JEE Chemistry Hydrocarbons MCQs Set 18

Answer: (a) 120 degrees

Explanation:
1. The planar ring with equal intermediate C-C bonds is benzene
2. Its carbons are sp2 hybridised in a regular hexagon, so the angle is 120 degrees

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: (a) \( \ce{CnH2n} \)

Explanation:
1. Alkenes have one double bond
2. Two hydrogens fewer than the alkane gives \( \ce{CnH2n} \)

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: (a) 3,3-Dimethylbutan-2-ol

Explanation:
1. Oxymercuration-demercuration gives Markovnikov hydration without rearrangement
2. OH goes to C-2 and the skeleton stays unchanged
3. The product is 3,3-dimethylbutan-2-ol, whereas acid hydration would rearrange to 2,3-dimethylbutan-2-ol

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) Olefinic hydrocarbons

Explanation:
1. Alkenes reduce permanganate to brown \( \ce{MnO2} \)
2. The purple colour disappears, which is the test for a double bond

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) Sabatier and Senderens' reaction

Explanation:
1. Catalytic hydrogenation of alkenes over finely divided nickel is the Sabatier-Senderens reaction
2. Wurtz and Kolbe form C-C bonds

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) \( \ce{(CH3)2C=C(CH3)2} \)

Explanation:
1. Acetone, \( \ce{(CH3)2C=O} \), comes from a \( \ce{=C(CH3)2} \) end
2. Only \( \ce{(CH3)2C=C(CH3)2} \) has such ends on both sides

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) 1 < 3 < 2

Explanation:
1. Electrophilic addition is faster when the carbocation formed is more stable
2. Isobutene gives a tertiary cation, but-2-ene a secondary cation and ethene only a primary cation, so the order is 1 < 3 < 2

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) Ammoniacal \( \ce{AgNO3} \)

Explanation:
1. Propyne is a terminal alkyne with an acidic hydrogen and gives a white precipitate of silver propynide
2. Propene gives no precipitate, while bromine, permanganate and ozone react with both

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) 2-Chloro-2-methylpropane

Explanation:
1. H adds to the \( \ce{CH2} \) carbon to form the tertiary carbocation
2. Chloride then attacks the tertiary carbon, giving \( \ce{(CH3)3C-Cl} \)

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) Ozonolysis

Explanation:
1. Ozonolysis cleaves the C=C bond into carbonyl compounds
2. The structures of these fragments show where the double bond was

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) Ethene

Explanation:
1. Ethene is unsaturated and decolourises Baeyer's reagent
2. Ethyne reacts with Tollens' reagent through its acidic hydrogen, while ethane and methane do not decolourise Baeyer's reagent

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: (b) Saytzeff's rule

Explanation:
1. Saytzeff's rule says the more substituted alkene is the major elimination product
2. But-2-ene is more substituted than but-1-ene

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) 2-Butene

Explanation:
1. Only a symmetrical alkene with a \( \ce{CH3-CH=} \) unit on each side gives just ethanal
2. But-2-ene gives two molecules of \( \ce{CH3CHO} \)

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: (a)

Explanation:
1. The acylium ion \( \ce{CH3CO+} \) attacks the terminal carbon to give the stable tertiary carbocation
2. Chloride captures it, giving 4-chloro-4-methylpentan-2-one

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: (a) Raney nickel and \( \ce{H2} \)

Explanation:
1. Alkenes add \( \ce{H2} \) readily over a finely divided metal catalyst such as Raney nickel
2. The other reducing agents do not reduce an isolated C=C bond

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: (c) Ethene

Explanation:
1. The succinate ion \( \ce{-OOC-CH2-CH2-COO-} \) loses two \( \ce{CO2} \) at the anode (Kolbe electrolysis)
2. The two \( \ce{CH2} \) radicals join as \( \ce{CH2=CH2} \)

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) Ethylene

Explanation:
1. Hydrogenation needs an unsaturated compound, so ethane is ruled out
2. Acetylene would give a precipitate with ammoniacal \( \ce{AgNO3} \), so ethene is the answer

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: (c) Electrophilic addition

Explanation:
1. The proton, an electrophile, attacks the pi bond first
2. The bromide then adds to the carbocation, so it is electrophilic addition

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) \( \ce{CH2=C=C(CH3)-CH=CH2} \)

Explanation:
1. The fragments must join as \( \ce{CH2=} \), \( \ce{=C=} \) (gives \( \ce{CO2} \)), \( \ce{=C(CH3)-CH=} \) (gives \( \ce{CH3COCHO} \)) and \( \ce{=CH2} \)
2. This is \( \ce{CH2=C=C(CH3)-CH=CH2} \)
3. Option (c) would give \( \ce{CO2} \) and acetone instead

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) 2-Butene

Explanation:
1. \( \ce{CH3-CH=CH-CH3} \) is cleaved at the double bond
2. Both halves are oxidised to \( \ce{CH3COOH} \)

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: (b) Pd/\( \ce{H2} \)

Explanation:
1. Catalytic hydrogenation over palladium adds \( \ce{H2} \) to the C=C bond
2. Metal-acid reductions do not reduce an isolated alkene

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: (a) n-Propyl bromide

Explanation:
1. Peroxide causes a free radical chain addition
2. Br adds to the terminal carbon, giving 1-bromopropane (n-propyl bromide)

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: (d) All of these

Explanation:
1. Every alkene adds bromine across its C=C bond
2. The red-brown colour of bromine disappears in each case

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: (a) A more stable carbonium ion

Explanation:
1. Proton addition to the terminal carbon gives the secondary carbocation
2. This is more stable than the primary one, so iodide attacks the middle carbon

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) \( \ce{C2H4} \)

Explanation:
1. \( \ce{CH2=CH2} \) is split at the double bond into two \( \ce{CH2=O} \)
2. Ethyne would give glyoxal and ethane gives nothing

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.