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Ostrovityanka [42]
3 years ago
13

In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?

Chemistry
1 answer:
klio [65]3 years ago
4 0

Answer:

a. Cyclohexanone

Explanation:

The principle of IR technique is based on the <u>vibration of the bonds</u> by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is <em>a specific energy that generates a specific vibration</em>. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.

Now, we must remember that the <u>lower the wavenumber we will have less energy</u>. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.

If we look at the structure of all the molecules we will find that in the last three we have <u>heteroatoms</u> (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of <u>resonance structures</u> which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.

The molecule that fulfills this condition is the <u>cyclohexanone.</u>

See figure 1

I hope it helps!

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Answer:

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Explanation:

This a neutralisation reaction. The first thing to note is that this type of reaction forms salt and water.

Note that the formula of water is H2O. So write this at the right hand side.

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Next note that the salt should contain the acid radical (NO3) and the metallic part of the base (Ca) written well following the rule for writing chemical equation; thus

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The next step is to write the reactants at the right hand side of the equation;

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In the above equation, H atoms are three at the left hand side add 2 to to H- containig compound on both side as to balance up. Thus

2HNO3 + Ca(OH)2 ===> Ca(NO3)2 + 2H2O

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