CBSE Class 12 Chemistry Organic Chemistry Questions

SOME NOTEWORTHY POINTS

Question 2. If there are two or more functional groups, the order of decreasing priority is : \(-\text{COOH} > -\text{SO}_3\text{H} > \text{anhydride} > \text{Ester} > \text{Acid halide} > \text{nitrile} > \text{aldehyde} > \text{ketone} > \text{alcohol} > \text{amine} \gg \text{C}=\text{C} < > -\text{C}\equiv\text{C}-\)
Answer: Priority order dictates the principal functional group used for suffix naming in IUPAC nomenclature. Groups appearing earlier in the sequence take precedence as the suffix, while others are treated as prefixes.
In simple words: The highest priority functional group determines the main suffix of the chemical name.

Exam Tip: Memorize the top four priority groups (\(-\text{COOH}, -\text{SO}_3\text{H}, \text{anhydride}, \text{ester}\)) as they frequently appear in multi-functional IUPAC naming questions.

Question 3. Anti Markownikov's addition takes place only with \text{HBr} in presence of a peroxide.
Answer: The peroxide effect (free radical mechanism) is only applicable to hydrogen bromide (\text{HBr}). For \text{HCl}, the \(\text{H}-\text{Cl}\) bond is too strong to be homolytically cleaved by a free radical. For \text{HI}, the iodine free radicals (\(\text{I}^{\bullet}\)) readily combine with each other to form iodine molecules (\(\text{I}_2\)) rather than adding across the double bond.
In simple words: Only \text{HBr} undergoes anti-Markownikov addition with peroxides because chlorine and iodine reactions are energetically unfavorable.

Exam Tip: Always check for the presence of organic peroxide (\(\text{R}-\text{O}-\text{O}-\text{R}\)) and \text{HBr} before applying anti-Markownikov's rule.

Question 4. Order of reactivity of hydrohalic acid : \text{HI} > \text{HBr} > \text{HCl}.
Answer: The bond dissociation enthalpy decreases down the group as the size of the halogen atom increases (\(\text{H}-\text{I}\) bond is weakest and longest, while \(\text{H}-\text{Cl}\) is shortest and strongest). Therefore, \text{HI} releases protons most easily.
In simple words: Larger halogen atoms form weaker bonds with hydrogen, making them more reactive.

Exam Tip: Bond strength decreases as bond length increases down the halogen group.

Question 5. Order of reactivity of alcohols with Lucas reagent : \(3^{\circ} > 2^{\circ} > 1^{\circ}\).
Answer: Lucas reagent is anhydrous \text{ZnCl}_2\) and concentrated \text{HCl}. The reaction proceeds via the formation of a carbocation intermediate. Since \(3^{\circ}\) carbocations are the most stable, \(3^{\circ}\) alcohols react the fastest (instant turbidity), followed by \(2^{\circ}\) (turbidity in 5 minutes) and \(1^{\circ}\) (no turbidity at room temperature).
In simple words: Tertiary alcohols react fastest with Lucas reagent because they form stable tertiary carbocations.

Exam Tip: Remember the turbidity times: Tertiary = immediate, Secondary = 5 minutes, Primary = no reaction at room temperature.

Question 6. \(\text{R}-\text{H} + \text{X}_2 \xrightarrow{h\nu} \text{R}-\text{X} + \text{HX}\). The order of reactivity of halogens is \text{Cl}_2 > \text{Br}_2 > \text{I}_2\). Order of reactivity of hydrogen atom substituted is \(3^{\circ} > 2^{\circ} > 1^{\circ}\).
Answer: Halogenation of alkanes is a free-radical substitution reaction. Fluorination is too violent and explosive, chlorination is rapid, bromination is moderately fast and selective, and iodination is reversible. Tertiary hydrogen atoms are replaced most easily because the resulting free radical is most stable.
In simple words: Chlorine reacts faster than bromine, and tertiary hydrogens are replaced more easily than primary hydrogens.

Exam Tip: Fluorination is explosive and iodination requires an oxidizing agent like \text{HNO}_3\(or \text{HIO}_3\) to remove the \text{HI} formed.

Question 7. Benzene ring reacting with \text{X}_2\) in presence of \text{Fe}/\text{dark} gives aryl halide plus \text{HX} (Electrophilic Aromatic Substitution).
Answer: The Lewis acid catalyst (\text{Fe} or \text{FeX}_3\)) polarizes the halogen molecule to generate a strong electrophile (\(\text{X}^{+}\)), which then substitutes a hydrogen atom on the aromatic benzene ring.
In simple words: Lewis acids help halogens act as electrophiles to substitute hydrogens on benzene rings.

