JEE Chemistry Hydrocarbons MCQs Set 41

Answer: (d) F₃C–CH=CH₂

Explanation:
1. A +R or +I group on the double bond stabilises the adjacent carbocation, so Markovnikov addition is followed (options a, b, c).
2. The strongly electron-withdrawing CF₃ group (–I) destabilises the carbocation next to it, so H⁺ adds to the carbon bearing CF₃ and Cl goes to the terminal carbon: F₃C–CH₂–CH₂Cl, the anti-Markovnikov product.

Teacher's Note:
1. Electron-withdrawing groups reverse the direction of addition even without peroxide.
2. Compare the stability of the two possible carbocations.

Answer: (d) 4-chlorobenzaldehyde

Explanation:
1. Light promotes free-radical chlorination at the benzylic CH₃; two equivalents give the dichloromethyl compound ArCHCl₂.
2. Hydrolysis of ArCHCl₂ gives an unstable gem-diol ArCH(OH)₂, which loses water to give the aldehyde.
3. Product: 4-chlorobenzaldehyde.

Teacher's Note:
1. The ring chlorine is untouched in these conditions.
2. gem-Dihalides hydrolyse to aldehydes or ketones.

Answer: (a) Friedel–Crafts alkylation

Explanation:
1. In Friedel–Crafts alkylation, the alkyl group introduced activates the ring, so the product is more reactive than benzene and further alkylation occurs.
2. In acylation, the acyl group deactivates the ring, so substitution stops after one step.

Teacher's Note:
1. Activating products lead to polysubstitution.
2. Acylation is preferred for clean monosubstitution.

Answer: (c) D < A < C < B

Explanation:
1. Electron-withdrawing groups deactivate the ring and electron-releasing groups activate it.
2. CN is strongly deactivating (–I, –R), Cl is weakly deactivating (–I > +R), CH₃ activates by +I and hyperconjugation, and OCH₃ activates strongly by +R.
3. Order: D < A < C < B.

Teacher's Note:
1. A group with a lone pair on the ring atom (OCH₃) is a strong activator.
2. Halogens are deactivating but ortho/para directing.

Answer: (d) CH₃C(I)(Cl)CHD₂

Explanation:
1. Addition to an unsymmetrical alkyne follows Markovnikov's rule: D adds to the terminal carbon and Cl to the internal carbon, giving CH₃C(Cl)=CHD.
2. The second addition (DI) also follows Markovnikov's rule: D goes to the CHD carbon, I to the carbon already bearing Cl.
3. Product: CH₃C(I)(Cl)CHD₂.

Teacher's Note:
1. The halide adds to the more substituted carbon each time.
2. The second equivalent adds to the carbon that already has a halogen (stabilised cation).

Answer: (a) III < I < II

Explanation:
1. Acetophenone has a strongly deactivating –R acetyl group, chlorobenzene is weakly deactivated, and toluene is activated.
2. Order of reactivity: acetophenone < chlorobenzene < toluene.

Teacher's Note:
1. –R groups deactivate more strongly than halogens.
2. Alkyl groups activate the ring.

Answer: (c) angle strain

Explanation:
1. Angle strain arises when bond angles in a ring deviate from the ideal, so it is absent in open-chain compounds.
2. Torsional strain, steric hindrance and electrostatic forces all affect the stability of acyclic conformers.

Teacher's Note:
1. Angle strain is a feature of rings (Baeyer strain).
2. Think of the staggered and eclipsed ethane conformations: torsional strain.

Answer: (c) 2 molecules of CH₃COOH

Explanation:
1. Hot alkaline KMnO₄ followed by acid oxidatively cleaves the C=C bond.
2. Each CH= carbon bears a CH₃ and an H, so each fragment is oxidised to a carboxylic acid.
3. Product: 2 molecules of acetic acid.

Teacher's Note:
1. Cold dilute KMnO₄ gives the diol; hot, concentrated KMnO₄ cleaves the double bond.
2. A carbon with an H ends up as –COOH.

