Question

Organic chemistry. Draw mechanisms for the following.

Scheme III. Preparation of the Lactone Portione O OAC AC WHOL 756% Moo MOO DOWN АС Moon MOON Med OR OPh 23 RST 26 R:H 24 REH 27 R = SitBule, • (a) NaoMe, MeOH; (b) Hg(OAc),, MeOH; (c) NaCl; (d) NaB- Hq; (e) (Ph),CCI, pyr; (f) Nah, ArSO,C1; (g) LiAlH.; (h) Li, NH3 (liquid); (i) Phocsci, pyr; (j) CH2=CHCH,SnBus, hv; (k) t- BuSiMe,cl, imidazole, DMF. n

Answer 1

DAL DA •C-CH₃ a NGOME Com Jesus ad name Steps Jeg har sme & Meot (step-2) 애 ACOM (stets | HG (GAL) 2 OAL OH step-y. 1th Homel gtrek from below На РТ- W oll Thamton step Thans. Hon state for step-u OAL at the Tharoition state the SN' type intenmidiate fohmed at stabilised by the 4R'(Rensonances effect of oxygen so, Meol chill attack to this carbon which is adjacent to oxygen atom ACEJ Hg LOH cauta no into H me that & Hace (steps OH a-Hgh ot thaOAL Me on & Na BHq (step-6. OH of ou a secondary COH oovito wo (step-7) ocph e h v (she mallit o olt Meowik, Phimany c pha In step 7 the only phimary - chet with react because it is less hindened.

OH Meo -o-c(PH) Meo Meo (stek at Ansta Hd 1 Ar Cantinetterk om - Absos (steb-12) Meo quy Men ото I Li Altu (de protection of oth groug Step-13 Mega мео Litte [R=H] OR Meoko MOOR step 14 야애 o-H 2 py i MO 8TH (stek is ontotoo (Step-16 L (Phimany of group mone heactive than Secondaney Stet 16 co is betten leaving group -oph Med oph) than

J CH₂=CH-CH2-Sn Buz oph MOOK . CH2=CH-CH2 Snuz to - CH₂=CH-CH2 + sh Buz otoph (step-H) Lolita Moth ... meo s-sn Buz c- oph 애 CH₂=CH-CH2 MAK Meow 10 t-Bu-sime, a t imidazole, DMF stekk ме Si-me 11- (Imidazole) ме ме ST +-Bu 0-Situme Fol trou Mora Meonik.