Exam 1 Flashcards

1
Q
A

Fluorobenzene

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2
Q
A

Chlorobenzene

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3
Q
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Bromobenzene

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4
Q
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Iodobenzene

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5
Q
A

Toulene

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6
Q
A

Cumene

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7
Q
A

Ethylbenzene

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8
Q
A

Styrene

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9
Q
A

Ortho-xylene

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10
Q
A

Meta-xylene

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11
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Para-xylene

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12
Q
A

Phenol

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13
Q
A

Benzoic acid

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14
Q
A

Benzaldehyde

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15
Q
A

Acetophenone

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16
Q
A

Methyl benzoate

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17
Q
A

Anisole

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18
Q
A

Aniline

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19
Q
A

Nitrobenzene

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20
Q
A

Benzonitrile

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21
Q
A

Benzamide

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22
Q
A

Benzenesulfonic acid

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23
Q
A

Naphthalene

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24
Q
A

Anthracene

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25
Q
A

Toluidine (o-toluidine)

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26
Q
A

p-cresol and m-cresol

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27
Q
A

Salicylic acid

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28
Q
A

Acetyl salicylate

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29
Q

Priority of naming disubstituted groups

A
  1. Ester
  2. Acid (-COOH, -SO2H)
  3. Aldehyde and Ketone (-CHO, -C- w. double bond to O)
  4. Alcohol and Thiol (-OH, -SH)
  5. Amino (-NH2)
  6. Everything else (ether, halogen, -R, -OR (alkoxy) -NO2)
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30
Q

Naming Tri/Poly Substituted Benzene Derivative

A
  1. Find derivative with unique name and make that 1
  2. The second group will be whatever is closest, regardless of group attachment.
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31
Q

Criteria for Aromaticity

A
  1. Molecule is cyclic
  2. Molecule is planar (sp2 hybridized 120)
  3. Each atom in the ring is a conjugated p-orbital
  4. In this cyclic conjugated system there are (4n+2) pi electrons
    Huckel #’s = 2,6,10,14,18,etc.
    n= any integer
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32
Q
A

Furan / aromatic

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33
Q
A

Pyridine

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34
Q
A

2-phenylbutane / sec-Butylbenzene
Aromatic

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35
Q
A

Indole - Aromatic

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36
Q
37
Q
A

Cyclopentadienyl anion

38
Q

Rules for signals in 1H NMR

A
  1. Identify Unique Proton Environments: Each unique chemical environment in a molecule produces a separate signal in the NMR spectrum. Identical protons (chemically equivalent) will contribute to the same signal.
  2. Consider Symmetry: Molecules with symmetry often reduce the number of NMR signals because multiple protons may be in equivalent environments. For example:
    * In ethanol (CH₃CH₂OH), the three methyl (CH₃) protons are equivalent and give one signal.
    * In 1,2-dichloroethane (ClCH₂CH₂Cl), the two CH₂ groups are in an identical environment due to symmetry, so they give one signal.
  3. Check for Exchangeable Protons (OH, NH, COOH, etc.)
    * Hydroxyl (-OH) and amine (-NH) protons can exchange with solvent and sometimes appear as broad peaks or disappear in certain solvents.
    * Carboxylic acid (-COOH) protons usually appear as broad peaks in the 10-12 ppm range.
  4. Look for Neighboring Groups (Deshielding Effects)
    * Electronegative atoms (O, N, Cl, etc.) pull electron density away from nearby protons, shifting their signals downfield (higher ppm).
    * Alkyl groups (-CH₃, -CH₂-) shield protons, moving signals upfield (lower ppm).
  5. Use the n + 1 Rule to Determine Splitting
    * Multiplicity (splitting patterns) follow the n + 1 rule, where n is the number of equivalent protons on adjacent carbons.
    o Singlet (s) → No neighboring protons (n = 0)
    o Doublet (d) → One adjacent proton (n = 1)
    o Triplet (t) → Two adjacent protons (n = 2)
    o Quartet (q) → Three adjacent protons (n = 3)
    o Multiplet (m) → Complex splitting from multiple neighbors
  6. Check for Equivalent Groups in Ring Systems or Conjugation
    * Aromatic rings have characteristic splitting patterns due to coupling between ortho, meta, and para protons.
    * Allylic and vinylic protons in conjugated systems appear in the 4-7 ppm range.
  7. Remember that Integration Reflects Proton Count
    * The relative area under each peak corresponds to the number of protons contributing to that signal. Example: A peak with twice the integration of another means twice as many protons are in that environment.
39
Q

