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Blababa [14]
3 years ago
15

A ____________ alkene is more stable than a ____________ alkene because they have fewer steric interactions. In an elimination r

eaction, a geometry where the β hydrogen and the leaving group are on opposite sides of the molecule is called ____________ periplanar. In an ____________ mechanism, a nucleophile attacks the carbocation, forming a substitution product, while in an ____________ mechanism, a base removes a β hydrogen from the carbocation, forming a new π-bond. CH3CH2Br and NaOH react by an ____________ mechanism. Stronger bases, like hydroxide, favor ____________ reactions, whereas weaker bases, like water favor, ____________ reactions. Disubstituted alkenes always have the possibility to exist as two different ____________ . ____________ reactions are regioselective, favoring formation of the more substituted and more stable alkene
Chemistry
1 answer:
Vesna [10]3 years ago
6 0

Answer:

See explanation

Explanation:

A <u>trans</u> alkene is more stable than a <u>cis alkene</u> because they have fewer steric interactions.

<em>⇒ In a cis alkene there is steric hindrance, because the methyl groups are on the same side of the double bond. </em>

<em>Because of this steric crowding, there are van der Waals repulsive forces between the electron clouds of the groups. </em>

<em> </em>

<em>This decreases the stability of the cis alkene.</em>

<em />

In an elimination reaction, a geometry where the β hydrogen and the leaving group are on opposite sides of the molecule is called <u>anti</u> periplanar.

<em> ⇒ 'Anti’ refers to the two functional groups lying on opposite sides of the plane of the bond</em>

In an <u>SN1 </u>mechanism, a nucleophile attacks the carbocation, forming a substitution product,

<em> ⇒ The SN1 reaction is a substitution reaction, and means "nucleophilic substitution".The "1" says that the rate-determining step is unimolecular. Thus, the rate equation is often shown as having first-order dependence on electrophile and zero-order dependence on nucleophile.</em>

while in an <u>E1</u><u> </u>mechanism, a base removes a β hydrogen from the carbocation, forming a new π-bond.

<em> ⇒ E1 indicates a elimination, unimolecular reaction</em>

<em>This implies that the rate determining step of the mechanism depends on the decomposition of a single molecular species.</em>

<em>.This is a classic elimination reaction – forming a new C–C(π) bond, and breaking a C–H and C–leaving group bond.</em>

CH3CH2Br and NaOH react by an <u>SN2</u><u> </u>mechanism.

<em> ⇒  It's a type of reaction mechanism that is common in organic chemistry, where one bond is broken and one bond is formed, synchronously, (in one step.) </em>

Stronger bases, like hydroxide, favor<u> E2</u> reactions, whereas weaker bases, like water favor, <u>E1</u> reactions

Disubstituted alkenes always have the possibility to exist as two different <u>Diastereomer.</u>

<em>Diastereomer are stereoisomers that are not mirror images of one another and are non-superimposable on one another. They exist (always) in 2 forms.</em>

<u>Elimination reactions</u> are regioselective, favoring formation of the more substituted and more stable alkene.

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<em></em>

where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).

[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).

  • At the time needed to be determined:

<em>8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.</em>

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∴ The remaining concentration of Potassium-40 ([A] = 88.88%).

and that of argon-40 produced from potassium-40 decayed = 11.11%.

  • That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).

∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.

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