Organic Chemistry Hydrocarbons: Alkenes, Alkynes, and Aromatics Syllabus

Hydrocarbons: Alkenes, Alkynes, and Aromatics syllabus — full chapter and topic list

Expand any chapter to see its topics and sub-topics. This is the whole examinable outline for Hydrocarbons: Alkenes, Alkynes, and Aromatics in Organic Chemistry, not a summary of it.

1. 1

   Alkenes and Addition Reactions

   6 topics

   • Structure, Stability, and E/Z Nomenclature
   • Electrophilic Addition of HX
   • Acid-Catalyzed Hydration
   • Halogenation and Halohydrin Formation
   • Hydroboration-Oxidation
   • Oxidation: Epoxidation, Dihydroxylation, Ozonolysis
2. 2

   Alkynes

   5 topics

   • Acidity of Terminal Alkynes
   • Addition Reactions of Alkynes
   • Hydration to Ketones and Aldehydes
   • Reduction to Cis and Trans Alkenes
   • Alkylation of Acetylide Anions
3. 3

   Conjugation and Dienes

   3 topics

   • Conjugated vs Isolated Dienes
   • 1,2- vs 1,4-Addition
   • Diels-Alder Cycloaddition
4. 4

   Aromatic Compounds

   3 topics

   • Aromaticity and Huckel's Rule
   • Nomenclature of Benzene Derivatives
   • Aromatic, Antiaromatic, and Nonaromatic Compounds
5. 5

   Reactions of Aromatic Compounds

   4 topics

   • Electrophilic Aromatic Substitution
   • Substituent Effects on Reactivity
   • Nucleophilic Aromatic Substitution
   • Reactions of Side Chains

Hydrocarbons: Alkenes, Alkynes, and Aromatics flashcards for Organic Chemistry

21 of 50 cards from the Hydrocarbons: Alkenes, Alkynes, and Aromatics deck — real questions with worked answers.

1. What is the degree of unsaturation (index of hydrogen deficiency) and how is it calculated for a hydrocarbon $\ce{C_nH_m}$?

   It counts rings plus $\pi$ bonds. For a hydrocarbon: $\text{DoU} = \frac{2n + 2 - m}{2}$. Each double bond or ring = 1; each triple bond = 2.

2. Describe the hybridization, bond angle, and geometry of the carbons in a C=C double bond of an alkene.

   Each alkene carbon is $sp^{2}$ hybridized, trigonal planar, with bond angles of approximately $120^{\circ}$. The $\pi$ bond forms from side-on overlap of unhybridized $p$ orbitals.

3. In the E/Z (Cahn–Ingold–Prelog) system, when is an alkene designated Z versus E?

   Assign priority to the two groups on each double-bond carbon by atomic number. Z (zusammen) = the two higher-priority groups are on the same side; E (entgegen) = they are on opposite sides.

4. State Zaitsev's rule as it relates to alkene stability.

   The more substituted (more highly alkyl-substituted) alkene is generally the more stable and is typically the major product of elimination. Stability order: tetra > tri > di > mono-substituted.

5. Why are trans (E) alkenes generally more stable than cis (Z) alkenes?

   Trans alkenes have less steric strain because the bulky substituents are on opposite sides, minimizing van der Waals repulsion, whereas cis alkenes force them onto the same side.

6. What is Markovnikov's rule for the electrophilic addition of HX to an alkene?

   The hydrogen adds to the carbon that already has more hydrogens, and the halogen (X) adds to the more substituted carbon, so as to form the more stable (more substituted) carbocation intermediate.

7. Give the mechanism steps and regiochemistry for the addition of $\ce{HBr}$ to propene.

   Step 1: the alkene $\pi$ bond attacks $\ce{H+}$ to form the more stable secondary carbocation on C2. Step 2: $\ce{Br-}$ attacks the carbocation. Product: $\ce{CH3CHBrCH3}$ (2-bromopropane, Markovnikov).

8. What conditions cause anti-Markovnikov addition of HBr to an alkene, and why?

   Peroxides (ROOR) with $\ce{HBr}$ trigger a radical mechanism. Br$\cdot$ adds first to the less substituted carbon to form the more stable (more substituted) carbon radical, giving anti-Markovnikov product. Only works for HBr, not HCl/HI.

