Hydrodefluorination method of fluoro aromatic hydrocarbon
Through the combination of palladium catalyst and alcohol solvent, the efficient, safe and economical hydrodefluorination reaction of fluorinated aromatics is achieved, and the problem of fluorinated aromatics in the prior art is solved. It is suitable for fluorinated aromatics of various functional groups and is suitable for industrial applications.
Patent Information
- Application Number
- CN202510450920.1
- Authority / Receiving Office
- CN · China
- Patent Type
- Applications(China)
- Current Assignee / Owner
- Filing Date
- 2025-04-11
- Publication Date
- 2025-07-11
AI Technical Summary
The prior art is difficult to efficiently, safely and economically realize the hydrodefluorination reaction of fluoroaromatic hydrocarbons, especially when compatible with functional groups containing active hydrogen such as hydroxyl groups and amino groups, and commonly used reducing agents are expensive or have safety risks.
The defluorogenic hydrocarbons are defluorogenic by hydrodehalogen reaction using a combination of palladium catalyst, alcohol solvent and alkali. The alcohols are used as both solvents and reducing agents, and the reaction is carried out at room temperature and pressure, which is suitable for fluorogenic hydrocarbons of various functional groups.
It achieves high yield (up to 99%) fluoroaromatic defluorination, is compatible with a variety of functional groups, has mild reaction conditions, is suitable for industrial applications, avoiding the use of dangerous reducing agents and complex anaerobic operations.
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Figure CN120289263A_ABST
Abstract
Description
Technical Field
[0001] The present invention relates to the technical field of organic synthesis, and particularly to a method for hydrodefluorination of fluoroarenes. Background Art
[0002] Debromination reactions have attracted attention due to their extensive applications in organic synthesis and environmental treatment. At present, the existing technologies mainly focus on the hydrodebromination reactions of carbon-iodine bonds, carbon-bromine bonds and carbon-chlorine bonds. Commonly used reducing agents include hydrogen, aluminum amalgam, borane, sodium borohydride, triethylhydrogensilane, tert-butylmagnesium chloride, triethylamine, and tributyltin hydride, etc. These reagents are expensive and pose safety hazards, thus limiting their large-scale industrial applications.
[0003] In recent years, the debromination reaction methods using green and safe alcohols (such as n-butanol, isopropanol) as reducing agents and hydrogen sources have gradually attracted attention. However, such methods usually rely on organophosphine or carbene ligands and need to be operated under strict anaerobic conditions, and the process is relatively cumbersome.
[0004] In addition, due to the relatively large bond energy of the carbon(sp 2 )–fluorine bond, the hydrodefluorination reaction of fluoroarenes is more challenging. Currently, the literature reports mainly focus on the defluorination degradation of fluoroalkanes, and their products are mostly low-value substances such as fluoride ions, carbon dioxide and water, etc.; for fluoroarenes, the existing methods often have difficulty in achieving complete defluorination or easily generating complex products, and it is difficult to be compatible with functional groups containing active hydrogen such as hydroxyl groups and amino groups.
[0005] Therefore, there is an urgent need to develop a new method for the hydrodefluorination reaction of fluoroarenes that has advantages in terms of substrate generality, reaction selectivity and environmental friendliness. Summary of the Invention
[0006] Based on this, it is necessary to provide a method for hydrodefluorination of fluoroarenes to solve the problems existing in the prior art.
[0007] To achieve the above object, the present invention provides a technical solution:
[0008] A method for hydrodefluorination of fluoroarenes, comprising the steps of:
[0009] Mixing a fluoroarene, a palladium catalyst, a base and an alcohol, and carrying out a hydrodehalogenation reaction to prepare an aromatic compound;
[0010] Wherein, the structural formula of the fluoroarene is:
[0011] In the formula, Ar is an aromatic ring, and the aromatic ring includes at least one of a benzene ring, a heterocyclic ring and a fused ring;
[0012] R includes the following functional groups and any combination thereof:
[0013]
[0014] The alcohol includes ethanol and / or isobutanol.
[0015] Specifically, the reaction formula of the hydrodehalogenation reaction is as follows:
[0016]
[0017] Preferably, the structural formula of the aromatic ring includes one of the following functional groups:
[0018]
[0019] In some embodiments, the reaction temperature of the hydrodehalogenation reaction is 60°C to 140°C.
[0020] In some embodiments, the palladium catalyst includes:
[0021] a palladium compound catalyst; and / or
[0022] a catalyst formed by combining palladium metal with a carrier; and / or
[0023] a catalyst formed by combining a palladium compound with a carrier.
