7. (10 pts) Given the IR and NMR spectra of Compound X...

Transcript

00:01 Hi now we will discuss about what is the acceptable structure for d6 h10o here we have provided with ir and nmrdta from ir spectrum we have all the stretching frequencies for the bonds stretching wavelengths are given 2 ,995 2 ,895 cm inverse the stretching is observed this is responsible for the th h stretch th h stretch and this is so for alkali group because of the presence of the alkyl group we observed the air spectrum at 3 ,300 centimeter inverse we have a stretching and this is responsible for the oh h stretch o bond h stretch this is because of the alcohol group and also we have observed at 2 ,240 centimeter inverse also stretching is observed it is responsible because of the c triple bonds alkyne group so these groups are present in the compound from the ir spectrum now we will see the nmr data from nmr data we have to here we have the one two three four four type four signals are there so four types of hydrogen are there if we predict the four types of hydrogen based upon this information i are data we may have like one triple bond one alcohol and alkyl group so here proton a proton b proton here also protons and here proton d four types of four types of protons we can observe the proton nmr signals now we will see the proton nmr signals for proton a we know that we can find the multiplicity multiplicity can bound by using 2 n to i plus here i is equal to half n equal to number of proton so for each type of proton we have to find multiplicity for proton a it is 2 to 5 half plus 1 is equal to 6 so here the multiplicities form with one proton and the signal is at 3 .90 ppm .m.

03:01 Now we will see for proton b.

03:05 For proton b, 2 into 0 half plus 1 gives as to 1.

03:12 That makes it singlet.

03:14 It is singlet with 1 proton.

03:18 Signal is at 3 .88 ppm.

03:24 Now we will see for protons say 2 into 1 half plus 1.

03:31 It is a 2 to doublet with 2 protons and signal is at 2 .30 ppm now for proton b 2 into 0 0 1 is equal to 1 it is a singlet the signal is singlet with 3 proton and the the cell is at 1 .80 of for proton 2 into 1 .80 now for proton 2 into 2 to 1 half plus 1 2...

Question

7. (10 pts) Given the IR and NMR spectra of Compound X (C6H10O), propose an acceptable structure. 3300 2995, 2895 2240 ignore 3H, s 3H, d 2H, d 1H, m 1H, b ppm

Question

7. (10 pts) Given the IR and NMR spectra of Compound X (C6H10O), propose an acceptable structure. 3300 2995, 2895 2240 ignore 3H, s 3H, d 2H, d 1H, m 1H, b ppm

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