Aromatic Compounds flash cards

1. 1.What are the four conditions for a molecule (or ion) to be aromatic?

   (1) Cyclic, (2) planar, (3) every ring atom has a $p$-orbital allowing continuous conjugation (fully conjugated / delocalized), and (4) the cyclic $\pi$ system contains $(4n+2)$ $\pi$ electrons, $n = 0,1,2,\dots$ (Huckel's rule).

   Hint: Cyclic, planar, conjugated, $4n+2$.

2. 2.State Huckel's rule and give the allowed electron counts.

   A planar monocyclic fully conjugated ring is aromatic if it has $(4n+2)$ delocalized $\pi$ electrons. Allowed counts: $2, 6, 10, 14, \dots$ (for $n = 0,1,2,3$).

   Hint: $4n+2$ with $n$ a non-negative integer.

3. 3.What is antiaromaticity and how many $\pi$ electrons characterize it?

   A cyclic, planar, fully conjugated system with $4n$ $\pi$ electrons ($4,8,12,\dots$) is antiaromatic — it is destabilized (higher energy) relative to the open-chain analog. Example: cyclobutadiene ($4\pi$).

   Hint: $4n$ electrons; less stable than expected.

4. 4.Why is benzene aromatic?

   Benzene is cyclic, planar, and every C is $sp^2$ with a $p$-orbital; the six $p$-orbitals overlap continuously to give $6$ delocalized $\pi$ electrons $= 4n+2$ with $n=1$. So it satisfies Huckel's rule.

   Hint: $6\pi$ electrons, $n=1$.

5. 5.How many $\pi$ electrons does benzene have and how are they distributed?

   $6$ $\pi$ electrons delocalized over all six carbons — not localized in three double bonds. All six C–C bonds are equivalent (bond order $1.5$).

   Hint: Six, fully delocalized.

6. 6.What is the C–C bond length in benzene and why is it uniform?

   All C–C bonds are $1.39$ Å — intermediate between a single bond ($1.54$ Å) and a double bond ($1.34$ Å). Uniformity arises from delocalization giving each bond order $1.5$.

   Hint: $1.39$ Å, between single and double.

7. 7.Describe the hybridization and geometry of benzene carbons.

   Each carbon is $sp^2$ hybridized with bond angles of $120^\circ$. The molecule is planar and hexagonal; the unhybridized $p$-orbitals (perpendicular to the plane) form the $\pi$ system.

   Hint: $sp^2$, $120^\circ$, planar hexagon.

8. 8.What is resonance energy (stabilization energy) of benzene and what does it signify?

   About $150$ kJ/mol ($\approx 36$ kcal/mol) — the difference between the actual energy of benzene and that of the hypothetical 1,3,5-cyclohexatriene. It measures the extra stability from delocalization.

   Hint: $\sim 150$ kJ/mol; delocalization stability.

9. 9.How does the heat of hydrogenation of benzene reveal its stability?

   Expected $\Delta H$ for 3 double bonds $\approx 3 \times (-120) = -360$ kJ/mol, but benzene's actual $\Delta H_{hyd} \approx -208$ kJ/mol. The $\sim 150$ kJ/mol difference is the resonance/stabilization energy.

   Hint: Less exothermic than 3 isolated C=C.

10. 10.Why does benzene undergo substitution rather than addition reactions?

    Addition would destroy the aromatic $6\pi$ delocalized system and its large resonance stabilization. Substitution preserves the aromatic ring, so it is thermodynamically favored.

    Hint: Keeping the ring aromatic is favorable.

11. 11.How many resonance (Kekule) structures does benzene have and what does resonance imply?

    Two equivalent Kekule structures (alternating double bonds) that interconvert. The real molecule is a resonance hybrid — a single delocalized structure more stable than either canonical form.

    Hint: Two Kekule forms; hybrid is real.

12. 12.Is the cyclopentadienyl anion aromatic? Explain.

    Yes. The 5-membered ring is planar, fully conjugated, and the carbanion contributes a lone pair to the $\pi$ system giving $6$ $\pi$ electrons $= 4n+2$. Hence aromatic and unusually stable.

    Hint: $6\pi$, cyclopentadienyl minus.

13. 13.Is the cycloheptatrienyl (tropylium) cation aromatic?

    Yes. The 7-membered ring cation is planar and fully conjugated with $6$ $\pi$ electrons ($4n+2$, $n=1$). Tropylium is aromatic and remarkably stable for a carbocation.

    Hint: $C_7H_7^+$, $6\pi$.

14. 14.Is cyclopentadienyl cation aromatic or antiaromatic?

