Transcript Alkane ∆Hf values (kcal/mol)

CEM 850, Fall 2004
Some notes on Thermochemistry,
Bond Strengths, and Strain energies
Ned Jackson
Alkane ∆Hf values (kcal/mol)
Branching
Carbon number n
1
2
3
4
5
6
7
8
-17.9
-20.04
-25.02
-30.37
-32.07
-35.08
-36.73
-40.14
-39.96
...
-44.35
-44.89
...
-48.96
-49.82
...
-53.99
Octane s :
Octane
-49.82
2-methylheptane
3-methylheptane
4-methylheptane
3-ethylhexane
-51.50
-50.82
-50.69
-50.40
2,5-dimethylhexane
2,4-dimethylhexane
2,3-dimethylhexane
3,4-dimethylhexane
3-ethyl-2-methylpentane
-53.21
-52.44
-51.13
-50.91
-50.48
2,2-dimethylhexane
3,3-dimethylhexane
3-ethyl-3-methylpentane
-53.71
-52.61
-51.38
2,3,4-trimethylpentane
-51.97
2,2,3-trimethylpentane
2,2,4-trimethylpentane
2,3,3-trimethylpentane
-52.61
-53.57
-51.73
2,2,3,3-tetramethylbutane
-53.99
Alkane ∆Hf values show
Systematic Patterns
• Can we estimate ∆Hf by summing energy
equivalents for transferable molecular
“building blocks?”
– Bond Equivalents
– Group Equivalents
• Fragment transferability in comparisons
between compounds implies deep similarity
Bond Equivalents
• Estimate ∆Hf values from C-H, C-C bonds:
–
–
–
–
–
–
Ethane ∆Hf = -20.04; 6 C-H, 1 C-C
Propane ∆Hf = -25.02; 8 C-H, 2 C-C
\C-H = -3.765; C-C = +2.55
Predict ∆Hf(C5H12) = 4C-C + 12C-H = -34.98
N-pentane -35.08 but isopentane = -40.14
\Bond equivalents fail for branching.
Group Equivalents
• All alkanes can be expressed in terms of
four building blocks: (CH3)i(CH2)j(CH)k(C)l
(nonspecified bonds implicitly to C)
• Enthalpy Equivalents:
–
–
–
–
CH3
CH2
CH
C
-10.08
-4.95
-1.90
+0.50
Group Equivalents (cont’d)
• Analogous equivalents for alkenes and
aromatics can be similarly derived, with
value for CH3 held at -10.08 kcal/mol no
matter what it’s attached to.
• This method defines a “strainless ideal” for
hydrocarbons of arbitrary formula, and
allows the definition of “strain.”
Strain Energies
• Cycloalkanes (CH2)n
Ring Size n
² Hf
² Hf Hypothetical
Strainless (CH2)n
Strain E.
2
3
4
5
6
12.54
12.74
6.78
-18.26
-29.43
2x(-4.95) 3x(-4.95) 4x(-4.95) 5x(-4.95) 6x(-4.95)
= -9.9 = -14.85 = -19.8 = -24.75 = -29.7
22.44
27.59
26.58
6.49
0.27
Thermochemistry--why care?
• Besides simple reaction ∆H and ∆G values,
detailed energetics define reaction direction
• Combined with bond strengths and kinetics
of the reactions of interest, even imperfect
energetic ideas put limits on mechanistic
possibilities
• Lead in to tools for comparing reactions!
Bond Strengths
• An X-Y bond, as defined by its atoms X and
Y, is not a uniform (thus transferable)
molecule building block
• The bond equivalent approach did not lead
to a reliable method for ∆Hf estimation
• A group equivalent approach was required
• Some bond strengths allow development of
group equivalent ideas for reactions
R-H BDEs worth remembering
•
•
•
•
•
•
•
H-H
CH3-H
CH3CH2-H
(CH3)2CH-H
(CH3)3C-H
H2C=CHCH2-H
PhCH2-H
104.2 kcal/mol
105.1
100.5
99.1
95.2
88.1
89.6
An ordinary C-C s bond
• Generic C-C bond strengths in R-R’
– Use group equivalents to estimate ∆Hf values
for R-R’, R-H, and R’-H
– Get ∆Hf of R•, R’• radicals from R-H, R’-H via
C-H BDEs + BDE(H2) = 104.2 kcal/mol
– Calculate R-R’ BDE
Cracking of Butane: 1-2 vs 2-3
•
•
•
•
•
•
•
∆Hf(butane) = 2(-10.08 -4.95) = -30.06 est.
