. Write the two resonance hybrids for the carbocation that would be formed by protonation at C-1 of 2-methyl-1,3-pentadiene. Wit
1 answer:
Answer:
Answer is explained in the explanation section below.
Explanation:
This question is requires the diagram, which I have drawn and it is attached in the attachment below. Please for this answer, refer to the diagram attached in the attachment below.
Referring to the diagram, attached. As we know that, Allylic secondary carbocations are more stable than Allylic tertiary carbocations.
Hence,
C2 will have a more positive charge since a tertiary carbocation (C2) is more stable than a secondary carbocation (C4). Therefore, the resonance structure will favor the positive charge at C2.
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Answer:
The answer is quartet 2.40 ppm.
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Explanation:
Solution
Multiplicity or (n+1) rule:
It helps in determination of multiplicity of an individual proton or individual types of proton which are available in the molecule.
Multiplicity =(n+1)
Thus
The non equivalent protons which are attached from adjacent atom is denoted by n.
Now because there are three non-equivalent protons are present at adjacent carbon of methylene group, hence the multiplicity of methylene hydrogen is given as follows:
The multiplicity will be the same for the two hydrogen's. thus we compute multiplicity only for one hydrogen atom stated below:
Non- equivalent = 3
Multiplicity = (3 +1)
= 4
= Quartet for 2H
A quartet for 2H indicates that the hydrogen atoms attached from the carbon, which is attached one side from a methyl group and the other side form an atom that have no any hydrogens.
Now due +I effect of carbonyl group, chemical shift value is high for these two hydrogens which is exactly at 2.40 ppm or 2.40 Quartet.
