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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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2 years ago
How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr
pickupchik [31]

Answer: There are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = 87.5^{o}C = (87.5 + 273) K = 360.5 K

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

Thus, we can conclude that there are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

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kakasveta [241]
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⋅
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formula units of barium nitrate is the quotient multiplied by the molar mass:
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m
o
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×
16.6*10^23
⋅
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