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Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet SpectroscopyProblem 23
Chapter 15, Problem 23

Phenolphthalein is an acid–base indicator that is colorless below pH 8 and red above pH 8. Explain briefly why the first structure is colorless and the second structure is colored.
Chemical structures showing phenolphthalein; left structure labeled colorless, right structure labeled red.

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1
Step 1: Analyze the chemical structures provided in the image. Structure A (colorless) has a lactone ring, which is a cyclic ester. Structure B (red) has undergone a reaction with hydroxide ions (HO⁻), leading to the opening of the lactone ring and the formation of a conjugated system.
Step 2: Understand the role of conjugation in color. In organic molecules, conjugation refers to the overlap of p-orbitals across adjacent atoms, allowing delocalization of electrons. This delocalization lowers the energy gap between the molecule's highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO), enabling absorption of visible light and imparting color.
Step 3: Explain why structure A is colorless. In structure A, the lactone ring prevents extensive conjugation. The lack of a fully conjugated system means the molecule absorbs light in the ultraviolet region, which is not visible to the human eye, making it appear colorless.
Step 4: Explain why structure B is red. In structure B, the lactone ring is opened, creating a highly conjugated system that extends across the molecule. This conjugation allows the molecule to absorb light in the visible spectrum, specifically in the blue-green region, which results in the complementary color red being observed.
Step 5: Relate the pH dependency to the structural change. At pH below 8, the lactone ring remains intact (structure A), and the molecule is colorless. Above pH 8, hydroxide ions deprotonate the molecule, opening the lactone ring and forming the conjugated system (structure B), which is red.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Acid-Base Indicators

Acid-base indicators are substances that change color in response to changes in pH. They typically exist in two forms: one that is protonated (often colorless) and one that is deprotonated (often colored). The transition between these forms occurs at a specific pH range, allowing indicators like phenolphthalein to signal the acidity or basicity of a solution.
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Structural Changes in Phenolphthalein

Phenolphthalein has different structural forms depending on the pH of the solution. Below pH 8, it exists predominantly in a protonated form, which is colorless. Above pH 8, it loses a proton and transitions to a deprotonated form, which has a conjugated double bond system that absorbs visible light, resulting in a red color.
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Conjugation and Color

The color of a compound is often related to its electronic structure, particularly the presence of conjugated systems. In phenolphthalein, the deprotonated form features a conjugated system of alternating double bonds, which allows for the absorption of specific wavelengths of light, leading to the observed color change. This phenomenon is a key principle in understanding how molecular structure influences color in organic compounds.
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Related Practice
Textbook Question

Match four of the following UV absorption maxima (λmax) with the corresponding compounds: (1) 232 nm; (2) 256 nm; (3) 273 nm; (4) 292 nm; (5) 313 nm; (6) 353 nm.

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Textbook Question

Predict the products of the following reactions.

(b) cyclopentadiene + anhydrous HCl

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Textbook Question

There is a different, thermally allowed cycloaddition of two butadiene molecules. Show this reaction, and explain why it is thermally allowed. (Hint: Consider the dimerization of cyclopentadiene.)

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Textbook Question

One milligram of a compound of molecular weight 160 is dissolved in 10 mL of ethanol, and the solution is poured into a 1-cm UV cell. The UV spectrum is taken, and there is an absorption at λmax = 247 nm. The maximum absorbance at 247 nm is 0.50. Calculate the value of e for this absorption.

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Textbook Question

Classify the following dienes and polyenes as isolated, conjugated, cumulated, or some combination of these classifications.

(a) cycloocta-1,4-diene

(b) cycloocta-1,3-diene

(c) cyclodeca-1,2-diene

(d) cycloocta-1,3,5,7-tetraene

(e) cyclohexa-1,3,5-triene (benzene)

(f) penta-1,2,4-triene

Textbook Question

Predict the products of the following reactions.

(a) allyl bromide + cyclohexyl magnesium bromide

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