Transcript I. Alkynes

Chapter 8
Introduction
• An alkyne
is a hydrocarbon that contains a
carbon-carbon triple bond
• Acetylene is the simplest alkyne.
• It is produced industrially from methane by high
temperature decomposition (pyrolysis)
• It is the starting material of many organic
molecules
I. Alkynes: An Overview
A.
Electronic Structure
B.
Naming Alkynes
A.
Electronic Structure
Carbon-carbon triple bond results from:
– sp hybrid orbital on each C forming a s bond and
– unhybridized py and pz orbitals forming a p bond
– The remaining sp orbitals form bonds to other atoms
at 180º to C-C triple bond.
– The triple bond is shorter and stronger than single or
double bond
– The triple bond is shorter and stronger than single
or double bond
• Breaking a p bond in acetylene (HCCH) requires
318 kJ/mole (in ethylene it is 268 kJ/mole)
B.
Naming Alkynes
• Like alkanes and alkenes, alkynes are named according
to the system devised by the International Union of Pure
and Applied Chemistry (IUPAC).
Steps to naming alkynes1
a. Find the longest continuous carbon chain containing the
triple bond
b. Name using the suffix –yne to indicate an alkyne.
– The position of the triple bond is indicated by giving the
number of the first alkyne carbon
c. Numbering of chain with triple bond is set so that the
smallest number possible includes the triple bond
Steps to naming alkynes2
d. If more than one triple bond is present:
– Indicate the position of each and use the suffixes diyne, -triyne, …
– A compound with two triple bonds is a diyne
– A triyne has three triple bonds
Steps to naming alkynes3
e. If a triple and a double bond are present:
– Name the compound an enyne
– An enyne has a double bond and triple bond
– Number from chain that ends nearest a double or
triple bond - double bond is preferred if both are
present in the same relative position
• Alkynes as substituents are called “alkynyl”
Practice Problem: Name the following compounds:
Practice Problem: There are seven isomeric alkynes with the
formula C6H10. Draw and name them
II. Synthesis of Alkynes
A.
Elimination Reactions of
Dihalides
B.
Alkylation of Acetylide Anions
A.
Elimination Reactions of Dihalides
• Treatment of a 1,2 dihaloalkane (vicinal dihalide) with
KOH or NaNH2 produces a two-fold elimination of HX
and formation of an alkyne
– Intermediate is a vinyl halide
– Vicinal dihalides are available from addition of
bromine or chlorine to an alkene
– Vinyl halides give alkynes when treated with strong
base.
Dehydrohalogenation of Vicinal dihalides
Dehydrohalogenation of vicinal dihalides results in twofold
elimination (loss) of HX and formation of an alkyne
– uses strong base (KOH or NaNH2)
– Intermediate is a vinyl halide
B.
Alkylation of Acetylide Anions
III. Reactions of Alkynes
A.
Addition of HX and X2
B.
Hydration of Alkynes
C.
Reduction of Alkynes
D.
Oxidative Cleavage of Alkynes
III. Reactions of Alkynes
E.
Alkyne Acidity: Formation of
Acetylide Anions
F.
Alkylation of Acetylide Anions
G.
An Introduction to Organic
Synthesis
Introduction
• Addition reactions of alkynes are similar to those
of alkenes
• Alkynes react with many electrophiles to give
useful products by addition
A.
Addition of HX and X2
• Addition of excess H-X to an alkyne gives a dihalide
product
– Regiochemistry is Markovnikov (X attaches to the more
highly substituted sp carbon and H adds to the less highly
substituted side)
Addition of X2 (where X = Br or Cl)
• Addition of excess X2 to an alkyne gives a tetrahalide
product.
• Initial addition of X2 (Br2 or Cl2) to an alkyne gives
trans intermediate
Addition of HX to Alkynes Involves Vinylic Carbocations
• Addition of H-X to alkyne should produce a vinylic
carbocation intermediate
– Secondary vinyl carbocations form less readily than primary
alkyl carbocations
– Primary vinyl carbocations probably do not form at all
• Nonethelss, H-Br can add to an alkyne to give a vinyl
bromide if the Br is not on a primary carbon
• Vinylic carbocations are less stable than similarly
substituted alkyl carbocations.
– Vinyl carbocations have sp-hybridized carbons and thus
lack stabilizing hyperconjugative interactions
– Stability of carbocations:
3º alkyl > 2º alkyl > 1º alkyl ~ 2º vinyl > +CH3 ~ 1º vinyl
Addition of HX and X2: Summary
• Addition of excess H-X to an alkyne gives a dihalide
– Markovnikov Regiochemistry
• Addition of excess X2 to an alkyne gives a tetrahalide
product.
