Solutions to test set of exam exercises E2017 T2 in General and organic chemistry PDF

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Test set of exam exercises in General and organic chemistry, E2017 (T2) General chemistry 1) a) Which has the more stable 2s orbital, a potassium atom or a K+ cation? Since The 2s orbital is not affected by the reaction of K-atom to the K-cation but the 4s orbital. Therefore have the same stable 2s orbital. b) What would be the three next orbitals to fill after the 3d orbital?

After filling the 3d orbitals the 4p, then 5s and 4d-orbitals are filled (se figur 4.38, s. 146)

c) Arrange the following molecules in order of increasing bond polarity: HF, H2S and SbH3 Compare the electronegativities (figure 5.5, p. 174): - F(4.0) and H(2.1), difference HF = 1.9 - H(2.1)and S(1.8), difference H2S = 0.3 - Sb(1.9) and H(2.1), difference SbH3 = 0.2 The ranking: HF > H2S > SbH3

2) Determine the Lewis structures of NH3 and BF4-. Use the procedure given on pp. 177–8 of the text to determine Lewis structures.

NH3 1) 2) 3) 4)

There are 8 valence electrons. H-atoms are bonded to N and each N—H bond requires two electrons. There are no outer atoms other than H. Place remaining electrons on the N atom: 5) Optimise electron configurations of the inner atoms: N has an octet in each structure, so it is the optimal structure (Formal charge)

BF41) There are 3 + (4 x 7) + 1 = 32 valence electrons. 2) F-atoms are bonded to B and each B-F bond requires two electrons (=8 electrons, leaving 24 valence electrons). 3) Place 3 electron pairs around each outer F atom, leaving 24 – (4 x 6) = 0 electrons 4) Both F and B have an octet, so it is the correct Lewis structure.

F F B F F

3) Balance the following two equations: (a) in acidic aqueous solution and ( b) in basic aqueous solution: a) C2O42- + HNO2 → CO2 + NO C2O42– 2CO2 + 2e– (HNO2 + H+ + e– NO + H2O) 2 Net: C2O42– + 2HNO2 + 2H+ 2CO2 + 2NO + 2H2O

b) NiO2 + Mn(OH)2 → Mn2O3 + Ni(OH)2 NiO2 + 2H+ + 2e–  Ni(OH)2 2Mn(OH)2  Mn2O3 + H2O + 2H+ + 2e– NiO2 + 2Mn(OH)2  Ni(OH)2 + Mn2O3 + H2O

4) Propose a structural formula for each of the following compounds: a) C5H11Br, 2 signal in 1H-NMR 11−11 2

= 0, indicative of an alkane structure

-

IHD ( CnH2n+1) =

-

2 signals is indicative of two sets of equivalent H -atoms

All three methyl groups are equivalent:  1.1

CH3

CH3 3.2 C CH2Br CH3

b) C4H8O, δ=1.0(triplet,3H), δ=2.1(singlet, 3H), δ=2.4 (quartet, 2H)   

Index of hydrogen deficiency is 1, indicating a carbonyl group. The triplet (3 H) and quartet (2 H) are diagnostic of an ethyl group. The singlet at  2.1 (3 H) is typical of a methyl group bonded to a carbonyl group.

O These data correspond to butanone:

5) Based on the following IR spectrum and mass spectrum of an unknown compound, propose at least one possible structure. -

-

The IR spectrum indicates that that the compound is a ketone. The mass spectrum indicates a molecular weight of 86. We can calculate a molecular formulae, i.e. 86 – 16 (O-atom molecular weight) = 70. Trying to identify no. of C-atoms by dividing 70 with 12 (C molecular weight) = 5.8. This suggest 5 carbon atoms, leaving 10 mass unit for H-atoms. This is indicative of a molecular formulae, C5H10O The peak at m/z=43, indicate the loss of a propyl group. The compound likely has a three carbon chain (either a propyl group or isopropyl group) on one side of the ketone:

Annex 1. IR spectrum and mass spectrum of an unknown compound....