CBSE Class 11 Chemistry Hydrocarbons MCQs Set 06

Answer: (c) 1-Butyne

Explanation:
1. Terminal alkynes (R–C≡C–H) have an acidic hydrogen on an sp carbon.
2. This hydrogen is replaced by Ag to give a white precipitate of silver alkynide, e.g. CH₃CH₂C≡CAg.
3. 2-Pentyne is an internal alkyne, so it has no acidic hydrogen on the triple-bonded carbon.

Teacher's Note:
1. Only terminal alkynes (with ≡C–H) give this test.
2. Cuprous chloride gives a red precipitate in the same way.

Answer: (a) AgNO₃ in NH₄OH

Explanation:
1. Both propyne and propene decolourise Br₂/CCl₄ and dilute KMnO₄.
2. Only propyne, a terminal alkyne, forms a white precipitate (CH₃C≡CAg) with ammoniacal silver nitrate.
3. Propene gives no precipitate.

Teacher's Note:
1. Choose the test that depends on a difference, not on a common property.
2. A precipitate is a clear observation.

Answer: (b) H₂ with Lindlar's catalyst

Explanation:
1. Lindlar's catalyst (Pd on CaCO₃ poisoned with quinoline or sulphur compounds) stops hydrogenation at the alkene stage.
2. Both hydrogen atoms add from the same side, giving the cis alkene.
3. Na in liquid NH₃ would give the trans isomer.

Teacher's Note:
1. Lindlar gives cis; Na/liq. NH₃ gives trans.
2. The catalyst is poisoned so it does not reduce the alkene further.

Answer: (d) 2,2-Dichlorobutane

Explanation:
1. The first HCl adds by Markovnikov's rule: CH₃CH₂C≡CH → CH₃CH₂CCl=CH₂ (2-chlorobut-1-ene).
2. The second HCl adds to the same carbon again (Cl is already on it and stabilises the cation).
3. The product is a gem-dihalide, CH₃CH₂CCl₂CH₃ (2,2-dichlorobutane).

Teacher's Note:
1. Alkynes give gem-dihalides with excess HX.
2. Both chlorine atoms end up on the carbon that carries the alkyl group.

Answer: (b) Benzene

Explanation:
1. Calcium carbide reacts with water to give ethyne: CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂.
2. Ethyne polymerises (cyclic trimerisation) when passed through a red hot tube at about 873 K: 3C₂H₂ → C₆H₆.
3. So B is benzene.

Teacher's Note:
1. Ethyne to benzene is a standard NCERT conversion.
2. Three ethyne molecules make one benzene ring.

Answer: (c) Benzene

Explanation:
1. Soda lime (NaOH + CaO) causes decarboxylation of the sodium salt.
2. C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃.
3. The COONa group is replaced by H.

Teacher's Note:
1. Decarboxylation removes CO₂ (as carbonate) from the carboxylate salt.
2. The same method gives methane from sodium acetate.

Answer: (b) Toluene

Explanation:
1. Alkanes with six or more carbon atoms form aromatic hydrocarbons at high temperature and pressure with a catalyst such as Cr₂O₃ on alumina.
2. n-Hexane gives benzene; n-heptane cyclises and loses 4H₂ to give toluene (methylbenzene).

Teacher's Note:
1. Aromatisation means ring closure with loss of hydrogen.
2. Seven carbons: a six-membered ring plus a CH₃ group.

Answer: (a) Toluene

Explanation:
1. The CH₃ group in toluene is electron-releasing (+I and hyperconjugation), so the ring has higher electron density and is activated.
2. –CHO is strongly deactivating and –Cl is weakly deactivating.
3. So toluene reacts faster than benzene and the other two.

Teacher's Note:
1. Electron-releasing groups activate the ring towards electrophiles.
2. Electron-withdrawing groups deactivate it.

Answer: (c) SO₃

Explanation:
1. Fuming sulphuric acid (oleum) contains SO₃.
2. SO₃ is electron-deficient at sulphur and acts as the electrophile in sulphonation.
3. Attack on the ring gives benzenesulphonic acid, C₆H₅SO₃H.

Teacher's Note:
1. Nitration uses NO₂⁺ and sulphonation uses SO₃.
2. An electrophile is electron-deficient.

Answer: (d) Base

Explanation:
1. The stronger acid, H₂SO₄, donates a proton to HNO₃.
2. HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻.
3. Since HNO₃ accepts the proton, it behaves as a base, and the nitronium ion is generated.

Teacher's Note:
1. Identify who gives and who takes the proton.
2. The nitronium ion NO₂⁺ is the real electrophile.

