Exam Analysis: Organic Chemistry, Midterm 1

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1 Exam Analysis: Organic Chemistry, Midterm 1 1) TEST BREAK DOWN: There are three independent topics covered in the first midterm, which are hybridization, structure and isomerism, and resonance. The test is usually 60 minutes, and worth 100 marks. All of the questions will be multiple choice questions, and in order to master the term test, you will need to have good knowledge of each topic. You need to a solid understanding of isomerism which covers constitutional, conformational and stereoisomers. It also carries most of the points along with resonance as the next major topic on term test 1. For the detailed importance of each topic, please refer to the question statistics as follows. 2) TEST STATISTICS: Topics/Years Oct Oct 2011 Oct 2012 Total Hybridization Structure/ Isomerism Resonance Total Hybridization Structure/Isomerism Resonance 0

2 3) KNOWLEDGE SUMMARY: Hybridization: Organic Chemistry involves studying of molecules that are primarily made up of carbon atom and hydrogen atom. These atoms are linked via a covalent bond which involves sharing of electrons. To form a covalent bond, each atomic orbital overlaps with neighboring atom s atomic orbital. o The overlap of these orbitals results in the formation of molecular hybridized orbitals. There are four types of atomic orbitals: s- orbital, p-orbital, d- orbital and f-orbital. o Overlap of 1s and 1p orbital results in 2 sp hybridized orbitals, where the total number of atomic orbitals equals the number of total hybridized orbitals. o Overlap of 1s and 2p orbital results in 3 sp2 hybridized orbitals, overlap of 1s and 3p orbitals results in 4 sp3 hybridized orbitals. To find the hybridization of any atom in a molecule, we simply count the number of total electron groups around the atom. The total number of electron groups equals the number of attached atoms and number of lone pairs on the atom in question. o For example HCN, which has a single covalent bond between hydrogen atom and carbon atom, and triple covalent bond between carbon and nitrogen atom has sp hybridized carbon atom and sp hybridized nitrogen atom. Carbon has 2 electron groups around it in the form of hydrogen atom and nitrogen atom while no lone pairs. Nitrogen atom has 1 lone pair and a carbon atom thus again total 2 electron groups. 2 electron groups means that total of 2 hybridized orbitals will be present on each atom.. Hybridized orbitals formed by association of atomic orbitals show directionality and particular bond angles. Sp hybridized orbitals have bond angle of 180 because that allows the lowest energy states by having 2 electron groups as far as possible. Sp2 with 3 electron groups are 120 degrees apart while sp3 hybridized atoms have degree bond angles between each atom. SUMMARY OF QUESTIONS TO BE ASKED: 1. Hybridization of particular atom in a molecule o Count the number of electron groups around the atom in question o Check for lone pairs on atoms such as nitrogen and oxygen etc. RELATED PAST TEST QUESTIONS: Oct-2012 Midterm 1 Questions 1, 2, 3, 4,5, 6, 7,8,9, 10

3 Answer: 1) The hybridization of Nitrogen atom in lumigan involves counting the number of electron groups around the nitrogen atom. Since nitrogen atom has a valence of 5, it has a lone pair or electrons and three bonds meaning it is trivalent. Next we simply count the electron groups and since nitrogen is neutral, we see it has 3 bonds and 1 lone pair, which means a total of four electron groups and thus nitrogen atom must be sp3 hybridized because it requires 4 atomic orbitals to make up 4 hybridized sp3 orbitals. (C) 2) In order to assign priority around chiral compounds, we look at Cahn ignold rules. We prioritize according to atomic number. When the lowest group points at the back, and the priority from highest to lowest leads to clockwise direction, the stereo chemical orientation is R, and if its anticlockwise then it is S. 3) Across a double bond the stereochemical configuration is in terms of cis/trans or E/Z isomerism. Cis or Z isomers are those that have two largest groups on the same side while E and trans isomers have two groups on the opposite side of the double bond. Between C4 and C5, we see that two smallest groups face the same side which means that the two largest groups face the same side as well. This results in Z isomers. 4) Amide oxygen atom has lone pairs which allows it to be either lewis base or Bronsted Lowry base. The lone pair of electrons can attack any electrophilic carbon or electrophilic hydrogen. Since Bronsted Lowry base is a proton acceptor and it accepts it by its lone pair of electrons, the answer is C. 5) When counting carbon atoms that are sp2 hybridized, we should look for carbon atoms that are double bonded to other atoms, carbocation and alkyl radicle carbon atoms. A sp2 hybridized carbon atom means that an unhybridized p orbital is present, and this unhybridized p orbital holds the radicle electron, pi bond electrons or it is empty in the case of carbocation. Answer (B) 6) The alcohol functional group is described as secondary since the carbon bonded to alcohol is bonded to two other carbon atoms. 8) The bond on C3 has 3 bonds and a hydrogen atom which is not shown. Since carbon is tetravalent, and thus four electron groups we know that its hybridization would be sp3. Since C2 has a double bond, it would have 3 groups around it which means its hybridization would be sp2. (D) Oct-2011 Midterm 1 Questions 2-4 Answer: 2) The question requires us to find out the number of carbon atoms that are sp2 hybridized. This means that the carbon atom must have a double bond or unhybridized p oribital. From the structure provided we locate carbon atoms that have double bond attached to them. Since there are only 2 carbon atoms that qualify the criteria, the answer is 2 or (C)

