ORGANIC CHEMISTRY (LL)-PACKAGE
ORGANIC CHEMISTRY (LL)-PACKAGE
8th Edition
ISBN: 9781319316389
Author: VOLLHARDT
Publisher: MAC HIGHER
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Chapter 1, Problem 52P

(a)

Interpretation Introduction

Interpretation:The highly covalent polarized single bond should be located in the compound with biological activity against cells of prostate cancer.

Concept introduction:The difference in electronegativity gives rise to partial positive and negative charges on two atoms. The extent of polarity is quantitatively expressed as dipole moment. It is quantitatively defined as product of charges of two bonded atoms and the separation between them. The electronegativity difference decides the direction of dipole moment. It is indicated by an arrow head from positive end towards negative end. For example, the carbonyl bond is polar with partial postive charge on carbon and partial negative charge on oxygen as illustrated below.

  ORGANIC CHEMISTRY (LL)-PACKAGE, Chapter 1, Problem 52P , additional homework tip  1

The difference in electronegativity gives rise to partial positive and negative charges on two atoms. The extent of polarity is quantitatively expressed as dipole moment. It is quantitatively defined as product of charges of two bonded atoms and the separation between them.

(b)

Interpretation Introduction

Interpretation: The highly polarized covalent double bond should be located in the compound with biological activity against cells of prostate cancer.

Concept introduction:The difference in electronegativity gives rise to partial positive and negative charges on two atoms. The extent of polarity is quantitatively expressed as dipole moment. It is quantitatively defined as product of charges of two bonded atoms and the separation between them. The electronegativity difference decides the direction of dipole moment. It is indicated by an arrow head from positive end towards negative end. For example, the carbonyl bond is polar with partial postive charge on carbon and partial negative charge on oxygen as illustrated below.

  ORGANIC CHEMISTRY (LL)-PACKAGE, Chapter 1, Problem 52P , additional homework tip  2

The difference in electronegativity gives rise to partial positive and negative charges on two atoms. The extent of polarity is quantitatively expressed as dipole moment. It is quantitatively defined as product of charges of two bonded atoms and the separation between them.

(c)

Interpretation Introduction

Interpretation: The nearly nonpolar covalent bond should be located in the compound with biological activity against cells of prostate cancer.

Concept introduction:The difference in electronegativity gives rise to partial positive and negative charges on two atoms. The extent of polarity is quantitatively expressed as dipole moment. It is quantitatively defined as product of charges of two bonded atoms and the separation between them. The electronegativity difference decides the direction of dipole moment. It is indicated by an arrow head from positive end towards negative end. For example, the carbonyl bond is polar with partial postive charge on carbon and partial negative charge on oxygen as illustrated below.

  ORGANIC CHEMISTRY (LL)-PACKAGE, Chapter 1, Problem 52P , additional homework tip  3

The difference in electronegativity gives rise to partial positive and negative charges on two atoms. The extent of polarity is quantitatively expressed as dipole moment. It is quantitatively defined as product of charges of two bonded atoms and the separation between them.

(d)

Interpretation Introduction

Interpretation: The sp -hybridized carbon atom should be located in the compound.

Concept introduction:In accordance with VSEPR model, if four bonds are present around the central atom the hybridization associated is sp3 and theses four orbitals tend to stay as far as 109.5 ° from one another so as to have minimum repulsions.

Similarly, if three bonds are present around the central atom the hybridization associated is sp2 and theses orbitals tend to stay as far as 120 ° from one another so as to have minimum repulsions.

Sequence that determines hybridization state of an atom is as follows:

  • A single bond is always associated with sigma bonds. Molecules that have all atoms connected by sigma bond exclusively are stated to be in sp3 hybridization.
  • A double bond is always associated sp2 hybridization.
  • A triple bond is always associated sp hybridization.

(e)

Interpretation Introduction

Interpretation:The sp2 -hybridized carbon atom should be located in the compound with biological activity against cells of prostate cancer.

Concept introduction:In accordance with VSEPR model, if four bonds are present around the central atom the hybridization associated is sp3 and theses four orbitals tend to stay as far as 109.5 ° from one another so as to have minimum repulsions.

Similarly, if three bonds are present around the central atom the hybridization associated is sp2 and theses orbitals tend to stay as far as 120 ° from one another so as to have minimum repulsions.

Sequence that determines hybridization state of an atom is as follows:

  • A single bond is always associated with sigma bonds. Molecules that have all atoms connected by sigma bond exclusively are stated to be in sp3 hybridization.
  • A double bond is always associated sp2hybridization.
  • A triple bond is always associated sp hybridization.

(f)

Interpretation Introduction

Interpretation:The sp3 -hybridized carbon atom should be located in the compound with biological activity against cells of prostate cancer.

Concept introduction:Sequence that determines hybridization state of an atom is as follows:

  • A single bond is always associated with sigma bonds. Molecules that have all atoms connected by sigma bond exclusively are stated to be in sp3 hybridization.
  • A double bond is always associated sp2 hybridization.
  • A triple bond is always associated sp hybridization.

(g)

Interpretation Introduction

Interpretation:Longest bond in the molecule should be located in the compound.

Concept introduction:Bond strength varies inversely to bond length. Stronger bonds have small bond length and vice-versa. The bond length varies inversely to bond order.

Compound with biological activity against cells of prostate cancer is as follows:

  ORGANIC CHEMISTRY (LL)-PACKAGE, Chapter 1, Problem 52P , additional homework tip  4

(h)

Interpretation Introduction

Interpretation:Shortest bond in the molecule should be located in the compound with biological activity against cells of prostate cancer.

Concept introduction:Bond strength varies inversely to bond length. Stronger bonds have small bond length and vice-versa. The bond length varies inversely to bond order.

Compound with biological activity against cells of prostate cancer is as follows:

  ORGANIC CHEMISTRY (LL)-PACKAGE, Chapter 1, Problem 52P , additional homework tip  5

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Transmitance 3. Which one of the following compounds corresponds to this IR spectrum? Point out the absorption band(s) that helped you decide. OH H3C OH H₂C CH3 H3C CH3 H3C INFRARED SPECTRUM 0.8- 0.6 0.4- 0.2 3000 2000 1000 Wavenumber (cm-1) 4. Consider this compound: H3C On the structure above, label the different types of H's as A, B, C, etc. In table form, list the labeled signals, and for each one state the number of hydrogens, their shifts, and the splitting you would observe for these hydrogens in the ¹H NMR spectrum. Label # of hydrogens splitting Shift (2)
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