Chapter 11: Spectroscopy Flashcards

1
Q

infrared (IR) spectroscopy

A

infrared light passed through a sample, and absorbance is measured.

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2
Q

useful absorptions in sectroscopy are:

A

2500 to 25,000 nm which corresponds to 4000 to 400 cm-1 in wavenumber

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3
Q

what does not show up in a infrared (IR) spectroscopy

A

symmetric stretches beacuse they involve no net change in dipole moment

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4
Q

O-H

A

broad around 3300 cm-1 for alcohol and 3000 cm-1 for carboxylic acid

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5
Q

N-H

A

sharp around 3300 cm-1

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6
Q

C=O

A

sharp around 1750 cm-1

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7
Q

untraviolet (UV) spectroscopy

A

passing ultraviolet light through a sample tht is usually dissolved in an inert, nonabsorbing solvent and recording the absorbance. The absorbance is plotted against wavelenght.

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8
Q

electron transition in UV spectroscopy

A

molecus with pi-electrons or nonbonding electrons can be excited by ultraviolet light to higher energy antibolding orbitals. Lower energy gap between highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) are more easily excited and can absorb longer wavelenghts with lower energy.

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9
Q

conjugated systems in UV spectroscopy

A

shifts absorption spectrum, resulting in higher maximun wavelenght.

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10
Q

nuclear magnetic resonance spectroscopy (NMR)

A

certian amount nuceli have mahnetic moemnts that are orinted ar ramdon.

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11
Q

apha-state in NMR

A

lower energy

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12
Q

betha-state in NMR

A

higher energy

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13
Q

TMS in NMR

A

at =0 is used as calibration standard

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14
Q

hydrogen of an aromatic ring are

A

6.0-8.5 ppm

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15
Q

alkyl group

A

0-3 ppm

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16
Q

alkynes

A

2-3 ppm

16
Q

spliting rule

A

n + 1

n is the number of adjacent hydrogens

17
Q

alkenes

A

4.6-6 ppm

18
Q

aldehydes

A

9-10 ppm

19
Q

carboxylic acid

A

10.5-12 ppm

20
Q

sp hybridized

A

2-3 ppm

21
Q

sp2 hybridized

A

4.6-6 ppm

22
Q

sp3 hybridized

A

0-3 ppm