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Identifying Aromatic Compounds in a Mixture Using 1H NMR Spectroscopy
Question
"The 1H NMR spectrum of a solution containing a 1:1 mixture of two aromatic compounds with the molecular formula C10H14 exhibits two signals (two singlets) at around 6.7 ppm in a 1:1 ratio. There are also three signals at around 2.2 ppm (all singlets) in a 2:1:1 ratio. Which two compounds are present in the mixture from the following options: 1,2,4,5-tetramethylbenzene, 1,2,3,5-tetramethylbenzene, 1,2,3,4-tetramethylbenzene, and 1,4-diethylbenzene.
Explain which one of the following is correct, out of A, B, C, and D:
A) 1,2,4,5-tetramethylbenzene and 1,2,3,4-tetramethylbenzene
B) 1,2,4,5-tetramethylbenzene and 1,2,3,5-tetramethylbenzene
C) 1,2,3,4-tetramethylbenzene and 1,2,3,5-tetramethylbenzene
D) 1,2,3,5-tetramethylbenzene and 1,4-diethylbenzene
"
Perplexity / DeepSeek (Pro)
<p><think>
Okay, so I have this NMR problem to solve, and I'm a bit new to this, but I'll try to work through it step by step. The question is about identifying two aromatic compounds in a 1:1 mixture based on their 1H NMR spectra. The molecular formula for both is C10H14, and the options given are various tetramethylbenzenes and 1,4-diethylbenzene.</p>
<p>First, I remember that in 1H NMR, the aromatic region is usually between 6.5-8 ppm. The problem states that there are two singlets around 6.7 ppm in a 1:1 ratio. Since they're singlets, that means the aromatic protons don't have any neighboring protons to split the signal. So, each compound must have equivalent aromatic protons that aren't next to any others.</p>
<p>Looking at the options, all the tetramethylbenzenes have four methyl groups attached to a benzene ring. The positions of these methyl groups will determine the symmetry of the molecule and thus the NMR signals.</p>
<p>Let me consider each compound:</p>
<ol>
<li><p><strong>1,2,4,5-Tetramethylbenzene (Durene)</strong>: This has methyl groups at positions 1,2,4,5. The ring is symmetric, so the aromatic protons are in equivalent positions. Since each proton is adjacent to a methyl group, they might be deshielded, but because of symmetry, there would be only one set of equivalent protons. So, this would show a singlet in the aromatic region. Also, the methyl groups are all equivalent, so they would show as a single peak in the aliphatic region.</p></li>
<li><p><strong>1,2,3,5-Tetramethylbenzene (Isodurene)</strong>: The methyl groups are at positions 1,2,3,5. This structure is less symmetric than durene. The aromatic protons might not all be equivalent, leading to more than one signal. Alternatively, if the symmetry is such that the protons are equivalent, it might still show a singlet, but I'm not sure. The methyl groups might also be in different environments, leading to multiple signals in the aliphatic region.</p></li>
<li><p><strong>1,2,3,4-Tetramethylbenzene (Prehnitene)</strong>: This has methyl groups at positions 1,2,3,4. This structure is similar to isodurene but arranged differently. The symmetry here might be lower, leading to non-equivalent aromatic protons and possibly multiple signals. The methyl groups could also be in different environments, causing multiple peaks.</p></li>
<li><p><strong>1,4-Diethylbenzene</strong>: This compound has two ethyl groups at the para positions. The aromatic protons would be equivalent due to the symmetry, giving a singlet. However, the ethyl groups would split into a quartet and a triplet in the aliphatic region, not singlets. Since the problem mentions singlets at around 2.2 ppm, this might not fit because ethyl groups typically show up as multiplets, not singlets.</p></li>
</ol>
<p>Now, looking at the NMR data: the aromatic region has two singlets in a 1:1 ratio. This suggests that each compound contributes one singlet, meaning each has equivalent aromatic protons. So, both compounds must have high symmetry.</p>
<p>From the options, 1,2,4,5-tetramethylbenzene (durene) is known to have a high degree of symmetry, giving a single aromatic singlet. If paired with another compound that also has a single aromatic singlet, it would fit the data.</p>
<p>Looking at the aliphatic region, there are three singlets in a 2:1:1 ratio. This suggests that one compound has two equivalent methyl groups (integral 2) and the other has two sets of equivalent methyl groups (each integral 1). </p>
