Question

1. For the following substrates, predict whether a carbocation rearrangement will take place in an SN1/E1 mechanism by first drawing the carbocation intermediate, then deciding if the carbocation will rearrange (can it form a more stable carbocation by hydride or methyl shift?): Carbocation: Sturcture Rearrange? If yes, draw the rearranged product. If no, write No н.с Br Н,с, Cн, 2. For each pair of nucle ophiles, circle which will react faster with CHsl in an Sx2 reaction. Briefly explain why. a) CH30 or CH3CO2 b) NH3 or PH3 3. Draw all possible E2 products using CH3ONA as the base, and circle the major product: CH, оноNа OTs
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Answer 1

& Carbocatian ntabilite is increased with greater number of hypercang ugable H. ė H3 CH3 Brite Carboealian No rearra Parmatian [ Now this carboealian wan't occur rearrangement I Reasan I ) Initially the carbocalian has 4 hypercangugable H-celaim to dabiline it, at the two adjaccent eralam e-lande-3. CHH But if the rearrangement occurs to the e-3 by hydrite shift no aftietianal dabilisation is gained, rather same 4 hypercan- -gugable H-atain of e.2 and e-u Ha fint . A t XH

11) Naw sceppone the rearrangeenent occurs to the e-l, Then there are hypercoingugable H atam at e-6 and 2 hupeneangigable H-alam at e-2." 13H (4 (W CH3 LOG S4 ingable floalan decrease Henee, the no ot dabilising hypercangugal and indeed it us a less stable Carbocation. It H H H3 CHE (CHS) P ACH 6 H Rearrange- Carbocalan Parmalan ricent by Me shift o Here the cartocalian ean bertabilised by the rearrangement. • Because cinitially there were anly a hypercargigable H-citam of e-3 o But after rearrangement nia methyl shifo there are o hyper canguegable H-atam atabilising the earbocation. CHICHI F CH3 e rearrange -ment G-hypercangagable H-dan. o Mechanizam (SN) CH B r CH Nu (Muckaphile) det exis NU o This will fallaw SMP mechanisan. o Reasan • The Nu ean easily approach the carbocatian as there are nonci- -gbawring bulky graup, unlike the other molecule. o Honec will allow the SN mechanisam, which have the 1st Vep as the Ropes (rate determining veo ... CH3 0 Mechanisam CEARD E 1). CAS CAB CH3 CH3 CHI Minar ChinCH3 Al rearrange Rodos mcent

Reason • Here unlike before the approach at the nucleakale is rather fiffcieellt due to the eangented enuiseanmeent. • The post will also be highly Cangerted if the SNI pathway in fallawed, ound due to istorie ean gestian it will be cerintable • The alkene however, its not causing adfcitianal congestian. by Lindseducing aftitianal guzaup. .. Between the two alkene, the one which is more s bebocê tuled, has greater no ar hypercangugable t is major product i s ..2...18 -> CHTO do CH3O- we -I → eff- --4 - 1 TS2 5- ecarteinated T-S 7.51 AEL a reactian coordinate For eflson reaction eoantinata Far e H3 cool1 00 1 -CHC00- TeI H -E-0.... -H Togn sN2 reactian at the ts, the no af graups around the central e-alan increases, as the e-atambecaines 5- Caarteinated • Thus the pos becames mare storically congested if the graund ware ar nuelcophile vill has tulky gracipis and thus Tos becames marc cunvable cand Aç increases. O CHO being smaller than effy cool causes lesser Cange- astian ints, and A&, < Alz. o Thus CH3 Or will react faster In Another reason far toores to react faster is because it has localised -ve charge over oxygenotaim, where els amplifics it by ti cefleet.. CH3 CHCEO) CH-01-)

Brut the eflseoot due to the canajugatioun, has de localesed -ve change and this is bad nucleophile; campare to choora o This is also areasan why cHz or will act faster an 6) NHz ₂ valanec electronie Canfiguration at Nin 2P [ îîl (ox | 1x boy) PH₂ → Cantiguratian at p Valance eleckranie 3P Vacant orbital where the lane painean ocelut Kesada debralisa o The lane pairat Nisi localised but the p has the lane paire dela tian. realysing through the vacenta arbital and thus it is not much available andalheis is not as good as NHS. Hence NH3 will react faster Oil CHION 0 H30CH3 OTS Mayar Minar P-2 pol Mechanisam CH3 wilt eformat 015 Sets OCH3 OTS CH381-1) o OTS is a good nucleophile • The structural criteria for the Ezis the antiperiplanar arrangement and thus the Hound oTS un anti with respect to each other.

o The pris majar because it is fnare stable due to the Cangugalian with o, but that is absent in the P-2. o The Tos being product like the pal has mare stable Trs and thus par ciltbe major product.