Chemical Principles
Chemical Principles
8th Edition
ISBN: 9781337247269
Author: Steven S. Zumdahl; Donald J. DeCoste
Publisher: Cengage Learning US
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Chapter 21, Problem 40E

(a)

Interpretation Introduction

Interpretation: The solubility of minoxidil in acidic and basic aqueous solution should be determined.

  Chemical Principles, Chapter 21, Problem 40E , additional homework tip  1

Concept Introduction: The Lewis structure of an organic compound represents the bonding of atoms with lone pairs (if any). It indicates the bonds with atoms and also arrangement of atoms in molecule.

Hybridization of any atom indicates the molecular geometry of molecule. The formula to check the hybridization can be written as:

Hybridization = Number of sigma bonds + Number of lone pair

(a)

Expert Solution
Check Mark

Answer to Problem 40E

Minoxidil must be soluble in acidic aqueous solution.

Explanation of Solution

Minoxidil contains two −NH2 groups. The amine groups are basic in nature as N atom of this functional group contains one lone pair. Thus it must be soluble in acidic aqueous solution.

As in acidic solution, it will form salt and therefore will be soluble.

(b)

Interpretation Introduction

Interpretation: The hybridization of five nitrogen atoms in minoxidil should be determined.

  Chemical Principles, Chapter 21, Problem 40E , additional homework tip  2

Concept Introduction: The Lewis structure of an organic compound represents the bonding of atoms with lone pairs (if any). It indicates the bonds with atoms and also arrangement of atoms in molecule.

Hybridization of any atom indicates the molecular geometry of molecule. The formula to check the hybridization can be written as:

Hybridization = Number of sigma bonds + Number of lone pair

(b)

Expert Solution
Check Mark

Answer to Problem 40E

  -NH2; hybridization= 3 sigma bonds+1 lone pair = 4 = sp3=N-;  hybridization= 2 sigma bonds+1 lone pair = 3 = sp2

Explanation of Solution

  Chemical Principles, Chapter 21, Problem 40E , additional homework tip  3

Minoxidil contains two −NH2 groups. There are three N atoms in the ring. Thus the hybridization of N atoms are:

  -NH2; hybridization= 3 sigma bonds+1 lone pair = 4 = sp3=N-;  hybridization= 2 sigma bonds+1 lone pair = 3 = sp2

(c)

Interpretation Introduction

Interpretation: The hybridization of each carbon atoms in minoxidil should be determined.

  Chemical Principles, Chapter 21, Problem 40E , additional homework tip  4

Concept Introduction: The Lewis structure of an organic compound represents the bonding of atoms with lone pairs (if any). It indicates the bonds with atoms and also arrangement of atoms in molecule.

Hybridization of any atom indicates the molecular geometry of molecule. The formula to check the hybridization can be written as:

Hybridization = Number of sigma bonds + Number of lone pair

(c)

Expert Solution
Check Mark

Answer to Problem 40E

  >CH2; hybridization= 4 sigma bonds+0 lone pair = 4 = sp3=CH-;  hybridization= 3 sigma bonds+0 lone pair = 3 = sp2

Explanation of Solution

  Chemical Principles, Chapter 21, Problem 40E , additional homework tip  5

Minoxidil contains two types of C atoms. The double bonded carbon atoms are part of one ring whereas another ring contains only single bonded C atom.

  >CH2; hybridization= 4 sigma bonds+0 lone pair = 4 = sp3=CH-;  hybridization= 3 sigma bonds+0 lone pair = 3 = sp2

(d)

Interpretation Introduction

Interpretation: The bond angles marked as a, b, c, d, e and f in the given molecule of minoxidil should be determined.

  Chemical Principles, Chapter 21, Problem 40E , additional homework tip  6

Concept Introduction: The Lewis structure of an organic compound represents the bonding of atoms with lone pairs (if any). It indicates the bonds with atoms and also arrangement of atoms in molecule.

Hybridization of any atom indicates the molecular geometry of molecule. The formula to check the hybridization can be written as:

Hybridization = Number of sigma bonds + Number of lone pair

(d)

Expert Solution
Check Mark

Answer to Problem 40E

  1. 109°
  2. 109°
  3. 120°
  4. 120°
  5. 109°
  6. 120°

Explanation of Solution

The hybridization of bonded atoms determines the bond angle.