[0024] Wherein, the palladium compound includes at least one of palladium acetate, palladium chloride, and ammonium chloropalladite, and the carrier includes at least one of silica, carbon, calcium carbonate, calcium phosphate, and alumina, preferably alumina.
[0025] In some embodiments, the base includes at least one of sodium hydroxide, sodium ethoxide, sodium tert-butoxide, potassium tert-butoxide, and sodium metaaluminate, preferably sodium tert-butoxide.
[0026] In some embodiments, the ethanol includes ultra-dry ethanol, analytical grade anhydrous ethanol, or 95 wt.% ethanol.
[0027] In some embodiments, the molar ratio of palladium to fluoroarene in the palladium catalyst is (0.5 - 10):100, preferably 7:100. Herein, the meaning of "palladium" refers to the total molar amount of Pd in the overall catalyst. In some embodiments, the molar ratio of base to fluoroarene is (1 - 5):1, preferably 4:1.
[0028] Advantages of the present invention:
[0029] In this reaction, ethanol and / or isobutanol serve as both a solvent and a reducing agent simultaneously, eliminating the need for additional addition of a reducing agent and avoiding the use of dangerous reducing agents such as hydrogen and lithium aluminum hydride.
[0030] This catalyst system has excellent substrate applicability and can be well compatible with common electron-donating groups, electron-withdrawing groups, and heterocycles, with a maximum yield of up to 99%.
[0031] In addition, this reaction system can be carried out in analytical pure ethanol and air without complex anaerobic operations, which is convenient for large-scale industrial applications. Description of the Drawings
[0032] Figure 1 It is the reaction formula for the hydrodehalogenation reaction. Detailed Embodiments
[0033] In order to better illustrate the purpose, technical solutions, and advantages of the present invention, the present invention will be further described below in conjunction with specific embodiments.
[0034] In the embodiments, unless otherwise specified, the test methods used are all conventional methods, and the materials, reagents, etc. used, unless otherwise specified, can all be obtained from commercial channels.
[0035] Example 1: Catalyst Optimization
[0036] Add a suitable stir bar into a 10 mL pressure-resistant sealed tube, palladium catalyst (3.3 mol% in terms of palladium relative to the molar percentage of 2-fluorobiphenyl), 2-dicyclohexylphosphino-2',4',6'-triisopropylbiphenyl (X-Phos, 2.0 mol% relative to the molar percentage of 2-fluorobiphenyl), 2-fluorobiphenyl (0.2 mmol), sodium hydroxide (0.4 mmol). After evacuating and replacing with argon three times, add dodecane (0.1 mmol, as a gas-phase internal standard) and 1 mL of ultra-dry ethanol, and react at 120 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction, and take the upper organic phase to calculate the gas-phase yield. The results are shown in Table 1.
[0037] Among them, the reaction formula is as follows:
[0038]
[0039] Table 1 Optimization of the Catalyst
[0040]
[0041] Example 2: Optimization of the Base
[0042] Add a magnetic stir bar into a 10 mL pressure-resistant sealed tube, catalyst (palladium / aluminum oxide, 3.3 mol% in terms of palladium based on the molar percentage relative to 2-fluorobiphenyl), X-Phos (2.0 mol% in terms of the molar percentage relative to 2-fluorobiphenyl), 2-fluorobiphenyl (0.2 mmol), base (0.4 - 0.8 mmol). After evacuating and replacing with argon three times, add dodecane (0.1 mmol, as a gas-phase internal standard) and 1 mL of ultra-dry ethanol, and react at 120 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction. Take the upper organic phase to calculate the gas-phase yield, and the results are shown in Table 2.
[0043] The reaction equation is as follows:
[0044]
[0045] Table 2 Optimization of the base
[0046]
[0047] As can be seen from Table 1, after a series of screenings of the base in the reaction system, we selected sodium tert-butoxide (t-BuONa) as the optimal base, and the target product can be obtained with a gas-phase yield of 58%. Further increasing the amount of t-BuONa, the reaction yield can be increased to 68%.
[0048] Example 3: Optimization of water
[0049] Add a magnetic stir bar of appropriate size into a 10 mL pressure-resistant sealed tube, palladium / aluminum oxide (3.3 mol% in terms of palladium based on the molar percentage relative to 2-fluorobiphenyl), X-Phos (2.0 mol% in terms of the molar percentage relative to 2-fluorobiphenyl), 2-fluorobiphenyl (0.2 mmol), sodium tert-butoxide (0.8 mmol). After evacuating and replacing with argon three times, add water (0 - 4.0 mmol), dodecane (0.1 mmol, as a gas-phase internal standard), and 1 mL of ultra-dry ethanol, and react at 120 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction. Take the upper organic phase to calculate the gas-phase yield.
