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Question 15 Marks
An aromatic compound $'A\ ’$ on treatment with $\ce{CHCl_3/ KOH}$ gives two compounds $'B\ ’$ and $'C\ ’.$ Both $B$ and $C$ give the same product $'D\ ’$ when distilled with zinc dust. Oxidation of $D$ gives $E$ having molecular formula $\ce{C_7H_6O_2}.$ The sodium salt of $E$ on heating with sodalime gives $F$ which may also be obtained by distilling $A$ with zinc dust. Identify $A$ to $F.$
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Question 25 Marks
Answer the following question: Write mechanism of the reaction of $HI$ with methoxymethane.
Answer
The cleavage of methoxymethane with $HI$ follows $SN_2$ mechanism.
Step $I:$

Step $II:$​​​​​​​

If excess of $HI$ is used, methyl alcohol formed in step $II$ further reacts with another molecule of $HI$ to form $\ce{CH_3I}.$
Step $III:$​​​​​​​
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Question 45 Marks

Identify A and B in the following reactions:
  1.  
  1. $\text{C}_2\text{H}_5\text{OH}\xrightarrow{\text{Cu},573\text{k}}\text{A}\xrightarrow{\text{CH}_3\text{MgBr}}\text{B}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}_2\text{O}/\text{H}^{+}$
Answer
  1.  
  1.  
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Question 55 Marks
  1. The acid strength of alcohols decreases in the order
  1. Phenol is more easily nitrated than benzene.
Answer
  1. The acidic character of alcohols is due to the polar nature of $O-H$ bond. As alkyl groups have $+I$ effect, therefore, electron density on oxygen increases from $1$ to $3^\circ$ alcohols. In other words, the $O-H$ bond in $3^\circ$ alcohols is most polar whereas it is least polar in $1^\circ$ alcohols. Thus, the acid strength of alcohols follows the order
  1. Nitration involves attack of electrophile nitronium $\ce{(NO_2)}$ ion on benzene ring. Due to $+R$ effect of $O-H$ group electron density increases at ortho and para position in phenol. Therefore, phenol is more easily nitrated than benzene.
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