Chem 2114 Unit 1 Worksheet - answers PDF

Preview

Summary

Download Chem 2114 Unit 1 Worksheet - answers PDF

Description

CHEM 2114 - Vahlberg

Unit 1 Worksheet Answers

The worksheet is a supplement for the homework. Doing only the worksheet problems will not give you enough practice to do well on the test. Do the homework first, THEN do the worksheet. 1. a) Write the equation for the reaction between methylamine (CH3NH2) and an aqueous solution of hydrochloric acid. Methylamine is a base, so this is an acid-base reaction. The proton moves from the acid to the base. CH3NH2 (aq) + HCl(aq)  CH3NH3+ (aq) + Cl-(aq) b) Now, draw the Lewis structure of each reactant and use curved arrows to show the movement of electrons in the reaction.

c) Write the formula of the conjugate acid of methylamine. CH3NH3+ d) Write the formula of the conjugate base of methylamine. CH3NHe) Methylamine is a 1° / 2° / 3° amine. It is a primary 1° amine. f) Identify all polar bonds in the two reactants and the electronegative atom in each. C-N, N-H (there are 2), and H-Cl N and Cl are the more electronegative atoms in the bonds. 2. Write and balance the equation for the reaction between acetylene and sodium amide. Sodium amide is a 1° / 2° / 3° / does not apply amide. The clue to the type of reaction lies in the fact that sodium amide is a REALLY strong base. This is an acid-base reaction. Clearly, acetylene is the acid, and both hydrogens are equivalent. H-C≡C-H + NaNH2  H-C≡C- Na+ + NH3 Sodium amide is an ionic compound containing the amide ion. The amide ion is not the amide functional group, so the 1° / 2° / 3° classification does not apply. 3. a) Draw the Lewis structure for nitromethane, CH3NO2.

b) Draw the Lewis structure for the conjugate base of nitromethane and any resonance structures. Use curved arrows to show the electron pair movement among resonance structures.

c) Compare the pKa’s of nitromethane and methane: which is the stronger acid? pKa of nitromethane is 10.2 (about the same as phenol) and pKa of methane is 50, so nitromethane is the stronger acid. d) Explain why nitromethane is able to act as an acid. The loss of a proton results in a conjugate base stabilized by resonance. 4. a) Write the full electronic configurations for carbon, oxygen, chlorine, and hydrogen. C: 1s22s22p2 O: 1s22s22p4 Cl: 1s22s22p63s23p5 H: 1s1 b) Then identify the number of valence electrons each has. For main group elements, the number of valence electrons is the number of s and p electrons in the highest shell, so C has 4 valence electrons, O 6, Cl 7, and H 1. These are also the last digit in their respective group numbers. 5. 2Na(s) + Cl2(g)  2NaCl a) What is the physical state of NaCl at room temperature? solid b) The physical state is dictated by the type of bond present in NaCl. What type of bond is present in NaCl? ionic (Ionic compounds are characterized by high melting points.) c) What type of bond is present in Cl2? How do you know this? 100% covalent. A polar covalent bond would result if there were a significant difference in electronegativities of the two elements in the bond. Since both elements are the same, there is no difference in electronegativity. d) Use your knowledge of intermolecular forces to explain why NaCl is soluble in water. Water is polar. Ionic compounds dissolve in water because ion-dipole attractions between the ions and the water molecules are even stronger than the hydrogen bonding that holds water molecules together. e) Use your knowledge of intermolecular forces to explain why NaCl is insoluble in hexane, C6H14. Hexane is nonpolar and held together by dispersion forces. There is no appreciable attraction between the nonpolar hexane and the ions of NaCl, therefore there is no driving force to break up the NaCl crystal. 6. For CO2, SO2, H2O, and XeF2: a) Draw the Lewis structure

b) Give the hybridization on the central atom C has two electron domains, so the hybridization is sp. S has three electron domains, so the hybridization is sp2. O has four electron domains, so the hybridization is sp3. Xe has five electron domains, so the hybridization is sp3d. c) Give the bond angle The hybridization dictates the bond angle sp means a bond angle of 180°. sp2 means a bond angle of 120°. sp3 means a bond angle of 109.5°. sp3d means a bond angle of 180°, since the nonbonding electrons go to the equatorial positions of the trigonal bipyramid. d) Show the dipole moments, overall dipole moment, and identify the molecule as polar or nonpolar. Polar molecules have polar bonds arranged such that the dipole moments do not cancel. All four molecules have polar bonds. Bond dipole moments are in blue. The overall dipole moment, if nonzero, is red. CO2 : The dipole moments cancel because the molecule is linear, so CO2 is nonpolar.

SO2 : The dipole moments do not cancel because the molecule is nonlinear, so SO2 is polar.

H2O: The dipole moments do not cancel because the molecule is nonlinear, so H2O is polar.

XeF2: The dipole moments cancel because the molecule is linear, so XeF2 is nonpolar.

e) Name the intermolecular forces present in each. CO2 has dispersion attractions. SO2 has dispersion attractions and dipole-dipole attractions. H2O has dispersion attractions, dipole-dipole attractions, and hydrogen bonding. XeF2 has dispersion attractions.

7. For the following compounds, locate and identify the functional groups.

8. a) For the following compound, give all bond angles and hybridizations.

O CH 3C CCH 2CCH 2CH(OH)CH 2CHO From left to right, 109.5° and sp3, then 180° and sp for the next two C’s, then 109.5° and sp3, then 120° and sp2, 109.5° and sp3 for the next 3 C’s and the O in the hydroxyl group, and 120° and sp2 for the final C.

b) Draw the 3-D structure of the compound.

O

HO H

O H

H HH

HH

HH

HH

c) Which bonds are polar? The two carbonyls and the CO and OH bonds in the alcohol. 9. Arrange in order of increasing acid strength: HCO3-, HSO4-, CH3CHO, CH3CO2H, C6H5OH, CH3OH, CH3C≡CH, CH4, NH3

CH4, NH3, CH3C≡CH, CH3CHO, CH3OH, (HCO3- ≈ C6H5OH), CH3CO2H, HSO4-...