Question

IR spectrum of a compound \( C_5H_{10}O \) shows a band at 1715 cm\(^{-1}\). The same compound showed two signals, a triplet and a quartet, in its NMR spectrum. Identify the compound from the following. 

 

• A. (A)
• B. (B)
• C. (C)
• D. (D)

Hint:

In IR spectroscopy, a band at around 1715 cm\(^{-1}\) typically corresponds to a carbonyl group. In NMR, the triplet-quartet pattern is characteristic of an ethyl group (-CH\(_2\)-CH\(_3\)).

Answer 1

The Correct Option is B

1. Molecular Formula and Unsaturation

The molecular formula is $\text{C}_5\text{H}_{10}\text{O}$.

The Index of Hydrogen Deficiency (IHD) is calculated as:

$$\text{IHD} = \frac{2C + 2 + N - H - X}{2}$$

For $\text{C}_5\text{H}_{10}\text{O}$ (where $C=5, H=10, N=0, X=0$):

$$\text{IHD} = \frac{2(5) + 2 + 0 - 10 - 0}{2} = \frac{12 - 10}{2} = 1$$

This indicates one ring or one double bond. Since all options are linear ketones, the IHD corresponds to the $\text{C}=\text{O}$ double bond (a ketone).

2. Infrared (IR) Spectrum

The IR spectrum shows a band at $\mathbf{1715\ \text{cm}^{-1}}$.

This strong absorption is characteristic of a ketone carbonyl stretching vibration ($\text{C}=\text{O}$), which typically appears between 1705 and $1725\ \text{cm}^{-1}$. This confirms that the compound is one of the given ketones.

3. Nuclear Magnetic Resonance (NMR) Spectrum

The ${}^1\text{H NMR}$ spectrum shows two signals: a triplet and a quartet.

The multiplicity of a signal is determined by the $n+1$ rule, where $n$ is the number of equivalent protons on adjacent carbon atoms. The ratio of the integrals (not given, but implied by the structure) of the two signals will be $4:6$ or $6:4$.

Triplet ($n+1=3 \implies n=2$ adjacent $\text{H}$'s): This signal comes from $\text{CH}_3$ groups coupled to a $\text{CH}_2$ group.

Quartet ($n+1=4 \implies n=3$ adjacent $\text{H}$'s): This signal comes from a $\text{CH}_2$ group coupled to a $\text{CH}_3$ group.

Now, let's analyze the symmetry and signals of the given options:

Compound	Structure	Number of Unique 1H Signals	Predicted Multiplicity
(A) Pentan-2-one	$\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CH}_3$	4	$\text{CH}_3$ (s), $\text{CH}_2$ (t), $\text{CH}_2$ (sextet), $\text{CH}_3$ (t)
(B) Pentan-3-one	$\text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_2-\text{CH}_3$	2	$\mathbf{\text{CH}_3}$ (t) and $\mathbf{\text{CH}_2}$ (q)
(C) Hexan-2-one	$\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$	5	$\text{CH}_3$ (s), $\text{CH}_2$ (t), $\text{CH}_2$ (quintet), $\text{CH}_2$ (sextet), $\text{CH}_3$ (t)
(D) 3-Methylbutan-2-one	$\text{CH}_3-\text{CO}-\text{CH}(\text{CH}_3)_2$	3	$\text{CH}_3$ (s), $\text{CH}_3$ (d), $\text{CH}$ (septet)

Compound (B), $\text{Pentan-3-one}$, is a symmetrical ketone:

 

$$\text{CH}_3^a-\text{CH}_2^b-\text{CO}-\text{CH}_2^b-\text{CH}_3^a$$

$\text{CH}_3^a$ protons (6H): These are equivalent and adjacent to the $\text{CH}_2^b$ group (2 $\text{H}$'s). This gives a triplet.

$\text{CH}_2^b$ protons (4H): These are equivalent and adjacent to the $\text{CH}_3^a$ group (3 $\text{H}$'s). This gives a quartet.

Compound (B) is the only option that is consistent with the observation of exactly two signals, one being a triplet and the other a quartet, in the ${}^1\text{H NMR}$ spectrum.