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kicyunya [14]
3 years ago
6

The tosylate of (2R,3S)-3-phenylbutan-2-ol undergoes an E2 elimination on treatment with sodium ethoxide. Draw the structure of

the alkene that is produced.

Chemistry
1 answer:
andrezito [222]3 years ago
6 0

Answer:

(R)-but-3-en-2-ylbenzene

Explanation:

In this reaction, we have a very <u>strong base</u> (<em>sodium ethoxide</em>). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an <u>E2 mechanism</u>, therefore, the hydrogen that is removed must have an <u>angle of 180º</u> with respect to the leaving group (the "OH"). This is known as the <u>anti-periplanar configuration</u>.

The hydrogen that has this configuration is the one that placed with the <u>dashed bond</u> (<em>red hydrogen</em>). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.

See figure 1

I hope it helps!

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If the pKa of Lysine is 5, then at a pH of 7, it is most likely going to be a) Neutral. b) Negatively charged. c) Positively cha
Sladkaya [172]
C. Positively charged
7 0
3 years ago
Some fats are called "saturated" fats because __________. Select one: a. their carbon skeleton is bent to form four fused rings
marin [14]

Answer:

All carbons in the carbon skeleton contain the maximum number of hydrogen atoms

Explanation:

Saturated fats are class of compounds having all the fatty acids containing only single bonds. In other words, carbon skeleton has only single bonds.

Unsaturated compounds undergoes hydrogenation to form saturated fats.

In hydrogenation, hydrogen atoms are added to the carbon attached to double bond. After become saturated, no more hydrogen atoms can be added.

Therefore, it can be said that in saturated fats all carbons in the carbon skeleton contain the maximum number of hydrogen atoms.

6 0
3 years ago
A boy swings a rubber ball attached to a string over his head in a horizontal, circular path. The piece of string is 1.15 m long
gregori [183]

Answer:

v = 16.49 m/s

Explanation:

Given that,

Length of the string, l = 1.15 m

The ball makes 137 complete turns each minute.

We know that, 1 turn = 6.28 rad

137 turns = 860.79 rad

1 min = 60 s

\omega=\dfrac{860.79\ rad}{60\ s}\\\\=14.34\ rad/s

We need to find the tangential velocity of the ball. It can be given by

v=r\omega\\\\=1.15\times 14.34\\\\v=16.49\ m/s

So, the tangential velocity of the ball is 16.49 m/s.

5 0
3 years ago
The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.
harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})

= 70molCO_2

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})

= 87.5molO_2

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

84.0molO_2(\frac{4molO_2}{5molO_2})

= 67.2molCO_2

7 0
3 years ago
Read 2 more answers
A sample of gas in a 14.6 L flexible container is at 25.0oC and 1.00atm. What is the volume of the sample when heated to 220.0oC
inn [45]

Answer: 24.1 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=14.6L\\T_1=25.0^oC=(25+273)K=298K\\V_2=?\\T_2=220.0^0C=(220+273)K=493K

Putting values in above equation, we get:

\frac{14.6}{298K}=\frac{V_2}{493}\\\\V_2=24.1L

Thus the volume of the sample when heated to 220.0oC and the pressure is constant is 24.1 L

7 0
3 years ago
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