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Chapter 1: Problem 19

Draw the two constitutional isomers with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10}\), showing all atoms and their bonds.

Short Answer

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CH3-CH2-CH2-CH3 (Butane)and(CH3)CH-(CH3)CH3 (Isobutane)

Step by step solution

01

- Understand Constitutional Isomers

Constitutional isomers have the same molecular formula but differ in the connectivity of their atoms. For \(\text{C}_{4}\text{H}_{10}\), determine the different ways to arrange the carbon atoms.
02

- Draw the Straight-Chain Isomer

Draw the isomer with a straight chain of four carbon atoms (butane): C-C-C-C Then add the hydrogen atoms to ensure each carbon has four bonds.

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Key Concepts

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

Constitutional Isomers
Constitutional isomers are a type of isomer where molecules have the same molecular formula, but the atoms are connected in different ways. This means that while they have the same number of each type of atom, the structure of how these atoms are linked together varies. For example, in the case of \(\text{C}_{4}\text{H}_{10}\), which is butane, there are multiple ways to connect the four carbon atoms to generate different structures. Understanding this is essential because it affects the physical and chemical properties of the molecules.
Molecular Formula
A molecular formula gives us the number and types of atoms in a molecule. It does not tell us anything about how these atoms are connected. For example, \(\text{C}_{4}\text{H}_{10}\) tells us that there are four carbon atoms and ten hydrogen atoms. This formula can correspond to more than one structure, which is why we can have different isomers with the same molecular formula. This is a critical concept because knowing the molecular formula alone isn't enough to understand a molecule's structure or properties. The connectivity of atoms is necessary to get the full picture.
Carbon Connectivity
Carbon connectivity refers to the way carbon atoms are linked (or bonded) to each other and to other atoms. In the context of constitutional isomers, changing which carbon atoms are connected to each other can lead to different isomers. For example, the \(\text{C}_{4}\text{H}_{10}\) molecule can have a straight-chain arrangement (n-butane) where all carbon atoms form a continuous chain, or a branched arrangement (isobutane) where one carbon atom is connected to three others, forming a branch. Understanding carbon connectivity is crucial for determining the structure and classification of different organic molecules.
Butane
Butane is a hydrocarbon with the molecular formula \(\text{C}_{4}\text{H}_{10}\). It exists as two constitutional isomers: n-butane and isobutane. n-Butane has a straight chain of four carbon atoms: C-C-C-C. In n-butane, all carbon atoms are connected in a straight line, and the hydrogens are arranged to ensure each carbon has four bonds. Isobutane, the other isomer, has a branched structure: (CH鈧�)鈧僀-CH鈧�. In this form, three carbon atoms are connected to the central carbon atom, forming a T-shaped branch. These distinct structures show how butane can exist in different forms, affecting its properties such as boiling point and reactivity.

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Most popular questions from this chapter

(a) Draw two resonance forms for nitrite ion, \(\mathrm{NO}_{2}^{-}\). What can you say about the geometry of this molecule (linear or bent)? (Hint: Consider the effect of electron repulsion exerted by the lone pair on nitrogen.) (b) The possibility of valence-shell expansion increases the number of feasible resonance forms, and it is often difficult to decide on one that is "best." One criterion that is used is whether the Lewis structure predicts bond lengths and bond angles with reasonable accuracy. Draw Lewis octet and valence-shell-expanded resonance forms for \(\mathrm{SO}_{2}\) (OSO). Considering the Lewis structure for SO (Exercise \(1-8)\), its experimental bond length of \(1.48 \mathrm{~A}\), and the measured \(\mathrm{S}-\mathrm{O}\) distance in \(\mathrm{SO}_{2}\) of \(1.43 \mathrm{~A}\), which one of the various structures would you consider "best"?

Draw the Lewis structure of \(\mathrm{HClO}_{2}\) (HOCIO), including the assignment of any charges to atoms. Strategy To solve such a problem, it is best to follow one by one the rules given in this section for drawing Lewis structures. Solution \- Rule 1: The molecular skeleton is given as unbranched, as shown. \- Rule 2: Count the number of valence electrons: $$ \mathrm{H}=1,2 \mathrm{O}=12, \mathrm{Cl}=7, \text { total }=20 $$ \- Rule 3: How many bonds (shared electron pairs) do we need? The supply of electrons is 20 ; the electron requirement is 2 for \(\mathrm{H}\) and \(3 \times 8=24\) electrons for the other three atoms, for a total of 26 electrons. Thus we need \((26-20) / 2=3\) bonds. To distribute all valence electrons according to the octet rule, we first connect all atoms by two-electron bonds, H:O:Cl:O, using up 6 electrons. Second, we distribute the remaining 14 electrons to provide octets for all nonhydrogen atoms, (arbitrarily) starting at the left oxygen. This process requires in turn 4,4, and 6 electrons, resulting in octet structures without needing additional electron sharing: \- Rule 4: We determine any formal charges by noting any discrepancies between the "effective" valence electron count around each atom in the molecule we have found and its outer-shell count when isolated. For \(\mathrm{H}\) in \(\mathrm{HOClO}\), the valence electron count is 1 , which is the same as in the \(\mathrm{H}\) atom, so it is neutral in the molecule. For the neighboring oxygen, the two values are again the same, 6 . For \(\mathrm{Cl}\), the effective electron count is 6 , but the neutral atom requires 7 . Therefore, Cl bears a positive charge. For the terminal \(\mathrm{O}\), the electron counts are 7 (in the molecule) and 6 (neutral atom), giving it a negative charge. The final result is

Draw a Lewis structure for each of the following molecules and assign charges where appropriate. The order in which the atoms are connected is given in parentheses. (a) CIF (b) \(\mathrm{BrCN}\) (c) \(\mathrm{SOCl}_{2}(\mathrm{CISCl})\) (d) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (e) \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) (f) \(\mathrm{N}_{2} \mathrm{H}_{2}\) (HNNH) (g) \(\mathrm{CH}_{2} \mathrm{CO}\) (h) \(\mathrm{HN}_{3}\) (HNNN) (i) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{NNO})\)

Draw Lewis structures for the following molecules: \(\mathrm{HI}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{OH}, \mathrm{HSSH}, \mathrm{SiO}_{2}(\mathrm{OSiO})\), \(\mathrm{O}_{2}, \mathrm{CS}_{2}(\mathrm{SCS})\)

Ammonia, \(\mathrm{NH}_{3}\), is not trigonal but pyramidal, with bond angles of \(107.3\). Water, \(\mathrm{H}_{2} \mathrm{O}\), is not linear but bent \(\left(104.5^{\circ}\right)\). Why? (Hint: Consider the effect of the nonbonding electron pairs.)
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