JEE Chemistry Hydrocarbons MCQs Set 32

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

Explanation:
1. The rate depends on the stability of the carbocation formed on protonation.
2. (IV) gives Ph₂C⁺-CH₂CH₃, a tertiary carbocation stabilised by two phenyl groups; (III) gives a tertiary benzylic cation; (II) gives a tertiary cation; (I) gives only a secondary cation.
3. So IV > III > II > I.

Teacher's Note:
1. Carbocation stability: diaryl tertiary > aryl tertiary > tertiary > secondary.
2. The more stable the carbocation, the faster the addition.

Answer: (c) deactivating and meta directing

Explanation:
1. The nitro group withdraws electrons by both -I and -M effects, so it deactivates the ring.
2. The sigma complex from meta attack is the least destabilised, so the group is meta directing.

Teacher's Note:
1. -M groups direct meta.
2. The ortho-para directing deactivators are the halogens.

Answer: (c) CH₃-CHCl-CH₂-NO

Explanation:
1. In NOCl the nitrosonium ion NO⁺ is the electrophile, which adds to the terminal CH₂ carbon to give the secondary carbocation CH₃CH⁺CH₂NO.
2. Chloride then adds to the secondary carbon: CH₃CHClCH₂NO (Markovnikov).

Teacher's Note:
1. The positive part of the reagent is the electrophile.
2. The negative part ends up on the more substituted carbon.

Answer: (b) 2-chloro-2-methylpropane (tert-butyl chloride)

Explanation:
1. H⁺ adds to the CH₂ carbon to give the tertiary carbocation (CH₃)₃C⁺ (Markovnikov), which is captured by Cl^-.

Teacher's Note:
1. Markovnikov: H goes to the carbon with more hydrogens.
2. Tertiary cation is the most stable.

Answer: (b) 2,2-dichloropentane

Explanation:
1. The first HCl adds by Markovnikov addition to give 2-chloropent-1-ene (H on the terminal carbon, Cl on C-2).
2. The second HCl again adds with Cl going to C-2, giving the gem-dihalide CH₃CCl₂CH₂CH₂CH₃.

Teacher's Note:
1. Two moles of HX with a terminal alkyne give a gem-dihalide at the internal carbon.
2. The chlorine already present on C-2 directs the second Cl to C-2.

Answer: (a) 5

Explanation:
1. Substitution at C-1, C-2, C-3, C-4 and C-5 gives 1,2-, 2,2-, 2,3-, 2,4- and 2,5-dibromopentane, with the original stereocentre at C-2 retained.
2. Chiral products: 1,2 (1), 2,3 (2 diastereomers), 2,4 (the (S,S) isomer only; the (S,R) form is meso) (1), 2,5 (1). The 2,2-dibromide is achiral.
3. Total = 1 + 2 + 1 + 1 = 5 (RDKit check).

Teacher's Note:
1. Use the pure (S) configuration at C-2 as the fixed starting point.
2. A meso compound is achiral.

Answer: (d) The major product, 2-bromobutane, forms through the more stable secondary radical

Explanation:
1. Bromine radicals are selective: they abstract the secondary hydrogen, giving the more stable secondary radical, so 2-bromobutane is the major product (about 98 percent).
2. The reaction is a radical chain reaction (no carbocation), and bromine is less reactive than chlorine, so the other statements are false.

Teacher's Note:
1. Selectivity: Br· > Cl·, reactivity: Cl· > Br·.
2. Radical stability controls the site of halogenation.

Answer: (c) Isobutane, n-butane

Explanation:
1. Isobutane has a tertiary C-H bond, which gives a tertiary radical that is more stable than the secondary or primary radicals from n-butane.
2. Ease of bromination follows the stability of the radical: isobutane > n-butane. Ethane is easier than methane too.

Teacher's Note:
1. Reactivity of C-H: 3 degree > 2 degree > 1 degree > CH₄.
2. Compare the most reactive C-H in each molecule.

Answer: (d) the +I effect and hyperconjugation of the methyl group

Explanation:
1. The methyl group releases electrons by the inductive effect and by hyperconjugation (C-H sigma bonds overlapping with the ring pi system).
2. This stabilises the sigma complexes formed by ortho or para attack, where the positive charge sits on the carbon bearing the methyl group.

Teacher's Note:
1. Methyl has no lone pair, so it has no +M effect.
2. Alkyl groups direct ortho-para and activate.

Answer: (d) 1,2-dibromo-1-phenylpropane (anti addition)

Explanation:
1. Bromine adds across the C=C bond through a bromonium ion; bromide attacks from the opposite face (anti addition).
2. The product has Br on each of the two former alkene carbons: C₆H₅CHBr-CHBrCH₃.

Teacher's Note:
1. Halogen addition to alkenes gives vicinal dihalides.
2. In CCl₄ there is no radical or allylic substitution.

Answer: (c) all three give the same major product, 2-methylbutan-2-ol

Explanation:
1. Each alkene is protonated to a carbocation: x and y give the tertiary 2-methylbutan-2-yl cation directly, and z gives a secondary cation that undergoes a 1,2-hydride shift to the same tertiary cation.
2. Water captures this cation, so all three give 2-methylbutan-2-ol.

Teacher's Note:
1. Rearrangement to the most stable cation makes different alkenes converge to one product.
2. Both HCl and H₃O⁺ addition follow Markovnikov regioselectivity.

Answer: (a) CH₃-CHBr-CH₂-C≡CH

Explanation:
1. The double bond is more reactive than the triple bond towards electrophiles, because protonation of an alkene gives an ordinary secondary carbocation whereas protonation of an alkyne gives a less stable vinyl cation.
2. Markovnikov addition of HBr to the C=C gives CH₃CHBrCH₂C≡CH.

Teacher's Note:
1. Alkene reacts faster than alkyne in electrophilic addition.
2. Markovnikov rule applies to the alkene end.

Answer: (a) 2-bromo-2-methylbutane

Explanation:
1. The tertiary C-H (at C-2) gives the most stable (tertiary) radical, and bromine is very selective for it.

Teacher's Note:
1. Bromine selectively replaces tertiary hydrogen.
2. Count radicals: tertiary > secondary > primary.

Answer: (c) substitution keeps the aromatic stabilisation (resonance energy) of the ring, while addition would destroy it

Explanation:
1. Benzene has about 150 kJ per mole of extra aromatic stability. Addition would convert the ring into a non-aromatic product and lose this stability, so it is strongly disfavoured.
2. In substitution a proton is lost from the sigma complex and the aromatic ring is regenerated.

Teacher's Note:
1. Aromatic stabilisation drives the choice between substitution and addition.
2. The pi electrons of benzene are available but are in a stable closed cloud.

Answer: (b) (A) and (C) only

Explanation:
1. (A) correct: NO₂ is meta directing. (C) correct: the bulky tert-butyl group goes para to the ethyl group (ortho is hindered).
2. (B) is wrong: toluene gives a mixture of ortho and para (not only ortho). (D) is wrong: benzaldehyde is nitrated at the meta position to give 3-nitrobenzaldehyde.

Teacher's Note:
1. Bulky electrophiles prefer the para position.
2. CHO is meta directing.

Answer: (a) 1-bromo-2-methylpropane

Explanation:
1. In the presence of peroxide Br· adds to the CH₂ carbon to give the tertiary radical (CH₃)₂C·-CH₂Br.
2. This takes H from HBr, giving (CH₃)₂CH-CH₂Br (anti-Markovnikov).

Teacher's Note:
1. Peroxide effect puts Br on the less substituted carbon.
2. The radical formed is the more stable tertiary one.

Answer: (d) +M effect of chlorine

Explanation:
1. Chlorine has lone pairs that are donated into the ring (+M), which puts negative charge on the ortho and para carbons and stabilises the sigma complex from ortho or para attack.
2. The -I effect decides the rate (deactivation) and the +M effect decides the orientation.

Teacher's Note:
1. Chlorine: -I wins for reactivity, +M wins for orientation.
2. Resonance donation is felt only at ortho and para positions.

Answer: (a) 2-bromo-5-hydroxybenzaldehyde (Br para to OH)

Explanation:
1. OH is a strong activator and directs to its ortho and para positions; CHO directs meta to itself.
2. The position para to OH (which is also ortho to CHO) is both activated by OH and not crowded, so 2-bromo-5-hydroxybenzaldehyde is the major product; the position meta to OH is not activated.

Teacher's Note:
1. The strongest activating group controls orientation.
2. Para to the stronger donor is usually preferred to the hindered position between two groups.

Answer: (a) iodobenzene

Explanation:
1. In ICl chlorine is more electronegative, so iodine carries the partial positive charge: I^δ+-Cl^δ-.
2. AlCl₃ removes Cl^- to give I⁺, the electrophile, which attacks benzene to give C₆H₅I.

Teacher's Note:
1. The less electronegative atom of an interhalogen is the electrophile.
2. Only one substituent is introduced.

Answer: (d) 2-(bromomethyl)-2-methyltetrahydropyran, formed by intramolecular attack of OH on the bromonium ion

Explanation:
1. Br₂ forms a bromonium ion on the double bond.
2. The OH group four carbons away attacks the more substituted (tertiary) carbon, closing a six-membered ring containing oxygen (a better nucleophile than the bromide present in low concentration).
3. The product is 2-(bromomethyl)-2-methyloxane.

Teacher's Note:
1. Intramolecular nucleophiles give cyclic ethers (halo-etherification) when a 5- or 6-membered ring can form.
2. The nucleophile attacks the more substituted carbon of the bromonium ion.

Answer: (d) (CH₃)₂C(Cl)-CH₂I

Explanation:
1. In ICl iodine is the electrophilic end (I^δ+), so I⁺ adds first to the CH₂ carbon to give the tertiary cation.
2. Chloride then attacks the tertiary carbon, giving (CH₃)₂CCl-CH₂I.

Teacher's Note:
1. Electrophile is the less electronegative atom.
2. Nucleophile goes to the more substituted carbon (Markovnikov).

