133rd one Solved... (With 95% assistance frm smone called govind...(Someone! :D) , not to forget 5% of my own ( i am writing the sol.n [3]))
I- Friedel Craft ALkylation (Ethyl group on benzene)
II- Attach -NH2 at para position.(preferably by para-niration followed by reduction)
III- Convert -NH2 to Diazonium.
IV- Remove Diazonium group by D3PO2/D2O...!!
P.S. Not to forget the commendable presence of memory shown by Mr.Pritish fr his ability to even think of never-b4-dreamt/heard stuff like Deuteration & Mozingo Synthesis (i think Bingo Synthesis wud've been better :P)
13
Avik
·2010-04-01 12:15:28
Here goes the 1st one...
I- Benzene to Toluene (Friedel Cr.)
II- NBS (to form BromoMethyl Benzene)
III- Replacing -Br by -CN.
IV- Stephen's ReaG. or 1eq H2/Pd (to form Ph-CH2.CH=NH)
V- 1eq D2/Pd
VI- NaNO2/HCl (to form Ph-CH2-CHD-OH)
VII-SOCl2
VIII-TPH (TriPhenyl TinHydride---To reduce halide to alkane)
Solution- 85% Govind, 15%Me (5% the usual typing part :P,,, 10% 'coz of a bit of my own mental efforts :) )
106second one
I - benzene to ethyl benzene (friedel crafts)
II - Br2/hv OR NBS (so that one benzylic H gets substituted by Br)
III - Mg/ether (formation of grig. reagent)
IV - D2O
I hope correct
third one:
I - benzene to ethyl benzene (friedel crafts)
II - Br2/Fe (para bromination)
III - Mg/ether (grig. reagent)
IV - D2O
13
Avik
·2010-04-04 11:24:31
Nice ones Asish [1] ...GR is indeed very useful.
An alternate pathway fr second-
I-Ethyl Benzene (FC)
II- NBS
III- LiAlD4
(Govind's suggestion)
