CBSE Class 11 Chemistry Organic Chemistry Assignment

ONE MARK QUESTIONS

1 Write the structural formula of the following (1 mark each)

Question 1(a). 2,3-Dibromo-1-phenylpentane
Answer: \(\text{C}_6\text{H}_5-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}(\text{Br})-\text{CH}_2-\text{CH}_3\)
In simple words: A five-carbon alkane chain with a phenyl ring at carbon 1 and bromine atoms at carbons 2 and 3.

Exam Tip: Number the pentane chain starting from the carbon attached to the phenyl group.

Question 1(b). 6-Hydroxyheptanal
Answer: \(\text{CH}_3-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CHO}\)
In simple words: A seven-carbon aldehyde containing an -OH group at carbon 6.

Exam Tip: The aldehyde carbon gets priority as carbon 1.

Question 1(c). p-Nitroaniline
Answer: Benzene ring substituted with an \(-\text{NH}_2\) group and a \(-\text{NO}_2\) group at the para position to each other.
In simple words: A benzene ring with amino and nitro groups directly opposite to each other.

Exam Tip: 'p' stands for para (1,4-positions on the benzene ring).

Question 1(d). Cyclohex-2-en-1-ol
Answer: A six-membered cyclic ring with a double bond between carbons 2 and 3, and an \(-\text{OH}\) group at carbon 1.
In simple words: A cyclohexene ring carrying an alcohol group adjacent to the double bond.

Exam Tip: The principal functional group \(-\text{OH}\) determines carbon 1.

Question 1(e). 4-Ethyl-1-fluoro-2-nitrobenzene
Answer: Benzene ring with fluorine at carbon 1, nitro at carbon 2, and an ethyl group at carbon 4.
In simple words: A benzene ring substituted with fluoro, nitro, and ethyl groups following lowest locant rules.

Exam Tip: Numbering starts from the principal substituent or alphabetical priority if halo substituent is present.

Question 1(f). Pent-4-en-2-ol
Answer: \(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}_3\)
In simple words: A five-carbon chain with a double bond at carbon 4 and an alcohol group at carbon 2.

Exam Tip: Alcohol suffix \(-\text{ol}\) takes precedence over the alkene for principal functional group status.

Question 1(g). 3,4-Dimethylphenol
Answer: Benzene ring with an \(-\text{OH}\) group at carbon 1, and methyl groups at carbons 3 and 4.
In simple words: Phenol substituted with two methyl groups at the meta and para positions relative to the hydroxyl group.

Exam Tip: Phenol carbon attached to \(-\text{OH}\) is always carbon 1.

Question 1(h). 3-Nitrocyclohexene
Answer: A six-membered cyclic ring with a double bond between carbons 1 and 2, and a nitro group at carbon 3.
In simple words: A cyclohexene ring with a nitro group next to the double bond carbon.

Exam Tip: Double bond carbons get lower locants (1 and 2) than the substituent.

Question 1(i). m-Nitrophenol
Answer: Benzene ring with an \(-\text{OH}\) group at carbon 1 and a nitro group at carbon 3.
In simple words: A benzene ring with hydroxyl and nitro groups at meta positions to each other.

Exam Tip: 'm' stands for meta (1,3-positions).

Question 2(a). 2,2-dimethyl pentane
Answer: Incorrect.
Correct IUPAC name: 2,2-Dimethylpentane
In simple words: Corrected spacing and capitalization in the IUPAC name.

Exam Tip: IUPAC names are written as a single continuous word with numbers separated by commas and letters by hyphens.

Question 2(b). 2,5,7-trimethyloctane
Answer: Incorrect.
Correct IUPAC name: 2,4,7-Trimethyloctane
In simple words: The numbering must follow the lowest locant rule giving substituents the lowest possible numbers (2,4,7 instead of 2,5,7).

Exam Tip: Always test numbering from both ends of the longest carbon chain to get the lowest set of locants.

Question 2(c). 4-chloro-2-methylpentane
Answer: Incorrect.
Correct IUPAC name: 2-Chloro-4-methylpentane
In simple words: Numbering from the other end gives locants 2 and 4, which is lower than 4 and 2.

Exam Tip: Lowest locant rule takes precedence over alphabetical order when choosing numbering direction.

Question 2(d). but-4-ol-1-yne
Answer: Incorrect.
Correct IUPAC name: But-1-yn-4-ol (or But-3-yn-1-ol depending on orientation; standard numbering gives But-3-yn-1-ol)
In simple words: The principal functional group \(-\text{OH}\) must get the lowest possible number (carbon 1).