Answer: (a) the gamma hydrogen (on the CH₂ next to the double bond, the allylic carbon)

Explanation:
1. Under radical conditions the most easily abstracted hydrogen is the allylic one, since it gives a resonance-stabilised allyl radical.
2. In but-1-ene the allylic hydrogens are on the CH₂ next to the C=C, which is the gamma carbon in the labelling given.

Teacher's Note:
1. Vinylic hydrogens are the hardest to abstract.
2. Allylic and benzylic hydrogens are the easiest.

Answer: (b) I > II > III

Explanation:
1. The more rings are fused, the smaller the loss of aromatic stabilisation when the electrophile attacks, so the sigma complex is more stable (the remaining rings keep their aromaticity).
2. Reactivity: anthracene > naphthalene > benzene.

Teacher's Note:
1. Fused rings are more reactive than benzene.
2. Resonance energy lost per ring is smaller in anthracene than in benzene.

Answer: (a) a chlorine radical

Explanation:
1. Initiation is homolysis of the Cl-Cl bond by light or heat: Cl₂ gives 2Cl·.

Teacher's Note:
1. Initiation makes radicals from molecules.
2. Propagation consumes and regenerates them.

Answer: (a) DDT: 1,1,1-trichloro-2,2-bis(4-chlorophenyl)ethane

Explanation:
1. Protonated chloral is an electrophile and attacks the para position of chlorobenzene to give an alcohol.
2. The alcohol is protonated, loses water to give a carbocation, and a second chlorobenzene attacks para, giving CCl₃CH(C₆H₄Cl)₂ (DDT).

Teacher's Note:
1. Two successive electrophilic substitutions occur.
2. Chlorobenzene reacts at the para position.

Answer: (b) II > III > I

Explanation:
1. The phenoxide ion (-O^-) is a very strongly activating group, the most reactive.
2. Fluorine is only weakly deactivating (+M partly offsets -I).
3. The trimethylammonium group carries a full positive charge and strongly deactivates the ring.

Teacher's Note:
1. Rank groups: -O^- (very strong activator) > H > -F > -NR₃⁺.
2. Halogens deactivate only moderately.

Answer: (c) the carbocations they would form (vinyl and phenyl cations) are very unstable, and the C-Cl bond has partial double-bond character

Explanation:
1. Friedel-Crafts alkylation needs an alkyl halide that can form a carbocation (or a polarised complex) with AlCl₃.
2. In vinyl and aryl halides the chlorine is on an sp² carbon; the C-Cl bond is shorter and stronger (lone-pair conjugation), and the vinyl or phenyl cation is extremely unstable.

Teacher's Note:
1. Aryl and vinyl halides are inert to SN1-type ionisation.
2. Carbocation stability decides which halides work.

Answer: (d) chain termination step: two methyl radicals combine

Explanation:
1. Two CH₃· radicals can couple to form a C-C bond: CH₃· + CH₃· gives CH₃CH₃.
2. Radical-radical combination is a termination step because it consumes radicals and ends the chain.

Teacher's Note:
1. Termination steps remove radicals.
2. Traces of ethane and chloroethane are evidence for the radical mechanism.

Answer: (c) para to the CH₂ group on the unsubstituted phenyl ring

Explanation:
1. The ring that already carries NO₂ is strongly deactivated; the other ring carries only an alkyl group (CH₂Ar), which is activating.
2. So nitration takes place on the unsubstituted ring, para (less hindered) to the CH₂ group.

Teacher's Note:
1. Attack occurs on the more reactive ring.
2. Alkyl groups direct ortho-para.

Answer: (a) 4

Explanation:
1. Electron-releasing groups activate: -OCOCH₃ (weakly, through oxygen lone pair), -C₆H₅, -CH₃ and -NH₂.
2. Electron-withdrawing groups deactivate: -Br (-I outweighs +M for the rate), -NO₂, -CHF₂ and -CCl₃ (strong -I effect of the halogens).
3. That is four deactivating groups.

Teacher's Note:
1. Halogens deactivate but direct ortho-para.
2. Groups carrying several electronegative atoms on the first carbon are electron-withdrawing.

Answer: (c) CH₃-CH(D)-C·(CH₃)₂

Explanation:
1. Br· abstracts the weakest and most reactive bond. The C-D bond is stronger than C-H (lower zero-point energy), so it is not broken.
2. The tertiary C-H hydrogen gives the most stable (tertiary) radical, so the radical is at the carbon bearing the two CH₃ groups; deuterium stays on the adjacent carbon.

Teacher's Note:
1. Kinetic isotope effect: C-H breaks faster than C-D.
2. Tertiary radical is the most stable.

Answer: (d) a racemic pair of enantiomers (2R,3R and 2S,3S) from the cis alkene and the meso compound from the trans alkene

Explanation:
1. Bromine adds anti through a bromonium ion (stereospecific).
2. Anti addition to the cis alkene gives the chiral (2R,3R)/(2S,3S) pair formed in equal amounts (racemate).
3. Anti addition to the trans alkene gives (2R,3S), which is meso.

Teacher's Note:
1. Cis alkene + anti addition gives racemate; trans alkene + anti addition gives meso.
2. Meso compounds have a plane of symmetry.