CBSE Class 11 Chemistry Organic Chemistry Worksheet

Q1. Give IUPAC name of the followings:-

Question 1. \(\text{CH}_3\text{CH}_2\text{CH} = \text{CH}_2\)
Answer: But-1-ene
In simple words: A four-carbon straight-chain alkene with a double bond starting at position 1.

Exam Tip: Number the carbon chain from the end closer to the double bond.

Question 2. \(\text{CH}_3\text{C} \equiv \text{C} - \text{CH}_3\)
Answer: But-2-yne
In simple words: A four-carbon alkyne with a triple bond between carbons 2 and 3.

Exam Tip: Lowest locant rule gives the triple bond position as 2.

Question 3. \(\text{CH}_3 - \text{CH}(\text{I}) - \text{CH}_3\)
Answer: 2-Iodopropane
In simple words: A three-carbon propane chain with an iodine atom attached to the middle carbon.

Exam Tip: Alphabetical order applies if multiple different substituents are present, but here only iodo is present.

Question 4. \(\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2-\text{CH}_3\)
Answer: Ethoxyethane
In simple words: An ether containing two ethyl groups bonded to an oxygen atom (diethyl ether).

Exam Tip: Name ethers as alkoxyalkanes where the larger alkyl group forms the alkane root.

Question 5. \(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{COOH}\)
Answer: Butanoic acid
In simple words: A four-carbon carboxylic acid.

Exam Tip: The carbon of the carboxylic acid group is always carbon-1.

Question 6. \(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CHO}\)
Answer: Butanal
In simple words: A four-carbon aldehyde.

Exam Tip: Aldehyde carbon receives number 1 in numbering the principal chain.

Question 7. \(\text{CH}_3\text{COCH}_2-\text{CH}_2-\text{CH}_3\)
Answer: Pentan-2-one
In simple words: A five-carbon ketone with the carbonyl oxygen on carbon 2.

Exam Tip: Number from the end nearest to the carbonyl group (\(\text{C}=\text{O}\)).

Question 8. \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COCl}\)
Answer: Butanoyl chloride
In simple words: An acyl chloride derived from butanoic acid.

Question 9. \(\text{CH}_3 - \text{CONH}_2\)
Answer: Ethanamide
In simple words: A two-carbon acid amide (acetamide).

Exam Tip: Amide carbon is counted as carbon-1.

Question 10. \(\text{CH}_3\text{COOC}_2\text{H}_5\)
Answer: Ethyl ethanoate
In simple words: An ester formed from ethanol and ethanoic acid.

Exam Tip: The alkyl group attached to oxygen is named first as a separate word, followed by the carboxylate name.

Question 11. \(\text{CH}_3\text{CH}_2-\text{NO}_2\)
Answer: Nitroethane
In simple words: An ethane molecule with a nitro substituent.

Exam Tip: Nitro groups are always treated as substituents in IUPAC nomenclature.

Question 12. \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}\)
Answer: Butanenitrile
In simple words: A four-carbon nitrile containing the \(-\text{C}\equiv\text{N}\) group.

Exam Tip: The carbon of the nitrile group is included in the main chain numbering.

Question 13. \(\text{CH}_3-\text{CH}(\text{C}_2\text{H}_5)-\text{C}(\text{CH}_3)_2-\text{CH}_2-\text{CH}_3\)
Answer: 3,3-Dimethyl-2-ethylpentane (Correct IUPAC: 3-ethyl-2,2-dimethylpentane based on longest chain selection giving 3,3,4-trimethylhexane? Let us trace longest chain: pentane chain with substituents or hexane chain: 3-ethyl-2,2-dimethylpentane)
In simple words: A pentane main chain substituted with one ethyl and two methyl groups.

Exam Tip: Always select the longest continuous carbon chain as the parent hydrocarbon.

Question 14. \(\text{CH}_3-\text{CH}_2-\text{CH}(\text{C}_2\text{H}_5)-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3\)
Answer: 4-Ethyl-3-methylhexane
In simple words: A six-carbon hexane chain with methyl and ethyl branches.

Exam Tip: Number from the end that gives lower locant numbers to substituents.

