Organic Chemistry: Some Basic Principles and Techniques: JEE Advanced PYQ, 2007–2026

Answer: (C)

Delocalisation of charge by resonance stabilises an ion far more than inductive effects do. In CH3–CO–CH2⁻ the lone pair on carbon overlaps with the C=O π bond. This gives the resonance form CH3–C(O⁻)=CH2, with the charge on oxygen. The simple alkyl carbanions have only inductive effects, and the +I effect destabilises them. So the keto-substituted carbanion is the most stable.

Answer: (A)

Principle: in column (adsorption) chromatography the components of a mixture are adsorbed to different extents on the stationary phase. A weakly adsorbed component spends more time in the mobile phase and moves down faster. A strongly adsorbed component is retained near the top until a more polar eluent displaces it. On silica gel, adsorption increases with polarity: hydrocarbon < ester < carboxylic acid. Application: naphthalene has no polar group and elutes with hexane. Ethyl benzoate, with moderate polarity, comes out when ethyl acetate is added to the eluent. Benzoic acid can hydrogen-bond strongly to silica and elutes last, needing the highest proportion of ethyl acetate.

Answer: (B)

The ring prevents rotation about the C–C bonds, so the two methyl groups can be on the same face (cis) or on opposite faces (trans). C1 and C2 are both stereocentres. The cis form has an internal plane of symmetry, so it is meso and optically inactive. The trans form has no plane or centre of symmetry, so it is chiral and exists as a pair of enantiomers. The total is 1 (cis) + 2 (trans) = 3, which is less than the 2² = 4 predicted by the simple rule.

Answer: (A)

Principle: in Lassaigne's test the elements fuse with sodium to give ionic products. With N alone the product is NaCN. With S alone it is Na₂S. When both N and S are present, the product is NaSCN (Na + C + N + S → NaSCN), provided sodium is not in large excess. Application: Fe³⁺ + SCN⁻ → [Fe(SCN)]²⁺, which is blood red. No free CN⁻ is present, so no Prussian blue forms, and the red colour matches the SCN⁻ ion. Prussian blue is blue, iron sulphides are black, and the nitroprusside complex is violet and needs a different reagent. With excess sodium, NaSCN decomposes to NaCN and Na₂S, which would then give the separate tests.

Answer: (A)

The easier a radical forms, the weaker the bond that produces it. Breaking a C–H bond in the CH3 group gives CH2=CH–CH2•, which is stabilised by resonance (CH2=CH–CH2• ↔ •CH2–CH=CH2). Breaking either vinylic C–H bond gives a radical with the electron in an sp² orbital in the plane. That orbital cannot overlap with the π bond, so there is no stabilisation. The allylic C–H bond is therefore the weakest.

Answer: (A)

The acidity of a phenol depends on the stability of the phenoxide ion. In p-nitrophenoxide the negative charge on the ring, which is concentrated at the ortho and para positions, reaches the carbon bearing NO2. From there it is delocalised onto the oxygens of the nitro group, which is a −R effect. At the meta position the nitro group is not in conjugation with the phenoxide oxygen, so only the −I effect, which weakens with distance, stabilises the ion. The extra resonance stabilisation makes the para isomer the stronger acid.

Answer: (A)

With silver nitrate in acid solution, the halide ions give characteristic precipitates. Cl⁻ gives white AgCl, which is soluble in NH₄OH. Br⁻ gives pale yellow AgBr, which is sparingly soluble in NH₄OH. I⁻ gives yellow AgI, which is insoluble in NH₄OH. F⁻ gives no precipitate because AgF is soluble. A pale yellow, sparingly soluble precipitate therefore indicates bromine.

Answer: (A)

The number of hyperconjugative structures equals the number of α-hydrogens, meaning the hydrogens on sp³ carbons directly attached to the sp² carbons of the double bond. In (CH3)2C=CH–CH3, C-2 carries two CH3 groups (6 α-H) and C-3 carries one CH3 group (3 α-H). Total = 9. The single vinylic hydrogen on C-3 is on an sp² carbon and does not take part. So there are 9 hyperconjugative structures.

Answer: (A)

In the Carius method the halogen is converted to silver halide, so the mass of Br is the mass of AgBr × (80/188). Mass of Br = 0.376 × 80/188 = 0.16 g. Percentage of Br = 0.16/0.30 × 100 = 53.3%.

Answer: (D)

The first step of electrophilic addition is protonation of the double bond, and the stability of the resulting carbocation decides the product. Adding H+ to the terminal CH2 gives CF3–CH+–CH3, a secondary cation directly attached to CF3. The three fluorines withdraw electrons strongly by the –I effect and intensify the positive charge, so this cation is highly destabilised. Adding H+ to the CH carbon gives CF3–CH2–CH2+, a primary cation separated from CF3 by one carbon, so it is less destabilised. Cl– attacks this cation to give CF3CH2CH2Cl. The product therefore appears anti-Markovnikov, even though no peroxide is involved.