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General Chemistry II

by: Ansel Ankunding PhD

General Chemistry II CHEM 1120

Ansel Ankunding PhD
Utah State University
GPA 3.99

Cheng-Wei Chang

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Cheng-Wei Chang
Class Notes
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This 57 page Class Notes was uploaded by Ansel Ankunding PhD on Wednesday October 28, 2015. The Class Notes belongs to CHEM 1120 at Utah State University taught by Cheng-Wei Chang in Fall. Since its upload, it has received 33 views. For similar materials see /class/230507/chem-1120-utah-state-university in Chemistry and Biochemistry at Utah State University.

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Date Created: 10/28/15
Chapter 18 Amines and Amides Learning Objectives 1 Provide both IUPAC and common names for amines 2 Differentiate primary secondary and tertiary amines 3 Recognize the physical properties of amines 4 Recognize heterocyclic amines 5 Provide both IUPAC and common names for amides and write the equations for amidation from amines 6 Write the equations for the hydrolysis of amides Sections 181 182 183 184 185 Amines Properties of Amines Heterocyclic Amines and Alkaloids Amides Hydrolysis of Amides Recommended additional problems 181 8 1841 1843 1845 1848 1851 Amines A Classi cation of amines R N U 39z39 I I 73 2 3 2 amine 3 amine I 639 R lrl R R 4 ammonium B Nomenclature of amines 39 39 I VT NH NH2 NH2 HN H2N A NH2 39 NHZ 39 HO y i HOk NH NH2 1 o O 182 Properties of Amines A Boiling points B Solubility in water C Amines react as bases D Ionization of an amine in water E Basicity of amines F Amine salts G Properties of amine salts 183 Heterocyclic Amines and Alkaloids A Heterocyclic amines O U U H W B Alkaloids amines in plants 184 Amides A Preparation of amides B Nomenclature of amides i in VH0 C Physical properties of amides 185 Hydrolysis of Amides A Acid hydrolysis of amides B Base hydrolysis of amides Chapter 21 Nucleic Acid and Protein Synthesis Learning Objectives Describe the components that make up the nucleic acids Describe the primary structures of RNA and DNA Describe the double helix of DNA Describe the process of DNA replication Identify the different RNAs and describe the synthesis of mRNA Describe the function of codon in genetic code Describe the process of protein synthesis from mRNA Describe the correlation between altered DNA and the sequential mutation 10 Describe the preparation and uses of recombinant DNA 11 Describe the methods by which a Virus infect sa cell pwaaewwe Sections 211 Components of Nucleic Acids 212 Primary Structure of Nucleic Acids 213 DNA Double Helix 214 DNA Replication 215 RNA and Transcription 216 The Genetic Code 217 Protein Synthesis Translation 218 Genetic Mutations 219 Recombinant DNA 21 1 O Viruses Recommended additional problems 2179 2181 2183 2192 2199 21101 211 Components of Nucleic Acids N Pyrimidine purine Pyrimidine NH2 N O H Cytosine C Purine NH2 N N lt I j Q N Adenine A Ribose O 0 CH3 NH 1 L A N N O39 H H Thymine T Uracil U 0 N NH lt 39 A N H N NH2 Guanine G Nucleosides and Nucleotides NH2 N NH2 O Bases Nucleosides Nucleotides Adenine A Adenosine A Adenosine 539monophosphate AMP Guanine G Guanosine G Guanosine RNA 539monophosphate GMP Cytosine C Cytidine C Cytidine 539 1non0phosphate CMP Uracil U Uridine U Uridine 539m0nophosphate UMP Adenine A Deoxyadenosine A Deoxyadenosine 539monophosphate dAMP Guanine G Deoxyguanosine G Deoxyguanosine DNA 539monophosphate dGMP Cytosine C Deoxycytidine