Which isomer would you expect to undergo E2 elimination faster, trans-1-bromo-4-tert-butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its more stable chair conformation, and explain your answer.

Answers

Answer 1
Answer:

Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.

Explanation:

The two primary requirements for an E-2 elimination reaction are:

1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.

2.The hydrogen and leaving group must have a anti-periplanar position .

Any substrate which would follow the above two requirements can give elimination reactions.

For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane  to be stable it  must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.

Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.

The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.

The trans-1-bromo-4-tert-butylcyclohexane  does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.

so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.

Answer 2
Answer:

Final answer:

Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.

Explanation:

In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.

In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.

In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.

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Essentially all of the mass of an atom is due to the ______.(A) electrons.(B) neutrons.(C) nucleons.(D) protons.

Answers

Answer:

Atom is made up of NUCLEUS and electrons revolving around nucleus.

Nucleus itself contains protons and neutrons. Protons and neutrons are approximately 1680 times heavier than the electrons. So the major contribution to the mass of an atom comes from the nucleus.

Final answer:

The mass of an atom is mostly carried by the nucleons, protons, and neutrons, in its nucleus. Electrons contribute very little to the overall mass of an atom because of their small mass.

Explanation:

Essentially, all of the mass of an atom is due to the nucleons. An atom is primarily composed of protons, neutrons, and electrons. However, the mass of an electron is so small that it contributes very little to the overall mass of an atom. The term 'nucleons' refers to the particles in the nucleus of an atom, namely the protons and neutrons. These particles carry most of the atom's mass given their relatively large mass compared to electrons.

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Which of the following species is amphoteric? A)HCO32- B) HF O C) NH4+ D) P042-

Answers

Answer:

HCO_(3)^(-)

Explanation:

Chemical species which can behave as both acids and bases are known as amphoteric species.

HCO_(3)^(-) can behave as both acid and base.

HCO_(3)^(-) can donate H+:

HCO_3^(-) \rightarrow CO_3^(2-) + H^+

HCO_(3)^(-) can accept H+ as well:

HCO_3^(-) + H^+ \rightarrow H_2CO_3

HF can only behave as acid, as it can only donate H+.

HF(aq) \rightarrow H^(+)(aq) + F^(-)(aq)

NH_4^(+) is a conjugate acid of NH_3.

PO_4^(2-) is a conjugate base.

Final answer:

In chemistry, an amphoteric species can act as both a base and an acid. Among the given options, HCO3- (bicarbonate ion) is amphoteric because it can either donate or accept protons.

Explanation:

Among the provided options, the species that are amphoteric are HCO3- (bicarbonate ion). The term amphoteric refers to substances that can act both as an acid and a base. In other words, they can either donate or accept protons. Let's take HCO3- as an example. This ion can act as a base by accepting H+, forming H2CO3, or it can act as an acid by donating H+, forming CO32-. This dual behavior makes it an amphoteric species.

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How did Mendeleev come up with the first periodic table of the elements? (1 point)A He determined the mass of atoms of each element.
B He estimated the number of electrons in atoms of each element.
C He arranged the elements by different properties to find a pattern.
D He organized the elements by their atomic number.​

Answers

Mendeleeve's periodic table was based on the atomic mass of each element.

The correct answer is option A: He determined the mass of atoms of each element.

The periodic table can be described as a regular arrangement of elements based on some periodic properties. The periodic table has evolved through the ages. Its patter has been changed several times before it arrived at its present form.

The  Mendeleev periodic table was based on the mass of atoms of the elements. This table was published in Russia in 1869. His table even predicted elements that were yet to be discovered.

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Answer:

1. He arrange the elements by different properties to find a pattern!

Here are the rest!

2.. atoms are indivisible

3.  He determined the amount of charge on an electron

4. Like charges repel and opposite charges attract

5.Alpha particle deflection

Hope this helps the ballers!

The material the start all reactions are ______ and the materials that are formed are called _____.

Answers

Reactant and then product

Answer:

I think the second space is chemical bonds

Calculate the pH of a titration at the point when 15.0 mL of 0.15 M NaOH is added to 30.0 mL 0f 0.20 M HNO.

Answers

Answer:

The correct answer is 1.10.

Explanation:

Based on the given information, the molarity of the NaOH is 0.15 M, that is, 0.15 moles per liter of the solution.

Now the moles present in the 15 ml of the solution will be,  

0.015 × 0.15 = 2.25 × 10⁻³ moles of NaOH or 0.0025 moles of NaOH

Now, molarity of the HNO₃ given is 0.20 M, which means 0.2 moles per liter of the solution.  

Now the moles present in the 30 ml of the solution will be,  

0.030 × 0.2 = 0.006 moles of HNO₃

Now the complete disintegration of acid and base will be,  

NaOH (aq) (0.025 moles) ⇔ Na⁺ (aq) (0.025) + OH⁻ (aq) (0.025 moles)

HNO₃ (aq) (0.006 moles) ⇔ H⁺ (0.006 moles) + NO₃⁻ (aq) (0.006 moles)

Now the additional Hydrogen ions at titration point is,  

= 0.006 - 0.0025 = 0.0035 moles of H+

Now the concentration of H+ ions in the 45 ml of the solution will be,  

= 0.0035/45 × 1000

= 0.078 M

pH = -log[H⁺] = -log [0.078]

= 1.10