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Hydrocarbons MCQs

Class 11 Chemistry — questions with answers and worked explanations.

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These questions are drawn from the Pariksha Sutra question bank for Hydrocarbons, part of the Class 11 Chemistry syllabus. Each one shows the correct answer and, where a method helps, the working behind it.

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Questions

Question 1
The IUPAC name of (CH3)3C-CH2-CH3 is:
  1. 3,3-dimethylbutane
  2. 2-methylpentane
  3. 2,2-dimethylbutane
  4. neopentane
Answer: 2,2-dimethylbutane
  1. The IUPAC name of (CH3)3C-CH2-CH3 is 2,2-dimethylbutane.
  2. Longest continuous carbon chain: 4 carbons (butane)
  3. Numbering: Numbering from the left gives the lowest locant numbers (2,2 vs. 3,3)
  4. Substituents: Two methyl groups attached at carbon-2 (2,2-dimethyl)
Question 2
In the free-radical chlorination of methane, the chain-initiation step is:
  1. CH4 + Cl. -> CH3. + HCl
  2. Cl2 -> 2 Cl.
  3. CH3. + Cl2 -> CH3Cl + Cl.
  4. CH3. + Cl. -> CH3Cl
Answer: Cl2 -> 2 Cl.
  1. Homolytic cleavage of a diatomic molecule by heat or light generates radicals that start the chain.
  2. In free‑radical chlorination, the first radicals must be chlorine atoms.
  3. Thus the initiation step is the homolysis of chlorine gas into two chlorine radicals.
  4. Cl2 -> 2 Cl.
Question 3
The IUPAC name of CH3-CH=CH-CH2-CH3 is:
  1. 2-Methylbut-2-ene
  2. Pent-3-ene
  3. Pent-2-ene
  4. Pent-1-ene
Answer: Pent-2-ene
  1. Identify the longest continuous carbon chain – it contains five carbons, so the parent hydrocarbon is a pentene.
  2. Number the chain to give the double bond the lowest possible locant; numbering from the left gives the C=C between C‑2 and C‑3, i.e., the double bond at carbon 2.
  3. Hence the IUPAC name is Pent-2-ene.
Question 4
According to Saytzeff's rule, dehydrohalogenation of 2-bromobutane mainly gives:
  1. Buta-1,3-diene
  2. 2-Methylpropene
  3. But-2-ene
  4. But-1-ene
Answer: But-2-ene
  1. Identify the β‑hydrogen atoms on C‑1 and C‑3 of 2‑bromobutane; removal with base can give a C1=C2 or C2=C3 double bond.
  2. The C2=C3 alkene (but‑2‑ene) is disubstituted, whereas the C1=C2 alkene (but‑1‑ene) is only monosubstituted.
  3. Saytzeff’s rule favors the more substituted alkene, so the major product is but‑2‑ene.
Question 5
Acid-catalysed hydration (H2O/H2SO4) of propene gives mainly:
  1. Propan-1-ol
  2. Propan-2-ol
  3. Propane
  4. Propanal
Answer: Propan-2-ol
  1. In acid‑catalysed hydration the double bond is protonated to give the more stable carbocation; for propene this is the secondary carbocation at C‑2. Water attacks this carbocation and after deprotonation yields the corresponding alcohol. Hence the product formed is Propan‑2‑ol.
Question 6
The degree of unsaturation (number of rings plus pi bonds) for a compound with molecular formula C6H12 is:
  1. 2
  2. 3
  3. 1
  4. 0
Answer: 1
  1. Calculate DU = (2C + 2 − H)/2 for a hydrocarbon.
  2. Substitute C = 6 and H = 12: (2·6 + 2 − 12)/2 = (12 + 2 − 12)/2 = 2/2.
  3. The degree of unsaturation equals 1, meaning one ring or one pi bond.
  4. Answer: 1
Question 7
Which reagent is used in the second dehydrohalogenation step to convert a haloalkene (vinyl halide) into an alkyne?
  1. Dilute HCl
  2. Sodium amide (NaNH2)
  3. Aqueous NaCl
  4. Water
Answer: Sodium amide (NaNH2)
  1. The first elimination with a base gives a vinyl halide (haloalkene).
