Organic Chemistry Class 1 Flashcards
A physical organic chemist argued that the formation of the axial alcohol is favored with bulky reducing agents solely because H experiences minimal steric hindrance when attacking from the equatorial direction. Therefore, this argument is based upon:
A. the stability of the product.
B. ΔH of the reaction.
C. thermodynamic considerations.
D. kinetic considerations.
D. kinetic considerations.
The stability of the products is used to determine ΔHrxn, a thermodynamic principle. Thus choices A, B, and C are essentially the same. Steric considerations in the transition state control the rate of a chemical reaction. Rate means you’re thinking of kinetics (choice D).
Hydride reducing reagents are reactive because they contain at least one nucleophilic hydrogen atom. Under what conditions should hydrogen be nucleophilic?
A. When they are bonded to more electronegative atoms
B. When they have a partial negative charge
C. When they are acidic
D. When they are sterically unhindered
B. When they have a partial negative charge
Nucleophile is a code word for an atom that has some degree of negative charge (choice B), and nucleophilic hydrogens are actually strong bases. Hydrogens bound to more electronegative atoms have positive character and are electrophilic, not nucleophilic (eliminate choice A); if the electronegativity of the element is great enough, it may make the hydrogen acidic (eliminate choice C). Steric factors play no role in the electronic nature of the hydrogen (eliminate choice D).
The passage indicates that benzene is susceptible to electrophilic attack.
Benzene could be used as an inert solvent in all of the following reactions EXCEPT:
A. Grignard reactions with CH3MgBr.
B. SN2 reactions with NaI.
C. ester hydrolysis with KOH.
D. bromination reactions with Br2/FeBr3.
D. bromination reactions with Br2/FeBr3.
D. Recall that an electrophile is an atom with some degree of positive charge. When molecular bromine interacts with the Lewis acid catalyst FeBr3, a very strong electrophile (effectively Br+) is produced, which can be used to brominate aromatic compounds like benzene. The carbons in Grignard reagents are nucleophilic because carbon is more electronegative than Mg (eliminate choice A). SN2 reactions use iodide because it is also a good nucleophile (eliminate choice B). Finally, hydroxide with its negative charge is clearly nucleophilic (eliminate choice C).
Which one of the following reagents is a good Lewis acid?
A. AlCl4–
B. BF3
C. HF
D. NH3
A Lewis acid is a molecule that can accept a pair of nonbonding electrons. These molecules can be recognized easily because they always have a positive charge and/or an incomplete octet. None of the species in choice A, C, or D fit this criterion, but the boron atom in boron trifluoride (choice B) lacks a complete octet.
The stereoconfiguration around the migrating α carbon remains intact during a Baeyer-Villiger oxidation. Therefore, it can be postulated that during the key migration step, this α carbon is:
A. a carbocation.
B. a radical.
C. π bonded to another atom
D. sp3 hybridized having four σ bonds
D. Unchanged stereoconfiguration implies that the carbon remained chiral throughout the reaction. Choice D is correct because only sp3 hybridized carbons may be chiral.
What does an acetate group look like
Which one of the following compounds is considered to be the weakest oxidizing agent?
A. HNO3
B. HClO4
C. H3PO4
D. O3
C. An oxidizing agent is a molecule with two or more highly electronegative atoms directly bonded together. Since phosphorus is not highly electronegative, phosphoric acid (H3PO4) is not an effective oxidizing agent.
What are the hybridization states of N-1 and N-2 in thalidomide?
At first glance, both nitrogen atoms appear to be best described as sp3hybridized because both have three bonds and one lone pair. However, resonance structures can be drawn with the lone pairs on each of the nitrogen atoms being in resonance with neighboring carbonyls. In order to accommodate the resonance structures, the lone pairs must be in an unhybridized p orbital, which requires each nitrogen atom to be sp2 hybridized.
How many p electrons does thalidomide possess?
A. 7
B. 12
C. 14
D. 18
D. Thalidomide has a total of 18 π electrons; 14 from double bonds, and 4 from the two nitrogen lone pairs. Choice A (7) is the number of double bonds. Choice C is the number of π electrons from the double bonds alone.
Treatment of thalidomide with KOH followed by PhCH2Br will produce which of the following functional groups?
A. Carboxylic acid
B. Alcohol
C. Amide
D. N-alkylated imide
D. An imide is a nitrogen flanked by two carbonyl groups (such as the one found in thalidomide), and is acidic if the third group on nitrogen is a proton. Therefore, when treated with a base (KOH), the imide is more likely to react than the carbonyl carbon (eliminating choice A). When the anion forms, it is readily alkylated by the benzyl bromide (via an SN2 mechanism) to give an N-alkylated imide (see figure below). An amide (choice C) would not result because neither C—N bond of the imide is broken. Alcohols are not possible products from any reaction of this carboxylic acid derivative.
A low retention time in gas chromatography is associated with what type of compound
a volatile (low boiling point) compound
Rf in thin-layer chromatography is associated with what type of compound?
a nonpolar compound
(because non polars travel farther on the mobile phase while the stationary phase is polar and holds on to polar molecules more tightly)
Retention time in gas chromatography is the time it takes a compound to be eluted and detected after injection into the system. Of the following compounds, which will display both a short retention time and a high Rf in analysis by thin-layer chromatography?
A. I
B. II
C. III
D. IV
III
because it’s non polar with lowest molecular mass
