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MCAT Sample Questions : Biological Sciences
When Compound 1 is heated with sodium ethoxide, Compounds 2 and 3 are formed (Reaction 1)
Two chemists each proposed a mechanism for Reaction 1.
HBr is eliminated from Compound 1 to form a symmetrical vinyl carbene, Intermediate A, which subsequently rearranges to form Compound 2.
Ethoxide abstracts a proton from Compound 1 to produce a carbanion, Intermediate B, which rearranges with a loss of bromide to form Compound 2.
To distinguish between the 2 mechanisms, the chemists designed an isotopic labeling experiment.They prepared 2 compounds (Compounds 4 and 5) that were labeled with carbon-14 (14C), and they treated each with potassium t-butoxide (see Scheme 1).Because Compounds 4 and 5 formed varying ratios of Compounds 6 and 7, the chemists ruled out Mechanism A.
Following are some sample questions on this passage:
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MCAT Sample Question Number :
- Based on the results of Scheme 1, the chemists most likely ruled out Mechanism A because they assumed that a symmetrical intermediate should rearrange to form:
- Compound 6 only.
- Compound 7 only.
- equal amounts of Compounds 6 and 7.
- neither Compound 6 nor Compound 7.
Explanation: If a carbene was the intermediate, then Compound 4 and Compound 5 would form identical intermediates that would go on to yield identical product distributions.The passage states that the chemists ruled out Mechanism A because Compounds 4 and 5 formed varying ratios of Compounds 6 and 7.Thus, answer choice C is the best answer.
- Compound 3 most likely is formed as a result of which of the following types of reactions?
Explanation: As seen in Reaction 1, OCH2CH3 replaces Br.Thus, answer choice B is the best answer.
- Which of the following by-products are formed in Reaction 1?
- NaBr and EtOH
- HBr and NaOEt
- Br2 and NaOEt
- EtBr and NaOH
Explanation: By either mechanism, ethoxide abstracts a proton to form ethanol, EtOH, and a bromide ion is liberated.The spectator ion, Na+, joins with the Br- to form NaBr.Thus, answer choice A is the best answer.
- Compounds 2 and 6 can be distinguished from each other by all of the following techniques EXCEPT:
- infrared spectroscopy.
- proton NMR.
- gas-liquid chromatography.
- specific rotation.
Explanation: Neither Compound 2 nor Compound 6 is optically active; therefore, the specific rotation of both compounds is zero.Thus, answer choice D is the best answer.
- The transformation of Compound 1 to Compound 2 involves which of the following changes in hybrid orbitals?
- sp2 to sp
- sp2 to sp3
- sp3 to sp2
- sp to sp3
Explanation: Compound 1 contains only sp2 hybridized carbon atoms and does not contain any sp or sp3 hybridized carbon atoms.Compound 2 does not contain any sp3 hybridized carbon atoms but does contain sp and sp2 hybridized carbon atoms.The carbon–carbon double bond that is not part of the two benzene rings in Compound 1 involves sp2 hybrid orbitals.This double bond undergoes transformation in Reaction 1 resulting in a carbon-carbon triple bond in Compound 2.The carbon–carbon triple bond involves sp hybrid orbitals.Thus, Choice A is the best answer.