Question 15. \(\text{CH}_3-\text{CH}(\text{C}_2\text{H}_5)-\text{C}(\text{CH}_3)_2-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_3\)
Answer: 2,3,5,5-Tetramethylhexane (or analogous branched structure)
In simple words: A hexane skeleton heavily methylated.

Exam Tip: Alphabetize substituent prefixes when writing out the name.

Question 16. \(\text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}_2-\text{CH}(\text{CH}_3)_2\)
Answer: 2,2,4-Trimethylpentane (Isooctane)
In simple words: A five-carbon pentane chain with three methyl branches.

Exam Tip: Use prefixes like di-, tri-, tetra- for identical substituents.

Question 17. \(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{C}(\text{CH}_3)=\text{CH}_2\)
Answer: 2,3-Dimethylbut-1-ene
In simple words: A four-carbon alkene with two methyl groups and a terminal double bond.

Exam Tip: Double bond takes precedence over alkyl substituents for lowest locant numbering.

Question 18. \(\text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{C}\equiv\text{C}-\text{CH}_3\)
Answer: 4-Methylhex-2-yne
In simple words: A six-carbon alkyne with a methyl substituent at position 4.

Exam Tip: Numbering starts from the end closer to the triple bond.

Question 19. \(\text{CH}_3-\text{CH}_2-\text{CH}(\text{C}_2\text{H}_5)-\text{COO}-\text{CH}_3\)
Answer: Methyl 2-ethylbutanoate
In simple words: Methyl ester of 2-ethylbutanoic acid.

Exam Tip: Ester alkyl part is named first, followed by the alkanoate name.

Question 20. \(\text{CH}_3-\text{CH}=\text{CH}-\text{C}(=\text{O})-\text{COOH}\)
Answer: 2-Oxopent-3-enoic acid (or similar keto-acid structure)
In simple words: An unsaturated keto-acid.

Exam Tip: Carboxylic acid has highest priority and dictates principal functional group suffix.

Question 21. \(\text{CH}_2=\text{C}(\text{CH}_3)-\text{C}\equiv\text{C}-\text{OC}_2\text{H}_5\)
Answer: 4-Ethoxy-2-methylbut-1-en-3-yne
In simple words: A hydrocarbon chain containing both double and triple bonds and an alkoxy group.

Exam Tip: En and yn take alphabetical priority when numbering from equivalent ends, but unsaturation takes precedence over alkoxy.

Question 22. \(\text{CH}_3-\text{CH}=\text{C}(\text{OCH}_3)-\text{CH}_3\)
Answer: 2-Methoxybut-2-ene
In simple words: A four-carbon alkene with a methoxy group at position 2.

Exam Tip: Double bond gets lower locant (2 over 3).

Question 23. \(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\text{CH}_2\text{NH}_2\)
Answer: 3,4-Dimethylpentan-1-amine
In simple words: A five-carbon primary amine with two methyl substituents.

Exam Tip: Amino group takes suffix '-amine' attached to the longest chain containing the nitrogen-bearing carbon.

Question 24. \(\text{CH}_3-\text{C}(=\text{O})-\text{CH}(\text{C}_2\text{H}_5)-\text{CH}(\text{CH}_3)-\text{CH}_3\)
Answer: 3-Ethyl-4-methylpentan-2-one
In simple words: A five-carbon ketone with ethyl and methyl branches.

Exam Tip: Ketone carbonyl oxygen gets lower locant number (2).

Question 25. \(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_2\text{Br}\)
Answer: 1-Bromo-4-methylpentan-2-one
In simple words: A bromo-substituted methylpentanone.

Exam Tip: Principal functional group (ketone) determines suffix '-one', numbering gives it position 2.

Question 26. \(\text{CH}_3-\text{CH}(\text{C}_6\text{H}_5)-\text{CH}_2\text{OH}\)
Answer: 2-Phenylpropan-1-ol
In simple words: Propan-1-ol with a phenyl ring at carbon 2.

Exam Tip: Hydroxyl group has highest priority, giving carbon-1 to \(-\text{CH}_2\text{OH}\).

Question 27. \((\text{CH}_3\text{CH}_2)_3\text{COH}\)
Answer: 3-Ethylpentan-3-ol
In simple words: A tertiary alcohol with three ethyl groups attached to the carbinol carbon.

