Study Guide For Exam 3
Study Guide For Exam 3 CHEM 2321
Popular in Organic Chemistry I
verified elite notetaker
Popular in Chemistry
This 10 page Study Guide was uploaded by Hayley Lecker on Thursday November 12, 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 226 views. For similar materials see Organic Chemistry I in Chemistry at University of Texas at El Paso.
Reviews for Study Guide For Exam 3
The glitch where the document didn't show was fixed, apparently Study Soup had an error. This document is going to be a life saver! The thorough detail is amazing.
Something went wrong notification popped up.
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 11/12/15
CDMMDM FUNCTIlGl llAL GHDUPS Organic Chemistry Exam 3 Study Guide Important Information Professor s Email isalutepedu Class Website organicutepeducourses2324 Functional Groups This will important on the exam since you need to what reacts with what and how to identify them Tipe m autumn Wallalulu Elimp e Funmm mp T5111 11 nu Bulimia mu Enid i 31 m FIFE WEE nal 3 13 Eli Almhul Fl19H llgll 43H minimise A E 42 Wm M and H EH maturitygm HES lii 11 I l li uo g n 3 H m p EEtH Heart Emil Fl llHIE FL ll lgEIlg Em Fl lFl Churn mg my 33 If 05 a v s r x t 130 Hikene g z a Hf EH demurehm Haema HJLH BM gum Ia int i EHl quhide rm m5 Ir Mb I lmin H tH urnLE tH rm E Hit Hi triplel3th Sul de H S H mfg EH EH 2 II I Hi r r EH MlIE H2 WWW H ENE FEMlam Ellllg Il39slIEIE ailii nineser ner Hawvi H EH EH E E Eh So let s practice this From Practice Exam 3 2015 15 Match each functional group centered on a given numbered carbon in the structure below to a best functional group name on the right Answers may be repeated 1 C1 e a aldehyde 2 C2 d b enol ether 3 C3 a c ester 4 C4 c d ketal 5 C5 b e ketone Okay looking at the first molecule we can see Cl has only a double bonded oxygen to that that can be either a ketone or an aldehyde but from the chart above we know it is a ketone because it is not at the end of a carbon change Once you get to C5 the only choose left over is enol ether which looks like RC o R R2 R3 6lO Match each functional group centered on a given numbered carbon in the structure below to a best functional group name on the right Answers may be repeated Interestingly this compounds looks a lot like a common sedative 3 ngH 6 C6 c a nitrile N C7 8CQ 7 C7 e b imine N N 8 C8 d c enam1ne HClg 9 C9 a d cyanohydrin HO C10 b e amide Now onto a nitrogencarbon based molecule The first one that is easily identified is the nitrile at C9 because nitriles are the only functional group with a carbonnitrogen triple bond just with nitrogen and nothing less Now let s look at the rest of the molecule well there is a double bond carbonoxygen at C7 that is a characteristic of amide s so that is taken off Now the let me post images of what imine cyanohydrin and enamines look like lmine Cyanohydrin Enamine 3 NR R5NR4 R 1J 2 1 R3 R R R2 So all three have different characteristics we can see C6 looks like an enamine and the C8 has the characteristics of a cyanohydrin so that less C10 as an imine Now it s important to know mechanism s because that will be the whole test basically Let s first look at figuring out mechanisms for this I will use two practice tests 2014 and 2015 for exam 3 to give more coverage ll15 Order the reactants intermediates and final product of the reaction of methyl ethanoate and excess Grignard reagent of iodomethane Counter ions byproducts nonbonding electrons and charges are not shown So this is between an ester and a Grignard reagent so CMgX is a Grignard reagent So first find the ester if it is available to see That would be D you can see the ester and the Grignard reagent it doesn t have the MgI but that isn t important for the nal product Now let s find the nal product I like to nd the first and last because it makes it easier to fill in Grignard reagents with the general structure RMgBr are great nucleophiles They add to ketones aldehydes esters twice acid halides twice epoxides and a number of other carbonyl containing compounds In this reaction we are using an ester so it will reaction twice and not need a hydronium work up The general reaction is 0 lil39 MQE II Fquot So an alcohol will be formed which is e Now let s think about the rst so we know the first d is the starting material now let s look at the others since we know it not e it must be either abor c If we look at a that doesn t seem right because one oxygen is gone and a ketone formed so lets look at b B is the correct answer because the Grignard reagent comes in and attacks the oxygen ends with a negative charge this goes to A because the OCH3 group leaves forming a ketone Because the ketone is more reactive than the ester it can rereact to make an alcohol so it goes to C and finally E 11 d 12 b 13 a 14 C 15 6 CH3 CH3 CH3 0 CH3 n CH3 0O CH3 CH3O CHsJ l CH3 a O b CH3 c CH3 d CH3 O e OH CH3CH3 CH3 2327 Order the intermediates in the conversion of methanal to an acetal For simplicity one step was skipped Charges and the starting materials and products are not shown Let me first put the mechanism for methanol to an acetal H H DfH 39 H 39 39 H 39339 CHg H I H C H quot H c H H H H 0 H CH3 H DH CH T 3 hemiacetal H EI H HI H acetau DE 0 CH3 HI H3cng HageH H H H I H H H H C H I Uh Uh FID 0 an 393 CH3 CH3 2 c39 CH3 EC So as you read in the instructions there is no ending or starting material so we have look at what could be the second step and could be the before the end step So looking at that general mechanism we can see C look like before protonation to make an acetal the extra is what comes off which is water Now lets find the starting second step well B look like what is in the general mechanism with two alcohol groups also in the picture now to figure out the intermediates After B either d or e looks like it could come next but e makes more since one oxygen would come in and protonate the carboncation Now for the third step A would make since because the porton would shift so water could leave in step D as a leaVing group that makes is so the final alcohol could come in and attack which lead to C 23 B 24 E 25 A 26 D 27C OH H H O I OH CI H O a H b H c H d H e OH H OACP H Now on to the mechanisms I am going to include Dr Salvador s tables on here because they can be difficult to understand at first but once you use them they become very helpful They are turned sideway for quick printing in the next three and third pages Table ef erhexylie Aei Derivatives and Alth Substitution Diegenell Cempelulm Tuequetiee D Dtegem Reagent New Diagonal Ceimpeundy Reed Cleekwise Fer example A C l h l t Acid Thienyl Clitmtide All Aeid er Atty Halide and Anhydrtdte Atmth An Ester Alcohol Acid C t l t t Etieetlet Amine Heat m Beee er Acid Cate at Heat Diemmethene Amine DEC ml EIE tlquot hilie A i en Thiem l Elmi ride Pyridine CetalA eie Phee here113 T hrumide Almnmlitmq HIth Aeidq Water quotHelen l Cl mide heat Acid Water Beee Acid Water Ease Se ifl f M39C39 39h39 l or Water Base Alk xide twilth Ease gluing M39C39 39h39 l err llemtide er Cerbeevlate Salt Tame If C edlugc mn nf CHI h lj liit Arid Derivatives Additim and S uh hlti a m li 13th 112331 Fangs quot ipmam tim r 1f aw machamigm if E h ma a11 Net Substitm uns Mai 4 Emma 3 Kamila 2 Midade 1 mcnhn1 131 11 C fh f uid 2 vszi Li l 1EE Earh yjlat Salt R m f qh yh l LiA I a 141239 C j lh m i f fj yh l Liam 314 EMg39K ma R39Li CH2CHCHHH I LIEUIa J17 0139 H7 tranaitinn meta catalysis Li L I 1339 m rm1t f m I2CHCH MH ER39MEH39X 3 Halide E3CM Li Lmr39 i a LEE 1a 1Q C cyanideevanehvd nes 31110 lithium er alk maEneeimm Griguard l eldelwdes rge te Seeendan aleelilels i ketenes ge t0 er ianf m i earhen diexide geee t0 earhexvlie aei e H intennediate efmah39 g ijme delive vee is ealledl e hemiamj39nallg geminal amine aleelml er earhii nl amine emmeuia ijmines er Shi quot base primer amj39nespiminesz hirdrenrlamine eximes hirdragiue hvdrammes seeminder eminreE iminiumsa enemines water i aldehydes fever n rate germinal M i metDues fava kete fem 111E equivalent efe alienhula i alIilelilydes gen 10 hemiaeetals 139 metDues Eat to hemiketals twe equ veleute ef e printerF mleehel acid eatellyeie l aldehytles gen In aee alls l metDues gun to ketals prawn I eIImiiin inn i emu beee enellate alleelhu and acid 1139 These eatellyeie i ellell ether H 53 hydride two equivalents of 3 11113111313 hiall with uid 1 aldehvdes 39 o to gammygi primary 311311015 with NHEI LL our 3 ldElllF EE go to illliinl i hammers In to t ms seconds1W alcohols with k g 1 him NaEILL or warm twin 15 l imi girl to amines with NaEfCMI h and So let s look at problems from Exam 3 2015 Match each set of reagents to a major product on the right Answers may be repeated Assume any necessary workup between steps Now for these problems you need to use salvador s first table I included If you see the trend everything is on there except 20 which isn t on the table 16 butanoic acid thionyl chloride pyridine catalysis followed by methanol d a butanamide 17 sodium butanoate hydrochloric acid followed by methanol acid catalysis d b butanenitrile 18 methyl butanoate ammonia followed by thionyl chloride heat b c butanoyl chloride 19 butanamide water acid catalysis followed by thionyl chloride pyridine catalysis c d methyl butanoate 20 butanoic acid heat followed by excess ammonia e e not ad Match each set of reagents to a major product below Answers are repeated Assume any necessary workup between steps Extra space is provided so that you can draw your intermediates Now these problems deal with both substitution and reduction So the rst problem is a carboxylic acid diazomethane that would turn it into an ester now looking at the second part that s a Grignard reagent and as I said above grignard s reaction with ester twice first to make a ketone and secondly to make an alcohol Girgnard s make ester s into tertiary alcohols the only tertiary alcohol is D so one down 21 ethanoic acid diazomethane followed by excess methyl magnesium chloride d 22 butanoic acid dimethylamine dicyclohexylcarbodiimide followed by diisobutylaluminum hydride followed by sodium borohydride For this problem you will need three tables rst is substitution An amine and DCC was added so the carboxylic acid turned to an amide now looking at the reaction table we know that amides with diisobutylaluminum CH32CHCH22A1H turns to an aldehyde now looking at the final table we go under H since sodium borohydride is a hydride and it says that aldehydes go to primary alcohols There is TWO primary alcohol in the answers however the methyl would add to the end of chain not in the middle so it is A 23 24 25 26 27 propanoic acid sodium hydroxide followed by methyllithium followed by sodium borohydride Now this one is a little easier it is 3 charts but simple steps A carboxylic acid base NaOH is a base gives carboxylic salt this is basically an ionic bond between the Na and the last OH turns to 0 Now a reduction occurs which turns it into a ketone Now let s look at the third table for a hydride with a ketone the table says that ketones turn to secondary alcohols the only secondary alcohol is C so that was easy C ethanoic acid thionyl chloride pyridine catalysis followed by tritert butoxyaluminum hydride followed by ethyl magnesium bromide So now an carboxylic acid with those two things turns into a acyl halide so we have step one So now we have to figure out what happens with tri tertbutooxyalumnium hydride look to the right So the thing to know about acyl halides and that compound is it will reduce acid chlorides to aldehydes and stop there So we have an aldehyde now so now it is going to reaction with methyl magnesium bromide a Grignard reagent so look at the third table it says that aldehydes go to SECONDARY ALCOHOLS well there is only one which is C Lithium trLwatoxy alunmlinum hydride 2methylpropanoic thionyl chloride pyridine catalysis followed by sodium 2methylpropanoate followed by lithium aluminum hydride So this first would turn into an acyl halide and then it is followed by a carboxylate salt This would turn it into an anhydride this would happen because the salt part of the carboxylate salt would draw off the Cl and the negatively charged oxygen would come in for the attach creating the anhydride Now a reduction occurs anhydride will turn to an alcohol and a carboxylic acid we are interested in the alcohol if you draw out the structure it will be a propane with a methyl then an OH group at the end which is B propanoic acid ammonia heat followed by thionyl chloride heat followed by methyl magnesium chloride followed by water acid catalysis followed by sodium borohydride A carboxylic acid NH3 and heat will give an amide now we are doing a SECOND substitution which turns it into a nitrile The Grignard reagent with water and acid if you saw the mechanism above turns this into a ketone In the last step we know ketones go to secondary alcohols from the third page table The only secondary alcohol is C butanoic acid ammonia followed by heat followed by lithium aluminum hydride Now at rst you may think this goes to an amide but no heat is added so it goes a carboxylic salt With LiAlH3 Carboxylic salts make PRIMARY alcohols This would end up in A because the lithium would attack the oxygen double bond and give a carbocation an hydrogen That would end in A Alcohols we count the number of carbons attached to the carbon condemn the OH OH OH OH H o 5 63 39L a 3 I 3 A c d OH 0 carbons 1 carbon 2 carbons 3 carbons e not ad Wale39 Primary 391 39 Secmdary l l Ternary 3 l alcohol alcohol alcohol
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'