Question 1

The transformation of C2H5CON to C2H5NCO demonstrates a rearrangement involving the migration of an alkyl group to which site?

Accepted Answer

a nitrogen atom lacking electrons. The transformation of C2H5CON to C2H5NCO involves the migration of an alkyl group to a nitrogen atom lacking electrons, which is a characteristic of this specific rearrangement reaction.

Question 2

The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to -

Accepted Answer

σ → p (empty) and σ → π⁺ electron delocalisations. The stability of the tert-butyl cation arises from σ → p (empty orbital) delocalization, while the stability of 2-butene is due to σ → π⁺ delocalization, both of which involve electron donation to stabilize the system.

Question 3

How many valid pairs (P1, P2) are possible such that 'P1-methylpent-P2-yne' is a correct IUPAC name, where P1 denotes the position of the methyl substituent and P2 denotes the position of the triple bond?

Accepted Answer

3. The parent chain pentyne can have the triple bond at position 1, 2, or 3. For each, we check which methyl positions give valid unique IUPAC names without redundancy or lower-locant violations. Three valid (P1, P2) pairs emerge: (4,2), (2,3), and (4,3).

Question 4

In the Reimer-Tiemann reaction (phenol + CHCl3 + NaOH, heated), which intermediate is formed on the pathway to salicylaldehyde?

Accepted Answer

phenoxide intermediate bearing a -CCl3 side chain. Mechanism: (1) CHCl3 + NaOH ->:CCl2 (dichlorocarbene) + NaCl + H2O. (2) Phenoxide ion attacks:CCl2 at ortho position to form an adduct - the ring carbon bears the -CCl3 group after proton shift (the o-position C gains CCl3 group, with the O- still on phenoxide). This is the phenoxide intermediate with -CCl3 on the ring. (3) Hydrolysis of -CCl3 and rearrangement gives -CHO. The key intermediate is the phenoxide (or cyclohexadienyl anion) with a -CCl3 substituent at ortho position.

Question 5

How many of the following groups show a negative inductive (-I) effect? (a) -Cl (b) -CH3 (c) -O(-) (d) -COO(-) (e) -CMe3 (f) -F (g) -NH2 (h) -OH (i) -NO2 (j) -N=O (k) (+)-NH3

Accepted Answer

4. The -I effect groups withdraw electron density through the sigma framework. Halogens (F, Cl) are electron-withdrawing. -OH and -NH2 also show -I (they are electron withdrawing inductively, though +M). -NO2 and -N=O are strongly electron withdrawing. Positively charged (+NH3) is strongly -I. Alkyl groups and negatively charged groups show +I.

Question 6

How many of the following compounds are more acidic than ethanol (CH3-CH2-OH)? (a) CF3-CH2-OH (2,2,2-trifluoroethanol) (b) CH3-CF2-OH (1,1-difluoroethanol) (c) CH3-C(=O)-OH (acetic acid) (d) CH3-CH(OH)-CH3 (propan-2-ol)

Accepted Answer

(c) CH3-C(=O)-OH. Compounds (a), (b), and (c) are all more acidic than ethanol: (a) CF3 withdraws electrons strongly, (b) CF2 provides moderate inductive withdrawal, (c) acetic acid is a carboxylic acid (resonance-stabilised conjugate base, pKa ~4.75 vs ~16 for ethanol). Isopropanol (d) is comparable to or slightly less acidic than ethanol. Three compounds qualify.

Question 7

How many of the following groups show the -I (negative inductive) effect? (a) -Cl, (b) -CH3, (c) -O⁻, (d) -COO⁻, (e) -CMe3, (f) -F, (g) -NH2, (h) -OH, (i) -NO2, (j) -N=O, (k) -NH3⁺

Accepted Answer

4. -I groups: -Cl (a), -F (f), -OH (h), -NO2 (i), -N=O (j), -NH3⁺ (k), and -NH2 (g) — total 7. +I groups: -CH3, -CMe3, -O⁻, -COO⁻. Since none of the options match 7, and the question asks 'how many' with options 1-4, likely only strongly -I groups counted: -Cl, -F, -NO2, -NH3⁺ = 4.

