All categories

2 years ago

Which one of the following is not true concerning Diels-Alder reactions?

Chemistry

1 answer:

zloy xaker [14]2 years ago

6 0

Answer:Option c is incorrect

Explanation:

Diels-Alder reaction is a 4+2 cycloaddition reaction .

The reaction occurs between a diene and a dieneophile .

Generally In the Diels-alder reaction HOMO of the diene and LUMO of the dienophile react with proper orbital symmetry. But vice-versa can also be used.

HOMO-Highest occupied molecular orbital

LUMO-Lowest unoccupied molecular orbital

The primary driving force for the diels alder reaction is is the conversion of 2pi bonds into 2 sigma bonds. The sigma bonds are energetically more favorable than pi bonds.

The diels alder reaction happens through orbital interaction and hence the substituents on either the diene or dienophile do not change there stereochemistry in the product so it is a stereospecific reaction.Since predominantly only isomer would be produced. so a is correct.

The Diels alder reaction is a concerted( more bonds form at a time) reaction which means it is just a one step reaction. so statement b is correct.

The option c is incorrect as diels alder reaction occurs through orbital interaction in a pericyclic manner. Since Diels alder reactions are pericyclic in nature and occur through orbital symmetry they do not involve polar intermediates like carbocation or radicals.

The dienes must be conjugated as on account of conjugation stability of a diene also increases. Also since we know that due to conjugation the energy of LUMO decreases and that of HOMO increases and so HOMO is more reactive and generally we involve HOMO of the diene and LUMO of the dienophile. So conjugated dienes are important.

So for a bond formation to take place in a Diels-Alder reaction HOMO and LUMO with proper symmetry must overlap.

You might be interested in

In a hospital laboratory, a 10.0 mL sample of gastric juice (predominantly HCl), obtained several hours after a meal, was titrat

LiRa [457]

Answer: 1.14

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL$

Putting values in above equation, we get:

$1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M$

To calculate pH of gastric juice:

molarity of $H^+$ = 0.072

$pH=-log[H^+]$

$pH=-log(0.072)=1.14$

Thus the pH of the gastric juice is 1.14

3 0

2 years ago

A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f

Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

8 0

2 years ago

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c

Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

$18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M$

Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

$42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M$