Most organic acids can be represented as \(\mathrm{RCOOH}\) where \(\mathrm{COOH}\) is the carboxyl group and \(\mathrm{R}\) is the rest of the molecule. (For example, \(\mathrm{R}\) is \(\mathrm{CH}_{3}\) in acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH} .\) ) (a) Draw a Lewis structure for the carboxyl group. (b) Upon ionization, the carboxyl group is converted to the carboxylate group, \(\mathrm{COO}^{-}\). Draw resonance structures for the carboxylate group.

Short Answer

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The Lewis structure for a carboxyl group includes a \(\mathrm{C}\) atom bonded to an \(\mathrm{O}\) atom through a double bond and to a \(\mathrm{OH}\) through a single bond. Upon ionization, it is converted to carboxylate ion (\(\mathrm{COO}^{-}\)), in the Lewis structure of which, \(\mathrm{C}\) is bonded to two \(\mathrm{O}\) atoms, while the negative charge can be on either \(\mathrm{O}\). The two structures arising due to charge distribution form the resonance structures.

Step by step solution

01

Drawing Lewis Structure for Carboxyl Group

The carboxyl group \(\mathrm{COOH}\) contains elements carbon (C), oxygen (O), and hydrogen (H). The valency of these elements are 4, 2, and 1 respectively. Draw the Lewis structure ensuring each atom fulfills these requirements. C is connected to one O by a double bond (since O requires two bonds), and to the other O by a single bond. This second O is also bonded to H.
02

Ionization of the Carboxyl Group

When \(\mathrm{COOH}\) is ionised, it loses a hydrogen ion \(\mathrm{H}^{+}\), and the leftover \(\mathrm{O}\) carries a negative charge represented as \(\mathrm{COO}^{-}\). The negatively charged \(\mathrm{O}\) and the other \(\mathrm{O}\) are both bonded to the \(\mathrm{C}\) atom.
03

Drawing Resonance Structures for the Carboxylate Group

In the carboxylate ion, the negative charge on the \(\mathrm{O}\) atom can be delocalized (shared) between the two \(\mathrm{O}\) atoms through a process called resonance, giving rise to two equivalent structures. These structures are drawn such that in one, the double bond is between \(\mathrm{C}\) and first \(\mathrm{O}\) (with the second \(\mathrm{O}\) carrying the negative charge), while in the second structure, the double bond is between \(\mathrm{C}\) and second \(\mathrm{O}\) (with the first \(\mathrm{O}\) carrying the negative charge).

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Most popular questions from this chapter

Compare single, double, and triple bonds in a molecule, and give an example of each. For the same bonding atoms, how does the bond length change from single bond to triple bond?

Which of the following molecules has the shortest nitrogen-to-nitrogen bond: \(\mathrm{N}_{2} \mathrm{H}_{4}, \mathrm{~N}_{2} \mathrm{O}, \mathrm{N}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4} ?\) Explain.

In the vapor phase, beryllium chloride consists of discrete \(\mathrm{BeCl}_{2}\) molecules. Is the octet rule satisfied for Be in this compound? If not, can you form an octet around Be by drawing another resonance structure? How plausible is this structure?

Nitroglycerin, one of the most commonly used explosives, has the structure The decomposition reaction is $$ \begin{array}{r} 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}(l) \longrightarrow \\ 12 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \end{array} $$ The explosive action is the result of the heat released and the large increase in gaseous volume. (a) Calculate the \(\Delta H^{\circ}\) for the decomposition of one mole of nitroglycerin using both standard enthalpy of formation values and bond enthalpies. Assume that the two \(\mathrm{O}\) atoms in the \(\mathrm{NO}_{2}\) groups are attached to \(\mathrm{N}\) with one single bond and one double bond. (b) Calculate the combined volume of the gases at STP. (c) Assuming an initial explosion temperature of \(3000 \mathrm{~K}\), estimate the pressure exerted by the gases using the result from (b). (The standard enthalpy of formation of nitroglycerin is \(-371.1 \mathrm{~kJ} / \mathrm{mol} .\) )

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