as there is pi bond in alkene that has to be break for addition reaction and due to bond breaking one carbon acquire more electron density and gets negative charge and other 1 get positive charged due to electron deficiency so here nucleophilic and electrophiic both reactions are possible
The positive species which may attack on carbon or a pi bonds are electrophile as Cl+, NO2+ etc. but H+ and Positive metallic ions are not electrophile .
Alkenes have pi bonds that are readily available to react because the strength of a pi bond isn't as strong as a sigma bond. Pi electrons will attack the nucleophile to form the respective carbocation. Alkanes only contain sigma bonds and have no pi electrons to attack a nucleophile. In order for an alkane to become a strong enough nucleophile it must not be sterically hindered (primary carbons prefered to tertiary) and most likely deprotenated by a very strong base ( likely stronger than sodium amide ).
when an electrophile attacks 2nd position in pyridine the +ve charge goes on the electronegative nitrogen atom and destabilizes the intermediate ,in case of 3rd position attack the +ve charge goes on the ring which is more stable than the intermediate formed during the 2nd position attack
in sn1 reaction the electrophile leaves the substrate forming a carboncation.afterwards the nucleophile while attack the carboncation and usually recimes may be formed in sn1 reaction depending on whether the carboncation experienced a front of backside attack. in sn2 reaction the departing and attacking proccess occurs at the same time. these is pule rampai from the university of johannesburg
This is a really great question that I often drill into my students heads. During the attack of an alkyl halide by a nucleophile, we see the electrons from the Nucleophile emptying into the anti-bonding orbital present at the rear of the carbon atom. It is at this position..... backside.....that the large-lobed anti-bonding orbital is located. This is seen in the SN2 process. In an SN1 process, a carbocation can form,,,,, resulting in an sp2 intermediate....which can readily be attacked by a nucleophile. Halogens such as I- are excellent leaving groups since the anion is stable. F- is a poor leaving group, since the huge negative charge is so concentrated that is results in an unstablized anion.
Alkenes have pi bonds that are readily available to react because the strength of a pi bond isn't as strong as a sigma bond. Pi electrons will attack the nucleophile to form the respective carbocation. Alkanes only contain sigma bonds and have no pi electrons to attack a nucleophile. In order for an alkane to become a strong enough nucleophile it must not be sterically hindered (primary carbons prefered to tertiary) and most likely deprotenated by a very strong base ( likely stronger than sodium amide ).
The positive species which may attack on carbon or a pi bonds are electrophile as Cl+, NO2+ etc. but H+ and Positive metallic ions are not electrophile .
Alkenes have pi bonds that are readily available to react because the strength of a pi bond isn't as strong as a sigma bond. Pi electrons will attack the nucleophile to form the respective carbocation. Alkanes only contain sigma bonds and have no pi electrons to attack a nucleophile. In order for an alkane to become a strong enough nucleophile it must not be sterically hindered (primary carbons prefered to tertiary) and most likely deprotenated by a very strong base ( likely stronger than sodium amide ).
When cyclohexene(C6H10) reacts with bromine (Br2), trans-1,2-cyclohexane.This stereochemistry is obtained because bromine acts as both an electrophile and a nucleophile creating a cyclic bromonium ion intermediate. This means the second bromine, which acts as a nucleophile, can only attack the partially positive carbon from the opposite side of the side that is a part of the cyclic bromonium ring.
Nucleophilic substitution occurs when a nucleophile (some species with an affinity for positive charge) attacks an atom that is electron rich (electrophile). The atom under attack cannot form additional bonds so must release a bond to another atom or side group that has less affinity for it. This often happens simultaneously in an SN2 type reaction but can also occur in the slower 2 step SN1 reaction. Now the nucleophile has taken the place of (substituted itself for) the atom or group with less affinity.
when an electrophile attacks 2nd position in pyridine the +ve charge goes on the electronegative nitrogen atom and destabilizes the intermediate ,in case of 3rd position attack the +ve charge goes on the ring which is more stable than the intermediate formed during the 2nd position attack
in sn1 reaction the electrophile leaves the substrate forming a carboncation.afterwards the nucleophile while attack the carboncation and usually recimes may be formed in sn1 reaction depending on whether the carboncation experienced a front of backside attack. in sn2 reaction the departing and attacking proccess occurs at the same time. these is pule rampai from the university of johannesburg
when an electrophile attacks at 2 or 5 position we get an intermediate in which + charge is on the ring but when the attack is on 3 or 4 position the + charge get on nitrogen atom which is the strongly electronegative in the ring and get destabilised. and hence the electrophile prefer to attack on 2 or 5 position
nucleophle means nucleus loving.it can attack at the electron deficient center if it is not hindered.
The given reaction is:The given reaction is an SN2 reaction. In this reaction, CN−acts as the nucleophile and attacks the carbon atom to which Br is attached. CN−ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.
because the bond between the halogen and the carbon in the benzene ring (aryl halide) or a carbon participating in a double bond (vinylic halide) is much too strong--stronger than that of an alkyl halide--to be broken by a nucleophile (Sn2). Also the electrons of the double bond or benzene ring repel the approach of a nucleophile from the backside. They do not undergo Sn1 reactions because the carbocation intermediate they would produce is unstable and does not readily form.
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