Chapter 21, Problem 21.55AP

Interpretation Introduction

(a)

Interpretation:

The compound from given spectra is to be identified.

Concept Introduction:

In the IR spectroscopy, change in the dipole moment produces absorption of energy. All the functional group have different absorption frequency based on which they can be differentiated. For carbonyl $\text{C}=\text{O}$, the absorption takes place at $1725\text{c}{\text{m}}^{-1}$ which is different for carboxylic acid and esters. In proton NMR, the proton resonance is taken into consideration for different functional groups. Due to change in surroundings of the proton, the chemical shift changes.

Answer to Problem 21.55AP

The structure of compound is shown below.

Explanation of Solution

The odd molecular mass of the compound indicates that it contains odd number of nitrogen atoms. Hydroxamate test is given by esters. So, compound contains ester group. The IR band at $1733\text{c}{\text{m}}^{\text{-1}}$ is for ester, at $1200\text{c}{\text{m}}^{-1}$ is for $-\text{OH}$ functional group and at $2237\text{c}{\text{m}}^{\text{-1}}$ is for nitrile group. Hence, the compound contains both ester and nitrile functional group. The triplet and quartet pattern in NMR indicates the presence of ethyl ester group. The singlet of $2H$ group indicates presence of methylene group in between carbonyl and nitrile group. The structure of the compound is shown below.

Figure 1

Conclusion

The given compound is ethyl cyanoacetate.

Interpretation Introduction

(b)

Interpretation:

The compound from given spectra is to be identified.

Concept Introduction:

In the IR spectroscopy, change in the dipole moment produces absorption of energy. All the functional group have different absorption frequency based on which they can be differentiated. For carbonyl $\text{C}=\text{O}$, the absorption takes place at $1725\text{c}{\text{m}}^{-1}$ which is different for carboxylic acid and esters. In proton NMR, the proton resonance is taken into consideration for different functional groups. Due to change in surroundings of the proton, the chemical shift changes.

Answer to Problem 21.55AP

The structure of given compound is shown below.

Explanation of Solution

The given compound gives IR absorption peak at $1743\text{c}{\text{m}}^{\text{-1}}$ which means it contains ester group. The presence of triplet –quartet pattern in NMR indicates that ethyl group is present. The resonance at $\text{δ}1.6\left(2H\right)$ is coupled with both $\text{δ}1.0\left(3H\right)\text{and}\text{δ}4.0\left(2H\right)$ indicates propyl group adjacent to the carboxylate oxygen atom. Hence, the compound is propyl propanoate.

Figure 2

Conclusion

The given compound is propyl propanaote.

Interpretation Introduction

(c)

Interpretation:

The compound from given spectra is to be identified.

Concept Introduction:

In the IR spectroscopy, change in the dipole moment produces absorption of energy. All the functional group have different absorption frequency based on which they can be differentiated. For carbonyl $\text{C}=\text{O}$, the absorption takes place at $1725\text{c}{\text{m}}^{-1}$ which is different for carboxylic acid and esters. In proton NMR, the proton resonance is taken into consideration for different functional groups. Due to change in surroundings of the proton, the chemical shift changes.

Answer to Problem 21.55AP

The structure of given compound is shown below.

Explanation of Solution

The strong absorption peak in IR at $3200{\text{cm}}^{-1}$ indicates the presence of hydroxyl group. The absorption at $2250{\text{cm}}^{-1}$ indicates the presence of nitrile group. The triplet-triplet pattern of two hydrogen atoms having same coupling constant indicates the presence of $-{\text{CH}}_{\text{2}}-{\text{CH}}_{\text{2}}$ group in between the nitrile and hydroxyl group. So, the given compound is $3-\text{hydroxypropanenitrile}$ as shown below.

Figure 3

Conclusion

The given compound is $3-\text{hydroxypropanenitrile}$.

Interpretation Introduction

(d)

Interpretation:

The compound from given spectra is to be identified.

Concept Introduction:

In the IR spectroscopy, change in the dipole moment produces absorption of energy. All the functional group have different absorption frequency based on which they can be differentiated. For carbonyl $\text{C}=\text{O}$, the absorption takes place at $1725\text{c}{\text{m}}^{-1}$ which is different for carboxylic acid and esters. In proton NMR, the proton resonance is taken into consideration for different functional groups. Due to change in surroundings of the proton, the chemical shift changes.

Answer to Problem 21.55AP

The structure of given compound is shown below.

Explanation of Solution

The two molecular ion peaks at $\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{z}$}\right.=\text{180}$ and $182$ indicates that bromine in present in the compound. The IR absorption peak at $1743\text{c}{\text{m}}^{-1}$ indicates that ester group is present in the compound. The triplet-quartet pattern at $\text{δ}1.30\left(3H,\text{t},J=7\text{Hz}\right)$ $\text{δ}4.23\left(2H,\text{q},J=7\text{Hz}\right)$ indicates the presence of ethyl group attached to the carboxylate oxygen. The quartet of one hydrogen atom at $\text{δ}4.37\left(1H,\text{q},J=7\text{Hz}\right)$ indicates the presence of $-\text{CH}$ group attached with the methyl group.

