Question

Which one of the following groups shows maximum -I effect?

• A. $-\stackrel{+}{N}(C{H}_3{)}_3$
• B. $-\stackrel{+}{N}{H}_3$
• C. $-I$
• D. $-F$

Answer 1

The correct option is A $-\stackrel{+}{N}(C{H}_3{)}_3$

Permanent shift in electron density away from the carbon chain to the substituent is called -I effect.
Electron withdrawing groups show negative inductive effect. Electron withdrawing groups means substituent which withdraws electron pair towards itself. So the correct order of -I effect will be
$-\stackrel{+}{N}(C{H}_3{)}_3>-\stackrel{+}{N}{H}_3>F>I.$

The $-\stackrel{+}{N}(C{H}_3{)}_3$ and $-\stackrel{+}{N}{H}_3$ shows maximum electron withdrawing effect as compared to $-F$ and $-I$ due to presence of positive charge. Steric Repulsion of alkyl group causes the bond length(N-R) to increase, hence the electron donation by -R group becomes difficult, leading to $-\stackrel{+}{N}{R}_3$ better EWG as compared to $-\stackrel{+}{N{H}_3}$.

Among F and I more electronegative element has better withdrawing ability of electron hence F>I
Hence $-\stackrel{+}{N}(C{H}_3{)}_3$ will show maximum -I effect.