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3 years ago

Draw the primary alcohol with molecular formula C5H12O that cannot be synthesized by the hydroboration–oxidation of an alkene.

Chemistry

1 answer:

alexandr402 [8]3 years ago

3 0

Answer:

The answer is neopentyl alcohol. (CH₃)₃C-CH₂-OH

Explanation:

This alcohol cannot be synthesized by the hydroboration–oxidation of an alkene since the reaction of hydroboration–oxidation go through steps that involved the addition of BH₃ into the original structure. In the case of neopentyl alcohol, if we carry out retro synthetic by the hydroboration–oxidation, we will find the original alkene is (CH₃)₃C=CH₂ which is impossible since the center carbon atom has 5 bonds.

The condition for hydroboration–oxidation reaction is that the alkene must have the structure of R-CH=CH₂

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If you are given 96.0 grams of O2, how many grams of H20 are made?

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2 years ago

Aqueous solutions of barium nitrate and potassium phosphate are mixed. What is the precipitate and how many molecules are formed

kykrilka [37]

Aqueous solutions of barium nitrate and potassium phosphate are mixed. What is the precipitate and how many molecules are formed?

Barium nitrate has a chemical symbol of Ba(NO3)2 and potassium phosphate has a chemical symbol K2PO4. The reaction between these two is a double replacement reaction yielding barium phosphate and potassium nitrate.

The chemical equation representing the reaction is,

Ba(NO3)2 + K2PO4 à KNO3 + BaPO4

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2 years ago

Read 2 more answers

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

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1 year ago

How many moles are in 117.30 grams of KCLO3?

eduard

0.0081599680129254

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2 years ago