alkenes and alkynes.ppt
TRANSCRIPT
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Unsaturated Hydrocarbons
Lecture Outline Class odds and ends Defining unsaturation in hydrocarbons Naming alkenes and alkynes Cis and trans isomers Reactions of alkenes and alkynes
o Hydrogenationo Halogenationo Hydrohalogenationo hydration
Polymers Aromatic compounds Properties of aromatic compounds
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Compare the following structures of ethane, ethene and ethyne.
1. What difference do you notice among the first and the last two structures? (Ethane is saturated and ethene and ethyne are unsaturated?
2. How would you define unsaturated hydrocarbons?
Unsaturated Hydrocarbons
Ethane
C C
H
H
H H
H
H
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Bond Angles in Alkenes and AlkynesAccording to VSEPR theory: Three groups in a double
bond are bonded at 120° angles.
Alkenes are flat, because the atoms in a double bond lie in the same plane.
The groups attached to a triple bond are at 180° angles. Copyright © 2007 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Naming Alkenes
Look and study the names of the following alkenes
Alkene IUPAC Common
H2C=CH2 ethene ethylene
H2C=CH─CH3 propene propylene
cyclohexene
CH2=CH─CH2─CH3 1-butene
CH3─CH=CH─CH3 2-butene
How is the naming of alkenes done? Talk over with your group members
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Naming Alkynes
Now, let’s look at the case of alkynes. Naming is done in
a fashion similar to that of alkenes.
Alkyne IUPAC Common
HC≡CH ethyne acetylene
HC≡C─CH3 propyne
H3C CH C C CH3
Br
2 13454-Bromo-2-pentyne
cycloheptyne
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Naming Alkenes and Alkynes with Substituents
Now what if the alkene or alkyne has a substituent? Here is an example. Write the IUPAC name for
CH3
│
CH3─CH─CH=CH─CH3
STEP 1 Name the longest carbon chain pentene
STEP 2 Number the chain from the double bond
STEP 3 Give the location of each substituent: 4-methyl- 2-pentene
H3CCH
CH
HC
CH3
CH3
21
34
52-pentene
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Naming Alkynes with Substituents
Write the IUPAC name for CH3
│
HC≡C─CH─CH3
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Summary: Naming Alkenes and Alkynes
Okay, so let’s summarize the rules for naming alkenes and alkynes. Here is what we do:
Name the longest carbon chain with a double or triple bond
Indicate the location of the double or triple bond in the main chain by number, starting at the end closer to the double or triple bond
Cycloalkenes do not require the numerical prefix but the bonds are given numbers 1 and 2
Give the location and name of each substituent (alphabetical order) as a prefix to the name
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Learning Check
Write the IUPAC name for each of the following:
1. CH2=CH─CH2─CH3
2. CH3─CH=CH─CH3
CH3
|3. CH3─CH=C─CH3
4. CH3─CC─CH3
Solution
1) ______________
2) ______________
3) ______________
4) ______________
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Learning Check
Write the structural formula for each of the following:
A. 2-pentyne
B. 3-methyl-2-pentene
Solution
a) _______________
b) ________________
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Cis and Trans Isomers
In an alkene, cis and trans isomers are possible
because the double bond Is rigid.
Cannot rotate.
Has groups attached to the carbons of the double bond that are fixed relative to each other.
CH3 CH3 CH3
CH = CH CH = CH
cis trans CH3
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Cis-trans isomers occur when
different groups are attached to
the double bond.
In a cis isomer, groups are attached on the same side of the double bond.
In the trans isomer, the groups are attached on opposite sides.
Cis-Trans Isomers
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
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Cis-Trans Isomerism
Cis-trans isomers do not occur if a carbon atom in the double bond is attached to identical groups.
