Question

Propose skeletal structures for compounds that satisfy the following molecular formulas. There is more than one possibility in each case.

(a) ${\mathbf{C}}_5{\mathbf{H}}_{12}$

(b) ${\mathbf{C}}_2{\mathbf{H}}_7\mathbf{N}$

(c)${\mathbf{C}}_3{\mathbf{H}}_6\mathbf{O}$

(d)${\mathbf{C}}_4{\mathbf{H}}_9\mathbf{Cl}$

Short answer

a.

b.

c.

d.

Step-by-step solution

Skeletal Structures

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.

Step 2: Proposing the skeletal structures for the following compounds

(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)