Question

Identify the pair of alkanes which can exhibit chain isomerism and have a neo isomer respectively?
A. ${{\text{C}}_{3}}{{\text{H}}_{8}}\text{, }{{\text{C}}_{4}}{{\text{H}}_{10}}$
B. ${{\text{C}}_{4}}{{\text{H}}_{10}}\text{, }{{\text{C}}_{5}}{{\text{H}}_{12}}$
C. ${{\text{C}}_{5}}{{\text{H}}_{12}}\text{, }{{\text{C}}_{6}}{{\text{H}}_{14}}$
D. ${{\text{C}}_{6}}{{\text{H}}_{14}},{{\text{C}}_{7}}{{\text{H}}_{16}}$

Answer 1

The general configuration of alkanes is ${{\text{C}}_{\text{n}}}{{\text{H}}_{2\text{n}+\text{2}}}$, where n is the natural number. Draw the isomers of the given alkanes to check whether they show chain and neo isomerism or not. Also, the minimum requirement of carbon atoms for this isomerism should be known, which is 4 in chain and 5 in neo isomerism.

Complete answer:
Let us first discuss what chain and neo isomers are:
Chain isomers- The carbon compounds with the same molecular formula but have different atomic arrangements, the one with branches. The minimum number of carbon atoms required for chain isomerism are 4.
Neo isomers- In this, a carbon atom is attached to four other carbon atoms. Neo-isomers are the type of chain isomers only. The minimum number of carbon atoms required for neo isomers are 5.
Let us discuss the options one-by-one along with their chain and neo isomers:
A) ${{\text{C}}_{3}}{{\text{H}}_{8}}\text{, }{{\text{C}}_{4}}{{\text{H}}_{10}}$- The compound ${{\text{C}}_{3}}{{\text{H}}_{8}}$ is propane. As, the number of carbon atoms are less than 4, so it cannot have chain isomerism. ${{\text{C}}_{4}}{{\text{H}}_{10}}$ is butane. It has four carbon atoms, so it can form chain isomers. But, it cannot form neo isomers. The only isomer of butane is 2-methyl propane. The only isomer is

B) ${{\text{C}}_{4}}{{\text{H}}_{10}}\text{, }{{\text{C}}_{5}}{{\text{H}}_{12}}$- ${{\text{C}}_{4}}{{\text{H}}_{10}}$is butane. It has four carbon atoms, so it can form chain isomers. But, it cannot form neo isomers. As given in the above part the only isomer of butane is 2-methyl propane.
The compound ${{\text{C}}_{5}}{{\text{H}}_{12}}$ is pentane. It has both chain and neo-isomer, as it has 5 carbon atoms. The chain isomer is 2-methyl butane. The neo isomer is 2,2-dimethylpropane.

C)${{\text{C}}_{5}}{{\text{H}}_{12}}\text{, }{{\text{C}}_{6}}{{\text{H}}_{14}}$: As discussed in the above solution part, the compound ${{\text{C}}_{5}}{{\text{H}}_{12}}$ is pentane. It has both chain and neo-isomer. The chain isomer is 2-methyl butane. The neo isomer is 2, 2-dimethylpropane.
The compound ${{\text{C}}_{6}}{{\text{H}}_{14}}$ is hexane. It has both chain and neo-isomer, as it has more than 5 carbon atoms. The chain isomers are 2-methyl pentane and 2,3-dimethyl butane. The neo isomers are 2,2-dimethylbutane.

D)${{\text{C}}_{6}}{{\text{H}}_{14}},{{\text{C}}_{7}}{{\text{H}}_{16}}$: As given in the above solution part we know that hexane has both chain and neo-isomers. The chain isomers are 2-methyl pentane and 2,3-dimethyl butane. The neo isomers are 2, 2-dimethylbutane.
The compound ${{\text{C}}_{7}}{{\text{H}}_{16}}$ is heptane. It has both chain and neo-isomer, as it has more than 5 carbon atoms. The chain isomers are 2-methyl hexane, 2,3-dimethyl pentane and 2,4-dimethyl pentane.

The neo isomers are 2,2-dimethyl-3-methyl butane, 2,2-dimethyl pentane and 3,3-dimethyl pentane.
${{\text{C}}_{5}}{{\text{H}}_{12}}\text{, }{{\text{C}}_{6}}{{\text{H}}_{14}}$ and ${{\text{C}}_{6}}{{\text{H}}_{14}},{{\text{C}}_{7}}{{\text{H}}_{16}}$ are the pair of alkanes which can exhibit chain isomerism and has a neo isomer respectively.

The correct options are (c) and (d).

Note:
As the branching in the carbon chain increases, the surface area decreases and vander waal forces of attraction decrease. The force of attraction decreases, thus, the boiling point of alkane is also decreased. The neo isomers have less boiling point than chain isomers.