Chem/Phys II Flashcards
(286 cards)
bond angles to know: trigonal planar
120 degrees
bond angles to know: linear
180 degrees
VSEPR theory
Valence-shell electron-pair repulsion theory; because electron pairs repel, molecules adjust their shapes so that valence electron pairs are as far apart as possible
trigonal planar
has 3 binding regions (no electron pairs)
aluminum
is an exception to the quartet rule, makes 3 bonds
bond length is determined by
the sizes (radii) of the two bonding atoms and how many electron pairs they share
SN1 stereochemistry
In an SN1 reaction, the original stereochemistry of the molecule is lost when the carbocation intermediate (a planar structure) is formed. Thus, the product is present as a racemic mixture. Such a mixture includes each enantiomer in a 1:1 ratio, meaning that it promotes no net rotation of polarized light.
epimer
These molecules differ at a single stereogenic center, classifying them as epimers.
anomeric carbon
anomeric carbon (the carbon atom bound to two separate oxygen groups)
E1 reaction-DONT NEED TO KNOW
E1 reactions are unimolecular eliminations. These reactions are analogous to SN1 mechanisms, as both are first-order and involve the formation of a carbocation. As such, both are heavily favored by tertiary reagents. Additionally, use of a mild or weak base tends to push a reaction toward E1, as E2 reactions (which are bimolecular) require moderately strong basic species. Finally, like SN1 processes, E1 reactions prefer protic solvents for carbocation stabilization.
UV-visible spectroscopy
-An analytical technique involving measurement of the UV-visible light absorbed by a substance. -mainly used to analyze conjugated systems
What would PCC oxide an alcohol to?
As a weak oxidizing agent, PCC will oxidize a primary alcohol to an aldehyde. According to IUPAC naming convention, this aldehyde group will be given the highest priority, so our numbering should begin at that end of the chain. The parent alkane is five carbons long, and our high-priority aldehyde is given the suffix “-al,” yielding “pentanal.” From there, the methyl substituents are numbered based on the carbons to which they are attached.
What would CrO3 and K2Cr2O7 turn a primary alcohol into?
strong oxidants and will thus oxidize a primary alcohol directly to a carboxylic acid
addition of pyridine (ring with an N in it)
(a weak base) will decrease the chance of this protonation, inhibiting carbocation formation
which is a better acid ethanol, or ethanthiol
Since sulfur is much larger than oxygen, the conjugate base of ethanethiol is better able to delocalize negative charge, increasing its stability. Remember, the more stable the conjugate base, the stronger the acid.
SN2
These processes, formally known as bimolecular nucleophilic substitution reactions, happen in one step and involve a “backside attack” by a strong nucleophile. Since the attacking atom must bind at the same time as the leaving group is removed, these reactions require an unhindered substrate.
good leaving groups are
weak bases
Can acetals form under basic conditions?
No, acetals cannot form under basic conditions. Under basic conditions, only hemiacetals can be formed. Acetals, which result from the reaction of a hemiacetal with an alcohol, require acidic conditions.
Hemiacetals and hemiketals exist in equilibrium with
aldehydes and ketones.
alpha hydrogen
An alpha hydrogen is one that is bound to the carbon immediately adjacent to a carbonyl carbon
chromic acid
Chromic acid (H2CrO4), like many chromium-containing reagents, is a strong oxidizing agent. Reaction of a primary alcohol with such a compound will oxidize it as thoroughly as possible. Since a primary alcohol has only one bond to carbon, it possesses the ability to form three bonds to oxygen, creating a carboxylic acid.
reducing agent NaBH4 turns an aldehyde into?
could turn an aldehyde into an alcohol
kinetic v thermal product
The kinetic product is the one that is less thermodynamically stable, but is easier to synthesize due to a lower activation energy. In general, kinetic enolates are less substituted than their thermodynamic counterparts. Here, the double bond has formed in the position that is less sterically hindered (between carbons 1 and 6).If this were the thermodynamic enolate, it would have the double bond between carbons 1 and 2 (the more substituted position adjacent to what formerly was the carbonyl carbon).
reducing agents
-Chemically, deuterium and hydrogen behave identically. Thus, lithium aluminum deuteride may be considered to be equivalent to lithium aluminum hydride (LAH). LAH is a strong reducing agent and is certainly capable of reducing both carboxylic acids and aldehydes. It would therefore be the most appropriate choice. -Sodium borohydride would be perfectly adequate for selective reduction of the aldehyde, but it lacks the reducing power necessary to reduce the carboxylic acid. -Sodium hydride is a strong base, but is not a hydride source and thus is not a reducing agent. -Potassium cyanide is a nucleophile and a weak base. It is not an appropriate choice for any reduction reactions.