Organic 2 Review Problem 472

Organic 2 Review Problem 472 - 456 P RACTICAL ORGANIC...

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456 PRACTICAL ORGANIC CHEMISTRY Example of Calculation. Weight of triacetin taken — 1-150 g. Flask A requires ... . 9-15 ml. M. HCl solution Flask B (Control) requires . . . 24-85 ml. ^1/.HCl solution Difference in volume of M. HCl required = 15*70 ml. iooo ml. M.HC1 EE iooo ml. M.NaOH EE i G. MoI. NaOH EE i Acetyl group .'. 15-7 ml. A r .HCl = 1570 ml. M.NaOH = 1570/1000 G. MoI. NaOH = 15*70/1000 Acetyl groups Thus 1-150 g. triacetin contain 15-70/1000 Acetyl groups, 15*70 218 218 g. (i G. MoI.) triacetin contain X Acetyl groups iooo 1-150 = 2-97 Acetyl groups. 2. DETERMINATION OF THE NUMBER OF ACETYL GROUPS IN: (ii) HEXACETYLMANNITOL. C 6 H 8 (O-COCH 3 ) 6 . Molecular Weight, 434. Continue precisely as for triacetin, but since hexacetyl-mannitol (p. 142) is a crystalline compound, weigh out 1-0-1-2 g. of the finely powdered substance, either from a weighing-bottle, or by direct weighing in the flask A. The results are excellent. (iii) PENTACETYLGLUCOSE. C 6 H 7 O(O-COCHg) 5 . Molecular Weight, 390. A modification of the above method is required for acetylated sugars,
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