Preparation of Alkyl Halides - NEET Chemistry Class 12 Free MCQ Test

The correct answer is (b) 3° > 2° > 1° bromides.

Explanation:

This reaction is the Hunsdiecker reaction, where the dry silver salt of a carboxylic acid is heated with bromine (Br₂) in CCl₄ to yield an alkyl bromide. The mechanism involves free-radical decarboxylation, and the yield of alkyl bromide depends on the stability of the intermediate alkyl radical.

Key Points:

1. Reactivity Order:
   The stability of the alkyl radical formed during decarboxylation follows the order:
   3° > 2° > 1°.

   • Tertiary (3°) radicals are most stable due to hyperconjugation and inductive effects.

   • Primary (1°) radicals are least stable.

2. Yield of Alkyl Bromide:
   Since the reaction proceeds via a radical intermediate, the yield of alkyl bromide correlates with radical stability:

   • Highest yield for 3° bromide (most stable radical).

   • Lowest yield for 1° bromide (least stable radical).

3. Mechanism:

   • The silver carboxylate (RCOOAg) reacts with Br₂ to form RCOOBr.

   • Homolytic cleavage produces R• (alkyl radical) and CO₂.

   • The alkyl radical then reacts with Br• to form R-Br.

Why Not Other Options?

• (a) & (c): Incorrect because 1° bromides do not form in higher yields than 2° or 3°.

• (d): Incorrect because 2° bromides form in higher yields than 1° (but lower than 3°).

Final Answer:

(b) 3° > 2° > 1° bromides (due to radical stability).

Additional Note:

The Hunsdiecker reaction is limited to stable radicals, so 3° and 2° substrates work best, while 1° often gives poor yields or side products.

Except F₂, almost all halogens react with RCOOAg giving alkyl halide via Hunsdiecker reaction.

With l₂ if RCOOAg is in excess, R— I formed in the first step reacts further with unreacted salt to give ester as
R—COOAg + R—I → R—COOR + AgI

If free radical halogenation generate a chiral carbon, racemic mixture of halides are always formed due to equal probability of halogenation from both sides of planar free radical.

The gaseous byproducts escape out on its own continuously driving the reaction in forward direction.

Nucleophilic attack in step-ll occur at the carbon atom which can better accommodate the positive charge. Hence, attack of Br^- or Cl^- in second step occur at 2° carbon rather that at 1° carbon.

Both SO₂CI₂ and Cl₂ undergo homolytic bond fission when heated or irradiated with light.

Answer: (b) The order of SN₂ reaction of the alkyl halide is RI > RBr > RCl > RF.

Explanation: Iodine is a good nucleophile and a good leaving group. Thus, it eliminates easily from an alkyl halide favouring SN₂ elimination reaction

NBS in CCI₄ brings about allylic brom ination by free radical mechanism:

In free radical halogenation of alkane, the first step of propagation is exothermic when Cl₂ is used while it is endothermic when Br₂ is used. Also chlorination occur at very fast rate, hence very less selective while bromination occur at very slow rate, occurs selectively where most stable free radical is formed.

As mentioned in mechanism, reaction proceed via carbocation intermediate. Hence, alcohol forming most stable carbocation reacts most easily.

• In cyclopropanol (a and c), removal of OH⁻ leads to a carbocation on a highly strained 3-membered ring. Carbocations on such strained small rings are generally less stable because the ring strain is high, and the carbocation cannot delocalize effectively.

• In cyclopentenol (b and d), removal of OH⁻ generates a carbocation on a 5-membered ring. This ring is less strained, and the carbocation can be stabilized through hyperconjugation and potentially some resonance with the ring pi system if any double bonds are present (especially in compound b, which seems to have a double bond in the ring).

Structure b) shows a double bond in the ring (cyclopentene), so the carbocation formed can be resonance stabilized (allylic carbocation).

Conclusion:

Alcohol (b) formsthe most stable carbocation, hence most reactive.

ZnCI₂ is a Lewis acid, helps in the form ation of carbocation intermediate, hence catalyse the reaction.

The order of stability of carbocation intermediates formed follows the order of reactivity.

Only three pairs of enantiomers are formed.

(I) has two optically active enantiomers and (II) has two geometrical isomers.

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S- configuration at C-2 would not change because the change will only takes place where the new bond is formed.

The presence of anhydrous ZnCl₂ is to be break the C-O bond in alcohols. ZnCl₂ is a Lewis acid and reacts with the oxygen of the alcohol group.