(a)
Interpretation:
The structure for the eight constitutional isomers of molecular formula C4H11N should be drawn.
Concept Introduction:
There are three types of

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

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.

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
ALEKS 360 ACCESS CARD F/GEN. ORG.CHEM
- Use the reaction coordinate diagram to answer the below questions. Type your answers into the answer box for each question. (Watch your spelling) Energy A B C D Reaction coordinate E A) Is the reaction step going from D to F endothermic or exothermic? A F G B) Does point D represent a reactant, product, intermediate or transition state? A/ C) Which step (step 1 or step 2) is the rate determining step? Aarrow_forward1. Using radii from Resource section 1 (p.901) and Born-Lande equation, calculate the lattice energy for PbS, which crystallizes in the NaCl structure. Then, use the Born-Haber cycle to obtain the value of lattice energy for PbS. You will need the following data following data: AH Pb(g) = 196 kJ/mol; AHƒ PbS = −98 kJ/mol; electron affinities for S(g)→S¯(g) is -201 kJ/mol; S¯(g) (g) is 640kJ/mol. Ionization energies for Pb are listed in Resource section 2, p.903. Remember that enthalpies of formation are calculated beginning with the elements in their standard states (S8 for sulfur). The formation of S2, AHF: S2 (g) = 535 kJ/mol. Compare the two values, and explain the difference. (8 points)arrow_forwardIn the answer box, type the number of maximum stereoisomers possible for the following compound. A H H COH OH = H C Br H.C OH CHarrow_forward
- 7. Magnesium is found in nature in the form of carbonates and sulfates. One of the major natural sources of zinc is zinc blende (ZnS). Use relevant concepts of acid-base theory to explain this combination of cations and anions in these minerals. (2 points)arrow_forward6. AlF3 is insoluble in liquid HF but dissolves if NaF is present. When BF3 is added to the solution, AlF3 precipitates. Write out chemical processes and explain them using the principles of Lewis acid-base theory. (6 points)arrow_forward5. Zinc oxide is amphoteric. Write out chemical reactions for dissolution of ZnO in HCl(aq) and in NaOH(aq). (3 points)arrow_forward
- Draw the product(s) formed when alkene A is reacted with ozone, followed by Zn and H₂O. If no second product is formed, do not draw a structure in the second box. Higher Molecular Weight Product A Lower Molecular Weight Product draw structure ... draw structure ...arrow_forwardRank A - D in order of increasing rate of reaction with H2 and Pd/C. ب ب ب ب A B с Which option correctly ranks the alkenes in order of increasing rate of reaction with H₂ and Pd/C? О Barrow_forwardDraw the product of the following Sharpless epoxidation, including stereochemistry. Click the "draw structure" button to launch the drawing utility. -OH (CH3)3C-OOH Ti[OCH(CH3)2]4 (+)-DET draw structure ... Guidarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- 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