C Deoxyguanosine 539mon0phosphate dCMP Thymine T Deoxythymidine T Deoxythymidine 539monopho sphate dTMP ADP and ATP 212 Primary Structure of Nucleic Acids Frag quotWilma39xa39rialazre 135139me Ema and ha gf i apw ghmi fxa Wng Edmat inn inc uhlishing as Benjamin 213 DNA Double Helix Hbose riboseNNI E0 N G C bose 1 Nlt gt N N H 0 ribose A I Nf H V H Ci l ia Ha MeVa J NW5 9quot quotfo x 393 if mH r a my rhfmmg sf a x 2quot quotsugar 39 39 Sugar Hm 393 DNA chain Adenine A E D Edwina thymine Gymsmegc 539 E Guaaazme enxyrihusef sugar Sugarphgsphaze P a gham gba kmm x nnnnnnn y mgen hand 11nrn x If u M asquot I39M L H a H VRW quotH W 139 1 My 1 5111331 Sugar WK w er a in DNA In mm M chain chain H 53 35I Guanine Evinsine Timbe aka enaysax await and intagfcaf hemfsw Cany ght Fearsoza Edmuatinn inn gpublia hing 55 Hemamin nmmings 214 DNA Replication Useful online animation httpVcellndsunodakeduanimationshomehtm During DNA replication An enzyme helicase unwinds the parent DNA at several sections At each open DNA section called a replication fork DNA polymerase catalyzes the formation of 539 339 539 to 339 ester bonds of the leading strand The lagging strand growing in the 339539 339 to 539 direction is synthesized in short sections called Okazaki fragments The Okazaki fragments are joined by DNA ligase to give a single 339 539 DNA strand Template strand on on Pymphcsgihate re eased V or naleaside triphasphate inwieding DNA V 3quot DNA i 39 39f 3 5 heading strand M Replication fork Okazakifragments 3 5 215 RNA and Transcription Transcription 1nRNA is formed from a gene on a DNA strand Translation tRNA molecules bring amino acids to lnRNA to build a protein Replication Transcription 7ransla on Amino acids Timberlukc wcmtmwn 39 39 I w s H Inc quotHiW39s 39 Cummings Maj or function Percentage of total RNAs Messenger RNA mRN A Carries genetic information from 5 1 0 DNA to the ribosomes Transfer RNA tRNA Brings amino acids to the 39 1015 ribosome to make the protein Ribosomal RNA rRNA Makes up 23 of ribosomes 13 75 i ribosomal proteins Where protein synthesis takes place A tRNA 0 ma aquot a omcmwmw Sarina Accepmr stem r V B rRNA and mRNA in translation C RNA Polymerase During transcription RNA polymerase moves along the DNA template in the 3 5 direction to synthesize the corresponding mRNA The mRNA is released at the termination point Termination site 39 ontamplate 5 Strand 1 2 a mmptate I strand S Imhesis of myHNA 5quot Rim go ymerase Tlmbarlaka Ganeral mm and Biafag cai C mismv Copyrigth Pearson Educatian lnc publishing as Benjamin Cummings D Processing of mRNA The DNA of eukaryotes contains exons that code for proteins along with introns that do not code for proteins Initiation site I 39farminaiinn Transcription PremeNA Processing spacing m fl Hummus b mi Timberlakar enemr manic and iamgrm nemrsrrjc Gapymight f aersm Emailsn mm gu lla hing as Benjamin Bumm mga E Regulation of transcription Operon consiSts of control site including promoter and operator and genes cluster that are usually coded for proteins involved in a serious of reactions Promoter promote the transcription of genes Repressor coded byregulatory gene that produces protein to repress the transcription of the corresponding genes Example Lac operon a The lamist apemn w egul39aimy gene i f h Cammi SHE b Transcriptiun turnad RNA puiymeme Razpiressdr Q anm p mn 39 tiumed an Hm egsm Lam353 Mad We ga mwsi a a inducer rapmssmr 39 10 216 The Genetic Code A Codons sets of three bases triplet that coded for a speci c amino acid start function