Answer: (b) cyclopentadienyl cation

Explanation:
1. Aromaticity needs a planar, cyclic, fully conjugated system with (4n + 2) π electrons.
2. Pyrrole (6 π electrons including the N lone pair), the cyclopentadienyl anion (6) and pyridine (6) are aromatic.
3. The cyclopentadienyl cation has only 4 π electrons (4n, n = 1): it is antiaromatic.

Teacher's Note:
1. Count π electrons: 2, 6, 10 are aromatic; 4, 8 are antiaromatic.
2. Cyclopentadiene itself is not aromatic because of its sp³ carbon.

Answer: (a) B, C and D

Explanation:
1. Cyclopropenyl cation: planar, cyclic, conjugated with 2 π electrons (4n + 2, n = 0), so aromatic.
2. Cyclopentadienyl cation has 4 π electrons: antiaromatic. Cycloheptatrienyl anion has 8 π electrons: antiaromatic.
3. Cycloheptatriene has an sp³ CH₂, so it is not fully conjugated or planar: non-aromatic.
4. Not aromatic: B, C and D.

Teacher's Note:
1. Huckel's rule needs 2, 6, 10 … π electrons.
2. An sp³ carbon in the ring breaks the cyclic conjugation.

Answer: (d) CH₃CH₂C≡CH

Explanation:
1. Alcoholic KOH removes one HBr to give a vinyl bromide, CH₃CH₂CH=CHBr or CH₃CH₂CBr=CH₂.
2. NaNH₂ in liquid NH₃ is a stronger dehydrohalogenating agent and removes the second HBr to give the terminal alkyne, which exists as its sodium salt CH₃CH₂C≡C^–Na⁺.
3. Water work-up gives but-1-yne.

Teacher's Note:
1. Two successive eliminations convert a vicinal dihalide to an alkyne.
2. Terminal alkynes are acidic and form sodium salts with NaNH₂.

Answer: (c) Na/liquid NH₃

Explanation:
1. Sodium in liquid ammonia (dissolving-metal reduction) adds electrons and protons in an anti fashion: the intermediate vinylic radical is more stable in the trans configuration, giving trans-alkenes.
2. H₂–Pd/BaSO₄ (Lindlar) gives cis-alkenes; NaBH₄ does not reduce alkynes.

Teacher's Note:
1. Lindlar gives cis; Na/NH₃ gives trans.
2. The mechanism passes through a vinylic anion.

Answer: (d) four

Explanation:
1. Peroxide effect: anti-Markovnikov addition gives 2-bromo-3-methylpentane, CH₃CHBr–CH(CH₃)–CH₂CH₃.
2. It has two different chiral centres (C-2 and C-3), so the number of stereoisomers = 2² = 4.

Teacher's Note:
1. For n different chiral centres, the maximum number of stereoisomers is 2ⁿ.
2. Check that the molecule has no symmetry before applying 2ⁿ.

Answer: (a) 1-phenylbut-2-ene

Explanation:
1. Geometrical isomerism in an alkene needs two different groups on each doubly bonded carbon.
2. 1-Phenylbut-2-ene, C₆H₅CH₂CH=CHCH₃, satisfies this.
3. The terminal alkenes have two H atoms on one double-bond carbon, and 1,1-diphenylpropane has no double bond.

Teacher's Note:
1. A terminal =CH₂ carbon always has two identical groups.
2. Saturated compounds do not show cis–trans isomerism.

Answer: (b) neopentane

Explanation:
1. The hydrocarbon with M = 72 is C₅H₁₂ (12 × 5 + 12 = 72).
2. Among its isomers, neopentane (CH₃)₄C has all twelve hydrogens equivalent, so it gives only one monohalide, (CH₃)₃CCH₂Cl.

Teacher's Note:
1. Count the types of equivalent hydrogens to count monohalides.
2. The other options are not C₅H₁₂ hydrocarbons.

