Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free