In the chemical synthesis of DNA and RNA, hydroxyl groups are normally converted to triphenylmethyl (trityl) ethers to protect the hydroxyl group from reaction with other reagents. Triphenylmethyl ethers are stable to aqueous base but are rapidly cleaved in aqueous acid. $$ \mathrm{RCH}_{2} \mathrm{OCPh}_{3}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{RCH}_{2} \mathrm{OH}+\mathrm{Ph}_{3} \mathrm{COH} $$ (a) Why are triphenylmethyl ethers so readily hydrolyzed by aqueous acid? (b) How might the structure of the triphenylmethyl group be modified to increase or decrease its acid sensitivity?

Short Answer

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Question: Explain the reasons for the ready hydrolysis of triphenylmethyl ethers by aqueous acid and how modifications to the structure of the triphenylmethyl group can affect its acid sensitivity. Answer: Triphenylmethyl ethers are readily hydrolyzed by aqueous acids due to the steric hindrance and electron repulsion provided by the three phenyl groups attached to the central carbon atom. This destabilizes the positive charge on the central carbon during the hydrolysis process, making it more susceptible to attack by the nucleophile (water). Modifications to the structure of the triphenylmethyl group can alter its acid sensitivity by changing either the steric hindrance or the electron distribution. Introducing electron-donating groups on the phenyl rings will increase acid sensitivity, while introducing electron-withdrawing groups or decreasing the size of substituents will decrease acid sensitivity.

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01

(a) Reasons for ready hydrolysis of triphenylmethyl ethers by aqueous acid

Triphenylmethyl ethers are readily hydrolyzed by aqueous acids due to the steric hindrance and electron repulsion provided by the three phenyl groups attached to the central carbon atom. This destabilizes the positive charge on the central carbon during the hydrolysis process, making it more susceptible to attack by the nucleophile (water). The overall process can be explained in the following steps. Step 1: Protonation $$ \mathrm{RCH}_{2} \mathrm{OCPh}_{3} + \mathrm{H}^{+} \longrightarrow \mathrm{RCH}_{2} \mathrm{OCPh}_{3}^{+} \mathrm{H} $$ The trityl ether reacts with a proton (from an aqueous acid) which results in the protonation of the oxygen atom and a positive charge formation. Step 2: Nucleophilic attack $$ \mathrm{RCH}_{2} \mathrm{OCPh}_{3}^{+} \mathrm{H} + \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{RCH}_{2} \mathrm{OH} + \mathrm{Ph}_{3} \mathrm{COH} $$ Due to the destabilization caused by steric hindrance and electron repulsion, the positively charged carbon atom is readily attacked by a nucleophile (water in this case) which leads to the cleavage of the bond between the oxygen and carbon atoms. The trityl group loses a proton and forms triphenylmethanol, and the RCH2 group forms an alcohol, which was the goal of protection in the first place.
02

(b) Modifying the structure of the triphenylmethyl group

To modify the structure of the triphenylmethyl group and alter its acid sensitivity, we can make changes to either the steric hindrance or the electron distribution. 1. Increase acid sensitivity: The acid sensitivity can be increased by introducing electron-donating groups (-OH, -OCH3, or -NH2) on the phenyl rings. This will increase the electron density on the central carbon, making it more susceptible to the protonation and nucleophilic attack. 2. Decrease acid sensitivity: On the other hand, decreasing acid sensitivity can be achieved by introducing electron-withdrawing groups (-NO2, -CN, or -CF3) on the phenyl rings. This will decrease the electron density on the central carbon and make it less susceptible to protonation and nucleophilic attack. Additionally, decreasing the size of the substituents will reduce steric hindrance and make the trityl ether less reactive.

