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Preparation, properties, reactions and nature of carbon-halogen bond: NEET MCQs with answers

49 questions on this topic of Organic Compounds Containing Halogens (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Arrange the following leaving groups in increasing order of their leaving ability in nucleophilic substitution reactions of alkyl derivatives: (I) F⁻ (II) Cl⁻ (III) Br⁻ (IV) I⁻

  1. I < II < III < IV
  2. IV < III < II < I
  3. I < III < II < IV
  4. II < I < III < IV
Show answer and explanation

Answer: (A) I < II < III < IV

Leaving group ability is inversely related to base strength. As we go down group 17, the size of the halide ion increases and the C-X bond strength decreases, making I⁻ the weakest conjugate base and the best leaving group. Hence, the order is F⁻ < Cl⁻ < Br⁻ < I⁻.

medium

2. Identify product 'B' in the given sequence of reactions: CH₃-CH=CH₂ + Cl₂ (500 °C) → A A + H₂O (NaOH, Δ) → B

  1. CH₃-CH(OH)-CH₂Cl
  2. CH₃-CH(OH)-CH₃
  3. CH₂=CH-CH₂OH
  4. CH₂(OH)-CH(OH)-CH₃
Show answer and explanation

Answer: (C) CH₂=CH-CH₂OH

At high temperature (500 °C), propene reacts with chlorine via allylic chlorination to yield allyl chloride (CH₂=CH-CH₂Cl) as product A. Alkaline hydrolysis of allyl chloride with aqueous NaOH substitutes the chloride with a hydroxyl group to yield allyl alcohol (CH₂=CH-CH₂OH) as product B.

hard

3. Which of the following compounds reacts fastest with silver nitrate solution (AgNO₃/EtOH) to give a precipitate of AgX?

  1. Allyl chloride
  2. Chlorobenzene
  3. Vinyl chloride
  4. 2-Chloropropane
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Answer: (A) Allyl chloride

Reaction with AgNO₃ involves the formation of a carbocation in an SN1-like process where Ag⁺ helps in the departure of the halide ion. Allyl chloride ionizes readily because the resulting allyl carbocation (CH₂=CH-CH₂⁺) is resonance-stabilized. In vinyl chloride and chlorobenzene, the C-Cl bond possesses partial double bond character due to resonance and cannot ionize, while 2-chloropropane forms a less stable secondary carbocation.

medium

4. Which of the following alkyl halides undergoes dehydrohalogenation with alcoholic KOH at the highest rate?

  1. CH₃-CH₂-CH₂-Br
  2. (CH₃)₂CH-Br
  3. CH₃-CH₂-Br
  4. (CH₃)₃C-Br
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Answer: (D) (CH₃)₃C-Br

Dehydrohalogenation of alkyl halides by base involves an E2 mechanism where the transition state has double bond character. The stability of the resulting alkene determines the rate: tertiary alkyl halides give more substituted, more stable alkenes, and their transition states are more stable. The rate order for dehydrohalogenation is 3° > 2° > 1°.

easy

5. Which among the following chlorohydrocarbons has the shortest carbon-chlorine bond length?

  1. CH₂=CH-Cl
  2. CH₃-CH₂-Cl
  3. CH₂=CH-CH₂-Cl
  4. (CH₃)₃C-Cl
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Answer: (A) CH₂=CH-Cl

In vinyl chloride (CH₂=CH-Cl), the unshared pair of electrons on chlorine is conjugated with the carbon-carbon double bond, giving the C-Cl bond partial double-bond character. Additionally, the carbon attached to chlorine is sp² hybridised, making the bond shorter than sp³ C-Cl bonds.

hard

6. Given below are two statements: Statement I: The dipole moment of chlorobenzene (1.69 D) is lower than that of cyclohexyl chloride (1.86 D). Statement II: The C-Cl bond in chlorobenzene possesses partial double bond character due to resonance and the sp² hybridized carbon of the ring is more electronegative than the sp³ carbon of cyclohexane.