Multiplicity/splitting in 1H NMR

A

Use the n + 1 Rule to Determine Splitting
* Multiplicity (splitting patterns) follow the n + 1 rule, where n is the number of equivalent protons on adjacent carbons.
o Singlet (s) → No neighboring protons (n = 0)
o Doublet (d) → One adjacent proton (n = 1)
o Triplet (t) → Two adjacent protons (n = 2)
o Quartet (q) → Three adjacent protons (n = 3)
o Multiplet (m) → Complex splitting from multiple neighbors

40
Q

Rules for signaling in 13C NMR

A
  1. Each Unique Carbon Gives a Separate Signal
    - Every chemically distinct carbon in a molecule produces one signal in the ¹³C NMR spectrum.
    - Equivalent (symmetrical) carbons appear as a single peak because they have the same chemical environment.
  2. Consider Molecular Symmetry
    - Symmetrical molecules reduce the number of unique carbon signals.
    Example:
    Butane (CH₃CH₂CH₂CH₃) has two signals: one for CH₃ and one for CH₂.
    2-Methylpropane (isobutane, (CH₃)₂CHCH₃) has three signals due to symmetry.
  3. Carbons Directly Attached to Electronegative Atoms Shift Downfield (Higher ppm)
    - Sp³ carbons (alkanes, CH₃, CH₂, CH) appear upfield (0-50 ppm).
    - Sp² carbons (alkenes, benzene rings, carbonyls) appear downfield (100-220 ppm).
    - Carbons attached to oxygen, nitrogen, or halogens shift downfield (higher ppm).
  4. Double and Triple Bonds Shift Carbon Signals
    - Alkene (C=C) carbons appear around 100-150 ppm.
    - Aromatic (benzene ring) carbons appear in the 110-160 ppm range.
    - Alkyne (C≡C) carbons are upfield (~70-100 ppm) due to shielding effects.
  5. Carbonyl Carbons Have the Highest Chemical Shifts
    - Ketones and Aldehydes (C=O): 190-220 ppm
    - Carboxylic acids, Esters, and Amides (C=O): 160-185 ppm
  6. DEPT Spectra Help Differentiate Carbon Types
    - ¹³C NMR alone does not show splitting (due to low natural abundance of ¹³C and decoupling), but DEPT (Distortionless Enhancement by Polarization Transfer) helps distinguish types:
    DEPT-135:
    CH₃ and CH carbons → Positive signals
    CH₂ carbons → Negative signals
    DEPT-90:
    Only CH carbons appear
    Regular ¹³C NMR:
    All carbons appear, including quaternary (C without H)
  7. Carbon Coupling Effects (Multiplicity)
    Typically, ¹³C NMR spectra are recorded as decoupled spectra (broad peaks without splitting), but if ¹H-¹³C coupling is observed:
    CH₃ (methyl) → Quartet
    CH₂ (methylene) → Triplet
    CH (methine) → Doublet
    Quaternary C (no H) → Singlet
  8. Check for Equivalent Carbon Atoms in Rings and Conjugated Systems
    Benzene rings and symmetrical cyclic structures reduce the number of signals due to equivalent carbons.
41
Q

¹³C NMR Chemical Shift Chart

42
Q

______ protons experience a net magnetic field strength that is smaller than the applied magnetic field

43
Q

______ protons experience a net magnetic field strength that is greater than the applied magnetic field

A

Deshielded

44
Q

In the presence of unpaired electrons, ______ nuclei experience extreme downfield shifts due to strong local magnetic fields.

A

Paramagnetic

45
Q

______ protons are surrounded by electron density that generates an opposing magnetic field, reducing the effective field they experience.

A

Diamagnetic

46
Q

Which of the following compounds displays two triplets and a singlet in its 1H NMR spectrum?
A. CH3CH(OCH3)2
B. CH3OCH2CH2OCH3
C. CH3OCH2CH(OH)CH3
D. CH3OCH2CH2CH2CH2OCH3
E. CH3CH2OCH2CH3

A

CH3OCH2CH2CH2CH2OCH3

47
Q

What is a conjugated diene

A

A conjugated diene is a system in which two double bonds are separated by a single bond (i.e., alternating single and double bonds). This allows for electron delocalization across the π system, which significantly impacts the molecule’s stability, reactivity, and spectroscopic properties.

48
Q

Rules to Determining MO Notation for HOMO & LUMO

A
  1. Determine the Number of π Molecular Orbitals

For a conjugated diene or polyene with nn π-electrons, there will be nn molecular orbitals.
Each π-bond contributes 2 electrons to the system.

Example: 1,3-Butadiene (4 π-electrons)

2 π-bonds → 4 π-electrons
This system will have 4 molecular orbitals labeled π₁, π₂, π₃, π₄** (in increasing energy).
  1. Assign π Molecular Orbitals (Energy Levels)

The π molecular orbitals are filled following Aufbau’s principle, meaning the lowest energy orbitals fill first.

