Chlorination of Other Alkanes  Can you write the complete mechanism for the chlorination of cyclopentane? initiation propagation step 1 propagation step 2 termination overall reaction

Chlorination of Propane - Selectivity  With propane, there is a mix of monochlorinated products…but the percentages show that 2-chloropropane is preferred. What would the percentages be if all H atoms were equally reactive?

Chlorination of Propane - Selectivity  There are more propagation steps for this reaction, because two different propyl free radicals can form: BDE = 410 kJ BDE = 397 kJ

Chlorination of Propane - Selectivity  E A for the 1° radical CH 3 CH 2 CH 2 ▪ is 4kJ higher than E A for the 2 ° radical. The 2 ° radical (CH 3 ) 2 CH ▪ forms more quickly.  BDE’s show that breaking the bond for the 1° radical requires more energy (410 kJ) than for the 2° radical (397 kJ).

Chlorination of Propane - Selectivity  ∆H° for 1-chloropropane is -118 kJ, and for 2-chloropropane is -127 kJ. 2-chloropropane is thermodynamically favored, too.

Stability of Free Radicals  Table 4-2 shows the stability of the different types of free radicals:

Bromination of Propane  With propane, there is also a mix of monobrominated products… Why must the system be heated? Why is 2-bromopropane so much more preferred than 2-chloropropane?

Bromination of Propane  Why must the system be heated?  Here, the formation of the free radical is endothermic instead of exothermic because forming the HBr bond does not give as great an energy payback (368 kJ) as does the formation of the HCl bond (431 kJ).  (The ∆H for the two radicals is the same as for the chlorination rxn.)

Bromination of Propane  Why is 2-bromopropane so much more preferred than 2-chloropropane?  The difference in E A between the 1° and 2° free radicals is larger (9 kJ) than seen in the chlorination reaction (4 kJ) making the formation of the 2° free radical much faster for bromination than for chlorination.

The Hammond Postulate  In an endothermic reaction step, the energy of the transition state is closer to that of the “products.”  This means the structure of the transition state is close to that of the products of that step.  For this reaction, it means that the transition state will be closer in structure to the 2° free radical.

The Hammond Postulate  Related species that are closer in energy are also closer in structure. The structure of a transition state resembles the structure of the closest stable species. In an endothermic process, the transition state will resemble products in energy and in structure. In an exothermic process, the transition state will resemble reactants in energy and in structure.  So exothermic processes are less selective.

Chlorination of Propane Why is 2- chloropropane less preferred than 2- bromopropane? Because the differences in E A that are present apply to something that is very akin to propane, so there is little to select for.

Reactive Intermediates  Reactive intermediates are short- lived species that are never present in high concentrations.  Most reaction mechanisms involve reactive intermediates. carbocations free radicals carbanions carbenes

Carbocations  aka carbonium or carbenium ions  C has a positive charge. sp 2 hybridized, so the bonds to C + are planar. powerful electrophile (Lewis acid)

Carbocations - Stability  The positive charge on the C is stabilized by any alkyl groups bonded to it through induction. The C-H bond of an alkyl group is a little bit polar, with the C being more electronegative. These electronegative C atoms donate electron density through the sigma bonds to the C atom with the + charge.

Carbocations - Stability  The positive charge on the C is also stabilized by any alkyl groups bonded to it through hyperconjugation. This is the sharing of electron density by the partial overlap of the empty 2p orbital with one of the three sp 3 orbitals on the alkyl C.

Carbocations - Stability  induction  hyperconjugation  (resonance for unsaturated carbocations) most stable

Free Radicals  No charge, but very reactive. The C atom is sp 2 hybridized, so the bonds to the C are planar.

Free Radicals - Stability  induction  hyperconjugation  (resonance for unsaturated carbocations) Exactly the same as for the carbocations…and for the same reasons.

Free Radicals  resonance stabilization Similar structures (+ instead of an unpaired electron) can be drawn for the resonance-stabilized carbocation. allylic radicals

Carbanions  Negatively charged, strong nucleophile The C atom is sp 3 hybridized, so the bonds to the C are tetrahedral…just like ammonia. Carbanions are stronger bases than amines due to their negative charge.

Carbenes  No charge, but very reactive. The C atom is sp 2 hybridized.