Organic Chemistry Flashcards

Question

Diazo

Answer 1

**Decreases** bond length **Increases** bond energy

Answer 2

Same molecular formula, different structural formula Same in writing, different in drawing

Answer 3

1. Structural Isomers: same molecular formula, different connectivity 1. Positional Isomers: Structural Isomers that have the same functional groups positioned differently 2. Functional Isomers: Structural Isomers that have different functional groups

Answer 4

Same molecular formula, same connectivity, but different orientation across a double bond. * Cis = same side, Trans = opposite side * E/Z when different groups are attached to either side * Z = high priority groups on same side * E = high priority groups on opposite side

Answer 5

Same molecular formula, same connectivity, but different 3D arrangements across one or more chiral centers

Answer 6

2^#chiral- (#meso)

Answer 7

Mirror images. ALL chiral centers are reversed

Answer 8

More than one chiral center, but not all chiral centers are inverted

Answer 9

They have chiral centers, but as a molecule they are achiral and optically inactive

Answer 10

1. the same 2. the same 3. different

Answer 11

* Same molecular formula * Same connectivity * Same stereochemistry * Can rotate about a single bond

Answer 12

* Eclipsed * Syn-periplanar: highest torsional strain, unstable, bulky groups eclipse eachother * Anticlinal eclipsed: high torsional strain, unstable, bulky groups eclipse hydrogens * Staggered * Gauche: low torsional strain, stable, bulky groups 60° staggered * Anti: lowest torsional strain, most stable, bulky groups 180° staggered

Answer 13

* Chair: most stable, everything staggered * Twist boat: less stable, things not completely eclipsed * Boat: least stable, everything is eclipsed

Answer 14

EM fields in one direction

Answer 15

* Specific rotation: chiral molecules containing a single enantiomer will rotate polarized light ("optically active") * *l* = levorotatory * *d* = dextrorotatory

Answer 16

1. Assign priorities 2. Turn molecule so lowest priority group is at the back 3. Order 1st, 2nd, and 3rd priority groups 4. go right- R, go left- L

Answer 17

* Start with atoms directly bonded to chiral carbon * Atom with the higher MW has greater priority * If atoms are the same, look at the next shell * Atom with higher MW on second shell has greater priority

Answer 18

Mixtures that contain equal amounts of both enantiomers. (Racemate) Racemic mixtures do not rotate polarized light- optically inactive

Answer 19

* Convers enantiomers to diastereomers * Separation of diastereomers * Convert diastereomers back to enantiomers

Answer 20

Proteins are made of L-amino acids

Answer 21

* hydrophobic * LD forces present only * lower BP than compounds the same size with functional groups * Heptane has same BP as water

Answer 22

Alkane + O₂ → CO₂ + H₂O

Answer 23

Alkane + halogen + free radical initiator → alkyl halide * UV light and peroxides are free radical initiators

Answer 24

* Cyclopropane has the highest ring strain * Cyclobutane has the second highest ring strain * Cyclohexane has the lowest ring strain

Answer 25

* Angle: caused by deviation from the ideal sp³ tetrahedral bond angle of 109.5° * Torsional: molecule aving eclipsed conformations rather than staggered

Answer 26

Prefix: hydroxyl, hydroxy Suffix: -ol, alcohol

Answer 27

* Hydrogen bonding * Higher boiling point than the same compound without the alcohol group * water soluble * IR: 3300 cm^-1 and broad due to hydrogen bonding

Answer 28

R-OH + HX \<--\> R-X + H₂O * SN1: stable carbocation, 3° carbon center, protic solvent * SN2: unstable carbocation, primary carbon center, aprotic (but polar) solvent * need good LG

Answer 29

* unimolecular, intermediate carbocation formed

Answer 30

* bimolecular, passes through transition state

Answer 31

* Primary: Oxidized to carboxylic acids by KMnO₄ and CrO₃, Oxidized to aldehyde by PCC * Secondary: always oxidized to ketone * Tertiary: do not oxidize

Answer 32

* Protect by adding Cl-SiMe₃ to R-OH * Alcohol is "capped" into R-O-SiMe₃ * Deprotect: add F^-

Answer 33

R-OH + SOCl₂ --\> R-Cl (by products: SO₂ + HCl) R-OH + PBr₃ --\> R-Br (by products: H₃PO₃, R₃PO₃, HBr)

Answer 34

R-SO₃^- (good leaving groups) * R = Methane : methanesulfonate * R = Toluene : tosylate * R = trifluoromethane: triflate