For 1,3-Butadiene (4 π-electrons, 4 orbitals):

π₁ (Lowest energy, bonding orbital) → Occupied
π₂ (Second lowest, bonding orbital) → Occupied
π₃ (First antibonding orbital) → Unoccupied*
π₄ (Highest energy, strongly antibonding) → Unoccupied*

HOMO = Highest Occupied Molecular Orbital = π₂
LUMO = Lowest Unoccupied Molecular Orbital = π₃*
49
Q

Wavelength vs Conjugation

A

Larger conjugated systems have smaller HOMO-LUMO gaps, shifting absorption to longer wavelengths (red shift)

Non-conjugated systems will have the shortest wavelengths

50
Q

When an α-spin state nucleus absorbs radio frequency radiation of appropriate frequency, it undergoes what is called __________ .
(A) Chemical shift
(B) Electronic excitation
(C) Spin-flip
(D) Relaxation
(E) Economic inflation
(F) Crying for help

51
Q

Which of the following is true about the relationship between the energy gap (ΔE) between the spin states for a 1H nucleus and the strength of the external magnetic field?
A. they are inversely proportional
B. they are directly proportional
C. there is no relationship
D. the magnetic field is slightly less
E. none of these

A

they are directly proportional

52
Q

Consider the molecular orbital of benzene. In the ground state how many molecular
orbitals are filled with electrons? (2 points)
a. 1
b. 2
c. 3
d. 4
e. 6

A

3

Remember: 3 pi bonds = 6 p-orbitals with 3 filled and 3 unfilled (anti)

53
Q

What happens when a conjugated polyene absorbs UV-Vis radiation?

A

The electrons become excited from the radiation, absorb the energy, and move from HOMO to LUMO.

54
Q

AE aromatic

A

addition-elimination aromatic where an electrophilic aromatic substitution (EAS) reaction occurs not through the typical arenium ion intermediate, but instead via an initial addition step followed by elimination, essentially adding a substituent to an aromatic ring through a non-classical pathway

55
Q

Which of the following is true about the location of signals in a 1H NMR spectrum?
a. It indicates the number of neighboring protons
b. It indicates the electronic environment of neighboring protons
c. It indicates the number of different protons
d. It indicates the electronic environment of absorbing protons
e. It indicates the number of protons in the signal

A

It indicates the electronic environment of absorbing protons

56
Q

Which of the following compounds displays two triplets and a singlet in its H NMR
spectrum?
a. CH3CH(OCH3)2
b. CH3OCH2CH2OCH3
c. CH3OCH2CH(OH)CH3
d. CH3OCH2CH2CH2CH2OCH3
e. CH3CH2OCH2CH3

A

CH3OCH2CH2CH2CH2OCH3

57
Q

Nomenclature for Ethers

58
Q

Acid catalyzed reaction of alcohols (Synthesis of Ethers)

59
Q

Williamson Ether Synthesis (Synthesis of Ethers)

60
Q

Alkoxymercuration of Alkenes (Synthesis of Ethers)

61
Q

Acidic Cleavage (Rxn of Ethers)

62
Q

E1/SN1 (Rxn of Ethers)

63
Q

What are Epoxides

64
Q

Alkene & Peroxy acid (Synthesis of Epoxides)

65
Q

Synthesis of Epoxides from Halohydrins

66
Q

Acid-catalyzed epoxide opening

67
Q

Base catalyzed epoxide opening

68
Q

NMR Magnetic field

69
Q

Chemical shift

70
Q

Characteristics of chemical shifts

71
Q

Number of signals

72
Q

Intensity

73
Q

Splitting/spin-spin coupling

74
Q

Complex coupling

75
Q

13C NMR differences

76
Q

Conjugated Pi System Types

77
Q

Dienes (Conjugated Pi System)

78
Q

Electrophilic Addition Rxn (Conjugated Pi Systems)

79
Q

Diels Adler Rxn (Conjugated Pi System)

80
Q

Sigmatropic Rearrangements (Conjugated Pi Systems)

81
Q

Aromatic Compounds

82
Q

Aromatic Nomenclature - Monosubstituted benzene derivatives

83
Q

Aromatic Nomenclature - Disubstituted benzene derivatives

84
Q

Aromatic Nomenclature - Tri/Polysubstituted benzene derivatives

85
Q

What makes something aromatic

86
Q

Aromatic Radical Rxns

87
Q

Oxidation of alkyl benzene

88
Q

Birch Reduction (Benzene)