CBSE Class 11 Chemistry HOTs Hydrocarbons Set 02

Higher Order Thinking Questions

Question 1. Assertion (A): Ethyne is more acidic than ethene.
Reason (R): The carbon atoms of ethyne are sp hybridised, and an sp orbital has more s-character (50%) than an sp² orbital (33%).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) A greater s-character makes the carbon more electronegative, so it holds the C–H bonding electrons more tightly and the hydrogen is released more easily as H⁺. Hence R correctly explains A.
In simple words: More s-character means higher electronegativity of carbon and a more acidic C–H bond.

Teacher's Note:
1. Percentage s-character: sp 50%, sp² 33%, sp³ 25%.
2. Cite the hybridisation to explain acidity.

Question 2. Assertion (A): Benzene undergoes electrophilic substitution reactions rather than addition reactions.
Reason (R): Substitution keeps the delocalised π electron system, and so the stability of the aromatic ring, intact.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) Addition would destroy the delocalised six-π-electron system and the resonance stabilisation. In substitution the ring is restored after the loss of H⁺, so substitution is preferred. R correctly explains A.
In simple words: Benzene keeps its special stability by swapping a hydrogen instead of adding to the ring.

Teacher's Note:
1. Resonance energy is the stability that benzene would lose in addition.
2. Addition happens only under drastic conditions (H₂/Ni, Cl₂/sunlight).

Question 3. Assertion (A): HCl adds to propene in the presence of peroxide according to Markovnikov's rule, just as it does in its absence.
Reason (R): The H–Cl bond is too strong (high bond enthalpy) to be broken by the free radical, so the radical chain does not occur with HCl.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) The peroxide effect is shown only by HBr. With HCl, the step Cl–H + radical is energetically unfavourable (strong H–Cl bond), so the ionic mechanism continues and the Markovnikov product, 2-chloropropane, is obtained. Hence A is true and R explains it.
With HI the I· radical is not reactive enough to add to the double bond, so HI also does not show the effect.
In simple words: Only HBr has a bond that is weak enough for the radical path and a bromine radical reactive enough to add.

Teacher's Note:
1. Peroxide effect: HBr only.
2. HCl: bond too strong. HI: iodine radical too unreactive.

Question 4. Assertion (A): The staggered conformation of ethane is more stable than the eclipsed conformation.
Reason (R): In the staggered form the electron clouds of the C–H bonds of the two carbons are farthest apart and repel each other least.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) In the staggered form the dihedral angle is 60°, so the bond-pair repulsions (torsional strain) are minimum. In the eclipsed form they are at maximum, and the energy is about 12.5 kJ/mol higher. R explains A.
In simple words: Bonds that sit farther apart repel less, so the staggered form has lower energy.

Teacher's Note:
1. Staggered: least torsional strain; eclipsed: most torsional strain.
2. The two forms interconvert freely at room temperature.

Question 5. Assertion (A): cis-But-2-ene has a higher boiling point than trans-but-2-ene.
Reason (R): The cis isomer has a net dipole moment, while in the trans isomer the bond dipoles cancel.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) In cis-but-2-ene the two CH₃ groups are on the same side, so the bond dipoles add up to a small net dipole moment, giving dipole–dipole attractions (boiling point 277 K). In trans-but-2-ene the dipoles cancel (boiling point 274 K). R explains A.
In simple words: A polar molecule has stronger attractions between molecules, so it boils at a higher temperature.

Teacher's Note:
1. Compare the symmetry of the two isomers.
2. The trans isomer packs better in a crystal, so it has the higher melting point.

Question 6. Consider the statements about electrophilic aromatic substitution. I: Hyperconjugation increases the electron density at the ortho and para positions in alkylbenzenes. II: Electron-donating groups are activators and o/p-directors because they stabilise the arenium ion. III: Electron-withdrawing groups are activators and m-directors. IV: Halogens are activators and o/p-directors. Select the correct statements.
(a) I and II only
(b) I, II and III
(c) II, III and IV
(d) I and IV only
Answer:
Ans. (a) Statements I and II are correct.
III is wrong: electron-withdrawing groups deactivate the ring, so they are deactivators and m-directors. IV is wrong: halogens withdraw electrons inductively and so are deactivating, though they direct o/p through their lone pairs.
In simple words: Donors activate and send the new group ortho/para; withdrawers deactivate and send it meta; halogens deactivate but still go ortho/para.

Teacher's Note:
1. Test each statement separately.
2. Halogens are the special case.

Question 7. Why does nitrobenzene not undergo Friedel–Crafts alkylation or acylation, whereas chlorobenzene does?
Answer:
The –NO₂ group is strongly electron-withdrawing (–I and –M), so it reduces the electron density of the ring so much that the ring cannot attack the weak carbocation or acylium electrophile. The nitro group can also complex with the AlCl₃ catalyst.
The chlorine atom is only weakly deactivating, because its +M effect (lone pair donation) partly offsets its –I effect, so chlorobenzene is still reactive enough to give o- and p-products.
In simple words: A strongly deactivated ring is too electron-poor for Friedel–Crafts reactions; a weakly deactivated ring can still react.