Identical Identical
2-bromopropene 1,1-dibromoethene (not cis or trans) (not cis or trans)
C C
H Br
H CH3
C C
H Br
BrH
H
H
H Br
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Naming Cis-Trans Isomers
The prefixes cis or trans are placed in front of the alkene name when there are cis-trans isomers.
cis trans
cis-1,2-dibromoethene trans-1,2-dibromoethene
C C
Br H
BrH
C C
Br Br
H H
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Learning Check
Name each, using cis-trans prefixes when needed.
C C
CH3 H
CH3H
C C
Br Br
H H
A.
B.
C C
CH3 Cl
ClH
C.
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Lecture 3 Unsaturated Hydrocarbons
Reactions of Alkenes and Alkynes
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
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Addition Reactions
There are four addition reactions we will study, summarized in table 12.2
TABLE 12.2
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Question: what happens in hydrogenation? Study the examples given below to answer this question?(Side note: a catalyst such as Pt or Ni is used to speed up the reaction)
Hydrogenation
HC CH + 2H2Ni
HC CH
H H
H H
H2C CH2
H HPt
H2H2C CH2 +
Your conclusion about hydrogenation:
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Adding H2 to double
bonds in vegetableoils produces Compounds with
higher melting points. Solids at room
temperature such as margarine,
soft margarine,
and shortening.
Hydrogenation of Oils
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
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Learning Check
Write an equation for the hydrogenation of 1-butene using a platinum catalyst.
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Trans Fats
In vegetable oils, the unsaturated fats usually contain
cis double bonds.
During hydrogenation, some cis double bonds are converted to trans double bonds (more stable) causing a change in the fatty acid structure
If a label states “partially” or “fully hydrogenated”, the fats contain trans fatty acids.
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Learning Check
Write the product of each the following reactions:
Pt
CH3─CH=CH─CH3 + H2
Pt
+ H2
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Halogenation
Question: What happens in halogenation reaction? Think about the following reactions.
+ CH3CHC 2Cl2CH3
ClCl
ClCl
CHC
BrBr
CH2H2C Br2CH2 +H2C
Your conclusion about halogenation of alkenes and alkynes is that …
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Write the product of the following addition reactions: 1. CH3─CH=CH─CH3 + Cl2
2. + Br2
Learning Check
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Testing for Unsaturation
When bromine (Br2) is added to an alkane, the red color of bromine persists.
When bromine (Br2) is added to an alkene or alkyne, the red color of bromine disappears immediately.
Br2
Br2
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
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Hydrohalogenation
What happens in hydrohalogenation reactions? Again, think about where atoms of a hydrogen halide end up.
ClH
CH3 CHCHCH3+ HClCH3CHCHCH3
Br
H + HBr
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Markovnikov’s Rule
When an unsymmetrical alkene undergoes hydrohalogenation, the H in HX adds to the carbon in the double bond that has the greater number of H atoms .
HCl
CH2CHCH3
CH2CHCH3 + HCl
ClH
CH2CHCH3 Does not form
C with the most H
Product that forms
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Hydration
In the addition reaction called hydration An acid H+ catalyst is required. Water (HOH) adds to a double bond. An H atom bonds to one C in the double bond. An OH bonds to the other C.
H OH
H+ │ │CH3-CH3─CH=CH─CH3 + H─OH
CH3─CH─CH─CH3
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Hydration
When hydration occurs with a double bond that has an
unequal number of H atoms, The H atom bonds to the C in the double bond with
the most H. The OH bonds to the C in the double bond with the
fewest H atoms.
OH H H+ │ │CH3─CH=CH2 + H─OH CH3─CH─CH2
(1H) (2H)
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Learning Check
Write the product for the hydration of each of the
following:
H+
1. CH3─CH2─CH=CH─CH2─CH3 + HOH
CH3
│ H+
2. CH3─C=CH─CH2─CH3 + HOH
H+
3. + HOH
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Learning Check
Write the products of each reaction
C.
B.
A.
HOH +CH3CHCHCH3
Pt
+ Cl2 CH2CHCH3
+ H2
H+
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In a regioselective reaction, one constitutional isomer is the major or the only product.
