• $C_6H_6$: 12 $\sigma$, 3 $\pi$.
• $C_6H_{12}$ (Cyclohexane): 18 $\sigma$, 0 $\pi$.
• $CH_2Cl_2$: 4 $\sigma$, 0 $\pi$.
• $CH_2=C=CH_2$: 6 $\sigma$, 2 $\pi$.
• $CH_3NO_2$: 6 $\sigma$, 1 $\pi$.
• $HCONHCH_3$: 8 $\sigma$, 1 $\pi$.

Isopropyl Alcohol: Y-shape with OH at the center joint.

2,3-Dimethylbutanal: 4-carbon chain. CHO at end. Methyls at C2 and C3.

Heptan-4-one: 7-carbon zigzag chain. Double bond O at the central peak (C4).

(Based on standard NCERT diagrams for these labels)

(a) Propylbenzene.
(b) 3-Methylpentanenitrile.
(c) 2,5-Dimethylheptane.
(d) 3-Bromo-3-chloroheptane.
(e) 3-Chloropropanal.
(f) 2,2-Dichloroethanol.

(a) 2,2-Dimethylpentane: Correct (Locants 2,2 vs 2).
(b) 2,4,7-Trimethyloctane: Correct (Set 2,4,7 < 2,5,7).
(c) 2-Chloro-4-methylpentane: Correct (Alphabetical priority for equivalent positions).
(d) But-3-yn-1-ol: Correct (Functional group OH gets priority over triple bond).

(a) Acid: Methanoic, Ethanoic, Propanoic, Butanoic, Pentanoic acid.
(b) Ketone: Propanone, Butanone, Pentan-2-one, Hexan-2-one, Heptan-2-one.
(c) Alkene: Ethene, Propene, But-1-ene, Pent-1-ene, Hex-1-ene.

(a) 2,2,4-Trimethylpentane: Alkane. No functional group.
(b) Citric Acid: 3 -COOH groups, 1 -OH group.
(c) Hexanedial: 2 -CHO (Aldehyde) groups.

(Based on standard figures)
(a) Aldehyde (-CHO), Hydroxyl (-OH), Methoxy (-OMe).
(b) Amino (-NH2), Ketone (C=O), Double Bond (-C=C-).
(c) Nitro (-NO2), Double Bond (-C=C-).

$O_2NCH_2CH_2O^-$ is more stable.
The nitro group ($-NO_2$) exerts a strong electron-withdrawing inductive effect (-I), which disperses the negative charge on the oxygen, stabilizing the ion. In contrast, the ethyl group in the other ion exerts a +I effect, destabilizing the negative charge.

Alkyl groups attached to a $\pi$ system act as electron donors primarily due to Hyperconjugation. The bonding pair of electrons in the C-H sigma bond delocalizes into the adjacent empty or partially filled p-orbital or $\pi^*$ orbital, donating electron density.

(a) Phenol: +R effect. O pushes electrons into the ring. Negative charge at ortho/para.
(b) Nitrobenzene: -R effect. NO2 pulls electrons. Positive charge at ortho/para.
(c) Benzaldehyde: -R effect. C=O pulls electrons from ring.
(e) Benzyl Carbocation: Positive charge delocalized over the ring.

Electrophiles ($E^+$): Electron-loving. Electron deficient. Accept electron pair. (e.g., $H^+, BF_3$).
Nucleophiles ($Nu^-$): Nucleus-loving. Electron rich. Donate electron pair. (e.g., $OH^-, NH_3$).

(a) $HO^-$: Nucleophile.
(b) $CN^-$: Nucleophile.
(c) $CH_3C^+O$: Electrophile.

(a) Nucleophilic Substitution ($S_N$).
(b) Electrophilic Addition.
(c) Elimination.
(d) Nucleophilic Substitution ($S_N1$).

(a) Position Isomers: Functional group/substituent position differs.
(b) Geometrical Isomers: Cis-Trans arrangement.
(c) Resonance Contributors: Same atomic arrangement, different electron distribution.

Homolysis: Curved arrows (fish-hook) to each atom. Produces Free Radicals.
Heterolysis: Curved arrow to more electronegative atom. Produces Carbocation and Carbanion.

Inductive (I): Permanent shift of $\sigma$ electrons due to electronegativity difference.
Electromeric (E): Temporary shift of $\pi$ electrons in presence of reagent.
Acidity Order Explanation:
(a) $-I$ effect of Cl stabilises anion. More Cl = Stronger Acid.
(b) $+I$ effect of alkyl groups destabilises anion. More/Larger alkyl = Weaker Acid.

(a) Crystallisation: Solubility difference in a solvent.
(b) Distillation: Boiling point difference of volatile liquids.
(c) Chromatography: Differential adsorption/partition on stationary phase.

Fractional Crystallisation: Used when two compounds have different solubilities in the same solvent S. Repeated crystallisation separates the less soluble component first.

Simple Distillation: For liquids with large B.P. difference, stable at B.P.
Reduced Pressure: For liquids that decompose at their normal B.P. Lowers B.P.
Steam Distillation: For substances insoluble in water but volatile in steam. Lowers total pressure required to boil.

Principle: Organic compounds are fused with metallic sodium to convert covalent elements (N, S, Halogens) into water-soluble ionic sodium salts.

This sodium extract is then used for wet tests to detect these elements using specific reagents (like $FeSO_4$ for N, Lead Acetate for S, $AgNO_3$ for X).

Dumas Method: Based on the conversion of nitrogen in the compound to molecular nitrogen gas ($N_2$) by heating with Copper(II) oxide ($CuO$). The volume of $N_2$ collected is measured.
Kjeldahl’s Method: Based on the conversion of nitrogen into ammonium sulphate $(NH_4)_2SO_4$ by digestion with conc. $H_2SO_4$. Ammonia is liberated by alkali and absorbed in acid for titration. (Not applicable to nitro/diazo compounds).

