Chapter 15, Problem 15.19P

Interpretation Introduction

(a)

Interpretation:

The CN stretching mode that absorbs the higher-frequency IR photons is to be indicated for given pair of compounds. The reason for it is to be explained.

Concept introduction:

We simplify the picture of molecular vibrations by considering the ball-and-spring model, which treats bonds as simple springs that connect atoms together. According to Hooke’s law, the spring vibrates at a particular frequency (${\text{ν}}_{\text{spring}}$) that depends on the mass (m) of the object connected to the spring (in this case, an atom) and the spring’s force constant (k), which is often thought of as the stiffness of the spring. The stiffer the spring, the more difficult it is to compress and extend it. A stronger and stiffer bond tends to lead to a higher vibrational frequency. A stiffer spring exerts a greater force on the attached atoms, causing them to move faster. Triple bonds tend to be stronger and stiffer than double bonds, just as double bonds tend to be stronger and stiffer than single bonds.

Answer to Problem 15.19P

The CN stretching mode that absorbs the higher-frequency IR photons due to strong and stiffer bond is indicated below,

Explanation of Solution

The given pair of compounds is,

The triple bonds tend to be stronger and stiffer than double bonds. A stronger and stiffer bond tends to lead to a higher vibrational frequency. In given pair of compounds, with the faster vibration, the $\text{C }\equiv \text{N}$ stronger and stiffer than $\text{C = N}$ bond. Therefore $\text{C }\equiv \text{N}$ bond absorbs the higher-frequency IR photons.

The CN stretching mode that absorbs the higher-frequency IR photons is shown below,

Conclusion

The CN stretching mode that absorbs the higher-frequency IR photons is indicated on the basis of the relationship between strength and stiffness of the bond and vibrational frequency.

Interpretation Introduction

(b)

Interpretation:

The CN stretching mode that absorbs the higher-frequency IR photons is to be indicated for given pair of compounds. The reason for it is to be explained.

Concept introduction:

We simplify the picture of molecular vibrations by considering the ball-and-spring model, which treats bonds as simple springs that connect atoms together. According to Hooke’s law, the spring vibrates at a particular frequency (${\text{ν}}_{\text{spring}}$) that depends on the mass (m) of the object connected to the spring (in this case, an atom) and the spring’s force constant (k), which is often thought of as the stiffness of the spring. The stiffer the spring, the more difficult it is to compress and extend it. A stronger and stiffer bond tends to lead to a higher vibrational frequency. A stiffer spring exerts a greater force on the attached atoms, causing them to move faster. Triple bonds tend to be stronger and stiffer than double bonds, just as double bonds tend to be stronger and stiffer than single bonds.

Answer to Problem 15.19P

The CN stretching mode that absorbs the higher-frequency IR photons due to strong and stiffer bond is indicated below,

Explanation of Solution

The given pair of compounds is,

The double bonds tend to be stronger and stiffer than single bonds. A stronger and stiffer bond tends to lead to a higher vibrational frequency. In given pair of compounds, with the faster vibration, the $\text{C }=\text{N}$ stronger and stiffer than $\text{C - N}$ bond. Therefore $\text{C }=\text{N}$ bond absorbs the higher-frequency IR photons.

The CN stretching mode that absorbs the higher-frequency IR photons is shown below,

Conclusion

The CN stretching mode that absorbs the higher-frequency IR photons is indicated on the basis of the relationship between strength and stiffness of the bond and vibrational frequency.

Interpretation Introduction

(c)

Interpretation:

The CN stretching mode that absorbs the higher-frequency IR photons is to be indicated for given pair of compounds. The reason for it is to be explained.

Concept introduction:

We simplify the picture of molecular vibrations by considering the ball-and-spring model, which treats bonds as simple springs that connect atoms together. According to Hooke’s law, the spring vibrates at a particular frequency (${\text{ν}}_{\text{spring}}$) that depends on the mass (m) of the object connected to the spring (in this case, an atom) and the spring’s force constant (k), which is often thought of as the stiffness of the spring. The stiffer the spring, the more difficult it is to compress and extend it. A decrease in the mass of an atom leads to an increase in vibrational frequency. A stronger and stiffer bond tends to lead to a higher vibrational frequency.

Answer to Problem 15.19P

The CN stretching mode that absorbs the higher-frequency IR photons due to lower mass is indicated below,

Explanation of Solution

The given pair of compounds is,

In both compounds there is $\text{C }\equiv \text{ N}$ bond so both has essentially the same strength and therefore each bond’s spring stiffness must be about the same too. Therefore spring stiffness does not come into the play.

But ${}^{13}\text{C}$ has higher mass than ${}^{12}\text{C}$ so lower mass of ${}^{12}\text{C}$ contributes to a higher frequency of vibration. Because the frequency of vibration is the same as the photon frequency, a ${}^{12}\text{C }\equiv \text{ N}$ stretch must absorb higher-frequency photons than ${}^{13}\text{C }\equiv \text{ N}$ stretch.

The CN stretching mode that absorbs the higher-frequency IR photons is shown below,

Conclusion

The vibrational mode that absorbs at a higher frequency in the IR region is determined on the basis of the relationship between mass and vibrational frequency.