🇺🇦 nmr_spectroscopy / organic chemistry ⌬

It’s #NMRweekend time! 🧲⁉️ Can you identify the correct structure based on the spectrum? Take part in the poll below 👇. I’ll post the full explanation in a few days. 😉 #nmrchallenge #quiz 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

The hidden order behind a complex multiplet 🧩 At first glance, this ¹⁹F NMR signal might look like a chaotic forest of peaks, but it’s actually a beautiful demonstration of a higher-order XX'AA'MM'NN'R spin system. Why does a simple molecule produce such a complicated pattern? It all comes down to the crucial difference between chemical and magnetic equivalence. Thanks to the molecule's plane of symmetry, the two fluorine atoms (and their neighboring protons) are chemically equivalent. However, because they are locked in a rigid cyclic structure, each fluorine interacts (couples) differently with the other protons across the ring depending on their spatial distance and geometry. This magnetic non-equivalence, combined with the diastereotopic CH2 protons in the ring, turns what would be a simple signal into this intricate, perfectly calculable pattern. As you can see, the experimental spectrum matches the calculated one flawlessly! Find this and other building blocks at enaminestore.com #NMR #FluorineNMR #spinsystem 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

It’s #NMRweekend time! 🧲⁉️ Can you identify the correct structure based on the spectrum? Take part in the poll below 👇. • UPD ✅. The correct answer is B! Here is how we rule out the other options: Why not A? The spectrum only shows two signals in the aromatic region. Structure A is trisubstituted and would display three aromatic protons. Why not C? In Structure C, the two aromatic protons are adjacent (ortho) and would split each other, meaning they would share the same coupling constant. Instead, we see two doublets with different coupling constants (J = 8.6 Hz and J = 5.8 Hz). This proves they are not coupled to each other, but are instead coupling individually to the fluorine atom! The OH Signal: That far-downfield peak at 12.01 ppm is a classic signature of a phenolic OH proton participating in strong intramolecular hydrogen bonding with the adjacent carbonyl oxygen! All of this perfectly matches the arrangement in Structure B. • #nmrchallenge #quiz 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

How Much Sample Do You Need for NMR? A Practical Guide From standard 1D proton spectra to low-concentration 2D experiments: sample requirements for optimal signal-to-noise. 👉 full post on my Substack! 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

It’s #NMRweekend time! 🧲⁉️ Can you identify the correct structure based on the spectrum? Take part in the poll below 👇. I’ll post the full explanation in a few days. 😉 UPD ✅. The correct answer is C! Why not A? The methyl group in Structure A should appear as a triplet. Here, we have a doublet. Why not B? A phenolic OH should appear at a higher chemical shift (>8 ppm in CDCl3). Here, we have an aliphatic alcohol signal at 1.47 ppm. #nmr #nmrchat #chemistry #quiz

In today’s #NMRMultiplet we’re going to consider this complex 1H NMR signal from a norbornane derivative (centered at 2.23 ppm, labeled (tdd). 🔍 Feature Highlights: 1️⃣ Distinct W-Coupling (4J): Highlighted with the red curved line in the picture is a clear 4J(H,H) W-coupling (approximately 3.5 Hz) to a bridge syn-proton. The unique rigid skeleton provides the ideal geometry for this strong long-range “W” interaction. 2️⃣ Negligible Bridgehead Coupling (3J): Notice that there is no 3J coupling to the adjacent bridgehead proton. This is a classic Karplus relationship in action: the dihedral angle between this target proton and the bridgehead H is approximately 90°, predicting a vicinal constant near zero. 3️⃣ Decoding the rest: The other three couplings in our splitting tree are: 14.67 Hz, 14.67 Hz, and 11.13 Hz. They come from vicinal 3J(H,F) interactions and a geminal 2J(H,H) pairing. #coupling #norbornane 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

It’s #NMRweekend time! 🧲🧩 Can you identify the correct structure based on the 1D NOESY spectrum? (Honestly, the ¹H NMR alone is enough to crack this one 😉). Take part in the poll below 👇. I’ll post the full explanation in a few days! UPD ✅ The correct answer is Structure B! Why not A? The aliphatic splitting pattern simply doesn't fit. In Structure A, the methyl group is adjacent to a CH proton, meaning it would appear as a doublet. Our spectrum clearly shows a 3H singlet at 1.45 ppm. Why not C? To choose between B and C, we look at the 1D NOESY experiment. Irradiating the methyl group reveals a clear nOe correlation to the meta-coupled aromatic doublet (J = 2.5 Hz). If Structure C were correct, the proton closest to the methyl group would have an adjacent ortho neighbor, and we would see an nOe to an ortho-coupled doublet instead! All evidence points perfectly to Structure B. #nmr #nmrchat #chemistry #quiz 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