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Most popular questions from this chapter

Following are \({ }^{1} \mathrm{H}-\mathrm{NMR}\) and \({ }^{13} \mathrm{C}-\mathrm{NMR}\) spectral data for compound \(\mathrm{F}\left(\mathrm{C}_{12} \mathrm{H}_{16} \mathrm{O}\right)\). From this information, deduce the structure of compound \(F\). $$ \begin{array}{lll} \hline{ }^{1} \text { H-NMR } & \multicolumn{2}{c}{{ }^{13} \text { C-NMR }} \\\ \hline 0.83(\mathrm{~d}, 6 \mathrm{H}) & 207.82 & 50.88 \\ 2.11(\mathrm{~m}, 1 \mathrm{H}) & 134.24 & 50.57 \\ 2.30(\mathrm{~d}, 2 \mathrm{H}) & 129.36 & 24.43 \\ 3.64(\mathrm{~s}, 2 \mathrm{H}) & 128.60 & 22.48 \\ 7.2-7.4(\mathrm{~m}, 5 \mathrm{H}) & 126.86 & \\ \hline \end{array} $$

Account for the fact that water-insoluble carboxylic acids ( \(\mathrm{p} K_{\mathrm{a}} 4-5\) ) dissolve in \(10 \%\) aqueous sodium bicarbonate \((\mathrm{pH} 8.5)\) with the evolution of a gas but that waterinsoluble phenols ( \(\mathrm{p} K_{\mathrm{a}} 9.5-10.5\) ) do not dissolve in \(10 \%\) sodium bicarbonate.

Draw a structural formula for each compound. (a) 1-Bromo-2-chloro-4-ethylbenzene (b) \(m\)-Nitrocumene (c) 4-Chloro- 1,2 -dimethylbenzene (d) 3,5 -Dinitrotoluene (e) \(2,4,6\)-Trinitrotoluene (f) \((2 S, 4 R)\)-4-Phenyl-2-pentanol (g) \(p\)-Cresol (h) Pentachlorophenol (i) 1-Phenylcyclopropanol (j) Triphenylmethane (k) Phenylethylene (styrene) (I) Benzyl bromide (m) 1-Phenyl-1-butyne (n) \((E)\)-3-Phenyl-2-propen-1-ol

Given here are \({ }^{1}\) H-NMR and \({ }^{13}\) C-NMR spectral data for two compounds. Each shows strong, sharp absorption between 1700 and \(1720 \mathrm{~cm}^{-1}\), and strong, broad absorption over the region \(2500-3000 \mathrm{~cm}^{-1}\). Propose a structural formula for each compound. (a) \(\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{3}\) (b) \(\mathrm{C}_{10} \mathrm{H}_{10} \mathrm{O}_{2}\) $$ \begin{array}{cc} \hline{ }^{1} \mathbf{H}-\mathrm{NMR} & { }^{13} \mathrm{C}-\mathrm{NMR} \\ \hline 2.49(\mathrm{t}, 2 \mathrm{H}) & 173.89 \\ 2.80(\mathrm{t}, 2 \mathrm{H}) & 157.57 \\ 3.72(\mathrm{~s}, 3 \mathrm{H}) & 132.62 \\ 6.78(\mathrm{~d}, 2 \mathrm{H}) & 128.99 \\ 7.11(\mathrm{~d}, 2 \mathrm{H}) & 113.55 \\ 12.4(\mathrm{~s}, 1 \mathrm{H}) & 54.84 \\ & 35.75 \\ & 29.20 \\ \hline \end{array} $$ $$ \begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 2.34(\mathrm{~s}, 3 \mathrm{H}) & 167.82 \\ 6.38(\mathrm{~d}, 1 \mathrm{H}) & 143.82 \\ 7.18(\mathrm{~d}, 1 \mathrm{H}) & 139.96 \\ 7.44(\mathrm{~d}, 2 \mathrm{H}) & 131.45 \\ 7.56(\mathrm{~d}, 2 \mathrm{H}) & 129.37 \\ 12.0(\mathrm{~s}, 1 \mathrm{H}) & 127.83 \\ & 111.89 \\ & 21.13 \\ \hline \end{array} $$

Account for the fact that \(p\)-nitrophenol \(\left(K_{\mathrm{a}} 7.0 \times 10^{-8}\right)\) is a stronger acid than phenol $$ \left(K_{\mathrm{a}} 1.1 \times 10^{-10}\right) $$

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