Transcript Chapter 1 Structure and Bonding

Chapter 11 Alkenes and IR Lect. #2
I.
Infrared Spectroscopy
A.
Energy Absorption and Vibration
1) IR electromagnetic radiation is just less energetic than visible light
2) This energy is sufficient to cause excitation of vibrational energy levels
3) Wavelength (l) = 2.5-16.7 x 10-6 m
1
~
   600 - 4000 cm 1  1  10 kcal/mol
l
4)
5)
n = wavenumbers. Larger n = higher energy
Excitation depends on atomic mass and how tightly they are bound
a) Hooke’s Law for 2 masses connected by a spring
(m  m2 )
~  k f 1
m1m2
b)
k = constant
f = force constant = bond strength
m-term = reduced mass
Many possible absorptions per molecule exist: stretching, bending,…
Vibrational modes leading to IR absorptions:
B.
Using IR in Organic Chemistry
1. Functional Groups have characteristic IR absorptions
2.
Fingerprint Region (600-1500 cm-1) is unique for every molecule and lets
us match an unknown with a known spectrum
4.
IR of Alkenes
a. Alkene C—H absorbs at higher energy than alkanes because the
force constant is stronger than alkanes (sp2 hybridization)
b.
Substitution pattern of alkenes give characteristic absorptions
i. Terminal alkenes give 915, 995 cm-1
H
H
C
C
H
ii.
R
Geminal disubstituted gives 890 cm-1
H
R
C
C
H
R
iii. trans disubstituted gives 970 cm-1
H
R
C C
R
H
II.
Stability of Alkenes
A.
Heat of Hydrogenation
1) DH for the hydrogenation of an alkene tells us how stable it was
C
2)
C
H2 / Pt
H
H
C
C
All alkene isomers give the same saturated product, so we can compare
B.
R
Stability of alkene isomers
R
C C
R
R
>
R
1.
2.
3.
>
C C
H
R
R
R
H
C C
H
R
>
R
R
>
C C
H
R
H
H
>
C C
H
H
H
H
C C
H
H
Hyperconjugation makes substituted alkenes more stable by stabilizing
p-orbitals
cis alkenes are less stable than trans alkenes because of steric crowding
cis cycloalkenes are more stable than trans for the small rings
III. Preparation of Alkenes
A.
E2 Elimination
CH3
CH3CH2CCH3
Br
CH3CH2O-Na+, EtOH, 70 oC
CH3
+
C C
-HBr
CH3CH2
CH3
H
CH3
C C
CH3
70%
1)
2)
3)
H
Regioselectivity = selective reaction at one of multiple sites
T.S. is lower energy for the major product
Saytzev Rule = most highly substituted product will be favored
H
30%
4)
A bulky base will lead to the other product as major (Hoffman Rule)
CH3
CH3
(CH3)3CO-K+, iBuOH, 70 oC
CH3CH2CCH3
Br
+
C C
-HBr
CH3CH2
CH3
H
C C
CH3
CH3
H
CH3CH2CH2CCH3
Br
trans product more stable than cis product
CH3CH2O-Na+, EtOH
CH3CH2
H
CH3CH2
+
C C
-HBr
H
H
CH3
CH3CH2CH2CH=CH2
H
31%
18%
S,R
H
CH3CH2
CH3
H
CH3
Br
-HBr
H3C S,S
Br
CH3CH2
H
+
Some eliminations are stereospecific due to anti elimination
R,R
H
CH3
C C
51%
6)
H
73%
23%
5)
H
H
CH3CH2
CH3
H
CH3
E-product
H
CH3
CH3CH2
CH3
Br
-HBr
R,S
CH3CH2
CH3CH2
H3C
Br
CH3
H
CH3
H
CH3
H
CH3
Z-product
B.
Dehydration of Alcohols
1. Mineral acid and heat will dehydrate an alcohol to an alkene
H+, D
C C
2.
3.
4.
C C
H OH
Reaction is easier for more substituted products
Primary alcohols by E2
Secondary and Tertiary by E1 (with rearrangement to most substituted
product)
conc H2SO4
CH3CH2OH
CH2CH2
o
170 C
OH
CH3CHCH2CH3
(CH3)3COH
CH3
OH
CH3C CH2 CCH3
H
H
CH3CH=CHCH3
80%
o
100 C
dil H2SO4
50% H2SO4
100 oC
50% H2SO4
H2C=C(CH3)2
100%
o
50 C
CH3
CH3
H
C C
CH3
+
CH2CH3
54%
CH3CCH=CHCH3
H
8%