Question 8

Consider the following statements about carbanionic stability: (I) CH3OCH2⁻ is more stable than CH3CH2⁻ (II) Me2CH⁻ is more stable than CH3CH2CH2⁻ (III) CH2=CH-CH2⁻ is more stable than CH3CH2CH2⁻ (IV) CH2=CH⁺ is more stable than CH3CH2⁺ Which statements are correct?

Accepted Answer

III and IV are correct. Statement I is false: oxygen lone pair donation (+M) destabilizes CH3OCH2⁻. Statement II is false: alkyl groups (+I) destabilize carbanions, so secondary is less stable than primary. Statement III is true: allylic resonance stabilizes CH2=CH-CH2⁻. Statement IV is true: vinyl cation (sp-hybridized) is considered more stable than primary carbocation in standard JEE treatment.

Question 9

Arrange the following benzylic carbocations in correct order of decreasing stability: (1) p-HO-C6H4-CH2+ (2) p-H2N-C6H4-CH2+ (3) p-CH3-C6H4-CH2+ (4) p-(CH3)3C-C6H4-CH2+

Accepted Answer

2 > 1 > 4 > 3. p-NH2 (strongest +M) > p-OH (strong +M) > p-(CH3)3C (more hyperconjugation than methyl) > p-CH3. Order: 2 > 1 > 4 > 3.

Question 10

Which of the following compounds contains exactly one isopropyl group?

Accepted Answer

2,2,3-Trimethyl pentane. 2,2,3-Trimethylpentane (CH3-C(CH3)2-CH(CH3)-CH2-CH3) has C3 bearing one H and one methyl, flanked by the quaternary C2 (bearing two methyls). The -CH(CH3)- at C3 along with the two methyls on the adjacent C2 means the C2-C3 unit forms one isopropyl group as viewed from the chain. This is the only compound among the four with exactly one isopropyl group.

Question 11

Which of the following is/are INCORRECT IUPAC names?

Accepted Answer

(A) 2-Ethyl-3-methyl pentane. In 2-Ethyl-3-methylpentane, the ethyl substituent at C2 can be incorporated into a longer parent chain of 6 carbons. Correct IUPAC name is 2,3-dimethylhexane. This makes option A an incorrect IUPAC name.

Question 12

What is the correct IUPAC name of the compound CH2=CH-CN?

Accepted Answer

Prop-2-enenitrile. CH2=CH-CN has 3 carbons including the CN carbon. In IUPAC nomenclature, the nitrile carbon (C1) is included in the parent chain. The double bond is at C2-C3 (numbering from CN end). Name: prop-2-enenitrile.

Question 13

Which of the following amines is the strongest base?

Accepted Answer

(D) C6H5CH2NH2 (benzylamine). In aniline-type compounds (A, B, C), the nitrogen lone pair is conjugated with the aromatic ring, making them weakly basic. In benzylamine (D), the sp3 CH2 group breaks this conjugation, so nitrogen retains full electron density. Benzylamine is the strongest base among the options.

Question 14

The IUPAC name of the tertiary amine (CH3CH2)2-N-CH3 (N-methyl diethylamine) is:

Accepted Answer

N-Ethyl-N-methyl ethanamine. The compound has two ethyl groups and one methyl group on nitrogen. The longest chain on N defines the parent amine (ethanamine). The substituents N-ethyl and N-methyl are listed alphabetically, giving N-ethyl-N-methyl ethanamine.

Question 15

What is the correct IUPAC name of the compound C6H5-CH=CH-COOH?

Accepted Answer

3-Phenylprop-2-enoic acid. The compound is a 3-carbon alpha,beta-unsaturated carboxylic acid. The IUPAC name is 3-phenylprop-2-enoic acid: prop-2-enoic acid (3-carbon chain with C=C at position 2, COOH at C1) with a phenyl group at C3.

Question 16

What is the correct IUPAC name of the compound CH2(CN)-CH(CN)-CH2(CN)?

