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# Oxidation and Reduction-Oxidising and Reducing agents - definitions, examples, Practice problems

Our teeth can become stained over time from things like smoking, drinking coffee, red wine, and ageing.

Who wouldn't desire a bright, white smile?

One of the most common dental procedures each year is tooth whitening. There are numerous options available if you want to enhance the beauty of your smile. The terms "bleaching" and "whitening" are frequently used interchangeably.

Bleaching products contain peroxide (hydrogen peroxide or carbamide peroxide), an oxidising agent that eliminates stains from teeth's surface and interior. These products may make teeth appear lighter than they are naturally.

I hear you asking, "Wait a minute, what exactly are oxidising agents? Are there also reducing agents if there are oxidising agents? So what exactly are reducing agents?

Take a chill pill. Because the information on this content page has the answers to these queries.

So let's explore oxidising and reducing agents in further detail.

• Oxidising Agents
• Examples of Oxidising Agents
• Reducing Agents
• Examples of Reducing Agents
• Understanding the Terms Involved
• Practice Problems
• Frequently Asked Questions - FAQ

### Oxidising Agents

• An oxidising agent (also known as an oxidizer or an oxidant) is a chemical species that tend to oxidise other substances, causing them to lose electrons and thus increase their oxidation state.
• Halogens, oxygen, and hydrogen peroxide are common oxidising agents.
• Oxidising agents typically exist in the highest possible oxidation states.
• Ions, atoms and molecules with a high affinity for electrons act as effective oxidising agents.
• The power of the oxidising agent increases with the increase in their electron affinity.
• Oxidising agents are electron acceptors. They have a great tendency to accept electrons and themselves undergo reduction.

### Examples of Oxidising Agents

1. There are numerous oxidising agents that are often utilised in both industrial settings and daily life. Examples include sulphuric acid (H2SO4), potassium nitrate (KNO3), and household bleach (NaClO).

Example: $Zn\left(s\right)+{H}_{2}{SO}_{4}\left(aq\right)\to Zn{SO}_{4}\left(aq\right)+{H}_{2}\left(g\right)$

In this equation, H2SO4 act as an oxidising agent. It oxidises Zn and itself gets reduced.

- - Oxidation

- Reduction

1. Oxygen is a highly reactive non-metal with effective oxidising characteristics that is a member of the periodic table's chalcogen group. Due to the high oxidising properties of oxygen in the air, metals typically react with oxygen to produce metal oxides.

Example: $2Cu\left(s\right)+{O}_{2}\left(g\right)\to 2CuO\left(s\right)$

In this equation, O2 acts as an oxidising agent. It oxidises Cu and itself gets reduced.

- Oxidation

- Reduction

1. Halogens have higher electronegativities than the other elements in their respective period and are hence said to have a greater ability to gain electrons. Halogens like iodine, bromine, chlorine, and fluorine are a few examples of effective oxidising agents. Fluorine is thought to have the highest electronegativity of all the elements, making it the most powerful oxidising agent.

Example:$2NaCl\left(aq\right)+{F}_{2}\left(g\right)\to 2NaF\left(aq\right)+{Cl}_{2}\left(g\right)$

In this equation, F2 act as an oxidising agent. It oxidises NaCl and itself gets reduced.

- Oxidation

- Reduction

1. Hydrogen Peroxide functions as a mild oxidising, cleaning, and bleaching agent.

Example: ${4H}_{2}{O}_{2}\left(aq\right)+PbS\left(s\right)\to {4H}_{2}O\left(aq\right)+PbS{O}_{4}\left(aq\right)$

In this equation, H2O2 act as an oxidising agent. It oxidises PbS and itself gets reduced.

- Oxidation

- Reduction

### Reducing Agents

• A reducing agent (also known as a reducer or a reducant) is a chemical species that tends to reduce other substances, causing them to gain electrons and thus decrease their oxidation state.
• Elements of groups 1 and 2 are common reducing agents.
• Reducing agents typically exist in the lowest possible oxidation states.
• Ions, atoms and molecules with a low affinity for electrons act as effective reducing agents.
• The power of the reducing agent decreases with the increase in their electron affinity.
• Reducing agents are electron donators. They have a great tendency to donate electrons and themselves undergo oxidation.

### Examples of Reducing Agents

1. Metals like Na, Fe, Zn, and Al, as well as non-metals like C, S, and H2 are examples of common reducing agents. Some substances, including hydracids, like HCl, HI, HBr, and H2S, act well as reducing agents.

Example: $2HCl\left(aq\right)+{F}_{2}\left(g\right)\to 2HF\left(aq\right)+{Cl}_{2}\left(g\right)$

In this equation, HCl act as a reducing agent. It reduces F2 and itself gets oxidised

- Oxidation

- Reduction

1. When added to solutions, lithium is believed to be a powerful reducing agent.

Example: $2Li\left(s\right)+{F}_{2}\left(g\right)\to 2LiF\left(s\right)$

In this equation, Li acts as a reducing agent. It reduces F2 and itself gets oxidised.

- Oxidation

- Reduction

1. Iodide salts are thought to be mild reducing agents. Iodine is produced when they interact with oxygen. These also have a variety of antioxidant qualities.

