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Alcohols, Phenols, and Ethers — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryAlcohols, Phenols, and Ethers2 Marks
Question
Match the starting materials given in Column I with the products formed by these (Column II) in the reaction with II.
 
Column I
 
Column II
(i)
$CH_3-o-CH_3$
(a)
(ii)
(b)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{CH}_3-\text{C}-\text{I}+\text{CH}_3\text{OH} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
(iii)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{H}_3\text{C}-\text{C}-\text{O}-\text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
(c)
(iv)
(d)
$CH_3-OH + CH_3-I$
 
 
(e)
 
 
(f)
 
 
(g)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{CH}_3-\text{C}-\text{OH}+\text{CH}_3\text{I} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
✓

Answer

  Column I   Column I
(i) $CH3-O-CH3$ (d) $CH_3-OH + CH_3-I$
(ii) (e)
(iii) $\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{H}_3\text{C}-\text{C}-\text{O}-\text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$ (b) $\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{CH}_3-\text{C}-\text{I}+\text{CH}_3\text{OH} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
(iv) (a)
Explanation:
  1. $CH_3-O-CH_3$ is a symmetrical ether so the products are $CH_3I$ and $CH_2OH.$
  2. In $(CH_3)_2CH-O-CH_3$ unsymmetrical ether, one alkyl group is primary while another is secondary. So, it follows ${S_N}^2$ mechanism. Thus, the halide ion attacks the smaller alkyl group and the products are
  1. In this case, one of the alkyl group is tertiary and the other is primary. It follows ${S_N}^1$ mechanism and halide ion attacks the tertiary alkyl group and the products are $(CH_3)_3\ C-I$ and $CH_3OH.$
  2. Here, the unsymmetrical ether is alkyl aryl ether. In this ether $0-CH_3$ bond is weaker than $0-C_6H_5$ bond which has partial double bond character due to resonance. So, the halide ion attacks on alkyl group and the products are $C_6H_5-OH$ and $CH_3I.$

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