[nertil1]

can somebody help me? I need help with finding the structures from the nmr spectra.

here's the pdf
http://www.filelodge.com/files/room32/890019/Assignment%202.pdf"

[Albert]

Show us your opinions/attempts, and we'll be pleased to help you. Otherwise we'll both break forum rules...

[Albert]

...however, since I've already finished to work, now, here is a list of useful hints.

A:  What kind of molecule gives a positive result when carrying out a test with Tollen's reagent AND has a peak (for 1 hydrogen) at more than 9.5 ppm?

B: If you've realized what A is, you should think about the reaction between it and a Grignard reagent. Moreover, Grignard reagents don't yield a vast spectrum of products, do they?

C: B gets oxidized and the product has no peaks above 2.5 ppm. Besides, there's a singlet for 3 hydrogens and there's no hydrogen on its own, hence, I sincerely doubt that C is an alcohol...

D: Well, what reacts with iodine yielding a yellow precipitate of which is iodoform...? 

To sum up, you may even not look at the spectra at all in order to work out what A,B,C and D are.

[nertil1]

here's what i have so far?

A)CH3CH2CH2CH=O (ALDEHYDE)

B)CH3CH2CH2CH(OH)CH3 (SECONDARY ALOCOHOL)

C)CH3CH2CH2C=O(CH3) (KETONE)

D)CH3CH2C(I2)C=O(CH3)

(the numers between the letters are supposed to be subscrips by the way)

Can anybody check these and tell me if I'm right or not?
it's due soon
thanks for your help
-nertil

[Albert]

D)CH3CH2C(I2)C=O(CH3)

This is obviously incorrect, while the others are all correct.

You have a methyl ketone, right? It reacts with iodine and a base. You yield two products: one is iodoform (yellow).

Now, you can just write the chemical equation and try to find a plausible molecule to complete it, or you can have a look at this: http://www.wellesley.edu/Chemistry/chem211lab/Orgo_Lab_Manual/Appendix/ClassificationTests/aldehyde_ketone.html#Iodoform

[swati]

You can also cross-check your answer with the given NMR spectra

For D NMR spectra is given as
1H   broadened singlet     ~ 10.5
2H   triplet                        ~  2.1
2H   sextet                       ~  1.6
3H   triplet                        ~  1.0   

C on reaction with I₂ and base will give iodoform CHI₃ (yellow product) and carboxylic acid

So D is CH₃CH₂CH₂COOH

*  For 1H of COOH we get broadened singlet at 10.5 .The COOH proton get exchanged with  the solvent H₂O so rapidly that the protons of the neighbouring carbon wont effect the splitting pattern of COOH proton , therefore we get only singlet.

*  For 2H of carbon no. 2 we get triplet at 2.1 .The COOH proton get exchanged with the solvent H2O so rapidly that the protons of the neighbouring carbon wont feel its presence , so only 2 neighbouring protons on carbon no.3 will effect the splitting pattern of protons on carbon no.2 . So splitting pattern will be 2+1=3

*  For 2H of carbon no. 3 we get sextet at 1.6 . This is due to 5 protons on neighbouring carbon atoms. So splitting pattern will be 5+1=6   

*  For 3H of carbon no. 4 we get triplet at 1.0 . This is due to 2 protons on neighbouring carbon atoms. So splitting pattern will be 2+1=3

[nertil1]

thanks a lot guys for your help