Given below are two statements:
Statement - I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow S_N ² mechanism.
Statement - II: A secondary alkyl halide when treated with a large excess of ethanol follows S_N ¹ mechanism.
In the light of the above statements, choose the most appropriate from the options given below:

Solution

Chemistry ma’am · 4 steps

For a secondary halide, the reagent decides: strong and concentrated means SN2, weak and protic means SN1.

1. Secondary halides. A secondary carbon is not too crowded for a backside attack, and its carbocation is fairly stable. So it can react either way, and the conditions choose the mechanism.

   Condition	Statement I	Statement II
   Nucleophile	strong, high concentration	ethanol, weak and neutral
   Crowding at carbon	no bulky groups, backside attack is easy	not the deciding factor
   Solvent role	nucleophile attacks in one step	polar protic solvent that also acts as nucleophile (solvolysis)
   Mechanism	S_N2	S_N1

2. Statement I. S_N2 rate = k[RX][Nu⁻]. A strong nucleophile at high concentration speeds up this one-step attack, and with no bulky groups nothing blocks the back side. So Statement I is true.
3. Statement II. Ethanol is a weak nucleophile and cannot push out the halide by itself. As a polar protic solvent it stabilises the carbocation and the leaving halide ion, so the halide ionises first and ethanol then attacks the carbocation. This is S_N1 solvolysis, so Statement II is true.
4. So. Both statements are true, which is option (a).

The idea to keep. Primary halides lean SN2, tertiary lean SN1. For secondary, read the reagent: a strong nucleophile at high concentration gives SN2, a weak one in a polar protic solvent gives SN1.

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