Which has the highest BP? The lowest?
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Transcript of Which has the highest BP? The lowest?
Which has the highest BP? The lowest?
Which will be miscible with each other?
HO
H
O
Alkenes
Addition Reactions
Synthesis
Which is the Nucleophile? The Electrophile?
Addition Reaction is a Two-Step Mechanism
Markovnikov’s Rule
The addition of H-X across a double bond results in the more
highly substituted alkyl halide as the major product.
Addition of HBr or HClMarkovnikov Addition
Markovnikov
Br
C CH
H
HCH3
CH3
HBr
CH3
CH3
C
H
H
C
CH3
CH3C
H
H
Br
C
H
not formed
Regiochemistry Determined by Stability of Intermediate
H-BrBr
H
HBr Br H
3o carbocation
Hyperconjugation
Carbocation Stabilitymore highly substituted, lower energy
3o Carbocation forms Preferentially
What Alkenes are Needed to form these Alkyl Halides?
HBr Addition with RO-ORAnti-Markovnikov
HBr
RO-OR, h
H Br
Free-Radical Mechanism
Initiation:
Propagation: i)
ii)
RO-OR 2 RO
RO + HBr Br + ROH
+ BrBr
Br+ H-Br
H Br+ Br
.
. .
. .
. .
HydrationAddition of H2O
C CH
H
CH3
CH3
H2SO4 (aq)
CH3
CH3C
H
H
H
C
OH
H
C CH
H
HCH3
CH3
H2O
OH
H
C CH
H
HCH3
CH3
H2O
-H
catalytic
Addition of Br2
Br
Br
Br2
A Similar Mechanism to the Addition of HBr?
Trans is formed exclusively(No cis is formed)
Bromonium ion intermediate is formed
Bromonium Ion is Opened Equally from Both Sides
Br2
Br Br Br Br
Br
Br
Br
Br
Br Br + -
Bromonium Ion Intermediate
Bromonium ion
- Br
Br
Br
BrBr
+ BrBr
Catalytic Hydrogenationsyn addition
Mechanism – Syn Addition
Syn Addition of H2
CH3CH3
H2, 1% Pt/C
75 psi
CH3 CH3H H
cis only
Hydrogenationall alkene bonds are reduced
H2, Pt/C
Three mol equivalents of hydrogen gas are consumed.
Hydrogenated Vegetable Oil
Oxymercuration HydrationMarkovnikov additionRegiospecific Reaction
1) Hg(OAc)2 inTHF/H2O
2) NaBH4
OH
H
Oxymercuration Mechanism
OH H
(H )NaBH4
organomercurial alcohol
HgOAc
OH
OAc
HgOAc
OH
H
OAcHgOAc
in H2O
OH2
+
+
Hg
OAc
Hg(OAc)2
Predict the Oxymercuration Hydration Products
What Alkene would you use to Make These Alcohols?
Hydroboration Hydration Anti-Markovnikov
Syn addition
31) BH3-THF
2) H2O2, NaOH
H OH3
BH3 adds to Alkene
Hydroboration
BH3H BH2 H B
HH2
B
H
H H
Trialkylborane
H2O2
OH-
H OH3
3
Alkylborane
Regiochemistry is Anti-Markovnikov
2 Complementary Hydration Reactions
CH3
1) BH3, THF
2) H2O2, NaOH
OH
CH3
H
1) Hg(OAc)2, H2O-THF
2) NaBH4
OH
CH3
Oxymercuration – the more highly substituted alcohol forms
Hydroboration – the less highly substituted alcohol forms
Predict Both Oxymercuration and Hydroboration Products
Epoxidation of Alkenes
Mechanism of Epoxide Formation
Epoxide opens to form a trans-1,2-diol
Alkenes react with basic KMnO4 or OsO4 to form cis-1,2-diols
Oxidation of Alkenes: Draw the major product(s)
KMnO4, NaOH
MCPBA H3O+
Alkenes can be cleaved with acidic KMnO4
Alkenes can be cleaved with acidic KMnO4
Olefinic carbons with H are converted to carboxylic acids.
Olefinic carbons w/out H are converted to ketones.
Terminal olefinic carbons are converted to CO2.
KMnO4, H3O+
O + O
OH
KMnO4, H3O+
O + O C O
Fill in the missing compounds.KMnO4, H3O+
A + B
KMnO4, H3O+
KMnO4, H3O+
D (with a cyclohexyl ring)
O
O
O
OH
+ CO2
C
Problem
The sex attractant of the common housefly is a hydrocarbon named muscalure, with the formula C23H46. On treatment with aqueous acidic KMnO4, two products are obtained, CH3(CH2)12CO2H and CH3(CH2)7CO2H. Propose a structure for muscalure.
Molecular Form ula = C 23H 46
Form ula W eight = 322.61134Com position = C(85.63%) H(14.37%)
OH
O
OH
O
KMnO4, H3O+
Problem
A symmetrical unknown compound A, C8H16, reacts with H2 on a 1% Pt/C catalyst
to form B (C8H18). Treatment of A with
acidic permanganate affords butanone only. Identify A and B.
CH3CCH2CH3
O
butanone
O
butanone
O
HH
H2
O3
A
B
KMnO4, H3O+
Conjugated Dienes
Which of these compounds are conjugated alkenes?
B and C only.
Looking at Buta-1,3-diene
Addition to Conjugated Dienes
resonance forms of the allylic carbocation
The 1,2-addition product is typical Markovnikov.
The 1,4-addition product is due to the resonance forms of the allylic carbocation intermediate.