Please written by hand Br2 (1.0 equiv.),HOAC NaOAcH2O2Br (aq., 30 wt-%,5.5 equiv.)HOHONaOH(1.5 equiv.), 20°CÓMeÓMe(aq., 2 wt-%)2223Но.BrBrCH2 (1.5 еquiv.), К2СО,(1.5 equiv.), DMF, 100°C, 1 hНоÓMeÓMe24 (91% from 22)25 (74%)

Answered: Image /qna-images/answer/2ad09515-b78e-448a-947c-da7cf09688d4.jpg

Br2 (1.0 equiv.), HOAC NaOAc H2O2 Br (aq., 30 wt-%, 5.5 equiv.) HO HO NaOH (1.5 equiv.), 20°C OMe ÓMe (aq., 2 wt-%) 22 23 НО. Br Br CH212 (1.5 equiv.), K2CO3 (1.5 equiv.), DMF, 100°C, 1 h HO ÓMe ÓMe 24 (91% from 22) 25 (74%)

Br2 (1.0 equiv.), HOAC NaOAc H2O2 Br (aq., 30 wt-%, 5.5 equiv.) HO HO NaOH (1.5 equiv.), 20°C OMe ÓMe (aq., 2 wt-%) 22 23 НО. Br Br CH212 (1.5 equiv.), K2CO3 (1.5 equiv.), DMF, 100°C, 1 h HO ÓMe ÓMe 24 (91% from 22) 25 (74%)

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.10QAP

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Question

Please written by hand

Br2 (1.0 equiv.),
HOAC NaOAc
H2O2
Br (aq., 30 wt-%,
5.5 equiv.)
HO
HO
NaOH
(1.5 equiv.), 20°C
ÓMe
ÓMe
(aq., 2 wt-%)
22
23
Но.
Br
Br
CH2 (1.5 еquiv.), К2СО,
(1.5 equiv.), DMF, 100°C, 1 h
Но
ÓMe
ÓMe
24 (91% from 22)
25 (74%)

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