JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 16

Answer: (c) 1 > 2 > 3

Explanation:
1. Alkyl groups donate electrons (+I) and also hinder solvation of the alkoxide ion, so acidity falls as the alcohol becomes more branched.
2. Cyclohexylmethanol is primary and the most acidic, so it has the lowest pKa.
3. Tricyclohexylmethanol is tertiary and the least acidic, so it has the highest pKa. Order of pKa: 1 > 2 > 3.

Teacher's Note:
1. A higher pKa means a weaker acid.
2. Acidity of alcohols: primary > secondary > tertiary.

Answer: (c) 4 > 2 > 3 > 1

Explanation:
1. An electron-withdrawing group stabilises the carboxylate ion and increases acidity. The formyl group (-M, -I) does this most, so 4 is the strongest acid.
2. Benzoic acid has no substituent.
3. The ethyl group donates electrons (+I) and weakens the acid, and the hydroxyl group donates by +M from the para position, weakening it most. Order: 4 > 2 > 3 > 1.

Teacher's Note:
1. Para donors (+M) lower the acidity of benzoic acid.
2. Para acceptors (-M) raise it.

Answer: (c) \(\ce{-CN}\)

Explanation:
1. A group raises acidity if it withdraws electrons and stabilises the benzoate ion.
2. The cyano group is strongly electron withdrawing (-I and -M).
3. The methoxy, methyl and amino groups all donate electrons and decrease the acidity.

Teacher's Note:
1. Electron-withdrawing groups increase Ka of acids.
2. Electron-donating groups decrease it.

Answer: (c) 1 > 2 > 3 > 4

Explanation:
1. The two halogens on the ring exert a -I effect, which stabilises the carboxylate ion.
2. The -I effect falls in the order F > Cl > Br > I because electronegativity falls.
3. So acidity falls in the same order: 1 > 2 > 3 > 4.

Teacher's Note:
1. Inductive effect depends on the electronegativity of the atom.
2. Order of -I: F > Cl > Br > I.

Answer: (c) ii > iii > i

Explanation:
1. The formyl group is strongly electron withdrawing (-I), so (ii) is the strongest acid.
2. The phenyl group is also mildly electron withdrawing by -I (sp2 carbons), so (iii) is stronger than (i).
3. The ethyl group donates electrons, so propanoic acid is the weakest. Order: ii > iii > i.

Teacher's Note:
1. Compare the groups attached to the CH2 next to COOH.
2. CHO is a stronger -I group than phenyl.

Answer: (d) 3 > 2 > 1

Explanation:
1. Basic strength of a carbanion is higher if the lone pair is in an orbital of lower s-character.
2. The lone pair of (3) is in an sp3 orbital (25 percent s), of (2) in sp2 (33 percent) and of (1) in sp (50 percent).
3. The more s-character, the more tightly held the pair, so the order of basicity is 3 > 2 > 1.

Teacher's Note:
1. More s-character means more stable carbanion and weaker base.
2. Acidity of terminal alkynes is explained the same way.

Answer: (b) B > D > A > C

Explanation:
1. The p-hydroxy group donates its lone pair into the ring by +M and stabilises the benzylic cation best, so B is the most stable.
2. Benzyl cation (D) has no substituent.
3. A meta hydroxy group cannot donate by resonance and only withdraws by -I, so A is less stable than D, and the p-formyl group withdraws by -M and -I, making C the least stable.

Teacher's Note:
1. Only ortho and para groups can donate by resonance to a benzylic centre.
2. Withdrawing groups destabilise carbocations.

Answer: (b) 1 and 3 only

Explanation:
1. Hyperconjugation needs a C - H bond on the carbon next to a pi bond, an empty p orbital or a radical centre.
2. The decalin radical has adjacent C - H bonds, and the \(\ce{CH2}\) of diphenylmethane has C - H bonds next to the rings.
3. In tert-butylbenzene the carbon attached to the ring has no hydrogen, so there is no hyperconjugation.

Teacher's Note:
1. The carbon next to the unsaturated centre must carry a hydrogen.
2. Tertiary butyl has no alpha hydrogen.

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

Explanation:
1. Basic strength falls as pKb rises.
2. Ethylamine is the strongest base (lone pair on a nitrogen attached to an alkyl group), benzylamine is a little weaker because of the phenyl -I effect.
3. Aniline is far weaker because its lone pair is delocalised into the ring, and diformylamine is the weakest because the lone pair is delocalised into two carbonyl groups. Order of pKb: iii > iv > ii > i.

Teacher's Note:
1. Delocalised lone pairs make weaker bases.
2. A larger pKb means a weaker base.

Answer: (d) pyrrolidin-2-yl cation

Explanation:
1. The positive carbon in (d) is attached to nitrogen, whose lone pair is donated to the empty p orbital (an iminium ion), giving every atom a complete octet.
2. The cations in (a) and (c) are stabilised only by allylic delocalisation, and the phenyl cation has its empty orbital in the plane of the ring, so it is very unstable.

Teacher's Note:
1. A lone pair on the atom next to the cation gives the strongest stabilisation.
2. Phenyl cations cannot be stabilised by the ring pi system.

Answer: (c) succinimide

Explanation:
1. In succinimide the lone pair on nitrogen is delocalised over two carbonyl groups, so it is hardly basic at all (the N - H is in fact acidic).
2. Naphthylamine is an aromatic amine of low basicity, while benzylamine and cyclohexylamine are aliphatic amines and are strongly basic.

Teacher's Note:
1. Amides and imides are very weak bases.
2. Aliphatic amines are the strongest simple bases.

