Draw a structure for a compound that meets the following description: An optically active compound, C5H10O with an IR absorption
at 1730 cm-1.
1 answer:
Question: A optically active compound, C5H10O, exhibits IR absorption at 1730 cm-1.
Its carbon NMR shifts are given below. The number of hydrogen's at each carbon, determined by DEPT, is given in parentheses after the chemical shift.
13C NMR: δ 22.6 (3), 23.6 (1), 52.8 (2), 202.4 (1)
Draw the structure of this compound in the window below
Explanation:
3-methylbutanal is a butanal substituted by the methyl group at the 3rd position. It is a volatile constituent in the olive. Also, it is used as a flavoring agent and a plant metabolite, it is also a Saccharomyces cerevisiae metabolite. It is also called as the Isovaleraldehyde organic compound. The liquid is colorless at STP, and also found in very low concentrations. It is also seen to be produced commercially for different use. Mostly used compound reagent in the preparation of pharmaceuticals and pesticides.
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Answer:
Z* = 3.55
Explanation:
Slater rule says that:
Z*= Z - S
Z* be the nuclear effective charge
Z is the nuclear charge
S is the shielding constant
First we write the electronic configuration of platinum:
The first Slater rule says that we need to group:
The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.
For the third rule we have two considerations, if is ns or np and if is nd or nf:
For our case, we have an electro that is in ns, so the rule says that
-electrons within same group shield 0.35, except the 1s which shield 0.30
-electrons within the n-1 group shield 0.85
-electrons within the n-2 or lower groups shield 1.00
Now we can proceed with the calculation:
The first consideration in the third rule does not apply as we only have one electron on this level.
The second consideration will be as follow for the level 5, where we have 17 electrons.
Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.
So the result for S=(60*1.00 + 17*0.85) = 74.45
And the equation is: Z* = 78 - 74.45
So Z* = 3.55