MCAT Organic Chemistry Review

Card Set Information

MCAT Organic Chemistry Review
2013-09-23 17:04:28

Bonding, stereochemistry, hydrocarbons, alkylhalides and alcohols, aldehydes and keytones, carboxylic acids and carboxylic acid derivatives, amines, amino acids, and proteins, carbohydrates, separation and identification of organic compounds
Show Answers:

  1. How many bonds is a carbon atom allowed to form? What about the oxygen atom? The hydrogen atom? The nitrogen atom? The halides?
    • A carbon atom has one s and three p orbitals in its outmost shell, allowing it to form 4 single bonds, or a double bond, where two electrons are shared, or a triple bond, where three electrons pairs are shared.
    • An oxygen atom may form 2 single bonds, or one double bond and has 2 unshared (lone) electron pairs.
    • A hydrogen atom will form only one single bond.
    • A nitrogen atom may form 3 single bonds, and is capable of double and triple bonds. It has one unshared electron pair.
    • The halides are all able to form only one (single) bond. Halides all have three unshared electron pairs.
  2. What is the difference between composition and constitution in organic chemistry? What is isomerism?
    • Composition: the molecular formula
    • Constitution: the order and kinds of bonds
    • Isomerism: different compounds with the same composition.
  3. What is a hybrid orbital?
    In the carbon atom, what is the result of mixing one s and three p orbitals?
    Give the geometry of the hybridzed orbitals:  one s + three p's 
    one s + two p's 
    one s + one p
    • In organic molecules, the orbitals of atoms are combined to form hybrid orbitals, consisting of a mixture of the s and p orbitals. In a carbon atom, if the one s and three p orbitals are mixed, the result is four hybrid sp3 orbitals. (Three hybridized sp2 orbitals result from the mixing of one s and two p orbitals, and two hybridized sp orbitals result from the mixing of one s and one p.)
  4. What are alkanes?
    • fully saturated (no double or triple bonds) hydrocarbon molecules containing only sp3 hybridized carbon atoms (single bonds) and are, therefore, unreactive except when exposed to heat or light.
    • They may be unbranched, branched, or cyclic.
    • General formula is CnH2n+2 for straight chain
    •    2 hydrogen atoms subtracted for each ring
  5. How does the number of carbons determine the name of a compound?
    Name the following:
    C-C-C-             C-C-C-C- 
     H3C                             CH3
          CH-                 H3C-C-        
          /                             |
     H3C                             CH3
    CH3CH2CH-      H3C-C-CH2-
                |                  |
                CH3             CH3
    What would a square represent?
    What would a hexagon represent?
    • The root of the compound is name according to the number of carbons in the longest carbon chain
    • C1=meth    C6=hex
    • C2=eth       C7=hept
    • C3=prop     C8=oct
    • C4=but       C9=non
    • C5=pent     C10=dec
    • When naming these as fragments, (alkyl fragments are the alkane minus one H atom with symbol R), the suffix '-yl' is used. If naming the alkane, the suffix '-ane' is use.
    • Some prefixes results from a carbon with one R group attached is a primary (normal or n-) carbon, two is secondary (sec) and three it is tertiary (tert or t-) carbon.
    • Some alkyl groups have special names like
    •  C-C-C- n-propyl   C-C-C-C- n-butyl
    •  H3C                            CH3
    •                                   |
    •       CH- isopropyl   H3C-C-   tert-butyl
    •      /                             |
    •  H3C                             CH3
    •                                       CH3
    •                                       |
    • CH3CH2CH- sec-butyl H3C-C-CH2- neopentyl
    •            |                          |
    •            CH3                      CH3
    • cyclobutane
    • cyclohexane
  6. How would one determine the nomenclature for branched-chain alkanes?
    • by determining the longest straight chain (i.e. the highest number of carbons attached in a row).
    • The groups attached to the straight or main chain are numbered so as to achieve the lowest set of numbers.
    • Groups are cited in alphabetical order.
    • If a groups appears more than once, the prefixes di-(2), tri-(3), tetra-(4) are used.
    • If two chains of equal length compete for selection as the main chain, choose the chain with the most substituents.
  7. Give the physical properties of alkanes at room temperature and one atmosphere of pressure.
    • Straight chain alkane state
    •   1-4 carbons - gases (i.e. CH4, CH3CH3...)
