QuestionIdentify the compound classes for the formula $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}_{2}$ . Check all that apply: carboxylic acid, alcohol, phenol, amide, ether, ester.

Studdy Solution

STEP 1

Assumptions1. The molecular formula of the compound is $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{}_{}$ . The possible functional groups are carboxylic acid, alcohol, phenol, amide, ether, ester

STEP 2

First, let's consider the functional group of carboxylic acid. Carboxylic acids have the general formula $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{COOH}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{}_{2}$ .

STEP 3

Comparing the general formula of carboxylic acid with the given molecular formula, we can see that they match. Therefore, the compound could belong to the carboxylic acid class.

STEP 4

Next, let's consider the functional group of alcohol. Alcohols have the general formula $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{OH}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+2} \mathrm{}$ .

STEP 5

Comparing the general formula of alcohol with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the alcohol class.

STEP 6

Next, let's consider the functional group of phenol. Phenols have the general formula $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{}$ .

STEP 7

Comparing the general formula of phenol with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the phenol class.

STEP 8

Next, let's consider the functional group of amide. Amides have the general formula $\mathrm{R-C-NH}_{2}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{}$ .

STEP 9

Comparing the general formula of amide with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the amide class.

STEP 10

Next, let's consider the functional group of ether.thers have the general formula $\mathrm{R-O-R'}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+2} \mathrm{}$ .

STEP 11

Comparing the general formula of ether with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the ether class.

STEP 12

Finally, let's consider the functional group of ester. Esters have the general formula $\mathrm{R-COO-R'}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n} \mathrm{}_{2}$ .

STEP 13

Comparing the general formula of ester with the given molecular formula, we can see that they match. Therefore, the compound could belong to the ester class.
The compound with the molecular formula $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{}_{2}$ could belong to the carboxylic acid class and the ester class.

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QuestionIdentify the compound classes for the formula $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}_{2}$ . Check all that apply: carboxylic acid, alcohol, phenol, amide, ether, ester.

Studdy Solution

STEP 1

Assumptions1. The molecular formula of the compound is $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{}_{}$ . The possible functional groups are carboxylic acid, alcohol, phenol, amide, ether, ester

STEP 2

First, let's consider the functional group of carboxylic acid. Carboxylic acids have the general formula $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{COOH}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{}_{2}$ .

STEP 3

Comparing the general formula of carboxylic acid with the given molecular formula, we can see that they match. Therefore, the compound could belong to the carboxylic acid class.

STEP 4

Next, let's consider the functional group of alcohol. Alcohols have the general formula $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{OH}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+2} \mathrm{}$ .

STEP 5

Comparing the general formula of alcohol with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the alcohol class.

STEP 6

Next, let's consider the functional group of phenol. Phenols have the general formula $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{}$ .

STEP 7

Comparing the general formula of phenol with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the phenol class.

STEP 8

Next, let's consider the functional group of amide. Amides have the general formula $\mathrm{R-C-NH}_{2}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{}$ .

STEP 9

Comparing the general formula of amide with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the amide class.

STEP 10

Next, let's consider the functional group of ether.thers have the general formula $\mathrm{R-O-R'}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+2} \mathrm{}$ .

STEP 11

Comparing the general formula of ether with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the ether class.

STEP 12

Finally, let's consider the functional group of ester. Esters have the general formula $\mathrm{R-COO-R'}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n} \mathrm{}_{2}$ .

STEP 13

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QuestionIdentify the compound classes for the formula C5H12O2\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}_{2}C5​H12​O2​. Check all that apply: carboxylic acid, alcohol, phenol, amide, ether, ester.

Studdy Solution

STEP 1

Assumptions1. The molecular formula of the compound is C5H12\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{}_{}C5​H12​​ . The possible functional groups are carboxylic acid, alcohol, phenol, amide, ether, ester

STEP 2

STEP 3

Comparing the general formula of carboxylic acid with the given molecular formula, we can see that they match. Therefore, the compound could belong to the carboxylic acid class.

STEP 4

Next, let's consider the functional group of alcohol. Alcohols have the general formula CnH2n+1OH\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{OH}Cn​H2n+1​OH. If we convert this to a more general form, it becomes CnH2n+2\mathrm{C}_{n} \mathrm{H}_{2n+2} \mathrm{}Cn​H2n+2​.

STEP 5

Comparing the general formula of alcohol with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the alcohol class.

STEP 6

Next, let's consider the functional group of phenol. Phenols have the general formula C6H5OH\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}C6​H5​OH. If we convert this to a more general form, it becomes C6H6\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{}C6​H6​.

STEP 7

Comparing the general formula of phenol with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the phenol class.

STEP 8

Next, let's consider the functional group of amide. Amides have the general formula R−C−NH2\mathrm{R-C-NH}_{2}R−C−NH2​. If we convert this to a more general form, it becomes CnH2n+1\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{}Cn​H2n+1​.

STEP 9

Comparing the general formula of amide with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the amide class.

STEP 10

Next, let's consider the functional group of ether.thers have the general formula R−O−R′\mathrm{R-O-R'}R−O−R′. If we convert this to a more general form, it becomes CnH2n+2\mathrm{C}_{n} \mathrm{H}_{2n+2} \mathrm{}Cn​H2n+2​.

STEP 11

Comparing the general formula of ether with the given molecular formula, we can see that they do not match. Therefore, the compound could not belong to the ether class.

STEP 12

Finally, let's consider the functional group of ester. Esters have the general formula R−COO−R′\mathrm{R-COO-R'}R−COO−R′. If we convert this to a more general form, it becomes CnH2n2\mathrm{C}_{n} \mathrm{H}_{2n} \mathrm{}_{2}Cn​H2n​2​.

STEP 13

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QuestionIdentify the compound classes for the formula $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}_{2}$ . Check all that apply: carboxylic acid, alcohol, phenol, amide, ether, ester.

Assumptions1. The molecular formula of the compound is $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{}_{}$ . The possible functional groups are carboxylic acid, alcohol, phenol, amide, ether, ester

Next, let's consider the functional group of alcohol. Alcohols have the general formula $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{OH}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+2} \mathrm{}$ .

Next, let's consider the functional group of phenol. Phenols have the general formula $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{}$ .

Next, let's consider the functional group of amide. Amides have the general formula $\mathrm{R-C-NH}_{2}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{}$ .

Next, let's consider the functional group of ether.thers have the general formula $\mathrm{R-O-R'}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n+2} \mathrm{}$ .

Finally, let's consider the functional group of ester. Esters have the general formula $\mathrm{R-COO-R'}$ . If we convert this to a more general form, it becomes $\mathrm{C}_{n} \mathrm{H}_{2n} \mathrm{}_{2}$ .