Answer: (b) Benzophenone

Explanation:
1. Friedel–Crafts reactions need a ring that is not strongly deactivated.
2. Benzophenone has a C=O group attached to the ring, which strongly withdraws electrons.
3. Toluene, p-xylene and chlorobenzene (weakly deactivated) can all be acylated.

Teacher's Note:
1. Strongly deactivated rings (–NO₂, –COR) fail in Friedel–Crafts reactions.
2. Halogens only deactivate weakly.

Answer: (b) C₆H₆ + C₂H₅Cl → C₆H₅C₂H₅ + HCl (anhydrous AlCl₃)

Explanation:
1. Friedel–Crafts alkylation introduces an alkyl group into the ring using an alkyl halide and anhydrous AlCl₃.
2. C₆H₆ + C₂H₅Cl gives ethylbenzene and HCl.
3. The second is a substitution of OH, the third is not balanced (Cl₂ is not released), and the fourth is Grignard reagent formation.

Teacher's Note:
1. Look for an aromatic ring, an R–X or RCOX and AlCl₃.
2. The by-product is HX, not X₂.

Answer: (d) –F

Explanation:
1. –NHCOR, –OCOR and –CH=CH₂ all donate electrons to the ring (by resonance or hyperconjugation) and activate it.
2. Fluorine withdraws electrons by the inductive effect more strongly than it donates by resonance, so it deactivates the ring.
3. It is, however, ortho/para directing.

Teacher's Note:
1. Halogens are deactivating but o,p-directing.
2. Activating groups usually carry a lone pair or alkyl character.

Answer: (c) p-Xylene

Explanation:
1. In p-xylene (1,4-dimethylbenzene) all four ring hydrogens are equivalent because of symmetry.
2. So only one ring-substituted monochloro product, 2-chloro-1,4-dimethylbenzene, is possible.
3. o-Xylene gives two and m-xylene gives three ring-substituted monochloro isomers.

Teacher's Note:
1. Count the number of different kinds of ring hydrogens.
2. Symmetry reduces the number of products.

Answer: (a) Aniline > acetanilide > benzene > acetophenone

Explanation:
1. –NH₂ is a very strong activator.
2. In acetanilide the N lone pair is partly delocalised onto the C=O group, so –NHCOCH₃ activates less than –NH₂ but still activates.
3. Benzene has no substituent, and –COCH₃ deactivates the ring, so acetophenone is the least reactive.

Teacher's Note:
1. Activators first, then benzene, then deactivators.
2. Acylation of the nitrogen weakens its activating power.

Answer: (a) 5 < 1 < 2 < 4 < 3

Explanation:
1. –N⁺(CH₃)₃ is the strongest electron-withdrawing group, then –NO₂, so these deactivate most.
2. Benzene is the reference.
3. –CH₃ activates weakly, and –OH activates strongly through resonance.

Teacher's Note:
1. Place groups on a scale from strongly withdrawing to strongly donating.
2. Charge on the substituent makes it strongly deactivating.

Answer: (d) 1-Bromo-3-nitrobenzene

Explanation:
1. Nitration gives nitrobenzene.
2. The –NO₂ group is deactivating and meta-directing, so the incoming Br goes to the meta position.
3. Monobromination thus gives 1-bromo-3-nitrobenzene (m-bromonitrobenzene).

Teacher's Note:
1. Nitro is a meta director.
2. Bromination of a deactivated ring is slow but follows the meta rule.

Answer: (b) Ethanal

Explanation:
1. Water adds across the triple bond to give an unstable enol, CH₂=CHOH (vinyl alcohol).
2. The enol rearranges at once to the more stable carbonyl compound, CH₃CHO (ethanal).
3. Higher alkynes give ketones in the same way, e.g. propyne gives propanone.

Teacher's Note:
1. Hydration of an alkyne gives a carbonyl compound, not an alcohol.
2. Ethyne is the only alkyne that gives an aldehyde.

Answer: (c) CH₃CO⁺

Explanation:
1. AlCl₃ removes Cl⁻ from CH₃COCl: CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻.
2. The acylium ion CH₃CO⁺ is the electrophile and attacks the ring to give acetophenone.

Teacher's Note:
1. Acylation uses the acylium ion RCO⁺; alkylation uses R⁺.
2. AlCl₃ acts as a Lewis acid by accepting Cl⁻.

Answer: (c) Hexachlorocyclohexane (C₆H₆Cl₆)

Explanation:
1. In sunlight, chlorine adds to benzene by a free-radical mechanism, not by substitution.
2. C₆H₆ + 3Cl₂ → C₆H₆Cl₆ (benzene hexachloride, BHC or gammexane).
3. Substitution products form with FeCl₃ in the dark.

Teacher's Note:
1. Light means addition; a Lewis acid catalyst means substitution.
2. Three Cl₂ molecules add to the three double bonds.