  -NH2; hybridization= 3 sigma bonds+1 lone pair = 4 = sp3=N-;  hybridization= 2 sigma bonds+1 lone pair = 3 = sp2

  >CH2; hybridization= 4 sigma bonds+0 lone pair = 4 = sp3=CH-;  hybridization= 3 sigma bonds+0 lone pair = 3 = sp2

A sp3 hybrid C atom has 109° bond angle with tetrahedral geometry whereas a sp2 hybrid C atom has 120 ° bond angle with trigonal planer geometry. Similarly the single bonded N atom has 109° bond angle whereas a double bonded N atom has 120° bond angle. Thus, the marked bond angles must be:

  1. 109°
  2. 109°
  3. 120°
  4. 120°
  5. 109°
  6. 120°
(e)

Interpretation Introduction

Interpretation: Interpret the number of sigma bonds in the given molecule including hydrogen atoms.

  Chemical Principles, Chapter 21, Problem 40E , additional homework tip  7

Concept Introduction: The Lewis structure of an organic compound represents the bonding of atoms with lone pairs (if any). It indicates the bonds with atoms and also arrangement of atoms in molecule.

Hybridization of any atom indicates the molecular geometry of molecule. The formula to check the hybridization can be written as:

Hybridization = Number of sigma bonds + Number of lone pair

(e)

Expert Solution
Check Mark

Answer to Problem 40E

There are total 30 sigma bonds in the molecule.

Explanation of Solution

A single covalent bond is formed by one sigma bond whereas one double bond is composed of one sigma bond and one pi bond.

Each C atom must be bonded with four other atoms. Therefore in the bond line formula, each C atom must be bonded with other C or N atom and the tetravalency will be completed with H atoms.

Thus there are total 30 sigma bonds in the given molecule.

(f)

Interpretation Introduction

Interpretation: Interpret the number of p-bonds in the given molecule.

  Chemical Principles, Chapter 21, Problem 40E , additional homework tip  8

Concept Introduction: The Lewis structure of an organic compound represents the bonding of atoms with lone pairs (if any). It indicates the bonds with atoms and also arrangement of atoms in molecule.

Hybridization of any atom indicates the molecular geometry of molecule. The formula to check the hybridization can be written as:

Hybridization = Number of sigma bonds + Number of lone pair

(f)

Expert Solution
Check Mark

Answer to Problem 40E

There are total 3 p-bonds in the molecule.

Explanation of Solution

A single covalent bond is formed by one sigma bond whereas one double bond is composed of one sigma bond and one pi bond.

Each C atom must be bonded with four other atoms. Therefore, in the bond line formula, each C atom must be bonded with other C or N atom and the tetravalency will be completed with H atoms.

Thus, there are total 3 p-bonds in the given molecule.

