Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-chlorobutane. Based on this information, the major organic product(s) of this reaction are expected to be _____.

Answers

Answer 1

Answer:

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

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Zeros laced at the end of the significant number are...

Answers

Answer:

Zeros located at the end of significant figures are significant.

Explanation:

Hope it will help :)

What's another name for potential energy? 1. kinetic

2. stored

3. mechanical

4. moving

Answers

Answer: Stored energy

Explanation:

Answer:

2. stored energy It is 100% right

Explanation:

A compound is found to contain 18.28 % phosphorus , 18.93 % sulfur , and 62.78 % chlorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 169.4 g/mol. The molecular formula for this compound is

Answers

Answer:

1. EF = PSCl₃; 2. MF = PSCl₃

Explanation:

1. Empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our first job is to calculate the molar ratio of P:S:Cl.

Assume 100 g of the compound.

(a) Calculate the mass of each element.

Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.

(b) Calculate the moles of each element

$\text{Moles of P} = \text{18.28 g C} * \frac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\n\n\text{Moles of S} = \text{18.93 g S} * \frac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\n\n\text{Moles of Cl} = \text{62.78 g Cl} * \frac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}$

(c) Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3

(d) Write the empirical formula

EF = PSCl₃

The empirical formula for this compound is PSCl₃.

2. Molecular formula

(a) Calculate the ratio of the molecular and empirical formula masses

n = (169.4 u)/(169.40 u) = 1.000 ≈ 1

(b) Calculate the molecular formula

MF = (EF)ₙ = (EF)₁ = PSCl₃

The molecular formula for this compound is PSCl₃.

The empirical formula of the compound is PSCl₃.

To find the empirical formula, we first need to find the moles of each element in the compound. We can do this by dividing the mass of each element by its molar mass. The molar masses of the elements are:

P = 30.97 g/mol

S = 32.06 g/mol

Cl = 35.45 g/mol

The mass percentages given are for 100 g of the compound. So, the mass of each element in 100 g of the compound is:

P = 18.28 g

S = 18.93 g

Cl = 62.78 g

The moles of each element are then:

P = 18.28 g / 30.97 g/mol = 0.590 mol

S = 18.93 g / 32.06 g/mol = 0.590 mol

Cl = 62.78 g / 35.45 g/mol = 1.770 mol

The smallest whole number ratio of the moles of each element is 1:1:3. So, the empirical formula of the compound is PSCl3.

The molecular formula of the compound can be the same as the empirical formula, or it can be a multiple of the empirical formula. The molecular formula is not given, so we cannot say for sure what it is. However, we can say that the molecular formula must be a whole number multiple of the empirical formula PSCl3.

Learn more about empirical formula,here:

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Please help me i’m in need!

Answers

Answer:

balanced............

Draw a line-bond structure for CBrN. Explicitly draw all H atoms. You do not have to include lone pairs in your answer. In cases where there is more than one answer, just draw one.

Answers

Answer:

Br - C ≡ N

Explanation:

To draw the Lewis line-bond structure we need to bear in mind the octet rule, which states that in order to gain stability each atom tends to share electrons until it has 8 electrons in its valence shell.

C has 4 e⁻ in its valence shell so it will form 4 covalent bonds.

Br has 7 e⁻ in its valence shell so it will form 1 covalent bond.

N has 5 e⁻ in its valence shell so it will form 3 covalent bonds.

The most stable structure that respects these premises is:

Br - C ≡ N

It does not have any H atom.

2.67 grams of butane (C4H10) is combusted in a bomb calorimeter. The temperature increases from 25.68 C to 36.2C. What is the change in the internal energy (deltaE) in KJ/mol for the reaction if the heat capacity of the bomb calorimeter is 5.73 kJ/C?

Answers

The internal energy : 1310.43 kJ/mol

Further explanation

Internal energy (ΔE) can be formulated for Calorimeter :

$\tt \Delta E=C*.\Delta t$

C= the heat capacity of the calorimeter

Δt=36.2-25.68=10.52°C

$\tt \Delta E=5.73 kJ/^oC* 10.52^oC=60.2796~kJ$

mol butane(MW=58,12 g/mol)

$\tt (2.67)/(58,12 )=0.046$

the internal energy (ΔE) in KJ/mol

$\tt (60.2796)/(0.046)=1310.43~kJ/mol$

Final answer:

The change in internal energy when 2.67 grams of butane is combusted in a bomb calorimeter, given a temperature increase from 25.68 C to 36.2C and a heat capacity of 5.73 kJ/C for the calorimeter, is approximately 1308 kJ/mol.

Explanation:

To solve the problem of calculating the changes in internal energy when 2.67 grams of butane (C4H10) is combusted in a bomb calorimeter, it is necessary to understand calorimeter's heat capacity and how a bomb calorimeter works.

The first step will be to calculate the change in temperature which here is the final temperature subtracted from the initial temperature: 36.2 C - 25.68 C = 10.52 C.

Then, we multiply this temperature change by the heat capacity of the calorimeter to find the total heat produced by the reaction in kJ: 10.52 C * 5.73 kJ/C = 60.18 kJ.

The final step is to convert grams of butane to moles, because we are asked to find the energy change in kJ/mol. The molar mass of butane (C4H10) is approximately 58.12 g/mol. So we have approximately 2.67 g / 58.12 g/mol = 0.046 mol.

Finally, we divide the heat produced by the number of moles to get the energy change per mole of butane: 60.18 kJ / 0.046 mol = approximately 1308 kJ/mol.

Learn more about Bomb Calorimetry here:

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