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# Study Guide for Exam 1 CHEM 2321

UTEP

GPA 3.42

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## Popular in Organic Chemistry I

## Popular in Chemistry

This 4 page Study Guide was uploaded by Hayley Lecker on Wednesday September 16, 2015. The Study Guide belongs to CHEM 2321 at University of Texas at El Paso taught by Dr. James Salvador in Fall 2015. Since its upload, it has received 170 views. For similar materials see Organic Chemistry I in Chemistry at University of Texas at El Paso.

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Date Created: 09/16/15

Organic Chemistry Exam 1 Study Guide Important Information Professor s Email isalutepedu Class Website organicutepeducourses2324 Class Code Ebook utep232Xfall2015 1 How many carbons are in a given structure a Inn 39 Now in the example 2 to the right there are 25 carbons wheinO Every vertex is a carbon unless shown otherwise if it is not a carbon is will tell you Examples In the example 1 to the left if you count the bertexareas it tells you are carbons you get 29 carbons CH3 you count the vertexes 2 How many methyls methylenes or methines are in a given chemical structure a Methane is CH4 Methyl is CH3 Methylene is CH2 and finally Methine is CH So you need to imagine the carbons and there hydrogens around the structure Lets look at example 1 this will be used throughout the study guide as an example it is the chemical structure to the left and count the Methyls Methylenes and Methines Methyls 4 They are the CH3 noted on the structure Methylenes 7 The trick is to look at vertex and imagine the hydrogens are there so if it has 2 bonds already that carbon can only make two more and it will be the hydrogens not noted Methines 7 3 How many primary seconday tertiary and quaternary carbonsnitrogens are in a given structure a A carbon or nitrogen with one bond to another carbon is primary If bonded to two other carbons it is secondary if three other carbons tertiary if four other carbons quaternary So lets look at example 1 again Primary 2 Secondary 14 Tertiary 8 Quaternary 1 4 How many primary methyls methylenes or methines seconday methylenes or methines or tertiary methines are in a given chemical structure a We need to look at the bonds around the central carbon and the hydrogens on the central carbon Lets look for secondary methylenes in example 1 so we are looking for CH2 bonded to two other carbons Secondary Methylenes 7 5 How many pi bonds or unsaturation s are in a given structure a Any bond with more than one line is a pi pi bond it it has two lines it is one pi bond if three it has two pi bonds So let s count in example 1 There are 8 you may have counted 7 at first but that s because you forgot the oxygen which is double bonded to a carbon 6 How many unsaturated atoms are in a given structures a A carbon may have four bonds but if it is not bonded to four different atoms it is called unsaturated Nitrogen must be bonded to 3 different atoms to be called saturated oxygen must have two to be saturated So lets look at example 1 and count the unsaturation s There are 13 where there is a multiple bond it effects both carbons on the other side the reason it isn t 2 times pi bonds is because in the case of the double bonded oxygen it is only affecting 1 carbon 7 How many tetrahedral trigonal or linear atoms are in a given structure a You may remember the flower method each atom around a central atom is a petal So with linear atoms they will have two petals drawn around them with trigonal three and with tetrahedral 4 This is including any lone pairs So let s look at example 1 again for tetrahedral atoms There are 18 you must look at the nitrogen for the tetrahedral atom it is bonded to 3 other atoms with one lone pair 8 How many s sp sp2 or sp3 hybridized atoms does a given structure have a For hybridizations this is the same as the flower trick now it is each petal gets a letter the first petal is always s the for every other petal up until 3 more is p So let s find sp3 in example 1 There will be 18 9 How many carboncarbons bonds are 12 13 14 or 15 A long C l bond length is 11 JET carbon hydrogen C bond length is 15151 carbon carbon single bond length is 141 151 this is in cases of resonance such as in a ring where double bonds and single bonds which places very fast so its length is the avg of a single and double bond bond length is 13113 carbon carbon double bond length is 12 carbon carbon triple So let s look for 13 long in example 1 There should only be 5 since there is only 5 carbon carbon bonds that are double in that structure 10 How many rings are in a given structure a You look at the minimum number of bonds you need to break so there will be no rings we give rings mono bi tri tetra etc So lets look at example 1 again it