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Alkenes: geometrical isomerism and electrophilic additions: NEET MCQs with answers

61 questions on this topic of Hydrocarbons (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Given below are two statements: Statement I: Addition of HBr to but-1-ene in the presence of peroxides proceeds via a free radical mechanism and yields 1-bromobutane. Statement II: The peroxide effect is also observed when HCl or HI is added to but-1-ene. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct as the Kharasch effect operates via free radicals, giving anti-Markovnikov addition of HBr. Statement II is incorrect because the peroxide effect is not observed with HCl (the H-Cl bond is too strong to be broken homolytically by alkoxy radicals) or HI (iodine radicals combine to form I₂ instead of adding to the double bond).

medium

2. Arrange the following alkenes in order of decreasing rate of electrophilic addition of HBr: (I) Ethene (II) Propene (III) 2-Methylpropene (IV) 2,3-Dimethylbut-2-ene

  1. III > IV > II > I
  2. I > II > III > IV
  3. IV > III > II > I
  4. IV > II > III > I
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Answer: (C) IV > III > II > I

The rate of electrophilic addition depends directly on the electron density of the double bond and the stability of the intermediate carbocation formed during the rate-determining step. 2,3-Dimethylbut-2-ene forms a stable tertiary carbocation and has four electron-donating methyl groups (most reactive), followed by 2-methylpropene, propene, and ethene (least reactive).

hard

3. An alkene (X) having molecular formula C₅H₁₀ reacts with Br₂ in CCl₄ to form a vicinal dibromide (Y). Compound (X) exhibits geometrical isomerism. What is the IUPAC name of (X)?

  1. Pent-2-ene
  2. Pent-1-ene
  3. 2-Methylbut-2-ene
  4. 2-Methylbut-1-ene
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Answer: (A) Pent-2-ene

The acyclic alkene isomers of C₅H₁₀ are pent-1-ene, pent-2-ene, 2-methylbut-1-ene, 2-methylbut-2-ene, and 3-methylbut-1-ene. Among all these isomers, only pent-2-ene has two different groups attached to each sp² carbon (H and CH₃ on C2; H and C₂H₅ on C3) and can therefore exhibit cis-trans geometrical isomerism.

medium

4. An alkene (A) on ozonolysis gives acetone and propionaldehyde in equimolar amounts. What is the major product obtained when alkene (A) reacts with hydrogen chloride (HCl)?

  1. 3-Chloro-3-methylpentane
  2. 3-Chloro-2-methylpentane
  3. 2-Chloro-3-methylpentane
  4. 2-Chloro-2-methylpentane
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Answer: (D) 2-Chloro-2-methylpentane

Ozonolysis of (A) gives acetone ((CH₃)₂C=O) and propionaldehyde (CH₃CH₂CHO), identifying (A) as 2-methylpent-2-ene, (CH₃)₂C=CH-CH₂-CH₃. Addition of HCl follows Markovnikov's rule: H⁺ adds to C3 to generate the more stable tertiary carbocation at C2, which is attacked by Cl⁻ to yield 2-chloro-2-methylpentane.

easy

5. Given below are two statements: Statement I: Addition of HBr to propene in the presence of peroxides gives 1-bromopropane via a free radical mechanism. Statement II: Addition of HI to propene in the presence of peroxides also gives 1-iodopropane as the major product. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Peroxide effect (Kharasch effect) operates only with HBr. For HI, the H-I bond is weak, but iodine free radicals combine with each other to form I₂ rather than adding to the alkene double bond; hence, HI undergoes normal electrophilic addition following Markovnikov's rule to give 2-iodopropane.

hard

6. What is the major product obtained when 4-methylpent-1-ene is treated with hydrogen bromide (HBr) in the absence of peroxides?