<p>Wait, no, the aliphatic signals are three singlets in a 2:1:1 ratio. So, perhaps one compound contributes two singlets (maybe two different methyl environments) and the other contributes one singlet (all methyl groups equivalent). </p>
<p>Wait, no, the total ratio is 2:1:1, which sums to 4. Since the mixture is 1:1, each compound contributes two protons worth of signals. So, perhaps one compound has two sets of methyl groups, each contributing a singlet, and the other has one set contributing a singlet. But I'm getting a bit confused here.</p>
<p>Alternatively, considering the number of methyl groups: each tetramethylbenzene has four methyl groups. If a compound has all four methyl groups equivalent, it would show as a single singlet in the aliphatic region. If the methyl groups are in different environments, they would split into multiple singlets.</p>
<p>So, 1,2,4,5-tetramethylbenzene (durene) has all four methyl groups equivalent due to its symmetry, so it would show as a single singlet around 2.2 ppm. </p>
<p>1,2,3,5-tetramethylbenzene (isodurene) has methyl groups in different positions, so maybe two sets of equivalent methyl groups. For example, two methyl groups in one environment and two in another, leading to two singlets. But wait, the aliphatic region shows three singlets. Hmm.</p>
<p>Wait, the problem states that in the aliphatic region, there are three signals, all singlets, in a 2:1:1 ratio. So, perhaps one compound contributes two singlets (maybe two different methyl groups, each as singlets) and the other contributes one singlet. Or maybe one compound has two equivalent methyl groups (integral 2) and two singlets from two different methyl groups (each integral 1), but that would sum to 4, which is the total for two compounds.</p>
<p>Wait, the mixture is 1:1, so each compound contributes half the total integral. The aliphatic region has three singlets with integrals 2,1,1. So, perhaps one compound contributes the 2 and one of the 1s, and the other contributes the other 1. But that doesn't quite add up because each compound has four methyl groups, so each should contribute two singlets or one, depending on equivalence.</p>
<p>Alternatively, maybe one compound has two sets of equivalent methyl groups (each contributing a singlet) and the other has one set of equivalent methyl groups (contributing another singlet), but that would give three singlets in total. For example, compound A has two singlets (each integral 1) and compound B has one singlet (integral 2), but the ratio is 2:1:1, so that doesn't fit.</p>
<p>Wait, perhaps I'm overcomplicating. Let me think again. The aliphatic region has three singlets: 2,1,1. Since the mixture is 1:1, each compound contributes half the total integral. So, the total integral is 4 (2+1+1), so each compound contributes 2.</p>
<p>So, one compound contributes 2 (maybe one singlet of integral 2) and the other contributes 1 and 1 (two singlets each of integral 1). Or, one compound contributes two singlets (1 and 1) and the other contributes one singlet (2).</p>
<p>Looking at the options:</p>
<ul>
<li>1,2,4,5-tetramethylbenzene (durene) has all four methyl groups equivalent, so it would contribute a single singlet at around 2.2 ppm with integral 4 (but wait, in the mixture, each compound is present in 1:1, so the integral would be 2 for each compound's methyl groups). Wait, no, the total integral for the aliphatic region is 4 (2+1+1), so each compound contributes 2.</li>
</ul>
<p>So, durene has four equivalent methyl groups, so in the mixture, it would contribute a singlet with integral 2 (since it's 1:1, each contributes half). The other compound must contribute the remaining two integrals as 1 and 1, meaning it has two sets of equivalent methyl groups, each contributing a singlet of integral 1.</p>
<p>Looking at the options, 1,2,3,5-tetramethylbenzene (isodurene) has methyl groups in positions 1,2,3,5. Due to its symmetry, perhaps it has two sets of equivalent methyl groups. For example, the methyl groups at positions 1 and 3 might be equivalent, and those at 2 and 5 might be equivalent, leading to two singlets. Each would have integral 1 in the mixture, contributing to the 1:1 ratio.</p>
<p>So, the two compounds would be durene (1,2,4,5-tetramethylbenzene) and isodurene (1,2,3,5-tetramethylbenzene). Durene contributes a singlet at 2.2 ppm with integral 2, and isodurene contributes two singlets, each with integral 1. This would give the three singlets in a 2:1:1 ratio as described.</p>
<p>Therefore, the correct answer is option B: 1,2,4,5-tetramethylbenzene and 1,2,3,5-tetramethylbenzene.
