carry out the conversion: p-nitrotoluene to 2-bromobenzoic acid
Answers
Answered by
0
Sorry but I would like to answer your questions from last to first,
(1)Yes,2-bromo-4-nitrotoluene is major product as -CH3 is ortho-para orienting group and preference of -CH3(activating) OVER -NO2(deactivating) will take place
(2) -COOH is deactivating meta directing group (due to presence of >C=O) so friedel-crafts bromination would take place at meta position.
(3)Yes,as -NH2 is better activator than -CH3 so bromination would take place ortho with respect to -NH2
(4)I think it is mistake of your book for not showing any catalysts like AlCl3 but It should be friedel-crafts bromination as -Br replaced one of the -H at ortho (to -CH3) of benzene.
(5) As it is friedel-crafts bromination,Br+ ion will be easily satisfied(to be octet) by pi bond of benzene than sigma bond -C-H of-CH3.(since sigma bond is more stronger than pi bond so pi bond can be easily broken and form meta-stable sigma complex)
(1)Yes,2-bromo-4-nitrotoluene is major product as -CH3 is ortho-para orienting group and preference of -CH3(activating) OVER -NO2(deactivating) will take place
(2) -COOH is deactivating meta directing group (due to presence of >C=O) so friedel-crafts bromination would take place at meta position.
(3)Yes,as -NH2 is better activator than -CH3 so bromination would take place ortho with respect to -NH2
(4)I think it is mistake of your book for not showing any catalysts like AlCl3 but It should be friedel-crafts bromination as -Br replaced one of the -H at ortho (to -CH3) of benzene.
(5) As it is friedel-crafts bromination,Br+ ion will be easily satisfied(to be octet) by pi bond of benzene than sigma bond -C-H of-CH3.(since sigma bond is more stronger than pi bond so pi bond can be easily broken and form meta-stable sigma complex)
Similar questions