Chem Question

[Red_x_Lily]

This is from a practice exam we did in school recently –

Q) IR spectroscopy is an analytical tool often used in organic chemistry. Which of the following could be distinguished from the other molecules due to the absence of a particularly distinct and characteristic peak?

A)   CH₃CH₂CH₂CH₂OH
B)   CH₃CH₂CH₂COOH
C)   CH₃CH₂CH₂CH₂NH₂
D)   CH₃CH₂CH₂COOCH₃

I got the answer to this question right, according to the solutions, but there's no an explanation for it and I’m not sure if what I thought was right. Some of the students in my class, including the teacher, thought it was C…and I just wanted to know what everyone else thinks. 

The answer is: D

[jfaure]

A C=O bond, displayed in answer D, has more distinct wave number and can't be mistaken for another type of bond. However a N-H bond, as is the case in answer C, has a wave number that overlaps with an O-H (alcohols) bond. Therefore D is the correct answer.

[Chocolate]

A C=O bond, displayed in answer D, has more distinct wave number and can't be mistaken for another type of bond. However a N-H bond, as is the case in answer C, has a wave number that overlaps with an O-H (alcohols) bond. Therefore D is the correct answer.

Which of the following could be distinguished from the other molecules due to the absence of a particularly distinct and characteristic peak?
The question is asking for absence.

A)   CH3CH2CH2CH2OH     -Distinct peak at OH (of an alkanol)
B)   CH3CH2CH2COOH       -Distinct OH of an acid along side the C=O
C)   CH3CH2CH2CH2NH2    -Distinct NH2
D)   CH3CH2CH2COOCH3   -Have a look at it. Its got an ester linkage. Its got a C=O BUT so does B. Theres is nothing in D that differentiates it from B but theres the OH of the ACID in B that differentiates it from D.

Thats my take on it.

a N-H bond, as is the case in answer C, has a wave number that overlaps with an O-H (alcohols) bond.

Just to ask..why are you drawing comparison between the NH2 and the OH bonds? Sure, they overlap..but only when a compund contains these two functional groups. Neither of these 4 do..C) has only the NH2 so theres no OH that will affect it as you are suggesting. These compounds have their own distinct IR spectrum..I'm assuming theyre separate compunds run through the IR spectrophotometer at different times, not jumbled up.

[Red_x_Lily]

Which of the following could be distinguished from the other molecules due to the absence of a particularly distinct and characteristic peak?
The question is asking for absence.

A)   CH3CH2CH2CH2OH     -Distinct peak at OH (of an alkanol)
B)   CH3CH2CH2COOH       -Distinct OH of an acid along side the C=O
C)   CH3CH2CH2CH2NH2    -Distinct NH2
D)   CH3CH2CH2COOCH3   -Have a look at it. Its got an ester linkage. Its got a C=O BUT so does B. Theres is nothing in D that differentiates it from B but theres the OH of the ACID in B that differentiates it from D.

Thats my take on it.

Thank you so much! It makes much more sense to me now. =)

[beezy4eva]

Why does the high resolution NMR spectrum of ethylene glycol (HOCH2CH2OH) show two singlet peaks? I know why there is two peaks, coz there's two hydrogen environments, but i dont understand why the Hs on the CH2s wouldn't cause peak splitting.

[chem-nerd]

i think this was an error in the answers. the CH2s should be split into triplets.

edit: sorry didnt read q properly. CH2s are equivalent => no splitting

[beezy4eva]

Nope i asked my teacher so he looked up the high resolution NMR spectrum and it shows it as two singlets

[ed_saifa]

maybe because it is one envirnment so it doesn't get split?

[shinny]

yeh peak splitting only occurs if they're in different chemical environments since the n+1 specifically states that the number of peaks within a peak set is equal to the number of hydrogens in a NON-EQUIVALENT environment, plus one. since ethylene glycol is symmetrical, it wont split since the H's on the CH2- are in the same environment as the other hydrogens on the other CH2-

[kj_]

i think this was an error in the answers. the CH2s should be split into triplets.

answers are correct in this case, though

edit: what shinjitsuzx said