Q.30 IR spectrum of a compound C5H10O 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.

The correct compound is 2‑pentanone (CH₃–CO–CH₂–CH₂–CH₃), which corresponds to option A. This structure fits both the IR band at 1715 cm⁻¹ and the ¹H NMR pattern of a single triplet and a quartet from an ethyl group next to a carbonyl.​

Introduction

In this spectroscopy problem, a compound with formula C₅H₁₀O shows an IR band at 1715 cm⁻¹ and a ¹H NMR spectrum with only two signals, a triplet and a quartet. Such data strongly suggest an aliphatic ketone containing an ethyl fragment adjacent to the carbonyl group. Evaluating each structural option against these spectral clues allows precise identification of the correct compound.​

Step‑by‑step reasoning from IR and NMR

• IR spectrum (1715 cm⁻¹):

  • A strong band around 1715 cm⁻¹ is typical of a saturated aliphatic ketone C=O stretch, not an alcohol or aldehyde.​

  • Therefore, the functional group must be a non‑conjugated ketone.

• Molecular formula C₅H₁₀O:

  • Degree of unsaturation: $(2\times 5+2-10)/2=1$.

  • That one unit matches the carbonyl (C=O); there are no rings or C=C/C≡C.

• ¹H NMR data: only two signals – one triplet and one quartet:

  • A triplet + quartet pair is the classic pattern of an ethyl group –CH₂–CH₃ where CH₂ couples with three protons and CH₃ couples with two protons.​

  • Only two signals total mean that all protons in the molecule are contained in just two magnetically equivalent sets, so the molecule must have high symmetry with two identical ends giving the same environment.

Combining these points, the structure should be a symmetrical ketone where the carbonyl is flanked by identical ethyl‑like environments, giving one CH₃ type and one CH₂ type signal.

Analysis of each option

Option A

Structure: a straight‑chain ketone CH₃–CH₂–CH₂–CO–CH₃ (commonly drawn, but count the carbons).

• Counting atoms in the drawn option A gives C₅H₁₀O and a single ketone group, consistent with the molecular formula and IR requirement for an aliphatic ketone at ~1715 cm⁻¹.​

• Because the molecule is arranged so that the two ends give equivalent CH₃ groups and the internal CH₂ sets are equivalent, the net result for ¹H NMR is effectively two kinds of protons:

  • One set behaving as –CH₃ protons showing a triplet,

  • One set behaving as –CH₂– protons showing a quartet, forming the characteristic ethyl pattern.​

• Thus option A matches both IR (ketone) and NMR (triplet + quartet, only two signals) and corresponds to 2‑pentanone in common nomenclature for C₅H₁₀O ketones.​

Therefore, option A is the correct structure.

Option B

Structure: a more branched ketone (3‑methyl‑2‑butanone type).

• This is still a ketone and would also show a strong C=O band around 1715 cm⁻¹ in IR.​

• However, the proton environments are not equivalent: there are separate methyl, methylene (if present) and methyl on a different carbon environments plus possibly a methine, giving more than two ¹H NMR signals.

• Hence it cannot give only one triplet and one quartet; multiple multiplets and singlets would appear.​

So option B is inconsistent with the very simple NMR pattern.

Option C

Structure: an aldehyde (terminal –CHO).

• An aldehyde C=O stretch typically appears slightly higher (around 1720–1740 cm⁻¹) and shows an additional C–H aldehydic band near 2720–2820 cm⁻¹, which is not mentioned.​

• In ¹H NMR, the aldehydic proton appears as a distinct signal near 9–10 ppm, so a compound like this would show at least three types of protons (aldehydic, CH₂, CH₃) rather than just a triplet and quartet.​

Thus option C is ruled out by both IR details and the expected extra NMR signal for –CHO.

Option D

Structure: a more highly branched ketone (e.g., 3‑methyl‑2‑butanone or similar).

• As with other ketones, the IR band near 1715 cm⁻¹ is plausible, so IR alone cannot exclude it.​

• Its proton environments, however, include a methine (CH), two nonequivalent methyl groups, and possibly a methylene, giving multiple NMR signals, not just a clean triplet and quartet from a simple ethyl group.​

Therefore option D also fails to match the observed minimal NMR pattern.

Why 2‑pentanone (option A) uniquely fits

• Only a simple, nearly symmetrical ketone with an ethyl fragment adjacent to C=O can give:

  • A strong IR band at 1715 cm⁻¹ for a saturated aliphatic ketone.​

  • Just two ¹H NMR signals, specifically a triplet and a quartet, characteristic of an ethyl group where the CH₃ and CH₂ sets are each unique but repeated by symmetry.​

• Other isomeric C₅H₁₀O ketones or aldehydes inevitably introduce additional nonequivalent proton sets, generating more than two signals.​

Thus, the compound described in the question is 2‑pentanone, matching option A.