JEE Chemistry Organic Chemistry Some Basic Principles And Techniques MCQs Set 13

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

Explanation:
1. The rate depends on the stability of the carbocation formed
2. III gives a tertiary carbocation stabilised by two phenyl rings, II gives a tertiary benzylic cation, and I a secondary benzylic cation
3. The nitro group in IV destabilises the cation, so IV is least reactive
4. Order: III > II > I > IV. Answer: (b)

Teacher's Note:
1. A more stable carbocation means a faster electrophilic addition.
2. Electron-withdrawing groups such as -NO2 reduce reactivity.

Answer: (b) CH₃–CH⁺–CH₃

Explanation:
1. The peroxide effect operates for HBr only, not for HCl or HI
2. So HCl adds by the ordinary ionic (Markovnikov) mechanism
3. H⁺ adds to the terminal carbon giving the secondary carbocation CH₃–CH⁺–CH₃. Answer: (b)

Teacher's Note:
1. Peroxide effect is shown only by HBr.
2. The ionic pathway goes through the more stable carbocation.

Answer: (c) isobutyl alcohol

Explanation:
1. Hydroboration-oxidation gives anti-Markovnikov hydration
2. Boron joins the less substituted carbon of (CH₃)₂C=CH₂
3. The product is (CH₃)₂CH–CH₂OH, isobutyl alcohol (2-methylpropan-1-ol). Answer: (c)

Teacher's Note:
1. Hydroboration-oxidation places the OH on the less substituted carbon.
2. The product is a primary alcohol here.

Answer: (b) 2

Explanation:
1. Bromine adds across the double bond giving Ph–CHBr–CHBr–COOH
2. Both carbons carrying Br have four different groups
3. There are 2 chiral carbons. Answer: (b)

Teacher's Note:
1. A carbon with four different groups is chiral.

Answer: (c) A and B show chain isomerism

Explanation:
1. A is propan-2-ol and B is propan-1-ol
2. Both are alcohols and differ in the position of the OH group, so they are position isomers, not chain isomers
3. Methyl ethyl ether (C₃H₈O) is a functional isomer of both. The wrong statement is (c). Answer: (c)

Teacher's Note:
1. Same carbon skeleton with a different position of the group means position isomerism.

Answer: (d) 2-Methylpropene

Explanation:
1. Acid-catalysed hydration goes through a carbocation
2. 2-Methylpropene gives a tertiary carbocation, the most stable
3. So it reacts fastest. Answer: (d)

Teacher's Note:
1. More stable carbocation means faster hydration.

Answer: (c) Ph–CH(OH)–CH₂–CH₃

Explanation:
1. Oxymercuration-demercuration follows Markovnikov addition of water without rearrangement
2. The OH attaches to the carbon that better stabilises positive charge, the benzylic carbon
3. Product: Ph–CH(OH)–CH₂–CH₃. Answer: (c)

Teacher's Note:
1. Oxymercuration gives Markovnikov alcohols.
2. The benzylic carbon is the more stabilised site.

Answer: (a) BH₃/H₂O₂/OH⁻

Explanation:
1. The OH ends up on the less substituted alkene carbon, which is anti-Markovnikov hydration
2. This is done by hydroboration-oxidation
3. The other methods put OH on the more substituted (tertiary) carbon. Answer: (a)

Teacher's Note:
1. Anti-Markovnikov hydration needs BH3 followed by H2O2/OH-.

Answer: (b) Isobutylene

Explanation:
1. Isobutylene, (CH₃)₂C=CH₂, gives the tertiary carbocation (CH₃)₃C⁺
2. Water adds to give (CH₃)₃C–OH, a tertiary alcohol. Answer: (b)

Teacher's Note:
1. Only an alkene that can give a tertiary carbocation gives a tertiary alcohol.

Answer: (d) CH₃CH₂–C(CH₃)(OH)–CHCl–CH₃

Explanation:
1. HOCl adds as Cl⁺ and OH⁻
2. The positive charge develops on the more substituted (tertiary) carbon, so OH attaches there and Cl goes to the other carbon
3. Product: CH₃CH₂–C(CH₃)(OH)–CHCl–CH₃. Answer: (d)

Teacher's Note:
1. In halohydrin formation, OH goes to the more substituted carbon.

Answer: (a) CH₃–CHI–CH₃

Explanation:
1. Markovnikov's rule: H goes to the carbon with more hydrogens
2. The product is 2-iodopropane, CH₃–CHI–CH₃. Answer: (a)

Teacher's Note:
1. H adds to the carbon already holding more hydrogens.

