# Why is tertiary carbon atom more reactive towards sn1 reaction even though it is the most stable one??

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
anor277 Share
Feb 7, 2018

BECAUSE the formal cation that results from the tertiary carbon is intrinsically the MOST stable...

#### Explanation:

Now ${S}_{N} 1$ reactions are essentially bond-breaking, and these are conceived to occur when a carbon-halogen bond is BROKEN to give a carbocation intermediate....

Consider....

${H}_{3} C - C H X - C {H}_{3} \stackrel{\text{rate determining step}}{\rightarrow} {H}_{3} C - \stackrel{+}{C} H - C {H}_{3} + {X}^{-}$

The fast step is the reaction of some nucleophile, $X {'}^{-}$,...which could be the solvent or it could be even the initial leaving group, to give a new species....

${H}_{3} C - \stackrel{+}{C} H - C {H}_{3} + + X {'}^{-} \stackrel{\text{fast step}}{\rightarrow} {H}_{3} C - C H X ' - C {H}_{3}$

Now because secondary carbocations are stabilized with respect to methyl, and primary carbocations, and tertiary carbocations are stabilized with respect to secondary carbocations, the order of reactivity in terms of ${S}_{N} 1$ mechanisms should be....

${\underbrace{\text{Methyl, primary, secondary, tertiary}}}_{\rightarrow}$
$\text{RATE OF REACTIVITY IN BOND-BREAKING REACTION}$

And this order reflects the stability, the lifetime if you like, of the reactive carbocation intermediate. Sterically encumbered carbocations thus have faster rates of reactions in bond-breaking scenarios.

• 15 minutes ago
• 16 minutes ago
• 17 minutes ago
• 18 minutes ago
• A minute ago
• 4 minutes ago
• 10 minutes ago
• 11 minutes ago
• 12 minutes ago
• 14 minutes ago
• 15 minutes ago
• 16 minutes ago
• 17 minutes ago
• 18 minutes ago