Chemistry, asked by advait17, 10 months ago

What happens when CL2 is treated with propane?​

Answers

Answered by Anonymous
5

Answer:

The major product would be 2-chloropropane:

CH3-CH2-CH3 + Cl2 → CH3-CHCl-CH3 + HCl.

   ⇒The CCl4 is just a solvent that doesn’t react with chlorine.

   ⇒We also expect 1-chloropropane. It would be the minor product, despite there being 6 primary hydrogens and 2 secondary hydrogens, because the intermediate radical for 2-chloropropane formation (isopropyl) is more stable than the intermediate radical for 1-chloropropane formation (propyl).

Answered by Amazonalexa
3

Answer:

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The major product would be 2-chloropropane:

CH3-CH2-CH3 + Cl2 → CH3-CHCl-CH3 + HCl.

Notes:

The CCl4 is just a solvent that doesn’t react with chlorine.

We also expect 1-chloropropane. It would be the minor product, despite there being 6 primary hydrogens and 2 secondary hydrogens, because the intermediate radical for 2-chloropropane formation (isopropyl) is more stable than the intermediate radical for 1-chloropropane formation (propyl).

What are the products when chlorine reacts with propane in the presence of UV light?

What are the products when reacting propane and chlorine?

Which product is obtained when propane reacts with Cl2 in the presence of sunlight?

You could get the following, if the CCl4 does not participate in the reaction (meaning no external heat applied):

C3H8(g) + Cl2(g) + CCl4(g)= C3H6(1barg) + 2HCl(g) + CCl4(g)

ΔG(20C) = -91.5kJ (negative, so the reaction runs)

ΔH(20C) = -59.8kJ (negative, so the reaction is exothermic)

However, applying external heat, thus forcing the CCl4 to participate in the reaction, results in:

C3H8(g) + Cl2(g) + CCl4(g)= C4H2(g) + 6HCl(g)

ΔG(20C) = -61.0kJ (negative, so the reaction runs)

ΔH(20C) = +105.4kJ (positive, so the reaction is endothermic)

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Explanation:

When propane react with chlorine in presence of diffused sunlight then two products are formed

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