Exam 2 Study Guide
Exam 2 Study Guide CHEM 2321
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This 11 page Study Guide was uploaded by Hayley Lecker on Thursday October 15, 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 114 views. For similar materials see Organic Chemistry I in Chemistry at University of Texas at El Paso.
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Date Created: 10/15/15
Organic Chemistry Exam 2 Study Guide Important Information Professor s Email isalutepedu Class Website organicutepeducourses2324 Factors that Effect Boiling Points 1 The relative strength of the four intermolecular forces Ionic gt Hydrogen bondinggt diploe 2 3 dipolegt Van der Waals dispersion forces These are influenced by the function group present Boiling points increases as the number of carbons increase Branching decreases boiling point Examples of the intermolecular forces affecting boiling point If you look ionic has the higher boiling point while the dipole dipole has the least Boiling Poi nta Re ect Intermolecular Attractive Forces Example Alcohol derlvatlvea of elmllar molecular welght G G O WOH W0 Na Mol weight 7412 7412 961 N Sodium ame Duethyl ether nbutanol nbutoxide Strongest intermolecular D39POLE39D39POLE ggaNGGEN IONIC force Boiling point 35 C 117 C gt 260 C Now let s think about if there are no polar function groups such as butane C4H10 the only forces going on are Van der Waals so this would have a very low boiling point So let s look at amine and carboxylic acids as well so you can see the trend Amine derivatives of similar molecular weight I O K Wm ri CI Mol weight 731 731 1096 Tetramethyl Name NNodimethyl nbutylamme ammonium ethylamine chloride Stron est inta dwm DIPOLEDlPOLE Egagf NGGEquot IONIC force Boiling point 38 C 77 C gt 260 C Carboxyllc acid derivatives roughly similar molecular weight 0 0 O AO MO H NLOGN 89 Mol weight 881 881 1111 Name Methyl propionate Butyric acid Sodium butanoate Stron est interngoiecular D39POLE D39POLE 39ON39C lorce Boiling point 80 C 163 C gt 280 C BOTTOM LINE Intermolecular forces go in the order ionic gt Hydrogen bonding gt Dipoledipole gt Van der Wesls dispersion forces London forces Now the second factor affecting boiling point is increasing carbons this also is for the molecular weight the higher it is the higher the boiling point In a given series boiling point increases with molecular weight Why Increasing surface area gives riee to increased Van Der Weeis lnterectione As you can see as all of the examples get bigger the boiling points increase dramatically Now the final reason symmetry branching If something is symmetrical the likely hood of it lining up and bond is better when it can t its boiling point is higher OH X H Pentane 2239d39melhylpfopane 1 pentanol 3penianoi b 33 0c neopenIane P mp 9 c bp 137 C hp 115 o Branching decreases boiling point and polar functional groups that are most quotexposedquot will elevate boiling points to the greatest extent If we look at neopentane it is symmetrical so it can stack better it is highly quotbranchedquot so its boiling point is very low Which of those three amino isomers will have the himost boiling point and why NH2 N N H K nbutylamine methylamine NNdimethyl ethylarnine CHN MW a 7314 CHN MW 7314 CHN MW a 7314 If we follow the trends the first one will have the highest its unbranched a large number of carbons and hydrogen bonding the second highest will be NH then N on the right it is highly branched which leads to a low boiling point Factors that Effect Melting Points 1 9 Increases with size and weight Like boiling points the bigger the molecule the bigger the melting point This also includes if a CI vs N is on a 5carbon chain the Cl has a higher weight and will have a higher melting point Increase with bonds this is the same as with boiling point Ionic gt Hydrogen bondinggt diploe dipolegt Van der Waals dispersion forces These are influenced by the function group present Melting point decreases with branching Increases with better crystal packing if the molecules can fit nicely together they will resist melting Increases with higher rotational symmetry Rotational symmetry means you can turn the molecule and it still will look the same interactien nature ef attractien strength types at in nctien al greups galFinn he a rgmign m a very streng melecciies with charges such pesitiyeiy and negatively 35 quotENE SPEE39ES charged iena interactien at a streng an esamele the interactien Wampum melecuie with a ei water that has streng charge and a digeie with eesitiye and melecuie with a dipeie negatiye iens IiltE Ihlat er IEir39 wizang nab n special type at dieele medianquot eiceheis earlsesylie acids dipeie interactien quotIt and 2 amides thiels 1quotquot between an acidic and 2 amines any hydregen fair and a melecuie with a Iene pair and Iene pair a an acidic hydreagen dipeiediieele stireng melecuilar wealt heteines esters 12 dipeies generate siltyl halides any meiecule pa rtiaI charges with a dieele er a atrengly as and a peiarieed bend dispersien attractien ef wealt yery weal ailtenes elitenes hensenes Lendein tenses tandem dieeles termed en melecuilar rfaces larger the surface area the better shape and size matters eyery rnelecuile can term these but they are weal se it there are ether interacliens these interactiens will he deminantg Let s look at the