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Answers  253
344. (A) Base removes hydrogen ions, leaving an overall negative charge, while acid adds 
hydrogen ions, leaving an overall positive charge. Positive ions move toward the negative 
plate, and negative ions move toward the positive plate.
345. (B) The less polar compound will migrate farther (larger Rf value) than the more polar 
compound. Molecular mass is much less important to TLC.
346. (D) The change in position indicates a change in the carbonyl region of the com-
pound. HCl hydrolysis will convert an ester into a carboxylic acid with retention of a 
carbonyl peak. The ether has no carbonyl peak. Neither a ketone nor a carboxylic acid will 
change under these conditions.
347. (A) This is the iodoform test for a methyl ketone.
348. (D) Lithium aluminum hydride converts carbonyl groups to alcohols. Therefore, the 
increase by four hydrogen atoms indicates that the two carbonyl groups underwent reduc-
tion. No C=C underwent reduction. There is one degree of unsaturation that could be a 
double bond or a ring. The question asked for the maximum number of rings; therefore, 
there is one ring.
349. (A) Homolytic cleavage produces two free radicals. This is the initiation step in a 
free radical mechanism. A propagation step involves the formation of one free radical from 
another free radical. Instigation and proliferation steps do not exist.
350. (D) There are two types of carbons present here: six secondary and two tertiary. All of 
the secondary carbon atoms are equivalent, as are the tertiary carbons. Therefore, the NMR 
spectra should yield two peaks.
351. (C) Spin-spin coupling occurs when hydrogen atoms on one atom interact magneti-
cally with different hydrogen atoms on an adjacent atom. In Compounds A and D, there 
are no hydrogen atoms on adjacent atoms. In Compound B, all the hydrogen atoms are 
equivalent, so there will be no coupling. In Compound C, there are hydrogen atoms on 
adjacent carbon atoms, so coupling should take place.
352. (B) Amines (answer D) show absorption in the 3250–3500 range; alcohols (answer 
C) in the 3200–3650 range; aldehydes (answer A) in the 1690–1750 range; and alkynes 
(answer B) in the 2100–2,260 and 3260–3600 ranges. The correct answer is B, an 
alkyne.
353. (A) Amines are bases; therefore, treatment with a strong acid will protonate the 
amine, forming a cation (ammonium) and making it water soluble.
354. (B) Isobutane has the formula CH(CH3)3. Highly selective bromination produces 
primarily CBr(CH3)3. The methyl hydrogen atoms are equivalent, and there are no hydro-
gen atoms on the adjacent carbon atom to cause splitting.