JEE Chemistry Hydrocarbons MCQs Set 16

Answer: (b) Ethyl alcohol

Explanation:
1. Conc. \( \ce{H2SO4} \) at 170 degrees C removes water from ethanol to give ethene
2. Methanol has no adjacent carbon and cannot give an alkene

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: (c) Propene only

Explanation:
1. Both halides lose HCl by dehydrohalogenation
2. Each gives the same alkene, \( \ce{CH3CH=CH2} \)

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: (d) Ethylene glycol

Explanation:
1. Cold alkaline permanganate adds two OH groups across the double bond
2. Ethylene glycol is obtained

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) Ethene

Explanation:
1. Ethene is unsaturated and decolourises permanganate
2. It has no acidic hydrogen, so ammoniacal cuprous chloride gives no precipitate
3. Acetylene would give a red precipitate

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) Ethylene

Explanation:
1. \( \ce{CH2=CH2 + HOCl -> ClCH2CH2OH} \)
2. Ethylene chlorohydrin is the product

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) Polyethylene

Explanation:
1. Ethene molecules join in a polymerisation under heat and pressure
2. The product is polyethylene (polythene)

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) trans-2-Butene

Explanation:
1. Debromination is an anti elimination
2. Anti loss of the two Br atoms from the meso isomer places the two methyl groups on opposite sides, so trans-2-butene forms

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) 1-Butene

Explanation:
1. A primary halide can only lose a beta hydrogen from C-2
2. The alkene is \( \ce{CH2=CH-CH2CH3} \)

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: (d) 2-Pentene

Explanation:
1. Joining the carbonyl carbons through a double bond gives \( \ce{CH3CH=CHCH2CH3} \)
2. That is pent-2-ene

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) 1.35 Angstrom

Explanation:
1. C=C in ethene is about 1.34-1.35 Angstrom
2. Ethane is 1.54 and ethyne is 1.20

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) Propadiene

Explanation:
1. In \( \ce{CH2=C=CH2} \) the two end carbons are sp2 and the central carbon is sp
2. Propene has sp2 and sp3 carbons, and propyne has sp and sp3

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) Chlorine

Explanation:
1. Reactivity of halogens decreases down the group
2. Chlorine is the most reactive and iodine the least

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) 2-Chloropropane

Explanation:
1. The peroxide effect is not shown by HCl, because the chain step in which Cl radical adds is slow and the H abstraction from HCl is endothermic
2. So ordinary Markovnikov addition gives 2-chloropropane

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) 2-Iodopropane

Explanation:
1. HI does not show the peroxide effect because the H abstraction by I radical is endothermic and I radical adds poorly
2. So normal Markovnikov addition gives 2-iodopropane (isopropyl iodide)

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) \( \ce{C2H5-CHX-CH3} \)

Explanation:
1. Following Markovnikov's rule, H goes to the carbon with more hydrogens
2. X goes to the second carbon, giving 2-halobutane

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) Alcoholic KOH with heating

Explanation:
1. Elimination is favoured by a strong base in alcohol with heat
2. Aqueous alkali favours substitution to give alcohols

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) 2-Butene

Explanation:
1. Elimination follows Saytzeff's rule
2. The more substituted alkene but-2-ene predominates

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) \( \ce{CH3CH=CHCH2CH3} \)

Explanation:
1. Permanganate is decolourised by C=C bonds
2. Ethane and neopentane are saturated, and naphthalene is aromatic and resistant

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: (a) HCN

Explanation:
1. The Wacker process oxidises ethene to ethanal (\( \ce{CH3CHO} \))
2. HCN adds readily across the polar C=O of an aldehyde to give a cyanohydrin
3. The other reagents do not add readily to a saturated carbonyl

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) Its bond energy is high

Explanation:
1. The H-Cl bond is strong (about 431 kJ/mol)
2. Its homolysis and the H-abstraction step in the radical chain are unfavourable, so no 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: (d) All of these

Explanation:
1. \( \ce{CH2=CH-CH=CH2} \) has alternate double and single bonds, which makes it conjugated
2. It adds HBr and undergoes polymerisation to polybutadiene

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) \( \ce{BrCH2-CHBr-CH2-C#CH} \)

Explanation:
1. Electrophilic addition of bromine is faster for the C=C than for the C#C
2. The first mole of bromine is added to the double bond, leaving the alkyne unchanged

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) \( \ce{R-CH2-CH2-OH} \)

Explanation:
1. Hydroboration puts boron on the less substituted carbon
2. Oxidation replaces B by OH, so the primary alcohol is formed

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) An unsaturated bond

Explanation:
1. Cold dilute alkaline permanganate is decolourised by C=C and C#C bonds
2. So it is a test for unsaturation

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: (a) \( \ce{CH2Cl-CH=CH2} \)

Explanation:
1. At high temperature the radical substitution at the allylic hydrogen dominates over addition
2. Allyl chloride is the main product

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.