JEE Chemistry Hydrocarbons MCQs Set 14

Answer: (d) Gasoline

Explanation:
1. Gasoline contains about \( \ce{C6} \) to \( \ce{C10} \) hydrocarbons
2. Its boiling range matches 70-200 degrees C

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: (c) \( \ce{CO} \) and \( \ce{N2} \)

Explanation:
1. Producer gas is made by passing air over red hot coke
2. It is \( \ce{CO} \) with \( \ce{N2} \)

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: (c) n-Octane

Explanation:
1. Boiling point rises with molar mass and falls with branching
2. n-Octane (126 degrees C) is higher than iso-octane (99 degrees C) and 2,2,3,3-tetramethylbutane (106 degrees C)

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) Isobutane

Explanation:
1. Isobutane (about -12 degrees C) is branched and boils lower than n-butane (-0.5), 1-butene (-6) and 1-butyne (+8)
2. Branching weakens van der Waals forces

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) All of these

Explanation:
1. Reforming processes convert straight chain alkanes into branched or cyclic ones
2. Isomerisation, alkylation and cyclisation all raise the octane number

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: (c) \( \ce{C6-C11} \)

Explanation:
1. Gasoline is the fraction containing roughly 6 to 11 carbon atoms
2. The other ranges correspond to other fractions

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

Explanation:
1. \( \ce{CH4 + 2 O2 -> CO2 + 2 H2O} \)
2. Excess oxygen gives only carbon dioxide and water

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: (b) Straight chain olefins

Explanation:
1. Straight chain compounds ignite easily and knock most
2. Branching and aromatic rings resist knocking, so the straight chain olefin is worst here

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: (c) \( \ce{Al4C3} \)

Explanation:
1. \( \ce{Al4C3 + 12 H2O -> 4 Al(OH)3 + 3 CH4} \)
2. Calcium carbide gives ethyne, while SiC and \( \ce{B4C} \) are inert to water

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: (b) 2-Bromo-2-methylbutane

Explanation:
1. Bromination is very selective for the most stable radical
2. The tertiary hydrogen at C-2 is replaced, giving the tertiary bromide

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: (b) 2,3-Dimethylbutane

Explanation:
1. 2,3-Dimethylbutane has only two kinds of hydrogen (CH3 and CH)
2. n-Hexane and 2,2-dimethylbutane give three monochloro products and 2-methylpentane gives five

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) \( \ce{C2H6} \)

Explanation:
1. Catalytic hydrogenolysis replaces Cl by H: \( \ce{C2H5Cl + H2 -> C2H6 + HCl} \)
2. Ethane is formed

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: (c) Both 1,2 and 1,4 addition products

Explanation:
1. The allylic cation intermediate can be attacked at C-2 or C-4
2. Both 1,2-dibromobut-3-ene and 1,4-dibromobut-2-ene form

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) An alkene

Explanation:
1. Zinc removes the two vicinal bromines: \( \ce{BrCH2CH2Br + Zn -> CH2=CH2 + ZnBr2} \)
2. An alkene is formed

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: (c) \( \ce{CH2Cl-CH2OH} \) and aqueous \( \ce{NaHCO3} \)

Explanation:
1. \( \ce{CH2=CH2 + HOCl -> ClCH2CH2OH} \) (ethylene chlorohydrin)
2. Warming with aqueous sodium bicarbonate replaces Cl by OH to give \( \ce{HOCH2CH2OH} \)

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: (c) Addition

Explanation:
1. The pi bond is electron rich and easily attacked by electrophiles
2. So addition is characteristic

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) 2-Bromopropane

Explanation:
1. Markovnikov addition puts Br on the carbon with fewer hydrogens
2. The secondary carbocation gives 2-bromopropane

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: (b) \( \ce{C2H5OH} \)

Explanation:
1. \( \ce{CH3CH2OH -> CH2=CH2 + H2O} \) on heating with conc. \( \ce{H2SO4} \)
2. The other alcohols give different alkenes or none

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: (b) Two molecules of acetone

Explanation:
1. The alkene is \( \ce{(CH3)2C=C(CH3)2} \), symmetrical about the double bond
2. Cleavage gives two molecules of \( \ce{(CH3)2C=O} \)

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: (c) \( \ce{C2H4} \)

Explanation:
1. \( \ce{C2H5I + KOH(alc.) -> C2H4 + KI + H2O} \)
2. Ethene has a double bond and decolourises permanganate

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: (a) \( \ce{CH2=CH2} \)

Explanation:
1. Alcoholic KOH brings about dehydrohalogenation
2. Ethene is formed

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) Ethylene

Explanation:
1. One molecule of HBr is eliminated to give ethene
2. Ethene has no halogen left to eliminate further

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: (c) 2,3-Dimethyl-2-butene

Explanation:
1. Loss of water gives a secondary carbocation, which rearranges by a methyl shift to a tertiary carbocation
2. Loss of H+ from it gives the most substituted alkene, 2,3-dimethylbut-2-ene, which has no cis-trans forms

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: (a) \( \ce{CH3-CH^.-CH2Br} \)

Explanation:
1. In the presence of peroxide the reaction is a free radical chain
2. Br radical adds to the terminal carbon to give the more stable secondary radical \( \ce{CH3-CH^.-CH2Br} \)

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.