For the following reactions
CH₃CH₂CH₂Br + Z⊖ →(kₛ) CH₃CH₂CH₂Z + Br⊖; →(kₑ) CH₃CH=CH₂ + HZ + Br⊖; where,; Z⊖ = CH₃CH₂O⊖ (A) or H₃C-C(CH₃ |)-O⊖ (B)
Kₛ and Kₑ, are respectively, the rate constants for substitution and elimination and μ = kₛ/kₑ the correct options is

Solution

Chemistry ma’am · 4 steps

Small base substitutes, bulky base eliminates: tert-butoxide cannot reach carbon, so it grabs a hydrogen.

1. Two paths compete. On a primary bromide, the base can attack the carbon bearing Br, which is substitution (S_N2, kₛ). It can also pull off a β-hydrogen, which is elimination (E2, kₑ). The size of the base decides which path wins.
2. Ethoxide (A). CH₃CH₂O⁻ is small, so it reaches the back side of the CH₂Br carbon easily. Substitution dominates, so kₛ is large compared with kₑ, and μ_A = kₛ/kₑ is large.
3. Tert-butoxide (B). (CH₃)₃CO⁻ is very bulky and is also a stronger base. Its three methyl groups block backside attack, so kₛ falls. Plucking an exposed β-hydrogen is easy, so kₑ rises. That makes μ_B small and kₑ(B) > kₑ(A).

   Base	Size	Main path	μ = kₛ/kₑ
   CH₃CH₂O⁻ (A)	small	substitution	large
   (CH₃)₃CO⁻ (B)	bulky	elimination, larger kₑ	small

4. So. μ_A > μ_B and kₑ(B) > kₑ(A), which is option (b).

The idea to keep. With a primary halide, a small strong base like ethoxide gives mostly substitution. A bulky base like tert-butoxide cannot reach the carbon, so it switches the reaction to elimination.

Answer from the NTA final answer key. Solution by PrepWiser.