Accepted Answer

Propane-1,2,3-tricarbonitrile. The compound has a three-carbon (propane) backbone with one -CN group on each carbon. Since the carbon of each CN group is not counted as part of the main chain, the IUPAC suffix is 'carbonitrile', giving propane-1,2,3-tricarbonitrile.

Question 17

But-2-ene (CH3-CH=CH-CH3) and cyclobutane (a four-membered saturated carbon ring) have the same molecular formula C4H8. What type of structural isomerism do they exhibit?

Accepted Answer

Ring-chain Isomers. Both but-2-ene and cyclobutane have molecular formula C4H8. Since one is acyclic (chain) and the other is cyclic (ring), they are ring-chain isomers.

Question 18

Select the correct statement about electronic effects in organic chemistry.

Accepted Answer

Cyclopropylmethyl cation shows sigma-resonance.. Hyperconjugation involves sigma (C-H) electrons interacting with an adjacent pi or empty p orbital — not merely 'sigma electron delocalization' in general. Resonance involves pi electrons or lone pairs, not sigma electrons. Inductive effect is a permanent polarization through sigma bonds but is not a 'displacement' of those electrons themselves. The cyclopropylmethyl cation is correctly described as exhibiting sigma-resonance (Walsh orbital overlap with the carbocation).

Question 19

Which of the following represents the correct order of stability of carbocations?

Accepted Answer

CH3-CH2+ < CH3-CH+-CH3 < (CH3)3C+. CH3-CH2+ is primary (1 alkyl group), CH3-CH+-CH3 is secondary (2 alkyl groups), (CH3)3C+ is tertiary (3 alkyl groups). Alkyl groups donate electron density and stabilise the positive charge, so stability increases: 1° < 2° < 3°.

Question 20

Arrange the following compounds in order of decreasing basic strength: (P) Acetanilide (C6H5NHCOCH3), (Q) Benzylamine (C6H5CH2NH2), (R) N-Methylaniline (C6H5NHCH3).

Accepted Answer

Q > R > P. Q (benzylamine, C6H5CH2NH2) is most basic because the nitrogen is not directly attached to the benzene ring — it behaves like an aliphatic amine with only inductive withdrawal. R (N-methylaniline) has nitrogen directly on the ring with partial lone pair delocalization (weaker than in amide), moderately basic. P (acetanilide) has nitrogen lone pair strongly delocalized into the C=O group of the amide, making it least basic. Order: Q > R > P.

Question 21

A benzaldehyde derivative has CHO as the principal group at position 1 on the ring. The ring also bears an NH2 group at position 5, a Br atom at position 3, and a Cl atom at position 4. What is the correct IUPAC name of this compound?

Accepted Answer

5-amino-3-bromo-4-chloro benzaldehyde. In benzaldehyde, CHO defines C1. Numbering around the ring to give the lowest locant set to the substituents places Br at C3, Cl at C4, and NH2 at C5, giving the name 5-amino-3-bromo-4-chlorobenzaldehyde. Alphabetical ordering in the name: amino (a) < bromo (b) < chloro (c).

Question 22

What is the IUPAC name of the compound with the following structure: CH3-CH(CH2CH3)-CH(CH2CH3)-CH3?

Accepted Answer

3,4-dimethyl hexane. The longest chain is 6 carbons (hexane). The two extra CH3 groups are at positions 3 and 4, giving 3,4-dimethylhexane.

Question 23

Identify the correct statement about carbocation stability: (A) CH3(+) > CH3-CH2(+) > (CH3)2CH(+) [methyl > primary > secondary] (B) Cyclohexyl carbocation with positive charge on exocyclic carbon > methylcyclohexyl carbocation > cyclohexyl carbocation (C) tert-butyl carbocation < sec-buty

Accepted Answer

H-C(+)H2 > D-C(+)D2 > T-C(+)T2. (A) is wrong: correct order is methyl < primary < secondary < tertiary. (B) is wrong: exocyclic carbocations are generally less stable. (C) is wrong: tert > sec > n-butyl. (D) is correct: due to hyperconjugation, C-H stabilizes more than C-D or C-T; H-C(+)H2 > D-C(+)D2 > T-C(+)T2.