  2. A second elimination requires a very strong, non‑nucleophilic base to remove the remaining β‑hydrogen from the sp² carbon.
  3. Sodium amide (NaNH₂) in liquid ammonia provides the needed strong base, effecting the second dehydrohalogenation to form the alkyne. Sodium amide (NaNH₂)
Question 8
Hydration of propyne (CH3-C(triple)CH) with dilute H2SO4/HgSO4 gives:
  1. Propanoic acid
  2. Propanal
  3. Propan-1-ol
  4. Propanone (acetone)
Answer: Propanone (acetone)
  1. Propyne undergoes electrophilic addition of H⁺ (from H₂SO₄) in the presence of Hg²⁺ to give the more substituted vinyl cation, which after water attack forms an enol (CH₃‑C(OH)=CH₂).
  2. The enol rapidly tautomerises to the corresponding carbonyl compound, shifting the double bond and a proton to give the more stable ketone.
  3. Thus the product of hydration of CH₃‑C≡CH is propanone (acetone).
Question 9
The monochlorination of methane (CH4 + Cl2 under UV light) proceeds by a mechanism that is:
  1. Free radical substitution
  2. Electrophilic addition
  3. Electrophilic substitution
  4. Nucleophilic substitution
Answer: Free radical substitution
  1. UV light causes homolytic cleavage of Cl₂ giving two Cl· radicals.
  2. The chlorine radical abstracts H from CH₄ forming CH₃· and HCl, propagating the radical chain.
  3. Both propagation steps involve radicals, characteristic of a free‑radical substitution mechanism.
  4. Free radical substitution
Question 10
An open-chain hydrocarbon with one carbon-carbon double bond and 5 carbons has the molecular formula:
  1. C5H8
  2. C5H12
  3. C5H14
  4. C5H10
Answer: C5H10
  1. Open‑chain hydrocarbon with one C=C double bond is an alkene, whose general formula is CₙH₂ₙ.
  2. For five carbon atoms, n = 5, so the hydrogen count is 2×5 = 10, giving C₅H₁₀.
  3. Thus the molecular formula is C5H10.
Question 11
Dehydrohalogenation of an alkyl halide to give an alkene requires:
  1. Na metal
  2. Aqueous KOH
  3. Alcoholic KOH
  4. Dilute H2SO4
Answer: Alcoholic KOH
  1. Strong base removes the β‑hydrogen while the leaving group departs in a concerted E2 step.
  2. A non‑aqueous medium prevents competing SN1/SN2 substitution, so KOH dissolved in ethanol supplies OH⁻ without water.
  3. Hence dehydrohalogenation of an alkyl halide to form an alkene requires alcoholic KOH.
Question 12
Markovnikov addition is governed by the formation of the more stable intermediate, which is a:
  1. free radical
  2. carbene
  3. carbocation
  4. carbanion
Answer: carbocation
  1. Proton adds to the double bond giving the more substituted carbon a positive charge.
  2. The intermediate formed is a positively charged carbon species.
  3. Stability increases with greater substitution, favoring the more substituted carbocation.
  4. Thus the governing intermediate in Markovnikov addition is a carbocation.
Question 13
At STP, what volume is occupied by 0.25 mol of ethene gas? (molar volume = 22.4 L/mol)
  1. 5.6 L
  2. 2.24 L
  3. 22.4 L
  4. 11.2 L
Answer: 5.6 L
  1. Moles of ethene given = 0.25 mol.
  2. Molar volume at STP = 22.4 L per mole.
  3. Volume = moles × molar volume = 0.25 × 22.4 L = 5.6 L.
  4. 5.6 L
Question 14
Which reaction converts a vicinal dihalide into an alkyne?
  1. Dehydrohalogenation with alcoholic KOH then NaNH2
  2. Hydrogenation with Pt
  3. Ozonolysis
  4. Halogenation with Br2
Answer: Dehydrohalogenation with alcoholic KOH then NaNH2
