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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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That will be great!!!
Some topics need reinforcement however or a slightly more elaborate explanation. But keep up the good work!
HOPE THE ABOVE SITES HELPED ... :) VOTE ME UP
2) the more polarizable the lone pairs of the nucleophile the better. so RS- is better than RO- because the S has larger and more diffuse orbitals due to the 3s and 3p vs the 2s and 2p of O. so I->Br->Cl->F-
3) also, as a side note, charged nucleophiles are better than uncharged ones, so for example: RS->ROH or H2O
hope this helps, have a nice day
Also there is no such thing as pure OH-....hydroxide is always in salt form. (you need lithium, sodium, or potassium (a group IA metal) to balance the charge by ion stabilization (Coulombic interaction).
It is as if you were choking on a piece of meat. If I hit you behind the shoulders hard enough, the offending piece if meet will come flying out of your mouth in front.
Basicity is the ability to form strong bonds. If you are a relatively strong base, you form stronger bonds than weak bases. Therefore, it's harder for you to leave.
I don't know if this question has been answered, and that I've just missed it. But what exactly is "partial negativity"?
When you have a polar covalent bond then one end is more positive and one end more negative. They are not charged enough to be ionic and have a full charge. So these ends are called "partial" negative charges.
I mean, whats in the name?! - partial negative, meaning slightly negative -so, thats what it is!
Similarly, the weakest bases are least likely to form bonds by sharing electrons (i.e., they are good leaving groups). The strongest bases are most likely to form bonds by sharing electrons (i.e., they are poor leaving groups).
R-Br + KOH → R-OH + KBr (the "molecular" equation)
But we must remember that compounds like KOH and KBr are ionic compounds.So we really should write the reaction as
RBr + K+ + OH- → ROH + K+ + Br- (the "ionic" equation)
We see that the K+ appears on both sides of the equation. It does not really participate in the reaction. Its only function is to balance the charges on the OH- and Br- ions. We call it a _spectator_ ion.Since it does not participate in the reaction, we can cancel it from each side of the equation and get
R-Br + OH- → R-OH + Br- (the "net ionic" equation).
We often omit spectator ions from equations.
Hope this helps.
In an SN1 reaction, the leaving group leaves all by itself in a rate-determining step. The nucleophile doesn’t attack until later..Since only the substrate is involved in the rate determining step, this is a first-order (SN1) nucleophilic substitution.
The two mechanisms are quite different.
All molecules or ions with a lone pair of electrons or at least one π bond can act as nucleophiles. Because nucleophiles donate electrons, they are by definition Lewis bases.
The oxygen in water has a partial negative charge and can behave as a nucleophile
The chloride ion has a whole negative charge and it also can behave as a nucleophile
The norbornyl cation is a non-classical ion. Its structure involves the sharing of two valence electrons by three atoms: C6, C1, and C2.
It is not easy to explain the stability of the norbornyl cation. A clue may come from the structure itself: the C6-C1 and C6-C2 bonds are equal in length, but the C1-C2 bond is remarkably short.
If the original positive charge was on C2, we could re-hybridize C1 and C6 to get two p orbitals. These, plus the vacant p orbital on C2, give us essentially a cyclopropenium cation, without two of the frame-work σ bonds. That may be why the cation is so stable: under this interpretation, we could consider it as a Hückel aromatic π system with n = 0. It is as if we had a π bond (as well as a σ bond) between C1 and C2, with a lobe of a p orbital from C6 sticking into the π system.
Method 1: Bond strength,
The Carbon to Hydrogen bond is one of the strongest bonds in chemistry at roughly 420kJ per mol. As the bond strength decreases the more likely the group is to leave. Bromine a much weaker bond with hydrogen at roughly 300kJ per mol, (the weakest possible is around 200kJ per mol). So the process that requires has the lowest activation, or input energy, will happen.
Method 2; pKa of Conj. Base.
The pKa of the conjugate base of the leaving group is another more complex method for working out the best leaving group. I tend to use this one as I have a better picture in my head of pKa's rather than Bond strengths.
The conjugate base of Bromine is Hydrogen Bromide (HBr). This has a pKa value of -9. The lower the pKa value of the Conj. Base the more stable the leaving group will be. So Br- is a very good leaving group.
Use whichever method you find easier. Hope I've helped.
