(A) `overset(HCI+ZnCI_(2))larr` `Me_(3)C-CH_(2)OH overset(SOCI_(2))underset(PCI_(5))underset((or))rarr` (B)
(A) and (B) are:

Options

A. A. `Me_(3)C-CH_(2)CI` B. `Me_(3)C-CH_(2)CI`
B. B. `Me-overset(Me)overset(|)underset(CI)underset(|)C-CH_(2)Me` B. `Me_(3)C-CH_(2)CI`
C. C. `Me_(3)C-CH_(2)CI` B. `Me-overset(Me)overset(|)underset(CI)underset(|)C-CH_(2)-Me`
D. D. `Me-overset(Me)overset(|)underset(CI)underset(|)C-CH_(2)Me` B. `Me-overset(Me)overset(|)underset(CI)underset(|)C-CH_(2)-Me`

Explanation

To solve the given question, we need to identify the compounds A and B based on the reactions provided. Let's break down the steps involved in the reactions.

### Step-by-Step Solution:

1. **Identify Compound A:**
- The starting compound is 3-methyl-2-butanol, which can be represented as Me₃C-CH₂OH (tert-butyl alcohol).
- When this compound reacts with HCl and ZnCl₂, the H⁺ from HCl will protonate the hydroxyl (-OH) group, leading to the formation of a positively charged oxygen (oxonium ion).
- The structure at this point is:
\[ \text{Me}_3C-CH_2OH \xrightarrow{HCl, ZnCl_2} \text{Me}_3C-CH_2^+O^H \]

2. **Formation of Carbocation:**
- The positively charged oxygen is unstable, so it will lose a water molecule (H₂O) to form a carbocation.
- The resulting structure will be a secondary carbocation:
\[ \text{Me}_3C-CH_2^+ \]

3. **Methyl Shift:**
- To stabilize the carbocation, a methyl shift occurs. One of the methyl groups from the tert-butyl group shifts to the carbocation center.
- This results in a more stable tertiary carbocation:
\[ \text{Me}_3C^+-CH_2 \quad \text{(after methyl shift)} \]

4. **Formation of Compound A:**
- The carbocation is then attacked by a chloride ion (Cl⁻) from the HCl, leading to the formation of compound A, which is 2-chloro-3-methylbutane:
\[ \text{Me}_3C-CH_2^+ \xrightarrow{Cl^-} \text{Me}_3C-CH_2Cl \]

5. **Identify Compound B:**
- Now, for compound B, we start again with 3-methyl-2-butanol (Me₃C-CH₂OH).
- When this compound reacts with SO₂Cl₂ or PCl₅, the lone pair of electrons from the oxygen atom attacks the sulfur atom, forming a sulfonium ion.
- The structure at this point becomes:
\[ \text{Me}_3C-CH_2OH \xrightarrow{SOCl_2} \text{Me}_3C-CH_2^+O-SO_2Cl \]

6. **Formation of Compound B:**
- In this reaction, the chloride ion (Cl⁻) attacks the carbon atom in a single step (SN2 mechanism), leading to the formation of compound B, which is 2-chloro-3-methylbutane:
\[ \text{Me}_3C-CH_2^+O-SO_2Cl \xrightarrow{Cl^-} \text{Me}_3C-CH_2Cl \]

### Final Compounds:
- **Compound A:** 2-chloro-3-methylbutane (Me₃C-CH₂Cl)
- **Compound B:** 2-chloro-3-methylbutane (Me₃C-CH₂Cl)

### Conclusion:
Both compounds A and B are the same, which is 2-chloro-3-methylbutane.