Lithium → HCID 87142787, compound
Reaction equation
[F:1][CH:2]([O:7][CH:8](F)[C:9]([F:12])(F)[F:10])[C:3]([F:6])([F:5])[F:4].[CH3:14][Li].[Li]>C(OCC)OCC>[C:9](=[C:8]([O:7][CH:2]([C:3]([F:6])([F:5])[F:4])[F:1])[CH3:14])([F:12])[F:10]Participants
5
Reactant 3
Product 1
Reagents, catalysts & solvents 1
Procedure
Another reaction was performed under similar conditions using a 1:2 molar ratio of bis-(1,2,2,2-tetrafluoroethyl)ether (32.5 grams, 0.15 mols) and CH3Li (105 mL of 3.0 M solution of CH3Li in CH2(OC2H5)2, 0.32 mols of the base). 43.1 grams of a crude product liquid was collected. GC analysis demonstrated that the liquid contained 1.5% bis-olefin, 0.4% mono-olefin, 0.1% unreacted starting ether, 4.3% unknown by-product, 7.3% CF2═C(CH3)OCHFCF3 (major by-product discussed above) and 84.5% CH2(OC2H5)2. These reaction conditions demonstrate that doubling the amount of the lithium reagent produces more of the by-product CF2═C(CH3)OCHFCF3 at the expense of the desired mono-olefin.
Workup
- custom43.1 grams of a crude product liquid was collected
- customThese reaction conditions
Sources & evidence
- ORD record
- ord-7e98da2aad5f4b10b6e90bc0bdad117d
- Source dataset
- uspto-grants-2016_05
- Patent number
- US09353038B2
Reaction notes
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