JEE Chemistry 2025 Question with Solution

Given: Compound 'X' absorbs $2$ moles of hydrogen, indicating the presence of a triple bond that is converted into a single bond. Upon oxidation with $\mathrm{KMnO_4}$ in acidic medium, the products obtained are:

• $\mathrm{CH_3{-}C(=O){-}CH_3}$
• $\mathrm{CH_3{-}C(=O){-}OH}$
• $\mathrm{CH_3{-}C(=O){-}CH_2{-}C(=O){-}OH}$

Find: The total number of $\sigma$ bonds present in compound 'X'.

From the given oxidation products, count the $\sigma$ bonds in each part as stated in the solution:

• $\mathrm{CH_3{-}C(=O){-}CH_3}$ has $9$ $\sigma$ bonds
• $\mathrm{CH_3{-}C(=O){-}OH}$ has $7$ $\sigma$ bonds
• $\mathrm{CH_3{-}C(=O){-}CH_2{-}C(=O){-}OH}$ has $11$ $\sigma$ bonds

So,

$$9 + 7 + 11 = 27$$

Therefore, the total number of $\sigma$ bonds in compound 'X' is $27$. Hence, the correct option is A.

Given: 'X' absorbs $2$ moles of hydrogen and on oxidation gives ketone and acid fragments.

Find: Total $\sigma$ bonds in 'X'.

Absorption of $2$ moles of hydrogen indicates one triple bond in 'X'. Oxidative cleavage with acidic $\mathrm{KMnO_4}$ gives the listed fragments, and the provided working concludes the total count by summing the $\sigma$ bonds contributed by those parts.

$$\begin{aligned} \mathrm{CH_3{-}C(=O){-}CH_3} &\rightarrow 9 \, \sigma\text{ bonds} \\ \mathrm{CH_3{-}C(=O){-}OH} &\rightarrow 7 \, \sigma\text{ bonds} \\ \mathrm{CH_3{-}C(=O){-}CH_2{-}C(=O){-}OH} &\rightarrow 11 \, \sigma\text{ bonds} \end{aligned}$$

Adding them,

$$27 = 9 + 7 + 11$$

Thus, the total number of $\sigma$ bonds present in compound 'X' is $27$.