JEE Chemistry Hydrocarbons MCQs Set 15

Answer: (d) All of these

Explanation:
1. Acrylonitrile is \( \ce{CH2=CH-CN} \)
2. It is called vinyl cyanide, cyanoethene or prop-2-enenitrile

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: (d) Both (a) and (b)

Explanation:
1. \( \ce{HC#CH + AsCl3 -> ClCH=CH-AsCl2} \)
2. The product is the war gas lewisite, whose chemical name is beta-chlorovinyl dichloroarsine, so both describe it

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) 2-Butene

Explanation:
1. \( \ce{CH3-CH=CH-CH3} \) is symmetrical
2. Cleavage at the double bond 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) But-2-ene

Explanation:
1. But-2-ene is symmetrical, so both Markovnikov and anti-Markovnikov addition give the same product
2. The peroxide effect cannot be detected for it

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) Decolourise Baeyer's reagent giving a brown precipitate

Explanation:
1. Both have pi bonds and are oxidised by permanganate, giving brown \( \ce{MnO2} \)
2. Only ethyne gives a precipitate with silver nitrate
3. Ethene reacts with chlorine by addition, not explosively

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) \( \ce{RCHOH-CHOHR} \)

Explanation:
1. Cold dilute permanganate does syn hydroxylation of the double bond
2. The product is a vicinal diol

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) 2-Methylbutan-2-ol

Explanation:
1. Protonation follows Markovnikov's rule to give the tertiary carbocation
2. Water then adds to it giving the tertiary alcohol 2-methylbutan-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: (b) By dehydration with conc. \( \ce{H2SO4} \) at 170 degrees C

Explanation:
1. Loss of water from ethanol with conc. \( \ce{H2SO4} \) at 170 degrees C gives ethene
2. Dehydrogenation gives ethanal and HI reduction gives ethane

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) Detection of double bonds

Explanation:
1. Baeyer's reagent is cold dilute alkaline \( \ce{KMnO4} \)
2. Its purple colour is lost with a C=C bond, so it is used to detect unsaturation

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) Propylene

Explanation:
1. Markovnikov hydration of propene gives the secondary alcohol \( \ce{CH3CH(OH)CH3} \)
2. Ethylene gives ethanol and 2-methylpropene gives tert-butyl alcohol

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) \( \ce{BrCH2CH2CH2CH2COOH} \)

Explanation:
1. With peroxide HBr adds in the anti-Markovnikov way
2. Br goes to the terminal carbon, giving 5-bromopentanoic acid

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) Ammoniacal \( \ce{Cu2Cl2} \)

Explanation:
1. Bromine and permanganate are decolourised by ethene but not by ethane
2. Ammoniacal cuprous chloride reacts only with terminal alkynes, so neither ethene nor ethane reacts and it cannot distinguish them

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) \( \ce{CH3CH2CH=CHCH2CH3} \)

Explanation:
1. Two moles of the same acid means the alkene is symmetrical
2. \( \ce{CH3CH2CH=CHCH2CH3} \) (hex-3-ene) is cleaved to two moles of propanoic acid

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) Dehydrohalogenation

Explanation:
1. Alcoholic KOH removes H and Br from adjacent carbons
2. This elimination is dehydrohalogenation

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) Cyclohexene

Explanation:
1. This is a Diels-Alder reaction, a [4+2] cycloaddition
2. The diene and dienophile give a six-membered ring with one double bond

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) A cyclic ozonide (1,2,4-trioxolane)

Explanation:
1. Ozone adds across the double bond to form a cyclic ozonide
2. Formaldehyde is obtained only after reductive work-up

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) Hydrogenation

Explanation:
1. Addition of \( \ce{H2} \) over a catalyst saturates the double bond
2. An alkene becomes an alkane

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: (c) \( \ce{C3H6} \)

Explanation:
1. A saturated cycloalkane has the formula \( \ce{CnH2n} \)
2. \( \ce{C3H6} \) is cyclopropane, while \( \ce{C3H8} \) is an open chain alkane

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: (c) Addition

Explanation:
1. Breaking the pi part of a C=C and forming two new sigma bonds is addition
2. The molecule becomes saturated

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) Both \( \ce{CH2D-CHCl-CH(CH3)2} \) and the rearranged \( \ce{CH2D-CH2-CCl(CH3)2} \)

Explanation:
1. D+ adds to the terminal carbon to give a secondary carbocation, which Cl- can capture directly
2. A hydride shift also converts it into the more stable tertiary cation, which gives the rearranged chloride
3. So both products form

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) But-2-ene

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

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: (a) \( \ce{R2C=CR2} \)

Explanation:
1. Alkyl groups stabilise a double bond by hyperconjugation
2. The tetrasubstituted alkene has the most alkyl groups and is most stable

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) Boiling point increases

Explanation:
1. Larger molecules have stronger van der Waals forces
2. So the boiling point and melting point rise and the vapour pressure falls

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: (b) Propene

Explanation:
1. Markovnikov's rule is needed when an unsymmetrical alkene adds an unsymmetrical reagent
2. Propene is unsymmetrical, whereas ethene and but-2-ene are symmetrical and ethane does not add

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) Alcoholic KOH

Explanation:
1. Alcoholic KOH removes HBr by dehydrohalogenation
2. Aqueous KOH would instead give ethanol by substitution

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).