hi assalamu alaikum in the very morning so now we're gonna do chapter seven which is hello alkene we're gonna focus on the subtopic of 7.2 chemical properties focusing on part one of the video first so in this video we're gonna look into the description of halo alkene my hollow alkene will have a general formula of r x where x here refers to a halogen so we're going to look into the polar bond and where the polar bond gonna be susceptible for the nucleophilic attack so susceptible here means that it's going to be an easy target or the b muda and they go nucleophilic attack so on the next part we're going to explain the nucleophilic substitution direction where we're going to look into two types which is sn1 and sn2 mechanism on last part we're going to compare the relative reactivities of the primary secondary and tertiary hydroalkene towards hydrolysis and hypothesis so hydro disease happens when the nucleophile gonna be a water or it gonna have a hydroxide ion so water here is a neutral one neutral molecule and hydroxide ion here are going to be the negative charge or anion for alcoholicist it's going to be happening when the nucleophile will have o r minus here so r here can be och3 or it can be o c h2 ch3 or any alkyl chain okay so this have this come from the alcohol group which is ch3oh okay it's gonna be ch3o minus when it is becoming an ion and this one comes from base and this comes from the water all right so without any further ado let us start so halo again which containing a carbon halogen bond is a very polar bond because the halogen here is going to be more electronegative than the carbon so the electron density gonna move towards the halogen through so due to the electronegativity of the carbon atom this carbon here will carry a partially positive charge and this halogen group here gonna carry a partially negative charge and because it carries a positive a partially positive charge it will be an easy target for the nucleophile to attack the carbon okay and this refers to the electrophilic site and in a reaction they're gonna be a nucleophile and they're gonna be an electrophile so due to these properties uh the halo alkane can undergo nucleophilic substitution reaction and in the latter video you're going to see that the halo alkene can undergo elimination and it can be used to synthesis the green reagent however in this video we're going to focus on the nucleophilic substitution reaction first which is in the part one of the video for this two reaction letter you're going to look at it at part two okay so for the nucleophilic substitution reaction there are two types and the mechanism going to be depending on the structure of the hollow alkene so hollow again as you have looked in the previous video halogen can be divided into primary secondary or tertiary so this classification is important in order to determine the type of mechanism or the type of nucleophilic substitution reaction that are taking place so there are two types which is we have the unimolecular nucleophilic substitution and second we have the bimolecular nucleophilic substitution so substitution here refers to the letter s nucleophilic refers to the letter n and uni refer to one okay same goes to here where s refers to the substitution nucleophilic refer to the n and by refer compatible right so it's going to be sn2 here okay and in order to determine whether the reaction or the hollow alkene undergo sn1 or sn2 we're going to look into the class of the halo alkyl so for the uh for the primary and methyl halo alkene it will usually undergo sn2 where it will it is the usual case where it has a low steric hindrance so low steric hindrance means that it is not it is not uh protected by big group okay um static entrance for example if we have ch3 ch2 br okay so we're going to look into this mechanism in the lattice line however when their halo alkene has a secondary classification it will be depending on the strength of the nucleophile so if the nucleophile used in the reaction is a strong nucleophile uh for example hydroxide ion methoxide ion carboxylation or cyanide it will undergo sn2 however when the nucleophile use is a weak nucleophile it will undergo sn1 so the example of the weak nucleophile is water alcohol for example the ethanol here and also the amino group which is an amine or ammonia okay so as what you can see here the strong nucleophile you usually will have a negative charge meanwhile for the weak nucleophile usually it is it existing as a neutral molecule for example water where it has a lone pair but the overall charge here is neutral same goes to ammonia when it has a three nh-bond and one lone pair so this is known as a weak nucleophile so for the secondary hydroalkene when a strong nucleophile is used it will undergo sn2 if a weak nucleophile is used it will undergo sm1 meanwhile for the sharing halo alkene we will usually undergo sn1 and we must use the weak nucleophile in order for the sn1 reaction to occur okay so this flow chart here is very very important and going to be used along the process here so