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17 Cards in this Set

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What is this?
What is the hybridization of the carbonyl carbon in N,N-dimethylformamide?
- A carbonyl carbon is one that is double bonded to an oxygen. Since the carbon contains one π bond and three σ bonds (one double bond and two single bonds), it must be sp2 hybridized.
- Note that we do not need to know the structure of N, N-dimethylformamide to answer this question.
- The anion formed is stabilized by aromaticity.
- When 1,3-cyclopentadiene reacts with sodium, it goes from a nonaromatic to an aromatic compound.
- A planar cyclic compound is aromatic if it satisfies Huckel's rule—that is, it has (4n+2) π electrons, where n is any integer. The product of the reaction has 6 π electrons, so is therefore aromatic.
- An aromatic compound is very stable because the π electrons are delocalized throughout the ring. In this reaction the products are much more stable, so the reaction will be spontaneous.
What is the BEST explanation for this observation?
When aminobenzoic acid loses a proton, the conjugate base assumes a charge of –1. Therefore, any resonance structure must also posses the same net charge.
Why is this an appropriate resonsnce from of the conjugate base of p-aminobenzoic acid?
What are the characteristics of acetone; and why is it commonly used as a solvent to promote SN2 and E2 reactions?
- It is aprotic and polar.
- Since it cannot form strong intermolecular bonds and is of low molecular weight it has a low boiling point.
- Acetone exists in two forms, with the keto form predominating by 99%. This is because the carbon-oxygen double bond (the carbonyl bond) is much stronger than the carbon-carbon double bond.
How many distinct organic compounds have the molecular formula C5H12?
Three. Distinct compounds arise because of the different degrees of branching that can occur in the carbon chain.
For cyclic hexanes, what is the most favorable position for the substituents.
More favorable for substituents to occupy equatorial positions since they are more “out of the way.”
(T/F) Tautomers are a specific type of structural isomer.
- True.
- This reaction results in the formal migration of a hydrogen atom accompanied by a switch of adjacent conjugated double bonds.
What is the approximate C-N-C bond angle in diethylamine; and what is it's shape?
A. 107 deg
B. 110 deg
- The normal bond angle for an sp3 hybridized atom is 109.5 degrees.
- By looking at the drawing of diethylamine, you will notice that nitrogen has a free electron pair.
- Nonbonded (or lone) electrons pairs exert stronger repulsive effects, forcing the atoms attached to nitrogen down, and closer together.
- VSEPR predicts that the bond angle MUST BE LESS than 109.5°.
- Bromine: axial; t-butyl: equatorial
- The t-butyl is the bulkiest substituent, so it must assume the equatorial position
- Since the bromine is trans to the t-butyl and is two positions away, it must assume the opposite configuration as the t-butyl group, so is therefore axial.
in terms of axial and equatorial for Bromine and t-butyl group?
- In order to identify the product, we start by recognizing and classifying the reactants.
Methyl chloride: primary alkyl halide
Sodium ethoxide: strong nucleophile.
- What reaction mechanism does this scream out? SN2 : the ethoxide attacks while the chloride leaves, forming dimethyl ether, choice C. (The sodium is just a spectator ion here).
- This reaction type goes by the name of the Williamson ether synthesis.
What is the reasoning behind this? What kind of reaction is this?
There are three chiral centers so 8 stereoisomers.
*** In general, be on the look out for meso compounds (not present in this scenario)!
How many different stereoisomers does the following compound have?
Which of the following compounds will exhibit the greatest dipole moment: (E)-1,2-Dichloro-1,2,-diphenylethene AND (Z)-1,2-Dichloro-1,2-diphenylethene?
How many structural isomers of C3H6Br2 are capable of exhibiting optical activity?
see above.
see above.
- This is a SN2 reaction made to look difficult.
- SOCl2 reacts with primary and secondary alcohols to generate something with a very good leaving group, ClSO2–. (OH– is a poor leaving group.)
- The compound formed is reactive in nucleophilic substitution reactions.
- The Grignard reagent is a good nucleophile. To get the question correct, you have to identify R. Since our product has 8 carbons, and 2 came from the ethyl magnesium bromide, our starting compound must have 6 carbons attached to a leaving group; answer D.
See above.
- KOH dissociates in to K+ and OH– ions, the latter of which is a strong base. The elimination will therefore proceed via the E2 mechanism. Since the hydroxide ion is not a bulky base, and the substrate is a secondary alkyl halide (rather than a more sterically hindered tertiary alkyl halide), the resulting alkene will be the more thermodynamically stable, more substituted one depicted above.
- I.e. the double bond forms between the second and third carbon atoms.
*** (there is no such thing as racemic and inversion when it comes to elimination reactions!!!)
- When either of the alkyl chlorides undergoes elimination, the same alkene will be formed: propene.
- Since this is an E1 reaction, the substrate that can form the more stable carbocation will be most reactive.
- Isopropyl chloride, a secondary alkyl halide, can form a more stable carbocation than n-propyl chloride, a primary alkyl halide.
==> Therefore isopropyl chloride will react the most and n-propyl chloride will be present in excess.
see above