Econ 101 - Week 1 lecture notes
Econ 101 - Week 1 lecture notes 101
Popular in Economics
verified elite notetaker
One Day of Notes
verified elite notetaker
One Day of Notes
verified elite notetaker
verified elite notetaker
One Day of Notes
verified elite notetaker
verified elite notetaker
Popular in Economcs
This 10 page Class Notes was uploaded by Sieva Kozinsky on Thursday October 30, 2014. The Class Notes belongs to 101 at University of California Santa Barbara taught by Dr. Pavlov in Fall2014. Since its upload, it has received 51 views. For similar materials see Economics in Economcs at University of California Santa Barbara.
Reviews for Econ 101 - Week 1 lecture notes
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: 10/30/14
Prep101 http wwwprep101com mcat Passage I Questions 14 Cow s milk is a nutritionally complete food source which contains water vitamins minerals proteins carbohydrates and lipids An important protein in milk is casein This typically exists in milk as its calcium salt calcium caseinate Calcium caseinate is soluble at the normal pH of milk 66 but it becomes an insoluble form of casein in solutions that have a lower pH of 46 the isoelectric point of calcium caseinate Milk contains two other important types of proteins which can be denatured and precipitated by heat namely lactalbumins and lactoglobulins Lactose shown below is the main carbohydrate in milk It is a disaccharide which can be hydrolyzed with acid to give two monosaccharides galactose and glucose In the body the enzyme lactase digests lactose to galactose and glucose Adding a dilute solution of acetic acid to nonfat milk is the rst step in isolating several of the components of milk The solid material that precipitates Solid 1 is removed Care must be taken to avoid adding an excess of acetic acid The remaining acidic solution is then treated with powdered CaCO3 and heated to boiling after which time another solid Solid 2 forms Solid 2 is removed by ltration and hot ethanol is added to the aqueous ltrate The solution is Page 1 of 10 allowed to cool overnight during which time lactose crystallizes in the ask l A The passage states that an insoluble form of casein forms as the pH decreases from 66 to 46 Sour milk is acidic 2 A The passage describes three speci c types of compounds 1 Calcium caseinate which precipitates at low pH as casein 2 Proteins lactalbumins and lactoglobulins which precipitate when denatured by heat and 3 Lactose which hydrolyzes to galactose and glucose In the method mentioned in the passage the powdered calcium carbonate is probably added to neutralize any remaining acetic acid Solid 1 casein forms on the addition of acetic acid and is removed by filtration so it is not present when the CaCO3 is added Solid 2 denatured and precipitated proteinaceous material is insoluble in the CaCO3 3 C H20 From the equation it can be seen that the hydrolysis of lactose results in the net addition of H20 during the cleavage of the disaccharide 4 C Solid 2 forms after the solution is heated to boiling a condition that denatures and precipitates the lactalbumins and lactoglobulins Prep101 http wwwprep101com mcat Passage 11 Questions 59 Carvone shown below can exist as one of two enantiomers One of them carvone is responsible for the fragrance of spearmint oil and the other carvone provides the characteristic odor in caraway seed oil carvone A chemist attempted to isolate carvone from caraway seed oil which consists almost entirely of carvone and limonene shown below limonene Because the two compounds have different boiling points vacuum fractional distillation was chosen as the method of separation The apparatus in Figure l was assembled and a sample of caraway seed oil was placed in the distillation ask A bubbler or ebuliator was lowered into the distillation ask to introduce small air bubbles into the system The fractionating column and distillation head were wrapped with glass wool and the apparatus was connected with thick tubing to a vacuum source The contents of the distillation ask were heated Page 2 of 10 and two fractions were collected in the receiving ask 5 6 7 thermometer an in tiistillantion healli wantEr out 2 air control mmmgr K to 3939auun1 water in fraIiiunatiirig column Jr REEEi i g flask ii5 ilavtion headi Figure 1 Vacuum distillation apparatus The carbonyl group CO in carvone makes it more polar and it thus has a higher boiling point due to nonbonding interactions than the hydrocarbon limonene which contains no polar groups C The passage mentions that the bubbler ebuliator was added to introduce small air bubbles into the system This is the same function provided by a boiling chip at atmospheric pressure The air bubbles prevent superheating and bumping A The separation of limonene and carvone can be improved by modifying either the apparatus shown in Figure l or the method The separation of the two liquids takes place in the fractionating column as the two liquids vaporize and condense the lower