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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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Sometimes, especially in introductory courses the instructor will try to keep things "focused" in order to promote learning. Still, it's unfortunate that the instructor couldn't respond in a more positive and stimulating way to your question. These reactions do occur at $\ce{sp^2}$ hybridized carbon atoms, they are often just energetically more costly, ...

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It is well known that SN1 reactions often give incomplete racemisation: Although many first-order substitutions do give complete racemization, many others do not. Typically there is 5–20% inversion, although in a few cases, a small amount of retention of configuration has been found. These and other results have led to the conclusion that in many SN1 ...

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In general, yes, $\mathrm{S_N2}$ reactions are reversible. But not in this particular case. Rates of $\mathrm{S_N2}$ reactions depend on several factors: the nucleophile, the leaving group, the alkyl group undergoing substitution, and so on. In the context of your question, the leaving group ability is quite possibly the most important. Iodide, $\ce{I-}$, ...

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As we know, $\ce{S_N1}$ reaction is stereochemically non specific. This means, that it does not force the formation of one stereoisomer, as $\ce{S_N2}$ does. In $\ce{S_N1}$ reaction, presence of carbenium ion (=carbocation) as intermediate, results in the feasibility of attack from both sides of the ion, resulting in the formation of both stereoisomers(R and ...

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Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more ...

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In the second compound, steric repulsions (denoted by the curly lines) between the N-methyl groups and the methyl groups on the ring occur: This forces the $\ce{-NMe2}$ group to rotate such that the nitrogen lone pair is no longer parallel to the ring's $\pi$ system. As a result, the nitrogen lone pair can no longer contribute to the mesomeric stabilisation ...

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There are a number of factors that can influence the rate of an $\mathrm{S_{N}2}$ reaction. Solvent, leaving group stability, attacking group nucleophilicity, steric factors and electronic factors. In the series of compounds you've presented, all of these parameters are held constant except for the steric and electronic factors. Considering only steric ...

11

Yes, the haloalkane product will have the inverse stereochemistry of the alcohol that it was produced form. $\ce{PBr3}$ does this by converting the $\ce{-OH}$ into a good leaving group and providing a nucleophile all in one. This is can be seen in the two step mechanism: $\hspace{2.4cm}$ When $\ce{Br-}$ ejects $\ce{PBr2OH}$ in an $\mathrm{S_N2}$ reaction, ...

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This is a quote from Solomons and Frylhe Organic Chemistry 10th Ed, Chapter 13, p.614 Reactions of this type are quite general with other conjugated dienes. Conjugated trienes often show 1,6-addition. An example is the 1,6-addition of bromine to cycloocta-1,3,5-triene. A plausible mechanism for the bromination step would be the initial formation of a ...