∆Hf(methane) = -17.9
∆Hf(ethane) = 2(-10.08) = -20.16 est.
∆Hf(propane = 2(-10.08) -4.95 = -25.11 est.
∆Hf(Me•) = -17.9 +105.1 -52.1 = 35.1 est.
∆Hf(Et•) = -20.16 +100.5 -52.1 = 28.2 est.
∆Hf(Pr•) = -25.11 +100.5 -52.1 = 23.3 est.
Some Heats of Formation
H
CH3
CH3CH2
(CH3)2CH
(CH3)3C
H2C=CHCH2
PhCH2
52.1
365.7
35.1
262.0
28.4
215.6
22.0
191.5
11.0
165.5
40.9
229.5
49.5
216.5
H
52.1
34.7
0.0
0.0
-17.9
-17.9
-20.0
-20.0
-25.0
-25.0
-32.1
-32.1
4.9
4.9
12.0
12.0
F
19.0
-59.5
-65.3
-65.3
-56.0
-56.0
-70.1
-70.1
-30.2
-30.2
Cl
29.0
-54.4
-22.1
-22.1
-20.0
-20.0
-26.8
-26.8
-34.7
-34.7
-43.0
-43.0
-1.3
-1.3
4.5
4.5
Br
26.7
-50.9
-8.7
-8.7
-8.2
-8.2
-15.2
-15.2
-22.9
-22.9
-31.6
-31.6
11.4
11.4
20.0
20.0
I
25.5
-45.1
6.3
6.3
3.4
3.4
-1.7
-1.7
-9.5
-9.5
-17.2
-17.2
23.8
23.8
30.4
30.4
OH
9.3
-32.8
-57.8
-57.8
-48.0
-48.0
-56.2
-56.2
-65.2
-65.2
-74.7
-74.7
-29.6
-29.6
-22.6
-22.6
NH2
45.5
27.1
-11.0
-11.0
-5.5
-5.5
-13.0
-13.0
-20.0
-20.0
-28.8
-28.8
21.0
21.0
CH3
35.1
33.3
-17.9
-17.9
-20.0
-20.0
-25.0
-25.0
-32.1
-32.1
-40.1
-40.1
-0.2
-0.2
7.1
7.1
All energies in kcal/mol
-30.06 est.
-30.37 exp.
BDEs
Cleave 1,2
88.5 est.
89.4 exp.
CH3• + •
35.1 + 23.3 est.
35.1 + 23.9 exp.
Cleave 2,3
• + •
2 * 28.2 est.
2 * 28.4 exp.
BDEs
86.5 est.
87.2 exp.
Bond Diss’n Energies (BDEs)
H
CH3
CH3CH2
(CH3)2CH
(CH3)3C
H2C=CHCH2
PhCH2
52.1
365.7
35.1
262.0
28.4
215.6
22.0
192.0
11.0
165.5
40.9
229.5
49.5
216.5
H
52.1
34.7
104.2
400.4
105.1
314.6
100.5
270.3
99.1
251.2
95.2
232.3
88.1
259.3
89.6
239.2
F
19.0
-59.5
136.4
371.5
110.1
258.5
111.1
202.1
61.2
-17.3
Cl
29.0
-54.4
103.2
333.4
84.1
227.6
84.2
188.0
85.7
171.8
83.0
154.1
71.2
176.4
74.0
157.6
Br
26.7
-50.9
87.5
323.5
70.0
219.3
70.3
179.9
71.6
163.5
69.3
146.2
56.2
167.2
56.2
145.6
I
25.5
-45.1
71.3
314.3
57.2
213.5
55.6
172.2
57.0
155.9
53.7
137.6
42.6
160.6
44.6
141.0
OH
9.3
-32.8
119.2
390.7
92.4
277.2
93.9
239.0
96.5
223.9
95.0
207.4
79.8
226.3
81.4
206.3
NH2
45.5
27.1
108.6
403.8
86.1
294.6
86.9
255.7
87.5
238.6
85.3
221.4
74.0
222.6
CH3
35.1
33.3
105.1
416.9
90.2
315.3
88.5
273.9
89.2
256.9
86.2
238.9
76.2
263.0
77.5
242.7
The strength of a π bond
• Breaking ethylene’s π bond doesn’t lead to
two well-defined fragments. How can we
define a separate “bond strength” for it?
