Chapter 16.SE, Problem 24VC

Interpretation Introduction

a)

Interpretation:

Draw the products obtained when the compound shown (methoxybenzene) reacts with (i) Br₂, FeBr₃ and (2) CH₃COCl, AlCl₃.

Concept introduction:

Aromatic rings can be brominated using Br₂ in the presence of FeBr₃. The electrophile is Br⁺. Friedal-Crafts acylation reaction occurs when an aromatic compound is treated with an acyl chloride in the presence of AlCl₃. The electrophile is RCO⁺. In aromatic electrophilic substitution reactions methoxy group is an ortho and para directing and activating group.

To draw:

The products obtained when the compound shown (methoxybenzene) reacts with (i) Br₂, FeBr₃ and (2) CH₃COCl, AlCl₃.

Answer to Problem 24VC

(i) The products obtained when the compound shown (methoxybenzene) reacts with Br₂/FeBr₃ are p-bromomethoxybenzene and o-bromomethoxybenzene.

(ii) The products obtained when the compound shown (methoxybenzene) reacts with CH₃COCl/AlCl₃ are p-acetylmethoxybenzene and o-acetylmethoxybenzene.

Explanation of Solution

The methoxy group attached to the benzene ring is an ortho & para directing group. When methoxybenzene is brominated and alkylated it directs the incoming electrophiles, Br⁺ and CH₃CO⁺ to ortho & para positions.

Conclusion

(i) The products obtained when the compound shown (methoxybenzene) reacts with Br₂/FeBr₃ are p-bromomethoxybenzene and o-bromomethoxybenzene.

(ii) The products obtained when the compound shown (methoxybenzene) reacts with CH₃COCl/AlCl₃ are p-acetylmethoxybenzene and o-acetylmethoxybenzene.

Interpretation Introduction

b)

Interpretation:

Draw the structures of the products obtained when the compound shown, p-methylbenzaldehyde reacts with (i) Br₂, FeBr₃ and (2) CH₃COCl, AlCl₃.

Concept introduction:

Electrophilic substitution of disubstituted benzenes follows three simple rules. (i) If the directing influence of both the substituents reinforce each other, a single product results. (ii) If the directing influences of both the substituent groups oppose each other, the most powerful activating group among them has the dominant influence but usually a mixture of products results. (iii) In meta disubstituted compounds, further substitution in between the groups occurs only rarely, due to steric reasons.

To show:

The major product(s) produced when p-methylbenzaldehyde reacts with (i) Br₂, FeBr₃ and (2) CH₃COCl, AlCl₃.

Answer to Problem 24VC

(i) The major product produced when p-methylbenzaldehyde reacts with Br₂/FeBr₃ is 3-bromo-4-methylbenzaldehyde.

(ii) The major product produced when p-methylbenzaldehyde reacts with CH₃COCl/AlCl₃ is 3-acetyl-4-methylbenzaldehyde.

Explanation of Solution

The aldehydic group is a meta directing deactivating group while the methyl group is an ortho and para directing activating group. Thus both methyl and aldehydic groups can direct the entering electrophile into the same position, ortho to methyl and meta to aldehydic group. Thus the Br atom and –COCH₃ group get substituted at a position ortho to methyl and meta to aldehydic group.

Conclusion

(i) The major product produced when p-methylbenzaldehyde reacts with Br₂/FeBr₃ is 3-bromo-4-methylbenzaldehyde.

(ii)The major product produced when p-methylbenzaldehyde reacts with CH₃COCl/AlCl₃ is 3-acetyl-4-methylbenzaldehyde.