the topic of this video will be the e1 reaction the e1 reaction is a first order elimination reaction the e stands for elimination the one is for you know molecular just like the sn1 reaction only one molecule is involved in the rate determining step and just like sn1 the one molecule that is involved in the rate determining step is a substrate let's look at the general reaction our substrate will have a good leaving group just like it did for a substitution reaction the difference when we're talking about elimination reactions however is that there must be a proton on the carbon next to the carbon with a leaving group therefore the general structure of a substrate for an elimination reaction looks like this instead of a nucleophile we'll use a base because this is a deprotonation reaction this is a bronsted-lowry base the product will be an alkene plus the conjugate acid of the base plus the leaving group the mechanism of an e1 reaction starts very much like the mechanism of an sn1 reaction in the first step the leaving group leaves producing a carbo cation in the second step of an e1 mechanism the base removes the proton adjacent to the carbo cation the electrons from the carbon hydrogen bond form the PI bond between the carbon and the carbo cation carbon forming the alkene product now let's look at an example reaction we look at the reaction of bromo cyclohexane with methanol this reaction requires heat to form the cyclohexane product now let's look at the mechanism the first step of the mechanism the leaving group leaves forming the carbo cation you may find it helpful to draw in all of the hydrogens at the carbo cation center as well as the neighboring carbon centers the next step the base D protonates one of the hydrogens next to the carbon that is the carbo cation center because of the symmetry in this molecule it doesn't matter if the proton that is removed comes from the top or the bottom the methanol removes the proton and the electrons involved in the bond to that proton form the PI bond to the carbo cation this produces our cyclohexane product just like in an owl reaction it is the first step of the mechanism that is slow it is the rate determining step therefore the rate of an e1 reaction is dependent only on the concentration of the substrate the structure of the base has no bearing on the rate of an e1 mechanism we often see that the base is weak and it's often the solvent the structure of the substrate does affect the rate of the e1 reaction and typically what we see is that the tertiary substrates react most quickly followed by secondary followed by primary this follows the stability of the carbo cation that is formed in the rate determining step just like it did with the sn1 reaction methyl substrates cannot react in the e1 reaction because there is no proton on a neighboring carbon to the carbon with a leaving group hopefully at this point you'll know you're noticing that there are some similarities between the reaction and the sn1 reaction both of these reactions involve the formation of a carbo-cation and they involve a weak base or a weak nucleophile which are often the same thing what this means then is that the e1 and the sn1 almost always compete and you often get both products in such a reaction let's look at an example reacting this substrate which is tertiary with a weak nucleophile / weak base such as ethanol we will get a mixture of products we will get both the e1 product as well as the sn1 product take a moment and pause the video and work through the mechanism and make sure that you can understand how both the e1 and the sn1 products are formed again because e 1 and sn1 both involve the formation of a carbo-cation u1 reactions can experience rearrangements just like sn1 reaction let's look at an example if we treat this substrate with a weak base you'll get 2 products both of these products occur from a rearrangement of the carbo cation that is formed in the first step let's look at the mechanism the first step of an e1 mechanism is the leaving group leaves forming the carbo cation this is a secondary carbo cation ax but a hydride shift can give us the more stable tertiary carbo cation this garbo cannon can eliminate in one of two ways we can remove a proton from the secondary carbon next to the carbonyl cation Center to get a left product or we can remove a proton from the primary carbon next to the carbonyl iron Center to give the right product we get a mixture of these two products however the product on the left is the major product and the product on the right is the minor product this difference in the amount of the two products can be explained using zaitsev's rule zaitsev's rule states than in elimination reactions the most substituted alkene usually predominates because it is more stable