    •   5-17 carbons - liquids
    •   18+ carbons - solid
    • Boiling points of straight chain alkanes (aliphatic) increase with increasing number of carbons. (This is because they are nonpolar molecules and have weak intermolecular forces). Branching of alkanes leads to a dramatic decrease in boiling point.
    • As a rule, as the number of carbons increase, the melting points also increase.
    • Alkanes are soluble in nonpolar solvents (i.e. benzene, CCl4...) and not in aqueous solvents due to their low polarity and their inability to hydrogen bond.
    • Alkanes are the least dense of all classes of organic compounds. Thus petroleum, a mixture of hydrocarbons rich in alkanes, floats on water.
  8. In complete combustion, what is the hydrocarbon converted to?
    What is the combustion is incomplete?
    Summarize a strongly exothermic reaction.
    • Complete: Carbon dioxide and water
    • Incomplete: other products, such as carbon monoxide and soot can form
    • CnH2n+2 + excess O2 -> nCO2 + (n+1)H2O
  9. Radical substitution reactions with halogens may be summarized as
      RH + X2 + uv light(hf) or heat --> RX + HX
    Where the halogen X2 may be F2, Cl2, or Br2.
    Summarize the mechanism of halogenation using Cl2 as an example.
    • 1. Initiation: This involves the formation of free radicals (highly reactive substances which contain an unpaired electron, which is symbolized by a single dot).
    •     Cl:Cl + uv light or heat --> 2Cl.
    • 2. Propagation: The chlorine free radical begins a series of reactions that form new free radicals.
    •     CH4 + Cl. --> . CH3 + HCl
    •    .CH3 + Cl2 --> CH3Cl + Cl.
    • 3. Termination: These reactions end the radical propagation steps. Termination reactions destroy the free radicals (coupling).
    •     Cl. + . CH3 --> CH3Cl
    •     .CH3 + . CH3 --> CH3CH3
    •     Cl. + Cl. --> Cl2
    • Note: Radical substitution reactions can also occur with halide acids (i.e. HCl, HBr) and peroxides (i.e. HOOH - hydrogen peroxide). Chain propagation (step 2) can destroy many organic compounds fairly quick. This step can be inhibited by using a resonance stabilized free radical to "mop up" (termination) other destructive free radicals in the medium. The stability of a free radical depends on the ability of the compound to stabilize the unpaired electron. Thus, a tertiary compound is more stable than a secondary, which, in turn, is more stable than a primary compound.
  10. Why are cyclic alkanes strained compounds? Why is cyclohexane, in the chair conformation, stable?
    • The ring strain results from the bending of the bond angles in greater amounts than normal. The usual angle between bonds in an sp3 hybridized carbon is 109.5° (the normal tetrahedral angle.
    • It has the normal bond angles of 109.5° and while at any given moment, 99% of the cyclohexane molecules would be found in the chair conformation because of that stability, it can be found in any conformation in between the chair and boat formation.
  11. What is an ether group?
    • A group consisting of two alkyl groups on either side of an oxygen.
    • So, CH3OCH3 is di-methyl-ether
    • and CH3OCH2 is methyl-ethyl-ether
  12. What is conformation in terms of organic chemistry?
    • The shape due to free rotation about sigma bonds. Sigma bonds are single bonds.
    • Eclipsed conformation is when the atoms are in front of another (when looking at a Newman projection)
    • Staggered conformation is when the atoms behind are twisted so that they can be seen between the atoms in the front
    • In terms of room temperature, knowing that opposites attract and like repels (there is electron shell repulsion) a staggered conformation is more stable, and is therefore preferred. You will get all the different conformations at room temperature, but at any given moment most molecules will tend toward the more stable conformation.
  13. What is configuration in terms of chemistry?
    • The permanent, 3-dimentional shape of a molecule.
    • Trans: when the main substituents (like methyl groups) are on opposite sites of a double bond.
    • Cis: when the main substituents are on the same side of the double bond.
    • The more stable is trans since the groups will want to separate as much as possible. The double bond, however, prevents free rotation so it cannot spontaneous change shapes and will have different properties and are therefore two isomers.
