27

In alcoholic solution, the $\ce{KOH}$ is basic enough ($\mathrm{p}K_{\mathrm{a}} =15.74$) to deprotonate a small amount of the alcohol molecules ($\mathrm{p}K_{\mathrm{a}}= 16–17$), thus forming alkoxide salts ($\ce{ROK}$). The alkoxide anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the ...


14

Assuming your Product A is correct then product B should be benzene. Selenium act as a mild oxidizing agent at temperature 220 °C to 330 °C. In the above reaction selenium oxidizes cyclohexadiene to form benzene. Below is a reaction in which 1-methyl-1,3-cyclohexadiene is converted into toluene. you can also go to this link for more info


12

Wow, this is an amazing question. The expected reactivity is strongly dependent on the exact structure. For starters, trimethylammonium is about the size of a tert-butyl group. So, expect an A value around 4.9, which is very large. So, put that group equatorial on the ring. Once you do this, you see that there are only 2 hydrogen atoms that are anti-...


9

Aqueous $\ce{KOH}$ is alkaline in nature i.e. it dissociates to produce a hydroxide ion. These hydroxide ions act as a strong nucleophile and replace the halogen atom in an alkyl halide. $$\ce{RCl + KOH (aq) -> ROH + KCl}$$ This results in the formation of alcohol molecules and the reaction is known as nucleophilic substitution reaction. Alcoholic, $\...


9

I think the final product will also contain some Bromine atoms attached to it. and the no. of double bonds will be one less than that of the compound given in the picture. I may be wrong but with all of my known possibility of Organic reactions, I can actually reach to the final product somewhat different from that given in the picture and I don't think the ...


8

$\ce{OH-}$ acts as a nucleophile. Reactions carried out in alcohol tend to be elimination reactions, and reactions carried out in water (aqueous) tend to be substitution reactions. If water were used as a solvent in an elimination reaction involving $\ce{KOH}$, the equilibrium would be shifted towards the reactants (water reacting with product), so ...


8

Your assessment of which groups are trans to each other is correct but you have not taken into account that the molecule can adopt other conformers. The classic confomation for E2 elimination is when the hydrogen and the leaving group are trans. This is because it maximises the orbital overlap between the $\ce{C-X \sigma ^*}$ and the $\ce{C-H \sigma}$ ...


7

Conjugated dienes are significantly favorable to non-conjugated dienes (because of delocalization and hybridization, etc.), so this will be the major product. It is not to say that the other reaction cannot happen, but that a combination of factors leads to that product being major. All of chemistry occurs in an equilibrium of some degree, and the reaction ...


7

In chemistry, you don't solve things like that; you just recall the right class of reactions (Hofmann elimination in this case) and apply that knowledge. You can hardly expect to rediscover that from the first principles. Knowledge about other classes of organic compounds and their behavior in the presence of a base is not particularly helpful either. It is ...


7

The best explanation I was able to find is given by Avery A. Morton , John B. Davidson , Barton L. Hakan, J. Am. Chem. Soc., 1942, 64 (10), 2242–2247. If I understand it correctly, this boils down exactly to that you already propose. Disproportionation has been so universally accepted as the criterion of a free radical that there has been no attempt to ...


7

I believe you were right to conclude E1cB would occur. Both of your considerations are exactly on point. The amount of ring strain is minimal -- if we think that it's OK to deprotonate to form a sp2-like carbon in the first place, then we've already decided that the added ring-strain is not a huge issue. Of course, the added conjugation and the entropic ...


6

These are both reasonable mechanisms, and the question outlines well the factors favoring each. In favor of mechanism I: Low temperature suggests kinetic deprotonation Statistically more terminal hydrogens than internal hydrogens In favor of mechanism II: Small base suggests thermodynamic deprotonation In these cases where there are conflicting factors, ...


6

The reason is quite straightforward- $\ce{OH^-}$ is a weak base and a stronger nucleophile specially under polar protic conditions. Hence substitution occurs. $\ce{RO^-}$ is a strong base owing to the inductive effect (+ I effect) of the R group. Hence under alcoholic conditions, $\ce{RO^-}$ extracts the $\beta$ hydrogen of the halides and gives an alkene.


6

The above reactions belong to Williamson Synthesis of Ethers, probably the best of the alternatives for prepartion of ethers! The Williamson Ether synthesis reactions follow SN2 mechanism. Since the SN2 mechanism proceeds through a single step where the nucleophile performs a “backside attack” on the alkyl halide, the only thing stopping this, is steric ...