Answer 35

Acid + Alcohol = Ester

Answer 36

Replace C of esters with a different atom

Answer 37

more acidic

Answer 38

More branching = higher freezing/melting point, lower boiling point

Answer 39

1. COOH **5** 2. ArOH **10** 3. H2O **16** 4. ROH **16** 5. -CH2(CO)-R **20** 6. -CH2(CO)-OR **25**

Answer 40

-al, -aldehyde

Answer 41

* Prefix: keto- , oxo- * Suffix: -one , ketone

Answer 42

* Polar C=O bond * Dipole-dipole interactions give high boiling points (higher than alkanes, lower than alcohols and carboxylic acids)

Answer 43

Aldehydes and ketones react with 1 equivalent of alcohol to make a hemiacetal. They react w/ 2 equivalents of alcohol to make acetals

Answer 44

Like hemiacetal and acetal but ketone is starting reactant

Answer 45

Primary amine + aldehyde or ketone = imine Secondary amine + aldehyde or ketone = enamine

Answer 46

1. Ketone + halogen = **halogenation of alpha C **(adjacent to C=O) 2. Methyl ketone + halogen = **haloform + carboxylate** 3. Trihalogenated methyl is a good **leaving group**

Answer 47

occurs because of acidic alpha proton

Answer 48

Aldehydes oxidize to **carboxylic acids**. Ketones do not oxidize further.

Answer 49

Enol has alcohol, keto has ketone (keto is more stable and predominates)

Answer 50

* make R^-, which attacks C=O to make R-C-OH * Make C-C bonds * R-X + Li → R-Li * R-X + BuLi → R-Li * R-Li + C=O → R-C-OH

Answer 51

Reduces C=O to -CH₂- C=O + NH₂NH₂ → -CH₂- + N₂

Answer 52

* Grignard reagents produce R^- * R-X + Mg → R-Mg-X * R-Mg-X + C=O → R-C-OH

Answer 53

Steric hinderance- blocks access to electrophilic carbon- reactivity goes down

Answer 54

Alpha proton is acidic because resulting carbanion is stabilized by resonance

Answer 55

α,β-unsaturated carbonyl + nucleophile → **addition of the nucleophile at the β position** * Nucleophiles attack the β H

Answer 56

-oic acid, carboxylic acid, -dioic acid

Answer 57

High BP due to hydrogen bonding soluble in water

Answer 58

Occurs on electrophilic C of C=O

Answer 59

Nucleophilic attack occurs by nucleophilic O of COOH

Answer 60

LiAlH4: COOH → alcohol

Answer 61

COOH + ROH under acidic conditions = ester

Answer 62

1. Convert to enolizable form (w/ PBr₃, make an acyl halide) 2. Acyl halide enolizes 3. Halogenate the enol w/ Br₂ 4. Revert back to carboxylic acid (or hydrolysis)

Answer 63

1. Carboxylic acid converted to acyl halide, which can enolize 2. Acyl halide tautomerizes to its enol form by abstractino of acidic alpha hydrogen 3. Halogen (or some other E+) gets attacked by the alpha position 4. Revert back to a carboxylic acid. The net effect is that the alpha H gets substituted by an electrophile

Answer 64

H bonding causes dimerization of carboxylic acids

Answer 65

pKa of COOH = **5** pKa of H+ = **0** pKa of water = **16** (COOH is a weak acid)

Answer 66

EWG makes the acid stronger (helps distribute charge of COO- and stabilize it)

Answer 67

Conjugate base (carboxylate ion) is stabilized by resonance

Answer 68

* suffix = -oyl chloride

Answer 69

-oic anhydride

Answer 70

Yes- amides can hydrogen bond because of the N-H group. Hydrogen bonding involving the amide backbone of polypeptides form the secondary structure of proteins.

Answer 71

C=O shows up greater than 1700 cm^-1, close to 1800 cm^-1

Answer 72

2 bands between 1700-1800 cm^-1

Answer 73

C=O shows up at 1700 cm^-1 C-O ether stretch shows up at 1200 cm^-1

Answer 74

Acid Chloride

Answer 75

Carboxylic Acid

Answer 76

Nucleophile attacks C of C=O

Answer 77

Takes away the C=O of an amide.