  14. Does a trans compound tend to have higher or lower melting points than a cis? Why?
    What about the boiling point? Why?
    • Trans compounds tend to have higher melting points due to better symmetry.
    • Trans compounds tend to have lower boiling points due to less polarity.
    • Note: The greater number of attached alkyl groups, the greater the alkene's stability due to the alkyl groups being somewhat electron donating and stabilizing the double bond.
  15. Alkyl groups are somewhat _______ donating.
    • electron
    • They are not very electron donating. Just somewhat so.
    • A carbon that is positive needs three alkyl groups surrounding to create a more stable C. Then, if there were an OH- (like sodium hydroxide, because Na+ is a spectator ion) it would be happy to attack the carbon. And this would be a nucleophilic substitution.
  16. The atoms in most organic compounds are held together by what?
    Covalent bonds (the sharing of an electron pair between two atoms. Some ionic bonding (the transfer of electrons from one atom to another) does exist.
  17. What would be the first reaction to take place in the following scenario?
    R-CH2-X + OH-
    • An SN2 reaction:
    • The carbon would be partially positive (the halide group is very electronegative), so the OH- group would want to attack it first. Since the hydrogens and the R group are fairly equal, it is the halide group bond that is the most unstable and will break and be the leaving group next.
  18. What does it mean for a reaction to end with the molecule inverted?
    • If the new group attacks from the opposite side of the leaving group it is inverted and if there is an equal amount of each, which can happen in an SN1 reaction, it is a racemic mixture.
    • In an SN2 reaction, the nucleophile will always attach from behind and will be inverted.
  19. Consider the formation of beryllium chloride (BeCl2):
    How does the beryllium atom form two bonds with the atom of chlorine?
    • Chlorine's atomic number is 17
    •     outer shell configuration 3s23p5; featuring one unpaired electron in the third p orbital
    • Beryllium's atomic number is 4
    •     outer shell configuration 1s22s2; featuring no unpaired electrons.
    • The beryllium atom redistributes its electrons in the following manner:
    • 1. one electron is "promoted" from a 2s orbital into an empty 2p orbital, leaving two unpaired electrons - one in the 2s orbital and one in the 2p orbital
    • 2. the 2s orbital and the 2p orbital then "hybridize" to form two sp orbitals, each of which contains an unpaired electron
    • 3. each of these two hybridized sp orbitals is able to donate an electron to a covalent bond.
    • E↑ __ __ __ 2p        __ __ 2p
    • n|                            ↑   ↑  2sp
    • e|   2s            ⇒    
    • r|
    • g|  ⇅   1s                  ⇅  1s  
    • y| an s and one p form two equivalent
    •     orbital mix...        sp hybrid orbitals
    •         Orbital Energy Diagram for Be
  20. Describe the sigma and pi bonds.
    What is a triple bond?
    What happens when there are multiple bonds to rigidity, bond length, and bond strength?
    • Sigma: single bonds in which the electron density is between the nuclei. They are symmetric about the axis, can freely rotate, and are formed when orbitals (regular or hybridized) overlap directly. They are characterized by the fact that they are circular when a cross section is taken and the bond is viewed along the bond axis.
    • Pi bonds: double bonds in which the electron density overlaps both above and below the plane of atoms.
    • Triple bond: one sigma and two pi bonds.
    • The pi bonds in doubly and triply bonded molecules created a barrier to free rotation about the axis of the bond and are much more rigid than single bonded molecules.
    • The length of a bond decreases with multiple bonds.
    • Bond strength (and the amount of energy to break a bond) is larger in a sigma bond than in a pi bond, and the multiple bonds have greater bond strength than single bonds.
  21. Delocalization of charges in the pi bonds is possible when there are hybridized orbitals in adjacent atoms. This delocalization may be represented in two different ways. What are they?
    • molecular orbital (MO) approach - takes a linear combination of atomic orbitals to form molecular orbitals, in which electrons form the bonds.
    • resonance (valence bond) approach - a linear-combination of different structures with localized pi bonds and electrons, which (only) together depict the true molecule. It does NOT mean that the two draw structures are representations of the molecule resonating between the different structures. The true form is something in between the representation where the electron(s) are delocalized.
    • When you can draw multiple structures (when there are equivalent resonance structures) it increased the stability of the molecule.
