Organic Chemistry Part II Sections V-VIII Section V Carbonyls and Alcohols Section VI Carbohydrates Section VII Nitrogen Compounds Section VIII Organic Chemistry Laboratory Techniques BERKELEY Specializing in MCAT Preparation ERKELEY ® E • V • I • E • W P.O Box 40140, Berkeley, California 94704-0140 Phone: (800) 622-8827 Internet: NCATprep@berkeleyreview.com (80 0) M CAT-TBR http://www.berkeleyreview.com The Berkeley Review and The Berkeley Review logo are registered trademarks of The Berkeley Review This publication for The Berkeley Review® was written, edited, and composed on a desktop publishing system using Apple Macintosh® computers and Microsoft® Word Pages were created on theApple LaserWrite® Pro Line art was created using numerous graphics programs designed tor use on Macintosh computers The majority of the text type and display type was set in Times Roman and Palatine Cover Design by MacGraphics Copyright © 2012, 2010, 2007, 2005, 2004, 2002, 2000, 1995, 1994, 1993, 1992 by The Berkeley Review All rights reserved No partof this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the copyright owner Oxygen Containing Compounds a) b) c) d) Section V Alcohol Properties Alcohol Reactivity Alcohol Spectroscopy Aldehyde and Ketone Properties e) Aldehyde and Ketone Reactivity f) Aldehyde and Ketone Spectroscopy Carbonyls g) Ketals and Acetals and i Protecting Groups h) Carboxylic Acids and Derivatives i Carboxylic Acids Alcohols ii iii by Todd Bennett Esters Lactones iv Acid Anhydrides v O Off RX XCH3 vi /C\0 R Carbonyl Reactivity CH2 a) Attack at Carbonyl Carbon O O b) Deprotonation of a-Protons C R R CH2 +1 CH2I O 2more times Off ^5~" a) Qrignard Reaction || /C\ CI3 R O c) Oxidation-Reduction Reactions Name Reactions O /C\'CH2I R' Acid Halides Amides b) Aldol Condensation c) Claisen Condensation d) Transesterification O e) Wittig Reaction Pinacol Rearrangement g Iodoform Reaction h) Wolff-Kishner Reaction f) Off R O o C R /c\ OH + cl3 + CI3 ^f=^ OH C R Iodoform Test Synthetic Logic + HCI3 O yeUowoil a) b) c) d) Reactions of Acetoacetic Ester Reactions of Malonic Ester Decarboxylation Protecting Groups Carbonyl Biochemistry a) Biological Oxidation-Reduction b) Biochemical Reagents BERKELEY Ur-E-V.KE'W® Specializing in MCAT Preparation Carbonyls & Alcohols Section Goals Recognize the carbonyl functional groups and types of compounds °!> especially ketones and aldehydes You must know which compounds are most reactive towards You must be able to recognize functional groups such as amides, anhydrides, acid halides, and substitution (which isbasedon the leaving group strength), and mostelectrophilic (which is based on the electron withdrawing or donating capacity of the functional group) Be able to identify infrared peaks for carbonyl compounds Carbonyl compounds will have apeak in the infrared spectrum in the area of 1700± cm'1 This will mostlikely be useful whencomparing twocarbonyl compounds, or identifying an unknown carbonyl compound You may wish toknow roughly where esters, aldehydes, and ketones fall inthe IR range © Be able to identify common name reactions involving carbonyl compounds * You must recognize common name reactions from carbonyl chemistry Included inthis group should be the Aldolcondensation, Grienardreaction, Wittie reaction, and the Claisen reaction TheAldol and Claisen reactions have biological significance, because they play a role in select biochemical pathways such as glycolysis and beta oxidation Be able to recognize the acidity of alpha carbons The carbon alpha to thecarbonyl can bedeprotonated, ifit hasa proton bonded to it The pKa ofa standard ketone is the range 1/± Once deprotonated, an enolate is formed The enolate has an equilibrium of its own with the enol structure The conversion of a ketone into an enol is referred to as tautomerization * Be able to identify ketals, hemiketals, acetals and hemiacetals Although current accepted nomenclature does notdistinguish between ketals andacetals, youshould be aware of the functional group Acetals and ketals are best described as "double ethers." They play a major role in sugar chemistry and protecting groups in carbonyl synthesis Understand the difference between thermodynamic and kinetic enolates Thethermodynamic enolate is formed under conditions ofhighertemperature wherethe pathway of greateractivation energymaybe chosen Thethermodynamic enolateis the moresubstitutedand thus more stable intermediate which will lead tothe more stable final product The kinetic enolate is formed under conditions oflower temperature and greater steric hindrance where the pathway of lower activation energy must be chosen The kinetic enolate is the less substituted and thus less stable intermediate which will lead to the less stablefinal product Know the mechanisms for acidic and basic carbonyl reactions The mechanism for transesterification ispresent inbiochemistry and organic chemistry, soit important that you recognize the steps Also recognize what catalyst is necessary to carry out the process © Recognize common oxidizing and reducing agents n and/or the gain of bonds to hydrogen a nutshell, oxidation is defined as thegainofbondsto oxygen and/or the lossofbonds to hydrogen Oxidizing agentsincludeKMn04and K2Cr207 Reduction is definedas the lossofbonds to oxygen Reducing agents include LiAlH4 and NaBI-14 Know common reactions by both name and reagents The AAMC guide lists a series ofreactions that they expect you toknow byname Iris a eood idea tonot only know the general reaction, butalso the mechanism and reaction conditions Highlights of this list include the Aldol reaction, the Grignard reaction, the Witting reaction, the iodoform reaction, transesterification, and the Wolff-Kishner reduction Organic Chemistry Carbonyls and Alcohols Introduction Carbonyls and Alcohols The carbon-oxygen bond is a major part oforganic and biological chemistry A significant part of organic chemistry on the MCAT involves compounds that contain carbon-oxygen bonds In the case of carbonyl compounds, the carbonoxygen 7C-bond is easily broken to form new bonds to the carbonyl carbon and subsequently form a new compound The carbon-oxygen a-bond found in alcohols and sugars can undergo several reactions, but it is generally not as reactive as the carbon-oxygen rc-bond Our goalis to organize the vast multitude of reactions involving carbonyl compounds and alcohols Figure 5-1 shows several types of carbonyl compounds and carbonyl derivatives with which you should be familiar Types of Carbonyl Compounds O R R R R'Ov OR* R'Ov \/ R OH R'O \/ H Acetal R H Aldehyde Ketone H Hemiacetal OR' R'O \/ R/CVR R/CVR Ketal Hemiketal O O O O II II II II OH[ R R R OR' Carboxylic Acid O If If R NH2 R / R N R' NHPh R Phenylhydrzine derivative Lactone if H Imide Amide Acid halide R Acid anhydride Ester it 'N OH \/ ? Lactam If R R /c\ CHn Enolate Resonance Forms Figure 5-1 Copyright © by The Berkeley Review Exclusive MCAT Preparation OrgSlIllC ChCllllStry Carbonyls and Alcohols Oxygen Containing Compounds Oxygen containing compounds, because of the highly electronegative nature of oxygen, are very reactive Much of organic chemistry revolves around alcohols and carbonyls, so it is imperative to get a fundamental understanding of their properties, reactivity,and spectroscopicevidence that supports their existence Alcohol Properties Because of their ability to form hydrogen bonds, alcohols typically have high boiling points and are generally miscible in water Alcohols make good solvents as they are often liquids at room temperature and they have a large range between their melting and boiling points Alcohols are hydrophilic, polar molecules that become less hydrophilic (more lipophilic) as their carbon chain length increases The smaller alcohols (three carbons or less) are highly water soluble, but as the size of the alkyl group increases, their water solubility decreases As with all compounds, their physical properties vary with mass and branching, as well as the position of the hydroxyl group As the molecular mass increases, the boiling point increases, but the effect on the melting point is less clear As the branching increases, the boiling point decreases Table 5-1 shows the physical properties of several alcohols, from which the effects of mass, branching, and positioning of the hydroxyl group on the physical properties can be ascertained Isomer IUPAC Name Boiling (Common Name) Point Melting Density Point (g/mL) Water Solubility (g/lOOmL) CH3OH Methanol 64.6°C -98°C 0.791 High H3CCH2OH Ethanol 78.4°C -115°C 0.789 High H3CCH2CH2OH 1-Propanol (n-Propanol) 97.2°C -127°C 0.804 High (H3Q2CHOH 2-Propanol (i-Propanol) 82.3°C -90°C 0.786 High H3C(CH2)3OH 1-Butanol (w-Butanol) 117.3C -90°C 0.810 8.2 H3CCH(OH)CH2CH3 2-Butanol (sec-Butanol) 99.6°C -115°C 0.806 12.8 (H3C)2CHCH2OH 2-Methyl-l-propanol (f-Butanol) 107.7°C -122°C 0.802 11.3 (H3Q3COH 2-Methyl-2-propanol (f-Butanol) 82.0°C 24°C 0.789 High H3C(CH2)4OH 1-Pentanol (n-Pentanol) 137.6°C -79°C 0.814 2.1 H3CCH(OH)CH2CH2CH3 2-Pentanol 119.3°C 0.809 5.0 (H3CCH2)2CHOH 3-Pentanol 115.9°C 0.815 5.6 H3C(CH2)4CH3 1-Hexanol (n-Hexanol) 157.5°C 0.814 0.8 C6HnOH Cyclohexanol 161.5°C 0.956 2.1 H3C(CH2)6CH2OH n-Octanol 194.7°C 0.817 0.3 T able 5-1 Copyright©by The Berkeley Review The Berkeley Review OrgaiUC ChemiStry Carbonyls and Alcohols Oxygen Containing Compounds Example 5.1 Whattype of alcohol is the following molecule? HiC CH„ HO- A Primary alcohol B Secondary alcohol C Tertiary alcohol D Phenol Solution The compound has the alcohol functional group attached to a carbon that is bonded to two other carbons This is defined as a secondary alcohol, so the best answer is choice B Alcohol Reactivity Alcohols are nucleophilic reagents in organic chemistry They are not good nucleophiles in their protonated (neutral)state, but they can be deprotonated and converted into their anion (alkoxide) form under basic conditions Because alkoxides (the deprotonated form of the alcohol) are strong bases, they are not the ideal nucleophile, but they are generally better than alcohols Alcohol chemistry also involves oxidation into a carbonyl as we shall see later in this section Alcohols are commonly formed from the reduction of carbonyls, which we shall also postpone for the moment The common reactions to form and consume alcohols that not involve carbonyl compounds center around nucleophilic substitution Figure 5-2 shows nucleophilic substitution reactions that convert alkyl halides into alcohols Figure 5-3 shows nucleophilic substitution reactions that convert alcohols into alkyl halides Alkyl halides to alcohols R R h*%/ x ~2~~^ H*y H OH + X H R' \ R»';?-Br R' RCO," / 2.0H-(ac,)» H°~V"R + H „^ TT + H R" „„t.^—X Rv / „ R" • acetone t,ii»»^— OH Rx / R' +HX R' Figure 5-2 Copyright © by The Berkeley Review Exclusive MCAT Preparation OrCJ£UllG ChCHllStry Carbonyls and Alcohols Oxygen Containing Compounds Alcohols to alkyl halides O R ,^-oh S « R /s^ Cl ^—Cl H^ H Retention H R' R' h r^; PBr, OH Br -i H Inversion H R" R" h R^ OH ^L^ h Br r^; K Racemization R' Figure 5-3 Spectroscopic Evidence for Alcohols Alcohols can be detected using either infrared or NMR spectroscopy In IR spectra, hydroxyl groups present a distinct absorbance between 3200 and 3500 cm"1 that is medium in intensity and broad due to hydrogen bonding In XHNMR spectra, hydroxyl groups present asignal between 1and 5ppm that is broad due to hydrogen bonding, although the broadness varies with the solvent They have no definite 6-value (it varies with concentration and solvent) The peak slowly disappears with the addition of D2O to the NMR tube The OH group does notcouple well, so werarely consider splitting patterns foralcohols Figure 5-4 shows the 1HNMR spectrum for 2-propanol in carbon tetrachloride solvent Figure 5-5 shows the IR spectrum for 2-propanol obtained neat onsalt plates H H,C OH 6H CH, 1H 1H 3.0 ppm 2.0 1.0 Figure 5-4 Copyright ©by The Berkeley Review The Berkeley Review Ur£J£UllC t/llCllllStry Carbonyls and Alcohols Oxygen Containing Compounds PTVl 1392 cm-' 1455 cm-1 & 1365 cm-' 2984 cm Figure 5-5 Aldehyde and Ketone Properties Because aldehydes and ketones not form hydrogen bonds, they typically have boiling points only slightly higher than alkanes of equal mass Because of the polarity of the carbonyl bond, they are slightly miscible in water Aldehydes and ketones are aprotic, polar molecules that become less hydrophilic as their carbon chain length increases The smaller aldehydes and ketones (three carbons or less) are generally water soluble but as the size of the alkyl group increases, their water solubility decreases Table 5-2 shows the physical properties of several aldehydes and ketones, from which the effects of mass, branching, and positioning of the carbonyl group on the physical properties can be ascertained Isomer IUPAC Name (Common Name) Boiling Melting Point Point Water Solubility (g/lOOmL) HCHO Methanal (Formaldehyde) -21°C -92°C High H3CCHO Ethanal (Acetaldehyde) 21°C -121°C Infinite H3CCH2CHO Propanal (Propionaldehyde) 49°C -81°C 16.3 H3C(CH2)2CHO Butanal (n-Butyraldehyde) 76°C -99°C 6.8 H3C(CH2)3CHO Pentanal 103°C -92°C 3.3 H3C(CH2)4CHO Hexanal 128°C -56°C 2.1 C6H5CHO Benzaldehyde 178°C -26°C 0.3 H3CCOCH3 Propanone (Acetone) 56°C -94°C Infinite H3CCOCH2CH3 Butanone (Ethyl methyl ketone) 80°C -86°C 25.6 H3CCO(CH2)2CH3 2-Pentanone 102°C -78°C 5.7 (H3CCH2)2CO 3-Pentanone 1018C -41'C 4.9 H3CCO(CH2)3CH3 2-Hexanone 128°C -55°C 1.6 H3CCH2CO(CH2)2CH3 3-Hexanone 124°C H3CCOCH2CH(CH3)2 4-Methyl-2-Pentanone 119°C -85°C 1.9 C6Hi0O Cyclohexanone 156°C °C 2.2 C6H5COCH3 Acetophenone 202°C 21°C Insoluble 1.3 Table 5-2 Copyright © by The Berkeley Review Exclusive MCAT Preparation OrCJcllllC l^IlCllllStry Carbonyls and Alcohols Oxygen Containing Compounds Example 5.2 What is the IUPACname for the following compound? OH r H O A B C D l-Aldo-4-pentanol 4-Hydroxypentanal 5-Oxo-2-pentanol 2-Hydroxypentaldehyde Solution The longest chain isfive carbons and the highest priority functional group is the aldehyde The functional group with the most oxidized carbon receives the highest priority according toIUPAC convention For naming aldehydes, the"e" is dropped from the alkane chain of the samelengthand an "al" suffix is added This makes the compound pentanal, which makes choice B correct The OH is named hydroxy as a substituent Aldehyde and Ketone Reactivity Aldehydes consist ofa carbonyl with a hydrogen bonded to the carbonyl carbon along with either an alkyl group or in the case of formaldehyde, a second hydrogen Ketones consist of a carbonyl group with two alkyl substituents attached The chemistry occurs primarily at the electrophilic carbonyl center Aldehydes and ketones are reactive with most nucleophiles, but not by a traditional nucleophilic substitution mechanism Once a nucleophile attacks a carbonyl carbon, it forms a four-ligand intermediate with a negative charge on oxygen known as a tetrahedral intermediate This intermediate will be shown in the mechanism of many carbonyl reactions in this section The chemistry of aldehydes is similar to the chemistry of ketones except that analdehyde can be oxidized into a carboxylic acid while ketones cannot be oxidized easily Oxidation in carbonyl chemistry can be viewed as either the gain ofbonds to oxygen orthe loss of bonds to hydrogen We shall thoroughly address carbonyl reactions throughout this section Spectroscopic Evidence for Aldehydes and Ketones Aldehydes have infrared absorbances inthe 1720 cm"1 to1740 cm-1 range They are unique in the IR from other carbonyls due to two medium C-H stretches around2700 cm-1 and 2900 cm-1 Ketones haveinfrared absorbances in the 1710 cm-1 to 1725 cm-1 range In *HNMR, aldehyde hydrogens are found between and 10 ppm, which makes aldehydes easy to identify via *HNMR Ketones and aldehydes each have alpha protons which fall in the 2.0 to 2.5 ppm range in !HNMR Figure 5-6 shows the *HNMR spectrum for butanone in carbon tetrachloride solvent Figure 5-7 shows the IR spectrum for butanal obtained neat on salt plates Copyright ©by The Berkeley Review The Berkeley Review 28 Choice A is correct Sublimation is the physical process of converting a solid directly into a gas, so choices C and D are not valid answers As a laboratory technique, it is used to convert one solid into a gas, while leaving behind the other solids The gas can migrate away from the other solids and then be collected on a cold surface as a pure sample of the solid Purifying a solid from a liquid does not work, because the liquid can vaporize, resulting in more than one compound in the gas phase This eliminates choice B The best answer choice is A Passage V (Questions 29 - 35) Partition Coefficient Experiment 29 Choice B is correct Exactly 0.75 grams of the original 1.0 gram sample was isolated from the organic layer It is assumed that the other 0.25 grams of sample were dissolved into the aqueous layer The partition coefficient is the amount that dissolves into the organic layer divided by the amount that dissolves into the aqueous layer, which in this case is 0.75/0.25 = 3.0 This makes choice B the best answer 30 Choice D is correct As stated in the passage, the solubility of4-methylaniline in water at 25°C is 3.8 grams/100 mL and in diethyl ether at 25°C is 22.8 grams/100 mL The partition coefficient is the solubility of the sample in diethyl ether divided by the solubility of the sample in water The partition coefficient is found as follows: 22.8 3.8 100 100 _ 22.8 -228-114 - 6xl9 3.8 38 19 = 19 Choice D is the best answer 31 Choice D is correct As stated in the passage, the solubility of 4-methylaniline at 25°C in water is 3.8 grams/100 mL and the solubility of 4-methylaniline at 25°C in diethyl ether is 22.8 grams/100 mL This means that the maximum amount of 4-methylaniline that can dissolve into 10 mL diethyl ether is 2.28 grams The maximum amount of 4-methylaniline that can dissolve into 10 mL water is 0.38 grams The maximum total mass between the two solutions is 2.66 grams Adding 3.00 grams exceeds this amount, so not all of the 3.00 grams of 4methylaniline can dissolve into the two solutions combined The undissolved portion will not partition between the two solvents, but instead forms a precipitate at the bottom of the flask This precipitate makes the lower layer appear to have more 4-methylaniline than it actually has dissolved into solution This makes choice D the best choice Had the compound been fully soluble, the partition coefficient should have been the same, assuming any errors remains the same in magnitude 32 Choice Bis correct Because 4-methylaniline is more soluble in the diethyl ether than water, the pipette could be rinsed with diethyl ether to wash off any residue of 4-methylaniline Heating and cooling the pipette is pointless once the compound has adhered to the walls of the pipette Water is a poor choice to remove the adhered solid, because 4-methylaniline is relatively insoluble in water The residue on the glass was lost from the diethyl ether layer, so if it is recovered, it is best recovered into the diethyl ether layer Pick choice Bfor best results 33 Choice Ais correct If the ether were not anhydrous, itwould contain some water impurity This implies that in the 10 mL of ether, there is actually a small portion of water and therefore less than 10 mL of ether This eliminates choices C and D, because they indicate an excess (a value greater than 10 mL) of ether With less than 10 mL of ether, less 4-methylaniline than expected would dissolve into the ether layer This means that a smaller amount of 4-methylaniline would be isolated from the ether layer This would lower the numerator, and thus lower the value for the calculated partition coefficient This makes choice A the best answer 34 Choice Biscorrect Diethyl ether is used because itis immiscible in water and the organic compounds are soluble in it Any solvent that is used in place of diethyl ether must have the same properties Tetrahydrofuran, THF, is a cyclic ether, so it has the same properties as diethyl ether This eliminates choice A Cyclohexane and methylene chloride are both organic solvents that are immiscible in water, so choices C and D are eliminated Ethanol is miscible in water, so it cannot be used in lieu ofdiethyl ether This makes choice B the best answer 35 Choice Cis correct For two layers to form, the organic solvent must be immiscible in water, eliminating choice A The density is not important, as long as its not the same density as water As long as it has a density different than water, higher or lower, the system will have a layer on top and on bottom This eliminates choice B The solute should be soluble in the organic layer, so that itcan be removed from the aqueous layer Choice C is the best answer If its boiling point is less than room temperature, then it will exist as a gas at room temperature To be useful as a solvent, it needs tobea liquid at room temperature This eliminates choice D Copyright © by The Berkeley Review( 292 LAB & SPECTROSCOPY EXPLANATIONS Passage VI Acid/Base Extraction Experiment (Questions 36 - 42) 36 Choice D is correct Ether is chosen as a solvent, because it is immiscible in water and it is a liquid at room temperature The organic solvent used in lieu of ether must also be immiscible in water and a liquid at room temperature Choices B and C are eliminated, because they are not liquids Choice A (ethanol) dissolves into water, so choice A is eliminated Only choice D remains Chloroform is an immiscible liquid, so choose D 37 Choice B is correct To extract the contents of Tube from the organic layer, a basic aqueous solution was used, so Tube must be at a high pH This means that the solute is likely in its anionic form To remove the solute (dissolved anion) from solution, the solution must be neutralized To neutralize a basic solution, an acid is added This eliminates choices A and D The acid should be strong, so it can fully react with the base in solution The best answer is choice B, hydrochloric acid 38 Choice D is correct Tubes and contain the component that does not react with either the strong base or the weak base to become more water soluble Dinitrobenzene has no acidic protons, given that all of its hydrogens are bonded to carbon, so it will remain neutral and organic soluble during the entire separation It is not possible to convert dinitrobenzene into a charged species with just acid and base reactions The best answer is dinitrobenzene for both Tube and Tube 5, so choice D is the best answer 39 Choice C is correct The second proton is by definition less acidic than the first proton, so the second pKa is greater than the first pKa Choices A and B are eliminated The value of pKb can be found by subtracting the pKa for the conjugate acid from 14 Because the pKa for benzoic acid is lower than the pKa for the first proton of resorcinol, the pKb for the conjugate base of benzoic acid (benzoate) must be greater than the pKb for the conjugate base of the first acidic proton of resorcinol (pKb2)- The first proton lost by a diprotic acid is the second one gained by the fully deprotonated form of the conjugate base This makes choice C the best answer 40 Choice A is correct Because the melting points for the compounds isolated by Student II showed a narrower range than the compounds isolated by Student I, the productsfrom thescheme ofStudent II must have been purer than the compounds from the scheme ofStudent I Thus, the scheme for Student II must have worked better than the scheme for Student I The scheme for Student II isolated the stronger acid first (Tube 4), the weaker acid second (Tube 6), and the neutral species in the organic layer (Tube 5) It can be concluded that the scheme for Student II worked while the scheme for Student I did not The best answer is choice A 41 Choice D is correct 3-Methylaniline is an amine, making it a basic compound Basic compounds dissolve into water under acidic conditions However, in the reaction scheme for Student II, only base is added to the water and not acid As such, the amine is not protonated, and remains in the organic layer for both extraction steps The result is that it starts and finishes in the organic layer, which finishes in Tube A small amount may dissolve into water, so the best answer is choice D, predominantly in Tube 42 Choice D is correct A compound can beseparated using extraction techniques as long as it is not reactive in any of the mediums 4-Ethylcyclohexanone is a large ketone, which is not reactive with the ether layer, acidic water, or basic water 4-Ethylcyclohexanone can be isolated using acid-base extraction techniques, so choice A is eliminated 3-Bromophenol is a mild organic acid, which is not reactive with the ether layer or acidic water, and undergoes reversible deprotonation in basic water 3-Bromophenol can be isolated using acid-base extraction techniques, so choice B is eliminated 3-Nitrobenzoic acid is a mild organic acid, which is not reactive with the ether layer or acidic water, and undergoes reversible deprotonation in basic water 3-Nitrobenzoic acid can be isolated using acid-base extraction techniques, so choice C is eliminated Ethyl benzoate is an ester Esters can undergo hydrolysis under acidic aqueous conditions or saponification (base catalyzed hydrolysis) under basic aqueous conditions Because esters react under the conditions used in acid-base extraction, ethyl benzoate cannot be isolated using acid-base extraction techniques Choice D is the best answer Passage VII (Questions 43 - 49) Extraction and Thin Layer Chromatography Experiment 43 Choice C is correct An Rf value of 0.92 implies that the compound is highly soluble in the solvent and that the compound has a very low affinity for the surface of the TLC plate A change in solvent from nonpolar to polar would show a reduction in solubility, and thus a reduction in the Rf value The Rf cannot be negative, so choice D is eliminated The Rf value cannot be greater than 1, so choice A is eliminated The only value greater than zero but less than 0.92 is choice C, making it the best answer Copyright © by The Berkeley Review® 293 LAB & SPECTROSCOPY EXPLANATIONS 44' Sfvinl Be,CaUSe 3SUglr 3largGThis nUmber °f hydr0Xy{ most solub1^ solvent ?SICT? with a hydroxyl group, such C°ntainS as an alcohol is based on the SrouPs' idea that*wiI1 "like be dissolves like." inAna alkane, ether, and ketone nothave a hydroxyl group, so the best answer is choice C 45' plate r^ttZ^f 'T'1; ?rind?le thin kyer chromatography is having the spotsto migrate the at different rates^by ^^ exploiting theirbGhind differences in attraction to the solvent and affinity the TLCupplate Spotting should be done in away that allows the spots to migrate If the spots are placed below the top of the solvent (submerged into the solvent), they can dissolve into the solvent This makes choice Dthe best answer Choice Ashould be eliminated, because the rate of migration does not depend on the placing of the solvent and spots The spots are not going to become saturated, so they will not rise up the plate too rapidly Choice Bis eliminated The spots will dissolve in all directions, so radial migration is an incorrect answer Choice Cis eliminated Choice D is better than the other pooranswers 46 Choice Cis correct For column chromatography, the best separation results from the solvent that shows the greatest difference in Rf values in terms of ratio In hexane, the Rf values are different by a factor of while in ethanol, the Rf values are different by afactor of Do not subtract the Rf values to determine the separation, the Rf values must be divided A better separation is observed with ethanol than hexane, so choices Aand B should be eliminated Because the Rf values are small, the column should be short, so that less solvent is required and the components not remain on the column for any unnecessary time after separation has been estab ished A mixed-solvent of hexane and ethanol would not work well, because the solutes will travel faster and the ratio of the Rf values will decrease only slightly Choice C is the best answer 47 Choice Dis correct Carbon tetrachloride is anonpolar solvent, so it shows similar solubility properties (and thus similar TLC results) as other nonpolar solvents Acetone and propanal both contain the carbonyl functional group, which makes them polar This eliminates both choices Aand B Ethanol is not only polar, but it also is capable of forming hydrogen bonds with solute particles Choice Ccan thus be eliminated Because pentane is a hydrocarbon, it is nonpolar Pentane will behave most like carbon tetrachloride, so the best answer ischoice D 48 Choice Dis correct The Rf value is defined as the distance that aspot travels up aTLC plate (dspot) divided by the distance that the solvent travels up aTLC plate (dsolvent)- Because the spot migrates due to its affinity for the solvent, the spot can never migrate a greater distance than the solvent migrates This means that the dsolvent will always be agreater value than the dspot This consequently means that the Rf value can never be greater than or equal to 1.00 Choice D, 1.05, is not possible An Rf value of 0.0 simply means that the spot does not dissolve mto that particular solvent, and therefore it does not move An Rf value of 0.0 is possible 49 Choice Ais correct Trimyristin is a fatty acid triglyceride, meaning it is a triester Long chain esters are hydrophobic, because they cannot form hydrogen bonds and they are relatively nonpolar This eliminates choices Band D THF is an ether, making it hydrophobic Because of the oxygen in the ring, THF is a mildly polar solvent, although it is aprotic Tiie best answer is choice A Passage VIII (Questions 50 - 56) Thin Layer and Column Chromatography 50 Choice Cis correct Alumina is a polar adsorbent, so it hinders polar solutes more than nonpolar solutes Nonpolar solutes come off the column first, resulting in a fast elution time Benzoic acid is hydrophilic and polar, eliminating choice A Glycerol is also polar, so choice Bis eliminated Glycine is polar and hydrophilic, so despite the presence of ethanol as the eluting solvent, it is still hindered by the alumina Choice D is eliminated Anonpolar compound like o-xylene exhibits little affinity for the alumina, so it travels quickly down the column Using anonpolar solvent like ligroin speeds itup even more, so choice Cis the best answer 51 Choice Dis correct The compound with the larger Rf value climbs aTLC plate faster, so it travels through a The Rf values differ by afactor of two, so the relative migration distance in a given time differs by a factor of two This means that their rates also differ by a factor of two, meaning that their times for traveling a given column faster as well This means that it should have a smaller elution time This eliminates choices A and B distance shall also differ by a factor of two This makes choice D the best answer 52 Choice Cis correct Salicylic acid is highly polar, as stated in the passage The greatest Rf value corresponds to the greatest migration, which is observed in the most polar solvent Hexane and ligroin are nonpolar, so choices Aand Bare eliminated Methanol is more polar than pentanol, sochoice C is the best answer Copyright © by The Berkeley Review® 294 LAB & SPECTROSCOPY EXPLANATIONS 53 Choice D is correct The Brockmann Activity rating for alumina has to with the amount of water bound to the column According to the passage, a rating of III refers to a column that is 6% water by mass A rating of IV has more water than a column with a rating of III, so choice A is eliminated Because there is so much water bound to alumina, there is less binding of solute to the column Having more water bound with a rating of IV than a rating of III, a rating of IV represents a lower affinity than a rating of III This eliminates choice B Alumina is made of aluminum oxide, so there is no silicon present This eliminates choice C A Brockmann Activity rating of III is caused by 6% water by mass, so a rating greater than III results from more than six percent water by mass Given that a rating of I is perfectly anhydrous, a rating of IV requires more than grams of water per 100 grams of column, so choice D is the best answer 54 Choice B is correct Methanol has a high affinity for alumina, so the binding of methanol to the alumina is exothermic This eliminates choice A When methanol is added to the column, heat pockets can form, which results in the formation of vapor pockets in the column These vapor pockets can crack the gel, resulting in a break in uniformity for the pathway of solute through the column Choices B is the best answer Methanol is miscible in hexane, but even without knowing such a fact, there is no reason to assume that rate at which methanol migrates through the column has any dependence on the miscibility of the two solvents This eliminates choices C and D 55 Choice A is correct A higher Brockmann Activity rating represents a greater amount of water on the alumina, which corresponds to a lower affinity for polar molecules The lower affinity is due to the fact that water is taking up space that the polar molecule would bind if it was open This makes choice A the best answer Water does not cause the contraction of alumina If anything, it may dissolve alumina with such a high affinity for water Choice B is eliminated Alumina is polar, so although the addition of water may affect the polarity, choice C is incorrect, because alumina is polar to begin with, not nonpolar Water does not oxidize alumina, given that aluminum is fully oxidized in AI2O3 This eliminates choice D 56 Choice B is correct Alumina is polar, so the solute for which it has the highest affinity must also be polar Choice A, 1,3-cyclohexadiene, is nonpolar and hydrophobic, so it is eliminated Choice C, ethyl butanoate, is an ester, so while it has a dipole, it is classified as nonpolar species and hydrophobic Choice C is eliminated Choice B, cyclohexanol, is polar and able to form hydrogen bonds, so it has a high affinity for alumina Choice D, 3-methylpentanal, is an aldehyde, so it polar and should also exhibit an affinity for alumina A protic species has a greater affinity for alumina than an aprotic species, so choice B is a better answer than choice D Column Chromatography Passage IX (Questions 57 - 63) 57 Choice A is correct Because like dissolves like, fluorenone should be most soluble in another ketone As solubility increases in the solvent, the affinity is said to be higher The only ketone in the answer choices is acetone, so choice A is the best answer 58 Choice C is correct Fluorenone is more polar than fluorene, so independent of solvent, fluorene should elute from the column first The fact that a nonpolar, hydrophobic solvent is being used makes the difference in migration rate even greater Fluorene is a nonpolar species, so it has a low affinity for silica gel This eliminates choice A Fluorene is in fact lighter than fluorenone, but in a column, much like free fall, the mass of the species does not factor into its migration rate except for the steric hindrance associated with larger, more massive molecules Choice B is eliminated Fluorene does in fact have a higher affinity for cyclohexane then fluorenone does, so choice C is a valid statement Fluorene is smaller than fluorenone, but there are no pores in silica gel that separate by molecular size (molecular sieves that), so choice D is eliminated Choice C is the best answer 59 Choice B is correct The best solvent is the one that allows the two species to separate by the most This is determined by comparing the Rf values of fluorene and fluorenone in each of the solvents The greatest ratio represents the best solvent For Liquid II, the ratio is 0.33 to 0.24, which is less than For Liquid III, the ratio is 0.51 to 0.19, which is slightly greater than 2.5 For Liquid IV, the ratio is 0.55 to 0.41, which is less than For THF, the ratio is 0.42 to 0.31, which is slightly over Liquid II is best, so choice B is the best answer 60 Choice A is correct Alumina is a polar adsorbent and ligroin is a nonpolar solvent, so nonpolar species will exit the column before polar species Of the three compounds, biphenyl is nonpolar and hydrophobic, so it will elute from the column first This eliminates choices B and C Toluic acid is polar and hydrophilic, so it will elute from the column last This makes choice A the best answer Copyright © by The Berkeley Review® 295 LAB & SPECTROSCOPY EXPLANATIONS 61 Choice C is correct When the solvent is flush with the top of the silica gel, all of the solute is also flush with the top of the silica gel This means that the solute will enter the gel at the same time, resulting in sharper bands going down the column Choice C is the best answer The sample must dissolve into the solvent to be effective, so it cannot float, making choice Aincorrect Whether the solvent is flush with the top of the silica gel does not impact whether or not the solute adheres to the glass If that is an issue, it will result in problems as the sample migrates down the column, no matter where the solvent level starts Choice Bis eliminated The silica gel should not dry out, so choice D is incorrect The best answer is in fact choice C 62 Choice C is correct Although the world might be a better place with words like "sandophilic" in it, there is no such term This unfortunately makes choice Dincorrect The sample must reach the silica gel, so choice A is eliminated The sand does not interact with the silica gel, so it has no impact on the polarity of the gel Choice Bis eliminated The role of the sand at the top of the silica gel is ensure that the top of the silica gel stays flat, so all solute travels the same distance through the column no matter where itstarts Without the sand, pouring solvent into the system could disturb the flat surface Choice C is the best answer 63 Choice Bis correct Silica gel is a polar adsorbent, so its affinity for polar species is high while its affinity for nonpolar species is low The question states that when using silica gel, nonpolar species elute first, no matter what solvent is used This simply means that the affinity of the adsorbent must outweigh the affinity for the solvent Polar compounds follow the "like dissolves like" rule, so choice Ais eliminated Nonpolar species remain nonpolar when polar solvent is added, so choice C is eliminated Polar compounds not form large molecular networks any more than nonpolar molecules would, so choice D is eliminated Choice B is the best answer, because silica gel is highly polar and it thereby hinders polar species by binding them as the travel down a column Passage X (Questions 64 - 71) Gas Chromatography 64 Choice A is correct According to the data in Table 1, the reactant is in the lowest concentration in Trial III (16%), so Peak must be due to reactant In GC graphs, the peak with the smallest area correlates to the lowest concentration Upon adding reactant, spiking the mixture with reactant, Peak should be the one to grow This is done to confirm the identity of a peak The best answer is choice A 65 Choice Bis correct If the column ispacked with a nonpolar material, then nonpolar compounds will bind to the column more so than the polar compounds This allows the polar compounds to travel through the column with less resistance Polar compounds will thus have shorter elution times The best answer is choice B 66 Choice D is correct Figure represents the output for Trail III at 15°C According to Table 1, the relative concentrations at 15°C are Product A > Product B> Reactant Peak is the largest of area, so Peak must be associated with Product A Peak is the second largest,so Peak is associated with Product B Peak must be due to the reactant The first peak to appear has the shortest retention time, so the reactant has the lowest retention time This eliminates choices Band C Product Bcomes off last, so Product Bhas the longest retention time The best answer is choice D 67 Choice A is correct By definition, an increase in temperature favors the formation of the thermodynamic product, so choice Bis eliminated Choices C and D are the same answer, but worded differently There cannot be two best answer choices, so choices C and Dare both eliminated The MCAT writers love to place the same answer twice with slightly different wording From Table in the passage, it can be seen that the amount of reactant decreases with increasing temperature, so the reaction is proceeding more towards products This implies that the reaction is more favorable with increasing temperature The best answer is choice A 68 Choice D is correct Gas chromatography separates by differences in migratory rates of vaporized organic compounds Although size and mass are variables that affect the migratory rate of a gas, the primary differences in retention times in this example can be attributed to the attractive forces between the packing material in the column and the compounds If mass and size were the primary factors, then good separation would not be possible for the two products from the reaction in the passage The charge ofthe compound would have aneffect ifcharged compounds could easily vaporize Because charged compounds not readily vaporize, the gas chromatography apparatus does not distinguish charged species Although the passage doesn't mention it, background knowledge should tell you that ionic compounds (like NaCl) not vaporize easily The best answer is choice D Copyright © by The Berkeley Review® 296 LAB & SPECTROSCOPY EXPLANATIONS 69 Choice C is correct The best gas to propel through the gas chromatography apparatus is an inert gas that does not react with any of the compounds Nucleophiles can react with carbon dioxide, so choice A is eliminated Hydrogen gas can react with air, so hydrogen gas should not be used This eliminates choice B Helium is the only inert gas of the choices, sohelium should be chosen The best answer is choice C Water is too reactive 70 Choice A is correct The major product from Trial II is the less hindered compound It has no acid-base properties, so acid-base extraction is not viable Choice Dis eliminated The boiling point of the major product is likely to be close to the boiling point of the minor product, so only fractional distillation would work, not simple or vacuum distillation Choices Band C are eliminated Choice A is the best answer by default 71 Choice D is correct Astrong base isused to deprotonate a very weak acid This eliminates choices A andB The pKa for a hydrogen on a carbon that is alpha to a carbonyl is around 19, so the best answer is choice D Sometimes trivia is necessary to solve a problem Recrystallization Passage XI (Questions 72 - 79) 72 Choice C is correct Longer crystals are indicative of higher purity Impurities disrupt the crystal lattice, so crystals cannot grow as large This eliminates choices A and B Slow gradual formation of crystals yields the finest, purest crystals This makes choice C the best answer 73 Choice D is correct The ideal solvent should not bind to the lattice structure of the crystals, otherwise it could incorporate as an impurity in the crystal lattice as it precipitates from solution This makes Statement I valid The ideal solvent for recrystallization should have the solid insoluble at low temperatures and highly soluble at elevated temperatures This eliminates Statement II The impurities should be highly soluble in the solvent so that they not precipitate out and possibly get trapped in the crystals This makes Statement III valid Statements I and III are both valid, so choice D is the best answer 74 Choice Cis correct Physical properties can be used to determine the purity ofa compound If the density of the crystal is precise, it implies that there are no impurities in the crystal The melting point of the compound (crystalline ornot) can be used to determine the purity Abroad melting point range implies that the compound is not pure The mass of the crystal has nothing to with its shape and purity This makes choice Can invalid statement, and therefore the best answer The index of refraction is another of the physical properties that depends on the lattice structure The index of refraction varies with impurities The best answer is choice C 75 Choice Bis correct By keeping the solvent and glassware hot, the original solid remains in solution during the filtration The purpose of filtering at this point is to remove any solid (insoluble) impurities that are present, so the target compound should remain in solution for highest yield This makes choice Bthe best answer choice 76 Choice C is correct The ice bath serves to cool the solution to 0°C The solubility of a solute decreases as the temperature of the solution decreases, causing more solid to crystallize (precipitate) out of solution Overall, the purpose of the ice bath is to cool the solvent to decrease the solubility and thus increase the amount of crystals formed Pick C for correctness 77 Choice Bis correct If crystals form too rapidly, then impurities (such as solvent molecules) can get trapped in the crystal lattice If an impurity happens to collide with the surface of the crystal, it has time to escape if the crystals are formed slowly However, if the next layer of the crystal forms too quickly, the impurity cannot escape The best answer is choice B 78 Choice C is correct Ideally, the crystals formed from recrystallization are purer than the original solid, otherwise recrystallization would be pointless Choice A is a valid statement Considering that the solvent must be heated to itsboiling point to maximize solubility (and thus minimize the amount ofsolvent required), it is important to know the boiling point of the solvent This makes choice B valid If the solvent is highly volatile, then it cannot be raised to a very high temperature (the more volatile the liquid, the lower its boiling point), thus more solvent is required to fully dissolve the solid The more solvent required at maximum temperature, the more solvent present when the solution is cooled The more solvent present when recrystallizing, the less solid that will crystallize out from solution This reduces the percent recovery, making choice C the incorrect statement It is never possible to recover 100% of a material after it has been dissolved into solution A small portion always remains soluble The best answer is choice C Copyright ©by The Berkeley Review® 297 LAB & SPECTROSCOPY EXPLANATIONS 79 Choice C is correct Charcoal binds colored organic molecules If impurities are colored, they can bind the charcoal and the charcoal can be filtered away, taking the bound impurities with it This means that charcoal is employed when there are colored impurities This eliminates choices A and B If the target molecule is colored, then charcoal can bind it, and using charcoal would decrease the yield This means that the desired compound must be colorless, making choice Cthe best answer To be achemistry warrior supreme, pick C Passage XII (Questions 80 - 86) Qualitative Analysis 80 Choice A is correct Because Compound A is oxidized by chromic acid, it must be either an alcohol or an aldehyde Dinitrobenzoylchloride reacts readily with nucleophiles (such as alcohols) to form a derivative that is a solid at room temperature The formation of a precipitate therefore confirms that Compound A is a nucleophile, so the aldehyde choice is not possible Although you may not know what these reagents specifically, you should be able to deduce the purpose of each test The molecular formula for Compound A (C6H12O) indicates that the compound must be either a carbonyl, ether, or alcohol The ether choice is eliminated, because it would not be oxidized by the chromic acid The carbonyl choice is eliminated for the reasons stated above The compound must be an alcohol An O-H stretch is found at3500 cm'1 Tiie peak isbroad due to hydrogen bonding Because Compound Bdoes not change color in CrC>3/H+, it must eitherbe an etheror ketone Because it shows a positive iodoform test, it is a methyl ketone If you did not know that, you can deduce that the ether possibility is out, because ethers are inert and will not test positive with I2 and KOH The key peakin a ketone would be theC=0 peak at 1710 cm"1 This makes choice A the best answer 81 Choice Ais correct This question has more depth than meets the eye According to the formula, the compound has one degree of unsaturation, which can be either a ketone 7t-bond, an alkene 7t-bond, or a ring The reagent added (Br2 in CCI4) reacts with an alkene 7t-bond, therefore an alkene functionality must be present in the molecule The ketone is not possible for the structure, because the C=0 ofa ketone would require a second degree of unsaturation and given that the one unit of unsaturation present in the compound has already been used (by the alkene) There can be no carbonyl present in the compound The positive test confirms that there is an alkene present, so it does not show that it cannot be an alkene Choice B is thus eliminated The ether and alcohol functionalities are possible, because the molecule could be an alkene (for which the bromine in CCI4 test is positive) with an ether or alcohol functionality The positive test only shows that the compound cannot be a ketone, thus you mustpick A you correctness trooper you! 82 Choice Ais correct An acidic hydrogen (protic hydrogen) will exchange a proton for deuterium in the presence of D2O This results in the loss of one signal (broad peak) in the proton NMR spectrum The only compound with a protic hydrogen is the alcohol, choice A An ether should have been eliminated immediately, because ethers have no acidic protons Aketone and aldehyde have alpha hydrogens that can be removed by strong base, but they not readily exchange in water without the presence of a strong base or strong acid This question is best solved by selecting the most acidic compound of the choices Aproton on the alcohol oxygen is more acidic than the hydrogen on an alpha carbon No matter what your path to wisdom, finish off beautifully by choosing answer choice A 83 Choice Ciscorrect Because Compound Awas oxidized by the chromic acid, itcan be oxidized by the KMnC>4 as well Both chromic acid and potassium permanganate are strong oxidizing agents The other compound of interest isCompound C, which turned the bromine solution clear (confirming the presence analkene double bond) Alkenes also react with permanganate solution, forming a vicinal diol with syn orientation This is stated in the passage Both Compound Aand Compound Ccan react with aqueous potassium permanganate (KMnC>4(aq)), resulting in the purple permanganate solution changing color upon reaction This makes choice C the best answer 84 Choice C is correct Acid chlorides are highly reactive electrophiles (because the chloride is a good leaving group and the carbonyl carbon iselectron poor) Acid chlorides can react with even the weakest ofnucleophiles Alcohols, amines, and thiols are all nucleophiles, so they all will react with the dinitrobenzoyl chloride ((N02)2C6H3C0C1) kicking out the chloride leaving group Thus the best choice for a compound that will not react with (N02)2C6H3C0C1 is the ether The best answer is choice C Based on the inert nature of ethers in general, it is safe to pick the ether, even if you know nothing about acid halide reactivity In other words, for questions that ask "Which will not react?", they are really asking "Which is the least reactive species?" Copyright © by The Berkeley Review® 298 LAB & SPECTROSCOPY EXPLANATIONS 85 Choice B is correct Choices C and D can be eliminated, because tertiary alcohols and ethers cannot be oxidized by chromic acid (Cr03 in H+) The color change observed from the treatment of Compound Awith chromic acid shows that the compound can be oxidized A primary alcohol (choice A) can be eliminated, because when primary alcohols are oxidized, they form carboxylic acids which would turn litmus paper red when added to the paper For this reason, the secondary alcohol is the best choice Feel correctness euphoria with choice B Be euphoric and pick it 86 Choice A is correct This is perhaps a question of trivial knowledge to a point If youhave your chemical tests memorized, then this question is a no brainer The MCAT provides a great deal of information, therefore we recommend that you focus on extracting information via critical analysis of the passages and tables, rather than memorization The color of the precipitate is given in the passage as yellow This eliminates the choices C and D which list the precipitate as being black The passage also states that the test confirms the presence of a methyl ketone, which outright gives choice Aas the correct answer This is a free question for just reading the passage If you did not read the passage, the aldehyde should be eliminated, because aldehydes can be oxidized, and Compound Bdid not react with chromic acid If Compound Bwere an aldehyde, it would have showed a positive test with the chromic acid (a color change from orange to green) From the information listed in Table and knowledge that the precipitate is yellow, the best choice is answer A Synthesis and Extraction Passage XIII (Questions 87 - 93) 87 Choice Bis correct The limiting reagent is 1,4-dimethoxybenzene, not nitric acid, so the percent yield should be calculated based on 1,4-dimethoxybenzene The percent yield for the reaction is the actual yield (91.5 mg 2,5- dimethoxynitrobenzene) divided by the theoretical yield (183 mg 2,5-dimethoxynitrobenzene) based on the 1.0 mmole of the limiting reagent (1,4-dimethoxybenzene) Because the actual yield is only half of the theoretical yield, the percent yield is fifty percent The best answer is thus choice B The math is shown below moles product % yield = moles limiting reagent 88 91.5 mg, 138mg 183 g/mole _ 138g/mole 0.5 mmolesproduct = 50% 1.0 mmoles limiting reagent Choice B is correct The melting point of the product has a range of 1.5 °C, which is a small enough range to deduce that the product is relatively pure Because the melting point is sharp (a sharp melting point range is generally regarded as two degrees or less) and close to that of the literature value for the product, the best answer is choice B Be a Bsupporter! There isno mention ofa broad OH peak in the passage, so it is unlikely that it is wet Choice A is eliminated 89 90 Choice Bis correct Filtering collects (and thus removes) the solid particles from a solution ormixture Filtering is most commonly used inorganic lab for the separation of solids from liquids The best answer ischoice B Choice Cis correct Both nitrogen gas and carbon monoxide gas are colorless, so choices Aand Bare eliminated You should know the color of these two gases from your everyday experiences inlife Nitrogen gas makes up the majority of our air, and air is transparent Carbon monoxide is given off with the exhaust from combustion engines If you are aware of your trivia, sulfur dioxide is colorless to faint yellow in color, which eliminates choice D If you didn't know this, you hopefully thought of smog, a brown gas containing a nitrogen/oxygen molecule of some kind (nitrogen dioxide) Because smog is comprised of anitrogen/oxygen compound, and smog is brown, it is a logical conclusion to select NO2, choice C Choice C is the bestanswer 91 Choice B is correct Theerror in the thermometer resulted in a reading that was too low (the experimental value of 72.5 to 74 is less than the literature value of 74 to 75) This means that the mercury did not climb as high as it should have in the inner column The presence ofsome air above the mercury column (resulting from a reduced vacuum) hinders the ability of the mercury column to rise, thus Statement I results in a reading lower than expected A divot present between 45°C and 55°C results in the mercury column filling the divot at temperatures where the top of the mercury column is above the divot (temperatures greater than 55°C in Statement II) This causes the mercury column to not rise as high This results in a reading that is too low If the divot is above the mercury level, then it will have no effect, because the mercury has not risen high enough to fill the divot This means that Statement II is valid but Statement III is invalid Choice B is the best answer Copyright © by The Berkeley Review*1 299 LAB & SPECTROSCOPY EXPLANATIONS 92 Choice Ais correct Filtering the compound again would not purify the crystals, it would simply reduce the yield, so choice Bis eliminated Distilling the liquid (formed from melting) would be impractical given the high temperature and probability of deposition of the gas into solid throughout the distilling column, so choice Cis eliminated Addition of more sulfuric acid and nitric acid could further react with the product rather than the starting material, so choice Dis eliminated Recrystallization is the ideal laboratory technique to employ in the purification of a solid The best answer is choice A, so you better pick that A 93 Choice Bis correct The addition of cold water causes the crashing out (rapid crystallization) of product from solution This means that solubility decreased, not increased, eliminating choice Aand making choice Bthe best answer The solution already has two strong acids, a mixture of nitric and sulfuric, so the addition of water will not increase its acidity of the solution This eliminates choice C Water will not protonate a species that is deprotonated in the presence of strong acid, so choice D is eliminated You should recall that lower temperatures result in reduced solubility, so choice B is the best answer choice Not Based on a Passage Questions 94 -100 94 Organic Chemistry Lab Techniques Choice C is correct If the S-enantiomer has a specific rotation of -55°, then the R-enantiomer has a specific rotation of +55° The net rotation is positive, so there must be more of the R-enantiomer than the S-enantiomer This eliminates choice A Solving this is going to require math, so it is easier to start with an answer choice and work back to the specific rotation For plugging in the numbers, choice C is best, because only one calculation needs to be done The value is either going to be too high (implying that choice Bis correct), right on the mark (making choice C correct), or too low (implying that choice D is correct) A mixture of 30% of the S-enantiomer and 70% of the R-enantiomer has a 40% excess of the R-enantiomer A100% R-enantiomer solution has aspecific rotation of +55°, so 40% excess of R yields a rotation of 40% of +55°, which is 22° Choice C is the best answer 95 Choice A is correct The solute that elutes first from the column is the fastest In a TLC experiment, the fastest solute climbs the plate the most, resulting in the highest spot This means that the dark