Opening the book…
Additional Problems
Visualizing Chemistry
PROBLEM 1-16 Show the structures of all possible adducts of the following diene with 1 equivalent of HCl:
PROBLEM 1-17 Show the product of the Diels–Alder reaction of the following diene with 3-buten-2-one, H2CCHCOCH3. Make sure you show the full stereochemistry of the reaction product.
PROBLEM 1-18 The following diene does not undergo Diels–Alder reactions. Explain.
PROBLEM 1-19 The following model is that of an allylic carbocation intermediate formed by protonation of a conjugated diene with HBr. Show the structure of the diene and the structures of the final reaction products.
Mechanism Problems
PROBLEM 1-20 Predict the major product(s) from the addition of 1 equivalent of HX and show the mechanism for each of the following reactions.
(a) (b)
(c)
PROBLEM 1-21 We’ve seen that the Diels–Alder cycloaddition reaction is a one-step, pericyclic process that occurs through a cyclic transition state. Propose a mechanism for the following reaction:
PROBLEM 1-22 In light of your answer to Problem 21 propose mechanisms for the following reactions.
(a)
(b)
PROBLEM 1-23 Luminol, which is used by forensic scientists to find blood, fluoresces as a result of Diels–Alder-like process. The dianion of luminol reacts with O2 to form an unstable peroxide intermediate that then loses nitrogen to form a dicarboxylate and emit light. The process is similar to that in Problems 21 and 22. Propose a mechanism for this process.
PROBLEM 1-24 A useful diene in the synthesis of many naturally occurring substances is known as Danishefsky’s diene. It’s useful because after the Diels–Alder reaction it can be converted into a product that can't be accessed by a typical Diels–Alder reaction. Show the Diels–Alder adduct and propose a mechanism that accounts for the final products.
Conjugated Dienes
PROBLEM 1-25 Name the following compounds:
(a) (b)
(c) (d)
(a) 1 mol Br2 in CH2Cl2 (b) O3 followed by Zn (c) 1 mol HCl in ether (d) 1 mol DCl in ether
(e) 3-Buten-2-one (H2C=CHCOCH3) (f) Excess OsO4, followed by NaHSO3
PROBLEM 1-28 Electrophilic addition of Br2 to isoprene (2-methyl-1,3-butadiene) yields the following product mixture:
Of the 1,2-addition products, explain why 3,4-dibromo-3-methyl-1-butene (21%) predominates over 3,4-dibromo-2-methyl-1-butene (3%).
PROBLEM 1-29 Propose a structure for a conjugated diene that gives the same product from both 1,2 and 1,4-addition of HBr.
PROBLEM 1-30 Draw the possible products resulting from addition of 1 equivalent of HCl to 1-phenyl-1,3-butadiene. Which would you expect to predominate, and why?
Diels–Alder Reactions
PROBLEM 1-31 Predict the products of the following Diels–Alder reactions:
(a) (b)
PROBLEM 1-32 2,3-Di-tert-butyl-1,3-butadiene does not undergo Diels–Alder reactions. Explain.
PROBLEM 1-33 Show the structure, including stereochemistry, of the product from the following Diels–Alder reaction:
PROBLEM 1-34 How can you account for the fact that cis-1,3-pentadiene is much less reactive than trans-1,3-pentadiene in the Diels–Alder reaction?
PROBLEM 1-35 Would you expect a conjugated diyne such as 1,3-butadiyne to undergo Diels–Alder reaction with a dienophile? Explain.
PROBLEM 1-36 Reaction of isoprene (2-methyl-1,3-butadiene) with ethyl propenoate gives a mixture of two Diels–Alder adducts. Show the structure of both, and explain why a mixture is formed.
PROBLEM 1-37 Rank the following dienophiles in order of their expected reactivity in the Diels–Alder reaction.
PROBLEM 1-38 1,3-Cyclopentadiene is very reactive in Diels–Alder cycloaddition reactions, but 1,3-cyclohexadiene is less reactive and 1,3-cycloheptadiene is nearly inert. Explain. (Molecular models are helpful.)
PROBLEM 1-39 1,3-Pentadiene is much more reactive in Diels–Alder reactions than 2,4-pentadienal. Why might this be?
PROBLEM 1-40 How could you use Diels–Alder reactions to prepare the following products? Show the starting diene and dienophile in each case.
(a) (b)
(c)
(d)
PROBLEM 1-41 Show the product of the following reaction.
Diene Polymers
PROBLEM 1-42 Tires whose sidewalls are made of natural rubber tend to crack and weather rapidly in areas around cities where high levels of ozone and other industrial pollutants are found. Explain.
PROBLEM 1-43 1,3-Cyclopentadiene polymerizes slowly at room temperature to yield a polymer that has no double bonds except on the ends. On heating, the polymer breaks down to regenerate 1,3-cyclopentadiene. Propose a structure for the product.
UV Spectroscopy
PROBLEM 1-44 Arrange the molecules in each of the following sets according to where you would expect to find their wavelength of maximum absorption in UV spectroscopy, from shortest to longest wavelength.
(a)
(b)
(c)
PROBLEM 1-47 The following ultraviolet absorption maxima have been measured:
| Molecule | ultraviolet absorption maxima |
| 1,3-Butadiene | 217 nm |
| 2-Methyl-1,3-butadiene | 220 nm |
| 1,3-Pentadiene | 223 nm |
| 2,3-Dimethyl-1,3-butadiene | 226 nm |
| 2,4-Hexadiene | 227 nm |
| 2,4-Dimethyl-1,3-pentadiene | 232 nm |
| 2,5-Dimethyl-2,4-hexadiene | 240 nm |
What conclusion can you draw about the effect of alkyl substitution on UV absorption maxima? Approximately what effect does each added alkyl group have?
