Understanding sp2 hybridization in alkenes is essential for B. Pharm students studying organic chemistry and medicinal chemistry. This introduction explains how carbon atoms in alkenes form three sp2 hybrid orbitals, adopt a trigonal planar geometry, and create a sigma bond framework plus a perpendicular p orbital that forms the pi bond. Key concepts include bond angles (~120°), pi-bond electron density, restricted rotation, conjugation, resonance stabilization, and how hybridization influences reactivity, acidity, and spectroscopy. Mastery of these topics helps predict reaction mechanisms and molecular properties of drug-like molecules. Now let’s test your knowledge with 50 MCQs on this topic.
Q1. Which hybridization state does each carbon atom in ethene (C2H4) adopt?
- sp
- sp2
- sp3
- unhybridized
Correct Answer: sp2
Q2. How many sp2 hybrid orbitals does a carbon atom form in an alkene?
- One
- Two
- Three
- Four
Correct Answer: Three
Q3. What is the ideal bond angle around an sp2-hybridized carbon?
- 109.5°
- 120°
- 180°
- 90°
Correct Answer: 120°
Q4. Which orbital overlap forms the pi bond in a carbon–carbon double bond?
- sp2–sp2 head-on overlap
- sp2–s overlap
- unhybridized p–p sideways overlap
- sp–sp head-on overlap
Correct Answer: unhybridized p–p sideways overlap
Q5. In ethene, how many pi bonds are present?
- Zero
- One
- Two
- Three
Correct Answer: One
Q6. Which of the following describes the geometry of atoms bonded to an sp2 carbon?
- Tetrahedral
- Linear
- Trigonal planar
- Bent
Correct Answer: Trigonal planar
Q7. Why is rotation about a carbon–carbon double bond restricted?
- Because of steric hindrance only
- Because pi bond formation requires parallel unhybridized p orbitals
- Because sigma bonds are rigid
- Because of lone pair repulsion
Correct Answer: Because pi bond formation requires parallel unhybridized p orbitals
Q8. Which of the following is true about the sigma bond between the two carbons in an alkene?
- It is formed by sideways overlap of p orbitals
- It is formed by head-on overlap of sp2 orbitals
- It is weaker than a pi bond
- It does not contribute to the bond framework
Correct Answer: It is formed by head-on overlap of sp2 orbitals
Q9. Compared to a C–C single bond, a C=C double bond is generally:
- Longer and weaker
- Shorter and stronger
- The same length and strength
- Longer but stronger
Correct Answer: Shorter and stronger
Q10. How many sigma bonds does each carbon atom in ethene form?
- One sigma bond
- Two sigma bonds
- Three sigma bonds
- Four sigma bonds
Correct Answer: Three sigma bonds
Q11. What is the unhybridized orbital remaining on an sp2 carbon used for?
- Forming an additional sigma bond
- Holding a lone pair
- Forming the pi bond
- Pointless and non-reactive
Correct Answer: Forming the pi bond
Q12. Which of the following statements about s-character is correct?
- sp3 has more s-character than sp2
- sp2 has more s-character than sp3
- sp has less s-character than sp2
- All hybrids have equal s-character
Correct Answer: sp2 has more s-character than sp3
Q13. Which hydrogen is most acidic among these?
- Hydrogen on sp3 carbon (alkane)
- Hydrogen on sp2 carbon (alkene)
- Hydrogen on sp carbon (alkyne)
- All have equal acidity
Correct Answer: Hydrogen on sp carbon (alkyne)
Q14. Conjugation in a polyene system requires which orbital arrangement?
- Alternating sp3 and sp bonds
- Consecutive sp2 centers with overlapping p orbitals
- All carbons being sp3
- Only sigma bonds in sequence
Correct Answer: Consecutive sp2 centers with overlapping p orbitals
Q15. Which of the following best explains resonance stabilization in allylic systems?
- Localization of electrons in sigma bonds
- Delocalization of pi electrons across sp2 centers
- Increased steric strain
- Formation of additional sigma bonds
Correct Answer: Delocalization of pi electrons across sp2 centers
Q16. The carbonyl carbon in a ketone is hybridized as:
- sp
- sp2
- sp3
- unhybridized
Correct Answer: sp2
Q17. Which spectroscopic band region is most indicative of a C=C stretch in an alkene?
- Around 3300 cm−1
- Around 1700 cm−1
- Around 1600–1650 cm−1
- Around 2100–2200 cm−1
Correct Answer: Around 1600–1650 cm−1
Q18. Vinyl hydrogens (directly attached to sp2 carbon) typically appear in 1H NMR at what approximate chemical shift?
- 0.5–1.5 ppm
- 1.5–2.5 ppm
- 4.5–6.5 ppm
- 8.0–10.0 ppm
Correct Answer: 4.5–6.5 ppm
Q19. Which of these statements about benzene carbons is true?
