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Chapter 17: Problem 49

Make line drawings for all the isomers of \(\mathrm{C}_{6} \mathrm{H}_{14}\).

Short Answer

Expert verified
The isomers of C6H14 are as follows: 1. \(n-Hexane: CH_3-CH_2-CH_2-CH_2-CH_2-CH_3\) 2. \(2-Methylpentane: CH_3-CH(CH_3)-CH_2-CH_2-CH_2-CH_3\) 3. \(3-Methylpentane: CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3\) 4. \(2,2-Dimethylbutane: CH_3-CH_2-CH_2-CH(CH_3)-CH_2-CH_3\) 5. \(2,3-Dimethylbutane: CH_3-CH_2-CH(CH_3)-CH(CH_3)-CH_2-CH_3\)

Step by step solution

01

Identify the longest possible carbon chain

In this step, we will find the longest possible carbon chain that can be formed, which will be our parent structure. As the molecule has 6 carbon atoms, we will arrange them in a straight chain. \(CH_3-CH_2-CH_2-CH_2-CH_2-CH_3\)
02

Create branched isomers

Now, we will look for the branched isomers that can be formed by C6H14. We start by moving one of the terminal CH3 groups to the 2nd carbon atom, thus forming a branched structure. \(CH_3-CH(CH_3)-CH_2-CH_2-CH_2-CH_3\)
03

Create more branched isomers

Next, we will continue changing the position of the CH3 group to form new isomers. In this step, we will move the CH3 group to the 3rd carbon atom. \(CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3\)
04

Chain shortening

Now, let's shorten the main carbon chain and investigate the new branching possibilities. \(CH_3-CH_2-CH_2-CH(CH_3)-CH_2-CH_3\)
05

Create the last branched isomer

In this final step, we will form the last isomer by moving the CH3 group to the central carbon atom. \(CH_3-CH_2-CH(CH_3)-CH(CH_3)-CH_2-CH_3\) These are all the isomers of C6H14: 1. \(n- Hexane:\, CH_3-CH_2-CH_2-CH_2-CH_2-CH_3\) 2. \(2- Methylpentane:\, CH_3-CH(CH_3)-CH_2-CH_2-CH_2-CH_3\) 3. \(3- Methylpentane:\, CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3\) 4. \(2,2- Dimethylbutane:\, CH_3-CH_2-CH_2-CH(CH_3)-CH_2-CH_3\) 5. \(2,3- Dimethylbutane:\, CH_3-CH_2-CH(CH_3)-CH(CH_3)-CH_2-CH_3\)

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Most popular questions from this chapter

What do the letters in the acronym IUPAC stand for?

A hydrocarbon molecule consists of a chain of four carbon atoms. In this chain there are one \(\mathrm{C}=\mathrm{C}\) double bond and two \(\mathrm{C}-\mathrm{C}\) single bonds. What is the molecular formula? Draw a possible structure.

The geometry about each carbon in a particular hydrocarbon molecule is tetrahedral. What does this say about the possibility of there being \(C=C\) or \(C \equiv C\) bonds in the molecule? Explain.

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Which of the molecules in Problem \(17.72\) represent a pair of isomers? Discuss their differences in terms of bonds and/or linear versus branched.
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