Assessment
Question — of 15
Assessment

Assessment

Answer all 15 questions to test your understanding. Your results will be shown at the end.

15
Questions
5
Objectives
80%
Pass mark
Summary

Key Takeaways

  • Review the core ideas.
  • Connect concepts to practice.
  • Prepare for assessment.
1
Select All That Apply

Which species can exhibit resonance because they contain a continuous conjugated system with delocalizable electrons or lone pairs? Select all that apply.

  • CH2=CH-CH=CH2
  • CH2=CH-CH2-CH=CH2
  • CH2=CH-OCH3
  • CH3-CH2-CH3
✓ Answer recorded — results shown at the end
2
Single Choice

Which molecule does NOT allow resonance between the two double bonds because conjugation is interrupted?

  • ACH2=CH-CH=O
  • BCH2=CH-CH2-CH=CH2
  • CPh-NH2
  • DCH2=CH-C+H2
✓ Answer recorded — results shown at the end
3
True / False

A lone pair on an atom directly adjacent to a C=C bond can participate in resonance if it can occupy a p orbital aligned with the π system.

  • ATrue
  • BFalse
✓ Answer recorded — results shown at the end
4
Sequencing

Arrange the steps for generating a valid resonance form of the allylic cation CH2=CH-CH2+.

⠿Move the π electrons of the C1=C2 bond toward the C2-C3 region
⠿Draw the new double bond between C2 and C3
⠿Keep the positions of all atoms fixed
⠿Place the positive charge on C1 in the new contributor
✓ Answer recorded — results shown at the end
5
Select All That Apply

Which actions are allowed when drawing resonance structures of a conjugated molecule? Select all that apply.

  • Move a π bond to an adjacent bond position within the conjugated system
  • Move a lone pair into an adjacent π bond system
  • Rearrange the sigma-bond skeleton by shifting atoms
  • Keep the total formal charge unchanged across contributors
✓ Answer recorded — results shown at the end
6
Fill in the Blank

When drawing resonance structures, the positions of ___ must remain unchanged.

When drawing resonance structures, the positions of   must remain unchanged.
✓ Answer recorded — results shown at the end
7
Assertion–Reason

Assertion: A delocalized resonance hybrid is usually more stable than any single localized contributor. Reason: Delocalization spreads electron density over multiple atoms and lowers the overall energy of the molecule.

  • ABoth A and R are true and R is the correct explanation of A
  • BBoth A and R are true but R is NOT the correct explanation of A
  • CA is true but R is false
  • DA is false but R is true
✓ Answer recorded — results shown at the end
8
Single Choice

Which species is expected to be more stable due to resonance delocalization?

  • ACH3-CH2+
  • BCH2=CH-CH2+
  • CCH3-CH2-CH2-CH3
  • DCH3-CH2O-
✓ Answer recorded — results shown at the end
9
Select All That Apply

Which statements correctly compare localized and delocalized electron arrangements? Select all that apply.

  • Delocalization in a conjugated system can lower the total energy of the molecule
  • Aromatic rings are often unusually stable because π electrons are delocalized around the ring
  • A localized drawing with alternating single and double bonds is always the real structure of benzene
  • Resonance-stabilized ions are often more stable than similar ions without delocalization
✓ Answer recorded — results shown at the end
10
Fill in the Blank

In benzene, the observed C-C bond length is about ___ Å, intermediate between a typical C-C single bond (~1.54 Å) and a C=C double bond (~1.34 Å).

In benzene, the observed C-C bond length is about   Å, intermediate between a typical C-C single bond (~1.54 Å) and a C=C double bond (~1.34 Å).
✓ Answer recorded — results shown at the end
11
True / False

If benzene truly had three isolated C=C bonds and three isolated C-C bonds, two different C-C bond lengths would be expected instead of one common value.

  • ATrue
  • BFalse
✓ Answer recorded — results shown at the end
12
Match the Pairs

Match each bond description to its approximate C-C bond length or interpretation in benzene chemistry.

Concept
Typical C-C single bond
Typical C=C double bond
Observed C-C bond in benzene
Reason all benzene C-C bonds are equal
Definition
π electrons are delocalized around the ring
About 1.39 Å
About 1.34 Å
About 1.54 Å
0 of 4 pairs matched
✓ Answer recorded — results shown at the end
13
Match the Pairs

Match each structure description to the type of conjugation present.

Concept
CH2=CH-CH=CH2
CH2=CH-CH2+
CH2=CH-NH2
Ph-OCH3
Definition
π-positive conjugation
π-lone pair conjugation
π-lone pair conjugation
π-π conjugation
0 of 4 pairs matched
✓ Answer recorded — results shown at the end
14
Select All That Apply

Which species show π-lone pair conjugation? Select all that apply.

  • CH2=CH-OCH3
  • CH2=CH-CH=CH2
  • Ph-NH2
  • CH2=CH-CH2+
✓ Answer recorded — results shown at the end
15
Assertion–Reason

Assertion: In an allylic carbocation, the conjugation is classified as π-positive. Reason: The positively charged carbon provides an empty p orbital that can overlap with the adjacent π bond.

  • ABoth A and R are true and R is the correct explanation of A
  • BBoth A and R are true but R is NOT the correct explanation of A
  • CA is true but R is false
  • DA is false but R is true
✓ Answer recorded — results shown at the end
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