In: Chemistry
Indicate whether the statements below are CORRECT or INCORRECT. ONLY pick correct for statements that are COMPLETELY correct. For statements that are partially correct or sometimes correct but not always, explain your answer in ONE sentence:
(a) In 13C DEPT NMR spectra, a carbon in a methylene (CH2) group always has the same phase as one in a quaternary group.
(b) In 1H NMR spectra, two diastereomers will give different spectra.
(c) The 1H NMR spectrum of 3-methoxypentane consists of 4 signals including a singlet, a triplet and two pentets.
(d) The 1H and 13C NMR spectra of 1-bromo-4-methylbenzene consists of 4 and 6 aromatic signals, respectively.
(e) If a reaction is exothermic and has a positive entropy change, then the equilibrium constant will be greater than 1.
So far, I have incorrect, sometimes correct, incorrect (except for that it is not called a pentet), incorrect, correct. Can I please have some help on these? Thank you!!
(A) A carbon in a methylene (CH2) group will not have the same phase as one in a quaternary group. so, False
(B) In 1H NMR spectra, two diastereomers will give different spectra. Diastereoisomers are, essentially, different compounds. They have different physical properties (solubility, m.p., b.p., NMR, IR spectra etc.) so, True
(C) The 1H NMR spectrum of 3-methoxypentane consists of 4 signals including a singlet, a triplet and two pentets. True
(D) The 1H and 13C NMR spectra of 1-bromo-4-methylbenzene consists of 2 aromatic signals and 1 methyl (-Ch3) signal, respectively. so False
(E) Enthalpy of exothermic reaction ΔH is nagative, ΔH =
-ve , ΔS=+ve
ΔG = ΔH - TΔS = -ve so the reaction is favourable
hence If a reaction is exothermic and has a positive
entropy change, then the equilibrium constant will be greater than
1.
So, True