Transcript Alkenes and Alkynes Ch#3

Alkenes and Alkynes
Chapter #3
Alkene Introduction
• Hydrocarbon with carbon-carbon double bonds
• Sometimes called olefins, “oil-forming gas”
• General formula CnH2n n≥2
• Examples
n=2 C2H4
Common Names
Usually used for small molecules.
Examples:
CH2=CH2
ethylene
CH2=CH-CH3
propylene
CH3
CH2=C-CH3
isobutylene
Vinyl carbons are the carbons sharing a double bond
in blue
Vinyl hydrogens are the hydrogens bonded to vinyl
carbons in red
IUPAC Nomenclature
• Parent is longest chain containing the double or
triple bond.
• -ane changes to –ene (or -diene, -triene) for
double bonds, or –yne (or –diyne, -triyne).
• Number the chain so that the double bond, or
triple bond has the lowest possible number.
• In a ring, the double bond is assumed to be
between carbon 1 and carbon 2.
Name These Alkenes
CH2
CH CH2
CH3
CHCH2CH3
H3C
CH3
C CH CH3
CH3
CH3
Name These Alkenes
CH2
CH CH2
1-butene
CH3
C CH CH3
CH3
CH3
CH3
CHCH2CH3
H3C
Name These Alkenes
CH2
CH CH2
CH3
1-butene
CH3
C CH CH3
CH3
2-methyl-2-butene
CH3
CHCH2CH3
H3C
Name These Alkenes
CH2
CH CH2
CH3
1-butene
CH3
C CH CH3
CH3
2-methyl-2-butene
CH3
3-methylcyclopentene
CHCH2CH3
H3C
Name These Alkenes
CH2
CH CH2
CH3
1-butene
CHCH2CH3
H3C
2-sec-butyl-1,3-cyclohexadiene
CH3
C CH CH3
CH3
2-methyl-2-butene
CH3
3-methylcyclopentene
Name These Alkenes
CH2
CH CH2
CH3
1-butene
CHCH2CH3
H3C
2-sec-butyl-1,3-cyclohexadiene
CH3
C CH CH3
CH3
2-methyl-2-butene
CH3
3-methylcyclopentene
3-n-propyl-1-heptene
Alkene Substituents
= CH2
- CH = CH2
methylene
vinyl
Name = ?
- CH2 - CH = CH2
allyl
- CH2 - CH = CH2
allyl
Alkene Substituents
= CH2
- CH = CH2
methylene
vinyl
Name = Methylenecyclohexane
- CH2 - CH = CH2
allyl
Name =
- CH2 - CH = CH2
allyl
Alkene Substituents
= CH2
- CH = CH2
methylene
vinyl
Name = Methylenecyclohexane
- CH2 - CH = CH2
allyl
Name = vinylcyclohexane
Alkyne Common Names
• Acetylene is the common name for the two
carbon alkyne.
• To give common names to alkynes having
more than two carbons, give alkyl names to
the carbon groups attached to the vinyl
carbons followed by acetylene.
Alkyne Examples
Alkyne Examples
Isopropyl methyl acetylene
Alkyne Examples
Isopropyl methyl acetylene
sec-butyl Cyclopropyl acetylene
Cis-trans Isomerism
• Similar groups on same side of double bond, alkene
is cis.
• Similar groups on opposite sides of double bond,
alkene is trans.
• Cycloalkenes are assumed to be cis.
• Trans cycloalkenes are not stable unless the ring has
at least 8 carbons.
Name these:
H
CH3
C C
CH3CH2
H
Name these:
H
CH3
C C
CH3CH2
trans-2-pentene
H
Name these:
H
CH3
Br
C C
C C
CH3CH2
trans-2-pentene
Br
H
H
H
Name these:
H
CH3
Br
C C
CH3CH2
trans-2-pentene
Br
C C
H
H
H
cis-1,2-dibromoethene
Which of the following show cis/trans
isomers?
a. 1-pentene
b. 2-pentene
c. 1-chloro-1-pentene
d. 2-chloro-1-pentene
e. 2-chloro-2-pentene
Solution to the Question
Solution to the Question
Which of the following show cis/trans
isomers?
a. 1-pentene-No
b. 2-pentene- Yes
c. 1-chloro-1-pentene- Yes
d. 2-chloro-1-pentene- No
e. 2-chloro-2-pentene- yes
E-Z Nomenclature
• Use the Cahn-Ingold-Prelog rules to assign priorities
to groups attached to each carbon in the double
bond. Highest priority is #1 and is the element with
the largest atomic number.
