Question

Which of the following does not belong to the same homologous series?

• A. ${\mathrm{CH}}_4$
• B. ${\mathrm{C}}_2{\mathrm{H}}_6$
• C. ${\mathrm{C}}_3{\mathrm{H}}_8$
• D. ${\mathrm{C}}_4{\mathrm{H}}_8$

Answer 1

The correct option is D

${\mathrm{C}}_4{\mathrm{H}}_8$

The explanation for the correct answer

(D) ${\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{8}}$

Homologous series

• It is a series of compounds with similar functional groups and chemical properties differing from the successive member by${\mathrm{CH}}_2$.
• Homologous series of Hydrocarbons are of Three types
• Alkanes - ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$
• Alkenes - ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}$
• Alkynes - ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-2}$
• Where, $\mathrm{n}=$the number of Carbon atoms.
• Here, ${\mathrm{C}}_4{\mathrm{H}}_8$ belongs to the ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}$series.
• Rest of the options belongs to the ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$series.

Hence, Option (D) ${\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{8}}\mathbf{}$is correct

The explanation for the incorrect answer

(A) ${\mathbf{CH}}_{\mathbf{4}}$

• Here $\mathrm{n}=1$
• ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}={\mathrm{CH}}_4$
• Therefore, ${\mathrm{CH}}_4$belongs to the ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$series.

(B) ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}$

• Here $\mathrm{n}=2$
• ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}={\mathrm{C}}_2{\mathrm{H}}_6$
• Therefore, ${\mathrm{C}}_2{\mathrm{H}}_6$belongs to the ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$series.

(C) ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}$

• Here $\mathrm{n}=3$
• ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}={\mathrm{C}}_3{\mathrm{H}}_8$
• Therefore, ${\mathrm{C}}_3{\mathrm{H}}_8$belongs to the ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$series.

Option (A) (B) (C) belongs to the same series.

Therefore, Option (D) ${\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{8}}\mathbf{}$is correct