Method for directly performing heptafluoroisopropylation on aromatic hydrocarbon by utilizing electron donor-acceptor complex strategy

Abstract

The invention discloses a method for directly carrying out heptafluoroisopropylation on aromatic hydrocarbon by utilizing an electron donor-acceptor compound strategy, which comprises the following steps: by taking aromatic hydrocarbon and easily available heptafluoroisopropyl iodide as raw materials, carrying out the heptafluoroisopropylation on the aromatic hydrocarbon by utilizing the electron donor-acceptor compound strategy under illumination and alkaline conditions, thereby obtaining the heptafluoroisopropylation aromatic hydrocarbon. A series of compounds containing heptafluoroisopropyl are synthesized with high yield and high selectivity through a one-step reaction; the method is also suitable for other types of perfluoroalkylating reagents; the synthesized heptafluoroisopropyl substituted aromatic hydrocarbon has wide application prospects in the fields of bioactive molecules and functional materials, the reaction conditions are mild, the raw materials are simple and easy to obtain, the atom economy is high, and three wastes are few; # imgabs0 #

Description

Technical Field

[0001] The invention relates to a heptafluoroisopropylation reaction of aromatic hydrocarbon CH, and in particular to a method for directly carrying out heptafluoroisopropylation of aromatic hydrocarbon by utilizing an electron donor-acceptor complex strategy. Background Art

[0002] Heptafluoroisopropyl 3 F 7 ), often referred to as "Super CF 3 ", due to its unique properties (such as high lipophilicity, strong electron-withdrawing ability and large steric hindrance, etc.), it is different from other straight-chain perfluoroalkyl groups [J.Med.Chem., 1973, 16, 1207-1216; Chem.Rev., 1991, 91, 165-195.], and has been widely used in the fields of bioactive molecules and functional materials. In addition, studies have shown that most bioactive compounds containing heptafluoroisopropyl groups exist in aromatic compounds, such as the six commercially available insecticides containing heptafluoroisopropyl groups: Flubendiamide, the new phenylpyrazole nicotinamide fluorine-containing insecticide (Nicofluprole), Broflanilide, Cyhalodiamide, etc., all have good insecticidal activity (Formula 1). Therefore, it is of great research value to explore the method of introducing heptafluoroisopropyl groups into aromatic hydrocarbons.

[0003]

[0004] However, despite the wide application of heptafluoroisopropyl-containing aromatic compounds, practical methods for effectively incorporating heptafluoroisopropyl groups into various aromatics have not been fully explored. Generally, the methods for introducing heptafluoroisopropyl groups into aromatic substrates can be divided into two categories: 1) heptafluoroisopropylation of prefunctionalized aromatics; 2) CH heptafluoroisopropylation of aromatics. The first method is to react the pre-functionalized substrate with heptafluoroisopropyl iodide / bromine (under the action of a transition metal or a photosensitizer) or directly with a heptafluoroisopropyl metal reagent to obtain heptafluoroisopropylated aromatic hydrocarbons, such as boric acid, amino group, iodine atom, carboxylic acid, etc. in the substrate are directly converted into heptafluoroisopropyl [Chem. Rev., 2002, 102, 1359-1469; J. Fluorine Chem., 2016, 189, 59-67; Org. Chem. Front., 2016, 3, 304-308; Chem. Commun., 2016, 52, 796-799; Org. Lett., 2019, 21, 1093-1097; Org. Lett., 2022, 24, 7622-7626.] (Formula 2).

[0005]

[0006] In contrast, the CH heptafluoroisopropylation reaction of aromatics is a more atom-economical method for constructing heptafluoroisopropyl-containing molecules. The most common CH heptafluoroisopropylation reaction is that electron-rich aromatics are subjected to a sulfinating dehalogenation reagent, such as sodium dithionite, through a free radical process to achieve the heptafluoroisopropylation reaction of the substrate [Huang Weiyuan, Sulfinating Dehalogenation Reaction and Its Application, Hebei Education Press, 2003] (Formula 3). However, due to the instability of sodium dithionite, a large amount of difficult-to-treat sodium salt wastewater will be produced after the reaction. In addition, the CH heptafluoroisopropylation reaction in modern methods also relies on the participation of transition metals or photosensitizers [Org. Chem. Front., 2022, 9, 4435-4440; Commun. Chem., 2023, 6, 177.] (Formula 4).

[0007]

[0008] In summary, although heptafluoroisopropyl iodide can introduce heptafluoroisopropyl substituents into aromatic hydrocarbons through sulfinating dehalogenation reaction and the action of transition metals (or photosensitizers), the sulfinating dehalogenation reaction will produce a large amount of wastewater that is difficult to treat. Most other methods require the participation of transition metals or photosensitizers, or a series of heptafluoroisopropyl metal reagents that are not easy to obtain under the action of transition metals. Therefore, it is of great application value to develop a more economical, green and efficient method for synthesizing CH heptafluoroisopropylation products. Summary of the invention

[0009] The present invention aims to provide a heptafluoroisopropylation reaction method for aromatic hydrocarbon CH, so as to realize the efficient synthesis of various heptafluoroisopropylation products of aromatic hydrocarbons.

[0010] The invention uses aromatic hydrocarbons and readily available heptafluoroisopropyl iodide as raw materials, and synthesizes a series of heptafluoroisopropyl-containing compounds with high yield and high selectivity through a one-step reaction under light and alkaline conditions through an electron donor-acceptor (EDA) complex (electron donor-heptafluoroisopropyl iodide) strategy.