</think></p>
<p>The correct answer is <strong>B) 1,2,4,5-tetramethylbenzene and 1,2,3,5-tetramethylbenzene</strong>.</p>
<p><strong>Explanation:</strong></p>
<ul>
<li><strong>Aromatic Region (6.7 ppm):</strong> Both compounds exhibit a singlet due to their symmetric structures, resulting in two singlets in a 1:1 ratio.</li>
<li><strong>Aliphatic Region (2.2 ppm):</strong> 1,2,4,5-tetramethylbenzene contributes a singlet (integral 2), while 1,2,3,5-tetramethylbenzene contributes two singlets (each integral 1), matching the observed 2:1:1 ratio.</li>
</ul>
<p>Thus, the mixture contains these two compounds.</p>
<hr />
<h2>References</h2>
<p>[1] https://chem.libretexts.org/Bookshelves/Organic<em>Chemistry/Organic</em>Chemistry<em>(Morsch</em>et<em>al.)/15:</em>Benzene<em>and</em>Aromaticity/15.07:<em>Spectroscopy</em>of<em>Aromatic</em>Compounds</p>
<p>[2] https://en.wikipedia.org/wiki/Tetramethylbenzene</p>
<p>[3] https://chem.libretexts.org/Bookshelves/Organic<em>Chemistry/Organic</em>Chemistry<em>I</em>(Liu)/06:<em>Structural</em>Identification<em>of</em>Organic<em>Compounds-</em>IR<em>and</em>NMR<em>Spectroscopy/6.07:</em>H<em>NMR</em>Spectra<em>and</em>Interpretation<em>(Part</em>II)</p>
<p>[4] https://www2.chem.wisc.edu/deptfiles/OrgLab/handouts/Estimating%20the%201H%20shifts%20of%20aryl%20H-atoms.pdf</p>
<p>[5] https://www.chemicalbook.com/SpectrumEN<em>95-93-2</em>1HNMR.htm</p>
<p>To determine the two compounds present in the mixture based on the provided proton NMR data, let's carefully analyze the information:</p>
<hr />
<h3><strong>Analysis</strong></h3>
<ol>
<li><p><strong>Aromatic Region (6.7 ppm, two singlets, 1:1 ratio)</strong>:</p>
<ul>
<li>The aromatic hydrogens in aromatic compounds exhibit chemical shifts in the range of ~6.5-8 ppm.</li>
<li>The observation of two singlets in a 1:1 ratio suggests that the aromatic region of the two compounds is symmetric and does not show splitting due to coupling with other protons.</li>
</ul></li>
<li><p><strong>Aliphatic Region (2.2 ppm, three singlets, 2:1:1 ratio)</strong>:</p>
<ul>
<li>This indicates there are three distinct types of aliphatic proton groups contributing to the signal intensity.</li>
<li>Singlets in the aliphatic region correspond to isolated methyl groups or similar types of protons that do not couple (no neighbors).</li>
</ul></li>
<li><p><strong>Symmetry Effect</strong>:</p>
<ul>
<li>The number and splitting pattern of NMR signals depend on the symmetry of the chemical structures. More symmetrical compounds have fewer unique signals.</li>
</ul></li>
<li><p><strong>Possible Compounds</strong>:</p>
<ul>
<li><strong>1,2,4,5-Tetramethylbenzene</strong>:
<ul>
<li>Highly symmetrical, will show two singlets in the aromatic region (6.7 ppm) corresponding to two types of equivalent aromatic hydrogens.</li>
<li>Three singlets at ~2.2 ppm for three types of methyl groups: two equivalent at 2:1:1 intensity ratio. Matches data.</li>
</ul></li>
<li><strong>1,2,3,5-Tetramethylbenzene</strong>:
<ul>
<li>Less symmetrical than 1,2,4,5-tetramethylbenzene, but retains sufficient symmetry to have two singlets in the aromatic region.</li>
<li>Also has three distinct methyl groups giving singlets in the aliphatic region, in a 2:1:1 ratio. Matches data.</li>
</ul></li>
<li><strong>1,2,3,4-Tetramethylbenzene</strong>:
<ul>
<li>Asymmetrical. Would not exhibit two singlets (1:1) in the aromatic region, but rather more complex splitting.</li>
<li>This does not match the observed NMR data.</li>
</ul></li>
<li><strong>1,4-Diethylbenzene</strong>:
<ul>
<li>Would show only one aromatic singlet (not two) due to high symmetry.</li>
<li>Also, two signals in the aliphatic region (not three) for the ethyl groups. Does not match the data.</li>
</ul></li>
</ul></li>
</ol>
<hr />
<h3><strong>Conclusion</strong></h3>
<p>The correct pair of compounds is:
<strong>B) 1,2,4,5-Tetramethylbenzene and 1,2,3,5-Tetramethylbenzene</strong></p>
<ul>
<li>Both have symmetrical aromatic regions that generate two singlets in a 1:1 ratio.</li>
<li>Their aliphatic methyl regions produce three singlets with a 2:1:1 intensity ratio.</li>
</ul>
<hr />
<h3><strong>References</strong></h3>
<p>Relevant information from the search results:
