Image Matters o-Hydroxyacetophenone has a melting point of 4-6 degrees C. p-Hydroxyacetophenone has a melting point of 109-111 degrees C! Explain the reason for the considerable difference between the melting points of these two compounds. When performing TLC under the following conditions, you get suboptimal results. Consider the situation in each case and suggest a correction: you run TLC on a mixture of two unknown halogenated alkenes, but you see only one point with an Rf of 0.91. The solvent used was ethyl acetate. You run TLC on a mixture of thiol and amine and again only get a single point with an Rf of 0.15. The solvent was a mixture of petroleum ether and dichloromethane. When you place your TLC plate in the development chamber, the solvent in the chamber covers the dots on the baseline of the TLC plate. You run a TLC plate stained with three compounds: naphthalene, o-toluic acid, and fluorenol. Predict relative Rf values. For each of the following pairs, predict which compound will have the highest Rf value if both are run on a TLC SiO2 plate in 10% acetone/hexane: 4-decanone or 4-decanol; xylene or benzoic acid; cycloheptane or cycloheptanone. Arrange the following solvents in order of increasing polarity: ethyl acetate – dichloromethane – n-propanol – ethanol – toluene – heptanes Name three important things to remember when placing TLC plates.
1) Since the -OH group and the carbonyl group (CO) of orthohydroxyacetophenone can form intramolecular hydrogen bonds, its melting point is lower. The hydrogen bond between the –OH group and the carbonyl group (CO) of orthohydroxyacetophenone is a stable six-membered ring. However, parahydroxyacetophenone cannot form intramolecular hydrogen bond and can form intermolecular hydrogen bond. Thus, parahydroxyacetophenone molecules interact with other molecules through hydrogen bonding and require greater energy to break free into the liquid (i.e., higher melting point) stationary phase, and because differences in solubility in solvents (a) Halogenated alkenes have the same polarities, so they will show the same Rf value. i.e.: 0.91(b) Similarly, thiols and amines have more polarity (compared to halogenated alkenes), therefore they will have the same Rf values, i.e. say; 0.15 (c) Identify the compound above the eluent level. 3) Rf values in descending order он он naphthalene fluorenol Ó-toluic acid 4) 4-decanone > 4-decanol; benzoic acid > Xylene; Cycloheptanone >Cycloheptane 5) Ethanol > n-propanol >ethyl acetate >dichloromethane >toluene >heptane 6) The dye must not touch the eluent Staining too much solution. The capillary diameter should be around 1/32″ (0.7 to 1 mm). The initial stitch diameter should be 1/8″ to 3/16″ (3 to 5 mm) immediately after application. Too concentrated or inhomogeneous spots of a solution. The solution to be colored should be free of precipitates and insoluble oils and should not be too concentrated. 0.5 to 5% by weight is sufficient. The right dilution is found by trial and error. A volatile solvent is used. There is no need to stain using solutions prepared in DMF, dimethylacetamide, DMSO, or pyridine, to name a few. The staining solvent must be completely removed from the TLC before development. For DMF and pyridine, it is possible to place the colored TLC plate for “5min under vacuum before the run to evaporate the solvent.
1)
ortho has fewer hydrogen bonds due to steric hindrance
minus the melting point.. while for has many hydrogen bonds
then no more melting point.
in case of melting point you only need to see hydrogen
obligation..
5)
ethanol > propanol > toulene > ethyl acetate
> methane
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