Organic Chemistry Questions. Help 1. Could you predict the sign of the rotation

Question

Organic Chemistry Questions. Help 1. Could you predict the sign of the rotation for the benzamide derivative of (-)-sec-butylamine? Explain. 3. What would happen if you attempted to crystallize the salts formed by mixing (±)-sec-butylamine and (±)-tartaric acid? 4. Give all the equations for all thechemical reactions that occur in the resolution of (±)-s-butylamine w/ pure (+)-tartaric acid 5. Explain why each step in the purification of the (+)-sec-butylamine. In particular, indicate why the distillate is saturated with potssium hydroxide (KOH) before the amine layer is separated. Mass of (+)-tartarci acid used:50g Mass of water used:44g Mass of (+/-)-s-butylamine:23.75g Calculate the percent yield of the resolved amine

Explanation / Answer

he present study deals with the acidbase reaction of three solid-state forms of the nonsteroidal antiinflammatory drug indomethacin with ammonia gas. X-ray powder diffraction, optical microscopy, gravimetry, and spectroscopic methods were employed to establish the extent of the reaction as well as the lattice changes of the crystal forms. The glassy amorphous form readily reacts with ammonia gas to yield a corresponding amorphous ammonium salt. In addition, the metastable crystal form of indomethacin (the -form) also reacts with ammonia gas, but produces the corresponding microcrystalline ammonium salt. This reaction is anisotropic and propagates along the a-axis of the crystals. The stable crystal form (the -form), however, is inert to ammonia gas. Amorphous indomethacin can react with ammonia gas because it has more molecular mobility and free volume. The reactivity differences between the - and -forms are dictated by the arrangement of the molecules within the respective crystal lattices. The recently determined crystal structure of the metastable -form of indomethacin (monoclinic P21 with Z = 6, V = 2501.8 Å3, Dc = 1.42 g·cm-3) has three molecules of indomethacin in the asymmetric unit. Two molecules form a mutually hydrogen-bonded carboxylic acid dimer, while the carboxylic acid of the third molecule is hydrogen bonded to one of the amide carbonyls of the dimer. The carboxylic acid groups of the -form are exposed on the {100} faces and are access

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