  1. Vicinal dihalides undergo successive elimination of HX; the first equivalent of alcoholic KOH removes one halogen and a β‑hydrogen to give a vinyl halide (alkene).
  2. A second elimination is required to remove the remaining HX; a strong base such as NaNH₂ abstracts the acidic alkyne hydrogen and eliminates the second halide, furnishing the alkyne.
  3. Thus the conversion is achieved by dehydrohalogenation with alcoholic KOH followed by NaNH₂.
Question 15
Hydration of ethyne with dilute H2SO4 and HgSO4 gives:
  1. Ethanol
  2. Ethanal (acetaldehyde)
  3. Ethene
  4. Acetic acid
Answer: Ethanal (acetaldehyde)
  1. Dilute H₂SO₄/HgSO₄ adds water across the C≡C bond of ethyne to form the enol CH₃CH=OH.
  2. The enol is unstable and rapidly undergoes keto‑enol tautomerism, converting the –C=OH group to a carbonyl.
  3. Thus the product is ethanal (acetaldehyde).
Question 16
All carbon-carbon bond lengths in benzene are equal (about 139 pm) because of:
  1. Hydrogen bonding
  2. sp3 hybridisation
  3. Presence of sigma bonds only
  4. Delocalisation of pi electrons (resonance)
Answer: Delocalisation of pi electrons (resonance)
  1. Each carbon in benzene is sp² hybridised, forming three σ‑bonds (two C–C and one C–H).
  2. The remaining unhybridised p‑orbitals overlap to give a continuous π‑electron cloud over the ring.
  3. This delocalisation makes every C–C bond have a bond order of 1.5, giving equal lengths of about 139 pm – Delocalisation of pi electrons (resonance).
Question 17
The IUPAC name of the compound CH3-CH(C2H5)-CH2-CH2-CH3 is:
  1. 3-methylhexane
  2. 3-ethylpentane
  3. 2-ethylpentane
  4. 2-methylhexane
Answer: 3-methylhexane
  1. Identify the longest continuous carbon chain: six carbons → hexane backbone.
  2. Number the chain from the end that gives the substituent the lowest possible locant; numbering from the left gives a methyl group on carbon 3.
  3. Thus the compound is named 3‑methylhexane.
Question 18
Controlled combustion/oxidation makes alkanes burn completely to give:
  1. CO and H2
  2. CO2 and H2O
  3. CO2 and H2
  4. C and H2O
Answer: CO2 and H2O
  1. Complete oxidation of an alkane requires enough O₂ to convert every C atom to CO₂ and every H atom to H₂O.
  2. For CₙH₂ₙ₊₂, the balanced equation is CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → n CO₂ + (n+1) H₂O.
  3. Since all carbon and hydrogen are fully oxidized, the products are carbon dioxide and water.
  4. CO₂ and H₂O.
Question 19
The IUPAC name of (CH3)2C=CH2 is:
  1. 2-Methylprop-2-ene
  2. But-2-ene
  3. But-1-ene
  4. 2-Methylprop-1-ene
Answer: 2-Methylprop-1-ene
  1. Identify the longest carbon chain containing the C=C bond; it has three carbons → propene backbone.
  2. Number the chain so the double bond gets the lowest possible locant; numbering from the end nearer the C=C gives the double bond at carbon 1.
  3. The remaining methyl group is attached to carbon 2, giving a 2‑methyl substituent.
  4. Hence the IUPAC name is 2-Methylprop-1-ene.
Question 20
Controlled monohalogenation aside, dehalogenation of a vicinal dihalide such as 1,2-dibromoethane with zinc dust gives:
  1. Ethanol
  2. Ethyne
  3. Ethene
  4. Ethane
Answer: Ethene
  1. Zn dust reduces the C–Br bonds, forming ZnBr₂ and generating a carbanion intermediate at each carbon.
  2. The adjacent carbanions eliminate the two bromide ions, forming a C=C double bond between the two carbons.
  3. Thus 1,2‑dibromoethane is converted to ethene.

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