Exam Tip: Select longest chain containing the hydroxyl carbon (pentane chain).

Question 28. \(\text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3\)
Answer: 2,3,4,2-tetramethyl... -> 2,3,4,4-Tetramethylhexane
In simple words: A hexane core substituted with four methyl groups.

Question 29. \(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CO}-\text{CH}(\text{I})-\text{CH}_3\)
Answer: 4-Iodo-2-methylpentan-3-one
In simple words: A pentanone derivative with an iodo group and a methyl group.

Exam Tip: Number from the end giving the carbonyl carbon the lowest possible number (3).

Question 30. \(\text{CH}_3-\text{CH}_2-\text{CH}(\text{Cl})-\text{CH}(\text{CH}_3)-\text{C}(\text{CH}_3)(\text{OH})-\text{CH}_3\)
Answer: 3-Chloro-2,4-dimethylhexan-2-ol
In simple words: A hexanol derivative with chloro and methyl substituents.

Exam Tip: Hydroxyl group takes highest priority, numbering starts from the end nearest to \(-\text{OH}\).

Question 31. \(\text{Cyclohexyl}-\text{CH}(\text{CH}_3)-\text{CHO}\)
Answer: 2-Cyclohexyl-2-methylpropanal
In simple words: An aldehyde with a cyclohexyl ring and a methyl group at carbon 2.

Exam Tip: Aldehyde carbon is carbon-1.

Question 32. \(\text{CH}_3-\text{CO}-\text{CH}(\text{CH}_3)_2\)
Answer: 3-Methylbutan-2-one
In simple words: An isopropyl methyl ketone.

Exam Tip: Number from right to left to give carbonyl carbon position 2.

Question 33. \(\text{CH}_3-\text{CO}-\text{O}-\text{CO}-\text{CH}_3\) (represented as acid anhydride structure)
Answer: Ethanoic anhydride (Acetic anhydride)
In simple words: An acid anhydride derived from acetic acid.

Exam Tip: Named by replacing 'acid' in the corresponding carboxylic acid with 'anhydride'.

Question 34. \(\text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{COOC}_2\text{H}_5\)
Answer: Ethyl 2-methylbutanoate
In simple words: Ethyl ester of 2-methylbutanoic acid.

Exam Tip: Ester alkyl group named first as 'ethyl'.

Question 35. \(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\text{CHO}\)
Answer: 2,3-Dimethylbutanal
In simple words: A four-carbon aldehyde with two methyl branches.

Exam Tip: Aldehyde carbon is position 1.

Question 36. \(\text{CH}_3-\text{C}(\text{C}_2\text{H}_5)_2-\text{CH}_2-\text{C}(\text{CH}_3)_2-\text{CH}_3\) (or similar)
Answer: 2,2,4-Trimethyl-4-ethylpentane (or correct systematic equivalent like 2,2-dimethyl-4,4-diethylpentane)
In simple words: A heavily branched pentane hydrocarbon.

Exam Tip: Longest continuous carbon chain determines root name.

Question 37. \(\text{CH}_3-\text{C}\equiv\text{C}-\text{CO}-\text{CH}_2-\text{CH}_3\)
Answer: Hex-4-yn-3-one
In simple words: A conjugated keto-alkyne.

Exam Tip: Ketone group has priority over alkyne for lowest locant in suffix assignment.

Question 38. \(\text{CH}_3-\text{CH}(\text{Br})-\text{CH}(\text{CH}_3)-\text{CH}_3\)
Answer: 2-Bromo-3-methylbutane
In simple words: A butane derivative with bromo and methyl substituents.

Exam Tip: Number from the end closer to the substituent with alphabetical priority (bromo before methyl) if locants are equivalent, but here 2-bromo gives lower locant.

Question 39. \(\text{CH}_3-\text{CH}(\text{OH})-\text{CH}(\text{C}_2\text{H}_5)-\text{CH}_2\text{OH}\)
Answer: 3-Ethylbutane-1,2-diol
In simple words: A butane diol with an ethyl group at position 3.

Question 40. \(\text{CH}_3-\text{C}(\text{OH})(\text{CO})-\text{C}(\text{OH})(\text{CO})-\text{CH}_3\) (or similar alpha-diketone / diol)
Answer: 2,3-Dihydroxybutane-2,3-dione (or analogous structure)
In simple words: A compound containing both alcohol and ketone functionalities.

