For the following reactions

where:

k$$_s$$ and k$$_e$$ are respectively the rate constants for substitution and elimination, and $$\mu = \frac{k_s}{k_e}$$, the correct option is

We need to compare the relative rates of substitution ($$k_s$$) and elimination ($$k_e$$) for a primary alkyl halide reacting with two different alkoxide bases: Ethoxide ion (A) and tert-Butoxide ion (B).

Core Chemical Principles:

• Ethoxide ion ($$\text{CH}_3\text{CH}_2\text{O}^\ominus$$ - A): A small, sterically unhindered nucleophile/base. It easily attacks the carbon center, making substitution ($$\text{S}_\text{N}2$$) the major pathway. Thus, $$k_s$$ is high and $$\mu_A = \frac{k_s}{k_e}$$ is large.
• tert-Butoxide ion ($$(\text{CH}_3)_3\text{C--O}^\ominus$$ - B): A bulky, highly sterically hindered base. The steric crowding prevents it from attacking the carbon center for substitution, forcing it to abstract a peripheral proton to undergo elimination ($$\text{E}2$$). Thus, elimination is strongly favored, meaning $$k_e(B) > k_e(A)$$ and $$\mu_B = \frac{k_s}{k_e}$$ is very small.

Comparison Results:

1. Since substitution is favored with the unhindered base A and elimination dominates with the bulky base B, we have: $$\mu_A > \mu_B$$
2. Because tert-butoxide is a stronger, bulkier base designed specifically to drive elimination pathways, its elimination rate constant is significantly higher: $$k_e(B) > k_e(A)$$

Conclusion:

Steric hindrance in the tert-butoxide base suppresses substitution while accelerating elimination compared to the unhindered ethoxide base.

Answer: Option B — $$\mu_A > \mu_B$$ and $$k_e(B) > k_e(A)$$