Question

Reactivity of $PhMgBr$ with the given in the decreasing order is :

• A. $\left(i\right)>\left(iv\right)>\left(ii\right)>\left(iii\right)$
• B. $\left(iii\right)>\left(ii\right)>\left(iv\right)>\left(i\right)$
• C. $\left(i\right)>\left(iii\right)>\left(ii\right)>\left(iv\right)$
• D. $\left(i\right)>\left(ii\right)>\left(iii\right)>\left(iv\right)$

Answer 1

The correct option is A $\left(i\right)>\left(iv\right)>\left(ii\right)>\left(iii\right)$

At $o$-position, electron donating power to $(Me-)$ group is more than $p$ and $m$ (due to the $I$ effect, but equal hyperconjugative efect)

So, the order of electron-donating power of $(Me-)$ group is $o>p>m$. Nucleophilic addition is favoured by less-substituted EDG or more substituted EWG.

So, the order of reactivity of $PhMgBr$ with aromatic aldehyde is $\left(i\right)>\left(iv\right)>\left(ii\right)>\left(iii\right)$

Hence, the correct option is A.