# Tag Info

## Hot answers tagged alcohols

27

The problem arises from the metabolized products of methanol. Methanol oxidizes in the liver by an enzyme called alcohol dehydrogenase to formaldehyde which is further metabolized to formic acid by another enzyme called aldehyde dehydrogenase. This formic acid is the source for acute toxicity associated with methanol poisoning. Accumulation of this chemical ...

21

IPA has a different carbon:hydrogen ratio than ethanol. There is more incomplete combustion occurring with IPA, hence the smoky orange flame and smell of soot. Ethanol combusts more completely, leading to a blue (soot-free) flame and no smell. In response to your second question, ethanol likely has a lower latent heat of vaporisation than IPA, resulting in ...

17

Interesting observation. The blue flame color of all hydrocarbon fuels is due to the emission small diatomic carbon species such $C_2$ or CH. There is nothing magical about IPA having a yellow flame. The yellow flame originates from incomplete combustion. There is more carbon per mole of IPA as compared to ethanol. Yellow flames are called reducing flames ...

15

The term "Essigäther" is actually the German name for ethyl acetate i.e Essig = "vinegar" + Äther = "ether". "vinegar" becomes acetic, hence ethyl acetate becomes "acetic ether". (etymoline.com) But, ethyl acetate is an ester. So, why is it named "acetic ether"? According to Leopold Gmelin, ester ...

12

Seeing the reaction given here, I assume that is an on-paper reaction that takes place and that all three alcoholic groups are dehydrated. The mechanism (on paper) seems to be as follows Dehydration $1$ For the first dehydration, the carbocation formed is stabilised via ring expansion and then forms a double bond as follows: Dehydration $2$ Similar to the ...

10

The boiling point of non-ionic compounds are highly depend on their H-bonding abilities. For example, boiling point of water (molar mass: $\pu{18.02 g/mol}$) is $\pu{100 ^\circ C}$ at $\pu{1 atm}$ while that of ethanol (molar mass: $\pu{46.07 g/mol}$) is $\pu{78.4 ^\circ C}$ at $\pu{1 atm}$, even though ethanol is heavier and have more other intermolecular ...

10

Nilay's answer, as well as the comments, do a great job of explaining why methanol is toxic (and ethanol is comparatively less toxic). I'll take a different approach here: I'll explain why ethanol is the major component of alcoholic drinks. To start off, ethanol is not the only alcohol in alcoholic drinks1. Other alcohols such as Glycerol, Tryptophol, Tert-...

9

Mathew Mahindaratne has provided analysis based on experimental values of the boiling points of the two compounds. I would like to offer a different view using bonding analysis. Before I begin tackling the question, we shall first clarify the concept of the hydrogen bond. While the popular view of the hydrogen bond is as a particularly strong type of "...

9

The given answer is wrong in many ways: 4 did not satisfy both criteria C and E, because it is not a chiral compound to begin with. 10 did not satisfy the criteria D, because it does give a precipitate with iodine in presence of $\ce{NaOH}$. That gives only 5 in given answer to satisfy all criteria. Let's eliminate given compounds in systematic order: ...

9

tl;dr A slightly anthropomorphic approach with some basis in general organic chemistry: Step 1: The protonation of the carbonyl (An acid-base reaction) Here, we have three options – we can protonate the nitrogen (deactivated by the resonance with oxygen), the alcohol group ($\mathrm{sp^3}$) or the carbonyl group ($\mathrm{sp^2}$). As you can see this is ...

7

Expanding on @PeterMortensen's comments (1, 2) here is some other discussion of how a small contamination of sodium can lead to orange flames: From Why does the humidifier make a stove's flame orange?: From this answer to it: OK, I've managed to measure some spectra using my Amadeus spectrometer with custom driver. I used 15 s integration time with the ...

7

I think that correct answer should be 5 and 6 only, because: 12 will produce terephthalic acid, instead of benzoic acid as it possess two benzylic-H (also given by 7). 4 and 10 doesn't satisfy statement C and D, as you've mentioned.

7

A carbonyl group normally destabilizes carbocations but stabilizes carbanions. In this particular case, the rate of dehydration isn't guarded by carbocation stability, but by CH- acidity. In acidic conditions, OH group in alcohols is relatively easily protonated. Then, if there is a relatively acidic hydrogen in $\beta$-position it immediately dissociates, ...

7

The major product is the Zaitsev one i.e 1-methylcyclohexene. The OH needs to be protonated first in order for it to leave and moreover it is a very fast leaving group (also we have the case that $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ are very weak bases and considering that only elimination reactions occur tells us that it must follow an E1 elimination). When ...

