Select which option is the correct IUPAC name for the compound shown below?

The longest chain has 6 carbons (hexane).

Numbering from left to right:
$\text{C}_1\text{H}_3-\text{C}_2\text{H}(\text{CH}_3)-\text{C}_3\text{H}(\text{CH}_2\text{CH}_3)-\text{C}_4\text{H}(\text{CH}_3)-\text{C}_5\text{H}_2-\text{C}_6\text{H}_3$
Substituents are at positions 2 (methyl), 3 (ethyl), and 4 (methyl). The sum of locants is $2+3+4=9$.

Numbering from right to left:
$\text{C}_6\text{H}_3-\text{C}_5\text{H}_2-\text{C}_4\text{H}(\text{CH}_3)-\text{C}_3\text{H}(\text{CH}_2\text{CH}_3)-\text{C}_2\text{H}(\text{CH}_3)-\text{C}_1\text{H}_3$
Substituents are at positions 2 (methyl), 3 (ethyl), and 4 (methyl). The sum of locants is $2+3+4=9$.
Both numbering schemes give the same set of locants.

Ethyl comes before methyl. There are two methyl groups, so "dimethyl".
The name is 3-ethyl-2,4-dimethylhexane.

The correct option is A.

12. How many primary carbon atoms are contained in the hydrocarbon chain provided below?
A primary carbon atom (1°) is bonded to only one other carbon atom. Let's identify each carbon type in the given structure:

      CH3 (1°)
      |
CH3(1°)-C(4°)-CH2(2°)-CH(3°)-CH2(2°)-CH3(1°)
      |         |
      CH3(1°)   CH3(1°)

1. The leftmost $\text{CH}_3$ group: Primary (bonded to one carbon).
2. The top $\text{CH}_3$ group (attached to the quaternary carbon): Primary (bonded to one carbon).
3. The bottom-left $\text{CH}_3$ group (attached to the quaternary carbon): Primary (bonded to one carbon).
4. The bottom-right $\text{CH}_3$ group (attached to the tertiary carbon): Primary (bonded to one carbon).
5. The rightmost $\text{CH}_3$ group: Primary (bonded to one carbon).

Counting these, there are 5 primary carbon atoms.
Since 5 is not among the given options (A. n=1, B. n=3, C. n=6, D. n=2), there might be an error in the question's options. However, if forced to choose the closest option, 6 is the closest. Given the options, and the clear structure, the most accurate count is 5. If there is a single correct answer among the options, the question or options may be flawed. Assuming there might be a slight misinterpretation or a common error leading to one of the options, let's re-evaluate if any common mistake could lead to 6. If the leftmost carbon was part of a longer chain and the first methyl was not a terminal methyl, it could change. But as drawn, it's 5.

Let's assume there's a typo in the question and it meant to ask for the total number of methyl groups, which is 5. Or if one of the secondary carbons was somehow misidentified.
Given the options, and the possibility of a slight error in the question, I will state the correct count.

There are 5 primary carbon atoms. Since this is not an option, I cannot select one of the provided choices.

13. An Organic compound contains 20% Carbon, 6.66% Hydrogen, 47.33% Nitrogen and the rest was oxygen. Its molar mass is $60 \text{gmol}^{-1}$. The molecular formula of the compound is?

Percentage of Oxygen = $100\% - (20\% \text{ C} + 6.66\% \text{ H} + 47.33\% \text{ N})$
Percentage of Oxygen = $100\% - 73.99\% = 26.01\%$

Carbon (C): $20 \text{ g}$
Hydrogen (H): $6.66 \text{ g}$
Nitrogen (N): $47.33 \text{ g}$
Oxygen (O): $26.01 \text{ g}$

Moles of C = $\frac{20 \text{ g}}{12.01 \text{ g/mol}} \approx 1.665 \text{ mol}$
Moles of H = $\frac{6.66 \text{ g}}{1.008 \text{ g/mol}} \approx 6.607 \text{ mol}$
Moles of N = $\frac{47.33 \text{ g}}{14.01 \text{ g/mol}} \approx 3.378 \text{ mol}$
Moles of O = $\frac{26.01 \text{ g}}{16.00 \text{ g/mol}} \approx 1.626 \text{ mol}$

Smallest number of moles is approximately $1.626 \text{ mol}$ (for Oxygen).
C: $\frac{1.665}{1.626} \approx 1.02 \approx 1$
H: $\frac{6.607}{1.626} \approx 4.06 \approx 4$
N: $\frac{3.378}{1.626} \approx 2.07 \approx 2$
O: $\frac{1.626}{1.626} = 1$

The empirical formula is $\text{CH}_4\text{N}_2\text{O}$.

Empirical formula mass = $(1 \times 12.01) + (4 \times 1.008) + (2 \times 14.01) + (1 \times 16.00)$
Empirical formula mass = $12.01 + 4.032 + 28.02 + 16.00 = 60.062 \text{ g/mol}$

Molar mass = $60 \text{ g/mol}$
Empirical formula mass $\approx 60 \text{ g/mol}$
Since the molar mass is approximately equal to the empirical formula mass, the molecular formula is the same as the empirical formula.

The molecular formula is $\boxed{\text{CH}_4\text{N}_2\text{O}}$.

The correct option is B.

14. Which of the following is not an isomeric form of the alkene $\text{C}_5\text{H}_{10}$?
An alkene with the formula $\text{C}_5\text{H}_{10}$ has one double bond or is a cycloalkane. We need to check the molecular formula for each given option.

• A. 3-methyl-2-butene:
  • Butene (4 carbons) + methyl (1 carbon) = 5 carbons.
  • It's an alkene, so $\text{C}_5\text{H}_{10}$. (e.g., $\text{CH}_3-\text{CH}=\text{C}(\text{CH}_3)-\text{CH}_3$)
  • This is an isomer of $\text{C}_5\text{H}_{10}$.
• B. 3-methyl-1-butene:
  • Butene (4 carbons) + methyl (1 carbon) = 5 carbons.
  • It's an alkene, so $\text{C}_5\text{H}_{10}$. (e.g., $\text{CH}_2=\text{CH}-\text{CH}(\text{CH}_3)-\text{CH}_3$)
  • This is an isomer of $\text{C}_5\text{H}_{10}$.
• C. 2-methyl-2-butene:
  • Butene (4 carbons) + methyl (1 carbon) = 5 carbons.
  • It's an alkene, so $\text{C}_5\text{H}_{10}$. (e.g., $\text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3$)
  • This is an isomer of $\text{C}_5\text{H}_{10}$.
• D. 2-methyl-1-butene:
  • Butene (4 carbons) + methyl (1 carbon) = 5 carbons.
  • It's an alkene, so $\text{C}_5\text{H}_{10}$. (e.g., $\text{CH}_2=\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_3$)
  • This is an isomer of $\text{C}_5\text{H}_{10}$.

All the given options are isomers of $\text{C}_5\text{H}_{10}$. This suggests there might be an error in the question or options, as it asks "Which of the following is not an isomeric form". All listed compounds are indeed isomers of $\text{C}_5\text{H}_{10}$ (pentenes).

If I must choose one, and assuming there's a subtle error in the question's premise, I would re-examine if any of these names are invalid or refer to a different formula. All names are valid for $\text{C}_5\text{H}_{10}$ isomers.

Given that all options are valid isomers of $\text{C}_5\text{H}_{10}$, there is no correct answer among the choices for "not an isomeric form".