Class 11 Chemistry Chapter 8 Organic Chemistry – Some Basic Principles and Techniques – Extra Questions with Answers

Extra practice questions for Class 11 Chemistry Chapter 8 (Organic Chemistry – Some Basic Principles and Techniques), beyond the textbook. These Class 11 Chemistry Chapter 8 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. What is a carbanion?
Ans: A negatively charged carbon species with a lone pair of electrons.

Q2. Name the electron displacement effect transmitted through pi bonds/lone pairs via delocalization.
Ans: Resonance (mesomeric) effect.

Q3. What is a free radical?
Ans: A species containing an atom with an unpaired electron.

Q4. Give an example of a heterocyclic compound.
Ans: Pyridine (or furan, thiophene).

Q5. What is meant by chain isomerism?
Ans: Isomers differing in the arrangement/branching of the carbon skeleton, with the same molecular formula.

Short Answer Questions (2–3 marks)

Q6. Write the IUPAC name of CH₃-CH₂-CH₂-OH.
Ans: Propan-1-ol.

Q7. Draw and name a chain isomer of butane (C₄H₁₀).
Ans: 2-methylpropane (isobutane), CH₃-CH(CH₃)-CH₃, differing from n-butane by having a branched chain instead of a straight chain.

Q8. Which is more stable: a tertiary carbocation or a primary carbocation, and why?
Ans: A tertiary carbocation is more stable due to greater hyperconjugation and positive inductive (+I) effect from the three attached alkyl groups, which help disperse the positive charge.

Higher-Order Thinking / Application Questions

Q9. Explain, using the inductive effect, why trichloroacetic acid (Cl₃CCOOH) is a much stronger acid than acetic acid (CH₃COOH), connecting this to the stability of the conjugate base formed.
Ans: Chlorine is strongly electronegative and exerts a negative inductive (−I) effect, withdrawing electron density through the sigma bonds of the carbon chain. In trichloroacetic acid, the three electronegative chlorine atoms attached to the alpha carbon pull electron density away from the O-H bond and, more importantly, help stabilize the negative charge on the conjugate base (trichloroacetate ion) formed after the acid donates its proton, by dispersing that negative charge through the strong −I effect of the three chlorines. Since a more stabilized (lower energy) conjugate base corresponds to a stronger acid (easier to lose the proton), trichloroacetic acid is a considerably stronger acid than acetic acid, which lacks these electron-withdrawing chlorine substituents and thus cannot stabilize its conjugate base (acetate ion) nearly as effectively.

Q10. Explain why a tertiary carbocation (like (CH₃)₃C⁺) is more stable than a primary carbocation (like CH₃CH₂⁺), using both the inductive effect and hyperconjugation as reasoning.
Ans: A tertiary carbocation has three alkyl (methyl) groups directly attached to the positively charged carbon, compared to a primary carbocation which has only one alkyl group. Alkyl groups are electron-donating through the positive inductive (+I) effect, pushing electron density toward the electron-deficient positively charged carbon and partially neutralizing the positive charge, with three alkyl groups providing much more of this stabilizing donation than one. Additionally, tertiary carbocations benefit from far greater hyperconjugation: with three alkyl groups, there are many more adjacent C-H sigma bonds available to donate electron density into the empty p-orbital of the carbocation, further stabilizing the positive charge. Together, the combined effect of greater +I donation and more extensive hyperconjugative stabilization makes the tertiary carbocation significantly more stable than the primary carbocation.

Written by Satish

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