it is important for you to at least memorize it at first and try to understand it as you go along okay so now this is the the differences between the sn1 and sn2 mechanism so for sn1 because we're referring to one so it is focusing on the uni molecular nucleophilic substitution reaction and because it won we are saying that it is a first order reaction and the rate gonna be dependent on the concentration of the halo okay the other name of halo alkene gonna be okay and for usually for sn1 the weak nucleophile is being used which is the neutral molecule i have doctors now for example water and ammonia so for sn1 it involves the carbocation as intermediate and the rearrangement can occur so for sn1 you're gonna be having two step mechanism and for sn1 the the chili hello alkene gonna be um gonna be the one that will undergo more reactive towards sn1 mechanism and here you can see that the steric factor is not quite important okay so for sn2 um it is a bimolecular nucleophile substitution and because it refers to number two it is the second order reaction where it gonna be depending on the concentration of the halo alkene and the concentration of the nucleophile so in sn2 we're gonna be depending on the uh nucleophile the stronger profile which is hydroxide methoxide and cyanide and so on and here in this step it involved the activated complex and it refers to the one step mechanism and for this sn2 the metal headlight going to be more familiar to the sn2 mechanism so primary emitter highlight going to be the best one for sn2 and here the static factor is important so uh this is the differences between sn1 s and sn2 and so that whenever you got confused you can always refer back to the properties of sn1 and sn2 here okay so for sn1 it's going to be a first order reaction so sn1 it will have two step mechanism and it involves a tertiary carbocation better so in that saturday or by molecular we are referring to sn2 so it gonna be two involve one step and primary hollow arcane will be better for sn2 so subtle vertigo and sn2 gonna be two one one okay so these are the key point idea so now we're going to look into the sn1 mechanism first magen situator sn1 um gonna have two steps ion so ignites are two word antigua so you know that it consists of two steps okay so first uh we're gonna have for the sn1 reaction to occur first we need to identify whether it is a tertiary carbocation or not so this carbon is attached to halogen the carbon here gonna have one two and three alkyl group attached to it so it is a tesseri carbocation so for the step number one the the bond gonna be broken between the c x bond and is gonna uh produce a partially posit uh a positive charge on the carbon and because electron is being transferred to halogen it's gonna carry a negative charge and because it carries a partially positive charge okay so the partially positive charge going to be attacked by the nucleophile into here and as a result the nucleophile going to be attached to the carbon uh to the carbon bond and the nucleophile here can be for example water or hydroxide ion okay and this can be represented as in the diagram here so for sn1 reaction it involves two steps first you need to remove your halogen first and then barrel nucleophile which is the red the orange gap to enter the empty spot okay meanwhile for sn2 as what you can see here sn2 it will involve only one step okay so the strong nucleophile going to be used here and it will directly attack the electrophilic site and then this causes when the nucleophile enters or attack the carbon the nucleophile gona almost forming the bond with the carbon at the same time the cx bond here almost broken down so this is why we are denoting it with s with a dashed line here garrison put this produce and this process here is known as the activated complex and at the end you will directly form your nucleophile here your nucleophile muscle for a mechanism you need to show that it form a can activated complex first then only forming a product here all right so this is the representation where a strong nucleophile gonna change away there the halogen group where it will form an activated complex then the nucleophile and the halogen go gonna go up all right so remember that for the nucleophile attack it's going to be a backside attack masonry hydrogen and the intermediate which is activated complex here need to be shown all right and at the end of the mechanism you can see that the line here gonna be having an in inverted configuration so it was started up as like this okay so first they are like this at the end of the reaction it's gonna be like that initial and this is the final okay now we're going to look into the specific example here so let's say if we have rx which is the halo alkene reacting with water which is our weak nucleophile under reflux so you know that our oh is going to enter here and the halogen group going to be removed so water here act as a weak nucleophile so let's say if you are given a situation here and this process is known as hydrolysis using water which is a weak