  22. What are Lewis dot structures?
    The outer shell (or valence) electrons are those that form chemical bonds. Lewis dot structures are a method of showing the valence electrons and how they form bonds. These electrons, along with the octet rule (which states that a maximum of eight electrons are allowed in the outermost shell of an atom) holds only for the elements in the second row of the periodic table (C, N, O, F). The elements of the third row (Si, P, S, Cl) use d orbitals, and thus can have more than eight electrons in their outer shell.
  23. What causes a charge separation?
    If the chemical bond is made up of atoms of different electronegativity, there is a charge separation. There is a light pulling of the electron density by the more electronegative atom from the less electronegative atom. This results in the bond having a partial ionic character (i.e. a polar bond). The charge separation also causes an electrical dipole to be set up in the direction of partial positive to partial negative. A dipole has a positive end and a negative end and will line up an electric field.
  24. Put the following elements in order from most electronegative to least:
    A. fluorine, nitrogen, chlorine, oxygen
    B. fluorine, oxygen, nitrogen, chlorine
    C. nitrogen, chlorine, oxygen, fluorine
    D. nitrogen, oxygen, chlorine, fluorine
    B. is correct. Fluorine (4.0), oxygen (3.5), nitrogen (3.0), and chlorine (3.0). These elements are often paired with hydrogen (2.1) and carbon (2.5), resulting in bonds with partial ionic character. The dipole moment is a measure of the charge separation and thus, the electronegativities of the elements that make up the bond; the larger the dipole moment, the larger the charge separation. No dipole moment is found in molecules with no charge separation between atoms (i.e. Cl2, Br2), or, when the charge separation is symmetric resulting in a cancellation of bond polarity like vector addition in physics (i.e. CH4, CO2).
    (this multiple choice question has been scrambled)
  25. How does electronegativity affect the dipole moment?
    • Electronegativities in brackets: Fluorine (4.0), oxygen (3.5), nitrogen (3.0), and chlorine (3.0) are often paired with hydrogen (2.1) and carbon (2.5), resulting in bonds with partial ionic character.
    • The dipole moment is a measure of the charge separation and thus, the electronegativities of the elements that make up the bond; the larger the dipole moment, the larger the charge separation.
    • No dipole moment is found in molecules with no charge separation between atoms (i.e. Cl2, Br2), or, when the charge separation is symmetric resulting in a cancellation of bond polarity like vector addition in physics (i.e. CH4, CO2).
    • A molecule where the charge separation between atoms is not symmetric will have a non-zero dipole moment (i.e. CH3F, H2O, NH3).
  26. Which is stronger, non-polar or polar bonds?
    Non-polar bonds are generally stronger than polar covalent and ionic bonds, with ionic bonds being the weakest. However, in compounds with ionic bonding, there is generally a large number of bonds between molecules and this makes the compound as a whole very strong. For instance, although the ionic bonds in one compound are weaker than the non-polar covalent bonds in another compound, the ionic compound's melting point will be higher than the melting point of the covalent compound. Polar covalent bonds have a partially ionic character, and thus the bond strength is usually intermediate between that of ionic and that of non-polar covalent bonds. The strength of bonds generally decreases with increasing ionic character.
  27. A substance which has a formal positive charge or a partial positive charge is attracted to a substance with a formal negative charge or a partial negative charge. In general, a substance with a formal charge would have a greater force of attraction than one with a partial charge when faced with an oppositely charged specials. Give the exception to this statement.
    Spectator ions: Ions formed by elements in the first two groups of the periodic table (i.e. Na+, K+, Ca++) do not actively engage in reactions in organic chemistry. They simply watch the reaction occur then, at the very end, they associate with the negatively charged product.
  28. What two categories of compounds can created a carbon-carbon bond?
    • A) alkyl lithiums (RLi)
    • B) Grignard reagents (RMgBr),
    • because they have a δ- carbon; note that the carbon is δ- since lithium is to the left of carbon on the periodic table
    • ** looking at the charges:
    •    R   Mg  Br
    •    -    +2  -1
    • So this R group would be attracted to the Carbon nucleus.
  29. What is a nucleophile? An electrophile?
    • Nucleophile: a molecule with a free pair of electrons, and sometimes a negative charge, that seeks out partially or completely positively charged species (i.e. a carbon nucleus).
