Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE 1) IR so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one double bond equivalent. That leaves 3 carbons and 7 hydrogens. 2) NMR Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when adding electron withdrawing groups nearby. It\'s a pretty good guess that the triplet at ~1.1ppm is a methyl group. The fact that it\'s a triplet indicates the carbon next to it has two protons attached to it. That gives a CH3-CH2- (as well as the -COOH group) all that\'s left now is one carbon and two hydrogens (CH2) Looking at the other two proton peaks in the nmr ... there\'s one triplet and one sextet. The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it must be a CH2 next to the COOH Now you would have two pieces: CH3-CH2- and -CH2-COOH put them together and check the coupling of the protons... you should end up with two triplets (proton split by two others), one sextet (proton split by 5 others), and a small peak somewhere around 9-13ppm. (that\'s the peak indicated by the offset of 2.0ppm on the spectrum) So the structure would be: CH3-CH2-CH2-COOH (C4H8O2) Solution Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE 1) IR so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one double bond equivalent. That leaves 3 carbons and 7 hydrogens. 2) NMR Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when adding electron withdrawing groups nearby. It\'s a pretty good guess that the triplet at ~1.1ppm is a methyl group. The fact that it\'s a triplet indicates the carbon next to it has two protons attached to it. That gives a CH3-CH2- (as well as the -COOH group) all that\'s left now is one carbon and two hydrogens (CH2) Looking at the other two proton peaks in the nmr ... there\'s one triplet and one sextet. The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it must be a CH2 next to the COOH Now you would have two pieces: CH3-CH2- and -CH2-COOH put them together and check the coupling of the protons... you should end up with two triplets (proton split by two others), one sextet (proton split by 5 others), and a small peak somewhere around 9-13ppm. (that\'s the peak indicated by the offset of 2.0ppm on the spectrum) So the structure would be: CH3-CH2-CH2-COOH (C4H8O2).

1. Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE
1) IR
so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an
acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one
double bond equivalent.
That leaves 3 carbons and 7 hydrogens.
2) NMR
Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when
adding electron withdrawing groups nearby. It's a pretty good guess that the triplet at ~1.1ppm is
a methyl group. The fact that it's a triplet indicates the carbon next to it has two protons attached
to it.
That gives a CH3-CH2- (as well as the -COOH group)
all that's left now is one carbon and two hydrogens (CH2)
Looking at the other two proton peaks in the nmr ... there's one triplet and one sextet.
The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it
must be a CH2 next to the COOH
Now you would have two pieces:
CH3-CH2- and -CH2-COOH
put them together and check the coupling of the protons...
you should end up with two triplets (proton split by two others), one sextet (proton split by 5
others), and a small peak somewhere around 9-13ppm. (that's the peak indicated by the offset of
2.0ppm on the spectrum)
So the structure would be:
CH3-CH2-CH2-COOH (C4H8O2)
Solution
Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE
1) IR
so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an
acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one
double bond equivalent.
That leaves 3 carbons and 7 hydrogens.
2) NMR
Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when

2. adding electron withdrawing groups nearby. It's a pretty good guess that the triplet at ~1.1ppm is
a methyl group. The fact that it's a triplet indicates the carbon next to it has two protons attached
to it.
That gives a CH3-CH2- (as well as the -COOH group)
all that's left now is one carbon and two hydrogens (CH2)
Looking at the other two proton peaks in the nmr ... there's one triplet and one sextet.
The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it
must be a CH2 next to the COOH
Now you would have two pieces:
CH3-CH2- and -CH2-COOH
put them together and check the coupling of the protons...
you should end up with two triplets (proton split by two others), one sextet (proton split by 5
others), and a small peak somewhere around 9-13ppm. (that's the peak indicated by the offset of
2.0ppm on the spectrum)
So the structure would be:
CH3-CH2-CH2-COOH (C4H8O2)