following three examples for BOILING and MELTING points A E E Compound A has a strong polarized functional group an aldehyde This is creating a strong molecular dipolar partial and charges Thus it will form dipoledipole interactions This will have both the higher melting and boiling point B and C are alkanes B is a straight chain while C has lots of branching In both boiling and melting points branching decreases the change so B will have a greater melting and boiling point than C Let s look at a few other examples Molecule A has an ion or ionic Because of this if we look at the chart it has the strongest bonds so it will have the highest melting AND boiling point Molecule B has an OH group this hydrogen bonding and will have the second highest of the boiling and melting points Molecule C does have polar functional group but none ofthe oxygens bear an acidic proton so they only form weak dipoledipole interactions Factors that Effect Acidity 1 Charge Acidity increases with increasing positive charge on an atom 1 Acidity Increases with Increasing positive charge on an atom 6 H09 lt H20 lt H3O pKa 157 17 Increasing acidity C9 NH lt NH3 lt NH4 pKa 38 92 2 The role of the atom This can be confusing because acidity increases as we go across a ROW in the periodic table electronegativity plays a role in this However going down the periodic table increases acidity Think of this because as the molecule gets bigger there is size mismatch 2 Across the periodic table acidity increases with electronegativlty CH4 lt NH3 lt H20 lt HF pKa 50 38 157 32 Bearmegativity 25 30 34 40 but down the periodic table acidity increases with size HF lt HCI lt HBr lt Hi pKa 32 80 90 10 Electrmegativity 40 30 28 25 Ionic radius 133 181 196 220 picometres Also holds for oxygen versus eulttlr H20 lt H25 pKa 157 70 HaCOli lt H305H pKa 155 10 Resonance if the conjugate base is stable the acidity will increase 3 Resonance Remember any structural feature that increases the stability of the conjugate base will Increase acidity Key question can the lone pair of the conjugate base participate in resonance with an adjacent 1 bond Resonance will increase the dabih ty of the conjugate base th eretore in creas39ng aa oity because the negative diarge can be delocalized Examplel o contrast methanol versus phen pKa 15 lt 10 The anion of phenol can be stabilized through resonance more acidic Example 2 alcohols eg methanol versus carboxylic acids Ii e CH CH lt gt 3 H3O 0 chU oe Hacko pKa 15 lt 4 more acidic Question Which proton in red would you expect to be more acidic H H H 0 Mac rH ch CH3 H The answer to previous question would be the first molecule because it will have a resonance structure Inductive Effects electronegative atoms draw negative charges to themselves this leads to a stable conjugate base 4 Elcctrcttcgatittity an inductive effects T wci princictcc clccttcnwittttlrcwinc cuttctitucntc can ncrcccc acidity ct a nearby ctcm wlti t icmccccc with cicctrccccaticity an ticccccccc with icctcccing dictcnce tc the ctctc EIEE lr t IE atittilty incrcaccc irt the crcr F 2 El c Bar 1 c c c c w H t Ht cl F HEAR 1 HDMX HDJJV ft HU39JJX quotE HillUV H H H H H H H H H H pitta 435 31 5 SEE E l EEE c quot393 JL JL iItZi El IEEHEEHEEE li cHEcHE HD JLGHEEI pHa 459 at ENE As you can see when the Br in the second example is CLOSER to the oxygen the compound is more acidic because the Br can draw in the oxygen Orbitals spgtsp2gtsp3 As the orbital is bigger the compound is less acidic 5 Orbltelo The higher the echaractet of a bond to hydrogen the more acidic it will be 893 39quot SP 25 lt 33 lt 50 scharacter scharacter scharaoter H HqC Hs H H pKa 50 42 25 mdeashgaa39oi39ty Oueetlone What do you think about the acidity oi the following protons in red H R G 1 HNCH9R Nlt HNz R R 2 H 8 HO CH R O 39 2 69 R In Number 1 the first compound will have the lowest acidity followed by the second from the left the then the compound on the right Now for the second problem we need to think about charges the first compound will be more acidic because it bears a positive charge Nucleophile donate electrons Electrophiles receive electrons To find the number of moles an reaction will be at a certain point use the following equationthis can be used for Molarity as well x2 10 pKafpKab K 11 3002 35 Example pKa 151 H H N pKalOS h I2 N3 WM 1H H r H H O O 05 mole 500 mole 1 Given the number of moles of reactants underneath the reaction to the nearest tenth how many moles of the first reactant will be left at equilibrium a 01 b 02 c 03 d 04 e not ad We have the pKa forward and backward so 1039153911 39850E395K x2 5 x500 x From here you should plug in the x numbers and if it returns the K of 50E395 it is your number The answer ends up being A ioPKafPKab 5013 5 Partition eaef tients Tlhie it used in extratth and multiple extra ttiena Whe eq uatiena fer tlhie is I TF3 Detan fl elfwater partitien teieffient is the llgil t wklagp the leg P muat be between l and 5 tall he eptimal Example ef equatien given egPl l imlL 11 Ell mL r El initial material 1g Ea Iaigpr2l f1 k 13 quot1393 1 39 l iim How to calculate density by specific gravity 1 Find the molecular weight of the compound Divide by the total number of atom in the molecule 3 Divide by water s specific gravity 183 6 Then round to two significant figures Example C7H50 Iquot The molecular weight will be 