PROBLEM 1-48 1,3,5-Hexatriene has λmax = 258 nm. In light of your answer to Problem 14-47, approximately where would you expect 2,3-dimethyl-1,3,5-hexatriene to absorb?
PROBLEM 1-49 β-Ocimene is a pleasant-smelling hydrocarbon found in the leaves of certain herbs. It has the molecular formula C10H16 and a UV absorption maximum at 232 nm. On hydrogenation with a palladium catalyst, 2,6-dimethyloctane is obtained. Ozonolysis of β-ocimene, followed by treatment with zinc and acetic acid, produces the following four fragments:
- (a) How many double bonds does β-ocimene have?
- (b) Is β-ocimene conjugated or nonconjugated?
- (c) Propose a structure for β-ocimene.
- (d) Write the reactions, showing starting material and products.
General Problems
PROBLEM 1-50 Draw the resonance forms that result when the following dienes are protonated. If the resonance forms differ in energy, identify the most stable one.
(a) (b)
(c)
PROBLEM 1-51 Answer the following questions for 1,3,5-cycloheptatriene.
- (a) How many p atomic orbitals are in the conjugated system?
- (b) How many molecular orbitals describe the conjugated system?
- (c) How many molecular orbitals are bonding molecular orbitals?
- (d) How many molecular orbitals are anti-bonding molecular orbitals?
- (e) Which molecular orbitals are filled with electrons?
- (f) If this molecule were to absorb a photon of UV light an electron would move between which two molecular orbitals (be specific)?
PROBLEM 1-52 Treatment of 3,4-dibromohexane with strong base leads to loss of 2 equivalents of HBr and formation of a product with formula C6H10. Three products are possible. Name each of the three, and tell how you would use 1H and 13C NMR spectroscopy to help identify them. How would you use UV spectroscopy?
PROBLEM 1-53 Addition of HCl to 1-methoxycyclohexene yields 1-chloro-1-methoxycyclohexane as a sole product. Use resonance structures to explain why none of the other regioisomer is formed.
PROBLEM 1-54 Aldrin, a chlorinated insecticide now banned from use in most countries since 1990, can be made by Diels–Alder reaction of hexachloro-1,3-cyclopentadiene with norbornadiene. What is the structure of aldrin?
PROBLEM 1-55 Norbornadiene (Problem 54) can be prepared by reaction of chloroethylene with 1,3-cyclopentadiene, followed by treatment of the product with sodium ethoxide. Write the overall scheme, and identify the two kinds of reactions.
PROBLEM 1-56 The triene shown here reacts with 2 equivalents of maleic anhydride to yield a product with the formula C17H16O6. Predict a structure for the product.
PROBLEM 1-57 Myrcene, C10H16, is found in oil of bay leaves and is isomeric with β-ocimene (Problem 49). It has an ultraviolet absorption at 226 nm and can be hydrogenated to yield 2,6-dimethyloctane. On ozonolysis followed by zinc/acetic acid treatment, myrcene yields formaldehyde, acetone, and 2-oxopentanedial:
Propose a structure for myrcene, and write the reactions, showing starting material and products.
PROBLEM 1-58 Hydrocarbon A, C10H14, has a UV absorption at λmax = 236 nm and gives hydrocarbon B, C10H18, on hydrogenation. Ozonolysis of A, followed by zinc/acetic acid treatment, yields the following diketo dialdehyde:
An illustration shows the structure of diketo dialdehyde. It shows two carbonyl groups single bonded to each other. Each carbonyl group is bonded to a chain of three methylene groups and then to an aldehyde group.
- (a) Propose two possible structures for A.
- (b) Hydrocarbon A reacts with maleic anhydride to yield a Diels–Alder adduct. Which of your structures for A is correct?
- (c) Write the reactions, showing the starting material and products.
PROBLEM 1-59 Adiponitrile, a starting material used in the manufacture of nylon, can be prepared in three steps from 1,3-butadiene. How would you carry out this synthesis?
PROBLEM 1-60 Ergosterol, a precursor of vitamin D, has λmax = 282 nm and molar absorptivity ϵ = 11,900. What is the concentration of ergosterol in a solution whose absorbance A = 0.065 with a sample pathlength l = 1.00 cm?
PROBLEM 1-61 Dimethyl butynedioate undergoes a Diels–Alder reaction with (2E,4E)-2,4-hexadiene. Show the structure and stereochemistry of the product.
PROBLEM 1-62 Dimethyl butynedioate also undergoes a Diels–Alder reaction with (2E,4Z)-2,4-hexadiene, but the stereochemistry of the product is different from that of the (2E,4E) isomer (Problem 61). Explain.
PROBLEM 1-63 How would you carry out the following synthesis (more than one step is required)? What stereochemical relationship between the –CO2CH3 group attached to the cyclohexane ring and the –CHO groups would your synthesis produce?
PROBLEM 1-64 The double bond of an enamine (alkene + amine) is much more nucleophilic than a typical alkene double bond. Assuming that the nitrogen atom in an enamine is sp2-hybridized, draw an orbital picture of an enamine, and explain why the double bond is electron-rich.
PROBLEM 1-65 Benzene has an ultraviolet absorption at λmax = 204 nm, and para-toluidine has λmax = 235 nm. How do you account for this difference?
This page titled 1.9: Additional Problems was last modified on Mon, 17 Aug 2026 23:03:13 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Sol Parajon Puenzo (Cañada College) via source content that was edited to the style and standards of the LibreTexts platform.
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