JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 30

Answer: (a) (i) > (ii) > (iii) > (iv)

Explanation:
1. The cyclopentadienyl anion is aromatic (6 pi electrons) and the most stable.
2. The benzyl anion is stabilised by resonance with the ring.
3. The phenyl anion has its lone pair in an sp2 orbital in the plane and is not delocalised, but is better than an sp3 anion.
4. The cyclohexyl anion is a plain secondary sp3 carbanion, the least stable.

Teacher's Note:
1. Aromatic anion first, then resonance-stabilised, then s-character.
2. Alkyl anions are the least stable.

Answer: (d) III > I > II > IV

Explanation:
1. The p-nitro group stabilises the carbanion strongly (–M, –I).
2. The donor groups destabilise it; OCH3 (+M) does so more than CH3.
3. Order: III > I > II > IV.

Teacher's Note:
1. Electron-withdrawing groups stabilise anions.
2. The stronger the donor, the less stable the carbanion.

Answer: (d) triphenylacetaldehyde

Explanation:
1. Keto–enol tautomerism requires an alpha hydrogen.
2. The first three each have alpha hydrogens.
3. In triphenylacetaldehyde the alpha carbon carries three phenyl groups and no hydrogen.

Teacher's Note:
1. Check for an alpha hydrogen on the carbon next to C=O.
2. Without it, an enol cannot form.

Answer: (b) 2-nitroethanal, O2N–CH2–CHO

Explanation:
1. Tautomerism needs an alpha hydrogen.
2. Benzoquinone has no sp3 carbon, benzaldehyde and 2,2-dimethylpropanal have no alpha hydrogen.
3. In O2N–CH2–CHO the CH2 carries two very acidic hydrogens, flanked by NO2 and CHO.

Teacher's Note:
1. Look for sp3 C–H next to C=O or NO2.
2. Strong electron-withdrawing neighbours make the hydrogen very acidic.

Answer: (d) RCH2NO2

Explanation:
1. Nitroalkanes show nitro–aci-nitro tautomerism only when there is an alpha hydrogen.
2. RCH2NO2 has alpha hydrogens, but R3C–NO2 and (CH3)3C–NO2 have none.
3. (CH3)2NH has no group for such tautomerism.

Teacher's Note:
1. The nitro group makes the alpha hydrogens acidic.
2. Tertiary nitro compounds cannot form the aci form.

Answer: (a) 9 sigma bonds, 1 pi bond and 2 lone pairs

Explanation:
1. C–H bonds: 2 + 3 = 5; C=C sigma 1; C–C 1; C–O 1; O–H 1. Total sigma = 9.
2. One pi bond (C=C).
3. The oxygen carries two lone pairs.

Teacher's Note:
1. Draw the structure fully before counting.
2. A double bond is one sigma plus one pi.

Answer: (a) i > ii > iii > iv

Explanation:
1. The enol of cyclohexa-2,5-dienone is phenol, which is aromatic, so its enol content is essentially 100 %.
2. In the 1,3-dione the enol is conjugated and hydrogen bonded in part, so it has a large enol content.
3. The 1,4-dione has a much smaller enol content and simple cyclohexanone is least.

Teacher's Note:
1. Aromatic stabilisation of the enol dominates.
2. 1,3-Dicarbonyls enolise readily.

Answer: (d) If hyperconjugation increases, Ka increases.

Explanation:
1. Electron-donating effects destabilise the conjugate base and so reduce acidity.
2. Electron-withdrawing effects stabilise the base and raise Ka.
3. A more stable conjugate base means a stronger acid.
4. Hyperconjugation is electron donation, so it decreases Ka; statement 4 is wrong.

Teacher's Note:
1. Acid strength is decided by the stability of the conjugate base.
2. Donors lower acidity; acceptors raise it.

Answer: (d) 10⁻²

Explanation:
1. A larger Ka means a stronger acid.
2. 10⁻² is the largest value among the four.

Teacher's Note:
1. Larger Ka and smaller pKa mean stronger acid.
2. Compare exponents carefully.

Answer: (d) CH3OH

Explanation:
1. Oxygen is more electronegative than carbon, so the conjugate base CH3O⁻ is the best stabilised.
2. pKa values are about 16 (methanol), 25 (ethyne), 43 (benzene) and 50 (ethane).

Teacher's Note:
1. Hybridisation helps carbon acids but electronegativity of the atom bearing the charge matters more.
2. Alcohols are far more acidic than alkynes.

Answer: (b) propiolic acid

Explanation:
1. The sp-hybridised carbon of the alkyne is more electron-withdrawing than the sp2 carbon of the alkene, which in turn is more than an sp3 carbon.
2. So propiolic acid is the strongest acid (pKa about 1.9), then acrylic acid (4.25), then the saturated acids (about 4.8).

Teacher's Note:
1. Greater s-character on the carbon attached to COOH means a stronger –I effect.
2. Saturated alkyl groups are electron donating and weaken the acid.

Answer: (a) CF3COOH

Explanation:
1. Fluorine has the strongest –I effect.
2. Three fluorines stabilise the carboxylate best, so CF3COOH is the strongest.

Teacher's Note:
1. Acid strength follows the –I power of the substituents: F > Cl > Br.
2. Carboxylates with more withdrawing groups are more stable.