or stop function along the 1nRNA Sacand Letter mm Third Lamar quot mm mm 39 H H W nanim star was sans mas arias we if A AUG lays Glam amt Gin than t am EEG quotan ma a Ei r 39 Qi t m t f that uh walnut signal the and u peptide chain 539 CCU AGC GGA CUU 339 39 Pro S er Gly Leu 217 Protein Synthesis Translation A tRNA activation I Occurs when a synthetase uses energy of ATP hydrolysis to attach an amino acid to a speci c tRNA I Prepares each tRNA to use a triplet called an anticodon to complement a codon on 1nRNA ll o cnzou OMGMCH MNH SetIna Acceptor stem I Cosion an mRNA 113 c A Brxming glmlmmgyvtidu chum As 11 39r immnm nim w Isltmg u mim p ml a r 39 midmrmmmmma Immmwmi Mm 11 But gmwiaagv w hmmgw minds gamma air 5 H tnmpmnm m 55 K r E t mgiamj 39 2 ready ma mah rg d with a raw mains acid mh cmume already Martian at 111343543sz 12 218 Genetic Mutations A Substitution mutation A different base substitutes for the proper base in DNA changing the corresponding codon in the mRNA which leads to the incorporation of wrong amino acid in a polypeptide Eubstitutim at t by T limiting strand m tm Amine acid eequenue 1B Frame shift mutation An extra base adds to or is deleted from the normal DNA sequence causing all the codons in mRN A translated incorrectly Wrong amino acids are incorrectly incorporated v EDNA awning Strand C Effect of mutation Defective proteins will be produced 13 D Genetic diseases Genetic Disease Result Gaimwscn a The mmsfcxrase required for the membo sm of galacmsclvphosphaw is absent Mcumulntion of GabiP Beads tn cataract and mental retardation Cystic brosis The most common inile ted ime Thick mucus secretions make39brea aing dif cult and blmk pancreatic function Down syndrome The leading cause of mnmi retardation occurring in about I of every 3130 iivo39births Manta turd phyniwl problems minding heart mi cyedcfccis an 23w result of the variation of lter chrmm mines1 usualin chromosome2391 instead of apair Muscuiar dymophy One of 10 forms ofMD A mutation in this X chromarm results in the how or abnormal Duehmn production of dysrrapmn by hex gene This musciiwicsmymg disease agpcairs at about age 5 with by ago 26 and ocean in about I of 10008 males Huntington s disease Hm Appearing in middle age Iii meets the nervous system tingling in stolen physicai impaimncm It is the result of a mutation in a gene on stagnatiome 4 winchczm new in mapped to test peopie fanniies with HT Sicidmcc anemia A infective 31613103150511 Emma mutation in a gemrm chromosome 11 neoxygenn mng ab ity of mi bloat vihich on a sicklcd shape causing maintain plugng capillaries from red brood mail aggregation riemxph ia One or marewdefacrive mood aiming footers iead to mor39coaguiai mn woman s bieeding and internal henmhagm TnymSachs disuse chnsmir dase A is M 397 Mining m 39 39 m M J wilt a in mm madman Jess of motor comm anxi early Emilx 219 Recombinant DNA A DNA fragment from one organism is Combined with DNA in another A Preparing recombinant DNA Plasma membrane nissoives Restriction enzyme am plasmas and p asmids isolated ems p asmids Ligase jams sticky ends Gene obtained from another am using same restriction enzyme cloned malts produee new protein sum as enzymes or hormones Example of restriction enzyme EcoR1 339 539 SGAATTC G AATTC 3CTTAAGy PTTAA B DNA ngerprinting The ngerprint or pattern is unique to each individual DNA Typical procedure i Isolating the DNA from the rest of the cellular material in the nucleus This can be done either chemically by using a detergent to wash the extra material from the DNA or mechanically by applying a large amount of pressure in order to quotsqueeze outquot the DNA ii Restriction enzymes cut a DNA sample into smaller agments RFLPs iii The fragments are sorted by size which can be done by gel electrophoresis The DNA is poured into a gel such as agarose and an electrical charge is applied to the gel with the positive charge at the bottom and the negative charge at the top Because DNA has a slightly negative charge the pieces of DNA will be attracted towards the bottom of the gel the smaller pieces however will be able to move more quickly and thus further towards the bottom than the larger pieces 39 The differentsized pieces of DNA will therefore be separated by size with the smaller pieces towards the bottom and the larger pieces towards the top iv Denaturing the DNA so that the entire DNA is rendered singlestranded This can be done either by heating or chemically treating the DNA in the gel v Blotting the DNA The gel with the sizefractionated DNA is applied to a sheet of nitrocellulose paper and then baked to permanently attach the DNA to the sheet The Southern Blot is now ready to be analyzed xi In order to analyze a Southern Blot a radioactive isotope that adheres to certain base sequences in the fragments produces a pattern on X ray lm which is the ngerprint 15 C Polymerase chain reaction PCR i Produces multiple copies of a DNA in a short time ii Separates the sample DNA strands by heating iii Mixes the separated strands with enzymes polymerase and nucleotides to form complementary strands iv Can be repeated for many times to produce a large sample of the DNA 2110 Viruses A Types of Viruses Type of Viruses Representative Viruses Approximate number of genes 39 Single stranded DNA le 74 phage 5 Doublestranded DNA Adenovirus 2 30 Tiphage 1 50 Singlestranded RNA Tobacco mosaic Virus 6 Poliovirus 8 In uenza Virus 12 Doublestranded RNA Reovirus 22 16 B Reversed transcription i A retrovirus which contains Viral RNA but no viral DNA enters a cell ii The viral RNA uses reverse transcriptase to produce a viral DNA strand iii The viral DNA strand forms a complementary DNA strand iv The new DNA uses the nucleotides and enzymes in the host cell to synthesize new virus particles lt Ember 592 general rgam e 3311f asawg mmhemw epyright sf V Em Edeaazien Inn publishing as Eta amin Cummings C HIV viruses and AIDS i Is a retrovirus that infects T4 lymphocyte cells ii Decreases the T4 level making the immune system unable to destroy harmful organisms 39 iii Causes pneumonia and skin cancer associated with AIDS RNA Core Reverse transcripfase Lipid membrane Envelope motels Ylmbmlaxe 351mm Irrwrw 39 a 39N p O O N H O Thymine Azi dothymine 18 Chapter 16 Carboxylic Acids and Esters Learning Objectives 1 Provide both IUPAC and common names for carboxylic acids and esters 2 Recognize the physical properties of carboxylic acids and esters 3 Write the equations for esteri cation and hydrolysis of esters Sections 161 Carboxylic Acids 162 Properties of Carboxylic Acids 163 Esters 164 Naming Esters 165 Properties of Esters Recommended additional problems 1643 1644 1647 1648 1651 1655 1656 1659 A 161 Carboxylic Acids R 23H T quotOquot RCOOH RCOgH A Nomenclature of carboxylic acids M OH A OH OH Cl C B Preparation of carboxylic acids 162 Properties ofCarboxylic Acids A Boiling points B Solubility in water OH C Acidity of carboxylic acids a ll D Neutralization of carb oxylic acids 163 Esters R O R39 Tquot quotOquot RCOOR39 RCOZR39 A Esteri cation General reaction R 6 H H R 6 RI quot catalyst