• If high priority groups are on the same side, the
name is Z (for zusammen).
• If high priority groups are on opposite sides, the
name is E (for entgegen).
Example, E-Z
1
H3C
Cl
C C
H
2Z
1
H
CH2
2
2
Cl
2
1
CH CH3
C C
H
2
1
5E
Example, E-Z
1
H3C
Cl
C C
H
2Z
1
H
CH2
2
2
Cl
2
1
CH CH3
C C
H
2
1
5E
3,7-dichloro-(2Z, 5E)-2,5-octadiene
Physical Properties
• Low boiling points, increasing with mass.
• Branched alkenes have lower boiling points.
• Less dense than water.
•Nonpolar (Hydrophobic)
Alkene Synthesis
Elimination Reactions:
• Dehydrohalogenation (-HX)
• Dehydration of alcohols (-H2O)
Examples:
Cl
+
+ NaCl + HOH
+
NaOH
major
minor
OH
H
+ H2O
minor
major
Zaitsev’s rule: The major product contains the most
substituted double bond
Alkene Reactions
I. Addition Reactions
a. Hydration
H O-H
H+
C=C
+ H-O-H
H H
Catalyst
+ H-H
C-C
Alkane
c. Halogenation
C=C
+ X-X
X = Cl, Br, I
Follows Markovnikov’s Rule
Alcohol
b. Hydrogenation
C=C
C-C
X X
C-C
Dihalide
Catalyst = Ni, Pt, Pd
Regiospecificity
Markovnikov’s Rule: The proton (H+) of an acid adds to the
carbon in the double bond that already has the most H’s.
“Rich get richer.”
H O-H
Examples:
H
H
H+
+ H-O-H
C=C
H
CH3
H
H
C=C
H
+ H-Cl
CH3
H
C-C
H
H CH3
Major Products
H Cl
H
C-C
H
H CH3
Alkene Reactions (2)
I. Addition Reactions (cont.)
d. Hydrohalogenation
C=C
+ H-X
H X
Follows Markovnikov’s Rule
C-C
Alkyl halide
e. Glycol Formation
C=C
+ H-O-O-H
H-O
O-H
C-C
Glycol
Alkene Reactions
Step 1: Pi electrons attack the electrophile.
Step 2: Nucleophile attacks the carbocation
Terpenes
• Composed of 5-carbon isopentyl groups.
• Isolated from plants’ essential oils.
• C:H ratio of 5:8, or close to that.
• Pleasant taste or fragrant aroma.
• Examples:
Myrcene (From bay or myrcia plants)
α-Pinene (From pine trees)
Β-Selinene (From celery)
Menthol (From peppermint oil)
Camphor (From evergreen trees)
R-Carvone (From spearmint)
Classification
• Terpenes are classified by the number of
carbons they contain, in groups of 10.
• A monoterpene has 10 C’s, 2 isoprenes.
• A diterpene has 20 C’s, 4 isoprenes.
• A sesquiterpene has 15 C’s, 3 isoprenes.
Terpenes
head
tail
head
head
tail
tail
2-methyl-1,3-butadiene
Isoprene
OH
tail
head
Geraniol (roses)
Head to tail link
of two isoprenes
Called diterpene
head
Menthol (pepermint)
Head to tail link of
two isoprenes
another diterpene
Structure of Terpenes
Two or more isoprene units, 2-methyl-1,3-butadiene with
some modification of the double bonds.
myrcene, from
bay leaves
ALKENE REVIEW
Describe the geometry around the
carbon–carbon double bond.
a. Tetrahedral
b.Trigonal pyramidal
c. Trigonal planar
d.Bent
e.Linear
Answer
a. Tetrahedral
b.Trigonal pyramidal
c. Trigonal planar
d.Bent
e.Linear
Give the formula for an alkene.
a. CnH2n-4
b.CnH2n-2
c. CnH2n
d.CnH2n+2
e.CnH2n+4
Answer
a. CnH2n-4
b.CnH2n-2
c. CnH2n
d.CnH2n+2
e.CnH2n+4
Name CH3CH=CHCH=CH2.
a. 2,4-butadiene
b.1,3-butadiene
c. 2,4-pentadiene
d.1,3-pentadiene
e.1,4-pentadiene
Answer
a. 2,4-butadiene
b.1,3-butadiene
c. 2,4-pentadiene
d.1,3-pentadiene
e.1,4-pentadiene
Calculate the unsaturation number for
C6H10BrCl.
a. 0
b.1
c. 2
d.3
Answer
a. 0
b.1
c. 2
d.3
U = 0.5 [2(6) + 2 – (12)] = 1
Name
H3C
.
