CHEM 2081 Exam 2

Question 1

Markovnikov’s Rule

Answer 1

Electrophilic addition favors the formation of the more stable carbocation, which is where the electrophile will add

Carbocation rearrangements occur via 1,2-hydride shift or 1,2-methyl shift

Question 2

Cyclic Transition State

Answer 2

When the electrophile has a lone pair, two new E-C bonds form simultaneously from either side of the alkene bond to form a ring, preventing rearrangements

Question 3

Formation of Carbenes

Answer 3

A carbene (two equivalent (R/X/H) groups connected to a carbon with a lone pair - highly reactive) forms via two mechanisms:

1. H2CN₂ (diazomethane) and heat
2. CHCl₃ (chloroform) and base (NaOH)

Adds to molecule via cyclopropanation, pay attention to stereochemistry

Question 4

Formation of Dihalides (Addition of X₂)

Answer 4

Molecular halides first add and form a ring before breaking to add a second anti across the former alkene/alkyne bond, can add twice if 2 equivalents are used on an alkyne

Question 5

Formation of Halohydrins (X and OH)

Answer 5

X₂ and H₂O are used and X is first added to form a ring before breaking to add H₂O to other side, which deprotonates to leave OH and X anti to each other

Question 6

Oxy-Mercuration Reaction

Answer 6

1. Hg(OAc)₂
2. NaBH₄

Follows Markovnikov’s rule, adds OH to more-subbed side of alkene/alkyne

For Alkene: Forms alchol
For Alkyne: Forms Ketone after two tautomerizations

Question 7

H-X Addition

Answer 7

Adds to one side, no stereochemical control if chiral center forms
+enantiomer

Question 8

Epoxidation

Answer 8

Takes place with peroxyacids (e.g., MCPBA) which provide O to form ring, which then treated with either H₂O/H⁺ or H₂O/H^- will open and add two OH groups anti to each other

Question 9

Hydroboration-Oxidation

Answer 9

1. BH₃, THF (acid)
2. H₂O₂, NaOH

anti-Markovikov (adds OH to least-subbed C of alkene, forms aldehyde when alkyne)

Question 10

Terpene Biosynthesis

Answer 10

Occurs by adding isoprene (5-carbon) units together

Question 11

Addition to Conjugated Dienes

Answer 11

Occurs via 1,2 (kinetic control - cold) or 1,4 (thermodynamic control - heat) mechanism with resonance

(major product determined by rxn conditions)

Question 12

Conjugated pi Systems

Answer 12

Resonance structures disperse electron density across 3 carbons, all p orbitals must be parallel (sp²)

Thermodynamically stabilizes molecule with pi system

Question 13

Hückel’s Rules for Aromaticity

Answer 13

1. An aromatic compound is: cyclical, planar (all carbons sp²), has a completely conjugated pi bond system, meets pi electron requirement of 4n+2
2. An antiaromatic species has a pi electron count of 4n
3. All other species are nonaromatic

8+ membered rings are often nonaromatic due to folding (not planar)

Question 14

Aromaticity of Large Rings and Heterocycles

Answer 14

Large/Multiple rings can be aromatic as long as they follow Hückel’s rules, specifically about pi electrons

Heterocycles (e.g., pyridine) can also sometimes be aromatic long as they too follow Hückel’s rules, and the lone pairs of the distinct atom are in the p orbital plane

Question 15

Conjugation and Maximum Wavelength (Lambda) for UV-Vis Spectroscopy

Answer 15

The more conjugated a molecule (more pi bonds it possesses), the greater the maximum wavelength will be

Question 16

Acid-Catalyzed Hydration Reactions

Answer 16

H₂O, H₂SO₄ (cat.) leads to OH adding on more-subbed carbon

CH₃OH, H₂SO₄ (cat.) leads to ester adding on more-subbed carbon

Question 17

Aromatic Rings in SN₁ Reactions

Answer 17

Strong LG increases stability, strengthens aromaticity of molecule, makes it ideal for substitution