Markovnikov’s RuleThe electrophile adds to the sp2 carbon that is bonded to the greater number of hydrogens:
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The positive part of the reagent becomes attached to the double bond carbon which bears the greatest number of hydrogen atoms
Markovnikov’s Rule
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Markovnikov’s Rule
The acid proton will bond to carbon 3 in order to produce the most stable carbocation possible.
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Example
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Show how you would accomplish the following synthetic conversions.(a) Convert 1-methylcyclohexene to 1-bromo-1-methylcyclohexane.
This synthesis requires the addition of HBr to an alkene with Markovnikov orientation. Ionic addition of HBr gives the correct product.
Solved Problem 1
Solution
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Anti-Markovnikov’s Rule
Free Radical Additon of HBr:
In the presence of peroxides, (ROOR), HBr adds to an alkene to form the “anti-Markovnikov” product.
Peroxides produce free radicals. Only HBr has just the right reactivity for each step of the
free-radical chain reaction to take place. The peroxide effect is not seen with HCl or HI because the
reaction of an alkyl radical with HCl or HI is strongly endothermic.
Atom H attached to C atom with fewer hydrogen atom.
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OXIDATION REACTION OF ALKENES
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Ozone, O3, adds to alkenes to form molozonide Molozonideis converted to ozonide that may be reduced to obtain ketones and/or aldehydes
E.g:
1) OZONOLYSIS
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2) HYDROXYLATION
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Example
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3) OXIDATION WITH HOT KMnO4 Alkene react with hot/concentrated KMnO4 to give compounds
containing –C=O group – (form ketone and/or acids) In the reaction, C=C is cleavage and bond to oxygen
KMnO4R1 – C = C – R4
R2 R3
R1 – C = O
R2
O = C – R4
R3
+
R1 – C = C – H
R2 R3
KMnO4 R1 – C = O
R2
+ O = C – OH
R3
R1 – C = C – H
R2 H
KMnO4 R1 – C = O
R2
+ CO2 + H2O
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4) OXYMERCURATION–DEMERCURATION REACTION
Reagent is mercury(II) acetate, which dissociates slightly to form +Hg(OAc).
+Hg(OAc) is the electrophile that adds to the pi bond. The intermediate is a three-membered ring called the
mercurinium ion. Overall, the addition of water follows Markovnikov’s rule.
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5) HYDROBORATION OF ALKENES
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SYNTHESIS OF ALKENES - ELIMINATION RXNS
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1) DEHYDRATION OF ALCOHOL
Alkenes are also generally prepared by the dehydration of alcohols in the presence of a strong acid.
heatH+
C
H
C
OH
+ H2O
Sulfuric and phosphoric acids are often used for this reaction. Industrial processes use flow reactors with the alcohol in the gas phase passing over a solid Lewis acid such as alumina, Al2O3, with heating.
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Characteristics of the Dehydration Reaction
.(1) The conditions (acid strength and temperature) required for the dehydration depend on the structure of the alcohol
primary alcohol
CH3CH2OH conc. H2SO4
180o CCH2=CH2 + H2O
20% H2SO4
85o CCH2=C + H2OCH3
CH3tertiary alcohol
CH3COHCH3
CH3
General Order of Reactivity
R-C-OH
tertiary alcohol
R
R> R-C-OH
secondary alcohol
R
HR-C-OH>
primary alcohol
H
H
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2) DEHYDROGENATIONS OF ALKYL HALIDES
Elimination reaction of hydrogen halide (HX) from alkyl halide (RX)
Reagents used commonly KOH dissolved in alcohol and NaOH dissolved in alcohol
E.g:
– C – C –
H X
KOH,
alcohol – C = C –
+ H2O + KX
Br
KOHBr
KOH
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Polymers
Polymers are Large, long-chain molecules. Found in nature, including cellulose in plants,
starches in food, proteins and DNA in the body.
Also synthetic such as polyethylene and polystyrene, Teflon, and nylon.
Made up of small repeating units called monomers.
Made by reaction of small alkenes.
