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sashaice [31]
2 years ago
11

Draw the alkyl bromide(s) you would use in a malonic ester synthesis of ethyl 2-methyl-4-pentenoate

Chemistry
2 answers:
Rudik [331]2 years ago
6 0

The structure of the alkyl bromides used in a malonic ester synthesis of ethyl 2-methyl-4-pentenoate are as drawn in the attached image.

<h3>Ethyl 2-methyl-4-pentenoate by Malonic ester synthesis.</h3>

The malonic ester synthesis is a chemical reaction characterized by the alkylation of diethyl malonate or a similar ester of malonic acid at the carbon alpha (directly adjacent) to both carbonyl groups, and subsequently converted to a substituted acetic acid.

Hence, it follows from the structure of Ethyl 2-methyl-4-pentenoate that the alkyl bromides used are Ethyl bromide and methyl bromide.

Read more on Malonic ester synthesis;

brainly.com/question/17237043

crimeas [40]2 years ago
5 0

The structure of alkyl bromide that would be used in a malonic ester synthesis of ethyl 2-methyl-4-pentenoate are ethyl bromide and methyl bromide.

<h3>What is alkyl bromide?</h3>

When halogens such as Br, Cl, I, attched to the sp3 carbon atom of alkyl group, they called alkyl halides.

Here, bromide is attached to an alkyl group.

Ethyl bromide and methyl bromide are the alkyl halides used in synthesis of ethyl 2-methyl-4-pentenoate

Thus, ethyl bromide and methyl bromide are the alkyl bromides.

Learn more about alkyl bromide

brainly.com/question/14126879

#SPJ4

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For the decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 °C 2 H2O2(aq)2 H2O(l) + O2(g) the following data hav
saul85 [17]

Answer:

K= 0.06611

Explanation: The rate of reaction is defined as the change in concentration of any of reactant or products per unit time. From the given reaction, the rate of reaction may be equal to the rate of disappearance of reactant which is equal to the rate of appearance of products.

The average rate of disappearance of H2O2 over the time period from t=0 min at 8.92×10^-2 to t=9.63min at 4.72×10^-2 is given as -4.36×10^-3Mmin-1.

We can say:

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•The concentration at time t. [H2O2]t is 4.72×10^-2

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The expression of rate constant for a first order reaction is shown as

K=2.303/t log[H2O2]o/ [H2O2]t

Substitute the values of t, [H2O2]o and [H2O2]t in the equation of rate constant.

K=2.303/9.63 log [8.92×10^-2]/ [4.72×10^-2]

K= 0.2391 (log 8.92×10^-2 - log 4.72×10^-2)

K= 0.2391 [-1.0496-(-1.3261)]

K= 0.2391 (-1.0496+1.3261)

K= 0.2391 (0.2765)

K= 0.06611

Since the value of k is almost constant, the decomposition of H2O2 is a first order reaction.

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Answer:

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Explanation:

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First energy level is n=1 has 1s2 so total 2 electrons


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Third has 3s2 3p5 electrons just 7 total... p can contain 6 electrons but only 5 are present. So the third level has lesser number of electrons than that can be filled


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