When alcohol reacts with HBr then substitution reaction takes place and the OH group is substituted by the Br atom. In the first step, the OH group is converted into an H₂O group, and a secondary carbocation is formed as a product. Then rearrangement of carbocation takes place and more stable tertiary carbocation is obtained. 
In the first step, tertiary carbocation reacts with bromide ion, and an alkyl halide is obtained as a product. 

The reaction with mechanism is as follows:

Thus, option second is the correct answer.

Explanation:

(1) “Lucas’ reagent” is a solution of anhydrous zinc chloride in concentrated hydrochloric acid. This solution is used to classify alcohols of low molecular weight. The reaction is a substitution in which the chloride replaces a hydroxyl group.

(2) Lucas reagent converts alcohols to alkyl chlorides: tertiary alcohols give an immediate reaction, indicated when the alcohol solution turns cloudy; secondary alcohols usually show evidence of reacting within five minutes; primary alcohols do not react to any significant extent. The order of reactivity of alcohols is: 

Among the given alcohols,  is a tertiary alcohol. So, it will react instantaneously with Lucas reagent. 

Thus, option 3 is the correct answer.

Regioselective

 Anti-Markovnikov 

(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

The phthalein dye test is commonly used to identify the presence of certain functional groups in organic compounds. This test uses phthalein dyes, which are specific for certain groups, especially phenolic compounds.

Phthalein Dye Test:

• Phenolphthalein and other phthalein dyes are often used in acid-base titrations. They undergo color changes depending on the pH of the solution.
• The phthalein dye test is specifically used to detect phenolic compounds. In the test:
  • If a phenolic compound is present, the phthalein dye will turn a pink color in an alkaline medium. This is because the hydroxyl group (-OH) in phenolic compounds reacts with the dye, causing the color change.

Explanation of the Options:

1. Phenolic: Correct. Phthalein dye (especially phenolphthalein) is commonly used to identify phenolic compounds. The color change in the test indicates the presence of a phenolic hydroxyl group (-OH).
2. Alcohol: Incorrect. While alcohols also contain hydroxyl groups, phthalein dye is more specific for phenols, not alcohols. Alcohols typically do not cause the same color change with phthalein dyes.
3. Aldehyde: Incorrect. The phthalein dye test is not used to identify aldehydes. Aldehydes are typically detected using other reagents, like Tollens’ reagent or Fehling’s reagent.
4. Carboxylic acid: Incorrect. Carboxylic acids can be detected using other tests (e.g., reacting with sodium bicarbonate to release CO₂), but phthalein dyes are not specific for carboxylic acids.

A. H₃C-CH₃OH (ethanol) → C₆H₆ (benzene):

• Reaction Name: Dehydration of ethanol (or Decarboxylation in some cases).
• This transformation involves the removal of water (dehydration) from ethanol to form ethene (C₂H₄), followed by further dehydration to form benzene (C₆H₆).
• Intermediate: Ethene (C₂H₄), which then undergoes a second step to form benzene.
• Reagent/Condition: The reaction is typically carried out with conc. H₂SO₄ (sulfuric acid) and heat (Δ).

Correct Match: A → IV (Conc. H₂SO₄, Δ)

B. CH₃COOH (acetic acid) → CH₃CH₂OH (ethanol):

• Reaction Name: Reduction of acetic acid to ethanol.
• The reduction of acetic acid (CH₃COOH) to ethanol (CH₃CH₂OH) typically involves the addition of hydrogen (H₂).
• Intermediate: Acetyl radical (CH₃CO·) or Hydride transfer.
• Reagent/Condition: This is typically achieved by the use of NaOH (base) and heat (Δ), which assists in the reduction.

Correct Match: B → II (NaOH, Δ)

C. CH₃CH₂OH (ethanol) → C₃H₇OH (propanol):

• Reaction Name: Hydrogenation of ethanol to propanol.
• This involves the hydrogenation of ethanol (C₂H₅OH) to form propanol (C₃H₇OH).
• Intermediate: Ethanol as the starting material, and hydrogen is added to the system.
• Reagent/Condition: The reaction is carried out using H₂ (hydrogen) and a catalyst.

Correct Match: C → III (H₂, catalyst)

D. C₆H₆OH (phenol) → C₆H₆ (benzene):

• Reaction Name: Dehydration of phenol.
• Phenol (C₆H₆OH) is converted to benzene (C₆H₆) by the elimination of water (dehydration).
• Intermediate: Phenoxide ion (C₆H₅O⁻), which is formed when phenol loses a proton.
• Reagent/Condition: Conc. H₂SO₄ (sulfuric acid) and heat.

Correct Match: D → I (Oleum)
 

Explanation: 

Statement I: Image formation needs regular reflection and/or refraction.
This statement is correct. Image formation relies on the interaction of light with surfaces. Here’s how:
Regular Reflection: This occurs when parallel rays of light strike a smooth surface and reflect back in a parallel manner. Mirrors are a prime example of regular reflection. This type of reflection allows for clear, focused images.
Refraction: This occurs when light passes from one medium to another (like from air to water or air to glass). The change in speed causes the light to bend, resulting in image formation. This is how lenses in cameras, telescopes, and our own eyes work.
Statement II: The variety in colour of objects we see around us is due to the constituent colours of the light incident on them.
This statement is also correct. The colors we perceive are based on the interaction between light and objects:
White Light: Sunlight or light from a bulb contains all the colors of the rainbow (the visible spectrum).
Selective Absorption: Objects absorb certain colors of light and reflect others. For instance, a red apple absorbs most colors except red, which it reflects back to our eyes.
Pigments: Many objects, especially those we paint or dye, contain pigments that absorb specific colors and reflect others.
Hence, option (3) is the correct answer.

Step: Find the time taken by the sunlight to pass through a glass slab.
The time taken by the sunlight to pass through a glass slab is given by; 
$t = \frac{\text{optical path length}}{\text{speed of light}}$
$\Rightarrow t = \frac{\mu \times L}{c}= \frac{1.5\times 5\times 10^{-3}}{3\times 10^{8}}= \left(\frac{5}{2}\right)\times 10^{-11}~\text{s}~~~[L = 5~\text{mm}]$
Hence, option (2) is the correct answer. 
 

Step: Find the correct variation of $\dfrac{1}{v}$ with $\dfrac{1}{u}$ for a concave mirror.
According to the mirror equation:
$\dfrac{1}{v}+\dfrac{1}{u}= \dfrac{1}{f}$
$\Rightarrow \dfrac{1}{v}= -\dfrac{1}{u}+\dfrac{1}{f}~~~...(1)~~~~[f = -ve,~\text{for concave mirror}]$
Equation $(1)$ represents a straight line having a negative slope and negative intercept. 
Therefore, the correct graph is shown below;
               
Hence, option (2) is the correct answer.