OCHEM 1 QS missed

How many lone pairs of electrons are found on the central atom in the chlorate ion?

A) 0
B) 1
C) 2
D) 3

Which of the following best describes the shape of an SF? and a BrF? molecule, respectively?

A) See-saw and T-shaped
B) T-shaped and tetrahedral
C)See-saw and trigonal planar
D) Tetrahedral and T-shaped

No conceptual stuff here; this is one of the few things you have to memorize. You need to memorize VSEPR shapes for all combinations of bonds and lone pairs. Again, you’ll need to draw the Lewis structure to determine the number of lone pairs. SF? has four bonded substituents plus one lone pair, this is always a see-saw shape. BrF? has three bonded substituents and two lone pairs, which will always be T-shaped. Which makes A the best answer.
Cis-1-Bromo-2-Chlorocyclohexane is most stable in which of the following conformations?

a) Chair conformation, with both substituents in the equatorial position
b) Chair conformation, with Bromine in the equatorial position
c) Boat conformation, with both substituents in the axial position
d) Chair conformation, with both substituents in the axial position.

As always BE CAREFUL. It is actually not possible to have both Br and Cl in the equatorial position when they are cis to oen another and on adjacent carbons. On these questions you need to draw a chair with axial and equatorial substituents shown. What combinantions are possible depends entirely on whee they are on the ring compared to each other and whether they are cis/trans; and the various combinations aren’t worth memorizing. In this case, when Br and Cl are cis to each other, onet must be axial and one equatorial. Bromine should be in the equatorial because it is the largest, so B is correct.
Utilizing the Cahn-Ingold-Prelog rules, which of the following correctly lists the substituents of the chiral carbon in the compound below in order or decreasing priority.

A) hydroxyl group, aldehyde group, ethanol group, hydrogen
B) hydroxyl group, ethanol group, aldehyde group, hydrogen
C) hydroxyl group, aldehyde group, methanol group hydrogen
D) hydroxyl group, methanol group, aldehyde group hydrogen

C; These are the simple rules we use when assigning R and S and they are based solely on molecular weight with one important caveat: when comparing identical atoms you look at the atoms attached to each atom with a double or triple bond counting the same as if that atom were bonded to the other atom two or three different times. C=O, for example, is like carbon being bonded to two oxygens and would thus be higher priority than C-OH. Here, the hydroxyl is first because oxygen has a higher molecular weight than the other two carbons. The next in line is the aldehyde group because it has a double bond to its oxygen. Then the methanol group and finally the hydrogen; making Answer C correct.
Which of the following is NOT true of the nitrate ion?
A) The molecule bears three formal charges
B) the molecule contains three oxygen atoms
C) the nitrogen atom bears one lone pair
D) Each oxygen bears two or more lone pairs.
Carefully note that you are lookign for the NOT true statement. First, you’ll need to draw the lewis structure for nitrate, NO3-. This should show you that the molecule does contain three separate formal charges, one on the nitrogen and one on each of two oxygens. The nitrogen, however, does NOT contain a lone pair. All of the oxygens do bear At LEAST two lone pairs, so C is the correct answer.
Many organic chemistry reactions begin with abstraction of an alpha proton. The reactivity of alpha protons is best explained by the:

A) electron withdrawing effect of the neighboring carbonly
B) electron donating effect of the neighboring carbonyl
C) stabilization of the resulting carbocation
D) resonance of the resulting carbanion

Answer A is true, the carbonyl does have a partial positive charge on the carbonyl carbon, which does withdraw electrons, but A is not the best answer. Remember not to get too sold on answers too quickly. B is false because of what we just said regarding A. C is false because abstraction of a proton leaves a carbanion, not a carbocation. D is correct; it is the ability of the unstable carbanion to resonate up onto the carbonyl that makes the conjugate base quite stable.
What is the formal charge on the nitrogen atom in the nitrite ion?
A) o
B) +1
C) -1
D) +2
Another question that basically tests you ability to draw lewis structures; hopefully you’re getting that this is very important. The Nitrite ion has two oxygens bound to the central nitrogen atom with one having a formal negative charge and the other being couble bond. This gives nitrogen 3 bonds and thus no formal charge.
According to VSPER theory, the hybridization and shape of the chlorite ion are best described as?

