A compound 'X' absorbs 2 moles of hydrogen and 'X' upon oxidation with $$KMnO_{4}|H^{+}$$ gives

The total number of $$\sigma$$ bonds present in the compound 'X' is ________.

The reagent $$H_2/Pd$$ adds across a carbon-carbon multiple bond. Therefore, when a compound absorbs 2 moles of $$H_2$$, it must contain exactly two carbon-carbon double bonds (each $$C=C$$ consumes one mole of $$H_2$$).

Hot, acidic $$KMnO_4$$ cleaves every $$C=C$$ completely, converting each carbon of the double bond into a carbonyl carbon that finally appears as $$-COOH$$. If a carbon of the original double bond was $$\;CH_2=$$, that terminal carbon is oxidised further to $$CO_2$$.

Let the oxidation product obtained from ‘X’ be $$HOOC\,(CH_2)_6\,COOH$$ (suberic acid).
Tracing the process backwards:
  • Replace each carboxyl carbon of the acid by the corresponding alkene carbon (because each $$-COOH$$ arose from a double-bond carbon).
  • Add back the $$CH_2$$ that was lost as $$CO_2$$ at both ends.

Working in this way we recover the precursor:

$$CH_2=CH\;-\;(CH_2)_6\;-\;CH=CH_2$$

This structure is $$\textbf{deca-1,9-diene}$$. It indeed contains two isolated $$C=C$$ bonds and therefore absorbs exactly two moles of $$H_2$$, matching the data given.

Now count the $$\sigma$$ bonds in deca-1,9-diene.

Number of carbon atoms, $$n = 10$$, hence the carbon skeleton has $$n-1 = 9$$ $$C-C$$ $$\sigma$$ bonds.

Hydrogen count (obtained from the structure):
  • Two terminal $$CH_2=$$ groups ⇒ $$2 \times 2 = 4$$ H
  • Two $$CH=$$ carbons (one in each $$CH=CH_2$$) ⇒ $$2 \times 1 = 2$$ H
  • Six internal $$CH_2$$ groups ⇒ $$6 \times 2 = 12$$ H
Total hydrogens $$= 4 + 2 + 12 = 18$$.

Therefore, $$C-H$$ $$\sigma$$ bonds $$= 18$$.

Total $$\sigma$$ bonds in the molecule:
$$C-C: 9 \;+\; C-H: 18 = 27$$.

Hence, the compound ‘X’ has $$\boxed{27}$$ $$\sigma$$ bonds.