The incorrect statement regarding ethyne (C₂H₂) is:
AThe carbon–carbon bond in ethyne is weaker than that in ethene
BBoth carbon atoms in ethyne are sp hybridised
CThe C–C bond length in ethyne is shorter than that in ethene
DEthyne molecule is linear
✅ Correct: A
The C≡C triple bond in ethyne is stronger (not weaker) than the C=C double bond in ethene. Bond energies: C≡C (837 kJ/mol) > C=C (614 kJ/mol) > C−C (347 kJ/mol). Bond lengths decrease: C≡C (120 pm) < C=C (134 pm) < C−C (154 pm). Statement A is incorrect.
Q2
Which of the following catalysts is used for partial reduction of alkynes to give cis-alkenes?
CSodium in liquid ammonia (Birch reduction conditions)
DCold dilute KMnO₄
✅ Correct: A
Lindlar's catalyst (Pd on CaCO₃ + quinoline, partially poisoned) selectively reduces alkynes to cis-alkenes via syn addition. Sodium in liquid ammonia gives trans-alkenes (anti addition). Cold KMnO₄ is for hydroxylation of alkenes.
Q3
The number of different bromo derivatives formed when ethane reacts with excess bromine in the presence of diffused sunlight is:
A5
B7
C9
D11
✅ Correct: C
Ethane (C₂H₆) + excess Br₂ → polybromo derivatives. Possible derivatives: C₂H₅Br, C₂H₄Br₂ (1,1- and 1,2-), C₂H₃Br₃ (1,1,1- and 1,1,2-), C₂H₂Br₄ (1,1,1,2- and 1,1,2,2-), C₂HBr₅, C₂Br₆. Total = 1+2+2+2+1+1 = 9 derivatives.
Q4
Statement I: Neopentane (2,2-dimethylpropane) forms only one monosubstituted derivative. Statement II: The melting point of neopentane is higher than that of n-pentane. Which statements are correct?
AOnly Statement I is correct
BOnly Statement II is correct
CBoth statements are correct
DBoth statements are incorrect
✅ Correct: C
Statement I: Correct — neopentane has 12 equivalent H atoms (all on 4 equivalent CH₃ groups) → only 1 monosubstituted product. Statement II: Correct — neopentane (mp −16.6°C) > n-pentane (mp −129.7°C) because the compact spherical shape allows close packing in crystal lattice. Both correct.
Q5
The main product of acid-catalysed hydration of propene (CH₃CH=CH₂) in the presence of H₂SO₄/H₂O is:
A1-propanol (CH₃CH₂CH₂OH)
B2-propanol ((CH₃)₂CHOH)
CPropanoic acid
DPropanal
✅ Correct: B
Acid-catalysed hydration follows Markovnikov's rule: H adds to carbon with more H atoms, OH adds to carbon with fewer H atoms. For propene: H⁺ adds to CH₂ (C1) → more stable secondary carbocation at C2 → OH⁻ attacks → 2-propanol (isopropanol).
Q6
Ozonolysis of 2-butene (CH₃CH=CHCH₃) followed by reductive workup with Zn/H₂O gives:
AButanal and butanal
BTwo molecules of ethanal (acetaldehyde)
COne molecule of butanone
DMethanal and propanal
✅ Correct: B
Ozonolysis cleaves the C=C double bond. 2-butene: CH₃CH=CHCH₃. Cleaving at C=C gives two CH₃CHO (ethanal/acetaldehyde) molecules. Reductive workup (Zn/H₂O) gives aldehydes. So: two molecules of ethanal (CH₃CHO).
BThe π electron cloud makes benzene ring nucleophilic, attracting electrophiles
CBenzene has low electron density due to aromaticity
DBenzene's σ bonds are very reactive
✅ Correct: B
The delocalized π electron cloud above and below the benzene ring makes it electron-rich (nucleophilic) — it attracts electrophiles. EAS preserves aromaticity (substitution rather than addition), which is thermodynamically favorable due to the resonance energy (~150 kJ/mol) of benzene.
Q8
The correct order of boiling points of the following alkanes is: methane (CH₄), ethane (C₂H₆), propane (C₃H₈), butane (C₄H₁₀):
ACH₄ > C₂H₆ > C₃H₈ > C₄H₁₀
BC₄H₁₀ > C₃H₈ > C₂H₆ > CH₄
CCH₄ > C₄H₁₀ > C₃H₈ > C₂H₆
DAll have same boiling point
✅ Correct: B
Boiling points of alkanes increase with molecular mass (more electrons → stronger London dispersion forces). Order: C₄H₁₀ (−0.5°C) > C₃H₈ (−42°C) > C₂H₆ (−89°C) > CH₄ (−161°C).
Q9
In the free radical halogenation of alkanes, the relative reactivity of C−H bonds is:
A1° > 2° > 3°
B3° > 2° > 1°
CAll C−H bonds react equally
D2° > 3° > 1°
✅ Correct: B
Free radical stability determines reactivity of C−H bond toward halogenation. More stable radical → easier to form → faster reaction. Stability: 3° > 2° > 1°. So 3° C−H bonds are most reactive, followed by 2° and 1°. Selectivity: fluorination < chlorination < bromination.
Q10
Which of the following reacts with Na metal to liberate H₂ gas?
AEthane
BEthene
CEthyne
DBenzene
✅ Correct: C
Ethyne (acetylene) reacts with Na: C₂H₂ + 2Na → 2NaC₂H + H₂↑ (sodium acetylide). Terminal alkynes (−C≡CH) are acidic (pKa ≈ 25) because sp carbon is more electronegative than sp³ or sp² — they donate H⁺ to Na. Alkanes, alkenes, and benzene do not react with Na.
