Structural formula

structural formula

source : melayunak al.

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.


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


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.


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


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 .



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.


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.


Compound is CH3COCH3 ( acetone or propanone -2).

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