The two explanations are that the eclipsed state is less stable (van der Waals repulsions) or that the staggered state is more stable (hyperconjugation).
We have learned that the electron clouds surrounding H atoms repel each other. Indeed, at distances of less than 210 pm, two H atoms do repel. At longer distances, they attract. The maximum attraction occurs at about 240 pm. The distance between eclipsed hydrogen atoms in ethane is about 230 pm, so the H atoms should be attracting rather than repelling each other.
Another explanation is that the rotational barrier arises because hyperconjugation makes the staggered conformation more stable. It arises because in the antiperiplanar staggered conformation, the bonding σ orbital of one C-H bond is oriented perfectly to overlap with the σ* orbital of an antiperiplanar C-H bond on the adjacent carbon atom. This interaction leads to a lower energy situation.
We would represent this as H-CH₂-CH₂-H <—> H⁺ CH2=CH2 H⁻
In staggered ethane, there are six (equivalent) antiperiplanar interactions of this type. They are not major contributors, but they do contribute. This makes the staggered form more stable than the eclipsed form.
So you are correct. It is the partially positively charged carbon that becomes a target for the electrons of the nucleophile.
If the nucleophile is weak to attack the partially positive α carbon, the leaving group will leave first (SN1). Then the nucleophile can attack the fully positive α carbon.
Every one of these can undergo a separate reaction mechanism depending on the reagents.. It depends on the reagents used....Do you know? or are you supposed to synthesize this molecule?
Then you can consider mechanism.
As we know from electronegativity (Br has 2.96 C has 2.55 and H has 2.20) partial charges are being induced in the molecule so that Br has partial negative charge relative to others and H atoms partially positive charge.
So why does that nucleophile essentially "attack" the nucleus namely the Carbon atom? Is that a statistical phenomena? (meaning on average the attack is mainly towards nucleus)
Are not positively charged H-atoms "attacked" as well by the nucleophile because opposites attract.
It seems really abstract that we can just say hey, this will be a Sn2-reaction, just believe it or not.
In elimination reactions, you're eliminating a particular leaving group, and it results in a double bond.
However, i do not understand which reaction mechanism has to be followed when for what solvent, reagent or substrate...
Does it need to be mugged up?
CH3CH-Br-CH2CH3 + OH- follows SN2 mechanism , i.e. it first reaches a transition state and then we get the product which is CH3CH-OH-CH2CH3OH + Br-
But i cannot understand why does not it follow SN1 mechanism i.e. it first becomes a carbocation and then OH- bonds with it.
Similarly, i cannot understand by seeing the solvent, reagent or substrate which reaction mechanism will be followed...
I'd elaborate....but there are errors in your product...you have a diol apparently...is that just a typo? Are you sure you understand how substitution occurs?
when i came across this question i thought it would first undergo halogenation......
then it would react with NaOH to give phenol....
i dont know if it is right or wrong.... can someone tell me please?
it's rather enjoyable. My own organic teacher taught us the same way, to make teaching more fun.
Example - Rate: [ ? ] [ ? ]
ps: thank you so so much for all these awesome videos!!!
IMMEDITAE ANSWER REQUESTED plZ
This could be in the form of lone pairs (i.e., the oxygen in methanol) or a partial negative charge (i.e., one of the oxygens in nitrate)
i didn't get why the reaction will be fast...
The proton transfer to generate a hydroxide ion should be drawn with an oxygen lone pair of electrons from one water molecule attacking one of the protons on another water molecule. Simultaneously, both electrons in the breaking O-H bond should be shown as forming the new lone pair on the hydroxide ion product. The SN2 mechanism should be drawn by showing one of the oxygen lone pairs attacking the electrophillic carbon and both electrons in the C-Br bond leaving with the bromide.
This is NOT a common convention in the videos, in fact I would say it causes more confusion, is inaccurate, and will mislead you from the actual convention for writing mechanisms that organic chemists follow. Just stating a fact.
if yes then see,
the hydrogens are not actually at the place they should be.
because as the Br- leaves , and the OH- comes and joins with the carbon , the two hydrogens are pushed towards the space created as the Br- leaves.
atleast i think of it like that.
hope i didn't confuse you.
just think about it in 3-Dimension and you will get it better.
Great video as always...
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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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At 2:33, Sal said "single bonds" but meant "covalent bonds."
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