JEE Chemistry Hydrocarbons MCQs Set 27

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Peroxide generates bromine radicals, which add to the terminal carbon to give the more stable secondary radical
2. So anti-Markovnikov product 1-bromobutane forms, and the radical mechanism is the reason

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The product is 1,2-dibromobutane, \( \ce{CH2Br-CHBr-CH2CH3} \)
2. C-2 carries four different groups
3. The bromonium ion is attacked from either face, giving both enantiomers, so a racemic mixture of two optical isomers forms

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Chlorination is a radical chain reaction that must be started by breaking \( \ce{Cl-Cl} \)
2. Sunlight supplies the energy that homolyses \( \ce{Cl2} \), so without light there is no reaction

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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Elimination follows Saytzeff's rule
2. The more substituted alkene but-2-ene is more stable and is the major product

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) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The HI formed reduces the alkyl iodide back to alkane, so the reaction is reversible
2. Iodic acid oxidises HI and is used to push the reaction forward, but this is a remedy rather than the reason for reversibility

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) A is true, but R is false.

Explanation:
1. The tertiary C-H bond of isobutane is oxidised by permanganate to give \( \ce{(CH3)3C-OH} \), so the assertion is true
2. Strong oxidising agents do attack alkanes (tertiary C-H and combustion), so the reason is false

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) A is true, but R is false.

Explanation:
1. All twelve hydrogens of neopentane are equivalent, so only one monosubstituted product forms
2. Its C-H bond energy is not unusually high and does not explain this, so the reason is false

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: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Knocking causes premature, uneven combustion and loss of power
2. Fuels that resist knocking are indeed preferred, but that is a consequence of the assertion rather than its explanation

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) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( \ce{Ag+} \) forms a pi complex with the C=C bond, which increases solubility
2. Alkenes are weakly polar, but this is not what causes the complex formation

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Dehydration of 2-butanol gives both alkenes, with but-2-ene as the major product
2. Saytzeff's rule (more substituted alkene predominates) accounts for this product distribution

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: (c) A is true, but R is false.

Explanation:
1. The cyclic bromonium ion is opened by \( \ce{Br-} \) or \( \ce{Cl-} \), giving both products, so the assertion is true
2. The mechanism is ionic and not free radical, so the reason is false

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) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. n-Heptane has an octane number of zero and straight chain alkanes knock easily
2. The octane scale does measure gasoline quality, but this does not explain why straight chains rate low

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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The central sp carbon forms two pi bonds in perpendicular planes
2. So the two \( \ce{CH2} \) groups lie in perpendicular planes and the hydrogens are not coplanar
3. The hybridisation of the carbons explains this

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Peroxide initiates a radical chain addition, and the bromine radical adds to the terminal carbon
2. This gives 1-bromopropane (anti-Markovnikov), and the radical mechanism is the reason

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( \ce{HC#CH + NaNH2 -> HC#CNa + NH3} \)
2. Acetylene (pKa about 25) is a stronger acid than ammonia (pKa about 35), so it protonates the amide ion to give ammonia

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) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The C-Cl bond in aryl halides has partial double bond character from delocalisation, so substitution is difficult
2. Halogens are o,p-directing, but that effect concerns electrophilic ring substitution and does not explain the low reactivity of the halogen

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) A is false, but R is true.

Explanation:
1. Benzene is itself alkylated under these conditions, so it cannot be used as an inert solvent
2. The reason statement about Friedel-Crafts is true

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The methyl group donates electrons by induction and hyperconjugation
2. This stabilises the intermediate arenium ion and so activates the ring

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Fuming sulphuric acid contains free \( \ce{SO3} \), a strong electrophile
2. It attacks the ring, so \( \ce{SO3} \) as the attacking species explains the formation of the sulphonic acid

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: (c) A is true, but R is false.

Explanation:
1. Activating groups donate electrons to the ring, so the assertion is true
2. The nitroso group (\( \ce{-NO} \)) withdraws electrons and is deactivating (though o,p-directing), so the reason is false

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) A is false, but R is true.

Explanation:
1. With \( \ce{AlCl3} \) benzene gives acetophenone by acylation, not chlorobenzene, so the assertion is false
2. Chlorination of benzene is indeed electrophilic substitution, so the reason is true

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) A is false, but R is true.

Explanation:
1. Annulenes are monocyclic conjugated polyenes with an even number of carbons, so the assertion is false
2. Their general formula is \( \ce{(CH=CH)n} \), so the reason is true

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. [10]Annulene has 10 pi electrons (4n+2, n = 2)
2. The two internal hydrogens of a planar ring would clash, so the ring twists out of plane and loses aromaticity
3. Non-planarity explains the assertion

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The slow step is attack of \( \ce{NO2+} \) on the ring to form the arenium ion
2. Since the C-H or C-D bond is broken in a fast step, there is no kinetic isotope effect, which explains the equal rates

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The cyclopentadienyl anion is planar and cyclic with six pi electrons, so it is aromatic and highly stabilised
2. The allyl anion has only simple delocalisation, so aromaticity explains the greater stability

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.