Using the<br><br>Cahn-Ingold-Prelog

Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the d

Elina [12.6K]

Answer:

an increase in 1-butene was observed when t-butoxide was used

Explanation:

When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.

The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

8 0

2 years ago

Materials:

shusha [124]

Answer:

Nice experiment

Explanation:

8 0

2 years ago

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

2 years ago

A ground water tank has its height 2m. Calculate the pressure at its bottom when it is completely filled with water.

Law Incorporation [45]

The pressure at the bottom : 19600 N/m²

<h3>Further explanation</h3>

Given

A ground water tank has its height 2m

Required

The pressure at its bottom

Solution

Hydrostatic pressure is the pressure caused by the weight of a liquid.

The weight of a liquid is affected by the force of gravity.

The hydrostatic pressure of a liquid can be formulated:

$\large{\boxed{\bold {P_h ~ = ~ \rho.g.h}}$

Ph = hydrostatic pressure (N / m², Pa)

ρ = density of liquid (kg / m³)

g = acceleration due to gravity (m / s²)

h = height / depth of liquid surface (m)

ρ = density of water (kg / m³) = 1000

g = acceleration due to gravity = 9.8 m/ sec²

The pressure

$\tt P=1000\times 9.8\times 2=19600~N/m^2$

6 0

2 years ago

What is the coefficient for hydrogen in the balanced equation for the reaction of solid molybdenum(iv) oxide with gaseous hydrog

anastassius [24]

The coefficient for hydrogen in the balanced equation of solid molybdenum(iV) oxide with gaseous hydrogen is 2

Explanation

Coefficient is defined to as a number in front of a chemical formula in a balanced chemical equation.

The reaction of molybdenum (iv) oxide with gaseous hydrogen is as below,

MoO2 + 2 H2→ Mo +2 H2O

From balanced equation above the coefficient for H2 is 2 since the number in front of H2 is 2

3 0

3 years ago

Using theCahn-Ingold-Prelog