Question

Substituents on an aromatic ring can have several effects on electrophilic aromatic substitution reactions. Substituents can activate or deactivate the ring to substitution, donate or withdraw electrons inductively, donate or withdraw electrons through resonance, and direct substitution either to the ortho/para or to the meta positions. From the following lists, select the substituents that have the indicated property. The substituents are written as -XY, where X is the atom directly bound to the aromatic ring.

Substituents on an aromatic ring can have several effects on electrophilic aromatic substitution reactions. Substituents can activate or deactivate the ring to substitution, donate or withdraw electrons inductively, donate or withdraw electrons through resonance, and direct substitution either to the ortho/para or to the meta positions. From the following lists, select the substituents that have the indicated property. The substituents are written as -XY, where X is the atom directly bound to the aromatic ring.

Answer 1

Concepts and reason

The concept uses to solve this question is inductive effect. The electrophiles are the molecules carrying positive charge. The molecules or the ions carrying a positive charge, are attracted by the negative charge compounds.

Fundamentals

The inductive effect is an effect caused by the presence of the atoms having a greater electronegativity difference. The difference in the electronegativity either causes the donation of the atom in the carbon chain or withdrawal of the electron.

Part 1

The compounds, which are not able to donate the electron in the aromatic ring do not undergo the activation of the ring. The atoms attached to the having free lone pairs of electrons, which are attached (attachment other than carbon) to less electronegative molecule cannot donate electrons.

The donation of the electron makes the ring electron rich. The presence of the electrons makes the attachment of the electrons easy. The groups $- {\rm{C}}{{\rm{H}}_{\rm{3}}}\left( {{\rm{methyl}}} \right)$, ${\rm{ - R}}\left( {{\rm{alkyl}}} \right)$, $- {\rm{N}}{{\rm{H}}_{\rm{2}}}\left( {{\rm{amino}}} \right)$, $- {\rm{N}}{{\rm{R}}_{\rm{2}}}\left( {{\rm{amide}}} \right)$, $- {\rm{OH}}\left( {{\rm{hydroxyl}}} \right)$, $- {\rm{OR}}\left( {{\rm{ketone}}} \right)$are electron donating.

Part 2

The donation of the electron makes the ring electron rich. The-OR (ketone) group is capable of donating the electrons in the ring. The oxygen is attached to carbon on both side. Hence, can easily pass down the electron, as the carbon is less electronegative.

[Part 2]

Part 2

The $- {{\mathop{\rm OCOCH}\nolimits} _3}$and $- {\rm{NHCOC}}{{\rm{H}}_{\rm{3}}}$is not an electron withdrawing group. They tend to donate the electron in the chain.

The –CN group is highly electron withdrawing group due to the presence of electronegative nitrogen. It pulls electron from the ring.

The –COOH group is highly electron withdrawing group due to the presence of electronegative oxygen. It pulls electron from the ring.

In the resonating structure, the ortho para directing groups are the groups which are capable of donating the electrons as a lone pair. They transfer their lone pair of the electron to the ortho and para position of the ring. The fluorine is highly electronegative atom, it cannot lose its electron.  

The nitrogen is capable of donating the lone pair of electrons present in its outer most shell. After donating the electron, it gains a positive charge.

The oxygen is capable of donating the lone pair of electrons present in its outer most shell. After donating the electron, it gains a positive charge. The positive charge is stable on the oxygen.

Ans: Part 1

The compound causes activation of the ring towards substitution is $- {\rm{OH}}$.

Part 2

The compound causes activation of the ring towards substitution is $- {\rm{OC}}{{\rm{H}}_{\rm{3}}}$.

The electron withdrawing group in resonance is -CN.

The electron withdrawing group in resonance is -COOH.

The ortho para directing group would be $- {\rm{N}}{{\rm{H}}_{{\rm{2}}{\rm{.}}}}$

The ortho para directing group would be $- {\rm{N}}{{\rm{O}}_{\rm{2}}}$