Question

2. Predict the product CHE CH,SNa 3. Provide a curved-arrow mechanism, and explain the observed stereochemistry 1) NaOH(aq) (excess) 2) H' workup 4. Explain the difference in reactivity م - م . KOBU no reaction! 5. Explain the difference in reactivity conc. Heso no reaction!

Answer 1

CH₂s Na This is SN2 type of reach. Chasna is very strong nucleophile. ¢ inversion of configuration takes place here. SO SO CHgs Na • Cha dus....c... Transitionstate CH Inversion of configuran 3 NaOH (ay) excess ) xt workup- yo SH co NaOH HO In this step basse abstracts one protons hom SH f go formed

UDSERVATION INNERENCI which And - acts Altack as stoung nucleophile. from same side of met cheaple leaving group. & Br ukup WO- KO Bu 80 Fву In this the Br P t are anliperiplafar so the removal of leaving group takes place easily as - Br is axially situated bout M BP KOBU Np reac”. In this case the removal for of us aquationally situated so leaving group is not possible.

tonë HasOy H Dove No recich. Bredt's rule states that double bond at bridgehead- position is unstable f such product is not formed. double bond at bridged head position is not stable.