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Korvikt [17]
3 years ago
5

Hey please help man...

Chemistry
1 answer:
malfutka [58]3 years ago
4 0

Answer:

1. Cations are positively charged and have less electrons than protons.

2. Anions are negatively charged and have more electrons than protons.

3. Columns on the periodic table are known as groups.

4. The elements in the first group are called alkalis metals and have one valence electron.

Explanation:

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Answer:

See explanation below

Explanation:

First, we need to understand that the monochlorination of an alkane like this one, involves substitution of one of the atoms of hydrogen of the molecule for an atom of chlorine.

This reaction takes place when the alkane reacts with Cl₂ in presence of light or heat.

When this happens, the first step involves the breaking of the double bond of the chlorine to form the ion Cl⁻.

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The third step, the alkane with the lone pair of electron substract a chlorine for the beggining and form the mono chlorinated product.

The final step involves forming the remaining products with the remaining reagents there.

In the picture attached you have the mechanism and product for this reaction:

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Need help with chemistry question
luda_lava [24]

Answer:

See explanation

Explanation:

In this case, we have to check two variables:

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2) The carbon bonded to the leaving group.

Let's check one by one:

<u>2-chloro-3-methylbutane</u>

<u />

In this molecule, the leaving group is "Cl", the carbon bonded to the leaving group has two neighbors. Therefore, we have a <u>secondary substrate.</u>

<u>1-phenylpropan-1-ol</u>

<u />

In this molecule, the leaving group is "OH", the carbon bonded to the hydroxyl group has two neighbors also. So, we have a <u>secondary substrate.</u>

<u>(E)-pent-3-en-2-yl 4-methylbenzenesulfonate</u>

<u />

In this case, the leaving group is "OTs" (Tosylate), the carbon bonded to the tosylate group has as a neighbor a double bond. Therefore, we have an <u>allylic substrate.</u>

<u>3a-bromooctahydro-1H-indene</u>

<u />

In this molecule, the leaving group is "Br", the carbon bonded to the bromine has three neighbors. So, we have a <u>tertiary substrate.</u>

<u>1-iodo-3-methylbutane</u>

<u />

In this molecule, the leaving group is "I", the carbon bonded to the iodide has only one neighbor. So, we have a <u>primary substrate.</u>

<u />

See figure 1

I hope it helps!

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