1. Isodesmic Reactions Isodesmic Reactions: chemical changes in which there is a net formal retention of the number of bonds (groups) but a change in their formal relation to each other. Consider the following reaction: CH 3 CH 2 CH 2 CH 3 + CH 3 CH 3  2 CH 3 CH 2 CH 3 -125.5±0.7 -83.7±0.4 -104.7±0.5  H r (298)= 2(-104.7 ±0.5) - (-125.5±0.7) - (83.7 ±0.4)  H r (298)= (-0.2 ±1.3) kJ mol -1

3. 3CH 3 CH 2 CH 2 CH 3  3 CH 3 CH 3 + cyclohexane -125.5±0.7 -83.7±0.4 -123.4 ±0.8  H r (298)= (-123.4 ±0.8) + 3(-83.7 ±0.4) -3 (-125.5 ±0.7 )  H r (298)= (2.0 ± 2.5) kJ mol -1

4. 2CH 3 CH 2 CH 2 CH 3  2 CH 3 CH 3 + cyclobutane -125.5 ±0.7 -83.7±0.4 28.4±0.6  H r (298) = (28.4±0.6) +2(-83.7±0.4)-2(-125.5±0.7 )  H r (298) = (112±1.7) kJ mol -1 strain energy Please read: Fuchs, R. “The Evaluation of Strain and Stabilization in Molecules Using Isodesmic Reactions,” J. Chem Educ. 1984, 61, 133-136.

5. List the following olefins in terms of their stability 1. Ethylene 2. Propene 3. Tetramethylethylene 4. Trimethylethylene 5. cis 2-butene 6. trans-butene

6. Heats of Hydrogenation of Alkenes / kJ mol -1 H 2 + CH 2 =CH 2  CH 3 CH 3 ;  H r (298) = -136.4 H 2 + CH 3 CH=CH 2  CH 3 CH 2 CH 3 ;  H r (298) = -125.1 H 2 + trans CH 3 CH=CHCH 3  CH 3 CH 2 CH 2 CH 3 ;  H r (298) = -114.6 H 2 + cis CH 3 CH=CHCH 3  CH 3 CH 2 CH 2 CH 3 ;  H r (298) = -118.4 H 2 + CH 3 C=CHCH 3  CH 3 CHCH 2 CH 3 ; | | CH 3 CH 3  H r (298) = -112.5 H 2 + CH 3 C=CCH 3  CH 3 CHCHCH 3 ; | | | | CH 3 CH 3 CH 3 CH 3  H r (298) = -111.3

7. Isodesmic Reactions / kJ mol -1 CH 2 =CH 2 + CH 3 CH 3  CH 3 CH 2 CH 2 CH 3 ; 52.3±0.4 -83.7±0.4 -125.5±0.7  H r (298) = -125.5±0.7-(-83.7±0.4 )-(52.3±0.4) = -94.1±0.9 CH 3 CH=CH 2 + 2CH 3 CH 3  CH 3 CH 2 CH 3 + CH 3 CHCH 3 20.1±0.8 -83.7±0.4 -104.6±0.5 | -134.3 ±0.7 CH 3  H r (298) = (-134.3±0.7)-(104.6±0.5)-2(-83.7±0.4)-(20.1±0.8)  H r (298) = (-91.6±1.2)

8. Isodesmic Reactions trans CH 3 CH=CHCH 3 +2CH 3 CH 3  2CH 3 CHCH 3 ; -11.4±1.0 -83.7±0.4 | CH 3 -134.3±0.7  H r (298) = 2(-134.3±0.7) - 2(-83.7±0.4)-(-11.4 ±1.0)  H r (298) = (-89.9±1.3) kJ mol -1 cis CH 3 CH=CHCH 3 +2CH 3 CH 3  2CH 3 CHCH 3 ; | CH 3  H r (298) = 2(-134.3±0.7) - 2(-83.7±0.4)- (-7.1±1.0)  H r (298) = (-94.1±1.3) kJ mol -1