Example: ${Cr}_{2}{O}_{7}^{2-}+6{I}^{-}+{14H}^{+}\to 2{Cr}^{3+}+{3I}_{2}+7{H}_{2}O$

In this equation, ${I}^{-}$ acts as a reducing agent. It reduces ${Cr}_{2}{O}_{7}^{2-}$ and itself gets oxidised.

- Oxidation

- Reduction

1. Due to the presence of a free ketone group or a free aldehyde group, reducing sugars function similarly to reducing agents. The term "reducing sugars" refers to all monosaccharides as well as disaccharides, polysaccharides, and oligosaccharides.

Example:

### Understanding the Terms Involved

Observe the given reaction. When lead (Pb) reacts with copper sulphate solution, lead (Pb) is converted to lead sulphate (PbSO4) and copper sulphate (CuSO4) to copper (Cu). There is an increase in the oxidation state of lead, confirming its oxidation, and a decrease in the oxidation state of copper, confirming its reduction.

In a redox reaction, the oxidising agent undergoes reduction i.e. it gains electrons, and the reducing agent undergoes oxidation i.e. it loses electrons.

In the given reaction, the oxidising agent is copper sulphate (CuSO4) as copper (Cu) goes from +2 to 0 oxidation state i.e it accepts 2 electrons from lead (Pb) and undergoes reduction.

Similarly, the reducing agent in this reaction is lead (Pb) as it goes from 0 to +2 oxidation state i.e it donates 2 electrons to copper (Cu) and undergoes oxidation.

### Practice Problems

Q1. Which will behave as an oxidising agent in the given reaction?

$\mathbit{P}\mathbit{b}{\mathbit{S}\mathbit{O}}_{4}\left(\mathbit{a}\mathbit{q}\right)+\mathbit{Z}\mathbit{n}\left(\mathbit{s}\right)\to \mathbit{Z}\mathbit{n}{\mathbit{S}\mathbit{O}}_{4}\left(\mathbit{a}\mathbit{q}\right)+\mathbit{P}\mathbit{b}\left(\mathbit{s}\right)$

A. PbSO4(aq)
B. Zn(s)
C. Pb(s)
D. Both (A) and (C)

Solution: The oxidation state of Pb in PbSO4(aq) is +2, and the oxidation state of Pb in Pb(s) is 0 i.e. Pb2+ accepts two electrons from Zn and becomes Pb. Therefore Pb2+ (PbSO4) undergoes reduction, or in other words, it acts as an oxidising agent.

(reduced/oxidising agent)

The oxidation state of Zn in Zn(s) is 0 and the oxidation state of Zn in ZnSO4(aq) is +2 i.e. Zn donates two electrons to Pb2+ and becomes Zn2+. Therefore, Zn undergoes oxidation, or in other words, it acts as a reducing agent.

(oxidised/redducing agent)

Hence, PbSO4(aq) is the oxidising agent in the given reaction.

So, option A) is the correct answer.

Q2. Which of the following will act as a reducing agent in the given reaction?

$4\mathbit{H}\mathbit{N}{\mathbit{O}}_{3}\left(\mathbit{a}\mathbit{q}\right)+\mathbit{C}\mathbit{u}\left(\mathbit{s}\right)\to \mathbit{C}\mathbit{u}\left(\mathbit{N}{\mathbit{O}}_{3}{\right)}_{2}\left(\mathbit{a}\mathbit{q}\right)+{2\mathbit{N}\mathbit{O}}_{2}\left(\mathbit{g}\right)+{2\mathbit{H}}_{2}\mathbit{O}\left(\mathbit{a}\mathbit{q}\right)$

A. HNO3(aq)
B. Cu(s)
C. Cu(NO3)2
D. 2H2O(aq)

Solution: The oxidation state of N in HNO3(aq) is +5, and the oxidation state of N in NO2(g) is +4 i.e. N5+ accepts an electron from Cu and becomes N4+. Therefore N5+ (HNO3) undergoes reduction, or in other words, it acts as an oxidising agent.

(reduced/oxidising agent)

The oxidation state of Cu in Cu(s) is 0 and the oxidation state of Cu in Cu(NO3)2(aq) is +2 i.e. Cu donates two electrons and becomes Cu2+. Therefore, Cu undergoes oxidation, or in other words, it acts as a reducing agent.

(oxidised/reducing agent)

Hence, Cu (s) is the reducing agent in the given reaction.

So, option B) is the correct answer.

Q3. For the reaction $\mathbit{C}\mathbit{u}{\mathbit{C}\mathbit{l}}_{2}+{\mathbit{H}}_{2}\to 2\mathbit{H}\mathbit{C}\mathbit{l}+\mathbit{C}\mathbit{u}$, pick out the correct statement.

A. CuCl2 is reduced and H2 is oxidised.
B. CuCl2 is oxidised and H2 is reduced.
C. CuCl2 is both oxidised and reduced.
D. H2 is both oxidised and reduced.