Answer: (b) 4 > 3 > 2 > 1

Explanation:
1. Reactivity depends on the electron density of the ring.
2. Two methoxy groups (+M) activate most, in 1,3-dimethoxybenzene.
3. Methyl (+I) activates the anisole slightly, chlorine (-I) deactivates a little, and the formyl group (-M) deactivates most. Order: 4 > 3 > 2 > 1.

Teacher's Note:
1. Donors raise and acceptors lower ring reactivity.
2. Two strong donors beat one donor and one weak donor.

Answer: (d) \(\ce{-OCH3}\) (methoxy)

Explanation:
1. The methoxy group donates its lone pair strongly into the ring (+M), which makes the ring most electron rich.
2. In acetamido and acetoxy groups the lone pair on N or O is also pulled toward the carbonyl, so they activate less.
3. The hydroxymethyl group has no lone pair on the carbon attached to the ring.

Teacher's Note:
1. A lone pair that is shared with a carbonyl donates less.
2. Methoxy is a strong activator.

Answer: (c) phenyl acetate

Explanation:
1. The acetoxy group of phenyl acetate has a lone pair on oxygen that can be donated to the ring (+M), so the ring is activated.
2. Chlorine deactivates through -I, and the nitro group deactivates strongly through -M and -I.
3. So phenyl acetate is more reactive than benzene and the others.

Teacher's Note:
1. A group with a lone pair on the atom attached to the ring is activating.
2. Halogens are the exception: they are deactivating but ortho-para directing.

Answer: (b) triphenylmethyl chloride

Explanation:
1. The most stable carbocation is the one with the greatest delocalisation of the positive charge.
2. The triphenylmethyl (trityl) cation has three phenyl rings sharing the charge and so is most stable.
3. Diphenylmethyl has two rings, benzyl has one, and the cycloheptadienyl cation is not aromatic.

Teacher's Note:
1. More aryl groups mean more resonance.
2. The cation of a non-aromatic cycloheptadiene is not special.

Answer: (a) methylamine

Explanation:
1. In methylamine the lone pair on nitrogen is attached to an alkyl group and is fully available for protonation.
2. In the three anilines the lone pair is delocalised into the ring, and the effect is greatest with the nitro group.
3. So methylamine is the most basic in the list.

Teacher's Note:
1. Aliphatic amines are stronger bases than aromatic amines.
2. A -M group in the ring lowers basicity further.

Answer: (c) 1-methyl-3-phenylcyclohex-2-en-1-yl radical

Explanation:
1. The radical in (c) is tertiary, allylic, and the allyl system is further conjugated with a phenyl group on its far end.
2. That gives the most delocalisation.
3. The others are secondary (a), secondary allylic (b) and primary (d) with no delocalisation.

Teacher's Note:
1. Radical stability: more delocalisation and more alkyl groups both help.
2. Allylic and benzylic radicals are especially stable.

Answer: (c) o-nitrophenol

Explanation:
1. Sodium bicarbonate dissolves only acids stronger than carbonic acid (\(\ce{H2CO3}\)).
2. Benzoic acid, benzenesulphonic acid and picric acid are all strong enough and evolve \(\ce{CO2}\).
3. o-Nitrophenol (pKa about 7.2) is weaker than carbonic acid and does not react.

Teacher's Note:
1. Test: carboxylic acids dissolve in bicarbonate, most phenols do not.
2. Picric acid is a very strong phenol.

Answer: (b) \(\ce{(CH3)2NH > CH3NH2 > (CH3)3N > NH3}\)

Explanation:
1. In water, basicity depends on +I effect and on solvation of the ammonium ion.
2. For methyl-substituted amines the order is secondary > primary > tertiary > ammonia.
3. The tertiary amine is poorly solvated, which cancels its extra +I effect.

Teacher's Note:
1. For methylamines: (CH3)2NH > CH3NH2 > (CH3)3N in water.
2. In the gas phase tertiary is the strongest.

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

Explanation:
1. The nitro group at ortho and para positions withdraws electrons by both -M and -I, so those nitrophenols are much more acidic than phenol.
2. At the meta position only -I operates, so 3-nitrophenol is weaker.
3. 2-Nitrophenol has intramolecular hydrogen bonding that holds its H, so it is a little less acidic than the para isomer. Order: III > II > IV > I.

Teacher's Note:
1. Ortho nitrophenol is a bit weaker than para because of hydrogen bonding.
2. Meta nitrophenol is acidified only by -I.

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

Explanation:
1. In the gas phase there is no solvation, so only the +I effect and polarisability of the alkyl groups matter.
2. Each methyl group adds electron density and stabilises the ammonium ion.
3. So basicity rises with the number of methyl groups: III > II > I > IV.

Teacher's Note:
1. Solvation is absent in the gas phase.
2. Basic strength in the gas phase follows the number of alkyl groups.

Answer: (a) \(\ce{-CH3}\)

Explanation:
1. Hyperconjugation depends on the number of alpha hydrogen atoms. The methyl group has three, the ethyl group two, and the tert-butyl group none.
2. Inductive effect increases with the size of the alkyl group, so methyl has the smallest.
3. The deuterated methyl group has weaker hyperconjugation than methyl because the C - D bond is stronger.

Teacher's Note:
1. Count the alpha hydrogens for hyperconjugation.
2. Methyl: most hyperconjugation, least +I among the alkyl groups.

Answer: (a) II > I > III

Explanation:
1. In (II) the oxygen lone pair is donated to the cationic carbon by +M effect, which is the strongest stabilisation.
2. (I) is stabilised by two methyl groups (+I and hyperconjugation).
3. In (III) the oxygen is one carbon away, so it only withdraws by -I effect and destabilises the cation. Order: II > I > III.

Teacher's Note:
1. A lone pair on the atom next to a cation stabilises it by resonance.
2. An electronegative atom one carbon away destabilises a cation.