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SN2 Reaction Mechanism

Which would u...

Question

Which would undergo $S_{N} 1$ reaction faster in the following pair and why?
${C H}_{3} - C H_{2} - B r$ and $\begin{matrix} C H_{3} | C H_{3} & - & C & - & C H_{3} | B r \end{matrix}$

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Solution

$S N_{1}$ reaction
Substitution Nucleophilic Unimolecular reaction:
It proceeds via the formation of a carbocation.
Reactivity towards $S N_{1}$ reaction depends upon the stability of carbocation and since tertiary carbocation is the most stable i.e, why tertiary halides show high reactivity towards $S N_{1}$ reaction.
Reactivity order of carbocation $3^{\circ} > 2^{\circ} > 1^{\circ}$ .
Therefore, this would undergo $S N_{1}$ reaction faster as it forms a $3^{\circ}$ carbocation.
Thus, the tert-butyl bromide will go $S_{N} 1$ reaction faster than ethyl bromide.

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S_{N} 1

C H_{3} - C H_{2} - B r and C H_{3} - C H_{3} | C | B r - C H_{3}

S_{N} 1

C H_{3} - C H_{2} - C H_{2} - B r

C H_{3} - C H - C H_{3}

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Which would undergo $S_{N} 1$ reaction faster in the following pair and why?

$C H_{3} | C H_{3} - C H_{2} - B r a n d C H_{3} - C - C H_{3} | B r$

S_{N} 2

K I

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H ∣ C H_{3} - C H_{2} - C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{2} - C H_{2} - B r

(I) (I I)

C H_{3} - C H ∣ C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{3} ∣ C ∣ C H_{3} - C H_{2} C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - B r ∣ C H - C H_{3} a n d C H_{3} - C H_{3} ∣ C H - C H_{2} - B r ∣ C H - C H_{3}

(I) (I I)

C H_{3} - C H_{2} - B r + N a O H_{1} C H_{3} - C H_{2} - O H + N a B r \to

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C H_{3} - C H_{3} | C | C H_{3} - C H_{2} - B r + N a O H_{2} (C H_{3})_{3} C - C H_{2} - O H + N a B r \to

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K_{1}

K_{2}

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