Q11
The product formed when propyne (CH₃C≡CH) reacts with one mole of HBr following Markovnikov's rule is:
A1-bromopropene
B2-bromopropene (CH₃CBr=CH₂)
C1,2-dibromopropane
D2-bromopropyne
✅ Correct: B
Addition of HBr to propyne (CH₃C≡CH): H adds to terminal carbon (C3, more H), Br adds to C2 (fewer H). Product: CH₃CBr=CH₂ = 2-bromopropene (2-bromoprop-1-ene). If excess HBr: adds again → 2,2-dibromopropane (Markovnikov twice).
Q12
Birch reduction (Na/liquid NH₃/alcohol) of benzene gives:
ACyclohexane
BCyclohexadiene
C1,4-cyclohexadiene
DCyclohexene
✅ Correct: C
Birch reduction: Na metal in liquid NH₃ with alcohol reduces benzene to 1,4-cyclohexadiene (cyclohexa-2,5-diene). The electrons add to positions 1 and 4, giving a non-conjugated diene. Substituents on benzene influence which positions are reduced.
Q13
Which reagent is used for the conversion of alkene to alkane (hydrogenation)?
AH₂/Ni or Pd (Raney Ni)
BNaBH₄
CLiAlH₄
DH₂SO₄
✅ Correct: A
Catalytic hydrogenation: alkene + H₂ → alkane, with Ni (Raney Ni), Pd, or Pt as catalyst. The catalyst adsorbs H₂ and alkene → syn addition of H₂ across double bond. NaBH₄ and LiAlH₄ are used for reducing carbonyl compounds.
Q14
Dehydrohalogenation of 2-bromo-2-methylbutane with alcoholic KOH gives the major product as:
A2-methylbut-1-ene
B2-methylbut-2-ene
C3-methylbut-1-ene
D3-methylbut-2-ene
✅ Correct: B
Zaitsev's rule: elimination preferentially gives the more substituted (more stable) alkene. 2-bromo-2-methylbutane: (CH₃)₂CBrCH₂CH₃. Possible alkenes: 2-methylbut-2-ene (trisubstituted) or 2-methylbut-1-ene (disubstituted). More substituted = 2-methylbut-2-ene (major product, Zaitsev).
Q15
The reaction of benzene with Cl₂ in presence of AlCl₃ (Lewis acid catalyst) gives:
ABenzene hexachloride (BHC)
BChlorobenzene (C₆H₅Cl)
CDichlorobenzene
DCyclohexyl chloride
✅ Correct: B
Friedel-Crafts halogenation: Cl₂ + AlCl₃ → Cl⁺ (electrophile) + AlCl₄⁻. Cl⁺ attacks benzene ring → electrophilic substitution → chlorobenzene (C₆H₅Cl). Without catalyst, Cl₂ + benzene (UV light) gives benzene hexachloride (BHC) by radical addition — different reaction.
Q16
When acetylene (C₂H₂) is passed through dilute H₂SO₄ in presence of HgSO₄ catalyst at 60°C, the product formed is:
AEthanol
BAcetic acid
CAcetaldehyde (ethanal)
DVinyl alcohol
✅ Correct: C
Hydration of acetylene (Markovnikov addition of water): C₂H₂ + H₂O → CH₂=CHOH (vinyl alcohol/enol, unstable) → tautomerises → CH₃CHO (acetaldehyde/ethanal). Catalyst: HgSO₄ in dilute H₂SO₄. This is an example of keto-enol tautomerism.
Q17
The reaction of toluene with concentrated HNO₃/H₂SO₄ (mixed acid) at 60°C gives mainly:
Ao-nitrotoluene and m-nitrotoluene
Bp-nitrotoluene and m-nitrotoluene
Co-nitrotoluene and p-nitrotoluene
Dm-nitrotoluene only
✅ Correct: C
CH₃ group on benzene is an ortho-para director (+I and hyperconjugation). Nitration of toluene gives predominantly o-nitrotoluene and p-nitrotoluene (ratio approx 58:40, with ~2% meta). The methyl group activates ortho and para positions toward electrophilic attack.
Q18
Kolbe's reaction involves the reaction of sodium phenoxide with CO₂ under pressure and heat to give:
APhenol and carbonic acid
BSodium salicylate (sodium 2-hydroxybenzoate)
CBenzoic acid
DSodium carbonate and phenol
✅ Correct: B
Kolbe's reaction (Kolbe-Schmitt synthesis): sodium phenoxide + CO₂ (high pressure, 125°C) → sodium salicylate (sodium 2-hydroxybenzoate). On acidification: salicylic acid. This is used industrially to produce aspirin precursor.
Q19
Cracking of high molecular weight hydrocarbons gives:
AOnly alkanes
BAlkanes and alkenes
COnly alkenes
DAromatic hydrocarbons
✅ Correct: B
Cracking (thermal or catalytic) of high molecular weight hydrocarbons (from petroleum) gives a mixture of alkanes and alkenes of lower molecular weight. Example: C₁₆H₃₄ → C₈H₁₈ + C₈H₁₆. Alkenes formed are important feedstocks for polymer industry.
Q20
The reaction of benzene with CH₃Cl in presence of anhydrous AlCl₃ is called:
AClemmensen reduction
BFriedel-Crafts alkylation
CBirch reduction
DKolbe's reaction
✅ Correct: B
Friedel-Crafts alkylation: benzene + alkyl halide (CH₃Cl) + AlCl₃ → toluene (methylbenzene) + HCl. AlCl₃ generates carbocation CH₃⁺ (electrophile) which attacks the benzene ring. Problems: polyalkylation and rearrangement. Named after Charles Friedel and James Crafts (1877).
R
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Expert · 5 Years Experience
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