This preview has intentionally blurred parts. Sign up to view the full document

View Full Document

Unformatted Document Excerpt

the 2.1Give direction of the dipole moment (if any ) for each of the following molecules: (a) HF, Answer: (b) IBr, (c) Br2, (d) F2. H (a) F ; (b) I Br 2.2 Boron trifluoride (BF3) has no dipole moment (=0). Explain how this observation confirms the geometry of predicted by VSEPR theory. Register to View Answeris SP2-hybridized, so all atoms on the same plane, the center of positive charge and the center of negative charge coincide, and the molecule has no net dipole moment 2.3 Tetrachloroethene doesnt have a dipole moment. Explain this fact on the basis of the shape of Cl2C=CCl2. Answer: Cl Cl Cl Cl The shape of the tetrachloroethene is a planar, the vector sum of the dipole moment of each C-Cl bonds is zero. 2.4 Sulfur dioxide has a dipole moment, on the other had, carbon dioxide has no dipole moment. What do these facts indicate about the geometries of the two molecules? Answer: The fact that SO2 has a dipole moment indicates that the molecule is angular, not linear: S O O The fact that CO2 has no dipole moment indicates that its shape is linear, not angular: O C O 2.5 Using a three-dimensional formula, show the direction of the dipole moment of CH3OH. net dipole O H CH3 2.6 Trichloromethane (CHCl3, also called chloroform) has a larger dipole moment than CFCl3. Use three dimensional structures and bond moments to explain this fact. Answer: H C Cl Cl Cl Cl F C Cl Cl 2.7 Indicate the direction of the important bond moments in each of the following compounds (neglect C-H bonds). You should also give the direction of the net dipole moment for the molecule. If there is not net dipole moment, state that =0. (a) cis-CHF=CHF (b) trans-CHF=CHF Answer: (a) cis-CHF=CHF H C F C F H (c) CH2=CF2 (d)CF2=CH2 net dipole moment (b) trans-CHF=CHF H C F C H F (c) CH2=CF2 H C H C F F F =0. net dipole moment (d) CF2=CH2 F C F C F =0. 2.8 Write structural formulas for all of the alkenes (a) with the formula C2H2Br2 and (b) with the formulaC2Br2Cl2. In each instance designate compounds that are cis, trans isomers of each other. Predict the dipole moment of each one. Answer: (a) Br Br Br H Br H C H C H H C C Br =0 Br C C H cis-1,2-Dibromoethene (b) Br trans-1,2-Dibromoethene 1,1-Dibromoethene Br Br Cl C Cl C Cl Cl C C Br =0 cis-1,2-Dibromo-1,2-dichloroethene trans-1,2-Dibromo-1,2-dichloroethene Br Cl C Br C Cl 1,1-Dibromo-2,2-dichloroethene 2.9 Write structural formulas (a) for two constitutional isomeric primary alkyl bromides with the formula C4H9Br, (b) for a secondary alkyl bromide ,and (c) for a tertiary alkyl bromide with the same formula. Answer: Br (a) Br Br (b) Br (c) 2.10 Although we shall discuss the naming of organic compounds later when we discuss the individual families in detail, one method of naming alkyl halides is so straightforward that it is worth describing here. We simply name the alkyl group attached to the halogen and add the word fluoride, chloride, bromide, or iodide. Write formulas for (a) ethyl fluoride and (b) isopropyl chloride. What are names for (c) CH3CH2CH2Br, (d) CH3CHFCH3, and (e) C6H5I? (a) CH3CH2F (b) CH3CHCH3 Cl (c) propyl br omide (d) isopropyl fluoride (e) phenyl iodide 2.11 Write structural formulas for (a) two primary alcohols , tertiary alcohol all having the molecular formula C4H10O. Answer: (b) (a) CH3CH2CH2CH2OH & (CH3)2CHCH2OH; (b) a secondary alcohol, and (c) a CH3CH2CH(CH3)OH; (c) (CH3)3COH. 2.12 One way of naming alcohols is to name the alkyl group that is attached to the OH and add the word alcohol. Write the structures of (a) propyl alcohol and (b) isopropyl alcohol. Answer: propyl alcohol isopropyl alcohol