It’s #NMRweekend time! 🧲⁉️ Can you identify the correct structure based on the spectrum? Take part in the poll below 👇. I’ll post the full explanation in a few days. UPD ✅ The correct answer is Structure B. Ruling out A and C: Both of these structures have adjacent aromatic protons. If either were correct, we would see strong ortho couplings in the spectrum (typically around 8.0 Hz). Confirming B: Our spectrum shows no ortho coupling at all! Instead, we see two doublets and a triplet that all share a small coupling constant of J = 2.1 Hz. This indicates that none of the aromatic protons are directly next to each other, perfectly matching the 3,5-disubstituted pattern of Structure B, where only meta couplings are observed. #nmrchallenge #quiz 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

Negative chemical shifts in ¹³C NMR? Yes, they exist! 🧲📉 Take a look at picture, showing the spectrum for diiodomethane CH₂I₂. Normally, bonding a carbon to highly electronegative halogens strips away electron density, deshielding the nucleus and pushing the signal downfield. But iodine plays by a completely different set of rules! As you can see, the ¹³C signal sits way upfield at a staggering -66.6 ppm. Why does this happen? It’s all thanks to the Heavy Atom Effect, specifically, the HALA (Heavy Atom on Light Atom effect). Because iodine is so massive, relativistic effects—primarily spin-orbit coupling—kick in. This induces a massive shielding environment at the directly attached carbon nucleus, overpowering the standard inductive effect. Fun fact: This shielding is highly localized! Notice how the ¹H shift for the CH₂ group stays perfectly normal at 3.87 ppm. The spin-orbit coupling diminishes rapidly with distance, leaving the protons largely unaffected. Have you ever worked with heavily iodinated compounds like iodoform CHI₃ or carbon tetraiodide CI₄? The shifts go even further into the negative! Drop your favorite NMR anomalies in the comments on my Substack 👇 #iodine #heavyatomeffect 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

It’s #NMRweekend time! 🧲⁉️ Can you identify the correct structure based on the spectrum? Take part in the poll below 👇. #nmrchallenge #quiz UPD ✅. The correct answer is Structure A. Structure C can be ruled out due to an incorrect aromatic proton spin coupling pattern: this structure wouldn’t have doublet with only one small meta coupling of 1.7 Hz. Structure B would have a benzylic methylene group around 2.8–3.5 ppm; additionally, it would appear as a multiplet due to coupling with the adjacent S-CH2 group. 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

UPD ✅ The correct answer to this #NMRweekend challenge is Structure C! Here is the quick breakdown of how we get there: •Why not A? We would expect to see 3 distinct methyl signals (an N-Me at ~3 ppm and two C-Me at ~1.8–1.9 ppm). Our spectrum only has 2 signals! •Why not B? Methylene protons attached to an iodine (CH2-I) usually show up downfield around 3.5 ppm. The spectrum is completely empty there. •Why C? The chemical shifts are a perfect match, and the spectrum reveals a classic feature of rigid bicyclic systems: strong 4-bond couplings. If you check the expanded region, the ¹³C satellites of the CH2 signal are beautifully split by a coupling constant of ⁴J = 9.3 Hz! 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

It’s #NMRweekend time! 🧲🧩 Can you identify the correct structure based on the spectrum? Take part in the poll below 👇. #nmrchallenge #quiz 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!

What does sour look like on an NMR spectrometer? 🍋‍🟩📉 Here’s the 600 MHz ¹H NMR spectrum of lime juice. Those massive, intense peaks between 2.6 and 2.8 ppm? That’s the defining feature of lime: citric acid, showing off a textbook AB spin system thanks to its diastereotopic protons. You can also clearly see the complex sugar region (glucose and fructose) from 3.2–4.2 ppm, and the alpha-glucose anomeric proton doublet sitting at 5.22 ppm. Unlike oranges, limes are very low in sucrose, which is perfectly reflected in the missing high-intensity sucrose signals! More details on my Substack. #naturalproducts #lime 📱Telegram | 📱Instagram | 📱Twitter | 📱LinkedIn | ☕️BuyMeACoffee | Substack!