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Chapter 24: Problem 45

Draw the molecular structure for \((\mathbf{a})\) a carboxylic acid that is an isomer of ethyl ethanoate, \((\mathbf{b})\) an alcohol that is an isomer of dimethylether.

Short Answer

Expert verified
The required molecular structures are: 1. Butanoic acid (\(CH_3CH_2CH_2COOH\)) as an isomer of ethyl ethanoate: ``` H H H H | | | | H - C - C - C - C - O - H | | | H H OH ``` 2. Ethanol (\(CH_3CH_2OH\)) as an isomer of dimethyl ether: ``` H H | | H - C - C - O - H | H ```

Step by step solution

01

Calculate the molecular formula for ethyl ethanoate

Ethyl ethanoate (ethyl acetate) has an ethyl group (C2H5) attached to an ethanoate group (CH3COO). Thus, the molecular formula for ethyl ethanoate is \(C_4H_8O_2\).
02

Calculate the molecular formula for dimethyl ether

Dimethyl ether has two methyl groups (CH3) attached to an oxygen atom. Therefore, the molecular formula for dimethyl ether is \(C_2H_6O\).
03

Determine the carboxylic acid isomer of ethyl ethanoate

Now, we need to find a carboxylic acid with the molecular formula \(C_4H_8O_2\). The simplest carboxylic acid with this formula is butanoic acid. It consists of a 4-carbon chain with a carboxylic acid group (-COOH) at one end. Its structure is: \(CH_3CH_2CH_2COOH\).
04

Draw the molecular structure for butanoic acid

The molecular structure for butanoic acid is: ``` H H H H | | | | H - C - C - C - C - O - H | | | H H OH ```
05

Determine the alcohol isomer of dimethyl ether

Next, we need to find an alcohol with the molecular formula \(C_2H_6O\). The simplest alcohol with this formula is ethanol. It consists of a 2-carbon chain with a hydroxyl group (-OH) at one end. Its structure is: \(CH_3CH_2OH\).
06

Draw the molecular structure for ethanol

The molecular structure for ethanol is: ``` H H | | H - C - C - O - H | H ``` So the required structures are: 1. Butanoic acid (\(CH_3CH_2CH_2COOH\)) as an isomer of ethyl ethanoate. 2. Ethanol (\(CH_3CH_2OH\)) as an isomer of dimethyl ether.

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Most popular questions from this chapter

Adenine and guanine are members of a class of molecules known as purines; they have two rings in their structure. Thymine and cytosine, on the other hand, are pyrimidines, and have only one ring in their structure. Predict which have larger dispersion forces in aqueous solution, the purines or the pyrimidines.

(a) What is the empirical formula of cellulose? (b) What is the monomer that forms the basis of the cellulose polymer? (c) What bond connects the monomer units in cellulose: amide, acid, ether, ester, or alcohol?

Give the molecular formula of a hydrocarbon containing six carbon atoms that is (a) a cyclic alkane, (b) a cyclic alkene, \((\mathbf{c})\) a linear alkyne, \((\mathbf{d})\) an aromatic hydrocarbon.

Write a condensed structural formula for each of the following: \((\mathbf{a})\) an amide with the formula $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{NO},(\mathbf{b})\( a cyclic ether with the formula \)\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O},(\mathbf{c})$ a diketone compound with the formula \(\mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}_{2}\), (d) a phenyl carboxylic acid with the formula \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{2}\)

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