Question

2. Using the arrow-pushing formalism, draw the mechanism for the transformation of your phosphine into the phosphonium salt you observed by HP NMR. 3. A brief written component answering the following questions. a. Based on your collected data, is 1-bromobutane or 2-bromobutane a more active electrophile? How do you explain this observation? b. Based on your collected data, is PPh or P(Bu) a more active nucleophile? How do you explain this observation?

Answer 1

2.2 Reaction between triphenyl phosphine with alkyl halide gives phosphonium salt. I ph w ph + RyX ph triphenyl phosphine phosphonium salt Above compound can be easily determined by 31p NMR specta. Chemical shift is obserred In ppm ors. Above product shows peak at &- 62 ppm.

2.3 a 1 - bromobutane is more active electrophile, because in I- bromobutane -Br is attached to 1 (primary) carbon while in a-bromo butane Br. is attached to secondary carbon atom 1 carbon atom is more electon deficient than 20 due to +I effect T-bromobutane Chess I effect) More reactive 2-bromo Duca (more t I effect) Jess reactive pph is less active elechophile than PBUZ because size of Ppha is bigger than p Buz There are three phenyl ring present in pph which makes it less nucleophilic I than P Buz.