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Anastaziya [24]
3 years ago
6

The conjugation of the trans/cis product is likely diminished due to steric interactions between the cis-phenyl and the nearby v

inyl hydrogen. This interaction likely causes the phenyl ring to twist out of the plane, and thus no longer to be fully conjugated with the adjacent double bond. The trans-trans isomer has no such steric interactions, and thus it is expected to be more fully conjugated. The two isomers are thus may have different light absorbing properties.
Considering all this, which product stereoisomer do you expect to have the highest lambda max value in the UV-V is spectrum- the trans/trans or the cis/trans? Briefly explain your answer.
Chemistry
1 answer:
max2010maxim [7]3 years ago
8 0

Answer:

The trans/trans stereoisomer.

Explanation:

So, we are going to have two stereoisomers or arrangements for this chemical compound, that is the Cis/trans, the trans/trans.

For a compound to be seen with UV, the main or primary condition is conjugation or how conjugated the chemical specie or compound is.

Thus, the more the conjugation, the more or the higher the wavelength or lands value.

=> The Trans/trans stereoisomer will have the highest lamda value or wavelength because the energy gap in its π - π* Transitioning deceases due to the STERIC interactions, and the more the decrease in the energy gap, the more or the higher the lamda value or wavelength.

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Determine the empirical formula of a compound containing 40.6 grams of carbon, 5.1 grams of hydrogen, and 54.2 grams of oxygen.
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The Empirical formula of compound is C₁H₂O₁. The Molecular Formula of the compound is 4 (C₁H₂O₁).

<h3>What is Empirical Formula ?</h3>

Empirical formula is the simplest whole number ratio of atoms present in given compound.

Element   %   Atomic mass   Relative no. of atoms  Simplest whole ratio

C          40.6       12                   \frac{40.6}{12} = 3.3                        \frac{3.3}{3.3} = 1

H          5.1          1                      \frac{5.1}{1} = 5.1                          \frac{5.1}{3.1} = 2

O         54.2       16                    \frac{54.2}{16} = 3.3                        \frac{3.3}{3.3} = 1

The Empirical formula of compound is C₁H₂O₁ or CH₂O

<h3>How to find the Molecular formula of compound ?</h3>

Molecular formula = Empirical formula × n

n = \frac{\text{Molecular weight}}{\text{Empirical Formula weight}}

   = \frac{118.084}{30}

   = 4

Molecular formula = n × Empirical formula

                              = 4 (C₁H₂O₁)

Thus from the above conclusion we can say that The Empirical formula of compound is C₁H₂O₁. The Molecular Formula of the compound is 4 (C₁H₂O₁).

Learn more about the Empirical Formula here: brainly.com/question/1603500

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