A hydrocarbon A has the molecular formula C8H14. Its NMRspectrum consists of a triplet at 1.2 ppm (relative peak area of2), a triplet at 1.9 ppm (relative peak area of 2) and a singlet at2.0 ppm (relative peak area of 3). When compound A is treated withBr2 in CCl4 (no light), a new compound B is formed with a molecularformula of C8H14Br2. The NMR spectrum of B looks the same as thatof A, except that the positions of the triplets and the singlet areall shifted downfield. When compound B is treated with H2O,compound C is formed slowly (molecular formula C8H16O2). The NMRspectrum of C consists of a triplet at 2.1 ppm (relative peak areaof 2), a triplet at 3.4 ppm (relative peak area of 2), a singlet at3.5 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 thecompound has been shaken with D2O. When compound B is treated withhot KOH, a new hydrocarbon D is formed as a major product, alongwith some minor geometric isomers. Compound D has a molecularformula of C8H12. The NMR spectrum of compound D consists of adoublet at 2.0 ppm (relative peak area of 2) which is coupled to atriplet at 4.2 ppm (relative peak area of 1) and a singlet at 2.1ppm (relative peak area of 3). Â
Write clear structural formulas for compounds A, B, C and D whichare consistent with these observations and briefly explain yourreasoning.
