[Nekromantis]

Hi,

I must write the simpliest method synthesis 2-methylocyklohexanone from cyklohexanone. My idea is:

http://img839.imageshack.us/img839/8895/synteza.png

Correct?

[Dan]

Theoretically fine, but a couple of points:

1. I would be inclined to use a stronger base to irreversibly generate the enolate and then quench it with an alkyl halide. Can you see why?

2. MeI is a better choice than MeBr. MeI is a liquid at room temperature and 1 atm whereas MeBr is a gas - so MeI is much easier to handle.

[Nekromantis]

Ok, thanks for your help

1. I would be inclined to use a stronger base to irreversibly generate the enolate and then quench it with an alkyl halide. Can you see why?

no, why ?

[nox]

There's more than 1 electrophile in your reaction.

[orgopete]

Aldol?

[Dan]

Overalkylation and aldol condensations would probably be major side reactions.

The issue with using NaOEt is that the ketone-enolate equilibrium lies towards the ketone side. This means that there are always a lot of ketones around that could catch any enolates in an aldol reaction, and also that any 2-methylcyclohexanone that is formed will be exposed to base, enolise, and participate in side reactions.

If a very strong base is used, the enolate will form irreversibly. This would normally be achieved by slowly adding the ketone to the base, ensuring that the base is always in excess so that all the ketone is immediately enolised before any appreciable aldol reactions can take place. This enolate solution can then be added to an excess of MeI.

Side reactions are avoided because you don't have base and MeI present at the same time.

[orgopete]

If enolate alkylation is unfamiliar, then perhaps you have learned about enamine chemistry. 

[Nekromantis]

Thanks for help

But I need your help once againe

I have 2 synthesis and I would like you to check my synthesis.

http://imageshack.us/photo/my-images/809/synteza2.gif/

http://imageshack.us/f/12/synteza3.gif/

[Dan]

http://imageshack.us/photo/my-images/809/synteza2.gif/

A few points:

1. NaOH then H⁺ will not reduce anything, there is no reducing agent there. Can you draw a mechanism for it?

2. LiAlH₄ will probably take out the double bond as well via conjugate reduction. In any case it will probably be a mess. The reduction you propose is normally carried out with a Luche reduction, but...

3. You can't alkylate an allylic alcohol with a cuprate, but you can alkylate an a,b-unsaturated ketone with a cuprate. Given this, it would be more sensible to alkylate first, then reduce.

http://imageshack.us/f/12/synteza3.gif/

Right idea, but there are problems:

1. Your arrows in the first step. You have attacked at one end of the epoxide, but your arrows then show the C-O epoxide bond at the other end cleaving.

2. You have installed a methyl group, but the Grignard reagent you used is ethylmagnesium bromide - with this reagent your product would be 1-phenylbutan-1-ol, not 1-phenylpropan-1-ol as drawn. Keep count of your carbons.

[Nekromantis]

1. NaOH then H+ will not reduce anything, there is no reducing agent there. Can you draw a mechanism for it?

It's my mistake. Probably i saw product where isc group OH not CO and i want enolan ion.

2. LiAlH4 will probably take out the double bond as well via conjugate reduction. In any case it will probably be a mess. The reduction you propose is normally carried out with a Luche reduction, but...

I guess H2/Pt?

1. Your arrows in the first step. You have attacked at one end of the epoxide, but your arrows then show the C-O epoxide bond at the other end cleaving.

I readjusted

2. You have installed a methyl group, but the Grignard reagent you used is ethylmagnesium bromide - with this reagent your product would be 1-phenylbutan-1-ol, not 1-phenylpropan-1-ol as drawn. Keep count of your carbons.

My mistake

http://imageshack.us/photo/my-images/143/synteza3.gif/