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Chem 104 Homework #
1. What are valence electrons? Electrons in the outermost shell (the valence shell). For the following elements: (1) write the number of valence electrons each has (2) draw the Lewis structure for each atom C __ 4 __ S __ 6 __ H __ 1 __ Cl 7 (halogen) O __ 6 __ Br 7 (halogen) N __ 5 __ Based on the Lewis structures you drew above, how many covalent bonds can each of the following atoms form? C __ 4 __ O __ 2 __ H __ 1 ___ N __ 3 __ 2. What is a halogen? After your explanation, list all of them. Halogens are the elements in column VIIA of the periodic table (column 7 of the main-group elements-- the taller columns). F, Cl, Br, I, At 3. Below is the structural formula of ethane, a gas which is a component of natural gas and petroleum. Draw the Lewis structure of ethane using dots to show the location of all valence electrons. C C H H H H H H
C H O N S Cl Br
How many valence electrons does each carbon of ethane have (including electrons that are being shared with other atoms)? ___ 8 ___ Why is this number significant? Because the valence shell is full when it contains 8 electrons (an atom is more stable when its valence shell is full, so this is a favorable situation).
4. Read the last page of the syllabus. Name one thing you can do to ensure that you do well in Chem 104 this semester. 5. Look at the first page of the syllabus. Where can you go if you need help understanding the material covered in Chem 104? C C H H H H H H
4. Which of the following structures are not possible? If a structure is not possible, explain why. CH 3 CH 2 CH 2 CH 3 CH (^3) CH CH 3 CHCHCH 3 2 CH 2 CH 2 Br Cl No, because the 2nd^ No, because the 3rd^ carbon No, because the carbon from the left has from the left needs one first carbon on too many H’s. more H. the left needs one more H. 5. What requirement must be met for two compounds to be constitutional isomers? They must have the same molecular formula (same number of each type of atom). For example: suppose a certain compound has the molecular formula C 5 H 12. All the constitutional isomers of this compound must have 5 carbons and 12 hydrogens (so they will have the molecular formula C 5 H 12 , but the carbons and hydrogens will be bonded together differently.) 6. Use a model set to build the pair of molecules shown below. Do the two structures represent constitutional isomers or are they the same molecule? CH 3 CH 2 CH 2 CH 3 CH 2
CH 3 CH 2 CHCH 3
CH 3
They are constitutional isomers because the first molecule is just a simple chain of 5 carbons, but the second molecule is a chain of 4 carbons with the fifth carbon branching off. (Circle the longest carbon chain in each in order to visualize this.) Now build the pair of molecules below. Do the two structures represent constitutional isomers or are they the same molecule? CH 3 CH 2 CH 2 CH 3
CH 3 CH 2 CH 2 CH 3
These are just two different representations of the same molecule—they are NOT constitutional isomers. Both of them are composed of a simple chain of 4 carbons (circle the longest carbon chain to convince yourself). The molecule on the left is simple drawn in such a way that the 4 carbons aren’t all lined up in one row.
Chem 104 Homework #
1. Draw ALL the constitutional isomers with the molecular formula C 5 H 12. It may be helpful to circle the longest carbon chain of each structure you’ve drawn, to be sure that none of them are actually the same molecule. When you think you have them all, build each of the structures using your model kit to be sure that none of them are the same molecule. CH 3 CH 2 CH 2 CH 2 CH 3 CH 3 CH 2 CHCH 3 CH 3
CH 3 CCH 3
CH 3
CH 3
2. Draw a full structural formula for this cycloalkane: CH 2 CH 2 CH 2 CH 2 CH CH 2 CH 3 C H H H C C C C C C H H H H H H H H H H H Draw a line formula for the cycloalkane above. (Here is an example of another cycloalkane to guide you: CH 2 CH 2 CH 2 CH 2 CH 2 = 3. Draw the full structural formula for a cycloalkane composed of 4 carbons, with an iodine atom bonded to any two of the carbons—draw the iodine atoms so that the molecule has the trans configuration. C C C C H I I H H H H H C C C C H H H H I H I H OR MORE on next page…
Below is a line formula for a 5-carbon cycloalkane. Assuming that only one chlorine atom becomes attached to each cycloalkane molecule, draw the halogenation product for this reaction. Draw the product using a line formula.
