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CAMRYN > Lab Final > Flashcards

Lab Final Flashcards

1
Q

what was lab 1

A

synthesis of [Ni(NH3)x]Br2

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2
Q

what was the point of lab 1

A

find out how many NH3 in complex and find Ni coordination number

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3
Q

reagents in lab 1

A

NiCl2-6H2O
conc NH3
KBr

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4
Q

how did we verify we made the right thing in lab 1

A

test tube tests (pdt in water)

  1. NaOH – ph paper for basic gas as Ni(OH)2 and NH3 g
  2. dimethyl glyoxime – replaces NH3 in complex, change in color to red
  3. HCl, DCM, NaOCl – DCM forms second layer (NP). NaOCl oxidizes the bromide to bromine (NP) and goes into DCM
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5
Q

how did we find out how much ammonia

A

back titration with h2so4
complex + H2SO4 -> (NH4)2SO4 + Ni(OH)6

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6
Q

color of product from lab 1

A

purple solid

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7
Q

how did we make product from lab 1

A

NiCl2-6H2O in water Ni will complex w water
sub h2o with NH3
add KBr to switch anions

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8
Q

in test tube check, lab 1 product with dimethylglyoxime makes what

A
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9
Q

what was lab 2

A

synthesis of [VO(acac)2]

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10
Q

what is interesting about [VO(acac)2]

A

used as a catalyst in the oxidation of anthracene to 1,8-anthraquinone bc it is square pyramidal, leaving an open site below the vanadyl bond

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11
Q

how did we make VO(acac)2

A
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12
Q

how did we verify we made VOacac

A

FT-IR presence of new peaks

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13
Q

What was the qualitative assessment of VOacac

A

6 test tubes, pdt in

  1. dcm - none
  2. meoh
  3. pyridine
  4. thf
  5. nh3
  6. sat sodium carb

tested physical interaction of effect of solvation - verified vacant site

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14
Q

How does TLC work in lab 2

A

less polar moves up higher

more polar moves less

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15
Q

TON & TOF

A

TON = moles pdt/moles catalyst (no unit)

refers to the moles of product formed in relation to the moles of catalyst used

A higher turnover number correlates to a higher amount of product formed and thus more effective catalysis

TOF = TON/time (s^-1)

number of turnovers per time, usually seconds

A higher turnover frequency correlates to more product in a short amount of time, again meaning more effective catalysis

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16
Q

how did we purify the 1,8-anthraquinone

A

extraction

solid is washed with toluene bc product is not soluble but starting material is

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17
Q

what is VO(acac)2

A

acac = acetylacetonate

vanadyl acetylacetonate

bis(acetylacetonato)oxovanadium (IV)

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18
Q

what does adding Na2CO3 to reaction in lab 2 do

A

removes H from acac

makes the double bonded O become a c=c and a o-, making the o reactive

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19
Q

What does FTIR measure

A

%transmittance vs wavenumber

streching/bending of functional groups

20
Q

color of VO(acac)2

A

aqua

21
Q

color of 1,8-anthraquinone

A

orange

22
Q

what was something significant about the reaction for the catalyst VOacac

A

needed significant stirring since it was a biphasic system - water & pdt in the EtAc

23
Q

how is voacac used as a catalyst (mechanism with h2o2)

A

This reaction uses H2O2 as a reagent, which supplies the oxygen needed for the reaction. The vanadium catalyst is necessary to make the hydrogen peroxide reactive to the alkene. The vanadium center reacts with the hydrogen peroxide to make HO•.7 This radical reacts with the alkene to give an alcohol. The second equivalent of hydrogen peroxide reacts again with the vanadium and alkene to give the second OH bond. The final equivalent of hydrogen peroxide reacts with the protons to give a carbonyl bond and water.7 This reaction would not be possible without the use of 1 as the catalyst. The final equivalent of hydrogen peroxide is necessary to create the ketone.

24
Q

some possible changes to test effectiveness of catalyst

A

Some changes include increasing and decreasing the amount of hydrogen peroxide, catalyst, the reaction time, and the reaction temperature. It would also be beneficial to test the yield, TON, and TOF without heat, catalyst, and hydrogen peroxide separately. More catalyst and a longer reaction time would likely increase yield and TON

25
Q

goal of experiment 3

A

synthesize 3 Mo(CO)4L2 isomers

26
Q

how could we tell isomers from each other

A

IR!!

cis - 4 peaks

trans - 2 peaks

When a molecule is more symmetrical, there are generally fewer IR active peaks.

27
Q

what three isomers in lab 3

A

cis-tetracarbonyl-bis(piperidine)-molybdenum(0), [Mo(CO)4(pip)2] (pip = piperidine)

tetracarbonyl-bis(triphenylphosphine)-molybdenum(0), [Mo(CO)4(PPh3)2] (cis & trans)

28
Q

Important aspect about lab 3 synthesis

A

performed under nitrogen using a Schlenk line. It is essential to complete these reactions without oxygen as the molybdenum complexes can react with oxygen to form MoO2 and MoO3. this is because in the complex it is Mo(0)

29
Q

how did we make the three isomers

A

First, molybdenum hexacarbonyl undergoes a substitution reaction with piperidine in toluene. This reaction requires a two-hour reflux to synthesize a tetracarbonyl-bis(piperidine)-molybdenum(0) isomer, (1). Then, a substitution reaction with triphenyl phosphene gives the first isomer of tetracarbonyl-bis(triphenylphosphine)-molybdenum(0) (2). Finally, the other stereoisomer of tetracarbonyl-bis(triphenylphosphine)-molybdenum(0) is produced by thermal isomerization to give 3.