Exam Tip: Ketones have higher priority than alcohols in IUPAC suffix ranking.

Question 41. \(\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_3\) (Note: transcribed as \(\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2\))
Answer: Buta-1,3-diene
In simple words: A conjugated diene with four carbon atoms.

Exam Tip: Use 'diene' suffix with position locants 1,3.

Question 42. \(\text{CH}_2(\text{COOH})-\text{CH}_2-\text{CH}(\text{CN})-\text{COOH}\) (polyfunctional)
Answer: 2-Cyanobutane-1,4-dioic acid
In simple words: A dicarboxylic acid derivative containing a cyano group.

Exam Tip: Carboxylic acid takes absolute priority over nitrile.

Question 43. \(\text{CH}_2(\text{CHO})-\text{CH}=\text{CH}-\text{CHO}\)
Answer: Pent-2-enedial
In simple words: A five-carbon dialdehyde with a double bond at carbon 2.

Exam Tip: Two terminal aldehyde groups are designated by the suffix '-dial'.

Question 44. \(\text{CH}_2(\text{COOH})-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}(\text{OH})-\text{CH}(\text{CH}_3)-\text{COOH}\) (or polyfunctional acid)
Answer: 4-Hydroxy-3,5-dimethylhexanedioic acid
In simple words: A substituted dicarboxylic acid.

Exam Tip: Carboxylic acid carbons are numbered 1 and 6.

Question 45. \(\text{CH}_2(\text{NH}_2)-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_2\text{NH}_2\)
Answer: 2-Methylbutane-1,4-diamine
In simple words: A diamine with a methyl branch at carbon 2.

Exam Tip: Numbering gives amino groups lowest possible locants (1,4).

Question 46. \(\text{CH}_2(\text{Cl})-\text{CH}(\text{Cl})-\text{CH}_2\text{Cl}\)
Answer: 1,2,3-Trichloropropane
In simple words: Glycerol trichloride, a halogenated alkane.

Exam Tip: Symmetric molecule; numbering from either end gives 1,2,3.

Question 47. \(\text{CH}_3-\text{CH}(\text{OH})-\text{CH}(\text{CH}_3)-\text{CH}(\text{OH})-\text{CH}_3\)
Answer: 3-Methylpentane-2,4-diol
In simple words: A pentane diol with a methyl substituent.

Exam Tip: Number from the end giving lower locant to the first hydroxyl group.

Question 48. \(\text{HOOC}-\text{CH}=\text{CH}-\text{COOH}\)
Answer: But-2-enedioic acid (Fumaric acid / Maleic acid)
In simple words: An unsaturated dicarboxylic acid.

Exam Tip: Specify cis/trans (or E/Z) configuration if stereochemistry is specified.

Question 49. \(\text{HOOC}-\text{CH}_2-\text{CH}(\text{COOH})-\text{CH}_2-\text{CH}(\text{COOH})-\text{CH}_2\text{COOH}\) (Propane-1,2,3-tricarboxylic acid derivative)
Answer: Hexane-1,3,5-tricarboxylic acid (or analogous polycarboxylic acid)
In simple words: An aliphatic polycarboxylic acid.

Exam Tip: When more than two carboxylic acid groups are present, use the 'tricarboxylic acid' nomenclature suffix.

Question 50. \(\text{CH}_3-\text{C}(\text{OH})=\text{C}(\text{CH}_3)-\text{CO}-\text{CH}_3\)
Answer: 4-Hydroxy-3-methylpent-3-en-2-one
In simple words: An enolic ketone derivative.

Exam Tip: Ketone carbonyl takes suffix priority over alcohol hydroxyl.

Question 51. \(\text{HOOC}-\text{CH}_2-\text{CH}(\text{COOH})-\text{CH}_2-\text{CH}(\text{COOH})-\text{COOH}\)
Answer: Butane-1,2,3,4-tetracarboxylic acid (or similar polyacid)
In simple words: A tetracarboxylic acid.

Exam Tip: Use multiplicative suffixes like 'tetracarboxylic acid' when four or more equivalent acid groups are attached.

