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3 years ago

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Which reactant should be used to convert propene to 1,2-dichloropropane?

2 answers:

raketka [301]3 years ago

3 0

Propene +cl2 -> 1,2 dichloropropane

babunello [35]3 years ago

3 0

Answer:

Reactant is shown below

Explanation:

To convert propene to 1,2-dichloropropane, $Cl_{2}$ should be used in a polar aprotic slovent like $CH_{2}Cl_{2}$ .
Here, $Cl_{2}$ adds onto double bond and forms a chloronium cation followed by nucleophilic attack by a chloride ion.
Chloronium ion intermediate has transient carbocationic character, Therefore chloride ion attack at more substituted position.
Reaction mechanism has been shown below.

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What alcohol is formed when the alkene is treated with h2o in the presence of h2so4? qs14?

pentagon [3]

Here we have to get the alcohols which are formed in the reaction of an alkene with H₂SO₄ and H₂O.

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Nutka1998 [239]

Q = mCpΔT

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3 years ago

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3 years ago

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2 years ago

4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. T

andre [41]

<u>Answer:</u> The enthalpy change of the reaction is -243 J/mol

<u>Explanation:</u>

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

$q=c\times \Delta T$

where,

c = heat capacity of calorimeter = 1620 J/K

$\Delta T$ = change in temperature = $0.150^oC=0.150K$ (Change remains same)

Putting values in above equation, we get:

$q=1620J/K\times 0.15K=243J$

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

$S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}$

We are given:

Moles of monoclinic sulfur = 1 mole

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -243 J

n = number of moles = 1 mole

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol$

Hence, the enthalpy change of the reaction is -243 J/mol

8 0

3 years ago