Answer: (b) Li/NH₃

Explanation:
1. Li in liquid ammonia (Birch-type reduction) converts an internal alkyne to the trans-alkene.
2. Pt/H₂ gives hexane, Pd/BaSO₄ (Lindlar) gives the cis-alkene and LiAlH₄ does not reduce a simple alkyne.

Teacher's Note:
1. Same logic as Na/NH₃.
2. Catalytic hydrogenation with Pt goes all the way to the alkane.

Answer: (b) a vinyl group

Explanation:
1. Formaldehyde (HCHO) comes from a terminal =CH₂ unit in the alkene.
2. A terminal =CH₂ attached to the rest of the molecule is a vinyl (or vinylidene) group.

Teacher's Note:
1. Ozonolysis cuts C=C and puts C=O on each carbon.
2. =CH₂ becomes H₂C=O.

Answer: (d) 2-methyl-2-pentene

Explanation:
1. Join the two carbonyl fragments with a C=C in place of the oxygens: (CH₃)₂C=O + O=CHCH₂CH₃ gives (CH₃)₂C=CHCH₂CH₃.
2. That is 2-methyl-2-pentene.

Teacher's Note:
1. Reverse the ozonolysis: remove O atoms and join the carbons.
2. Check that the alkene has the right chain length.

Answer: (a) cyclopenta-1,3-diene

Explanation:
1. Cyclopenta-1,3-diene has an sp³ CH₂ in the ring, so it is not fully conjugated and not aromatic.
2. Furan (6 π electrons), naphthalene (10) and benzene (6) are planar, cyclic and fully conjugated.

Teacher's Note:
1. Look for an sp³ carbon in the ring.
2. Furan uses an oxygen lone pair to complete the sextet.

Answer: (c) 2-butene

Explanation:
1. An aldehyde C_nH_2nO of mass 44: 14n + 16 = 44, so n = 2, giving CH₃CHO.
2. Joining two CH₃CHO units through C=C gives CH₃CH=CHCH₃, that is 2-butene (symmetrical).

Teacher's Note:
1. Work out the carbonyl compound's formula from its mass.
2. A symmetrical alkene gives two identical carbonyl molecules.

Answer: (d) CH₃CHO

Explanation:
1. Ozone adds to 2-butene to form an ozonide.
2. Reductive work-up (Zn/H₂O) splits the ozonide into two molecules of acetaldehyde, CH₃CHO.

Teacher's Note:
1. Zn prevents over-oxidation of aldehydes to acids.
2. A symmetrical alkene gives two identical products.

Answer: (b) CH₃CH₂C≡CH

Explanation:
1. Sodium amide formed in liquid NH₃ removes the acidic hydrogen from a terminal alkyne to form a sodium alkynide.
2. Only but-1-yne (CH₃CH₂C≡CH) has a terminal ≡C–H; the internal alkynes are only reduced (to trans-alkenes) under these conditions.

Teacher's Note:
1. Terminal alkyne C–H is acidic (sp carbon).
2. Internal alkynes have no acidic hydrogen.

Answer: (d) CH₄

Explanation:
1. The terminal alkyne hydrogen is acidic enough to protonate the Grignard reagent.
2. CH₃MgX + CH₃C≡CH gives CH₄ + CH₃C≡CMgX.

Teacher's Note:
1. Grignard reagents act as strong bases towards acidic hydrogens.
2. The gas evolved is methane.

Answer: (a) CH₃C≡CH + 2HBr

Explanation:
1. Propyne adds HBr twice by Markovnikov's rule: first to give 2-bromopropene, then CH₃CBr₂CH₃ (both bromines on C-2).
2. (b) gives 1,1-dibromopropane type, (c) gives 1,1-dibromoethane and (d) gives 2-bromopropane.

Teacher's Note:
1. Both halogens end up on the same carbon in the addition to alkynes.
2. 2,2-Dibromopropane is a gem-dihalide.