The tosylate of (2R,3S)-3-phenylbutan-2-ol undergoes an E2 elimination on treatment with sodium ethoxide. Draw the structure of
the alkene that is produced.
1 answer:
Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very <u>strong base</u> (<em>sodium ethoxide</em>). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an <u>E2 mechanism</u>, therefore, the hydrogen that is removed must have an <u>angle of 180º</u> with respect to the leaving group (the "OH"). This is known as the <u>anti-periplanar configuration</u>.
The hydrogen that has this configuration is the one that placed with the <u>dashed bond</u> (<em>red hydrogen</em>). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
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<u>Explanation:</u>
A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.
- <u>For 1:</u>
When potassium phosphate reacts with silver nitrate, it leads to the formation of silver phosphate and potassium nitrate.
The balanced chemical equation follows:
- <u>For 2:</u>
When barium bicarbonate reacts with calcium hydroxide, it leads to the formation of barium carbonate, calcium carbonate and water
The balanced chemical equation follows:
Answer:
The activation energy is
Explanation:
The gas phase reaction is as follows.
The rate law of the reaction is as follows.
The reaction is carried out first in the plug flow reactor with feed as pure reactant.
From the given,
Volume "V" =
Temperature "T" = 300 K
Volumetric flow rate of the reaction
Conversion of the reaction "X" = 0.8
The rate constant of the reaction can be calculate by the following formua.
Rearrange the formula is as follows.
The feed has Pure A, mole fraction of A in feed is 1.
= change in total number of moles per mole of A reacte.
Substitute the all given values in equation (1)
Therefore, the rate constant in case of the plug flow reacor at 300K is
The rate constant in case of the CSTR can be calculated by using the formula.
The feed has 50% A and 50% inerts.
Hence, the mole fraction of A in feed is 0.5
= change in total number of moles per mole of A reacted.
Substitute the all values in formula (2)
Therefore, the rate constant in case of CSTR comes out to be
The activation energy of the reaction can be calculated by using formula
In the above reaction rate constant at the two different temperatures.
Rearrange the above formula is as follows.
Substitute the all values.
Therefore, the activation energy is