exam 1 (Orgo 2)

Question 1

1/2 amine/amide N-H stretches (IR)

Answer 1

primary: 2 peaks
secondary: 1 peak

Question 2

overtone (IR)

Answer 2

secondary absorptions that are generally twice the frequency
(carbonyl (C=O))

Question 3

aldehyde peak

Answer 3

2700-2830 cm-1 small doublet

Question 4

higher bond energy

Answer 4

higher frequency (higher energy), higher cm-1

Question 5

bond energy trend increase

Answer 5

single, double, triple
more EN

Question 6

s character

Answer 6

stronger bond, higher frequency

Question 7

DoU

Answer 7

(2+2C-H+N-X)/2

Question 8

alcohol fragmentation

Answer 8

alpha + dehydration

Question 9

amine fragmentation

Answer 9

alpha

Question 10

ketone + aldehyde fragmentation

Answer 10

alpha + McLafferty (4 C!)

Question 11

alpha mech

Answer 11

……………..⏺+……………………………….
C -❌- C - OH…………..➡️……..C = OH +
…………….- NH2………..➡️…….C = NH2 +
…………….= O……………➡️…….C ≡ O +

Question 12

dehydration mech

Answer 12

…………❌……………………
C - C - OH…..➡️…..C=C
|………………………………..
H ❌…………………………..

Question 13

McLafferty

Answer 13

H is 4 C away
……………………..⏺+
………………………O
………↙️⬅️⬅️⬅️||
……H………………C
……|⤵️……….↱/..\
R - C - C -❌- C…..R/H

⬇️

……..⏺+
……..O-H
……..|
……..C
…..//..\
… C….R/H

+ R-C=C

Question 14

-15

Answer 14

*CH3

Question 15

-18

Answer 15

H2O

Question 16

-29

Answer 16

CH2CH3 or HCO

Question 17

-35/37

Answer 17

Cl

Question 18

-43

Answer 18

C3H7

/\

Question 19

-79/81

Answer 19

Br

Question 20

cm-1, m/z, ppm

Answer 20

IR, mass spec, NMR

Question 21

M:M+2 = 1:1 (mass spec)

Answer 21

Br

Question 22

M:M+2 = 3:1 (mass spec)

Answer 22

Cl

Question 23

IR: aldehyde > ketone

Answer 23

less e- donation, so more double bond character, making for a higher frequency

Question 24

angle strain in rings (IR)

Answer 24

more angle strain (less Cs), more energy, higher wavelength

Question 25

mass spec instrument range

Answer 25

can range from 60-1200 MHz, but usually 300-400 MHz

Question 26

anti-parallel spin state (NMR)

Answer 26

higher in energy, against B0, resonance

Question 27

as B0 increases in strength

Answer 27

applied resonance frequency strength increases (to bring the nuclei into resonance)

Question 28

induced field

Answer 28

Every nuclei in existence is surrounded by e-. The orbiting and circulating e- generate their own, tiny magnetic field. It shields the nuclei, making it require less radiofrequency (UPFIELD)

Question 29

downfield

Answer 29

deshielded, left, more radiofrequency feels more of the applied field

Question 30

upfield

Answer 30

shielded, right, less radiofrequency feels less of the applied field

Question 31

delta scale

Answer 31

used to standardize observed frequencies for NMR remember resonance frequency is dependent on the strength of the applied B0 The x-axis of an NMR spectrum is called the delta scale. It shows the position of resonance of each nucleus relative to a standard (TMS) and has units of ppm.

Question 32

resonance frequency

Answer 32

the frequency that brings the NMR-active nuclei into resonance, dependent on nuclei itself and B0. Directly proportional to B0.

Question 33

applied field

Answer 33

B0

Question 34

field anisotropy

Answer 34

the orientation of nuclei in the magnetic field determines the chemical shift ⬆️..........➡️........🔄 .......⬅️ ⬆️......⬆️....⬇️............⬇️....⬆️ ⬆️..........⬅️........🔄........➡️ B0 ⬇️ shielding / ⬆️ deshielding 🔄 pi e- circulating

Question 35

field anisotropy - double bond

Answer 35

Pi bonds have electron current above and below the plane. The direction of the induced magnetic field by the pi-electrons is ADDED to the applied field through the hydrogen, therefore the hydrogen is DESHIELDED ⬆️..........➡️........🔄 .........⬅️ ⬆️......⬆️....⬇️.....=.......⬇️....⬆️ ⬆️..........⬅️........🔄..........➡️ B0 ⬇️ shielding / ⬆️ deshielding 🔄 pi e- circulating

Question 36

field anisotropy - triple bond

Answer 36

the triple bond is like a cylinder so the pi-electrons completely circulate. Now the induced field is OPPOSING the applied field through the proton, making the proton SHIELDED from applied field ⬆️..........➡️........🔄 .......⬅️ ⬆️......⬆️....⬇️...|||...⬇️....⬆️ ⬆️..........⬅️........🔄........➡️ B0 ⬇️ shielding / ⬆️ deshielding 🔄 pi e- circulating

Question 37

field anisotropy - aromatics (rings)

Answer 37

aromatics have a lot of pi-electron density known as ring current induced field is ADDED to the applied field (even more now) through the protons, they are DESHIELDED ⬆️..........➡️........🔄 .........⬅️ ⬆️......⬆️....⬇️.....♨......⬇️....⬆️ ⬆️..........⬅️.........🔄..........➡️ B0 ⬇️ shielding / ⬆️ deshielding 🔄 pi e- circulating

Question 38

dihedral angles

Answer 38

J values descend from it the angle between two bonds originating from different atoms in a Newman projection greater dihedral angles = greater coupling constant

Question 39

cis/trans

Answer 39

Z/E

Question 40

diastereotopic protons

Answer 40

chemically non equivalent protons that each produce distinct chemical shifts. Z/E and cis/trans if there's a stereocenter (SC) in a molecule, any CH2 group's protons will also be diastereotopic... 1. the effect is felt most strongly when the CH2 is directly next to the SC 2. the proton signal may or may not be resolved on the NMR spectrum A stereocenter is an atom, typically carbon, that has four attachments that are different from each other.

Question 41

axial and equatorial Hs interconversion in NMR is...

Answer 41

...very fast at room temperature, it is faster than the NMR time scale can detect, so they average to a SINGLET

Question 42

acidic protons (O-H, N-H) exchange...

Answer 42

...faster than the NMR timescale. This is why alcohols and amines are typically contaminated with H2O or acidic or basic impurities. This can lead to proton exchange, which is very fast and results in an averaged uncoupled signal

Question 43

13 C NMR

Answer 43

same principles as 1H NMR! NO SPLITTING due to low abundace of 13C R4C singlet R3C-H doublet R2CH2 triplet RCH3 quartet **a proton decoupled 13C NMR tells you how many different carbons a molecule has!! shielding and symmetry are same

Question 44

integration in 13C NMR...

Answer 44

means nothing, so no correlation between # of C and peak height