NMR questions
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Eicosane (C20H42) can be cracked by heating to 700 K in the presence of a catalyst. The products are * an aromatic hydrocarbon C8H10 * an alkane C6H14 * another alkane. C20H42 ⟶ C8H10 + C6H14 + _______ Complete the equation for this reaction.
3 C2H6
A sample of butane has a volume of 20 cm3 at room temperature and pressure. The sample is burned completely in 1350 cm3 of air. The final mixture is cooled to room temperature and pressure. C4H10 + 6 O2 ⟶ 4 CO2 + 5 H2O Calculate the total volume of gas in the final mixture. Assume that air contains 21% by volume of oxygen.
M1 Initial volume O2 = 0.21 × 1350 = 283.5 (cm3) Alternative route: M1 Vol Air decreases by 6.5 × 20 = 130 cm3 M2 Volume of O2 remaining = M1 - (6.5 × 20) = 153.5 cm3 M2 = 1220 cm3 M3 Volume of CO2 formed = 20 × 4 = 80 cm3 M3 Vol CO2 produced = 4 × 20 = 80 cm3 M4 Total volume of gas left = M2 + M3 + (0.79 × 1350) = 1300 cm3 M4 Total Vol Air + CO2 = 1220 + 80 = 1300 cm3
4. This question is about compound X with the empirical formula C2H4O Figure 1 shows the infrared spectrum of X. Figure 2 shows the 13C NMR spectrum of X. The 1H NMR spectrum of X shows four peaks with different chemical shift values. The table below gives data for these peaks. Show how information from Figure 1, Figure 2 and the table can be used to deduce the structure of compound X. Just read it beacsue it will show you have to answer querstions like this
Stage 1: infrared 1a) (broad peak) at 3400 cm–1 (any value from 3230-3550) indicates OH in alcohols 1b) peak at 1720 cm–1 (any value from 1680-1750) indicates C=O Stage 2: 1H nmr 2a) peak at 3.9 ppm integration 1 so 1 H-C-O AND quartet so adjacent to CH3 (stated or shown) 2b) peak at 3.7 ppm integration 1 so HO-C-(stated or shown) 2c) peak at 2.1 ppm integration 3 so H3C-C=O AND singlet so no adjacent H (stated or shown) 2d) peak at 1.2 ppm integration 3 so H3C- AND doublet so adjacent to CH (stated or shown) 2e) sum of integration values = 8 Hence C4H8O2 Stage 3: 13C nmr 3a) peak at 210 ppm C=O aldehydes or ketones 3b) peak at 75 ppm C−O (alcohols, ethers or esters) 3c) peak at 25 ppm 3d) peak at 20 ppm 3e) structure
Suggest why it would be difficult to determine the structure of Z using the spectrum in Figure 3, without the information in the table in part (d). Legit just hsows a specturm with chemical shift and nothing else
(f) Cannot deduce splitting patterns of peaks (at about ẟ = 2.60) Or No integration values
Explain what can be deduced from the splitting patterns and chemical shift values for the peaks at ẟ = 4.10 ppm and at ẟ = 1.26 ppm Deduce the part of the structure of Z that causes the peaks at ẟ = 4.10 ppm and ẟ = 1.26 ppm Use Table B in the Data Booklet to help you answer this question.
M1 (Quartet) because neighbouring C has 3H M2 (At ẟ = 4.1 ppm) because connected to single bonded O of ester or M3 (Triplet) because neighbouring C has 2H M4 (At ẟ = 1.26 ppm) because R2CH2 or RCH3
How many different carbon environments are there in a molecule of Z? there are 6 peaks but two of thwm are ver clsoe toegther (You cnounted this as a isngle peak but it was 2) so you got this q wrong the answer was 7
In c NMR even if theres two peaks near the same spot they atre sperare peaks
Explain why the peaks in the 1H NMR spectrum are singlets
Adjacent C has no (non-equivalent) H attached (so no splitting/spin-spin coupling takes place)