Exam Tip: Alcohol suffix \(-\text{ol}\) has higher seniority than the triple bond \(-\text{yne}\).

Question 2(e). 1-hydroxyethanoic acid
Answer: Incorrect.
Correct IUPAC name: 2-Hydroxyethanoic acid (commonly known as glycolic acid)
In simple words: The carboxylic acid carbon is always carbon 1, so the hydroxyl group is at carbon 2.

Exam Tip: Carboxylic acid carbon is mandatorily carbon 1 in naming.

Question 2(f). Propan-2-ol
Answer: Correct.
In simple words: The name accurately describes a three-carbon chain with an alcohol group at carbon 2.

Exam Tip: Propan-2-ol is also commonly known as isopropyl alcohol.

Question 2(g). 2-Ethylpent-2-ene
Answer: Incorrect.
Correct IUPAC name: 3-Methylhex-2-ene
In simple words: The longest continuous carbon chain has 6 carbons (hexane), not 5.

Exam Tip: Always identify the absolute longest carbon chain containing the principal functional group or multiple bond.

Question 2(h). Hexa-1,6-diene
Answer: Incorrect.
Correct IUPAC name: Hexa-1,5-diene
In simple words: Numbering should provide the lowest possible locants for the double bonds (1 and 5 instead of 1 and 6).

Exam Tip: Number from the end that gives the lowest locant at the first point of difference.

Question 2(i). 4-ethyl-3-methylhexane
Answer: Correct.
In simple words: The numbering and alphabetical ordering of substituents are correct.

Exam Tip: Ethyl is written before methyl alphabetically (E before M).

Question 2(j). Pent-4-ene
Answer: Incorrect.
Correct IUPAC name: Pent-1-ene
In simple words: Numbering must start from the end closer to the double bond to give it the lowest locant (1 instead of 4).

Exam Tip: Double bonds take precedence over alkyl substituents for lowest locants.

3 Write IUPAC names for the following (1 mark each)

Question 3(a). [Structure: Isopropyl group / Propane derivative or branched chain]
Answer: 2-Methylpropane (or Propane if simple representation, standard branched structure gives 2-methylpropane)
In simple words: A three-carbon main chain with a methyl group at carbon 2.

Exam Tip: Commonly known as isobutane.

Question 3(b). [Structure: Cyclohexane ring]
Answer: Cyclohexane
In simple words: A saturated six-membered carbocyclic ring.

Exam Tip: Cyclic saturated hydrocarbons use the prefix 'cyclo-' followed by the alkane name.

Question 3(c). [Structure: Pentane derivative / branched chain]
Answer: 2-Methylbutane
In simple words: A four-carbon chain with a methyl substituent at carbon 2.

Exam Tip: Also known as isopentane.

Question 3(d). [Structure: Pent-2-ene]
Answer: Pent-2-ene
In simple words: A five-carbon chain with a double bond starting at carbon 2.

Exam Tip: Number from the end nearest to the double bond.

Question 3(e). [Structure: Methylenecyclohexane or substituted ring]
Answer: Methylenecyclohexane (or Cyclohexylidenemethane)
In simple words: A cyclohexane ring connected via a double bond to a \(=\text{CH}_2\) group.

Exam Tip: Exocyclic double bonds are named using the 'ylidene' or 'methylene' prefix.

Question 3(f). (CH3)3C-CH2-CHO
Answer: 3,3-Dimethylbutanal
In simple words: A four-carbon aldehyde chain with two methyl groups at carbon 3.

Exam Tip: Aldehyde carbon is carbon 1.

Question 3(g). HO-CH2-CH2-COOH
Answer: 3-Hydroxypropanoic acid
In simple words: A three-carbon carboxylic acid with an alcohol group at carbon 3.

Exam Tip: Carboxylic acid has higher priority than alcohol.

Question 3(h). CH3 CO CH2 CHO
Answer: 3-Oxobutanal
In simple words: A four-carbon chain containing both an aldehyde group and a ketone carbonyl group.

Exam Tip: Aldehyde has higher priority than ketone; ketone is prefixed as 'oxo'.

Question 3(i). (CH3)3C CH (C2H5) CH3
Answer: 2,2,3-Trimethylpentane
In simple words: A five-carbon main chain with methyl groups at carbons 2, 2, and 3.

Exam Tip: Longest chain selection must be carefully checked; pentane is the longest chain here.