Practice Problem: What products would you expect from the
following reactions?
B.
Hydration of Alkynes
• Hydration of alkynes is the addition H-OH to an alkyne
– Mercury (II) catalyzes Markovnikov oriented
addition
– Hydroboration-oxidation gives the non-Markovnikov
product
Mercury(II)-Catalyzed Hydration of Alkynes
• Mercuric ion (as the sulfate) is a Lewis acid catalyst
that promotes addition of water in Markovnikov
orientation
– The immediate product is a vinylic alcohol, or enol,
which spontaneously rearranges to a ketone
– Alkynes do not react with aqueous protic acids
Keto-enol Tautomerism
• Isomeric compounds that can rapidily interconvert by
the movement of a proton are called tautomers and the
phenomenon is called tautomerism
• Enols rearrange to the isomeric ketone by the rapid transfer
of a proton from the hydroxyl to the alkene carbon
• The keto form is usually so stable compared to the enol that
only the keto form can be observed
Mechanism of Mercury (II)
catalyzed Hydration
• Addition of Hg(II) to
alkyne gives a vinylic
carbocation intermediate
• Water adds and loses a
proton
• A proton from aqueous
acid replaces Hg(II)
Hydration of Unsymmetrical Alkynes
• Unsymmetrically substituted internal alkyne:
– If the alkyl groups at either end of the C-C triple bond are
not the same, both products can form and this is not
normally useful
Hydration of Unsymmetrical Alkynes
Terminal alkyne
– If the triple bond is at the first carbon of the chain (then H
is what is attached to one side) this is called a terminal
alkyne
– Hydration of a terminal alkyne always gives the methyl
ketone, which is useful
Mercury(II)-Catalyzed Hydration: Summary
H-OH adds to an alkyne to give an enol which converts to
ketone
– Markovnikov regiochemistry
– Mercury-containing vinylic carbocation intermediate
Practice Problem: What product would you obtain by hydration
of the following alkynes?
Practice Problem: What alkynes would you start with to prepare
the following ketones?
Hydroboration/Oxidation of Alkynes
• BH3 (borane) adds to alkynes to give a vinylic borane
• Oxidation with H2O2 produces an enol that converts to
the ketone or aldehyde
• Process converts alkyne to ketone or aldehyde with
Non-Markovnikov orientation
• This is opposite to mercuric ion catalyzed hydration
Comparison of Hydration of Terminal Alkynes
• Hydroboration/oxidation converts terminal alkynes to
aldehydes because addition of water is non-Markovnikov
• Mercury(II) catalyzed hydration converts terminal alkynes
to methyl ketones
Hydroboration-oxidation: Summary
Borane adds to an alkyne to give a vinylic borane which is
then oxidized by alkaline H2O2 to give an enol which
converts to ketone or aldehyde
– Non-Markovnikov regiochemistry
Practice Problem: What alkyne would you start with to prepare
each of the following compounds by a
hydroboration/oxidation reaction?
C.
Reduction of Alkynes
• Reduction of alkynes is the addition H-H to an alkyne
– Complete Catalytic hydrogenation using a metal
catalyst (Pd/C) or
– Partial Catalytic hydrogenation using Lindlar
catalyst
– Reduction with Lithium/ammonia
Catalytic Hydrogenation
• Addition of H2 over a metal catalyst (such as palladium
on carbon, Pd/C) converts alkynes to alkanes (complete
reduction)
– The addition of the first equivalent of H2 produces an
alkene intermediate, which is more reactive than the
alkyne so the alkene is not observed
Conversion of Alkynes to cis-Alkenes
• Addition of H2 using the Lindlar catalyst (chemically
deactivated palladium on calcium carbonate) produces
a cis alkene
– The two hydrogens add syn (from the same side of the
triple bond)
Conversion of Alkynes to trans-Alkenes
• Alkynes are reduced to trans alkenes with sodium or
lithium in liquid ammonia
– Anhydrous ammonia (NH3) is liquid below -33 ºC
– Alkali metals dissolve in liquid ammonia and function
as reducing agents
• The reaction involves a radical anion intermediate
Reduction of Alkynes : Summary
• Addition of H2 over a metal catalyst (Pd/C) to an alkyne
produces an alkane
• Addition of H2 using the Lindlar catalyst produces a cis
alkene
– syn stereochemistry
• Alkynes are reduced to trans alkenes with sodium or
lithium in liquid ammonia
– a radical anion intermediate
– anti stereochemistry
Practice Problem: Using any alkyne needed, how would you
prepare the following alkenes?
a) trans-2-Octene
b) cis-3-Heptene
c) 3-Methyl-1-pentene
D.
