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CHM138 Term Test 1 Exam Analysis

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University of Toronto St. George

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 02 2003 2004 2005 2006 2007 2008 2009 Oct 2011 Oct 2012 Total Hybridization 5 7 8 8 4 4 2 5 3 2 48 Structure/ 7 11 17 16 22 20 9 10 Isomerism 20 20 152 Resonance 1 - 2 3 2 2 2 2 1 1 16 Total 15 20 29 29 30 29 28 28 14 14 235 25 20 15 Hybridization 10 Structure/Isomerism Resonance 5 0 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 109.5 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 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) 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
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