Question

Racemic total synthesis of arcutinine OTMS 17 TMSCN MeLi OMe ОМe OMe OMe Me2SO OMOM оМом OTMS OMOM 78% 85% 1'CN OMe Me 84% Me 18 20 15 19 CICO2Allyl 84% Me Me 1) LIAIH4 2) TsC OR PdCl2(MeCN)2 OMe Pd(PPha)4 OMe OMe OMe OMOM then MOMC омом then MOMCI TCN Me NH OMOM Me Ts t"CN 84%, two steps OO TCN Ме Me 21 75% 55% Me 13 22 OAllyl 23, R H TMSOTf 96% 26, R TMS ОMe Мe DAF ГОМе н Pd(OAc)2, DAF 76%, 2.5:1 d.r Ts 25 (68% from S1b) TMS н Me Me Me O OMe Me OMe OMe 1) TFA OMe ТВАF DMP OMe На N' Ts Мом Ts Ts МОм Ts Мом 2) Phl(OAc)2 Н 28a (a-H20, 54% from 26) 24а (а-Н20, 93%) S1a (а-Н20, 95%) S1b (B-H20, 95%) 10 (67% from S1a) 24b (B-H20, 94%) 28b (B-H20, 22% from 26) Sml2 85% Me но н Me но н Me но н Me но н Me он Me PrCO2H CH2(NMe2)2 NaBH(OMe)з Н Ни N Ts Н. Н 77% 70% 82% N Ts N н 34 Н 33 Н Ts H Ts 32 S2 Li/naphthalene 93% Me Me C Me но н Ме нон Me но н Me Он о нон Me Me Me С Me PhIO NaOMe Н. 85% 70% н H N H H H H 2, (-arcutinine (71%) 3, ()-arcutinidine 35 36 (14%)

Can you please answer these two questions?

1- count the total number of steps in the synthesis

2- calculate an overall yield for the longest linear sequence in the synthesis

Answer 1

The which one steps are taken into consideration leads a molecule to another in a synthetic manner . total steps 2 : of synthesis . so, are thus 18 there. In ease of Overall yield linear synthesis is calcuated like the folloiting- : A B C D . gield = 80% : 40% 30% so, overal yield = (80%)x (40%) x30%)xico X SO 80 To X99 v 0 100 TeX 100 - =0.8 x 0.4x30 Overall yield. = 9.6% Similary,

Lome TOMO Om OMOM MBS 04 34% 78% (A) 857, 76) BA7) (8 557 757 ) 34% 76% e Not-linear More som Total 'B' steps are here can be considred as the longest linear synltiesis. so; Overall yield = (84%)*(78°;)* (857) x (84%) *$54)*(75%)*(847.)x26-) 100 :. = 844.78.85 *184x155x:75 x.84 4.576 X100 . Overall yield = 12.319 ~ 12.32%