Solution: The oxidation state of Cu in CuCl2 is +2, and the oxidation state of Cu in Cu is 0 i.e. Cu2+ accepts two electrons and becomes Cu. Therefore, Cu2+ (CuCl2) undergoes reduction, or in other words, it acts as an oxidisng agent.

(reduced/oxidising agent)

The oxidation state of H in H2 is 0, and the oxidation state of H in HCl is +1 i.e. H donates an electron and becomes H+. Therefore H (H2) undergoes oxidation, or in other words, it acts as a reducing agent.

(oxidised/reducing agent)

Thus, CuCl2 is reduced and H2 is oxidised in the given reaction.

So, option A) is the correct answer.

Q4. Which of the following will act as both a reducing agent and an oxidising agent in the given reaction?

$4{\mathbit{C}\mathbit{l}}_{2}\left(\mathbit{g}\right)+8{\mathbit{O}\mathbit{H}}^{-}\left(\mathbit{a}\mathbit{q}\right)\to {7\mathbit{C}\mathbit{l}}^{-}\left(\mathbit{a}\mathbit{q}\right)+\mathbit{C}\mathbit{l}{\mathbit{O}}_{4}^{-}\left(\mathbit{a}\mathbit{q}\right)+4{\mathbit{H}}_{2}\mathbit{O}\left(\mathbit{a}\mathbit{q}\right)$

A. Cl2(g)
B. Cl-(aq)
C. $Cl{O}_{4}^{-}\left(aq\right)$
D. All of these

Solution: The oxidation state of Cl in Cl2(g) is 0, and the oxidation state of Cl in Cl-(aq) is -1 i.e Cl accepts an electron and becomes Cl-(aq). Therefore, Cl undergoes reduction, or in other words, it acts as an oxidising agent.

(reduced/oxidising agent)

The oxidation state of Cl in Cl2(g) is 0, and the oxidation state of Cl in $\mathbit{C}\mathbit{l}{{\mathbit{O}}_{4}}^{-}\left(\mathbit{a}\mathbit{q}\right)$ is +7 i.e. Cl donates 7 electrons and becomes Cl+7(aq). Therefore, Cl undergoes oxidation, or in other words, it acts as a reducing agent.

(oxidised/reducing agent)

Hence, Cl2(g) gets reduced and oxidised, and can therefore act as both a reducing and an oxidising agent.

So, option A) is the correct answer.

## Frequently Asked Questions - FAQ

Q1. Is there a difference between a reducing agent and an anti-oxidant?
Antioxidant agents can stop a chain reaction by getting rid of the free radical intermediates. Antioxidants can be thought of as reducing agents since they are oxidised while acting as antioxidants. Examples include polyphenols, thiols and ascorbic acid.

Q2. What biological effects do oxidising agents have?
The harmful effects of oxidants include damage to lipids, proteins and DNA, as well as the acceleration of atherosclerosis, rheumatoid arthritis, and cataract development. Additionally, oxidants are to blame for the ischemia-reperfusion syndrome.

Q3. Why are metals referred to as reducing agents?
Because all metals are relatively electropositive and have low ionisation energies, they lose electrons relatively quickly. As a result of their tendency to donate electrons and become oxidised, metals act as reducing agents.

Q4.  What effect do oxidants have on respiration?
High doses of oxidants like ozone and nitrogen dioxide result in death from lung oedema. Worsening of asthma (acute effects) and effects on lung growth are well-established effects on the respiratory system (chronic effects). Long-term exposure to oxidising agents such as ozone has recently been linked to negative cardiovascular outcomes, including increased mortality.


text:
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Iwasawa theory, projective modules, and modular representations / [electronic resource] Ralph Greenberg.

Material type: TextPublisher: Providence, R.I. : American Mathematical Society, c2010Description: 1 online resource (v, 185 p. : ill.)ISBN: 9781470406097 (online)Subject(s): Iwasawa theory | Curves, EllipticAdditional physical formats: Iwasawa theory, projective modules, and modular representations /DDC classification: 512.7/4 LOC classification: QA247 | .G74 2010Online resources: Contents
Contents:
Chapter 1. Introduction. Chapter 2. Projective and quasi-projective modules. Chapter 3. Projectivity or quasi-projectivity of $X^{\Sigma _0}_E(K_\infty )$. Chapter 4. Selmer atoms. Chapter 5. The structure of $\mathcal {H}_v(K_\infty , E)$. Chapter 6. The case where $\Delta$ is a $p$-group. Chapter 7. Other specific groups. Chapter 8. Some arithmetic illustrations. Chapter 9. Self-dual representations. Chapter 10. A duality theorem. Chapter 11. $p$-modular functions. Chapter 12. Parity. Chapter 13. More arithmetic illustrations.
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"May 2011, volume 211, number 992 (second of 5 numbers )." -- T.p.

Includes bibliographical references.

Chapter 1. Introduction. Chapter 2. Projective and quasi-projective modules. Chapter 3. Projectivity or quasi-projectivity of $X^{\Sigma _0}_E(K_\infty )$. Chapter 4. Selmer atoms. Chapter 5. The structure of $\mathcal {H}_v(K_\infty , E)$. Chapter 6. The case where $\Delta$ is a $p$-group. Chapter 7. Other specific groups. Chapter 8. Some arithmetic illustrations. Chapter 9. Self-dual representations. Chapter 10. A duality theorem. Chapter 11. $p$-modular functions. Chapter 12. Parity. Chapter 13. More arithmetic illustrations.

