(R)-2-octyl tosylate is solvolyzed in water u | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In an ideal $S_{N}1$ reaction,the leaving group departs completely to form a free carbocation,which is planar. Nucleophilic attack from both sides

Vedclass · Accepted Answer

$(R)-2-$octanol and $(S)-2-$octanol in a $1.5 : 1$ ratio

Vedclass · Answer

(B) In an ideal $S_{N}1$ reaction,the leaving group departs completely to form a free carbocation,which is planar. Nucleophilic attack from both sides should lead to a racemic mixture ($1:1$ ratio of enantiomers).However,in practice,the leaving group (tosylate) remains close to the carbocation as an 'intimate ion pair' for a short time.This shields the front side,making the back-side attack by water more favorable.Therefore,inversion of configuration is slightly favored over retention of configuration.This results in a mixture where the $(S)-2-$octanol (inversion product) is formed in greater quantity than $(R)-2-$octanol (retention product).Thus,the ratio of $(S)-2-$octanol to $(R)-2-$octanol is greater than $1:1$,such as $1.5:1$.

$(R)-2-$octyl tosylate is solvolyzed in water under ideal $S_{N}1$ conditions. The product$(s)$ will be

Similar Questions

The rate of $\beta$-dehydrohalogenation in the following halides under treatment with a strong base follows the order:

Which of the following halides will give $S_N 1$ reaction fastest?

Monohalides on reacting with alcoholic $KOH$ give

What is the product formed by the addition of $HCl$ to $1$-phenylpropene?

The reaction $CH_3-CH(C_6H_5)-CH(Br)-C_6H_5 \xrightarrow{alc.KOH} A$. Identify the product $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module