Oxidative Cleavage of Alkynes
• Strong oxidizing reagents (O3 or KMnO4) cleave
– internal alkynes, producing two carboxylic acids
– terminal alkynes, producing a carboxylic acid and
carbon dioxide
• Neither process is useful in modern synthesis.
– Historically, these were used to elucidate structures
because the products indicate the structure of the alkyne
precursor
Oxidative Cleavage of Alkynes: Summary
Practice Problem: Propose structures for alkynes that give the
following products on oxidative cleavage by
KMnO4:
E.
Alkyne Acidity: Formation of
Acetylide Anions
• Reaction of strong anhydrous bases (Na+-NH2) with a
terminal alkyne produces an acetylide ion
– Terminal alkynes are relatively acidic
• Terminal alkynes are weak Brønsted acids (pKa ~ 25)
– Alkenes and alkanes are much less acidic.
• Acetylide anions are more stable than either alkyl
anions or vinylic anions because:
– They have sp-hybridized carbon and their negative charge is in
a hybrid orbital with 50% s character, allowing the charge to be
closer to the nucleus.
Formation of Acetylide Anions: Summary
• Reaction of strong anhydrous bases (Na+-NH2) with
a terminal alkyne produces an acetylide ion
Practice Problem: The pKa of acetone, CH3COCH3, is 19.3.
Which of the following bases is strong enough
to deprotonate acetone?
a) KOH (pKa of H2O =15.7)
b) Na+ -CΞCH (pKa of C2H2 =25)
c) NaHCO3 (pKa of H2CO3 = 6.4)
d) NaOCH3 (pKa of CH3OH = 15.6)
F.
Alkylation of Acetylide Anions
• Reaction of an acetylide anion with a primary alkyl halide
produces a larger alkyne
– Acetylide anions can react as nucleophiles as well as bases
– Acetylide anions can displace a halide ion from a 1o alkyl
halide
Limitations of Alkylation of Acetylide Ions
Reactions only are efficient with 1º alkyl bromides and
alkyl iodides
•
•
Acetylide anions can behave as bases as well as
nucelophiles
Reactions with 2º and 3º alkyl halides gives
dehydrohalogenation, converting alkyl halide to alkene
Alkylation of Acetylide Anions: Summary
• Reaction of an acetylide anion with a primary alkyl halide
produces a larger alkyne
Practice Problem: Show the terminal alkyne and alkyl halide
from which the following products can be
obtained. If two routes look feasible, list both:
Practice Problem: How would you prepare cis-2-butene starting
from propyne, an alkyl halide, and any other
reagents needed? This problem can’t be
worked in a single step. You’ll have to carry
out more than one reaction.
G. An Introduction to Organic Synthesis
Organic synthesis may be used to
– produce new molecules that are needed as drugs
or materials
– design, test and improve efficiency and safety for
making known molecules
– test ideas and methods, answering challenges
Synthesis as a Tool for Learning Organic Chemistry
• In order to propose a synthesis, one must be
familiar with reactions:
–
–
–
–
What they begin with
What they lead to
How they are accomplished
What the limitations are
• A synthesis combines a series of proposed
steps to go from a defined set of reactants to
a specified product
Strategies for Synthesis
1. Compare the target and the starting material
2. Consider reactions that efficiently produce the
outcome.
3. Look at the product and think of what can lead to it
Example
– Problem: prepare octane from 1-pentyne
– Strategy: use acetylide coupling
Practice Problem: Beginning with 4-octyne as your only source
of carbon and using any inorganic reagents
necessary, how would you synthesize the
following compounds?
a) Butanoic acid
b) cis-4-Octene
c) 4-Bromooctane
d) 4-Octanol (4-hydroxyoctane)
e) 4,5-Dichlorooctane
Practice Problem: Beginning with acetylene and any alkyl
halides needed, how would you synthesize
the following compounds?
a) Decane
b) 2,2-Dimethylhexane
c) Hexanal
d) 2-Heptanone
Chapter 8
Addition of HX and X2: Summary
Hydration of Alkynes
Mercury (II) catalyzes Markovnikov oriented addition
Hydroboration-oxidation gives the non-Markovnikov
product
Reduction of Alkynes : Summary
Oxidative Cleavage of Alkynes: Summary
Formation of Acetylide Anions: Summary
Alkylation of Acetylide Anions: Summary
• Reaction of an acetylide anion with a primary alkyl halide
produces a larger alkyne
Alkylation of Acetylide Anions: Summary
• Reaction of an acetylide anion with a primary alkyl halide
produces a larger alkyne