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Vitamin E acetate, or α-tocopheryl acetate, exists in two forms: D-form (solid/highly viscous at 20 °C) and a racemate (viscous liquid at 20 °C). Interestingly enough, the difference in physical properties is well denoted in German Wikipedia article, whereas English Wikipedia omits this. Also, I noticed many authoritative sources arbitrarily listing a ...

6

The Wharton rearrangement presents an opportunity to convert one α,β-unsaturated ketone to its regioisomeric allylic alcohol. To this end Ohloff, et al.,1,2 explored the transformation of α-ionone 1 into α-damascone 2. Epoxide 3 was exposed to hydrazine in methanol to afford bicyclic allylic alcohol 4 (30%) and the uncyclized allylic ...

4

Your question doesn't sound like you care about exact splitting, so I'm going to ignore magnetic equivalence/inequivalence. If you do care, you need to take that into account. For the purposes of this discussion, equivalent means chemically equivalent. You also need to keep in mind that NMR is achiral. In other words, opposite enantiomers will produce the ...

3

The thermometer that records the temperature is in the air high above the solution (usually above a Vigreux column). Your distillation was under ambient conditions, so there was air in contact with everything. The thermometer can only records a rising temperature if the temperature around it rises. During the distillation, this is due to methanol vapour (...

3

Your analysis is basically right. The three protons within a rapidly rotationaly averaged methyl group are magnetically and chemically equivalent, whereas the aromatic protons related by reflection symmetry are chemically but not magnetically equivalent. The ethylenic protons are not chemically equivalent and therefore not magnetically equivalent. The key to ...

3

According to given data, you may consider expected peak patterns are not complicated such as $\mathrm{ABX}$ or $\mathrm{A_2B}$ patterns (more reading look here. This is simply a $\mathrm{A_2X}$ pattern ($\delta_X = \pu{6 ppm}$ and $\delta_A = \pu{1 ppm}$ without geminal coupling of two $\ce{H_b}$ protons), and hence, peak is a $dt$ for $\ce{H_b}$ as ...

3

Hyperconjugation is indeed very difficult to practically in cases involving bridgehead protons. The underlying reason is essentially the same that also explains why these protons cannot be enolised if there is a carbonyl group adjacent to them. This is best explained by drawing out the compound in a three-dimensional manner, which I have done for you in the ...

3

More resonance structures does not necessarily mean more stabilization. Phenolate has a negative charge at an oxygen atom which cannot be shared without "disturbing" the aromaticity of the ring. This is in part because in the resonance structures drawn, there are negative charged carbons within the aromatic ring, which either implies that: The ring is ...

3

This is an interesting question. However, disadvantage is having only $\ce{^1H}$-$\mathrm{NMR}$ to deal with. Yet, it has pretty good resolution (probably using $\pu{400 MHz}$ machine) so we can resolve the splitting pattern very easily and predict the structure. First, OP has correctly assigned two degree of unsaturation. For molecular formula, $\ce{... 3 Finding a structure with the correct formula that matches the chemical shifts reasonably well is not so difficult. The multiplet integrals, normalized such that the area corresponding to one H is$\approx 33\$ suggests the following number of H (moving upfield): 1,1,2,2,3. The upfield singlet suggests an uncoupled terminal methyl group, so we begin with ...

1

The second one will be more stable. We usually first check for resonance as it is more effective. But here it has to be explained on basis of hyper conjugation because the resonance due to the ring is same in both. But the second one will have more hyper conjugating structures as greater number of alpha hydrogens are present. More about hyper conjugation ...

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Cross conjugation is a phenomena due to which (in simple words) resonance from two groups in opposite direction cancel out. It is a destabilising effect. It is something like this: These are some examples which will help you understand why this is a destabilising effect: In the above two examples the stability of carbanion decreases drastically as the two ...

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