Sodium Iodide → HCID 14977896, compound
Reaction equation
[CH3:1][O:2][C:3]1[CH:8]=[CH:7][C:6]([CH2:9][CH2:10][CH2:11][CH2:12][CH2:13][CH2:14]Br)=[CH:5][CH:4]=1.[I-:16].[Na+]>CC(=O)CC>[CH3:1][O:2][C:3]1[CH:8]=[CH:7][C:6]([CH2:9][CH2:10][CH2:11][CH2:12][CH2:13][CH2:14][I:16])=[CH:5][CH:4]=1Participants
4
Reactant 2
Product 1
Reagents, catalysts & solvents 1
Procedure
6-(4-Methoxyphenyl)hexyl iodide [VIa; Ar is 4-CH3OC6H4, R is H] was prepared from 16.3 g. of 6-(4-methoxyphenyl)hexyl bromide (Preparation C10) and 9.93 g. of sodium iodide in 325 ml. of 2-butanone, to give 19.3 g. of product as a pale yellow oil.
Sources & evidence
- ORD record
- ord-2ef42bbcdbfb4602849a60a6b9b9ceea
- Source dataset
- uspto-grants-1978_11
- Patent number
- US04125541
Reaction notes
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