CHEM110 Lab 5 imine synthesis PDF

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IMINE SYNTHESIS and REDUCTION

Experimental Observations (imine formation): -

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When 4-ethoxyaniline is added to the clear solution of ethyl lactate, there is a colour change from a colourless solution to a light yellow solution. When the salicylaldehyde is added to the mixture (of ethylactate and 4-ethoxyaniline), there is a colour change from a light yellow solution, to a light orange solution. After swirling the mixture a couple of times, a solid with a light yellow colour forms. After adding the 10 mL of ice cold brine, the solid changes from a light yellow to a light orange colour. After the mixture has been added to the vacuum filtration apparatus, and washed with two portions of ice-cold deionised water, the colour fades slightly as the colourful impurities wash out in the flask underneath the mixture. At this point, some solid starts to form. After vacuum filtration is finished, the solution of the impurities left in the flask has a light orange colour, and the filtrate/ solid (the i light pale orange colour.

Experimental Observations (amine formation): -

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6.2

After the stirrer bar has been added to the vial and 5 mls of methanol have been added, the solution mixture has a yellow colour. After the sodium borohydride is added to the vial on the magnetic hot plate, there is vigorous bubbling (it foams up). after adding a second portion of the NaBH4, the yellow solution colour starts to lose colour over time (bubbling is still occurring) After adding a third portion of the NaBH4 , the solution colour turns to a very pale yellow colour after some time (foaming and bubbling is still occuring). After adding 10 mL of 5% sodium bicarbonate solution to the reaction mixture, the solution foams up quite rapidly. The solution mixture now is a white cloudy colour (most of the yellow colour has faded). After adding the reaction mixture to the vacuum filtration apparatus, the resulting solid is a very fine white powder. After vacuum filtration and transferring the product into a conical flask, adding 3 mLs of ethanol, and standing a boiling stick in the flask, and then placing the flask on a hot plate; the solid starts to dissolve, and the resulting solution is a very light pale yellow. After using a pipette to transfer 1-2 drops to a vial, and leaving the remaining solution to stand undisturbed while cooling to room temperature, crystallization starts to occur. (medium sized white crystals start to form). After collecting the recrystallized product by vacuum filtration, white crystals remain as the final product. T.l.c Analysis of Reaction

Calculate the Rf for the imine (Im) and the amine (Am): Rf Imine: distance travelled by the spot / distance travelled by the solvent = 3/4.2 =0.714=0.71 Rf Amine: distance travelled by the spot / distance travelled by the solvent = 1.8/4.2 = 0.429=0.43 © School of Chemical Sciences, The University of Auckland

CHEM 110 Laboratory Assignment 5 [2020]

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At the time your sample was taken, had the reaction gone to completion? YES / NO At the time the sample was taken, the reaction had not gone to completion. This is because there are two spots in the tlc for amine, which means that the spot corresponding to the starting material (the smaller spot) is also present in the reaction mixture. Therefore, it can be inferred that the reaction had not gone to completion. Furthermore, the smaller point for the tlc of amine, travelled the same distance as the tlc spot point for imine, this means some of the imine has not reacted fully to form the amine. Therefore, imine is still present in the reaction mixture and hasn’t been completely reduced so the reaction had not gone completion at the time the sample was taken.

7 SPECTROSCOPY EXERCISES 7.1 Imine and Amine Draw the imine and amine products you would expect to obtain from reacting p-ethoxybenzaldehyde with the aniline assigned to your group (see Canvas for data).

7.2

UV-VIS Spectroscopy

What property of the imine and secondary amine allowed them to be viewed under the UV lamp in the t.l.c exercise? This is due to the conjugation pattern of the double-bonds present in the secondary amine and imine, where compounds containing only sigma (single) bonds are transparent to the UV-VIS light, and since the secondary imine and the amine have conjugated double bonds, this makes them visible under the UV lamp. © School of Chemical Sciences, The University of Auckland

CHEM 110 Laboratory Assignment 5 [2020]

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After isolation by filtration the imine was ___ORANGE___ in colour and the amine was ___WHITE (colourless)_________ in colour. UV-Vis spectrometry shows that the imine has a λmax at a much longer wavelength than the amine. Use the structure of the two compounds to explain these observations. ➔ Imine has λmax at a much longer wavelength than the amine because the imine is much more conjugated than the amine. In UV-VIS, compounds with more conjugation absorb at longer wavelengths, where the more extensive the conjugation within a molecule, the longer the wavelength for λmax. Conjugation refers to double bonds that are linked together (double bonds are conjugated when there is one single bond between the double bonds.) Imine is more conjugated because of the carbon-nitrogen double bond (C=N) bond in the imine, whereas the amine has a single carbon- nitrogen bond (C-N) which means that it is less conjugated than imine. Therefore the additional C=N bond in imine makes it more conjugated which is why the imine has λmax at a much longer wavelength. Furthermore, The more conjugated a compound is, the more highly coloured it is as it absorbs at higher wavelengths, which is why imine is orange (absorbs at a longer wavelength) while amine is white (absorbs at a much lower wavelength). 1 7.2 H NMR Your group supervisor will provide you with four 1H NMR spectra, corresponding to your aldehyde, your aniline, your secondary amine and a fourth compound not used in this experiment. Match the spectra to the corresponding compounds, giving at least 2 pieces of data that support your choice.

