review-questions-for-MCAT-OC-BC-pdf_pag263

Prévia do material em texto

254  Answers
355. (D) Filtration is out; both compounds are liquids. Extraction with either acid or 
base is out; it is difficult to form ions with an alcohol or an ether. Fractional distillation 
will work, since the ether is polar with dipole-dipole forces, but the alcohol has a stronger 
intermolecular hydrogen bond. The difference in the intermolecular forces will make a 
significant difference in their boiling points, with the alcohol having the higher.
356. (A) Since there are no H atoms on the adjacent C, there will be no splitting, and a 
singlet will be observed.
357. (D) The formula CnH2n indicates that the compound has either one ring or one 
double bond. All compounds of more than three carbons with a double bond would show 
more than one NMR peak. Therefore, the compound must be cyclobutane, since all the 
hydrogen atoms would have the same chemical environment.
358. (A) In structure I, there are two types of carbons (those that are attached to an OH 
group and those that are not attached), so there would be two 13C NMR signals. In structure 
II, there are four types of carbons (four signals). In structure III, there are four types of carbons 
(four signals). In structure IV, all the carbons are equivalent, so there would be only one signal.
359. (C) The reaction shown is the oxidation of a secondary alcohol, which yields a ketone. 
The absence of the O-H stretch would indicate that all the alcohol had been converted to 
the ketone.
360. (D) The amine group on III is basic, so extractions with dilute HCl will protonate 
the amine group, forming a water-soluble ion. The other materials do not react with acid, 
so they remain in the organic layer.
361. (A) Titration is not a separation technique, eliminating answer D. There would not 
be enough difference in the properties of the products to use solvent extraction. Paper 
chromatography is useful only for very small amounts of sample. Since there is 100 mL of 
mixture, fractional distillation would be the most logical method.
362. (D) IR absorption in the 1700 cm-1 area is indicative of a C=O bond, which all three 
compounds have.
363. (C) The 3600 cm–1 absorption in the IR spectrum is indicative of an O-H stretch. 
The base (OH–) would be left behind in the separated water layer.
364. (A) Since there are hydrogen atoms with different chemical environments, 1H NMR 
spectroscopy would be the best choice for identifying the various products.
365. (B) The aromatic system has three different hydrogen environments; therefore, there 
will be three resonance signals. (There are no hydrogen atoms on the carbon atoms with 
methyl groups attached.)
366. (A) The average of the two pKa values is 5.4 (= pI). At the pI, the overall charge is 0; 
therefore, the amino acid will not migrate.