CH 2C CCH3
2HBr
Br CH3CH 2 C CH2CH 3 + Br
Br CH3CH 2CH2 C CH3 Br
b. Addition of X2 Br2 and Cl2 also add to alkynes, with trans stereochemistry in first step.
CH3CH 2C CH
Br2
Br
H C C
CCl4
Br2 CCl4
CH3CH 2CBr2CHBr 2
CH3CH 2 Br Mechanism of HX addition is similar to HX addition to alkene. Goew through a vinylic carbocation. R
C
C
H H
Vinylic carbocation
C
sp hybridization- stabilized through hyperconjugation
Show mechanisms:
Problem 8.3, p. 250 for practice, e.g. 8.3(c) CH3CH 2CH2CH 2C CCH3
1HBr
?
2. Hydration of Alkynes: This is the addition of the elements of H-OH to the triple bond. There are two different ways of accomplishing this, different reagents, different mechanisms AND different products. These are: 5
a) Mercury (II) catalyzed hydration - leads to Markovinkov product b) Hydroboration/oxidation - leads to non-Markovnikov product But with BOTH of these reactions, the product is the result of the addition of ONE equivalent of H-OH. a) Mercury (II) Catalyzed Hydration (Mercuration-Demercuration) H3C H3C
CHCH2CH2 C CH
OH (CH3)2CHCH2CH2 C CH2
H2SO 4, H2O HgSO4
enol
Enol not isolated, but rearranges to a ketone
O (CH3)2CHCH2CH2 C CH3 ketone
The intermediate enol in brackets is not isolated but rearranges to a ketone because enols are basically unstable. These are not resonance structures. This rearrangement is called keto-enol tautomerism. Mechanism: R
C
C
H
Hg++SO4=
H RC
C
OH2 +
=
OH2 R
Hg SO4
C
H C
Hg+SO4=
-H+ O R C CH3
rearrangement HO
H C
C H
R
6
H3O+ R
OH H C C Hg+SO4=
Symmetrical
O
H3O+
RC CR
RCH2 C R
HgSO4
R's the same
Unsymmetrical
RCH2 C R'
HgSO4
R's different
Terminal
R C CH2R same cmpd
O
+
R C CH2R'
may be the same or different, depnds on R,R'
H3O+
RC CH
+
O
H3O+
RC CR'
O
O R C CH3
HgSO4
Always a methyl ketone adds by Markovnikov's rule
Examples
b) Hydroboration/oxidation (anti- or non-Markovnikov)
RC CR
BH3 THF
H C R
C
BH2 R
H2O2 OH
H C R
C
OH rearr. R
O RCH2 C R Ketone
The H of the borane adds as a hydride (H-). The final product or products depend on the R groups which may be the same or different. Terminal alkynes yield aldehydes RC CH
BH3 THF
BR2 RCH=CH
OH RCH=CH
H2O2 NaOH - H2O
7
Tautomerize
RCH2CHO
Aldehyde
Since terminal alkynes are less sterically hindered - may get doubly hydroborated intermediare, but after oxidation, product is still an aldehyde.
R C C H
BH2 RCH2 C H
BH3
H2O2,HO_
O RCH2 C
BH2
H
Even if the dihydroxy compound is formed in the second step, an equilibrium exists between that and the ketone (or aldehyde product and water). The equilibrium generally lies in favor of the formation of the keto product. OH R C R' OH
O R
C
+ H 2O R'
3. Reduction of Alkynes H2 (xs) CH3CH 3 catalyst As with alkenes, alkynes can be hydrogenated with H2 in the presnce of a catalyst under pressure. Under the conditions (pressure and catalyst) as the reduction of alkenes, it is difficult to stop after the addition of one equivalent of H2 but the process is a stepwise addition of hydrogen. The product of the first step is an alkene HC CH
Step 1:
HC CH
H2 catalyst
H2C CH2
DH˚ = -176 J/mol (-42 kcal/mol)
H2 CH3CH 3 DH˚ = -137 J/mol (-33 kcal/mol) catalyst Alkynes are easier to reduce than alkenes; so the first step proceeds under milder conditions than the second. The reaction can be controlled and stopped at the frist step (formation of the alkene) if a less active catalyst is used. The catalyst employed is called the Lindlar catalyst which is palladium on calcium carbonate then "poisoned" with lead acetate or quinoline. Step 2:
H2C CH2
Lindlar = Pd/CaCO3/Pb(OAc)2
8
The product is always the cis alkene because the hydrogen is delivered from the suface of the metal catalyst, both H atoms on the same side of the alkyne. (Review the reduction of alkenes mechanism in chapter 7). CH3C CCH3
+
H3C
H2 Lindlar
H
C
C
CH3
get cis product
H
A complementary reaction using metallic sodium (Na) or lithium (Li) in liquid ammonia (NH3) gives the trans addition alkene. Li NH3 (l )
RC CR
R H
C
C
H
get trans product
R
Mechanism: R
C
Li one-electron transfer
C R
R
C
H
R C
C
H2N
+ Li
R H NH2
R C
+ NH2
C
C
R
+
NH2
H Li (another one-electron transfer) H
R C C
H R trans alkene
R
H
4. Oxidative Cleavage As with alkenes, strong oxidizing agents will cleave the C-C bond of the triple bond.
9
R
R
C
C
C R'
C H
O
KMnO4 or ozonolysis
R
C
O +
'R
C
OH
OH O
KMnO4 or ozonolysis
R
C
OH
+
CO2
Problem 8.8 (p. 286) Propose structure of the alkyne that gives the products shown on oxidative cleavage.
Alkyne Acidity Brønsted Lowry theory - an acid is a species that donates a proton. Depending on conditions most any compound with a hydrogen can be an acid. O O CH3CH 2O + CH3C CH3CH 2OH + CH3C OH OpKa = 16 pKa = 4.75 CH3CH 2OH
+
CH3CH 2O-
NH2
pKa = 16 Remember pKa values!!!!!!
+
NH3 pKa = 35
10
Compound pKa Alkane Alkene NH3 Alkyne ROH HOH
>60 44 35 25 16-18 15.75
OH RC
5
O
10
OH
A terminal alkyne, in the presence of the conjugate base of a weaker acid, will result in the formation of an acetylide anion. RC CH
+ NH2
RC C:
pKa = 25
Hybridization of C:
+ NH3 pKa = 35
Alkane sp3
Alkene sp2
Alkyne sp
In acetylide anions the lone pair electrons are closer to nucleus thus more stabilized than a carbocation. sp orbital shorter. Acetylide anions are nucleophiles - seek out positive (=) centers.
RC C:
+ CH3-Br d+ d-
RC CCH3
+ Br-
This is an alkylation reaction. Notice - reaction results in a new carboncarbon bond (a larger carbon skeleton). Limited to 1˚ alkyl bromides. Thus
11
RCH2-Br will react in alkylation reactions (a substituition reaction) but R2CH-Br and R3C-Br will react differently. Br + 2˚ halide
RC C: acts as base
elimination product not substitution product
Will learn about this mechanism in Clapter 11
Synthesis We have now learned enough reactions including forming a new C-C bond so we can put them together in a series of steps to synthesize other molecules. Below are some of the transformations we will go over in class. 1. CH3CH 2CH2C CH
CH3CH 2CH2CH 2CH2CH 2CH2CH 3
CH3CH 2CH2
2. CH3CH 2CH2C CH
H
12
CH3 (Remember stereochemistry) H
3.
4.
Acetylene and any R-X to
Br CH3CH 2CH2 CCH3
13