Match the organic compounds with their formulas.

Matching:

Here's the correct matching between the names of organic compounds and their formulas:

• 1. пропановая кислота (Propanoic acid) corresponds to 3. \( C_2H_5COOH \)
• 2. пентаналь (Pentanal) corresponds to 4. \( C_4H_9CHO \)
• 3. сахароза (Sucrose) corresponds to 1. \( C_{12}H_{22}O_{11} \)
• 4. метиловый эфир бутановой кислоты (Methyl butanoate) corresponds to 2. \( C_3H_7COOCH_3 \)
• 5. пропанол (Propanol) corresponds to 5. \( C_3H_7OH \)

Explanation:

1. Пропановая кислота (Propanoic acid) is a carboxylic acid with 3 carbon atoms. Its general formula is \( RCOOH \). If R is \( C_2H_5 \), the formula is \( C_2H_5COOH \).

2. Пентаналь (Pentanal) is an aldehyde with 5 carbon atoms. Its general formula is \( RCHO \). For pentanal, R has 4 carbon atoms, so the formula is \( C_4H_9CHO \).

3. Сахароза (Sucrose) is a disaccharide, a common sugar, with the molecular formula \( C_{12}H_{22}O_{11} \).

4. Метиловый эфир бутановой кислоты (Methyl butanoate) is an ester. Butanoic acid is \( C_3H_7COOH \). When it forms a methyl ester, the hydrogen of the carboxyl group is replaced by a methyl group \( CH_3 \). So the formula is \( C_3H_7COOCH_3 \).

5. Пропанол (Propanol) is an alcohol with 3 carbon atoms. Its general formula is \( C_nH_{2n+1}OH \). For propanol, \( n=3 \), so the formula is \( C_3H_{2 \cdot 3 + 1}OH = C_3H_7OH \).

Answer: 1-3, 2-4, 3-1, 4-2, 5-5.