Chapter 15, Problem 15.41P

Interpretation Introduction

(a)

Interpretation:

The differences in the IR spectra of the reactant and product that would enable you to tell that the given reaction has taken place are to be determined.

Concept introduction:

The frequencies of the stretching vibrations are estimated from the type of bond and the frequency range given for that bond. Also, the frequency range for the given type of bond is identified by the functional group to which that bond resembles in the molecule.

Answer to Problem 15.41P

An OH stretch with stretching frequency $\sim \text{340}0{\text{ cm}}^{-1}$ and a $\text{C - O}$ stretch with stretching frequency $\sim {\text{1100 cm}}^{\text{-1}}$ would disappear from the reactant. The $\text{C = C}$ stretch having stretching frequency $\sim {\text{1650 cm}}^{-1}$ and $\text{C - H}$ bond attached to $\text{C = C}$ bond having stretching frequency $\sim {\text{3050 cm}}^{-1}$ would appear in the product.

Explanation of Solution

The given reaction is

In the above reaction, the OH group, that is the alcohol functional group, is present in the reactant, and C=C bond of alkene functional group is present in the product. The $\text{C - O}$ bond in the reactant is attached to the alcohol functional group. It’s stretching frequency is $\sim {\text{1100 cm}}^{\text{-1}}$. The $\text{O - H}$ bond is a characteristic of an alcohol. It’s stretching frequency is $\sim \text{340}0{\text{ cm}}^{-1}$. These two stretches will disappear from alcohol. The characteristic IR band for alkene i.e., for $\text{C = C}$ bond having stretching frequency $\sim {\text{1650 cm}}^{-1}$ and $\text{C - H}$ bond attached to $\text{C = C}$ bond having stretching frequency $\sim {\text{3050 cm}}^{-1}$ will appear in the product.

Conclusion

The differences in the IR spectra that would enable to tell that the given reaction has taken place are determined on the basis of bonds and the functional groups present in the reactant and the product.

Interpretation Introduction

(b)

Interpretation:

The differences in the IR spectra of the reactant and product that would enable you to tell that the given reaction has taken place are to be determined.

Concept introduction:

The frequencies of the stretching vibrations are estimated from the type of bond and the frequency range given for that bond. Also, frequency range for the given type of bond is identified by the functional group to which that bond resembles in the molecule.

Answer to Problem 15.41P

A very broad OH stretch having stretching frequency $\sim {\text{3000-2500 cm}}^{-1}$ from carboxylic acid of the reactant would disappear and the OH stretching frequency of an ester would appear in the product.

Explanation of Solution

The given reaction is

The $\text{R-COOH}$ i.e. carboxylic acid functional group is present in the reactant whereas the $\text{R-COOR'}$ i.e ester functional group is present in the product. Therefore, a very broad OH stretch having stretching frequency $\sim {\text{3000-2500 cm}}^{-1}$ from carboxylic acid would disappear, and the OH stretching frequency of an ester would appear in the product.

Conclusion

The differences in the IR spectra that would enable to tell that the given reaction has taken place are determined on the basis of bonds and the functional groups present in the reactant and the product.

Interpretation Introduction

(c)

Interpretation:

The differences in the IR spectra of the reactant and product that would enable you to tell that the given reaction has taken place are to be determined.

Concept introduction:

The frequencies of the stretching vibrations are estimated from the type of bond and the frequency range given for that bond. Also, frequency range for the given type of bond is identified by the functional group to which that bond resembles in the molecule.

Answer to Problem 15.41P

Two $\text{N-H}$ stretching bands from the primary amine would disappear from the reactant and only one N-H single band would appear in the product.

Explanation of Solution

The given reaction is,

Two $\text{N-H}$ stretching bands of stretching frequency $\sim {\text{3400 cm}}^{-1}$ are from the reactant having the primary amine group. The secondary amine is present in the product. Therefore, two $\text{N-H}$ stretching bands from the primary amine would disappear and only one $\text{N-H}$ single band would appear in the product.

Conclusion

The differences in the IR spectra that would enable to tell that the given reaction has taken place are determined on the basis of bonds and the functional groups present in the reactant and the product.

Interpretation Introduction

(d)

Interpretation:

The differences in the IR spectra of the reactant and product that would enable you to tell that the given reaction has taken place are to be determined.

Concept introduction:

The frequencies of the stretching vibrations are estimated from the type of bond and the frequency range given for that bond. Also, frequency range for the given type of bond is identified by the functional group to which that bond resembles in the molecule.

Answer to Problem 15.41P

The single N-H band having stretching frequency $\sim {\text{3400 cm}}^{-1}$ would disappear from the reactant and tertiary amine having no characteristic IR absorption bands above $\sim {\text{3000 cm}}^{-1}$ would appear in the product.

Explanation of Solution

The given reaction is

The single N-H band having stretching frequency $\sim {\text{3400 cm}}^{-1}$ present in the reactant would disappear. Tertiary amine which has no characteristic IR absorption bands above $\sim {\text{3000 cm}}^{-1}$ would appear in the product.

Conclusion

The differences in the IR spectra that would enable to tell that the given reaction has taken place are determined on the basis of bonds and the functional groups present in the reactant and the product.

Interpretation Introduction

(e)

Interpretation:

The differences in the IR spectra of the reactant and product that would enable you to tell that the given reaction has taken place are to be determined.

Concept introduction:

The frequencies of the stretching vibrations are estimated from the type of bond and the frequency range given for that bond. Also, frequency range for the given type of bond is identified by the functional group to which that bond resembles in the molecule.

Answer to Problem 15.41P

The epoxide having ether functional group would disappear from the reactant. An OH band having stretching frequency $\sim {\text{3400 cm}}^{-1}$ would appear for the alcohol functional group in the product.

