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Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 8
Keywords: leaving groups, nucleophilicity, SN1/SN2/E1/E2
Description: Goes through substitution and elimination competition reactions:
Total Problems: 10
Rank the following anions in order of decreasing stability (1 = most stable)
Each of these ions has a charge of -1, and none of them has any resonance forms. So the two things to consider are electronegativity and size. Electronegativity is relevant as we go from left to right across the periodic table. But here we are going up to down on the periodic table, so size is relavant. Larger ions are more stable than smaller ions (due to a smaller charge:size ratio), so I- is the most stable of the group.
Rank the following electrophiles in order of decreasing reactivity with NaN3 in DMF. (1 = most reactive)
This problem is similar to problem 536. The only difference among these compound is their leaving group, so the compound with the best leaving group will undergo substitution reactions most rapidly.
Good leaving groups are stable. Larger ions tend to be more stable than smaller atoms (due to a smaller charge:size ratio. See problem 288), so when going down the periodic table, stability increases. I- is more stable than Br-, which is more stable than Cl-, etc. So I- is the best leaving group, and 2-iodobutane will react the fastest with a nucleophile.
The only differences among these three molecules is their leaving groups, so whichever has the best leaving group will react fastest with the nucleophile (azide, N3-). Stable molecules make good leaving groups, so the best leaving group of the three is SO3R-, which has has three resonance forms (see problem 535). So the third compound will react the fastest with NaN3.