Question 8. Order of reactivity in \(\text{S}_{\text{N}}1\) and \(\text{S}_{\text{N}}2\) mechanism are as follows: For \(\text{S}_{\text{N}}1\) reactions: \(3^{\circ} > 2^{\circ} > 1^{\circ} > \text{CH}_3\text{X}\). For \(\text{S}_{\text{N}}2\) reactions: \(\text{R}-\text{I} > \text{R}-\text{Br} > \text{R}-\text{Cl}\).
Answer: \(\text{S}_{\text{N}}1\) proceeds via carbocation formation, making tertiary halides most reactive. \(\text{S}_{\text{N}}2\) proceeds via a concerted backside attack mechanism where steric hindrance hinders the attack, making methyl and primary halides most reactive. For leaving groups, iodide is the best leaving group due to the weakest carbon-iodine bond.
In simple words: \(\text{S}_{\text{N}}1\) favors bulky tertiary substrates due to carbocation stability, while \(\text{S}_{\text{N}}2\) favors unhindered primary substrates.

Exam Tip: Steric hindrance is the primary factor controlling \(\text{S}_{\text{N}}2\) rates, whereas electronic stability controls \(\text{S}_{\text{N}}1\) rates.

Question 9. In case of optically active alkyl halide, \(\text{S}_{\text{N}}2\) mechanism results in the inversion of configuration while \(\text{S}_{\text{N}}1\) mechanism in racemisation.
Answer: In an \(\text{S}_{\text{N}}2\) reaction, the nucleophile attacks from the side opposite to the leaving group, causing Walden inversion. In an \(\text{S}_{\text{N}}1\) reaction, the planar carbocation intermediate can be attacked by the nucleophile from either side with equal probability, yielding a racemic mixture (50% inversion + 50% retention).
In simple words: \(\text{S}_{\text{N}}2\) flips the molecule inside out like an umbrella, while \(\text{S}_{\text{N}}1\) attacks from both sides equally.

Exam Tip: Always specify optical inversion for \(\text{S}_{\text{N}}2\) and racemic mixture formation for \(\text{S}_{\text{N}}1\).

Question 10. Aryl halides are much less reactive towards nucleophilic substitution reactions. Presence of electron withdrawing groups (like \(-\text{NO}_2, -\text{CN}, -\text{COOH}\), etc.) at o- and/or p-position to halogen increases the rate of reaction.
Answer: Aryl halides exhibit partial double bond character in the \(\text{C}-\text{X}\) bond due to resonance, making cleavage difficult. Electron-withdrawing groups at ortho and para positions withdraw electron density from the benzene ring and stabilize the carbanion intermediate formed during nucleophilic aromatic substitution.
In simple words: Resonance makes aryl halides unreactive, but electron-withdrawing groups at ortho/para positions speed up substitution.

Exam Tip: Meta-nitro groups do not activate aryl halides towards nucleophilic substitution nearly as effectively as ortho or para nitro groups.

Question 12. All the three types of monohydric alcohols (\(1^{\circ}, 2^{\circ}\) or \(3^{\circ}\)) except methanol can be prepared from Grignard Reagent.
Answer: Formaldehyde (\(\text{HCHO}\)) reacts with Grignard reagents to yield primary alcohols. Other aldehydes yield secondary alcohols, and ketones yield tertiary alcohols. Since formaldehyde is the simplest aldehyde with only one carbon, methanol (which has one carbon) can only be prepared using formaldehyde.
In simple words: Grignard reagents react with aldehydes and ketones to form all alcohols, but methanol requires formaldehyde.

Exam Tip: Remember: Formaldehyde gives \(1^{\circ}\) alcohol, higher aldehydes give \(2^{\circ}\) alcohol, and ketones give \(3^{\circ}\) alcohol.

