cem 850, fall 2004

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CEM 850, Fall 2004 Some notes on Thermochemistry, Bond Strengths, and Strain energies Ned Jackson

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CEM 850, Fall 2004. Some notes on Thermochemistry, Bond Strengths, and Strain energies Ned Jackson. Alkane ∆H f values (kcal/mol). Alkane ∆H f values show Systematic Patterns. Can we estimate ∆H f by summing energy equivalents for transferable molecular “building blocks?” Bond Equivalents - PowerPoint PPT Presentation

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Page 1: CEM 850, Fall 2004

CEM 850, Fall 2004

Some notes on Thermochemistry, Bond Strengths, and Strain energies

Ned Jackson

Page 2: CEM 850, Fall 2004

Alkane ∆Hf values (kcal/mol)1

-17.9

2

-20.04

3

-25.02

4

-30.37-32.07

5

-35.08-36.73-40.14

6

-39.96...

-44.35

Carbon number nB

ranc

hing

Octanes:

7

-44.89...

-48.96

8

-49.82...

-53.99

Octane

2-methylheptane3-methylheptane4-methylheptane3-ethylhexane

2,5-dimethylhexane2,4-dimethylhexane2,3-dimethylhexane3,4-dimethylhexane3-ethyl-2-methylpentane

-49.82

-51.50-50.82-50.69-50.40

-53.21-52.44-51.13-50.91-50.48

2,2-dimethylhexane3,3-dimethylhexane3-ethyl-3-methylpentane

2,3,4-trimethylpentane

2,2,3-trimethylpentane2,2,4-trimethylpentane2,3,3-trimethylpentane

2,2,3,3-tetramethylbutane

-53.71-52.61-51.38

-51.97

-52.61-53.57-51.73

-53.99

Page 3: CEM 850, Fall 2004

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

Page 4: CEM 850, Fall 2004

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.

Page 5: CEM 850, Fall 2004

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 -10.08

– CH2 -4.95– CH -1.90– C +0.50

Page 6: CEM 850, Fall 2004

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.”

Page 7: CEM 850, Fall 2004

Strain Energies

• Cycloalkanes (CH2)n

2

12.54

2x(-4.95)= -9.9

22.44

3

12.74

3x(-4.95)= -14.85

27.59

4

6.78

4x(-4.95)= -19.8

26.58

5

-18.26

5x(-4.95)= -24.75

6.49

6

-29.43

6x(-4.95)= -29.7

0.27

Ring Size n

² Hf

² Hf HypotheticalStrainless (CH2)n

Strain E.

Page 8: CEM 850, Fall 2004

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!

Page 9: CEM 850, Fall 2004

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

Page 10: CEM 850, Fall 2004

R-H BDEs worth remembering

• H-H 104.2 kcal/mol• CH3-H 105.1

• CH3CH2-H 100.5

• (CH3)2CH-H 99.1

• (CH3)3C-H 95.2

• H2C=CHCH2-H 88.1

• PhCH2-H 89.6

Page 11: CEM 850, Fall 2004

An ordinary C-C 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

Page 12: CEM 850, Fall 2004

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.

Page 13: CEM 850, Fall 2004

Some Heats of FormationH F Cl Br I OH NH2 CH3

52.1 19.0 29.0 26.7 25.5 9.3 45.5 35.134.7 -59.5 -54.4 -50.9 -45.1 -32.8 27.1 33.3

H 52.1 0.0 -65.3 -22.1 -8.7 6.3 -57.8 -11.0 -17.9365.7 0.0 -65.3 -22.1 -8.7 6.3 -57.8 -11.0 -17.9

CH3 35.1 -17.9 -56.0 -20.0 -8.2 3.4 -48.0 -5.5 -20.0262.0 -17.9 -56.0 -20.0 -8.2 3.4 -48.0 -5.5 -20.0

CH3CH2 28.4 -20.0 -26.8 -15.2 -1.7 -56.2 -13.0 -25.0215.6 -20.0 -26.8 -15.2 -1.7 -56.2 -13.0 -25.0

(CH3)2CH 22.0 -25.0 -70.1 -34.7 -22.9 -9.5 -65.2 -20.0 -32.1191.5 -25.0 -70.1 -34.7 -22.9 -9.5 -65.2 -20.0 -32.1

(CH3)3C 11.0 -32.1 -43.0 -31.6 -17.2 -74.7 -28.8 -40.1165.5 -32.1 -43.0 -31.6 -17.2 -74.7 -28.8 -40.1

H2C=CHCH2 40.9 4.9 -1.3 11.4 23.8 -29.6 -0.2229.5 4.9 -1.3 11.4 23.8 -29.6 -0.2

PhCH2 49.5 12.0 -30.2 4.5 20.0 30.4 -22.6 21.0 7.1216.5 12.0 -30.2 4.5 20.0 30.4 -22.6 21.0 7.1

Page 14: CEM 850, Fall 2004

All energies in kcal/mol

Cleave 1,2 Cleave 2,3

CH3• + • • + •

-30.06 est.-30.37 exp.

