Question

12. Identify which ammonium ion has the lower pka. Use aromaticity concepts to explain your answers.(3pts) 13. What is the Hückel number of pi electrons for the molecule shown below? How many pi electron pairs does it have? Would you expect this molecule to be aromatic, antiaromatic or non aromatic? Explain. (3pts) 14. The rate of solvolysis of iodide 1 is 100 times faster than the rate of solvolysis of iodide 3. What explanation (with structures) can you provide for the difference in reaction rates. Use aromaticity concepts!!(5 pts) 15. Rank the reactivity of the compounds shown below when using a Friedel Crafts alkylation reaction. Make sure you label which is least reactive and most reactive. (Spts) Como

Answer 1

12)

Compound 2 has a lower pKa value than compound 1. As the compound 1 after losing one proton will become a cyclic five member ring with all sp2 hybridized atoms but it will have eight pi electrons (4 from two double bonds and four from one lone pair each of the two N atoms) which is a 4n pi electronic system so according to Huckel's rule it will be anti aromatic so unstable. But compound 2 after losing a proton becomes a non aromatic system because though it is cyclic system it has one sp3 hybrid carbon atom which makes it non planner so non aromatic. As non aromatic compounds are comparatively more stable than anti aromatic compounds so for compound 2 the release of proton will be easier than compound 1 so its acidity will be more hence its pKa value will be less.