91Ó°ÊÓ

Skeletal structures are simpler than condensed structures.Hydrogen atoms connected to the carbon atoms are not shown in skeletal structures. The atoms other than carbon and hydrogen are shown.

(a) The degree of unsaturation for the compound ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}$can be found as:

$\begin{array}{l}\text{Degreeofunsaturation=}\frac{\text{2C+2+N-H-X}}{\text{2}}\\ \text{=}\frac{\text{2×5+2+0-12-0}}{\text{2}}\\ \text{=0}\end{array}$

The degree of unsaturation is 0. Hence the compound has no double bond or ring. The structure of the possible compounds with the formula can be given as:

Skeletal structure of compound (a)

(b)The degree of unsaturation for the compound ${\text{C}}_{\text{2}}{\text{H}}_{\text{7}}\text{N}$can be found as:

$\begin{array}{l}\text{Degreeofunsaturation=}\frac{\text{2C+2+N-H-X}}{\text{2}}\\ \text{=}\frac{\text{2×2+2+1-7-0}}{\text{2}}\\ \text{=0}\end{array}$

The degree of unsaturation is 0. Hence the compound has no double bond or ring. The structure of the possible compounds with the formula can be given as:

Skeletal structure of compound (b)

(c) The degree of unsaturation for the compound${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}$ can be found as:

The degree of unsaturation is 1. Hence the compound has one double bond or ring. The structure of the possible compounds with the formula can be given as:

Skeletal structure of compound (c)

(d)The degree of unsaturation for the compound ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}\text{Cl}$can be found as:

$\begin{array}{c}\text{Degreeofunsaturation=}\frac{\text{2C+2+N-H-X}}{\text{2}}\\ \text{=}\frac{\text{2×4+2+0-9-1}}{\text{2}}\\ \text{=0}\end{array}$

The degree of unsaturation is 0. Hence the compound has 0 double bonds or rings. The structure of the possible compounds with the formula can be given as:

Skeletal structures of compound (d)