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3 years ago

What is the role of a strong acid catalyst in an elimination reaction?

Chemistry

1 answer:

Ivan3 years ago

8 0

Explanation:

An elimination reaction is a type of organic reaction in which two substituents are removed from a molecule in either a one or two-step mechanism. The one-step mechanism is known as the E2 reaction, and the two-step mechanism is known as the E1 reaction.

Another definiton is;

Elimination reaction, any of a class of organic chemical reactions in which a pair of atoms or groups of atoms are removed from a molecule, usually through the action of acids, bases, or metals and, in some cases, by heating to a high temperature. It is the principal process by which organic compounds containing only single carbon-carbon bonds (saturated compounds) are transformed to compounds containing double or triple carbon-carbon bonds (unsaturated compounds).

Strong Acids play two key roles in elimination reactions:

1. Without the acid catalyst, the reaction is painfully slow.

2. Acid greatly facilitates elimination of the leaving group.

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2 years ago

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4) the one that is reduced, which is the oxidizing agent

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2 years ago

Sanjay is learning about the elements that make up Earth’s crust. Ms. Richards, his Earth science teacher, gives him an element

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2 years ago

The measured value 24.00 g has an uncertainty of?

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Answer:

±0.005 g

Explanation:

The uncertainty depends on whether the measurement was obtained manually or digitally.

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The minimum uncertainty is ±0.01 g.

It may be greater, depending on random or personal errors

2. Digital

Most measurements of mass are now made on digital scales.

A digital device must always round off the measurement it displays.

For example, if the display reads 20.00, the measurement must be between 20.005 and 19.995 (±0.005).

If the measured value were 20.006, the display would round up to 20.01.

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3 years ago

Give regents and mechanism

Kryger [21]

Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a hydration of alkenes reaction. But, in this example, we have an anti-Markovnikov reaction. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: "hydroboration".

The first step of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the second step, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the third step, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In step number 4 we have the migration of the C-B bond to oxygen. Then in step number 5, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will protonate the molecule to produce the alcohol.

See figure 1

I hope it helps!

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3 years ago