Tquot u H9 R39 H O H o o v Example if 1 gt O H20 cat H2804 164 Naming Esters 165 Properties of Esters A Solubility in water B Acid hydrolysis of esters General reaction a H I o R o R39 R O H catalyst 39 H B H 3919 0 0 C Base hydrolysis of esters saponi cation quot9 e R O R39 R H M H M63 b H H H O R 00 I 00 GB M Na K most common Examples Chapter 17 Lipids Learning Objectives 1 Describe the classes of lipids 2 Write the structures of fatty acids and identify as saturated or unsaturated 3 Write the structural formula of a wax fat or oil produced by the reaction of a fatty acid and an alcohol or glycerol 4 Draw the structure of products from hydrogenation hydrolysis and oxidation of tn39acyl glycerol 5 Know the properties of glycerophospholipids 6 Know the types of lipids that contain sphingosine 7 Know the general structures of steroids 8 Know the composition and function of the lipid bilayer in cell membranes Sections 171 Lipids 172 Fatty Acids 173 Waxes Fats and Oils 174 Chemical Properties of Triacylglycerols 175 Glycerophospholipids 176 Sphingolipids 177 Steroids Cholesterol Bile Salts and Steroid Hormones 1 78 Cell Membranes Recommended additional problems 1777 1779 1781 1783 1785 1786 1788 171 Lipids Basic building block fatty acids Derivatives from fatty acids include prostaglandins waxes triacylglycerol Gheerophospholipids sphingolipids glycosphingolipids steroids Know the information in Figure 171 172 Fatty Acids A Properties of fatty acids hydrophobic 39 hydrophilic Stability of saturated and unsaturated double bonds Signi cance of cis double bond j A Prostaglandins HO OH 173 Waxes Fats and Oils A General structure of waxes 0 HO HO OH COZH N B Fats and oils Triacylglycerol 0 HOW OH O W HOW W OiVVVW O 0 j U 39 0 C Melting points of fats and oils 174 Chemical Properties of Triacylglycerols A Hydrogenation B Oxidation W C Hydrolysis acidic or neutral conditions D Saponi cation basic condition 175 Glycerophospholipids A General structure of glycerophospholipids O Phosphate amino alcohol Phosphate monoester Phosphate diester Amino alcohol 176 Sphingolipids A Sphingosine CHs CH212 CH CH CH OH CH NH2 CHz OH B General structure CH3 CH212 CH CH CH OH H CH N Fatty aCIds CH2 O Phosphate amino alcohol sphingomyelin C Glycosphingolipids H CH N Fatty adds 0 CH2 0 carbohydrates 177 Steroids Cholesterol Bile Salts and Steroid Hormones A Cholesterol m z oo 2N 20 I all em 2 D O 2 3 mam m C Lipoproteins transporting lipids Lipids are transferred in the forms of triacylglycerol or cholesteryl ester C H3 CH214 39Proteins apoproteins Cholesterol chylomicrons verylowdensity lipoprotein VLDL lowdensity lipoprotein LDL highdensity lipoprotein HDL 10 178 Cell Membranes Chapter 15 Carbohydrates Learning Objectives 1 Recognize monosaccharide as aldoses and ketoses with respect to the number of carbon atoms 2 Draw the D and L con guration of glucose galactose and fructose 3 Draw and identify the cyclic structures of monosaccharides 4 Recognizethe products from oxidation and reduction of monosaccharides 5 Recognize the monosaccharide units and linkages in oligosaccharides Sections 151 Carbohydrates 152 Structures of Monosaccharides 153 Cyclic Structures of Monosaccharides 154 Chemical Properties of Monosaccharides 155 Disaccharides 39 156 Polysaccharides Recommended additional problems 1547 1549 1559 1555 151 Carbohydrates A Types of carbohydrates Monosaccharides Disaccharides Oligosaccharides Polysaccharides B Monosaccharides i Based on the functional groups aldose ketose ii Based on the number of carbon atoms