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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, ...


24

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

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 ...


13

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 ...


13

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 ...


11

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 ...


10

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 ...


10

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 ...


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

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

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 ...


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

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

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 ...


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

OK, so let's look at your starting material, 3-iodopentane: $\ce{(CH3CH2)2CH-I}$ So it's SN1, so the $\ce{I}$ is going to leave spontaneously and give you the carbocation: $\ce{(CH3CH2)2CH+}$ The oxygen is in the ethanol so that will attack the carbocation giving the oxonium ion: $\ce{(CH3CH2)2CH-O+} ~ \ce{H-CH2CH3}$ Which will then be deprotanated by ...


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

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 ...


7

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....


7

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 ...


7

Reaction of alcohols with PCl5 and PCl3 (for reaction with alcohol) Reaction of cyclohexyl methyl ketone 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 them shows ...


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

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 ...


6

The first step of this reaction is always a proton transfer: Then, the substitution follows either the SN2 or SN1 pathway depending on whether R is methyl, primary, secondary, or tertiary. Methyl and primary ethers undergo acidic cleavage by the SN2 mechanism: Tertiary ethers undergo cleavage by the SN1 mechanism: Secondary ethers probably undergo ...


6

The reaction rate of SN2 depends primarily on the leaving group involved. Since in your question, $\ce{-Cl}$ is common in all 4 options we move on to the next important criteria. The partial positive charge on carbon (or the electrophilicity of the carbon) is considered more important than the steric effect that you mentioned. Look at Wikipedia's first line....


6

The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this: Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the ...


6

The acetone carbonyl is attacked by the CN$^-$ that comes from the NaCN. This tetrahedral intermediate has a oxygen anion, which then grabs a proton from the HCN, regenerating the CN$^-$.


6

I am assuming that you want the alcohol to react with the $\ce{OH}$ group and the alkyl halides to react with the $\ce{Cl}$ group. Both groups like to withdraw electrons, so they tend to be negative, and if we take them away, the resulting ions would be positive so they cannot react. This leaves us to the possibility of the $\ce{H+}$ being cleaved off from ...


6

A) is $\mathrm{S_N2}$ due to the secondary $\ce{C}$ and the sufficiently good nucleophile $\ce{I-}$. So inversion takes place. B) is $\mathrm{S_N1}$ due to the tertiary $\ce{C}$. The solvent is also more polar than in A). So the stereochemical information would be lost.


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