    • Electrophile: a substance that seeks electrons (H+ is an important electrophile).
  30. What is the general trend for nucleophile strength?
    • The general trend is that the stronger the nucleophile, the stronger the base it is.
    • For example
    •      RO>HO>>RCOO>ROH>H2O
  31. Consider the reaction
    NaOCH3 + CH3Br→
    • As you take the MCAT and you see things that maybe you don't recall or seem intimidating, start by finding things similar to what you do know.
    •    For instance NaOH (Sodium Hydroxide) is similar to NaOCH3 and you know that when you put NaOH in water you get Na+ OH- so you know you will get Na+ OCH3-.
    •    You see CH3Br and know that the C will be partially positive and the Br will be partially negative.
    •    You know the spectator ions, like Na+, are the 1st and 2nd groups of the periodic table.
    •    Look for the most charged thing... you know the OCH3- is it and it is a full formal charge and will be the most reactive and attracted to something positive. So it wants the C but C can only be bonded 4 times. It already has 3H and a Br. C-H bonds are almost equally shared, but the Br bond is not shared equally and is therefore weak. The weakest bond must break!
    •    You can identify a nucleophile, like OCH3, (something that is attracted to the C nucleus) and the leaving group, which is the Br.
    • So you get:
    • CH3OCH3 + Na+Br-
  32. Which of these molecules has the lowest boiling point?
    A. HO//
    B. /OH
    C. HO//
    D. HO///
    E. HO///
    B. is correct. All the molecules have hydrogen bonding capability. For compounds with the same functional group the lower molecular weight compounds generally have a lower boiling point due to the reduced opportunities for intermolecular bonding through van der Waals interactions.
    (this multiple choice question has been scrambled)
  33. Which of the following molecules are polarized?
         I. H-H
        II. F-Cl
       III. Br-Br
       IV. H-F
    A. I and III
    B. II and IV
    C. I and II
    D. II and III
    E. III and IV
    B. is correct. The polarization of a bond is determined by the relative electronegativity of each particular atom. The fluorine atom is more electronegative than a hydrogen atom and bromine atom, so they have polarized bonds, whereas the symmetrical molecules have no polarization.
    (this multiple choice question has been scrambled)
  34. Which of these molecules have the highest heat of combustion?
    A. H3-C-C-H3
    B. H2-C=C-H2
    C. H-C=C-H
    D. All have the same value.
    E. H-CH3
    C. is correct. Ethyne C has a very strong carbon-carbon triple bond, and will release the most energy when combusted (=burning).
    (this multiple choice question has been scrambled)
  35. The hybridization state of a tetrahedral carbon is:
    A. s
    B. p
    C. sp
    D. sp2
    E. sp3
    • E. is correct. When carbon has
    • 4 substituents = sp3 & tetrahedral (109.5°)
    • 3 substituents = sp2 & trigonal planar (120°)
    • 2 substituents = sp & linear (180°)
    • There are no carbon-carbon multiple bonds in tetrahedral molecules (which would be made from overlapping p orbitals). So all the orbitals (2s and the 2p orbitals in the x/y/z directions: 2px, 2py and 2pz) are available to form four sp3 hybrids for the sigma-framework.
  36. In the ammonium ion, NH4+, the nitrogen-hydrogen bonding is due to which overlapping orbitals?
    A. sp3-sp3
    B. sp2-sp2
    C. sp3-s
    D. sp-sp
    E. sp2-sp
    C. is correct. Nitrogen's atomic orbitals as well as the lone-pair of electrons are all sp3 hybridized and each of the nitrogen hybridized orbital's overlap to form a bond with the s orbital's of the four individual hydrogen atoms. The bonding between the nitrogen and each of hydrogens is therefore due to the overlapping of the sp3 from the nitrogen and s orbital from hydrogen to give sp3-s type bonding (similar to the central carbon in an alkane like methane, CH4)
    (this multiple choice question has been scrambled)
  37. Which of the labeled carbon atom(s) in the following skeletal diagram of cholesterol is (are) sp3-hybridized?

    A.C2 and C6
    B. C1, C3, C4, C5, C7, C8 and C9
    C. None of them
    D. All of them
    E. C1 only
    D. is correct. All of the numbered (in red) carbons (C1 to C9, inclusive) are all sp3 hybridized as they all have four single bonds attached to each of the highlighted carbons. Note: the hydrogen's are not all shown as this is a skeletal diagram of the cholesterol molecule.