–
–
–
–
Cis-trans isomerization of HDC=CHD?
Hydrogenation energies?
Spectroscopic measurements?
Others (full disassembly of molecule)?
Ethylene isomerization
• Heat cis or trans DHC=CHD and measure
the rate of isomerization as a function of T.
• From kinetic analysis, obtain ∆Hact for c-t
isomerization: ~66 kcal/mol.
• Problems: at high enough T, lots of other
chemistry can happen; some may catalyze
isomerization, making barrier appear too
low. Or, isomerization might not go via
rotation!? How to get a check on this value?
Hydrogenation Strategy
• H2C=CH2 + H2 —> H-H2C-CH2-H
12.5 + 0 —> -20.0; ∆Hrxn = -32.5 kcal/mol
• Broken: C-C π bond, [email protected] kcal/mol;
Formed: Two ethane C-H bonds @100.5
kcal/mol each
• BDE(π) = 201. -32.5 -104.2 = 64.3 kcal/mol
!Looks good!
Spectroscopic approaches?
• π—>π* Excited state has no π bonding, but
lmax = 171 nm = ~167 kcal/mol!? Pretty far
from 66!
• ∆IE (ethylene - ethyl)
(Electron’s energy-drop
from non- to π-bonding
= 10.51 - 8.12 eV
= 55 kcal/mol per e–
=> 110 kcal/mol!?
Energetics of Full Disassembly
of Ethylene
• Try to make a prediction:
– C-C s BDE is ~90 kcal/mol
– the π bond is ~65 kcal/mol
– Predict ~155 kcal/mol ∆H for C2H4 —> 2CH2
• ∆Hf(ethylene) = 12.5; ∆Hf(CH2) = 92.3; 184.612.5 = 172.1, almost 20 kcal/mol “too large”
--what’s going on?
• C-H bond strengths increase from C2H4 and CH2
Cyclopropane Stereomutation
• How strong is a C-C bond in cyclopropane?
– Look at isomerization via isotopic labeling
– Directly analogous to ethylene cis-trans
isomerization
– Should go via “real” open-chain biradical
•H C-CH -CH •
2
2
2
– What about hydrogenation energies?
– Can Strain E’s help?
Thermal Stereomutation
• Measured ∆Hact for c-t isomerization:
– 63.7 kcal/mol (1958); 59.8 kcal/mol (1972)
– ∆Hf of cyclopropane = 12.7 kcal/mol
– \biradical ∆Hf should be ca. 72.5 kcal/mol
• Primary C-H BDE back then was thought to
be ca. 97 kcal/mol, instead of 100.5
• Propane = -25 + 2(97-52) = 65…huh?
The propanediyl disaster
• Thermochem looked like biradical must rest
in a 5-9 kcal/mol well between c,t-isomers
CHD •
• CHD
D
D
D
D
Why don’t we expect a barrier
• General radical dimerization barrierless
• Conceptual reason: there’s no stabilization
to lose as bond formation begins.
• “Hammond postulate” and/or Bell-EvansPolanyi principle--the more exothermic the
process, the lower its barrier will be.
Review with current values
• We calculated the 2-3 cleavage barrier for
butane; cyclopropane should have the same
number, lowered by its strain energy, which
is released upon ring opening.
• So 87.2 -27.5 kcal/mol directly predicts a
barrier of 59.7, near the 1972 ∆Hact value.
