$\underset{A}{\mathop{PhF}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{B}{\mathop{PhCl}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{C}{\mathop{PhBr}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{D}{\mathop{PhI}}\,$
Despite the increase in size of the substituent \(Y\) from \(F \to I\), the proportion of \(o-\) isomer increases. An increasing steric effect will, as with the alkyl benzenes, be operating to inhibit \(o-\) attack, but this must here be outweighed by the electron-withdrawing inductive/field effect exerted by the halogen atom \((Y)\). This effect will tend to decrease with distance from \(Y\), being exerted somewhat less strongly on the distant \(p-\) position compared with the adjacent \(o-\) position. Electronegative \(F,\) and relatively little \(o-\) attack thus takes place on \(C_6 H_5F\), despite the small size of \(F.\) The electron- withdrawing effect of the halogen \((Y)\) decreases considerably from \(F\) to \(I\) (the biggest change being between \(F\) and \(Cl\)), resulting in increasing attack at the \(o-\) position despite the increasing bulk of \(Y.\)
\(Increase\,\,in\,\,size\,\,of\,\,Y\) \(\begin{gathered}
\downarrow \hfill \\
\downarrow \hfill \\
\downarrow \hfill \\
\downarrow \hfill \\
\end{gathered} \) \(\begin{array}{*{20}{c}}
Y&{\% \,o\, - }&{\% \,p\, - } \\
F&{12}&{88} \\
{Cl}&{30}&{69} \\
{Br}&{37}&{62} \\
I&{38}&{60}
\end{array}\)

$R - X \rightarrow R ^{\oplus} X ^{\ominus} \rightarrow R ^{\oplus} \| X ^{\ominus} \stackrel{ Y^\ominus }{\rightarrow} R - Y + X ^{\ominus}$
Ion pair Solvent separated ion pair
કોઈ વિદ્યાર્થી આપેલ પદ્ધતિના આધારે સામાન્ય લાક્ષણિકતાઓ આ રીતે લખે છે:
$(a)$ પ્રક્રિયા નબળા કેન્દ્રાનુરાગી દ્વારા તરફેણમાં છે
$(b)$ $R^ \oplus$ સરળતાથી પ્રક્રિયા કરી મોટુ સંયોજન આપે છે
$(c)$ પ્રક્રિયા રેસેમાઇઝેશન દ્વારા પૂર્ણ થાય છે
$(d)$ પ્રક્રિયા બિન-ધ્રુવીય દ્રાવક દ્વારા તરફેણ કરવામાં આવે છે.
કયા અવલોકનો યોગ્ય છે $?$
$(I)$ $C{H_2} = CH\mathop C\limits^ + HC{H_3}$
$\begin{array}{*{20}{c}}
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\
{(II)\,\,\,\,\,\,\,\,\,\,C{H_2} = C - \mathop {{\text{ }}C}\limits^ + {H_2}}
\end{array}$
$(III)$ $C{H_3}CH = CH\mathop C\limits^ + {H_2}$