[0050] The reaction equation is as follows:
[0051]
[0052] Table 3 Optimization of water
[0053]
[0054] As can be seen from Table 3, if the amount of water is too high or too low, the yield will decrease. When 6 equivalents of water are added, the highest reaction yield can reach 79%.
[0055] Example 4: Optimization of Alcohols
[0056] Add a magnetic stir bar of appropriate size into a 10 mL pressure-resistant sealed tube, palladium / aluminum oxide (3.3 mol% in terms of palladium relative to the molar percentage of 2-fluorobiphenyl), X-Phos (2.0 mol% relative to the molar percentage of 2-fluorobiphenyl), 2-fluorobiphenyl (0.2 mmol), sodium tert-butoxide (0.8 mmol). After evacuating and replacing with argon three times, add water (1.2 mmol), n-dodecane (0.1 mmol, as a gas-phase internal standard), 1 mL of alcohol, and react at 120 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction. Take the upper organic phase to calculate the gas-phase yield, and the results are shown in Table 4.
[0057] The reaction equation is as follows:
[0058]
[0059] Table 4 Optimization of Alcohols
[0060]
[0061] As can be seen from Table 4, a relatively high yield is obtained only in ethanol for this reaction. When using methanol, isopropanol, 2-butanol or tert-butanol, only trace amounts of products (<5%) are formed because ethanol has strong reducibility.
[0062] Example 5: Optimization of Temperature
[0063] Add a magnetic stir bar of appropriate size into a 10 mL pressure-resistant sealed tube, palladium / aluminum oxide (3.3 mol% in terms of palladium relative to the molar percentage of 2-fluorobiphenyl), X-Phos (2.0 mol% relative to the molar percentage of 2-fluorobiphenyl), 2-fluorobiphenyl (0.2 mmol), sodium tert-butoxide (0.8 mmol). After evacuating and replacing with argon three times, add water (1.2 mmol), n-dodecane (0.1 mmol, as a gas-phase internal standard), 1 mL of ultra-dry ethanol, and react at 100 - 140 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction. Take the upper organic phase to calculate the gas-phase yield, and the results are shown in Table 5.
[0064] The reaction equation is as follows:
[0065]
[0066] Table 5 Optimization of Temperature
[0067]
[0068] As can be seen from Table 5, lowering or raising the temperature results in a decrease in the reaction yield, and 120 °C is the optimal reaction temperature.
[0069] Example 6: Optimization of the Ligand
[0070] Add a suitable-sized magnetic stir bar to a 10 mL pressure-resistant sealed tube, palladium / aluminum oxide (3.3 mol% in terms of palladium based on the molar percentage of 2-fluorobiphenyl), X-Phos (2.0 mol% in terms of the molar percentage of 2-fluorobiphenyl or not added), 2-fluorobiphenyl (0.2 mmol), sodium tert-butoxide (0.8 mmol). After evacuating and replacing with argon three times, add water (1.2 mmol), n-dodecane (0.1 mmol, as a gas-phase internal standard), 1 mL of ultra-dry ethanol, and react at 120 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction, and take the upper organic phase to calculate the gas-phase yield.
[0071] The reaction equation is as follows:
[0072]
[0073] Table 6 Ligand Optimization
[0074]
[0075] As can be seen from Table 6, the reaction yield decreases after adding the ligand. Under the condition of no ligand, the target product can be obtained with a yield of 98%.
[0076] Example 7: Optimization of the Catalyst Dosage
[0077] Add a suitable-sized magnetic stir bar to a 10 mL pressure-resistant sealed tube, palladium / aluminum oxide (3.3 - 6.6 mol% in terms of palladium based on the molar percentage of 2-fluorobiphenyl), 2-fluorobiphenyl (0.2 mmol), sodium tert-butoxide (0.8 mmol). After evacuating and replacing with argon three times, add water (1.2 mmol), n-dodecane (0.1 mmol, as a gas-phase internal standard), 1 mL of ultra-dry ethanol, and react at 120 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction, and take the upper organic phase to calculate the gas-phase yield.
[0078] The reaction equation is as follows:
[0079]
[0080] Table 7 Catalyst Dosage Optimization
[0081]
[0082] As can be seen from Table 7, moderately increasing the catalyst dosage can completely convert the reaction raw materials, and the yield is further increased to 99%. However, trace amounts of over-hydrogenated products (cyclohexylbenzene) are formed. Therefore, the preferred catalyst dosage is 6.6 mol%.
