![General, Organic, & Biological Chemistry](https://www.bartleby.com/isbn_cover_images/9780073511245/9780073511245_largeCoverImage.gif)
(a)
Interpretation:
The structure for the eight constitutional isomers of molecular formula C4H11N should be drawn.
Concept Introduction:
There are three types of
![Check Mark](/static/check-mark.png)
Answer to Problem 18.48P
The structures for the eight constitutional isomers of molecular formula C4H11N are represented as follows:
Explanation of Solution
Four structures of primary amines can be drawn with the formula C4H11N.
Three structures of secondary amines can also be drawn.
A tertiary structure can also be drawn as follows:
(b)
Interpretation:
The systematic name for each amine should be given.
Concept Introduction:
In nomenclature of primary amine, the longest carbon chain bonded to nitrogen is determined and the −e ending of the parent
![Check Mark](/static/check-mark.png)
Answer to Problem 18.48P
The name of amines are as follows:
1-butanamine 2-methylpropan-1-amine butan-2-amine
2-methylpropan-2-amine N-ethylethanamine N-methylpropan-1-amine
N-methylpropan-2-amine N,N-dimethylethanamine
Explanation of Solution
The longest carbon chain has four carbons. So the alkane name is butane. N is attached to C-1. Therefore, the systematic name of the amine is butanamine.
The longest carbon chain has three carbons. There is a methyl group at C-2. So the parent name is 2-methylpropanamine. The N atom is bonded to C-1. Therefore, the name become 2-methylpropan-1-amine.
The longest carbon chain bonded to amine group has four carbons. The parent name is butanamine. The N atom is bonded to C-2. Therefore, the systematic name of the amine is butan-2-amine.
The longest carbon chain bonded to amine group has three carbons. There is a methyl group at C-2. The parent name is 2-methylpropanamine. The N atom is bonded to C-2. Therefore, the systematic name of the amine is 2-methylpropan-2-amine.
The secondary amine has the longest carbon chain with 2 carbons. So, the parent name is ethanamine. The N atom has bonded to C-1 and has 1 ethyl group as a substituent. Therefore, the systematic name become N-ethylethanamine.
The secondary amine has the longest carbon chain with 3 carbons. So, the parent name is propanamine. The N atom has bonded to C-1 and has 1 methyl group as a substituent. Therefore, the systematic name become N-methylpropan-1-amine.
The secondary amine has the longest carbon chain with 3 carbons. So, the parent name is propanamine. The N atom has bonded to C-2 and has 1 methyl group as a substituent. Therefore, the systematic name become N-methylpropan-2-amine.
The tertiary amine has the longest carbon chain with 2 carbons. So the parent name is ethanamine. N atom has bonded to C-1 and has two methyl groups and 1 ethyl group as substituents. So, the systematic name of the amine is N,N-dimethylethanamine.
(c)
Interpretation:
The chirality center present in one of the amines should be identified.
Concept Introduction:
An atom that has four different groups bonded to it is referred to as chirality center. A chiral molecule has a non-superimposable mirror image.
![Check Mark](/static/check-mark.png)
Answer to Problem 18.48P
Explanation of Solution
Butan-2-amine has long carbon chain with 4 carbons and amine group is bonded to C-2. This C-2 carbon has four different groups bonded to it as 1 ethyl group, 1 methyl group, 1 amine group and a hydrogen. So, C-2 carbon is a chirality center.
Want to see more full solutions like this?
Chapter 18 Solutions
General, Organic, & Biological Chemistry
- Please answer the question and provide a detailed drawing of the structure. If there will not be a new C – C bond, then the box under the drawing area will be checked. Will the following reaction make a molecule with a new C – C bond as its major product: Draw the major organic product or products, if the reaction will work. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry.arrow_forwardPlease do not use AI. AI cannot "see" the molecules properly, and it therefore gives the wrong answer while giving incorrect descriptions of the visual images we're looking at. All of these compounds would be produced (I think). In my book, I don't see any rules about yield in this case, like explaining that one product would be present in less yield for this reason or that reason. Please explain why some of these produce less yield than others.arrow_forwardPlease answer the question and provide detailed explanations.arrow_forward
- All of these compounds would be produced (I think). In my book, I don't see any rules about yield in this case, like explaining that one product would be present in less yield for this reason or that reason. Please explain why some of these produce less yield than others.arrow_forward5. Fill in the missing molecules in the following reaction pathway. TMSO Heat + CI then HF O₂N (1.0 equiv) AICI 3 OMearrow_forwarde. O₂N NO2 1. excess H2, Pd/C 2. excess NaNO2, HCI 3. excess CuCNarrow_forward
- Help with a periodic table task.' Procedure Part 1: Customizing a Periodic Table Use a textbook or other valid source to determine which elements are metals, nonmetals, metalloids (called semimetals in some texts), alkali metals, alkaline earth metals, transition metals, halogens, and noble gases. Download and print a copy of the Periodic Table of Elements. Use colored pencils, colorful highlighters, or computer drawing tools to devise a schematic for designating each of the following on the periodic table: Group numbers Period number Labels for these groups: alkali metals, alkaline earth metals, transition metals, inner transition metals (lanthanides and actinides), other metals, metalloids (semimetals), other nonmetals, halogens, and noble gases Metals, nonmetals, and metalloids Note: Write the group and period numbers and color/highlight each element for categorization. Be sure to include a key for the schematic. Take a photo of the completed periodic table and upload the…arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forwardCan you explain these two problems for mearrow_forward
- 个 ^ Blackboard x Organic Chemistry II Lecture (m x Aktiv Learning App x → C app.aktiv.com ← Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. Problem 28 of 35 :OH H HH KO Select to Edit Arrows CH CH₂OK, CH CH2OH 5+ H :0: Donearrow_forwardCan you explain those two problems for me please.arrow_forwardDo we need to draw the "ethyne" first for this problem? im confusedarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305580350/9781305580350_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305079373/9781305079373_smallCoverImage.gif)