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Chapter 11: Problem 71

What is the degree of unsaturation in cyclobutane? (a) Zero; (b) one; (c) two; (d) three

Short Answer

Expert verified
The degree of unsaturation in cyclobutane is one.

Step by step solution

01

- Identify the Molecular Formula

Cyclobutane has the molecular formula \( C_4H_8 \).
02

- Calculate the Saturated Formula

For an alkane (fully saturated hydrocarbon) with 4 carbon atoms, the formula is \( C_4H_{2n+2} \), where \(n\text{ equals the number of carbon atoms} \). Therefore, \(C_4H_{2(4) + 2} = C_4H_{10} \).
03

- Apply the Degree of Unsaturation Formula

The degree of unsaturation (DOU) is calculated using the formula: \[ \text{DOU} = \frac{2C + 2 - H}{2} \]. Plug in the values from Step 1 and Step 2: \[ \text{DOU} = \frac{2(4) + 2 - 8}{2} = \frac{10 - 8}{2} = 1 \].

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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
Cyclobutane is a cyclic hydrocarbon consisting of four carbon atoms arranged in a square structure. This cyclic compound has a molecular formula of \(C_4H_8\). In terms of geometry, the carbon atoms in cyclobutane form a ring, making it a member of the cycloalkanes. The ring structure introduces interesting characteristics: the bonds are slightly bent due to ring strain caused by the 90-degree angles, which is uncommon for carbon atoms preferring tetrahedral geometry with 109.5-degree angles.
However, understanding its properties also involves knowing the molecular formula and concept of degree of unsaturation.
headline of the respective core concept
The molecular formula represents the exact number of each type of atom in a molecule. For cyclobutane, the molecular formula is \(C_4H_8\). This means the molecule contains four carbon atoms and eight hydrogen atoms. Each carbon atom typically forms four bonds, and each hydrogen atom forms one bond.
The molecular formula helps in determining the nature of the compound, such as whether it is saturated, unsaturated, or aromatic. The formula is essential for calculating other important chemical properties, including degree of unsaturation.
headline of the respective core concept
A saturated hydrocarbon, also known as an alkane, contains only single bonds between carbon atoms and the maximum number of hydrogen atoms possible. The general formula for a saturated hydrocarbon with 'n' carbon atoms is \(C_nH_{2n+2}\). For example, butane (a straight-chain alkane) has a molecular formula of \(C_4H_{10}\), which fits the formula \(C_nH_{2n+2}\) for four carbon atoms.
Saturated hydrocarbons exhibit no degrees of unsaturation as they have the maximum hydrogen possible for their carbon skeleton, unlike cyclobutane.
headline of the respective core concept
The degree of unsaturation (DOU) provides insight into the level of hydrogen deficiency in a molecule. It helps identify whether a molecule contains double bonds, triple bonds, or rings. The DOU formula is: \[ \text{DOU} = \frac{2C + 2 - H}{2} \], where 'C' is the number of carbons, and 'H' is the number of hydrogens.
For cyclobutane (\(C_4H_8\)), the DOU calculation is: \[ \text{DOU} = \frac{2(4) + 2 - 8}{2} = \frac{10 - 8}{2} = 1 \].
This single DOU indicates one ring structure in cyclobutane, reflecting its cyclic nature and difference from a saturated hydrocarbon.

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Most popular questions from this chapter

Write the most likely major product(s) of each of the following haloalkanes with sodium ethoxide in ethanol or potassium tert-butoxide in 2-methyl-2-propanol (tert-butyl alcohol). (a) Chloromethane; (b) 1-bromopentane; (c) 2-bromopentane; (d) 1-chloro-1-methylcyclohexane; (e) (1-bromoethyl)-cyclopentane; (f) \((2 R, 3 R)-2\) -chloro-3-ethylhexane; (g) \((2 R, 3 S)-2\) -chloro-3ethylhexane; (h) \((2 S, 3 R)-2\) -chloro- 3 -ethylhexane.

Spectroscopic data for three compounds with the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{8}\) are given below; \(\mathrm{m}\) denotes a complex multiplet. Assign a structure to each compound. (Hint: One is acyclic; the others each contain one ring.) (a) IR \(910,1000,1650,3100 \mathrm{~cm}^{-1} ;{ }^{1} \mathrm{H}\) NMR \(\delta=2.79(\mathrm{t}, J=8 \mathrm{~Hz})\), 4.8-6.2 (m) ppm, integrated intensity ratio of the signals \(=1: 3\). (b) IR \(900,995,1650,3050 \mathrm{~cm}^{-1}\); \({ }^{1} \mathrm{H}\) NMR \(\delta=0.5 1.5(\mathrm{~m}), 4.8-6.0(\mathrm{~m}) \mathrm{ppm}\), integrated intensity ratio of the signals \(=5: 3\). (c) IR \(1611,3065 \mathrm{~cm}^{-1} ;{ }^{1} \mathrm{H}\) NMR \(\delta=1.5-2.5(\mathrm{~m}), 5.7(\mathrm{~m}) \mathrm{ppm}\), integrated intensity ratio of the signals \(=3: 1\). Is there more than one possibility?

You have just entered the chemistry stockroom to look for several isomeric bromopentanes. There are three bottles on the shelf marked \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br}\), but their labels have fallen off. The NMR machine is broken, so you devise the following experiment in an attempt to determine which isomer is in which bottle: You first treat a sample of the contents in each bottle with \(\mathrm{NaOH}\) in aqueous ethanol, and then you determine the IR spectrum of each product or product mixture. Here are the results: (i) \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br}\) isomer in bottle \(\mathrm{A} \stackrel{\mathrm{NaOH}}{\longrightarrow} \mathrm{IR}\) bands at \(1660,2850-3020\), and \(3350 \mathrm{~cm}^{-1}\) (ii) \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br}\) isomer in bottle \(\mathrm{B} \stackrel{\mathrm{NaOH}}{\longrightarrow} \mathrm{IR}\) bands at 1670 and \(2850-3020 \mathrm{~cm}^{-1}\) (iii) \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br}\) isomer in bottle \(\mathrm{C} \stackrel{\mathrm{NaOH}}{\longrightarrow} \mathrm{IR}\) bands at \(2850-2960\) and \(3350 \mathrm{~cm}^{-1}\) (a) What do the data tell you about each product or product mixture? (b) Suggest possible structures for the contents of each bottle.

Convert each of the following IR frequencies into micrometers. (a) \(1720 \mathrm{~cm}^{-1}(\mathrm{C}=\mathrm{O})\) (b) \(1650 \mathrm{~cm}^{-1}(\mathrm{C}=\mathrm{C})\) (c) \(3300 \mathrm{~cm}^{-1}(\mathrm{O}-\mathrm{H})\) (d) \(890 \mathrm{~cm}^{-1}\) (alkene bend) (e) \(1100 \mathrm{~cm}^{-1}(\mathrm{C}-\mathrm{O})\) (f) \(2260 \mathrm{~cm}^{-1}(\mathrm{C} \equiv \mathrm{N})\)

The molecular formulas and \({ }^{13} \mathrm{C}\) NMR data (in ppm) for several compounds are given here. The type of carbon, as revealed from DEPT spectra, is specified in each case. Deduce a structure for each compound. (a) \(\mathrm{C}_{4} \mathrm{H}_{6}: 30.2\left(\mathrm{CH}_{2}\right), 136.0(\mathrm{CH}) ;\) (b) \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}: 18.2\left(\mathrm{CH}_{3}\right), 134.9(\mathrm{CH}), 153.7(\mathrm{CH})\), \(193.4(\mathrm{CH}) ;(\mathrm{c}) \mathrm{C}_{4} \mathrm{H}_{8}: 13.6\left(\mathrm{CH}_{3}\right), 25.8\left(\mathrm{CH}_{2}\right), 112.1\left(\mathrm{CH}_{2}\right), 139.0(\mathrm{CH}) ;(\mathrm{d}) \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}: 17.6\left(\mathrm{CH}_{3}\right)\) \(25.4\left(\mathrm{CH}_{3}\right), 58.8\left(\mathrm{CH}_{2}\right), 125.7(\mathrm{CH}), 133.7\left(\mathrm{C}_{\text {quatemary }}\right) ;(\mathrm{e}) \mathrm{C}_{5} \mathrm{H}_{8}: 15.8\left(\mathrm{CH}_{2}\right), 31.1\left(\mathrm{CH}_{2}\right), 103.9\left(\mathrm{CH}_{2}\right)\) \(149.2\left(\mathrm{C}_{\text {quaternary }}\right) ;\) (f) \(\mathrm{C}_{7} \mathrm{H}_{10}: 25.2\left(\mathrm{CH}_{2}\right), 41.9(\mathrm{CH}), 48.5\left(\mathrm{CH}_{2}\right), 135.2(\mathrm{CH}) .\) (Hint: This one is difficult. The molecule has one double bond. How many rings must it have?)
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