6

Ethanol does the same. It is the consequence of deviation from the Raoult law. The other consequences are existence of azeotropes ( affects distillation ) and mixing volume deviations. If you mix 1 L of ethanol and 1 L of water, it warms up. When it cools down back, its total volume is not 2 L, but about 1.9 L. It is closely related to total energy of ...

6

It's better to check the resonance hybrid whenever we have a confusion, so, in this case resonance takes place between lone pair of oxygen atom and the double bond and the electrons shift from oxygen towards the alkene. This creates a partial positive character on the oxygen atom and partial negative character on the alkene carbon which is farthest from the ...

6

I have commented previously that the success of this reaction depends on the protonation of the free amino group of answer (a). In the absence of acid, heating amino ester (a) would give the starting material, the hydroxyamide because esters are more labile than amides. The reason this reaction can occur is that the amino group is protonated in acid which ...

6

Just to add another resource besides the excellent find by Nilay, in Organic Chemistry, The Name Game, it mentions ester from German Essigäther (acetic ether) an early name for ethyl acetate The unabridged Oxford English Dictionary (by subscription only) also mentions that Gmelin called these compounds as napthas but later he changed it to ester. H. Watts ...

5

If you really want to figure what the “purity” Or rather the volume of isopropyl to water by weight aka $70\%$, $91\%$. And you don’t have a gas chromatograph, this is about the closest your going to get to an accurate Volumetric representation. This is a fractional distillation column same basic principle as distilling drinking alcohol but much different ...

5

You need to understand something in organic chemistry. Not every question is going to be part of some named reaction that you just have to apply. For example in this question, the basic concepts of organic chemistry is used nothing more. First step, you look at the molecule and think, hmmm what can $\ce{H+}$ do? There are two possible things: The nitrogen ...

5

As I mentioned in my Comment the only serious issue was the formation of an allylic anion in acid medium. You seem to know how to form enols from ketones and how to form ketones from enols. I will ignore those steps in the diagram. Protonation of the carbonyl in 1 can cause ring fragmentation to form transient enol 2 that rapidly tautomerizes to diketone 3. ...

4

Your question: Given a sample of an alcoholic beverage (be it beer, wine, etc..), what are possible methods to find out the ABV (alcohol by volume) of it with moderate accuracy? There are a few method used in winery industry as Karl mentioned a comment in elsewhere. I assume you want to find out better methods to find the alcohol content: The best ...

4

In order to understand this, you need to have this basic knowledge of the kinetics of multistep reactions In chemical kinetics, there are two ways to deal with multiple-step mechanisms Rate determining step method- Here one specific step is the slowest. So we consider all steps after this step to be equally fast. The rate-determining step is the slowest ...

4

For this question, given the lack of initial information, I'm making the following assumptions: Butane-2,3-diol dehydrates to give butan-2-one 2-butanol dehydrates to give but-2-ene Butane-2,3-diol dehydration is nicely described in this paper from Emerson1: Aqueous 2,3-butanediol was shown to react in a pseudo-first-order reaction in the presence of ...

4

The mechanism should provide for the successive processes of carbocation formation, rearrangement and proton loss. Observations(Points to consider): The tendency to attain stablity by aromaticity. The high ring strain in the cyclobutane ring. The first step step should involve protonation of each hydroxy group attached. Protonation produces hydronium ion ...

3

Although this was an MD study (and might be said to be somewhat dated), quoting from Matsumoto and Gubbins (Ref. 1): Most of the molecules are in the two-bonding state, but a small fraction of molecules are in the one- or three-bonding state, as already pointed out in previous studies. 7 ,8 According to this study most of the methanol molecules ...

3

Both the reaction mechanisms are equally correct. Reaction takes place at numerous sites in most of the organic reactions, many kinds of products are formed, but the thing which actually give the major product is one of the intermediate, which is relatively stable than its peers. We know, that 3° carbocation is a more stable intermediate than an allyl ...

3

Unfortunately I did not find a high resolution spectrum of this compound, which would have quickly answered your question. Nevertheless, ionized alcohols and even more primary alcohols have a main fragmentation pathway which is the loss of water to yield an ionized alkene. In this case, the corresponding alkene would be 4-methyl-1-pentene, at m/z 84. The ...

3

If you leave it to E1, you will not get the selectivity you need. The difference between the methyl group and the methylene unit is steric bulk. You can select for deprotonation of the methyl for elimination via a bulkier base in an E2 reaction. In addition, you will need to convert the substrate so that it has a good leaving group. For an alcohol, this is ...

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