Question

Double bond equivalent in A is \(x\) and in \(\mathrm{B}\) is \(y\), then \((x-y)\) is \(\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{~N}-\mathrm{Ph}\) 'A' 'B' (A) 9 (B) 8 (C) 7 (D) 6

Short answer

The difference between the double bond equivalences of A and B is 1.5, which is not present in the given options. There might be an error in the given exercise information or answer options.

Step-by-step solution

Determine the molecular formula for A and B

First, we need to find the molecular formula for A and B. A = (\(\mathrm{CH_3CO}_2\mathrm{N}\) - Ph) = \(\mathrm{C_4H_6N_2O_2} - \mathrm{C_6H_5}\) B = \(\mathrm{Ph}\) = \(\mathrm{C_6H_5}\) Now we have the molecular formula for A and B, we can compute their DBE values.

Calculate the DBE for A and B

The DBE formula is defined as DBE = 1 + \(\frac{1}{2} (n_\mathrm{C} - n_\mathrm{H} + 2n_\mathrm{N})\), where \(n_\mathrm{C}\), \(n_\mathrm{H}\), and \(n_\mathrm{N}\) are the numbers of carbon, hydrogen, and nitrogen atoms, respectively. For A: DBE = 1 + \(\frac{1}{2} (n_\mathrm{C} - n_\mathrm{H} + 2n_\mathrm{N})\) = 1 + \(\frac{1}{2} (4 - 6 + 2 \times 2)\) = 1 + \(\frac{1}{2} (4)\) = 1 + 2 = 3 For B: DBE = 1 + \(\frac{1}{2} (n_\mathrm{C} - n_\mathrm{H} + 2n_\mathrm{N})\) = 1 + \(\frac{1}{2} (6 - 5 + 2 \times 0)\) = 1 + \(\frac{1}{2} (1)\) = 1 + 0.5 = 1.5 Now that we have the values of x (DBE for A) and y (DBE for B), we can compute the difference \((x - y)\).

Compute the difference (x-y)

We have found that x = 3 and y = 1.5, so we can compute the difference: (x - y) = 3 - 1.5 = 1.5 The difference between the double bond equivalences of A and B is 1.5, which is not present in the given options. There might be an error in the given exercise information or answer options. Please cross-check the instructions or consult your teacher.

Key concepts

Molecular Formula

Understanding the molecular formula is a crucial aspect of chemistry, especially when studying organic compounds. The molecular formula provides vital information about the number and types of atoms present in a molecule. For instance, in a compound like glucose, which has the molecular formula \( C_6H_{12}O_6 \), it indicates that each molecule contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.

The formula thus reflects the exact number of atoms of each element in a molecule, and it's the foundational step in determining the structure and properties of the compound. When approaching an organic chemistry problem, always begin by identifying and understanding the molecular formula of the compounds involved, as done in the step-by-step solution for finding the DBE of compound A and B in the given exercise.

DBE Calculation

The Double Bond Equivalent (DBE) calculation is essential for predicting the unsaturation level within an organic molecule—meaning the number of double bonds, triple bonds, and rings that a compound may contain. The formula for calculating DBE is \( DBE = 1 + \frac{1}{2} (n_C - n_H + 2n_N) \) where \( n_C \) is the number of carbon atoms, \( n_H \) is the number of hydrogen atoms, and \( n_N \) is the number of nitrogen atoms in the formula.

Remember, other elements like halogens and oxygen do not directly affect the DBE calculation. Halogens are treated as hydrogen atoms, while oxygen is omitted as it doesn't alter the DBE. Applying this formula, you can assess the potential complexity of the molecular structure and better predict its physical and chemical properties. DBE calculation is a fundamental skill to master for efficient organic chemistry problem solving.

Organic Chemistry Problem Solving

Problem solving in organic chemistry often involves several key steps: understanding the molecular formulas involved, knowing the relevant chemical rules and principles, and applying mathematical calculations correctly. Issues such as finding the DBE of molecules require students to combine these elements effectively.

When faced with an apparent discrepancy in the results, like in the solved exercise where the calculated DBE does not match the given options, it is important to reassess each step critically. Check the molecular formulas for any potential errors, reapply the DBE formula, and make sure all mathematical operations are done precisely. If the problem persists, it could indicate an error in the given data or exercise itself, as suggested at the end of the sample solution. This standpoint assists in reinforcing the understanding of organic chemistry concepts and developing a systematic approach to problem solving.