1. <a href="https://www2.chemistry.msu.edu/faculty/reusch/OrgPage/nmr.htm">Proton NMR Table - Michigan State University</a>
2. <a href="https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Liu)/06:_Structural_Identification_of_Organic_Compounds-_IR_and_NMR_Spectroscopy/6.06:_H_NMR_Spectra_and_Interpretation_(Part_I)">6.6: ¹H NMR Spectra and Interpretation (Part I) - LibreTexts</a></p>
OpenAI
<p>A concise way to see why the correct pair must be 1,2,4,5‐tetramethylbenzene and 1,2,3,4‐tetramethylbenzene (option A) is as follows:</p>
<p>1) Aromatic region (two singlets at ~ 6.7 ppm, 1:1 ratio).<br />
Each of the tetramethylbenzene isomers in the list (other than 1,4‐diethylbenzene) has only two ring‐hydrogens total, and in each isomer those two hydrogens turn out to be chemically equivalent.<br />
– 1,2,4,5‐Tetramethylbenzene (Durene) has its two ring‐H at carbons 3 and 6, which are equivalent by a symmetry plane.<br />
– 1,2,3,4‐Tetramethylbenzene also has its two ring‐H chemically equivalent (positions 5 and 6).<br />
– 1,2,3,5‐Tetramethylbenzene likewise has a plane mapping C4 ↔ C6.<br />
Hence each of these three tetramethylbenzene isomers gives “one” aromatic singlet (integrating to 2H). </p>
<p>Because the 1 : 1 mixture shows only two aromatic singlets total (rather than three or more), we immediately exclude 1,4‐diethylbenzene (it would give two ring‐signals by itself) and focus on pairing two tetramethylbenzenes. </p>
<p>2) Aliphatic/methyl region (three singlets near 2.2 ppm, in a 2 : 1 : 1 ratio).<br />
• 1,2,3,5‐Tetramethylbenzene by itself is known to give three distinct methyl singlets in a 2 : 1 : 1 ratio (because two of its four methyls are equivalent to each other, while the other two methyls are each unique).<br />
• Both 1,2,4,5‐ and 1,2,3,4‐tetramethylbenzene, by contrast, each give only two distinct methyl environments in a 1 : 1 ratio (two pairs of equivalent methyls). </p>
<p>When you mix 1,2,4,5‐ and 1,2,3,4‐tetramethylbenzene in a 1 : 1 molar ratio, each compound on its own would have shown two singlets (each 1 : 1). However, in the actual mixture, one methyl‐singlet from one isomer happens to coincide (chemically overlap) with one methyl‐singlet from the other isomer. That collapse/overlap reduces the total number of observable methyl signals from “2 + 2 = 4” down to “3.” </p>
<p>Moreover, the integrated intensities end up in a global 2 : 1 : 1 ratio (one big overlapped singlet for 12 protons total, and two smaller singlets for 6 protons each). This matches exactly what is reported: three singlets at 2.2 ppm in a 2 : 1 : 1 area ratio.</p>
<p>All other pairwise combinations either yield too many methyl signals (because 1,2,3,5‐tetramethylbenzene alone already has three non‐overlapping methyl sets) or do not produce just two singlets in the aromatic region (as would be the case if 1,4‐diethylbenzene were present). </p>
<p>Hence the only consistent choice is:</p>
<p>A) 1,2,4,5‐Tetramethylbenzene and 1,2,3,4‐Tetramethylbenzene.</p>
xAI
<p>Let me analyze the 1H NMR data given in the question and determine which pair of compounds matches the spectral information:</p>
<p>The key spectral features are:
1. Two singlets at ~6.7 ppm in 1:1 ratio
2. Three singlets at ~2.2 ppm in 2:1:1 ratio</p>
<p>Let's analyze what this means:
- The signals at 6.7 ppm correspond to aromatic protons
- The signals at 2.2 ppm correspond to methyl groups attached to the aromatic ring
- The 1:1 ratio of aromatic signals means equal numbers of equivalent aromatic protons
- The 2:1:1 ratio for methyl signals indicates three different types of methyl environments</p>
<p>Looking at the given options and their structures:</p>
<p>Among the options given, only 1,2,4,5-tetramethylbenzene and 1,2,3,5-tetramethylbenzene would give:
- Two equivalent aromatic protons (at 6.7 ppm)
- Three different types of methyl groups (at 2.2 ppm) in the required 2:1:1 ratio due to their symmetry</p>
<p>Therefore, the correct answer is B) 1,2,4,5-tetramethylbenzene and 1,2,3,5-tetramethylbenzene.</p>
<p>The other combinations would not match this spectral pattern because they would either have different numbers of aromatic protons or different ratios of methyl group environments.</p>