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Ochem Passage Questions

Discussion in 'MCAT Study Question Q&A' started by mscjulia, May 7, 2012.

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  1. mscjulia


    Jan 30, 2012
    I'm having difficulties in understanding two questions of the following passage. Could someone here please help me with it?

    The passage is from Princeton Review MCAT Science Workbook Ochem - passage 4 (advanced passage)

    "A research chemist synthesized Compound A, which was shown to be a powerful analgesic upon clinical testing. Further evidence indicated only one stereoisomer of A was the acitve agent. The researcher was instructed to separate Compound A into its different stereoisomers to help determine which one was the analgesic.

    Compound A, which showed an optical rotation of 0 degree, was treated with the R stereoisomer of Compound B (melting point 55C). Two easily separated solids, C and D, whose melting points were 101C and 90C, respectively, were isolated from the reaction mixture. Compound C and D were then independently treated with warm aqueous acid. After workup, Compound C gave Compounds E and B, which Compound D yielded F and B. Testing of Compounds E and F determined that E was the active analgesic, while F was biologically inactive."

    1. Compound A can be described as:

    a. a meso compound
    b. a racemic mixture
    c. an equal mixture of two diastereomers
    d. an optically pure compound

    Answer is B. The explanation given is "from the information given, Compound A is a mixture of two stereoisomers. With an optical rotation of 0 degree, it must be a mixture of enantiomers, which is racemic mixture".

    My concern is, I could tell only from the text that it is a kinda mixture. However, if consider calling it "compound A", shouldn't it be a pure chemical?

    Compound C and D are best described as:

    a. enantiomers
    b. tautomers
    c. diastereomers
    d. rotamers

    Answer is C. Explanation: "Compound C has configuration R, R while compound D has configuration S, R. They are different at only one center and are therefore diastereomers.

    My concern is that I couldn't tell how to figure out the configuration of C and D just based on the passage. I could only guess they are probably diastereomers.

    Thank you very much!
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  3. Morsetlis

    Morsetlis I wish I were a dentist 7+ Year Member

    Jan 22, 2010
    The "Garden" State
    Just because you all it "dopamine" doesn't mean it doesn't have both L- and R-DOPA.

    The question is basically: You react A with R-B, then some further reactions, and yield E, B, and F. How did you start with only A but ended up getting both E and F, which are different? If A were a mesocompound (0 rotation amongst different stereoisomers with non-0 rotations), treatment with only 1 other reagent (R-B) would not have yielded 2 different reagents at the end. Therefore, A must be a racemic mixture. If A were diastereomers, the optical rotation wouldn't be 0. Optically pure means it's 1 stereoisomer, not a mixture of two different ones.

    C was obtained by mixing one enantiomer of A with R-B. D was obtained by mixing the opposite enantiomer of A with R-B. Unless B had the same optical rotation as one of A, adding a non-180/360 rotation to your two enantiomers will yield diastereomers. Rotamers involve rotation around a single bond and are not stereoisomers.
  4. mscjulia


    Jan 30, 2012
    Thank you so much! Really appreciate your help!
  5. Hellovv14


    Apr 23, 2011
    I am concerning the first question: base on the solution, the answer is B, because by definition, an equal mixture of two diastereomers is racemic mixture. But isn't that is option C?! So why not c?

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  6. Nezeru

    Nezeru 2+ Year Member

    May 23, 2012
    A racemic mixture is a mixture of equal amounts of L and R enantiomers of a chiral molecule (non-superimposable mirror images), not a mixture of diastereomers, which differ at one or more chiral center and are not non-superimposable mirror images of each other.

    I think the only way you can tell between choices A and B is by seeing that you get two different products later on (for the first question)
  7. gettheleadout

    gettheleadout MS-3 Moderator Emeritus 7+ Year Member

    Jun 22, 2010
    As to why choice A is incorrect on the first question, Morsetlis' explanation is correct. If Compound A were meso, then it would not be possible to separate it into different enantiomers. In fact, it is impossible to have a racemic mixture of a meso compound because no enantiomers exist, so A and B are mutually exclusive. B is the only choice that satisfies any of the criteria other than optical rotation of 0.

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