Organic Chemistry Quiz: Spectroscopy Ir And 1h Nmr Recognition

What this quiz covers

This quiz focuses on Spectroscopy Ir And 1h Nmr Recognition, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 1-Butanol and 2-butanol
2. 2-Pentanone and 3-pentanone
3. Diethyl ether and methyl propyl ether
4. Cyclohexanone and hex-5-en-1-ol (correct answer)

Explanation: To be reliably distinguished by IR, the two isomers should have different functional groups that give rise to distinct, strong absorptions in the diagnostic region. Cyclohexanone has a strong C=O stretch (~1715 cm⁻¹) but no O-H or C=C stretch. Hex-5-en-1-ol has a strong, broad O-H stretch (~3300 cm⁻¹) and a medium C=C stretch (~1650 cm⁻¹) but no C=O stretch. These differences are unambiguous. The other pairs consist of isomers with the same primary functional group (alcohol, ketone, ether), whose IR spectra would be very similar, with differences mainly in the complex fingerprint region.

Question 2

1. Disappearance of a peak around 1650 cm⁻¹ and appearance of a strong peak around 1725 cm⁻¹. (correct answer)
2. Appearance of a strong peak at 1710 cm⁻¹ and a very broad peak from 2500–3300 cm⁻¹.
3. Disappearance of a peak around 3050 cm⁻¹ and appearance of a broad peak around 3300 cm⁻¹.
4. Appearance of a sharp peak around 2250 cm⁻¹ and disappearance of peaks around 2950 cm⁻¹.

Explanation: The reaction is ozonolysis with a reductive workup (DMS), which cleaves the double bond of cyclohexene to form hexane-1,6-dial (adipaldehyde). Spectroscopically, this corresponds to the loss of the alkene C=C bond (disappearance of the IR peak at ~1650 cm⁻¹) and the formation of two aldehyde C=O bonds (appearance of a strong IR peak at ~1725 cm⁻¹). Choice B describes the product of an oxidative workup (a dicarboxylic acid). Choice C incorrectly identifies the starting material peak and suggests alcohol formation. Choice D suggests nitrile formation, which is incorrect.

Question 3

1. Acetone
2. Cyclohexanone
3. Cyclopentanone
4. Cyclobutanone (correct answer)

Explanation: The frequency of the C=O stretch in a cyclic ketone is influenced by ring strain. As ring size decreases from six carbons, the bond angles are compressed, leading to increased s-character in the C=O bond and a higher stretching frequency. A typical acyclic ketone like acetone or a strain-free cyclic ketone like cyclohexanone absorbs around 1715 cm⁻¹. Cyclopentanone (~1750 cm⁻¹) is more strained and absorbs at a higher frequency. Cyclobutanone (~1780 cm⁻¹) is even more strained and absorbs at an even higher frequency.

Question 4

1. cis-1,3-Dimethylcyclohexane and trans-1,3-dimethylcyclohexane
2. cis-2-Butene and trans-2-butene
3. (R)-2-Butanol and (S)-2-butanol (correct answer)
4. 2-Pentanone and 3-pentanone

Explanation: This question tests your understanding of what structural features IR and ¹H NMR can detect. Both techniques are sensitive to differences in connectivity, functional groups, and molecular environments, but they have important limitations when it comes to stereochemistry. IR spectroscopy detects functional groups through characteristic bond vibrations, while ¹H NMR reveals hydrogen environments, coupling patterns, and integration ratios. Crucially, neither technique can distinguish between enantiomers (non-superimposable mirror images) when measured in achiral environments, since enantiomers have identical physical properties except for how they interact with plane-polarized light. The correct answer is C because (R)-2-butanol and (S)-2-butanol are enantiomers. They have identical IR spectra (same O-H, C-H, and C-O stretches) and identical ¹H NMR spectra (same chemical shifts, coupling patterns, and integrations). Only techniques involving chiral environments or optical activity can distinguish them. Option A is wrong because cis and trans-1,3-dimethylcyclohexane have different conformational preferences, leading to distinct ¹H NMR coupling patterns and potentially different IR fingerprint regions. Option B is incorrect because cis and trans-2-butene show different ¹H NMR patterns—the cis isomer has equivalent methyl groups while the trans isomer places them in different environments. Option D is wrong because 2-pentanone and 3-pentanone are constitutional isomers with different connectivity, producing clearly different ¹H NMR spectra with distinct integration patterns and chemical shifts. Remember: enantiomers are "spectroscopically invisible" to achiral techniques—only their interactions with chiral environments reveal their differences.

Question 5

1. 1-Propanol
2. 2-Propanol (correct answer)
3. Methyl ethyl ether
4. Propanal

Explanation: The disappearance of a signal upon addition of D₂O indicates the presence of a labile proton, such as that in an -OH or -NH group. This rules out methyl ethyl ether. The formula C₃H₈O corresponds to an alcohol or ether. The two possible alcohols are 1-propanol and 2-propanol. 1-Propanol (CH₃CH₂CH₂OH) would have four unique proton signals (for the OH, and the three different CH₂/CH₃ groups). 2-Propanol ((CH₃)₂CHOH) has three unique proton signals (for the OH, the CH group, and the two equivalent CH₃ groups). Since the spectrum shows three signals, the compound must be 2-propanol.