Chapter 1 Homework E.O.Chem PDF

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Manmeet Kaur Element Organic Chemistry Chapter 1 Homework: 32-37, 44 ,49, 54

32. Use$the$relative$positions$of$the$elements$in$the$periodic$table$(Table$1.3$or$inside$back$cover)$ to$classify$the$following$substances$as$ionic$or$covalent:$

a. F2 : covalent

b. O2: covalent

c. NaBr: ionic

d. SiF4: covalent

e. CH4: covalent

f. CaCl2: ionic

g. NaI: ionic

h. ClF: covalent

33. When$a$solution$of$salt$(sodium$chloride)$in$water$is$treated$with$a$silver$nitrate$solution,$a$ white$precipitate$form$immediately.$When$tetrachloromethane$is$shaken$with$aqueous$silver$ nitrate,$no$such$precipitate$is$produced.$Explain$these$facts$in$terms$of$the$types$of$bonds$present$in$ the$two$chlorides.$$ Sodium$chloride$is$an$ionic$compound,$which$is$made$of$Na+$and$Cl-$ions.$When$NaCl$is$dissolved$in$ water,$ions$are$separated.$Now$Cl-$reacts$with$silver$nitrate$to$form$white$precipitate$of$silver$ chloride.$Cl-$(aq)$+$AgNO3$(aq)$à$AgCl$$$$$$$(s)$+$NO3-$(aq).$Tetrachloromethane$is$a$covalent$ molecule$where$no$Cl-$ion$is$free$to$move$in$solution.$Therefore,$it$does$not$give$any$precipitate$of$ silver$chloride$when$shaken$with$silver$nitrate$solution$$

34. For$each$of$the$following$elements,$determine$(1)$how$many$valence$electrons$it$has$and$(2)$ what$its$common$valence$is:$ a.$N:$5$ve$

$b.$C$:$4$ve$

c:$F:$1$ve$

d:$O:$6$ve$$

e:$P:$5$ve.$$$$f.$S:$6$ve$

35.$Write$a$structural$formula$for$each$of$the$following$compounds,$using$a$line$to$represent$each$ single$bond$and$dots$for$any$unshared$electron$pairs:$ a.#CH3OH$$$

$b.#CH3CH2Cl$ $ $ $c.#C3H8$z$

d.#CH3CH2NH2$$ # # # e.#C2H5F$

$ $ $ $f.#CH2O$

36. Draw$a$structural$formula$for$each$of$the$following$covalent$molecules.$Which$bonds$are$polar?$ Indicate$the$polarity$by$proper$placement$of$the$symbols$d1$and$d2.$ a.#BF3$

$

$

$

$

$

#b.#CH3F$

# $

#

#

Fluorine is more electronegative than boron. So, the electrons of B-F bond will be shifted towards fluorine. So, fluorine has partial negative charge and boron has partial positive charge. Therefore, B-F bonds are polar. # # c.#CO2$$

$

$

$

$

Fluorine is more electronegative than carbon. So, the electrons of C-F bond will be shifted towards fluorine. So, fluorine has partial negative charge and carbon has partial positive charge. Therefore, C-F bond is polar. Electronegatively difference between carbon and hydrogen is not very high. Therefore, C-H bond is not so polar.

$

$

d.#Cl2$#

# #

Oxygen is more electronegative than carbon. So, the# electrons of C-O bond will be shifted towards oxygen. So, oxygen has partial negative charge and carbon has partial positive charge. Therefore, C-O bond is polar.

The molecule Cl2 is a homonuclear molecule. So, difference in electronegativity between two chlorine atoms is zero. Therefore, it is a nonpolar molecule.

e.#SF6$$

$

$

$

$

$

$

f.#CH4$$

$

# # #

Fluorine is more electronegative than sulfur. So, # # the electrons of S-F bond will be shifted towards # # fluorine. So, Fluorine has partial negative charge and sulfur has partial positive charge. Therefore, # S-F bonds are polar.

Carbon is more electronegative than hydrogen. So, the electrons of C-H bond will be shifted towards carbon. So, carbon has partial negative charge and hydrogen has partial positive charge. Therefore, C-H bonds are polar.

# #g.#SO2$

Oxygen is more electronegative than sulfur. So, the electrons of S-O will be shifted towards oxygen. So, oxygen has partial negative charge and sulfur has partial positive charge. Therefore, S-O bond is polar.

$

h.#CH3OH$

Electronegativity difference between carbon and hydrogen is not very high. Therefore, C-H bond is not so polar. Oxygen is more electronegative than carbon. Therefore, the electrons of C-O bond will be shifted towards oxygen. So, oxygen has partial negative charge and carbon has partial positive. Oxygen is more electronegative than hydrogen, therefore, electron of O-H shift towards oxygen. And C-O bond and O-H bonds are polar.

37. Consider$the$X!H$bond,$in$which$X$is$an$atom$other$than$H.$The$H$in$a$polar$bond$is$more$acidic$ (more$easily$removed)$than$the$H$in$a$nonpolar$bond.$Considering$bond$polarity,$which$ hydrogen$in$acetic$acid,$ $ CH3C$OOH,$do$you$expect$to$be$most$acidic?$Write$an$equation$for$the$reaction$between$acetic$acid$

and$potassium$carbonate.$$

When acetic acid is reacted with potassium carbonate the acidic hydrogen of acetic acid will be replaced by potassium.

44. What$is$the$molecular)formula)for$each$of$the$following$compounds?$Consult$Figures$1.13$and$ 1.14$for$the$abbreviated$structural$formulas.$ a.#nicotine$:$C10H14N2$ b.#adenine:$C5H5N5$ c.#limonene:$C10H16$

$d.#coumarin:$C9H6O2$ $e.#benzene$:$C6H6

49. #Add$curved$arrows$to$the$following$structures$to$show$how$electron$pairs$must$be$moved$to$ interconvert$the$structures,$and$locate$any$formal$charges.$

50. Add$curved$arrows$to$show$how$electrons$must$move$to$form$the$product$from$the$reactants$in$ the$following$equation,$and$locate$any$formal$charges.$

54. Use$lines,$dashed$wedges,$and$solid$wedges$to$show$the$geometry$of$CF4$and$CH3SH.$...