Hint: In this reaction, both products are alkyl iodide.

Explanation:

The reaction is as follows:

When ether reacts with HI then alcohol and alkyl halide is formed as a product. In this reaction, HI is taken in excess,

Hence, the alcohol formed in the first step also reacts with HI and converts into alkyl halide.

Option 1 is the correct answer.

Hint: Hydrolysis of phenolic ether takes place

Explanation:

The reaction follows partially ${\mathrm{S}}_{\mathrm{N}}2$ reaction mechanism in which iodide ion attacks at the alkyl group with less steric hindrance and weak $C-O$ bond. The reaction is as follows:

 + HI $\to$+CH₃I

When ether reacts with HI then cleavage of the ether bond takes place and alcohol with alkyl halide is obtained as a product. 

The reaction follows partially S_N1 reaction. The reaction is as follows:

When ether reacts with HI then cleavage of the ether bond takes place. The mechanism of the given compound is as follows:

Here, phenol and benzyl iodide are formed as products. 

The correct answer is option fourth.

(1) Five-membered ring

1. Tetrahydrofuran (THF)

• Type: Structural
• Achiral
  
  Count = 1

(2) Four-membered ring (Oxetanes)

2–3. 2-Methyloxetane

• Type: Optical isomers
• (R) and (S) forms
  
  Count = 2

4–5. 3-Methyloxetane

• Type: Optical isomers
• (R) and (S) forms
  
  Count = 2

(3) Three-membered ring (Epoxides / Oxiranes)

6–7. 1,2-Epoxybutane (Butylene oxide)

• Type: Optical isomers
• (R) and (S) forms
  
  Count = 2

8. 2,2-Dimethyloxirane

• Type: Structural
• Achiral
  
  Count = 1

9–10. 2,3-Dimethyloxirane

• Type: Geometrical + optical
• cis (meso) = 1
• trans (enantiomeric pair) = 2
  
  Count = 3

Final total:

Total = 10