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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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Answer:

(3) 5.36

Explanation:

Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.

The reaction is:

HAc + NaOH ⇒ NaAc + H₂O

V NaOH = 40 mL x 1 L/1000 mL = 0.040 L

mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol

mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol

mol HAc left after reaction = 0.0025 - 0.002 = 0.0005

Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation

pH = pKa + log ((Ac⁻)/(HAc))

(Notice we do not have to calculate the molarities of  Ac⁻ and HAc because the volumes cancel in the quotient)

pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36

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5 0
3 years ago
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Answer:

<h2>1.505 × 10²⁴ particles</h2>

Explanation:

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From the question we have

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