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It takes more energy for bromine to leave than it does for water to bond to the carbocation.
after dat , the unstable carbocation would be very eager to stabilise nd wont take noticible tym 4 da nxt step
the overall rate of the rxn (which depends upon da tym taken by da rxn 4 cumpletion) would deped upon the slower step which requires a considerable amount of tym , not da faster one
hope dat i ws able 2 make da thing clear :)
so it needs to wait till the leaving group leaves and when it does, it can then give its lone pair
Solvent: Sn1: protic, Sn2: aprotic
Stereochemistry: Sn1: racemic mixture product, Sn2: switch from (R) to (S)
Steps: Sn1: two steps, Sn2: one step
Nucleophile strength: Sn1: does not matter, Sn2: strong nuc.
Substrates: Sn1: tertiary/secondary/allylic/benzylic, Sn2: methyl/primary/secondary
Think of you and a friend washing dishes. You are washing, your friend is drying them. It takes you 5 seconds to wash a plate, and it takes your friend 10 seconds to dry the plate. Drying the plate is the rate determining step, because it determines how quickly you and your friend can finish the job!
1)it proceeds through a transition state i.e. intermediate of nucleophile and substrate, which is usually a penta-ordinate system.
2)It takes place in a single step i.e. it is a concerted reaction
3) and nucleophile usually attacks from opposite side of leaving group
And the reaction is Sn1 if reaction proceeds through a carbocation.
A reaction is USUALLY SN1 if the inner carbon is connected to 3 other methyl or multi-atom groups due to steric hindrance.
A reaction is USUALLY SN2 if the inner carbon is connected to no multi-atom group or 1 multi-atom group.
A reaction is SN1 if it proceeds through two steps.
A reaction is SN2 with a strong, usually charged nucleophile.
A reaction is SN1 with a weak, usually neutral nucleophile.
Why doesn't the bromine anion takes the H atom from the water molecule that attached to the substract? Why does it have to be another water molecule instead? Shouldn't it be more stable in terms of electron charges that the bromine anion attaches to a H, forming HBr?
The 1 and 2 refer to whether the reaction is unimolecular or bimolecular. That is, does the rate of the reaction depend on the concentrations of one or both components?
Let's say that you are carrying out a reaction between NaOH and an alkyl halide RX. Is the reaction SN1 or SN2?
You might first do an experiment in which [RX] and [OH-] both equal 0.1 mol/L and measure the rate of reaction r₁.
Then you would do a second experiment in which the only change is that [RX] = 0.2 mol/L. If the new rate r₂ is twice r₁, then the rate is _first order_ in [RX].
Then you would do a third experiment in which [RX] = 0.1 mol/L and [OH-] = 0.2 mol/L.
If the new rate r₃ is the same as r₁,you know that the rate of reaction depends _only_ on [RX]: the reaction depends on the concentration of only one component, the RX. The reaction is unimolecular, so we label it SN1.
Ir r₃ had been twice the value if r₁, the reaction would be first order in [OH-]. The rate of the reaction is found to depend on the concentrations of _two_ components, so this is a bimolecular reaction, and we label it SN2.
Hope this helps.
Oh but then yeah, the H+ wouldn't be attracted to the carbocation. Nevermind lol thanks
What you mean by "why a water molecule as a weak nucleophile would attack the positively charged substituent"......this is a two step reaction. What positively charged substituent are you referring to...the carbocation?
Water is in a much higher concentration, that is certainly true. However, Br- is a much weaker base/nucleophile than water. Once it's left, it's surrounded by a hydration shell of water molecules making it unavailable for reaction. It has nothing to to with electronegativity here....the determining factor is nucleophilicity which follows the same trend as basicity.
In general, SN2 reactions are favoured whenever you have:
a strong nucleophile
a poor leaving group
a 1° or 2° substrate
a polar aprotic solvent
SN1 reactions are favoured by:
a weak nucleophile
a good leaving group
a 3° substrate
a polar protic solvent.
You have to weigh these factors against each other and then decide which ones best fit your case.
What do strong acids do in aqueous solution? ...dissociate 100% into hydronium and anion.