CH3CH2CH2OH CH3CHOHCH3 2.13 Write structural formulas for the materials below. (a) diethyl ether CH3CH2OCH2CH3 (b) ethyl propyl ether CH3CH2OCH2CH2CH3 (c) ethyl isopropyl ether CH3CH2OCH(CH3)2 What name would you give to (d), (e) and (f)? (d) CH3OCH2CH2CH3 methyl propyl ether (e) (CH3)2CHOCH(CH3)2 diisopropyl ether (f) CH3OC6H5 methyl phenyl ether 2.14 One way of naming amines is to name in alphabetical order the alkyl groups attached to the nitrogen atom, using the prefixes di- and tri- if the groups are the same. An example is isopropylamine for (CH3)2CHNH2. Write formulas for (a) propylamine, (b) trimethylamine, and (c) ethylisopropylmethylamine. What are names for (d) CH3CH2CH2NHCH(CH3)2, (e) (CH3CH2CH2)3N, (f) C6H5NHCH3, (g) C6H5N(CH3)2. Answer: (a) (b) (c) CH3CH2CH2NH2 (CH3)3N H3CH2C N CH3 H C CH3 CH3 (d) (e) (f) (g) isopropylpropylamine tripropylamine methylphenylamine dimethylphenylamine 2.15 Which amines in Problem 2.14 are (a) primary amines, (b) secondary amines, and (c) tertiary amines? Answer: Among the amines in Problem 2.14, (a). (b). (c). propylamine CH3CH2CH2-NH2 is a primary amine; isopropylpropylamine CH3CH2CH2NHCH(CH3)2 and methylphenylamine C6H5NHCH3 are secondary amines; trimethylamine (CH3)3N,ethylisopropylmethylamine CH3(CH3CH2)NCH(CH3)2, tripropylamine (CH3CH2CH2)3N, and dimethylphenylamine C6H5N(CH3)2 are all tertiary amines. 2.16 Amines are like ammonia in being weal bases. They do this by using their unshared electron pair to accept a proton. (a) Show the reaction that would take place between trimethylamine and HCl. (b) What hybridization state would you expect for the nitrogen atom in the product of this reaction? (a) (CH3)3N + HCl [(CH3)3NH]+Cl(b) SP3 hybridization. 2.17 The compounds in each part below have the same (or similar) molecularweights. Which compound in each part would you expect to have the higher boiling point? Explain your answers. (a) CH3CH2CH2CH2OH or CH3CH2OCH2CH3. (b) (CH3)3N or CH3CH2NHCH3 (d) CH3CH2CH2CH2OH or HOCH2CH2CH2OH. Answer: In group (a) I think CH3CH2CH2CH2OH have the higher boiling point .In group (b) is CH3CH2NHCH3. In group (c) is HOCH2CH2CH2OH. The reason is that those compounds all have a strong hydrogen bonding which is limited to molecules having a hydrogen atom attached to an O, N, or F atom. 2.18 Which compound would you expect to have the higher melting point, propane or cyclopropane? Explain your answer? Answer: The cyclopropane has the higher melting point because it is more symmetrical and more compact. 2.19 Classify each of the following compounds as an alkane, alkene, alkyne, alcohol, aldehyde, amine, and so forth. (a) O (b) H 3C (c) OH C CH (d) O H (e) (f) OH CH3(CH2)7 (CH2 )12CH3 H H Answer: (a) (d) Ketone Aldehyde (b) (e) Alkyne Alcohol (c) (f) Alcohol Alkene 2.20 Identify all of the fuctional groups in each of the following compounds: (a) OH functional groups: HO- ,aromatic ring (b) O OH C CH2 C H2N O C N H H CH2 CH C O O CH3 O C O C O CH3 functional groups: OH (c) C- , -O- , -NH2, aromatic ring, H2 C H C CH3 NH2 functional groups: aromatic ring , (d) NH 2 HO functional groups:-OH ,double bond (e) O C OCH2CH2 N CH3 O functional groups: C (f) OCH2CH2 ,aromatic ring , O H O functional groups: double bond , H (g) O O O O O functional groups: O C ,double bond 2.21 There are four alkyl bromides with the formula C4H9Br. Write their structure formulas and classify each as to whether it is a primary, secondary or tertiary alkyl bromide. Br primary alkyl bromide secondary alkyl bromide Br Br primary alkyl bromide tertial alkyl bromide Br 2.22 There are seven isomeric compounds with the formula C4H10O. Write their structures and classify each compound according to its functional group. Answer: (a) OH Alcohol (b) OH Alcohol (c) OH Alcohol (f) O Ether (d) O Ether (e) O Ether (g) OH Alcohol 2.23 Write structural formulas for four compounds with the formula C3H6O and classify each according to its functional group. H H C H H C H C H H C H O H Alkene&Alcohol O C H C H H Aldehyde H C H C H H Alcohol C H O H H C H C H O H C H H Ether & Alkene 2.24 Classify the following alcohols as primary, secondary or tertiary (a) (CH3)3CCH2OH (b) CH3CH(OH)CH(CH3)2 (c) (CH3)2C(OH)CH2CH3 primary secondary tertiary OH (d) tertiary OH (e) secondary Classify the following amines as primary ,secondary,or tertiary: 2.25 (a) CH3NHCH(CH3)2 (b) CH3CH2CH(CH3)CH2NH2 (c) (CH3CH2)3N (d) (C6H5)2CHCH2NHCH3 (e) HN (f) N Answer: (b) is primary, (a), (d), (e) are secondary and (c), (f) are tertiary amine. 2.26 Write structural formulas for each of the following: (a) Three ethers with the formula C4H10O. (b) Three primary alcohols with the formula C4H8O. (c) A secondary alcohol with the formula C3H6O. (d) A tertiary alcohol with the formula C4H8O. (e) Two esters with the formula C3H6O2. (f) Four primary alkyl halides with the formula C5H11Br. (g) Three secondary alkyl halides with the formula C5H11Br. (h) A tertiary alkyl halide with the formula C5H11Br. (i) Three aldehydes with the formula C5H10O. (j) Three ketones with the formula C5H10O. (k) Two primary amines with the formula C3H9N. (l) A secondary amine with the formula C3H9N. (m) A tertiary amine with the formula C3H9N. (n) Two amides with the formula C2H5NO. Answer: (a) O O O (b) OH OH OH (c) OH (d) OH CH3 (e) O O O O (f) Br Br Br Br Br (g) Br Br Br (h) Br (i) O O O O (j) O O O (k) NH2 NH2 (l) N H (m) N (n) O NH2 NH O 2.27 Which compound in each of the following pairs would have the higher boiling point? Explain your answers. (a) CH3CH2CH2OH or CH3CH2OCH3 (b) CH3CH2CH2OH or HOCH2CH2OH O OH (c) O OH or (d) or NH N CH3 (e) F or F F (f) F or O O (g) OH or O (h) Hexane CH3(CH2)4CH3 or nonane CH3(CH2)7CH3 O (i) or Answer: (a) CH3CH2CH2OH because it has a hydrogen bonding. (b) HOCH2CH2OH because it has a stronger hydrogen bonding. (c) OH OH Because it has a hydrogen bonding. (d) Because it has hydrogen bonding. NH (e) F Because it has hydrogen bonding. F (f) Because it has a Dipole-Dipole Forces. O (g) OH Because it has a hydrogen bonding. (h) nonane CH3(CH2)7CH3 because it has larger molecular weight and size. O (i) Because it has a stronger van der Waals Forces. 2.28 Predict the key IR absorption bands whose presence would allow each compound in pairs a, c, d, e, g and i from Problem 2.27 to be distinguished from each other. (a) CH3CH2CH2OH or CH3CH2OCH3 O OH (c) O (d) or or OH (e) NH or N CH3 O O (g) OH or O (i) or O Answer: (a) The typical OH adsorption would be around 3200-3500 cm-1; (c) The typical ketone adsorption wpuld be around 1700 cm-1; (d) Same reason for (a); (e) The secondary amine would have the adsorption around 3300-3500 cm-1; while the tertiary amine wouldnt have the adsorption around this region; (g)The carboxylic acid would have the adsorption around 2500-3500 cm-1 due to the OH stretch; (i) The ketone would have a strong adsorption around 1700 cm-1. 2.30 Cyclic compounds of the general type shown here are called lactones. What functional group dose lactone contain? O O Answer: Ester group. 2.31 Hydrogen fluoride has a dipole moment of 1.82 D; its boiling point is 19.34C. Ethyl fluoride (CH3CH2F) has an almost identical dipole moment and has a larger molecular weight, yet its boiling point is -37.7C.Explain. Answer: The two molecules have almost the same dipole moment. That means they have almost the same London force. However, the major molecular interaction between hydrogen fluoride is the hydrogen bonding, it makes hydrogen fluoride a higher boiling point. 