• Just need to revise primary C-H BDE up by
3.5 kcal/mol (x2 = ~the 7.5 kcal/mol error)
The Methane Activation Problem
• Methane combustion is very exothermic
– CH4 + 2O2 --> CO2 + 2H2O
– ∆Hcomb = -17.9 + 0 --> -94.1 + 2(-57.8)
= -191.8 kcal/mol (plenty exothermic)
– It’s a great fuel, but…it isn’t liquid
– BP(CH4) = -162 ˚C = 111 K
– \ Not practical for automotive use
– (similar issues surround H2)
Partial oxidation to liquify CH4?
• Oxidation to methanol would be exothermic
– CH4 + 1/2O2 --> CH3OH
– ∆H = -17.9 + 0 --> -48.0 = -30.1 kcal/mol
• Energy from CH3OH combustion?
– CH3OH + O2 --> CO2 + 2H2O
– ∆Hcomb = -48.0 --> -209.7 = -161.7 kcal/mol
Hydrocarbon vs. Methanol Fuels:
Energy Densities
• Typical hydrocarbon “(CH2)n”
– Mass = 14 g/mol
– ∆Hcomb= -5 --> -94.1+(-57.8) = -146.9 kcal/mol
– = 10.5 kcal/mol•gram
• Methanol CH3OH
– Mass = 30 g/mol
– ∆Hcomb = -161.7 kcal/mol
– = 5.4 kcal/mol•gram
Challenge: CH4 --> CH3OH
105.1
(314.6)
98.1
(254.3)
H
H
H
O
H
H
²H f = -17.9 kcal/mol
IE = 12.61 eV (290.8 kcal/mol)
PA = 129.9 kcal/mol
Reagents?
Radicals
Acids
Bases
92.4 (277.2)
H
H
H
H
H
H
H
104.2
(382)
²H f = -48.0 kcal/mol
IE = 10.84 eV (250.0 kcal/mol)
PA = 180.3 kcal/mol
110
90.2
(259.4)
130.6 (66.6)
H
117 (180)
O
H
²H f = 137.4 kcal/mol
O
H
²H f = -27.7 kcal/mol
IE = 10.88 eV (250.9 kcal/mol)
PA = 170.4 kcal/mol
Protection of Methanol?
• The C-H bond strengths in methanol are
increased from 98.1 to 110 kcal/mol by
methanol protonation, becoming stronger
than those in methane (105.1 kcal/mol). Is
this enough to control selectivity?
• The key is the attacking species,
presumably either HO• or CH3O• radicals
here.
Radical selectivities
• Isobutane halogenation
– Bond strengths matter!
BDE's (in kcal/mol)
Primary C-H: 100.5
Tertiary C-H: 95.2
Cl-Cl: 58.0
H-Cl: 103.2
Primary C-Cl: 84.2
Tertiary C-Cl: 83.0
Br-Br 46.0
H-Br: 87.5
Primary C-Br: 70.3
Tertiary C-Br: 69.3
Selectivity
H3C H
H3C
Cl2/h
CH3
Br2/h
²H rxn (kcal/mol)
Primary: 64%
Tertiary: 36%
-28.9
-33.0
Primary: 2%
Tertiary: 98%
-11.3
-15.6
Radical Reactions:
Selectivity vs. Exothermicity
• The 103.2 kcal/mol H-Cl bond means that
H-abstraction from any simple alkyl R-H is
exothermic.
• H-Br bond strength is just 87.5 kcal/mol so
all H abstractions are endothermic. The
relative barriers differ by nearly the whole
energy difference between primary and
tertiary radicals.
How to obtain reaction barrier
i.e. ∆Hact values?
• Kinetics…for a later discussion
– Measure reaction rates as a function of T
– Extract rate constants for various T values
– Arrhenius or Eyring plots to obtain ∆Eact
and/or ∆Hact + ∆Sact
Reaction Mechanisms
•
•
•
•
•
•
•
•
•
How many particles (intra- vs. intermolecular)?
Activation energies
What parts end up where?
Symmetries of TSs/Intermediates
What bonding changes happen, and when?
Concerted or stepwise?
Ionic or radical?
Catalyzed or direct?
Energy inputs (∆, h, others?)?