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# Practice Test Quesitons

by AznBoi
Tags: practice, quesitons, test
 P: 471 My work, my answers are wrong obviously. 18. $$U_{E}=\frac{kq1q2}{d}$$ -- I substituted the electric force with the work equation to get this. 25. Q=mL, Q=mcT . I combined these to get mL=mcT, stuck from there. 47. F=q(v x B), v=F/qB, F=mv^2/r I combined these equations to get: mv/r=qB and solved for v and I then I just guessed. 48. Why is the answer D?? I know that it has something to do with lenz's law but I only know how to apply that law for magnets moving with either North or South pole. (If a north pole of a magnet moves towards the wire the induced current is counter clockwise right? If the south pole moves into the wire it is clock wise?)-- Is this right? Thanks for your help guys! =]
 HW Helper P: 508 Sorry we are not giving answers as such. You may get hints for your doubts and confusions only. Show your approach in case not getting the correct answers.
 P: 471 Oh, I just posted the picture first to see if it worked lol. Anyways, my attempts are on the board! Do you want me to post the answers too? I have the answer key btw, just not the solutions.
HW Helper
P: 508

## Practice Test Quesitons

Is the diagram is given? Should be, otherwise what is Q? what about the position of the point P?
P: 471
 Quote by mukundpa Is the diagram is given? Should be, otherwise what is Q? what about the position of the point P?
Yeah, I realized that lol. I forgot to include the picture because I got 17 right. So for 17, all I did was add the two electric field vectors (one to the left and one down) to get the resultant vector which is (c) right? Thanks.
 HW Helper P: 508 Now it is clear. There are two charges Q both at a distance d from the third corner of the square. Find potential V at that point due to both Q charges (Potential is a scalar quantity) then multiply q and V.
 HW Helper P: 508 For the next heat is required to first increase the temperature of ice up to the melting point then latent heat to melt and then to increase temperature of water.
 HW Helper P: 508 For the next you got the relation for v. Put the value of e/m for proton and probable values of B and r and will get the order of the velocity.
P: 471
 Quote by mukundpa Now it is clear. There are two charges Q both at a distance d from the third corner of the square. Find potential V at that point due to both Q charges (Potential is a scalar quantity) then multiply q and V.
So how do you find the potential v at the point? I know that $$V=\frac{U_{e}}{q} =Ed$$
I don't get what you mean by then multiply q and v.
 HW Helper P: 508 The potential at that point will be KQ/d due to each charge Q so what is the total potential? and what is the potential energy?
P: 471
 Quote by mukundpa The potential at that point will be KQ/d due to each charge Q so what is the total potential? and what is the potential energy?
OKay, so $$\frac{kQ}{d}+\frac{kQ}{d}=2\frac{kQ}{d} *q= 2k\frac{qQ}{d}$$? So how do you know that you need to add the potential differences rather than just the potential energies? You can do it either way right? So do you just add them because they are a scalar? Is there any time where you need to subtract them? (mabye when the charges are opposite?) So if the quantities are scalar you just add them and if they are vectors you need to use pythagorean or trig right?
 HW Helper P: 508 Going in the right direction.

 Related Discussions Introductory Physics Homework 5 Introductory Physics Homework 4 Linear & Abstract Algebra 3 Introductory Physics Homework 2


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# help!!!- optimization problem

• December 13th 2008, 05:21 PM
52090
help!!!- optimization problem
I dont know how to start on this probem. It would nice if someone helps me.

A water line runs east-west. A town wants to connect two new housing developments to the line by running lines from a single point on the existing line to the developments. One is 3 miles south of the existing line; the other is 4 miles south of the existing line and 5 miles east of the first development. Find the place on the existing line, relative to the 2 developments, to make the connection that minimizes the total length of the new line.

• December 13th 2008, 05:26 PM
Jhevon
Quote:

Originally Posted by 52090
I dont know how to start on this probem. It would nice if someone helps me.

A water line runs east-west. A town wants to connect two new housing developments to the line by running lines from a single point on the existing line to the developments. One is 3 miles south of the existing line; the other is 4 miles south of the existing line and 5 miles east of the first development. Find the place on the existing line, relative to the 2 developments, to make the connection that minimizes the total length of the new line.

By Pythagoras' theorem, the length of the line connecting the town 3 miles south of the existing line is $\sqrt{x^2 + 3^2}$ and the length of the line connecting the town 4 miles south of the existing line is $\sqrt{4^2 + (5 - x)^2}$. you want to minimize the sum of those

hint: the square roots aren't needed, hope you know why
• December 13th 2008, 05:59 PM
Soroban
Hello, 52090!

If we must use Calculus, the approach and set-up is quite ugly.

Quote:

A water line runs east-west. A town wants to connect two new housing developments
to the line by running lines from a single point on the existing line to the developments.
One is 3 miles north of the existing line; the other is 4 miles north of the existing line
and 5 miles east of the first development.

