[yvblake]

Please help me to get this problem. I am in undergraduate first semester O-Chem.

We have learned alkyl halides; SN/E; basic alcohols, epoxides, and ethers; alkenes; alkynes; and just started reduction and oxidation.

Here is a screenshot of the problem.



Thanks in advance for helping. If you solve it, can you please put your thought process for solving this problem too?

[sjb]

Please help me to get this problem. I am in undergraduate first semester O-Chem.

We have learned alkyl halides; SN/E; basic alcohols, epoxides, and ethers; alkenes; alkynes; and just started reduction and oxidation.

Here is a screenshot of the problem.
to


Thanks in advance for helping. If you solve it, can you please put your thought process for solving this problem too?

What are your thoughts? What are the key reactions you'll need to do?

[yvblake]

Thanks for your reply.


My first thought was like this: 

1. Deprotinate with NaH
2. Add 2 Carbons by nucleophilic substitution  
3. Deprotinate with NaH again 
4. Add 2 Carbons by nucleophilic substitution  

And hydrohalogenaiton with 2 equivalents of HBr would equal products of 3,3-Dibromohexane and 4,4-Dibromohexane

I just cannot figure out how to form a reaction with 3,3-Dibromohexane as a major product.

Second attempt, thoughts were like this-

Twice Halogenation of Ethyne followed by twice Elimination will not work because I can't think of any useful reactions with Dibromoethyne

Halogenation of Ethyne followed by Hydrohalogenation of 1,1-Dibromoethene will produce 1,1,2-Tribromoethane. Then twice elimination will produce Bromoethyne. Again this product does not seem useful to me because even after deprotinating, adding four carbons, having a bromine on the terminal end does not do me any more good than not having it there.

If I knew of some way to assemble carbon chains with alkenes and could produce 3-bromo-3-hexene, then a bromine could easily be added to the third carbon through hydrohalogenation.

So these are my thoughts so far.

Thanks again for reading and helping.

[sjb]

Can you draw 3,3-dibromohexane, and 4,4-dibromohexane?

[yvblake]

Oh, no you can't because they are actually the same! Wow, I was trying to make that so much harder than it actually was. Thanks for your help, sjb!

[kriggy]

You have mistake in step 4. Missing one carbon