Potassium Iodide → Hydrochloric Acid

HRID 1639099Notes

Reaction equation

HRID 1639099 reaction equation
[N:1]1[CH:6]=[CH:5][CH:4]=[CH:3][C:2]=1[N:7]1[CH2:12][CH2:11][NH:10][CH2:9][CH2:8]1.[Cl:13][CH2:14][CH2:15][CH:16]([OH:27])[CH2:17][O:18][C:19]1[CH:24]=[CH:23][C:22]([Cl:25])=[C:21]([CH3:26])[CH:20]=1.C(=O)([O-])[O-].[Na+].[Na+].[I-].[K+]>CC(O)C.C(O)CCC>[ClH:13].[ClH:13].[Cl:25][C:22]1[CH:23]=[CH:24][C:19]([O:18][CH2:17][CH:16]([OH:27])[CH2:15][CH2:14][N:10]2[CH2:9][CH2:8][N:7]([C:2]3[CH:3]=[CH:4][CH:5]=[CH:6][N:1]=3)[CH2:12][CH2:11]2)=[CH:20][C:21]=1[CH3:26].[ClH:13]

Participants

8

Reactant 4

HCID 4875Potassium IodideHCID 10340Sodium CarbonateHCID 944591-(2-Pyridyl)piperazineHCID 133239084-Chloro-1-(4-chloro-3-methylphenoxy)-2-butanol

Product 2

HCID 313Hydrochloric AcidHCID 14422856HCID 14422856, compound

Reagents, catalysts & solvents 2

HCID 2631-ButanolSolventHCID 3776IsopropanolSolvent

Procedure

This compound was prepared according to the procedure of Example 25. A mixture of 1.6 g (0.01 mole) of 1-(2-pyridyl)piperazine, 2.5 g (0.01 mole) of 4-chloro-1-(4-chloro-3-methylphenoxy)-2-butanol, 5.2 g (0.05 mole) of anhydrous sodium carbonate and 0.1 g of potassium iodide in a total volume of 200 ml of 1-butanol gave a golden oil as residue. The hydrochloride was formed in 2-propanol saturated with hydrogen chloride and the collected solid was recrystallized from methanol-water-ethyl ether to give 2.9 g (65%) of white solid, m.p. >250° C.

Workup

  1. customThis compound was prepared
  2. customgave a golden oil as residue

Sources & evidence

ORD record
ord-2a0066d872a54fe786b640cc06c7e2fb
Source dataset
uspto-grants-1989_11
Patent number
US04882330

Reaction notes

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