12Carbon s weight times 7 16oxygen s weight time 1 E H 1hydrogen s weight x 6 This equals 106 now divide that by the total number of atoms in the molecule which is 14 This equals 757 now divide that by water s specific gravity which is 6 this gives a final density of 126 and rounded to 13 If the number is less than 1 the compound will float If the number is equal to 1 or close to 1 the compound will emulsify If the number is greater than 1 the compound will sink Addition reactions is a reaction where a molecule reacts with another molecule having one or more multiple covalent bonds so it forms a molecule whose molecular mass is the sum of the molecular masses of the reacting molecules Examples H H W on H2 Ir HEEH I In I I H H 39 Br HEEE H 2 BrE I HIllIllH H Elr In both of these reactions something was added the double bond goes away because of the addition Substitution reactions are reactions in which an atom or group of atoms in a molecule is replaced by another atom or group of atoms Examples 39739 39739 39739 39739 HooH Elr2 h HIIIIIIBr HEr In this reaction Br replaces the H 39739 39739 39739 39739 HIIIIIIDH HI i HIIIIIIl H20 H H In this reaction the OH is replaced by I If you lookjust because two atoms were removed it doesn t mean two must be replaced the Iodine was place to sub for the OH Elimination reactions are reactions where a multiple covalent bond is formed in a molecule by the removal of another usually smaller molecule Examples H H H H I I H3334 x I H39lll39lII39UH i39 IFEmEWIr H20 H H H H leads to its elimination as water In this reaction the acid pronates the OH group and this H1 IBr ICCx l HGEE H HEir H H In this reaction both the H and Br are eliminated to form HBr and the carbon goes from a single to triple bond Oxidation and Reduction Oiddotiono and roductiono O AHltH PCC g bond broken C H f 04 gt A H bond iormed C O oxidation state 0 C 1 oxidation state of C 1 oxidation state of carbon has become more positive by 2 this is an oxidation H H Cl h H H x I c r en C H CH gt Ca bond formed 00 739 2 oxidation state of C 43 oxidation state of C 1 oxidation state of carbon has become more posmve by 2 this is an oxidation 0 H c o 1 D39BAL39H C H bond broken co band formed CH 2 H20 oxidation state of C 3 oxidation state a C 1 Oxidalion Stale 039 cafbon has become more negative by 2 this is a reduction 1 in PdIC H i 2 H gt H bond broken 0 6 H bond Iormed CH 2 oxidation state of carbons have 4 2 each become more negative by 1 this is also a reduction Reduction 3 do nitiono Oxidation 3 de nitions Gain of electron Loss oi electron 39 Formation of a 0H bond Loss oi a 0H bond Loss of a 00 bond or its equivalent Formation of a 00 bond or its equivalent such as C N CCI etc So a reductive will result in a net increase in the number of CH bonds or a decrease in number of C0 bonds this applies for CCl CBr etc An oxidation will result in net decrease in the number of CH bonds or an increase in the number of C0 bonds or CCl CBr etc Anti Syn Exo and Endo Exo and Endo are only used with the methyl is on the two largest bridged Endo looks like As you can see there is a bridge formed when you line the bridgehead together and if the methyl is pointing downward in relation to the cup it is Endo Please note this only applies if the bridgehead s hydrogens are cis Exo Is up in relation to the bridge this again only is true if the bridgehead hydrogens are cis Syn and Anti are used when the methyl is on the smallest bridge this is usually on a bridge that is in the middle of the bridgehead Syn means the methyl points to the largest bridge Anti means the methyl points away from the largest bridge Cis and trans are used when the methyl is on the bridgehead Endo Rule Alder Rule As the pi bonds of the electron withdrawing group approach the C2 and C3 atoms of the diene the proximity of the electron withdrawing group to these carbons results in a secondary overlap between orbitals of the carbons and orbitals of the electron withdrawing group E or Z nomenclature Z and E is a general system to describe cis and trans relationships The EZ method of naming is based on the CahnIngold Prelog system of substituent priorities To determine whether E or Z assign each group bonded to doubled bonded carbons are priority number Assign groups to the left a 1 and 2 and right 1 and 2 Assign 1 to group with the atom having a higher atomic number if 1 1 Tl Ell HECEC CECH chk CEH EH HE H 3 E E Erutteme EE Imme og Bum Hamel Em As you can see in the above image 2 makes a quotboatquot and E looks more like a quotchairquot if you don t look at the hydrogens and just look at the methyls So if 1 groups are on the same side it is Z isomer if 1 groups are on opposite sides it is E isomer R or S F 8 Br 0 F 39 39 OH A a c OH R or S can be tricky but you don t have to redraw it every time to do it Instead leave the hydrogen where it is and still rank the others You will end up either going counterclockwise or clockwise This table will help position of Direction of 4m ranked substituent 1st 2nd 3rd ranked Assignment substituents Back cw 9 equivalent drawmg e BaCk cow 8 H39 Me Front CW 8 HO From COW R clockwise So if you are looking at the molecule backwards if you redrew to where the fourth ranked was in the back it would be opposite of what you originally thought
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