A) sp³, tetrahedral
B) sp², trigonal planar
C) sp, linear
D) sp³, bent

9) D; Draw the Lewis structure and you should find that chlorine has two lone pairs around it plus two bonded oxygens. That is a hybridization of sp3 and a shape of bent; exactly like water. Tetrahedral may be tempting, but this is only the shape if the four substituents are all bonded atoms with no lone pairs.
10) A; Decarboxylation will occur, which means that the carboxylic acid will break off to form CO2. As this happens, the bond between carbons 3 and 4 will break and the electrons will go onto carbon 3, making it a carbanion with a formal charge of -1. Carbon 2 will be unaffected, so it remains uncharged at zero. This product also has a resonance form with the negative charge on the oxygen, and no formal charge on either carbon 2 or 3. However, because none of the answer choices reflect this option, this is of no concern.
Which of the following bonds has the highest bond energy.

A) C=O
B) S=O
C) P=O
D) N=O

The bond with the highest bond energy is the one that is the most stable. Because the options are all double bonded to an oxygen, we can just look at the effect of the first species on the double bond. These species all vary in size, with the smallest atom being capable of getting the closest to the oxygen and thus forming the shortest, strongest double bond with the most pi orbital overlap. D is thus the best answer.
In the followign molecule
14) B; This is another good test of where you’re at. Are you mastering every single line of each lesson? Are you ensuring that you have EXCELLENT conceptual definitions of all principles and vocabulary? You should know that enantiomers must have opposite R and S at every chiral center. At the Cl and Br carbons this is not the case because Cl is in the same position on both. They are not anomers because these only occur on cyclic hemiacetals such as glucose. They ARE diastereomers, but answer B is the best answer because an epimer is a type of diastereomer and is exactly what they are. Epimers, if you don’t recall, are pairs of molecules with one or more chiral center that differ at ONLY by their R/S configuration at ONE of those chiral centers.
This is an excellent way to test your R/S abilities. You should get R for the ethanol bearing carbon on the lower right side and S for the one on the upper left side. The compound is NOT a meso compound, because both the chlorine bearing carbons are R. The position of the ethanol groups is actually appropriate for a meso compound
Which of the following is true of the absolute configurations fo the two ethanol-bearing carbons in the molecule shown above?
A) Both are R
B) Both are S
C) One is R and one is S
D) Neither is chiral
Which molecule is expected to have the larger dipole moment, AsCl? or AsCl??

A) AsCl?, because AsCl? is trigonal planar and has no net dipole
B) AsCl?, because it is trigonal pyramidal
C) AsCl?, because it has two more polar As-Cl bonds than does AsCl?
D) AsCl?, because it is symmetrical

17) B; To answer this correctly, you must be able to draw the Lewis structures, confidently decide how many, if any, lone pairs exist, and have memorized all your VSEPR shapes. AsCl3 will be trigonal pyramidal and the lone pair cannot cancel the three polar As-Cl bonds, resulting in a net dipole. AsCl5, on the other hand, will be trigonal bipyramidal and all the dipoles will exactly cancel each other out. This makes B the only plausible answer.
Which of the following statements must be true concerning a molecule containing one ore more polar bonds?

I. The molecule will have a net dipole moment
II. The molecule must contain atoms of varying electronegativities
III. The molecule will have a higher boiling point than molecules with a similar molecular weight but no polar bonds.

A) I only
B) II only
C) II and III
D) I and II

Statement I is false because the geometry could be such that the individual dipoles cancel. Statement II is true because if it were not true there would be no dipoles. Finally, statement III could be false. If two molecules had similar molecular weights and one had dipoles, but no net dipole, and the other had no dipoles at all, they would be expected to have similar boiling points. In other words, a molecule wiht no net dipole would not have intermolecular forces caused by its individual dipoles.
Which of the following isomers is most effectively resolved via reaction with a chiral reactant?

