Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 596

(a) Hydroquinone is used in photographic developers to aid in the conversion of silver ion into free silver. What property of hydroquinone is being taken advantage of here? (b) p-Benzoquinone can be used to convert iodide ion into iodine. What property of the quinone is being taken advantage of here?

Short Answer

Expert verified
(a) Hydroquinone is being used as a reducing agent in photographic developers to convert silver ions (Ag+) to free silver (Ag) by donating electrons to them. (b) The property of p-Benzoquinone being taken advantage of to convert iodide ions (I-) into iodine (I2) is its ability to act as an oxidizing agent, accepting electrons from iodide ions while getting reduced itself.

Step by step solution

01

Discuss hydroquinone

Hydroquinone is an aromatic compound that has the chemical formula C6H6O2. It has two alcohol (hydroxyl) groups attached to a benzene ring. This compound has a significant property called "reducing agent". A reducing agent, as the name suggests, has the ability to reduce other substances by donating electrons to them. For part (b):
02

Discuss p-Benzoquinone

p-Benzoquinone (also called 1,4-benzoquinone) is an organic compound with the chemical formula C6H4O2. It has a cyclic structure with a benzene ring, with two carbonyl (C=O) groups attached to it. p-Benzoquinone is an oxidizing agent, which has the ability to oxidize other substances by accepting electrons from them. Now, let's answer each part of the exercise: (a)
03

Property of hydroquinone being used

The property of hydroquinone that is being taken advantage of in photographic developers is its ability to act as a "reducing agent". It donates electrons to the silver ions (Ag+) in order to reduce them to free silver (Ag) which then forms the image on the photographic film or paper. (b)
04

Property of p-Benzoquinone being used

The property of p-Benzoquinone that is being taken advantage of when it is used to convert iodide ions (I-) into iodine (I2) is its ability to act as an "oxidizing agent". By accepting electrons from the iodide ions, it oxidizes them into iodine, while the p-Benzoquinone itself gets reduced.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Why is phenol \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) a stronger acid than an alcohol? (b) Would para-acetylphenol be a stronger or weaker acid than phenol? (c) Would 2, 4 -dinitrophenol be a stronger or weaker acid than \(p\) -nitrophenol? Draw contributing structures to support the answers.

How do you account for the fact that, unlike most phenols, \(2,4-\) dinitrophenol and \(2,4,6\) -trinitrophenol are soluble in aqueous sodium bicarbonate?

Give structures of all compounds below: (a) \(p\) -nitrophenol \(+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{A}\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{3} \mathrm{~N}\right)\) \(\mathrm{A}+\mathrm{Sn}+\mathrm{HCL} \rightarrow \mathrm{B}\left(\mathrm{C}_{8} \mathrm{~B}_{11} \mathrm{ON}\right)\) \(\mathrm{B}+\mathrm{NaNO}_{2}+\mathrm{HCL}\), then phenol \(\rightarrow \mathrm{C}\left(\mathrm{C}_{14} \mathrm{H}_{14} \mathrm{O}_{2} \mathrm{~N}_{2}\right)\) \(\mathrm{C}+\) ethyl sulfate \(+\mathrm{NaOH}\) (aq) \(\rightarrow \mathrm{D}\left(\mathrm{C}_{16} \mathrm{H}_{18} \mathrm{O}_{2} \mathrm{~N}_{2}\right)\) \(\mathrm{D}+\mathrm{SnCL}_{2} \rightarrow \mathrm{E}\left(\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{ON}\right)\) \(E+\) acetyl chloride phenacetin \(\left(\mathrm{C}_{10} \mathrm{H}_{13} 2 \mathrm{~N}\right)\), an analgesic ("pain-killer") and antypyretic ("fever- killer") (b) \(\beta-(-0-\) hy droxyphenyl \()\) ethyl alcohol \(+\mathrm{HBr} \rightarrow \mathrm{F}\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{OBr}\right)\) \(\mathrm{F}+\mathrm{KOH} \rightarrow\) coumarane \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}\right)\), insoluble in \(\mathrm{NaOH}\).

4-n-Hexylresorcinol is used in certain antiseptics. Outline its preparation starting with resorcinol and any aliphatic reagents.

When phloroglucinol, \(1,3,5\) -trihydroxybenzene, is dissolved in concentrated \(\mathrm{HClO}_{4}\), its nmr spectrum shows two peaks of equal area at \(86.12\) and 8 4.15. Similar solutions of \(1,3,5-\) trimethoxybenzene and 1,3 , 5-triethoxybenzene- show similar nmr peaks. On dilution, the original compounds are recovered unchanged. Solutions of these compounds in \(\mathrm{D}_{2} \mathrm{SO}_{4}\) also show these peaks, but on standing the peaks gradually disappear. How do you account for these observations? What is formed in the acidic solutions? What would you expect to recover from the solution of \(1,3,5\) -trimethoxybenzene in \(\mathrm{D}_{2} \mathrm{SO}_{4} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free