反应详情
EQUATION
反应方程式
REACTANTS
反应物
PRODUCTS
生成物
AUXILIARIES
试剂、催化剂与溶剂
PROCEDURE
实验过程
To (R)-benzyl 3-amino-4-(3′-chlorobiphenyl-4-yl)butanoate (Intermediate 9-2: 87.6 mg, 0.183 mmol) is added a solution of HCl in 1,4-dioxane (0.456 mL, 1.825 mmol) at room temperature. After stirring for 3 hours, the reaction mixture is concentrated under reduced pressure to give (R)-benzyl 3-amino-4-(3′-chlorobiphenyl-4-yl)butanoate hydrochloride. A mixture of (R)-benzyl 3-amino-4-(3′-chlorobiphenyl-4-yl)butanoate hydrochloride, fumaric acid monoethyl ester (33.4 mg, 0.220 mmol), EDCl (63.3 mg, 0.330 mmol), DIPEA (0.058 ml, 0.330 mmol) and HOAt (44.9 mg, 0.330 mmol) in DMF (1.8 ml) is allowed to stir at room-temperature for 3 hour. The reaction mixture is diluted with water, and then the products are extracted with EtOAc. The organic layer is washed with NH4OH, 1M HClaq and brine, dried over Na2SO4, filtered, and concentrated to give crude. The obtained residue is purified by silica gel flash column chromatography (heptane/EtOAc=100:0 to 0:100) to give (R,E)-ethyl 4-(4-(benzyloxy)-1-(3′-chlorobiphenyl-4-yl)-4-oxobutan-2-ylamino)-4-oxobut-2-enoate (72.9 mg); HPLC retention time=1.40 minutes (condition B); MS (m+1)=506.3; 1H NMR (400 MHz, CHLOROFORM-d) δ ppm 1.31 (t, J=7.1 Hz, 3 H) 2.58 (A of ABX, Jab=16.4 Hz, Jax=5.3 Hz, 1 H) 2.6 (B of ABX, Jab=16.4 Hz, Jbx=5.1 Hz, 1 H) 2.88 (A of ABX, Jab=13.6 Hz, Jax=8.1 Hz, 1 H) 3.03 (B of ABX, Jab=13.6 Hz, Jbx=6.3 Hz, 1 H) 4.24 (q, J=7.1 Hz, 2 H) 4.56-4.64 (m, 1 H) 5.12 (A of AB, J=12.1 Hz, 1 H) 5.18 (B of AB, J=12.1 Hz, 1 H) 6.57 (br d, J=9.1 Hz, 1 H) 6.77 (A of AB, J=15.4 Hz, 1 H) 6.81 (B of AB, J=15.4 Hz, 1 H) 7.19 (br d, J=8.1 Hz, 2 H) 7.29-7.47 (m, 10 H) 7.53-7.54 (m, 1 H).
WORKUP
后处理
- extractionthe products are extracted with EtOAc
- washThe organic layer is washed with NH4OH, 1M HClaq and brine
- dry with materialdried over Na2SO4
- filtrationfiltered
- concentrationconcentrated
- customto give crude
- customThe obtained residue is purified by silica gel flash column chromatography (heptane/EtOAc=100:0 to 0:100)