9. CH 3 C=CHCH 3 +2CH 3 CH 3  CH 3 CHCH 3 + CH 3 | -83.7±0.4 | | CH 3 CH 3 CH 3 CCH 3 -41.8 ±1.1 -134.3±0.7 | CH 3 -168.2±0.8  H r (298) = (-168.2±0.8)-(134.3±0.7)-[-2(83.7±0.4)+(-41.8±1.1]  H r (298) = (-93.3±1.7) kJ mol -1

10. 2CH 3 CH 3 + CH 3 C = CCH 3  2 C(CH 3 ) 4 | | -83.7±0.4 H 3 C CH 3 -168.1±0.8 -68.2±1.1  H r (298) = 2(-168.1±0.8) - [(-68.2±1.1)+2(-83.7±0.4)]  H r (298) = (-100.6±3.9) kJ mol -1

11. Summary Hydrogenation  a Isodesmic  a ethylene -134.6 -94.1±0.9 propene -125.1 -9.5 -91.6±1.2 -2.5±1.5 cis 2-butene -114.6 -20.0 -94.1±1.3 0.0±1.6 trans 2-butene -118.4 -16.2 -89.9±1.3 -4.2±1.6 2-methylbutene -112.5 -22.1 -93.3±1.7 -0.8±1.9 2,3-dimethylbutene -111.3 -23.3 -100.8±3.9 +6.7±4.0 a relative to ethylene

12. CH 2 =CHCH 2 CH 2 CH 3 + 3CH 3 CH 3  CH 3 CH 2 CH 3 + CH 3 CHCH 3 -21.3±1.0 -83.7±0.4 -104.6±0.5 CH 3 + CH 3 CH 2 CH 2 CH 3 ) -134.2±0.7 -125.5±0.7  H r (298)=(-134.2±0.7)+(-104.6±0.6)+(-125.5±0.7)-[3(-83.7±0.4)] +(-21.3±1.0)]  H r (298)= (- 91.9±1.9) kJ mol -1 CH 3 CH=CHCH 2 CH 3 + 3 CH 3 CH 3  CH 3 CH 2 CH 3 +2 CH 3 CHCH 3 cis -27.6±1.0 CH 3 trans -31.9±1.1 cis  H r (298)=2(-134.2±0.7)+(-104.6±0.5)-[(-27.6±1.0)+3(-83.7±0.4)]  H r (298)= (-94.3±2.2) kJ mol -1 trans  H r (298)=2(-134.2±0.7)+(-104.6±0.5)-[(-31.9±1.1)+3(-83.7±0.4)]  H r (298)= (-90.0± 2.2) kJ mol -1

13. Aromaticity 82.6±0.7 -123.4±0.8 -5.0±0.7  H r (298)= 3 (-5.0±0.7) - 2(123.4±0.8)-(82.6±0.7)  H r (298)= (149.2±3.3) kJ mol -1

14. What’s the strain energy of cyclopropane? + CH 3 CH 3  CH 3 CH 2 CH 2 CH 2 CH 3 -83.7±0.4 -146.9±0.9 Cyclopropane  H f (298)= 53.3±0.6  H r (298)= (-146.9±0.9) - (-83.7±0.4) -(53.3±0.6)  H r (298)= -116.5±1.2  H s (298)= 116.5 kJ mol -1

15. What’s the strain energy of spiropentane?  H f (298)= 185.2±0.8 CH3 + 4CH 3 CH 3  2 CH 3 CH 2 CH 2 CH 3 + CH 3 CCH 3 CH 3 -83.7±0.4 -125.5±0.7 -168.1±0.8  H r (298)= (-168.1±0.8) + 2(-125.5±0.7) - 4(-83.7±0.4) -(185.2±0.8)  H r (298)= (-269.5±2.4) kJ mol -1  H s (298)= 269.5 - 2(116.5) = 36.5 kJ mol -1 cyclopropane  H s (298)= 116.5 ± 1.2