chemsample-NMR-discussion


 * __General Discussion of Results ~__**

//Conducted by: Robert Theis//
 * __NMR ~__**

The spectrum for **2 - Pentanone** yielded the following hydrogen peaks, and integration: Triplet @ 1.0 (3H) Multi(6)plet @ 1.6 (2H) Singlet @ 2.1 (3H) Triplet @ 2.4 (2H)

Given this information, and the molecular structure of the compound, all of the spectrum's integrals and peaks can be accounted for ... The triplet @ 1.0 corresponds to the -CH 3 on the far left of the mol., split twice by the -CH 2 to its right. The multiplet @ 1.6 corresponds to the left most -CH 2, split three times by the neighboring -CH 3 , and twice by its neighboring -CH 2. The triplet @ 2.4 corresponds to the right most -CH 2, split twice by the -CH 2 directly to its left. And, finally, the singlet @ 2.1 corresponds to the -CH 3 to the right of the carbonyl group (Note - that this group is unsplit).



Note also, that the degree of deshielding in the hydrogens can be accounted for ... As the groups move closer to the carbonyl, their associated hydrogens become more, and more deshielded.

The spectrum for my **Unknown** yielded the following hydrogen peaks, and integration: Singlet @ 2.1 (3H) Singlet @ 7.1 (2H)

What this information tells me is that I most likely have **A)** a -CH3 group near a carbonyl (resulting in the deshielding around 2.0 -2.2), and **B)** that I have a -CH2 group directly bonded to an Oxygen. However, playing with actual possible formula's, it is difficult to establish any real firm structure based on this spectral information. It may be, that the integration is actually 4H to 6H, whereby we would be looking at a Benzene ring with one stemming carbon.

//Conducted by: Julia Valentino and Liz Wasil//
 * __NMR ~__**

The spectrum for ethyl-4-methylbenzoate has the following hydrogen peaks, and integration:

Molecular formula: C10 H12 O2 : 5° unsaturations


 * Splitting of Hydrogen’s || d /ppm || Integration (#H) ||
 * doublet || 7.8 || 2 ||
 * doublet || 7.3 || 2 ||
 * quartet || 4.3 || 2 ||
 * singlet || 2.4 || 3 ||
 * triplet || 1.4 || 3 ||

The signal at 7.8ppm and 7.3ppm is the H’s within the aromatic ring. They split one another creating a doublet for both the integrations. The signal located at 4.3ppm is identified as –CH2, located within the branch next to–CH3. Since there is (3H) neighboring the -CH2 it is split four ways, this is following the (n+1) rule. The –CH3 obtains the signal at 1.4ppm, and is a triplet. Those two signals are deshielded by the ester. The –CH3 on the fourth carbon in the ring has its own signal because it’s attached to another carbon.



//Conducted by Sarah Vorpahl and Adrian Ruiz//
 * __NMR ~__**

=
__Ester Banana (3-methylbutyl ethoate):__ ====== Singlet 2.0 3H Triplet 4.0 2H Multiplet 2.5 2H Triplet 1.0 3H

__Ethyl Acetate__: Singlet 2.0 3H Quartet 4.0 2H Triplet 1.2 3H

__Unknown #11__: Quartet 5.0-5.8 Quartet 4.0-4.4 Quintet 1.8-2.5 Triplet 1.7-1.7

__Unknown #4__: Singlet 5.2-5.4 Doublet 2.8-3.1 Singlet 2.4 Singlet 2.0-2.1

__**Post Laboratory Discussion Questions ~**__

1) Propose a structure consistent with the molecular formula and H NMR spectrum of your **unknown**compound. Explain your answer.

2) Propose a structure consistent with the molecular formula, H, C, and DEPT spectra of your**unknown** compound. Explain you answer. __**Note**__ - that, the answers to questions //1 and 2// should be included in the general discussion of our results above.

3) Using the H, C, and DEPT spectra, along with chemical shift arguments and resonance theory, assign all of the peaks in the H and C NMR spectra of **4 - propoxybenzaldehyde**. Exlain your reasoning.



There are a total of 4 signals in the 4-propoxybenaldehyde molecule shown above. There is a signal for the aromatic ring around 6-7ppm -this will have splitting, and have a much smaller peak than the other signals. Then, the aldehyde gets its own signal around 2-3ppm. This peak is smaller than the signal of the terminal methyl group, which is between the interval of 1-2ppm. This signal is going to cause splitting of the two hydrogens that are on the carbon attached to the oxygen (CH 2 -O gives a signal around 4.8ppm). The splitting will give you a quartet and a triplet for the terminal methyl hydrogens ... which is also the tallest peak. NMR table used: [[[|http://nmr] table2|[] ]]

4) Using data presented in your NMR handout, assign the spectrum of **4 - allylanisole**.



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