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Chapter 22: Problem 528

The \(\mathrm{pK}_{\mathrm{a}}\) of \(\mathrm{NH}_{4}^{+}\) is \(9.24\), that of \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\) is \(10.64\). Is methylamine a stronger or a weaker base than ammonia? How many times stronger or weaker? What do you think this difference is due to?

Short Answer

Expert verified
Methylamine (CH3NH2) is a stronger base than ammonia (NH3), as it has a lower pKb value (3.36) compared to ammonia (4.76). It is approximately 25 times stronger as a base. The difference in basicity can be attributed to the electron-donating inductive effect of the methyl group (-CH3), which stabilizes the positively charged conjugate acid (CH3NH3+) and increases methylamine's propensity to accept a proton.

Step by step solution

01

Calculate the pKb values for NH4+ and CH3NH3+

Using the relationship between pKa and pKb, we can calculate their pKb values: pKb(NH4+) = 14 - pKa(NH4+) = 14 - 9.24 = 4.76 pKb(CH3NH3+) = 14 - pKa(CH3NH3+) = 14 - 10.64 = 3.36
02

Compare the pKb values to determine the stronger base

To determine the stronger base, we need to compare the calculated pKb values: pKb(NH4+) = 4.76 pKb(CH3NH3+) = 3.36 The lower the pKb value, the stronger the base. Therefore, methylamine (CH3NH2) is a stronger base than ammonia (NH3).
03

Calculate the strength ratio

To find the strength ratio, calculate the difference in pKb values and use the formula: Difference in pKb values = pKb(NH4+) - pKb(CH3NH3+) = 4.76 - 3.36 = 1.40 Ratio = 10^(Difference in pKb values) = 10^1.40 ≈ 25 Methylamine (CH3NH2) is approximately 25 times stronger as a base compared to ammonia (NH3).
04

Discuss the possible cause of the difference

The difference in basicity between ammonia (NH3) and methylamine (CH3NH2) can be attributed to the inductive effect of the methyl group (-CH3). The methyl group's electron-donating nature stabilizes the positively charged conjugate acid (CH3NH3+) by dispersing the positive charge. This stabilization effect results in a higher propensity for methylamine to accept a proton, making it a stronger base than ammonia.

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Most popular questions from this chapter

Novocaine, a local anesthetic, is a compound of formula \(\mathrm{C}_{13} \mathrm{H}_{20} \mathrm{O}_{2} \mathrm{~N}_{2}\). It is insoluble in water and dilute \(\mathrm{NaOH}\), but soluble in dilute HC1. Upon treatment with \(\mathrm{NaNO}_{2}\) and \(\mathrm{HCl}\) and then with \(\beta\) -naphthol, a highly colored solid is formed. When Novocaine is boiled with aqueous \(\mathrm{NaOH}\), it slowly dissolves. The alkaline solution is shaken with ether and the layers are separated. Acidification of the aqueous layer causes the precipitation of a white solid A; continued addition of acid causes \(A\) to redissolve. Upon isolation \(A\) is found to have a melting point of \(185-6^{\circ}\) and the formula \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{~N}\). Evaporation of the ether layer leaves a liquid \(\mathrm{B}\) of formula \(\mathrm{C}_{6} \mathrm{H}_{15} \mathrm{ON}\). B dissolves in water to give a solution that turns litmus blue. Treatment of \(B\) with acetic anhydride gives C, \(\mathrm{C}_{8} \mathrm{H}_{17} \mathrm{O}_{2} \mathrm{~N}\), which is insoluble in water and dilute base, but soluble in dilute HCl. \(\mathrm{B}\) is found to be identical with the compound formed by the action of diethylamine on ethylene oxide. (a) What is the structure of Novocaine? (b) Outline all steps in a complete synthesis of Novocaine from toluene and readily available aliphatic and inorganic reagents. $$ \begin{array}{|l|l|l|} \hline {\text { CARBOXYLIC ACIDS }} \\ \hline \text { Name } & \text { Formula } & \begin{array}{c} \text { M.p. } \\ { }^{\circ} \mathrm{C} \end{array} \\ \hline \text { o-Nitrobenzoic } & \circ-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 147 \\ \hline \text { m-Nitrobenzoic } & \mathrm{m}-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 141 \\ \hline \text { p-Nitrobenzoic } & \mathrm{p}-\mathrm{O}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 242 \\ \hline \text { Phthalic } & \mathrm{o}-\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2} & 231 \\ \hline \text { Isophthalic } & \mathrm{m}-\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2} & 348 \\ \hline \text { Terephthalic } & \mathrm{p}-\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2} & 300 \text { sub1. } \\ \hline \text { Salicylic } & \mathrm{o}-\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 159 \\ \hline \text { p-Hydroxybenzoic } & \mathrm{p}-\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 213 \\ \hline \text { Anthranilic } & \mathrm{o}-\mathrm{H}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 146 \\ \hline \text { m-Aminobenzoic } & \mathrm{m}-\mathrm{H}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 179 \\ \hline \text { p-Aminobenzoic } & \mathrm{p}-\mathrm{H}_{2} \mathrm{NC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 187 \\ \hline \text { o-Methoxybenzoic } & \mathrm{o}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 101 \\ \hline \text { m-Methoxybenzoic } & \mathrm{m}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 110 \\ \hline \text { p-Methoxybenzoic(Anisic) } & \mathrm{p}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH} & 184 \\ \hline \end{array} $$

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