Part 3 mechanism please Part IexcessexcessMCPBACH;MgBrCNaOHNaOCIH2SO4→E (C3H12)АBH2OНОАСethermixture ofdiastereomersPart 2(CH3),SH20NaBH,FH;O*H;SO4A.H (C;H¿O2)-78 °CCH;OHPart 3E (CH12) +H (C;H4O2)I (C„H1602)

Answered: Image /qna-images/answer/7f462786-6202-489c-8ee1-f7b4a9b0e7c6.jpg

Part 1 excess excess MCPBA CH,MgBr C NaOH NaOCI B H3O* → D H,SO, → E (CH12) H,0 НОАС ether mixture of diastereomers Part 2 O3 (CH3),S NaBH, F CH;OH H;SO4 G H;O* H (C;H,O2) -78°C H,O Part 3 E (C,H12) + H (C;H,O2) I (C„H1602)

Part 1 excess excess MCPBA CH,MgBr C NaOH NaOCI B H3O* → D H,SO, → E (CH12) H,0 НОАС ether mixture of diastereomers Part 2 O3 (CH3),S NaBH, F CH;OH H;SO4 G H;O* H (C;H,O2) -78°C H,O Part 3 E (C,H12) + H (C;H,O2) I (C„H1602)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Part 3 mechanism please

Part I
excess
excess
MCPBA
CH;MgBr
C
NaOH
NaOCI
H2SO4
→E (C3H12)
А
B
H2O
НОАС
ether
mixture of
diastereomers
Part 2
(CH3),S
H20
NaBH,
F
H;O*
H;SO4
A.
H (C;H¿O2)
-78 °C
CH;OH
Part 3
E (CH12) +
H (C;H4O2)
I (C„H1602)

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