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Organic Chemistry
Chemistry · Class 11NCERT pp. 256–2942 sets × 25 questions
Key facts and formulas for Organic Chemistry, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Bond count | Single bond = 1 σ; double = 1 σ + 1 π; triple = 1 σ + 2 π |
| s character | sp 50%, sp² 33%, sp³ 25%; more s means more electronegative |
| Carbocation / radical stability | 3° > 2° > 1° > methyl (hyperconjugation; allyl and benzyl also resonance stabilised) |
| Carbanion stability | CH₃⁻ > 1° > 2° > 3° (alkyl groups destabilise) |
| Hückel rule | Aromatic: planar, cyclic, conjugated, 4n+2 π electrons |
| Degree of unsaturation | DoU = (2C + 2 − H) ÷ 2 for a hydrocarbon |
| Rf value | Rf = distance moved by component ÷ distance moved by solvent front |
| Carius method | %Br = 80/188 × mass AgBr ÷ mass compound × 100; %S = 32/233 × mass BaSO₄ ÷ mass × 100 |
| Dumas method | %N = 28 × V(N₂ at STP, mL) ÷ 22400 ÷ mass of compound × 100 |
| Kjeldahl method | %N = 1.4 × N × V (acid used) ÷ mass; H₂SO₄ is dibasic so N = 2M |
| Combustion analysis | %C = 12/44 × mass CO₂ ÷ mass × 100; %H = 2/18 × mass H₂O ÷ mass × 100 |
Three sample questions
Q1. The number of σ and π bonds in propyne, CH₃–C≡CH, is
- 6 σ and 2 π
- 5 σ and 3 π
- 7 σ and 1 π
- 6 σ and 3 π
Hint and solution
Hint: Count every single bond as one sigma. A triple bond is one sigma plus two pi.
- C–H bonds: 3 + 1 = 4 σ — The methyl carbon has three C–H bonds and the terminal carbon has one. All are sigma bonds.
- C–C: 1 single + 1 triple — The single bond C1–C2 gives one sigma. The triple bond C2≡C3 gives one sigma and two pi.
- σ = 4 + 1 + 1 = 6; π = 2 — Add all the sigma bonds: 4 C–H, 1 single C–C and 1 from the triple bond. Pi bonds come only from the triple bond.
- Answer: 6 σ and 2 π — So propyne has 6 sigma and 2 pi bonds. Option A.
Answer: A. 4 C–H plus 2 C–C sigma bonds make 6 σ. The triple bond adds two π bonds.
Q2. Which hybridised carbon is the most electronegative?
- sp³
- sp²
- sp
- All are equal
Hint and solution
Hint: Electronegativity rises with s character. Compare the s percentage of sp³, sp² and sp.
- s character: sp³ 25%, sp² 33%, sp 50% — Each hybrid orbital mixes one s with some p orbitals. The more s, the higher the s percentage.
- s orbital is closer to the nucleus — Electrons in an s orbital are held nearer the nucleus than in a p orbital.
- More s character → more electronegative — So the hybrid with the highest s character attracts bonding electrons best.
- Answer: sp — The sp carbon is most electronegative. Option C.
Answer: C. sp has 50% s character, so its electrons are nearest to the nucleus.
Q3. The IUPAC name of (CH₃)₂CH–CH₂–CH(CH₃)₂ is
- 2,4-dimethylhexane
- 2,4-dimethylpentane
- 1,1,3,3-tetramethylpropane
- isohexane
Hint and solution
Hint: Write it out and find the longest continuous chain. Then place the methyl groups with the lowest numbers.
- CH₃–CH(CH₃)–CH₂–CH(CH₃)–CH₃ — Draw the full structure. Each end carbon carries a CH₃ branch, which can be taken into the chain.
- Longest chain = 5 C → pentane — Count the continuous carbons: CH₃, CH, CH₂, CH, CH₃. That is five, so the parent is pentane.
- Methyls on C2 and C4 — The two CH₃ branches sit on the second and fourth carbons from either end.
- Answer: 2,4-dimethylpentane — The name is 2,4-dimethylpentane. Option B.
Answer: B. The longest chain has five carbons, with methyl groups on C2 and C4.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.