Question 13. Alcohols Reactivity with \text{HX}: \(1^{\circ} < 2^{\circ} < 3^{\circ}\). Water solubility, acidic strength.
Answer: The reactivity of alcohols with hydrohalic acids depends on the cleavage of the \(\text{C}-\text{OH}\) bond, which forms a carbocation intermediate. Thus, tertiary alcohols react fastest. Water solubility decreases as the alkyl chain length increases due to hydrophobic effects. Acidic strength of alcohols follows the order \(1^{\circ} > 2^{\circ} > 3^{\circ}\) due to the electron-donating inductive effect (+I effect) of alkyl groups intensifying negative charge on oxygen.
In simple words: Tertiary alcohols are most reactive toward HX, while primary alcohols are most acidic among aliphatic alcohols.

Exam Tip: Alkyl groups are electron-donating (+I), which destabilizes alkoxide ions and decreases acidity.

Question 14. Presence of electron withdrawing group increases the acid strength of alcohol, phenol and carboxylic acid while presence of electron donating group decreases the acid strength. E.W.G. : \(\text{NO}_2, -\text{X}, -\text{CN}, -\text{COOH}\), etc., E.D.G. : \(-\text{R}, -\text{OR}, -\text{OH}, -\text{NH}_2\), etc.
Answer: Electron-withdrawing groups (EWG) stabilize the conjugate base (alkoxide, phenoxide, or carboxylate ion) by dispersing the negative charge through inductive and/or resonance effects, thus increasing acidity. Electron-donating groups (EDG) intensify the negative charge and destabilize the conjugate base, decreasing acidity.
In simple words: EWGs pull electron density away and make molecules more acidic, whereas EDGs push electron density and make them less acidic.

Exam Tip: Strong EWGs like \(-\text{NO}_2\) at ortho and para positions dramatically increase the acidity of phenols.

Question 15. In electrophilic aromatic substitution reaction, ring activating groups like \(-\text{OH}, -\text{NH}_2, -\text{OR}\), are o- and p-directing whereas ring deactivating groups like \(-\text{CHO}, >\text{C}=\text{O}, -\text{COOH}, -\text{NO}_2, -\text{SO}_3\text{H}\) are m-directing. Halogens (\(\text{F}, \text{Cl}, \text{Br}\) and \(\text{I}\)) are deactivating but are o- and p-directing groups.
Answer: Activating groups release electrons into the ring via resonance, increasing electron density particularly at ortho and para positions. Deactivating groups withdraw electron density. Halogens are unique because their -I effect withdraws electron density (deactivating), but their +M (resonance) effect directs incoming electrophiles to ortho and para positions.
In simple words: Groups with lone pairs activate and direct ortho/para, while carbonyl and nitro groups deactivate and direct meta. Halogens are deactivating ortho/para directors.

Exam Tip: Halogens are an important exception: they deactivate the benzene ring yet remain ortho/para directing due to lone pair resonance.

Question 16. \(3^{\circ}\) alcohols are resistant to oxidation due to lack of \(\alpha\)-hydrogen.
Answer: Oxidation of alcohols requires the removal of at least one hydrogen atom from the carbon atom bearing the hydroxyl group (\(\alpha\)-carbon) and the oxygen atom. Since tertiary alcohols lack an \(\alpha\)-hydrogen atom, they cannot undergo oxidation under normal conditions.
In simple words: Tertiary alcohols cannot be oxidized easily because they have no hydrogen on the carbon attached to the \(-\text{OH}\) group.

Exam Tip: Primary alcohols oxidize to aldehydes then carboxylic acids, secondary alcohols oxidize to ketones, and tertiary alcohols resist normal oxidation.

Question 17. Order of Acid strength : Alcohol < Phenol < Carboxylic acid, it is because of resonance stabilization of phenoxide and carboxylate ion. In carboxylate ion negative charge is delocalised over two oxygen atoms. while in phenoxide ion it is delocalised over one oxygen atom and the less electronegative C atom of benzene ring.
Answer: Carboxylic acids are the strongest organic acids because the negative charge in a carboxylate ion is equally delocalized over two highly electronegative oxygen atoms. Phenols are weaker because the negative charge is shared with less electronegative carbon atoms of the benzene ring. Alcohols are the weakest because their alkoxide ions have localized negative charge.
In simple words: Carboxylic acids form the most stable conjugate bases with charge shared over two oxygens, making them the most acidic.

Exam Tip: Greater delocalization of negative charge over electronegative atoms always increases acid strength.

Question 18. All organic compounds which form intermolecular H-bonds with water are soluble in water.
Answer: Solubility in water depends on the ability of solute molecules to form hydrogen bonds with water molecules. Lower alcohols, amines, and carboxylic acids are highly soluble due to strong hydrogen bonding, but solubility decreases as the hydrophobic alkyl chain length increases.
In simple words: Compounds that can form hydrogen bonds with water will dissolve in water.

