in this video I will go over some practice problems to help you prepare for your organic chemistry exam most of the practice problems will be centered on various addition reactions organic synthesis and mechanisms so let's begin with this question so here we're asked to predict the product of this reaction now if bromine reacts with this alkan what we have to do is to break the pi bond and attach each of the bromine atoms across the double bond of the of the alkan and this broine atom must be added in an anti-addition fashion such that the two bromine atoms have to be on opposite stereochemistry or opposite bonds and this should be our product let's go to the next question here we're asked to predict the product of this reaction now there are two reactions basically the first one is where the al alkan is reacting with bromine under light condition so this first reaction will give us alkalhalide in which the bromine atom will be attached to the more substituted carbon and this will be the major product other monroination products are possible but this is going to be the major product now in step two we're reacting this alkalihalide with sodium methoxide to predict the product of this reaction we need to ask ourself the question what type of alkhalide do we have so our alihalite is a tertiary alkhalide now for the nucleophile the sodium ethoxide is a strong nucleophile because there is a negative charge on the oxygen meaning that which is an indication of an excess electron density around the oxygen so that makes it a strong nucleophile now as a base the compound is unstable the methoxy group is unstable because the negative charge on the oxygen cannot be stabilized by resonance or by size and so since this is an unstable base it's quite reactive and as such we regard it as a strong base so the sodium metthtoxide is a strong nucleophile that also behave as a strong base now this will result in a competition between SN2 and E2 so since our alihalide is tertiary alihalide um and our our root nucleophile could also is a strong nucleifile that could also double as a strong base e2 is going to prevail with that being said we can go ahead and draw the E2 product so since what we have here is a is a regular base we will have the most substituted arcane as the major product you could also have a minor product the Hoffman product is going to be the Maya product so next question this question asks us to provide the reagent that will be able to convert the tertiary alkaalhalide to an ether so you should apply retroynthetic analysis to solving this problem and retro synthesis simply means working backwards so since the product is ether you want to ask yourself the question how can we make an ether right also another thing you have to look at is the only difference between our starting material and the product is that the chlorine atom in the starting material has now been replaced with a methoxy group so the question then becomes what reagent can we react with the tertiary alihalide to convert it to the ether so that reagent must have OC3 group the question is there are two types we could make this a strong nucleophile or we could make it a weak nucleifile this is why it's quite important for us to certain mechanism that this reaction will go through since our alihalide is a tertiary alkhalide this simply means that even though there's possibility of having SN2 and SN1 because it this is definitely a substitution product the mechanism can never be SN2 because we have a tertiary alkaalhide so it has to be SN1 mechanism if it's SM1 mechanism definitely a nucleophile has to be a weak nucleophile so the correct answer then has to be this alcohol so this question asks us to provide the product of this reaction so essentially we need to we need to ask ourel the question what type of group does these reagents add across the double bond of the alkan it has to be hydrogen and hydroxy group but this addition addition of these groups is usually done via antimacovnikov fashion So this simply means that I have to attach the hydro the hydrogen to the more substituted carbon and the O will go to the less substituted carbon so to draw the product I'll break the pi bond and I will simply attach the O group to the less substitute carbon and that will be the product of this of this reaction next question so here we're asked to predict the product of this reaction to solve this problem we have to chloride and purodine what does this reagent do tossy chloride and purodin simply converts a bad living group the O here is a bad living group so the toss chloride and purodin will convert that alcohol to a tossillate which is a good living group so essentially we would have we treat this with the first reaction which is toss chloride and puridin the alcohol will get converted to oscillates next we have to react this tossillate with potassium toxide in the second step so potassium 3xide is a bulky base and the tossulate is a primary tossulate you can treat the dosilate as the way you treat your alkaalhalide so when you react this primary calculate with potassium 3 oxide the prevailing mechanism is going to be E2 and as such your product will be such that we have if this is your alpha carbon and we have one beta carbon all you have to do is to place a double bond between the alpha and beta carbon so our product is going to be this alkan this question asks us to predict the product of this reaction now again the first thing you want to do is to ask yourself the question what type of group or groups does the does this reagent combination add across the double bond of the alkan and how does it add the groups so the groups that are being added by these reagents are or the groups that are being generated by these reagents are hydrogen and hydroxy groups and they will be added in a marovnikov fashion so this simply means that the hydroxy group will go to the more substituted carbon and the hydrogen will go to the less substituted carbon so to draw the product of this reaction simply break the pi bond and attach the hydroxy group to the more substituted carbon and that's all you have to do the hydrogen will go here but we don't have to show the hydrogen next question this question asks us to predict the product product of this reaction