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An alkene ‘A’ (molecular formula C5H10) on ozonolysis gives a mixture of two compounds ’B’ and ’C. Compound ‘8’ gives positive Fehling’s test and also forms iodoform on treatment with l2, and NaOH. Compound ‘C’ does not give Fehling’s test but forms iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C.


Molecular formula =C5H10

Degree of unsaturation = (Cn+1)-Hn2

where Cn = number of carbon atoms
Hn = number of hydrogen atoms

                      =(5+1)-102=1

Compound A will be either alkene or cyclic hydrocarbon. Since, A is undergoing ozonolysis hence A must be an alkene.

Possible structures of alkene are
l. CH3—CH2—CH2—CH=CH2

ll. CH3—CH2—CH=CH—CH3

lll.. CH3—CH—CH=CH2
            |
          CH3

lV. CH3—C—CH—CH2
           |
       CH3

Ozonolysis of structure I produces aldehyde only

CH3—CH2—CH2—CH=CH2→(ii) Zn/H2O(i) O3CH3—CH2—CH2—C=O
                                                                                               |
                                                                                              H

                                                                                                             

Ozonolysis of structure II produces aldehyde only

CH3—CH2—CH=CH—CH3→(ii) Zn/H2O(i) O3CH3—CH2—CHO+CH3CHO

Ozonolysis of structure III Produces aldehyde only
CH3—CH—CH=CH2→(ii) Zn/H2O(i) O3CH3—CH—CHO+HCHO
            |                                                   |
           CH3                                              CH3        
Ozonolysis of structure IV produces both aldehyde and ketone
CH—CH=CH—CH3→(ii) Zn/H2O(i) O3 CH3—C=O+CH3CHO
           |                                                  |
         CH3                                              CH3

After ozonolysis of each of structures I, II and III produces only aldehydes as both Components. But as given in the question one compound doesnt give Fehling test but must give iodoform test. Hence, compound must be a ketone with           O
          ||
CH3—C—  group. Hence, correct structure is IV.

CH3—CH=C—CH3 →(ii) Zn/H2O(i) O3H3C—CHO+O=C—CH3
                   |                                 Acetaldehyde     | 
                  CH3                                  [B]               CH3
            2methylbu-2-ene                                 Acetone
                                                                                 [C]
              [A]                                                              

Formation of iodoform from ‘B’ and ‘C’ may be explained as follows

CH3CHO+3l2+4NaOH→∆CHl3+HCOONa+3Nal+3H2O
Acetaldehyde                   lodoform     Sodium
    [B]                                                 formate
CH2COCH3+3l2+4NaOH→∆CHl3+CH3CPPNa+3Nal+3H2O
Acetone                              lodoform  Sodium   
   [C]                                                          acetate

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