[0011] The technical solution of the present invention is as follows:

[0012] A method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy, comprising:

[0013] In an inert atmosphere, aromatic hydrocarbons, heptafluoroisopropyl iodide, an electron donor, an alkaline substance and an organic solvent are mixed, and a substitution reaction is carried out under light and room temperature conditions. The reaction is monitored by TLC until the reaction is complete. The reaction solution is separated and purified to obtain a product, heptafluoroisopropyl substituted aromatic hydrocarbon;

[0014] in,

[0015] The inert atmosphere is nitrogen or a rare gas;

[0016] The molar ratio of aromatic hydrocarbon, heptafluoroisopropyl iodide and electron donor is 1:1.0-1.2:0.02-0.1, preferably 1:1.05:0.05;

[0017] The molar ratio of the alkaline substance to the aromatic hydrocarbon is 0.5 to 1:1, preferably 0.5:1;

[0018] The electron donor is selected from the group consisting of: triphenylphosphine, tri(4-methylphenyl)phosphine, tri(4-bromophenyl)phosphine, tri(4-trifluoromethylphenyl)phosphine, tri(3,5-bis(trifluoromethyl)phenyl)phosphine, tri(pentafluorophenyl)phosphine, trimethyl phosphite, triethyl phosphite, triisopropyl phosphite, triphenyl phosphite, and one or more thereof, preferably trimethyl phosphite;

[0019] The alkaline substance is selected from one or more of potassium carbonate, sodium carbonate, cesium carbonate, potassium phosphate, triethylamine, and triethylenediamine;

[0020] The organic solvent is selected from: one or more of DMF, DMA, NMP, DMSO, ethyl acetate, dichloromethane, acetonitrile, tetrahydrofuran, and methyl tert-butyl ether; the volume molar ratio of the organic solvent to the aromatic hydrocarbon is 1 to 20:1, mL / mmol;

[0021] The illumination may be one of the following light sources: white light, ultraviolet light (365nm), purple light (395-400nm), blue light (450-455nm), green light (530-535nm), yellow light (595-600nm), red light (665-695nm), preferably white light or ultraviolet light (365nm);

[0022] The method for separating and purifying the reaction liquid is as follows: after the reaction is completed, water is added to the reaction liquid to quench it, ethyl acetate is extracted, the organic phases are combined, and they are concentrated under reduced pressure. A mixed solution of petroleum ether / ethyl acetate in a volume ratio of 1 to 10:1 is used as an eluent for silica gel column chromatography separation, and the eluate containing the target compound is collected, the solvent is evaporated and dried to obtain the product heptafluoroisopropyl substituted aromatic hydrocarbon;

[0023] The general formula of the aromatic hydrocarbon is shown in formula (1), the structural formula of heptafluoroisopropyl iodide is shown in formula (2), and the general formula of the product heptafluoroisopropyl substituted aromatic hydrocarbon is shown in formula (3);

[0024] The reaction formula is as follows:

[0025]

[0026] In formula (1) or formula (3),

[0027] n = 0, 1;

[0028] Het / Ar represents an aryl or heteroaryl group; an aryl group is, for example, a phenyl group or a naphthyl group; a heteroaryl group is an azaaryl group or a thioaryl group, for example, a pyridyl group, a pyrazinyl group, a pyrrolyl group, a thienyl group, an indolyl group, a carbazolyl group or a quinolyl group;

[0029] m = 1, 2, 3;

[0030] R (m) represents 1 to 3 substituents R on the Het / Ar ring, each of which is independently selected from: H, alkyl (such as methyl, ethyl, tert-butyl, etc.), primary amino, secondary amino, tert-amino, phenyl, alkoxy (such as methoxy, cyclopentyloxy, etc.), benzyloxy, halogen (fluorine, chlorine, bromine, iodine), carbomethoxy, trifluoromethyl, trifluoromethylthio, (N-methylmorpholine) group, boric acid pinacol ester group, tert-butyloxycarbonyl or a substituent derived from Vitamin E as shown below:

[0031]

[0032] The method of the present invention is also applicable to other perfluoroalkyl iodides, and the above-mentioned "heptafluoroisopropyl iodide" can also be replaced by other perfluoroalkylating agents, such as ethyl iodide, ethyl difluoroiodide, perfluorocyclohexyl iodide, perfluorohexyl iodide, perfluorooctyl iodide, perfluorobutyl iodide, etc.

[0033] The heptafluoroisopropyl substituted aromatic hydrocarbons synthesized in the present invention have broad application prospects in the fields of biologically active molecules and functional materials. For example, the 4-heptafluoroisopropyl-2-trifluoromethylaniline synthesized in Example 9 of the present invention is an important intermediate for the industrial production of broflanilide insecticides. For specific applications, see Example 50.

[0034] Compared with the prior art, the present invention has the following beneficial effects:

[0035] The present invention realizes an efficient and practical heptafluoroisopropylation reaction strategy of various anilines and heterocyclic aromatic hydrocarbons through an electron donor acceptor (EDA) complex strategy, and synthesizes a series of heptafluoroisopropylated aromatic hydrocarbon compounds with excellent yields under visible light.

[0036] The advantages of the method of the present invention are that the raw materials are readily available, the atom economy is high, the three wastes are less, the use of metal catalysts or photosensitizers is avoided, and it has high selectivity and a wide substrate range, providing a green and practical way for efficiently synthesizing various types of heptafluoroisopropyl products using simple and readily available reagents. DETAILED DESCRIPTION

[0037] The present invention is described in detail below in conjunction with specific embodiments. The following embodiments will help those skilled in the art to further understand the present invention, but are not intended to limit the present invention in any form. It should be noted that, for those of ordinary skill in the art, several variations and improvements may be made without departing from the concept of the present invention. These all belong to the protection scope of the present invention.

[0038] Embodiment 1:

[0039]

[0040] In a glove box, add K 2 CO 3 (0.5mmol, 69.1mg), P(OMe) 3 (0.05mmol, 6.2mg), N,N-dimethylformamide (1ml), o-methylaniline (1mmol, 107.2mg) and heptafluoroisopropyl iodide (1.05mmol, 310.7mg). Then, the Schlenk tube was sealed with a screw cap with a polytetrafluoroethylene liner and the Schlenk tube was taken out of the glove box. The stirred reaction mixture was irradiated with 10W white light (about 3.0cm from the bulb), and the reaction temperature was kept at room temperature by a fan in an air-conditioned room at 25°C. The reaction progress was monitored by TLC. After 2h, the reaction was complete, and the reaction mixture was quenched with 10ml of water, and then extracted with ethyl acetate (3×10ml), and the organic phase was collected. The solvent was concentrated under reduced pressure. The crude product was purified by column chromatography using silica gel (petroleum ether / ethyl acetate = 10:1) to obtain a yellow oily heptafluoroisopropylation product (241.2mg, purity: 99%, yield: 94%). 1 HNMR (400MHz, CDCl 3 )δ7.27(s,1H),7.25(d,J=7.6Hz,1H),6.73(d,J=7.9Hz,1H),3.87(s,2H),2.21(s,3H).

[0041] The same reactants were used for the reaction under the same operation steps with 0.05 mmol or less of electron donors. The results are shown in Table 1 below:

[0042] Table 1

[0043]

[0044]

[0045] In Table 1, superscript a Denotes the yield determined by GC using diphenylmethane as internal standard.

[0046] The same reactants were used under the same operating steps, and DMF was replaced with 1.0 ml of the following solvents for the reaction. The results are shown in Table 2 below:

[0047] Table 2

[0048]

[0049] In Table 2, superscript a Denotes the yield determined by GC using diphenylmethane as internal standard.

[0050] Taking the same reactants and the same operating steps, the following 0.5 mmol or less base was used to replace potassium carbonate for the reaction. The results are shown in Table 3 below:

[0051] Table 3

[0052]

[0053] In Table 3, superscript a Denotes the yield determined by GC using diphenylmethane as internal standard.

[0054] Taking the same reactants and the same operating steps, different types of light sources were used to replace white light for the reaction. The results are shown in Table 4 below:

[0055] Table 4

[0056]

[0057]

[0058] In Table 4, superscript a Denotes the yield determined by GC using diphenylmethane as internal standard.

[0059] Embodiment 2:

[0060]

[0061] The difference from Example 1 is that the nitrogen-containing aromatic hydrocarbon substrate used is aniline (1 mmol, 93.1 mg), and the other reaction conditions and operation steps are the same as those of Example 1 (225.3 mg, purity: 99%, yield: 94%). 1 H NMR (400 MHz, CDCl 3 ) δ7.39 (d, J = 8.4Hz, 2H), 6.73 (d, J = 8.7Hz, 2H), 3.88 (s, 2H).

[0062] Embodiment 3:

[0063]

[0064] The difference from Example 1 is that the nitrogen-containing aromatic hydrocarbon substrate used is 2-isopropylaniline (1 mmol, 135.2 mg), and the other reaction conditions and operating steps are the same as those of Example 1 (276.7 mg, purity: 99%, yield: 91%). 1 HNMR (400MHz, CDCl 3 )δ7.42(d,J=2.5Hz,1H),7.30(dd,J=8.4,2.3Hz,1H),6.81-6.64(m,1H),3.88(s,2H),2.94(p,J=6.8Hz,1H),1.33(dd,J=6.9,0.9Hz,6H).

[0065] Embodiment 4:

[0066]

[0067] The difference from Example 1 is that the nitrogen-containing aromatic hydrocarbon substrate used is 2-tert-butylaniline (1 mmol, 149.2 mg), and the other reaction conditions and operating steps are the same as those of Example 1 (292.8 mg, purity: 99%, yield: 92%). 1 HNMR (400MHz, CDCl 3 )δ7.43(s,1H),7.22(d,J=9.9Hz,1H),6.67(d,J=8.5Hz,1H),4.08(s,2H),1.43(s,9H).

[0068] Embodiment 5:

[0069]

[0070] The difference from Example 1 is that the nitrogen-containing aromatic hydrocarbon substrate used is 2-chloroaniline (1 mmol, 127.6 mg), and the other reaction conditions and operating steps are the same as those of Example 1 (234.2 mg, purity: 99%, yield: 80%). 1 H NMR (400 MHz, CDCl 3 )δ7.48(s,1H),7.27(d,J=7.6Hz,1H),6.81(d,J=8.8Hz,1H),4.34(s,2H).

[0071] Embodiment 6:

[0072]

[0073] The difference from Example 1 is that the nitrogen-containing aromatic hydrocarbon substrate used is 2-bromoaniline (1 mmol, 172.0 mg), and the other reaction conditions and operating steps are the same as those of Example 1 (274.5 mg, purity: 99%, yield: 72%). 1 H NMR (400 MHz, CDCl 3 )δ7.64(d,J=2.2Hz,1H),7.35-7.28(m,1H),6.80(dd,J=8.7,0.9Hz,1H),4.19(s,2H).

[0074] Embodiment 7:

[0075]

[0076] The difference from Example 1 is that the substrate used is 2-phenylaniline (1 mmol, 169.1 mg), and the other reaction conditions and operating steps are the same as those of Example 1 (307.6 mg, purity: 99%, yield: 91%). 1 H NMR (400 MHz, CDCl 3 )δ7.55-7.45(m,4H),7.45-7.35(m,3H),6.83(d,J=8.3Hz,1H),4.00(s,2H).

[0077] Embodiment 8:

[0078]

[0079] The difference from Example 1 is that the substrate used is 2-methoxyaniline (1 mmol, 123.1 mg), and the other reaction conditions and operation steps are the same as those of Example 1 (272.0 mg, purity: 99%, yield: 93%). 1 H NMR (400 MHz, CDCl 3 )δ7.04(d,J=8.4Hz,1H),6.97(s,1H),6.74(d,J=8.3Hz,1H),4.04(s,2H),3.88(s,3H).

