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Chapter 3: Problem 23

There are three isomers of \(\mathrm{C}_{5} \mathrm{H}_{8}\) that are alkynes. Draw them. There are seven isomers of \(\mathrm{C}_{6} \mathrm{H}_{10}\) that are alkynes. Draw them.

Short Answer

Expert verified
The three isomers of \[ \text{C}_{5}\text{H}_{8} \] are 1-pentyne, 2-pentyne, and 3-methyl-1-butyne. The seven isomers of \[ \text{C}_{6}\text{H}_{10} \] are 1-hexyne, 2-hexyne, 3-hexyne, 3-methyl-1-pentyne, 4-methyl-2-pentyne, 3-methyl-2-pentyne, and 2-ethyl-1-butyne.

Step by step solution

01

Identify the alkynes for \[ \text{C}_{5}\text{H}_{8} \]

The molecular formula \[ \text{C}_{5}\text{H}_{8} \] suggests that there are three isomers of alkynes. These are molecules containing a triple bond. The structure of these isomers can vary by the position of the triple bond. Draw the structures: \[ \text{CH}_3-\text{C} \equiv\ \text{C}-\text{CH}_2-\text{CH}_3 \] (1-pentyne), \[ \text{CH}_3-\text{CH}_2-\text{C} \equiv\ \text{C}-\text{CH}_3 \] (2-pentyne), and \[ \text{CH}_3-\text{CH}(\text{C} \equiv\ \text{C})-\text{CH}_3 \] (3-methyl-1-butyne).
02

Identify the alkynes for \[ \text{C}_{6}\text{H}_{10} \]

The molecular formula \[ \text{C}_{6}\text{H}_{10} \] suggests that there are seven isomers of alkynes. These include variations of carbon chain length and the position of the triple bond. Draw the structures: \[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{C} \equiv\ \text{C}-\text{CH}_3 \] (1-hexyne), \[ \text{CH}_3-\text{CH}_2-\text{C} \equiv\ \text{C}-\text{CH}_2-\text{CH}_3 \] (2-hexyne), \[ \text{CH}_3-\text{CH}(\text{C} \equiv\ \text{C})-\text{CH}_2-\text{CH}_3 \] (3-hexyne), \[ (\text{CH}_3)_2\text{CH}-\text{C} \equiv\ \text{C}-\text{CH}_3 \] (3-methyl-1-pentyne), \[ (\text{CH}_3)_2\text{CH}-\text{C} \equiv\ \text{C}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \] (4-methyl-2-pentyne), \[ \text{CH}_3-\text{CH}_2-\text{C} \equiv\ \text{C}-\text{CH}(\text{CH}_3)-\text{CH}_3 \] (3-methyl-2-pentyne), and \[ \text{CH}_3-\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{C} \equiv\ \text{C}\text{CH}_3 \] (2-ethyl-1-butyne).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

headline of the respective core concept
Understanding alkyne isomers involves a few core concepts that are integral to grasp the topic fully. Let's break down the critical elements: Molecular Formula, Triple Bonds, Isomer Structures, and Structural Isomerism. We'll discuss each in detail to ensure a clear understanding of alkyne isomers.
Molecular Formula
A molecular formula provides essential information about the composition of a molecule. For example, in the exercise, we see two molecular formulas: \( \text{C}_5\text{H}_8 \) and \( \text{C}_6\text{H}_{10} \). The subscript numbers next to the carbon (C) and hydrogen (H) elements tell us how many atoms of each are present.
Identifying correct molecular formulas is crucial when determining possible isomers. The number of carbon and hydrogen atoms suggest different structures that maintain the same number of these atoms but differ in their arrangement. This leads us to the concept of isomerism.
Triple Bonds
Alkynes are hydrocarbons that feature at least one triple bond between carbon atoms. A triple bond consists of one sigma bond and two pi bonds, leading to a linear structure. This nature of triple bonds significantly affects the molecule's properties and possible isomeric forms.
When dealing with isomers of alkynes, the location of this triple bond in the carbon chain is crucial. Different placements of the triple bond result in different isomers even if the rest of the molecule remains unchanged. For instance, in \( \text{C}_5\text{H}_8 \), positioning the triple bond from the end (1-pentyne) versus near the middle (2-pentyne) creates different structures.
Isomer Structures
Isomers are compounds that have the same molecular formula but different structures. In alkyne isomers, the different structures arise from:
  • changing the position of the triple bond
  • varying the length of the carbon chain
  • adding or moving branches within the carbon chain
These structural variations don't affect the molecular formula but result in different compounds. For example, \( \text{C}_6\text{H}_{10} \) can form seven different isomers by varying the triple bond's position and altering branches in the carbon chain, as shown in the exercise.
Structural Isomerism
Structural isomerism refers to the phenomenon where molecules with the same molecular formula have different physical and chemical structures. In the context of alkynes, structural isomerism often involves shifts in the position of the triple bond and changes in the carbon backbone.
For \( \text{C}_5\text{H}_8 \), the three structures (1-pentyne, 2-pentyne, and 3-methyl-1-butyne) show how rearranging the carbon atoms or triple bond placement leads to distinct isomers. Similarly, \( \text{C}_6\text{H}_{10} \) contains seven different isomers displaying structural isomerism due to varying triple bond locations and different branching patterns in the carbon chain.
Understanding structural isomerism is key to appreciating the diversity of organic compounds possible with a given molecular formula.

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Most popular questions from this chapter

Find the compound of the formula \(\mathrm{C}_{10} \mathrm{H}_{16}\) that is composed of two six-membered rings and which has only three different carbons.

Calculate the degrees of unsaturation \((\Omega)\) for the following formulas. Give two possible structures for each one. (a) \(\mathrm{C}_{5} \mathrm{H}_{6}\) (b) \(\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}\) (c) \(\mathrm{C}_{7} \mathrm{H}_{10}\) (d) \(\mathrm{C}_{5} \mathrm{H}_{8} \mathrm{Br}_{2}\)

Devise syntheses of the following three compounds, starting in each case with any alkene that contains four carbons or fewer. You do not have to write mechanisms, although at this point it may be very helpful for you to do so. $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH} \quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3} $$

Develop a bonding scheme for the compound \(\mathrm{H}_{2} \mathrm{CO}\) in which both carbon and oxygen are hybridized \(s p^{2}\).

Draw structures for the following compounds: (a) \((Z)-3\)-fluoro-2-methyl-3-hexene (b) 2,6 -diethyl-1,7-octadien-4-yne (c) trans-3-bromo-7-isopropyl-5-decene (d) 5-chloro-4-iodo-6-methyl-1-heptyne
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