Question

A hydrocarbon A has the molecular formula C8H14. Its NMR spectrum consists of a triplet at 1.2 ppm (relative peak area of 2), a triplet at 1.9 ppm (relative peak area of 2) and a singlet at 2.0 ppm (relative peak area of 3). When compound A is treated with Br2 in CCl4 (no light), a new compound B is formed with a molecular formula of C8H14Br2. The NMR spectrum of B looks the same as that of A, except that the positions of the triplets and the singlet are all shifted downfield. When compound B is treated with H2O, compound C is formed slowly (molecular formula C8H16O2). The NMR spectrum of C consists of a triplet at 2.1 ppm (relative peak area of 2), a triplet at 3.4 ppm (relative peak area of 2), a singlet at 3.5 ppm (relative area of 3) and a broad singlet at 4.9 ppm (relative area of 1). The singlet at 4.9 ppm disappears after the compound has been shaken with D2O. When compound B is treated with hot KOH, a new hydrocarbon D is formed as a major product, along with some minor geometric isomers. Compound D has a molecular formula of C8H12. The NMR spectrum of compound D consists of a doublet at 2.0 ppm (relative peak area of 2) which is coupled to a triplet at 4.2 ppm (relative peak area of 1) and a singlet at 2.1 ppm (relative peak area of 3).  

Write clear structural formulas for compounds A, B, C and D which are consistent with these observations and briefly explain your reasoning.

Answer 1

1) For (A) M.F. C8H14

i) Unsaturation sites: General formula for saturated hydrocarbon is CnH2n+2 and with n=8 we will have C8H18 as the formula. A is short by 4 H i.e. there are 2 unsaturation sites. These unsaturation sites can account for 2 double bond or 1 double + 1 ring or 1 triple bond.

ii) PMR data and analysis: 3 PMR signals with peak integration(area) ratio as 2:2:3 for A ,it means 3 kinds of protons are present in A. Total integration 2 mm +2mm +3mm = 7mm total number of H is exactly double but hence 2 mm = 1 H. so,

(1.2, t, 2mm = 4H) indicates 2 –CH2-groups,

(1.9, t, 2mm = 4 H) indicates 2 –CH2- groups,

(2.0, s, 3mm = 6H) indicate 2 –CH3 groups,

No peak in the region 4-6ppm of olefinic protons so no olefinic proton present and all are sp3C-H protons may be allylic

Combining all pieces we can propose follows structure 1,2-Dimethylcyclohexene.

2) Given that A on Br2 in CCl4 (in absence of light) gives B with M.F. C8H14Br2.

Also given that PMR for B is having same patter as that of A only there is downfield shift for all signals.

For B M.F. C8H14Br2

i) Unsaturation sites: M.F. C8H14Br2 corresponding hydrocarbon will be C8H16 (Compensation of 1 H per Br removal). Still there is deficiency of 2H i.e. there is 1 unsaturation site.It account for ring.

Hence structure proposed is 1,2-Dibromo-1,2-dimethylcyclohexane.

3) B on treatment with H2O slowly forms C with M.F. C8H16O2.

i) Unsaturation sites: M.F. C8H16O2 hence corresponding hydrocarbon C8H16 (no compensation for O removal). Still deficiency of 2H i.e. 1 unsaturation site present it counts for ring

ii) PMR data and analysis:

(2.1, t, 2mm = 4H) indicate 2-CH2-

(3.4, t, 2mm = 4H) indicate 2 –CH-

(3.5, s, 3mm = 6H) indicate 2 –CH3

(4.9, s, 1mm = 2H exchangeable with D2O) indicate 2 –OH groups.

Hence the structure proposed is, 1,2-Dimethylcyclohexane-1,2-diol.

4) B on treatment with hot KOH gives 2 hydrocarbons.

D is the major one with M.F. C8H12

For D,

i) Unsaturation sites: M.F. C8H12 but corresponding saturated hydrocarbon formula will be C8H18 this means D is short by 6H i.e. 3 unsaturation sites. These unsaturation sites account for 2 C=C and 1 ring certainly as a bromo alkane undergo elimination reaction with KOH under hot conditions.

ii)PMR data and analysis:

(2.0, d, 2mm = 4H) indicate 2 allylic –CH2-

(4.2,t, 1mm = 2) indicate 2 olefinic spC-H i.e. C=C-H

(2.1, s, 3mm = 6H) indicate 2 allylic –CH3 groups

Hence structure proposed is 2,3-Dimethylcyclohexa-1,3-diene.

Geometrical Isomer of D say D’is also shown in image.

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