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3 years ago

Does methylenecyclohexane obey Zaitsev's rule? why or why not

Chemistry

1 answer:

kifflom [539]3 years ago

6 0

Answer:

Explanation: Zaitsev’s or Saytzev’s (anglicized spelling) rule is an empirical rule used to predict regioselectivity of 1,2-elimination reactions occurring via E1 mechanism or via E2 mechanism. It states that in a regioselective E1 or E2 reaction the major product is the more stable alkene, i.e., the alkene with the more highly substituted double bond.

E1 reaction always follow Zaitsev’s rule; with E2 reactions, there are exceptions (see antiperiplanar).

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Classify this chemical reaction: 2 HNO3 (aq) + Sr(OH)2 (aq) → Sr(NO3)2 (aq) +2 H2O (l)

ad-work [718]

Answer:

This is not an answer its just to get points

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3 years ago

Diamond and graphite are two crystalline forms of carbon. In which form are the bonds between the C atoms arranged at tetrahedra

Fynjy0 [20]

Answer:

The answer is letter A. Diamond

Explanation:

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2 years ago

Which period does this atom belong to?

GrogVix [38]

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2 years ago

What is the correct name for the compound named 1-chloro-3-pentyne

velikii [3]

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3 years ago

Which of the following reactions have a positive ΔSrxn? Check all that apply.

PolarNik [594]

Answer:

The reactions that have a positive ΔS rxn are the first and the fourth choices:

2A(g) + B(s) → 3C(g)

2A(g) + 2B(g) → 5C(g)

Explanation:

ΔS rxn is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

2A(g) + B(s) → 3C(g)

1) The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: ΔS rxn is positive.

2A(g) + B(g) → C(g)

Using the same reasoning, 3 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

A(g) + B(g) → C(g)

Again, 2 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

2A(g) + 2B(g) → 5C(g)

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: ΔS rxn is positive.

6 0

3 years ago