Question 52. \(\text{Br}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{I}\)
Answer: 1-Bromo-5-iodopentane
In simple words: A haloalkane with both bromo and iodo substituents.

Exam Tip: Alphabetical order determines prefix naming: bromo comes before iodo.

Question 53. \(\text{CH}_3-\text{CH}(\text{OCH}_3)-\dots\)
Answer: 2-Methoxypropane (or derivative depending on attached chain)
In simple words: An ether containing a methoxy group.

Exam Tip: Alkoxy substituent is treated alphabetically.

Q3. What do you mean by isomerism? Discuss its various types giving at least one e.g. in each case.

Question Q3. What do you mean by isomerism? Discuss its various types giving at least one e.g. in each case.
Answer: Compounds having the same molecular formula but different physical and chemical properties are known as isomers, and the phenomenon is called isomerism.
Types of Isomerism:
1. Constitutional (Structural) Isomerism: Compounds with the same molecular formula but different connectivity of atoms.
- Chain Isomerism: e.g., \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\) (butane) and \(\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_3\) (2-methylpropane).
- Position Isomerism: e.g., \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) (propan-1-ol) and \(\text{CH}_3\text{CH}(\text{OH})\text{CH}_3\) (propan-2-ol).
- Functional Isomerism: e.g., \(\text{CH}_3\text{CH}_2\text{OH}\) (ethanol) and \(\text{CH}_3-\text{O}-\text{CH}_3\) (methoxymethane).
- Metamerism: e.g., \(\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3\) (ethoxyethane) and \(\text{CH}_3-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_3\) (1-methoxypropane).
2. Stereoisomerism: Compounds with the same structural formula but different spatial arrangement of atoms.
- Geometrical Isomerism: e.g., cis-but-2-ene and trans-but-2-ene.
- Optical Isomerism: e.g., d-lactic acid and l-lactic acid.
In simple words: Isomerism occurs when two or more compounds share the exact same chemical formula but are arranged differently, leading to different behaviours.

Exam Tip: Clearly categorize isomerism into structural and stereoisomerism before diving into sub-types.

Q4. Explain the following with one e.g. in each case: - (i) Hemolytic Fission (ii) Heterolytic fission of covalent bonds

Question Q4. Explain the following with one e.g. in each case: (i) Hemolytic Fission (ii) Heterolytic fission of covalent bonds
Answer: (i) Hemolytic Fission: A covalent bond breaks symmetrically such that each bonded atom retains one of the shared electrons, resulting in the formation of free radicals. Example: Photochemical dissociation of chlorine into chlorine radicals: \(\text{Cl}-\text{Cl} \xrightarrow{h\nu} \dot{\text{Cl}} + \dot{\text{Cl}}\).
(ii) Heterolytic Fission: A covalent bond breaks unsymmetrically such that both shared electrons remain with one of the fragments, producing ions (a cation and an anion). Example: Ionization of chloromethane: \(\text{CH}_3\text{Cl} \rightarrow \text{CH}_3^+ + \text{Cl}^-\).
In simple words: Hemolytic fission splits a bond evenly to form radicals, while heterolytic fission splits it unevenly to form positive and negative ions.

Exam Tip: Use single-barbed fishhook arrows (\(\cap\)) for homolytic mechanisms and double-barbed curved arrows for heterolytic mechanisms.

Q5. What are carbocations? Discuss their various types.

Question Q5. What are carbocations? Discuss their various types.
Answer: A carbocation is an organic species containing a carbon atom bearing a positive charge and a sextet of electrons (6 electrons in its valence shell).
Types of Carbocations:
1. Primary (\(1^\circ\)) Carbocation: The positively charged carbon is attached to only one other carbon atom (e.g., \(\text{CH}_3\text{CH}_2^+\)).
2. Secondary (\(2^\circ\)) Carbocation: The positively charged carbon is attached to two other carbon atoms (e.g., \((\text{CH}_3)_2\text{CH}^+\)).
3. Tertiary (\(3^\circ\)) Carbocation: The positively charged carbon is attached to three other carbon atoms (e.g., \((\text{CH}_3)_3\text{C}^+\)).
Stability order: \(3^\circ > 2^\circ > 1^\circ\) due to inductive effect, hyperconjugation, and steric stabilization.
In simple words: Carbocations are carbon ions with a positive charge. Tertiary carbocations are the most stable.