has 5 rings 11 How many hydrogens are in a given structures l 22ClN X 2lrings 2lpi Pidouble or triple bonds if a double bond it has 1 pi bond if triple it has 2 pi If we use that formula for example 1 you will get 35 hydrogens 12 How many rings plus unsaturations are implied from a given molecular formula Rings Pi 1 C llZ lNz The above equation can be used to calculate the rings and pi bonds in example 1 which would be 13 We know from the above questions that there are 8pi bonds and 5 rings which is 13 13 Given molecular formulas before and after hydrogenations how many rings and how many unsaturations are in a given compound Example Cngg Using the first equation we can find out the number of rings and unsaturation is 5 After hydlrogenation the compound is now CgIlm to find the number of pi bonds take the new number of hydlrogens minus the old number then divide by 2 So 16 8824 pi bonds from there you can minus 4 from 5 to find out there was only 1 ring 14 How many sigma bonds are implied from a given molecular formula The following pasted text is for example 1 To find the number of sigma bonds in a molecule the following equation is used Sigma 1 C N 0 H rings C Carbon NNitrogen OOxygen HlHlydrogen Rings Number of Rings in molecule To further simplify this equation is it sigma Eatomsrings1 the symbol 2 means the sum of In wordls this equation is the sum of the total number of atoms in the equation the number of rings minus 1 So for mifepristone example on the above page It will be Sigma 2935125 1 71 This is 29 carbon 35 hydrogen 1 nitrogen 2 oxygen 5 rings andl finally minus 1 Hlallogens are not included when doing this equation 15 How many total bond are implied from a given molecular formula Total number of bondls equations The sum of sigma pi total bonds this simply means 2C32 N12Hl12X total bonds Where C is carbon N is nitrogen ll is hydrogen and X is the halogens If you have an oxygen in the compound the equation is 2C32N12Hl12XO The trend for this equation is taking the yaliancy of the atom divided by 2 times the number ofthat atom in the compound So for example 1 the total number of bonds is 79 that is reasonable since we have 71 sigma and 8 pi 16 What is the corresponding systematic name or line drawing for a given structure If we look at 16 on the practice exam 1 spring 2014 we can see how easy this actually can be The following Newman Projections represents which compound H H a cis13dimethylcyclohexane b trans13dimethylcyclohexane c cis14dimethylcyclohexane d trans14dimethylcyclohexane The answer is d you may be asking why Well every circle contains 2 carbons and lines that do not meet the center of the circle are the back carbon s substituents So looking at the first circle of the first newman projection the methyl is in front we will call that carbon 1 now count the shortest distance to the second methyl you can see it will be 4 because it is by the other circle but DOES NOT meet the circle s center Now to determine cis or trans look at the orientation of the methyls one points up and one points down even though they aren t on adjacent carbons they are still orientated as trans 17 What is the energy difference between the two given conformations to a tenth of kcalmol For this we can look at question 17 on exam 1 spring 2014 1 Given the following energy values what is the energy difference in kcalmole between the two conformations given in question 16 CH3CH3gauche 08 HH eclipsing 10 HCH3eclipsing 12 CH3CH3eclipsing 41 a 08 b 16 c 24 d 36 e not ad Well lets look at the first conformation in that conformation one methyl group is on the equatorial position and one in the axial if you look down the bonds at them you will see they are anti to each other giving a value of 0 for that conformation Now for the second one it is best explain by chegg In the diequitorial conformation of structure 39d39 each methyl in axial position will contribute 18kcal energy in the form of strain The hydrogens in the axial methyl group will interact with each of the 13 di axial hydrogens present on the ring Each of these steric interacgions is approximately equivalent to one gauche butane interaction of 09 kcalmole that means there are two axial hydrogens for one axial methyl hence 2 x 09 kcal for one axial methyl groups All that is saying is that hydrogens are interacting and methyl and hydrogens will interact That gives you 36 for the second conformation however because the first one is 0 360 will still be 36 for the energy difference 18 Another important concept to know is neo this means there is a central carbon bonded to four other carbons creating a cross kinda of Neo is NOT an IUPAC name so to rename something in IUPAC way you must first draw it out then look for the longest chain with the most substituents

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