Chem 210 - Sample Exam #5 Solutions

Sample Exams are designed to give you an idea of the types of questions you can expect.  Exam numbers may not exactly agree with currently assigned material, so you may need to examine other sample exams to see questions that might be similar to the upcoming exam. I have made every effort to show correct answers, but there might be an incorrect answer, so if in doubt, you are probably right.

1. Describe simple chemical tests that would distinguish between the following. Tell exactly what you would do and see.

1,3-pentadiene and n-pentane = Br₂ reacts with double bonds, but not with alkanes

1,3-pentadiene and 1-pentene = Br₂ reacts with both, but at different amount; diene reacts with AgNO₃

cyclohexanol and 1-bromocyclohexane = alcohol easily dissolves in sulfuric acid, not the haloalkane

2. When treated with 1 mole equivalent of HCl, 1,3-butadiene yields a mixture of 3-chloro-1-butene and 1-chloro-2-butene. Show why you get both of these two products.  3-chloro-1-butene is produced when one mole of HCl reacts using a carbocation intermediate.  However, 1-chloro-2-butene also forms because a carbocation rearrangment occurs to also produce a more stable substituted double bond (between carbon atoms 2 and 3).

3. Show the complete reaction for the dehydrohalogenation of 4-bromo-1-hexene and show only the major product.

Major product would the conjugated diene, due to stability compared to the non-conjugated diene.

4. Give structures of the chief product or products expected from addition of one mole of HCl to each of the following compounds.

3-cyclohexen-1-ol  = equal quantities of the 3- and 4-chlorocyclohexanol

3-methyl-1,3-butadiene C=C-C(CH₃)₂-Cl

1,4-pentadiene C=C-C-C(Cl)-C

5. Describe how an experiment can be set up to show that E2 elimination occurs anti not syn using a cyclic compound as the starting point.

6. Starting with cyclohexanol, prepare:

cyclohexene Use H₂SO₄ and heat, for elimination

3-bromocyclohexene Use acid to eliminate (dehydrate) then use Br^. free radical to add to allylic carbon

1,3-cyclohexadiene Use acid to dehydrate  the 3-bromocyclohexene above, and base to dehydrohalogenate

7. Starting with cyclopentanol, outline a synthesis protocol to produce stereochemically pure cis-1,2-cyclopentanediol or pure trans-1,2-cyclopentanediol.

-- Use H⁺ to dehydrate to the alkene.
-- For cis isomer, use alkaline KMnO₄
-- For trans isomer, use oxirane derivative (from alkene and peroxide) and treat with H⁺ and water 

8. Starting with ethene, show how you would produce the two-carbon epoxide (ethylene oxide). Show the product for reaction of this epoxide (H⁺ catalysis) with (a) water, (b) ethanol, and (c) HBr.

(a) ethene + peroxide, then water and H⁺ to produce ethanediol

(b) oxirane + ethanol to produce diethyl ether

(c)  oxirane + Br^- to produce 2-bromoethanol

9. Show the complete reaction mechanism for the reaction of an epoxide with water.

epoxide (oxirane) reacts with water in presence of H⁺ to produce a diol

10. Give the principal organic product for each of the following reactions:

cyclopentane + Cl₂ (300 ^oC) chlorocyclopentane

cyclopentene + Br₂ (in CCl₄) 1,2-dibromocyclopentane

cyclopentene + Br₂ (300 ^oC) 3-bromo-1-cyclopentene

cyclopentene + cold KMnO₄ 1,2-cyclopentanediol

11. Show the reaction products of propene with (a) HCl(aq), (b) HBr(aq) + peroxides, (c) Cl₂ (low temp, CCl₄), and (d) Cl₂ (500 ^oC, gas phase).

(a) 2-chloropropane

(b) 1-bromopropane

(c) 1,2-dichloropropane

(d) 3-chloro-1-propene

12. Explain why the allyllic free radical is much easier for form, and is more stable, than even the tertiary free radical. Conversely, why is the vinyl free radical so hard to form, even being harder to form than the methyl free radical.

-- Ease of formation resonance structures because the allyl free radical is extremely stable
-- Difficulty in forming any type of resonance structure because it becomes extremely electron rich

13. NBS (n-bromosuccinimide) is a reagent used for the specific purpose of brominating alkenes at the allylic position via free radical substitution. NBS maintains a constant, low concentration of free Br₂.

Show the product(s) you would get by treating 1-octene with NBS. 3-bromo-1-octene and 1-bromo-2-octene

Why do you get allylic rearrangement? because C-3 free radical (allylic free radical) can resonant with C-1.  Therefore Br^. can add to either C-1 or C-3 free radical to produce the 3-bromo- or 1-bromo- products listed.

14. Draw resonance structures (if they exist) for (a) the allyl free radical, (b) allyl carbocation, and (c) 1,4-pentadiene.

(a) C=C-C. <-> .C-C=C

(b) C=C-C+ <-> +C-C=C

(c) C=C-C-C=C (no resonance possible because it has isolated double bonds)

EXTRA CREDIT (4 points): Do either part A or part B. If you do both, you will be graded on part A.

A. Starting with isoprene, show the mechanism for free radical polymerization. Then, show the structure of the polymer, which is latex, or natural rubber. [-C-C(CH₃)=C-C-]_n

B. When 1,4-hexadien-3-ol is dissolved in H₂SO₄, it is converted completely into 3,5-hexadien-2-ol. How do you account for this? resonance stabilization of allylic free radical forms conjugated diene