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(i) Compound(A)is:...

(i)
Compound`(A)`is:

A

B

C

D

Text Solution

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a. In compound`(A)`, the `(-NH_(2))`group is removed and from the percentage of`(Br)` in(A) ,it is suggested that two`(Br)`groups are present in`(A)`.The formation of one mono-nitro derivation from`(B)`suggests that two`(Br)`grops are present at para -position `(OMP = 231 )` .
So compound (A) is
In compolund C , the `( - NH_2)` group inremoved and one more (Br) group is introduced in (C ) , this suggests that compound (D)ocntaisn three (Br ) group, which is confirmed from the percentage of ` Br ` in (D).
The formation of one mon-nitro derivative from (D) suggests that three `Br` groups are present at symetrical jposition is ` (D) ( AUS = 23 1)`.
So Comoud (D) is `
Reations :

` (##KSV_CHM_ORG_P2_C15_E01_036_S03.png" width="80%">.
Colculation fo mumber of Br in (A) , (B) and (D) In (A) : 100 gm of compound contains ` = 63 . gm of Br `
` 251 gm ` of compound contins ` = ( 63.7 xx 251)/(100) gm`
of `Br = ( 63 . 7 xx 25 1)/( 100 x 780) mol ` of ` Br = 99 = 2 mol Br`.
` Br = ( 63. 7 xx 25 1)/( 100 x 80) mol ` of ` Br = 1. 99 ~~2 mol Br`.
In (B) : molar mass of (B) ` ( C_6 H_4 Br _2 _) = 236 gm`
Moles of `Br` in ` (B) = ( 67. 80 xx 236) /( 1 00 xx 80) = 2 mol Br`
In (D : ) Molar mass of (D) `(C_6 H_3 Br_3) = 315 gm`
Moles of Br in `(D) = ( 76 . 4 xx 315 ) /( 100 xx 80 ) =3 mol Br`.
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