`Me-underset(Et)underset(|)(CH)-CH=CH-Phoverset(HCl)rarrW,W` is

Options

A. `Me-underset(Et)underset(|)(CH)-CH_(2)-underset(Cl)underset(|)(CH)-Ph`
B. `Me-underset(Et)underset(|)(CH)-underset(Cl)underset(|)(CH)-CH_(2)-Ph`
C. `Me-underset(Cl)underset(|)(CH)-underset(Et)underset(|)(CH)-underset(Ph)underset(|)(CH_(2))`
D. `Cl-underset(Et)underset(|)(CH)-underset(Me)underset(|)(CH)-CH_(2)-Ph`

Answer

To solve the question regarding the reaction of the compound with HCl, we can break it down step by step.

### Step-by-Step Solution:

1. **Identify the Structure**: The compound given is CH3-CH=CH-Ph (where Ph represents a phenyl group). This is an allylic system with a double bond between the second and third carbon atoms.

2. **Protonation of the Alkene**: When HCl is added to the reaction, the first step involves the protonation of the double bond. The double bond between the second and third carbon (C2=C3) will react with HCl. The H+ from HCl will add to one of the carbons of the double bond.

3. **Formation of Carbocation**: The addition of H+ leads to the formation of a carbocation. The most stable carbocation will form, which is typically at the more substituted carbon. In this case, the proton will add to the less substituted carbon (C2), resulting in a carbocation on C3.

4. **Resonance Stabilization**: The carbocation formed (C3) can be stabilized by resonance with the phenyl group (Ph). The positive charge can delocalize onto the phenyl ring, which stabilizes the carbocation.

5. **Nucleophilic Attack by Chloride Ion**: After the carbocation is formed and stabilized, the Cl- ion from HCl will attack the positively charged carbon (C3). This results in the formation of the final product.

6. **Final Product**: The final product will be CH3-CH(Cl)-CH2-Ph, where Cl is attached to the carbon that was initially part of the double bond.

### Final Answer:
The product W is CH3-CH(Cl)-CH2-Ph.

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