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Q9.5:
An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.

Solution :

Step 1: Analyzing Ozonolysis Products

Ozonolysis of an alkene involves the oxidative cleavage of a carbon-carbon double bond ($C=C$) by ozone ($O_3$), followed by reductive workup ($Zn/H_2O$). The reaction follows the general pattern:

$R_1R_2C=CR_3R_4 \xrightarrow{O_3, Zn/H_2O} R_1R_2C=O + R_3R_4C=O$

The given products are:

  • Ethanal: $CH_3CHO$ (derived from a fragment containing a terminal double bond with one hydrogen atom).
  • Pentan-3-one: $CH_3CH_2COCH_2CH_3$ (derived from a fragment containing a substituted double bond).

Step 2: Reconstruction of the Parent Alkene

To identify the structure of 'A', we remove the oxygen atoms from the carbonyl groups of both products and join the remaining carbon atoms via a double bond [Per the Criegee mechanism of ozonolysis].

  • Fragment 1: $CH_3CH=O \rightarrow CH_3CH=$
  • Fragment 2: $O=C(CH_2CH_3)_2 \rightarrow =C(CH_2CH_3)_2$

Connecting these two fragments results in the following structural skeleton:

$CH_3CH=C(CH_2CH_3)_2$

Step 3: IUPAC Nomenclature

To determine the systematic name, we follow the IUPAC guidelines:

  1. Longest Chain Selection: The longest chain containing the double bond consists of 5 carbons (pent-ene).
  2. Numbering: Numbering starts from the end nearest to the double bond to assign the lowest possible locant. Numbering from the left gives the double bond position at $C2$.
  3. Substituent Identification: There is an ethyl group ($-C_2H_5$) located at the $C3$ position.

Structure: $CH_3 - CH = C(C_2H_5) - CH_2 - CH_3$

IUPAC Name: 3-ethylpent-2-ene

Step 4: Visual Representation

CH3 CH C CH2CH3 CH2CH3
Fragment Structure
Product 1 $CH_3CHO$ (Ethanal)
Product 2 $CH_3CH_2COCH_2CH_3$ (Pentan-3-one)

Final Solution: The structure of alkene ‘A’ is $CH_3CH=C(C_2H_5)CH_2CH_3$, and its IUPAC name is 3-ethylpent-2-ene.


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