Find the place on the existing line, relative to the two developments,
to make the connection that minimizes the total length of the new line.

Code:

                          * B                         * |     A *              *  |       | *          *    | 4     3 |  *      *      |       |    *  *        |       * - - - * - - - - - *       C  x  P    5-x    D

Development $A$ is 3 miles from the line: . $AC = 3$
Development $B$ is 4 miles from the line: . $BD = 4$
$CD = 5$

Let $P$ be a point on $CD$.
Let $x \,=\, CP$, then $5-x \,=\, PD$

We want to minimize the distance: $AP + PB$

In right triangle $ACP\!:\;AP \:=\:\sqrt{x^2+3^2}$
In right triangle $BDP\!:\;PB \:=\:\sqrt{(5-x)^2 + 4^2}$

The total distance is: . $D(x) \;=\;\left(x^2+9\right)^{\frac{1}{2}} + \left(x^2-10x+41\right)^{\frac{1}{2}}$

. . and that is the function we must minimize.

I'll wait in the car . . .
.


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# Can anyone explain how to store a string as a variable?

breebreebran
I don't have any big programs. I just write programs to get me through my math classes.
But I want to have the output look pretty.
Like one program I have is for my finance class.

Code:
prompt p
if p = 1
then
Prompt b,r,n,t
r/100→x
b/(1+x/n)^(n*t)→l
disp l
else
Prompt b,r,n,t
r/100 → x
b*(x/n)/((1+x/n)^(n*t)-1) → m
disp m

But what I want to do is instead of prompting p, I want to prompt "present value?"
Then If present value = 1 then etc.
It just looks neater if it prompts an entire string instead of one letter.
So instead of prompting b,r,n,t it'll prompt
"Ending Balance?"
"Rate?"
"Number of compounds?"
"Time?"
And then I can save those strings as variables.
So ending balance → b
Rate → r
Number of compounds → n
And then just have the rest of the program work with those letters instead of having to type out (rate/numberofcompounds)
Every time I try to use strings instead of letters it always gives me a syntax error.
Any help?

## Answers and Replies

Mentor
Well you put a print statement in front of each prompt.

Mentor
breebreebran, you didn't say what kind of calculator you have, which would be useful information. It's possible that the prompt command has an optional first parameter of a string, like this:
prompt "Present value? ", p

If not, do as jedishrfu suggests, and print the string before executing the prompt.


text:
Sciencemadness Discussion Board » Fundamentals » Reagents and Apparatus Acquisition » tert-Butyl Alcohol Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues

Pages:  1
Author: Subject: tert-Butyl Alcohol
497
International Hazard

Posts: 778
Registered: 6-10-2007
Member Is Offline

Mood: HSbF6

tert-Butyl Alcohol

I've looked around and have been unable to find any sort of OTC source or synthesis of tert-butyl alcohol.

I would imagine it could be had by cleavage of tert-butyl methyl ether, but I haven't seen a source of that either. Are there any other easily accessible compounds that have a tert-butyl that could be cleaved? I can't think of anything off the top of my head.

