All categories

3 years ago

An alkene with the molecular formula C8H16 undergoes ozonolysis to yield a mixture of (CH3)2C=O and (CH3)3CCHO. The alkene is:

Chemistry

1 answer:

aalyn [17]3 years ago

8 0

Answer:

2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Explanation:

In ozonolysis (hydrolysis step involve a reducing agent such as Zn, $Me_{2}S$ etc.), a pi bond is broken to form ketone/aldehyde.

Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.

Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.

Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.

Here, 2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Reaction steps are shown below.

You might be interested in

The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

6 0

2 years ago

Can sedimentation and decantation be used for all types of mixtures? Explain

Bumek [7]

Explanation:

1. Sedimentation and decantation cannot be used for all types of mixtures.

Decantation is a separation technique in which is used to separate immiscible liquids or mixtures containing liquid and solids within them.

In decantation, gravity is used to bring the denser materials to settle at the bottom.

For homogenous mixtures, it is not possible to use decantation. A solution of sugar and water will not decant.

2. Yes, mass of an object reduces the settling time of such object in a mixture.

The higher the mass, the faster the rate of settling. Also, as we know, mass is directly proportional to density. A body with a high density will settle faster in solution.

4 0

3 years ago

what would be the value of 'G' on the surface of earth if it's mass was twice & its radius half of what it is how?

Katen [24]

Answer:

What would be the value of g on the surface of the earth if its mass was twice and its radius half of what it is now ? g2=8g1 Thus , the value of g on the surface of the earth would be eight times the present value.

Explanation:

Hope this helps! :)

8 0

3 years ago

Read 2 more answers

Hii

-BARSIC- [3]

Answer:

im interrested

Explanation:

4 0

2 years ago