JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 40

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Chapter-wise Objective Questions: Organic Chemistry Some Basic Principles and Techniques

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Multiple Choice Questions

Question 1: Which compound shows a dipole moment?
(a) 1,4-Dichlorobenzene
(b) 1,2-Dichlorobenzene
(c) trans-1,2-Dichloroethene
(d) trans-2-Butene
Show Answer & Explanation

Answer: (b) 1,2-Dichlorobenzene

Explanation:
1. In 1,2-dichlorobenzene the two C-Cl moments are at 60 degrees and add up to a resultant
2. The other three are symmetrical with cancelling moments

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Question 2: Which of the following is a polar compound?
(a) \( \ce{C2H6} \)
(b) \( \ce{CCl4} \)
(c) HCl
(d) \( \ce{CH4} \)
Show Answer & Explanation

Answer: (c) HCl

Explanation:
1. HCl has a large electronegativity difference and a linear shape, so the bond dipole is the molecular dipole
2. The others are symmetrical and non-polar

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Question 3: Resonance structures of a molecule do NOT have
(a) Identical arrangement of atoms
(b) Nearly the same energy content
(c) The same number of paired electrons
(d) Identical bonding
Show Answer & Explanation

Answer: (d) Identical bonding

Explanation:
1. Resonance structures differ only in the distribution of electrons, so the bonding pattern is not identical
2. They must have the same atomic arrangement, similar energy and the same number of paired electrons

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Question 4: All the bonds in benzene are equal due to
(a) Tautomerism
(b) Inductive effect
(c) Resonance
(d) Isomerism
Show Answer & Explanation

Answer: (c) Resonance

Explanation:
1. The pi electrons are delocalised over all six carbons
2. Every C-C bond has the same intermediate length (1.39 angstrom)

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Question 5: Aromatic properties of benzene are explained by
(a) Aromatic sextet theory
(b) Resonance theory
(c) Molecular orbital theory
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. The pi sextet, resonance between Kekule structures and molecular orbital delocalisation all account for benzene's stability

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Question 6: Which one of the following is correct regarding the inductive effect of the substituents?
(a) -NR2 < -OR > -F
(b) -NR2 > -OR > -F
(c) -NR2 < -OR < -F
(d) -NR2 > -OR < -F
Show Answer & Explanation

Answer: (c) -NR2 < -OR < -F

Explanation:
1. The -I effect increases with the electronegativity of the atom attached to carbon
2. N (3.0) < O (3.5) < F (4.0), so -NR2 < -OR < -F

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Question 7: Benzene is unreactive towards addition because
(a) It has double bonds
(b) It has carbon-carbon single bonds
(c) Carbon atoms are sp2 hybridised
(d) The pi electrons are delocalised
Show Answer & Explanation

Answer: (d) The pi electrons are delocalised

Explanation:
1. Delocalisation gives benzene extra resonance stability
2. Addition would destroy this aromatic stabilisation, so benzene prefers substitution

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Question 8: The C-C bond length in benzene lies between the single and the double bond lengths. The reason is
(a) Resonance
(b) Isomerism
(c) Metamerism
(d) Inductive effect
Show Answer & Explanation

Answer: (a) Resonance

Explanation:
1. Resonance between the Kekule structures gives every C-C bond partial double-bond character
2. The length is therefore intermediate

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Question 9: Credit for the ring structure of benzene goes to
(a) Wohler
(b) Faraday
(c) Kekule
(d) Baeyer
Show Answer & Explanation

Answer: (c) Kekule

Explanation:
1. Kekule proposed the cyclic structure with alternating double bonds in 1865
2. Faraday isolated benzene, and Wohler is known for the synthesis of urea

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Question 10: The polarisation of electrons in acrolein may be written as
(a) \( \ce{CH2^{δ-}=CH-CH^{δ+}=O} \)
(b) \( \ce{CH2^{δ-}=CH-CH=O^{δ+} \)}
(c) \( \ce{CH2^{δ+}-CH=CH-O^{δ-} \)}
(d) \( \ce{CH2^{δ+}=CH-CH=O^{δ-} \)}
Show Answer & Explanation

Answer: (d) \( \ce{CH2^{δ+}=CH-CH=O^{δ-} \)}

Explanation:
1. Oxygen pulls the electrons of C=O toward itself and the conjugated C=C is polarised in the same direction
2. The terminal carbon becomes electron-poor (delta plus) and oxygen is delta minus