boiling liquid distilling rst If the fractionating column is shortened Answer D the liquids will vaporize and condense fewer times ie there will 8 9 Prep101 be fewer theoretical plates and the level of separation will not be as effective Cooling the condenser with ice water Answer C will have no effect on the degree of separation because condensation takes place after the separation has occurred Creating a lower pressure inside the distilling apparatus Answer B will lower the boiling points of both liquids and reduce the differences in boiling point making it more dif cult to effect a separation of the two compounds Heating the distillation ask at a slower rate Answer A will allow both liquids more time in the fractionating column increase the number of theoretical plates allowing liquid and vapor to equilibrate and therefore effect a greater degree of separation of the two compounds B Carbon 5 is the only stereogenic carbon atom chiral center carbon attached to 4 different functional groups carbons 2 and 7 are not stereocenters A The boiling point of a liquid is the temperature at which the vapour pressure of the liquid equals the surface pressure The vapor pressure of a liquid increases with increasing temperature Hence the boiling point of a liquid decreases as the pressure on the surface of the liquid is decreased with applied vacuum If a leak develops in the apparatus the vacuum will decrease and the surface pressure will increase as will the boiling points of both liquids Page 3 of 10 http wwwprep101com mcat Passage III Questions 1014 When Compound 1 is heated with NaOEt Compounds 2 and 3 are formed Reaction 1 j Compound 2 maj or Br heat Compound 1 H 0C2Hs Compound 3 minor Two mechanisms were proposed for Reaction 1 Mechanism A HBr is eliminated from Compound 1 to form a symmetrical vinyl carbene Intermediate A which then rearranges to form Compound 2 Mechanism B Ethoxide abstracts a proton from Compound 1 to produce a carbanion Intermediate B This then rearranges with loss of bromide to form Compound 2 Prep101 http wwwprep101com mcat 10 C If a carbene was the intermediate intermediate A then Compound 4 and Compound 5 would form identical intermediates that would go on to yield 3 equal amounts of compounds 6 and 7 C097 O Br v H 11 B As seen in Reaction 1 OCHZCH3 Interrred1ateB C replaces Br in a substitution reaction Br l 12 A By either mechanism ethoxide abstracts a proton to form ethanol EtOH and a bromide ion is liberated The Nafrom NaOEt joins with the Brquot to form NaBr To distinguish between the 2 mechanisms an isotopic labeling experiment was designed 2 compounds Compounds 4 and 5 were labeled 13 D Neither Compound 2 nor Compound with carbonl4 MC and then each was treated 6 is optically active therefore the with potassium tbutoxide see Scheme 1 speci c rotation of both compounds is Because Compounds 4 and 5 formed Varying zero ratios of Compounds 6 and 7 Mechanism A was ruled out 14 A Compound 1 contains only spz hybridized carbon atoms Compound 2 only contains sp and spz hybridized carbon atoms The carbon carbon double bond that is not part of the two benzene rings in Compound 1 contains spz hybrid orbitals This double bond undergoes transformation in Reaction 1 resulting in a carbon carbon triple bond in Compound 2 The carbons of the carbon carbon triple bond contains sp hybrid orbitals Scheme 1 denotes C14 label Page 4 of 10 Prep101 http wwwprep101com mcat Passage IV Questions 1520 As part of an experiment a starting material was treated with NaOH in acetone solvent CH32CO however the starting material was recovered unreacted Instead a small amount of Product A shown below was isolated O OH Product A It was determined that Product A resulted from the aldol selfcondensation of acetone Product A was identi ed based on the following observations Observations about Product A Product A had a molecular weight of 116 Elemental analysis of Product A showed only carbon hydrogen and oxygen in the molecule 3 Signals in the infrared spectrum of Product A included a broad band at 3400 cm391 and an intense signal at 1720 cm391 4 Product A was a methyl ketone as it gave a positive iodoform test 5 When Product A was treated with Br in CCl4 the red bromine colour persisted because no carboncarbon double bonds were present to react with the bromine NC The structure of Product A was further confirmed when treatment with hot sulfuric acid resulted in the corresponding dehydration product Product B Page 5 of 10 15 B The question states that the product of the aldol selfcondensation is a 3 hydroxy ketone meaning that the condensation is not complete in that the product has not lost water Note An aldol addition reaction forms a 3 hydroxycarbonyl compound whereas an aldol condensation reaction implies that the addition product has lost water Because the addition product is a dimer of the starting compound the product has a molar mass that is twice that of the starting compound Thus the molecular weight of the