Why would you possibly make the side product (HBr) that is 0% in concentration? Water surrounds the Br- molecule stabilizing it (solvating it). You end up with hydronium as the regeneration of catalyst. There is something known as the "leveling effect" you cannot have a stronger acid in water than hydronium. You cannot have a stonger base in water than hydroxide. Hope that helps. Just because it has a negative charge doesn't mean it's a strong base. Often water is stronger than good leaving groups.
adnan humayun pakistan
my textbook says we can add NaOH to any alkyl halide to give an alcohol and NaX..
R-X + NaOH---> R-OH +NaX..
my question is ... can we add h20 instead of naoh? and which mechanism is this?
is it nucleophilic substituion rxn? if yes, then is it sn1 or sn2?
SN2 requires the forcing of a leaving group vs SN1 being a situation where an opportunistic nucleophile waits to jump in on say a carbocation which has formed as the result of a reactant leaving on its own accord?
it has two FREE PAIR electrons!!
You can also get SN1 reactions with strong nucleophiles, e.g., iodide ion with tert-butyl bromide.
Usually a 3° substrate goes SN1, and a 1° substrate goes SN2. The 2° substrates can go either way, depending on the other factors.
Keep in mind that BOTH reaction mechanisms proceed in most situations, they just proceed at very different rates. So yes, a weak nucleophile could react via SN2, but because it's weak, it will only manage to "kick out" the halide on the electrophile every once in a while. The halide spontaneously leaving to give a (stable) carbocation would happen much more frequently, and once the carbocation is formed (slow step), the weak nucleophile would react with it quickly. So both happen, but because in this situation SN1 proceeds so much more quickly than SN2, it is the dominant reaction mechanism.
In this example, there is no hydride or methyl shift that would produce a more stable carbocation, but carbocation rearrangements do happen in SN1 reactions, for the same reasons they do in E1 reactions.
For example, if your car is hit in the rear end by a car that is going fast enough, your front bumper might fly off.
Hope this helps.
Let's say you have 2-bromobutane. and above the arrow is NaOH/H2O. ...NaOH is stronger and will certainly react faster than H2O. Is HO- a good nucleophile? yes. it is negatively charged. Sn2. If the alkyl halide is tertiary, Sn2 is not possible. Then you must do an E2 reaction if possible.
In an Sn1, the first step is rate -limiting (solvolysis). It always is, (unless you are doing an acid catalyzed dehydration of and alcohol).
though the video on this topic has not been made but it comes under the topic of nucleophilic reactions.
i wanted a detailed explanation of Nucleophilic internal substitution or also known as Sni.
However, the solvent is practically infinite in concentration (water). Water is much more basic than Br- (1x10^4 times more). It would be highly improbably that Br- would remove the H. It's already surrounded by a cage of water that stabilizes the anion....it's not even able to break out of the cage to attack the hydrogen.
oxygen is a electronegative element so it must be much stable to hold that negative charge...
methyl carbon = bonded to no other carbons
primary carbon = bonded to 1 carbon
secondary carbon = bonded to 2 carbons
tertiary carbon = bonded to 3 carbons
quarternary carbon = bonded to 4 carbons
The simple answer...reactions are classified by the end result to the organic compound only (It's an electrophile). We do not consider what gets exchanged (substituted) on the inorganic nucleophile.
An SN2 reaction happens _in a single step_. The nucleophile attacks the central carbon _while the leaving group is leaving_. Since the nucleophile always attacks from the side exactly opposite of where the leaving group leaves, the stereochemistry of the central carbon is inverted.
You can help favour SN1 over SN2 or vise versa by changing the nucleophile, electrophile, and solvent.
that would seem more logical to me, since H3O+ would also probably instantly react with Br-, this also depends on the solubility of HBr acid..
or we can just choose.
If the bromine were bonded to any of the other (primary) carbons, it would be called isobutylbromide.
R-C-(CH3)3 is a tert-butyl group, whereas
R-CH2-CH-(CH3)2 is an isobutyl group.
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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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This is great, I finally understand quadratic functions!
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