4 00 4 10 br m 4h 5 55 d 1h 5 65 d 1h 6 05 m 1h 6 60 d 1h 6 65 d 1h 7 00 t 1h 9 55 s 1h 11 65 11 75 b 1h anal calcd for c 15 h 19 no 2 hcl 0 25h 2 o c 62 93 h 7 22 n 4 89 found c 62 53 h 7 48 n 4 71 u03b1 d u2212 115 4 c 0 5 dmso the procedure described for example 5a1 was followed starting from example 2c2 117 mg to give example 5c2 yield 22 mg as a white solid mp dec gt 280 u00b0 c lc ms method 25 rt 0 75 min elsd 99 wv 95 mh 246 3 1 h nmr data identical to the data reported for example 5c1 anal calcd for c 15 h 21 no 2 hcl 0 25h 2 o c 61 96 h 7 28 n 4 82 found c 62 26 h 7 51 n 4 57 u03b1 d 118 4 c 0 5 dmso the procedure described for example 5a1 was followed starting from example 2d1 107 mg to give example 5d1 yield 79 mg as a.