JEE Chemistry Organic Chemistry Some Basic Principles And Techniques MCQs Set 14

Answer: (b) 2-nitropropane

Explanation:
1. In free-radical nitration, the 2° hydrogen is replaced more readily than the 1° hydrogen
2. The major product is 2-nitropropane. Answer: (b)

Teacher's Note:
1. Radical stability: 3° > 2° > 1°.

Answer: (a) n-butane

Explanation:
1. n-Butane has two types of hydrogens (CH₃ and CH₂), so it gives 1-chlorobutane and 2-chlorobutane
2. 2,4-Dimethylpentane gives 3, benzene gives 1 and 2-methylbutane gives 4. Answer: (a)

Teacher's Note:
1. Count the number of different kinds of hydrogen atoms.

Answer: (b) CH₃–CH₂–CHBr–CH₃

Explanation:
1. Bromination is highly selective for the weaker secondary C–H bond
2. The chief product is 2-bromobutane. Answer: (b)

Teacher's Note:
1. Bromine radicals are very selective for the most substituted C-H.

Answer: (d) –CCl₃

Explanation:
1. –CCl₃ withdraws electrons strongly by the –I effect of three chlorines, so it deactivates the ring most
2. –OCH₃ and –CH₃ activate the ring. Answer: (d)

Teacher's Note:
1. Strong -I groups deactivate the ring.

Answer: (b) toluene

Explanation:
1. The ring is activated by hyperconjugation and the +I effect of the alkyl group
2. Toluene has three α-hydrogens, the most for hyperconjugation among the four
3. So toluene is the most reactive. Answer: (b)

Teacher's Note:
1. The number of alpha hydrogens decides the hyperconjugation effect (Baker-Nathan order).

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

Explanation:
1. Two methyl groups make p-xylene the most reactive, followed by toluene
2. The nitro group is strongly deactivating, so p-nitrotoluene is less reactive
3. p-Dinitrobenzene has two nitro groups and is the least reactive. Answer: (a)

Teacher's Note:
1. Count activating and deactivating groups on the ring.

Answer: (c) both inductive and hyperconjugative effects

Explanation:
1. The methyl group pushes electrons by the +I effect and by hyperconjugation
2. Both raise the electron density of the ring. Answer: (c)

Teacher's Note:
1. Alkyl groups activate by +I effect and hyperconjugation.

Answer: (c) electrophilic substitution

Explanation:
1. The electrophile NO₂⁺ attacks the benzene ring and replaces a hydrogen
2. This is electrophilic aromatic substitution. Answer: (c)

Teacher's Note:
1. Benzene prefers substitution to retain its aromaticity.

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

Explanation:
1. –OH is a strongly activating group, –CH₃ a weakly activating group, H is the reference and –NO₂ is strongly deactivating
2. Order: phenol > toluene > benzene > nitrobenzene. Answer: (c)

Teacher's Note:
1. Activating groups raise the rate and deactivating groups lower it.

Answer: (a) NO₂⁺

Explanation:
1. Conc. H₂SO₄ protonates HNO₃, which loses water to give the nitronium ion NO₂⁺, the electrophile. Answer: (a)

Teacher's Note:
1. The nitronium ion is generated in the nitrating mixture.

Answer: (c) to produce the electrophile

Explanation:
1. AlCl₃ is a Lewis acid that removes Cl⁻ from RCl or RCOCl
2. This generates the carbocation or acylium ion that attacks the ring. Answer: (c)

Teacher's Note:
1. A Lewis acid catalyst creates the electrophile.

Answer: (b) chlorobenzene

Explanation:
1. Chlorine is an ortho-para director in spite of being deactivating, because of its +M effect
2. –NO₂, –CCl₃ and –COOH are meta directors. Answer: (b)

Teacher's Note:
1. Halogens are deactivating but o,p-directing.

