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torisob [31]
4 years ago
11

A proton transfer reaction can occur when an aldehyde is placed in strong base, such as an alkoxide ion, producing an alcohol an

d a charged conjugate base that is resonance stabilized. In the left box, draw the curved arrows for the proton transfer. In the middle and right boxes, draw the two structures for the resonance-stabilized product as noted in the box-specific directions. Be sure to include all lone pairs and nonzero formal charges.

Chemistry
1 answer:
Pani-rosa [81]4 years ago
7 0

Hi, you have not provided structure of the aldehyde and alkoxide ion.

Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.

Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.

The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.

After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.

All the structures are shown below.

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Answer:

The edge of the length is \mathbf{L = 8.54 \times 10^{-10} \ m}

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From the given information:

The associated energy for a particle in three - dimensional box can be expressed as:

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h = planck's constant = 6.626 \times 10^{-34} \ Js

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The energy at the ground state can be calculated by using the formula:

E_1 =\dfrac{3h^2}{8mL^2}

At first excited energy level, one of the quantum values will be 2 and the others will be 1.

Thus, the first excited energy will be: 2,1,1

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E_2 =\dfrac{(2^2+1^2+1^2)h^2}{8mL^2}

E_2 =\dfrac{(4+1+1)h^2}{8mL^2}

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The transition energy needed to move from the ground to the excited state is:

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\Delta E= \dfrac{6h^2}{8mL^2}-  \dfrac{3h^2}{8mL^2}

\Delta E= \dfrac{3h^2}{8mL^2}} ----- (1)

Recall that:

the  wavelength identified with the electronic transition is: 800 nm

800 nm = 8.0  × 10⁻⁷ m

However, the energy-related with the electronic transition is:

\Delta E =\dfrac{hc}{\lambda}

\Delta E =\dfrac{6.626 \times 10^{-34} \times 2.99 \times 10^8}{8.0 \times 10^{-7} }

\Delta E =2.48 \times 10^{-19}  \ J

Replacing the value of \Delta E in (1); then:

2.48 \times 10^{-19}= \dfrac{3h^2}{8mL^2}}

Making the edge length L the subject of the formula; we have:

L = \sqrt{\dfrac{3h^2}{8m \times2.48 \times 10^{-19}} }

L = \sqrt{\dfrac{3\times (6.626 \times 10^{-34})^2}{8(9.1 \times 10^{-31} ) \times2.48 \times 10^{-19}} }

\mathbf{L = 8.54 \times 10^{-10} \ m}

Thus, the edge of the length is \mathbf{L = 8.54 \times 10^{-10} \ m}

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