2H-Azepin-2-one, hexahydro-1-(hydroxymethyl)- → Formaldehyde

HRID 622152Notes

Reaction equation

HRID 622152 reaction equation
[CH2:1]([N:3]1[CH2:9][CH2:8][CH2:7][CH2:6][CH2:5][C:4]1=[O:10])[OH:2]>[OH-].[Na+].C(O)C>[C:4]1(=[O:10])[NH:3][CH2:9][CH2:8][CH2:7][CH2:6][CH2:5]1.[CH2:1]=[O:2]

Participants

5

Reactant 1

HCID 831272H-Azepin-2-one, hexahydro-1-(hydroxymethyl)-

Product 2

HCID 712FormaldehydeYield 50%HCID 7768Caprolactam

Reagents, catalysts & solvents 2

HCID 14798Sodium HydroxideCatalystHCID 702EthanolSolvent

Procedure

It has already been disclosed that N-methylolcaprolactam can be prepared in a yield of 67% of the theoretical yield from caprolactam and excess paraformaldehyde (50% excess) in 95% strength ethanol in the presence of 4.2 mol % of sodium hydroxide as catalyst (J. Am. Chem. Soc. 70 (1948), 2115-2118). The reaction specified is carried out in approximately 50% strength solution.

Workup

  1. customcan be prepared in a yield of 67% of the

Sources & evidence

ORD record
ord-92168a9cefa441fdb1ef366cce53e80a
Source dataset
uspto-grants-1988_09
Patent number
US04769454

Reaction notes

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