Question

$\beta -$ Elimination or anti-elimination reaction is carried out with base $\left(B^{\ominus }\right)$ as shown.
The following bases are used,
$\left(I\right)\stackrel{\ominus }{O}H\left(II\right)R{O}^{\ominus }\left(III\right)RCO{O}^{\ominus }\left(IV\right)\stackrel{\ominus }{C}N\left(V\right)N{O}_3^{\ominus }$

The decreasing order of reactivity for the above elimination is :

• A. $\left(II\right)>\left(I\right)>\left(IV\right)>\left(III\right)>\left(V\right)$
• B. $\left(V\right)>\left(III\right)>\left(IV\right)>\left(I\right)>\left(II\right)$
• C. $\left(II\right)>\left(I\right)>\left(III\right)>\left(IV\right)>\left(V\right)$
• D. $\left(I\right)>\left(II\right)>\left(III\right)>\left(IV\right)>\left(V\right)$

Answer 1

The correct option is A $\left(II\right)>\left(I\right)>\left(IV\right)>\left(III\right)>\left(V\right)$

Rate of $E2$ reaction depends the strength of base.
Rate of reaction increases when strong bronsted base is used.
$\text{Acidic order :}HN{O}_3>RCOOH>HCN>H_{2}O>ROH$
$\text{Basic order :}N{O}_3^{\ominus }<RCO{O}^{\ominus }<C{N}^{\ominus }<\stackrel{\ominus }{OH}<R{O}^{\ominus }$
Hence, order of $\beta -$elimination is :$\left(II\right)>\left(I\right)>\left(IV\right)>\left(III\right)>\left(V\right)$