CBSE Class 11 Chemistry Hydrocarbons MCQs Set 05

Answer: (c) n-Butane

Explanation:
1. Wurtz reaction couples two alkyl groups from alkyl halides and sodium in dry ether.
2. Two molecules of ethyl bromide give n-butane only: 2CH₃CH₂Br + 2Na → CH₃CH₂CH₂CH₃ + 2NaBr.
3. Isobutane, n-pentane and isopentane would each need two different alkyl halides, which gives a mixture of alkanes.

Teacher's Note:
1. Wurtz reaction doubles the carbon chain of a single alkyl halide.
2. Use it only with one type of halide, otherwise a mixture forms.

Answer: (d) 3

Explanation:
1. 2-Methylbutane is CH₃CH(CH₃)CH₂CH₃.
2. Add the double bond in every possible position: 2-methylbut-1-ene, 2-methylbut-2-ene and 3-methylbut-1-ene.
3. All three have the same carbon skeleton, so all three give 2-methylbutane on hydrogenation.

Teacher's Note:
1. Keep the carbon skeleton fixed and move the double bond.
2. Check that each position is possible without exceeding four bonds on carbon.

Answer: (b) 3

Explanation:
1. n-Butane has a straight four-carbon chain, so the alkene must be unbranched.
2. The alkenes are but-1-ene, cis-but-2-ene and trans-but-2-ene.
3. Geometrical isomers are different compounds, so the count is 3.

Teacher's Note:
1. Count cis and trans forms separately.
2. Remember that the carbon skeleton does not change in hydrogenation.

Answer: (a) 1-Bromo-1-methylcyclopentane

Explanation:
1. Free radical bromination is highly selective and removes the hydrogen that gives the most stable radical.
2. Radical stability order is 3° > 2° > 1°.
3. Methylcyclopentane has one tertiary hydrogen (on the carbon carrying CH₃), so substitution occurs there.

Teacher's Note:
1. Bromine is more selective than chlorine.
2. Look for the tertiary C–H bond first.

Answer: (a) The alkane (A) must be added in excess

Explanation:
1. Chlorination is a chain reaction that can continue to di-, tri- and tetra-substituted products.
2. With excess alkane, a chlorine radical is far more likely to meet an alkane molecule than a chloroalkane.
3. So mainly the monochloro product is formed.

Teacher's Note:
1. Excess halogen gives polyhalogen compounds.
2. No light means no initiation, so no reaction.

Answer: (d) 3 and 5

Explanation:
1. Step (1) is initiation because radicals are formed from Cl₂ by light.
2. In propagation a radical reacts with a molecule and gives a new radical: Cl· + CH₄ → CH₃· + HCl, then CH₃· + Cl₂ → CH₃Cl + Cl·.
3. Steps (2) and (4) are not part of the accepted mechanism.

Teacher's Note:
1. A propagation step uses one radical and makes another radical.
2. Initiation makes radicals; termination removes them.

Answer: (c) Isomerisation

Explanation:
1. Both compounds have the formula C₄H₁₀.
2. Only the carbon skeleton changes from straight chain to branched chain, so this is isomerisation.

Teacher's Note:
1. Same molecular formula, different structure means isomers.
2. Anhydrous AlCl₃ with HCl is the usual catalyst.

Answer: (b) trans-But-2-ene

Explanation:
1. Acid-catalysed dehydration of an alcohol gives an alkene (elimination of water).
2. By Saytzeff's rule the more substituted alkene, but-2-ene, is the major product.
3. Of the two but-2-ene isomers, trans is more stable because the two methyl groups are far apart.

Teacher's Note:
1. Saytzeff: the more substituted alkene predominates.
2. Trans alkenes are generally more stable than cis.

Answer: (a) H₂ with finely divided Ni

Explanation:
1. Alkenes add dihydrogen in the presence of a metal catalyst such as Ni, Pt or Pd (Sabatier–Senderens reduction).
2. The C=C double bond becomes a C–C single bond.
3. Br₂ in CCl₄ gives a dibromide and cold KMnO₄ gives a diol.

Teacher's Note:
1. Remember Ni, Pd and Pt as hydrogenation catalysts.
2. Hydrogenation increases the number of C–H bonds.

Answer: (a) The hydrogen adds to the doubly bonded carbon that already has more hydrogen atoms, and the halide to the carbon with fewer hydrogen atoms

Explanation:
1. The negative part of the reagent goes to the carbon carrying fewer hydrogens.
2. For example, CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (2-bromopropane).

Teacher's Note:
1. The rich get richer: H goes where there are already more H atoms.
2. Do not confuse it with the peroxide effect.