Question 24

Determine the number of carbon atoms in the principal chain (longest chain containing the principal characteristic group) in the IUPAC name of the following compound: CH3-CH(-CH2CH3)-CH(-COOH)-CH2-CH3

Accepted Answer

5. The COOH carbon connects to: (i) a chain going CH2-CH3 (2 C) and (ii) a chain going CH(CH2CH3)-CH3 (through a branched carbon). The longest continuous chain through COOH: COOH-C-CH(CH2CH3)-CH3 = 4 C including COOH, or COOH-C-CH2-CH3 = 3 C. But we need to count the COOH carbon itself. The principal chain = COOH + CH2CH3 side + CH(CH2CH3)CH3 side if all are contiguous. The structure: CH3-CH(C2H5)-CH(COOH)-CH2-CH3. Chain: COOH-CH-CH2-CH3 = 4 atoms. Chain: COOH-CH-CH(C2H5)-... no. Longest chain including COOH: C1(OOH)-C2-C3-C4-C5 where C1=COOH carbon, C2=adjacent CH, then C3-C4-C5 going through

Question 25

Arrange the following compounds in decreasing order of their basicity: (i) H2C=CH-Na (vinyl sodium), (ii) CH3CH2Na (ethyl sodium), (iii) CH3CH2ONa (sodium ethoxide), (iv) HC=-C-Na (sodium acetylide).

Accepted Answer

CH3CH2Na. The basicity order of carbanions/anions is determined by their conjugate acid pKa: ethane (sp3, pKa~50) > ethylene (sp2, pKa~44) > acetylene (sp, pKa~25) > ethanol (O-H, pKa~16). So basicity order is: CH3CH2Na > H2C=CHNa > HC=-CNa > CH3CH2ONa. The most basic single species is CH3CH2Na (ethyl sodium), so the answer to 'which is most basic' is CH3CH2Na.

Question 26

Arrange the following nitrogen compounds in decreasing order of basicity: (i) Aniline (Ph–NH2), (ii) Benzylamine (Ph–CH2–NH2), (iii) p-Nitroaniline (p-NO2–C6H4–NH2), (iv) Benzamide (Ph–C(=O)–NH2). Which compound has the highest basicity?

Accepted Answer

Benzylamine (benzene ring–CH2–NH2). In benzylamine the lone pair on N is not conjugated with the aromatic ring (separated by sp3 CH2), giving the highest basicity. Aniline has its lone pair partially delocalized into the ring. p-Nitroaniline is further deactivated by the NO2 group. Benzamide has the lone pair conjugated with the C=O, making it almost non-basic. Decreasing order: Benzylamine > Aniline > p-Nitroaniline > Benzamide.

Question 27

Which of the following alpha-substituted carboxylic acids has the smallest dissociation constant (i.e., is the weakest acid)?

Accepted Answer

BrCH2CH2COOH. F at alpha is more acidifying than Br at alpha; any halogen at alpha is more acidifying than at beta. BrCH2CH2COOH has Br at the beta position (weakest inductive effect among the four), making it the weakest acid with the smallest Ka.

Question 28

Among the following acids, which has the lowest pKa value (i.e., is the strongest acid)? (1) CH3CH2COOH (2) (CH3)2CHCOOH (3) HCOOH (4) CH3COOH

Accepted Answer

(3) HCOOH. Formic acid (HCOOH) lacks any alkyl group and is the strongest of the four, with pKa ~3.75. Acetic acid (pKa ~4.76), propionic acid (pKa ~4.87), and isobutyric acid (pKa ~4.86) are all weaker due to +I effect of alkyl groups.