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In the presence of $\ce{AgBF4}$, diphenylsulfide is alkylated to 3-chloropropyl diphenylsulfonium tetrafluoroborate. The sulfonium salt is stable and can be isolated as a crystalline product. This sufonium salt can be deprotonated (e.g. with $\ce{NaH}$ in $\ce{THF}$) to the corresponding sulfonium ylide. In the latter, intramolecular displacement of $\ce{... 10 Secondary and tertiary alcohols react with strong acid (and bases) such$\ce{HX}$following SN1 mechanism. $$\ce{ROH <=> HX + ROH_2^+ + X^-} \tag{1}$$ $$\ce{ROH_2^+ <=> R^+ +H_2O} \tag{2}$$ $$\ce{R+ +X- \rightarrow RX} \tag{3}$$ In these reactions, the rate determining step is the formation of the carbocation (2). So the successive attack of ... 10 Agreed,$ \ce{Br-} $is a good leaving group, so one would expect that an SN1 mechanism is not too difficult to have in a polar protic solvent. But do not forget that the mechanism of any reaction is determined by not only the nature of the leaving group, but also the carbon skeleton. A primary carbocation is extremely difficult to obtain, and will not, in ... 10 I think the final product will also contain some Bromine atoms attached to it. and the no. of double bonds will be one less than that of the compound given in the picture. I may be wrong but with all of my known possibility of Organic reactions, I can actually reach to the final product somewhat different from that given in the picture and I don't think the ... 9 I have asked a question previously which concerns the same electrophilic system with the exception that nucleophilic attack happens on the carbonyl carbon and not on the α-carbon. From the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on pp. 890-891: The$\pi^{*}(\ce{C=O})$and$\sigma^{*}(\ce{C-X})$orbitals add together to form a new,... 9 The shorthand way that we draw structures in organic chemistry, with implicit hydrogens, leads us to often forget that the hydrogens are there, or to neglect considering them in our analyses. In electrophilic aromatic substitution (EAS) reactions, the generally accepted mechanism involves attack of the aromatic ring on the electrophile to give a Wheland ... 9 From Peter Sykes: Pyridine (62), like benzene, has six at electrons (one being supplied bynitrogen) in delocalised it orbitals but, unlike benzene, the orbitals will be deformed by being attracted towards the nitrogen atom because of the latter's being more electronegative than carbon. This is reflected in the dipole of pyridine, which has the ... 8$\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}$, the equilibrium would be shifted towards the reactants (water reacting with product), so ... 8 This is a rather intellectually-stimulating question and one that is also very difficult to answer. You have constructed a very good case for why the nucleophilicity order would not be expected to reverse in polar aprotic solvents, i.e. the order of intrinsic nucleophilicities of the halide ions should be$\ce {I^- > Br^- > Cl^- > F^-}$, based on ... 8 For what it's worth, this reaction does not appear in the literature. So, whether it actually forms the product or not is anybody's guess. However, if I had to suggest a mechanism... I would note first that an SN1 reaction on a primary centre, as you have proposed, is difficult. I think it more likely that the alkene assists in the expulsion of the chloride.... 8 TL;DR - Bond angle strain gets a lot worse in going from cyclpropyl chloride to the transition state for the 3-membered ring example, than for cyclopentyl chloride going to the corresponding transition state. The rate for each reaction will be determined by the free energy barrier in each case. Cyclopropyl chloride is strained because the carbons "want" to ... 8 Reaction of alcohols with PCl5 and PCl3 (for reaction with alcohol) Mechanism for conversion of ketone to dichloride with phosphorus pentachloride (for reaction with ketone) Now as per the question as it is not given in excess so it will react with any one of the alcohol or ketone. It's quite hard to tell which one of them will react first because both of ... 8 You are correct on your initial assumption. This hydrolysis is assisted by the neighboring thiomethoxy group ($\ce{SCH3}$) as depicted in the following scheme: The cyclic methylsulfonium intermediate, which is achiral in this case, is a mimic of bromonium ion in bromination of alkenes (also see Chemistry of Sulfur Mustard on releasing chlorine through ... 7 explain why substitution reactions with enolate anions don't take place at other sp2 carbon atoms As explained in this earlier answer,$\ce{S_{N}2}$reactions do occur at$\ce{sp^2}$carbons, they're just higher energy and consequently, less common. So the question becomes, "why is this pathway higher in energy?" Since the$\ce{S_{N}2}$reaction involves ... 7 The reason is quite straightforward-$\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$hydrogen of the halides and gives an alkene. 7 Not a bad question, but a poor answer. The person who mentioned$\mathrm{S_N2}$' is on a much better track. Nucleophilic attack on a carbonyl or 1,4-Michael acceptor is not an$\mathrm{S_N2}$reaction. It is also very misleading. It's like saying a chicken and a human are the same thing because they both have two legs.$\mathrm{S_N2}$reactions involve ... 7 The order of reactivity by substitution in these two reactions is difference because they have different mechanisms. The substitution of an alkyl halide by a strong nucleophile in a polar aprotic solvent is an SN2 mechanism. A$\ce{C-Nu}$bond forms and a$\ce{C-X}$bond breaks at the same time:$$\ce{CH3CH2Br + NaOH ->[\text{acetone}] Na}\bigg[\ce{... 7 Let's do this step by step and figure your questions out. First, what a nucleophile needs is positive terminal in the compound. The other carbon attached to oxygen is not as positive as the one where the attack has been done due to inductive effect of$\ce{-CH2}$adjacent to it. The product (1) is not formed as there is no source providing$\ce{H+}\$ to form ...

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A cursory look through several organic chemistry texts (Clayden, Carey, March, Vollhardt) as well as SciFinder doesn't reveal any instances of the same reaction used in the Khan Academy video - this would imply that either the reaction is made up to illustrate a point (albeit not very well) or comes from somewhere more obscure (historical literature in a ...

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There are several arguments. Firstly, five-membered ring formation is generally kinetically more favourable than 6-membered ring formation. As ring size increases, the entropy of activation also increases – this is explained in greater detail in Clayden 2ed, pp 805–807. It is true that six-membered rings are less strained, but that is generally more of a ...

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