Question 3(j). CH2 = CH CH2 CH = CH2
Answer: Penta-1,4-diene
In simple words: A five-carbon chain with two double bonds at positions 1 and 4.

Exam Tip: Use 'diene' suffix for two double bonds.

Question 3(k). HOOC CH (CH3) CH2 CH2 CH3
Answer: 2-Methylpentanoic acid
In simple words: A five-carbon carboxylic acid chain with a methyl group at carbon 2.

Exam Tip: Acid carbon is carbon 1.

Question 3(l). CH3 CH2 CH2 NO2
Answer: 1-Nitropropane
In simple words: A three-carbon alkane chain with a nitro substituent at carbon 1.

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

Question 3(m). HO CH (CH3) CH2 CHO
Answer: 3-Hydroxybutanal
In simple words: A four-carbon aldehyde chain with a hydroxyl group at carbon 3.

Exam Tip: Numbering starts from the aldehyde carbon.

Question 3(n). H2N CH=CH-COOH
Answer: 3-Aminoprop-2-enoic acid
In simple words: An acrylic acid derivative with an amino group at carbon 3 and a double bond at carbon 2.

Exam Tip: Carboxylic acid takes principal functional group priority.

Question 3(o). CH3 CO CO C2H5
Answer: Hexane-2,3-dione
Answer: Hexane-2,3-dione
In simple words: A six-carbon dicarbonyl chain with ketone groups at carbons 2 and 3.

Exam Tip: Number from the end closer to the carbonyl groups.

Question 3(p). CH3 CH2 CH2 COO C2H5
Answer: Ethyl butanoate
In simple words: An ester formed from butanoic acid and ethanol.

Exam Tip: The alkyl group attached to oxygen is named first as a separate word (ethyl), followed by the acid part with '-anoate'.

Question 3(q). CH3 CH Cl2
Answer: 1,1-Dichloroethane
Answer: 1,1-Dichloroethane
In simple words: A two-carbon ethane chain with two chlorine atoms on the same carbon.

Exam Tip: Geminal dihalides have halogens on the same carbon atom.

Question 3(r). HOOC-COOH
Answer: Ethanedioic acid (Commonly oxalic acid)
In simple words: A two-carbon dicarboxylic acid.

Exam Tip: Smallest dicarboxylic acid containing two carboxyl groups.

Question 3(s). CH2 = CH CH2 C\(\equiv\)C CH3
Answer: Hex-1-en-4-yne
In simple words: A six-carbon chain with a double bond at carbon 1 and a triple bond at carbon 4.

Exam Tip: When both double and triple bonds are present, numbering gives lowest locants to multiple bonds collectively, but if there's a tie, double bond gets lower locant.

Question 3(t). I-(CH2)3 COOH
Answer: 4-Iodobutanoic acid
In simple words: A four-carbon carboxylic acid with an iodine atom at carbon 4.

Exam Tip: Carboxylic acid carbon is carbon 1.

Question 3(u). CH3 CO COOH
Answer: 2-Oxopropanoic acid (Pyruvic acid)
In simple words: A three-carbon alpha-keto acid.

Exam Tip: Carboxylic acid takes priority over ketone.

Question 3(v). (CH3 )2 CH CH(C2H5) CH3
Answer: 2,3-Dimethylpentane
In simple words: A five-carbon chain with methyl substituents at carbons 2 and 3.

Exam Tip: Identify the longest continuous chain correctly (pentane, not hexane).

Question 3(w). CH3CH(CH3)CH2CH2CH=CH2
Answer: 5-Methylhex-1-ene
In simple words: A six-carbon alkene chain with a methyl group at carbon 5.

Exam Tip: Numbering starts from the end nearest to the double bond.

Question 3(x). CH3C(CH3)2CH2CH2CH2CH(C2H5)CH3
Answer: 2,2,7-Trimethyloctane
In simple words: An eight-carbon alkane chain with methyl groups at carbons 2, 2, and 7.

Exam Tip: Find the longest continuous carbon chain (octane).

TWO MARKS QUESTIONS

Question 1(a). Distinguish between Nucleophile and Electrophile with example (2 marks each)
Answer:
- Nucleophile (Nu-): An electron-rich species that donates an electron pair to form a covalent bond. Examples: \(\text{OH}^-, \text{NH}_3\).
- Electrophile (E+): An electron-deficient species that accepts an electron pair. Examples: \(\text{NO}_2^+, \text{CH}_3^+\).
In simple words: Nucleophiles seek nuclei (positive centers), while electrophiles seek electrons (negative centers).

Exam Tip: Mention both definition and suitable chemical examples to secure full marks.

Question 1(b). Distinguish between Heterolytic and Homolytic fissions with example (2 marks each)
Answer:
- Homolytic Fission: The bonding electrons are shared equally between the two atoms, forming free radicals. Example: \(\text{Cl}-\text{Cl} \xrightarrow{\text{UV}} \text{Cl}^\bullet + \text{Cl}^\bullet\).
- Heterolytic Fission: The bonding electrons are retained by one atom, forming ions (cations and anions). Example: \(\text{CH}_3-\text{Cl} \rightarrow \text{CH}_3^+ + \text{Cl}^-\).
In simple words: Homolytic splits evenly into radicals; heterolytic splits unevenly into ions.

Exam Tip: Use fish-hook arrows (\(\cap\)) for homolytic and full curved arrows for heterolytic fission.

Question 2(a). Give reason: Metallic sodium is used to prepare Lassaigne’s extract.
Answer: Metallic sodium is reactive and is used to convert covalently bonded elements (nitrogen, sulphur, halogens) in an organic compound into water-soluble ionic sodium salts (\(\text{NaCN}, \text{Na}_2\text{S}, \text{NaX}\)) which can be easily identified using qualitative reagents.
In simple words: Sodium converts covalent organic compounds into ionic compounds that dissolve in water for testing.

Exam Tip: Highlight the conversion of covalent bonds to ionic sodium salts.

Question 2(b). Give reason: Lassaigne’s extract is boiled with dil HNO3 before testing for halogens.
Answer: Lassaigne's extract is boiled with dilute \(\text{HNO}_3\) to decompose any sodium cyanide (\(\text{NaCN}\)) or sodium sulphide (\(\text{Na}_2\text{S}\)) present, which would otherwise interfere with the silver nitrate test by forming precipitates of silver cyanide or silver sulphide.
In simple words: Boiling with nitric acid destroys interfering cyanide and sulfide ions before adding silver nitrate.

Exam Tip: State clearly that \(\text{HCN}\) and \(\text{H}_2\text{S}\) gases are expelled during this boiling step.

Question 3(a). Arrange the following in the increasing order of their stabilities: \((CH_3)_3CCH_2^+, (CH_3)_3C^+, CH_3CH_2CH_2^+, CH_3CH^+CH_2CH_3\)
Answer:
Increasing order of stability:
\(\text{CH}_3\text{CH}_2\text{CH}_2^+ < (\text{CH}_3)_3\text{CCH}_2^+ < \text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3 < (\text{CH}_3)_3\text{C}^+\)
In simple words: Stability of carbocations increases with hyperconjugation and inductive effect (\(3^\circ > 2^\circ > 1^\circ\)). Neopentyl carbocation is primary but hindered; simple primary is least stable.

Exam Tip: Tertiary carbocations are most stable due to maximum alpha hydrogens and hyperconjugative structures.

Question 3(b). Arrange the following in the increasing order of their stabilities: \((CH_3)_2\bar{C}H, (CH_3)_3\bar{C}, CH_3\bar{C}H_2, \bar{C}H_3\)
Answer:
Increasing order of stability:
\((\text{CH}_3)_3\bar{\text{C}} < (\text{CH}_3)_2\bar{\text{C}}\text{H} < \text{CH}_3\bar{\text{C}}\text{H}_2 < \bar{\text{C}}\text{H}_3\)
In simple words: Carbanion stability decreases with an increasing number of electron-donating alkyl groups due to +I effect intensifying the negative charge.

Exam Tip: Carbanion stability order is opposite to carbocation stability order (\(1^\circ > 2^\circ > 3^\circ\)).

Question 4. Define isomerism. Draw the structural isomers of \(C_5H_{12}\) and \(C_4H_{10}\)
Answer:
Definition: Compounds having the same molecular formula but different physical and chemical properties due to different arrangements of atoms are called isomers, and the phenomenon is known as isomerism.
- Structural isomers of \(\text{C}_5\text{H}_{12}\) (3 isomers):
1. Pentane (\(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3\))
2. 2-Methylbutane (\(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3\))
3. 2,2-Dimethylpropane (\(\text{C}(\text{CH}_3)_4\))
- Structural isomers of \(\text{C}_4\text{H}_{10}\) (2 isomers):
1. Butane (\(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3\))
2. 2-Methylpropane (\((\text{CH}_3)_2\text{CH}-\text{CH}_3\))
In simple words: Isomers have identical formulas but different structural connectivity.

