Question

Name the carboxylic acid and alcohol from which each ester is derived. (a) COC(=O)C1CCCCC1 (b) CC(=O)OC1CCC(OC(C)=O)CC1 (c) CCC=CC(=O)OC(C)C (d) CCOC(=O)CCC(=O)OCC

Short answer

Question: Identify the carboxylic acid and alcohol precursors for each of the following esters given in SMILES notation: (a) COC(=O)C1CCCCC1 (b) CC(=O)OC1CCC(OC(C)=O)CC1 (c) CCC=CC(=O)OC(C)C (d) CCOC(=O)CCC(=O)OCC Answer: (a) Cyclohexanecarboxylic acid (C1CCCCC1C(=O)O) and methanol (CO) (b) Acetic acid (CC(=O)O) and dihydrofuran-2-ylmethanol (C1CCOC1CO) (c) 4-Pentenoic acid (CCC=CC(=O)O) and isopropanol (OC(C)C) (d) 4-Oxopentanoic acid (CCC(=O)CC(=O)O) and ethanol (CCO)

Step-by-step solution

(a) Breaking the ester linkage in the first molecule

First, let's examine the SMILES notation for the first molecule: COC(=O)C1CCCCC1. We can see the ester linkage by identifying the carbonyl group (C=O) and the adjacent oxygen (O). We need to break this linkage between the carbonyl carbon and the oxygen atom. This will give us the carboxylic acid and alcohol precursors.

(a) Identifying the carboxylic acid and alcohol precursors

After breaking the ester linkage, we can identify the carboxylic acid as cyclohexanecarboxylic acid, with SMILES notation: C1CCCCC1C(=O)O, and the alcohol precursor as methanol, with SMILES notation: CO.

(b) Breaking the ester linkage in the second molecule

Now, let's examine the SMILES notation for the second molecule: CC(=O)OC1CCC(OC(C)=O)CC1. In this case, there are two ester linkages. We need to break both linkages to find the carboxylic acid and alcohol precursors.

(b) Identifying the carboxylic acid and alcohol precursors

After breaking the ester linkages, we can identify the carboxylic acid as acetic acid, with SMILES notation: CC(=O)O, and the alcohol precursor as dihydrofuran-2-ylmethanol, with SMILES notation: C1CCOC1CO.

(c) Breaking the ester linkage in the third molecule

Next, let's examine the SMILES notation for the third molecule: CCC=CC(=O)OC(C)C. We can identify the ester linkage and break it to find the carboxylic acid and alcohol precursors.

(c) Identifying the carboxylic acid and alcohol precursors

After breaking the ester linkage, we can identify the carboxylic acid as 4-pentenoic acid, with SMILES notation: CCC=CC(=O)O, and the alcohol precursor as isopropanol, with SMILES notation: OC(C)C.

(d) Breaking the ester linkage in the fourth molecule

Finally, let's examine the SMILES notation for the fourth molecule: CCOC(=O)CCC(=O)OCC. We can identify the ester linkage and break it to find the carboxylic acid and alcohol precursors.

(d) Identifying the carboxylic acid and alcohol precursors

After breaking the ester linkage, we can identify the carboxylic acid as 4-oxopentanoic acid, with SMILES notation: CCC(=O)CC(=O)O, and the alcohol precursor as ethanol, with SMILES notation: CCO.