Answer 78

Ester + alcohol → new ester

Answer 79

Hydrolysis of fats and glycerides (esters) to a base

Answer 80

LG is a neutral amine

Answer 81

Acid chloride \> anhydride \> ester \> amide * Acid halides are the most reactive because halides are good LGs * Amides are the most stable derivatives because NR₂^- is a bad LG. The C-N also has partial double bond character (stable)

Answer 82

carboxylate ion (COO- can redistribute the negative charge via resonance and thus stabilize)

Answer 83

No- it has double bond character Rings with amides have higher strain beta-lactam: 4-membered ring with 1 amide

Answer 84

4-membered ring with 1 amide Has higher ring strain (double bond character of amide)

Answer 85

beta-keto esters → beta-keto acids → enols → ketos

Answer 86

1. acidic alpha proton comes off, resulting carbanion attacks new R group 2. Hydrolysis of ester turns it into a β-keto carboxylic acid. 3. β-keto acids undergo decarboxylation because β-keto group stabilizes the resulting carbanion via enol formation. 4. Net effect: R group is made to attach to the alpha carbon of acetone

Answer 87

H alpha to a carbonyl group is more acidic than a regular H

Answer 88

* Solutes are distributed between two immiscible solvents * Organic phase usually less dense than aqueous phase. EXCEPT chloroform, which is more dense.

Answer 89

* Separates liquids based on **boiling point** * Simple distillation: done with a normal column, separates 2 liquids with large difference in BP * Fractional distillation: done w/ fractionating column, separates liquids with smaller differences in BP * Vacuum distillation: under low pressure, lowers BP for all components, doesn't require increasing T as much

Answer 90

* Molecules move slower if more attracted to stationary phase and repelled by mobile phase * Molecules move faster if attracted to the mobile phase and repelled by stationary phase

Answer 91

* stationary phase is a liquid coated on inside of a column * Mobile phase is gas * Substrate equilibrates between mobile (gas) and stationary (liquid coat) phase * Substances w/ greater affinity for stationary phase come out of column slower. * Polar substrate has more affinity for polar stationary phase, hydrophobic substrate has more affinity for hydrophobic stationary phase

Answer 92

* solvent = mobile phase, paper = stationary phase * Separate pigment in dyes * Pigments stick to paper * R_f value = d_pigment/d_{solvent front} * R_f=0 : pigment has not moved * R_f=1 : pigment moved as far as solvent

Answer 93

* Like paper chromatography * Instead of paper, use a plate coated with a specific stationary phase of your choosing

Answer 94

* Barely dissolve compound, let it recrystalize out of solution, compound ends up more pure * Use just enough solvent to make a warm saturated solution * As solution cools, solubility decreases, compound comes out of solution * Choose a solvent that your compound is soluble in at a warm temp, but not at a cool temp * Impurities should be highly soluble in solvent at all temperatures

Answer 95

* Vibrations: bonds can stretch, compress, and bend like a spring * Rotations: molecules rotate, most produce waves in microwave region, which overlaps with vibration spectra

Answer 96

* Transmittance vs. decreasing wavenumber

Answer 97

Anything with a H atom (O-H, N-H, C-H)

Answer 98

Does not involve H. C=O, C=C, C-C, C-O

Answer 99

1700 cm^-1

Answer 100

O-H, N-H, or alkyne C-H * O-H is the broadest, N-H is slightly sharper, alkyne C-H is very sharp * Broad peaks are due to H-Bonding in O-H and N-H

Answer 101

Primary colors of pigments (yellow, magenta, cyan) is exactly the complementary colors of the primary colors of light

Answer 102

Red - very acidic Orange - acidic Yellow - weakly acidic Green - neutral Blue - basic Purple - very basic

Answer 103

Conjugated systems **decrease** the energy of electromagnetic radiation that is absorbed. The more conjugated double bonds there are the **longer** the wavelengths of absorbed radiation.

Answer 104

* bombard a molecule into electrons- it is fragmented into ions * faster electrons produce more fragmentation and smaller molecular ion peaks * relative abundance vs. m/z (mass to charge ratio) * Magnetic field resolves the different ions * Peaks clustered close together are isotopes * parent peak = molecular ion peak (depicts the ion without fragmentation) * Base peak is the tallest (most abundant)

Answer 105

* measuring molecular weight of a molecule * identify the molecule by fragmentation patters * identify heteroatoms by their characteristic isotope ratios

Answer 106

* Measures chemical shift relative to a standard called TMS (tetramethylsilane) in ppm * The more "different" two protons are, the farther their chemical shifts * H is shielded by C, because C is not so electronegative * H is deshielded by O, because O is electronegative

Answer 107

* More shielded protons appear **upfield** **(to the right)** * More deshielded protons appear **downfield (to the left)**

Answer 108

One signal the height n times the signal for a single proton

Answer 109

* Produced by neighboring protons (Attached to adjacent atoms) * Split into n+1 peaks * J value defines how far apart things get split * Protons across single and aromatic bonds get split approximately the same * Protons across double bonds get split farther apart

