Question

Which of the following statements are correct about the following reactions?

• A. Reaction $1$ proceeds by $S_{N}2$ and reaction $2$ by $S_{N}1$ mechanism
• B. Reaction $1$ proceeds by $S_{N}1$ and reaction $2$ by $S_{N}2$ mechanism
• C. (B)$\Rightarrow$(-) $Ph-\stackrel{OH}{\stackrel{|}{C}H}-CH=CH-Me$
  (C)$\Rightarrow$ $(\pm )$ $Ph-CH=CH-\underset{OH}{\underset{|}{C}H}-Me$The products $\left(B\right)$ and $\left(C\right)$ are respectively
• D. (B)$\Rightarrow$(-) $Ph-\stackrel{OH}{\stackrel{|}{C}H}-CH=CH-Me$
  
  (C)$\Rightarrow$ $(\pm )$ $Ph-\underset{OH}{\underset{|}{C}H}-CH=CH-Me$
  The products $\left(B\right)$ and $\left(C\right)$ are, respectively

Answer 1

Answer: (A), (C)

The correct options are
A Reaction $1$ proceeds by $S_{N}2$ and reaction $2$ by $S_{N}1$ mechanism
C

(B)$\Rightarrow$(-) $Ph-\stackrel{OH}{\stackrel{|}{C}H}-CH=CH-Me$

(C)$\Rightarrow$ $(\pm )$ $Ph-CH=CH-\underset{OH}{\underset{|}{C}H}-Me$

The products $\left(B\right)$ and $\left(C\right)$ are respectively

For Reaction 1 :

5N of $NaOH$ is added in $Ph-CH-CH=CH-Me$ thus it proceeds through $S_{N}2$ mechanism

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$OCOR$

Thus it produces a (-) product = $Ph-\stackrel{OH}{\stackrel{|}{C}H}-CH=CH-Me=\left(B\right)$

For Reaction 2 :

Addition of dip. $NaOH$ makes it follow a $S_{N}1$ mechanism.

Thus it forms 2 products

$(\pm )$ $Ph-CH=CH-\underset{OH}{\underset{|}{C}H}-Me=\left(C\right)$