Question

How many amide isomers are possible for $C_4H_9ON$?

• A. 4
• B. 5
• C. 6
• D. 8

Answer 1

Answer: (D)

8

Answer 2

The molecular formula given is $C_4H_9ON$. We are looking for the number of amide isomers with this formula. An amide functional group is $-CO-N<$. The formula $C_4H_9ON$ has a degree of unsaturation of 1, which is accounted for by the C=O double bond in the amide group.

The general structure of an amide is $R_1 - CO - N(R_2)R_3$, where $R_1$, $R_2$, and $R_3$ can be hydrogen atoms or alkyl groups. The carbon in the carbonyl group is already accounted for. We have 3 additional carbon atoms ($C_3$) and the remaining hydrogen atoms to distribute among $R_1$, $R_2$, and $R_3$. The total number of carbon atoms in $R_1$, $R_2$, and $R_3$ must sum up to 3.

We can classify amides based on the number of alkyl groups attached to the nitrogen atom:

1. Primary amides: $R_2 = H$, $R_3 = H$. The structure is $R_1 - CO - NH_2$. The group $R_1$ must contain 3 carbon atoms. The possible alkyl groups with 3 carbons ($C_3H_7$) are n-propyl ($CH_3CH_2CH_2-$) and isopropyl ($(CH_3)_2CH-$).

   • $CH_3CH_2CH_2-CO-NH_2$ (Butanamide)
   • $(CH_3)_2CH-CO-NH_2$ (2-Methylpropanamide)

   There are 2 primary amide isomers.

2. Secondary amides: One alkyl group ($R_2$) and one hydrogen ($R_3=H$) attached to nitrogen. The structure is $R_1 - CO - NH - R_2$, where $R_2$ is an alkyl group. The total number of carbon atoms in $R_1$ and $R_2$ must be 3, and $R_2$ must have at least one carbon.

   • Case 2a: $R_1 = H$. $R_2$ must contain 3 carbons ($C_3H_7$). Possible $R_2$ groups are n-propyl and isopropyl.

     • $H - CO - NH - CH_2CH_2CH_3$ (N-Propylformamide)
     • $H - CO - NH - CH(CH_3)_2$ (N-Isopropylformamide)

     (2 isomers)

   • Case 2b: $R_1$ contains 1 carbon ($CH_3-$). $R_2$ must contain 2 carbons ($C_2H_5-$).

     • $CH_3 - CO - NH - CH_2CH_3$ (N-Ethylacetamide)

     (1 isomer)

   • Case 2c: $R_1$ contains 2 carbons ($C_2H_5-$). $R_2$ must contain 1 carbon ($CH_3-$).

     • $CH_3CH_2 - CO - NH - CH_3$ (N-Methylpropanamide)

     (1 isomer)

   • Case 2d: $R_1$ contains 3 carbons ($C_3H_7-$). $R_2$ must contain 0 carbons, which is not an alkyl group. This case is not possible for secondary amides.

   There are $2 + 1 + 1 = 4$ secondary amide isomers.

3. Tertiary amides: Two alkyl groups ($R_2$ and $R_3$) attached to nitrogen. The structure is $R_1 - CO - N(R_2)R_3$, where $R_2$ and $R_3$ are alkyl groups. The total number of carbon atoms in $R_1$, $R_2$, and $R_3$ must be 3, and $R_2, R_3$ must each have at least one carbon.

   • Case 3a: $R_1 = H$. $R_2$ and $R_3$ must distribute the 3 carbons, with at least 1 carbon each. The only possible distribution is 1 carbon and 2 carbons.

     • $R_2 = CH_3-$ and $R_3 = C_2H_5-$ (or vice versa).
     • $H - CO - N(CH_3)(C_2H_5)$ (N-Ethyl-N-methylformamide)

     (1 isomer)

   • Case 3b: $R_1$ contains 1 carbon ($CH_3-$). $R_2$ and $R_3$ must distribute the remaining 2 carbons, with at least 1 carbon each. The only possible distribution is 1 carbon and 1 carbon.

     • $R_2 = CH_3-$ and $R_3 = CH_3-$.
     • $CH_3 - CO - N(CH_3)_2$ (N,N-Dimethylacetamide)

     (1 isomer)

   • Case 3c: $R_1$ contains 2 carbons ($C_2H_5-$). $R_2$ and $R_3$ must distribute the remaining 1 carbon, with at least 1 carbon each. Not possible.

   • Case 3d: $R_1$ contains 3 carbons ($C_3H_7-$). $R_2$ and $R_3$ must distribute the remaining 0 carbons, with at least 1 carbon each. Not possible.

   There are $1 + 1 = 2$ tertiary amide isomers.

Total number of amide isomers = Number of primary amides + Number of secondary amides + Number of tertiary amides Total isomers = 2 + 4 + 2 = 8.

The 8 isomers are:

1. Butanamide ($CH_3CH_2CH_2CONH_2$)
2. 2-Methylpropanamide ($(CH_3)_2CHCONH_2$)
3. N-Ethylacetamide ($CH_3CONHCH_2CH_3$)
4. N-Methylpropanamide ($CH_3CH_2CONHCH_3$)
5. N-Propylformamide ($HCONHCH_2CH_2CH_3$)
6. N-Isopropylformamide ($HCONHCH(CH_3)_2$)
7. N-Ethyl-N-methylformamide ($HCON(CH_3)(C_2H_5)$)
8. N,N-Dimethylacetamide ($CH_3CON(CH_3)_2$)