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stira [4]
2 years ago
10

Acid-catalyzed dehydration of 3,3-dimethyl-2-butanol gives three alkenes: 2,3-dimethyl-2-butene, 3,3-dimethyl-1-butene, and 2,3-

dimethyl-1-butene. Draw the structure of the carbocation intermediate leading to the formation of 2,3-dimethyl-2-butene.
Chemistry
1 answer:
Fynjy0 [20]2 years ago
5 0

1,1-hexene > 2,3,3-dimethyl-2-butene > 3-methyl-3-hexene > cis-3-hexene

The Saytzeff rule states that an alkene becomes more stable the more highly substituted it is. Since 2,3-dimethyl-2-butene is the most substituted alkene among the ones listed, it will thus be the most stable.

Ozone treatment of 2,3-dimethyl but-2-ene results in ozoinde, which with further reduction yields propanone and water. Trans-2-butene undergoes ozonolysis, producing the main ozonide. Acetaldehyde, syn- and anti-acetaldehyde oxide, and this main ozonide are the products of its breakdown.

Let's not forget that the alkenes' pi bonds are least stabilized by alkyl groups, making terminal alkenes the least stable of the group. This means that among the alkenes listed, 1-hexene is the least stable.

Learn more about 2,3-dimethyl-2-butene  here brainly.com/question/24146247

#SPJ4.

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Explanation:

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klasskru [66]

Answer:

There are 70 grams of KOH

Explanation:

First, we calculate the weight of 1 mol of KOH:

Weight 1 mol KOH: Weight K + Weight 0 + Weight H=  39g+ 1g+ 16g= 56 g/mol

1 mol-------56 g KOH

1,25 mol----x= (1,25 molx56 g KOH)/1 mol= <em>70 g KOH</em>

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4 years ago
How many grams are equivalent to 1.80 x 10^-4 tons? (English tons)
konstantin123 [22]

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163.44 g

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1.80 × 10∧-4 ton× 2000 lb/ton× 454 g/lb

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3 years ago
10 The graph below compares the density, in grams per cubic centimeter (g/cm'), of
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3 years ago
The rate law for the reaction 2A + B →C is –rA= kACA2CBwithkA= 25 (L/mol)2sec. What arekBandkC?
givi [52]

Explanation:

The given reaction equation is as follows.

             2A + B \rightarrow C

So, rate constants for different reactants and products written as follows.

             \frac{-k_{A}}{\text{stoichiometric coefficient of A}} = \frac{-k_{B}}{\text{stoichiometric coefficient of B}} = \frac{k_{C}}{\text{stoichiometric coefficient of C}}

As per the reaction equation, the stoichiometric coefficients of reactants and products are as follows.

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Therefore,

      \frac{-k_{A}}{\text{stoichiometric coefficient of A}} = \frac{-k_{B}}{\text{stoichiometric coefficient of B}} = \frac{k_{C}}{\text{stoichiometric coefficient of C}}

      \frac{-k_{A}}{-2} = \frac{-k_{B}}{-1} = \frac{k_{C}}{1}    

             \frac{-k_{A}}{-2} = k_{B} = k_{C}

Hence,          k_{B} = k_{C} = \frac{25}{2} (L/mol)^{2}

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Thus, we can conclude that k_{B} and k_{C} are 12.5 (L/mol)^{2}.

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