JEE Chemistry Hydrocarbons MCQs Set 12

Answer: (b) C5 - C10

Explanation:
1. Gasoline is the light liquid fraction of petroleum
2. Its molecules contain roughly 5 to 10 carbon atoms

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) \( \ce{H2O + CO2} \)

Explanation:
1. Complete combustion converts all carbon to \( \ce{CO2} \) and all hydrogen to \( \ce{H2O} \)
2. The other options are partial oxidation products

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: (d) Acetylene

Explanation:
1. Rotation is restricted by overlap of pi bonds in ethylene and by steric repulsion in hexachloroethane, and ethane has a small torsional barrier
2. In acetylene the cylindrical pi cloud has no preferred orientation, so rotation about the triple bond is effectively free

Teacher's Note:
1. Terminal alkynes have an acidic sp C-H bond (50% s-character) and give precipitates with ammoniacal AgNO3 and Cu2Cl2.
2. Mercuric hydration of an alkyne gives a ketone, except ethyne which gives ethanal.

Answer: (d) \( \ce{CO2} \)

Explanation:
1. Carbon in \( \ce{CO2} \) is in the +4 state, the highest
2. The other compounds still contain carbon in lower oxidation states

Teacher's Note:
1. Cold dilute alkaline KMnO4 gives a diol from an alkene, while hot acidic KMnO4 cleaves the double bond.
2. Terminal alkynes are separated from other hydrocarbons by their acidic hydrogen.

Answer: (d) Cracking

Explanation:
1. A large alkane breaks into a smaller alkene and a smaller alkane
2. This is cracking

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: (d) Neopentane

Explanation:
1. In neopentane all twelve hydrogens are equivalent
2. Only one monochloro derivative, 1-chloro-2,2-dimethylpropane, is possible

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) All of these

Explanation:
1. \( \ce{CH3COONa + NaOH -> CH4 + Na2CO3} \)
2. Reduction of \( \ce{CH3I} \) gives \( \ce{CH4} \)
3. \( \ce{Al4C3 + 12 H2O -> 4 Al(OH)3 + 3 CH4} \)

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) A mixture of methane and ethylene

Explanation:
1. Ammoniacal cuprous chloride reacts with the acidic hydrogen of terminal alkyne, forming a red precipitate of copper acetylide
2. Methane and ethylene do not react and pass out

Teacher's Note:
1. The more s-character, the shorter and stronger the bond and the more acidic the C-H.
2. Li or Na in liquid NH3 gives the trans alkene, while Lindlar's catalyst gives cis.

Answer: (d) \( \ce{C20H42} \)

Explanation:
1. Alkanes with about 18 or more carbon atoms are waxy solids
2. \( \ce{C20H42} \) is a solid paraffin

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: (d) Alkanes

Explanation:
1. Alkanes are saturated and have no pi bond or polar C=O group
2. They undergo only substitution, whereas the other classes give addition

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) \( \ce{CH3-CHBr2} \)

Explanation:
1. A gem-dihalide has both halogens on the same carbon
2. Only \( \ce{CH3CHBr2} \) (1,1-dibromoethane) fits
3. The others are vicinal

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) 2-Methylpentane

Explanation:
1. An isopropyl group is \( \ce{-CH(CH3)2} \) at a chain end
2. 2-Methylpentane, \( \ce{(CH3)2CH-CH2CH2CH3} \), has it
3. The others have tert-butyl or ethyl ends

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) \( \ce{CH4 + C2H6 + C3H8} \)

Explanation:
1. Natural gas is mainly methane with smaller amounts of ethane and propane
2. The other mixtures are not hydrocarbon gas

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) A mixture of the above three

Explanation:
1. The coupling of \( \ce{CH3-CH3} \), \( \ce{CH3-C2H5} \) and \( \ce{C2H5-C2H5} \) all occur
2. Ethane, propane and butane are formed together

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: (d) All of these

Explanation:
1. Vapour phase nitration breaks C-C bonds as well as C-H bonds
2. So nitromethane, nitroethane and nitropropanes are all obtained

Teacher's Note:
1. Use Huckel's rule (4n + 2 pi electrons in a planar cyclic conjugated ring) to judge aromaticity.
2. Side chain reactions in light go through radicals, while ring reactions with a Lewis acid go through cations.

Answer: (d) Homolysis

Explanation:
1. Light causes homolytic fission of the Cl-Cl bond to give chlorine radicals
2. This starts the radical chain

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: (d) Producer gas

Explanation:
1. Marsh gas, natural gas and coal gas all contain methane
2. Producer gas is \( \ce{CO + N2} \) and has no methane

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: (d) n-Propane

Explanation:
1. Coupling gives \( \ce{C2H5-C2H5} \) (butane), \( \ce{C2H5-C3H7} \) (pentane) and \( \ce{C3H7-C3H7} \) (hexane)
2. Propane has three carbons and cannot be made from an ethyl and a propyl group

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) Fractional distillation

Explanation:
1. The components of crude oil have different boiling points
2. So fractional distillation separates them

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: (d) Tetraethyl lead (TEL)

Explanation:
1. Tetraethyl lead, \( \ce{Pb(C2H5)4} \), is the classic antiknock additive
2. The other compounds are not used for this purpose

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) Cyclopropane

Explanation:
1. The three-membered ring has large angle strain
2. Bromine opens the ring to give 1,3-dibromopropane

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) Reacts with water

Explanation:
1. \( \ce{RMgX} \) behaves as a strong base and takes a proton from water
2. That destroys it giving the alkane \( \ce{R-H} \)

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) 70

Explanation:
1. Octane number equals the percentage of iso-octane
2. So it is 70

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: (d) \( \ce{Zn-Hg/HCl} \)

Explanation:
1. Clemmensen reduction converts \( \ce{C=O} \) into \( \ce{CH2} \) using zinc amalgam and concentrated HCl
2. The other reagents do not do this reduction under these conditions

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.