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Organic Chemistry: Some Basic Principles WBJEE MCQ & Practice Questions

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Organic Chemistry: Some Basic Principles Revision Guide for WBJEE Chemistry

Organic Chemistry: Some Basic Principles is part of WBJEE Chemistry preparation. A strong revision plan should combine concept review with exam-style questions so that definitions, relationships, calculations and exceptions are recalled accurately.

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Top 20 Organic Chemistry: Some Basic Principles MCQs with Answers

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WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

1. The hybridisation of each carbon atom in ethyne is:

  • A. sp3
  • B. dsp2
  • C. sp
  • D. sp2
Answer: C
Explanation: Each carbon has two electron domains and is sp hybridised.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

2. The hybridisation of carbon atoms in ethene is:

  • A. sp3d
  • B. sp2
  • C. sp
  • D. sp3
Answer: B
Explanation: Each alkene carbon has trigonal-planar sp2 hybridisation.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

3. The carbon atom in methane is:

  • A. sp3 hybridised
  • B. sp2 hybridised
  • C. sp hybridised
  • D. unhybridised
Answer: A
Explanation: Four sigma bonds give tetrahedral sp3 carbon.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

4. The approximate H-C-H bond angle in methane is:

  • A. 120°
  • B. 180°
  • C. 90°
  • D. 109.5°
Answer: D
Explanation: Tetrahedral sp3 geometry gives about 109.5°.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

5. The approximate bond angle around an sp2 carbon is:

  • A. 180°
  • B. 90°
  • C. 120°
  • D. 109.5°
Answer: C
Explanation: sp2 hybridisation gives trigonal planar geometry.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

6. The approximate bond angle around an sp carbon is:

  • A. 90°
  • B. 180°
  • C. 120°
  • D. 109.5°
Answer: B
Explanation: sp hybridisation gives linear geometry.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

7. A carbon-carbon double bond contains:

  • A. One sigma and one pi bond
  • B. Two sigma bonds
  • C. Two pi bonds
  • D. One sigma only
Answer: A
Explanation: A double bond consists of one σ and one π bond.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

8. A carbon-carbon triple bond contains:

  • A. Three sigma bonds
  • B. Two sigma and one pi bond
  • C. Three pi bonds
  • D. One sigma and two pi bonds
Answer: D
Explanation: A triple bond contains one σ and two π bonds.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

9. Homolytic bond cleavage produces:

  • A. Carbanions only
  • B. Ions of opposite charge only
  • C. Free radicals
  • D. Carbocations only
Answer: C
Explanation: Homolysis divides the bonding pair equally.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

10. Heterolytic bond cleavage produces:

  • A. Only carbenes
  • B. Ions
  • C. Only radicals
  • D. Only neutral atoms
Answer: B
Explanation: Heterolysis transfers both bonding electrons to one fragment.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

11. An electrophile is a species that:

  • A. Accepts an electron pair
  • B. Donates an electron pair
  • C. Always has negative charge
  • D. Contains no vacant orbital
Answer: A
Explanation: Electrophiles are electron-pair acceptors.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

12. A nucleophile is a species that:

  • A. Accepts an electron pair
  • B. Must be positively charged
  • C. Has no lone pair or pi electrons
  • D. Donates an electron pair
Answer: D
Explanation: Nucleophiles donate electron density.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

13. Which is a typical nucleophile?

  • A. BF3
  • B. AlCl3
  • C. OH-
  • D. H+
Answer: C
Explanation: Hydroxide has lone-pair electron density available for donation.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

14. Which is a typical electrophile?

  • A. NH3
  • B. NO2+
  • C. OH-
  • D. CN-
Answer: B
Explanation: Nitronium ion is strongly electron deficient.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

15. BF3 behaves as a Lewis acid because boron has:

  • A. An incomplete octet
  • B. A complete octet and negative charge
  • C. A lone pair to donate
  • D. Ten valence electrons
Answer: A
Explanation: Electron-deficient boron can accept an electron pair.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

16. The inductive effect is transmitted mainly through:

  • A. Pi bonds only
  • B. Hydrogen bonds
  • C. Ionic lattices
  • D. Sigma bonds
Answer: D
Explanation: Inductive polarization propagates through σ bonds.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

17. The resonance effect involves delocalisation through:

  • A. Nuclei
  • B. Hydrogen bonds
  • C. Conjugated p orbitals
  • D. Only isolated sigma bonds
Answer: C
Explanation: Resonance requires overlapping orbitals and electron delocalisation.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

18. Hyperconjugation is often described as delocalisation involving:

  • A. Hydrogen bonds
  • B. Sigma C-H/C-C electrons adjacent to a pi or vacant p system
  • C. Only lone pairs on halogens
  • D. Nuclear electrons
Answer: B
Explanation: Adjacent σ bonds can interact with a p/π system.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

19. A +I group generally:

  • A. Releases electron density through sigma bonds
  • B. Withdraws electron density strongly
  • C. Has no electronic effect
  • D. Accepts a proton only
Answer: A
Explanation: Positive inductive groups push electron density.

WBJEE Chemistry · Organic Chemistry: Some Basic Principles · Very Very Hard

20. The stability order of simple alkyl carbocations is generally:

  • A. 1° > 3° > 2° > CH3+
  • B. 2° > 1° > 3° > CH3+
  • C. 3° > 2° > 1° > CH3+
  • D. CH3+ > 1° > 2° > 3°
Answer: C
Explanation: Hyperconjugation and +I alkyl effects stabilise more substituted carbocations.

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