Organic Chemistry
Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 23, Problem 23.7P
Interpretation Introduction

(a)

Interpretation:

The increasing order of basicity of propylamine, ammonia, and dipropylamine in aqueous solution is to be arranged.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia. Amines are basic in nature because the nitrogen can donate its lone pairs and also the ability of the nitrogen to accept the proton in water.

Expert Solution
Check Mark

Answer to Problem 23.7P

The increasing order of basicity of propylamine, ammonia, and dipropylamine in aqueous solution is shown below.

Ammonia<propylamine<dipropylamine

Explanation of Solution

It is known that the order of basicity of amines in aqueous solution is NH3<3°amine<1°amine<2°amine. The dipropylamine amine is secondary amine. Therefore, it is most basic. The propylamine is primary amine. Therefore, it is second most basic. Ammonia is least basic. Therefore, the increasing order of basicity of propylamine, ammonia, and dipropylamine in aqueous solution is shown below.

Ammonia<propylamine<dipropylamine

Conclusion

The increasing order of basicity of propylamine, ammonia, and dipropylamine in aqueous solution is shown below.

Ammonia<propylamine<dipropylamine

Interpretation Introduction

(b)

Interpretation:

The increasing order of basicity of methyl-3-aminopropanoate, sec-butylamine, and H3N+CH2CH2NH2 in aqueous solution is to be arranged.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia. Amines are basic in nature because the nitrogen can donate its lone pairs and also the ability of the nitrogen to accept the proton in water.

Expert Solution
Check Mark

Answer to Problem 23.7P

The increasing order of basicity of methyl-3-aminopropanoate, sec-butylamine, and H3N+CH2CH2NH2 in aqueous solution is shown below.

H3N+CH2CH2NH2<methyl-3-aminopropanoate<sec-butylamine

Explanation of Solution

It is known that the order of basicity of amines in aqueous solution solution is NH3<3°amine<1°amine<2°amine. The sec-butylamine is secondary amine. Therefore, it is most basic. The methyl-3-aminopropanoate is primary amine. Therefore, it is second most basic. In case of H3N+CH2CH2NH2, the positive charge on nitrogen makes it least basic. The increasing order of basicity of methyl-3-aminopropanoate, sec-butylamine, and H3N+CH2CH2NH2 in aqueous solution is shown below.

H3N+CH2CH2NH2<methyl-3-aminopropanoate<sec-butylamine

Conclusion

The increasing order of basicity of methyl-3-aminopropanoate, sec-butylamine, and H3N+CH2CH2NH2 in aqueous solution is shown below.

H3N+CH2CH2NH2<methyl-3-aminopropanoate<sec-butylamine

Interpretation Introduction

(c)

Interpretation:

The increasing order of basicity of aniline, methyl m-aminobenzoate, and methyl p-aminobenzoate in aqueous solution is to be arranged.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia. Amines are basic in nature because the nitrogen can donate its lone pairs and also the ability of the nitrogen to accept the proton in water.

Expert Solution
Check Mark

Answer to Problem 23.7P

The increasing order of basicity of aniline, methyl m-aminobenzoate, and methyl p-aminobenzoate in aqueous solution is shown below.

Methylp-aminobenzoate<methylm-aminobenzoate<aniline

Explanation of Solution

The structure of aniline, methyl m-aminobenzoate, and methyl p-aminobenzoate is shown below.

Organic Chemistry, Chapter 23, Problem 23.7P , additional homework tip  1

Figure 1

Amines are basic in nature because the nitrogen can donate its lone pairs and also the ability of the nitrogen to accept the proton in water.

The basicity of aniline substituted compound depends on the group that is attached to benzene ring. If the group attached to benzene ring is electron withdrawing group, it will decrease the basicity of the compound. If the group attached to benzene ring is electron donating group, it will increase the basicity of the compound.

There is no group attached in case of aniline. Therefore, it is most basic. In case of methyl m-aminobenzoate and methyl p-aminobenzoate, the CH3OCO group is attached to benzene ring which is an electron withdrawing group. Therefore, this group decreases the electron density on nitrogen atom by I effect. This result the decreases in basicity. The effect is more at para position. Therefore, methyl p-aminobenzoate is less basic than methyl m-aminobenzoate. The increasing order of basicity of aniline, methyl m-aminobenzoate, and methyl p-aminobenzoate in aqueous solution is shown below.

Methylp-aminobenzoate<methylm-aminobenzoate<aniline

Conclusion

The increasing order of basicity of aniline, methyl m-aminobenzoate, and methyl p-aminobenzoate in aqueous solution is shown below.

Methylp-aminobenzoate<methylm-aminobenzoate<aniline

Interpretation Introduction

(d)

Interpretation:

The increasing order of basicity of benzylamine, p-nitrobenzylamine, cyclohexylamine, aniline in aqueous solution is to be arranged.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia. Amines are basic in nature because the nitrogen can donate its lone pairs and also the ability of the nitrogen to accept the proton in water.

