**60.** Given below are two statements: **Statement I**: CH₃–O–CH₂–Cl will undergo Sₙ1 reaction though it is a primary halide. **Statement II**: CH₃ | CH₃–C–CH₂–Cl | CH₃ will not undergo Sₙ2 reaction very easily though it is a primary halide. In the light of the above statements, choose the **most appropriate answer** from the options given below: - (1) **Statement I** is incorrect but **Statement II** is correct. - (2) Both **Statement I** and **Statement II** are incorrect - (3) **Statement I** is correct but **Statement II** is incorrect - (4) Both **Statement I** and **Statement II** are correct

Key concepts you must know

1. SN1 reaction requirements

   • Formation of a stable carbocation (positively charged carbon) after the leaving group departs.
   • Stabilisation can happen through resonance or by nearby atoms that can donate electron density.

2. SN2 reaction requirements

   • Back-side attack by the nucleophile occurs in one concerted step.
   • Reaction rate drops sharply if bulky groups crowd the attack site or its neighbourhood.

3. Neighbouring group/alpha-hetero effect

   • If the carbon bearing Cl is next to an electronegative atom like O, resonance can spread the positive charge over C and O, stabilising the carbocation formed in SN1.

Applying the theory to each statement

Statement I: $CH_3\! -\! O\! -\! CH_2\! -\! Cl$

• When Cl leaves, we get a primary carbocation $CH_3\! -\! O\! -\! CH_2^+$.
• Oxygen has lone pairs; it can donate electron density and form the oxonium resonance form $CH_3\! -\! O^+\!=\!CH_2$ which stabilises the positive charge.
• Because the cation is stabilised, SN1 is feasible even though the carbon is primary. ✔️

Statement II: $(CH_3)_3C\! -\! CH_2\! -\! Cl$

• The Cl is on a primary carbon, but right next door is a huge tert-butyl group $(CH_3)_3C\! -$.
• For SN2, a nucleophile must attack from the back of the C–Cl bond. The bulky tert-butyl group creates severe steric hindrance, making that backside approach extremely difficult.
• Therefore, the molecule reacts via SN2 very slowly, if at all. ✔️

What is the question asking?

We have two funny-looking molecules with a chlorine (Cl) atom attached. The question asks whether they like to do two different kinds of ‘swap’ reactions (called SN1 and SN2) even though each has the Cl on a primary carbon (a carbon attached to only one other carbon).

• SN1 is like waiting in line at the canteen: the Cl leaves first, the carbon becomes lonely (forms a positive charge), and then somebody else comes to keep it company. This works best if friends nearby can stabilise the lonely carbon.

• SN2 is like a fast karate-chop: a new friend attacks from the back at the same time the Cl leaves. It works only if there is enough space at the back (little crowding).

The two statements in the problem argue:

1. The first molecule will happily do the canteen line (SN1) even though it’s primary.
2. The second molecule does not like the karate-chop (SN2) because big bulky groups block the way.

The task: decide which statement(s) are right.

Step-by-step answer

1. Check Statement I

   • Leaving group departure: $CH_3\! -\! O\! -\! CH_2\! -\! Cl \xrightarrow{\text{Cl}^-\,\text{leaves}} CH_3\! -\! O\! -\! CH_2^+$
   • Resonance with oxygen: $\leftrightarrow CH_3\! -\! O^+\!=\!CH_2$
   • Carbocation is stabilised (\Rightarrow) SN1 favourable. Statement I is correct.

2. Check Statement II

   • For SN2 the nucleophile must approach the C bearing Cl from the backside.
   • The adjacent $(CH_3)_3C\! -$ group is bulky, causing strong steric hindrance.
   • Therefore reaction via SN2 is strongly disfavoured. Statement II is correct.

3. Selecting the option

   • Both statements are correct ⟹ Option (4).