Structural formula Part-3

Structural formula

source : melayunak al. esy.es

Structural formula

Q 1– Molecular formula of organic compd is C2H4O. It gives red ppt. with fehling solution &gives yellow ppts. with I2 & NaOH .Give name & formula of compd.

Ans.

Compd reduces Fehling solution, hence it is an aldehyde. So compd is CH3CHO ( Acetaldehyde).

It gives yellow ppt. of iodoform with I2 & NaOH.This is iodoform test.

CH3CHO +I2 +NaOH —-> CHI3 +NaI +HCOONa +H2O

iodoform

Compd is CH3CHO (acetaldehyde or ethanal).

Q 2— Molecular formula of organic compound is C2H6O. It reacts with Na & liberates H2 gas.It also gives iodoform test .Identify compd.

Ans.

compd reacts with Na & liberates H2 gas , hence compd is alcohol.

It gives iodoform test ,hence it is C2H5OH.

2C2H5OH + 2Na —–> 2C2H5ONa +H2 (hydrogen gas)

C2H5OH +I2 +NaOH—–> CHI3 +NaI +HCOONa +H2O

iodoform

Compd is C2H5OH ( ethyl alcohol or ethanol).

Q 3— Molecular formula of compound ‘A’ is C2H4Cl2.It reacts with aq. NaOH to form ‘B’. Compd ‘B’ forms Ag- mirror with Tollen’s reagent & gives pink colour with Schiff’s reagent. Identify A & B .

Ans.

Compd ‘A’ reacts with aq. NaOH to form ‘B’ which reduces Tollen’s reagent & pink colour with Schiff’s reagent ,hence compd ‘B’ is an aldehyde. Thus, in compd ‘A’ both the Cl-atoms are present on same C -atom.

Hence compd is CH3-CHCl2 (Ethylidene di chloride).

CH3CHCl2 +NaOH(aq.)—->CH3 CH(OH)2—–>CH3CHO +H2O

‘A’ ‘B’

CH3CHO +Ag2O——>CH3COOH +2Ag(silver mirror)

‘A’ is CH3CHCl2 (Ethylidene di chloride or 1,1-dichloro ethane) ; ‘B’ is CH3CHO (Acetaldehyde or ethanal).

Q 4– Molecular formula of compound is C3H6O . It gives addition reaction with HCN & NaHSO3 but it does not give red colour with Schiff’s reagent.Identify the compd.

Ans.

Compd C3H6O gives addition reaction with HCN & NaHSO3 therefore it may be aldehyde or ketone. But it does not reduce Schiff’s reagent , hence it is a ketone or Acetone.