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Question 11: In the mixture of conc. \( \ce{H2SO4} \) and \( \ce{HNO3} \), the nitrating species is
(a) \( \ce{N2O4} \)
(b) \( \ce{NO2^+} \)
(c) \( \ce{NO2} \)
(d) \( \ce{NO2^-} \)
Show Answer & Explanation

Answer: (b) \( \ce{NO2^+} \)

Explanation:
1. \( \ce{HNO3} \) is protonated by the stronger acid \( \ce{H2SO4} \) and loses water
2. This gives the nitronium ion \( \ce{NO2^+} \), the electrophile

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Question 12: The increasing order of the +I effect of the alkyl groups is
(a) \( \ce{(CH3)3C-} \) < \( \ce{(CH3)2CH-} \) < \( \ce{CH3CH2-} \)
(b) \( \ce{CH3CH2-} \) < \( \ce{(CH3)2CH-} \) < \( \ce{(CH3)3C-} \)
(c) \( \ce{(CH3)2CH-} \) < \( \ce{(CH3)3C-} \) < \( \ce{CH3CH2-} \)
(d) \( \ce{(CH3)3C-} \) < \( \ce{CH3CH2-} \) < \( \ce{(CH3)2CH-} \)
Show Answer & Explanation

Answer: (b) \( \ce{CH3CH2-} \) < \( \ce{(CH3)2CH-} \) < \( \ce{(CH3)3C-} \)

Explanation:
1. The +I effect grows with the number of alkyl groups on the carbon that is attached
2. The order is ethyl < isopropyl < tert-butyl

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Question 13: Which of the following effects is observed in an ethylene molecule?
(a) Electromeric effect
(b) Inductive effect
(c) Homolytic fission
(d) None of these
Show Answer & Explanation

Answer: (a) Electromeric effect

Explanation:
1. The electromeric effect is the complete transfer of pi electrons to one atom in the presence of an attacking reagent
2. It operates in a multiple bond like C=C, and ethylene has no permanent polarity for the inductive effect

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Question 14: The cyclopentadienyl anion is
(a) Aromatic
(b) Non-aromatic
(c) Non-planar
(d) Aliphatic
Show Answer & Explanation

Answer: (a) Aromatic

Explanation:
1. The ring is planar and fully conjugated with 6 pi electrons (4 from the double bonds and 2 from the carbanion lone pair)
2. This obeys the 4n + 2 rule

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Question 15: The orbital interaction between the sigma bonds of a substituent group and a neighbouring pi orbital is known as
(a) Hyperconjugation
(b) Inductive effect
(c) Steric effect
(d) Dipole-dipole interaction
Show Answer & Explanation

Answer: (a) Hyperconjugation

Explanation:
1. Overlap of a C-H (or C-C) sigma bond with an adjacent empty p or pi orbital delocalises electrons
2. This is hyperconjugation (no-bond resonance)

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Question 16: The reactivity towards nucleophilic addition of (I) HCHO, (II) \( \ce{CH3CHO} \) and (III) \( \ce{CH3COCH3} \) is
(a) II > III > I
(b) III > II > I
(c) I > II > III
(d) I > II < III
Show Answer & Explanation

Answer: (c) I > II > III

Explanation:
1. Alkyl groups donate electrons and crowd the carbonyl carbon
2. Fewer alkyl groups mean more reactivity, so HCHO > \( \ce{CH3CHO} \) > \( \ce{CH3COCH3} \)

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Question 17: Which of the following is the most stable carbocation?
(a) \( \ce{CH3-C^+H-CH3} \)
(b) \( \ce{(CH3)3C^+} \)
(c) \( \ce{CH3^+} \)
(d) \( \ce{CH3-CH2-C^+H2} \)
Show Answer & Explanation

Answer: (b) \( \ce{(CH3)3C^+} \)

Explanation:
1. The tert-butyl cation has nine C-H bonds for hyperconjugation and three +I alkyl groups
2. Its stability order is 3 degrees > 2 degrees > 1 degree > methyl

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Question 18: Which is the decreasing order of stability of the carbocations (i) \( \ce{CH3-C^+H-CH3} \), (ii) \( \ce{CH3-C^+H-O-CH3} \) and (iii) \( \ce{CH3-C^+H-CO-CH3} \)?
(a) (i) < (ii) < (iii)
(b) (i) > (ii) > (iii)
(c) (iii) > (ii) > (i)
(d) (ii) > (i) > (iii)
Show Answer & Explanation

Answer: (d) (ii) > (i) > (iii)

Explanation:
1. In (ii) the oxygen lone pair stabilises the cation by +M donation
2. In (iii) the electron-withdrawing C=O group destabilises it, so (ii) > (i) > (iii)

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Question 19: The order of decreasing stability of the carbanions (1) \( \ce{(CH3)3C^-} \), (2) \( \ce{(CH3)2CH^-} \), (3) \( \ce{CH3CH2^-} \) and (4) \( \ce{C6H5CH2^-} \) is
(a) 1 > 2 > 3 > 4
(b) 4 > 3 > 2 > 1
(c) 4 > 1 > 2 > 3
(d) 1 > 2 > 4 > 3
Show Answer & Explanation

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

Explanation:
1. The benzyl carbanion is resonance stabilised and is the most stable
2. Among alkyl carbanions, +I groups destabilise, so primary > secondary > tertiary

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Question 20: Choose the chain-terminating step from the following.
(a) \( \ce{H2 -> H. + H.} \)
(b) \( \ce{Br. + HBr -> H. + Br2} \)
(c) \( \ce{H. + Br2 -> HBr + Br.} \)
(d) \( \ce{Br. + Br. -> Br2} \)
Show Answer & Explanation

Answer: (d) \( \ce{Br. + Br. -> Br2} \)

Explanation:
1. A termination step removes radicals without forming new ones
2. Two bromine atoms combining to \( \ce{Br2} \) does this, while the other steps produce radicals

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Question 21: Which of the following reactions requires a radical intermediate?
(a) \( \ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3} \)
(b) \( \ce{CH3-CHO + HCN -> CH3-CH(OH)CN} \)
(c) \( \ce{CH3-CH=CH2 + HBr -> CH3-CH2-CH2Br} \) (in presence of peroxide)
(d) \( \ce{CH3CHO + NH2OH -> CH3-CH=N-OH} \) (acid catalysed)
Show Answer & Explanation

Answer: (c) \( \ce{CH3-CH=CH2 + HBr -> CH3-CH2-CH2Br} \) (in presence of peroxide)

Explanation:
1. In the presence of peroxide, HBr adds by a free radical chain (anti-Markovnikov) mechanism
2. The other three reactions are ionic (electrophilic or nucleophilic addition)

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Question 22: Which of the following species is paramagnetic in nature?
(a) Free radical
(b) Carbocation
(c) Carbanion
(d) All of the above
Show Answer & Explanation

Answer: (a) Free radical

Explanation:
1. A free radical has an unpaired electron, which makes it paramagnetic
2. Carbocations and carbanions have all electrons paired

Teacher's Note:
1. Decide the hybridisation, charge or electronic effect first, then compare the options on that basis.
2. Remember the trend: more s-character means shorter bonds and higher acidity; more alkyl groups mean more stable carbocations and radicals.

Question 23: In which of the following species is the central carbon atom negatively charged?
(a) Carbanion
(b) Carbonium ion
(c) Carbocation
(d) Free radical
Show Answer & Explanation

Answer: (a) Carbanion

Explanation:
1. A carbanion has a carbon with three bonds and a lone pair
2. It carries a negative charge

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Question 24: Which of the following free radicals is the most stable?
(a) Primary
(b) Methyl
(c) Secondary
(d) Tertiary
Show Answer & Explanation

Answer: (d) Tertiary

Explanation:
1. Alkyl groups stabilise a radical by hyperconjugation and +I effect
2. Stability order: tertiary > secondary > primary > methyl

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

Question 25: Which of the following carbanions is the most stable?
(a) Methyl
(b) Primary
(c) Secondary
(d) Tertiary
Show Answer & Explanation

Answer: (a) Methyl

Explanation:
1. Alkyl groups push electrons towards the negative carbon and destabilise it
2. So the order is methyl > primary > secondary > tertiary

Teacher's Note:
1. Draw the species with all lone pairs and charges before answering.
2. For stability questions, ask whether the charge can be delocalised by resonance or hyperconjugation.

Question 26: Among the given cations, the most stable carbonium ion is
(a) sec-Butyl
(b) tert-Butyl
(c) n-Butyl
(d) None of these
Show Answer & Explanation

Answer: (b) tert-Butyl

Explanation:
1. The tertiary carbocation is stabilised by nine hyperconjugative C-H bonds and three +I groups
2. So tert-butyl is the most stable

Teacher's Note:
1. Link each reaction to its intermediate: carbocation (SN1, E1), planar and racemising; concerted (SN2), inversion.
2. Directing effects follow from electron donation or withdrawal by the substituent.

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