I could probably buy off the internet for $30-40/ 500ml but really would rather not. panziandi National Hazard Posts: 490 Registered: 3-10-2006 Location: UK Member Is Offline Mood: Bored t-Butyl Alcohol should be easily obtained from a chemical supply house. Make some calls and ask if they are able to supply individuals. Some will and t-Butyl alcohol is not watched in any sense of the word and is not that dangerous either. If you want to prepare it however it would be possible... ...Grignard! Methyl Magnesium Halide + Acetone => t-Butyl alcohol That is assuming you can find a source of acetone, methanol + some halogenating agent, magnesium, some form of drying agent etc. Easily doable I'm sure. stoichiometric_steve International Hazard Posts: 818 Registered: 14-12-2005 Member Is Offline Mood: satyric  Quote: Originally posted by panziandi Easily doable I'm sure. Not sure if someone unable to acquire t-Butanol should attempt Grignards. panziandi National Hazard Posts: 490 Registered: 3-10-2006 Location: UK Member Is Offline Mood: Bored Possibly true! redox National Hazard Posts: 268 Registered: 22-2-2011 Location: The Land of Milk and Honey Member Is Offline Mood: Chalcogenetic Methyl t-butyl ether (MTBE) can be found in an auto store as an octane booster. I was in the store today and bought four cans (on sale for 99 cents each)! I got 1.5 liters of pure MTBE for 4 dollars! My quite small but growing Youtube Channel: http://www.youtube.com/user/RealChemLabs Newest video: Synthesis of Chloroform The difference between chemists and chemical engineers: Chemists use test tubes, chemical engineers use buckets. BromicAcid International Hazard Posts: 3155 Registered: 13-7-2003 Location: Wisconsin Member Is Offline Mood: Rock n' Roll Meanwhile I recently checked several auto stores and found none. The capriciousness of chemical availability. Shamelessly plugging my attempts at writing fiction: http://www.robvincent.org hkparker International Hazard Posts: 601 Registered: 15-10-2010 Location: California, United States Member Is Offline Mood: No Mood I too have looked around for quite a while for t-BuOH with no luck and considered opening a thread on it, so I'm glad to see one open. When I looked into it a grignard reaction seemed like the best way to make it at home (though grignards are very difficult to do). It was really expensive and hard to find online. My YouTube Channel "Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday mr.crow International Hazard Posts: 884 Registered: 9-9-2009 Location: Canada Member Is Offline Mood: 0xFF Wow MTBE! Extraction solvent. If you really wanted t-butanol you could reflux this with HBr to remove the methyl ether. Then you get t-butyl bromide, reflux with NaOH or something. We are talking about tert-butanol and 2-methyl-2-butanol in the Make Potassium thread as well Double, double toil and trouble; Fire burn, and caldron bubble Nicodem Super Moderator Posts: 4230 Registered: 28-12-2004 Member Is Offline Mood: No Mood  Quote: Originally posted by mr.crow Wow MTBE! Extraction solvent. If you really wanted t-butanol you could reflux this with HBr to remove the methyl ether. Then you get t-butyl bromide, reflux with NaOH or something. Reaction of NaOH with t-butyl bromide (or chloride) gives isobutene, a useful compound, but a gas at room temperature. t-Butyl halides form mainly t-butanol upon solvolysis in water with acetone as cosolvent (a typical SN1 reaction thought to students and an lab exercise in kinetics). Not much of a prospect for preparative use, but might be an interesting experience in separation techniques. Otherwise, MTBE can be cleaved directly to a mixture of methanol, t-BuOH and isobutene by dilute aq. non-nucleophilic acid catalysis (e.g., H2SO4, NaHSO4, etc.). The scientific literature is full of studies in this regard, but the preparative methods are mostly described in the patent literature. MTBE reacts already with HCl, no need to use HBr. Essentially, MTBE can be used instead of t-BuOH in most SN1 reactions. gsd International Hazard Posts: 847 Registered: 18-8-2005 Member Is Offline Mood: No Mood  Quote: Originally posted by Nicodem MTBE reacts already with HCl, no need to use HBr. Essentially, MTBE can be used instead of t-BuOH in most SN1 reactions. @ Nicodem (CH3)3.C.O.CH3 + 2HCl ------> (CH3)3CCl + CH3Cl + H2O Is this what you are implying? If yes then can it be done with aqueous HCl - say 30 % ? Can you please site a preparative reference? gsd mr.crow International Hazard Posts: 884 Registered: 9-9-2009 Location: Canada Member Is Offline Mood: 0xFF Quote: Originally posted by Nicodem  Quote: Originally posted by mr.crow Wow MTBE! Extraction solvent. If you really wanted t-butanol you could reflux this with HBr to remove the methyl ether. Then you get t-butyl bromide, reflux with NaOH or something. Reaction of NaOH with t-butyl bromide (or chloride) gives isobutene, a useful compound, but a gas at room temperature. t-Butyl halides form mainly t-butanol upon solvolysis in water with acetone as cosolvent (a typical SN1 reaction thought to students and an lab exercise in kinetics). Not much of a prospect for preparative use, but might be an interesting experience in separation techniques. Otherwise, MTBE can be cleaved directly to a mixture of methanol, t-BuOH and isobutene by dilute aq. non-nucleophilic acid catalysis (e.g., H2SO4, NaHSO4, etc.). The scientific literature is full of studies in this regard, but the preparative methods are mostly described in the patent literature. MTBE reacts already with HCl, no need to use HBr. Essentially, MTBE can be used instead of t-BuOH in most SN1 reactions. Thanks Nicodem, didn't know that about tert alcohols and ethers Double, double toil and trouble; Fire burn, and caldron bubble MagicJigPipe International Hazard Posts: 1554 Registered: 19-9-2007 Location: USA Member Is Offline Mood: Suspicious Contrary to what some have said it is extremely difficult, for some strange reason, to find t-butanol. Methyl t-butyl ether (MTBE) is, however, very easy to find so that is always an option. If anyone needs a source for ACS Reagent MTBE online I know of at least one: about 500 mL for$20.

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Yesterday, in passing, I saw a hand sanitizer that contained T-butanol. a msds google search lists several that do.

I don't imagine it would be easy to separate from the other alcohols it's commonly mixed with...

2-Butoxyethanol Can be obtained as whiteboard cleaner, if that helps.
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I fail to see how 2-butoxyethanol has anything to do with t-butanol. Thanks for the info (it is quite common in a variety of cleaning solvents) but I just don't understand.

I think the main reason people seek t-butanol is, not because it is a 4-carbon, saturated alcohol but, because it is a tertiary alcohol. That is why 2-butoxyethanol is useless to people seeking it for that purpose.

But I see no reason why it shouldn't be mentioned--it would be bad to post that info in a new thread I guess.

