The correct order of `S_(N)2//E2` ratio for the % yield of product of the following halide is,
P.`CH_(3)-CH_(2)-underset(I) underset(|) overset (Ph) overset(|) (C)-CH_(3)`
Q. `CH_(3)-underset (Ph) underset(|) (CH)-underset(I) underset(|) (CH)-CH_(3)`
R. `CH_(3)-CH_(2)-I`
S. `CH_(3)-CH-underset(I) underset(|) (CH)-CH_(3)`

Options

A. `RgtSgtQgtP`
B. `RgtQgtSgtP`
C. `PgtRgtSgtQ`
D. `QgtPgtRgtS`

Explanation

To determine the correct order of the `S_N2` to `E2` ratio for the percentage yield of products from the given halides, we will analyze each compound based on steric hindrance and the nature of the reactions involved.

### Step-by-Step Solution:

1. **Identify the Structure of Each Compound:**
- **P:** `CH₃-CH₂-CH(Ph)-C(CH₃)₂` (3° carbon with iodine)
- **Q:** `CH₃-CH(Ph)-CH(CH₃)-CH₃` (2° carbon with iodine)
- **R:** `CH₃-CH₂-I` (1° carbon with iodine)
- **S:** `CH₃-CH(I)-CH(CH₃)-CH₃` (2° carbon with iodine)

2. **Determine the Reactivity for `S_N2`:**
- The `S_N2` reaction is favored at less sterically hindered (1° > 2° > 3°) centers.
- **R** has a 1° carbon, thus it will have the highest `S_N2` reactivity.
- **Q** and **S** both have 2° carbons, but **S** has less steric hindrance due to the absence of a bulky group (phenyl) compared to **Q**.
- **P** has a 3° carbon, which is the least favorable for `S_N2`.

Therefore, the order of `S_N2` reactivity is:
- R > S > Q > P

3. **Determine the Reactivity for `E2`:**
- The `E2` reaction favors more substituted alkenes and stable products (due to hyperconjugation and resonance).
- **P** will favor `E2` because the product formed can be stabilized by resonance with the phenyl group.
- **R** will have the least `E2` reactivity because it forms a less stable alkene.
- **Q** and **S** will have moderate `E2` reactivity, but **Q** is less favored due to steric hindrance from the phenyl group.

Therefore, the order of `E2` reactivity is:
- P > Q > S > R

4. **Establish the Ratio of `S_N2` to `E2`:**
- Since `S_N2` is favored at 1° and less hindered 2° centers, while `E2` is favored at more substituted centers:
- The ratio of `S_N2` to `E2` will be highest for R, followed by S, then Q, and finally P.

Thus, the final order of `S_N2/E2` ratio is:
- R > S > Q > P

### Conclusion:
The correct order of `S_N2/E2` ratio for the percentage yield of the products is:
**R > S > Q > P**