Teacher's Note:
1. Compare –NO₂ (strong deactivator) with –Cl (weak deactivator).
2. Friedel–Crafts reactions generally fail with –NO₂, –CN, –COR and –SO₃H on the ring.

Question 8. A hydrocarbon A (molecular formula C₂H₂) is obtained from calcium carbide and water. A is passed through a red hot iron tube to give B. B reacts with CH₃Cl in the presence of anhydrous AlCl₃ to give C. Identify A, B and C and write the equations. Predict the main products when C is nitrated.
Answer:
A = ethyne: CaC₂ + 2H₂O → HC≡CH + Ca(OH)₂.
B = benzene: 3HC≡CH → C₆H₆ (red hot iron tube, 873 K).
C = toluene: C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl (anhydrous AlCl₃, Friedel–Crafts alkylation).
Nitration of toluene (conc. HNO₃ + conc. H₂SO₄) gives mainly o-nitrotoluene and p-nitrotoluene, since –CH₃ is an ortho/para-directing activating group.
In simple words: Ethyne forms benzene, benzene is methylated to toluene, and toluene is nitrated at the ortho and para positions.

Teacher's Note:
1. Write each step with the reagent over the arrow.
2. Methyl is an ortho/para director.

Question 9. Arrange 1-butene, butane, 1-butyne and NH₃ in increasing order of acidity and explain.
Answer:
Order: butane < 1-butene < NH₃ < 1-butyne.
The acidity of a C–H bond depends on the s-character of the carbon: sp³ (butane, 25%) < sp² (1-butene, 33%) < sp (1-butyne, 50%). The conjugate base of 1-butyne is the most stabilised carbanion.
NH₃ is a stronger acid than the alkene and the alkane but weaker than the terminal alkyne. This is why the alkyne can be converted into its sodium salt with NaNH₂, but an alkene cannot.
In simple words: More s-character on the carbon makes the anion more stable and the hydrogen more acidic.

Teacher's Note:
1. The acidic order is alkane < alkene < NH₃ < alkyne.
2. A stronger acid has a more stable conjugate base.

Question 10. The C–C bond length in benzene (139 pm) lies between that of ethane (154 pm) and ethene (134 pm). Explain this with the help of resonance, and say what would be expected if benzene had three fixed single and three fixed double bonds.
Answer:
Benzene is a resonance hybrid of two Kekulé structures. The six π electrons are delocalised over all six carbons, so each C–C bond has about 50% double-bond character (a bond order of 1.5), giving a length between those of C–C (154 pm) and C=C (134 pm).
If benzene had three fixed single (154 pm) and three fixed double (134 pm) bonds, the ring would be an irregular hexagon with alternating bond lengths, and 1,2-disubstituted derivatives with the substituents on a single bond and on a double bond would be two different compounds. Neither is observed: o-xylene exists as only one compound, and all bonds are 139 pm.
In simple words: Delocalisation evens out all six bonds, so benzene is a regular hexagon.

Teacher's Note:
1. Resonance structures differ only in electron positions, not in atom positions.
2. Mention the one o-disubstituted isomer as evidence.

Question 11. Ethyl bromide is treated with sodium in dry ether to give X. X is then treated with Cl₂ in the presence of light to give monochloro derivatives. Identify X and state how many different monochloro structural isomers can form.
Answer:
Wurtz reaction: 2CH₃CH₂Br + 2Na → CH₃CH₂CH₂CH₃ + 2NaBr, so X = n-butane.
n-Butane has two kinds of hydrogen: those on the CH₃ carbons (1°) and those on the CH₂ carbons (2°). Replacement gives 1-chlorobutane and 2-chlorobutane.
So two monochloro structural isomers are possible. At room temperature the 2° hydrogen is replaced more easily, so 2-chlorobutane is the major product.
In simple words: Count the different kinds of hydrogen atoms in butane: two kinds give two monochloro products.

Teacher's Note:
1. Chlorination is not very selective, so both products are obtained.
2. Radical stability: 3° > 2° > 1°.

Question 12. How can you distinguish between ethane, ethene and ethyne by simple chemical tests?
Answer:
Test 1 – Br₂ water or Br₂ in CCl₄: Ethene and ethyne decolourise it, and ethane does not. This separates ethane from the other two.
Test 2 – Ammoniacal AgNO₃ (Tollens' reagent): Ethyne gives a white precipitate of silver acetylide, and ethene gives none.
(An alternative to separate ethane: Baeyer's reagent, which is decolourised by ethene and ethyne but not by ethane.)
In simple words: Bromine separates the saturated from the unsaturated; the silver test then picks out ethyne.

Teacher's Note:
1. Ethyne with Cu₂Cl₂ in ammonia gives a red precipitate in the same way.
2. Mention the observation (colour change or precipitate) for each test.