  1. 1-Bromo-4-methylpentane
  2. 2-Bromo-2-methylpentane
  3. 3-Bromo-2-methylpentane
  4. 2-Bromo-4-methylpentane
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Answer: (B) 2-Bromo-2-methylpentane

Electrophilic addition of H⁺ gives a secondary carbocation (CH₃)₂CH-CH⁺-CH₃ at C-2. A 1,2-hydride shift from C-4 to C-3 produces a much more stable tertiary carbocation (CH₃)₂C⁺-CH₂-CH₃. Subsequent attack by Br⁻ gives 2-bromo-2-methylpentane.

easy

7. Which of the following compounds has the capability to show geometrical isomerism?

  1. 1,1-Dibromoprop-1-ene
  2. Pent-2-ene
  3. 2-Methylbut-2-ene
  4. 2,3-Dimethylbut-2-ene
Show answer and explanation

Answer: (B) Pent-2-ene

For an alkene to exhibit geometrical isomerism, each doubly bonded carbon atom must be attached to two different groups. In pent-2-ene (CH₃-CH=CH-CH₂-CH₃), C2 carries -H and -CH₃ while C3 carries -H and -CH₂CH₃, satisfying the condition. The others have at least one sp² carbon attached to two identical groups.

easy

8. Between cis-but-2-ene and trans-but-2-ene, trans-but-2-ene has a:

  1. Zero dipole moment and higher melting point
  2. Zero dipole moment and lower melting point
  3. Higher dipole moment and higher melting point
  4. Higher dipole moment and lower melting point
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Answer: (A) Zero dipole moment and higher melting point

Trans-but-2-ene is symmetrical, causing the opposing bond dipole moments to cancel out and giving it a dipole moment of zero. Its symmetrical structure allows it to pack more closely and efficiently into the crystal lattice, resulting in a higher melting point than the cis-isomer.

medium

9. The total number of geometrical isomers possible for 3,4-dimethylhex-3-ene is:

  1. 1
  2. 4
  3. 3
  4. 2
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Answer: (D) 2

In 3,4-dimethylhex-3-ene, each sp² carbon is bonded to a methyl group (-CH₃) and an ethyl group (-CH₂CH₃). Since both groups on each carbon are different, it exhibits geometrical isomerism, yielding exactly two isomers: cis (where both ethyl groups lie on the same side) and trans (where they lie on opposite sides).

medium

10. What is the major product formed when 3,3-dimethylbut-1-ene is treated with aqueous dilute sulfuric acid?

  1. 3,3-Dimethylbutan-2-ol
  2. 3,3-Dimethylbutan-1-ol
  3. 2,3-Dimethylbutan-1-ol
  4. 2,3-Dimethylbutan-2-ol
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Answer: (D) 2,3-Dimethylbutan-2-ol

Electrophilic addition of H⁺ forms a secondary carbocation [(CH₃)₃C-CH⁺-CH₃]. A 1,2-methide shift then takes place rapidly to give the more stable tertiary carbocation [(CH₃)₂C⁺-CH(CH₃)₂], which captures H₂O followed by loss of H⁺ to yield 2,3-dimethylbutan-2-ol.

hard

11. How many geometrical isomers are possible for octa-2,4,6-triene (CH₃-CH=CH-CH=CH-CH=CH-CH₃)?

  1. 8
  2. 3
  3. 4
  4. 6
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Answer: (D) 6

For a symmetrical polyene with n double bonds (n = 3, odd), the number of geometrical isomers is given by 2^(n-1) + 2^((n-1)/2). Here, n = 3, so number of isomers = 2^(3-1) + 2^((3-1)/2) = 2² + 2¹ = 4 + 2 = 6.

easy

12. Which of the following alkenes reacts fastest with cold, concentrated sulfuric acid?

  1. 2-Methylpropene
  2. Propene
  3. Ethene
  4. trans-But-2-ene
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Answer: (A) 2-Methylpropene

Electrophilic addition of H₂SO₄ involves the formation of a carbocation as the rate-determining step. 2-Methylpropene forms a stable tertiary carbocation, (CH₃)₃C⁺, which has the lowest activation energy among the choices. Hence, 2-methylpropene exhibits the highest rate of reaction.

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