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Chapter 21 Solutions

Chemical Principles

Ch. 21 - Prob. 11ECh. 21 - Name each of the following cyclic alkanes, and...Ch. 21 - Prob. 13ECh. 21 - Prob. 14ECh. 21 - Prob. 15ECh. 21 - Prob. 16ECh. 21 - Prob. 17ECh. 21 - Prob. 18ECh. 21 - Prob. 19ECh. 21 - Prob. 20ECh. 21 - Prob. 21ECh. 21 - Prob. 22ECh. 21 - Prob. 23ECh. 21 - Prob. 24ECh. 21 - Prob. 25ECh. 21 - Prob. 26ECh. 21 - Prob. 27ECh. 21 - Prob. 28ECh. 21 - Prob. 29ECh. 21 - Prob. 30ECh. 21 - Name the following compounds.Ch. 21 - Prob. 32ECh. 21 - Prob. 33ECh. 21 - Prob. 34ECh. 21 - Prob. 35ECh. 21 - Prob. 36ECh. 21 - Prob. 37ECh. 21 - Prob. 38ECh. 21 - Prob. 39ECh. 21 - Prob. 40ECh. 21 - Prob. 41ECh. 21 - Draw structural formulas for each of the following...Ch. 21 - Prob. 43ECh. 21 - Prob. 44ECh. 21 - Prob. 45ECh. 21 - Prob. 46ECh. 21 - Prob. 47ECh. 21 - Prob. 48ECh. 21 - Prob. 49ECh. 21 - Prob. 50ECh. 21 - Prob. 51ECh. 21 - Prob. 52ECh. 21 - Prob. 53ECh. 21 - Prob. 54ECh. 21 - Prob. 55ECh. 21 - Prob. 56ECh. 21 - Prob. 57ECh. 21 - Prob. 58ECh. 21 - Prob. 59ECh. 21 - Give an example reaction that would yield the...Ch. 21 - Prob. 61ECh. 21 - Prob. 62ECh. 21 - Prob. 63ECh. 21 - Prob. 64ECh. 21 - Prob. 65ECh. 21 - Prob. 66ECh. 21 - Prob. 67ECh. 21 - Prob. 68ECh. 21 - Prob. 69ECh. 21 - Prob. 70ECh. 21 - Prob. 71ECh. 21 - Prob. 72ECh. 21 - Prob. 73ECh. 21 - Prob. 74ECh. 21 - Prob. 75ECh. 21 - Prob. 76ECh. 21 - Prob. 77ECh. 21 - Prob. 78ECh. 21 - Prob. 79ECh. 21 - Prob. 80ECh. 21 - Prob. 81ECh. 21 - Prob. 82ECh. 21 - Prob. 83ECh. 21 - Prob. 84ECh. 21 - Prob. 85ECh. 21 - Prob. 86ECh. 21 - Prob. 87ECh. 21 - Prob. 88ECh. 21 - Prob. 89ECh. 21 - Prob. 90ECh. 21 - Prob. 91ECh. 21 - Prob. 92ECh. 21 - Prob. 93ECh. 21 - Prob. 94ECh. 21 - Prob. 95ECh. 21 - Draw the structures of the tripeptides gly-ala-ser...Ch. 21 - Prob. 97ECh. 21 - Prob. 98ECh. 21 - What types of interactions can occur between the...Ch. 21 - Prob. 100ECh. 21 - Prob. 101ECh. 21 - Prob. 102ECh. 21 - Prob. 103ECh. 21 - Prob. 104ECh. 21 - Prob. 105ECh. 21 - Prob. 106ECh. 21 - Prob. 107ECh. 21 - Prob. 108ECh. 21 - Prob. 109ECh. 21 - Prob. 110ECh. 21 - Prob. 111ECh. 21 - Prob. 112ECh. 21 - Prob. 113ECh. 21 - Prob. 114ECh. 21 - Prob. 115ECh. 21 - Prob. 116ECh. 21 - Prob. 117ECh. 21 - Prob. 118ECh. 21 - Prob. 119ECh. 21 - Prob. 120ECh. 21 - Prob. 121ECh. 21 - Prob. 122ECh. 21 - Prob. 123ECh. 21 - Prob. 124ECh. 21 - Prob. 125ECh. 21 - Prob. 126ECh. 21 - Prob. 127AECh. 21 - Prob. 128AECh. 21 - Prob. 129AECh. 21 - Prob. 130AECh. 21 - Prob. 131AECh. 21 - Prob. 132AECh. 21 - Prob. 133AECh. 21 - Prob. 134AECh. 21 - Prob. 135AECh. 21 - Prob. 136AECh. 21 - Prob. 137AECh. 21 - Prob. 138AECh. 21 - Prob. 139AECh. 21 - Prob. 140AECh. 21 - Prob. 141AECh. 21 - Prob. 142AECh. 21 - Prob. 143AECh. 21 - Prob. 144AECh. 21 - Prob. 145AECh. 21 - Prob. 146AECh. 21 - Prob. 147AECh. 21 - Prob. 148AECh. 21 - Prob. 149AECh. 21 - Prob. 150AECh. 21 - Prob. 151AECh. 21 - Prob. 152AECh. 21 - Prob. 153AECh. 21 - Prob. 154AECh. 21 - Prob. 155AECh. 21 - Prob. 156AECh. 21 - Prob. 157AECh. 21 - Prob. 158AECh. 21 - Prob. 159AECh. 21 - Prob. 160AECh. 21 - Prob. 161AECh. 21 - Name each of the following cyclic alkanes.Ch. 21 - Prob. 163AECh. 21 - Prob. 164AECh. 21 - Prob. 165AECh. 21 - Prob. 166AECh. 21 - Prob. 167AECh. 21 - Prob. 168AECh. 21 - Prob. 169CPCh. 21 - Prob. 170CPCh. 21 - Prob. 171CPCh. 21 - Prob. 172CPCh. 21 - Prob. 173CPCh. 21 - Prob. 174CP
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