Exam Tip: Draw all structures clearly and write their IUPAC names to ensure full credit.

Question 5. Give a pair of functional isomers of (a) \(C_2H_4O_2\) (b) \(C_3H_6\)
Answer:
(a) For \(\text{C}_2\text{H}_4\text{O}_2\):
- Acetic acid / Ethanoic acid (\(\text{CH}_3\text{COOH}\))
- Methyl formate / Methyl methanoate (\(\text{HCOOCH}_3\))
(b) For \(\text{C}_3\text{H}_6\):
- Propene (\(\text{CH}_3-\text{CH}=\text{CH}_2\))
- Cyclopropane (cyclic ring)
In simple words: Functional isomers belong to different homologous series while sharing the same molecular formula.

Exam Tip: Always double check that the total number of carbon, hydrogen, and oxygen atoms matches in both isomers.

Question 6. Write short note on the following with the help of an example: (a) Metamerism (b) Position isomerism
Answer:
(a) Metamerism: Arises due to different alkyl chains attached to the same polyvalent functional group (such as \(-\text{O}-, -\text{S}-, -\text{NH}-, -\text{C}(=\text{O})-\)).
Example: Methoxypropane (\(\text{CH}_3-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_3\)) and Ethoxyethane (\(\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3\)).
(b) Position Isomerism: Arises when compounds with the same molecular formula and carbon skeleton differ in the position of the functional group or multiple bond.
Example: But-1-ene (\(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\)) and But-2-ene (\(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\)).
In simple words: Metamers differ by alkyl groups around a heteroatom; position isomers differ by where a functional group sits on the chain.

Exam Tip: Provide clear structural formulas and IUPAC names for both examples.

THREE MARKS QUESTIONS

Question 1. Write the bond line formula for (a) Pentan-2-one (b) 3-Formyl hexane-1,6-dioic acid (c) 2,3-dimethyl butanal
Answer:
(a) Pentan-2-one: A zigzag line of 5 carbons with a double-bonded oxygen (carbonyl) on vertex 2.
(b) 3-Formyl hexane-1,6-dioic acid: A 6-carbon zigzag chain with carboxylic acid groups at both ends (1 and 6) and an aldehyde group (\(-\text{CHO}\)) branching off carbon 3.
(c) 2,3-Dimethyl butanal: A 4-carbon zigzag chain with an aldehyde group at carbon 1 and methyl groups branching off carbons 2 and 3.
In simple words: Bond line formulas represent carbon chains as lines and vertices without writing carbons and hydrogens explicitly.

Exam Tip: Ensure terminal lines and double-bonded oxygen atoms are clearly visible.

Question 2. Which is more stable and why? (a) \(NO_2CH_2CH_2O^-\) or \(CH_3CH_2CH_2O^-\) (b) 3,3-dimethyl-1-butene or 3-methyl-1-pentene (c) \(CH_3CH_2^+\) or \((CH_3)_2CH^+\)
Answer:
(a) \(\text{NO}_2\text{CH}_2\text{CH}_2\text{O}^-\) is more stable because the electron-withdrawing \(-\text{NO}_2\) group exerts a strong -I effect, dispersing the negative charge on oxygen.
(b) 3-Methyl-1-pentene is more stable because it has more alpha hydrogens (hyperconjugation) making the double bond more substituted and stable than 3,3-dimethyl-1-butene.
(c) \((\text{CH}_3)_2\text{CH}^+\) is more stable because it is a secondary carbocation stabilized by more hyperconjugative structures and +I effect than the primary \(\text{CH}_3\text{CH}_2^+\) carbocation.
In simple words: Charge dispersal increases stability in anions and carbocations; more alkyl substitution increases alkene stability.

Exam Tip: Always cite inductive effect, hyperconjugation, or resonance to justify stability.

Question 3. Explain with examples: (a) Resonance effect (b) Inductive effect (c) Electromeric effect
Answer:
(a) Resonance Effect: Polarity produced in a molecule by the interaction of two pi-bonds or a pi-bond and a lone pair of electrons. Example: Phenol, where the lone pair on oxygen delocalizes into the benzene ring (+R effect).
(b) Inductive Effect: Polarization of a sigma bond caused by the electronegativity difference of adjacent atoms. Example: Chloroethane (\(\text{CH}_3-\text{CH}_2-\text{Cl}\)), where chlorine pulls sigma electrons towards itself (-I effect).
(c) Electromeric Effect: Temporary complete transfer of a shared pair of pi-electrons to one of the atoms under the influence of an attacking reagent. Example: Addition of acids to alkenes.
In simple words: Resonance involves pi-electrons, inductive involves sigma bond polarization, and electromeric is a temporary transfer during reactions.

