[yankeekd25]

1. Draw the carbene formed from the above reagents, providing and labeling ALL orbitals present (including types of orbitals) and ALL electrons present (including types of electrons).
 
2.
Draw a possible multi‐ step mechanism by which a carbene is formed from potassium hydroxide and tribromomethane .
 
3.
Draw a possible mechanism by which the Product is formed and identify the product formed in this reaction. 

I don't want the answers right away. If possible, could someone please walk me through how to do this? Thank you.

[yankeekd25]

Can anyone at all please tell me how to get started?

[nj_bartel]

KOH and CHBr3 are the initial reacting species.  Br- is a very good leaving group.  Olefins are electron donating.  Carbenes are electron deficient.

[yankeekd25]

KOH and CHBr3 are the initial reacting species.  Br- is a very good leaving group.  Olefins are electron donating.  Carbenes are electron deficient.

Ok, well I figured out the carbene:

-:CBr2

and the major product formed.
I just do not know how to correctly write the mechanism or if the 2 Br's that are formed on the cyclo portion of the molecule will be cis or trans. I believe cis? Also, the carbonyl group and the ethyl groups remains in the final product right?

[nj_bartel]

1)  Carbenes are neutral molecules.

2) The two Br's are on the same carbon.

3) Ethyl substituent and carbonyl group are unreactive in this case.

[a student]

3) Ethyl substituent and carbonyl group are unreactive in this case.

as you know olefines can do insertion reaction with carbens but what about carbonyls?

[jinclean]

Well, in this reactions the KOH is a stong base to disproton to the tribromethane.then ,the anion will take place a α eliminate  to give a dibromocarbene.The carbene is a very  activity reagent to make an addition to the double bond of the carbon-carbon or carbon-oxide.So in this reaction,you will get two  product probable .See the Reimer–Tiemann reaction, you will know more properties about the carbenes and how to synthesis them