JEE Chemistry Hydrocarbons MCQs Set 13

Answer: (d) Sulphonation

Explanation:
1. The reaction replaces H by \( \ce{-SO3H} \)
2. This is sulphonation

Teacher's Note:
1. Aromatic rings keep their stability by electrophilic substitution, not addition.
2. Activating groups direct ortho and para, while strongly withdrawing groups direct meta.

Answer: (d) Lower

Explanation:
1. Branching makes the molecule more compact with less surface area
2. Weaker van der Waals forces give a lower boiling point

Teacher's Note:
1. Trends such as boiling point and stability depend on molecular size, branching and substituent effects.
2. Apply the trend directly to rank the options.

Answer: (d) Higher saturated hydrocarbons

Explanation:
1. Paraffin wax is a mixture of long chain alkanes, roughly \( \ce{C20} \) to \( \ce{C30} \)
2. They are saturated and solid at room temperature

Teacher's Note:
1. Fractions of crude oil come off in order of boiling point: gases, petroleum ether, gasoline, kerosene, diesel, lubricating oil.
2. Cracking breaks large molecules into smaller, more useful ones.

Answer: (b) Substitution reaction

Explanation:
1. A hydrogen atom is replaced by chlorine
2. This is free radical substitution

Teacher's Note:
1. Ozonolysis splits C=C into two carbonyl groups, so join the carbonyl carbons to find the alkene.
2. Hydroboration-oxidation gives the anti-Markovnikov alcohol, while oxymercuration gives the Markovnikov alcohol.

Answer: (d) \( \ce{AlCl3} \)

Explanation:
1. Anhydrous \( \ce{AlCl3} \) with HCl isomerises n-alkanes into branched alkanes
2. The other reagents are reducing agents

Teacher's Note:
1. Write the balanced equation first and track every carbon.
2. Check whether the reagent gives elimination (alcoholic KOH) or substitution (aqueous KOH).

Answer: (b) Dehydrohalogenation

Explanation:
1. Alcoholic KOH removes HX from alkyl halides to give alkenes
2. Aqueous KOH would instead give substitution

Teacher's Note:
1. Recall the key fact or reaction first, then eliminate options that contradict it.
2. Link each answer to the underlying bonding or mechanism rather than memorising it.

Answer: (d) Methane

Explanation:
1. \( \ce{Al4C3 + 12 H2O -> 4 Al(OH)3 + 3 CH4} \)
2. The carbide contains carbon as \( \ce{C^4-} \)

Teacher's Note:
1. Check the type of reaction involved and what it does to the carbon skeleton.
2. Compare the options one by one against that.

Answer: (b) Electrolysis of aqueous \( \ce{RCOOK} \)

Explanation:
1. Kolbe electrolysis gives the coupled alkane \( \ce{R-R} \) at the anode
2. \( \ce{RCOOAg} \) with iodine gives \( \ce{R-I} \), chlorination gives a mixture of chloroalkanes, and tert-butyl chloride with ethanol gives an ether or alkene

Teacher's Note:
1. Link each named reaction to its reagent and the change in carbon number.
2. Elimination gives alkenes, reduction gives alkanes, and coupling doubles the alkyl chain.

Answer: (b) Gasoline

Explanation:
1. Boiling point increases with the number of carbon atoms
2. Gasoline has the smallest molecules among these, so it distils first

Teacher's Note:
1. Remember the composition of common fuels: LPG is butane and propane, natural gas is mostly methane, producer gas is CO with N2.
2. Match each fuel to its main constituent.

Answer: (d) Difference in the rates of diffusion of gases

Explanation:
1. Methane (marsh gas, firedamp) diffuses through a porous pot at a different rate from air
2. This is Graham's law of diffusion

Teacher's Note:
1. Straight chain alkanes knock easily, while branched alkanes and aromatics resist knocking.
2. Octane number is the percentage of iso-octane in a mixture with n-heptane.

Answer: (b) Decarboxylation of sodium acetate with soda lime

Explanation:
1. \( \ce{CH3COONa + NaOH -> CH4 + Na2CO3} \)
2. Wurtz coupling always gives at least two carbons, and the other two reactions do not give alkanes

Teacher's Note:
1. Identify the functional group that is removed or formed and count the carbon atoms of the product.
2. Wurtz and Kolbe couple two alkyl groups, soda lime removes one carbon, and Grignard reagents pick up an acidic proton.

Answer: (b) Cyclopropane

Explanation:
1. The C-C-C angle of 60 degrees is far from 109.5 degrees
2. Cyclopropane has the highest angle strain

Teacher's Note:
1. Count sigma bonds and lone pairs to find hybridisation: 4 gives sp3, 3 gives sp2, 2 gives sp.
2. Bond length falls and bond strength rises with increasing s-character and bond order.

