Chemistry, 16.09.2019 21:00 kel3194

Be sure to answer all parts. draw the products formed when 2−propanol [(ch3)2choh], the main ingredient of rubbing alcohol, is treated with h2so4.

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Answer from: andrewschmitz4704

2 propanol if reacted with H₂SO₄ at 130 C will be obtained dipropyl ether

and if reacted at 180 C, propene will be obtained

$\boxed{\boxed{\bold{Further~explanation}}}$

Alcohol is an alkane derivative where one H atom is replaced by a OH⁻ group

2 propanol is a compound that has a molecular formula equal to 1 propanol but different in its functional group

Several alcoholic reactions occur including:

1. Reaction with active metals

The H atom from the OH group is substituted by the active metal to form alkoxide

2C₃H₇OH + 2Na ---> 2C₃H₇ONa + H₂

2. Substitution with Halogen

Cluster Oh can be substituted with halogens if reacted with concentrated HCl, PCl₃, PCl₅

C₃H₇OH + HCl ---> C₃H₇Cl + H₂O

3. Oxidation

Oxidation of primary alcohols forms aldehydes and carboxylic acids

Secondary alcohols form ketones

tertiary alcohols are not oxidized

4.Esterification

Alcohol reacts with carboxylic acids to form esters and water, better known as esterification reactions

5. Dehydration of alcohol

Alcohol heated with concentrated sulfuric acid dehydrates (release of water molecules) forms ether (at 130 C) or alkene (at 180 C)

2 propanol if reacted with H₂SO₄ at 130 C will be obtained dipropyl ether

and if reacted at 180 C, propene will be obtained

$\boxed{\boxed{\bold{Learn~more}}}$

substitution reaction

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oxidation of primary alcohol.

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the formation of ethyl acetate from ethanol and acetic acid

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the structure of the one tertiary (3 °) alcohol

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Keywords: alcohol, alkane derivative, 2 propanol, dehydration

Be sure to answer all parts. draw the products formed when 2−propanol [(ch3)2choh], the main ingredi

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

Ithink this is what you meant. the temperature at which a liquid begins to boil is called the boiling point. the boiling point of water is 212°f or 100°c. and at that point will start to turn into steam (gas)

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Be sure to answer all parts. draw the products formed when 2−propanol [(ch3)2choh], the main ingredi...

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