Question

Question 26 1 pts 1. a) For the following benzene compound: CH3-C.CH4-CH3 (para), the total number of non- equivalent hydrogens is 2: True/False True False Question 27 1 pts 1. b) CH3-C6H4-Cl (ortho), the total number of non-equivalent hydrogens is 3: True/False True False Question 28 1 pts 1. c) olefinic hydrogens (hydrogen attached to sp² hybridized carbon) give d values at about 3-4 ppm : True/False True False

Answer 1

summery: two methyl group having same Chemical envirnoment hence these having one group of non equivalent proton. And then another 4 Hydrogen on aromatic ring having also same chemical envrionoment hence there is 2 group of non equivalnent proton.

Question 26) for the following benzene compound CH3-CHE CHA (para) the total number of Non-equivalent Hydrogen is 2. to true o False Explination :- structure of given compound che DH THO who trºi Here an Ha type of proton of two methyl 9804 Shows ap 6H singlet which are non.eguivalent. Hur în Aromatic ging four pooton Shows 4H singlet Haring 7.20 8ppm chemical shift value.

summery: here All the proton having different chemical environment. Hense they split differntally and total number of non equivalent proton is 5.

- 9. 27) ethich-cl (ortho) the number of non- equivalent Hodoo qen is 3. o true to false . Explination:- stoucture of given compound. Ho a @ crka @ HY CL Here all the proton Having diffreut Chemical Envirnoment Hence total number of non-eqwvalent proton is 5.

05.128 Olefinc Hyd709en CH3d209en attached to sp? Hybridized carbon) given S value as about 3-40? o toue o false Ex plination :. Here amy Olefinic compound we have to consider like Ethene.(SHO) Here on 4 Hydounen give an sing let at 6.98 ppm to 7.9 Sppro this is an range of spa Hybridized carbon atom chemical shift but alue. proton are more deshielded.

Summery: as proton on Sp2 Hybridization get deshileded on applyed magnetic field hence it get higher chemical shift value in between 6 to 7.9 as range.

8.29) In a 300 MHz NMR 1.58 value is 4 50 Hz to true o false Explination:- Giren. Instrumenteel frequency = 300 MHz chemical shift value = 1.5 ppm Formwo for calcevaring an chemical Shift value is s= Frequency Cha) Instrumental frequency in MHZ i 1.5 = Re 300 MHZ 1.5300 = 2 i [23-450 Hz therefore the frequency is at 450 Hz. due to peat

Summery: its is an very simple to calculate an any things having this Formula.Summery : Sp2 hybridized carbon atom an aromatic ring having chemical shift value are at 7.25 ppm

are 8.30xd value for benzene Hydrogen at around 5-6 otore Lo False o False Explination:- Here all the proton GH sing let Having Hydrogen at 7.25 8ppm. Hydrogen attached to carbon are spa Hybridized Hence are get deshielded.

- 8.31) pattern of cth is quartet (9). CH-CH=CH-CHiCl : splitting otoue o false Explination: H H i-cl 4-5-5-5-5-01 3HCtriplet) 0.98ppm the three pooron on methyl couple with adjecent ch. get toiplet by Cht) rule. ore we can done by vain pascal triangle get intensity ratio.

Summery: Here an the methyl group 3 Hydrogen get couple with adjesent 2H get split into triplet.