Exam Tip: As the hydrocarbon chain length increases, water solubility decreases rapidly.

Question 19. Intermolecular H-bonds of p- and m-nitrophenol increases water solubility/Acid strength while intramolecular H-bonds in o-nitrophenol decreases these properties.
Answer: Ortho-nitrophenol forms stable intramolecular hydrogen bonding between the phenolic \(-\text{OH}\) group and the adjacent nitro group, preventing intermolecular hydrogen bonding with water and reducing volatility and water solubility. Para- and meta-nitrophenols form intermolecular hydrogen bonds with water, increasing solubility and acidity.
In simple words: Intramolecular hydrogen bonding in ortho-nitrophenol locks up the hydrogen, lowering its solubility and volatility compared to para-nitrophenol.

Exam Tip: Intramolecular hydrogen bonding causes chelation and decreases boiling point and water solubility (e.g., ortho-nitrophenol is steam volatile).

Question 20. In the reaction of alkyl aryl ether (Anisole) with \text{HI}, the products are always alkyl halide and phenol, because \(\text{O}-\text{R}\) bond is weaker than \(\text{O}-\text{Ar}\) bond.
Answer: The oxygen-carbon bond in aryl ethers (\(\text{O}-\text{Ar}\)) acquires partial double bond character due to resonance with the benzene ring, making it stronger. Therefore, cleavage occurs at the aliphatic \(\text{O}-\text{R}\) bond, yielding phenol and an alkyl iodide.
In simple words: Anisole reacts with HI to yield iodomethane and phenol because the phenyl-oxygen bond is strengthened by resonance.

Exam Tip: With mixed ethers containing an aryl group, the alkyl group always forms the alkyl halide.

Question 22. In reaction of toluene with \text{CrO}_3\) in presence of acetic anhydride is used to protect benzaldehyde as benzylidenediacetate to avoid further oxidation to benzoic acid.
Answer: Etard reaction uses chromium trioxide in acetic anhydride to convert toluene to benzylidenediacetate, which is then hydrolyzed to benzaldehyde. Acetic anhydride prevents over-oxidation of benzaldehyde into benzoic acid.
In simple words: Acetic anhydride traps the intermediate aldehyde so it does not get over-oxidized into an acid.

Exam Tip: This specific variation of the Etard reaction is a standard method for preparing benzaldehyde directly from toluene.

Question 23. Order of reactivity of aldehydes and ketones towards nucleophilic attack: \text{HCHO} > \text{CH}_3\text{CHO} > \text{CH}_3\text{CH}_2\text{CHO} \), \(\text{RCHO} > \text{RCOR}\); \(\text{ArCHO} > \text{ArCOR} > \text{ArCOAr}\). Because of (i) +I inductive effect of alkyl groups. (ii) steric hindrance of alkyl and aryl group reactivity of ketones is less than aldehydes.
Answer: Nucleophilic addition depends on both electronic and steric factors. Alkyl groups donate electrons (+I effect) which reduces the positive charge on the carbonyl carbon. Additionally, bulky alkyl or aryl groups create steric hindrance that blocks the approach of the incoming nucleophile.
In simple words: Ketones are less reactive than aldehydes because alkyl groups push electrons into the carbon and block incoming attackers.

Exam Tip: Formaldehyde is always the most reactive carbonyl compound in nucleophilic addition reactions.

Question 24. Boiling and melting points of various organic compounds depend on intermolecular forces of attraction which depend on the following: (a) Intermolecular/intramolecular H-bonding. (b) Dipole-dipole interaction. (c) Molecular size. (d) Surface area (branching decreases surface area of molecules).
Answer: Physical properties like boiling and melting points are directly related to the strength of intermolecular forces. Hydrogen bonding provides the strongest attraction, followed by dipole-dipole interactions and London dispersion forces (which increase with molecular weight and surface area). Branching reduces surface area, decreasing van der Waals forces and lowering boiling points.
In simple words: Stronger intermolecular forces, larger molecular sizes, and straight chains lead to higher boiling and melting points.

Exam Tip: Straight-chain isomers always have higher boiling points than branched-chain isomers due to greater surface contact.

