JEE Chemistry Hydrocarbons MCQs Set 19

Answer: (a) \( \ce{CH3-CH(OH)-CH2Cl} \)

Explanation:
1. The chloronium ion is attacked by water at the more substituted carbon
2. So OH goes to C-2 and Cl to C-1, giving 1-chloropropan-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: (c) Zero

Explanation:
1. The product is 2-methylbutane, \( \ce{(CH3)2CH-CH2CH3} \)
2. Its C-2 carries two methyl groups, so there is no chiral centre and no optical isomers

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) Isobutyl bromide

Explanation:
1. Under radical conditions Br adds to the less substituted carbon
2. The product is anti-Markovnikov, (CH3)2CHCH2Br, isobutyl 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: (b) \( \ce{H2C=CH-C#CH} \)

Explanation:
1. In vinylacetylene the two doubly bonded carbons are sp2 and the two triply bonded carbons are sp
2. The order from the left is sp2, sp2, sp, sp

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

Explanation:
1. At 165-170 degrees C the intermediate ethyl hydrogen sulphate decomposes
2. The product is ethene (dehydration)

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) 2-Phenylpropan-2-ol

Explanation:
1. Protonation gives the tertiary benzylic carbocation \( \ce{C6H5-C^+(CH3)2} \)
2. Water attacks it to give 2-phenylpropan-2-ol

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: (b) Solution of \( \ce{Br2} \) in \( \ce{CCl4} \)

Explanation:
1. Bromine in \( \ce{CCl4} \) loses its red-brown colour when it adds to a C=C bond
2. Ammoniacal cuprous or silver salts detect only terminal alkynes

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) Isopropyl alcohol

Explanation:
1. Markovnikov addition of \( \ce{H2SO4} \) gives isopropyl hydrogen sulphate
2. Hydrolysis gives propan-2-ol

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) Propene

Explanation:
1. Joining \( \ce{CH2=} \) and \( \ce{=CH-CH3} \) gives \( \ce{CH2=CH-CH3} \)
2. So the alkene is propene

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) \( \ce{H2C=CH2 + C6H6} \)

Explanation:
1. Friedel-Crafts alkylation of benzene with ethene and \( \ce{AlCl3} \) (with trace HCl) gives \( \ce{C6H5-C2H5} \)
2. Ethane is unreactive and propene gives cumene

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) n-Propyl bromide

Explanation:
1. The peroxide effect gives anti-Markovnikov addition of HBr
2. The product is 1-bromopropane (n-propyl 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: (a) \( \ce{C2H4} \)

Explanation:
1. Ethene is oxidised by permanganate but has no acidic hydrogen to give a silver salt
2. Ethane and methane do not react with \( \ce{KMnO4} \), while ethyne reacts with both

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) \( \ce{C6H5CH(Br)CH2CH3} \)

Explanation:
1. Protonation gives a secondary cation next to \( \ce{CH2Ph} \)
2. A hydride shift gives the more stable benzylic cation, which is captured by bromide to give 1-bromo-1-phenylpropane

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) 1-Bromo-2-butene under thermodynamically controlled conditions

Explanation:
1. At low temperature the 1,2 adduct (3-bromobut-1-ene) forms faster (kinetic control)
2. At 40 degrees C the more stable 1,4 adduct 1-bromobut-2-ene dominates (thermodynamic control)

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) Ethyl alcohol

Explanation:
1. Markovnikov hydration of any higher alkene gives a secondary or tertiary alcohol
2. Only ethene gives a primary alcohol, ethanol
3. Methanol would need methene, which does not exist

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) cis-Diol

Explanation:
1. \( \ce{OsO4} \) adds to the double bond as a cyclic osmate ester, delivering both oxygens to the same face
2. Reduction with \( \ce{NaHSO3} \) releases the cis-1,2-diol

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) Acetylene

Explanation:
1. Acetylene burns in oxygen with a very hot flame (about 3000 degrees C)
2. This oxy-acetylene flame is used for welding and cutting metals

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) Acetylene

Explanation:
1. \( \ce{CaC2 + 2 H2O -> Ca(OH)2 + C2H2} \)
2. Acetylene has an acidic hydrogen and gives a white precipitate of silver acetylide

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: (c) \( \ce{CH3CH2COOH + CO2} \)

Explanation:
1. Permanganate cleaves the triple bond
2. The ethyl end becomes \( \ce{CH3CH2COOH} \) and the terminal CH end is oxidised to \( \ce{CO2} \)

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) Acetylene

Explanation:
1. \( \ce{CH3COOH} \) is reduced to ethanol (A), which dehydrates to ethene (B)
2. Ethene adds \( \ce{Br2} \) to give 1,2-dibromoethane (C), and alcoholic KOH eliminates 2 HBr to give ethyne (D)

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) \( \ce{sp^2-sp} \)

Explanation:
1. The carbon of the triple bond is sp hybridised
2. The carbon of the double bond is sp2, so the single bond between them is sp2-sp

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: (a) Benzene

Explanation:
1. Three molecules of ethyne cyclotrimerise at about 400 degrees C
2. Benzene is formed

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) The sigma electron density of the C-H bond in acetylene is nearer to carbon, which has 50% s-character

Explanation:
1. An sp carbon has 50% s-character and is more electronegative
2. The C-H electrons are held closer to carbon, so the hydrogen is lost as \( \ce{H+} \) more easily
3. The acetylide ion is also stabilised

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: (c) Ammoniacal silver nitrate reagent

Explanation:
1. But-1-yne has an acidic terminal hydrogen and gives a white precipitate with ammoniacal \( \ce{AgNO3} \)
2. But-2-yne, butane and propene give none, while bromine and permanganate react with the unsaturated ones

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) Cyclohexene

Explanation:
1. \( \ce{C6H10} \) has two degrees of unsaturation, and one \( \ce{H2} \) uptake means one C=C and one ring
2. The dialdehyde with six carbons comes from cleaving cyclohexene

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.