Basic Organic Chemistry (GOC) — Top 20 Questions
Q1
Which of the following species is the most stable carbocation?
ACH₃⁺ (methyl carbocation)
BCH₃CH₂⁺ (primary carbocation)
C(CH₃)₂CH⁺ (secondary carbocation)
D(CH₃)₃C⁺ (tertiary carbocation)
✅ Correct: D
Stability of carbocations: 3° > 2° > 1° > CH₃⁺. (CH₃)₃C⁺ (tertiary) is most stable due to hyperconjugation — 9 C−H bonds in 3 methyl groups donate electron density to empty p-orbital. More hyperconjugation = more stability.
Q2
The correct order of inductive effect (−I effect) among the following groups is:
A−NR₂ > −OR > −F
B−F > −OR > −NR₂
C−OR > −F > −NR₂
D−NR₂ > −F > −OR
✅ Correct: B
Inductive effect (−I, electron withdrawing) depends on electronegativity: F(3.98) > O > N. Order: −F > −OR > −NR₂. Higher electronegativity = stronger electron withdrawal = stronger −I effect.
Q3
In the resonance structures of benzene, which of the following statements is correct?
AThe two structures of benzene contribute unequally
BThe actual structure of benzene is a resonance hybrid of all contributing structures
CBenzene exists as a mixture of the two Kekulé structures
DThe resonance structures of benzene can be separated
✅ Correct: B
The actual structure of benzene is a resonance hybrid — not a mixture of structures but a single structure that is a blend of all contributing resonance forms. All C−C bond lengths in benzene are equal (139 pm), intermediate between single (154 pm) and double (134 pm) bonds.
Q4
Which of the following is a nucleophile?
ABF₃
BAlCl₃
CH₂O
DH⁺
✅ Correct: C
Nucleophile: an electron-rich species that donates an electron pair to an electrophile. H₂O has two lone pairs on oxygen → it is a nucleophile. BF₃, AlCl₃ are Lewis acids (electrophiles). H⁺ is an electrophile (electron deficient).
Q5
The stability of carbanions follows the order:
A3° > 2° > 1° > CH₃⁻
BCH₃⁻ > 1° > 2° > 3°
C1° > 2° > 3° > CH₃⁻
D3° > CH₃⁻ > 1° > 2°
✅ Correct: B
Carbanion stability is opposite to carbocation: electron-donating groups (alkyl groups) destabilize carbanions. Methyl carbanion (CH₃⁻) is most stable and 3° is least stable. Order: CH₃⁻ > 1° > 2° > 3° (more alkyl groups = more electron density = less stable carbanion).
Q6
In which of the following reactions does an electrophilic substitution take place?
ANitration of benzene
BAddition of HBr to ethene
CElimination of HBr from ethyl bromide
DHydrolysis of ethyl chloride
✅ Correct: A
Nitration of benzene: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O. The electrophile is NO₂⁺ (nitronium ion). It substitutes H on benzene ring → electrophilic aromatic substitution (EAS). Addition to ethene is electrophilic addition; others are elimination and nucleophilic substitution.
Q7
Which of the following shows +M (positive mesomeric) effect?
A−NO₂
B−CN
C−COOH
D−NH₂
✅ Correct: D
+M effect: lone pair on substituent delocalizes into benzene ring → electron donation. Groups with lone pairs: −NH₂, −OH, −OR, −SH show +M. −NO₂, −CN, −COOH show −M effect (withdraw electrons from ring through resonance).
Q8
The hyperconjugation in propene (CH₃−CH=CH₂) involves:
AOverlap of σ(C−H) bond orbital with adjacent π bond
BOverlap of p-orbital with σ(C−C) bond
COverlap of lone pairs with π bond
DNo-bond resonance involving only carbon atoms
✅ Correct: A
Hyperconjugation: overlap of σ(C−H) bond orbital (of the methyl group) with the adjacent π bond (C=C). This delocalizes electron density from C−H into the π system, stabilizing the molecule. Also called ‘no-bond resonance’ — one C−H bond contributes to π system.
Q9
Which of the following is the most acidic compound?
AEthanol (C₂H₅OH)
BPhenol (C₆H₅OH)
CAcetic acid (CH₃COOH)
DCarbonic acid (H₂CO₃)
✅ Correct: C
Acidity order: CH₃COOH (pKa ≈ 4.75) > H₂CO₃ (pKa₁ ≈ 6.35) > C₆H₅OH (pKa ≈ 10) > C₂H₅OH (pKa ≈ 16). Acetic acid is most acidic — the carboxylate ion is stabilized by resonance.
Q10
Among the following, which is the strongest base?
AAniline (C₆H₅NH₂)
BMethylamine (CH₃NH₂)
CAmmonia (NH₃)
DDimethylamine ((CH₃)₂NH)
✅ Correct: D
Basicity of amines (in gas phase and in water, aliphatic > aromatic): (CH₃)₂NH > CH₃NH₂ > NH₃ >> C₆H₅NH₂. Dimethylamine is the strongest base — 2 electron-donating methyl groups increase electron density on N. In water, due to solvation, CH₃NH₂ ≈ (CH₃)₂NH but (CH₃)₂NH is generally stronger.