The aldehyde is spectrum W-3 1. In spectrum 3, there are a total of 10 hydrogens (3H+2H+2H+2H+1H=10 hydrogens) which aligns with p-benzaldehyde which has a total of 10 hydrogens in the molecule. This is the only spectra with a total of 10 hydrogens. This is a distinguishing feature for the aldehyde. 2. there is a singlet at 9.9 ppm in the spectrum. This signal corresponds to HCOR which has a range of 9.5-10 ppm (a hydrogen attached to a carbon that is also doubly bonded to an oxygen). This signal is present because of the hydrogen attached to the carbonyl (C=O), and since it is singlet, this means that there will be no vicinal hydrogens which is also true for p-benzaldehyde. This is a distinguishing feature for the aldehyde. 3. There are 2 doublets in the aromatic region (6.5-8 ppm) with an area of 2H each, which means that there are 4 hydrogens in the aromatic region which aligns with p-benzaldehyde. Since each signal is a doublet, that means there will be 1 vicinal hydrogen for a doublet. This also aligns with p-benzaldehyde. The splitting pattern (2 doublets, 2H each) also means that the ring is 1,4-disubstituted which is true for p-benzaldehyde. 4. There is a quartet signal at 4 ppm. This signal corresponds to R2CHO. The signal has an area of 2H, which means that there are two hydrogens attached to a carbon that is also bonded to an oxygen. Furthermore, since this is a quartet, this means there are 3 vicinal hydrogens (a CH3 vicinal fragment). This is true for the p-benzaldehyde that has the ethoxy substituent (OCH2CH3) attached to the ring, and this signal is due to the OCH2 fragment. 5. There is also a triplet signal at around 1.4 ppm. This signal corresponds to R2CHCZ (Z=O). The signal has an area of 3H which means that there are three hydrogens that are attached to a carbon that is attached to a carbon that is bonded to an oxygen. The signal is a triplet which means it has 2 vicinal hydrogens (a CH2 fragment). This is true for the p-benzaldehyde due to the ethoxy substituent attached to it (OCH2CH3).This signal is due to the CH3 fragment from the ethoxy substituent (OCH2CH3). © School of Chemical Sciences, The University of Auckland

CHEM 110 Laboratory Assignment 5 [2020]

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The aniline is spectrum W-4 1. In spectrum 4 there are a total of 9 hydrogens (3H+2H+2H+2H=9 hydrogens), which aligns with p-methoxyaniline which has a total of 9 hydrogens in the molecule. This is the only spectra with a total of 9 hydrogens. This is a distinguishing feature for the aniline. 2. There is a singlet at around 3.8 ppm with an area of 3H. This signal corresponds to R2CHO. This means that there are three hydrogens that are bonded to a carbon that is also bonded to an oxygen. The singlet also means that there are no vicinal hydrogens. This is true for the p-methoxyaniline due to the methoxy group (OCH3) attached to the ring. 3. There is a chemical shift (signal) at around 5.5 ppm with a signal area of 2H. The signal is broad and short. This is due to the NH2 group. The position of the signal (at 5.5 ppm) is because NH2 in the aniline is deshielded by both the nitrogen and the aromatic ring. The area of 2H means that there are two hydrogens attached to the nitrogen and this aligns with the structure of the aniline. This is also a distinguishing feature for the aniline. 4. There are 2 doublets in the aromatic region (6.5-8 ppm) with an area of 2H each, which means that there are 4 hydrogens in the aromatic region which aligns with p-methoxyaniline. Since each signal is a doublet, that means there will be 1 vicinal hydrogen for a doublet. This also aligns with p-methoxyaniline. The secondary amine is spectrum W-2 1. In spectrum 1, there are a total of 19 hydrogens (3H+3H+2H+2H+2H+2H+2H+2H +1H =17 hydrogens). This aligns with the secondary amine which has 19 hydrogens. This is the only spectra with 19 Hs. This is a distinguishing feature of the secondary amine. 2. There is also a triplet signal at around 1.4 ppm. This signal corresponds to R2CHCZ (Z=O). The signal has an area of 3H which means that there are three hydrogens that are attached to a carbon that is attached to a carbon that is bonded to an oxygen. The signal is a triplet which means it has 2 vicinal hydrogens (a CH2 fragment). This is true for the secondary amine due to the ethoxy substituent (OCH2CH3) attached to one of the rings. Therefore this signal is due to the CH3 fragment in the ethoxy (OCH2CH3) substituent. 3. There is a singlet at around 3.8 ppm with an area of 3H. This signal corresponds to R2CHO. This means that there are three hydrogens that are bonded to a carbon that is also bonded to an oxygen. This is true for the secondary amine due to the methoxy group substituent (OCH3) attached to one of the rings. 4. There is a quartet signal at 4 ppm. This signal corresponds to R2CHO. The signal has an area of 2H, which means that there are two hydrogens attached to a carbon that is also bonded to an oxygen. Since this is a quartet, this means there are 3 vicinal hydrogens (a CH3 vicinal fragment). The secondary amine has an ethoxy substituent (OCH2CH3) on one of the rings, and this signal is due to the OCH2 fragment. 5. There is a singlet signal at around 4.3 ppm that has an area of 2H. This would correspond to R2CHX (X=N). This is because there are two hydrogens that are attached to a carbon that is also bonded to a nitrogen. The position of the signal at 4.3 ppm is due to deshielding because the hydrogens are attached to the carbon which is attached to a benzene ring (which is an electron withdrawing group) and also an electronegative nitrogen. This is a distinguishing feature for the secondary amine. 6. There are 4 doublets in the aromatic region (6.5-8ppm). This means that there are 8 hydrogens in the aromatic region which aligns with the secondary amine structure. 7. There is also a broad short signal at around 7.4 ppm (that is separate from the four doublets for the aromatic hydrogens). This signal is due to the N-H bond in the secondary amine. It has an area of 1H which means that one hydrogen is attached to the nitrogen. This is a distinguishing feature for the secondary amine. © School of Chemical Sciences, The University of Auckland...