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Question

Which of the following has the highest rate of electophilic aromatic substitution reaction?

A
C6H6
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B
C6H5CH3
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C
C6H5NO2
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D
C6H5COOH
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Solution

The correct option is B C6H5CH3
Electron donating group will increase the electron density on benzene ring and favours the electrophilic attack at ortho and para position. These groups are also called ortho-para directing groups or activating groups.

Electron withdrawing group will decrease the electron density at ortho and para position and thus making difficult for electrophile to attack at ortho and para position.
Electrophilic attack occurs at meta position but at relatively slower rate as compared to compounds having activating groups.
Hence, these group are called meta directing or deactivating groups.

Compound (a) does not have any activating group.

Compound (b) has electron donating methyl group which shows +H effect. Thus, it increases electron density at ortho and para position making it favourable for electrophilic substitution reaction

In compound (c), NO2 group shows -R and -I effect which will reduces the electron density at ortho and para position Electrophile will attack at meta position but it is relatively slower than compound having electron donating group.

In compound (d), COOH show -R effect which reduce the electron density on benzene ring. Thus, reduce the rate of electrophilic substitution reaction.

Therefore, C6H5CH3 undergoes electrophilic substitution reaction most easily.

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