C
CH3
C
H
a.
b.
c.
d.
Trans-2-pentene
Cis-2-pentene
Trans-3-methyl-2-pentene
Cis-3-methyl-2-pentene
CH2CH3
Name
H3C
CH3
C
H
a.
b.
c.
d.
C
CH2CH3
Trans-2-pentene
Cis-2-pentene
Trans-3-methyl-2-pentene
Cis-3-methyl-2-pentene
Name
H3C
CH3
C
H
a.
b.
c.
d.
e.
C
CH2CH3
E-2-pentene
Z-2-pentene
E-3-methyl-2-pentene
Z-3-methyl-2-pentene
Z-2-methyl-2-pentene
Name
H3C
CH3
C
H
a.
b.
c.
d.
e.
C
CH2CH3
E-2-pentene
Z-2-pentene
E-3-methyl-2-pentene
Z-3-methyl-2-pentene
Z-2-methyl-2-pentene
H
H
Cl2
C
H
a. ClCH2CH2Cl
b.ClCH=CHCl
c. CH2=CH2
d.CH2=CHCl
C
H
NaOH
Answer
a. ClCH2CH2Cl
b.ClCH=CHCl
c. CH2=CH2
d.CH2=CHCl
Chlorine is added across the double
bond, then HCl is lost.
H
CH3
C
H2O
C
catalyst
H
a. (CH3)2CHOH
b.CH3CH2CH2OH
c. HOCH2CH2CH2OH
d.CH3CH(OH)CH2OH
H
Answer
a. (CH3)2CHOH
b.CH3CH2CH2OH
c. HOCH2CH2CH2OH
d.CH3CH(OH)CH2OH
Water adds by Markovnikov’s orientation
across the double bond.
Identify the product formed from the
polymerization of tetrafluoroethylene.
a. Polypropylene
b.Poly(vinyl chloride), (PVC)
c. Polyethylene
d.Poly(tetrafluoroethylene), Teflon
Answer
a. Polypropylene
b.Poly(vinyl chloride), (PVC)
c. Polyethylene
d.Poly(tetrafluoroethylene), Teflon
Teflon is formed from the polymerization
of tetrafluoroethylene.
H3C
CH3
C
H2
C
Pd
H
a. CH3CCCH3
b.CH2=CHCH=CH2
c. CH3CH=CHCH3
d.CH3CH2CH2CH3
H
Answer
a. CH3CCCH3
b.CH2=CHCH=CH2
c. CH3CH=CHCH3
d.CH3CH2CH2CH3
Hydrogen adds across the double bond
to form an alkane.
H3C
H
C
OH
a. (CH3)2CHOSO3H
b.CH3CH=CH2
c. (CH3)2C=O
d.CH3CH2COOH
H2SO4
CH3
heat
7.15 Answer
a. (CH3)2CHOSO3H
b.CH3CH=CH2
c. (CH3)2C=O
d.CH3CH2COOH
Acid dehydrates alcohols to form alkenes.
Give the products from the catalytic
cracking of alkanes.
a. Alkanes
b.Alkenes
c. Alkynes
d.Alkanes + alkenes
e.Alkanes + alkynes
Answer
a. Alkanes
b.Alkenes
c. Alkynes
d.Alkanes + alkenes
e.Alkanes + alkynes
Give the products from the
dehydrogenation of alkanes.
a. Alkanes
b.Alkenes
c. Alkynes
d.Alkanes + alkenes
e.Alkanes + alkynes
Give the products from the
dehydrogenation of alkanes.
a. Alkanes
b.Alkenes
c. Alkynes
d.Alkanes + alkenes
e.Alkanes + alkynes
End Chapter #3