[0080] Embodiment 9:

[0081]

[0082] The difference from Example 1 is that the substrate used is 2-trifluoromethylaniline (1 mmol, 161.0 mg), the light source used is a 10 W ultraviolet lamp (365 nm), the reaction time is 24 h, and the other reaction conditions and operating steps are the same as those of Example 1 (236.9 mg, purity: 98.8%, yield: 73%).1 H NMR (400 MHz, CDCl 3 )δ7.67(s,1H),7.50(d,J=8.8Hz,1H),6.81(d,J=8.7Hz,1H),4.47(s,2H).

[0083] Embodiment 10:

[0084]

[0085] The difference from Example 1 is that the substrate used is 3-methylaniline (1 mmol, 107.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (210.1 mg, purity: 97.1%, yield: 90%). 1 HNMR (400MHz, CDCl 3 )δ7.30-7.19(m,1H),6.53(dt,J=4.3,2.2Hz,2H),3.80(s,2H),2.42(d,J=8.5Hz,3H).

[0086] Embodiment 11:

[0087]

[0088] The difference from Example 1 is that the substrate used is 3-fluoroaniline (1 mmol, 111.0 mg), the reaction time is 24 h, and the other reaction conditions and operation steps are the same as those of Example 1 (234.0 mg, purity: 98.3%, yield: 87%). 1 H NMR (400 MHz, CDCl 3 )δ7.32(t,J=8.5Hz,1H),6.50(dd,J=8.7,2.4Hz,1H),6.42(ddd,J=13.5,2.3,1.0Hz,1H),4.05(s,2H).

[0089] Embodiment 12:

[0090]

[0091] The difference from Example 1 is that the substrate used is 2,6-diethylaniline (0.1 mmol, 149.1 mg), and the other reaction conditions and operation steps are the same as those of Example 1 (304.3 mg, purity: 98.7%, yield: 96%). 1 H NMR (400 MHz, CDCl 3)δ7.20(s,2H),3.91(s,2H),2.58(q,J=7.5Hz,4H),1.30(t,J=7.6Hz,6H).

[0092] Embodiment 13:

[0093]

[0094] The difference from Example 1 is that the substrate used is 2,6-dichloroaniline (1 mmol, 160.1 mg), the light source used is a 10 W ultraviolet lamp (365 nm), the reaction time is 24 h, and the other reaction conditions and operating steps are the same as those of Example 1 (164.5 mg, purity: 99.0%, yield: 50%). 1 H NMR (400 MHz, CDCl 3 )δ7.41(s,2H),4.77(s,2H).

[0095] Embodiment 14:

[0096]

[0097] The difference from Example 1 is that the substrate used is N-methylaniline (1 mmol, 107.1 mg), the reaction time is 12 h, and the other reaction conditions and operating steps are the same as those of Example 1 (258.5 mg, purity: 97.3%, yield: 94%). 1 HNMR (400MHz, CDCl 3 )δ7.39(d,J=8.6Hz,2H),6.64(dd,J=9.0Hz,0.8Hz,2H),4.03(s,1H),2.87(s,3H).

[0098] Embodiment 15:

[0099]

[0100] The difference from Example 1 is that the substrate used is N-cyclohexylaniline (1 mmol, 175.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (315.5 mg, purity: 97.1%, yield: 92%). 1 HNMR (400MHz, CDCl 3)δ7.37(d,J=8.5Hz,2H),6.62(d,J=8.5Hz,2H),3.87(s,1H),3.30(tt,J=9.8,3.9Hz,1H ), 2.07 (dq, J = 12.3, 3.9 Hz, 2H), 1.74 (ddt, J = 43.3, 12.7, 4.0 Hz, 3H), 1.48-1.13 (m, 5H).

[0101] Embodiment 16:

[0102]

[0103] The difference from Example 1 is that the substrate used is N-acetylcyanoaniline (1 mmol, 132.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (243.0 mg, purity: 97.3%, yield: 81%). 1 HNMR (400MHz, CDCl 3 )δ7.49(d,J=8.6Hz,2H),6.77(d,J=8.5Hz,2H),4.42(t,J=6.8Hz,1H),4.12(d,J=6.8Hz,2H).

[0104] Embodiment 17:

[0105]

[0106] The difference from Example 1 is that the substrate used is N-n-butyl-N-propionylaniline (1 mmol, 202.1 mg), the reaction time is 24 h, and the other reaction conditions and operating steps are the same as those in Example 1 (325.6 mg, purity: 98.6%, yield: 88%). 1 H NMR (400 MHz, CDCl 3 )δ7.45(d,J=8.7Hz,2H),6.71(d,J=8.9Hz,2H),3.70(dt,J=13.8,7.0Hz,2H),3.38(ddd,J=21.3,8.2,6. 7Hz, 2H), 2.62 (t, J=6.9Hz, 2H), 1.62 (tdd, J=9.3, 7.0, 5.3Hz, 2H), 1.50-1.31 (m, 2H), 1.08-0.94 (m, 3H).

[0107] Embodiment 18:

[0108]

[0109] The difference from Example 1 is that the substrate used is 4-tert-butylaniline (1 mmol, 149.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (285.3 mg, purity: 97.2%, yield: 90%). 1 HNMR (400MHz, CDCl 3 )δ7.31(d,J=7.0Hz,2H),6.68(dd,J=9.0,0.9Hz,1H),4.18(s,2H),1.28(d,J=1.1Hz,9H).

[0110] Embodiment 19:

[0111]

[0112] The difference from Example 1 is that the substrate used is 4-fluoroaniline (1 mmol, 111.0 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (235.2 mg, purity: 98.7%, yield: 84%). 1 H NMR (400 MHz, CDCl 3 )δ7.05-6.96(m,2H),6.67(dd,J=8.9,4.8Hz,1H),4.17(s,2H).