  1. Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
  2. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
  3. Statement I is correct but Statement II is incorrect
  4. Statement I is incorrect but Statement II is correct
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Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Both statements are correct. In chlorobenzene, delocalization of chlorine's lone pair into the ring introduces partial double bond character, shortening the bond and opposing the inductive dipole. Additionally, the sp² hybridized carbon is more electronegative than an sp³ carbon, reducing the bond polarity and overall dipole moment compared to cyclohexyl chloride.

easy

7. When chloroform is exposed to air and sunlight, it forms a poisonous gas (phosgene). To prevent this oxidation, which substance is added to chloroform?

  1. 1% Phenol
  2. 1% Ethanol
  3. 1% Acetic acid
  4. 1% Methanol
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Answer: (B) 1% Ethanol

Chloroform is stored in closed dark brown bottles completely filled to the brim. About 1% ethanol is added to act as a negative catalyst to suppress the oxidation and to convert any traces of toxic phosgene (COCl₂) formed into non-toxic diethyl carbonate.

easy

8. Which of the following reagents is best suited for the conversion of primary alcohols into alkyl chlorides in highest purity without requiring fractional distillation to separate side products?

  1. Thionyl chloride (SOCl₂) in presence of pyridine
  2. Phosphorus pentachloride (PCl₅)
  3. Phosphorus trichloride (PCl₃)
  4. Concentrated HCl in the presence of anhydrous ZnCl₂
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Answer: (A) Thionyl chloride (SOCl₂) in presence of pyridine

Reaction with thionyl chloride (SOCl₂) gives alkyl chloride along with SO₂ and HCl, both of which are gaseous and escape directly from the reaction mixture, leaving behind the pure alkyl halide.

medium

9. The reaction R-Cl + NaI → R-I + NaCl in dry acetone is called the:

  1. Swarts reaction
  2. Wurtz reaction
  3. Sandmeyer reaction
  4. Finkelstein reaction
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Answer: (D) Finkelstein reaction

In the Finkelstein reaction, NaCl (or NaBr) formed is insoluble in dry acetone and precipitates, driving the halogen exchange forward.

NCERT Class 12

easy

10. Match List-I with List-II to identify the correct synthetic transformation: List-I (A) 2 C₆H₅Cl + 2 Na / dry ether → Biphenyl (B) C₆H₅Cl + CH₃Cl + 2 Na / dry ether → Toluene (C) 2 CH₃CH₂Br + 2 Na / dry ether → Butane (D) C₆H₅N₂⁺Cl⁻ + Cu₂Br₂ / HBr → C₆H₅Br List-II (I) Wurtz reaction (II) Sandmeyer reaction (III) Fittig reaction (IV) Wurtz-Fittig reaction

  1. A-I, B-IV, C-III, D-II
  2. A-IV, B-III, C-I, D-II
  3. A-III, B-I, C-IV, D-II
  4. A-III, B-IV, C-I, D-II
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Answer: (D) A-III, B-IV, C-I, D-II

The coupling of two aryl halides with sodium is the Fittig reaction (A-III). The coupling of one aryl halide and one alkyl halide is the Wurtz-Fittig reaction (B-IV). The coupling of two alkyl halides is the Wurtz reaction (C-I). Replacement of a diazonium group by bromine using Cu₂Br₂ is the Sandmeyer reaction (D-II).

medium

11. For the same alkyl group, the boiling points of alkyl halides follow the order:

  1. RI > RBr > RCl > RF
  2. RF > RCl > RBr > RI
  3. RCl > RBr > RI > RF
  4. All are equal
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Answer: (A) RI > RBr > RCl > RF

Boiling point increases with molar mass and van der Waals forces, so iodides boil highest and fluorides lowest.

NCERT Class 12

medium

12. Given below are two statements: Statement I: Optically active 2-bromooctane undergoes complete inversion of configuration during an SN2 reaction with hydroxide ion. Statement II: An SN2 mechanism involves a carbocation intermediate where the nucleophile can attack equally from both faces.

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

Statement I is correct because an SN2 reaction proceeds via backside attack leading to Walden inversion of configuration. Statement II is incorrect because SN2 proceeds via a single-step concerted transition state without forming any carbocation intermediate.

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