Q.Which compound in each of the following pairs will react faster in $S_N2$ reaction with $^-OH$?

Concept: $S_N2$ reactivity here is governed by leaving group ability (pair i) and steric hindrance (pair ii) — not by any ambident-nucleophile behaviour. $^-OH$ is not an ambident nucleophile: it has only one nucleophilic atom (oxygen). Ambident nucleophiles like $CN^-$ or $NO_2^-$ have two different donor atoms and can give two different products; that isn't relevant here.

Reasoning:

1. In $S_N2$, the rate depends on how easily the leaving group departs. Iodide ($I^-$) is a better leaving group than bromide ($Br^-$) because the C–I bond is weaker and $I^-$ is more stable (larger, more polarizable). So $CH_3I$ reacts faster than $CH_3Br$. …

The Core Idea: What Makes an $S_N2$ Reaction Fast?

An $S_N2$ reaction is a single-step, bimolecular substitution. The nucleophile ($^-OH$ here) attacks the carbon from the side opposite the leaving group. This means two things matter enormously:

1. The leaving group must be able to depart easily. A good leaving group stabilises the negative charge it carries after leaving. In the halogens, this ability increases down the group: $I^- > Br^- > Cl^- > F^-$.
2. The carbon centre must be accessible. The nucleophile needs a clear path to the back of the carbon. Any bulky groups near that carbon physically block the attack — this is steric hindrance.

Let's apply these two principles to each pair.

Pair (i): $CH_3Br$ vs $CH_3I$

Both are primary alkyl halides with no branching at the reacting carbon. So steric hindrance is identical — the only difference is the leaving group.

Step 1: Compare leaving group ability.

The leaving group departs as a halide ion ($Br^-$ or $I^-$). The better the leaving group, the lower the activation energy for the $S_N2$ step.

Iodide ($I^-$) is a much better leaving group than bromide ($Br^-$). Why? Iodine is larger and more polarisable — its negative charge is spread over a bigger volume, making it more stable in solution. Also, the $C-I$ bond is weaker than the $C-Br$ bond, so it breaks more easily.

Step 2: Apply the rate effect.

Since the nucleophile and the carbon skeleton are identical, the reaction with the better leaving group will be faster.

Result for (i): $CH_3I$ reacts faster than $CH_3Br$.

Pair (ii): $(CH_3)_3CCl$ vs $CH_3Cl$

Here, the leaving group is the same (chloride) in both, but the carbon skeleton is drastically different.

Step 1: Examine the carbon centre.

• $CH_3Cl$ is methyl chloride — the carbon is attached to three hydrogens and one chlorine. There is almost no steric bulk around the backside.
• $(CH_3)_3CCl$ is tert-butyl chloride — the carbon is attached to three methyl groups and one chlorine. Those three methyl groups are large and stick out in all directions.

Step 2: Visualise the backside attack. …

Method: Steric Hindrance & Leaving Group Ability in $S_N2$

Concept-first understanding:

In $S_N2$ reactions, the nucleophile attacks from the backside of the carbon–leaving group bond. Two factors dominate the rate:

1. Leaving group ability – better leaving groups (weaker bases, more polarizable) leave faster.
2. Steric hindrance – bulky groups around the reaction centre block the backside attack, slowing the reaction.

(i) $CH_3Br$ vs $CH_3I$

Method: Compare leaving group ability (basicity & polarizability).

Steps:

1. Identify the leaving groups:

   • $Br^-$ (bromide)
   • $I^-$ (iodide)

2. Recall the trend:

   • Better leaving groups are weaker bases and more polarizable.
   • Basicity order: $F^- > Cl^- > Br^- > I^-$ (least basic = best leaving group).
   • Polarizability increases down the group: $I^-$ is largest and most polarizable.

3. Apply to the pair:

   • $I^-$ is a better leaving group than $Br^-$.
   • Both substrates are methyl halides (no steric difference).

Result:

$CH_3I$ reacts faster than $CH_3Br$ with $^-OH$.

(ii) $(CH_3)_3CCl$ vs $CH_3Cl$

Method: Compare steric hindrance around the reaction centre.

Steps:

1. Identify the substrate type:

   • $(CH_3)_3CCl$ = tertiary alkyl halide (3 bulky methyl groups).
   • $CH_3Cl$ = methyl halide (no bulky groups).

2. Recall the $S_N2$ steric requirement:

   • The nucleophile must approach the backside of the carbon. …

Common Mistakes: Ambident Nucleophile Reactivity & $S_N2$ Reaction Rates

Students often confuse nucleophile strength with leaving group ability when comparing $S_N2$ rates. Here are the most frequent errors and how to avoid them.

Mistake 1: Confusing Leaving Group Ability with Nucleophilicity

The error: Thinking that a stronger nucleophile (like $^-OH$) always reacts faster with a better nucleophile (like $CH_3I$ vs $CH_3Br$) — but the question is about the substrate, not the nucleophile.

Why it’s wrong: In $S_N2$, the rate depends on leaving group ability, not on how good the nucleophile is at attacking itself. The nucleophile ($^-OH$) is the same in both comparisons.

How to avoid: Always identify what is changing — here, it’s the halide leaving group (Br vs I) or the alkyl group (tertiary vs primary). The nucleophile is fixed.

Mistake 2: Forgetting the Leaving Group Trend in $S_N2$

The error: Saying $CH_3Br$ reacts faster than $CH_3I$ because Br is smaller or more electronegative.

Why it’s wrong: In $S_N2$, the better leaving group is the one that can stabilize the negative charge after departure. Iodide ($I^-$) is larger, more polarizable, and a weaker base than bromide ($Br^-$), so it leaves more easily.

Correct reasoning:

• Leaving group ability: $I^- > Br^- > Cl^- > F^-$
• Therefore, $CH_3I$ reacts faster than $CH_3Br$ with $^-OH$.

How to avoid: Memorize the leaving group trend: larger, weaker base = better leaving group. Use periodic trends: down the group, leaving ability increases.

Mistake 3: Ignoring Steric Hindrance in $S_N2$

The error: Thinking $(CH_3)_3CCl$ reacts faster because it has more alkyl groups (electron-donating) that stabilize the transition state.

Why it’s wrong: $S_N2$ is extremely sensitive to steric hindrance. The nucleophile must attack from the back side, and bulky groups block this approach. Tertiary carbons are so hindered that $S_N2$ is nearly impossible.

Correct reasoning:

• $CH_3Cl$ (primary) has no steric hindrance → fast $S_N2$
• $(CH_3)_3CCl$ (tertiary) is severely hindered → $S_N2$ is negligible; it prefers $S_N1$ or elimination

How to avoid: Remember the $S_N2$ reactivity order:

Methyl > Primary > Secondary > Tertiary (tertiary is essentially unreactive in $S_N2$).

Mistake 4: Misapplying “Ambident Nucleophile” Concept Here

The error: Thinking $^-OH$ is an ambident nucleophile (it can attack via O or H) and that this affects the rate comparison.

Why it’s wrong: $^-OH$ is not ambident — it has only one nucleophilic atom (oxygen). Ambident nucleophiles (like $^-CN$, $^-NO_2$) have two possible attack sites. This question is purely about substrate reactivity.

How to avoid: Only invoke ambident nucleophile behavior when the nucleophile itself has multiple nucleophilic atoms. Here, focus on the substrate (alkyl halide) differences.

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