Question

The 1^H NMR spectrum below corresponds to a hydrocarbon with the molecular formula C_6H_14. Deduce the structure of the molecule from the data below.

Answer 1

ANS.

The given compound is C₆H_14. It is an alkane as it follows the C_nH_2n+2 structural formula of alkanes.So the given compound is a HEXANE.

Now, hexane has 5 isomers . So the possible structures are:

• Now n-hexane has 2 sest of equivalent hydrogens. One set has 6 chemically equivalent Hs of CH₃ groups and another set has 6 equivalent hydrogens of CH₂ groups. NMR will show 2 peaks.
• 2-methylpentane has 5 sets of equivalent hydrogens, thus its NMR will show 5 peaks.
• 3-methylpentane has 4 sets 0f equivalent hydrogens, thus itsNMR will show 4 peaks.
• 2,3-dimethylbutane has 2 sets of equivalent hydrogens .One set has 12 hydrogens of methyl groups which have same chemical environment and another set will have 2 similar hydrogens of CH group. Its NMR will give 2 peaks of 12:2 ratio or 6:1.
• 2,2-dimethylbutane has 3 sets of equivalent hydrogens. Thus its NMR will have 3 peaks.

NOW, the given spectrum in question shows only two peaks of 12:2 ratio, so the compound referring to the given nmr is 2,3-dimethylbutane. Splitting of spectral line is spin spin coupling of both sets of equivalent hydrogens.

Hence the final answer is 2,3-dimethylbutane .