9. What are the reagents, regiochemistry, and a hallmark of acid-catalyzed hydration of an alkene?

   Reagents: $\ce{H2O}$ with catalytic $\ce{H2SO4}$ (or $\ce{H3O+}$). Markovnikov addition of $\ce{-OH}$ to the more substituted carbon via a carbocation. Because of the carbocation, rearrangements (hydride/methyl shifts) can occur.

10. Compare acid-catalyzed hydration, oxymercuration, and hydroboration-oxidation in terms of alkene regiochemistry.

    Acid-catalyzed hydration and oxymercuration–demercuration give Markovnikov alcohols. Hydroboration–oxidation gives the anti-Markovnikov alcohol ($\ce{-OH}$ on the less substituted carbon).

11. What is the stereochemistry and regiochemistry of halogenation (e.g., $\ce{Br2}$) addition to an alkene?

    Anti addition of the two halogens across the double bond, proceeding through a cyclic bridged halonium ion, giving a trans (vicinal) dihalide (racemic where applicable).

12. In halohydrin formation ($\ce{Br2}/\ce{H2O}$), where do the $\ce{-Br}$ and $\ce{-OH}$ end up?

    Anti addition: water (becoming $\ce{-OH}$) attacks the more substituted carbon of the bromonium ion (Markovnikov for OH), and Br ends up on the less substituted carbon, giving a trans halohydrin.

13. Give the reagents and the net stereochemistry/regiochemistry of hydroboration-oxidation of an alkene.

    Reagents: (1) $\ce{BH3}$ (or $\ce{BH3\cdot THF}$); (2) $\ce{H2O2}$, $\ce{NaOH}$. Result: syn addition of H and OH, with $\ce{-OH}$ on the less substituted carbon (anti-Markovnikov). No carbocation, so no rearrangements.

14. Why is hydroboration-oxidation both syn and anti-Markovnikov?

    $\ce{BH3}$ adds in one concerted step (syn) through a four-center transition state; boron (bulky, electrophilic center placement) goes to the less hindered carbon, and oxidation replaces B with OH with retention, giving anti-Markovnikov syn-OH.

15. What reagent and product result from epoxidation of an alkene, and what is the stereochemistry?

    A peroxyacid such as mCPBA converts the alkene to an epoxide (oxirane) via syn addition of a single oxygen across the double bond; cis/trans relationships of substituents are retained.

16. Contrast the products of syn dihydroxylation of an alkene using $\ce{OsO4}$ versus anti dihydroxylation via epoxide opening.

    $\ce{OsO4}$ (then $\ce{NaHSO3}$ or $\ce{H2O}$), or cold dilute $\ce{KMnO4}$, gives a syn (cis) vicinal diol. Epoxidation followed by aqueous acid/base ring opening gives an anti (trans) vicinal diol.

17. What are the products of ozonolysis of an alkene under reductive workup ($\ce{O3}$; then $\ce{Zn}/\ce{H2O}$ or $\ce{Me2S}$)?

    The C=C bond is cleaved to give two carbonyl compounds: aldehydes and/or ketones. Each doubly bonded carbon becomes a $\ce{C=O}$.

18. How do ozonolysis products differ under oxidative workup ($\ce{H2O2}$) versus reductive workup?

    Reductive workup ($\ce{Zn}$, $\ce{Me2S}$) stops at aldehydes/ketones. Oxidative workup ($\ce{H2O2}$) further oxidizes any aldehydes to carboxylic acids (ketones remain ketones).

19. Describe hybridization and geometry of the carbons in an alkyne triple bond.

    Each carbon is $sp$ hybridized, linear, with $180^{\circ}$ bond angles. The triple bond consists of one $\sigma$ bond and two mutually perpendicular $\pi$ bonds.

20. Why are terminal alkynes acidic, and what is the approximate $\mathrm{p}K_a$ of a terminal alkyne C–H?

    The conjugate base (acetylide) has its lone pair in an $sp$ orbital with high s-character (50%), holding electrons close to the nucleus and stabilizing the negative charge. $\mathrm{p}K_a \approx 25$.

21. Rank the acidity of a terminal alkyne, an alkene, and an alkane C–H, and explain the trend.

    Alkyne ($\mathrm{p}K_a \approx 25$) > alkene ($\approx 44$) > alkane ($\approx 50$). Higher s-character of the carbanion orbital ($sp > sp^{2} > sp^{3}$) increases stability of the conjugate base and thus acidity.

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