Step 2: alpha-bromo (o-chlorophenyl)-cyclopentyl ketone.
To 21.0 g of the above ketone is added 10.0 g of bromine in 80 ml of carbon tetrachloride dropwise at 0 °C. After all the Br2 has been added, an orange suspension forms. This is washed with a dilute aqueous solution of sodium bisulfite (~65 mL x2 of 10% solution) and evaporated to give 1-bromocyclopentyl-(o-chlorophenyl)-ketone, bp 111-114 °C (0.1 mm Hg). Yield is ~66%. This bromoketone is unstable and must be used immediately. Also, attempts to distill it at 0.1 mm Hg lead to some decomposition, so it should be used without further purification.
You got reaction mass after reaction and there is excess of Bromine. You add water solution of sodium bisulfite to reaction mass, shake it, divide layers. Repeat procedure. You have to discard water layer (with sodium bisulfite) after washing. NaHSO3 bounds excess bromine from reaction mixture after reaction. Do you understand?
I don't know how to explain more clearly. x2 Means two times as I said before.
you have to pour it in sink or waste water tank.
you have to pour it into a sink or into a waste water tank.
can Ethyl benzoate be used instead of decalin?
You can substitute decalin by cyclohexane, probably
Cyclohexane has a boiling point of 81 °C what would mean the thermal rearrangement would need 7 to 10 days at reflux. But then one could use IPA, same bp and it is known to work from literature (7 to 10 days at reflux).
Ethylbenzoate works.
Ethylene Glycol is another solvent which works and of better availability.
The yield is never near to quantitative, actually when using the Ketimine as base yields suffer, for decent yields the HCl salt must be used. It must also be done under strictly anhydrous conditions or yields suck. (the Imine will hydrolyze back to MeNH2, unsurprisingly regarding the rigorous conditions).
The rearrangement is catalyzed by Lewis Acids, MgCl2 And AlCl3 (anhydrous) work well. 0,1 mol equivalent are needed but AlCl3 can also take care of any water still present, the amount used must be adjusted of course, say X g needed to scavenge the water plus 0,1 mol to catalyze the reaction.
The use of AlCl3 reduces the temperature needed to ~130 °C (to take the same time as 180 °C would take un-catalyzed).
Another tip: Stay away from CCl4 the stuff is not just toxic but its mean and hard to get for good reasons. Gladly it is completely unnecessary for brominating the alpha position. The foolproof way to do this is by using CuBr2 what does the job quantitativly and without side-reactions.
For reaction conditions just Google "Bromination with CuBr2" and you will quickly find the standard conditions for this described and those work perfectly well in this case.
Step 4: 2-Methylamino-2-(o-chlorophenyl)-cyclohexanone (Ketamine).
, and the resulting acidic solution is made basic.
I don't understand your confuse. I already explained. Can you reformulate your question please?
Sir,In the final step,when the reaction boils enough for 3 hours,add HCI and DCM,and remove the solvent,will it crystallize when it becomes an oil?
11. Ketamine
10 g of methylketimine is dissolved in 100 mls undecane and boiled at 195 °C for 3-4 hrs. Ketamine is extracted with 20% HCl. Acidic extract is basified and extracted with DCM. Solvent is removed, giving the product as an oil that quickly crystallizes. It can be purified by recrystallization from pentane/ether or hexane/ether. The yields are close to quantitative.
What is written here?
It is simple acid/base extraction. Please, read several times topic Extraction, if you don't understand process. There is video explanation. You have to make acid space there, extract with ether (75 ml). Then, basify mixture and extract again (75 ml).
You got reaction mass after reaction and there is excess of Bromine. You add water solution of sodium bisulfite to reaction mass, shake it, divide layers. Repeat procedure. You have to discard water layer (with sodium bisulfite) after washing. NaHSO3 bounds excess bromine from reaction mixture after reaction. Do you understand?
10. 1-hydroxy-cyclopentyl-(o-chlorophenyl)-N-methylketimine
45 g of the above bromoketone is dissolved in 50 mls benzene, add therein 50mls triethylamine (17 g/23 mL is required for neutralization of HBr, but a 2x excess is used). The soln. is then saturated with 5 g methylamine, obtained by dripping a saturated soln of 15 g MeNH2·HCl onto 10 g NaOH, dried thru NaOH. The rxn is left for 1 day and the solvents are removed under aspirator vacuum, giving 30 g (80%) of methylketimine.
Sir,at this step,Is the product underneath?
Yes, you have to stir this reaction too without heating.Sir,Do I need to heat/stir in this part?
What do you mean?How to show saturation?
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