Exam Tip: Remember stability increases with alkyl substitution due to hyperconjugation and electron-releasing inductive effects.

Q6. What are electrophiles & nucleophiles? Explain with e.g.

Question Q6. What are electrophiles & nucleophiles? Explain with e.g.
Answer: - Electrophiles (Electron-loving species): Electron-deficient species that seek electron pairs. They can be positively charged or neutral molecules with vacant orbitals. Example: \(\text{NO}_2^+\), \(\text{BF}_3\).
- Nucleophiles (Nucleus-loving species): Electron-rich species having an unshared pair of electrons that donate electron density to electron-deficient centers. Example: \(\text{OH}^-, \text{NH}_3\).
In simple words: Electrophiles are electron seekers, while nucleophiles are electron donors.

Exam Tip: Lewis acids act as electrophiles, and Lewis bases act as nucleophiles.

Q7. How does hyper conjugation effect explain the stability of alkenes?

Question Q7. How does hyper conjugation effect explain the stability of alkenes?
Answer: Hyperconjugation involves the delocalization of \(\sigma\)-electrons of the \(\text{C}-\text{H}\) bond of an alkyl group directly attached to an unsaturated carbon atom into an adjacent unsaturated \(\pi\)-orbital. Greater the number of \(\alpha\)-hydrogens present in an alkene, greater is the number of contributing hyperconjugative structures, and hence greater is its stability. For example, 2-methylbut-2-ene is more stable than but-1-ene because it has more \(\alpha\)-hydrogens.
In simple words: More alkyl groups attached to the double bond mean more hyperconjugative structures, making the alkene more stable.

Question Q8. What is resonance? How does resonance explain that all carbon-carbon bond lengths in benzene are equal (139 pm)?
Answer: Resonance is the phenomenon where a single chemical structure cannot fully represent all the properties of a molecule, so two or more contributing structures are used, and the true structure is a resonance hybrid. In benzene (\(\text{C}_6\text{H}_6\)), two equivalent Kekule structures contribute equally to the hybrid. This delocalization of \(\pi\)-electrons over all six carbon atoms gives every carbon-carbon bond a partial double bond character, resulting in an intermediate bond length of 139 pm (between a pure single bond of 154 pm and a pure double bond of 134 pm).
In simple words: Resonance spreads electrons evenly around the benzene ring, making all carbon-carbon bonds identical in length.

Exam Tip: Emphasize that benzene does not rapidly oscillate between two structures; rather, it exists as a single hybrid with delocalized pi electrons.

Q9. What is resonance effect? What are its various types?

Question Q9. What is resonance effect? What are its various types?
Answer: The resonance effect (\(R\) effect) is defined as the polarity produced in a molecule by the interaction of two \(\pi\)-bonds or a \(\pi\)-bond and a lone pair of electrons present on an adjacent atom.
Types of Resonance Effect:
1. Positive Resonance Effect (\(+R\) effect): Groups release electrons toward the conjugated system through resonance (e.g., \(-\text{OH}, -\text{OR}, -\text{NH}_2\)).
2. Negative Resonance Effect (\(-R\) effect): Groups withdraw electrons away from the conjugated system through resonance (e.g., \(-\text{NO}_2, -\text{CHO}, -\text{CN}\)).
In simple words: The resonance effect describes how atoms push or pull electrons through conjugated bonds.

Exam Tip: Groups with lone pairs on the atom directly attached to the ring generally show \(+R\) effect.

Q10. Describe chemistry of Lassaigne’s test used for the detection of (i) Nitrogen (ii) Halogens (iii) Sulphur (iv) Phosphorus.

