OCHEM 1 QS missed Flashcard
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.
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.
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
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.
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.
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.
A) sp³, tetrahedral
B) sp², trigonal planar
C) sp, linear
D) sp³, bent
A) Both are R
B) Both are S
C) One is R and one is S
D) Neither is chiral
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
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.
B) conformational isomers
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.
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.
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) 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.
A) (CH3)3CCl and CH3Cl
B) CH?Br and CH?(CH?)?CH?
C) CH? and (CH?)?CCl
D) CH?OH and CH?OCH?