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

Explanation:
1. Allyl alcohol gives the delocalisation-stabilised allyl carbocation, so it is the most reactive
2. The 2° alcohol is more reactive than the 1° alcohol
3. Phenol does not undergo C–O cleavage with HX because the C–O bond has partial double bond character
4. Order: II > IV > III > I. Answer: (d)

Teacher's Note:
1. Reactivity of alcohols with HX depends on the stability of the carbocation or ease of SN2.

Answer: (c) t-pentyl chloride

Explanation:
1. Protonation and loss of water give the unstable primary neopentyl cation
2. A methyl shift gives the tertiary t-pentyl cation, (CH₃)₂C⁺–CH₂CH₃
3. Cl⁻ attacks it to give t-pentyl chloride. Answer: (c)

Teacher's Note:
1. Rearrangement to a more stable carbocation follows the SN1 pathway.

Answer: (c) a carbocation

Explanation:
1. SN1 is a two-step reaction in which the leaving group first departs to form a carbocation
2. This is the slow, rate-determining step. Answer: (c)

Teacher's Note:
1. SN1 means substitution nucleophilic unimolecular.

Answer: (b) AgCN

Explanation:
1. AgCN is covalent, so the nitrogen end (with the lone pair) attacks the alkyl halide
2. This gives R–N≡C (isocyanide). Answer: (b)

Teacher's Note:
1. KCN is ionic and gives alkyl cyanide (nitrile).

Answer: (c) PhCHCl(CH₃)

Explanation:
1. SN1 reactivity follows carbocation stability
2. PhCH⁺CH₃ is a secondary benzylic cation, more stable than the primary benzylic cation from PhCH₂Cl
3. The nitro group destabilises the cation in (d), and PhCl is unreactive. Answer: (c)

Teacher's Note:
1. Secondary benzylic carbocations are stabilised by the ring and the methyl group.

Answer: (b) Swarts reaction

Explanation:
1. The Swarts reaction converts alkyl chlorides or bromides to fluorides using AgF, Hg₂F₂ or SbF₃
2. Finkelstein makes iodides. Answer: (b)

Teacher's Note:
1. Swarts gives fluorides and Finkelstein gives iodides.

Answer: (d) nitroethane

Explanation:
1. AgNO₂ is covalent, so the nitrogen lone pair attacks the carbon
2. The main product is nitroethane, C₂H₅NO₂ (KNO₂ would give the nitrite ester). Answer: (d)

Teacher's Note:
1. AgNO2 gives nitroalkane and KNO2 gives alkyl nitrite.

Answer: (b) HI > HBr > HCl

Explanation:
1. Reactivity follows acid strength and nucleophilicity of X⁻
2. The order is HI > HBr > HCl. Answer: (b)

Teacher's Note:
1. The stronger the acid, the easier the protonation of the alcohol.

Answer: (b) SN2

Explanation:
1. The substrate is a methyl halide, which cannot form a stable carbocation, and OH⁻ is a strong nucleophile
2. The reaction goes through a one-step backside attack, SN2. Answer: (b)

Teacher's Note:
1. Methyl and primary halides react by SN2.

Answer: (d) propyl chloride with KCN

Explanation:
1. Butanenitrile has four carbons, of which one comes from CN⁻
2. So the alkyl halide must have three carbons: propyl chloride
3. CH₃CH₂CH₂Cl + KCN → CH₃CH₂CH₂CN + KCl. Answer: (d)

Teacher's Note:
1. Count the carbon of the cyanide group in the nitrile name.

Answer: (b) nucleophilic substitution

Explanation:
1. CN⁻ replaces Br⁻ at a saturated carbon
2. This is nucleophilic substitution. Answer: (b)

Teacher's Note:
1. The nucleophile replaces the leaving group.

Answer: (d) (C₆H₅)₃CCl

Explanation:
1. Hydrolysis goes through SN1
2. Triphenylmethyl chloride gives the trityl cation, which is stabilised by three phenyl rings
3. So it is hydrolysed most readily. Answer: (d)

Teacher's Note:
1. Greater carbocation stability gives faster SN1 hydrolysis.