• Halogens (Carius Method): Compound heated with fuming $HNO_3$ and $AgNO_3$. Halogen converts to $AgX$ precipitate. Mass of ppt determines %.
• Sulphur (Carius Method): Heated with fuming $HNO_3$. Sulphur oxidises to $H_2SO_4$, precipitated as $BaSO_4$ using $BaCl_2$. Mass of $BaSO_4$ determines %.
• Phosphorus: Oxidised to phosphate using fuming $HNO_3$. Precipitated as ammonium phosphomolybdate $(NH_4)_3PO_4 \cdot 12MoO_3$ or $MgNH_4PO_4$.

It is based on the principle of partition chromatography. The special paper contains water trapped in it, acting as the stationary phase. The solvent moving up the paper is the mobile phase. Components of the mixture separate due to their differing partition coefficients between the stationary and mobile phases.

The sodium extract may contain Sodium Cyanide ($NaCN$) and Sodium Sulphide ($Na_2S$) if Nitrogen and Sulphur are present. These ions react with Silver Nitrate to form white ($AgCN$) or black ($Ag_2S$) precipitates, interfering with the halogen test.
Boiling with conc. $HNO_3$ decomposes cyanide into $HCN$ gas and sulphide into $H_2S$ gas, removing them.

Organic compounds are covalent in nature and do not ionize in aqueous solution. Fusion with highly reactive metallic sodium converts these covalently bonded elements (N, S, X) into water-soluble ionic salts ($NaCN, Na_2S, NaX$), allowing for their detection via standard ionic wet tests.

Sublimation.
Camphor is a volatile solid that sublimes (turns directly to gas) on heating. Calcium sulphate is non-volatile. Heating the mixture allows separation of pure camphor on the cooler parts of the funnel.

In steam distillation, a mixture of organic liquid and water boils when the sum of their vapour pressures equals the atmospheric pressure ($P_{total} = p_{organic} + p_{water} = P_{atm}$).
Since $p_{water}$ contributes significantly, $p_{organic}$ is lower than $P_{atm}$. Consequently, the organic liquid boils at a temperature lower than its normal boiling point, preventing decomposition of heat-sensitive compounds.

No.
$CCl_4$ (Carbon tetrachloride) is a covalent compound. The C-Cl bond is strong and does not ionize in water to yield chloride ions ($Cl^-$). Without free $Cl^-$ ions, reaction with $AgNO_3$ to form $AgCl$ precipitate is not possible.

Potassium Hydroxide (KOH) is a strong alkali that readily absorbs Carbon Dioxide ($CO_2$) to form Potassium Carbonate ($K_2CO_3$). The increase in the weight of the KOH U-tube corresponds to the mass of $CO_2$ produced, allowing calculation of carbon percentage.

Lead acetate test detects sulphide ions ($S^{2-}$) as black $PbS$.
If sulphuric acid ($H_2SO_4$) were used, the lead ions ($Pb^{2+}$) would react with sulphate ions to form Lead Sulphate ($PbSO_4$), which is a white insoluble precipitate. This would interfere with the test. Acetic acid prevents this as Lead Acetate is soluble.

Mass of compound ($m$) = 0.20 g. %C = 69%. %H = 4.8%.

Mass of compound = 0.50 g.
Acid: 50 mL 0.5 M $H_2SO_4$. (Normality = $0.5 \times 2 = 1$ N).
Initial meq of acid = $50 \times 1 = 50$.
Base used: 60 mL 0.5 M NaOH. (Normality = 0.5 N).
meq of base used = $60 \times 0.5 = 30$.

Acid neutralized by $NH_3$ = $50 – 30 = 20$ meq.

Mass of sample = 0.3780 g. Mass of $AgCl$ = 0.5740 g.
Molar mass $AgCl = 108 + 35.5 = 143.5$.

Mass of sample = 0.468 g. Mass of $BaSO_4$ = 0.668 g.
Molar mass $BaSO_4 = 137 + 32 + 64 = 233$.

Numbering: Double bond gets priority over triple bond if locants are same from ends. Here, double bond at C1, triple at C5.
Structure: $C_1(H_2)=C_2(H)-C_3(H_2)-C_4(H_2)-C_5 \equiv C_6(H)$.
Question asks for $C_2-C_3$ bond.
$C_2$ is part of double bond $\rightarrow$ $sp^2$.
$C_3$ has single bonds $\rightarrow$ $sp^3$.
Answer: (c) $sp^2 – sp^3$.

The Prussian blue colour in Lassaigne’s test for nitrogen is due to the formation of Iron(III) hexacyanoferrate(II).
Formula: $Fe_4[Fe(CN)_6]_3$.
Answer: (b) $Fe_4[Fe(CN)_6]_3$.

Stability order: Tertiary ($3^\circ$) > Secondary ($2^\circ$) > Primary ($1^\circ$).
(a) $1^\circ$ (neopentyl).
(b) $(CH_3)_3C^+$ is $3^\circ$ (Tert-butyl). Most stable due to +I effect of 3 methyl groups and 9 hyperconjugative structures.
(c) $1^\circ$ (propyl).
(d) $2^\circ$ (sec-butyl).
Answer: (b).

Chromatography is considered the best and latest technique for separation, isolation, and purification, especially for complex mixtures and small quantities.
Answer: (d).

Reaction: $CH_3CH_2I + KOH(aq) \rightarrow CH_3CH_2OH + KI$.
Here, the nucleophile $OH^-$ replaces the leaving group $I^-$.
Answer: (b) Nucleophilic substitution.