# Consider the reactions :(i) ${\left({\mathrm{CH}}_{3}\right)}_{2}\mathrm{CH}-{\mathrm{CH}}_{2}\mathrm{Br}\stackrel{{\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{OH}}{⟶}$ ${\left({\mathrm{CH}}_{3}\right)}_{2}\mathrm{CH}-{\mathrm{CH}}_{2}{\mathrm{OC}}_{2}{\mathrm{H}}_{5}+\mathrm{HBr}$ (ii) ${\left({\mathrm{CH}}_{3}\right)}_{2}\mathrm{CH}-{\mathrm{CH}}_{2}\mathrm{Br}\stackrel{{\mathrm{C}}_{2}{\mathrm{H}}_{5}{\mathrm{O}}^{-}}{⟶}$ ${\left({\mathrm{CH}}_{3}\right)}_{2}\mathrm{CH}-{\mathrm{CH}}_{2}{\mathrm{OC}}_{2}{\mathrm{H}}_{5}+{\mathrm{Br}}^{-}$The mechanisms of reactions (i) and (ii) are respectively :

1. A

2. B

3. C

4. D

FREE Lve Classes, PDFs, Solved Questions, PYQ's, Mock Tests, Practice Tests, and Test Series!

+91

Verify OTP Code (required)

### Solution:

• First reaction is SN1 reaction because, ${\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{OH}$, is a polar protic solvent and a weak nucleophile. Moreover, in SN1 reaction, rearrangement takes place to form a more stable carbocation.
• Second reaction is SN2 reaction because, ${\mathrm{C}}_{2}{\mathrm{H}}_{5}{\mathrm{O}}^{-}$is a strong nucleophile and the reaction proceeds with the formation of transition state.

## Related content

Join Infinity Learn Regular Class Program!