Question

What is meant by homologous series of carbon compounds? Classify the following carbon compounds into two homologous series and name them.\[{{\mathbf{C}}_3}{{\mathbf{H}}_4}\], \[{{\mathbf{C}}_3}{{\mathbf{H}}_6}\], \[{{\mathbf{C}}_4}{{\mathbf{H}}_6}\], \[{{\mathbf{C}}_5}{{\mathbf{H}}_8}\],\[{{\mathbf{C}}_5}{{\mathbf{H}}_{10}}\].

Answer 1

Hint:In the above-given question, we are asked about the homologous series. A homologous series is a phenomenon which can be shown in the organic compounds. Hydrocarbons, which are compounds which are made from pure carbon and hydrogen, may include some other elements, Can form homologous series.

Complete step by step answer:
A homologous series is the order of organic compounds which belong to the same function group but increases gradually by a methylene group that is $C{H_2}$. The successor or predecessor of a compound can be found by adding or subtracting methylene respectively. And any series starts with the prefix meth, eth, prop, but, pent, hex, hept, oct, non, dec and so on meth stands for one carbon, eth for two carbon, prop for three and so on. There are various series of compounds in the organic chemistry of hydrocarbons, that is alkanes, alkenes and alkynes, alkanes consist of carbon-carbon single bond Alkenes consist of carbon-carbon double bond, and alkyne consists of carbon-carbon triple bond.
This series contains just a general formula. Each series has a different general formula. The general formula of alkanes, alkenes and alkynes are described below.
Alkanes-\[{{\mathbf{C}}_n}{{\mathbf{H}}_{{\mathbf{2n}} + {\mathbf{2}}}}\]
Alkenes-\[{{\mathbf{C}}_n}{{\mathbf{H}}_{{\mathbf{2n}}}}\]
Alkynes- ${C_n}{H_{2n - 2}}$
Now the compound given in the question may belong to an alkane, alkene or alkynes, we can find it using its general formula, taking the value of carbon and hydrogen from the compound, we will equate the number of hydrogen to the equation of number of hydrogen in the general formula;

Compound	Number of carbon	number of hydrogen	Alkane\[{\mathbf{2n}} + {\mathbf{2}}\]	Alkene\[{\mathbf{2n}}\]	Alkyne\[{\mathbf{2n}} - {\mathbf{2}}\]	Functional group present
\[{{\mathbf{C}}_3}{{\mathbf{H}}_4}\]	$3$	$4$	\[{\mathbf{2}}\left( {\mathbf{3}} \right) + {\mathbf{2}} = {\mathbf{4}}\]\[{\mathbf{6}} + {\mathbf{2}} = {\mathbf{4}}\]$8 \ne 4$	$2(3) = 4$$6 \ne 4$	$2(3) - 2 = 4$$6 - 2 = 4$$4 = 4$	alkynes
\[{{\mathbf{C}}_3}{{\mathbf{H}}_6}\]	$3$	$6$	$2(3) + 2 = 6$$6 + 2 = 6$$8 \ne 6$	$2(3) = 6$$6 = 6$	$2(3) - 2 = 6$$6 - 2 = 6$$4 \ne 6$	alkenes
\[{{\mathbf{C}}_4}{{\mathbf{H}}_6}\]	$4$	$6$	$2(4) + 2 = 6$$8 + 2 = 6$$10 \ne 6$	$2(4) = 6$$8 \ne 6$	$2(4) - 2 = 6$$8 - 2 = 6$$6 = 6$	alkynes
\[{{\mathbf{C}}_5}{{\mathbf{H}}_8}\]	$5$	$8$	$2(5) + 2 = 8$$10 + 2 = 8$$12 \ne 8$	$2(5) = 8$$10 \ne 8$	$2(5) - 2 = 8$$10 - 2 = 8$$8 = 8$	alkynes
\[{{\mathbf{C}}_5}{{\mathbf{H}}_{10}}\]	$5$	$10$	$2(5) + 2 = 10$$10 + 2 = 10$$12 \ne 10$	$2(5) = 10$$10 = 10$	$2(5) - 2 = 10$$10 - 2 = 10$$12 \ne 10$	alkenes

Note:
The general formula of a functional group can be used to find if a compound does belong to that functional group or does not belong to that functional group, you just need to equate the given values with the general equation.