Organic Chemisty 1 CHEM 307.002
Exam 7 Review (50 points maximum)
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).
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
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).© BrainMass Inc. brainmass.com October 10, 2019, 12:16 am ad1c9bdddf
The expert examines monochlorination products of the reaction.