JEE Chemistry Hydrocarbons MCQs Set 20

Answer: (a) \( \ce{CH3COOH + CH3CH2COOH} \)

Explanation:
1. Ozonolysis of an alkyne followed by hydrolysis gives carboxylic acids
2. Each carbon of the triple bond becomes a \( \ce{-COOH} \), giving ethanoic and propanoic acids

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) Mobil oil

Explanation:
1. The fractions of coal tar are light oil, middle (carbolic) oil, heavy oil, anthracene oil and pitch
2. Mobil oil is a petroleum lubricant

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) Tetrahedral, planar and linear

Explanation:
1. Methane is sp3 (tetrahedral), ethene is sp2 (planar) and ethyne is sp (linear)

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) Ethyne, iodomethane, iodoethane and sodamide

Explanation:
1. Ethyne is deprotonated by \( \ce{NaNH2} \) to give the acetylide, which is alkylated with \( \ce{CH3I} \)
2. The process is repeated with \( \ce{C2H5I} \), giving pent-2-yne

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) Propanone

Explanation:
1. Markovnikov hydration of propyne gives an enol \( \ce{CH3-C(OH)=CH2} \)
2. The enol tautomerises to acetone (propanone)

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) \( \ce{C2H5OH} \)

Explanation:
1. \( \ce{CaC2} \) and water give acetylene (A), hydration gives acetaldehyde (B)
2. Hydrogenation of the aldehyde over Ni gives ethanol (C)

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) KOH in \( \ce{C2H5OH} \)

Explanation:
1. A gem-dihalide loses two molecules of HCl with alcoholic KOH
2. This gives the alkyne

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. Kolbe electrolysis of potassium fumarate gives \( \ce{C2H2} \)
2. Calcium carbide and water give \( \ce{C2H2} \), and ethylene bromide loses 2 HBr with alcoholic KOH

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: (b) Potassium fumarate

Explanation:
1. Electrolysis of potassium fumarate (\( \ce{-OOC-CH=CH-COO-} \)) loses two \( \ce{CO2} \) at the anode
2. The two \( \ce{CH} \) radicals combine to give \( \ce{HC#CH} \), while succinate would give ethene

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) Ammoniacal silver nitrate

Explanation:
1. Bromine water and Baeyer's reagent react with any unsaturation
2. Only a terminal alkyne forms a precipitate with ammoniacal \( \ce{AgNO3} \)
3. So that reaction shows the triple bond in propyne

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) Cyclooctatetraene

Explanation:
1. Nickel cyanide catalyses tetramerisation of ethyne (Reppe synthesis)
2. The product is cyclooctatetraene, \( \ce{C8H8} \)

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) But-1-yne

Explanation:
1. \( \ce{C4H6} \) adding four Br atoms has a triple bond or two double bonds
2. A white precipitate with ammoniacal \( \ce{AgNO3} \) shows a terminal alkyne, so A is but-1-yne

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) \( \ce{C2H2} \)

Explanation:
1. Ethyne has acidic hydrogen atoms
2. \( \ce{2 HC#CH + 2 Na -> 2 HC#CNa + H2} \)

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) A white precipitate with \( \ce{AgNO3} \) (ammoniacal) and a red precipitate with \( \ce{Cu2Cl2} \) (ammoniacal)

Explanation:
1. Silver acetylide is white
2. Copper(I) acetylide is red

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: (d) Propyne

Explanation:
1. The C-C single bond shortens as the s-character of the other carbon increases
2. In propyne it joins an sp3 carbon to an sp carbon (50% s), which is the shortest

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) Formed by sp-sp overlap (as in alkynes)

Explanation:
1. Bond strength increases with s-character of the overlapping orbitals
2. The sp-sp sigma bond (and the triple bond overall) is the strongest

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) Ethylene and benzene

Explanation:
1. Ethylene and benzene both have sp2 carbons and bond angles of 120 degrees
2. Ethane is 109.5 and acetylene is 180 degrees

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) \( \ce{CH3CH2COCH3} \)

Explanation:
1. Markovnikov addition of water gives the enol \( \ce{CH3CH2C(OH)=CH2} \)
2. This tautomerises to the methyl ketone, butan-2-one

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) \( \ce{C2H2} \)

Explanation:
1. Sodamide is a very strong base that removes the acidic hydrogen of ethyne
2. The sodium salt is sodium acetylide

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. Decolourising bromine shows unsaturation
2. A precipitate with ammoniacal \( \ce{AgNO3} \) shows a terminal alkyne, so the gas is ethyne

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) (ii) and (iii)

Explanation:
1. Butane has only sp3 carbons and ethyne has only sp carbons
2. But-2-ene has sp3 and sp2, and vinylacetylene has sp2 and 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: (d) Vinyl chloride

Explanation:
1. \( \ce{HC#CH + HCl -> CH2=CHCl} \) with mercuric chloride catalyst
2. Vinyl chloride is the product, which is the monomer of PVC

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) Acetone

Explanation:
1. Hydration follows Markovnikov's rule
2. The enol \( \ce{CH3-C(OH)=CH2} \) tautomerises to acetone

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) Mesitylene

Explanation:
1. Three propyne molecules combine in a red hot tube
2. The ring formed is 1,3,5-trimethylbenzene (mesitylene)

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) \( \ce{C2H2} \)

Explanation:
1. Only ethyne has an acidic C-H bond
2. It gives red copper(I) acetylide

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.