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Use of CH 3 I forms C 6 H 5 —N + (CH 3 ) 3 I - as a by-product. While the use of CH 2 O and HCOOH forms the intermediate electrophile, CH 2 OH which in turn can attack the activated benzene ring to give p-NH 2 , C 6 H 4 , CH 2 OH.

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Amines

Short Answer Type

111.

Write the structure of the Zwitter ion of sulphanilic acid.

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112.

What happens when n-pentyl nitrite reacts with aniline in the presence of HCl in ice-cold solution?

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113.

Convert aniline into chlorobenzene.

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114.

Write the structures of A and B.

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115.

What is the structure of compound A?
$2 comma space 4 space nitro space benzene rightwards arrow from left parenthesis ii right parenthesis space anisole to left parenthesis straight i right parenthesis space NaNO subscript 2 plus HCl space 0 minus 4 degree straight C of straight A$

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116. How does the formation of 2° and 3° amines can be avoided during the preparation of 1° amine by alkylation?

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117.

Compare the reactions of MeN and Ph₃N with BF_3.

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118.
Why is it preferred to use methanol (and not methyl iodide HCHO or HCOOH) for preparing C₆H₅N(CH₃)₂ from aniline ?

Use of CH₃I forms C₆H₅—N⁺ (CH₃)₃I^- as a by-product. While the use of CH₂O and HCOOH forms the intermediate electrophile, CH₂OH which in turn can attack the activated benzene ring to give
p-NH₂, C₆H₄, CH₂OH.