[toadesque]

Let's say you want to have this as your final product:

      OH
       |
C-C-C-C-C-C-C-C
    |
    C


And you have to start with

C-C=O

What are the steps you need to take to get there? I know this question will be on my exam so I really need to understand it. I think you have to start with Grignard reagent and turn the C=O group into OH right? But how do you get those 6 extra carbons to get the 8 total since you only started with 2?

[toadesque]

anyone got something?

[lancenti]

Could I confirm whether it's an isopropyl group on the left / that it's a secondary alcohol you're after?

If it's a tertiary, then you'll want to use your grignard reagent twice - but remember that you have to 'reactivate' your functional group after the first time.

[orgopete]

This looks like an aldol condensation problem. So, start there.

The hard part is to make 3-methyl-1-butene. Take the aldol product and add (CH₃)₂CuLi to give isobutyraldehyde. Reduce, make bromide, eliminate to 3-methyl-1-butene, and make epoxide from it.

Reduce aldol product catalytically, reduce to alcohol, make bromide, then Grignard. Add to epoxide above and that should do it.

[Scatter]

Jesus.  This is really on an exam as an aldol problem?  Wow.  My Ochem course feels easy now.

[Orgchmartist]

Using acetaldehyde as starting material, two molecular acetaldehyde condensed in the presence of suitable base to yield 1,3 unsaturated aldehyde, then this aldehyde was reduced under HCONH4 Pd/C to obtain butyraldehyde. butyraldehyde obtained from former step was undergo above condensation and reduction again then react with isopropylmagnesium chloride to give your desired compound.

[orgopete]

Let me add a correction to my suggested route. I did not name the aldehyde (it isn't isobutyraldehyde, see scheme) correctly. I think that is what Orgchmartist was correcting.