JEE Chemistry Hydrocarbons MCQs Set 17

Answer: (d) Polymerisation

Explanation:
1. Many \( \ce{CH2=CHCl} \) units join to form polyvinyl chloride
2. This is addition polymerisation

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) Both (a) and (b)

Explanation:
1. The cyclic bromonium ion can be opened by \( \ce{Br^-} \) or by \( \ce{Cl^-} \)
2. Both 1,2-dibromoethane and 1-bromo-2-chloroethane form

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) \( \ce{CH3-CH2-CH2Br} \)

Explanation:
1. Peroxide gives bromine radicals
2. Br adds to the terminal carbon to form the more stable secondary radical, giving 1-bromopropane (anti-Markovnikov)

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) Both an aldehyde and a ketone

Explanation:
1. The alkene is \( \ce{(CH3)2C=CH-CH3} \)
2. Cleavage gives acetone (a ketone) and ethanal (an aldehyde)

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) Both (a) and (b)

Explanation:
1. Each doubly bonded carbon becomes a carbonyl carbon
2. The CHR end gives an aldehyde and the CR2 end gives a ketone

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) \( \ce{CH3-CH=CH-CH3} \)

Explanation:
1. Only but-2-ene has a carbon-carbon double bond
2. The others are saturated

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) \( \ce{C2H4} \)

Explanation:
1. Dehydration of ethanol gives ethene
2. Ethyl acetate would need acetic acid, and ethane and ethyne are not formed

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) Olefin

Explanation:
1. Alkenes \( \ce{CnH2n} \) are called olefins
2. Alkanes are paraffins and alkynes are acetylenes

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) One sigma and one pi bond

Explanation:
1. A C=C consists of one sigma bond from sp2 overlap
2. and one pi bond from side-on overlap of p orbitals

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: (b) Elimination

Explanation:
1. The halide loses H and X from adjacent carbons
2. A double bond forms, which is elimination

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) Propene

Explanation:
1. \( \ce{PCl3} \) converts the alcohol into 1-chloropropane (A)
2. Alcoholic KOH removes HCl to give propene (B)

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) Propene

Explanation:
1. Dehydrohalogenation removes HBr
2. \( \ce{CH3CH2CH2Br -> CH3CH=CH2} \)

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: (b) 2-Methyl-2-butene

Explanation:
1. Neopentyl bromide has no beta hydrogen, so direct elimination is impossible
2. Ionisation and a methyl shift give the tertiary cation, which loses H+ to give the most stable alkene, 2-methylbut-2-ene

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: (b) Markovnikov's rule

Explanation:
1. The electrophile adds so that the more stable carbocation is formed
2. This is the modern statement of Markovnikov's rule

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) \( \ce{CH3-CHCl2} \)

Explanation:
1. The Cl on the double bond stabilises the adjacent cation by lone pair donation
2. So H adds to \( \ce{CH2} \) and Cl to the carbon already holding Cl, giving the gem-dichloride

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) Peroxides

Explanation:
1. In the presence of peroxides HBr adds by a free radical chain
2. The product is anti-Markovnikov

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) HBr

Explanation:
1. Only HBr has the right bond energies for both radical chain steps to be exothermic
2. So only HBr shows the peroxide effect

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) \( \ce{Cu2Cl2 + NH4Cl} \)

Explanation:
1. Cuprous chloride with ammonium chloride dimerises ethyne to vinylacetylene
2. Adding HCl to it gives chloroprene

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: (a) Synthetic rubber

Explanation:
1. Polymerisation of chloroprene gives neoprene
2. It is a synthetic rubber

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) \( \ce{(CH3)2CH-CH3} \) and \( \ce{CH3CH2OMgBr} \)

Explanation:
1. The Grignard reagent takes the acidic proton of the alcohol O-H
2. This gives isobutane (\( \ce{(CH3)2CHCH3} \)) and \( \ce{C2H5OMgBr} \)

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)

Explanation:
1. The chloronium ion forms, and water attacks the carbon that can better carry positive charge, the tertiary C-3
2. Cl ends up on the less substituted C-2, so the product is 2-chloro-3-methylpentan-3-ol

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) Addition

Explanation:
1. The electron rich pi bond is easily attacked
2. Addition is the typical reaction of ethylene

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) 80

Explanation:
1. Moles of propene = 21/42 = 0.5
2. One mole of \( \ce{Br2} \) adds to one mole of the alkene, so 0.5 mol \( \ce{Br2} \) = 0.5 x 160 = 80 g

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) Catalytic dehydration of a tertiary alcohol

Explanation:
1. Over hot copper, primary alcohols give aldehydes and secondary alcohols give ketones by dehydrogenation
2. Tertiary alcohols cannot be dehydrogenated and lose water to give alkenes

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: (d) 5 sigma, 1 pi

Explanation:
1. \( \ce{CH2=CH2} \) has four C-H sigma bonds and one C-C sigma bond
2. The double bond adds one pi bond

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.