Answer: (d) \( \ce{CH4} \)

Explanation:
1. Marsh gas comes from decomposition of vegetation under water
2. It is mainly methane

Teacher's Note:
1. Fractions of crude oil come off in order of boiling point: gases, petroleum ether, gasoline, kerosene, diesel, lubricating oil.
2. Cracking breaks large molecules into smaller, more useful ones.

Answer: (b) Reduction of an alkyl halide

Explanation:
1. Reducing \( \ce{CH3-X} \) (for example with Zn and HCl) gives \( \ce{CH4} \)
2. Wurtz and Kolbe give alkanes with at least two carbons, and hydrogenation of an alkene also gives at least \( \ce{C2H6} \)

Teacher's Note:
1. Write the balanced equation first and track every carbon.
2. Check whether the reagent gives elimination (alcoholic KOH) or substitution (aqueous KOH).

Answer: (d) \( \ce{C2H6} \)

Explanation:
1. C-H bond length increases as the s character of the carbon decreases
2. sp3 carbon in ethane (109 pm) has the longest bond, then sp2 (107 pm) and sp (106 pm)

Teacher's Note:
1. Decide which functional groups can react with the reagent and which cannot.
2. A test that works for both compounds cannot distinguish them.

Answer: (c) Propane

Explanation:
1. Ozone adds across a C=C (or C#C) bond
2. Propane is saturated and does not react

Teacher's Note:
1. Draw the possible intermediates and pick the most stable one (tertiary, then secondary, then primary).
2. Only HBr shows the peroxide effect, not HCl or HI.

Answer: (d) \( \ce{sp^2} \)

Explanation:
1. Each carbon forms three sigma bonds
2. This uses three sp2 hybrid orbitals, and the leftover p orbitals form the pi bond

Teacher's Note:
1. Compare the structures directly: more substituted alkenes and carbocations are more stable.
2. Use hyperconjugation and the inductive effect to explain stability orders.

Answer: (b) Butane

Explanation:
1. Wurtz reaction: \( \ce{2 C2H5I + 2 Na -> C4H10 + 2 NaI} \)
2. Two ethyl groups join to give butane

Teacher's Note:
1. Link each named reaction to its reagent and the change in carbon number.
2. Elimination gives alkenes, reduction gives alkanes, and coupling doubles the alkyl chain.

Answer: (b) A molecular solid

Explanation:
1. Methane molecules are non-polar and held only by weak van der Waals forces
2. So solid methane is a molecular solid

Teacher's Note:
1. Trends such as boiling point and stability depend on molecular size, branching and substituent effects.
2. Apply the trend directly to rank the options.

Answer: (d) Both \( \ce{C2H5OH} \) and \( \ce{CH3CH2NH2} \)

Explanation:
1. A Grignard reagent takes a proton from compounds with acidic O-H or N-H and gives the alkane
2. Both ethanol and ethylamine release \( \ce{CH4} \), while acetone mainly undergoes addition to give a tertiary alcohol

Teacher's Note:
1. Identify the functional group that is removed or formed and count the carbon atoms of the product.
2. Wurtz and Kolbe couple two alkyl groups, soda lime removes one carbon, and Grignard reagents pick up an acidic proton.

Answer: (c) \( \ce{B2H6} \) followed by \( \ce{NaOH, H2O2} \)

Explanation:
1. Hydroboration-oxidation adds H and OH anti-Markovnikov
2. So propene gives the primary alcohol, propan-1-ol
3. \( \ce{H2SO4/H2O} \) would give propan-2-ol

Teacher's Note:
1. Electrophilic addition goes through the more stable carbocation, so H adds to the carbon with more hydrogens.
2. With HBr and a peroxide the radical path reverses the orientation.

Answer: (b) Cracking

Explanation:
1. Heating large molecules under controlled conditions breaks C-C bonds
2. This is cracking

Teacher's Note:
1. Remember the composition of common fuels: LPG is butane and propane, natural gas is mostly methane, producer gas is CO with N2.
2. Match each fuel to its main constituent.

Answer: (b) \( \ce{CnH2n+2} \)

Explanation:
1. Alkanes are saturated acyclic hydrocarbons
2. Their formula is \( \ce{CnH2n+2} \)

Teacher's Note:
1. Check the type of reaction involved and what it does to the carbon skeleton.
2. Compare the options one by one against that.

Answer: (d) \( \ce{CH3Br} \)

Explanation:
1. Reduction of \( \ce{CH3Br} \) gives methane
2. The Wurtz reaction of \( \ce{CH3Br} \) with sodium gives ethane
3. The other compounds cannot give both directly

Teacher's Note:
1. Write the balanced equation first and track every carbon.
2. Check whether the reagent gives elimination (alcoholic KOH) or substitution (aqueous KOH).