Answer: (b) y > x > z

Explanation:
1. Two chlorine atoms on the alpha carbon (y) have a large –I effect close to the carboxyl group.
2. In x and z the substituent is on the beta carbon, so the effect is weaker, and NO2 is stronger than Cl.
3. Order: y > x > z (pKa about 1.8, 3.6 and 4.0).

Teacher's Note:
1. Proximity matters: the –I effect falls rapidly with distance.
2. Then compare the strength of the groups.

Answer: (b) 1 > 2 > 3

Explanation:
1. The –I effect of bromine weakens with distance from the COOH group.
2. Bromine on the same carbon (1) has the strongest effect, then the adjacent carbon (2), then the next (3).

Teacher's Note:
1. Nearer electron-withdrawing group means stronger acid.
2. The –I effect is negligible beyond three carbons.

Answer: (d) PhSO3H > PhCO2H > PhOH > PhCH2OH

Explanation:
1. Sulphonic acids are stronger than carboxylic acids (pKa about −2 versus 4.2).
2. Phenol (pKa 10) is weaker than a carboxylic acid but stronger than benzyl alcohol (pKa about 15).

Teacher's Note:
1. Sulphonic > carboxylic > phenol > alcohol.
2. Resonance stabilisation of the conjugate base decides this order.

Answer: (a) phenol > water > ethyl alcohol

Explanation:
1. The phenoxide ion is resonance stabilised, so phenol is the strongest acid.
2. The ethyl group is electron-donating, which destabilises ethoxide, so ethanol is weaker than water.

Teacher's Note:
1. Resonance-stabilised bases come from stronger acids.
2. Alkyl groups reduce the acidity of alcohols relative to water.

Answer: (b) IV < II < I < III

Explanation:
1. Terminal alkyne (pKa about 25) is weaker than an alcohol (about 16 to 18).
2. Cyclohexanol is weaker than phenol (about 10).
3. The carboxylic acid (about 4.9) is the strongest.
4. Order of increasing acidity: IV < II < I < III.

Teacher's Note:
1. Acids ranked by pKa: carboxylic acid < phenol < alcohol < alkyne.
2. Smaller pKa means stronger acid.

Answer: (a) CH3OH

Explanation:
1. Alkyl groups donate electrons and destabilise the alkoxide ion, and solvation of the anion is also hindered.
2. Methanol has no alkyl group, so it is the most acidic.

Teacher's Note:
1. Acidity of alcohols: methyl > primary > secondary > tertiary.
2. This is the reverse of their basicity as bases in gas phase.

Answer: (a) the imino nitrogen (=NH)

Explanation:
1. Protonation at the imino nitrogen gives the guanidinium ion, \( \ce{H2N-C^{+}(NH2)-NH2} \), in which the charge is delocalised over three nitrogen atoms.
2. Protonation at an amino nitrogen would not give this symmetrical resonance-stabilised cation.

Teacher's Note:
1. Guanidine is one of the strongest organic bases because its conjugate acid is so stable.
2. Always protonate where the resulting cation is best stabilised.

Answer: (a) I > III > II > IV

Explanation:
1. Chlorine on the alpha carbon (I) has the strongest –I effect.
2. Chlorine on the beta carbon (III) has a weaker effect but still raises acidity over acetic acid (II).
3. Isobutyric acid (IV) has two electron-donating methyl groups and is the weakest.
4. Order: I > III > II > IV.

Teacher's Note:
1. Order the groups by –I and +I effects and by distance.
2. Branching in the alkyl chain reduces acidity.

Answer: (a) i > iii > ii > iv

Explanation:
1. Donor substituents stabilise the benzyl cation: OCH3 (+M) > CH3 (+I, hyperconjugation).
2. Unsubstituted benzyl comes next, and the nitro group, which withdraws electrons, destabilises the cation the most.
3. Order: i > iii > ii > iv.

Teacher's Note:
1. Electron donors stabilise cations; acceptors destabilise them.
2. Compare the strength of +M and +I effects.

Answer: (a) III > IV > I > II

Explanation:
1. Nitro groups at ortho and para positions stabilise the phenoxide ion by –M and –I effects.
2. 2,4-Dinitrophenol (pKa 4.1) is stronger than p-nitrophenol (7.15).
3. Phenol (10.0) is stronger than p-cresol (10.3), as the methyl group donates electrons.
4. Order: III > IV > I > II.

Teacher's Note:
1. More –M groups at o/p positions mean a stronger phenol.
2. Donor groups weaken phenols.

Answer: (b) sodium picrate, CO2 and H2O

Explanation:
1. Picric acid has pKa about 0.4, so it is a stronger acid than carbonic acid.
2. Hence it decomposes bicarbonate: picric acid + NaHCO3 gives sodium picrate + H2O + CO2.

Teacher's Note:
1. Phenol itself does not react with NaHCO3.
2. Three nitro groups make the phenoxide extremely stable.

Answer: (a) CH3(+) < CH3CH2(+) < (CH3)2CH(+) < (CH3)3C(+)

Explanation:
1. Each alkyl group adds +I effect and hyperconjugation (more alpha hydrogens).
2. Methyl has none, ethyl has 3 alpha hydrogens, isopropyl 6, and tert-butyl 9.
3. Hence stability rises from methyl to tertiary.

Teacher's Note:
1. Count alpha hydrogens to see the hyperconjugation effect.
2. The most substituted cation is the most stable.