starting compound is 1442 72 16 C This question involves a reaction at equilibrium Thus Le Chatelier39s principle is in effect and the position of equilibrium can be shifted toward Product A by removing it as it forms Catalysts Answers A and B do not affect the position of equilibrium and heating acetone to the boiling point Answer D would remove reactant and shift the equilibrium to the left 17 D is a constitutional isomer of Product A the other compounds are not 18 B Product B contains a double bond because it results from the dehydration loss of a molecule of water of Product A Thus the addition of a drop of Br2CCl4 to Product B will result in the red color of bromine disappearing as bromine adds across the double bond 19 A The six hydrogen atoms in acetone are magnetically equivalent and their 1H NMR signal appears as a singlet near 8 2 ppm Prep101 20 D All three compounds are methyl ketones The passage states that Product A gives a positive iodoform test so we can deduce that all three compounds give a positive iodoform test Passage V Questions 2125 A chemist wishes to make the products shown below by the following synthetic schemes Synthesis 1 NaNH NH CSHI IBT j gt A T B Lindlar39s Catalyst 4 C HBr ether Br Syn1hesis2 Brz CH2CH2 KMnO4 E D A E T FeBr3 H Br 21 D Acetylene has an acidic H atom which can be removed by the NaNH2 base to give acetylide anion 22 B Acetylide anion performs an SN2 attack on the alkylhalide to give 5methylheptlyne Looking at the final product gives the structure of the alkyl component 23 B Lindlars catalyst reduces the alkyne to an alkene Page 6 of 10 http wwwprep101com mcat 24 C The conditions employed effect an electrophilic aromatic substitution reaction a bromination of benzene 25 A The conditions employed create a carbocation from H addition to ethene The carbocation then effects an EAS reaction in a similar fashion to a Friedel Crafts alkylation at a position para to the bromine atom bromine is an op director Passage VI Questions 26 29 Three different methods of preparing alcohols from alkenes are shown below H H20 V Equation 1 Equation 1 shows the acid catalysed hydration of an alkene The reaction is regioselective and not stereoselective and rearragements may occur Stepll Equalion2 Equation 2 shows the oxymercurationdemercuration reaction This is a two step reaction and the final alcohol is formed via the oxymercurial alcohols shown after step I The mercury is then removed in the second step by NaBH4 Prep101 http wwwprep101com mcat to give the alcohol product Overall the reaction is regioselective and the rst step is also stereoselective and no molecular rearrangements 0 C C111 i B12 TI39FH I Egg uBh T H OI Sm iquotquotquot SBPH O E 3 39393939I Equation 3 shows a hydroboration reaction followed by a second step to remove the boron species to give the nal alcohol product The overall reaction is regioselective and is also stereoselective in both steps Molecular rearrangements do not take place 26 A The addition of H to the alkene attack of H by the alkene results in one end of the alkene bearing a positive charge This carbocation can rearrange under certain circumstances when a more stable carbocation can be formed from a rearrangement 27 C The alcohol shown in the scheme above comes from antiMarkownikov addition of water across the double bond via the organoborane species followed by oxidation by peroxide as shown in equation I above The other two equations describe Markownikov addition of water to an alkene 28 B NaBH4 in step II of equation 2 contains mostly H atoms not oxygen atoms Its name is sodium borohydride and it is a supplier of hydride ion Reduction is the addition of hydrogen to a molecule H202 in step II of equation 3 is an oxidizing agent it is hydrogen peroxide it supplies oxygen In this case the oxygen atoms bond to the molecule at the position where the boron atom was therefore oxygen is thus added to the molecule ie an oxidation Page 7 of 10 29 D All alcohols show a broad band OH stretch in the IR spectrum from 3300 3500 cm391 Passage VII Questions 30 33 A chemist was attempting to synthesise compound B from iodide A using MeOH as the nucleophile but the reaction was unsuccessful Instead of obtaining the desired product B an unknown molecule C was obtained Molecule C gave 4 signals in the 1H NMR spectrum one singlet integrating to 6H atoms one singlet integrating to 3H atoms a quartet integrating to 2H atoms and a triplet integrating to 3H atoms C is the product of a reaction mechanism involving a molecular rearrangement step y I y C2339i e L E 30 A OMe A is the only molecule which fits with the NMR data and also the fact that a molecular rearrangement has taken place The molecular rearrangement takes place because C is formed by an SNI reaction rather than the SN2 reaction desired by the chemist The SNI reaction gives a primary carbocation intermediate which then rearranges to a more stable tertiary carbocation prior to attack by methoxide nucleophile