nucleophile so as what you can see here the carbon here going to be a tertiary hydroalkene because it is attached with three alkyl group okay so it will undergo sn1 reaction so okay you can always refer back to the flow chart here the cherie i can undergo sn1 so now let us look in let us draw the mechanism okay so the you know that sn1 would involve two-step okay and it's gonna have and it gonna occurs for tesheri carbocation so this is the test so step number one is we're gonna break the uh br one here first and then this is known as a slow process or in other words it's going to be the red determining step okay and the removal the break of the bond between carbon and hydrogen group gonna transfer two electron to the bromine and this carries a partially positive charge on the carbon and the bromine gonna carry a partially negative charge okay because usually the bromine here gonna have a lone pair but now because two electrons are being transferred here you're gonna have eight electron in total however when we calculate the formal charge it's gonna be seven because bromine will initially have seven valence electron minus eight electron that they have as the lone pair electrons so you're gonna carry a negative charge here however for the carbon carbon will have four valence electrons and it is attached with three carbon minus three so we're gonna carry a partially positive charge which is plus one for the carbon and negative one for the problem okay now for step one is done now we're gonna do the step number two which is the nucleophilic attack so because we have a partially positive charge here our nucleophile here is now a water so water is the weak nucleophile so water will use its lone pair and use to attack the c plus here and this is a fast reaction and the o will attach to the carbon and the oxygen here will carry a partially positive charge because oxygen will have 6 valence electron minus 1 2 3 single bond and one two lone pair electron so we're gonna carry a partial uh positive one charge okay but now we're going to produce c ch3 c3 oh so we have to remove one hydrogen here so the another the another water molecule in the beaker for example will then be used to attack one of the hydrogen and this is going to break the bond in between the oh and transfer to electron to the oxygen and as a result you will get the oxygen here will receive two electrons and hence the oxygen will now have a neutral charge because oxygen will have six valence electron minus two uh two bonding pairs and one two three four on bad electrons so you're gonna be zero this one here is oxygen okay so we're gonna and we are basically just have shown our major product here and as a result oh gonna have a history of plus ion okay and you can continue to use your br to attack the hydrogen and here gonna attach to hears again in order to get hbr however this step is usually not required because we only need to show until the major product here all right now moving on to the next uh example here so now we're gonna do the hydrolysis with the hydroxide ion so let's say if you have sodium hydroxide so the hydrogen ion comes from the dissociation of the naoh so oh minus here going to be a strong nucleophile because it carries a negative charge so oh he's going to enter here and x is going to be remove up okay and this produce an alcohol so the first thing first we have to know our nucleophile which is a strong base and if we look into our reaction here it is a primary halo alkene because it is attached with one alkyl group okay so it is a primary halo alkane and then it is attached with only one alkyl group so one alkyl group will have low steric hindrance and then it will undergo sn2 [Music] so in this case it will undergo sn2 reaction as shown in the flowchart here okay now let us look into the mechanism so first we draw again the structure of the given halo alkene which is our primary halo again and we need to draw it in terms of the 3d structure so one two of them two of the atom is in plane one out of plane and one behind the plane okay and the hydroxide ion come from the naoh will then be used to attack the electrophilic site okay and this causes the breakdown of the cbr1 where the two electrons will be transferred into bromine and this is a slow step reaction and as mentioned we need to show our transition state so the transition state involves the partially bonding between the attacking nucleophile and the hollow arch nucleophile so this is going to be our transition state so bondings okay and at the end of the reaction okay we can say that it is a fast process so we're gonna plus vr minus here okay and as what you can see here gonna carry a partially negative charge because it is a nucleophile and here is the electrophilic side so you're gonna carry a positive charge and here is the new nucleophile which is a negative charge again so it's gonna be negative partially positive partially negative okay and as mentioned for sn2 mechanism okay at first you're gonna draw it this way but at the end of the reaction it's gonna be the