Question

The compound(s) which will have only two signals in the \( ^1 \text{H} \) NMR spectrum in 3:2 ratio is(are) 

 

• A. (A)
• B. (B)
• C. (C)
• D. (D)

Hint:

In \( ^1 \text{H} \) NMR, the signal intensity ratio corresponds to the number of protons in each distinct environment.

Answer 1

The Correct Option is C, D

Step 1: Analysis of NMR Requirements

For a compound to have only two signals in its ${}^1\text{H NMR}$ spectrum, it must have only two distinct sets of chemically equivalent protons.

For the ratio of the signals to be $3:2$, the number of protons in the two sets must be in the ratio $3x:2x$.

Since all options contain $\text{CH}_3\text{CO}$ groups (acetyl groups), the total number of protons is $6 + 4 = 10$.

The ratio of $3:2$ corresponds to $\frac{6 \text{ H}}{4 \text{ H}}$ or $\frac{3 \text{ H}}{2 \text{ H}}$.

The total number of $\text{H}$ atoms in the $\text{CH}_3\text{CO}$ groups is $3 \times 2 = 6$ protons.

This means the acetyl protons must form one signal (integrating to 6 $\text{H}$, or 3 parts), and the aromatic/alkane protons must form the second signal (integrating to 4 $\text{H}$, or 2 parts).

Total protons: $6 (\text{acetyl}) + 4 (\text{other}) = 10 \text{ H}$. Ratio is $6:4$, which simplifies to $3:2$.

The required compound must have:

Symmetry such that all six acetyl $\text{CH}_3$ protons are equivalent, giving $\mathbf{1}$ signal (integration 6 $\text{H}$).

Symmetry such that all four aromatic protons are equivalent, giving a second $\mathbf{1}$ signal (integration 4 $\text{H}$).

Step 2: Analysis of Options

(A) $\mathbf{o}\text{-diacetylbenzene}$

• Structure: $1,2\text{-diacetylbenzene}$
• Acetyl Protons: The two $\text{CH}_3\text{CO}$ groups are equivalent by symmetry. This gives 1 signal (integration 6 $\text{H}$).
• Aromatic Protons: The four aromatic protons are not all equivalent. Protons $3$ and $6$ are equivalent, and protons $4$ and $5$ are equivalent. This gives 2 signals.
• Total Signals: $1 (\text{acetyl}) + 2 (\text{aromatic}) = \mathbf{3}$ signals. Does not match.

(B) $\mathbf{m}\text{-diacetylbenzene}$

• Structure: $1,3\text{-diacetylbenzene}$
• Acetyl Protons: The two $\text{CH}_3\text{CO}$ groups are equivalent by symmetry. This gives 1 signal (integration 6 $\text{H}$).
• Aromatic Protons: The four aromatic protons are all non-equivalent ($\text{H}2, \text{H}4, \text{H}5, \text{H}6$). This gives 4 signals.
• Total Signals: $1 (\text{acetyl}) + 4 (\text{aromatic}) = \mathbf{5}$ signals. Does not match.

(C) $\mathbf{p}\text{-diacetylbenzene}$

• Structure: $1,4\text{-diacetylbenzene}$ * Acetyl Protons: The plane of symmetry passes through the $\text{C}2-\text{C}5$ bond and the $\text{C}3-\text{C}6$ bond. The two $\text{CH}_3\text{CO}$ groups are equivalent. This gives 1 signal (integration 6 $\text{H}$).
• Aromatic Protons: The four aromatic protons ($\text{H}2, \text{H}3, \text{H}5, \text{H}6$) are all equivalent by symmetry. This gives 1 signal (integration 4 $\text{H}$).
• Total Signals: $1 (\text{acetyl}) + 1 (\text{aromatic}) = \mathbf{2}$ signals.
• Integration Ratio: The ratio of the two signals is $6:4$, which simplifies to $\mathbf{3:2}$. Matches.

(D) Hexane-2,5-dione

• Structure: $\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CO}-\text{CH}_3$
• Acetyl Protons: The two terminal $\text{CH}_3\text{CO}$ groups are equivalent by symmetry. This gives 1 signal (integration 6 $\text{H}$).
• Methylene Protons: The two internal $\text{CH}_2$ groups are equivalent by symmetry. This gives 1 signal (integration 4 $\text{H}$).
• Total Signals: $1 (\text{acetyl}) + 1 (\text{methylene}) = \mathbf{2}$ signals.
• Integration Ratio: The ratio of the two signals is $6:4$, which simplifies to $\mathbf{3:2}$. Matches.

Step 3: Conclusion

Both (C) $\mathbf{p}\text{-diacetylbenzene}$ and (D) $\mathbf{\text{hexane-2,5-dione}}$ satisfy the criteria of having only two signals in a $3:2$ integration ratio.

Therefore, the correct options are (C) and (D).