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emmainna [20.7K]
3 years ago
15

. Write the two resonance hybrids for the carbocation that would be formed by protonation at C-1 of 2-methyl-1,3-pentadiene. Wit

hout doing a calculation, would you expect C-2 or C-4 (the two end carbons of the allylic cation) to have the most positive charge on it
Chemistry
1 answer:
never [62]3 years ago
8 0

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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Read 2 more answers
Which one of the following best represents the predicted approximate chemical shift and coupling for the hydrogen(s) indicated w
melomori [17]

Answer:

The answer is quartet 2.40 ppm.

Note: Kindly find an attached image below for the part of the solution to this question

Sources: The image was researched from Course hero platform

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

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= 4

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