hello everybody my name is Iman welcome back to my YouTube channel I'm really sorry about the video before this cutting abruptly my sound got cut off but I wanted to salvage that part so it ends abruptly we're just gonna go ahead here and pick up where we left off which was question eight the question is says one the following straight chain Fischer projection is converted to a chair or ring confirmation its structure will be blank all right so this is what we're starting off with all right we're going to form a ring structure which means that the last alcohol group is going to go ahead and attack the aldehyde that's going to form our anime carbon all right um and so we're gonna have to draw this in the ring structure now this is going to be oxygen all right we're going to draw it like so please forgive my drawing all right we're gonna go ahead and number our carbons here one two three four five in the ring structure what those are going to relate to is one is going to be that annumeric carbon that's formed right this new chiral carbon that's formed when we create this um ring structure all right this here is going to be two three four all right this is going to be five where our ch2oh is going to be this is going to point in the up Direction here so we're just going to go ahead and fill that in all right now our annumeric carbon can go in either position so we're not going to worry about that right now we're going to move ahead to two all right and what we know here is that um for the alcohol groups on the left they're going to go on the top position all right any alcohol on the left side goes in the top position any alcohols on the right go in the down position all right so starting off with two all right this is going to go in the down position for two all right so we're going to put that alcohol group here three has it on the left so it goes on the up position so up oh four is again on the right so it goes in the down position o h all right and our our um oh on that C1 carbon right the substituent can be really in either position on the anomeric carbon so we can skip that for now we already assigned C2 three and four all right um and just looking at the answer choices the only one that matches what we have for C two three four five that we know what it's going to be is answer Choice C and what it has for position one all right is the the up positions to where it assists with this so the beta and animer of this all right so that's how you can work through this right we know what's right we look for the answer twist that fits that all right and it really could go either way for the for for position one for our annumeric carbon um and so answer Choice C is the correct answer for a here 9 says why is the alpha animer of D cloak glucose less likely to form than the B Amber oh this is a good question well the hydroxyl group on the annumeric carbon of the B animer is going to be equatorial to be in that CIS in that CIS formation with the ch2oh that equatorial position is going to create less non-bonded strain than the alpha animal which has the hydroxyl group of the anomeric compound in an axial position all right and so essentially it's going to cause less strain so there's less electron repulsion and so what that means is answer Choice B is the correct one B animer is preferred for metabolism that has nothing to do with it Alpha anima is more stable known unnecessarily it forms l-glucose also not true all right so the correct answer for nine is B 10 says which two polysaccharides share all of their glycosidic linkage types in common alrighty so glycogen and amyl amylopectin are the only polysaccharide forms that demonstrate branching structure making them the most similar in terms of linkage both glycogen and amylopectin they use an alpha 1 4 and an alpha 1 6 linkage all right cellulose uses beta14 linkages and amylose does not contain Alpha One six linkages so which two polysaccharides share all of their glycosidic linkage types in common it is going to be glycogen and amylopectin so answer Choice d 11 says which of the following is digestible by humans and is made of only one type of monosaccharide while maltose and cellulo bios both have the same glucose subunits only maltose is digestible by humans because the beta glycosidic linkages in cellulo BIOS cannot be actually cleaved by the human body all right only maltose is digestible by humans fantastic and then last but not least the reaction below is an example of one step in blank all right let's notice something here all right we're forming ring structures all right we're forming a ring structure what is different between the ring structures oh look at this the anomeric carbon this has our high alcohol group in the down position that has this has our alcohol group in the up position here the alcohol in the ch2oh group are trans and here they're CIS we have already done a problem just like this where we describe this interconversion between these two as a Muto rotation and so the correct answer here is C all right phenomenal that's all I have for you I hope this was helpful let me know if you have any questions comments or concerns down below other than that good luck happy studying and have a beautiful beautiful day future doctors