[0083] Example 8: Effects of water and oxygen on the reaction yield
[0084] Add a suitable stir bar into a 10 mL pressure-resistant sealed tube, palladium / aluminum oxide (6.6 mol% based on the molar percentage of 2-fluorobiphenyl, calculated as palladium), 2-fluorobiphenyl (0.2 mmol), sodium tert-butoxide (0.8 mmol). After evacuating and replacing with argon three times (or under air conditions), add water (1.2 mmol or not), dodecane (0.1 mmol, as a gas-phase internal standard), and 1 mL of ethanol. React at 120 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction. Take the upper organic phase to calculate the gas-phase yield. The results are shown in Table 8.
[0085] The reaction equation is as follows:
[0086]
[0087] Table 8 Effects of water / oxygen on the reaction yield
[0088]
[0089] As can be seen from Table 8, when analytical grade ethanol (EtOH, Extra Dry) is used to replace ultra-dry ethanol (EtOH, AR), the reaction yield only slightly decreases (98% vs 99%, No. 2 vs No. 1).
[0090] When analytical grade ethanol is used as the solvent, no addition of water has no significant effect on the reaction yield (98% vs 98%, No. 3 vs No. 2), because the trace amount of water contained in analytical grade ethanol itself can promote the reaction.
[0091] In addition, this reaction can proceed smoothly in air to obtain the target product (98% vs 98%, No. 4 vs No. 3).
[0092] Since analytical grade solvents are used and the reaction is carried out under air conditions without the need for complex and strict anaerobic operations, therefore, it is preferred to use analytical grade ethanol as the solvent and air as the reaction gas atmosphere. For some sensitive substrates, anaerobic operation is selected.
[0093] Example 9: Isobutanol as the reaction solvent and reducing agent
[0094] Add a magnetic stir bar of appropriate size to a 10 mL reaction tube, along with palladium / aluminum oxide (6.6 mol% based on palladium, relative to the molar percentage of 2-fluorobiphenyl), 2-fluorobiphenyl (0.2 mmol), sodium tert-butoxide (0.8 mmol). Then add dodecane (0.1 mmol, as a gas-phase internal standard) and 1 mL of analytical grade isobutanol. React for 12 hours at 120 °C in air. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction. Take the upper organic phase to calculate the gas-phase yield, which is 99%.
[0095] The reaction equation is as follows:
[0096]
[0097] Table 9 Effect of isobutanol as a solvent and reducing agent on the yield
[0098]
[0099] As can be seen from Table 9, isobutanol can also be used as a solvent and reducing agent in the reaction to achieve the hydrodefluorination reaction of fluoroarenes and is insensitive to water and oxygen. Using a common test tube as the reactor and appropriately reducing the temperature, the target product can also be obtained with a yield of 99% under the reflux condition of isobutanol. Since the price of isobutanol (1071 RMB / 25 L) is more expensive than that of ethanol (510 RMB / 25 L), researchers can comprehensively consider the reactor cost (pressure-resistant reactor vs. atmospheric-pressure reactor) and solvent cost (isobutanol vs. ethanol) to select appropriate reaction conditions.
[0100] Example 10: Defluorination of p-methoxyfluorobenzene
[0101] Add a magnetic stir bar of appropriate size to a 10 mL pressure-resistant sealed tube, along with palladium / aluminum oxide (6.6 mol% based on palladium, relative to the molar percentage of 4-methoxyfluorobenzene), p-methoxyfluorobenzene (0.2 mmol), sodium tert-butoxide (0.8 mmol). Then add dodecane (0.1 mmol, as a gas-phase internal standard) and 1 mL of analytical grade ethanol. React for 12 hours at 120 °C. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction. Take the upper organic phase to calculate the gas-phase yield, which is 99%.
[0102] The reaction equation is as follows:
[0103]
[0104] MS(EI) m / z calcd for C7H8O [M] + : 108.1, found: 108.1.
[0105] Example 11: Defluorination of 8-fluoroquinoline
[0106] Add a stir bar of appropriate size to a 10 mL pressure-resistant sealed tube, palladium / aluminum oxide (1.65 mol% in terms of palladium based on the molar percentage relative to 8-fluoroquinoline), 8-fluoroquinoline (0.2 mmol), sodium tert-butoxide (0.8 mmol), add dodecane (0.1 mmol, as a gas-phase internal standard), 1 mL of analytical grade ethanol, 120 μL of water, and react at 140 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction. Take the upper organic phase to calculate the gas-phase yield, which is 93%.
