For the following reactions, ks and ke are respectively, the rate constants for substitution and elimination reactions…

Step 1: Identify the bases

Base A: $\ce{CH3CH2O-}$ (ethoxide) - small base

Base B: $\ce{(CH3)3C-O-}$ (tert-butoxide) - bulky base

Step 2: Understand substitution vs elimination

Small base (ethoxide):

• Can approach carbon easily
• Favors $S_N2$ substitution
• Some elimination also occurs
• $\mu_A = \frac{k_s}{k_e}$ is higher (more substitution)

Bulky base (tert-butoxide):

• Sterically hindered
• Cannot easily approach carbon for $S_N2$
• Acts as base instead → E2 elimination favored
• $\mu_B = \frac{k_s}{k_e}$ is lower (less substitution, more elimination)

Step 3: Compare μ values

$\mu = \frac{k_s}{k_e} = \frac{\text{substitution rate}}{\text{elimination rate}}$

• Higher μ: More substitution relative to elimination
• Lower μ: More elimination relative to substitution

Ethoxide (A): More substitution → $\mu_A$ is higher

tert-Butoxide (B): More elimination → $\mu_B$ is lower

Therefore: $\mu_A > \mu_B$ ✓

Step 4: Compare $k_e$ values

Elimination rate constant:

• Bulky bases are better at elimination
• tert-Butoxide is stronger base and bulkier
• $k_e$(B) > $k_e$(A) ✓

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

Key Principles:

Base size and selectivity:

| Base | Size | Favors | μ value | |------|------|--------|----------| | Small (OEt⁻, OH⁻) | Small | $S_N2$ | High | | Bulky (t-BuO⁻, LDA) | Large | E2 | Low |

Competition between $S_N2$ and E2:

• Same substrate can undergo both
• Base size determines selectivity
• Temperature also affects (heat favors elimination)
• Substrate structure matters (1° favors $S_N2$, 3° favors E2)