Question

Problem 7.6 Identify the nucleophile and leaving group and draw the products of each substitution reaction. + OCH,CH " . NAOH + NaOH - FL + N O . NACN + NaCN - Problem 77

Answer 1

Arue :: page:01 Given information, Identify the nucleophile and leaving group and draw the products of each substatection rachon. bet, . Nucleophiles are Electron euch species, • They contain po (tt) bond or alone paion, not necessarily with a negative change then, Leaving groups are Hie arboric or neutral species, which accepts the elechons from heterolytic bond cleavage and departs as more stable species, + OCH2CH3 B31 het, The freātiment of 3-boromiopentane with Ethoxide ion gives 3- Ethoxypentane.

Page:02. the saction fis, + OCH CH3 -> +6013 5:38: OCHYCH3 Here electropheleo the Ethoxide ion (nucleophole) atlacks the canton having a leaving group (Bor group). Therefore, est le nucleophile ps Foche CH3| the leaving group is [Bor] + NaOH ---> het, the treatmünt of chlorocyclohexane with sodium hydroxid gives cyclohexanol in the product.

Page:03 The guackon is L t NatioH --- + Nat:ci: Here, the hyckroxide ion (nucleophile) attacks thee glectrophile Carbon having a leaving group c-clgroup), and forms the required product. Therefore, : The nucleophile is TOH the leaving group is [cr] - i + 13 → het, The treatment of t- podobutane with azide ion giver -azidobutane in the product. The reachon is,

pageiou Herre; thie azede poncnucleophile) attacks this electrophilc carbon daving, a leaving group (-group) and forms the suquioud product, Therefore, The neckophile ps Na the leaving group is (2) + NOCN- het, cyanide gives the treatment of bromocyclohexane with secticum Cyclohexane carbonibile in-tie product. The ouaction is, CEN: + Na :CN + Na* 1897 Here, the cyanide. fon (nucle.ophile) attache the electrophilec carbon braving a leaving group (- Ba group) and -forms the required prodch. • Thenefore, The nucleophile is F-EN the leaving group is [Bro] (lansuer.