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Natali5045456 [20]
3 years ago
7

In an elimination reaction, the non-preferred geometry in which the β hydrogen and the leaving group are on the same side of the

molecule is called ____________ periplanar. According to the Zaitev rule, the major product in a β-elimination has the more substituted double bond. In an E1 reaction, the rate depends on only the alkyl halide concentration. The rate of the E2 reaction increases as the strength of the base increases because the reaction is biolecular and the base appears in the rate equation. Carbocation intermediates are involved in ____________ mechanisms. Polar aprotic solvents increase the rate of the ____________ reactions. Anti periplanar geometry is the preferred arrangement for any alkyl halide undergoing ____________ elimination, regardless of whether it is cyclic or acyclic.
Chemistry
1 answer:
Ratling [72]3 years ago
8 0

Answer:

a. syn

b. E1

c. E2

d. E2

Explanation:

In organic chemistry, the rule of Saytzeff, Saytzev or Zaitsev states that in an elimination reaction (β-elimination) in which more than one alkene can be formed, the most thermodynamically stable will be the majority.

In general, the most substituted alkene is the most stable, due to the electronic sharing properties of the alkyl groups with the double bond C = C (hyper-conjugation). Also, in some cases, another stabilizing effect may be incurred when establishing the regioselectivity such as the conjugation of the double bond with other groups.

This rule is valid except in bimolecular elimination reactions (E2) in which there is a significant steric hindrance, branched substrate and / or bulky base, and without the possibility of conjugation, then applying Hofmann's rule.

The elimination reaction E1 has a potential energy profile similar to that of  an SN1 reaction. The formation step of carbocation is very endothermic,  with a transition state which is what determines the speed of the  reaction. The second step is a rapid and exothermic deprotonation. Base  does not participate in the step that determines the speed of the process, so it  only depends on the concentration of alkyl halide.

The bimolecular elimination E2 takes place without intermediates and consists of a single ET, in which the base abstracts the proton, the leaving group leaves and the two carbons involved are rehybridized from sp3 to sp2.

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allsm [11]

The concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 M dextrose solution to 25 ml using a 25 ml volumetric flask is 0.56M.

Concentration is defined as the number of moles of a solute present in the specific volume of a solution.

According to the dilution law, the degree of ionization increases on a dilution and it is inversely proportional to the square root of concentration. The degree of dissociation of an acid is directly proportional to the square root of a volume.

M₁V₁=M₂V₂

Where, M₁=1.0M, V₁=14ml, M₂=?, V₂=25ml

Rearrange the formula for M₂

M₂=(M₁V₁/V₂)

Plug all the values in the formula

M₂=(1.0M×14 ml/25 ml)

M₂=14 M/25

M₂=0.56 M

Therefore, the concentration of a dextrose solution after the dilution is 0.56M.

To know more about dilution

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The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g
ra1l [238]

Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g)  ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ

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4 years ago
How many lone pairs does CO2 have? Does it have Single bonds or Double bonds? Does it obey the octet rule?
Aleks04 [339]
The molecule CO2 have 2 lone pairs located in the oxygen atoms. It has two double bonds also connecting carbon atom to the two oxygen atoms. It does obey the octet rule so the atoms need 8 electrons in each of them that is why electrons are shared.
3 0
3 years ago
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