Which of the following substances can exist in an optically active form?
b. ethyl cyclohexane
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Examples of determining optical active forms in a number of organic compounds are given
Monochlorination Products of this Reaction
Organic Chemisty 1 CHEM 307.002
Exam 7 Review (50 points maximum)
1. Please complete the following reactions by supplying the missing reactants or product(s) as indicated by a ? mark. Do not balance and do not show inorganic products or mechanisms. (22 points.)
2. Consider the reaction below to answer the following questions. (8 pts.)
a. Draw all the monochlorination products of this reaction.[Hint: there are four products]
b. Place a circle on the monochlorinated product(s) that have chiral carbon(s).
c. Briefly explain whether this radical chlorination is an oxidation or reduction process.
3. draw structures corresponding to the following names. Show proper number of hydrogens (condensed form, such as CH3 is fine) about each carbon atom except for cyclic structures that may be shown in polygons. (10 pts.)
(1) (E)-2,3-dibromo-2-butene (2) (S)-2-aminopropanoic acid
(3) (2R, 3S)-dibromobutane (Fischer Projection ) (4) 5-ethyl-3,4-dimethyloctane
(5) Determine whether the chiral carbon in the following structure is the R- or S-configuration. Show how priorities are determined.
4. A hydrocarbon A, which is optically active, has the formula C7H12. It absorbs two equivalents of hydrogen H2, gas on catalytic reduction over a palladium catalyst to give an optically active hydrocarbon B with formula C7H16. On partial hydrogenation with one equivalent of H2 gas, it gives optically active (Z)-4-methyl-2-hexene. Upon ozonolysis, of hydrocarbon A, only two products are formed: acetic acid (CH3CO2H), and (R)-2-mehtylbutanoic acid. Propose structures for compound A and B. (10 points).View Full Posting Details