Chapter 11, Problem 11.34E

(a)

Interpretation Introduction

Interpretation:

The Lewis structures, names and functional groups present of the given compounds have to be given.

Explanation of Solution

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\text{.}$

The name of the compound is hexane and it is alkane (only single bonds are present).

The number of valence electrons in carbon is four=$\left(\text{6}\right)\left(\text{4}\right)\text{=24}$

The number of valence electrons in hydrogen is one=$\left(\text{14}\right)\left(\text{1}\right)\text{=14}$

The total number of valence electrons in hexane is thirty-eight.

The Lewis structure of the hexane is,

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}{\text{CH=CH}}_{\text{2}}.$

The name of the compound is 1-hexene and it is alkene (one double bond is present).

The number of valence electrons in carbon is four=$\left(\text{6}\right)\left(\text{4}\right)\text{=24}$

The number of valence electrons in hydrogen is one=$\left(\text{12}\right)\left(\text{1}\right)\text{=12}$

The total number of valence electrons in 1-hexene is thirty-six.

The Lewis structure of the 1-hexene is,

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}{\text{CH}}_{\text{2}}\text{OH}\text{.}$

The name of the compound is 1-pentanol and it is an alcohol (hydroxyl group is present).

The number of valence electrons in carbon is four=$\left(5\right)\left(\text{4}\right)\text{=20}$

The number of valence electrons in hydrogen is one=$\left(\text{12}\right)\left(\text{1}\right)\text{=12}$

The number of valence electrons in oxygen is six=$\left(1\right)\left(6\right)\text{=6}$

The total number of valence electrons in 1-pentanol is thirty-eight.

The Lewis structure of the 1-pentanol is,

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}\text{CHO}\text{.}$

The name of the compound is pentanal and it is an aldehyde (carbonyl group is present).

The number of valence electrons in carbon is four=$\left(5\right)\left(\text{4}\right)\text{=20}$

The number of valence electrons in hydrogen is one=$\left(\text{10}\right)\left(\text{1}\right)\text{=10}$

The number of valence electrons in oxygen is six=$\left(1\right)\left(6\right)\text{=6}$

The total number of valence electrons in pentanal is thirty-six.

The Lewis structure of the pentanal is,

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COOH}\text{.}$

The name of the compound is butanoic acid and it is a carboxylic acid (carboxyl group is present).

The number of valence electrons in carbon is four=$\left(4\right)\left(\text{4}\right)\text{=16}$

The number of valence electrons in hydrogen is one=$\left(8\right)\left(\text{1}\right)\text{=8}$

The number of valence electrons in oxygen is six=$\left(2\right)\left(6\right)\text{=12}$

The total number of valence electrons in butanoic acid is twenty-six.

The Lewis structure of the butanoic acid is,

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COOCH}}_{\text{3}}.$

The name of the compound is methyl propanoate and it is an ester.

The number of valence electrons in carbon is four=$\left(4\right)\left(\text{4}\right)\text{=16}$

The number of valence electrons in hydrogen is one=$\left(8\right)\left(\text{1}\right)\text{=8}$

The number of valence electrons in oxygen is six=$\left(2\right)\left(6\right)\text{=12}$

The total number of valence electrons in methyl propanoate is twenty-six.

The Lewis structure of the methyl propanoate is,

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{3}}.$

The name of the compound is 3-pentanone and it is a ketone (carbonyl group is present).

The number of valence electrons in carbon is four=$\left(5\right)\left(\text{4}\right)\text{=20}$

The number of valence electrons in hydrogen is one=$\left(10\right)\left(\text{1}\right)\text{=10}$

The number of valence electrons in oxygen is six=$\left(1\right)\left(6\right)\text{=6}$

The total number of valence electrons in 3-pentanone is thirty-six.

The Lewis structure of the 3-pentanone is,

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{OCH}}_{\text{3}}\text{.}$

The name of the compound is 1-methoxybutane and it is an ether (oxygen is present).

The number of valence electrons in carbon is four=$\left(5\right)\left(\text{4}\right)\text{=20}$

The number of valence electrons in hydrogen is one=$\left(12\right)\left(\text{1}\right)\text{=12}$

The number of valence electrons in oxygen is six=$\left(1\right)\left(6\right)\text{=6}$

The total number of valence electrons in 1-methoxybutane is thirty-eight.

The Lewis structure of the 1-methoxybutane is,

(b)

Interpretation Introduction

Interpretation:

The molecules that are isomers to each other and the chirality have to be checked.

Answer to Problem 11.34E

The molecules that are isomers to each other are,

1. 1. ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{OH}$ and ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}{\text{OCH}}_{\text{3}}.$
2. 2. ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{3}}\text{CHO}$ and ${\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{CO}\text{.}$
3. 3. ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{COOH}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COOCH}}_{\text{3}}\text{.}$

Explanation of Solution

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}{\text{CH}}_{\text{2}}\text{OH}\text{.}$

The name of the compound is 1-pentanol and it is an alcohol (hydroxyl group is present).  The molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}\text{.}$

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{OCH}}_{\text{3}}\text{.}$

The name of the compound is 1-methoxybutane and it is an ether (oxygen is present).  The molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}\text{.}$

Both these compound have similar molecular formula but difference in spatial arrangement of atoms.  Hence, they are isomers to each other.

