Roadmap #3 question

[Jo07]

The reaction on the right side, one over from the start...
CH3
C C C C C The arrows that are going up and down from that reactant...
Br Are they both Elimination rxns? E1 because they are in polar protic solvents--alcohols? I'm confused as to the locations of the double bonds.

Can anyone clear this up for me?

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[Jo07]

The lower one is E2?

 

[TheWiredNerv]

If it is a very stericly hindered base it'll be an E2 reaction but it'll yield the less substituded alkene. If you use a non bulky base like CH3CH20- then this E2 reaction will yield the more substituded alkene.

 

[Jo07]

One more question...

On Roadmap 2, the middle right side... LiAlH4, Et2O is being used to reduce that compound. Why does it reduce the =O completely, instead of to an -OH? This is compared to Roadmap 1 where in the rxn, middle right side, it reduces COOH to CH3OH.

Any help is appreciated!

 

[Jo07]

anyone?

 

[TheWiredNerv]

In the reaction in question, LiAlH4 is reducing an amide. In reduction of amides with LiAlH4 you lose the carbony group as CO2 and the R group connects to the nitrogen group and you generate a primary amine.

In roadmap 1, LiAlH4 is reducing a carboxylic acid which yields alcohols.
In a nutshell, LiAlH4 reacts differently with amides than with pure carbonyl compounds.

 

[Jo07]

Thanks so much