Propose a structure for a molecule C14 H12 that has the following ^1 H NMR spectrum and has IR

step 1 Solving Problem 48 of Chapter 15. So propose a structure for the following empirical formula C14

Propose a structure for a molecule C14 H12 that has the...

Key Concepts

Infrared Spectroscopy

IR spectroscopy is used to identify functional groups by measuring the absorption of infrared light at specific frequencies. The characteristic absorption bands, such as those from out-of-plane C–H bending in aromatic rings, provide clues about the substitution pattern on the aromatic system.

Aromatic Ring Substitution Patterns

Understanding aromatic substitution patterns is essential when analyzing spectral data as it allows for the prediction of how substituents affect chemical shifts and characteristic absorption frequencies. The arrangement of substituents on the benzene ring influences both NMR splitting patterns and specific IR absorption peaks related to aromatic C–H vibrations.

Degrees of Unsaturation

This concept involves calculating the number of rings and multiple bonds present in a molecule using its molecular formula. It is critical for deducing whether a compound contains aromatic rings and/or other unsaturations, thereby serving as a foundational step in structure elucidation.

1H NMR Spectroscopy

1H NMR provides detailed information about the hydrogen atoms in a molecule, including their chemical environment, multiplicity, and integration. This technique helps in distinguishing different types of protons (e.g., aromatic versus aliphatic) and in understanding how atoms are connected within the molecule.

Transcript

00:01 Solving problem 48 of chapter 15.

00:06 So propose a structure for the following empirical formula c14, h12.

00:17 And from the nmr spectrum, we know that we have only basically two signals.

00:24 One, a five, around five, with integration as one.

00:29 So around five integration as one.

00:34 It's a singlet, basically.

00:39 And one between seven and eight, it's a singlet but correspond to five.

00:46 So let's write the integration more than the area, more than.

00:52 So this one is one area.

00:56 And this is the region.

00:59 So we need a total of 14 hydrogen, 14 carbon, sorry.

01:05 And 12 hydrogens.

01:07 So around 7, 8, we have the benzolic protons.

01:15 So let's start drawing one benzene ring.

01:19 So now the fact that we have so many carbons and so many protons, but other than the signal of the proteins of the benzene ring, we only have one other signal with integration one.

01:37 Suggests that probably we have two benzene rings connected somehow.

01:45 The molecules, i think it has to be symmetric so that we only have one type of signal for the benzene protons.

01:58 So, for example, in this way, we would have 12 carbons because they're six from this ring, six from this ring.

02:08 So let's suppose that they are connected to another carbon atom...

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