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Q9.6:
An alkene ‘$A$’ contains three $C – C$, eight $C – H$ $\sigma$ bonds and one $C – C$ $\pi$ bond. ‘$A$’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘$A$’.

Solution :

Step 1: Analyzing the Molecular Formula of Alkene 'A'

The total number of bonds provided is as follows:

  • $C–C$ $\sigma$ bonds: $3$
  • $C–H$ $\sigma$ bonds: $8$
  • $C=C$ $\pi$ bond: $1$

In an alkene with $n$ carbon atoms, the number of $C-C$ $\sigma$ bonds is $(n-1)$. Since we have $3$ $C-C$ $\sigma$ bonds, we determine the number of carbon atoms as follows:

$n - 1 = 3 \implies n = 4$

With $n=4$ carbons, the general formula for an alkene is $C_n H_{2n}$, which gives us $C_4 H_8$. We verify the $C-H$ $\sigma$ bonds: In a saturated alkane ($C_4 H_{10}$), there are $10$ $C-H$ bonds. The formation of one $\pi$ bond removes $2$ hydrogen atoms, resulting in $10 - 2 = 8$ $C-H$ bonds. This matches the given data [Per the general formula for alkenes].

Step 2: Analyzing Ozonolysis Products

The alkene '$A$' undergoes ozonolysis to produce two moles of an aldehyde with a molar mass of $44\text{ u}$.

Let the aldehyde be $R-CHO$. Its molar mass is calculated as:

$M_{aldehyde} = M_R + M_C + M_H + M_O$

$44 = M_R + 12 + 1 + 16$

$44 = M_R + 29 \implies M_R = 15$

An alkyl group with a molar mass of $15$ is a methyl group ($-CH_3$). Therefore, the aldehyde formed is $CH_3CHO$ (Ethanal or Acetaldehyde).

Step 3: Determining the Structure of Alkene 'A'

Ozonolysis of an alkene follows the cleavage of the $C=C$ double bond [Per the Criegee mechanism]. If the reaction yields two moles of acetaldehyde ($CH_3CHO$), the original alkene must have been symmetrical at the double bond:

$CH_3-CH=CH-CH_3 + O_3 \xrightarrow{Zn/H_2O} 2 CH_3CHO$

The structure is $But-2-ene$.

Step 4: Structural Visualization

CH₃ H H CH₃

Step 5: IUPAC Nomenclature

The carbon chain contains 4 carbon atoms with a double bond starting at the second position. Following the IUPAC rules for naming alkenes:

  1. Longest chain: Butane derivative.
  2. Positioning: Double bond gets the lowest locant ($2$).
  3. Suffix: $-ene$.

Result: $But-2-ene$.

Final Solution: The IUPAC name of alkene 'A' is But-2-ene.


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