- Cl 2 (uv light) Cl (HCl is also a product)HCl is also a product))
Chem 104 Homework #
1. Draw each of the following compounds using full structural formulas: 2,2-dibromobutane 1,2-dichloropentane C C Br C Br
H
H
H
C H
H H
H H
Cl C C C Cl H H
C C
H H
H H H
H
H
H
2. Draw each of the following compounds using condensed formulas: 1,3,5-trifluoropentane 2-iodononane CH 2 CH 2 CHCH 2 CH 2
F F F
CH 3 CHCH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3
I
3. Draw each of the following using line formulas: cis-1,3-dibromocyclobutane 1-bromo-2-methylpentane Br Br Br MORE on next page…
Chem 104 Homework #
1. Draw these molecules using full structural formulas: 1,2-dibromo-1,1,2-trifluoroethane cis-1,3-dichlorocyclohexane C C H Br F F Br F C C C
C
C C
Cl H H H
H
H
H
H Cl H
H
H
Draw these molecules using condensed formulas: 1,1,1-trichlorodecane trans-1,2-dimethylcyclopentane CCH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 Cl Cl Cl
CH 2
CH 2
CH 2
C
C
CH 3
H
H
CH 3
Draw this molecule using a line formula: 3,3,5-trimethylheptane
2. What is wrong with each of the following molecules? Draw an arrow to the “problem spot” and briefly state what’s wrong. CH 3 CH 2 CH 2 C CCH 3 CH 3 CH 3 CH 3 Too many bonds to carbon—carbon can only have a total of 4 bonds. Therefore, these two molecules are not possible. (Removing a methyl group from each would “fix the problem.”) MORE on next page…
3. For the four molecules named or pictured below, circle the molecules in which the carbon-carbon bonds can rotate freely: ethane cyclobutane (you should know (you should know the structure of the structure of ethane) cyclobutane) Ethane is CH 3 CH 3. The C-C Cyclobutane is shown above. bond CAN rotate freely. The C-C bonds can NOT rotate freely because they are constrained in a ring. C C I H H I H C C F The C-C double bond can NOT The C-C triple bond can NOT rotate rotate freely—double bonds never freely—triple bonds never rotate freely. rotate freely. 4. Draw the structure of an alkene that is composed of only two carbons. C C
H
H
H
H
What is the geometry (3D shape) of this molecule? planar (It is flat—all 6 atoms are in same plane.)
5. Draw the structure of an alkyne that is composed of only two carbons. H C C H What is the geometry of this molecule? _linear (All 4 atoms lie in a straight line.) C C C C H H H H H H H H
3. Draw each of the following compounds: 4-methyl-1-pentyne 2-methyl-2-hexene C
CH 3
CH 3
C
CH 2 CH 2 CH 3
CH 3 CHCH 2 C CH H
CH 3
trans-3-heptene cis-1-chloro-2-pentene C
H
CH 3 CH 2
C
CH 2 CH 2 CH 3
H
C
CH 2
H
C
CH 2 CH 3
H
Cl cis-2-chloro-2-methyl-3-heptene trans-5-bromo-2,6-dimethyl-3-octene C H CH 3 CH C CHCHCH 2 CH 3 H Br CH 3 CH 3 C CH 3 C H C CH 2 CH 2 CH 3 H Cl CH 3 2-hexyne propyne CH 3 CH 2 CH 2 C CCH 3 CH 3 C^ CH
4. Think carefully about the compound “ cyclopentene .” Based on its name, how many carbons does its structure contain? _5 (“pent”) Are there any double or triple bonds? yes, double bond (-ene ending) What does the prefix ‘cyclo’ tell you? the carbons are arranged in a ring Try to draw its structure below. You shouldn’t have to look it up—just use what you’ve learned about nomenclature to come up with a reasonable structure. H 2 C H 2 C CH
CH
CH 2
Chem 104 Homework #
1. Show the products for the hydrogenation and halogenation reactions below. H 2 C H 2 C CH 2 CH CH CH CH 3
- H 2 Ni^ H 2 C H 2 C CH 2 CH 2 CH CH 2 CH 3 H 2 C H 2 C CH 2 CH CH CH CH 3
- Cl 2 H 2 C H 2 C CH 2 CH CH CH CH 3 Cl Cl C + H 2 Pd CH 3 H C CH 2 CH 3 H CH 3 CH 2 CH 2 CH 2 CH 3 2. Show the two products of the hydration reaction below. LABEL which is major and which is minor.
- H 2 O C H+ H H C CH 2 CH 2 CH 3 CH 3 CH 2 CHCH 2 CH 2 CH 3 OH CH 3 CH 3 CCH 2 CH 2 CH 3 CH 3 OH major minor Show the two products of the hydration reaction below. Do you think there will be a major and a minor product, OR will the two products occur in about equal amounts? H 2 C H 2 C CH 2 CH CH CH CH 3
- H 2 O H+
H 2 C
H 2 C CH 2
CH
CH CH 2
CH 3
OH
H 2 C
H 2 C CH 2
CH 2
CH CH
CH 3 OH
about) equal amount)s of t)hese t)wo MORE on next page…
Chem 104 Homework #
1. Show the two products of each reaction below. For one reaction, there will be a major product and a minor product—label them accordingly. For the other reaction, there will be roughly equal amounts of each product—label them as ~equal. H 2 C H 2 C CH 2 CH CH CH CH 3
- HBr about) equal amount)s of t)hese t)wo H 2 C H 2 C CH 2 CH 2 CH CH CH 3 Br H 2 C H 2 C CH 2 CH CH CH 2 CH 3 Br CH 3 CH 2 CHCH=CH 2 CH 3
- HCl major (^) minor CH 3 CH 2 CHCH 2 CH 2 CH 3 CH 2 CHCHCH (^3) CH 3 Cl CH 3 Cl 2. There are two different alkenes (with their double bonds in different positions) that will produce the product shown for this reaction. Draw both of the alkenes that will lead to formation of this product. (It doesn’t matter whether each alkene is cis or trans—either isomer will react the same way.)
- HBr (^) CH 3 CH 2 CHCH 2 CH 2 CH 3 Br This alkene could react) wit)h HBr t)o produce t)he above compound as t)he product).
CH 3 CH 2 CH=CHCH 2 CH 3
CH 3 CH=CHCH 2 CH 2 CH 3
This alkene could also react) wit)h HBr t)o produce t)he above compound as one of t)he 2 possible product)s.
3. Why are trans fats called “trans fats”? Because they contain double bonds that are in the trans configuration (rather than the naturally occurring cis configuration). 4. Read the article on the next page about henna, a dye that is used to decorate the body (like tattoos). Henna’s skin-staining properties are due to a compound called lawsone. Is lawsone an aromatic compound? Explain why or why not. Yes, because it contains a benzene ring in its structure.
Henna: Dye derived from green henna leaves is used to decorate the body with intricate designs By Rachel Petkewich I first) met) Ruby Bansal at) school when we were nine years old. Through high school chemist)ry, college life, and budding science careers in different) part)s of t)he U.S., we kept) in close cont)act). In Sept)ember, I at)t)ended her wedding. In accordance wit)h her family's Indian t)radit)ion, Ruby's hands were adorned wit)h mehendi-designs drawn on t)he body using henna. The night) before t)he wedding, Deepal Vora, a Bost)on- based henna art)ist) born in Mumbai, India, decorat)ed Ruby and also her sist)er Neet)a and me. Vora applied t)he henna as a past)e by squeezing it) out) of a small plast)ic bag t)hrough a t)iny hole and drawing t)he pat)t)ern freehand. Each of her designs included a version of a peacock— a mot)if, she t)old us, t)hat) means good luck. Court)esy Of Deepal Vora As Vora expert)ly put) t)iny swirls and dot)s of t)he green-brown henna past)e on my hands, I felt) a cooling sensat)ion and not)iced a smell like t)hat) of fresh-cut) grass and spinach. As a rule of t)humb, t)he longer t)he henna st)ays on t)he skin, t)he longer and darker t)he st)ain will be. So t)o keep t)he past)e from flaking off t)oo quickly, Vora swabbed t)he pat)t)erns wit)h a mixt)ure of lemon juice and sugar. The next) morning, I scraped t)he dried past)e off wit)h t)he back of a but)t)er knife, revealing auburn-colored mehendi. The st)ains darkened t)o coffee color t)he day aft)er t)he wedding. Henna past)e is prepared from t)he green leaves of Lawsonia inermis, a small t)ree t)hat) grows in warm, arid regions of t)he world such as India, Pakist)an, and Nort)hern Africa. Numerous art)ifact)s found in Middle East)ern and Medit)erranean count)ries, dat)ing back t)o 1400 B.C., show women wit)h henna pat)t)erns on t)heir hands. The earliest) writ)ing on an art)ifact) t)hat) refers t)o t)he specific use of henna as an adornment) for a bride or a woman's special occasion is an inscript)ion on a t)ablet) from about) 2100 B.C. found in nort)hwest) Syria, says Cat)herine Cart)wright)-Jones, a Kent) St)at)e Universit)y Ph.D. st)udent) who is carrying out) research on t)he cult)ure and geography of henna. The use of henna is not) a religious pract)ice, she says. It) is commonly used as a hair dye and, increasingly in t)he West)ern world, for body art). The mat)erial can also reduce dandruff, kill ringworm and head lice, act) as a sunscreen, and, in some preparat)ions, be used t)o rust)proof t)he hulls of met)al ships. In India, especially in desert) areas where t)he t)emperat)ures are ext)remely high, henna was first) used not) t)o decorat)e t)he body but) t)o cool it), Vora says. Henna's characteristic staining properties stem from