giving compound C Prep101 http wwwprep101com mcat 31 B MeONa is the only MeO39 nucleophile in the list This nucleophile is much stronger than MeOH and is more likely to give the desired SN2 reaction to product B rather than the SNl reaction to product C 32 A Given the reaction conditions and that no promoting conditions for radical reactions are apparent answer A is the only sensible choice as carbocations are able to undergo rearrangements 33 C An SN2 reaction inverts the stereochemistry at the carbon that the nucleophile attacks and an SNI reaction destroys the chirality at the carbon containing the leaving group leading to the planar carbocation intermediate Passage VIII Questions 34 37 The rate of the baseinduced dehydrohalogenation reaction of cis l bromo4tertbutylcyclohexane is proportional to the concentration of both the bromide and the base however that of the trans isomer is proportional only to the concentration of the bromide 34 CCH33 BI C 35 A E2 Because the rate is proportional to both the concentration of both the bromide and the base and an elimination dehydrohalogenation takes place reaction must proceed by an E2 mechanism Page 8 of 10 36 B 37 A because rate is only proportional to the concentration of bromide reaction must proceed by an El mechanism Passage IX Questions 38 42 Chemists investigating the antibiotic agent penicillin G wished to prepare a number of analogues which might solve the problem of bacterial resistance to this class of drugs A method of production was required which was efficient and cheap Chemists decided to start from 6APA and following the scheme outlined in Reaction 1 they synthesized a number of analogues with a variety of groups at R H o X Fl iisma 0 Wiism Kgtlt1 Qlt Gmitqniallaicaid R11 min and egg GAIN Rmmi In order to obtain large enough amounts of 6APA for this synthetic scheme chemists used Penicillin G in an enzymatic cleavage reaction to form 6APA reaction 2 H I Me Penicillin acylase I ab 6APA O 0 J Penic1l1inG Read139on2 38 39 40 41 42 Prep101 C II is the only reaction above which will give an acid chloride B Friedel Crafts alkylation utilizes acid chlorides in reaction with an aromatic system and a Lewis acid catalyst to give an acylated aromatic product Alcohols in reaction with an acid chloride give esters Amines in reaction with an acid chloride give amides C The result of hydrolysis of an amide is a carboxylic acid C and an amine 6APA also has an amine group A This is the only route that will form a dipeptide D 4 Say we used Gly and Ala We would obtain GlyGly GlyAla AlaGly and AlaAla Page 9 of 10 http wwwprep101com mcat Passage X Questions 43 47 Salbutamol is an important molecule used for the treatment of asthma It is a B2 agonist and relaxes bronchial smooth muscle resulting in dilation of the airways In the laboratory Salbutamol can be synthesized according to the reactions shown in Scheme 1 M 1 E 32 CM 0 Bmirriim Oh we 0 PlHZIHB1 FHQCNtB1 042 N I quotJH 111 0 o4 CH2 P PdC CH2 l I quotb ta VI m a V Sdmcl Salbtlani 43 D The reaction of an alcohol with a carboxylic acid under acidic conditions HCl is an acid catalyzed esteri cation 44 D The PhCH2NHtBu is a nucleophilic amine and it is reacting as a nucleophile with a primary halide Primary halides most always react with nucleophiles in an SN2 reaction 45 B LiAlH4 is a reducing agent Lithium aluminum hydride supplies Hquot hydride ion which acts as a nucleophile at electrophilic centres Overall the reaction is addition of H2 to a molecule which is a definition of reduction Prep101 46 B LiAlD4 supplies Dquot ion as a nucleophile which attacks electrophilic carbonyl centres It is not responsible for the alcohol H atom which comes from either water or a protic solvent at the end of the reaction 47 B This is a catalytic hydrogenation and PdC is the catalyst Passage XI Questions 48 52 Non steroidal antiin ammatory agents N SAIDs are a successful class of compounds used for the treatment of pain and in ammation in a range of disorders and are especially useful in the treatment of arthritis One such drug is Ibuprofen In the laboratory chemists can prepare the racemic ibuprofen compound by many different routes One such synthetic route is shown in Scheme 1 below 48 B The reaction is indeed a Friedel Crafts acylation and the AlCl3 is a Lewis acid catalyst and so is usually present in less than equimolar amounts Page 10 of 10 http wwwprep101com mcat 49 C The reaction is a Friedel Crafts acylation which is an electrophilic aromatic substitution reaction The alkyl group of molecule 1 is an orthopara director 50 B The reaction is a reduction NaBH4 is sodium borohydride and Hquot is the reducing species 51 A SN2 reactions proceed with inversion 52 of con guration at chiral centres Walden inversion and primary substrates are better than secondary or tertiary C The conversion of a nitrile to a carboxylic acid is achieved by an acid catalyzed hydrolysis reaction
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'