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}\text{CHO}\text{.}$

The name of the compound is pentanal and it is an aldehyde (carbonyl group is present).  The molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}\text{.}$

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{3}}.$

The name of the compound is 3-pentanone and it is a ketone (carbonyl group is present).  The molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}\text{.}$

Both these compound have similar molecular formula but difference in spatial arrangement of atoms.  Hence, they are isomers to each other.

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COOH}\text{.}$

The name of the compound is butanoic acid and it is a carboxylic acid (carboxyl group is present).  The molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{18}}{\text{O}}_{\text{2}}\text{.}$

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COOCH}}_{\text{3}}$

The name of the compound is methyl propanoate and it is an ester.  The molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{18}}{\text{O}}_{\text{2}}\text{.}$

Both these compound have similar molecular formula but difference in spatial arrangement of atoms.  Hence, they are isomers to each other.

None of the compounds is chiral.

(c)

Interpretation Introduction

Interpretation:

The intermolecular forces present in the compounds have to be given.

Concept Introduction:

Intermolecular forces:  Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds.  Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions.  The three major types of intermolecular interactions are,

• Dipole-dipole interactions
• London dispersion forces
• Hydrogen bonds

Explanation of Solution

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\text{.}$

The name of the compound is hexane and it is alkane (only single bonds are present).  Hexane is a non-polar molecule and the polarity would cancel out since the molecule is symmetric.  The intermolecular force present in hexane is London dispersion forces.

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}{\text{CH=CH}}_{\text{2}}.$

The name of the compound is 1-hexene and it is alkene (one double bond is present).  Hexene is a non-polar molecule and the intermolecular force present in hexene is van der Waals dispersion forces.

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}{\text{CH}}_{\text{2}}\text{OH}\text{.}$

The name of the compound is 1-pentanol and it is an alcohol (hydroxyl group is present).  The molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}\text{.}$  1-Pentanol consists of both London dispersion force and hydrogen bonding, but the most dominant force will be London dispersion force because 1-pentanol is sparingly soluble in water.

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}{\text{2}}_{}\text{CHO}\text{.}$

The name of the compound is pentanal and it is an aldehyde (carbonyl group is present).  The molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}\text{.}$  Pentanal is polar molecule but it cannot participate in hydrogen bonding.  The most dominant intermolecular force in pentanal is London dispersion force (van der Waals forces).

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COOH}\text{.}$

The name of the compound is butanoic acid and it is a carboxylic acid (carboxyl group is present).  The molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{18}}{\text{O}}_{\text{2}}\text{.}$  Butanoic acid is a polar molecule and participates in hydrogen bond.  London dispersion force and hydrogen bonding will be present as the intermolecular force in butanoic acid, but the most dominant force present will be hydrogen bonding.

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COOCH}}_{\text{3}}$

The name of the compound is methyl propanoate and it is an ester.  The molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{18}}{\text{O}}_{\text{2}}\text{.}$  The intermolecular force present in methyl propanoate is London dispersion force.

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{3}}.$

The name of the compound is 3-pentanone and it is a ketone (carbonyl group is present).  The molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}\text{.}$  3-pentanone is polar due to the presence of carbonyl group.  The intermolecular force present in 3-pentanone is London dispersion force (van der Waals forces).

The formula of the given compound is identified as ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{OCH}}_{\text{3}}\text{.}$

The name of the compound is 1-methoxybutane and it is an ether (oxygen is present).  The molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}\text{.}$  The intermolecular force present in 1-methoxybutane is dipole-dipole attractions since it occurs in polar molecules with dipoles.

(d)

Interpretation Introduction

Interpretation:

The boiling points of the compounds have to be given from lowest to highest.

Answer to Problem 11.34E

The boiling points of the compounds from lowest to highest,

$\text{Hexene<Hexane<1-Methoxybutane<Methyl}\text{propanoate<3-pentanone<Pentanal<1-Pentanol<Butanoic}\text{acid}\text{.}$

Explanation of Solution

The greater the electronegativity differences between atoms in a bond, the polar the bond.  Partial negative charges are found on the most electronegative atoms, the others are partially positive.

The presence of oxygen is more polar because of its electronegativity.  The combination of carbons and hydrogen as in hydrocarbon or in the hydrocarbon portion of a molecule with functional group is always polar.

The greater the forces of attraction, the higher the boiling point (or) the greater the polarity, the higher the boiling point.

The polarity of the functional groups are,

$\text{Amide>Acid>Alcohol>Ketone}\approx \text{Aldehyde>Amine>Ester>Ether>Alkane}$

The boiling points of the compounds from lowest to highest,

$\text{Hexene<Hexane<1-Methoxybutane<Methyl}\text{propanoate<3-pentanone<Pentanal<1-Pentanol<Butanoic}\text{acid}\text{.}$