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Andrej [43]
2 years ago
7

Dehydration of 2-methyl-2-pentanol forms one major and one minor organic product. Draw the structures of the two organic product

s of this reaction.
Chemistry
1 answer:
Korvikt [17]2 years ago
7 0

Answer:

The major product is 2-methyl-2-pentene [ CH₃-CH₂-CH=C(CH₃)₂ ] and a minor product 2-methyl-1-pentene [ CH₃-CH₂-CH₂-C(CH₃)=CH₂ ].

Explanation:

Dehydration reaction is a reaction in which a molecule loses a water molecule in the presence of a dehydrating agent like sulfuric acid (H₂SO₄).

<u>Dehydration reaction of 2-methyl-2-pentanol</u> gives a major product 2-methyl-2-pentene and a minor product 2-methyl-1-pentene.

CH₃-CH₂-CH₂-C(CH₃)₂-OH (2-methyl-2-pentanol)→ CH₃-CH₂-CH=C(CH₃)₂ (2-methyl-2-pentene, major) + CH₃-CH₂-CH₂-C(CH₃)=CH₂ (2-methyl-1-pentene, minor)

<u>Since more substituted alkene is more stable than the less substituted alkene. So, the trisubstituted alkene, 2-methyl-2-pentene is more stable than the disubstituted alkene, 2-methyl-1-pentene.</u>

<u>Therefore, the trisubstituted alkene, 2-methyl-2-pentene is the major product and the disubstituted alkene, 2-methyl-1-pentene is the minor product.</u>

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grin007 [14]

Answer:

\%m/m=6.85\%

Explanation:

Hello,

In this case, we are asked to compute the by mass percent for the given 3.2 g of ethylene glycol in 43.5 g of water. In such a way, since the by mass percent is computed as follows:

\%m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%

Whereas the solute is the ethylene glycol and the solvent the water, therefore we obtain:

\%m/m=\frac{3.2g}{3.2g+43.5g} *100\%\\\\\%m/m=6.85\%

Best regards.

5 0
3 years ago
What is the mass of silver chlorate (191.32 g/mol) that decomposes to release 0.466L of oxygen gas at STP? AgC1036) _AgCl). _026
vesna_86 [32]

Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

2AgClO_3\rightarrow 2AgCl+3O_2

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in \frac{0.466}{22.4}=0.0208 mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from \frac{2}{3}\times 0.0208=0.01387 moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

\text{Mass of }AgClO_3=\text{Moles of }AgClO_3\times \text{Molar mass of }AgClO_3

Molar mass of silver chlorate = 191.32 g/mole

\text{Mass of }AgClO_3=0.01387mole\times 191.32g/mole=2.654g

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3 0
2 years ago
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Explanation:

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