so here the two groups will be hydrogen and Br and they will be added across the double bond of the of the across the double bond of the alken and they will be added via mnikov of addition so what this simply means that the Br will go to the more substituted carbon and the hydrogen will go to the less substituted carbon so to draw the product of this reaction simply break the pi bond and attach the Br to the more substituted carbon next we're asked to predict the product of this reaction to do this this is oonolysis oxidation so all we have to do is to divide this molecule into two fragments by drawing a bond through the al through the alkan so this is going to give us two fragments this and this next we attach oxygen to both of the alkanes so give this an oxygen and then give this an oxygen and that's going to be the product so this will be the product of this reaction in this question we're asked to provide the starting materials that will give give this product so we need to employ retroynthetic approach to solve this problem now since our product here is ether and the WilliamC ether synthesis require a strong nucleophile reacting with um a a an alkhalide via SN2 fashion so this means I want to deconstruct this product into two fragments that makes sense so what I could do would be to cleave the bone here if I did that then I will have two fragments basically i will have this and I will have the metal group so how can I turn these fragments into meaningful reactants that can undergo SN2 reaction so I need to make this into a strong nucleophile i need to make the aloxxy group into a strong nucleophile by giving that oxygen a negative charge so that makes that a strong nucleophile all all I have to do is to attach a good living group to the metal say bromine in this question we're asked to provide the product of this reaction now we need to recall that if we react an alken with hydrogen bromide using organic radical um using an organic peroxide which is also a radical initiator and if this reaction is taking place under light condition then the H and Br will be attached via will be added across the double bond of the alken via antimicarovnikov addition this simply means that the Br will go to the less substituted carbon and the hydrogen will go to the more substituted carbon so to draw our product we will simply have to attach bromine to the less substituted carbon here we're asked to convert to provide the reagent that will convert this alkan to the alkan so the reaction that involves conversion of alken to alkan is hydrogenation so just write hydrogen here and then you could include any catalyst which could be paladium catalyst or you could use nickel catalyst or you could use platinum catalyst you only need to use one catalyst alongside hydrogen now this question asks us to draw the mechanism for the for the reaction showing on the screen since we are reacting an alkan with bromine and um the reaction is taking place under light conditions then this has to go through radical mechanism and radical mechanism there are three stages initiation propagation and termination step so what happens in the initiation step is that the bromine molecule will be split into radicals so under light condition the bond between the bromine atom will break and then one electron will go to one broine atom the other electron will go to the other bromine and then you will have two bromine radicals so in the initiation steps in the initiation step two radicals are generated now the radicals generated in the initiation step will then convert the alkan into an alkal radical which will in turn convert neutral bromine reactants to radicals also so what happens essentially is that a bromine radical one of the bromine radicals that is generated in the initiation step would then react with the alkan and the way this happens is that the bond between the hydrogen and carbon will break one electron will go to the carbon the other electron will be used by this hydrogen to form a bond with the broine radical so hydrogen bromide is formed and we will also have an alkaalle radical the alkaal radical would then convert the bromine reactant neutral bromine reactant into a radical and in the process the product will be formed so a neutral bromine reactant then combine with the alkal radical and the way this happens is that the bond between the bromine atoms will break one electron will go to this bromine the other electron will be used by this bromine to form a bond with this alkal radical so a bromine radical is formed and our keyhide is also formed in the process and this is our product so most of the products are formed in the propagation step now the broine radical that's formed here will go on to conver to to convert more of the alkan into alk radical and the alk radical will again react with neutral bromine reactant and convert it to broine radical and more of the product is formed and this cycle will continue until all the starting materials the alkan and the bromine are all depleted or completely exhausted and when the reactants the neutral reactants are completely exhausted and now we're left with just radicals broine radical and alkal and alkal radical those radicals will then start combining in the termination step so in the termination step two broine radicals can combine to terminate so that forms a neutral broine molecule and the reaction gets terminated or we could have a bromine radical combining with alkal radical to form alide and this is a minus source of our product most of the products are actually formed in the propagation step another way we could have termination is the alkal radicals can also combine with itself which will form a byproduct so this side product can also be formed in the process this question asks us to draw the mechanism for this reaction to draw the mechanism we'll start by drawing out the alkan now the geometry of the alkan is such that it is flat as a result the bromine molecule can approach the alken from the top face or from the bottom face so let's take it that the bromine approaches the alken from the top face what's going to happen is that the pi electrons in the pi bond can attack this bromine atom and as such a partial positive charge will be developed on the more substituted carbon of the alkan the bromine can also donate electron simultaneously to this carbon of the alkan in the process forming a bridge head and then this broine will actually be kicked out so we are going to form a bremonium aon and this is going to be a bridge and here is the bromide aon that is kicked out now since the top face is already occupied by the bromonium aon the nucleophile can only attack from the opposite side or from the bottom from the bottom face this carbon has this carbon will develop partial positive charge as the bremonium ion will pull electrons from that carbon from the more substituted carbon now don't forget that this reaction also involved water um actually the bromine is diluted with water so we have less concentration of the bromide and we have more water molecules so this simply means that since the bromine is in very low concentration the bromide will also be in very low concentration and the water has better chance of getting to this carbon first so that water molecule will attack that carbon and then this bond between carbon and the broine brominium ion is going to break and the excess electron will go to the bromanium a thereby neutralizing the positive charge on the bronium aon so a product will be such that water molecule the oxygen will form a bond with the carbon and now we have bromine on the opposite side as the hydroxonium or hydrononeium ion since oxygen have three bonds it is going to get a positive charge to get rid of the positive charge on the oxygen we can then use the bromide ion for that deep protonation so so this is going to grab the proton the bond between hydrogen and oxygen breaks excess electron goes to the oxygen thereby neutralizing the positive charge on the oxygen eventually we will have hydroxy group on this side and we'll have broine on this other side and we'll also have a side product HBR question here ask us to pro propose a reasonable synthesis to transform the alkan into an ether so whenever you're solving synthesis problem it's good to apply retroynthetic analysis and retroynthetic analysis simply means walking backwards so try to work backward at some point try to walk forward and then eventually you connect the reactions so in working backwards what reactants are going to react in order to form this product retroynthesis demands that you try to deconstruct the product into simpler fragments so if I choose to break this bone here then I will have this fragment and I'll also have this fragment since I know that making ether require one of the good ways to make ether is to use a strong nucleifile and react that strong nucleifile with an alkalite via SN2 reaction i can make this into a strong nucleophile and then just add a living group to the methyl group here the next question would then be how can I make this aloxxide ion of course the aloxxide aon can be made simply by deeprotonating an alcohol using a strong base so the strong base here could be LDA or I could use sodium hydide or I could even use sodium so either of these reagent will deproinate will pull off the oxygen from the oxygen the the hydrogen from the oxygen and then it will confer a negative charge on the oxygen atom at this point we might choose to work forward going forward whenever you're starting a synthesis with an alkan it's pretty easy to convert the alkan into a compound that can do many reactions and one of such compounds is alkaalhalide so I I'll try to convert this alkan to an alkahalide and the way to do that will be radical bromination so if I react that alkan with bromine under light condition then I'll form this alihalide as the major product the next question would then be how can I connect the alkhalide to to the alcohol i know that I can think of how I can add a an O group to an alkan to form the alcohol so in other words I would need to make my alkhalite into an alkan and then treat that alkan with a reagent that can add hydrogen and O group across the double bond of the alkan say I want to convert this alkali helite to this type of alkan and then I can now add O and hydrogen across the top bond of the alkan to form my alcohol so in order to make this alkan I need to treat the alkhalide with a a strong base so I have to perform an E2 mechanism since my alihalide is a tertiary alkihalide can decide to treat it with sodomite irregular base of course we will have competition between substitution reaction and elimination reaction because this reagent is a strong nucleophile which could also function as a strong base however because our alihilite is tertiary E2 mechanism will prevail eventually I'll try to connect this alkan to the alcohol now I'm thinking what kind of reagent can I use that will add hydroxy group to the less substituted carbon so I'm going to use borine which is stabilized by the solvent THF and I will use sodium hydroxide in the second step of the reaction with hydrogen peroxide now we've successfully proposed a reasonable synthesis to transform the alkan to the ether here we're also asked to [Music] show syn synthesis a reasonable synthesis that could transform the alkan into an alcohol now as usual try working backwards whenever you have a synthesis problem like this try applying retroynthetic analysis and then the next thing will be to identify the functional group in the product so this basically has an alcohol now I'm going to be thinking how can I what reagents can I use to make this alcohol right so I can make an alcohol by treating an alkan with a reagent that can add hydrogen and hydroxy group across the double bond of the alkane this time around the O has to go to the more substituted carbon to make this happen I will need water and catalytic acid such as catalytic sulfuric acid this will add O group to the more substituted carbon giving us the alcohol at this point I can also choose to work forward now the starting material here is an alkan the easier thing to do would be to convert the alkan to an alkalhalide to make this happen I'll react the alkan with bromine under light conditions so the next thing would then be how can I convert the alkan the alka helite to this alkan of course we have to do elimination reaction now mind you in order to do it reaction I could use there are two types of bases we could use a