Naming, Substrate Rules, & Intro to Reactivity Flashcards
What is the IUPAC name for the following structure:
5-butyl-6-isobutyl-8-secbutyl-dodecane
First, find the longest carbon chain in the structure (the parent chain), and then find the substituents. Notice that the first substituents, on both the left and right side of the molecule, end up being five carbons away from the terminal carbon. If we count left to right (the incorrect way) we end up with the substituent numbering -5,6,8. Whereas, if we number the parent chain from right to left (the correct way), there is only a slight difference but we wind up with a lower number for the middle substituent, isobutyl, when counting this way.
What is the IUPAC name for the following structure:
1-iodomethane
The prefix “meth” = 1 carbon
The common name would just be iodomethane
What is the IUPAC name for the following structure:
1-bromoethane
The prefix “eth” = 2 carbons
The common name is just bromoethane
What is the IUPAC name for the following structure:
1-fluoropropane
The prefix “prop” = 3 carbons
The common name is just fluoropropane
What is the IUPAC name for the following structure:
1-chlorobutane
The prefix “but” = 4 carbons
The common name is just chlorobutane
What is the IUPAC name for the following structure:
2-chloropropane
The common name is isopropyl chloride (the structure is a chlorine atom covalently bonded to the #2 carbon of an isopropyl group)
What is the IUPAC name for the following structure:
2-bromobutane
The common name is secbutyl bromide (the structure is a butyl group with a bromine substituent covalently bonded to the second carbon)
What is the IUPAC name for the following structure:
1-chloro-2,2-dimethylpropane
The common name is neopentyl chloride (the structure is a neopentyl group covalently bonded to chlorine at a terminal carbon)
What is the IUPAC name for the following structure:
1-iodo-2-methylbutane
The common name is isopropyl iodide (the structure is an isopropyl group covalently bonded to iodine at the #2 carbon)
What is the molecular/electronic geometry for the following molecule: H2C=O
What are the hybridization states for each atom? (besides H)
Also, note the bond angles around the central atom
The molecular and electronic geometries of this molecule are trigonal planar due to the sp2 hybridization of the carbon atom, giving bond angles of 120 degrees.
What is the molecular/electronic geometry for the following molecule: CH3+, why is this important when considering the stereoselectivity of certain reactions?
What are the hybridization states of the atoms? (Besides H)
The molecular and electronic geometries of this molecular are trigonal planar. This is due to the sp2 hybridization state (remember that + charges DO NOT count as a group when determining the hybridization state) of the central carbon atom, giving bond angles of 120 degrees. This allows for a nucleophile to approach the localized charge from either face, resulting in the racemization of the products.
List the stability trend for carbocations
The alkyl groups around a carbocation will help stabilize the positive charge due to their positive inductive effect (increasing electron density at the carbocation center). Also, remember that mesomeric effects (resonance stabilization) is stronger than inductive effects.
How does a positive inductive effect impact the acidity of a carboxylic acid?
Will the pKa increase or decrease?
The pKa will increase, the positive inductive effect will increase the electron density of the O-H bond and consequently strengthen the bond (decreasing the acidity).
How does a positive inductive effect impact the acidity of a carboxylic acid?
Will the pKa increase or decrease?
The pKa will decrease, the negative inductive effect will decrease the elctron density of the O-H bond, weaknening the bond so that the proton is more likely to dissociate (becoming more acidic).
What is hydrohalogenation?
In a hydrohaolgenation addition, such as the case with HBr added to an alkene, the inductive effect of the alkyl groups on an asymmetric alkene will increase the electron density at one carbon and consequently decrease the electron density at the adjacent carbon, thus polarizing the pi electrons of the double bond. The proton, H+, will be added to the carbon with increased electron density (intuitively this makes sense because a proton has a positive charge and is attracted to the δ- carbon due to its increased electron density) and bromide to the carbon with decreased electron density.
Some use the rule that the proton will be added to the carbon with the most number of hydrogens and the bromide will be added to the carbon with the least number of bonded hydrogens (this rule does not always work so we must consider the inductive effects).