Answer: (b) 1-Bromopropane

Explanation:
1. In the presence of peroxide the addition of HBr proceeds by a free radical mechanism (peroxide effect, or Kharasch effect).
2. The bromine radical attacks the terminal carbon to give the more stable secondary radical.
3. The product is the anti-Markovnikov one: CH₃CH₂CH₂Br.

Teacher's Note:
1. The peroxide effect works only with HBr, not with HCl or HI.
2. Anti-Markovnikov product means Br goes to the carbon with more hydrogens.

Answer: (c) But-2-ene

Explanation:
1. Geometrical isomerism needs restricted rotation about the double bond and two different groups on each doubly bonded carbon.
2. In CH₃CH=CHCH₃ each double-bond carbon carries H and CH₃, so cis and trans forms exist.
3. In the others, one doubly bonded carbon carries two identical groups (two H atoms or two CH₃ groups).

Teacher's Note:
1. Check each C of the double bond for two different groups.
2. A =CH₂ or =C(CH₃)₂ end never gives cis-trans isomers.

Answer: (d) 8 and 1

Explanation:
1. C–H bonds: 3 + 1 + 2 = 6, all σ.
2. C–C single bond: 1 σ; C=C double bond: 1 σ + 1 π.
3. Total σ = 6 + 1 + 1 = 8 and π = 1.

Teacher's Note:
1. A single bond is one σ; a double bond is one σ and one π.
2. Count the C–H bonds first.

Answer: (a) Ethane > ethene > ethyne

Explanation:
1. More shared electron pairs between the carbons pull the nuclei closer.
2. The C–C bond lengths are about 154 pm (ethane), 134 pm (ethene) and 120 pm (ethyne).

Teacher's Note:
1. Higher bond order means a shorter bond.
2. Remember 154, 134 and 120 pm.

Answer: (b) 12.5 kJ/mol

Explanation:
1. The eclipsed conformation has repulsion between the electron clouds of C–H bonds on the two carbons.
2. NCERT gives the energy difference as about 12.5 kJ/mol.
3. This is small enough to be overcome at room temperature, so rotation about the C–C single bond is almost free.

Teacher's Note:
1. The barrier is much smaller than a normal bond energy.
2. Staggered is lower in energy than eclipsed.

Answer: (d) 2,4-Dimethylhexane

Explanation:
1. The longest chain has six carbons, so the parent is hexane.
2. The two CH₃ groups are on carbon 2 and carbon 4 when numbered from the end that gives the lowest locants.
3. Numbering from the other end would give 3,5, which is higher.

Teacher's Note:
1. Choose the longest chain first, then the lowest set of locants.
2. Do not select the ethyl group as part of a shorter main chain.

Answer: (a) But-2-ene

Explanation:
1. Alcoholic KOH removes HBr (dehydrohalogenation).
2. Two alkenes can form, but-1-ene and but-2-ene.
3. Saytzeff's rule says that the more substituted alkene, but-2-ene, is the major product.

Teacher's Note:
1. Alcoholic KOH gives elimination; aqueous KOH gives substitution.
2. The more substituted double bond is more stable.

Answer: (c) Ethene

Explanation:
1. Alkenes are oxidised by cold dilute KMnO₄ to diols, and KMnO₄ is reduced to brown MnO₂.
2. The pink colour disappears, which is a test for the C=C bond.
3. Alkanes and cycloalkanes do not react with Baeyer's reagent.

Teacher's Note:
1. Baeyer's test and bromine water test are for unsaturation.
2. Saturated hydrocarbons give no colour change.

Answer: (b) Ethane-1,2-diol

Explanation:
1. Cold dilute alkaline KMnO₄ adds two OH groups across the double bond.
2. CH₂=CH₂ + H₂O + [O] → HOCH₂CH₂OH (ethane-1,2-diol, glycol).

Teacher's Note:
1. Mild conditions give a diol; strong hot acidic KMnO₄ breaks the double bond.
2. KMnO₄ acts as the source of [O].

Answer: (d) Butane

Explanation:
1. At the anode the propanoate ion loses an electron and CO₂ to give an ethyl radical: CH₃CH₂COO⁻ → CH₃CH₂· + CO₂ + e⁻.
2. Two ethyl radicals combine: 2CH₃CH₂· → CH₃CH₂CH₂CH₃.
3. The alkane formed has twice the number of carbon atoms of the alkyl part of the acid salt.

Teacher's Note:
1. Kolbe's method doubles the alkyl group (R–R).
2. Hydrogen is liberated at the cathode.