Draw the ketone produced from the oxidation of 2-pentanol.

1 answer:

8 0

When 2-pentanol gets oxidized you end up with 2-pentanone. In this case, since we are oxidizing a secondary alcohol you end up with a ketone. They cannot be oxidized further than that. If it was a primary alcohol you would end up with an aldehyde.

You can find the mechanism as well the molecule formula of both 2pentanol and 2pentanone attached.

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Molodets [167]

The answer would be c as the cart is not in motion therefor ruling out kinetic and it is completely at rest making all of it energy potential

6 0

3 years ago

Which of the following is true for the equilibrium constant of a reaction? (5 pc

Slav-nsk [51]

The equilibrium constant of the reaction is represented by the symbol K. Thus, option C is the correct and accurate statement about the equilibrium constant.

<h3>What is the equilibrium constant?</h3>

The equilibrium constant is a representation of the concentration of the products and the reactants of the reaction that is raised to the powers through their stoichiometry coefficient.

Its value varies and changes at different temperatures and is not always less than 1. The equilibrium constant is the ratio of the coefficient of the products to reactants.

Therefore, option C. equilibrium constant is represented by K is true.

Learn more about the equilibrium constant here:

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2 years ago

A standard room temperature and pressure, most of the elements on the periodic table are

IgorC [24]

Gasses i’m pretty sure.

6 0

3 years ago

Read 2 more answers

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

LiRa [457]

Explanation:

(a) As the given chemical reaction equation is as follows.

$OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}$

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate = $K \times [OCl^{-}] \times [l^{-}]$