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1,2-dihydroxybenzene is obtained when two of the adjacent hydrogen atoms in benzene are replaced with an OH group. A skeleton of the molecule is shown here. (a) Complete a Lewis structure for the molecule using bonds and electron pairs as needed. (b) Are there any resonance structures for the molecule? If so, sketch them. (c) Are the resonance structures in (a) and (b) equivalent to one another as they are in benzene?

Short Answer

Expert verified
(a) Complete the structure with OH groups and lone pairs. (b) Yes, there are resonance structures. (c) They are not equivalent due to OH groups.

Step by step solution

01

Analyze the Molecular Structure

1,2-Dihydroxybenzene, also known as catechol, has two hydroxyl (-OH) groups attached to adjacent carbons on a benzene ring. Begin by drawing a benzene ring with alternating single and double bonds, then select any two adjacent carbon atoms to attach the OH groups.
02

Complete the Lewis Structure

Add the OH groups to the two neighboring carbon atoms identified in Step 1. Ensure each carbon has four total bonds, and each oxygen atom in the OH group has two lone pairs to complete their octet.
03

Determine Possible Resonance Structures

For dihydroxybenzene, consider the movement of double bonds in the benzene ring: the pi electrons can move around the ring. Draw alternative structures by changing the positions of the double bonds while keeping the OH groups fixed.
04

Evaluate the Resonance Structures

Compare the resonance structures to see if they all contribute equally to the description of the molecule. Note that while benzene itself has equivalent resonance structures, the presence of OH groups disrupts this symmetry.
05

Conclude on Resonance Structure Equivalence

The resonance structures of 1,2-dihydroxybenzene are not equivalent like those of pure benzene. The position of the OH groups causes electron distribution effects, making some resonance forms more stable than others.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis structure
Lewis structures are diagrams that represent the bonding of atoms within a molecule along with any lone pairs of electrons that may exist. To create a Lewis structure for 1,2-dihydroxybenzene, you start by drawing its basic skeleton:
  • A six-membered carbon ring representing benzene, typically shown with alternating single and double bonds.
  • Two hydroxyl groups (OH) added to two adjacent carbons.
Here's how to proceed:
  • Add OH groups to adjacent carbon atoms on the benzene ring.
  • Each carbon atom must have four bonds in total, considering the existing carbon-carbon bonds from the ring and the newly attached OH groups.
  • Every oxygen atom should exhibit two lone pairs to satisfy the octet rule.
This method ensures that each atom in catechol achieves a full valency, reflecting how molecules share electrons to form bonds.
resonance structures
Resonance structures illustrate different ways to represent the electron arrangement within a molecule, highlighting the possible configurations of pi electrons. In the context of 1,2-dihydroxybenzene, we can explore the resonance by moving around the double bonds located within the benzene ring. To identify resonance structures:
  • Keep the OH groups fixed; they're bonded to the same carbon atoms throughout each structure.
  • Focus on shifting the pi bonds around the ring.
  • This involves interchanging single and double bonds between the carbon atoms of the benzene ring.
The resonance forms illustrate that while the benzene core is conjugated with moving pi electrons, the presence of the hydroxyl groups affects the stabilization and appearance of these forms. Not all generated forms will contribute equally to the real structure; this is where analyzing stability and equivalence becomes crucial.
benzene ring
A benzene ring is a highly stable, cyclic hydrocarbon structure consisting of six carbon atoms connected in a hexagonal arrangement, with alternating single and double bonds. This creates what is called a conjugated system. In benzene:
  • The electrons in the pi bonds are delocalized, meaning they are shared across the entire ring.
  • This delocalization leads to a resonance-stabilized structure, giving benzene its well-known stability and symmetry.
The introduction of substituents, like hydroxyl groups in 1,2-dihydroxybenzene, modifies this perfect symmetry. The hydroxyl groups exert an electron-donating effect through resonance, affecting the electron cloud's uniformity and subsequently the resonance structures. Thus, while the base benzene ring exhibits interchangeable resonance forms, adding functional groups like OH disrupts this equivalence by drawing the electron density preferentially toward themselves, impacting the overall electron distribution of the molecule.

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Most popular questions from this chapter

Consider the collection of nonmetallic elements: \(\mathrm{B}\), As, \(\mathrm{O}\), and I. (a) Which two would form the most polar single bond? (b) Which two would form the longest single bond? (c) Which one would be likely to form a compound of formula \(\mathrm{XY}_{3}\) ? (d) Which element would likely to participate in two covalent bonds?

Some chemists believe that satisfaction of the octet rule should be the top criterion for choosing the dominant Lewis structure of a molecule or ion. Other chemists believe that achieving the best formal charges should be the top criterion. Consider the dihydrogen phosphate ion, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) in which the \(\mathrm{H}\) atoms are bonded to \(\mathrm{O}\) atoms. \((\mathbf{a})\) What is the predicted dominant Lewis structure if satisfying the octet rule is the top criterion? (b) What is the predicted dominant Lewis structure if achieving the best formal charges is the top criterion?

Consider the lattice energies of the following Group \(2 \mathrm{~A}\) compounds: \(\mathrm{BeH}_{2}, 3205 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{MgH}_{2}, 2791 \mathrm{~kJ} / \mathrm{mol} ;\) \(\mathrm{CaH}_{2}, 2410 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{SrH}_{2}, 2250 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{BaH}_{2}, 2121 \mathrm{~kJ} / \mathrm{mol}\) (a) What is the oxidation number of \(\mathrm{H}\) in these compounds? (b) Assuming that all of these compounds have the same three-dimensional arrangement of ions in the solid, which of these compounds has the shortest cation-anion distance? (c) Consider \(\mathrm{BeH}_{2}\). Does it require \(3205 \mathrm{~kJ}\) of energy to break one mole of the solid into its ions, or does breaking up one mole of solid into its ions release \(3205 \mathrm{~kJ}\) of energy? (d) The lattice energy of \(\mathrm{ZnH}_{2}\) is \(2870 \mathrm{~kJ} / \mathrm{mol}\). Considering the trend in lattice enthalpies in the Group 2 compounds, predict which Group 2 element is most similar in ionic radius to the \(\mathrm{Zn}^{2+}\) ion.

(a) Construct a Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), in which each atom achieves an octet of electrons. (b) How many bonding electrons are between the two oxygen atoms? (c) Do you expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{H}_{2} \mathrm{O}_{2}\) to be longer or shorter than the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) ? Explain.

The \(\mathrm{Ti}^{2+}\) ion is isoelectronic with the Ca atom. (a) Write the electron configurations of \(\mathrm{Ti}^{2+}\) and Ca. (b) Calculate the number of unpaired electrons for Ca and for \(\mathrm{Ti}^{2+} .(\mathbf{c})\) What charge would Ti have to be isoelectronic with \(\mathrm{Ca}^{2+}\) ?

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