Chapter 21, Problem 1Q

Interpretation Introduction

(a)

Interpretation:

The dodecane, $\text{1-hexyne}$, $\text{1-decene}$ and $\text{1,2-dimethylbenzene}$ should be matched with the correct set of IR frequencies.

Concept introduction:

IR spectroscopy is spectroscopic method to identify the structure of organic compounds. When IR is passed through the sample, different bonds vibrate at different frequencies of IR by absorbing IR light. In IR spectrum energy in terms of wavenumber( ${\text{cm}}^{\text{-1}}$ ) is the horizontal axis and intensity is the vertical axis.

As different molecules have different bonds of varying strength so they absorb IR light of different frequencies. This results in different absorption bands in the IR spectrum characteristic to that bond.

IR region is divided into 3 regions, near, mid and far IR region. The mid-IR region extends from $4000-600{\text{cm}}^{\text{-1}}$ and it is the most important region as approximately all organic compounds absorb in this region.

The allowed vibrational energy levels for IR transition is as follows:

  $E=h{\nu }_{\text{ο}}\left(\nu +\text{1}\right)$

Where,

• $E$ is IR energy absorbed by a molecule.
• $h$ is Planck's constant.
• ${\nu }_o$ is zero point vibrational level.
• $\nu$ is vibrational level with integer value of 1,2,3,4......

Explanation of Solution

The absorption by a particular bond vibration appears as peak at definite positions in the IR spectrum irrespective of the compound in which they take place.

Frequency of peaks and their intensity is majorly based on the factors as follows:

• Kind of vibration that is whether stretching or bending.
• Bond order
• Electronegativity difference
• Molecular weight

On the basis of peak intensity, $\text{w}$ is used to denote weakly intense peaks, $\text{s}$ is used to denote strongly intense peak, $\text{m}$ for medium intensity and $\text{b}$ for broad peaks.

The correct match of the frequencies with the proper compound is tabulated as follows:

  $\begin{array}{ccc}\text{Name}\text{of}\text{compound}& \text{Frequencies}\left({\text{cm}}^{\text{-1}}\right)& \text{Vibrations}\\ \text{Dodecane}& \begin{array}{l}\text{2924(s)}\\ \text{1467(m)}\end{array}& \begin{array}{c}\text{C}-\text{H}\text{stretch}\\ \text{C}-\text{H}\text{bend}\end{array}\\ \text{1-hexyne}& \begin{array}{l}\text{3311(s)}\\ \text{2961(s)}\\ \text{2119(m)}\end{array}& \begin{array}{c}\text{C}\equiv \text{C}-\text{H}\text{stretch}\\ \text{C}-\text{H}\text{stretch}\\ \text{C}\equiv \text{C}\text{stretch}\end{array}\\ \text{1-decene}& \begin{array}{l}\text{3049(w)}\\ \text{2951(m)}\\ \text{1642(m)}\end{array}& \begin{array}{c}\text{C=C-H}\text{stretch}\\ \text{C}-\text{H}\text{stretch}\\ \text{C=C}\text{stretch}\end{array}\\ \text{1,2-dimethylbenzene}& \begin{array}{l}\text{3020(s)}\\ \text{2940(s)}\\ \text{1606(s)}\\ \text{1495(s)}\\ \text{741(s)}\end{array}& \begin{array}{c}\text{aromatic}\text{C}-\text{H}\text{stretch}\\ \text{C}-\text{H}\text{stretch}\\ \text{C=C}\text{stretch}\\ \text{C}-\text{H}\text{bend}\\ \text{aromatic}\text{C}-\text{H}\text{bend}\end{array}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The phenol, benzyl alcohol, and methoxybenzene should be matched with the correct set of IR frequencies.

Concept introduction:

IR spectroscopy is spectroscopic method to identify the structure of organic compounds. When IR is passed through the sample, different bonds vibrate at different frequencies of IR by absorbing IR light. In IR spectrum energy in terms of wavenumber( ${\text{cm}}^{\text{-1}}$ ) is the horizontal axis and intensity is the vertical axis.

As different molecules have different bonds of varying strength, they absorb IR light of different frequencies. This results in different absorption bands in the IR spectrum characteristic to that bond.

IR region is divided into 3 regions, near, mid and far IR region. The mid-IR region extends from $4000-600{\text{cm}}^{\text{-1}}$ and it is the most important region as approximately all organic compounds absorb in this region.

The allowed vibrational energy levels for IR transition is as follows:

  $E=h{\nu }_{\text{ο}}\left(\nu +\text{1}\right)$

Where,

• $E$is IR energy absorbed by a molecule.
• $h$ is Planck's constant.
• ${\nu }_o$ is zero point vibrational level.
• $\nu$ is vibrational level with integer value of 1,2,3,4......

Explanation of Solution

The absorption by a particular bond vibration appears as peak at definite positions in the IR spectrum irrespective of the compound in which they take place.

Frequency of peaks and their intensity is majorly based on the factors as follows:

• Kind of vibration that is whether stretching or bending.
• Bond order
• Electronegativity difference
• Molecular weight

On the basis of peak intensity, $\text{w}$ is used to denote weakly intense peaks, $\text{s}$ is used to denote strongly intense peak, $\text{m}$ for medium intensity and $\text{b}$ for broad peaks.

The correct match of the frequencies with the proper compound is tabulated as follows:

  $\begin{array}{ccc}\text{Name}\text{of}\text{compound}& \text{Frequencies}\left({\text{cm}}^{\text{-1}}\right)& \text{Vibrations}\\ \text{methoxybenzene}& \begin{array}{l}\text{3060}\left(\text{m}\right)\\ \text{2835}\left(\text{m}\right)\\ \text{1498}\left(\text{s}\right)\\ \text{1247}\left(\text{s}\right)\\ \text{1040}\left(\text{s}\right)\end{array}& \begin{array}{l}\text{aromatic}\text{C}-\text{H}\text{stretch}\\ \text{alkyl C}-\text{H}\text{stretch}\\ \text{ aromatic}\text{C=C}\text{stretch}\\ \text{conjugatedC}-\text{O stretch}\\ \text{saturated C}-\text{O stretch}\end{array}\\ \text{phenol}& \begin{array}{l}\text{3370}\left(\text{s}\right)\\ \text{3045}\left(\text{m}\right)\\ \text{1595}\left(\text{s}\right)\\ \text{1224}\left(\text{s}\right)\end{array}& \begin{array}{l}\text{O}-\text{H stretch}\\ \text{aromatic}\text{C}-\text{H}\text{stretch}\\ \text{aromatic}\text{C=C}\text{stretch}\\ \text{conjugated C}-\text{O stretch}\end{array}\\ \text{benzyl alcohol}& \begin{array}{l}\text{3330}\left(\text{br,s}\right)\\ \text{3030}\left(\text{m}\right)\\ \text{2950}\left(\text{m}\right)\\ \text{1454}\left(\text{m}\right)\\ \text{1223}\left(\text{s}\right)\end{array}& \begin{array}{l}\text{O}-\text{H stretch}\\ \text{aromatic}\text{C}-\text{H}\text{stretch}\\ \text{alkyl C}-\text{H}\text{stretch}\\ \text{aromatic}\text{C=C}\text{stretch}\\ \text{C}-\text{O}\text{stretch }\end{array}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The $\text{2-phenyl propanal}$, $\text{1-amino octane}$, phenylacetate, heptanoic acid, $\text{2-methylpropanamide}$, $\text{2-pentanone}$ and acetophenone should be matched with the correct set of IR frequencies.

Concept introduction:

IR spectroscopy is spectroscopic method to identify the structure of organic compounds. When IR is passed through the sample, different bonds vibrate at different frequencies of IR by absorbing IR light. In IR spectrum energy in terms of wavenumber( ${\text{cm}}^{\text{-1}}$ ) is the horizontal axis and intensity is the vertical axis.

As different molecules have different bonds of varying strength so they absorb IR light of different frequencies. This results in different absorption bands in the IR spectrum characteristic to that bond.

IR region is divided into 3 regions, near, mid and far IR region. The mid-IR region extends from $4000-600{\text{cm}}^{\text{-1}}$ and it is the most important region as approximately all organic compounds absorb in this region.

The allowed vibrational energy levels for IR transition is as follows:

  $E=h{\nu }_{\text{ο}}\left(\nu +\text{1}\right)$

Where,

• $E$is IR energy absorbed by a molecule.
• $h$ is Planck's constant.
• ${\nu }_o$ is zero point vibrational level.
• $\nu$ is vibrational level with integer value of 1,2,3,4......

Explanation of Solution

The absorption by a particular bond vibration appears as peak at definite positions in the IR spectrum irrespective of the compound in which they take place.

Frequency of peaks and their intensity is majorly based on the factors as follows:

• Kind of vibration that is whether stretching or bending.
• Bond order
• Electronegativity difference
• Molecular weight

On the basis of peak intensity, $\text{w}$ is used to denote weakly intense peaks, $\text{s}$ is used to denote strongly intense peak, $\text{m}$ for medium intensity and $\text{b}$ for broad peaks.

The correct match of the frequencies with the proper compound is tabulated as follows: