Missed organic chemistry - U World
C.The product will distill first because hexanoic acid has a higher boiling point.
A majority of excess hexanoic acid is removed from the product in the extraction but a small amount remains, making the product impure. Which statement accurately describes the subsequent distillation of the product? A.Hexanoic acid will distill first because it has a lower boiling point than the product. B.Decreasing the pressure of the distillation will improve the purification of the product. C.The product will distill first because hexanoic acid has a higher boiling point. D.Increasing the length of the fractionating column will not give separation of the product and hexanoic acid
DF=VT1/ VF1× VT2 / VF2 Substituting values for the transfer volumes (VT1 and VT2) and final volumes (VF1 and VF2) yields DF=0.5 mL/ 10 mL × 0.5 mL/ 1.0 DF=F= 1/20×1/2=1/40 Therefore, the lipid extract was diluted by a factor of 40.
According to Experiment 2, the samples from the lipid extract were diluted by what factor to generate the NMR samples? A.2 B.20 C.40 D.80
D. Because this carbon atom does not have any alkyl groups, it is the least substituted and therefore the least sterically hindered nucleophile.
An alkoxide nucleophile is used instead of iodide in Experiment 1. Which of the alkoxides is the LEAST sterically hindered?
D.80 If a value is close to your answer but not exact... still go with closest! The amount of bacteria recovered is slightly less than 104 (or 10,000). Therefore, the number of bacteria recovered is slightly less than a 100-fold reduction. Of the options available, 80 is the closest to 100
By approximately what factor were the bacteria recovered from hands reduced when hands were washed with 3 mL of soap versus no soap? A.13 B.2 C.15 D.80
A.downfield because the ester deshields He and Hf. An electronegative substituent withdraws electrons and deshields neighboring protons from the magnetic field, causing a downfield signal (larger chemical shift).
How does the addition of the ester to Compound 1 affect the shift of He and Hf? The signals for He and Hf will move: A.downfield because the ester deshields He and Hf. B.downfield because the ester shields He and Hf. C.upfield because the ester deshields He and Hf. D.upfield because the ester shields He and Hf
The peptide bradykinin contains the following amino acids: arginine (Arg, R), proline (Pro, P), glycine (Gly, G), phenylalanine (Phe, F), and serine (Ser, S). -Arg has a positively charged side chain, which is basic - Ser has a polar uncharged side chain. Both residues are considered hydrophilic
How many hydrophilic amino acid residues are present in bradykinin? (Bradykinin, a vasodilator containing nine amino acids (Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg) A.0 B.3 C.5 D.6
A.A ketone and an aldehyde A retro-aldol reaction: **** breaks α and β carbons and H20 is added to form ketones or Aldehydes or both aldehydes form if the carbonyl and or β carbon has hydrogen subsituents, whereas ketones form if all substituents are carbon chains
If Compound 4 is heated and treated with a base, which of the following functional groups would result from this reaction? A.A ketone and an aldehyde B.A ketone and an alcohol C.Two aldehydes D.Two ketones
C.Hexanes "C18 hydrocarbon HPLC" = reverse HPLC reverse HPLC = nonpolar stationary phase C18 hydrocarbon & polar mobile phase (water, methanol, acetone)
If a C18 hydrocarbon HPLC column is used for analysis of the Scheme III reaction results, which of the following solvents would NOT be used for the mobile phase? A.Acetone B.Methanol C.Hexanes D.Water
C.Conversion of the alcohol to a mesylate The alcohol can be converted into a mesylate (-SO3CH3) to improve its leaving group ability. However, an alcohol's leaving group ability can be improved through the formation of a mesylate (-SO3CH3) or a tosylate (-SO3C6H4CH3). The mesylate and tosylate anions can stabilize their negative charges by the delocalization of electrons through resonance.
If an alcohol were to undergo a substitution reaction, which of the following modifications would improve the alcohol's leaving group ability? A.Protection of the alcohol B.Deprotonation of the alcohol C.Conversion of the alcohol to a mesylate D.Oxidation of the alcohol
C.Mannose **not glucose. Glucose differs from Compound 1 in the orientation of the substituent at C-2. Glucose and mannose are epimers.
If researchers needed to synthesize Compound 1 before beginning the synthesis of Compound 5, which of the following sugars would require the fewest steps if used as a starting material? A.Fructose B.Ribose C.Mannose D.Glucose
A.Potassium phthalimide is a starting material. gabriel synthesis = a method that uses potassium phthalimide and diethyl bromomalonate as starting materials to synthesize amino acids. The starting reagents of the Gabriel and Strecker syntheses are all planar; therefore, the product is a mixture of L- and D-amino acids. (The synthesis begins with an SN2 reaction,)
If researchers used the Gabriel synthesis to make the dansylalanine backbone, which of the following statements is true? A.Potassium phthalimide is a starting material. B.Diethyl malonate is a starting material. C.Potassium cyanide is a starting material. D.An aldehyde is a starting material
B.a tertiary alcohol cannot be oxidized. a tertiary alcohol can not be oxidized because it doesn't have any C-H bonds to lose!! oxidation of an organic molecule requires a decrease in the number of C-H bonds and an increase in the number of C-O bond
If the chromic acid reagent is added to Compound C, no reaction occurs because: A.a tertiary alcohol cannot be reduced. B.a tertiary alcohol cannot be oxidized. C.an aldehyde cannot be oxidized D.an aldehyde cannot be reduced.
B.n → π* when absorbing light, we will be exciting electrons from a lower energy to higher energy orbital A would be true (at 240 nm) but specified nonbonding + 280 band so we know we need to use nonbonding n = nonbonding and is a region between the bonding and antibonding MOs so still goes higher in energy C and D = EMISSION (high energy to low energy)
In the ultraviolet (UV) range, pyrimidine has maximum absorptions at 240 nm and 280 nm, and the absorption at 280 nm is due to nonbonding electron excitation. The nonbonding electrons on nitrogen in pyrimidine undergo which of the following transitions when absorbing UV light? A.π → π* B.n → π* C.π* → π D.π* → n
C.ketoses tautomerize to aldoses. Tollen's Test = looks for reducing sugars which are aldoses *most ketoses can tautomerize to aldoses so they can also give a (+)
Ketoses such as fructose are expected to give a positive Tollens test because: A.ketoses cannot mutarotate. B.ketoses are not reducing sugars. C.ketoses tautomerize to aldoses. D.ketoses are not hemiacetals
C.HCl protonates the carboxylate, and the fatty acid is found in the organic layer. Fatty acids sodium salts are normally soluble in water due to ion-dipole interactions of the charged carboxyl group and sodium ion with water. Protonation of fatty acid sodium salts renders the molecule insoluble in water because the fatty acid is no longer a charged species, and therefore cannot interact as readily with water. The long-chain hydrocarbon makes fatty acids hydrophobic, allowing for solubility in organic solvents.**** Hexanes are nonpolar organic molecules that can readily interact with the protonated fatty acids. Therefore, protonated fatty acids will enter the organic layer after extraction with the organic solvent hexanes.
Researchers separated fatty acid sodium salts from glycerol by adding HCl and then extracting with hexanes. Which statement correctly describes the extraction process? A.HCl protonates the carboxylate, and the fatty acid is found in the aqueous layer. B.The carboxyl group is deprotonated, and the fatty acid is dissolved in the polar solvent. C.HCl protonates the carboxylate, and the fatty acid is found in the organic layer. D.The carboxyl group is deprotonated, and the fatty acid is dissolved in the nonpolar solvent
A.CH3(CH2)2CH3, pKa = 50 higher pKa = less acidic = more basic = most nucleophilic
Suppose that Experiment 1 was done in diethyl ether instead of acetone, and NaI was replaced with another nucleophile. The conjugate base of which of the following compounds would be the most nucleophilic? A.CH3(CH2)2CH3, pKa = 50 B.(CH3)2NH, pKa = 40 C.CH3CH2OH, pKa = 15.9 D.HF, pKa = 3.2
D.Addition of a resolving agent in order to separate enantiomers, need to add chiral molecule which turns them into diastereomers which can be separated resolving agent = chiral molecule Once the diastereomers are separated, the resolving agent is removed, yielding the original molecules as single enantiomers.
The (R)-enantiomer of the antiasthma drug albuterol is the active isomer. If a researcher wants to separate a racemic mixture of albuterol, which of the following methods will most likely separate the enantiomers? A.Extraction with dilute base B.Thin-layer chromatography C.Fractional distillation D.Addition of a resolving agent
B less affinity will elute first (undesired product)
The HPLC analysis of the Shi epoxide revealed a 10:1 mixture of isomers in favor of the desired isomer. Based on the chromatogram shown in Figure 3, which of the following most likely depicts the chromatogram of the mixture? (A chiral column that had an affinity for the desired epoxide was used. A standard of the desired epoxide was obtained to optimize the conditions and determine the retention time of the epoxide.)
D.ensure that glycosidases do not degrade the acceptor substrate it is necessary to ensure that microsomes do not cause side reactions such as degradation of the reaction substrates. In the absence of donor substrate, glycosyltransferases in the microsome will not catalyze any reactions. This will allow for conclusions to be drawn about whether the new product in lane 5 is due to glycosidase activity or glycosyltransferase activity.
The assay in lane 4 most likely functions to: A.determine if the enzyme is necessary for glycosyl transfer. B.verify the identity of products A and B. C.demonstrate how HCT affects the donor substrate. D.ensure that glycosidases do not degrade the acceptor substrate
A.A reverse-phase HPLC column was used with a nonpolar stationary phase and polar mobile phase. tryptophan is more polar than compound 2** (look at O groups) therefore since tryptophan eluted first, it must be more attached to mobil phase and so mobile phase must be polar
What can be inferred about the HPLC column used in the passage if tryptophan has a shorter retention time than Compound 2? A.A reverse-phase HPLC column was used with a nonpolar stationary phase and polar mobile phase. B.A reverse-phase HPLC column was used with a polar stationary phase and nonpolar mobile phase. C.A normal-phase HPLC column was used with a polar stationary phase and nonpolar mobile phase. D.A normal-phase HPLC column was used with a nonpolar stationary phase and polar mobile phase.
D.The peaks are cis-trans isomers of docosenoic acid. **says no fragmentation in the passage!! must be cis/trans vs constitutional since they have the same name and #9 tells us the location of the pi bond
What can be inferred from Figure 1 about the relationship of Peaks A and B? A.Nothing can be inferred about the relationship of the peaks. B.The peaks are fragments of docosenoic acid. C.The peaks are constitutional isomers of docosenoic acid. D.The peaks are cis-trans isomers of docosenoic acid.
B.Conjugated pi bonds A conjugated system of alternating single and double bonds in a molecule causes electron delocalization. This delocalization allows the molecule to absorb a photon in the ultraviolet or visible region, exciting an electron to a higher energy state. (chirality does not contribute to a compounds ability to fluoresce)
What feature of dansylalanine allows for monitoring protein unfolding? A.Isomerization of double bonds B.Conjugated pi bonds C.Resonance structure at the C-terminus D.Chirality of the amino acid
B.α-D Highest number chiral carbon has R stereochemistry = D Sugar For D sugars, the anomeric carbon substituent is below plane of the ring, the sugar is the α-anomer
What is the configuration of the sugar in the coumarin derivatives shown in Figure 1? A.α-L B.α-D C.β-L D.β-D
A.(2R, 3S, 5S)-2-amino-5-hydroxy-3-methyl-4-oxohexanal **DONT FORGET METHYL* remember longest chain can include groups shown with stereochemistry
What is the correct systematic name for the compound shown below? A.(2R, 3S, 5S)-2-amino-5-hydroxy-3-methyl-4-oxohexanal B.(2S, 4S, 5R)-5-amino-2-hydroxy-4-methyl-6-oxohexan-3-one C.(1S, 3S, 4R)-4-amino-1,3-dimethyl-2,5-dioxopentan-1-ol D.(2R, 3S, 5S)-5-hydroxy-3,5-dimethyl-1,4-dioxopentan-2-amine
D.Add 0.05 M H2SO4(aq) to the organic layer followed by 0.01 M NaOH(aq). to move an amine into aqueous, protonate it with strong acid!!
Which additional extraction steps would cause phosphatidylethanolamine and 2,6-dimethoxyphenol to enter the aqueous layer consecutively? A.Add 0.05 M NaOH(aq) to the organic layer followed by 0.01 M H2SO4(aq). B.Add 0.01 M NaHCO3(aq) to the organic layer followed by 0.05 M HCl(aq). C.Add 0.01 M H2CO3(aq) to the organic layer followed by 0.05 M NaHCO3(aq). D.Add 0.05 M H2SO4(aq) to the organic layer followed by 0.01 M NaOH(aq).
B.2-ethyl-3-methylpentanoic acid
Which carboxylic acid would need to be used to form Compound A (shown below) based on the reaction described in the passage? A.2-sec-butylbutanoic acid B.2-ethyl-3-methylpentanoic acid C.2,3-diethylbutanoic acid D.4-ethyl-3-methylpentanoic acid
B.L3 in the B&D method and L1 in the GC method lipids are the least polar component so will end up in the organic*** organic depends on the densities of each!!
Which layer in the B&D and GC methods in Figure 2 is expected to contain sterols isolated from the Y. lipolytica extract? A.L1 in the B&D method and L2 in the GC method B.L3 in the B&D method and L1 in the GC method C.L3 in the GC method and L2 in the B&D method D.L3 in the GC method and L1 in the B&D method
B.I and II only Gas-liquid chromatography is a technique used to separate molecules in a mixture based on their boiling points The mobile phase is an inert gas such as helium or nitrogen (Number I), and the stationary phase is a liquid that coats a solid support on the inside of the column (Number II). The vapors then travel through the column in a heated oven to the detector. The column must be placed in a heated environment to allow volatile molecules to remain in the gas phase. Columns kept at room temperature would not separate the molecules.
Which of the following characterize gas-liquid chromatography? I. Gas mobile phase II. Liquid stationary phase III. Separation based on polarity IV. Room-temperature column conditions A.I only B.I and II only C.I and III only D.I, II, and IV only
B **base catalyzed aldol first requires enolate formation!!
Which of the following compounds would be a plausible intermediate in the base-catalyzed aldol reaction of the first step of the citric acid cycle shown below?
D.A single spot on a thin-layer chromatography plate Therefore, TLC can be used to assess product purity by comparing a standard of pure theobromine to the isolated product. If a single spot with a larger Rf than that of theobromine is visible on the product sample track on the TLC plate, then the isolated product is pure.
Which of the following observations would indicate the reaction product is pure? A.One absorbance peak from size-exclusion chromatography B.Two peaks in a high-performance liquid chromatogram C.Two unique −CH3 signals in the 1H NMR spectrum D.A single spot on a thin-layer chromatography plate
D.Reduction with LiAlH4 LiAlH4 reduces carboxylic acids and esters to primary alcohols NaBH4 is wrong because its a reducing agent like LiAlH4 not oxidizing agent.
Which of the following reactions will convert hexanoic acid and the carboxylic acid derivative described in the passage to the same product? A.Base-catalyzed aldol condensation B.Decarboxylation C.Oxidation with NaBH4 D.Reduction with LiAlH4
C.The product is a mixture of L- and D-alanine. the Strecker synthesis is not a stereospecific reaction and produces a mixture of L- and D-amino acids.
Which of the following statements accurately describes the stereochemical outcome of the Strecker synthesis of alanine? A.The product is L-alanine. B.The product is D-alanine. C.The product is a mixture of L- and D-alanine. D.Alanine does not contain a stereocenter.
B. First see if cis vs trans then do clockwise walking up and dow tells us cis vs trans (axial up, equatorial up, equatorial down, axial down) **axial vs equatorial does not
Which of the following structures depicts the same molecule?
C.High-performance liquid chromatography (Thin-layer chromatography (TLC) separates small sample sizes and includes a stationary phase, solvent mobile phase, and detector (UV). However, TLC is not done under pressure. - d wrong)
Which separation technique is optimal for purification of small sample sizes and employs a stationary phase, a solvent mobile phase (under pressure), and a detector to separate compound mixtures based on polarity? A.Gas chromatography B.Extraction C.High-performance liquid chromatography D.Thin-layer chromatography
C.No, because they are enantiomers. Enantiomers rotate plane-polarized light by the same magnitude in opposite directions
Would (R,R)- and (S,S)-labetalol be expected to rotate plane-polarized light in the same direction? A.Yes, because they are diastereomers. B.Yes, because the rotations would have the same magnitude. C.No, because they are enantiomers. D.No, because the rotations would cancel out each other.
B.a carboxyl group. ****The amino group of one amino acid reacts with the carboxyl group of another amino acid to form a peptide bond (An amide is formed during peptide bond formation rather than reacting with an amino group to form a peptide bond.-C)
A peptide bond is formed in bradykinin synthesis when the amino group of arginine reacts with: A.an anhydride group B.a carboxyl group. C.an amide group. D.a hydroxyl group.
C.LiOH need a strong base*
A reaction between a long-chain anhydride and phenylethylamine is done to produce a long-chain amide and a carboxylic acid. Which of the following aqueous solutions can separate the products of this reaction in an extraction? A.NaHSO4 B.HNO3 C.LiOH D.HClO4
B.II only Fractional distillation works best for compounds that have boiling points that are <25°C apart****** (Vacuum distillation is appropriate for compounds that have boiling points >150°C, which may decompose when heated beyond this temperature.) ( simple distillation. It is best suited for compounds with boiling points <150°C and >25°C apart)
A student needs to separate a mixture of chloroform (bp 61°C) and benzene (bp 80°C). What type(s) of distillation would be expected to give the best separation of the two compounds? I. Simple distillation II. Fractional distillation III. Vacuum distillation A.I only B.II only C.I and III only D.II and III only
C.2 SILLY MISTAKE 3 fatty acids will be formed, but only 2 are distinct
A triacylglyceride isolated from a plasma sample is shown below. When the triacylglyceride is hydrolyzed, how many different products are obtained other than glycerol? A.0 B.1 C.2 D.3
D.15 min carboxylic acids have high boiling points** essentially able to dimerize for extra stability so will elute last
Acetic acid most likely eluted from the gas chromatograph at what time point? A.1 min B.2 min C.5 min D.15 min
A.enhancing the nucleophilicity of alcohols.
Acid helps catalyze the Fischer esterification reaction by doing all of the following EXCEPT: A.enhancing the nucleophilicity of alcohols. B.forming a stable carbocation intermediate. C.creating a good leaving group. D.increasing carbonyl electrophilicity.
D be sure that you are looking at a ring from the correct perspective to identify whether it matches if OH goes under the ring it would hit CH3***** clockwise rather than Br, so D is opposite
All of the following are equivalent to Compound 1 EXCEPT:
A.Yes, because they are ketoses that differ in configuration at C2 only. For aldoses such as glucose, the anomeric carbon is C1, whereas for ketoses such as fructose, the anomeric carbon is C2. α-D-fructose and β-D-fructose are diastereomers that differ in orientation at C2 (the anomeric carbon) only. Therefore, α-D-fructose and β-D-fructose are anomers.
Can α-D-fructose and β-D-fructose be classified as anomers? A.Yes, because they are ketoses that differ in configuration at C2 only. B.Yes, because they differ in configuration at more than one carbon. C.Yes, because they are C1 epimers. D.No, because they are enantiomers
A. CrO3 = strong oxidizing agent*** -if given to an aldehyde, ketone, or alcohol they will oxidize them as far as possible**, but cannot change R groups so will turn primary alcohol into carboxylic acid!!! (first becomes aldehyde and then keeps going
Compound D gives a positive result for the chromic acid test. What is the structure of the product formed by the reaction of Compound D with the chromic acid reagent?
B.S and O Therefore, a bond between S and O is formed during the synthesis of a tosylate. Cl acts as the leaving group, breaking the bond between S and Cl.
During the synthesis of a tosylate, which of the following pairs of atoms form a covalent bond? A.C and Cl B.S and O C.S and Cl D.C and O
B.the chiral tertiary halide undergoes an SN1 reaction, which forms a carbocation and is not stereospecific tertiary halide can stabilize carbocation so will undergo SN1
If a primary alkyl halide and a chiral tertiary halide both undergo substitution reactions, the reaction with the tertiary halide will have more products because: A.the primary halide undergoes an SN2 reaction, which occurs in one step and is not stereospecific. B.the chiral tertiary halide undergoes an SN1 reaction, which forms a carbocation and is not stereospecific. C.the nucleophile can add to the electrophile from either side of the primary alkyl halide. D.the leaving group on a tertiary halide is not sterically hindered
C.H2O A glycosidic bond is the α- or β-linkage between a sugar and an -OH of another molecule. Hydrolysis of a glycosidic bond is cleavage of the linkage by addition of H2O, breaking the molecule into two smaller units.
In the second step, Glu35 creates a good nucleophile for the cleavage of the glycosidic bond between Mur2Ac and Asp52 by removing a proton from which of the following molecules? A.NH3 B.H3O+ C.H2O D.OH
A.protect the α-amino group from nucleophilic reaction. you dont want the N attached to Boc to do the Nucleophilic attack!! you want the other NH2 group. Therefore Boc is large and will prevent this Because there are two amines in Compound 1, the one bonded to the α-carbon needs to be protected so it will not react. The Boc group provides steric hindrance to prevent the amino group from acting as a nucleophile.