inverted way and you have to show the transition state so in the sn2 reaction the nucleophile attack from the backslide the reblocal of the electrophilic carbon that is from the site directly opposite bonded to the halogen and at the same time the backside attack causes the product form to have inverse configuration so here is the inverse configuration from the original configuration so from this y-shape all right so now we're going to look into the reactivity of various halo alkene towards the nucleophilic substitution reaction so antara ayudo alkene and bromo alkene plural alkene and fluoroalkyl which one is more reactive towards each other towards the nuclear substitution reaction so we can say that io and iodo are again going to be more reactive than from moroccan and it is more reactive than chloroalkine and it is more reactive than fluoroalkine this is because the strength of the carbon hydrogen bond increases from iodine into fluorine so our iodo alkene here will have the least strong one weaker spot stronger spot [Music] so weakest bone here gonna be the most reactive species this is because when the ri here which is the weakest bond it is easier to break the bond and hence easier for the molecule to form a carbocation and leave out the halogen group but however in this case is if r and f bond are strong it is very very difficult to break the bond hence it is less reactive okay and r here refers to the alkyl group so it can be ch3 or it can be ch2 ch3 and so on okay so you can imagine it as ch3i or ch3 br if they were to coming out in the equation and they asked you to compare so remember weakest bond is the most reactive point okay because it is easier to break the c and i bond apart okay now we're going to look into the steric effect okay for the stereo effect it basically means that it is the effect on the relative rate caused by the space filling properties of those part of the molecule attached at or near to the reacting side so here we're gonna have a primary halo alkene okay honey attached to one alkyl group so usually we're going to say that it will undergo sn1 reaction and the steric effect here means that it is the space filling properties where the carbon gonna attach with a small atom or not and as you know hydrogen is a small atom so there will be no stereo effect okay so if primary hello alkene it will be easier for the okay it will it will be easier to undergo sn2 mechanism however as you were to look into this case okay we have crr ch2br so if i were to expand the structure i'm gonna get ch ch and cb are here so we can say that the carbon here gonna be our primary halo alkene because the carbon here gonna attach with one alkyl group carbonyl attached to alkyl group however this primary hydroarcane with large alkyl group okay that the bond that the alkyl group is and this is because you do the steric hindrance so primary halo alkene be the high static hindrance for example ch3ch2vr it will undergo sn1 mechanism instead okay so because of the more standard effect it will not undergo sn2 but will undergo sn1 okay so let us look into the mechanism here and the example so let's say if we have a structure of c ch3 ch3 ch3 and this carbon here as mentioned is a primary halo alkene okay if you were to expand it this way so it is primary hello again so if you follow the chart usually the primary halo alkene will undergo sn2 however because of the large alkyl group here it will undergo sn1 okay and as mentioned sn1 will have two steps and usually it will undergo for the tesheri halo alkene however because of the primary halo alkene would that have a static hindrance it can undergo s and y so um the cbr one first going to be pre-applied and this causes the carbon here gonna have a partially positive charge due to the formal charge so you can count the formal charge what you want and the br gonna carry a br here gonna carry a negative structure and this causes the carbon here to be a primary carbocation so this is not good and hence what we're going to do is we're going to do the rearrangement so we're going to bring the ch3 group here to the right hand side okay and this process is known as one two metal shift and as we bring the metal group to the right hand side it will carry a positive charge here okay and the positive charge here will have one two three alkyl group so it's going to be a tertiary carbocation and now we can enter our nucleophile so our nucleophile here is kcn so if they were to dissociate and form ion it gonna be k plus n cn minus okay c and minus so the the carbon gonna carry a negative charge so the cn going to be used to attack the c plus which is the electrophile here okay and remember the carbon gonna carry the negative charge and as the result the carbon gonna be attached with the the carbon plus here we're going to attach with the carbon minus here and this is going to form the ccn bond here and this shows the product that we are forming so we have ch3 ch3 cn and ch2ch3 which is ch2 and ch3 okay so i think that's all for today's video see you again in some other time bye