In the above reaction o/p ratio will be highest when

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MCQ

In the above reaction $o/p$ ratio will be highest when

✓
$R = - CH_3$
B
$R = - CH_2 -CH_3$
C
$R = - CHMe_2$
D
$R = - CMe_3$

✓

Answer

Correct option: A.

$R = - CH_3$

a
$(a)$ The steric demand of $H^{\ominus}$ is, however extremely small, and when attack on $C_6H_5Y$ is by any other electrophile, $E^{\oplus}$ , which will necessarily be larger, there will be increasing interaction between $E$ and $Y$ in the transition state for attack at the position $o-$ to $Y (57 \,b, R = E)$ as attacking electrophile and substituent increase in size ; there can be no such interaction in the transition state for $p-$ attack $(57a, R = E)$. This will be reflected in an increasing $\Delta \,G^+$ for $o-$ attack, a consequently slower reaction, and the relative proportion of $o-$ product will thus fall as the size of $E$ and / or $Y$ increase. This is illustrated by the falling $f_{o^-} /f_{p^-}$ ratios which are observed for the nitration of alkylbenzenes $(Y ---CH_3 \to CMe_3)$ under comparable conditions ;

Increase in size of $Y$ $\begin{gathered}
\downarrow \hfill \\
\downarrow \hfill \\
\downarrow \hfill \\
\downarrow \hfill \\
\downarrow \hfill \\
\end{gathered} $ $\begin{array}{*{20}{c}}
Y&{\% \,\,o - }&{\% \,p\, - } \\
{C{H_3}}&{58}&{37} \\
{C{H_2}Me}&{45}&{49} \\
{CHM{e_2}}&{30}&{62} \\
{CM{e_3}}&{16}&{73}
\end{array}$

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