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Chapter 3: Problem 47

Find all the compounds of the formula \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{Br}_{2}\) in which the dipole moment is zero. It is easiest to divide this long problem into sections. First, work out the number of degrees of unsaturation so that you can see what kinds of molecules to look for. Then, find the acyclic molecules. Next, work out the compounds containing a ring.

Short Answer

Expert verified
The compounds are 2,3-dibromobutene and 1,3-dibromocyclobutane.

Step by step solution

01

Calculate the Degrees of Unsaturation

The molecular formula is \(\text{C}_{4}\text{H}_{6}\text{Br}_{2}\). To calculate the degrees of unsaturation, use the formula \[ \text{DU} = \frac{2C + 2 - H + N - X}{2} \] where \(C\) is the number of carbons, \(H\) is the number of hydrogens, \(N\) is the number of nitrogens, and \(X\) is the number of halogens. For \( \text{C}_{4}\text{H}_{6}\text{Br}_{2} \): \[ \text{DU} = \frac{2(4) + 2 - 6 + 0 - 2}{2} = \frac{8 + 2 - 6 - 2}{2} = 1 \] So, the degrees of unsaturation is 1.
02

Identify Acyclic Compounds

Next, determine the possible acyclic structures. Considering the molecular formula and the single degree of unsaturation, the compound must contain either a double bond or a triple bond. Possible acyclic structures include 1,4-dibromobutene, 2,3-dibromobutene, and others. However, only symmetric structures result in zero dipole moments. So, examine the acyclic structures to find symmetric ones.
03

Identify Cyclic Compounds

Now, consider the compounds containing rings. A cyclic compound can also account for the degree of unsaturation. Identify possible cyclic structures such as 1,2-dibromocyclobutene, 1,3-dibromocyclobutene, etc. Evaluate the dipole moments for these structures and determine which configurations produce zero dipole moments. Symmetric structures such as 1,3-dibromocyclobutane can lead to zero dipole moments.
04

Verify Symmetrical Structures

Evaluate the symmetry of each identified structure. For a compound to have a zero dipole moment, it must be symmetric so that any individual dipole moments cancel out each other. Confirm that structures like 2,3-dibromobutene and 1,3-dibromocyclobutane are symmetric and thus have zero dipole moments.

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Key Concepts

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

Degrees of Unsaturation
Degrees of unsaturation is an important concept in organic chemistry. It provides information about the presence of rings and multiple bonds in a molecule. One degree of unsaturation (DU) corresponds to either one double bond, one ring, or one triple bond. To calculate the degrees of unsaturation, we can use the formula:
In the case of , the degrees of unsaturation is calculated as follows: where . This tells us that the molecule has one degree of unsaturation, which means it could have either a double bond or a ring.
Dipole Moment
The dipole moment is a measure of the separation of positive and negative charges within a molecule. It is represented by a vector with both magnitude and direction. A molecule with a net dipole moment is polar, while a molecule with zero dipole moment is non-polar. For a molecule to have zero dipole moment, it must be symmetric, so that the dipole moments of any bonds cancel each other out. This happens when:
  • The molecule has a symmetrical structure.
  • The vector sum of the individual bond dipole moments is zero.
Thus, identifying symmetric structures is crucial for determining whether a molecule could have a zero dipole moment.
Acyclic Compounds
Acyclic compounds do not contain rings. For a molecule like , acyclic structures must be considered to find those with zero dipole moments. Acyclic structures can feature double bonds, given the degree of unsaturation of 1. A few possible acyclic isomers include 1,4-dibromobutene and 2,3-dibromobutene. However, only symmetric isomers will exhibit zero dipole moments. For example:
  • 2,3-Dibromobutene is symmetric along the midline, leading to zero dipole moment.
  • 1,4-Dibromobutene is not symmetric and thus does not have zero dipole moment.
Cyclic Compounds
Cyclic compounds feature rings, which contribute to degrees of unsaturation. We can consider cyclic isomers for . These compounds, like cyclic butenes, can account for the degree of unsaturation. Examples include:
  • 1,2-Dibromocyclobutene
  • 1,3-Dibromocyclobutene
    • However, only 1,3-dibromocyclobutane has the symmetry required to yield a zero dipole moment. The bromine atoms are positioned symmetrically, causing the dipole moments to cancel out.
Symmetry in Molecules
Symmetry is a key factor in determining the dipole moment of a molecule. For a compound to have a zero dipole moment, it must be symmetrical. Symmetry ensures that the dipole moments of individual bonds cancel each other out. Here’s what to keep in mind:
  • Molecules with a center of symmetry or a plane of symmetry are more likely to have zero dipole moments.
  • In our case, 2,3-dibromobutene and 1,3-dibromocyclobutane are symmetric, leading them to have zero net dipole moments.
  • Verifying symmetric structures involves checking the uniform distribution of identical substituents on opposite sides of the molecule.

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Most popular questions from this chapter

Alkanes combine with oxygen to produce carbon dioxide and water according to the following scheme: $$ \mathrm{C}_{n} \mathrm{H}_{2 n+2}+(3 n+1) / 2 \mathrm{O}_{2} \rightarrow(n+1) \mathrm{H}_{2} \mathrm{O}+n \mathrm{CO}_{2} $$ This process is generally referred to as combustion. An important use of this reaction is the quantitative determination of elemental composition (elemental analysis). Typically, a small sample of the compound is completely burned, and the water and carbon dioxide produced are collected and weighed. From the weight of water, the amount of hydrogen in the original compound can be determined. Similarly, the amount of carbon dioxide formed allows us to determine the amount of carbon in the original compound. Oxygen, if present, is usually determined by difference. The determination of the relative molar proportions of carbon and hydrogen in a compound is the first step in deriving its molecular formula. If combustion of \(5.00 \mathrm{mg}\) of a hydrocarbon gives \(16.90\) \(\mathrm{mg}\) of carbon dioxide and \(3.46 \mathrm{mg}\) of water, what are the weight percents of carbon and hydrogen in the sample?

A compound containing only carbon, hydrogen, and oxygen was found to contain \(70.58 \%\) carbon and \(5.92 \%\) hydrogen by weight. Calculate the empirical formula for this compound. If the compound has a molecular weight of approximately \(135 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula?

Although you should be able to draw structures for the following compounds, they are named incorrectly. Give the correct systematic name for each compound: (a) 5 -chloro-2-methyl-6-vinyldecane (b) 3 -isopropyl-1-pentyne (c) 4-methyl-6-hepten-1-yne (d) \((E)-3\)-propyl-2, 5 -hexadiene

Draw and write the systematic names for 10 of the acyclic isomers of 1 -heptene \(\left(\mathrm{C}_{7} \mathrm{H}_{14}\right)\). There are 36 isomers, including 1-heptene.

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. $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH} \quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3} $$
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