    (this multiple choice question has been scrambled)
  38. An "isolated diene" is defined as a hydrocarbon with two double bonds separated by more than one single bond. A "conjugated diene" however has its two double bonds separated by one single bond. As such, why is a conjugated diene considered more stable than an isolated diene?
    A. Due to decreased stability as a result of electron delocalization
    B. Due to electron delocalization and orbital hybridization of the C-C single bonds
    C. Due to sigma electron pair conjugation
    D. Due to the geometry of the cumulated d orbitals of the existing double bonds.
    E. All of the above
    B. is correct. The pi electrons in the double bonds of an isolated diene are localized between two carbons whereas the pi electrons in a conjugated diene are delocalized over a wider area which further stabilizes a molecule. If you were aware that delocalization of electrons (=pi electrons) resulted in greater stability, that would have gotten you the correct answer. In addition to resonance, hybridization energy affects stability. For example, in 1,3-butadiene the carbons with the single bond are sp2 hybridized unlike in nonconjugated dienes where the carbons with single bonds were sp3 hybridized. This difference in hybridization shows that the conjugated dienes have more 's' character and draw more of the pi electrons, thus making the single bond stronger and shorter than an ordinary alkane C-C bond.
    (this multiple choice question has been scrambled)
  39. What are isomers? Give the different types of isomers.
    • Two different molecules with the same number and type of atoms (the same molecular formula).
    • Structural isomers: different atoms and/or bonding patterns in relation to each other.
    •       CH3                        H
    •        |                            |
    • H3C-C-CH2CH2CH3   H3C-C-CH2CH3
    •        |                            |
    •        H                           CH2
    •                                     |
    •                                     CH3
    • Conformational isomers: differ only by the rotation about single bonds. As a result, substituents (=ligans=attached atoms or groups) can be maximally close (eclipsed conformation), maximally apart (anti or staggered conformation) or anywhere in between (i.e. gauche conformation). Though all conformations occur at room temperature, anti is most stable since it minimizes electron shell repulsion.
    • Geometric isomers: occur because carbons that are in a ring or double bond structure are unable to freely rotate. Geometric isomers occur only as alkenes and cyclic compounds. This results in cis and trans compounds.
    •     H     H                       H      Br
    •           /                              /
    •       C=C           and          C=C
    •      /                              /    
    •    Br      Br                    Br       H
    • cis-dibromoethene     trans-dibromoethene
    • Stereoisomers: different compounds with the same structure, differing only in the spatial orientation of the atom (=configuration) and may further divide into enanitomers and diastereomers.
  40. What are enantiomers and diastereomers?
    Not a common MCAT question.
    • The are subdivisions of stereoisomers.
    • Enantiomers: two non-superimposable molecules, which are mirror to each other. In order to have one, a molecule must be chiral (contain a carbon atom that has four different substituents attached to it). They have the same chemical and physical properties. The only difference is their interactions with other chiral molecules, and their rotation of plane polarized light. Examples: A and B, C and D.
    • Diastereomers: any pair of stereoisomers that are not enantiomers. They are both chemically and physically different from each other. Examples: A and C, B and D.
  41. Describe the steps of the R,S nomenclature.
    Not a common MCAT question.
    • 1. Identify and asymmetric carbon, and the four attached groups.2. Assign priorities to the four groups, using the following rules:   i.  Atoms of higher atomic number has higher priority.   ii. An isotope of higher atomic mass receives higher priority.   iii. The higher priority is assigned to the group with the atom of higher atomic number or mass at the first point of difference.   iv. If the difference between the two groups is due to the number of otherwise identical atoms, the higher priority is assigned to the group with the greater number of atoms of higher atomic number or mass.   v.  To assign priority of double or triple bonded groups, these atoms are replicated:
    • 3. View the molecule along the bond from the asymmetric carbon to the group of lowest priority (i.e. the asymmetric carbon is near, and the low priority group is far away).