Question 25. Benzaldehyde does not reduce Fehling's reagent.
Answer: Fehling's solution is a mild oxidizing agent that reduces aliphatic aldehydes to carboxylate ions, yielding a red precipitate of \(\text{Cu}_2\text{O}\). Aromatic aldehydes like benzaldehyde are less reactive and lack sufficient reducing power to reduce Fehling's solution.
In simple words: Aromatic aldehydes are not reactive enough to reduce Fehling's reagent, though they do reduce Tollens' reagent.

Exam Tip: Tollens' reagent is strong enough to oxidize both aliphatic and aromatic aldehydes, whereas Fehling's reagent only oxidizes aliphatic aldehydes.

Question 26. The more the \(K_a\) value, the lesser is the \(pK_a\), a stronger acid always has higher \(K_a\) but a lower \(pK_a\) value.
Answer: Acid dissociation constant \(K_a\) measures acid strength directly (larger \(K_a\) means more dissociation). Since \(pK_a = -\log K_a\), a higher \(K_a\) results in a mathematically smaller negative logarithm (\(pK_a\)).
In simple words: Stronger acids have higher \(K_a\) numbers and smaller \(pK_a\) numbers.

Exam Tip: Remember the inverse relationship: high \(K_a\) = low \(pK_a\) = strong acid.

Question 27. Order of Acid strength : \text{HCOOH} > \text{C}_6\text{H}_5\text{COOH} > \text{C}_6\text{H}_5\text{CH}_2\text{COOH} > \text{CH}_3\text{COOH}, \text{CH}_3\text{CH}_2\text{COOH}
Answer: Formic acid (\(\text{HCOOH}\)) is stronger than acetic acid because the hydrogen atom lacks the electron-donating +I effect present in the methyl group of acetic acid. Benzoic acid is stronger than aliphatic acids due to the electron-withdrawing resonance effect of the phenyl ring.
In simple words: Formic acid is stronger than acetic acid because it lacks an electron-pushing methyl group.

Question 28. \(-\text{C}-\text{OH}\) group in carboxylic acid is not a true carbonyl group because of resonance. Hence, carboxylic acids do not give addition reaction of aldehydes and ketones.
Answer: The lone pair on the hydroxyl oxygen of a carboxylic acid delocalizes into the carbonyl carbon via resonance, neutralizing the positive charge on the carbonyl carbon and preventing nucleophilic addition reactions typical of aldehydes and ketones.
In simple words: Resonance makes the carbonyl carbon in carboxylic acids much less positive, preventing addition reactions.

Exam Tip: Carboxylic acids react with bases to form salts rather than undergoing typical carbonyl addition reactions.

Question 29. (a) In aqueous solution order of basicity is : \((\text{C}_2\text{H}_5)_2\text{NH} > (\text{C}_2\text{H}_5)_3\text{N} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3\). \((\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3\). (b) In gaseous (vapour) state order of basicity is : \(\text{RNH}_2 < \text{R}_2\text{NH} < \text{R}_3\text{N}\).
Answer: In the gas phase, basicity increases with the number of electron-donating alkyl groups due to inductive stabilization (\(3^{\circ} > 2^{\circ} > 1^{\circ}\)). In aqueous solution, basicity is determined by a combination of inductive effect, steric hindrance, and hydration of the ammonium cation, making secondary amines the strongest bases.
In simple words: In water, secondary amines are the most basic due to a balance of steric hindrance and hydration effects.

Exam Tip: Memorize the distinct order for ethyl and methyl amines in aqueous solution: secondary > tertiary > primary for ethyl, and secondary > primary > tertiary for methyl.

Question 30. Basic character of aromatic amine
Answer: Aromatic amines are weaker bases than ammonia and aliphatic amines because the lone pair on the nitrogen atom is delocalized into the benzene ring via resonance, making it less available for protonation. Electron-donating groups increase basicity, while electron-withdrawing groups decrease basicity.
In simple words: Aniline is a weak base because its nitrogen lone pair is shared with the benzene ring.

Exam Tip: Electron-withdrawing groups like \(-\text{NO}_2\) at para positions decrease the basicity of aniline significantly.

Question 31. Hinsberg's reagent (\(\text{C}_6\text{H}_5\text{SO}_2\text{Cl}\)) is used to separate the mixtures of \(1^{\circ}, 2^{\circ} \) and \( 3^{\circ}\) amines.
Answer: Primary amines react with benzenesulfonyl chloride to form a sulfonamide soluble in alkali. Secondary amines form a sulfonamide insoluble in alkali. Tertiary amines do not react at all.
In simple words: Hinsberg's reagent distinguishes amines based on whether their sulfonamide derivatives dissolve in alkali.

