Bromine → Perfluorooctyl bromide

HRID 2209863

Reaction equation

[F:1][C:2]([F:28])([C:6]([F:27])([F:26])[C:7]([F:25])([F:24])[C:8]([F:23])([F:22])[C:9]([F:21])([F:20])[C:10]([F:19])([F:18])[C:11]([F:17])([F:16])[C:12]([F:15])([F:14])[F:13])C(O)=O.[Br:29]Br.FF>>[F:1][C:2]([Br:29])([F:28])[C:6]([F:27])([F:26])[C:7]([F:25])([F:24])[C:8]([F:23])([F:22])[C:9]([F:21])([F:20])[C:10]([F:19])([F:18])[C:11]([F:17])([F:16])[C:12]([F:15])([F:14])[F:13]

Participants

Reactant 3

HCID 24408Bromine

HCID 24524Fluorine2×

HCID 67821Perfluorononanoic acid

Product 1

HCID 9873Perfluorooctyl bromideYield 142.9%

Procedure

Perfluorononanoic acid 100 g (0.215 mole), bromine 21.0 g (0.129 mole) and Freon 113 (500 ml) were mixed in the plastic reactor. The mixture was treated with fluorine in the same manner as described in Example 1. Approximately 1.7 fold excess fluorine was used, producing 92 grams of perfluorooctyl bromide (86% yield). Thus, using approximately half the amount of bromine as used in Example 1 produced a lower yield of the perfluoroalkyl halide.

Sources & evidence

ORD record: ord-a8f75fceb2184feea80b206f32a1ed69
Source dataset: uspto-grants-1995_10
DOI: 10.6084/m9.figshare.5104873.v1
Patent number: US05455373

Linked notes

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