    • 4. Consider the clockwise or counterclockwise order of the priorities of the remaining groups. If they increase in a clockwise direction ,the asymmetric carbon is said to have the R configuration. If they decrease in a clockwise direction, the asymmetric carbon is said to have the S configuration.
  42. What is plane polarized light?
    Light is an electromagnetic wave that contains oscillating fields. In ordinary light, the electric field oscillates in all directions. However, it is possible to obtain light with an electric field that oscillates in only one plane. This type of light is known as plane polarized light.
  43. What are optical isomers? Describe how their nomenclature is defined.
    Not a common MCAT question.
    • Stereoisomers that differ by different spatial orientations about a chiral carbon atom. When plane polarized light is passed through a sample of chiral substance, it will emerge vibrating in a different plane than it started. Optical isomers differ only in this rotation.
    • If the light is rotated in a clockwise direction, the compound is dextroroatrary, and is designated by a D or (+).
    • If the light is rotated in a counterclockwise direction, the compound is levrorotary, and is designated by an L or (-). All L compounds have the same relative configuration as L-glyceraldehyde.
    • A racemic mixture will show no rotation of plane polarized light. This is a consequence of the fact that a racemate is a mixture with equal amounts of the D and L forms of a substance.
  44. How is specific rotation (α) defined?
    (Not a common MCAT question)
    • It is an inherent physical property of a molecule and is defined as:
    • α=          Observed rotation in degrees       
    •     (tube length in dm)(concentration in g/ml)
    • The observed rotation is the rotation of the light passed through the substance.
    • The tube length of the tube that contains the sample in question.
    • The specific rotation if dependent on the solvent used, the temperature of the sample, and the wavelength of the light.
    • Note: there is NO clear correlation between the absolute configuration (i.e. R, S) and the direction of rotation of plane polarized light.
  45. Which of the following pairs of compounds represent pairs of structural isomers?
    A. 2-methylbutaine and n-pentane
    B. 2-bromopentane and 3-bromopentane
    C. sec-butyl bromide and tert-butyl bromide
    D. n-propyl bromide and isopropyl bromide
    E. all of the above
    E. is correct. Structural (constitutional) isomers have the same molecular formula but a different arrangement of atom connectivity. Structural isomers can come in three sub-classes: (i) chain isomers (variation in the basic carbon skeleton); (ii) position isomers (position of a functional group in a chain), and; (iii) functional group isomers (where the re-arrangement of atoms also leads to a change in the functional group).
  46. Choose the molecules which exist as (E)- and (Z)- isomers:
    A. A and B
    B. A
    C. A and C
    D. B and C
    E. A and D
    C. is correct. Geometric isomers (E-/Z-) exist in molecules where there is restricted rotation about a multiple bond (e.g., C=C) and where there is no plane or axis of symmetry about the double bond. Terminal alkenes, those with the double bond at the end of a carbon chain, do not exhibit geometric isomerism (after all, since H is not considered as a substituent, there is no substituent attached to the terminal carbon if it is a pure alkene).
    (this multiple choice question has been scrambled)
  47. Which one of the following can exhibit geometric isomerism?
    A. 1-hexene
    B. 1-pentene
    C. 2,3-dimethyl-2-butene
    D. 2,3-dimethyl-1-butene
    E. 2-hexene
    E. is correct. Geometric isomers (E-/Z-) exist in molecules where there is restricted rotation about a multiple bond (e.g., C=C) and where there is no plane or axis of symmetry about the double bond. Terminal alkenes, those with the double bond at the end of a carbon chain, do not exhibit geometric isomerism.
    (this multiple choice question has been scrambled)
  48. Benzene reacts with bromine to form bromobenzene, but the reaction requires the used of an appropriate Lewis acid catalyst such as ferric bromide.

    It is reasonable to assume that Reaction I will proceed via which of the following mechanisms?
    A. Electrophilic substitution of the first order
    B. Nucleophilic substitution of the second order
    C. Nucleophilic substitution of the first order
    D. Electrophilic substitution of the second order.
    A. is correct. It may be recalled that a Lewis acid ACCEPTS electrons. In this way, the catalyst ferric bromide makes the reaction bromine quite positive. Bromine does not like to be positive!!!!!! Now Br really, really, really wants electrons (= powerful electrophile). That is why it can add to the electron dense benzene ring.
    (this multiple choice question has been scrambled)