    Antiaromatic. It has $4$ $\pi$ electrons ($4n$, $n=1$) in a planar conjugated 5-membered ring, making it highly unstable.

    Hint: $4\pi$ electrons.

15. 15.Why is cyclooctatetraene (COT) non-aromatic?

    COT has $8$ $\pi$ electrons ($4n$), and to avoid antiaromatic destabilization it adopts a non-planar tub shape with localized double bonds. It behaves like a normal polyene.

    Hint: Tub-shaped, $8\pi$, non-planar.

16. 16.What makes pyridine aromatic and where is its nitrogen lone pair?

    Pyridine is a 6-membered aromatic ring ($6\pi$). The N lone pair lies in an $sp^2$ orbital in the ring plane (NOT part of the $\pi$ system), so it is available for basicity/protonation.

    Hint: Lone pair in plane, not in $\pi$ cloud.

17. 17.What makes pyrrole aromatic and why is it a weak base?

    Pyrrole's N contributes its lone pair into the ring $\pi$ system to reach $6$ $\pi$ electrons. Because that lone pair is tied up in aromaticity, pyrrole is a very weak base.

    Hint: N lone pair is in the $\pi$ system.

18. 18.Is naphthalene aromatic? How many $\pi$ electrons?

    Yes. Naphthalene (two fused benzene rings) has $10$ $\pi$ electrons $= 4n+2$ with $n=2$, and is planar and fully conjugated — aromatic (though Huckel's rule is strictly for monocyclic systems, it works here).

    Hint: $10\pi$, fused bicyclic.

19. 19.Define electrophilic aromatic substitution (EAS).

    A reaction in which an electrophile $E^+$ replaces a hydrogen on an aromatic ring. The aromatic ring acts as a nucleophile (electron-rich $\pi$ cloud) attacking $E^+$; the ring's aromaticity is restored after loss of $H^+$.

    Hint: $E^+$ replaces H; ring is nucleophile.

20. 20.Outline the general two-step mechanism of EAS.

    Step 1 (slow, rate-determining): the $\pi$ system attacks $E^+$ forming a resonance-stabilized carbocation (arenium ion / sigma complex) — aromaticity temporarily lost. Step 2 (fast): a base removes $H^+$ from the $sp^3$ carbon, restoring aromaticity.

    Hint: Form arenium ion, then lose $H^+$.

21. 21.What is the arenium ion (sigma complex, Wheland intermediate)?

    The resonance-stabilized carbocation intermediate formed when $E^+$ bonds to a ring carbon. Positive charge is delocalized over three ring carbons (ortho, ortho, para to the $sp^3$ carbon); the ring is temporarily non-aromatic.

    Hint: Three-carbon delocalized cation.

22. 22.Why is the first step of EAS rate-determining?

    Forming the arenium ion destroys aromaticity (high energy transition state, large activation energy), whereas the second step (loss of $H^+$) regenerates the stable aromatic ring and is fast.

    Hint: Losing aromaticity costs energy.

23. 23.Give the reagents and electrophile for nitration of benzene.

    Reagents: conc. $HNO_3$ + conc. $H_2SO_4$ (nitrating mixture). Electrophile: the nitronium ion $NO_2^+$. Product: nitrobenzene.

    Hint: $HNO_3/H_2SO_4$; electrophile $NO_2^+$.

24. 24.How is the nitronium ion $NO_2^+$ generated in nitration?

    $H_2SO_4$ protonates $HNO_3$; loss of water gives $NO_2^+$: $HNO_3 + 2\,H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2\,HSO_4^-$. $H_2SO_4$ acts as a strong acid to generate the electrophile.

    Hint: $H_2SO_4$ protonates then dehydrates $HNO_3$.

25. 25.Give the reagents and electrophile for chlorination of benzene.

    Reagents: $Cl_2$ with a Lewis-acid catalyst ($AlCl_3$ or $FeCl_3$). The catalyst polarizes $Cl_2$ to generate the electrophile $Cl^+$ (as $Cl$–$Cl\cdots AlCl_3$). Product: chlorobenzene.

    Hint: $Cl_2 / FeCl_3$; electrophile $Cl^+$.

26. 26.Why is a Lewis acid (e.g., $FeBr_3$) needed for bromination of benzene?

    Benzene's aromatic cloud is not nucleophilic enough to react with $Br_2$ alone. $FeBr_3$ polarizes $Br$–$Br$, forming a stronger electrophile ($Br^+$ / $Br$–$Br\cdots FeBr_3$) that the ring can attack.

    Hint: To make $Br_2$ a strong enough electrophile.

Open the interactive deck for the other 70 cards, with self-grading so the ones you keep missing come back.