A) enantiomers
B) conformational isomers
C) diastereomers
D) structural isomers

Answer B is false because conformational isomers are not true isomers; they are the same molecule and therefore cannot be separated. Answer C is false because diastereomers are already non-identical mirror images and would not be easier to separate after reaction with a chiral molecule. Answer D is false because structural isomers are completely different molecules with different bond to bond connectivity and reaction with a chiral reactant would be meaningless. For enantiomers, however, reaction with a chiral molecule will produce two different substances. A is thus the best answer.
Which of the following alkly-substituted alkenes will have the highest heat of combustion per carbon carbon bond?

A) mono-substituted
B) cis-di-substituted
C) trans-di-substituted
D) tetra-substituted

22) A; A tetra-substituted alkene will be the most stable because alkenes are electron hungry and these four R groups will all be donating electron density via conjugation. Ironically, though, they will also be the most reactive. Because stable molecules have a LOW heat of combustion, the correct answer is the least substituted alkene, Answer A.
Specialized Machines can detect the relative energy of individual electrons before, during, and after a reaction. If such a machine is utilized to detect the relative energy of the s and p valence electrons of elemental carbon and compare it to the energy of these same carbon electrons in methane, which of the following best describes the observed energy differences?
A) Both s and p electrons will increase in energy from elemental carbon to methane

B) both s and p electrons will decrease in energy from elemental carbon to methane

C) The s electrons will increase in energy from elemental carbon to methane, but the p electrons will decrease

D) The s electrons will decrease in energy from elemental carbon to methane, but the p electrons will increase.

This question tests your understanding of how orbitals hybridize to form bonds. in methane, the s and p electronic orbitals of carbon hybridize to form a new orbital that is slightly higher in energy than the s orbitals were before, but slightly lower than the original p orbitals. This makes C the only possible answer.
Cis-1,2-dibromoethene has a higher boiling point, but a lower melting point, than trans-1,2-dibromoethene. which of the following provides the best explanation for this observation?

A) cis-1,2-dibromoethene has a net dipole moment, increasing boiling point, but is less symmetrical, decreasing melting point.

B) trans-1,2-dibromoethene has a net dipole moment, increasing boiling point, but is less symmetrical, decreasing melting point

C) cis-1,2-dibromoethene has no net dipole, decreasing boiling point, but is more symmetrical, increasing melting point

D) trans-1,2-dibromoethene has no net dipole, increasing boiling point, but is more symmetrical, decreasing melting point.

As we’ve said before, it is a VERY good idea to draw these two structures out, even if you know exactly what they are. Having them in front of you will make it easier on your brain and will help things “jog in your head” that wouldn’t otherwise. The cis version of dibromoethene does have a net dipole, while the trans version does not. A dipole creates greater intermolecular forces and increases boiling point. However, the shape of the trans version is much more conducive to stacking. Closer stacking means closer contact in the solid and thus higher melting points. Answer A exactly states these two facts.
A student in the lab attempts to react N, N- dimethylamine with Chlorocyclohexane. Assuming a reaction does occur, which mechanism is most likely?

A) A two-step reaction beginning with the spontaneous dissociation of the chlorine group to form a carbocation
B) A single step reaction involving back side attack of the chlorine bearing carbon
C) A multi-step nucleophilic aromatic substitution reaction
D) A single step reaction involving abstraction of a hydrogen.

Answer A is false because -as far as the MCAT is concerned–SN1 reactions only occur with tertiary substrates. Answer B is false because the dimethylamine is too sterically hindered, especially for attack from within a ring. Answer C is false because cyclohexane is not aromatic. Finally, Answer D is true. The reaction will most likely proceed via an E2 reaction with the dimethylamine acting as a base.
Which of the following pairs represent a molecule that is unlikely to undergo an SN2 reaction and a molecule that is unlikely to undergo an SN1 reaction respectively?

A) (CH3)3CCl and CH3Cl
B) CH?Br and CH?(CH?)?CH?
C) CH? and (CH?)?CCl
D) CH?OH and CH?OCH?

30) A: Answer B is false because methyl bromide is an almost ideal candidate for SN2 with bromine being an excellent leaving group and the methyl backbone providing almost no steric interference. Answer C is false because tert-butyl chloride is very likely to undergo SN1. Answer D is false because methanol can undergo an SN2 reaction as long as the alcohol gets protonated first.

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