"There must be no barriers to freedom of inquiry ... There is no place for dogma in science. The scientist is free, and must be free to ask any question, to doubt any assertion, to seek for any evidence, to correct any errors. ... We know that the only way to avoid error is to detect it and that the only way to detect it is to be free to inquire. And we know that as long as men are free to ask what they must, free to say what they think, free to think what they will, freedom can never be lost, and science can never regress." -J. Robert Oppenheimer
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I thought I saw it in the thread, I must have had a brain fart.
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 Quote: Originally posted by gsd @ Nicodem (CH3)3.C.O.CH3 + 2HCl ------> (CH3)3CCl + CH3Cl + H2O Is this what you are implying? If yes then can it be done with aqueous HCl - say 30 % ? Can you please site a preparative reference? gsd

The general SN1 substitutions of MTBE go like this (where HNu is the nucleophile):

tBu-O-Me + HNu ---(acid catalysis)---> tBuNu + MeOH

In the case of HNu = HCl:

tBu-O-Me + HCl -----> tBuCl + MeOH

Since methanol is a primary alcohol, it can not undergo SN1 reactions, though it can react under the same conditions in other ways, depending on the conditions and nature of the nucleophile and catalyst (e.g., if HNu is CH3COOH, then it can obviously get esterified, while if HNu is HI it can undergo SN2 substitution to yield methyl iodide, etc.).

A very good article describing various SN1 reactions of tBuOH, its ethers and acid catalysed reactions of other aliphatic ethers is:
J. Am. Chem. Soc., 54 (1932) 2088–2100 (DOI: 10.1021/ja01344a062). A few excerpts:
 Quote: When tert.-butyl ethyl ether and concentrated hydrochloric acid in the molecular ratio of one of the former to five of the latter are mixed the ether dissolves, but is rapidly converted into tert.-butyl chloride, which separates. When such a mixture was allowed to stand overnight 99% of the theoretical quantity of the chloride was obtained. ... There was an induction time before the chlorides separated. In order to have the results comparable zero time was taken when a measurable amount of chloride had separated-1 to 3% of the theoretical amount. ... The tertiary butyl ethers react so rapidly with concentrated hydrochloric acid at 25 °C that it was necessary to study them under different conditions.

When applying this reaction on MTBE, using a mol ratio 10 : 1 of 27% HCl(aq) vs. MTBE, the reaction reached 25% conversion in 3.5 hours at 0 °C (while ethyl t-butyl ether required only 0.83 hours and t-BuOH 1.6 hours). So the reactivity MTBE in this chlorination is twice less than that of t-BuOH, but still relatively fast and obviously very fast at room temperature.

PS: In context of another thread an advice for those that have no distillation apparatus: N-(t-butyl)acetamide is a solid at room temperature (mp 96-98°C). So, if you wander what else you can do with MTBE, do this using the Ritter reaction on acetonitrile. If you have no acetonitrile, you can use urea as the nucleophile (the patent number for Ritter reactions using urea is already posted in some other thread).

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@Nicodem

Thanks for the very detailed reply. The JACS paper is very interesting indeed.

The preparation of tBuCl from tBuOH and aqHCl is a very basic reaction one encounters in Organic Syntheses. However as anyone who has actually performed this procedure can tell, any significant yields are realized only when a) A very large excess of HCl is used or b) when the product is "salted out" of the reaction mixture by inorganic salts like CaCl2 or ZnCl2 etc.

In case of tBuCl preparation using MTBE and aqHCl will the same principle hold good? Here the matter is further complicated by presence of Methanol in aqueous phase.

gsd
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I wouldn't know, as I never performed the reaction myself. t-Butyl chloride should not have much of a solubility in aq. HCl and the fact that in the article they obtain a 99% yield with ethyl t-butyl ether which cleaves to ethanol (a better cosolvent than methanol), can only mean the solubility can not be much of an issue (or perhaps they just got an impure product, or maybe it was due to the 5-fold excess of HCl and overnight reaction time?).

The Org. Synth. note about the addition of CaCl2 giving a better yield due to salting out the product might also be an misinterpretation of the practical results. It could be that with a just 3-fold excess of HCl and 15-20 min is not enough to achieve full conversion. In such case the addition of CaCl2 would indeed drive the reaction to completion by increasing the chloride concentration and reducing the free [H2O] content. Since the additionally formed product phases out, it might wrongly appear as salting out the dissolved tBuCl, while in reality it is just converting the dissolved tBuOH to tBuCl.

Chances are the Org. Synth. procedure wasn't really optimized. Or it could be optimized only on the economical base, as it gives a cheap reagent (tBuCl) from another cheap reagent (tBuOH), but uses a severalfold excess of another cheap reagent (conc. HCl). So the economic optimal might not be using even more HCl, as it would be when starting from an expensive alcohol to give an expensive chloride.

…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)

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 Quote: Originally posted by 497 I've looked around and have been unable to find any sort of OTC source or synthesis of tert-butyl alcohol. I would imagine it could be had by cleavage of tert-butyl methyl ether, but I haven't seen a source of that either. Are there any other easily accessible compounds that have a tert-butyl that could be cleaved? I can't think of anything off the top of my head. I could probably buy off the internet for $30-40/ 500ml but really would rather not. Ullmann's 5th sez — 2-Methy-2-propanol ... In the literature a number of other methods for the preparation of butanol are cited which are generally of no significance other than for laboratory preparation. A comprehensive review is to be found in .... B Cornils & E Zilly Methodicum chimicum: a critical survey of proven methods and their application in chemistry, natural science, and medicine. Vol. 5 (1975) p. 84 Byda Ullmann also notes that — prod. of propylene oxide yields one ton per 1.2 t of propylene oxide. Must be a SL of it out there. Neil International Hazard Posts: 556 Registered: 19-3-2008 Member Is Offline Mood: No Mood This page http://www.fuel-testers.com/review_gas_treatment_products.ht... Lists "Turbo Octane Boost 180" which claims T-Butanol and methanol as the ingredients While http://www.chemcas.com/msds112/cas/1666/124-38-9_75-65-0_106... Lists a brake cleaner composed of "carbon dioxide tert-butyl alcohol, tert-butanol 1,2-butylene oxide methylal (dimethoxymethane) perchloroethylene (tetrachloroethylene) methyl chloroform" Squall181 Harmless Posts: 46 Registered: 21-2-2011 Member Is Offline Mood: No Mood So I found some octane booster(Spectrum Plus @ K-mart) that is made with methyl-t-butyl ether. So i decided to buy and try to distill the ether off. I set up for basic distillation and turned on the heat. To my surprise the liquid is refusing to boil. So after a while I decided to take the temp of the liquid, while the still head was reading 30C the liquid in the flask was at about 120C. Unfortunately I was unable to locate an MSDS for this product. Does anyone have any ideas on what's happening here, what other ingredient could be increasing the boiling point of the liquid. I'm about ready to give up on trying to make it boil. Takron Harmless Posts: 47 Registered: 19-5-2010 Member Is Offline Mood: No Mood  Quote: Originally posted by Nicodem Otherwise, MTBE can be cleaved directly to a mixture of methanol, t-BuOH and isobutene by dilute aq. non-nucleophilic acid catalysis (e.g., H2SO4, NaHSO4, etc.) If this cleavage was done, would one separate the t-BuOH from the rest by cooling it? The t-BuOH would crystalize and the rest cold be poured off right? Dr.Bob International Hazard Posts: 2506 Registered: 26-1-2011 Location: USA - NC Member Is Offline Mood: No Mood  Quote: Originally posted by Takron If this cleavage was done, would one separate the t-BuOH from the rest by cooling it? The t-BuOH would crystalize and the rest cold be poured off right? The t-butanol may not freeze into pure crystals from a mixture of solvents. While it readily freezes when pure, once you mix several liquids together, it is much harder to remove one by freezing it. Distilling them would likely work much better, then you could certainly freeze the crude t-BuOH distillate to try to repurify it by crystallization, and then pour off the non-solid material once ~90% has frozen. shadow Hazard to Self Posts: 52 Registered: 17-10-2007 Member Is Offline Mood: No Mood I don't get it. Carolina Biological------ 500ml -----$12.95

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 Quote: Originally posted by shadow I don't get it. Carolina Biological------ 500ml -----\$12.95 http://www.carolina.com/p2p/endecaSearch.do?pageName=Home+Page&keyword=tert+butyl+alcohol&Search.x=0&Search.y=0[/url]

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text:
# Abstract:

Measures of variability of the least-squares estimator $\hat{\theta}$ are essential to assess the quality of the estimation. In nonlinear regression, an accurate approximation of the covariance matrix of $\hat{\theta}$ is difficult to obtain (Clarke 80). In this paper, a second-order approximation of the entropy of the distribution of $\hat{\theta}$ is proposed, which is only slightly more complicated than the widely used bias approximation of Box (Box 71). It is based on the "flat" or "saddle-point approximation" of the density of $\hat{\theta}$. The neglected terms are of order ${\cal O}(\sigma^4)$, while the classical first order approximation neglects terms of order ${\cal O}(\sigma^2)$. Various illustrative examples are presented, including the use of the approximate entropy as a criterion for experimental design.

# Classification:

62K05, 62F12, 62E17, 62J02, 62E20


text:
# Axis and angle of rotation from 3x3 rotation matrix and properties of rotation matrix

A 3x3 rotation matrix is given. We know the properties of rotation matrix:

• The columns of rotation matrix are unit vectors perpendicular to each other.
• The column vectors indicate where the unit vectors along the principal axes are transformed.
• The rows of rotation matrix are unit vectors perpendicular to each other.
• The row vectors indicate the vectors that are transformed into the unit vectors along the principal axes.
• The inverse of rotation matrix is its transpose.

Suppose, we don't know anything about the Rodrigues formula, so we can't use Trace(R) = 1+2cos$\theta$. Is it possible to find the axis and angle of rotation from the given rotation matrix and the properties of rotation matrix? I need an answer with proper explanation.

If $A$ is your rotation matrix, then the axis can be found by solving the vector equation $Ax=x$. Take a unit vector $u$ along the axis, another unit vector $v$ perpendicular to $u$ (say by Gram-Schmidt) and $w=u\times v$ (vector product). Then $Av=(\cos\theta )v+(\sin\theta)w$ for some $\theta$ (that's the angle of rotation).
The right way to find $\theta$ is of course to use $\text{trace}(A)=1+2\cos\theta$.