Question

A compound 'X' absorbs 2 moles of hydrogen and 'X' upon oxidation with $KMnO_4$ | $H^+$ gives

$CH_3 - \underset{||}{C} - CH_3$, $CH_3 - \underset{||}{C} - OH$ and $CH_3 - \underset{||}{C} - CH_2CH_2 - \underset{||}{C} - OH$. $O$ $O$ $O$

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

Answer 1

30

Answer 2

We first “decode” the clues. The oxidation‐products obtained from X (namely,

$$\rm CH_3COCH_3,\quad CH_3COOH,\quad CH_3COCH_2CH_2COOH,$$

when written with each carbonyl shown as “$\ce{C=O}$”–$\ce{CH_3COCH_3}$ is acetone, $\ce{CH_3COOH}$ is acetic acid, and the third is a (methyl–substituted) dicarboxylic acid) indicate that every oxidizable “side–chain” in X has been cleaved cleanly to give these fragments. Adding their carbon counts (3 + 2 + 6) shows that X originally contained 11 carbon atoms.

Also we are told that “X absorbs 2 moles of hydrogen”. (Hydrogenation adds H₂ across isolated double bonds only.) Hence X must contain two C=C bonds. (Its two degrees of unsaturation come solely from these double bonds.) Thus the molecular formula of X is deduced from “11 carbons with 2 C=C” i.e.   Saturated alkane with 11 C would be $\ce{C_{11}H_{24}}$. Two double bonds remove 4 hydrogens so   X = $\ce{C_{11}H_{20}}$.

Now, to “count sigma bonds” we note that in any acyclic (tree–like) structure the connectivity involves a total of (number of C–atoms −1) C–C bonds; here that equals 10. (Whether a bond is single or double, each (C–C) multiple bond has 1 sigma component; the extra bonds are pi bonds.) Also each hydrogen is attached by a sigma bond. Thus the number of C–H sigma bonds is 20.

Therefore the total number of sigma bonds present in X is   $10$ (C–C sigma bonds) $+20$ (C–H sigma bonds) $= 30.$