In the synthesis of dansylalanine, the Boc group on Compound 1 most likely functions to: A.protect the α-amino group from nucleophilic reaction. B.protect the C-terminus from nucleophilic reaction C.make the N-terminus a better leaving group. D.make the carboxyl a better leaving group
A.S, because the configuration is inverted by the reaction draw starting product!!! then do inversion and see!!
Researchers produced ethylhexyl glycerin using a partially deuterated ethylhexyl halide and generated the product shown below. What was the configuration of the deuterated carbon prior to the reaction? A.S, because the configuration is inverted by the reaction B.R, because the configuration is inverted by the reaction C.S, because the configuration is preserved by the reaction D.R, because the configuration is preserved by the reaction
A.Transesterification new OR group formed!
The conversion of megestrol to megestrol acetate shown in Scheme 1 represents which kind of reaction? A.Transesterification B.Saponification C.Aldol condensation D.Oxidation
D.Carbon 4 only -amides require harsh conditions (strong acid or strong base and heat) to undergo hydrolysis -esters/carboxylic acid derivatives are more reactive and can hydrolyze under mild conditions
The precursor to Compounds 3-6 is shown below with carbonyl carbon atoms labeled. Which of the carbonyl carbon atoms is most likely to undergo hydrolysis when treated with a weak acid at room temperature? A.Carbon 1 only B.Carbons 1, 2, and 3 only C.Carbons 1 and 4 only D.Carbon 4 only
C.II and III only **be careful not to get confused with E1 vs E2 and Sn1 vs Sn2
The substrate tert-butylchloride reacts in Experiment 2 but not in Experiment 1 because: (experiment 1 = Sn2, experiment 2 = Sn1) I. the leaving group is bonded to a primary carbon. II. a stable carbocation forms in Experiment 2. III. the leaving group is bonded to a tertiary carbon. IV. an unstable carbocation forms in Experiment 2. A.I and II only B.I and IV only C.II and III only D.III and IV only
D.BH3 will selectively reduce carboxylic acids, and NaBH4 will selectively reduce ketones. Carboxylic acids can be selectively reduced to primary alcohols by the reducing agent BH3. They cannot be reduced by NaBH4, which is selective for the reactive carbonyls ketones and aldehydes.
Two separate reactions are conducted in which a compound containing a ketone, an ester, and a carboxylic acid is reacted with borane (BH3) in THF in one reaction and with NaBH4 in methanol in the other. Which of the following explains why different products are observed? A.BH3 will selectively reduce ketones, and NaBH4 will only reduce carboxylic acids. B.BH3 does not reduce ketones, and NaBH4 will selectively reduce esters. C.BH3 reduces carboxylic acids and esters, and NaBH4 will only reduce esters. D.BH3 will selectively reduce carboxylic acids, and NaBH4 will selectively reduce ketones.
A.An aldehyde and ammonia The Strecker synthesis is used to make α-amino acids from aldehydes using NH3 and potassium cyanide (KCN). The first step of the reaction proceeds with protonation of the carbonyl oxygen by H3O+, followed by nucleophilic attack of the carbonyl carbon by NH3, resulting in dehydration and imine formation. Therefore, an aldehyde and NH3 are used to form the imine intermediate in the Strecker synthesis.
What types of functional groups react to form the imine intermediate in the Strecker synthesis of amino acids? A.An aldehyde and ammonia B.An aldehyde and a secondary amine C.A ketone and a tertiary amine D.A ketone and a quaternary amine
A.2-chloro-2-(fluoromethyl)propane-1,3-diol SN2: primary >> secondary >>>>>>>>>>>>>tertiary (a lot faster on primary), therefore will happen at Br carbon and attach OH there
When the compound shown below undergoes an SN2 reaction with hydroxide, which of the following compounds will most likely form as the major product? A.2-chloro-2-(fluoromethyl)propane-1,3-diol B.2-(bromomethyl)-2-chloropropane-1,3-diol C.(2R)-3-bromo-2-(fluoromethyl)propane-1,2-diol D.(2S)-3-bromo-2-(fluoromethyl)propane-1,2-diol
D is what tert-butyl group looks like
Which structure is the product of the reaction of 4-tert-butyl-2-octanone and NaCN?
