cem 850, fall 2004

CEM 850, Fall 2004

Some notes on Thermochemistry, Bond Strengths, and Strain energies

Ned Jackson

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

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 -10.08

– CH2 -4.95– CH -1.90– C +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

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.

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 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

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

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 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

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.

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

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-H12.5 + 0 —> -20.0; ∆Hrxn = -32.5 kcal/mol

• Broken: C-C π bond, H-H@104.2 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

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!?

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

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?

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-Evans-

Polanyi 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

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

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!

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

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?)?