Question Q10. Describe chemistry of Lassaigne’s test used for the detection of (i) Nitrogen (ii) Halogens (iii) Sulphur (iv) Phosphorus.
Answer: Lassaigne's test converts covalently bonded elements in organic compounds into ionic forms by fusing the compound with sodium metal.
(i) Nitrogen: Sodium fusion extract reacts with iron(II) sulphate and acidified with concentrated sulphuric acid, yielding Prussian blue precipitate:
\(\text{Na} + \text{C} + \text{N} \rightarrow \text{NaCN}\)
\(6\text{NaCN} + \text{FeSO}_4 \rightarrow \text{Na}_4[\text{Fe}(\text{CN})_6]\)
\(3\text{Na}_4[\text{Fe}(\text{CN})_6] + 4\text{Fe}^{3+} \rightarrow \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \cdot x\text{H}_2\text{O}\) (Prussian blue).
(ii) Halogens: Sodium extract is acidified with nitric acid and treated with silver nitrate solution. A white precipitate soluble in ammonium hydroxide indicates chlorine; pale yellow precipitate sparingly soluble indicates bromine; yellow precipitate insoluble indicates iodine:
\(\text{NaX} + \text{AgNO}_3 \rightarrow \text{AgX}\downarrow + \text{NaNO}_3\).
(iii) Sulphur: Sodium extract treated with sodium nitroprusside gives a violet coloration:
\(\text{Na}_2\text{S} + \text{Na}_2[\text{Fe}(\text{CN})_5\text{NO}] \rightarrow \text{Na}_4[\text{Fe}(\text{CN})_5\text{NOS}]\) (Violet).
(iv) Phosphorus: The organic compound is heated with sodium peroxide, and the resulting solution is boiled with nitric acid and treated with ammonium molybdate, giving a canary yellow precipitate:
\(\text{H}_3\text{PO}_4 + 12(\text{NH}_4)_2\text{MoO}_4 + 21\text{HNO}_3 \rightarrow (\text{NH}_4)_3\text{PO}_4 \cdot 12\text{MoO}_3\downarrow + 21\text{NH}_4\text{NO}_3 + 12\text{H}_2\text{O}\).
In simple words: Lassaigne's test fuses organic substances with sodium metal to convert covalent elements into soluble ionic salts that can be easily identified using specific chemical reagents.

Exam Tip: Always mention that sodium fusion transforms covalent bonds into ionic salts before testing with reagents.

Instructions:

1. All questions are compulsory.

2. Please give the explanation for the answer where applicable.

Q1 - Define Organic chemistry.

Q2 - Organic chemistry is treated as a separate branch. Why?

Q3 - Write a short on the tetravalency of carbon. 

Q4 - What are the main characteristics of a Homologous series. 

Q5 - What are aliphatic hydrocarbons? Give their classification. 

Q6 - What are primary, secondary tertiary and quaternary carbon atoms? 

Q7 - What is the parent or principal chain in an organic compound? 

Q8 - Define primary suffix. Also give their names. 

Q9 - What is a Lassaigne's Extract? 

Instructions:

1. All questions are compulsory.

2. Please give the explanation for the answer where applicable.

Q1 - What is a homologous series? Explain with example. 

Q2 - What is an alkyl group? 

Q3 - Explain sp2 hybridization? 

Q4 - What do understand by sp hybridization? 

Q5 - Name the various systems in which organic compounds are named? 

Q6 - Write Resonance Structures of Nitrobenzene.

Q7 - Write a Short note on Electromeric Effect and Hyperconjucation. 

Q8 - How Organic Reaction can be classified. 

Q9 - Define Distillation and Differential Extraction. 

Q10 - Write test for Nitrogen and Sulphur. 

Q1 - What are isomers? Explain with example. (2 Marks)

Q2 - What is a sigma (s) bond?(1 Mark)

Q3 - What is the pi (p) bond? (1 Mark)

Q4 - Explain sp3 hybridisation in the terms of carbon atom. (1 Mark)

Q5 - What is the word root? Give example. (2 Marks)

Q6 - Define Metamerism and Position Isomerism. Give Examples. (3 Marks)

Q7 - What are Homolytic and Hetrolytic Fission? (3 Marks)

Q8 - Solve the following Problems:

A) During estimation of nitrogen present in an organic compound by Kjeldahl‘s method the ammonia evolved from 0.8 gm of nitrogen, neutralized 10 ml of 1 M . Find out the percentage of nitrogen in the compound.

B) In sulphur estimation, 0.160 gm of an organic compound gave 0.4820 gm of barium sulphate .What is the percentage of sulphur in the compound ? (5 Marks)

Q9 - Recognize the functional group present in the following compounds. (2 Marks)