The rate law for the subsitution reaction of 2-bromobutane and OH¯ in 75% ethanol and 25% H2O at 25°C is What % of the reaction take place by the SN2 mechanism when [OH¯] = 1.00 M If answer is x% then fill \frac{x}{10}

Engineering

Chemistry

Nucleophilic Substitution

Question

The rate law for the subsitution reaction of 2-bromobutane and OH¯ in 75% ethanol and 25% H₂O at 25°C is

What % of the reaction take place by the S_N2 mechanism when [OH¯] = 1.00 M
If answer is x% then fill $\frac{x}{10}$

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The rate law is: rate = k[2-bromobutane] + k'[2-bromobutane][OH⁻]. This shows the reaction proceeds via two simultaneous mechanisms: S_N1 (first term) and S_N2 (second term).

To find the percentage by S_N2, we calculate the fraction of the total rate due to the S_N2 path. At [OH⁻] = 1.00 M, the S_N2 rate is k'[RBr][1.00] = k'[RBr]. The S_N1 rate is k[RBr]. The total rate is (k + k')[RBr].

The fraction is: $\frac{k^{'} [RBr]}{(k + k^{'}) [RBr]} = \frac{k^{'}}{k + k^{'}}$

From the rate law, k = 0.0082 s⁻¹ and k' = 0.017 M⁻¹s⁻¹. Substituting the values:

$\frac{0.017}{0.0082 + 0.017} = \frac{0.017}{0.0252} \approx 0.6746$

Therefore, the percentage is 0.6746 × 100 = 67.46%. The final answer is 67.46/10 = 6.746.

Final Answer: $\frac{x}{10} = 6.746$