4 3) The hybridization of N1 is sp3 since nitrogen has 3 attached atoms and one lone pair of electrons (C) 4) The hybridization of carbonyl oxygen atom attached to C7 is sp2 since oxygen atoms has 2 lone pairs and 1 carbon atom attached to it, and three electron groups would mean a total 3 hybridized orbitals. (e) KNOWLEDGE SUMMARY Structure/ Isomerism Organic chemistry concerns with reactivity of atoms to form covalent bonds so that each atom can obtain a noble gas electron configuration or a complete octet. There are sigma bonds which are single covalent bonds formed between the atoms participating in the bonding. These bonds consist of 2 electrons and they are the always the first bonds formed between atoms. The electron density lies directly between the atomic nuclei. There are pi bonds which also comprise of 2 electrons but they have electron density above and below the plane of sigma bonds. They are formed due to parallel sideways overlap between unhybridized p orbitals. A double bond comprises of 1 sigma and 1 pi, where as a triple bond comprises of 1 sigma and 2 pi bonds. Organic molecules are said to be saturated if they don t contain a pi bond or a ring. A saturated compound has (2n + 2) number of hydrogen atoms where n is the number of carbon atoms. Degree of unsaturation is given by (2n + 2) x / 2 where x is hydrogen atoms or any monovalent atoms such as halogen. Oxygen atoms are ignored and each nitrogen is replaced by 1 carbon and 1 hydrogen atom. Bond length is the length between the atoms that are bonded to each other. The stronger the bond, the shorter the bond length. Bond dissociation energy is the energy required to dissociate a bond homolytically. This results in radicles and is different from heterolytic bond cleavage which results in ions. Isomers are compounds that have same molecular formula o Constitutional isomers or structural isomers differ by structural arrangement or connectivity of atoms in a molecule o Conformational isomers have same structure or connectivity but they differ by conformations around the sigma bond. Conformational isomers are same compounds but they are different due to variable ring strain between adjacent atoms. There are two main conformations, eclipsed and staggered. Eclipsed confirmations are high energy since in eclipsed conformations the adjacent carbons have their sigma bonds directly lining up with each other which results in electrostatic repulsion between sigma bonds and steric hindrance due to bulky groups in close proximity. o Cyclohexane assumes a chair conformation. This allows a lower energy state. The chair conformation in cyclohexane interconverts at room temperature. This interconversion converts axial atoms into equatorial position. Equatorial position is thermodynamically favorable because it allows bulky groups to be

5 as far as possible from other electron groups which reduces electrostatic repulsion o Stereoisomers are isomers which differ by spatial arrangement of atoms in space. These molecules are non-superimposable. Enantiomers are non-superimposable mirror images while diastereomers are non-superimposable non mirror images. Stereoisomers are compounds with stereo-centers also known as chiral carbons. These carbons are asymmetric in that they have four different groups attached to it. Absolute configuration involves prioritizing these groups around chiral carbon. Priority is given to larger atomic numbers. An R- absolute configuration is given to clockwise rotation of numbering while S- absolute configuration is given to counter clockwise numbering. (Note: Atom with the lowest atomic number should face at the back, otherwise the notation reverses). o Total number of stereoisomers is given by 2 n where n is the total number of chiral carbons. SUMMARY OF QUESTIONS TO BE ASKED: Determine degree of unsaturation o Using (2n + 2) x / 2, where n is number of carbon atoms in the molecule and x is number of hydrogen atoms, we calculate the degree of unsaturation Drawing bond line structures, and naming molecules Newman Projections Chair Conformations of Cyclohexane RELATED PAST TEST QUESTIONS: Answer: Feb 2009 Midterm 1 Questions 5 1) To calculate the degree of unsaturation in lisinopril, we count the number of pi bonds and rings present in our molecule. Since there are 6 pi bonds and 2 rings the total degree of unsaturation would be 8. We could also calculate using the formula (2n +2) x /2 and we should get the same answer. We will ignore any oxygen atom, and for each nitrogen we assume 1 carbon and 1 hydrogen in the formula. Oct 2004 Midterm 1 Part II Q1 1) I) To draw the structure in question, we use the bond line structure. The end part of the molecule corresponds to the base molecule length, whereas the prefixes are the substituents. (R)-4-bromo-3,3-dimethylhexane, we draw hexane structure and two methyl groups on carbon 3. On carbon 4, since bromine is assuming a position that makes carbon 4 absolute configuration (R) we draw bromine on a dashed line, and since lowest atomic number is pointing back, we reverse S- absolute configuration to make it R absolute