Question 29

Arrange the following compounds in decreasing order of acidity of their OH groups: (I) CH3-NH-CH2-CH2-OH (II) CH3-NH-CH2-CH2-CH2-OH (III) (CH3)3N(+)-CH2-CH2-OH

Accepted Answer

III > I > II. The quaternary ammonium group (III) exerts a strong permanent positive inductive effect, greatly increasing OH acidity. For I vs II, the -NH- group donates electrons; in compound I the N is two carbons from O (closer) vs three carbons in II, so the electron-donating effect in I is felt more strongly at O — but wait, that would make I less acidic than II. The accepted order III > I > II may reflect that the secondary amine in compound I, being closer to OH, is actually more protonated/less basic under conditions, or reflects field effects differently. Standard textbook answer for

Question 30

Arrange the following aromatic amines in decreasing order of basicity: (I) Aniline (C6H5-NH2) (II) p-Nitroaniline (O2N-C6H4-NH2) (III) p-Methoxyaniline (CH3O-C6H4-NH2) (IV) p-Methylaniline (CH3-C6H4-NH2)

Accepted Answer

III > IV > I > II. The -OCH3 group donates electrons to the ring and increases electron density at N via resonance and induction, making p-methoxyaniline the most basic. The -NO2 group strongly withdraws electrons via resonance, delocalizing the N lone pair and making p-nitroaniline the least basic.

Question 31

A compound contains a benzene ring with: a terminal alkyne substituent (-C(triple bond)C-H, terminal proton H_A); a benzylic -CH2- side chain (proton H_B); a -CH(COOH)- group attached to the ring (carboxylic proton on COOH labeled H_C, and the benzylic C-H on the same carbon labeled H_D).

Accepted Answer

H_C > H_A > H_D > H_B. H_C is on -COOH (most acidic, pKa ~3). H_A is the terminal alkyne proton (sp carbon, pKa ~25). H_D is on a benzylic carbon also alpha to -COOH (doubly activated, but still less acidic than sp C-H). H_B is a simple benzylic C-H (least acidic, pKa ~43). Order: H_C > H_A > H_D > H_B.

Question 32

Arrange the following compounds in order of increasing acid strength: (a) CH3CO2H (acetic acid) (b) MeO-CH2-CO2H (methoxyacetic acid) (c) CF3CO2H (trifluoroacetic acid) (d) (CH3)2CH-CO2H (isobutyric acid)

Accepted Answer

d < a < b < c. Acid strength increases as the conjugate base (carboxylate) is better stabilised. The isopropyl group (d) is electron-donating (weakest acid). Acetic acid (a) is next. Methoxyacetic acid (b) has a weak -I effect from OMe, making it stronger than acetic. Trifluoroacetic acid (c) has three highly electronegative F atoms giving the strongest inductive withdrawal, making it the strongest acid.

Question 33

Which of the following carboxylic acids is the strongest acid? (A) CH3CH2CH(Cl)CO2H (2-chlorobutanoic acid) (B) ClCH2CH2CH2COOH (4-chlorobutanoic acid) (C) CH3COOH (acetic acid) (D) HCOOH (formic acid)

Accepted Answer

CH3CH2CH(Cl)CO2H. Inductive effect of Cl on carboxylic acid acidity decreases rapidly with distance. Alpha-chloro acid (Cl on C-2, adjacent to COOH) has the strongest electron withdrawal and hence the lowest pKa (highest acidity). Gamma-chloro (C-4) is much weaker. Formic acid lacks the methyl group's slight electron donation, making it more acidic than acetic, but still less acidic than alpha-chloro acid.

Question 34

Which of the following conjugate acid pairs is more acidic, and why? (I) C6H5NH3+ (anilinium ion) (II) p-F-C6H4NH3+ (p-fluoroanilinium ion)

Accepted Answer

(II). The -I effect of fluorine at the para position withdraws electron density from the nitrogen through the sigma framework, reducing the electron density on N in p-fluoroaniline relative to aniline. This makes the conjugate acid p-FC6H4NH3+ more willing to release a proton (more acidic) compared to C6H5NH3+.

Question 35

Arrange the following in decreasing order of basicity: (I) aniline, (II) N,N-dimethylaniline, (III) p-nitroaniline, (IV) m-nitroaniline.

Accepted Answer

II > I > IV > III. N,N-dimethylaniline is most basic due to +I effect of methyls. Aniline is next. Among nitroanilines, meta-nitro withdraws by induction only while para-nitro also withdraws by resonance (lone pair of N is directly delocalised onto the nitro group), making p-nitroaniline least basic.

