Chapter 13, Problem 13.40P

How would each of the following pairs of compounds differ in their IR spectra?

a. d.

b. e. ${\text{CH}}_3\text{C}\equiv {\text{CCH}}_3$ and ${\text{CH}}_3{\text{CH}}_2\text{C}\equiv \text{CH}$

c. f. $\text{HC}\equiv {\text{CCH}}_2{\text{N(CH}}_2{\text{CH}}_3{)}_2$ and ${\text{CH}}_3{{\text{(CH}}_2\text{)}}_5\text{C}\equiv \text{N}$

Interpretation Introduction

(a)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR spectrum of cyclopentene is different from pentyne because the absorption of ${\text{C}}_{s{p}^2}={\text{C}}_{s{p}^2}$ bond is observed at ${\text{1650 cm}}^{-1}$ and absorption of ${\text{C}}_{sp}\equiv {\text{C}}_{sp}$ bond is observed at ${\text{2250 cm}}^{-1}$.

Explanation of Solution

The given compounds are cyclopentene and pentyne as shown below.

Figure 1

Cyclopentene contains ${\text{C}}_{s{p}^2}={\text{C}}_{s{p}^2}$ bond and shows absorption at ${\text{1650 cm}}^{-1}$ whereas 1-pentyne contains ${\text{C}}_{sp}\equiv {\text{C}}_{sp}$ and shows absorption at ${\text{2250 cm}}^{-1}$. Hence, the IR absorptions of cyclopentene is different from cyclopentane due to the bond strength.

Conclusion

The IR spectrum of cyclopentene is different from pentyne because the absorption of ${\text{C}}_{s{p}^2}={\text{C}}_{s{p}^2}$ bond is observed at ${\text{1650 cm}}^{-1}$ and absorption of ${\text{C}}_{sp}\equiv {\text{C}}_{sp}$ bond is observed at ${\text{2250 cm}}^{-1}$.

Interpretation Introduction

(b)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR absorption of propanoic acid is different from methyl acetate due to broad peak of $\text{O}-\text{H}$ bond is observed.

Explanation of Solution

The given compounds are propanoic acid and methyl acetate as shown below.

Figure 2

Propanoic acid contains $\text{O}-\text{H}$ and shows very broad peak at $\text{3400}-240{\text{0 cm}}^{-1}$ while there is no $\text{O}-\text{H}$ bond present in methyl acetate. Hence, the IR absorptions of propanoic acid is different from methyl acetate.

Conclusion

The IR absorptions of propanoic acid is different from methyl acetate due to broad peak of $\text{O}-\text{H}$ bond is observed.

Interpretation Introduction

(c)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR absorptions of butan-$2$-one is different from $2$-buten-$1$-ol due to peak at approximately at $\text{17}00{\text{ cm}}^{-1}$ because of $\text{C}=\text{O}$ peak in butan-$2$-one and broad peak of $\text{O}-\text{H}$ bond in $2$-buten-$1$-ol is observed.

Explanation of Solution

The given compounds are butan-$2$-one and $2$-buten-$1$-ol as shown below.

Figure 3

Butan-$2$-one contains $\text{C}=\text{O}$ and shows a peak approximately at $\text{17}00{\text{ cm}}^{-1}$ while there is no $\text{C}=\text{O}$ bond present in $2$-buten-$1$-ol. In the IR spectra of $2$-buten-$1$-ol, there is presence of very broad peak in the range of $\text{3600}-260{\text{0 cm}}^{-1}$ due to $\text{O}-\text{H}$ bond which is absent in butan-$2$-one. Hence, the IR absorptions of butan-$2$-one is different from $2$-buten-$1$-ol.

Conclusion

The IR absorptions of propanoic acid is different from methyl acetate due to broad peak of $\text{O}-\text{H}$ bond is observed.

Interpretation Introduction

(d)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR spectra of methyl hexanoate are different from $1,1$-dimethyloxycyclohexane due to the presence of ${\text{C}}_{s{p}^2}=\text{O}$ bond.

Explanation of Solution

The given compound are $1,1$-dimethyloxycyclohexane and methyl hexanoate as shown below.

Figure 4

Methyl hexanoate contains ${\text{C}}_{s{p}^2}=\text{O}$ bond and shows absorption around at ${\text{1730 cm}}^{-1}$ there is no ${\text{C}}_{s{p}^2}=\text{O}$ bond present in $1,1$-dimethyloxycyclohexane. Hence, the IR spectra of methyl hexanoate is different from $1,1$-dimethyloxycyclohexane due to the presence of ${\text{C}}_{s{p}^2}=\text{O}$ bond.

Conclusion

The IR spectra of methylhexanoate is different from $1,1$-dimethyloxycyclohexane due to the presence of ${\text{C}}_{s{p}^2}=\text{O}$ bond.

Interpretation Introduction

(e)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

But-$1$-yne contains ${\text{C}}_{sp}-\text{H}$ bond and shows absorption at ${\text{3300 cm}}^{-1}$ whereas but-$2$-yne does not contain this bond.

Explanation of Solution

But-$1$-yne (${\text{CH}}_3{\text{CH}}_2\text{C}\equiv \text{CH}$) contains ${\text{C}}_{sp}-\text{H}$ bond and shows absorption at ${\text{3300 cm}}^{-1}$ whereas but-$2$-yne (${\text{CH}}_3\text{C}\equiv {\text{CCH}}_3$)does not contain this bond. Hence, the IR absorptions of but-$1$-yne is different from but-$2$-yne.

Conclusion

But-$1$-yne contains ${\text{C}}_{sp}-\text{H}$ bond and shows absorption at ${\text{3300 cm}}^{-1}$ whereas but-$2$-yne does not contain this bond.

Interpretation Introduction

(f)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR spectra of N, N-diethylprop-$2$-yn-$1$-amine is different from heptanenitrile due to presence of ${\text{C}}_{sp}-\text{H}$ bond.

Explanation of Solution

The given compound are N, N-diethylprop-$2$-yn-$1$-amine (($\text{HC}\equiv {\text{CCH}}_2{\text{N(CH}}_2{\text{CH}}_3{)}_2$)) and heptanenitrile (${\text{CH}}_3{{\text{(CH}}_2\text{)}}_5\text{C}\equiv \text{N}$). N, N-diethylprop-$2$-yn-$1$-amine contain ${\text{C}}_{sp}-\text{H}$ bond and shows absorption at ${\text{3300 cm}}^{-1}$. There is no ${\text{C}}_{sp}-\text{H}$ bond present in heptanenitrile. Hence, the IR spectra of N, N-diethylprop-$2$-yn-$1$-amine is different from heptanenitrile due to the presence of ${\text{C}}_{sp}-\text{H}$ bond.

Conclusion

The IR spectra of N, N-diethylprop-$2$-yn-$1$-amine is different from heptanenitrile due to the presence of ${\text{C}}_{sp}-\text{H}$ bond.