Exam Tip: Primary amine derivative dissolves in alkali, secondary amine derivative is insoluble, and tertiary amine does not react.

Question 32. Sulphanilic acid exists as zwitter ion, therefore it is amphoteric in nature.
Answer: Sulphanilic acid contains both a basic amino group (\(-\text{NH}_2\)) and an acidic sulfonic acid group (\(-\text{SO}_3\text{H}\)). An intramolecular proton transfer creates a dipolar ion (zwitterion), allowing it to react with both acids and bases.
In simple words: Sulphanilic acid acts as both an acid and a base because internal proton transfer creates a zwitterion.

Exam Tip: Zwitterions possess both positive and negative charges within the same neutral molecule.

Question 33. Aryl diazonium salts are more stable than alkyl diazonium salts.
Answer: Aryl diazonium salts are stabilized by resonance involving the benzene ring, which disperses the positive charge. Alkyl diazonium salts lack this resonance stabilization and rapidly lose nitrogen gas to form carbocations.
In simple words: Benzene rings stabilize diazonium salts through resonance, whereas aliphatic diazonium salts break down immediately.

Exam Tip: Aryl diazonium salts are stable at low temperatures (0 to 5 degrees Celsius).

Question 34. Aniline, phenol and benzoic acid do not show Friedel-Crafts Reaction.
Answer: Aniline forms a complex with the anhydrous aluminum chloride (\(\text{AlCl}_3\)) Lewis acid catalyst, deactivating the ring. Phenol and benzoic acid similarly coordinate with the catalyst, preventing further electrophilic substitution.
In simple words: Aniline forms a complex with the catalyst and fails to undergo Friedel-Crafts alkylation or acylation.

Exam Tip: Aniline cannot be alkylated or acylated via Friedel-Crafts because the amino group bonds with the Lewis acid catalyst.

Question 35. The more the basic character, the more is the \(K_b\) value and lesser will be its \(pK_b\) value.
Answer: Base dissociation constant \(K_b\) measures basic strength directly. Since \(pK_b = -\log K_b\), a stronger base with a higher \(K_b\) value will have a lower \(pK_b\) value.
In simple words: Stronger bases have higher \(K_b\) values and smaller \(pK_b\) values.

Exam Tip: The \(pK_b\) scale is inversely proportional to basic strength, just like \(pK_a\) is inversely proportional to acid strength.

Question 35 (b). In phenol and aniline, electron donating or electron withdrawing groups when present in ortho position, always have acid strengthening effect. This is called ortho effect.
Answer: The ortho effect is primarily steric and electronic in origin. Substituents at the ortho position force coplanar functional groups out of alignment or stabilize the conjugate base through proximity effects, increasing acidity regardless of whether the group is electron-donating or electron-withdrawing.
In simple words: Groups at the ortho position in benzoic acid or phenol increase acidity due to steric and electronic interactions.

Exam Tip: Ortho-substituted benzoic acids are almost always stronger acids than their meta and para isomers.

USEFUL TIPS FOR IDENTIFICATION OF FUNCTIONAL GROUPS

Question 36. For identification of chloro, bromo or iodo alkanes, aq. \text{KOH} is added followed by \text{AgNO}_3\) solution then precipitate of \text{AgX} is formed.
Answer: Hydrolysis with aqueous KOH converts haloalkanes into alcohols and halide ions. Addition of silver nitrate precipitates silver halide (\(\text{AgCl}\) is white, \(\text{AgBr}\) is pale yellow, \(\text{AgI}\) is yellow).
In simple words: Halogen atoms in alkanes are detected by converting them to halide ions and precipitating them with silver nitrate.

Exam Tip: The color of the silver halide precipitate helps identify the specific halogen present.

Question 37. Iodoform Test is given by all organic compounds having \(\text{CH}_3-\text{C}(=\text{O})-\) or \(\text{CH}_3-\text{CH}(\text{OH})-\) when heated with \text{NaOH (aq)} and \text{I}_2 \), produces yellow iodoform (\(\text{CHI}_3\)).
Answer: Compounds containing a methyl ketone group or a secondary alcohol with an adjacent methyl group undergo halogenation followed by cleavage with sodium hydroxide to yield yellow iodoform crystals.
In simple words: Any compound with a \(\text{CH}_3\text{CH(OH)}-\) or \(\text{CH}_3\text{CO}-\) group gives a positive yellow iodoform test.

