36

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 ...


15

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, $\...


14

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-...


11

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 ...


10

As you have mentioned, the basic Fischer operations are: Vertical positions are below the plane of paper and horizontal positions are Above the plane of paper, thus you have already remember that when working with Fischer projection. What you have to remember about chiral compounds are: If you switch two groups, you get the epimer and if you switch two other ...


9

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.


9

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 ...


9

Pöytä: Your conclusion that D contains four bromine atoms symmetrically distributed is sound. The treatment of D with NaI serves to substitute one, if not two bromine atoms with iodine to form "J". Iodide acts as a reducing agent in a vinylogous fashion. (Compare this reaction with the reduction of a vicinal dibromide with iodide to form an alkene.)...


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 $\ce{C-H \sigma}$ ...


8

There are two possible eliminations that can happen (see the image below) As you can see, we then have two final products depending upon which $\ce{H}$ is eliminated. From the image 1a is cis and 2a is trans. Now, how do we decide which one is the major product? This can be based on looking at the stability of the transition state. This means we have to ...


7

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 ...


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 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.


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 ...


7

In this case, $\mathrm{E2}$ elimination is impossible regardless of condition used, because of lack of $\beta$-hydrogens. However, in acidic conditions, it is possible to have elimination reaction. Since 2,2-dimethylpropanol is a $1^\circ$-alcohol, initial carbocation formation is difficult. However, this formation of carbocations is accompanied by a ...


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

This question about the effect of the bulkiness of bases on the mode of E2 elimination has been studied and was published in 1956 by Brown, et al.[1] The elimination of 2-bromo-2-methylbutane was examined using solutions of potassium ethoxide, potassium t-butoxide, potassium t-amyloxide and potassium t-heptoxide in their respective alcohols. The clear result ...


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 ...


7

This was going to be a comment but it got too long. TL;DR: - This is not an answer, rather a justification for why the question is (probably) wrong. This paper (linked by @Rishi) gives us experimental evidence that under action of concentrated $\ce{H2SO4}$ hydrogens are exchanged from paraffins in the following fashion: $$ \begin{align} \ce{(CH3)2CHCH3 + ...


6

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/...


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

@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 ...


6

This excerpt is from an article in J Chem Ed (J. Chem. Educ., 1961, 38 (6), p 297 DOI: 10.1021/ed038p297) describing the contents of Markovnikov's 1870 paper in Liebig’s Annalen (translated into English, I guess): So this was before the discovery of the electron, but as you can see from the diagrams, the concept of chemical structure, atoms and valence was ...


6

Anindya Prithvi has said everything you needed to hear in two sentences. It is true that rate determining step for E1 mechanism is the carbocation formation. Thus, I present here the detail energy diagram for reaction progress of acid-catalyzed dehydration: The first step is the protonation of starting compound (SM), which is shown in the right hand top ...


6

Because of the steric hindrance of the two iPr groups, LDA is a very poor nucleophile. It is, as you note, a very strong base. The two products observed both arise from deprotonation of cyclohexene oxide. Deprotonation of one of the methylenes alpha to the epoxide ring opens it to give the allylic alcohol. Removal of one of the protons on the epoxide ring ...


6

$\ce{C-H}$ bond and $\ce{C-D}$ bonds are not the same in energy. $\ce{C-D}$ bond is stronger. So breaking $\ce{C-D}$ bond is more difficult and slower than breaking $\ce{C-H}$ bond. I hope you know the mechanisms of $\mathrm{E1}$ and $\mathrm{E2}$ reactions. In $\mathrm{E1}$ reaction, a carbocation is formed first, which is the rate-determining step (rds). ...


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 ...


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