Can somebody provide an arrow-pushing mechanism scheme for the acidic hydrolysis of 1-methoxy-1,4-cyclohexadiene to give 2-cyclohexanone?
Everyone in the high school group is confused about this one. Any kind of help will be greatly appreciated. We are not able to figure out how the double bond is shifting.
I am not expert in chemistry. I need to find a way to identify a $\ce{KNO3}$ and $\ce{KH2PO4}$ powder at home. I suspect the seller gave me the wrong powder. What is the easiest way to do it at home?
When 1-methoxycyclohexene is treated with dilute aqueous hydrochloric acid, methanol is produced along with another organic product. Identify this product. Explain why methoxycyclohexane is inert under these reaction conditions.
I don't understand why it isn't just a simple addition to...
According to this, the first step must be attack of H+ on the pi bond. But if I remember well, according to the previous problems I had solved, ether link broken before the H+ attacks the pi bond.
@Loong I guess the ether doesn't cleave in the first step because the solvent is polar protic which would promote SN1 mechanism and that would lead to a highly unstable methyl carbocation. Please correct me if I'm wrong.
What's going on here? I think that the neighbouring N is to blame
Will it be like, both the ketones will become alcohols and they'll form triple bond which will be reduced by LiAlH4 to double (because of the N nearby that could happen) and because that double bond is near an aromatic system it will get reduced again too
@PolarBear Well.. Out of the isolated ketone and amide,the amide is more nucleophilic,so H- should add there to form a coordinated organo-lithium complex(O-Li+) then,the NH2 could,via anchimeric assistance,remove the O-Li+ part to form an alpha lactam which will get hydrolyzed to an amide,and an amide with LiAlH4 will reduce to -CH2-NH2
Sorry.. I meant the ketone is more nucleophilic up there @PolarBear
@Mithoron So how do you suggest that the reduction should take place,to make the product? I thought that the lactam can be made because a 3 membered rings are common in anchimeric assistance.. And it will quickly hydrolyze and break.. That's what I thought, anyway
Formula
C13H8Cl3NO
SMILES
C1=C(C(=CC(=C1)Cl)Cl)N(C(C2=CC=CC=C2)=O)Cl
I found the diagram on the left in a book and drew the one on the right using
https://pubchem.ncbi.nlm.nih.gov/edit2/index.html
And got the SMILES description from that.
Any clues as to what might be an I...
@PolarBear the amide being reduced to an amine is normal with LiAlH4 and you should know the mechanism for that. The ketone being reduced to a CH2, though, is not normal. It has to do with the indole nitrogen participating in the reaction. There is no weird alpha lactam going on.
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I hope this is enough for you to figure out what's going on; if you need more elaboration, I suggest posting it as a question on main site.
I am not expert in chemistry. I need to find a way to identify a $\ce{KNO3}$ and $\ce{KH2PO4}$ powder at home. I suspect the seller gave me the wrong powder. What is the easiest way to do it at home?
When 1-methoxycyclohexene is treated with dilute aqueous hydrochloric acid, methanol is produced along with another organic product. Identify this product. Explain why methoxycyclohexane is inert under these reaction conditions.
I don't understand why it isn't just a simple addition to...
According to Chemspider structure here it is 3,4-Pentadien-2ylbenzene
Formula
C13H8Cl3NO
SMILES
C1=C(C(=CC(=C1)Cl)Cl)N(C(C2=CC=CC=C2)=O)Cl
I found the diagram on the left in a book and drew the one on the right using
https://pubchem.ncbi.nlm.nih.gov/edit2/index.html
And got the SMILES description from that.
Any clues as to what might be an I...
The Periodic Table
The Periodic Table