Diethyl Ether → Erlotinib Hydrochloride

HRID 1184635Notes

Reaction equation

HRID 1184635 reaction equation
[CH3:1][O:2][CH2:3][CH2:4][O:5][C:6]1[CH:7]=[C:8]2[C:20]([NH:21][C:22]3[CH:23]=[CH:24][CH:25]=[C:26]([C:28]#[CH:29])[CH:27]=3)=[N:19][CH:18]=[N:17][C:9]2=[CH:10][C:11]=1[O:12][CH2:13][CH2:14][O:15][CH3:16].C(OCC)C.[ClH:35].C(OCC)C>C(Cl)(Cl)Cl>[CH3:1][O:2][CH2:3][CH2:4][O:5][C:6]1[CH:7]=[C:8]2[C:20]([NH:21][C:22]3[CH:23]=[CH:24][CH:25]=[C:26]([C:28]#[CH:29])[CH:27]=3)=[N:19][CH:18]=[N:17][C:9]2=[CH:10][C:11]=1[O:12][CH2:13][CH2:14][O:15][CH3:16].[ClH:35]

Participants

5

Reactant 3

HCID 3283Diethyl EtherHCID 176870ErlotinibHCID 10953452Ether hydrochloride

Product 1

HCID 176871Erlotinib Hydrochloride

Reagents, catalysts & solvents 1

HCID 6212ChloroformSolvent

Procedure

Erlotinib free base (5 gm, obtained in reference example 3) is dissolved in chloroform (200 ml) at 25-30° C. to form a clear solution and then added diethyl ether (50 ml). To the resulting solution slowly added 15% diethyl ether HCl (5 ml) at 25-30° C. and stirred for 30 minutes at 25-30° C. Filtered the material, washed with a mixture of diethyl ether (10 ml) and chloroform (10 ml), and then dried at 60-65° C. under vacuum to give 4.9 gm of erlotinib hydrochloride polymorph form A [HPLC purity: 99.7%; Mean particle size (D50): 3.50 μm and 90 volume-% of the particles (D90): 4.61 μm].

Workup

  1. customto form a clear solution
  2. filtrationFiltered the material
  3. washwashed with a mixture of diethyl ether (10 ml) and chloroform (10 ml)
  4. customdried at 60-65° C. under vacuum

Sources & evidence

ORD record
ord-643d4aafd25d4d22b5cf876f527ab72c
Source dataset
uspto-grants-2014_03
Patent number
US08669265B2

Reaction notes

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