Q11
Homolytic fission of a covalent bond produces:
ATwo ions with opposite charges
BFree radicals with unpaired electrons
CA carbocation and hydride ion
DTwo nucleophiles
✅ Correct: B
Homolytic fission: each atom gets one electron from the shared pair → produces free radicals (neutral species with unpaired electrons). Heterolytic fission gives ions. Homolytic cleavage is initiated by UV light or heat (radical reactions).
Q12
The IUPAC name of CH₃CH(OH)CH₂COOH is:
A3-hydroxybutanoic acid
B2-methylhydroxyacetic acid
C3-hydroxy butanoic acid
Dβ-hydroxybutyric acid
✅ Correct: A
Chain: 4 carbons. COOH at C1, OH at C3. IUPAC: 3-hydroxybutanoic acid. Count from the COOH end: C1(COOH), C2(CH₂), C3(CH(OH)), C4(CH₃).
Q13
In which of the following is the +I (positive inductive) effect shown?
A−F
B−Cl
C−NO₂
D−CH₃ (alkyl group)
✅ Correct: D
+I effect: electron-donating through sigma bonds toward the reaction center. Alkyl groups (−CH₃, −C₂H₅, etc.) show +I effect — they push electron density toward the attached atom. All halogens, −NO₂, −CN show −I effect.
Q14
The number of σ bonds and π bonds in acetylene (C₂H₂) are respectively:
A2σ, 2π
B3σ, 2π
C2σ, 3π
D4σ, 2π
✅ Correct: B
Acetylene H−C≡C−H: H−C (1σ), C≡C (1σ + 2π), C−H (1σ). Total: 3σ bonds and 2π bonds. Each carbon is sp hybridized with 2 sp orbitals (for σ bonds) and 2 unhybridized p orbitals (for 2 π bonds in triple bond).
Q15
Which of the following correctly describes an SN2 reaction?
AUnimolecular, proceeds via carbocation intermediate, rate depends only on substrate
BBimolecular, concerted mechanism, inversion of configuration, rate depends on both substrate and nucleophile
CBimolecular, proceeds via carbocation, retention of configuration
DUnimolecular, proceeds via carbanion, inversion of configuration
✅ Correct: B
SN2 (Substitution Nucleophilic Bimolecular): concerted (one-step), backside attack by nucleophile, inversion of configuration (Walden inversion), bimolecular rate = k[substrate][nucleophile]. Favored by primary substrates and strong nucleophiles. No carbocation intermediate.
Q16
What is the IUPAC name of the compound: CH₃−CH=CH−CHO?
ABut-2-enal
BBut-3-enal
C2-butenal
DCrotonaldehyde (common name)
✅ Correct: A
CHO is at C1 (functional group suffix -al). C1=CHO, C2=CH, C3=CH, C4=CH₃. Double bond at C2-C3. IUPAC: but-2-enal. (The common name is crotonaldehyde but IUPAC is but-2-enal.)
Q17
Fries rearrangement involves conversion of:
APhenol ester to hydroxyaryl ketone
BNitrobenzene to aniline
CBenzene to phenol
DPhenol to benzene
✅ Correct: A
Fries rearrangement: phenol ester (ArOCOR) to hydroxyaryl ketone (ArOH with RCOA group at ortho or para position). In presence of Lewis acid (AlCl₃), acyl group migrates from O to the ring. Gives o- and p-hydroxyaryl ketones.
Q18
The term ‘aromaticity’ requires which of the following conditions (Hückel’s rule)?
APlanar, cyclic, conjugated, (4n+2) π electrons
BPlanar, cyclic, (4n) π electrons
CNon-planar, conjugated, (4n+2) π electrons
DCyclic only, any number of π electrons
✅ Correct: A
Hückel’s rule for aromaticity: (1) planar, (2) cyclic, (3) fully conjugated (alternating single and double bonds or lone pairs), (4) (4n+2) π electrons (n = 0,1,2…). Example: benzene has 6π electrons (n=1, 4×1+2=6). Cyclobutadiene (4π) is antiaromatic.
Q19
In the following reaction: CH₃Br + NaOH(aq) → CH₃OH + NaBr, the type of reaction is:
AElectrophilic substitution
BNucleophilic substitution (SN2)
CElimination
DRadical substitution
✅ Correct: B
CH₃Br + OH⁻ → CH₃OH + Br⁻. OH⁻ is a nucleophile attacking the carbon bearing Br (a leaving group). Since CH₃Br is primary (methyl), reaction is SN2 (bimolecular nucleophilic substitution). Backside attack, inversion (not applicable for CH₃ as it has no chiral center).
Q20
Which of the following compounds shows geometrical (cis-trans) isomerism?
ACH₂=CH₂ (ethene)
BCH₃CH=CHCH₃ (but-2-ene)
CCH₂=CHCl (vinyl chloride)
DCH₂=CH₂ (ethylene)
✅ Correct: B
Geometrical isomerism requires a C=C double bond with two different groups on each carbon. But-2-ene (CH₃CH=CHCH₃): each sp² carbon has CH₃ and H → different groups on each. Cis: both CH₃ on same side; trans: opposite sides. CH₂=CH₂ has two identical H on each carbon — no geometric isomerism.