For the reaction of 2-pentanol with NaBr/ H2SO4 that produces 2-Bromopentane and

Question

For the reaction of 2-pentanol with NaBr/ H2SO4 that produces 2-Bromopentane and 3-Bromopentane.

Can NMR distinguish between two compounds in a mixture? If so, why? Which isomer of SN1 reaction is the major product based on the NMR spectra? Explain whether or not it makes logical sense.

4.0 3.5 3,0 2.5 2.0 0.5 nt Paran Curr NAME SXAND PROCNO F2 on Parang rs ACO 20161202 Date Time NSTRUM 5 m Mu nu PROAHD 2g30 PULPFOG 65538 SOLVENT COC13 SMH 8273.146 H CAES 0.1263 3.9504243 sec 60 400 use: 6.0 see 300.0K 00000000 sec OHANNEL fi .ca...... NUC1 6.50 USec 0.00 dB SFO 400.13247 MHz F2 Processing paraneters 310 400.1300000 MHz NOH SS8 0.00 Hz. a NMR plot parameters CX 20.00 cm 5.000 2000.65 Hz F2P 0 000 pon F2 0.00 Hz PpHCN 0.25000 00.032

Explanation / Answer

The two SN1 products can be distinguished by 1H NMR. The chemical shift of CH proton in 2-bromopentane should be higher with greater multiplicity than the chemicha shift of CH proton in 3-bromopentane. So, the signal at 4.2 ppm) for the CH of 2-bromopentane while that at 4.0 ppm is for the CH of 3-bromopentane. Based on the intensity of these charcteristic signals we can find out which one is the major. The higher the intensity the higher will be the percentage. Since the intensity of the signal at 4.2 ppm is higher than the intensity of the signal at 4.0 ppm. This shows that 2-bromopentane is the major SN1 product.

We can calculate pertcentage of 2-bromopentane.

% of 2-bromopentane = 100 x intencity of CH at 4.2 ppm/ (intensity of CH at 4.2 ppm + intensity of CH at 4.0 ppm)

= 100 x 1/ (1.000 + 0.2976) = 77 %

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