Answer: (d) CH₃–CCl₂–CH₃

Explanation:
1. The first HCl gives CH₃–CCl=CH₂ (Markovnikov)
2. The second HCl adds to the same carbon, giving the gem-dihalide
3. Product: 2,2-dichloropropane, CH₃–CCl₂–CH₃. Answer: (d)

Teacher's Note:
1. With excess HX, an alkyne gives a geminal dihalide.

Answer: (c) B₂H₆ followed by H₂O₂

Explanation:
1. Propan-1-ol needs anti-Markovnikov hydration of propene
2. Hydroboration-oxidation (B₂H₆ then H₂O₂/OH⁻) does it
3. The other methods give propan-2-ol. Answer: (c)

Teacher's Note:
1. Hydroboration-oxidation gives the primary alcohol.

Answer: (b) C₆H₅CH(Br)CH₂CH₃

Explanation:
1. H⁺ adds to the terminal carbon giving the secondary carbocation C₆H₅CH₂–CH⁺–CH₃
2. A 1,2-hydride shift gives the more stable benzylic carbocation C₆H₅–CH⁺–CH₂CH₃
3. Br⁻ attacks it: C₆H₅CH(Br)CH₂CH₃. Answer: (b)

Teacher's Note:
1. Carbocations rearrange to become more stable.
2. Benzylic cations are stabilised by the ring.

Answer: (a) CH₃CH=N–NH–CO–NH₂

Explanation:
1. The NH₂ of semicarbazide, H₂N–NH–CO–NH₂, attacks the carbonyl carbon and water is lost
2. The product is the semicarbazone CH₃CH=N–NH–CO–NH₂. Answer: (a)

Teacher's Note:
1. Only the terminal NH2 attached to the other NH is nucleophilic.

Answer: (b) CH₃CHO

Explanation:
1. HCHO gives HOCH₂COOH (no chiral carbon) and acetone gives (CH₃)₂C(OH)COOH (no chiral carbon)
2. Acetaldehyde gives CH₃CH(OH)COOH, lactic acid, which has a chiral carbon. Answer: (b)

Teacher's Note:
1. The carbon must have four different groups to be chiral.

Answer: (c) 50% D + 50% L

Explanation:
1. The planar carbonyl group is attacked by CN⁻ from either face with equal probability
2. Equal amounts of both enantiomers form
3. The product is a racemic mixture. Answer: (c)

Teacher's Note:
1. Attack on a planar carbonyl from either side gives a racemic mixture.

Answer: (d) ketal

Explanation:
1. One mole of alcohol gives a hemiketal
2. A second mole of alcohol converts it to a ketal
3. So two moles of alcohol with a ketone give a ketal. Answer: (d)

Teacher's Note:
1. Acetals come from aldehydes and ketals from ketones.

Answer: (c) 3° alcohol

Explanation:
1. The Grignard reagent adds to the carbonyl carbon of the ketone
2. After hydrolysis the carbon bears OH, two methyl groups and the new R group
3. That is a tertiary alcohol. Answer: (c)

Teacher's Note:
1. Formaldehyde gives primary, other aldehydes secondary, ketones tertiary alcohols.

Answer: (c) n-propyl bromide

Explanation:
1. Peroxide makes the addition free-radical in nature (anti-Markovnikov)
2. The Br atom attaches to the terminal carbon
3. The product is n-propyl bromide. Answer: (c)

Teacher's Note:
1. The peroxide effect (Kharasch effect) reverses the orientation of HBr addition.

Answer: (a) isobutyl bromide

Explanation:
1. In the presence of peroxide, the bromine radical adds to the terminal carbon
2. This gives the more stable tertiary radical
3. The product is isobutyl bromide, (CH₃)₂CH–CH₂Br. Answer: (a)

Teacher's Note:
1. Free-radical addition forms the more stable radical.

Answer: (a) F₂

Explanation:
1. Fluorine is so reactive that it reacts with alkanes even in the dark
2. Chlorine and bromine need light or heat to form radicals, and iodine does not react appreciably. Answer: (a)

Teacher's Note:
1. Reactivity order of halogens: F2 > Cl2 > Br2 > I2.

Answer: (d) HNO₃ vapours at high temperature

Explanation:
1. Alkanes are nitrated in the vapour phase (400-500 °C) with HNO₃ vapour, by a free-radical process
2. The nitrating mixture is used for benzene. Answer: (d)

Teacher's Note:
1. Vapour-phase nitration of alkanes is a free-radical process.

Answer: (a) a fluorine free radical

Explanation:
1. Abstraction of hydrogen by F· is highly exothermic and has almost no activation energy
2. So the propagation step is fastest with fluorine. Answer: (a)

Teacher's Note:
1. Reactivity of halogen radicals decreases from F to I.