Figure 4

Conclusion

The given compound is $\text{ethyl-2-bromopropanoate}$.

Interpretation Introduction

(e)

Interpretation:

The compound from given spectra is to be identified.

Concept Introduction:

In the IR spectroscopy, change in the dipole moment produces absorption of energy. All the functional group have different absorption frequency based on which they can be differentiated. For carbonyl $\text{C}=\text{O}$, the absorption takes place at $1725\text{c}{\text{m}}^{-1}$ which is different for carboxylic acid and esters. In proton NMR, the proton resonance is taken into consideration for different functional groups. Due to change in surroundings of the proton, the chemical shift changes.

Answer to Problem 21.55AP

The structure of given compound is shown below.

Explanation of Solution

The compound has odd molecular mass which indicates the presence of nitrogen atom. The IR absorption peak at $\text{2200}\text{c}{\text{m}}^{-1}$ confirms the presence of nitrile group, as lowering inn absorption peak takes place that may be due to extended conjugation. The UV absorption confirms the extended conjugation. The NMR spectra shows peak at $\text{δ}7.4\left(5H,\text{apparent}\text{s}\right)$ which is for the benzene hydrogen atoms, indicating it is a monosubstituted benzene. The doublets at $\text{δ}5.85\left(1H,\text{d},J=17\text{Hz}\right)$, $\text{δ}7.35\left(1H,\text{d},J=17\text{Hz}\right)$; indicates the presence of alkene group and large difference in shift indicates that it is a trans alkene. The absorption at $970\text{c}{\text{m}}^{\text{-1}}$ confirms the presence of trans alkene. The given compound is $\text{trans-3-phenylpropenenitrile}$

Figure 5

Conclusion

The given compound is $\text{trans-3-phenylpropenenitrile}$.

Interpretation Introduction

(f)

Interpretation:

The compound from given spectra is to be identified.

Concept Introduction:

In the IR spectroscopy, change in the dipole moment produces absorption of energy. All the functional group have different absorption frequency based on which they can be differentiated. For carbonyl $\text{C}=\text{O}$, the absorption takes place at $1725\text{c}{\text{m}}^{-1}$ which is different for carboxylic acid and esters. In proton NMR, the proton resonance is taken into consideration for different functional groups. Due to change in surroundings of the proton, the chemical shift changes.

Answer to Problem 21.55AP

The structure of given compound is shown below.

Explanation of Solution

The absorption in IR at indicates the presence of amide group and at $1659{\text{cm}}^{-1}$ indicate the carbonyl group. The NMR peak at $\text{δ}7.6$ singlet shows the presence of hydrogen of amide, hence it is a secondary amide. The two doublets in NMR at $\text{δ}6.9\text{and}\text{δ}7.4$ indicates that it is a $1,4$ disubstituted benzene ring. The singlet at $\text{δ}2.0\left(3H\right)$ indicates the presence of methyl group adjacent to the carbonyl group indicating presence of acetamide. The triplet-quartet pattern indicates the presence of ethyl group adjacent to oxygen atom, thus ethoxy group is also present. Hence, the given compound is $N-\left(p\text{-ethoxyphenyl}\right)\text{acetamide}$.

Figure 6

Conclusion

The given compound is $N-\left(p\text{-ethoxyphenyl}\right)\text{acetamide}$

Interpretation Introduction

(g)

Interpretation:

The compound from given spectra is to be identified.

Concept Introduction:

In the IR spectroscopy, change in the dipole moment produces absorption of energy. All the functional group have different absorption frequency based on which they can be differentiated. For carbonyl $\text{C}=\text{O}$, the absorption takes place at $1725\text{c}{\text{m}}^{-1}$ which is different for carboxylic acid and esters. In proton NMR, the proton resonance is taken into consideration for different functional groups. Due to change in surroundings of the proton, the chemical shift changes.

Answer to Problem 21.55AP

The structure of given compound is shown below.

Explanation of Solution

The odd molecular mass of the compound indicates it contains nitrogen atom. The IR absorption peak at $3397$ and $3200\text{c}{\text{m}}^{-1}$ indicates the presence of primary amide . the absorption at $1655\text{c}{\text{m}}^{-1}$ is for $\text{C}=\text{O}$ stretching and at $1622\text{c}{\text{m}}^{-1}$ is for $\text{N}-\text{H}$ bending of amide. In the ${}^{13}\text{C}$ NMR of the compound peak at $\text{δ}180.5$ is for carbonyl carbon and at $\text{δ}38$, is for carbon adjacent to carbonyl carbon. Only one carbon peak is left but still more carbon atoms will be there as the molecular mass indicates so there will be three methyl group present on the $\alpha$ carbon of compound all having same chemical shift value. Hence the given compound is $2,2-\text{dimethylpropanamide}$.

Figure 7

Conclusion

The given compound is $2,2-\text{dimethylpropanamide}$.