circle corresponds to Band A This eliminates choices Band C By the same reasoning used to ascertain the dark spot, the hollow spot corresponds to Band C,making choice Athe best answer and thus, themost pickable answer 96 Choice D is correct Fractional distillation is used to separate liquids with boiling points that are close to one another This is achieved by increasing the surface area in the distilling column This makes choice A a valid statement, thereby eliminating it The boiling points are close to one another in fractional distillation, which makes choice B a valid statement, thereby eliminating it Fractional distillation is more selective, so the purity is high, but a great deal of residue is left behind in the distilling column, resulting in a lower yield with fractional distillation This makes choice D the best answer 97 Choice Ais correct The compound that gets extracted into sodium hydroxide solution is the organic compound that becomes an ion when treated with hydroxide Neutral acids become anions when they are deprotonated, so the answer is an organic acid The only organic acid is choice A, a carboxylic acid Choice A is the best answer 98 Choice B is correct Carboxylic acids are hydrophilic, although the aromatic ring helps with its lipid solubility It should be somewhat soluble in diethyl ether Despite the two hydrophilic groups on hydroxybenzoic acid, it is only slightly soluble in water, because of the benzene ring Carboxylic acids are rather easily deprotonated, so they can be converted into an anion and made to be more water soluble This makes a compound like benzoic acid highly solublein a basicenvironment The best answer is choice B 99 Choice C is correct Distillation involves boiling a liquid away from a mixture, so it works with two liquids that have different boiling points and don't form an azeotrope Choice A is eliminated Chromatography involves solutes migrating down a column at different rates, so it works for solutes Choice B is eliminated Crystallization dissolves and selectively precipitates a solid from a mixture of solids This eliminates choice D Extraction separates solids and liquids, but not gases Choice C is the bestanswer 100 Choice D is correct Butanone should be highly soluble in other ketones such as acetone, so choice A is eliminated Ketones don't oxidize with chromium (VI), so they show a negative Jone's test Choice Bis eliminated Butanone has a molecular mass of 72 grams per mole, so if it loses a methyl group during, it can form a fragment with a molecular mass of 57 amu Choice C is eliminated There are no chiral centers in butanone, so it cannot have optical activity This makes choice D the best answer Copyright © by The Berkeley Review® 300 LAB & SPECTROSCOPY EXPLANATIONS Writing a book of this nature ended up being a more difficult project than I first imagined back in 1990 For every one hundred questions you see in this book, about three hundred had to be written A great deal of time went into selecting which questions served the purpose of conceptual analysis, developing test-taking skills, honing table-reading skills, testing memory, and building confidence But as challenging as it was at times, it has been a wonderful experience This particular edition is the culmination of fourteen years of writing and rewriting Along the path I have incorporated the feedback and input of roughly 5000students I have tried my best to include many of the examples that have proven useful in lecture I have tried to keep in mind the sole purpose of this book in your life: to help you achieve a higher MCAT score in the biological sciences section and offer test-taking tips that are universally applicable It is imperative that you keep that in mind while working through our materials Your goal is to learn information and strategies that allow you to quickly pick the best of four choices when asked a multiple-choice question Completing this book would have never been possible without the support of many special people in my life I would like to thank: Cecile Santos: You have been an inspiration to my teaching I see the love you have in your eyes for the students you teach and it is awe-inspiring You have been my sounding board for teaching ideas and much of the text in this book Without you, I could never have done this Cambria and Madison: You fill my heart with so much love You keep me balanced at times I need it most Your laughter and smiles keep my spirits high and I am so lucky you came into my life You are the greatest children a father could ever want Our Wonderful Teachers: Over the years I have gleaned so many great ideas from you It has been an honor working alongside so many of you While so many of you have been helpful in my path, I would be remiss if I didn't specifically mention six very special teachers to me (although we have had many more.) Anthony Gopal showed me about the power of enthusiasm and how demonstrations and real life examples can drive classroom learning Moin Vera showed me how simple genius and fundamental concepts can make even the hardest topic seem easy Derek Welsbie showed me how simple logic, when applied with a little bit of information, can make every question easier to understand Sam Biggers showed me how bonding with the class and empathizing with their anxieties can build a bond that helps the entire team better Kim Gordy showed me the importance of silliness in gaining the students' interest and trust and how analogies can power learning Cecile Santos showed me the importance of engagement and open dialogue with her open forum discussion in lecture where every student is involved You are all incredible teachers and I learn something from you every time I watch you in action Thank you! Dale Schmidt: You have been incredibly supportive over the years, picking up the slack when it came Your work ethic and attention to detail is second to none I am always sure that if you did it, it was definitely done correctly Kimberly Gordy and Doug Coe: You spent so much time combing through my books to help them flow better I can't thank you enough, because I am not an easy person to work with when it comes to editing You managed to not only make this book better, but you made me realize the need for a different style and voice at many junctures Our Incredible Students: For the books I have written, I owe my biggest thank you to the thousands of students who have used them to help prepare for their MCAT Some of you are well into your practice and this is a distant memory Others of you have just recently finished your exam and are working your butt off to make it into medical school To each of you who offered some feedback, told me about a typo, asked me why a question was worded a certain way, or just simply said what you thought, I am deeply grateful This book, and the others I have written, is as much a part of you as it is a part of me This is our project and I am so happy you were a part of its development Copyright © by The Berkeley Review® 301 LAB & SPECTROSCOPY EXPLANATIONS Tfu PERKEJLEY rjR.F V.|^E-W'M PERIODIC TABLE OF THE ELEMENTS He H 4.0 1.0 Li Be 6.9 9.0 11 12 Na Mg 23.0 24.3 19 20 21 22 23 K Ca Sc Ti V 39.1 40.1 45.0 47.9 50.9 37 38 39 40 41 52.0 42 § 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 50 51 52 53 54 Sn Sb Te I Xe 118.7 121.8 127.6 126.9 131.3 84 85 86 Po At Rn (209) (210) (222) 43 44 45 46 47 48 49 In 114.8 75 76 77 78 79 Au Hf Ta Pt 180.9 183.9 186.2 Os 190.2 Ir 178.5 192.2 195.1 104 105 106 107 108 109 110 Rf Db 60 Ce Pr Nd 140.9 144.2 Pa Re 80 81 82 83 Hg Tl Pb Bi 197.0 200.6 111 204.4 207.2 209.0 112 Uuu Uub Uun Mt Hs Bh Sg (263) (262) (265) (266) (269) (272) (277) 140.1 Th 39.9 54.9 112.4 91 35.5 36 Ag 90 32.1 Kr 107.9 + 31.0 35 Pd 59 28.1 Br 106.4 58 27.0 34 Rh |(262) Ar Se 102.9 Ra 18 ci 33 Ru (223) 226.0 227.0 (261) 17 S As 101.1 89 R 16 P 32 Tc Ac§ 15 Ge (98) 88 14 Si 31 Mo 87 13 Al Ga 95.9 Fr 20.2 30 Nb W 19.0 Zn 92.9 138.9 16.0 29 Zr Ba 14.0 Cu 91.2 Cs 12.0 Ni Y 132.9 137.3 10.8 28 88.9 74 10 Ne 27 Sr 73 F Co 87.6 72 O Fe Rb 57t LaT N 26 85.5 56 C 25 Mn Cd 55 B 92 U 61 Pm 62 63 64 65 66 67 68 69 70 71 Sm Eu Gd Tb Dy Ho Er Tm Yb La 168.9 173.0 175.0 102 103 No (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 93 Np 232.0 (231) 238.0 (237) Specializing In MCAT Preparation 94 Pu (244) 95 Am 96 97 98 99 Cm Bk a Es (243) (247) (247) (251) 100 101 Fm Md (252) (257) (258) (259) Lr (260)