6 configuration II) To draw cis-1,2-dimethylcyclopropane, we draw the cyclopropane with C 3 H 6 structure, but we substitute carbon 1 and carbon 2 with one of their hydrogen with methyl groups. Since these methyl groups are cis, they will point the same direction in the form of wedges or dashes. Oct 2004 Midterm 1 Part II Q2 I) To draw the trichlorosubstituted cyclohexane through the two chair conformations we will draw four carbons in one plane while one carbon is above the plane and one carbon is below the plane. Since the two conformations interconvert at room temperature, axial chlorine atoms become equatorial and vice versa. Since carbon 1 and 2 are facing the same face, they will be axial and equatorial respectively while carbon 3 will be equatorial since it points on the opposite face. The reverse orientation will be for the second chair conformation II) Since there is one molecule with 2 equatorial positions and 1 axial versus other molecule with 2 axial and 1 equatorial position, we would expect the molecule with 2 equatorial and 1 axial chlorine atoms to be at lower energy than the other molecule. Oct 2004 Midterm 1 Q3 2) To draw the conformational isomers of 1-bromopropane from the given structure W, we turn the back side adjacent carbon by 60 degrees at a time. Turning it once towards left will bring us on fully eclipsed position with the two most bulky groups directly aligned. This gives us absolute maxima. If we rotate further 60 degrees to the left, again we get gauche conformation with the two bulky groups 60 degrees apart. This is analogous to structure in W and would be labeled as Y, relative minima. If we rotate further to the left, we have an eclipsed conformation but this time the hydrogen atom eclipses with bromine atom and not the methyl group. This has higher energy than previous gauche because the eclipsed conformation causes the groups to be too close to each other, and thus steric hindrance and electrostatic repulsion. If we rotate further to the left by 60 degrees, we will get a structure with methyl group exactly 180 degrees apart from bromine. This structure would be called anti-staggered conformation. It would be labeled as Z on the potential energy diagram. It represents absolute minima because the major bulky groups are furthest apart and none of the groups are eclipsed. KNOWLEDGE SUMMARY Resonance In organic chemistry often molecules with full charges stabilize themselves by distributing the charge over multiple atoms. This sharing of electron density or delocalization through unhybridized p orbitals or pi bonds is known as resonance. Resonance is shown by multiple contributors. None of the individual molecules represent the true structure of the molecule but it is the hybrid or mixture of

7 individual molecules that determines the molecule s actual structure For resonance to take place, the adjacent carbon must be sp2 hybridized or one with a double bond. Only lone pair of electrons that are delocalized and pi bond electrons can move. To draw appropriate resonance contributors, all atoms must have complete octet, must possess minimum charge separation, and if there is charge, it must be on an more electronegative atom. The total charge of the molecule must equal to the formal charge of an atom. The formal charge is an imaginary charge given to an atom in a molecule assuming that the atoms share the electron equally in a covalent bond. To calculate it we use the formula: Valence electrons lone pair of electrons ½ of bonding electrons SUMMARY OF QUESTIONS TO BE ASKED Draw resonance contributors of molecules in question RELATED PAST TEST QUESTIONS Fall 2009 Part II, Q2 2) To draw resonance structures including the structure given above, we move the pi bond electrons from the cyclohexadiene molecule towards the positive charge. This causes the positive charge to reside on the cyclohexene. Next, an adjacent sp2 hybridized carbon shares its pi bond electron density and takes on positive charge in the process. Last, the lone pairs of electrons present on adjacent nitrogen help transfer positive charge from carbon to nitrogen. To decide the major resonance contributor we look at the rules. The last structure is the major contributor because it satisfies the fundamental criteria that all atoms must satisfy octet rule. Since in all other structures, carbon atom has less than 8 electrons, last structure will complete octet will be much more stable and thus a major contributor. Oct 2012 Q10 pka is an indicator of the strength of an acid. A stronger acid has lower pka compared to a weaker acid. The strength of acid s conjugate base in terms of electronegativity, resonance and size determines the strength of an acid. If the base is weak, then it is a stronger acid. The basicity of the compound is lowered when the conjugate base can resonate, or has higher electronegative atom bearing the negative charge or its size is bigger resulting in greater stability of the base and thus stronger acid.

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