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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 ...

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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

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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-...

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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, $\... 10 I think the final product will also contain some Bromine atoms attached to it. and the no. of double bonds will be one less than that of the compound given in the picture. I may be wrong but with all of my known possibility of Organic reactions, I can actually reach to the final product somewhat different from that given in the picture and I don't think the ... 9 Pöytä: Your conclusion that D contains four bromine atoms symmetrically distributed is sound. The treatment of D with NaI serves to substitute one, if not two bromine atoms with iodine to form "J". Iodide acts as a reducing agent in a vinylogous fashion. (Compare this reaction with the reduction of a vicinal dibromide with iodide to form an alkene.)... 8$\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}$... 7 Conjugated dienes are significantly favorable to non-conjugated dienes (because of delocalization and hybridization, etc.), so this will be the major product. It is not to say that the other reaction cannot happen, but that a combination of factors leads to that product being major. All of chemistry occurs in an equilibrium of some degree, and the reaction ... 7 In chemistry, you don't solve things like that; you just recall the right class of reactions (Hofmann elimination in this case) and apply that knowledge. You can hardly expect to rediscover that from the first principles. Knowledge about other classes of organic compounds and their behavior in the presence of a base is not particularly helpful either. It is ... 7 The reason is quite straightforward-$\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$hydrogen of the halides and gives an alkene. 7 The best explanation I was able to find is given by Avery A. Morton , John B. Davidson , Barton L. Hakan, J. Am. Chem. Soc., 1942, 64 (10), 2242–2247. If I understand it correctly, this boils down exactly to that you already propose. Disproportionation has been so universally accepted as the criterion of a free radical that there has been no attempt to ... 7 I believe you were right to conclude E1cB would occur. Both of your considerations are exactly on point. The amount of ring strain is minimal -- if we think that it's OK to deprotonate to form a sp2-like carbon in the first place, then we've already decided that the added ring-strain is not a huge issue. Of course, the added conjugation and the entropic ... 7 A carbonyl group normally destabilizes carbocations but stabilizes carbanions. In this particular case, the rate of dehydration isn't guarded by carbocation stability, but by CH- acidity. In acidic conditions, OH group in alcohols is relatively easily protonated. Then, if there is a relatively acidic hydrogen in$\beta$-position it immediately dissociates, ... 6 Sodamide is bloody dangerous. If your bottle is too old, it likely has begun to oxidize some and you have potential azides or other polynitrogen compounds that may lead to an explosion. Even with good stuff, generally the reagent is not heated to avoid similar complications. This means normally this reagent works at low temperatures (often times dry ice/... 6 These are both reasonable mechanisms, and the question outlines well the factors favoring each. In favor of mechanism I: Low temperature suggests kinetic deprotonation Statistically more terminal hydrogens than internal hydrogens In favor of mechanism II: Small base suggests thermodynamic deprotonation In these cases where there are conflicting factors, ... 6 The above reactions belong to Williamson Synthesis of Ethers, probably the best of the alternatives for prepartion of ethers! The Williamson Ether synthesis reactions follow SN2 mechanism. Since the SN2 mechanism proceeds through a single step where the nucleophile performs a “backside attack” on the alkyl halide, the only thing stopping this, is steric ... 6 Here is a flowchart of what the commenters have stated. An α-elimination occurs stepwise 1 --> 2 --> 3. There is a phenyl migration (bridged?) to afford tolane 4. Look here for this chemistry. 6 @User6376297 raised a question: "Could the cyclopropane ring behave like a double bond, making this the analogue of an allyl bromide, and yielding$\ce{R2NCH2CH2CH=CH-tBu}$by attack on a cyclopropyl$\ce{CH2}$and ring opening?" Especially given the steric hindrance near the carbon bearing the leaving group. Indeed, that is the process going on in this ... 6 Anindya Prithvi has said everything you needed to hear in two sentences. It is true that rate determining step for E1 mechanism is the carbocation formation. Thus, I present here the detail energy diagram for reaction progress of acid-catalyzed dehydration: The first step is the protonation of starting compound (SM), which is shown in the right hand top ... 5 The difference between the different eliminations like E1, E2 or Hoffmann elimination is the proper choice of the base and proper solvent medium. Let's take your example i.e. 2-bromo-2-methylbutane, and choose different bases and reaction media for observing differences between E1, E2 and Hoffmann elimination. Why not E1? First let's consider the reaction ... 5 E2 elimination is stereospecific for anti elimination. In both variants of compound B that you have drawn, the OTs and the proton on the tertiary centre are syn to each other, meaning that the elimination is disfavoured. For this reason, the NaOMe deprotonates at the secondary centre, generating the products as drawn, despite the fact that this goes ... 5 I believe this is an example of a Cope elimination mechanism here The EtI is making the tertiary amine. The persulfate is taking the tertiary amine to the N-oxide. Heating this produces ethene and hydroxylamine of the starting amine. This is consistent with the Cope as "The sterically demanding amine oxide function reacts preferentially with the more ... 5 The process of elimination of two bromides which you are thinking is not the correct way how the reaction occurs in this case. First there will be a nucleophilic substitution (preferably SN2) by$\ce{I-}$on the two carbon atoms containing bromine. As iodide is a better nucleophile than bromide, this substitution is majorly driven forward, and after the ... 5 "A gets eliminated" or "the elimination of A" is just plain wrong, typical sloppy lab slang by many (too often native?) speakers. Very common, but still terrible. As you say, the small molecule X gets eliminated from substrate A. Chemical English is still just English, and that is the correct way to put the case in question into words in this language. "... 5 Here is what I think is going on:$Step1$: As the OP correctly identified is a Knoevenagel condensation to give A diethyl 4-cyclohexyl-benzalmalonate.$Step2$is an example of little-used ester hydrolysis using cyanide ion to give initially the acyl cyanide which is unstable in aq EtOH and gives the diacid B$Step3$is loss of one carboxy group by ... 5 In this case,$\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 ... 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. 4 @Waylander has provided a well-reasoned solution to the post. I offer a different interpretation. Knoevenagel product 1a has been converted in high yield to cyano ester 2a and subsequently hydrolyzed to phenylsuccinic acid (3a) 1. In the addition of cyanide to the double bond of 1a and protonation, one of the labile ester groups is cleaved by cyanide. (e.g.,... 4 There are two pathways that phenyllithium could follow - it could deprotonate, but there really not any reactive protons, or it could do lithium-halogen exchange with one of gem-bromines. This second pathway will lead to a carbene by loss of$\ce{Br-}$and the carbene (a highly reactive intermediate) will rearrange to give allene$\ce{H2C=C=CH2}\$. This ...

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