triose tetrose pentose hexose etc iii Examples CH0 CH0 CH0 H OH H OH H OH Ho H Ho H HO H H 0H H 0H HO H H 0H HO H H quot 0H CHZOH CHZOH CHZOH CHZOH CHO o H OH Ho H H 0H H OH H 0H H OH CHZOH CHZOH 152 Structures of Monosaccharides A Fischer projections CHO H OH CHZOH CHO H OH H 0H H OH CHZOH B Important monosaccharide Dglucose D galactose and Dfruc ose CHO HO H CHZOH CHO H OH H OH Ho H CHZOH 153 Cyclic Structures of Monosaccharides A Haworth structure CHZOH CHZOH H OH H H O H H H OH O OH OH OH OH H HO H HO CHZOH CHZOH H OH o H O HO H H OH I H OH OH OH H H HO H HO CH20H I I H OH H39 OH CHZOH CH20H HOHZC 0 OH H H OH CHZOH O H HO OH H OH CHZOH H CHZOH HO H HO O OH H H CHZOH H HO 154 Chemical Properties of Monosaccharides A Oxidation of monosaccharide B Rearrangement of monosaccharide C Reduction of monosaccharide 155 Disaccharides Knowhow to describe the types of glycosidic bonds Examples Maltose Lactose 156 Polysaccharides Amylopectin and glycogen Cellulose Chitin Chapter 13 Alcohols Phenols Thiols and Ethers Learning Objectives 1 Provide both IUPAC and common names for alcohols phenols ethers and thiols 2 Recognize the physical properties of alcohols phenols ethers and thiols 3 Write the dehydration equations of alcohols and predict the major product when applicable 4 Recognize the oxidation and reduction of alcohols and predict the products from oxidation of primary and secondary alcohols 5 Recognize the oxidation and reduction of thiols Sections 131 Alcohols Phenols and Thiols 132 Ethers 133 Physical properties of Alcohols Phenols and Ethers 134 Reactions of Alcohol and Thiols Recommended additional problems 13391341l34913531355l357 131 Alcohols Phenols and Thiols A Classi cation Alcohols molecules With OH group hydroxyl group Alkyl alcohol Aryl alcohol For alkyl alcohols ROII Primary 1 alcohol Secondary 2 alcohol Tertiary 3 alcohol Example OH Thiols mercapto molecules with SH group su1 1ydryl group B Nomenclature of alcohols Naming alcohols Step1 Identify longest carbon chain with OH group Step 2 Number the longest carbon chain from the carbon with OH Step 3 Apply appropriate name for the main carbon chain from alkane or cycloalkane but substitute 6 With 01 or thiol Remember to indicate the position of OH when necessary Step4 Name substituents when necessary Examples OH WOH OH CHg CHz CHz OH OH OH Br CH3 OH C Nomenclature of phenols OH OH OH CH3 Q Cl Dials OH HOWOH OH D Nomenclature of thiols Step1 Identify longest carbon chain with SH group Step 2 Number the longest carbon chain from the carbon with SH Step 3 Apply appropriate name for the main carbon chain from alkane or cycloalkane and add thiol at the end Remember to indicate the position of SH when necessary Step4 Name substituents when necessary 1ASH 39 SH 132 Ethers Ethers molecules contain oxygen atom that is single bonded to two alkyl or aryl groups 0 o A Nomenclature of ether Common name 0 IUPAC name use of alkoxy as substituent I Alkyl group Alkoxy group OH O O 0 WA 1 OCHZCH3 B Structural isomers of alcohols and others C Cyclic others 133 Physical properties of Alcohols Phenols and Ethers Know the effect of hydrogen bond A Boiling points B Solubility in water 134 Reactions of Alcohol and Thiols A Combustion B Dehydration of alcohols st04 A CC I I H20 l CC H OH Examples OH H heat gt OH heat gt SaytzefPs rule C Formation of ethers D Oxidation of alcohols 01 o CCH C H gt C C OH l I lt g lt 0 H1 H 0 oxidation H reduction T O I Cl3R quot C C R I OH H l 395 E Oxidation of thiols


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