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kogti [31]
3 years ago
14

Which of the following is an acceptable IUPAC name?(A) 3,6-dimethylheptane (B) 2-ethyl-3-methylheptane (C) 5-methyl-3-ethylhepta

ne (D) 4-ethyl-4-methylheptaneD

Chemistry
1 answer:
bearhunter [10]3 years ago
3 0

Answer:

D) 4-ethyl-4-methylheptane

Explanation:

The rules for naming of alkanes with substituents.

1. The carbon chain with the maximum number of carbon atoms must be selected as a parent chain.

2.Numbering should be done in such a way that each substituent gets the least number.

3. Substituents whose name comes before the another substituents's name in the English alphabet is written first.

4. Substituents are written first with their location in the chain and the name of the parent chain is done. Numbers are separated from numbers by comma and numbers are separated from letters by using hyphen.

Considering (a) 3,6-dimethylheptane

<u>Violation of Rule - 2 mentioned above</u>

The numbering of the parent chain is not done in a right way. Numbering must be done such that each substituent gets the least number. So, The correct name is 2,5-dimethylheptane

Considering (b) 2-ethyl-3-methylheptane

<u>Violation of Rule - 1 mentioned above.</u>

The carbon chain with the maximum number of carbon atoms must be selected as a parent chain. The parent chain selected is of 7 carbon atoms but the parent chain is of 8 carbon atoms. So, The correct name is 3,4-dimethyloctane

Considering (c) 5-methyl-3-ethylheptane

<u>Violation of Rule - 3 mentioned above.</u>

'e' comes before 'm' . So, ethyl- is written first than methyl- . So, The correct name is 3,6-dimethyloctane

Considering (d) 4-ethyl-4-methylheptane

<u>This is a IUPAC name and following all the rules mentioned above.</u>

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<u>Explanation:</u>

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1.) Calculation: If 9.02 x 1024 particles of vinegar (HC2H3O2)HC2H3O2) are added to 16.5 moles of eggshell (CaCO3) and 6.35 mole
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The theoretical yield of acetate is 2607 g. The actual yield of acetate is 1066.8 g. The percentage yield of acetate is 41%.

If 1 mole of vinegar contains 6.02  x 10^23 particles

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x = 1 mole x 9.02 x 10^24 /6.02 x 10^23

x = 15 moles of vinegar

The reaction is as follows;

2HC2H3O2 + CaCO3 -----> Ca(C2H3O2)2 + H2O + CO2

Since 2 moles of vinegar reacts with 1 mole of carbonate

x moles of vinegar reacts with 16.5 moles of carbonate

x =  2 moles x 16.5 moles/ 1 mole

x = 33 moles of vinegar

We can see that the vinegar is the reactant in excess hence the carbonate is the limiting reactant.

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Actual yield = 6.35 moles  x 158 g/mol = 1066.8 g

Percent yield = 1066.8 g/2607 g × 100/1

= 41%

Learn more: brainly.com/question/13440572?

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