In electrophilic aromatic substitution reactions, which of the following molecules are considered to be less reactive than benzene?

The following molecules are considered to be less reactive than benzene in electrophilic aromatic substitution reactions: Chlorobenzene, Fluorobenzene, Bromobenzene, Iodobenzene, and Nitrobenzene.

Explanation:

Electrophilic aromatic substitution (EAS) is where benzene acts as a nucleophile to replace a substituent with a new electrophile. That is, benzene needs to donate electrons from inside the ring. Electrophilic substitution relies on the electrophile being attracted to the delocalised electrons in the benzene ring. But chlorine is more electronegative than carbon, and so draws electrons in the ring towards itself. That makes the electron density around the ring rather less in chlorobenzene. It is less attractive for electrophiles, and so the reaction is slower.

Therefore all of the halogens (F,Cl,Br,F,Cl,Br,I) and nitrobenzene are more electronegative than hydrogen. Therefore they will inductively remove electron density from a benzene ring. However, you can also draw resonance structures for the various halobenzenes where the halogen substituent is donating electron density to the ortho and para positions of the benzene ring through resonance.

Therefore, in the case of the halobenzenes these two effects are working in opposite directions. In order to determine which effect controls the reaction, we need some data. The following table compares the relative rates of electrophilic aromatic nitration of the halobenzenes to the rate for benzene itself. If we assign benzene a relative rate of 1, we see that all of the halobenzenes react slower than benzene itself. This tells us that the inductive effect is stronger than the resonance effect in the halobenzene series.

Ar-X

Relative rates

Ar-H

1.0

Ar-F

0.11

Ar-Cl

0.02

Ar-Br

0.06

Ar-I

0.13

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answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole