Well, sorry for the delay, but here is some starting info.
You need to get your hands on this jounal reference first.
Jounal of Organic Chemistry, Vol 43 page 3249 (1978)
Read this and look-up the other references sited.
Basically, they coupled 1-Phenylethanol, C6H5CH(-OH)-CH3 with itself.
The 1-Phenylethanol, C6H5CH(-OH)-CH3 is refluxed in a mixture of titanium trichloride, TiCl3, and 0.33 mole equivalent of lithium aluminium hydride, LiAlH4, in a dimethoxyethane, (CH2OMe)2 solvent.
C6H5CH(-OH)-CH3 ==[LiAlH4/TiCl3 - (CH2OMe)2 solvent {reflux}]==>> C6H5-CH(-CH3)-CH(-CH3)-C6H5
Now, we want to couple benzylalcohol, C6H5CH2OH & ethylaminoalcohol, H2N-CH(-OH)-CH3
C6H5CH2OH + H2N-CH(-OH)-CH3 ==>> C6H5CH2CH(-NH2)-CH3 + 2H2O
ethylaminoalcohol, H2N-CH(-OH)-CH3 is made with acetaldehyde, CH3-CHO + ammonia, NH3 at 0-2 oC
The problem in using the [LiAlH4/TiCl3 - (CH2OMe)2 solvent {reflux}] reaction system:
C6H5CH2OH + H2N-CH(-OH)-CH3 ==[LiAlH4/TiCl3 - (CH2OMe)2 solvent {reflux}]==>> C6H5CH2CH(-NH2)-CH3 + 2H2O
is that ethylaminoalcohol, H2N-CH(-OH)-CH3 when heated looses water, H2O to form CH3CH=NH
H2N-CH(-OH)-CH3 ==HEAT==>> CH3CH=NH + H2O
The H2O reacts with the LiAlH4 causing the reaction to FAIL and the ethylaminoalcohol is no-longer present too couple with the benzylalcohol.
What is needed is [LiAlH4/XFactor - ether solvent {25 oC}]
The LiAlH4/XFactor reactions with the benzylalcohol, C6H5CH2OH & ethylaminoalcohol, H2N-CH(-OH)-CH3, reducing of the -OH and forming radicals.
(The mechanism is far more complex as stated above)
C6H5CH2. + H2N-CH.-CH3 ==>> C6H5CH2CH(-NH2)-CH3
Benzylic & sec-radicals are stable long enough to couple together.
Remember, that side-reaction will occur between C6H5CH2. and H2N-CH.-CH3, so the by-products MUST be remove.
Hope this helps. I don't want to post the entire synthesis, because it just makes it easier for the law to see new amphetamine synthesis at the Hive.
