JEE Chemistry Hydrocarbons MCQs Set 26

Answer: (b) \( \ce{O3} \)

Explanation:
1. Ozonolysis of an alkyne followed by reductive work-up with \( \ce{Zn/H2O} \) gives a 1,2-diketone
2. Ozone is X

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) \( \ce{C2H6} \) (excess) + \( \ce{Cl2} \) in UV light

Explanation:
1. A large excess of ethane makes a chlorine radical meet an ethane molecule rather than a product molecule
2. This keeps polychlorination low

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: (a) \( \ce{C2H6} \)

Explanation:
1. HI with red phosphorus is a strong reducing agent
2. It reduces \( \ce{C2H5OH} \) to ethane without changing the carbon skeleton

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: (a) \( \ce{H2} \); \( \ce{CO} \)

Explanation:
1. Water gas, a mixture of \( \ce{CO} \) and \( \ce{H2} \), is passed over a Co or Fe catalyst
2. This gives a mixture of hydrocarbons

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: (a) Wurtz-Fittig reaction

Explanation:
1. \( \ce{C6H5Br + n-C3H7Br + 2 Na -> C6H5-C3H7 + 2 NaBr} \)
2. Friedel-Crafts alkylation with n-propyl chloride rearranges to isopropylbenzene (cumene)

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) \( \ce{CH3CH=CH2} \) with \( \ce{B2H6} \), then alkaline \( \ce{H2O2} \)

Explanation:
1. Hydroboration followed by oxidation gives propan-1-ol, an alcohol
2. The other three reactions remove functional groups to give propane

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: (a) \( \ce{CH3COCH3 + CH3COOH} \)

Explanation:
1. The oxidative cleavage splits the double bond
2. The \( \ce{(CH3)2C=} \) end gives acetone and the \( \ce{=CH-CH3} \) end gives an aldehyde, which is oxidised to acetic acid

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) \( \ce{Ag2C2} \)

Explanation:
1. Both acidic hydrogens of ethyne are replaced by silver
2. The product is silver acetylide, \( \ce{Ag2C2} \)

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: (b) Molecular solid

Explanation:
1. Naphthalene molecules are held by weak van der Waals forces
2. It is a molecular solid and sublimes easily

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (d) 1,3,5-Heptatriene

Explanation:
1. 1,3,5-Heptatriene is an acyclic polyene with an \( \ce{sp3} \) carbon, so it is neither cyclic nor fully conjugated
2. The other three are planar cyclic systems with \( \ce{(4n+2)} \) pi electrons

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) 28 kg

Explanation:
1. Molar mass of \( \ce{CaC2} \) is 64.1 g/mol, so 64.1 kg is 1 kmol
2. 1 kmol \( \ce{CaC2} \) gives 1 kmol \( \ce{C2H2} \), then 1 kmol \( \ce{C2H4} \) (28 kg)
3. Polymerisation conserves mass, so 28 kg polyethylene

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) \( \ce{-Cl} \)

Explanation:
1. Chlorine withdraws electrons by the inductive effect and so deactivates the ring
2. Its lone pairs donate by delocalisation to the ortho and para positions, so it is o,p-directing

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (b) I > III > II > IV

Explanation:
1. The methyl group activates the ring, so toluene is most reactive
2. Benzene is next
3. Benzoic acid is deactivated by \( \ce{-COOH} \) and nitrobenzene is deactivated most strongly by \( \ce{-NO2} \)

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) Both (a) and (b)

Explanation:
1. Both permanganate and acidified dichromate oxidise the benzylic methyl group to \( \ce{-COOH} \)
2. Sulphuric acid alone is not an oxidising agent here

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) \( \ce{C2H2} \) and \( \ce{CH3CHO} \)

Explanation:
1. Calcium carbide gives acetylene (A)
2. Mercuric hydration of acetylene gives acetaldehyde (B)

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: (a) \( \ce{C2H2} \)

Explanation:
1. Ethyne is hydrated to ethanal (Y)
2. Ethanal is oxidised to acetic acid

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) \( \ce{(CH3)2C=CH2} \)

Explanation:
1. Electrophilic addition is fastest when the carbocation formed is the most stable
2. Isobutene gives a tertiary cation, which is more stable than the cations from the others

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) Acetic acid

Explanation:
1. Strong oxidative cleavage of \( \ce{CH3-CH=CH2} \) gives \( \ce{CH3COOH} \) and the terminal \( \ce{CH2} \) as formic acid
2. Acetic acid is X

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) Meta

Explanation:
1. Substituents on 1,2 are ortho, 1,3 are meta and 1,4 are para
2. So 1,3 is meta

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: (a) Ethylidene bromide

Explanation:
1. The first HBr gives vinyl bromide
2. The second HBr follows Markovnikov's rule to give \( \ce{CH3-CHBr2} \), ethylidene bromide

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) \( \ce{-X} \) (halogens)

Explanation:
1. Halogens withdraw by induction (deactivating) but donate by delocalisation (o,p-directing)
2. Amino, hydroxyl and alkyl groups are activating

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (a) Benzyl chloride

Explanation:
1. In light the substitution is a free radical reaction on the side chain
2. The benzylic radical gives benzyl chloride

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) \( \ce{C2H5COOH} \)

Explanation:
1. \( \ce{CH2=CH2 + CO + H2O -> CH3CH2COOH} \)
2. This is the Reppe carboxylation, which gives propanoic acid

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) \( \ce{CHI3} \)

Explanation:
1. Ethene gives bromoethane (X), which is hydrolysed to ethanol (Y)
2. Ethanol with \( \ce{I2} \) and base undergoes the iodoform reaction, giving yellow \( \ce{CHI3} \) (Z)

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) Vinyl bromide

Explanation:
1. One mole of HBr adds to give \( \ce{CH2=CHBr} \)
2. Further addition would give ethylidene 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.