Exam Tip: Ethanol is the only primary alcohol that gives a positive iodoform test.

Question 38. Tollens' reagent test is given by organic compound having \(-\text{CHO}\) group (formic acid \(\text{H}-\text{C}(=\text{O})-\text{H}\) also gives this test).
Answer: Tollens' reagent (ammoniacal silver nitrate) oxidizes aldehydes to carboxylate ions while reducing silver ions to metallic silver, forming a shining silver mirror on the inner walls of the test tube.
In simple words: Aldehydes reduce Tollens' reagent to form a shiny silver mirror.

Exam Tip: Formic acid acts as both an aldehyde and an acid, allowing it to reduce Tollens' reagent.

Question 39. Methanoic acid and ethanoic acid gives deep red colour solution with iron (III) chloride solution.
Answer: Reaction with neutral ferric chloride solution produces iron(III) carboxylate complexes that impart a deep red or brownish-red color to the solution.
In simple words: Lower carboxylic acids react with ferric chloride to form deep red complexes.

Exam Tip: Ferric chloride test is useful for identifying low molecular weight carboxylic acids.

Question 40. Benzoic acid gives buff coloured precipitate with iron (III) chloride solution.
Answer: Neutral ferric chloride reacts with sodium benzoate to precipitate iron(III) benzoate as a buff-colored (tan/brownish-yellow) solid.
In simple words: Benzoic acid forms a characteristic buff-colored precipitate with ferric chloride.

Exam Tip: This differentiates aromatic benzoic acid from lower aliphatic carboxylic acids.

Question 41. Acid chlorides and esters may be recognised after hydrolysis. The resultant acid and alcohol may be tested for.
Answer: Hydrolysis of acid chlorides yields carboxylic acids and hydrochloric acid. Hydrolysis of esters yields a carboxylic acid and an alcohol, which can be identified using standard functional group tests.
In simple words: Hydrolyze derivatives into their component acids and alcohols to identify them.

Exam Tip: Always break down carboxylic acid derivatives via hydrolysis before running qualitative identification tests.

Question 42. Amides are recognised by the liberation of ammonia on heating with alkali (\(\text{NaOH}\) or \(\text{KOH}\)) whereas ammonium salts liberate ammonia with alkali in the cold.
Answer: Heating amides with aqueous alkali hydrolyzes them to release ammonia gas, which turns moist red litmus blue. Ammonium salts liberate ammonia gas even at room temperature without heating.
In simple words: Amides need heat with alkali to release ammonia gas, whereas ammonium salts release it immediately at room temperature.

Exam Tip: Use the litmus paper test to confirm the basic ammonia gas released.

Question 43. \(1^{\circ}, 2^{\circ}\) and \(3^{\circ}\) alcohols are distinguished by Lucas test.
Answer: Lucas reagent (\(\text{ZnCl}_2/\text{HCl}\)) reacts with tertiary alcohols instantly to form cloudy turbidity, secondary alcohols react within 5 minutes, and primary alcohols show no reaction at room temperature.
In simple words: Lucas test uses turbidity timing to distinguish between primary, secondary, and tertiary alcohols.

Exam Tip: Speed of turbidity formation directly reflects carbocation stability.

Question 44. \(1^{\circ}, 2^{\circ}\) and \(3^{\circ}\) amines can be distinguished with the help of Hinsberg reagent.
Answer: Benzenesulfonyl chloride (\(\text{C}_6\text{H}_5\text{SO}_2\text{Cl}\)) forms alkali-soluble sulfonamides with primary amines, alkali-insoluble sulfonamides with secondary amines, and does not react with tertiary amines.
In simple words: Hinsberg's reagent separates and identifies amine types based on alkali solubility.

Question 45. \(1^{\circ}\) and \(2^{\circ}\) amines can be distinguished by Carbylamine test.
Answer: Primary amines heated with chloroform and ethanolic potassium hydroxide produce foul-smelling isocyanides (carbylamines). Secondary and tertiary amines do not give this test.
In simple words: Primary amines produce a foul-smelling isocyanide test with chloroform and alkali.

Exam Tip: The carbylamine test is specific for primary amines (both aliphatic and aromatic).