A substance`(X)` contains `41.37% C,6.89% H.0.166gm` of `(X)` gave`NH_(3)` which was absorbed in`50ml` of `N//10 H_(2)SO_(4)`.The excess of acid required `30ml`of `N//10 NaOH` for neutralisation.`(X)`on treatment with`HNO_(2)`gave succinc acid `(X)`on heating lost`NH_(3)` to give(A).(A) reacts with`Br_(2)`and `NAOH` to give(B) containing `41.02%C,5.88%H`, and `11.96%N.(B)`on further treatment with`Br_(2)`and `NaOH`gives `(C)(3`-amino propanoic acid).(C)reats with`NHO_(2)` to give `beta`-hydroxy-propanoic acid.
Compound (X) is:

Options

A. ` (##KSV_CHM_ORG_P2_C15_E01_056_O01.png" width="30%">
B. ` (##KSV_CHM_ORG_P2_C15_E01_056_O02.png" width="30%">
C.
D. ` (##KSV_CHM_ORG_P2_C15_E01_056_O04.png" width="30%">

Explanation

Percentage of `N=?`
Total acid `= 50 xx (1)/(10) = 5 mEq.`
Excess acid `= 30 xx (1)/(10) =3 mEq.`
Acid used `= 5 -3 = 3 mFq.`
Percentage of `N= (1. 4 xx "mEq. Of acid used" )/ ("weight of compound ")`
` = ( 1.4 xx 2)/( 0 . 116) = 24. 38%`
` C= 41 . 37 % H = 6. 89 % N= 24 . 38 %`
` O = 100 - ( 41. 37 + 6. 89 + 24 . 38 ) = 27 . 36`
Formula ` = C_4 H_8 O_2 N_2`, degeee fo undaturation `=2 ^@`.
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