Expert Solution
Check Mark

Answer to Problem 23.7P

The increasing order of basicity of benzylamine, p-nitrobenzylamine, cyclohexylamine, aniline in aqueous solution is shown below.

Aniline<p-nitrobenzylamine<benzylamine<cyclohexylamine

Explanation of Solution

The structure of benzylamine, p-nitrobenzylamine, cyclohexylamine, aniline is shown below.

Organic Chemistry, Chapter 23, Problem 23.7P , additional homework tip  2

Figure 2

Amines are basic in nature because the nitrogen can donate its lone pairs and also the ability of the nitrogen to accept the proton in water.

In case of aniline, the lone pair present at nitrogen atom is resonate with the benzene ring. Therefore, the electron density on nitrogen is decreased. Therefore, it is least basic.

In case of cyclohexylamine, there is no resonance. The cyclohexyl group is an electron donating group which increases the electron density on nitrogen atom. Therefore, the basicity of cyclohexylamine increased.

In case of benzylamine there is also no resonance because sp3 hybrid carbon is present in between the NH2 group and benzene ring. The phenyl group is an electron withdrawing group which decreases the electron density on nitrogen atom. Therefore, the basicity of benzylamine is less than cyclohexylamine.

In case of p-nitrobenzylamine there is also no resonance because sp3 hybrid carbon is present in between the NH2 group and benzene ring. The phenyl group is an electron withdrawing group which decreases the electron density on nitrogen atom. Due to presence of nitro group at para position the basicity of p-nitrobenzylamine is less than the benzylamine. Therefore, the increasing order of basicity of benzylamine, p-nitrobenzylamine, cyclohexylamine, aniline in aqueous solution is shown below.

Aniline<p-nitrobenzylamine<benzylamine<cyclohexylamine

Conclusion

The increasing order of basicity of benzylamine, p-nitrobenzylamine, cyclohexylamine, aniline in aqueous solution is shown below.

Aniline<p-nitrobenzylamine<benzylamine<cyclohexylamine

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Chapter 23 Solutions

Organic Chemistry

Ch. 23 - Prob. 23.11PCh. 23 - Prob. 23.12PCh. 23 - Prob. 23.13PCh. 23 - Prob. 23.14PCh. 23 - Prob. 23.15PCh. 23 - Prob. 23.16PCh. 23 - Prob. 23.17PCh. 23 - Prob. 23.18PCh. 23 - Prob. 23.19PCh. 23 - Prob. 23.20PCh. 23 - Prob. 23.21PCh. 23 - Prob. 23.22PCh. 23 - Prob. 23.23PCh. 23 - Prob. 23.24PCh. 23 - Prob. 23.25PCh. 23 - Prob. 23.26PCh. 23 - Prob. 23.27PCh. 23 - Prob. 23.28PCh. 23 - Prob. 23.29PCh. 23 - Prob. 23.30PCh. 23 - Prob. 23.31PCh. 23 - Prob. 23.32PCh. 23 - Prob. 23.33PCh. 23 - Prob. 23.34PCh. 23 - Prob. 23.35PCh. 23 - Prob. 23.36PCh. 23 - Prob. 23.37PCh. 23 - Prob. 23.38PCh. 23 - Prob. 23.39PCh. 23 - Prob. 23.40PCh. 23 - Prob. 23.41PCh. 23 - Prob. 23.42PCh. 23 - Prob. 23.43PCh. 23 - Prob. 23.44APCh. 23 - Prob. 23.45APCh. 23 - Prob. 23.46APCh. 23 - Prob. 23.47APCh. 23 - Prob. 23.48APCh. 23 - Prob. 23.49APCh. 23 - Prob. 23.50APCh. 23 - Prob. 23.51APCh. 23 - Prob. 23.52APCh. 23 - Prob. 23.53APCh. 23 - Prob. 23.54APCh. 23 - Prob. 23.55APCh. 23 - Prob. 23.56APCh. 23 - Prob. 23.57APCh. 23 - Prob. 23.58APCh. 23 - Prob. 23.59APCh. 23 - Prob. 23.60APCh. 23 - Prob. 23.61APCh. 23 - Prob. 23.62APCh. 23 - Prob. 23.63APCh. 23 - Prob. 23.64APCh. 23 - Prob. 23.65APCh. 23 - Prob. 23.66APCh. 23 - Prob. 23.67APCh. 23 - Prob. 23.68APCh. 23 - Prob. 23.69APCh. 23 - Prob. 23.70APCh. 23 - Prob. 23.71APCh. 23 - Prob. 23.72APCh. 23 - Prob. 23.73APCh. 23 - Prob. 23.74APCh. 23 - Prob. 23.75APCh. 23 - Prob. 23.76APCh. 23 - Prob. 23.77APCh. 23 - Prob. 23.78APCh. 23 - Prob. 23.79APCh. 23 - Prob. 23.80APCh. 23 - Prob. 23.81AP
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