6

@User6376297 raised a question: "Could the cyclopropane ring behave like a double bond, making this the analogue of an allyl bromide, and yielding $\ce{R2NCH2CH2CH=CH-tBu}$ by attack on a cyclopropyl $\ce{CH2}$ and ring opening?" Especially given the steric hindrance near the carbon bearing the leaving group. Indeed, that is the process going on in this ...


5

The difference between the different eliminations like E1, E2 or Hoffmann elimination is the proper choice of the base and proper solvent medium. Let's take your example i.e. 2-bromo-2-methylbutane, and choose different bases and reaction media for observing differences between E1, E2 and Hoffmann elimination. Why not E1? First let's consider the reaction ...


5

E2 elimination is stereospecific for anti elimination. In both variants of compound B that you have drawn, the OTs and the proton on the tertiary centre are syn to each other, meaning that the elimination is disfavoured. For this reason, the NaOMe deprotonates at the secondary centre, generating the products as drawn, despite the fact that this goes ...


5

Sodamide is bloody dangerous. If your bottle is too old, it likely has begun to oxidize some and you have potential azides or other polynitrogen compounds that may lead to an explosion. Even with good stuff, generally the reagent is not heated to avoid similar complications. This means normally this reagent works at low temperatures (often times dry ice/...


5

Here is a flowchart of what the commenters have stated. An α-elimination occurs stepwise 1 --> 2 --> 3. There is a phenyl migration (bridged?) to afford tolane 4. Look here for this chemistry.


5

I believe this is an example of a Cope elimination mechanism here The EtI is making the tertiary amine. The persulfate is taking the tertiary amine to the N-oxide. Heating this produces ethene and hydroxylamine of the starting amine. This is consistent with the Cope as "The sterically demanding amine oxide function reacts preferentially with the more ...


5

"A gets eliminated" or "the elimination of A" is just plain wrong, typical sloppy lab slang by many (too often native?) speakers. Very common, but still terrible. As you say, the small molecule X gets eliminated from substrate A. Chemical English is still just English, and that is the correct way to put the case in question into words in this language. "...


4

Liquid ammonia is cold, this means you are going to form the 'kinetic' product. Kinetic products are the ones that are most likely to occur via molecule collision, generally from the least sterically hindered deprotonations. There is some hinderance for forming the first intermediate on the way to product II. I would expect product I to form.


3

Well done to narrow the options down to E2 and SN2 and recognize that E2 is likely favored over SN2. As the comments suggest, in this case, it doesn't matter which one is deprotonated. The reason is that deprotonation at either beta-position gives the same product. If you are unsure, draw out both options. In future similar examples, you may come across ...


3

If the reagent is alcoholic KOH, then the reaction mechanism followed is E2 elimination (if steric/other conditions permit) However, if the reagent is aqueous KOH, it is simply a substitution (SN) type reaction in which the Br atoms are replaced by -OH groups (not to mention, if conditions permit, i.e. SN is actually possible in the given molecule*) *...


3

In a practical way, you would have to consider each particular case, there are tons of reasons that could make a reaction to yield mostly E1 or SN1 product (it will be always a mixture, that can vary from 50:50 to 0.01:99,9, I hope I'm making myself clear, they are always competing reactions). But there are several factors that contribute to make one of the ...


3

Example of a similar tertiary ammonium reaction with mechanism The thing is, it does react but there will be no cyclic product, in the end, I think that is the reason it is not the answer here.


3

It isn't necessarily the same situation as the haloalkane in water. In the sulfuric acid case, there is a large amount of acid present, and as you rightly say the alcohol and water are similar leaving groups, so the SN1 product is easily decomposed back to the tertiary carbocation. As a result, the mechanism tends towards the thermodynamic product, which ...


3

You have a tertiary halide. These dissociate rather easily into the relatively stable tertiary carbocation and the free anion. That in and by itself should be indication enough that the reaction proceeds via E1. Alcoholates are only moderately strong bases so the mechanism cannot be $\mathrm{E1_{cb}}$. You could consider E2 because the substrate is non-...


3

in this question, I think, we are not talking about priority of SN2 or E2, but according to the Clayden, (D) would go through E2 and (C) would choose SN2. But we know that reactivity of tertiary halides are much more than primary ones. So, I think (D) is the right choice, either.


3

The process of elimination of two bromides which you are thinking is not the correct way how the reaction occurs in this case. First there will be a nucleophilic substitution (preferably SN2) by $\ce{I-}$ on the two carbon atoms containing bromine. As iodide is a better nucleophile than bromide, this substitution is majorly driven forward, and after the ...


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