Question

The increasing order of reactivity of the following compounds in $S_N1$ reaction is:
Compounds:
A. $(CH_3)_3CBr$
B. $CH_3CH_2CH_2CH_2Br$
C. $(CH_3)_2CHCH_2Br$
D. $CH_3CH(Br)CH_2CH_3$

• A. $A<B<C<D$
• B. $B<C<D<A$
• C. $D<C<B<A$
• D. $B<D<A<C$

Hint:

$S_N1$ reactions proceed through carbocation formation. Stability of carbocation determines the reaction rate.

Answer 1

The Correct Option is B

Step 1: Mechanism of $S_N1$ Reaction
The $S_N1$ reaction proceeds in two steps. The rate-determining step involves the formation of a carbocation intermediate. Therefore, the reactivity of alkyl halides towards $S_N1$ reaction depends directly on the stability of the carbocation formed.
Step 2: Stability Order of Carbocations
The general order of carbocation stability is: \[ \text{Tertiary } (3^\circ)>\text{Secondary } (2^\circ)>\text{Primary } (1^\circ)>\text{Methyl} \] Step 3: Analyze the Compounds
A. $(CH_3)_3CBr$ (tert-Butyl bromide): Forms a Tertiary ($3^\circ$) carbocation. Most stable.
B. $CH_3CH_2CH_2CH_2Br$ (n-Butyl bromide): Forms a Primary ($1^\circ$) carbocation with a straight chain. Least stable.
C. $(CH_3)_2CHCH_2Br$ (Isobutyl bromide): Forms a Primary ($1^\circ$) carbocation. Although it is primary like B, the branching at the $\beta$-position makes it slightly more reactive than simple n-butyl in some contexts due to rearrangement potential, but in standard stability comparison, it is often grouped near B. However, for $S_N1$, strictly $1^\circ$ are very slow.
D. $CH_3CH(Br)CH_2CH_3$ (sec-Butyl bromide): Forms a Secondary ($2^\circ$) carbocation. Intermediate stability.
Step 4: Arranging in Increasing Order
Based on carbocation stability ($1^\circ<2^\circ<3^\circ$):
Lowest reactivity: Primary alkyl halides (B and C).
Intermediate reactivity: Secondary alkyl halide (D).
Highest reactivity: Tertiary alkyl halide (A).
Comparing B and C (both $1^\circ$): While both are primary and slow for $S_N1$, the order given in option (ii) places B (linear) as the least reactive and C (branched primary) slightly higher, followed significantly by D ($2^\circ$) and A ($3^\circ$). The clear distinction is: \[ \text{Primary} (B, C)<\text{Secondary} (D)<\text{Tertiary} (A) \] The option that matches this sequence ($...<D<A$) is (ii).