- They are sp-hybridized
- They are sp2-hybridized and aromatic
- They are sp3-hybridized and nonplanar
- They have no unhybridized p orbitals
Correct Answer: They are sp2-hybridized and aromatic
Q20. In propene (CH3–CH=CH2), which carbon(s) are sp2 hybridized?
- Only the terminal CH3 carbon
- Only the middle carbon
- Both carbons of the C=C double bond
- No carbon is sp2 hybridized
Correct Answer: Both carbons of the C=C double bond
Q21. Which phenomenon directly leads to cis–trans (E/Z) isomerism in alkenes?
- Free rotation about sigma bonds
- Restricted rotation due to the pi bond
- Rotation about C–H bonds
- Interconversion via sigma bond cleavage only
Correct Answer: Restricted rotation due to the pi bond
Q22. How many unhybridized p orbitals are present on each sp2 carbon in an alkene?
- Zero
- One
- Two
- Three
Correct Answer: One
Q23. Which statement about the electron density of a C=C pi bond is correct?
- Electron density is concentrated along the internuclear axis
- Electron density is above and below the plane of the sigma framework
- Electron density is localized inside the carbon nuclei
- Electron density is identical to that of a C–C single bond
Correct Answer: Electron density is above and below the plane of the sigma framework
Q24. Which is generally more reactive toward electrophilic addition, an isolated alkene or a conjugated diene?
- Isolated alkene because it has more electron density
- Conjugated diene due to delocalized pi electrons and resonance-stabilized intermediates
- Both are equally reactive
- Neither reacts
Correct Answer: Conjugated diene due to delocalized pi electrons and resonance-stabilized intermediates
Q25. Which orbital characteristic of sp2 hybridization increases s-character relative to sp3?
- Higher percentage of p-character
- Higher percentage of s-character (33%)
- No s-character
- Equal s- and p-character
Correct Answer: Higher percentage of s-character (33%)
Q26. Which of the following best explains why allylic carbocations are stabilized?
- Hyperconjugation only
- Resonance delocalization of the positive charge over sp2 centers
- Inductive donation from hydrogen atoms
- They are not stabilized
Correct Answer: Resonance delocalization of the positive charge over sp2 centers
Q27. In a conjugated polyene, increasing conjugation usually leads to:
- Higher HOMO–LUMO gap
- Lower HOMO–LUMO gap and absorption of longer wavelengths
- Complete loss of planarity
- More sp3 character
Correct Answer: Lower HOMO–LUMO gap and absorption of longer wavelengths
Q28. Which of the following is a direct consequence of increased s-character in an sp2 carbon compared to sp3?
- Lower electronegativity
- Shorter bond lengths and greater effective electronegativity
- More flexible bond rotation
- Increased number of unhybridized p orbitals
Correct Answer: Shorter bond lengths and greater effective electronegativity
Q29. Which reagent and condition typically converts an alkene into an alkane?
- Br2 addition in CCl4
- H2 with Pd/C (hydrogenation)
- O3 followed by reductive workup
- KMnO4 cold dilute
Correct Answer: H2 with Pd/C (hydrogenation)
Q30. In an sp2 carbon, what is the steric number used to determine hybridization?
- 1
- 2
- 3
- 4
Correct Answer: 3
Q31. Which of the following is true regarding bond rotation in conjugated systems?
- Rotation between conjugated sp2 centers is rapid and free
- Rotation is restricted because it breaks pi overlap and reduces conjugation
- Conjugation increases rotation freedom
- Rotation only occurs at high temperatures
Correct Answer: Rotation is restricted because it breaks pi overlap and reduces conjugation
Q32. Which molecular feature indicates partial double-bond character due to resonance?
- Bond length intermediate between typical single and double bonds
- Extremely long single bond lengths
- Complete loss of pi bonding
- Only sigma bond character present
Correct Answer: Bond length intermediate between typical single and double bonds
Q33. For a monosubstituted alkene, which isomer generally has higher dipole moment and often higher boiling point?
- Trans isomer
- Cis isomer
- Both identical
- Neither; dipole moment is zero
Correct Answer: Cis isomer
Q34. Which of these atoms in an imine (C=N) is typically sp2 hybridized?
- Carbon only
- Nitrogen only
- Both carbon and nitrogen
- Neither
Correct Answer: Both carbon and nitrogen
Q35. Which of the following reactions directly involves the pi electrons of an alkene attacking an electrophile?
