Pure cinnamic acid has a literature melting point of 132.5 -133.0 °C. Which of the following melting point data would not be an acceptable observed melting point for pure cinnamic acid in the lab?

Here are the solutions to the organic chemistry questions, focusing on the ones that had issues with the provided options:

12. How many primary carbon atoms are contained in the hydrocarbon chain provided below?
The structure is:

      CH3
      |
CH3-C-CH2-CH-CH2-CH3
      |         |
      CH3       CH3

Let's identify the primary carbon atoms (carbons bonded to only one other carbon atom):

1. The $\text{CH}_3$ group at the leftmost end of the main chain.
2. The $\text{CH}_3$ group attached above the quaternary carbon.
3. The $\text{CH}_3$ group attached below the quaternary carbon.
4. The $\text{CH}_3$ group attached below the tertiary carbon.
5. The $\text{CH}_3$ group at the rightmost end of the main chain.

Counting these, there are 5 primary carbon atoms in the given hydrocarbon chain.
The provided options are A. n=1, B. n=3, C. n=6, D. n=2. Since 5 is not among the given options, there appears to be an error in the question's choices.

14. Which of the following is not an isomeric form of the alkene $\text{C}_5\text{H}_{10}$?
An alkene with the molecular formula $\text{C}_5\text{H}_{10}$ has one double bond and a total of 5 carbon atoms. Let's determine the molecular formula for each given option:

• A. 3-methyl-2-butene: This compound has a 4-carbon chain (butene) and one methyl group, totaling 5 carbons. It has one double bond. Its molecular formula is $\text{C}_5\text{H}_{10}$.
• B. 3-methyl-1-butene: This compound has a 4-carbon chain (butene) and one methyl group, totaling 5 carbons. It has one double bond. Its molecular formula is $\text{C}_5\text{H}_{10}$.
• C. 2-methyl-2-butene: This compound has a 4-carbon chain (butene) and one methyl group, totaling 5 carbons. It has one double bond. Its molecular formula is $\text{C}_5\text{H}_{10}$.
• D. 2-methyl-1-butene: This compound has a 4-carbon chain (butene) and one methyl group, totaling 5 carbons. It has one double bond. Its molecular formula is $\text{C}_5\text{H}_{10}$.

All the listed options (3-methyl-2-butene, 3-methyl-1-butene, 2-methyl-2-butene, and 2-methyl-1-butene) are valid isomers of $\text{C}_5\text{H}_{10}$. Therefore, none of the options can be selected as "not an isomeric form" of $\text{C}_5\text{H}_{10}$. This question also appears to be flawed.

All other questions in the image have been addressed in previous messages.