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Polymerization
In polymerization, small repeating units called monomers join to form a long chain polymer.
monomer unit repeats
n
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Common Synthetic Polymers
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Polymers from Addition Reactions
TABLE 12.3
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More Monomers and Polymers
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Learning Check
What is the starting monomer for polyethylene?
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Learning Check
Name the monomer used to make Teflon and write a
portion of a Teflon polymer using four monomers.
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Recycling Plastics
Recycling is simplified by using codes found
on plastic items.
1 PETE Polyethyleneterephtalate
2 HDPE High-density polyethylene
3 PV Polyvinyl chloride
4 LDPE Low-density polyethylene
5 PP Polypropylene
6 PS Polystyrene Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
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Lecture 3 Unsaturated Hydrocarbons
Aromatic Compounds
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Benzene Has 6 electrons shared equally among the 6 C atoms.
Is also represented as a hexagon with a circle drawn inside.
Benzene Structure
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Aromatic Compounds in Nature and Health
Vanillin Aspirin
Ibuprofen Acetaminophen
COH
O
O
C O CH3
CH
O
OCH3
OH
CH
CH3
COH
O
CH2
CH3
CHH3COH
NH
O
C CH3
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Naming Aromatic Compounds
Aromatic compounds are named
With benzene as the parent chain.
With one side group named in front of benzene.
methylbenzene chlorobenzene
ClCH3
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Some Common Names
Some substituted benzene rings
Have common names used for many years.
With a single substituent use a common name or are named as a benzene derivative.
toluene aniline phenol
(methylbenzene) (benzenamine) (hydroxybenzene)
NH2 OHCH3
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Aromatic Compounds with Two Groups
Two naming systems are used when two groups are
attached to a benzene ring.
Number the ring to give the lowest numbers to the side groups.
Use prefixes to show the arrangement:
ortho(o-) for 1,2-
meta(m-) for 1,3-
para(p-) for 1,4-
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Aromatic Compounds with Two Groups
3-chlorotoluene 1,4-dichlorobenzene 2-chlorophenol
m-chlorotoluene p-dichlorobenzene o-chlorophenol
OHCH3
Cl
Cl
Cl
Cl
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Learning Check
Select the correct name for each compound:1) chlorocyclohexane2) chlorobenzene3) 1-chlorobenzene
1) 1,2-dimethylbenzene2) m-xylene3) 1,3-dimethylbenzene
CH3
CH3
Cl
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Learning Check
Write the structural formulas for each of the following:
A. 1,3-dichlorobenzene
B. o-chlorotoluene
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Learning Check
Identify the organic family for each:
A. CH3─CH2─CH=CH2
B.
C. CH3─C≡CH
D.
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Properties of Aromatic Compounds
Aromatic compounds Have a stable aromatic bonding system. Are resistant to many reactions. Undergo substitution reactions, which retain
the stability of the aromatic bonding system.
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Substitution Reactions
In a substitution reaction, a hydrogen atom on a benzene ring is replaced by an atom or group of atoms.
Type of substitution H on benzene replaced
by
Halogenation chlorine or bromine
atom
Nitration nitro group (—NO2)
Sulfonation —SO3H group
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Halogenation
In a halogenation An H atom of benzene is replaced by a chlorine or
bromine atom. A catalyst such as FeCl3 is needed in chlorination.
A catalyst such as FeBr3 is needed in bromination.
ChlorobenzeneBenzene
FeCl3 HCl+
Cl
Cl2+
H
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Nitration
In the nitration of benzene An H atom of benzene is replaced by a nitro (-NO2)
group from HNO3.
An acid catalyst such as H2SO4 is needed.
NitrobenzeneBenzene
H2SO4 HOH+
NO2
HNO3+
H
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Sulfonation
In a sulfonation An H atom on benzene is replaced by a —SO3H
group from SO3.
An acid catalyst such as H2SO4 is needed.
Benzenesulfonic acidBenzene
H2SO4
SO3H
SO3+
H
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Learning Check
Write the equation for the bromination of benzene including catalyst.