0.450 gm of an aromatic organic compound (A) on ignitin gives0.905 gm CO_(2) and 0.15gm H_(2)O 0.350gm of (A) on hydrolysis with Ca(OH)_(2) yields(B) which on mild reduction gives an opticallyactive compound(C). On heating(C) with I_(2) and NaOH, iodofrom is producedalong with (D). WithHCl, (D) gives a solid, which is more soluble in hotwater thanin cold. Idenify (A) to (D) with proper explanaion.

0.450 gm of an aromatic organic compound (A) on ignitin gives0.905 gm

1.	0.450 gm of an aromatic organic compound (A) on ignitin gives0.905 gm CO_(2) and 0.15gm H_(2)O 0.350gm of (A) on hydrolysis with Ca(OH)_(2) yields(B) which on mild reduction gives an opticallyactive compound(C). On heating(C) with I_(2) and NaOH, iodofrom is producedalong with (D). WithHCl, (D) gives a solid, which is more soluble in hotwater thanin cold. Idenify (A) to (D) with proper explanaion.
Answer» Solution :Percentage of `C = (12)/(44) xx ("Wt. of" CO_(2)xx100)/("Wt. of compound")` `= (12xx0.905xx100)/(44xx0.450) = 54.84%` Percentage of `H = (2)/(18) xx ("Wt. of" HO_(2)xx100)/("Wt. of compound")` `= (2xx0.15xx100)/(18xx0.450) = 3.7%` Percentage of `Cl = (35.5)/(143.5) xx ("Wt. of" AGCL xx 100)/("Wt. of compound")` `= (35.5xx0.574xx100)/(143.0.350) = 40.57%` Molecular mass of `(A) = 2 xx V.D. = 2xx87.5 = 175.0` Empirical formula`= C_(4) H_(4) Cl` `N = (MFW)/(EFw) = (175.0)/(87.5) = 2` Molecular formula `= C_(8) H_(8)Cl_(2)` `D.U` in `(A) = ((2n_(C) + 2) - (n_(Cl) + n_(H)))/(2)` `= ((2xx8+2)-(2+8))/(2) = 4^(@)` `4 D.U` in `(A)` suggests benzene ring with two `Cl` and two `C` atoms. REACTIONS with`Ca(OH)_(2)` of `(A)` suggests that two `Cl`atomare in the side chain becuasethere is no `EWG` present in the ring. Two `Cl` atoms can be vicinal or gemial lodoform reaction suggests that there must be`(overset(O)overset(\|\|)(-C-) CH_(3))` or `(CH_(3) - underset(\|)(C) - OH)` group. So ketone can be OBTAINED from `(A)` when two `Cl`atoms are in gemial postion. Reaction: