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Chapter 24: Problem 28

Draw all the possible noncyclic structural isomers of \(\mathrm{C}_{5} \mathrm{H}_{10}\) Name each compound.

Short Answer

Expert verified
The six possible noncyclic structural isomers of C5H10 are: 1. Pent-1-ene: C=C-C-C-C 2. Pent-2-ene: C-C=C-C-C 3. 3-methylbut-1-ene: C=C-C(C)-C 4. 2-methylbut-2-ene: C-C=C(C)-C 5. 2-methylbut-1-ene: C-C(C=C)-C 6. 3-methylbut-1-ene: C-C(C)-C=C

Step by step solution

01

Calculate the degree of unsaturation

To find the index of hydrogen deficiency/degree of unsaturation, we use the formula: IHD = (2 * Number_of_carbon + 2 - Number_of_hydrogen) / 2 For C5H10: IHD = (2 * 5 + 2 - 10) / 2 IHD = (10 + 2 - 10) / 2 IHD = 1 We have a degree of unsaturation of 1, which means there is one double bond or ring in each of our isomers. Since we are specifically looking for noncyclic structural isomers, each of our isomers will have one double bond.
02

Identify the initial structures

We'll start by identifying the initial structures of the isomers without including the double bond yet. We'll have a linear structure (5 carbons in a line) and several branched structures (branches starting from the 2nd or 3rd carbon in the chain). They are as follows: 1. Linear: C-C-C-C-C 2. Branched: (C-C(C)-C-C) 3. Branched: (C-C-C(C)-C)
03

Add the double bonds and draw the isomers

Now, we'll add the double bond to each structure, making sure we only have one double bond per isomer. 1. For the linear structure (C-C-C-C-C): Add the double bond at different positions: i. C=C-C-C-C ii. C-C=C-C-C 2. For the branched structure (C-C(C)-C-C): Add the double bond at different positions: i. C=C-C(C)-C ii. C-C=C(C)-C iii. C-C(C=C)-C iv. C-C(C)-C=C 3. For the branched structure (C-C-C(C)-C): Add the double bond at different positions: i. C=C-C(C)-C ii. C-C=C(C)-C In total, we have six different isomers of C5H10.
04

Name each compound

Now let's name each of the isomers using IUPAC rules: 1. C=C-C-C-C: Pent-1-ene 2. C-C=C-C-C: Pent-2-ene 3. C=C-C(C)-C: 3-methylbut-1-ene 4. C-C=C(C)-C: 2-methylbut-2-ene 5. C-C(C=C)-C: 2-methylbut-1-ene 6. C-C(C)-C=C: 3-methylbut-1-ene Thus, the six possible noncyclic structural isomers of C5H10 are: pent-1-ene, pent-2-ene, 3-methylbut-1-ene, 2-methylbut-2-ene, 2-methylbut-1-ene, and 3-methylbut-1-ene.

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Most popular questions from this chapter

Imagine a single DNA strand containing a section with the following base sequence: \(5^{\prime}\) -GCATTGGC-3'. What is the base sequence of the complementary strand? (The two strands of DNA will come together in an antiparallel fashion; that is, \(5^{\prime}-\mathrm{TAG}-3^{\prime}\) will bind to \(3^{\prime}-\) ATC-5'.)

(a) One test for the presence of an alkene is to add a small amount of bromine, a red-brown liquid, and look for the disappearance of the red-brown color. This test does not work for detecting the presence of an aromatic hydrocarbon. Explain. (b) Write a series of reactions leading to para-bromoethylbenzene, beginning with benzene and using other reagents as needed. What isomeric side products might also be formed?

Monosaccharides can be categorized in terms of the number of carbon atoms (pentoses have five carbons and hexoses have six carbons) and according to whether they contain an aldehyde (aldo- prefix, as in aldopentose) or ketone group (keto- prefix, as in ketopentose). Classify glucose and fructose in this way.

(a) What is the difference between a straight-chain and branched-chain alkane? (b) What is the difference between an alkane and an alkyl group?

(a) Give the empirical formula and structural formula for a cyclic ether containing four carbon atoms in the ring. (b) Write the structural formula for a straight-chain compound that is a structural isomer of your answer to part (a).
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