2.32 Which of the following solvents should be capable of dissolving ionic compounds? (a) liquid SO2 (b) liquid NH3 (c)benzene (d) CCl4 Register to View Answerand B are polar, and hence are capable of dissolving ionic compounds. 2.33 Write a three-dimensional formula for each of the following molecules using the wedge-dashed wedge-line formalism. If the molecule has a net dipole moment, you should so state. (You may ignore the small polarity of C-H bonds in working this and similar problems.) (a) CH3F (b) CH2F2 (c) CHF3 (d) CF4 (e) CH2FCl (f) BCl3 (g) BeF2 (h) CH3OCH3 (i) CH3OH (j) CH2O (a) F C H H (b) F H F C H H (c) H C F F (d) F F F C F F (e) F C Cl H (f) H Cl Cl Cl B (g) F Be F (h) H C H H H C O H H (i) H H O C H (j) O C H H H 2.35 Analyze the statement: For a molecule to be polar, the presence of polar bonds is necessary, but it is not a sufficient requirement. Answer: Its right. For a polar molecule polar bond is necessary, but its not sufficient. For instance CO2, it has polar bonds but it is not polar molecule. 2.36 Identify all of the functional groups in Crixivan (the structure of Crixivan is shown in the chapter-opening vignette). C6H5 N H N N H HN C(CH3)3 O O OH H H N OH H Answer: hydroxy, phenyl, carbonyl, amide, amine. 2.37 The IR spectrum of propanoic acid indicates that the absorption for the O-H stretch of the carboxylic acid functional group is due to a hydrogen bonded form. Draw the structure of two propanoic acid molecules showing how they could dimerize via hydrogen bonding. Answer: CH3 CH2 C O H O H O O C CH2 CH3 2.38 Tow isomers having molecular formula C4H6O are both symmetrical in structure. In their infrared spectra, neither isomer when in dilute solution in CCl4 (used because it is nonpolar) has absorption in the 3600-cm-1 region. Isomers A has absorption bands at approximately 3080,1620,and 700cm-1. Isomers B has bands in the 2900-cm-1 region and at 1780cm-1 .propose a structure for A and two possible structure for B Register to View Answer O B; O C O and H 2.39 When two substituents are on the same side of a ring skeleton they are said to be cis, and when on opposite sides, trans (analogous to use of those terms with 1,2-disubstituted alkene isomers). Consider stereoisomeric forms of 1,2-cyclopentanediol (compounds having a 5-membered ring and hydroxyl groups on two adjacent carbons that are cis in one isomer and trans in the other). At high dilution in CCl4 both isomers have an infrared absorption band at approximately 3626 cm-1 but only one isomer has a band at 3572 cm-1. (a) Assume for now that the cyclopentane ring is coplanar (the interesting actuality will be studied later) and then draw and label the two isomers using the wedge-dashed method of depicting the OH groups. (b) Designate which isomer will have the 3572-cm-1 band and explain its origin. ANSWER: (a) 1. 2. HO OH HO HO (b) The cis isomer will have the 3572 cm-1 band because only in it are the two hydroxyl groups close enough to form the intramolecular hydrogen bonding. The intermolecular hydrogen bonding couldnt be formed in the dilute carbon tetrachloride solution. 2.40 Compound C is asymmetric, has molecular formula C5H10O, and contains two methyl groups and a 30 functional group. It has a broad infrared absorption band in the 3200-3550-cm-1 region and no absorption in the 1620-1680-cm-1 region. (a) Propose a structure for C. (b) Is your suggested structure capable of stereoisomerism? If so, draw the stereo isomers using the wedge-dashed wedge method. Register to View AnswerH3 C H H OH H CH3 (b) Yes H3C H OH C H3 H OH H3C CH3 HO H3C H CH3 HO H3C H CH3 ... View Full Document

End of Preview

Sign up now to access the rest of the document