4-Bromobenzyl bromide → HCID 21716353, compound

HRID 1385908Notes

Reaction equation

HRID 1385908 reaction equation
[CH:1]1([O:6][C:7]2[CH:8]=[C:9]([CH:15]3[O:20][CH2:19][C:18](=[O:21])[NH:17][CH2:16]3)[CH:10]=[CH:11][C:12]=2[O:13][CH3:14])[CH2:5][CH2:4][CH2:3][CH2:2]1.[Br:22][C:23]1[CH:30]=[CH:29][C:26]([CH2:27]Br)=[CH:25][CH:24]=1>>[Br:22][C:23]1[CH:30]=[CH:29][C:26]([CH2:27][N:17]2[C:18](=[O:21])[CH2:19][O:20][CH:15]([C:9]3[CH:10]=[CH:11][C:12]([O:13][CH3:14])=[C:7]([O:6][CH:1]4[CH2:2][CH2:3][CH2:4][CH2:5]4)[CH:8]=3)[CH2:16]2)=[CH:25][CH:24]=1

Participants

3

Reactant 2

HCID 685274-Bromobenzyl bromideHCID 21716365HCID 21716365, compound

Product 1

HCID 21716353HCID 21716353, compoundYield 96.5%

Procedure

According to the same procedure as used in Example 12, using the 2-(3-cyclopentyloxy-4-methoxyphenyl)morpholin-5-one produced in Example 2 instead of 2-(3,4-dimethoxyphenyl)morpholin-5-one, and using 4-bromobenzyl bromide instead of methyl iodide, the above-described compound (yield 96.5%) was obtained as a brown oil.

Sources & evidence

ORD record
ord-8eabd5580c834c9f875287d363984f3b
Source dataset
uspto-grants-2001_07
Patent number
US06265402B1

Reaction notes

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