Explanation of Solution

The given reaction is

The epoxide having ether functional group would disappear from the reactant, and an OH band having stretching frequency $\sim {\text{3400 cm}}^{-1}$ would appear for the alcohol functional group present in the product.

Conclusion

The differences in the IR spectra that would enable to tell that the given reaction has taken place are determined on the basis of bonds and the functional groups present in the reactant and the product.

Interpretation Introduction

(f)

Interpretation:

The differences in the IR spectra of the reactant and product that would enable you to tell that the given reaction has taken place are to be determined.

Concept introduction:

The frequencies of the stretching vibrations are estimated from the type of bond and the frequency range given for that bond. Also, frequency range for the given type of bond is identified by the functional group to which that bond resembles in the molecule.

Answer to Problem 15.41P

The $\text{C = C}$ stretching band with frequency ${\text{1600 cm}}^{-1}$ and alkene $\text{C-H}$ band with stretching frequency ${\text{3050 cm}}^{-1}$ would disappear from the reactant. The $\text{C=O}$ band frequency will increase by about ${\text{30 cm}}^{-1}$ in the product.

Explanation of Solution

The given reaction is

In the above reaction, the ${\text{R}}_{\text{2}}\text{CO}$ group that is ketone functional group and $\text{C = C}$ bond of alkene functional group is present in the reactant. $\text{C=O}$ bond of ketone functional group is present in the product. The $\text{C = C}$ stretching band with frequency ${\text{1600 cm}}^{-1}$ and alkene $\text{C-H}$ band with stretching frequency ${\text{3050 cm}}^{-1}$ would disappear from the reactant. There is no conjugation present in the product so, the $\text{C=O}$ band frequency will increase by about ${\text{30 cm}}^{-1}$ in the product.

Conclusion

The differences in the IR spectra that would enable to tell that the given reaction has taken place are determined on the basis of bonds and the functional groups present in the reactant and the product.

Interpretation Introduction

(g)

Interpretation:

The differences in the IR spectra of the reactant and product that would enable you to tell that the given reaction has taken place are to be determined.

Concept introduction:

The frequencies of the stretching vibrations are estimated from the type of bond and the frequency range given for that bond. Also, frequency range for the given type of bond is identified by the functional group to which that bond resembles in the molecule.

Answer to Problem 15.41P

The $\text{C = C}$ stretching bands for an aromatic compound with stretching frequency $\sim {\text{1450-1550 cm}}^{-1}$ would disappear from the reactant. The $\text{C = C}$ stretching band for alkene with stretching frequency ${\text{1650 cm}}^{-1}$ and $\text{C-H}$ stretches with stretching frequency $\sim {\text{2800-3000 cm}}^{-1}$ would appear in the product.

Explanation of Solution

The given reaction is

The $\text{C = C}$ stretching bands for an aromatic compound with stretching frequency $\sim {\text{1450-1550 cm}}^{-1}$ would disappear from the given reactant. A simple $\text{C = C}$ stretching band for alkene having stretching frequency ${\text{1650 cm}}^{-1}$ and $\text{C-H}$ stretches with stretching frequency $\sim {\text{2800-3000 cm}}^{-1}$ would appear in the product.

Conclusion

The differences in the IR spectra that would enable to tell that the given reaction has taken place are determined on the basis of bonds and the functional groups present in the reactant and the product.

Interpretation Introduction

(h)

Interpretation:

The differences in the IR spectra of the reactant and product that would enable you to tell that the given reaction has taken place are to be determined.

Concept introduction:

The frequencies of the stretching vibrations are estimated from the type of bond and the frequency range given for that bond. Also, frequency range for the given type of bond is identified by the functional group to which that bond resembles in the molecule.

Answer to Problem 15.41P

The strong $\text{C-H}$ band which is attached to alkyne group with stretching frequency $\sim {\text{3300 cm}}^{-1}$ would disappear from the reactant. Also, the $\text{C}\equiv \text{C}$ stretch would become less intense in the product.

Explanation of Solution

The given reaction is

The strong $\text{C-H}$ band which is attached to the alkyne group with stretching frequency $\sim {\text{3300 cm}}^{-1}$ would disappear from the reactant. The two carbon atoms attached to $\text{C}\equiv \text{C}$ are more symmetric in the product. Therefore, the $\text{C}\equiv \text{C}$ stretching would become less intense in the product.

Conclusion

The differences in the IR spectra that would enable to tell that the given reaction has taken place are determined on the basis of bonds and the functional groups present in the reactant and the product.

Interpretation Introduction

(i)

Interpretation:

The differences in the IR spectra of the reactant and product that would enable you to tell that the given reaction has taken place are to be determined.

Concept introduction:

The frequencies of the stretching vibrations are estimated from the type of bond and the frequency range given for that bond. Also, frequency range for the given type of bond is identified by the functional group to which that bond resembles in the molecule.

Answer to Problem 15.41P

The $\text{C=O}$ band having stretching frequency ${\text{1680 cm}}^{-1}$ would disappear from the reactant. The OH band having stretching frequency ${\text{3400 cm}}^{-1}$ and $\text{C-O}$ band having stretching frequency ${\text{1100 cm}}^{-1}$ would appear in the product.

Explanation of Solution

The given reaction is

The reactant is conjugated ketone in the above reaction. The alcohol and alkene functional groups are present in the product. The $\text{C=O}$ band having stretching frequency ${\text{1680 cm}}^{-1}$ would disappear from the reactant. The OH band having stretching frequency ${\text{3400 cm}}^{-1}$ and $\text{C-O}$ band having stretching frequency ${\text{1100 cm}}^{-1}$ would appear in the product.

Conclusion

The differences in the IR spectra that would enable to tell that the given reaction has taken place are determined on the basis of bonds and the functional groups present in the reactant and the product.