[0113] Embodiment 20:

[0114]

[0115] The difference from Example 1 is that the substrate used is 4-chloroaniline (1 mmol, 127.0 mg), the reaction time is 24 h, and the other reaction conditions and operation steps are the same as those of Example 1 (265.5 mg, purity: 97.6%, yield: 90%). 1 H NMR (400 MHz, CDCl 3 )δ7.26(s,1H),7.20(dd,J=8.8,2.3Hz,1H),6.65(d,J=8.8Hz,1H),4.31(s,2H).

[0116] Embodiment 21:

[0117]

[0118] The difference from Example 1 is that the substrate used is 4-bromoaniline (1 mmol, 171.0 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (304.2 mg, purity: 97.4%, yield: 84%). 1H NMR (400 MHz, CDCl 3 )δ7.38(s,1H),7.33(dd,J=8.8,2.2Hz,1H),6.59(dd,J=8.8,0.9Hz,1H),4.51-4.09(m,2H).

[0119] Embodiment 22:

[0120]

[0121] The difference from Example 1 is that the substrate used is 4-iodoaniline (1 mmol, 219.0 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (346.5 mg, purity: 98.0%, yield: 90%). 1 H NMR (400 MHz, CDCl 3 )δ7.53-7.44(m,2H),6.49(d,J=9.1Hz,1H),4.35(s,2H).

[0122] Embodiment 23:

[0123]

[0124] The difference from Example 1 is that the substrate used is 4-methoxyaniline (1 mmol, 123.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (359.7 mg, purity: 97.7%, yield: 92%). 1 HNMR (400MHz, CDCl 3 )δ6.89(dd,J=8.9,2.7Hz,1H),6.83(s,1H),6.69(d,J=8.8Hz,1H),4.00(s,2H),3.74(s,3H).

[0125] Embodiment 24:

[0126]

[0127] The difference from Example 1 is that the substrate used is 4-benzyloxyaniline (1 mmol, 199.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (275.2 mg, purity: 99.3%, yield: 75%). 1 HNMR (400MHz, CDCl 3)δ7.51-7.31(m,5H),7.01-6.87(m,2H),6.68(d,J=8.8Hz,1H),4.99(s,2H),3.88(s,2H).

[0128] Embodiment 25:

[0129]

[0130] The difference from Example 1 is that the substrate used is 4-(N-methylmorpholinyl)aniline (1 mmol, 192.1 mg), the reaction time is 24 h, and the other reaction conditions and operation steps are the same as those of Example 1 (262.8 mg, purity: 97.1%, yield: 73%). 1 H NMR (400 MHz, CDCl 3 )δ7.22(d,J=7.6Hz,2H),6.68(d,J=8.6Hz,1H),4.25(s,2H),3.69(t,J=4.6Hz,4H),3.39(s,2H),2.40(t,J=4.6Hz,4H).

[0131] Embodiment 26:

[0132]

[0133] The difference from Example 1 is that the substrate used is methyl 4-aminobenzoate (1 mmol, 151.1 mg), the light source is a 10 W ultraviolet lamp (365 nm), the reaction time is 24 h, and the other reaction conditions and operating steps are the same as those in Example 1 (172.3 mg, purity: 98.0%, yield: 54%). 1 H NMR (400 MHz, CDCl 3 )δ8.03(s,1H),7.88(dd,J=8.7,1.9Hz,1H),6.70(dd,J=8.7,1.0Hz,1H),4.82(s,2H),3.85(s,3H).

[0134] Embodiment 27:

[0135]

[0136] The difference from Example 1 is that the substrate used is 4-cyclopentaneoxyaniline (1 mmol, 177.1 mg), the reaction time is 24 h, and the other reaction conditions and operation steps are the same as those of Example 1 (248.4 mg, purity: 97.9%, yield: 72%). 1 H NMR (400 MHz, CDCl 3)δ6.94-6.75(m,2H),6.67(d,J=8.8Hz,1H),4.61(tt,J=5.7,2.8Hz,1H),3.91(s,2H),1.94-1.70(m,6H),1.69-1.48(m,2H).

[0137] Embodiment 28:

[0138]

[0139] The difference from Example 1 is that the substrate used is 4-(trifluoromethylthio)aniline (1 mmol, 193.0 mg), the reaction time is 24 h, and the other reaction conditions and operation steps are the same as those of Example 1 (331.2 mg, purity: 97.2%, yield: 92%). 1 H NMR (400 MHz, CDCl 3 )δ7.59(s,1H),7.49(dd,J=8.6,2.0Hz,1H),6.71(d,J=8.6Hz,1H),4.64(s,2H).

[0140] Embodiment 29:

[0141]

[0142] The difference from Example 1 is that the substrate used is 4-aminophenylboronic acid pinacol ester (1 mmol, 219.1 mg), the reaction time is 24 h, and the other reaction conditions and operating steps are the same as those of Example 1 (325.1 mg, purity: 98.1%, yield: 84%). 1 H NMR (400 MHz, CDCl 3 )δ7.74(s,1H),7.69-7.58(m,1H),6.67(dd,J=8.2,1.1Hz,1H),4.50(s,2H),1.32(s,12H).

[0143] Embodiment 30:

[0144]

[0145] The difference from Example 1 is that the substrate used is (R)-2,5,7,8-tetramethyl-2-((4R,8R)-4,8,12-trimethyltridecyl)chroman-6-yl-4-aminobenzoic acid (1 mmol, 549.4 mg), the light source is a 10 W ultraviolet lamp (365 nm), the reaction time is 24 h, and the other reaction conditions and operating steps are the same as those in Example 1 (415.8 mg, purity: 97.4%, yield: 58%). 1H NMR (400 MHz, CDCl 3 )δ8.26(s,1H),8.09(dd,J=8.8,2.1Hz,1H),6.69(dd,J=50.7,8.7Hz,1H),4.83(s,2H),2.62(t,J=6.8Hz,2H ), 2.13 (s, 3H), 2.04 (dd, J = 16.9, 5.9Hz, 6H), 1.96-1.68 (m, 2H), 1.58-1.06 (m, 24H), 0.87 (t, J = 6.7Hz, 12H).

[0146] Embodiment 31:

[0147]

[0148] The difference from Example 1 is that the substrate used is dihydroindoline (1 mmol, 119.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (255.4 mg, purity: 98.0%, yield: 89%). 1 H NMR (400 MHz, CDCl 3 )δ7.93(s,1H),7.80(d,J=8.0Hz,1H),7.28(dt,J=4.2,2.2Hz,3H),2.51(d,J=3.7Hz,2H),2.43(s,2H).

[0149] Embodiment 32:

[0150]

[0151] The difference from Example 1 is that the substrate used is N-Boc pyrrole (1 mmol, 167.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (294.8 mg, purity: 98.0%, yield: 88%). 1 H NMR (400 MHz, CDCl 3 )δ7.50(dt,J=3.4,1.7Hz,1H),6.76(ddd,J=3.9,1.9,0.9Hz,1H),6.24(td,J=3.6,1.3Hz,1H),1.60(s,9H).

[0152] Embodiment 33:

[0153]

[0154] The difference from Example 1 is that the substrate used is N-phenylpyrrole (1 mmol, 143.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (223.9 mg, purity: 98.0%, yield: 72%). 1 HNMR (400MHz, CDCl 3 )δ7.46(qd,J=4.4,2.4Hz,3H),7.39(dt,J=7.8,1.9Hz,2H),6.88(dd,J=2.8,1.7Hz,1H),6.74(s,1H),6.43-6.34(m,1H).

[0155] Embodiment 34:

[0156]

[0157] The difference from Example 1 is that the substrate used is 2-methylindole (1 mmol, 131.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (258.0 mg, purity: 98.0%, yield: 86%). 1 HNMR (400MHz, CDCl 3 )δ8.26(s,1H),7.77(d,J=7.7Hz,1H),7.36-7.14(m,3H),2.54(d,J=3.7Hz,3H).

[0158] Embodiment 35:

[0159]

[0160] The difference from Example 1 is that the substrate used is 3-methylindole (1 mmol, 131.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (279.0 mg, purity: 98.0%, yield: 93%). 1 HNMR (400MHz, CDCl 3 )δ8.18(s,1H),7.75(dd,J=8.0,1.1Hz,1H),7.52-7.37(m,2H),7.30(ddd,J=8.1,6.7,1.4Hz,1H),2.55(d,J=2.4Hz,3H).

[0161] Embodiment 36:

[0162]

[0163] The difference from Example 1 is that the substrate used is 3,6-dibromocarbazole (1 mmol, 322.9 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (318.5 mg, purity: 98.0%, yield: 65%). 1 HNMR (400MHz, CDCl 3 )δ8.68(d,J=5.0Hz,1H),8.27(d,J=1.8Hz,1H),8.14(d,J=1.9Hz,1H),7.68(s,1H),7.58(dd,J=8.7,1.9Hz,1H),7.35(d,J=8.6Hz,1H).

[0164] Embodiment 37:

[0165]

[0166] The difference from Example 1 is that the substrate used is 2,6-diaminopyridine (1 mmol, 109.1 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (249.3 mg, purity: 98.0%, yield: 90%). 1 HNMR (400MHz, CDCl 3 )δ7.13(d,J=8.7Hz,1H),6.13(s,2H),5.83(dd,J=8.7,2.3Hz,1H),5.62(s,2H).

[0167] Embodiment 38:

[0168]

[0169] The difference from Example 1 is that the substrate used is 2-aminopyrazine (1 mmol, 95.0 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (215.7 mg, purity: 98.0%, yield: 82%). 1 H NMR (400 MHz, CDCl 3 )δ8.26(t,J=1.6Hz,1H),8.04-7.86(m,1H),7.24(s,2H).

[0170] Embodiment 39:

[0171]

[0172] The difference from Example 1 is that the substrate used is 2-iodothiophene (1 mmol, 209.9 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (343.1 mg, purity: 98.0%, yield: 91%). 1 H NMR (400 MHz, CDCl 3 )δ7.28(dd,J=3.9,2.5Hz,1H),7.03(d,J=4.8Hz,1H).

[0173] Embodiment 40:

[0174]

[0175] The difference from Example 1 is that the perfluoroalkylating agent used is perfluorobutyl iodide (1.05 mmol, 363.2 mg), the reaction time is 12 h, and the other reaction conditions and operating steps are the same as those of Example 1 (269.7 mg, purity: 98.0%, yield: 83%). 1 H NMR (400 MHz, CDCl 3 )δ7.23(d,J=6.9Hz,2H),6.70(d,J=8.7Hz,1H),3.91(s,2H),2.20(s,3H).

[0176] Embodiment 41:

[0177]

[0178] The difference from Example 1 is that the perfluoroalkylating agent used is perfluorohexyl iodide (1.05 mmol, 468.2 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as Example 1 (357.0 mg, purity: 98.0%, yield: 84%). 1 H NMR (400 MHz, CDCl 3 )δ7.23(d,J=6.5Hz,2H),6.71(d,J=8.9Hz,1H),3.96(s,2H),2.20(s,3H).

[0179] Embodiment 42:

[0180]

[0181] The difference from Example 1 is that the perfluoroalkylating agent used is perfluorooctyl iodide (1 mmol, 573.2 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as Example 1 (414.4 mg, purity: 98.0%, yield: 79%). 1H NMR (400 MHz, CDCl 3 )δ7.31-7.22(m,2H),6.65(d,J=8.5Hz,1H),4.31(s,2H),2.39(s,3H).