Exam Tip: Clearly distinguish permanent effects (Resonance, Inductive) from temporary effects (Electromeric).

Question 4. What is the difference between distillation, distillation under reduced pressure and steam distillation?
Answer:
- Simple Distillation: Used for liquids that boil without decomposition and have sufficiently different boiling points (e.g., chloroform and aniline).
- Distillation under Reduced Pressure: Used for liquids that decompose at or below their normal boiling points by lowering the pressure, which lowers the boiling point (e.g., glycerol).
- Steam Distillation: Used for organic compounds that are steam volatile, insoluble in water, and decompose near their boiling points (e.g., aniline in water).
In simple words: Different distillation methods are chosen based on whether a liquid boils normally, decomposes at high heat, or is steam volatile.

Exam Tip: Mention the specific property of the organic liquid for each technique.

Question 5. Discuss the principle behind the following techniques taking an example in each case: Crystallization, Chromatography, Sublimation
Answer:
- Crystallization: Based on the difference in solubilities of the compound and impurities in a suitable solvent. Example: Purification of impure copper sulfate.
- Chromatography: Based on the differential adsorption or partition of components of a mixture between a stationary phase and a mobile phase. Example: Separation of plant pigments using column or thin-layer chromatography.
- Sublimation: Based on the principle that certain solid substances convert directly from solid to vapor phase on heating without melting. Example: Separation of camphor from a mixture of salt and camphor.
In simple words: Crystallization uses solubility differences, chromatography uses phase distribution, and sublimation uses direct solid-to-vapor transitions.

Exam Tip: Clearly state the primary physical property exploited in each separation technique.

Question 6. Write one chemical test to detect the presence of the following elements in a given organic compound: (a) Nitrogen (b) Sulphur (c) Phosphorous
Answer:
(a) Nitrogen: Lassaigne's extract is treated with ferrous sulfate and acidified with concentrated sulfuric acid; formation of Prussian blue precipitate indicates nitrogen.
(b) Sulphur: Lassaigne's extract treated with sodium nitroprusside solution gives a deep violet coloration.
(c) Phosphorous: The organic compound is heated with sodium peroxide and boiled with nitric acid, followed by ammonium molybdate; a canary yellow precipitate confirms phosphorus.
In simple words: Specific chemical reagents produce characteristic colored precipitates or solutions for each element.

Exam Tip: Mention color changes (e.g., Prussian blue, violet, canary yellow) clearly in your answers.

Question 7. Draw the resonating structures of (a) Aniline (\(C_6H_5NH_2\)) (b) Nitrobenzene (\(C_6H_5NO_2\)) (c) Prop-2-enal
Answer:
(a) Aniline: The nitrogen lone pair delocalizes into the benzene ring, creating negative charge density at ortho and para positions with positive charge on nitrogen.
(b) Nitrobenzene: The electron-withdrawing nitro group withdraws pi electrons from the benzene ring via -R effect, creating positive charge density at ortho and para positions.
(c) Prop-2-enal (\(\text{CH}_2=\text{CH}-\text{CHO}\)): The carbonyl oxygen withdraws pi electrons through the conjugated system, giving structures like \(^+\text{CH}_2-\text{CH}=\text{CH}-\text{O}^-\).
In simple words: Resonating structures show how electron pairs shift across conjugated systems through curved arrows.

Exam Tip: Always use double-headed arrows (\(\leftrightarrow\)) between contributing structures.

Question 8. Explain the principle behind (a) Carius method of estimation of halogens (b) Kjeldahl’s method of estimation of nitrogen
Answer:
(a) Carius Method: A known mass of organic compound is heated with fuming nitric acid in the presence of silver nitrate in a Carius tube. Carbon and hydrogen are oxidized to carbon dioxide and water, while halogen is converted to silver halide (\text{AgX}), which is filtered, washed, and weighed.
(b) Kjeldahl’s Method: The organic compound is digested with concentrated sulfuric acid to convert nitrogen quantitatively into ammonium sulfate. The resulting solution is treated with excess sodium hydroxide, and the liberated ammonia gas is absorbed in a known volume of standard acid solution, the excess of which is determined by titration.
In simple words: Carius converts halogens to silver halide precipitates for weighing; Kjeldahl converts nitrogen to ammonia for acid-base titration.

Exam Tip: Mention quantitative conversion and formula calculation steps for full marks.