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Answers  243
correct about the rate-determining step. The rate law does not indicate anything about the 
number of steps in the mechanism.
228. (A) In aromatic substitution reactions, the aromatic ring acts as a nucleophile due to 
the presence of the double bonds. A dienophile has an electron-withdrawing group conju-
gated to the alkene. This is not the case with the aromatic ring.
229. (C) Conformers refers to rotation around a single bond, enantiomers must have a 
chiral center, and geometric isomers refers to orientation around a double bond or ring. The 
2-nitrotoluene and 4-nitrotoluene are constitutional (structural) isomers.
230. (B) The acid-catalyzed addition of water produces an alcohol. It starts with initial 
protonation of the double bonds and has a carbocation intermediate, with a Markovnikov 
product. It is anti-Markovnikov only if diborane/peroxide is used.
231. (D) The products do form a racemic mixture, since they are enantiomers. The reaction 
will be second-order: first-order in HBr and first-order in 1-butene. There is no carbocation 
intermediate formed. However, it is not possible to separate the enantiomers by fractional 
distillation, since they will have the exact same boiling point. Only some stereoselective 
means of separation may be used.
232. (A) Ozonolysis will cleave all the double bonds, yielding four pieces. However, three 
molecules (formaldehyde) are the same, so only two products will form.
233. (B) The HBr addition to 1-pentene will yield the Markovnikov product, II. Product 
I would be the result of anti-Markovnikov addition. Only product II will form, regardless 
of the temperature.
234. (D) In this reaction, one Br attacks one end of the C=C bond, and then the other Br 
attacks. The second attack can be from either side to give both enantiomers. Therefore, a 
racemic mixture will result.
235. (D) This is a hydrogenation (deuterium, D2, in this case) reaction, and these reactions 
produce a syn addition product.
236. (B) Hydrogenation using Na/ammonia produces a trans-alkene from an alkyne.
237. (A) This is a Friedel-Crafts alkylation to add an ethyl group. The isopropyl group is 
an o,p-director, but steric hindrance at the ortho position means that the para product will 
predominate.
238. (A) This is a Diels-Alder reaction involving a conjugated diene and a dienophile. The 
triple bond becomes a double bond, and the two double bonds lead to one double bond 
between their original positions. The final product should have two carbon-carbon double 
bonds.