[0182] Embodiment 43:

[0183]

[0184] The difference from Example 1 is that the substrate used is β-naphthylamine (1 mmol, 143.1 mg), the perfluoroalkylation reagent is perfluorohexyl iodide (1.05 mmol, 468.2 mg), the reaction time is 12 h, and the other reaction conditions and operating steps are the same as those in Example 1 (414.9 mg, purity: 98.0%, yield: 90%). 1 H NMR (400 MHz, CDCl 3 )δ8.05(d,J=8.5Hz,1H),7.82-7.61(m,2H),7.49(ddd,J=8.7,6.9,1.6Hz,1H),7.31(t,J=7.4Hz,1H),6.79(d,J=8.9Hz,1H),4.60(s,2H).

[0185] Embodiment 44:

[0186]

[0187] The difference from Example 1 is that the substrate used is tetrahydroquinoline (1 mmol, 133.1 mg), the perfluoroalkylation agent is perfluorohexyl iodide (1.05 mmol, 468.2 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those in Example 1 (294.8 mg, purity: 98.0%, yield: 88%). 1 H NMR (400 MHz, CDCl 3 )δ7.15(dd,J=8.1,1.5Hz,1H),7.08(d,J=7.3Hz,1H),6.61(t,J=7.7Hz,1H) ,4.40(s,1H),3.45-3.26(m,2H),2.82(t,J=6.4Hz,2H),2.04-1.83(m,2H).

[0188] Embodiment 45:

[0189]

[0190] The difference from Example 1 is that the substrate used is 2,6-diaminopyridine (1 mmol, 109.1 mg), the perfluoroalkylation reagent is perfluorohexyl iodide (1.05 mmol, 468.2 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those in Example 1 (375.7 mg, purity: 98.0%, yield: 88%). 1 H NMR (400 MHz, CDCl 3 )δ(400MHz, DMSO-d6)δ7.14(d,J=8.6Hz,1H),6.24(s,2H),5.81(d,J=8.5Hz,1H),5.65(s,2H).

[0191] Embodiment 46:

[0192]

[0193] The difference from Example 1 is that the perfluoroalkylating agent used is perfluorocyclohexyl iodide (1 mmol, 428.3 mg), the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (352.2 mg, purity: 98.0%, yield: 91%). 1 H NMR (400 MHz, CDCl 3 )δ7.31(dt,J=7.5,2.0Hz,2H),6.81-6.62(m,1H),3.89(s,2H),2.21(s,3H).

[0194] Embodiment 47:

[0195]

[0196] The difference from Example 1 is that the substrate used is N-phenylpyrrole (1 mmol, 143.1 mg), the perfluoroalkylation reagent is ethyl iododifluoroacetate, the reaction time is 12 h, and the other reaction conditions and operation steps are the same as those of Example 1 (227.9 mg, purity: 98.0%, yield: 86%). 1 H NMR (400 MHz, CDCl 3 )δ7.55-7.35(m,5H),6.90(ddt,J=2.7,1.7,0.9Hz,1H),6.69(dd,J=3.8,1.7Hz ,1H), 6.32(dd,J=3.8,2.8Hz,1H), 4.16(q,J=7.1Hz,2H), 1.23(t,J=7.2Hz,3H).

[0197] Embodiment 48:

[0198]

[0199] The difference from Example 1 is that different percentages of trimethyl phosphite were used for the reaction. The results are shown in Table 5 below:

[0200] Table 5

[0201] serial number Trimethyl phosphite percentage <![CDATA[Yield (%) a > 1 1mol% 87 2 2mol% 95 2 5mol% 99 3 10mol% 99

[0202] In Table 5, superscript a Denotes the yield determined by GC using diphenylmethane as internal standard.

[0203] Embodiment 49:

[0204]

[0205] The difference from Example 1 is that different proportions of potassium carbonate equivalents were used for the reaction. The results are shown in Table 6 below:

[0206] Table 6

[0207] serial number Potassium carbonate equivalent (equiv.) <![CDATA[Yield (%) a > 1 0.1 84 2 0.3 90 3 0.5 99 4 1.0 99

[0208] In Table 6, superscript a Denotes the yield determined by GC using diphenylmethane as internal standard.

[0209] Embodiment 50:

[0210] Brofenac was discovered by Mitsui Chemicals and is an efficient broad-spectrum diamine insecticide. The market size of bisamide insecticide products totaled US$2.283 billion (2019), with the top three being chlorantraniliprole, flubendiamide, and cyantraniliprole. In 2019, after the launch of brofenac, a rising star in the bisamide insecticide class, sales rose rapidly, and the compound's annual peak sales are expected to exceed US$800 million. Therefore, brofenac is one of the most promising insecticides in the past decade, and the synthesis method of brofenac also has important application value [C.-Y.Luo, Q.Xu, C.-Q.Huang, et al, Development of an efficient synthetic process for Broflanilide, Org.Process Res.Dev., 2020, 24, 1024-1031.].

[0211] Among the synthesized heptafluoroisopropyl products, 4-heptafluoroisopropyl-2-trifluoromethylaniline is an important synthetic intermediate for the industrial production of broflanilide insecticide. Further, broflanilide was synthesized using this intermediate as a raw material.

[0212]

[0213] Experimental process: Add 2-fluoro-3-(N-methylbenzylamine)benzoyl chloride (10mmol, 2.91g), 50mL xylene, and 4-heptafluoroisopropyl-2-trifluoromethylaniline (20mmol, 6.58g) to a 100mL clean dry flask equipped with a magnetic stirring bar. Stir the mixed reaction solution at 100℃ for 8 hours and condense and reflux. After the reaction is completed, slowly add 5wt% sodium carbonate (2.21g, 20mmol) aqueous solution to the reaction solution. Separate the liquid while hot, and the obtained organic phase is directly used for the next bromination reaction. Add hydrobromic acid (0.81g, 10mmol), water (4g), and sodium hydroxide (0.8g, 20mmol) to the organic phase of the previous step in sequence, then start stirring and heat the solution to 70℃, and slowly add 14wt% sodium hypochlorite (1.06g, 20mmol) aqueous solution to the reaction solution within 1h. Continue stirring at 70°C for 2 hours. After the reaction is completed, add 10wt% sodium sulfite (2g, 16mmol) aqueous solution, and then adjust the pH of the reaction system to 4 with concentrated hydrochloric acid to neutralize the reaction system. Separate the liquid while hot, separate and concentrate the organic layer under reduced pressure, and chromatograph the crude product on a silica gel column (eluent: petroleum ether / ethyl acetate = 3 / 1). The target product broflanilide (4.37g) was obtained with a yield of 66% (compared to 4-heptafluoroisopropyl-2-trifluoromethylaniline) by combining the two steps.