- Nucleophilic substitution (SN2)
- Electrophilic addition
- Free radical halogenation
- Elimination E2
Correct Answer: Electrophilic addition
Q36. In UV–Vis spectroscopy, a conjugated alkene system compared to an isolated double bond will typically show:
- Absorption at shorter wavelengths
- No absorption
- Absorption at longer wavelengths
- Only IR active modes
Correct Answer: Absorption at longer wavelengths
Q37. The presence of an sp2-hybridized carbon adjacent to a heteroatom (e.g., O or N) often results in:
- No effect on reactivity
- Enhanced resonance interaction and possible delocalization of lone pairs
- Complete loss of pi bonding
- Formation of sp hybridization
Correct Answer: Enhanced resonance interaction and possible delocalization of lone pairs
Q38. Which description best fits the shape of sp2 hybrid orbitals?
- Two lobes in a linear arrangement
- Four equivalent lobes in tetrahedral arrangement
- Three coplanar lobes arranged trigonal planar
- Spherical lobes only
Correct Answer: Three coplanar lobes arranged trigonal planar
Q39. What is the percent s-character of an sp2 hybrid orbital?
- 25%
- 33.3%
- 50%
- 0%
Correct Answer: 33.3%
Q40. Which property of an alkene is directly influenced by the presence of a pi bond?
- Ability to undergo free rotation about the C–C axis
- Planarity of the bonded atoms and reactivity toward addition reactions
- Total number of sigma bonds only
- Conversion to sp hybridization
Correct Answer: Planarity of the bonded atoms and reactivity toward addition reactions
Q41. How does conjugation affect the stability of an alkene?
- Conjugation decreases stability by localizing electrons
- Conjugation increases stability by delocalizing pi electrons
- Conjugation has no effect on stability
- Conjugation converts sp2 to sp3
Correct Answer: Conjugation increases stability by delocalizing pi electrons
Q42. Which of these correctly ranks hybridizations by increasing s-character?
- sp3 < sp2 < sp
- sp < sp2 < sp3
- sp2 < sp3 < sp
- All equal
Correct Answer: sp3 < sp2 < sp
Q43. In electrophilic addition of HBr to propene, the initial step involves:
- Nucleophilic attack by bromide on a saturated carbon
- Protonation of the pi bond to form a carbocation intermediate
- Radical abstraction of hydrogen
- Concerted insertion into C–H bonds
Correct Answer: Protonation of the pi bond to form a carbocation intermediate
Q44. Which of the following best describes the pi electron cloud in a cis-alkene versus its trans isomer?
- Pi cloud orientation is identical; only sigma framework differs
- Cis isomer may have a net dipole due to substituent orientation while trans often cancels dipoles
- Trans isomer is always more polar than cis
- Pi cloud disappears in cis isomer
Correct Answer: Cis isomer may have a net dipole due to substituent orientation while trans often cancels dipoles
Q45. Which of the following best predicts the site of electrophilic attack on a disubstituted alkene?
- Attack occurs at the more substituted carbon to form the more stable carbocation (Markovnikov rule)
- Attack always occurs at the less substituted carbon
- Attack is independent of substitution
- Attack never forms a carbocation intermediate
Correct Answer: Attack occurs at the more substituted carbon to form the more stable carbocation (Markovnikov rule)
Q46. Which statement is correct about the relationship between hybridization and electronegativity?
- Greater s-character makes the atom less electronegative
- Greater s-character increases effective electronegativity of the carbon
- Hybridization does not affect electronegativity
- sp3 is more electronegative than sp2
Correct Answer: Greater s-character increases effective electronegativity of the carbon
Q47. Which experimental observation supports that alkenes are planar at the double bond?
- Free rotation seen in NMR spectra
- X-ray/crystallographic structures showing trigonal planar geometry at double-bonded carbons
- Alkenes always being gases
- Presence of only sigma bonds
Correct Answer: X-ray/crystallographic structures showing trigonal planar geometry at double-bonded carbons
Q48. Which of the following is a consequence of pi-bond electron density being above and below the plane of the atoms?
- Enhanced ability for nucleophiles to attack from the plane only
- Facilitated sideways overlap leading to delocalization and conjugation
- No interaction with electrophiles
- Loss of planarity
Correct Answer: Facilitated sideways overlap leading to delocalization and conjugation
Q49. In a molecule exhibiting resonance with alternating double bonds, what happens to individual bond character?
- All bonds become pure single bonds
- Bonds become equivalent with partial double-bond character
- All bonds become pure double bonds
- Bond character is unchanged
Correct Answer: Bonds become equivalent with partial double-bond character
Q50. Why is an sp2-hybridized carbon often involved in key reaction centers of drug molecules?
- Because sp2 carbons never participate in reactions
- Because their planar geometry and pi systems enable specific stereoelectronic interactions, resonance stabilization, and predictable reactivity in addition and conjugation reactions
- Because they are too reactive to be useful
- Because sp2 carbons are always inert under physiological conditions
Correct Answer: Because their planar geometry and pi systems enable specific stereoelectronic interactions, resonance stabilization, and predictable reactivity in addition and conjugation reactions
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