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A compound of formula C5H12 gives one signal in the 1H NMR and two signals in the 13C NMR spectra. The compound is:
- a)Pentane
- b)2-methylbutane
- c)2, 2-dimethylpropane
- d)Cannot tell without more information
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A compound of formula C5H12 gives one signal in the 1H NMR and two sig...
Explanation:
The given compound has the formula C5H12, which suggests it is an alkane.
1H NMR:
The compound gives one signal in the 1H NMR. This indicates that there are no chemically distinct hydrogen atoms in the molecule.
13C NMR:
The compound gives two signals in the 13C NMR. This indicates that there are two different types of carbon atoms in the molecule.
Possible structures:
The compound can have the following structures:
a) Pentane: This compound has five carbon atoms and no branching. It would give only one signal in both 1H and 13C NMR spectra. Therefore, it is not the correct option.
b) 2-Methylbutane: This compound has a branched structure with four carbon atoms in the main chain and one methyl group attached to the second carbon atom. It would give one signal in the 1H NMR and two signals in the 13C NMR spectra. However, the signals in the 13C NMR spectra would be very close to each other, indicating that the carbon atoms are not very different in chemical environment. Therefore, this option is not the correct one.
c) 2,2-Dimethylpropane: This compound has a branched structure with three carbon atoms in the main chain and two methyl groups attached to the second and third carbon atoms. It would give one signal in the 1H NMR and two signals in the 13C NMR spectra. The signals in the 13C NMR spectra would be well separated, indicating that the carbon atoms are very different in chemical environment. Therefore, this option is the correct one.
d) Cannot tell without more information: This option can be eliminated based on the information provided in the question.
Conclusion:
The correct option is c) 2,2-Dimethylpropane, which has a branched structure with three carbon atoms in the main chain and two methyl groups attached to the second and third carbon atoms.
The given compound has the formula C5H12, which suggests it is an alkane.
1H NMR:
The compound gives one signal in the 1H NMR. This indicates that there are no chemically distinct hydrogen atoms in the molecule.
13C NMR:
The compound gives two signals in the 13C NMR. This indicates that there are two different types of carbon atoms in the molecule.
Possible structures:
The compound can have the following structures:
a) Pentane: This compound has five carbon atoms and no branching. It would give only one signal in both 1H and 13C NMR spectra. Therefore, it is not the correct option.
b) 2-Methylbutane: This compound has a branched structure with four carbon atoms in the main chain and one methyl group attached to the second carbon atom. It would give one signal in the 1H NMR and two signals in the 13C NMR spectra. However, the signals in the 13C NMR spectra would be very close to each other, indicating that the carbon atoms are not very different in chemical environment. Therefore, this option is not the correct one.
c) 2,2-Dimethylpropane: This compound has a branched structure with three carbon atoms in the main chain and two methyl groups attached to the second and third carbon atoms. It would give one signal in the 1H NMR and two signals in the 13C NMR spectra. The signals in the 13C NMR spectra would be well separated, indicating that the carbon atoms are very different in chemical environment. Therefore, this option is the correct one.
d) Cannot tell without more information: This option can be eliminated based on the information provided in the question.
Conclusion:
The correct option is c) 2,2-Dimethylpropane, which has a branched structure with three carbon atoms in the main chain and two methyl groups attached to the second and third carbon atoms.
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Community Answer
A compound of formula C5H12 gives one signal in the 1H NMR and two sig...
First calculate the degree of unsaturation. Here,
Degree of unsaturation= (2C+2 - H)/2= (2×5+2-12)/2=0
This compound has 0 degrees of unsaturation and has the general formula of alkanes. So, it has to be any one of the isomers of pentane.
It gives one signal in 1H NMR. So, all the hydrogens present are in equivalent positions(surrounded by same environment). Two signals in 13C NMR shows that there are two kinds of carbons that are not equivalent. If you draw the structures of all the isomers of pentane and check, then only 2,2-dimethylpropane will fit.
In n-pentane you will get 3 signals in 13C NMR, 3 signals in 1H NMR.
In 2-methyl butane, you will get 4 signals in 13C NMR, 4 signals in 1H NMR.
Degree of unsaturation= (2C+2 - H)/2= (2×5+2-12)/2=0
This compound has 0 degrees of unsaturation and has the general formula of alkanes. So, it has to be any one of the isomers of pentane.
It gives one signal in 1H NMR. So, all the hydrogens present are in equivalent positions(surrounded by same environment). Two signals in 13C NMR shows that there are two kinds of carbons that are not equivalent. If you draw the structures of all the isomers of pentane and check, then only 2,2-dimethylpropane will fit.
In n-pentane you will get 3 signals in 13C NMR, 3 signals in 1H NMR.
In 2-methyl butane, you will get 4 signals in 13C NMR, 4 signals in 1H NMR.
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