bulky base and a regular base a bulky base is going to form double bond between this alpha carbon and this beta carbon in which case your product will be this but this is not what we're trying to form we're trying to form the more substituted alkan we're trying to form the ZF product so I have to use a regular base a regular base that will create a double bond between this alpha carbon and this other beta carbon so we could use any regular base such as sodium oxide now we need to test the condition so this is a strong base that can also function as a strong nucleophile strong nucleophile which could also function as a strong base so there is always a competition between E2 and an SN2 however A2 will prevail since we have a tertiary alkhalide so these conditions are favorable for E2 mechanism and this will be the synthesis for for this problem here we're asked to predict the product of this reaction as well as circle the correct operative mechanism now um the substrates the tossillates should be treated like an alihalite so what type of tossillates do we have it's primary and tossate are very good living group now the nucleophile we have is a strong nucleophile because there's a negative charge on the oxygen suggesting excess electron density around the oxygen making it a strong nucleophile as a base whenever you want to check if a base is strong or weak look at how stable the base is if it's stable then it's a weak base but if it's unstable it's going to be reactive and as such will be considered a strong base so this in this base the negative charge on there's a negative charge on the oxygen and that negative charge cannot be stabilized by size or by resonance therefore it's unstable base making it a strong base so since we have a primary oxilate and a nucleophile that's a strong nucleophile that can also behave as a strong base SN2 will prevail essentially and in an SN2 mechanism the nucleophile attacks the electrophilic carbon thereby kicking out the living group and so our product will be such that the ethoxy group is attached to the electrophilic carbon in this problem first of all we need to predict the mechanism of this reaction and then decide what product is going to be formed to make this decision we need to try to find out what type of alkhalide we have so we have a secondary alkalhalide and a nucleophile is a weak nucleophile since it doesn't have there's no negative charge on the oxygen and this is a weak base because this molecule is stable as a base it's pretty stable so we have a weak nucleifile that could also function as a weak base the these conditions are favorable for SN1 and E1 so this means we're going to have two products we're going to have the SN1 product as well as the E1 product for the SN1 product all we have to do is to replace the bromine with the methoxy group so here we're going to form an ether for the E1 product you will need to you will need to place a double bond between the alpha and the beta carbon and these are the product of this reaction in this problem we are asked to predict the product of this reaction as well as choose the right mechanism that this reaction will go through i would like to draw the mechanism of this reaction so alcohol is reacting with hydrogen bromide the first step is going to be protonation the O attacks the proton and breaks the bond between the hydrogen and bromine and the excess electrons go to the broine atom so you would form hydrononeium and a bromide now this is an excellent leaving group and since oxygen has a positive charge that oxygen is electron deficient it's going to pull electrons from the nearby electron source the single bond so always draw your curved arrow from an electron source to an electron sink so you draw it from the bond from the center of the bond towards the oxygen and now the water molecule is going to leave and you will form a carocaton intermediate for the SN1 product the bromide ion would then attack the carocation again when drawing curved arrow draw it from electron source to electron sync the bromide the negatively charged bromine atom or the bromide is the electron source because it has lone pairs and then the electron sync is going to be the carbon because it has a positive charge that's how you can tell the electron and when this happens a bond is formed between the bromine and the carbon so I'll choose SN1 as the operative mechanism in this reaction next question we are reacting this alcohol with concentrated phosphoric acid and we're also asked to choose the operative mechanism let's try to draw the mechanism of this reaction we treat the alcohol with phosphoric acid the first step is going to be protonation so draw your draw your curved arrow from the electron source to electron sync the electron source is the lone pair on this oxygen and the electron sync is going to be the proton the bond between hydrogen and oxygen is going to break and the excess electron will go to the oxygen so a hydrononeium aon is formed and the conjugate base of the acid is also formed how should we draw the next curved arrow you need to identify your electron sink and electron source the electron sink is this oxygen because it's electron deficient the electron source is the bond between the oxygen and the carbon so the oxygen is going to pull an electron it's going to pull electrons from this bond towards itself thereby breaking the bond off and you will then have carocation and water molecule in the final step we will bring the conjugate base up here now this compound is not considered a nucleophile it is a base and this is because it tetrahedral structure makes it difficult to act as a nucleophile so what this is going to do essentially is that it will pull off a beta proton so if this is your alpha carbon the carbon with a positive charge then you could have we have a beta proton here we also have a beta proton here to form so there are two possible product that could be formed um but the major product will always be desired self product the product that will give you the more substituted alkan so since there's a a proton there the base will simply abstract that proton and the double bond is formed between the alpha and beta carbon and you would have you have this product of course the acid is regenerated