the compound 2-hydroxy-1,4-naphthoquinone, also known as lawsone, hennotannic acid, or natural orange 6. The structure of the compound is shown below. Henna leaves cont)ain up t)o 5% by weight) of t)he compound, which, in it)s pure form, is a yellow-orange powder t)hat) does not) dissolve in wat)er. Henna past)e is prepared by mixing crushed dry henna leaves wit)h a mild acidic ingredient). Vora uses lemon juice. The acid releases t)he lawsone from t)he plant). Various oils and herbs may also be added t)o enhance t)he scent) of t)he past)e. At) room t)emperat)ure, it) normally t)akes about) a day for t)he acid t)o act)ivat)e t)he dye and t)hree days for t)he past)e t)o lose it)s st)aining capabilit)ies. The process is fast)er in hot)t)er environment)s. Lawsone dye infuses skin, hair, and porous surfaces but) does not) permanent)ly or chemically alt)er t)hem. The dye molecules, which are about) t)he same size as amino acid molecules, migrat)e from t)he henna past)e int)o t)he st)rat)um corneum-t)hat) is, t)he out)ermost) layer of t)he skin, explains biophysicist) Boyan Bonev at) t)he Universit)y of Not)t)ingham, in England. The dye penet)rat)es down t)hrough t)he st)rat)um and does not) spread like ink on blot)t)ing paper. As a result), t)he st)ains init)ially appear darkest) on hands and feet) because t)hey have t)hicker st)rat)a cornea t)han ot)her part)s of t)he body. Various physiological fact)ors, such as skin t)ype and t)emperat)ure, hormone levels, and st)ress affect) t)he appearance of mehendi on people. Aft)er t)he dried past)e is scraped off t)he skin, air oxidat)ion or perspirat)ion can furt)her darken t)he st)ain over t)he next) 48 hours. Alt)hough henna may appear as different) shades, depending on t)hese physiological fact)ors, t)rue henna is a red dye, Vora says. Ot)her color product)s, market)ed as henna product)s, may not) act)ually cont)ain henna. "Black henna," for example, can cont)ain up t)o 40% of a synt)het)ic black hair dye called p-phenylenediamine. As I t)raveled home from Ruby's wedding, my hennaed hands at)t)ract)ed t)he at)t)ent)ion of many people, including complet)e st)rangers. Some recognized my mehendi wit)h compliment)s; ot)hers asked if my hands were t)at)t)ooed. Tat)t)oo ink has a different) composit)ion and is inject)ed int)o t)he dermal layer of skin. My henna disappeared before I finished t)his st)ory, and Ruby's henna wore off during her honeymoon. But) as wit)h many ot)her memorable t)imes in our friendship, we have pict)ures. Chemical & Engineering News ISSN 0009- Copyright) © 2006 American Chemical Societ)y O O Lawsone OH
5. Label each of the following compounds according to whether it will be soluble (S) in water or not soluble (N) in water: CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 __N____ C CH 3 H C CH 2 CH 3 H _N N CH 3 CHCH 3 OH ___S____ If an organic compound is not soluble in water, what would you observe when you try to mix it with water? If the organic compound is a liquid, you will observe two layers when you try to mix it with water. If the organic compound is a solid, it won’t dissolve in the water and you will observe particles of the compound in the water.
Chem 104 Homework #
1. The two compounds below have about the same molecular weight. Circle the one that will have a HIGHER boiling point. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OH (^) CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 This one is higher WHY does this one have a higher boiling point? Its boiling point is higher because the molecules form hydrogen bonds with each other. It takes energy to break these hydrogen bonds when the compound boils, so the temperature must higher for the compound to boil. The compound on the right does not form hydrogen bonds, so less energy is needed for the molecules to go from the liquid to the gas phase. Therefore, it boils at a lower temperature. 2. (a) Label the partial positive and partial negative charges on the alcohol below. (b) Draw a water molecule (H-O-H) below and label its partial positive and partial negative charges. (c) Show a hydrogen bond between the water molecule and the alcohol shown, using dots or dashes to represent the hydrogen bond. (Two possible hydrogen bonds are shown.)
C C C
H
O
H
H
H
H H
H
H
H
O H
+ +
3. Which is the WEAKEST interaction between two atoms? a. the carbon-carbon single bond in ethane b. the carbon-carbon double bond in ethene c. a hydrogen bond between two water molecules d. an ionic bond between Na+^ and Cl-^ in NaCl 4. Classify each of the alcohols below as 1o, 2o, or 3o. CH 3 CHCH 3 OH CH 3 CH 2 CHCH 2 CH 2 CH 2 OH Br C C C OH C H H H H H H H H Cl 2 o^2 o^1 o C CH 3 CH 3 CH 3 CH 2 OH CH 3 CH 2 CCH 2 CH 3 CH 2 CH 2 CH 2 CH 3 OH H 2 C H 2 C CH 2 C C CH 2 CH 3 H OH H 3 o^1 o^2 o
