[blueblueblue]

I need to come up with a synthesis of the picture below.  From the first structure to the last structure, I came up with this synthesis.  The difficult part was the formation of the double bond against Zaitsev's rule in the second structure.

I am thinking since tertiary butoxide is a hindered strong base, the tertiary alkyl halide will form a double bond to the beta carbon with the more hydrogen attached to it.

Please give me any feedback.

Thanks in advance for your help and taking the time to look at this.

[Vidya]

your approach is right
just one small error in second last step
it should be an enol (double bond with OH )

[OC pro]

The selectivity of the elimination will be problematic. There will be isomers of course. But the rest looks ok, though it would not be the preferred synthesis sequence to the target molecule.

[science123]

You can go from structure 2 to 5 using BH3.THF followed by H2O2/-OH and then use PCC to get to final structure.

[cundi]

Other posibility:
1.- Eliminations. As yours. (regioselectivity could be a problem)
2.- Hydroboration of the double bond
3.- Swern oxidation triggers the targeted molecule

[AndersHoveland]

forgive me,  but this seems suspiciously like you are trying to make an illicit ephedrine analogue.

However, can I suggest a different route? Phenyl nitromethane can be made by reacting toluene with nitrogen dioxide, at a temperature as low as 20degC (though the reaction proceeds more readily at 140degC, although then there is also phenyldinitromethane produced).

Phenyl nitromethane can then condense with a limited quantity of formalin CH2O in a Henry reaction (warmed with a base). Thus 1-nitro,1-phenyl,2-hydroxy-ethane will be obtained. Now protect the hydroxy group by forming an ether. The acidity will also cause the nitro group to undergo a Meyer reaction, becoming a ketone group.
Now use a selective reducing agent; you are no doubt familiar with several which would work.

Thus something similar to PhCH2CH2OCH3 should be obtained. Hydrolyze off that ether with some 20%HCl solution,
then oxidize the hydroxyl group with pyridium chlorochromate or 2-Iodoxybenzoic acid (many other common oxidizers could be used, but suffer from lower yields) to obtain the final aldehyde product.
PhCH2CH=O

Some information about 2-Iodoxybenzoic acid: it can oxidize methanol to formaldehyde in 94% yield, and can similarly oxidize ethylene glycol (vehicle anti-freeze) to glyoxal. However, dimethyl sulfoxide (DMSO) can not be used as a solvent for the latter, as its pressence will cause the ethylene glycol to be oxidized to formaldehyde instead. The 2-Iodoxybenzoic acid can then be re-oxidized and recycled after completion of the reaction.
2-Iodoxybenzoic acid can be prepared by the slow addition, over a half hour, of potassium bromate (76.0 g, 0.45 mol) to a vigorously stirred sulfuric acid mixture (0.73 M, 730 mL) containing 2-iodobenzoic acid (85.2 g, 0.34 mol).

Hopefully you find some of this helpful.