30
Q

list/draw out all reactions in lab 3

A

1 & 3 in minearal oil bath – hot bc toluene is solvent

2 doesnt need to get very hot bc dcm

31
Q

thermodynamic vs kinetic pdt in lab 3

A

The cis-isomer of a [Mo(CO)4(PPh3)2] complex, 2, formed in the second reaction and is believed to be kinetically favorable. Since PPh3 is a bulky ligand, it favors the trans isomer. This configuration reduces steric repulsion between the ligands.4 Therefore, the trans-isomer, 3, synthesized in the last step of the experiment is the thermodynamic product of the reaction and thus more thermodynamically stable.4

32
Q

20.65 mg of [Ni(NH3)x]Br2 was dissolved in 10.00 mL of 0.0496 M H2SO4, then the excess acid
was titrated with 0.0510 M NaOH solution. The average volume of NaOH was 12.80 mL.
Calculate the number of coordinated ammonia molecules (x) in the nickel complex.

A
33
Q

goal fo lab 4

A

synthesize ethylcobaloxime as a model for the vitamin b12 coenzyme

34
Q

synthetic scheme to make ethylcobaloxime

A

First, dimethylglyoxime and cobalt (II) bromide reacted to form 3.706 g of dibromo-(dimethylglyoxime)(dimethylglyoximato)cobalt(III)

Then, the addition of 4-tert-butylpyridine to the product of the first reaction gave bromo(4-tert-butylpyridine)cobaloxime

Finally, bromo(4-tert-butylpyridine)cobaloxime with sodium borohydride and ethyl iodide gave ethyl(4-tert-butylpyridine)cobaloxime

35
Q

what is vitamin b12

A

Vitamin B12 is one of the few naturally occurring organometallic molecules.1 It is bioactive and involved in cellular metabolism in two active coenzyme forms– methylcobalamin and 5-deoxyadenosylcobalamin. The vitamin also plays a role in biological processes, including DNA synthesis and regulation, nervous system function, and red blood cell formation, to name a few.2,3 Vitamin B12 is used in chemistry as it is a helpful catalyst.1 Reactions including isomerization, dehalogenation, and methyl transfer require the formation and cleaving of a Co-C bond.1 Vitamin B12 is used in synthesis as a catalyst for cobalt-mediated reactions.

36
Q

structure of vitamin b12

A

Vitamin B12 consists of a cobalt atom coordinated by a corrin, a tetrapyrrole macrocyclic ligand.3 One axial site is coordinated to a 5,6-dimethylbensimidazole nucleotide. The key to the vitamin’s bioactivity lies in the ability of its second axial site to form a metal-alkyl (Co-C) bond.3 This results in an octahedral complex with an essentially macrocyclic ligand. Vitamin B12 occurs in many forms called cobalamins.2 Cyanocobalamin is the principal form of the vitamin used in vitamin supplements and pharmaceuticals, where a second axial group is a cyano group.2

37
Q

First step in lab 4

A

dimethyl glyoxime + CoBr2 + O2 –> dibromo-(dimethylglyoxime)(dimethylglyoximato)cobalt(III) + 2H2O

o2 added as a direct stream of air… we can do this bc its Co(II)

product will crash out of acetone

38
Q

second step in lab 4

A

dibromo-(dimethylglyoxime)(dimethylglyoximato)cobalt(III) + 2 tertbutylpyridine —> bromo(4-tert-butylpyridine)cobaloxime

done in meoh bc more polar than acetone so dibromo complex will disolve

Co(III) so cannot be in air

39
Q

third step in lab 4

A

bromo(4-tert-butylpyridine)cobaloxime + NaBH4 + CH3CH2I –> ethyl(4-tert-butylpyridine)cobaloxime + NaBr + BH3 + HI

Co is the nucleohpile and CH3CH2I is the electrophile

Co-C bond is similar to B12

first the NaBH4 reacts with Br to give negative Co(I) and open site which reacts with the ethyl iodide

40
Q

why is ethylcobaloxime a good model

A

The ethylcobaloxime synthesized in this experiment is a good model for vitamin B12. On the cobaloxime, the N4 macrocycle, made from two monodeprotonated dioxime molecules, link at two points by hydrogen bonding to create an essentially planar cyclic ligand similar to the planar corrin in B12.3 Ethylcobaloxime also has an axial Co-N and Co-C bond, which is seen in vitamin B12. An ethylcobaloxime, like 3, is essentially a smaller model version of vitamin B12. The ease of preparation, purification, and characterization of this compound makes it a good model.4

41
Q

Drawbacks of ethylcobalozime as a model for b12

A

While the basic structure of 3 is similar to that of vitamin B12, there are some drawbacks to an ethylcobaloxime as a model. The difference in the model and the vitamin create a significant difference in the electrochemical properties of cobaloximes compared to vitamin B12.5 For a model to be sufficient, it should be able to reproduce the Co(III)/Co(II), Co(II)/Co(I) and RCo(III)/RCo(II) E1/2 values.5

42
Q

what are quantum dots

A

2-10 nm

semiconductor materials

physical properties change bc of size

smaller size = larger band gap

if small band gap like bulk then can slip down side

43
Q

lab 5 rxn

A

zn(c2h3o2) + Na2s ———> 2 na(c2h3o2) + ZnS

na(po3)n

h2o/rt

44
Q

why do we need sodium polyphosphate

A
45
Q

spectra for 5

A

uv - vis found the size from ½ the absorbance wavelength

photoluminscense ecplained the effects of the dopants based on if the wavelength of emission moved

theoretically dopants will lower band gap and increase wavelength but could go the other way if concentration is too high

CAMRYN (1 decks)

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