Question

Arrange the following in decreasing order of reactivity towards electrophilic aromatic substitution

• A. $\left(a\right)>\left(b\right)>\left(c\right)$
• B. $\left(c\right)>\left(b\right)>\left(a\right)$
• C. $\left(a\right)>\left(c\right)>\left(b\right)$
• D. $\left(c\right)>\left(a\right)>\left(b\right)$

Answer 1

The correct option is A $\left(a\right)>\left(b\right)>\left(c\right)$

In case of isotopes, the bond length follows the order :
$C-T<C-D<C-H$
So electron density becomes higher due to shorter bond length. Also, the bond becomes stronger for $C-T$ as compared to $C-D$ and $C-T$
Stronger the bond, more energy will be required to break the bond and lesser is the hyperconjugation shown.
$\therefore +H$ effect is more for $-C{H}_3$ as compared to $-C{D}_3$ and $-C{T}_3$

In electrophilic aromatc substitution reaction, carbocation intermediate is formed, which can be stabilized by $+H$ effect of these groups attached.
More the $+H$ effect shown more is the carbocation intermediate stability.
Carbocation intermediate stability follows the order :
$\left(a\right)>\left(b\right)>\left(c\right)$
Order of reactivity towards electrophilic aromatic substitution :
$\left(a\right)>\left(b\right)>\left(c\right)$