Aldehydes and ketones can be halogenated at their α-position by reaction with Cl2, Br2, or I2, under acidic conditions. Using Br
2 under acidic conditions, an intermediate enol is formed which adds bromine at the α-position. The reaction stops after the addition of one bromine because the electron-withdrawing halogen decreases the basicity of the carbonyl oxygen, making the protonation less favorable. Draw curved arrows to show the movement of electrons in this step of the mechanism.
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Answer:
the frequency of photons
Explanation:
Given: first ionization energy of 1000 kJ/mol.
No. of moles of sulfur = 1 mole
We know that plank's constant
Let the frequency of photons be ν
Also we know that ΔE = hν
this implies ν = ΔE/h
Hence, the frequency of photons
Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
