Learnosha CBSE 12
Chemistry Haloalkanes and Haloarenes 4 Marks
Appeared in: 2026 (3 sets)
Ethers are prepared by the dehydration of alcohols in presence of protic acids at $413\text{ K}$. Symmetrical and unsymmetrical ethers can also be prepared by Williamson synthesis. This reaction involves $\text{S}_\text{N}2$ attack of an alkoxide ion on primary alkyl halide. If a tertiary alkyl halide is used, elimination reaction occurs and an alkene is formed, no ether is formed. $\text{C} - \text{O}$ bond in ethers are cleaved under drastic conditions with excess of $\text{HI}$. When unsymmetrical ethers react with $\text{HI}$, the alkyl halide is formed from smaller alkyl group. If one of the alkyl group is tertiary, the alkyl halide is formed from the tertiary alkyl group because tertiary alkyl carbocation is more stable than the primary carbocation. Cleavage of alkyl aryl ethers takes place at the alkyl-oxygen bond due to more stable aryl-oxygen bond. The order of reactivity of hydrogen halides is $\text{HI} > \text{HBr} > \text{HCl}$. Aromatic ethers undergo electrophilic substitution reactions. The alkoxy group attached to the aromatic ring activates the ring towards electrophilic substitution and directs the incoming group to ortho- and para-positions.
✓ Verified CBSE Marking Scheme Official Evaluation Steps & EL10 Insight

CBSE Class 12 Chemistry: Marking Scheme Solution

Question: Predict the products of the following reactions:

  • (i) $CH_3-CH_2-O-CH_2-CH_3 + HI \xrightarrow{\Delta} ?$
  • (ii) $(CH_3)_3C-O-CH_3 + HI \xrightarrow{\Delta} ?$
  • (iii) $C_6H_5-O-CH_3 + HI \xrightarrow{\Delta} ?$
  • (iv) $CH_3-CH_2-CH_2-O-CH_3 + HBr \xrightarrow{\Delta} ?$

Step 1 (Governing Principles & Reaction Mechanism): [1 Mark]

The cleavage of ethers by hydrogen halides follows specific mechanistic pathways based on the nature of the alkyl groups:

  • For primary/secondary alkyl groups, the reaction follows the $S_N2$ mechanism, where the halide ion attacks the less sterically hindered alkyl group.
  • For tertiary alkyl groups, the reaction follows the $S_N1$ mechanism, forming a stable tertiary carbocation, resulting in the formation of a tertiary alkyl halide.
  • For alkyl aryl ethers, cleavage always occurs at the alkyl-oxygen bond because the aryl-oxygen bond has partial double bond character due to resonance.

Step 2 (Step-by-Step Reaction Analysis): [2 Marks]

(i) Symmetrical Ether: $CH_3CH_2-O-CH_2CH_3 + HI \rightarrow CH_3CH_2I + CH_3CH_2OH$

(ii) Tertiary Alkyl Group: $(CH_3)_3C-O-CH_3 + HI \rightarrow (CH_3)_3C-I + CH_3OH$ (Due to the stability of the tertiary carbocation).

(iii) Alkyl Aryl Ether: $C_6H_5-O-CH_3 + HI \rightarrow C_6H_5OH + CH_3I$ (Cleavage occurs at the $O-CH_3$ bond).

(iv) Unsymmetrical Ether: $CH_3CH_2CH_2-O-CH_3 + HBr \rightarrow CH_3CH_2CH_2OH + CH_3Br$ ($S_N2$ attack on the smaller methyl group).

Step 3 (Final Concluded Answer): [1 Mark]

The products are:

  1. $CH_3CH_2I$ (Iodoethane) and $CH_3CH_2OH$ (Ethanol)
  2. $(CH_3)_3CI$ (2-Iodo-2-methylpropane) and $CH_3OH$ (Methanol)
  3. $C_6H_5OH$ (Phenol) and $CH_3I$ (Iodomethane)
  4. $CH_3CH_2CH_2OH$ (Propan-1-ol) and $CH_3Br$ (Bromomethane)

💡 EL10 Concept Insight: Think of the halide ion as a hungry shopper; it prefers to attack the "smaller" or "easier" target (primary group) to avoid crowding, but if it sees a "super-stable" target (tertiary carbocation), it will always choose that one instead!

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