[0214]

[0215] Broflanilide. 1 H NMR (400 MHz, DMSO-d 6 )δ10.68(s,1H),8.41(d,J=2.3Hz,1H),8.00-

[0216] 7.88(m,1H),7.60(dt,J=14.1,7.2Hz,2H),7.42-7.23(m,6H),3.36(s,3H). 13 CNMR (101MHz, DMSO-d 6)δ170.63,163.12,154.74(d,J=262.2Hz),139.19,135.95,134.53(d,J=10.1Hz),133.20,131.52(q,J=31.3Hz),130.34,129.18,128.40,127.96,126.82(d,J=22.2Hz),125.34(d,J=4.0Hz),124.42(d,J=14.1Hz),123.65,123.17,120.93,120.25(qd,J=288.9Hz,28.3Hz),91.94-89.25(m),37.79. 19 F NMR(376MHz,DMSO-d 6 )δ-61.25,-75.21(d,J=7.4Hz),-122.97,-181.12--181.20(m).HRMS(ESI)m / z:[M+Na] + calcd.for C 14 H 13 F 7 NO 2 Na + 684.9955,found684.9959.

Claims

1. A method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy, characterized in that: The method comprises: In an inert atmosphere, aromatic hydrocarbons, heptafluoroisopropyl iodide, an electron donor, an alkaline substance and an organic solvent are mixed, and a substitution reaction is carried out under light and room temperature conditions. The reaction is monitored by TLC until the reaction is complete. The reaction solution is separated and purified to obtain a product, heptafluoroisopropyl substituted aromatic hydrocarbon; in, The electron donor is selected from: one or more of triphenylphosphine, tri(4-methylphenyl)phosphine, tri(4-bromophenyl)phosphine, tri(4-trifluoromethylphenyl)phosphine, tri(3,5-di(trifluoromethyl)phenyl)phosphine, tri(pentafluorophenyl)phosphine, trimethyl phosphite, triethyl phosphite, triisopropyl phosphite, and triphenyl phosphite; The general formula of the aromatic hydrocarbon is shown in formula (1), the structural formula of heptafluoroisopropyl iodide is shown in formula (2), and the general formula of the product heptafluoroisopropyl substituted aromatic hydrocarbon is shown in formula (3); The reaction formula is as follows: In formula (1) or formula (3), n＝0，1； Het / Ar represents an aryl or heteroaryl group; m＝1，2，3； R (m) represents 1 to 3 substituents R on the Het / Ar ring, each of which is independently selected from: H, alkyl, primary amino, secondary amino, tertiary amino, phenyl, alkoxy, benzyloxy, halogen, carbomethoxy, trifluoromethyl, trifluoromethylthio, (N-methylmorpholine) group, boric acid pinacol ester group, tert-butyloxycarbonyl or a substituent derived from Vitamin E as shown below:

2. The method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy as claimed in claim 1, characterized in that: Het / Ar represents phenyl, naphthyl, pyridyl, pyrazinyl, pyrrolyl, thienyl, indolyl, carbazolyl or quinolyl.

3. The method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy as claimed in claim 1, characterized in that: The molar ratio of aromatic hydrocarbon, heptafluoroisopropyl iodide and electron donor is 1:1.0-1.2:0.02-0.

1.

4. The method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy as claimed in claim 1, characterized in that: The molar ratio of the alkaline substance to the aromatic hydrocarbon is 0.5 to 1:

1.

5. The method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy as claimed in claim 1, characterized in that: The alkaline substance is selected from one or more of potassium carbonate, sodium carbonate, cesium carbonate, potassium phosphate, triethylamine and triethylenediamine.

6. The method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy as claimed in claim 1, characterized in that: The organic solvent is selected from one or more of DMF, DMA, NMP, DMSO, ethyl acetate, dichloromethane, acetonitrile, tetrahydrofuran, and methyl tert-butyl ether.

7. The method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy as claimed in claim 1, characterized in that: The volume molar ratio of the organic solvent to the aromatic hydrocarbon is 1 to 20:1, mL / mmol.

8. The method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy as claimed in claim 1, characterized in that: The illumination adopts one of the following light sources: white light, ultraviolet light, purple light, blue light, green light, yellow light, and red light.

9. The method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy as claimed in claim 1, characterized in that: The method for separating and purifying the reaction liquid is as follows: after the reaction is completed, water is added to the reaction liquid to quench it, ethyl acetate is extracted, the organic phases are combined, and they are concentrated under reduced pressure. A mixed solution of petroleum ether / ethyl acetate with a volume ratio of 1 to 10:1 is used as an eluent for silica gel column chromatography separation, and the eluate containing the target compound is collected, the solvent is evaporated and dried to obtain the product heptafluoroisopropyl substituted aromatic hydrocarbon.

10. The method for direct heptafluoroisopropylation of aromatic hydrocarbons using an electron donor-acceptor complex strategy as claimed in claim 1, characterized in that: The perfluoroalkylation reaction is carried out by replacing "heptafluoroisopropyl iodide" with ethyl difluoroiodide, perfluorocyclohexyl iodide, perfluorohexyl iodide, perfluorooctyl iodide or perfluorobutyl iodide.

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