[flying9412]

As showed below
IN scheme I :The reaction is smoothly completed in  6hs and a high yield is got.

While in schemeII: The same condition is processed,but what is surprise is that no disired product is got even trace amount.NMR-H shows the only product should be the oxidized reactant .So what's the matter with this reaction ?

PS:nitrogen is bubbled for 1h before the reactant is added.

[flying9412]

Can anybody see the picture?

[Borek]

Yes. Please wait patiently.

[jpg28]

I have not really done so much synthesis in the lab yet since I am still a fresh graduate, but I guess the effect of the t-butyl (I'm assuming it's a tert-butyl group, maybe the structure just lacks a carbon atom on both sides) group prevents a back-side attack of the lone pair of electrons on the oxygen atom of the hydroxyl group to the alkyl halide. That's just my input, though. Steric hindrance...

[flying9412]

I have not really done so much synthesis in the lab yet since I am still a fresh graduate, but I guess the effect of the t-butyl (I'm assuming it's a tert-butyl group, maybe the structure just lacks a carbon atom on both sides) group prevents a back-side attack of the lone pair of electrons on the oxygen atom of the hydroxyl group to the alkyl halide. That's just my input, though. Steric hindrance...

I'm sorry for my drawing mistake have made.It really contains two t-butyl groups . And some one else suggested me to change a basis ,and add trace mount of KI as catalyst .I don't know whether a stronger basis will help. I suggest that the electronic drawing of t-butyl groups will increase the charging density around O ,then it should be more easy to process this reaction unless the steric hindrance is too big to prevent attack any other groups.

[jpg28]

Don't worry about the drawing, all of us make mistakes. Still, I'm in no position to say that what I suggested was the correct answer to your problem. Our professor always told us that a reaction might look good on paper, but in reality (doing it in the lab), it may require more or may not even come out as planned.

[flying9412]

Of course，the increased eletronic destiny around O makes it more difficult for basis to capture proton ,so a stronger basis should help with my reaction .I will put it into practice. Thank u guys!

[orgopete]

This appears to be a very interesting result. Perhaps you could help us out by telling us what the product is that you are isolating. It appears that you are saying the alkylation reaction is slowing down and a slower competing reaction is now the preferred path. What is that reaction? I am unable to guess.

[flying9412]

Thank all of u for your attention to my problem .The NMR-H result :7.728 s 2H, 1.568 s 1H(I'm not sure about this signal ,it is sharp ,OH?or just some impurity ), 1.387 s 18H. Alough EI-MS is not detected ,the perchased reactant of Butylated hydroxytoluene from Acros also contains the impurity the same as my product,their TLC spolt and color(yellow) is identical.So I suggest the product should be the oxided reactant.

[jpg28]

Wouldn't the proton on the OH have a higher chemical shift (more downfield due to the electronegativity of the Oxygen)? I'm just curious about it, too; however, you did say that it was a sharp peak. This is puzzling. Could it really be coming from the OH?

[azmanam]

1.5 ppm in chloroform is water. Some potassium carbonate in your chloroform bottle will help some, but 1.5 is pretty common.

[orgopete]

The reaction seems like it should work.

Steric Hindrance as a Factor in the Alkylation of Ambident Anions: The Alkylation of Potassium 2,6-Di-t-butylphenoxide1,2
Nathan Kornblum, Raymond Seltzer
J. Am. Chem. Soc., 1961, 83 (17), pp 3668–3671

[azmanam]

And some one else suggested me to change a basis ,and add trace mount of KI as catalyst. I don't know whether a stronger basis will help.

catalytic iodide is a good idea, but not because it modulates the basicity of the reaction (I think you're saying adding KI will increase the basicity of the reaction).

Iodide is a good nucleophile.  It will displace bromide to make a primary iodide.  Iodide is also a better leaving group than bromide, so it will be displaced by phenoxide faster than the primary bromide.  This is why you add catalytic iodide - not because it modulates the basicity.

And if you're going to add catalytic iodide, you might get better solubility with tetrabutylammonium iodide (TBAI).