The reaction of (S)-2-bromobutane with OH- to produce (R)-butan-2-ol will be

Question

The reaction of (S)-2-bromobutane with OH- to produce (R)-butan-2-ol will be (A) first order in 2-bromobutane only (B) first order in OH- only (C) first order in 2-bromobutane and first order in OH- (D) second order in OH-

Tardigrade Admin · Accepted Answer

Configuration invert, means textS textN 2 reaction takes place Rate = K[Substrate] [.OH-].

Q. The reaction of (S)-2-bromobutane with $O H^{-}$ to produce (R)-butan-2-ol will be

first order in 2-bromobutane only

first order in $O H^{-}$ only

first order in 2-bromobutane and first order in $O H^{-}$

second order in $O H^{-}$

Configuration invert, means $S_{N}^{2}$ reaction takes place
Rate $=$ K[Substrate] $[O H^{-}]$

Q. The reaction of (S)-2-bromobutane with OH− to produce (R)-butan-2-ol will be

first order in 2-bromobutane only

first order in OH− only

first order in 2-bromobutane and first order in OH−

second order in OH−

Configuration invert, means SN2​ reaction takes place Rate = K[Substrate] [OH−]

Q. The reaction of (S)-2-bromobutane with $O H^{-}$ to produce (R)-butan-2-ol will be

first order in $O H^{-}$ only

first order in 2-bromobutane and first order in $O H^{-}$

second order in $O H^{-}$

Configuration invert, means $S_{N}^{2}$ reaction takes place
Rate $=$ K[Substrate] $[O H^{-}]$