[0107] The reaction equation is as follows:
[0108]
[0109] MS(EI) m / z calcd for C9H7N [M] + : 129.1, found: 129.1.
[0110] Example 12: Defluorination of p-fluorophenol
[0111] Add a stir bar of appropriate size to a 10 mL pressure-resistant sealed tube, palladium / calcium phosphate (3.3 mol% in terms of palladium based on the molar percentage relative to p-fluorophenol), p-fluorophenol (0.2 mmol), sodium hydroxide (0.6 mmol), add dodecane (0.1 mmol, as a gas-phase internal standard), 1 mL of analytical grade ethanol, and react at 120 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction. Take the upper organic phase to calculate the gas-phase yield, which is 99%.
[0112] The reaction equation is as follows:
[0113]
[0114] MS(EI) m / z calcd for C6H6O [M] + : 94.0, found: 94.1.
[0115] Example 13: Defluorination of 4-fluorobenzotrifluoride
[0116] Add a stir bar of appropriate size into a 10 mL pressure-resistant sealed tube, palladium / aluminum oxide (6.6 mol% in terms of palladium relative to the molar percentage of 4-fluorobenzotrifluoride), 4-fluorobenzotrifluoride (0.2 mmol), sodium tert-butoxide (0.8 mmol), add dodecane (0.1 mmol, as a gas-phase internal standard), 1 mL of analytical-grade ethanol, and react at 100 °C for 12 hours. After the reaction tube is cooled to room temperature, add 2 mL of ethyl acetate and 2 mL of saturated ammonium chloride to quench the reaction, and take the upper organic phase to calculate the gas-phase yield, which is 93%.
[0117] The reaction equation is as follows:
[0118]
[0119] MS(EI) m / z calcd for C7H5F3 [M] + : 146.0, found: 146.1.
[0120] It should be noted that the eq (equivalent) in the examples all refers to the equivalent relative to the fluoroarene in the examples.
[0121] It should be noted that the specific parameters or some reagents in the above examples are specific examples or preferred examples under the concept of the present invention, rather than limitations thereto; those skilled in the art can make adaptive adjustments within the concept and protection scope of the present invention.
Claims
1. A method for hydrodefluorination of fluorinated aromatic hydrocarbons, characterized in that, Comprising the steps of: Mixing a fluoroarene, a palladium catalyst, a base, and an alcohol to undergo a hydrodehalogenation reaction to prepare an aromatic hydrocarbon compound; Among them, the structural formula of the fluoroarene is as follows: In the formula, Ar is an aromatic ring, and the aromatic ring includes at least one of a benzene ring, a heterocyclic ring, and a fused ring; R includes the following functional groups and any combination thereof: The alcohol includes ethanol and / or isobutanol.
2. The hydrodefluorination method of a fluoroarene according to claim 1, characterized in that, The reaction temperature of the hydrodehalogenation reaction is 60°C to 140°C.
3. The hydrodefluorination method of fluorinated aromatic hydrocarbons according to claim 1, wherein The palladium catalyst includes: A palladium compound catalyst; and / or A catalyst formed by combining palladium metal with a support; and / or A catalyst formed by combining a palladium compound with a support.
4. The hydrodefluorination method of fluorinated aromatic hydrocarbons according to claim 1, characterized in that, The compound includes at least one of palladium acetate, palladium chloride, and ammonium chloropalladite.
5. The hydrodefluorination method of fluorinated aromatic hydrocarbons according to claim 1, characterized in that, The support includes at least one of silica, carbon, calcium carbonate, calcium phosphate, and alumina.
6. The hydrodefluorination method of the fluoroarene according to claim 1, characterized in that, The support is alumina.
7. The hydrodefluorination method of fluorinated aromatic hydrocarbons according to claim 1, characterized in that, The base includes at least one of sodium hydroxide, sodium ethoxide, sodium tert-butoxide, potassium tert-butoxide, and sodium metaaluminate.
8. The hydrodefluorination method of the fluoroarene according to claim 1, wherein The ethanol includes ultra-dry ethanol, analytical grade anhydrous ethanol, or 95 wt.% ethanol.
9. The hydrodefluorination method of fluorinated aromatic hydrocarbons according to claim 1, characterized in that, The molar ratio of palladium in the palladium catalyst to the fluoroarene is (0.5 to 10):
100.
10. The hydrodefluorination method of fluorinated aromatic hydrocarbons according to claim 1, characterized in that, The molar ratio of the base to the fluoroarene is (1 to 5):1.