[obliv]

Hello all!

This will be formed starting from benzene.

I am trying to figure out the reagents needed to get to this product, I was thinking of using friedel crafts but it does not seem to work, I know before the last reagent I will need to use PCC but then after that I do not know what to use to get rid of the OH to make it an aldehyde. Any help would be appreciated! thanks.

[nj_bartel]

Theoretical or real synthesis?

[obliv]

In this case theoretical.

[orgopete]

Well, if you are going to use PCC as the last reagent to your final product, what is the intermediate you need to make?

[Schrödinger]

I was thinking of a method, but I thought I'd get certain things clarified first.
When you use the Grignard reagent RMgX, is it okay to have other functional groups in the hydrocarbon part 'R'? Is there any problem associated with this?

[srihari]

@ schrodinger , what exactly do you want ?
I think it generally doesn't matter as long as you are thinking of attacking a
electrophilic carbon .
It matters if there are two sites of attack and you want one selectively
For ex : steric stuff may come into picture.

Is this what you were talking of??

[Dan]

I was thinking of a method, but I thought I'd get certain things clarified first.
When you use the Grignard reagent RMgX, is it okay to have other functional groups in the hydrocarbon part 'R'? Is there any problem associated with this?

If the R group contains a relatively acidic group it will tautomerise.

I like the organometallic strategy. I'd lithiate/metalate benzene and quench it with oxirane.

[Schrödinger]

I am extremely sorry. I was thinking of something completely different. However, thanks for the explanation Dan.

Here's my approach :
1. Benzene to Halobenzene
2. Halobenzene to Phenyl Magnesium halide
3. Addition of glyoxal
4. Selective reduction of alcohol

[srihari]

@ dan conjugation is also necessary na ?
without that I don't think it'll work

[Dan]

@ dan conjugation is also necessary na ?
without that I don't think it'll work

I don't understand your post, conjugation is necessary for what?

You should be able to deprotonate benzene (pKa 43) with MeLi (pKa of methane ~60), or even ^tBuLi, to give PhLi - quenching this organometallic with oxirane should give phenethyl alcohol.

You could also make the organometallic via a halobenzene (Schrödinger's post) by treatment with with the appropriate metal, or by metal-halogen exchange with something like BuLi.

[Schrödinger]

If the R group contains a relatively acidic group it will tautomerise.

Dan, can you please explain this part again? I don't think I understood 'relativley acidic group' and its 'tautomerization'. Can you please give an example so that I'd be able to understand better?

[Dan]

I don't think I understood 'relativley acidic group' and its 'tautomerization'. Can you please give an example so that I'd be able to understand better?

Sure, the point is that carbanions are very strong bases. Say for example you have BrCH₂CH₂CO₂^tBu.

And you want to treat it with Mg to make a Grignard. The proton alpha to the carbonyl is much more acidic than a proton beta to the carbonyl, so the problem is:
^-CH₂CH₂CO₂^tBu --tautomerisation--> CH₃CH^-CO₂^tBu <-> CH₃CH=C(O^-)O^tBu

Perhaps this is more obvious with alcohols, taking BrCH₂CH₂CH₂OH with Mg

^-CH₂CH₂CH₂OH ----> CH₃CH₂CH₂O^-

Does that clarify the point?

[Violet]

I am extremely sorry. I was thinking of something completely different. However, thanks for the explanation Dan.

Here's my approach :
1. Benzene to Halobenzene
2. Halobenzene to Phenyl Magnesium halide
3. Addition of glyoxal
4. Selective reduction of alcohol

If it's theoretical, why not just use epoxide in your step 3, and oxidize the OH in your stpe 4?

[srihari]

@ dan ,

I was thinking about the conjugation required to
have tautomeriztion in big molecules . Going from one
end to another , forgot the simple case
sorry and thanks for the example
my mistake
regards
Srihari

[Schrödinger]

I don't think I understood 'relativley acidic group' and its 'tautomerization'. Can you please give an example so that I'd be able to understand better?

Sure, the point is that carbanions are very strong bases. Say for example you have BrCH₂CH₂CO₂^tBu.

And you want to treat it with Mg to make a Grignard. The proton alpha to the carbonyl is much more acidic than a proton beta to the carbonyl, so the problem is:
^-CH₂CH₂CO₂^tBu --tautomerisation--> CH₃CH^-CO₂^tBu <-> CH₃CH=C(O^-)O^tBu

Perhaps this is more obvious with alcohols, taking BrCH₂CH₂CH₂OH with Mg

^-CH₂CH₂CH₂OH ----> CH₃CH₂CH₂O^-

Does that clarify the point?

Thanks for the explanation Dan.
Just to be sure that I have understood completely :
Suppose I react BrCH₂CHO with Mg, H₂C=CHOMgBr (the enolate form) will be formed predominantly. Am I right?