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3 years ago

Draw the major product of the reaction between 1-butanol and nah. include counterion in your answer.

Chemistry

2 answers:

BARSIC [14]3 years ago

7 0

Answer: Check explanation.

Explanation:

The reaction between 1-butanol and sodium hydride(NaH) is a very important reaction in industrial Chemistry in the Williamson Ether synthesis.

The 1-butanol is a primary alcohol, the Sodium(Na) in the sodium Hydride(NaH) is positively charged while the Hydrogen in the compound (that is, the sodium Hydride) is negatively charged. The Hydride is a very powerful base.

The reaction between 1-Butanol and sodium Hydride happens through the deprotonation of alcohol (1-butanol) by the sodium hydrides to give a sodium alkoxide and the evolution of Hydrogen gas.

The equation of reaction is given below:

C4H9OH + NaH ----------> CH3CH2CH2CH2 O^- Na^+ + H2.

NB: check attached file for the drawing.

Eduardwww [97]3 years ago

4 0

Answer:

The major product of the reaction between 1-butanol and NaH is attached in the drawing.

Explanation:

In this reaction a strong base, such as NaH, deprotes the alcohol to obtain an alkoxide ion. It is the first step of the method to prepare ethers by means of Williamson's synthesis, where subsequently the alkoxide ion attacks a haloalkane.

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What is the approximate mass of one mole of oxygen gas (O2)?

fenix001 [56]

The correct answer is B.

1 mol O2 x 15.999 O2/ 1 mol O2 = 15.999 O2

16 O2 when rounded.

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3 years ago

Assume that a raindrop has a volume of 3.48 cm3. If you had an Avogadro’s number of raindrops that just filled a cubic box. What

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V(cubic box) = 3,48 cm³ · 6,023·10²³.
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6 0

3 years ago

The number of neutrons in one atom of mercury with mass number 204 is:

Talja [164]

Number of neutrons is 121.

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3 years ago

what is an empirical formulaWhat is the percent composition by mass of nitrogen in (NH4)2CO3 (gram-formula mass = 96.0 g/mol)?

Margarita [4]

Answer:

Percentage composition = 14.583%

Explanation:

In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.

Percentage composition by mass of Nitrogen

Nitrogen = 14g/mol

In one mole of the compound;

Mass of Nitrogen = 1 mol * 14g/mol = 14g

Mass of compound = 1 mol * 96.0 g/mol = 96

Percentage composition of Nitrogen = (Mass of Nitrogen / Mass of compound) * 100

percentage composition = 14/96 * 100

Percentage composition = 0.14583 * 100

Percentage composition = 14.583%

3 0

3 years ago

A 1.93-mol sample of xenon gas is maintained in a 0.805-L container at 306 K. Calculate the pressure of the gas using both the i

Alex

Answer : The pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

Explanation :

First we have to calculate the pressure of gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $Xe$ gas = ?

V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

Now put all the given values in above equation, we get:

$P\times 0.805L=1.93mole\times (0.0821L.atm/mol.K)\times 306K$

$P=60.2atm$

Now we have to calculate the pressure of gas by using van der Waals equation.

$(P+\frac{an^2}{V^2})(V-nb)=nRT$

P = Pressure of $Xe$ gas = ?

V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

a = pressure constant = $4.19L^2atm/mol^2$

b = volume constant = $5.11\times 10^{-2}L/mol$

Now put all the given values in above equation, we get:

$(P+\frac{(4.19L^2atm/mol^2)\times (1.93mole)^2}{(0.805L)^2})[0.805L-(1.93mole)\times (5.11\times 10^{-2}L/mol)]=1.93mole\times (0.0821L.atm/mol.K)\times 306K$

$P=44.6atm$

Therefore, the pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

5 0

3 years ago