Question

Preceding and succeeding homologous of ${{C}_{10}}{{H}_{22}}$ respectively:
(A) ${{C}_{9}}{{H}_{20}},{{C}_{11}}{{H}_{22}}$
(B) ${{C}_{9}}{{H}_{20}},{{C}_{11}}{{H}_{24}}$
(C) ${{C}_{11}}{{H}_{24}},{{C}_{9}}{{H}_{20}}$
(D) ${{C}_{8}}{{H}_{18}},{{C}_{9}}{{H}_{20}}$

Answer 1

Before trying to solve the question, first see which type of compound is given in the question. Is it a saturated or unsaturated compound? If unsaturated, then we have to see if it is an alkene or an alkyne and then try to find the homologous series of the compound.

Complete step by step solution:
-There are 2 types of compounds- saturated and unsaturated compounds. Alkanes are termed as saturated compounds while alkenes and alkynes are unsaturated compounds.
-Alkanes are the open-chain aliphatic hydrocarbons. They are also called paraffin. Their general formula is ${{C}_{n}}{{H}_{2n+2}}$ where n is the natural number set.
-Alkenes are open-chain unsaturated aliphatic compounds. They are also called olefins. Their general formula is of the form ${{C}_{n}}{{H}_{2n}}$ where n is the natural number set.
-Alkynes are also open-chain unsaturated hydrocarbons which are less reactive than alkenes. They contain carbon-carbon triple bonds. Their general formula is denoted as ${{C}_{n}}{{H}_{2n}}_{-2}$ where n is the natural number set.
-Since n is a natural number, a very high number of compounds can be formed. To make the study of such compounds easier, they are combined together in the form of homologous series. Homologous series is a series of compounds of same functional groups, same chemical properties and also same general formula.
-Members of homologous series differ only in the $C{{H}_{2}}$ group. This makes it easy to study the type of reactions, dipole moment, boiling and melting points, reactivity, and all other properties of compounds of such types.
-Preceding homologous series will have 1$C{{H}_{2}}$ group less and succeeding series will have 1 group extra.
-Now observing the given compound, we find that it is of the form ${{C}_{n}}{{H}_{2n+2}}$. Thus the given compound is an alkane.
-Therefore the preceding compound of ${{C}_{10}}{{H}_{22}}$ can be found by directly reducing 1 $C{{H}_{2}}$group. So it will be ${{C}_{10-1}}{{H}_{22}}_{-2}$ ; i.e. ; ${{C}_{9}}{{H}_{20}}$ and the succeeding compound will be ${{C}_{10}}_{+1}{{H}_{22}}_{+2}$ which is ${{C}_{11}}{{H}_{24}}$.

Hence the correct answer is option B. ${{C}_{9}}{{H}_{20}},{{C}_{11}}{{H}_{24}}$.

Note: Always first verify if the compounds have the same functional group before considering them as homologous series. Check the general formula and then the functional groups. We may also observe that certain compounds will have the same general formula but are different homologous series as the functional groups contain ambidentate ions. Eg, ONO and $N{{O}_{2}}$.