Answer 110

* Prefix: amino- * Suffix: -amine

Answer 111

* 3° amines can be chiral, but the proton comes on and off. Always racemic because of spontaneous inversions. * 4° amines can be chiral and stay chiral

Answer 112

* Primary amines (R-NH₂) - 2 peaks around 3300 cm^-1 * Secondary amines (R₂-NH) - 1 peak around 3300 cm^-1 * Tertiary amines (R₃-N) - 0 peaks (no N-H bonds)

Answer 113

Amine + Acid derivative → amide

Answer 114

Peptide bond formation in protein synthesis (amine + carboxylic acid = amide)

Answer 115

Ar-NH₂ + HONO → Ar-N₂⁺ + H₂O + OH^- * Nitrous acid = HNO₂ = HONO * HONO → NO⁺ + OH^-

Answer 116

Amine + Methyl Iodide → exhaustive methylation of amine → elimination w/ methylated amine as LG * Unlike E1 reactions where more substituted double bond is formed (Zaitsev), Hoffman Elimination uses less substituted double bond (Hoffman)

Answer 117

Amines are basic. They like to gain a proton. R-NH₂ → R-NH₃+ * Difficult for neutral amines to lose a proton * An amide can lose a proton more easily

Answer 118

Electron **donating** groups increase the basicity of aromatic amines. Electron **withdrawing** groups decrease the basicity of aromatic amines. Anything ortho to the amine will **decrease** the basicity.

Answer 119

* Carb is a sugar, monosaccharide, disaccharide, polysaccharide * Prefix: deoxy * Suffix: -ose

Answer 120

1. Aldose: sugars with aldehyde group 2. Ketose: sugars with ketone group 3. Pyranose: sugars in 6-membered ring 4. Furanose: sugars in 5-membered ring 5. #ose: sugar with # carbon atoms

Answer 121

* At least a 3-Carbon backbone * an aldehyde or ketone group * At least 2 hydroxyl groups

Answer 122

glucose, fructose, galactose

Answer 123

Ribose makes up DNA. Deoxyribose makes up DNA.

Answer 124

disaccharide made from α-glucose and β-fructose joined at the hydroxyl groups on the anomeric carbons (making acetals)

Answer 125

Disaccharide made from β-galactose and α/β-glucose joined by a 1-4 linkage.

Answer 126

Glucose molecules joined by α1-4 linkage.

Answer 127

Same as starch. Glucose molecules joined by α1-4 linkage, but with additional α1-6 linkage for branching.

Answer 128

Glucose forms a pyranose when C5 attacks the carbonyl carbon

Answer 129

* -OH groups on the Left become Up * -OH groups on the Right become Down * -OH group on the anomeric carbon is up (beta) or down (alpha) * CH₂OH group on C5 points up for D, down for L

Answer 130

* Epimer: different configuration in just one chiral carbon * Anomer: different configuration in the chiral, anomeric C when the molecule is in cyclic form * Anomers are special types of epimers

Answer 131

* Acetal linkage involving the hydroxyl group of the anomeric carbon * Glycoside linkage can also mean the linkage between the sugar and the base in nucleotides * Hydrolized same way as acetal bond is hydrolyzed * glycoside + H2O + catalyst → hydrolysis. * Amylase is catalyst for starch * Glycosylase for nucleotide

Answer 132

L are more common in nature

Answer 133

* low pH: cationic * high pH: anionic * pH=pI : neutral zwitterion

Answer 134

Aspartic acid, glutamic acid | (R contains carboxylic acid)

Answer 135

Lysine, Arginine, Histidine | (R contains amine)

Answer 136

Requires a strong base or a biological enzyme

Answer 137

* Primary structure: sequence (N terminus to C terminus) * Secondary structure: * repetitive motifs from backbone interactions * H-Bonding between NH and C=O * alpha helices and beta sheets * alpha helix is right handed (R groups stick out) * Beta sheets: R groups stick out above and below sheet

Answer 138

Made from the cyclization of squalene (a terpene) * Cholesterol * Testosterone * Estrogen

Answer 139

* Polymerization of isoprene * Squalene has 6 isoprene units

Answer 140

Glycerol + 3 Fatty acids → Triacyl Glycerol.

Answer 141

Simplest phosphoric acid anhydride

Answer 142

Phosphodiester bonds link together the DNA and RNA backbone. Phosphoester bonds link the phosphates to sugar in ATP.

Answer 143

Carbonyl + Phosphorus Ylide → Alkene

Organic Chemistry Flashcards

(224 cards)