35.1 + 23.3 est.35.1 + 23.9 exp.

2 * 28.2 est.2 * 28.4 exp.

BDEs88.5 est.89.4 exp.

BDEs86.5 est.87.2 exp.

Page 15: CEM 850, Fall 2004

Bond Diss’n Energies (BDEs)H F Cl Br I OH NH2 CH3

52.1 19.0 29.0 26.7 25.5 9.3 45.5 35.134.7 -59.5 -54.4 -50.9 -45.1 -32.8 27.1 33.3

H 52.1 104.2 136.4 103.2 87.5 71.3 119.2 108.6 105.1365.7 400.4 371.5 333.4 323.5 314.3 390.7 403.8 416.9

CH3 35.1 105.1 110.1 84.1 70.0 57.2 92.4 86.1 90.2262.0 314.6 258.5 227.6 219.3 213.5 277.2 294.6 315.3

CH3CH2 28.4 100.5 84.2 70.3 55.6 93.9 86.9 88.5215.6 270.3 188.0 179.9 172.2 239.0 255.7 273.9

(CH3)2CH 22.0 99.1 111.1 85.7 71.6 57.0 96.5 87.5 89.2192.0 251.2 202.1 171.8 163.5 155.9 223.9 238.6 256.9

(CH3)3C 11.0 95.2 83.0 69.3 53.7 95.0 85.3 86.2165.5 232.3 154.1 146.2 137.6 207.4 221.4 238.9

H2C=CHCH2 40.9 88.1 71.2 56.2 42.6 79.8 76.2229.5 259.3 176.4 167.2 160.6 226.3 263.0

PhCH2 49.5 89.6 61.2 74.0 56.2 44.6 81.4 74.0 77.5216.5 239.2 -17.3 157.6 145.6 141.0 206.3 222.6 242.7

Page 16: CEM 850, Fall 2004

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)?

Page 17: CEM 850, Fall 2004

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?

Page 18: CEM 850, Fall 2004

Hydrogenation Strategy

• H2C=CH2 + H2 —> H-H2C-CH2-H12.5 + 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!

Page 19: CEM 850, Fall 2004

Spectroscopic approaches?• π—>π* Excited state has no π bonding, but

max = 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!?

Page 20: CEM 850, Fall 2004

Energetics of Full Disassembly of Ethylene

• Try to make a prediction: – C-C 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; 184.6-12.5 = 172.1, almost 20 kcal/mol “too large”--what’s going on?

• C-H bond strengths increase from C2H4 and CH2

Page 21: CEM 850, Fall 2004

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

•H2C-CH2-CH2•

– What about hydrogenation energies?– Can Strain E’s help?

Page 22: CEM 850, Fall 2004

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?

Page 23: CEM 850, Fall 2004

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

Page 24: CEM 850, Fall 2004

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-Evans-

Polanyi principle--the more exothermic the process, the lower its barrier will be.

Page 25: CEM 850, Fall 2004

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)

Page 26: CEM 850, Fall 2004

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)

Page 27: CEM 850, Fall 2004

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

Page 28: CEM 850, Fall 2004

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

Page 29: CEM 850, Fall 2004

Challenge: CH4 --> CH3OH

H

H

HH

O

H

HH

H

105.1 (314.6) 92.4 (277.2)

98.1 (254.3)

104.2 (382)

²H f = -17.9 kcal/molIE = 12.61 eV (290.8 kcal/mol)PA = 129.9 kcal/mol

²H f = -48.0 kcal/molIE = 10.84 eV (250.0 kcal/mol)PA = 180.3 kcal/mol

OH

H

90.2 (259.4)

²H f = -27.7 kcal/molIE = 10.88 eV (250.9 kcal/mol)PA = 170.4 kcal/mol

O

H

HH

H

130.6 (66.6)110

117 (180)H

²H f = 137.4 kcal/mol

Reagents?RadicalsAcidsBases

Page 30: CEM 850, Fall 2004

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.

Page 31: CEM 850, Fall 2004

Radical selectivities

• Isobutane halogenation– Bond strengths matter!

H3C CH3

H3C H

BDE's (in kcal/mol)Primary C-H: 100.5Tertiary C-H: 95.2

Cl-Cl: 58.0H-Cl: 103.2Primary C-Cl: 84.2Tertiary C-Cl: 83.0

Br-Br 46.0H-Br: 87.5Primary C-Br: 70.3Tertiary C-Br: 69.3

Cl2/hPrimary: 64%Tertiary: 36%

Br2/h Primary: 2%Tertiary: 98%

Selectivity

-28.9-33.0

²H rxn (kcal/mol)

-11.3-15.6

Page 32: CEM 850, Fall 2004

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.

Page 33: CEM 850, Fall 2004

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

Page 34: CEM 850, Fall 2004

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?)?