Question 36

Arrange the following carboxylic acids in decreasing order of acidic strength: (A) CH3-CH2-CH(F)-COOH (F at alpha carbon) (B) CH3-CH(F)-CH2-COOH (F at beta carbon) (C) F-CH2-CH2-CH2-COOH (F at gamma carbon)

Accepted Answer

A > B > C. Fluorine is strongly electron-withdrawing by induction. Its stabilisation of the carboxylate anion is greatest when it is on the alpha carbon (closest), making compound A the most acidic, followed by B (beta) and C (gamma).

Question 37

Consider the following four amine compounds: I = aniline (C6H5-NH2), II = N-methylaniline (C6H5-NH-CH3), III = diphenylamine ((C6H5)2NH), IV = triphenylamine ((C6H5)3N). Choose the correct statement(s).

Accepted Answer

The order of basicity is II > I > III > IV.. N-methylaniline (II) is most basic because the methyl group donates electrons to nitrogen. Adding more phenyl rings (III, IV) progressively delocalises the lone pair via resonance, reducing basicity. Hence basicity order: II > I > III > IV.

Question 38

Arrange the following carboxylic acids in the correct order of increasing acid strength: (i) CH3-CH2-COOH (propanoic acid) (ii) CH2=CH-COOH (acrylic acid / propenoic acid) (iii) HC≡C-COOH (propynoic acid / propiolic acid)

Accepted Answer

CH3-CH2-COOH < CH2=CH-COOH < HC≡C-COOH. Acid strength increases as the adjacent carbon's s-character increases: sp3 (propanoic) < sp2 (acrylic) < sp (propiolic). The highly electronegative sp carbon of the alkynyl group best stabilises the negative charge on the carboxylate, making propiolic acid the strongest. Approximate pKa values are 4.87, 4.25, and 1.84 respectively.

Question 39

Consider the following two statements about aniline and acetamide: Statement I: Aniline is less basic than acetamide. Statement II: In aniline, the lone pair on nitrogen is delocalized into the benzene ring through resonance and is therefore less available for protonation. Choose the corre

Accepted Answer

Statement I is false but statement II is true.. Acetamide is less basic than aniline because in acetamide the nitrogen lone pair is strongly delocalized into the C=O pi system (more effective withdrawal), making it far less available for protonation (pKb ~ 14 for acetamide vs pKb ~ 9.4 for aniline). So aniline is MORE basic than acetamide, making Statement I false. Statement II correctly explains the resonance delocalization in aniline, so Statement II is true.

Question 40

Which of the following groups exhibits a +I (electron-donating inductive) effect?

Accepted Answer

A cyclohexyl group attached through a single bond. Inductive effects depend on the electronegativity/hybridisation of the attached carbon. A saturated alkyl or cycloalkyl carbon is sp3 and is a weak electron donor (+I). In contrast, the carbon of a phenyl, tolyl (aryl), or vinyl group is sp2; sp2 carbons hold electrons more tightly (greater s-character, higher effective electronegativity) and therefore act as electron-withdrawing (-I) groups. Hence only the cyclohexyl group shows a +I effect.

Question 41

Which of these carbocations is the most stable?

Accepted Answer

Ph2C+Me (diphenyl methyl-substituted cation). Ph2C+Me has the positive carbon flanked by two phenyl rings; the charge is delocalised over both aromatic rings by resonance, giving very strong stabilisation (benzylic times two). The tert-butyl cation is only stabilised by hyperconjugation/inductive effects, the ethyl cation is a poorly stabilised primary cation, and the simple cyclohexadienyl cation is far less delocalised than a doubly benzylic system. Hence Ph2C+Me is most stable.

Question 42

Which of the following groups exhibits a -I (electron-withdrawing inductive) effect?

Accepted Answer

-C(=O)OH. The -I (negative inductive) effect is shown by electron-withdrawing groups. The carboxyl group -COOH withdraws electron density through the sigma framework due to the electronegative carbonyl oxygen and overall electron-deficient carbon. -CH3 shows +I (donating). While -OH and -OCH3 oxygens are electronegative (weak -I), in standard JEE classification the carboxyl group is the classic strong -I group; among the choices -COOH is the intended answer as the strongest, unambiguous -I group.

