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.
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.
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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Answer:
balanced............
Answer:
Yes.
Explanation:
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In this according to the attached file, we infer that the aniline can be nitrated by the addition of nitric acid and in presence of sulfuric acid that provides an acidic media. It leads to the formation of o-nitroaniline, m-nitroaniline and p-nitroaniline whereas the major products are the last two due to the steric hindrance.
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The correct option is B.
A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. The important parts of a voltaic cell: The anode is an electrode where oxidation occurs. The cathode is an electrode where reduction occurs.
A galvanic cell is an electrochemical cell that converts the free energy of a chemical process into electrical energy. A photogalvanic cell is one that generates species photochemically which react resulting in an electrical current through an external circuit.
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Answer:
B) chemical energy to electrical energy
Explanation:
b. 41 .4%
c. 80.5%
d. 0.805 %
Answer:
Option C. 80.5%
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
C3H8 + 5O2 —> 3CO2 + 4H2O
Next, we shall determine the mass of C3H8 that reacted and the mass of CO2 produced from the balanced equation.
This is illustrated below:
Molar mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44 g/mol
Mass of C3H8 from the balanced equation = 1 x 44 = 44 g
Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol
Mass of CO2 from the balanced equation = 3 x 44 = 132 g
From the balanced equation above,
44 g of C3H8 reacted to produce 132 g of CO2.
Next, we shall determine the theoretical yield of CO2.
This can be obtained as shown below:
From the balanced equation above,
44 g of C3H8 reacted to produce 132 g of CO2.
Therefore, 0.295 kg (i.e 295 g) will react to produce = (295 x 132)/44 = 885 g of CO2.
Therefore, the theoretical yield of CO2 is 885 g.
Finally, we shall determine the percentage yield of CO2 as follow:
Actual yield of CO2 = 712 g
Theoretical yield of CO2 = 885 g
Percentage yield of CO2 =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield of CO2 = 712/885 x 100
Percentage yield = 80.5%
Therefore, the percentage yield of CO2 is 80.5%.
Answer:
Non polar covlant
Explanation:
(B) H2S
(C) SO3
(D) PCl3
(E) CH2Cl2
(F) NO2
Answer:
D and E
Explanation:
D. PCl3
E. CH2Cl2