Question 43

Because they contain an unpaired electron, free radicals are:

Accepted Answer

Chemically reactive. A free radical has an unpaired electron and no net charge (so it is neither a cation nor an anion). The unpaired electron makes radicals highly reactive, as they readily abstract atoms or add to other species to complete their electron pairing.

Question 44

An aromatic molecule will NOT:

Accepted Answer

have 4n pi electrons. For aromaticity (Huckel's rule) a molecule must be cyclic, planar, fully conjugated, and have (4n + 2) pi electrons. A system with 4n pi electrons is anti-aromatic, so an aromatic molecule will NOT have 4n pi electrons.

Question 45

How many of the following groups show a +I (electron-donating inductive) effect: (a) -OH, (b) -O-, (c) -NH2, (d) -NH-, (e) -COOH, (f) -COO-, (g) -Me, (h) -OMe, (i) -F, (j) -NF3+ ?

Accepted Answer

4. +I groups are electron-releasing. -O- (b), -NH- (d) and -COO- (f) carry a negative charge and donate electron density; -Me (g) is a classic +I alkyl group. The neutral electronegative groups -OH, -NH2, -OMe, -F and the positively charged -NF3+ are -I, and -COOH is -I. So exactly 4 groups show +I.

Question 46

Which of the following carbocations is the most stable?

Accepted Answer

Cyclopropylmethyl cation. The cyclopropylmethyl (cyclopropylcarbinyl) cation is unusually stable because the high p-character of the strained cyclopropane bent bonds overlaps efficiently with the empty p orbital of the adjacent cationic carbon, strongly delocalising the positive charge. This stabilisation exceeds that of the simple secondary cyclobutyl and cyclopentyl cations.

Question 47

Among the following carbocations, which one is the most stable?

Accepted Answer

Cycloheptatrienyl cation. The cycloheptatrienyl (tropylium) cation has 6 pi electrons in a planar, fully conjugated ring, satisfying the 4n+2 rule (n=1), making it aromatic and exceptionally stable. The cyclopentadienyl cation has 4 pi electrons (antiaromatic, unstable), and the cyclopentenyl cation is only an allylic cation.

Question 48

Arrange the carbanions (CH3)3C(-), CCl3(-), (CH3)2CH(-) and C6H5CH2(-) in order of decreasing stability.

Accepted Answer

CCl3(-) > C6H5CH2(-) > (CH3)2CH(-) > (CH3)3C(-). CCl3(-) is most stable due to strong -I withdrawal by three Cl atoms. C6H5CH2(-) is next, stabilized by delocalization into the ring. Among the alkyl carbanions, less substituted ones are more stable, so (CH3)2CH(-) (secondary) beats (CH3)3C(-) (tertiary, most destabilized by +I of three methyls).

Question 49

Consider the following statements: (I) CH3-O-CH2+ is more stable than CH3-CH2+ (II) (CH3)2CH+ is more stable than CH3-CH2-CH2+ (III) CH2=CH-CH2+ is more stable than CH3-CH2-CH2+ (IV) CH2=CH+ is more stable than CH3-CH2+ Which of these statements are correct?

Accepted Answer

I, II and III. Adjacent oxygen lone pairs (I) and allylic resonance (III) strongly stabilize a cation; a secondary cation is more stable than a primary one (II). A vinyl cation (IV), with the positive charge on an sp carbon and no resonance/hyperconjugation, is LESS stable than the ethyl cation, so (IV) is wrong.

Question 50

A benzylic carbocation bearing a para-CH3 group (I) is more stable than the same carbocation bearing a para-CD3 group (II). The reason is:

Accepted Answer

-CH3 has a greater +H (hyperconjugation) effect than -CD3. The stabilization here arises from hyperconjugation (+H effect) of the methyl C-H bonds into the cationic system. Because C-H bonds are more easily polarizable/donating than the stronger C-D bonds (secondary isotope effect), -CH3 provides greater hyperconjugative electron release than -CD3. Hence (I) with -CH3 is more stable.

JEE Advanced Chemistry: Organic Chemistry – Basic Principles and Techniques questions with solutions

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