organic paper questions Flashcards
What is NaOAc
What sterics is the Hydrogen on here
What will the other sterics be when the I switches position
The Mg and Cl we define as both being on the top face (in same plane) and next to each other. When flipped things in same place alternate. If Mg goes axial, Cl must be equatorial
Which side would the equilibrium between ring flips lie
On the left. Cl is our priority group as its the biggest and so we want it in the most favourable position which is equatorial
Why couldn’t you do a ring flip for this compound
t-Bu will only sit in equatorial because axial would be far too unfavourable/ high in energy (conformational lock)
how do you know the reaction is Sn2- it cannot form a stable cation so must be Sn2
Draw a curly arrow mechanism for the following activation of an alcohol
no stereochemistry to worry about because its primary
An alcohol being activated by being turned into a sultanate ester. What are the steps in this reaction
Sultanate esters are good leaving groups. What happens in this substitution reaction
Alcohols can also be activated by being converted into acyl chlorides. What reactant is used to do this
Alcohols can also be activated by being converted into acyl chlorides. What reactant is used to do this
What is the mechanism of this reaction
Sn2 inversion where there is stereochemistry
Why would there be no reaction
The major product (most substituted alkene) could have two conformations because the sterics are not specified so could have two different transition states. So to work out the reaction profile decide what transition state had the lowest energy
Predict the major product and draw a curly arrow mechanism for the following reaction
heat favours elimination over substitution
elimination
- NaOMe is a small and not very sterically hindered base so can attack for a more substituted product.
- Secondary substrate cannot form a stabilised carbocation intermediate so must be E2
elimination
- KOCET3 is a big base/ sterically hindered so will go for least substituted product
2.Secondary substrate cannot form a stabilised carbocation intermediate so must be E2
- acetone- polar aprotic so SN2
- Top Br removed because its on sp3 hybridised carbon and other is sp2
Draw “curly arrow” mechanisms to determine the major products from the following reactions
How would this propagation step react and form
forms tertiary radical rather than primary as it is more stable/ transition state lower in energy
propagation step product
What would each of these termination steps look like
A-B; alkene + HBr= halogenolakane but under UV light so have to do initiation of the (t-BuO)2
What does an alcohol and H2SO4 make
alkene (elimination reaction)
halogenoalkane and CN
nitrile
What is the curly arrow mechanism for this (Br2 ring)
ring opening Br+, but instead of Br- attacking, water attacks
A bromohydrin alcohol can be treated with a base to get an epoxide. What Is the mechanism for this (deprotonation)
this reaction is fast
What is the mechanism for this
must form a stabilised tertiary carbocation intermediate
hyboronation has a ** four-membered transition state**; what would the this transition state
What is the mechanism for an alkene + mCPBA to form an epoxide
HOMO of alkene (πC-C) interacts with LUMO of the mCPBA (σ*O-O). The product will be racemic because the syn addition could add to either face
Alkene + mCPBA = epoxide (include transition state)
OsO4 makes 1,2 diol
syn addition
OsO4 makes 1,2 diol
dihydroxylation= making 1,2 diols
what is the mechanism for this
weak bases = strong acid = better leaving group. The most reactive will be the one with the stronger acid, and therefore the most stabilised anion. Sulfur at the end will be the most stabalised as it has the most electron withdrawing groups
Why is CH3- the strongest base
base- accepts proton
CH3- wants to accept the proton the most because it is the most unstable. F- is the weakest conj base because it is the most stable, therefore its conj acid HF will be the strongest acid in the table
Which would be a better base
base- accepts protons
O2- wants to accept protons the most because it is the most unstable, so O2 is the better base
Which will have the stronger conj acid
weak base= strong conj acid
so the weakest base which is water- conj acid H3O+
which are electron withdrawing group =O, NO2, F, Cl,
all
How do electron withdrawing groups improve an electrophiles ability
electrophile- electron withdrawing
electron-withdrawing effect makes any carbon center even more electron deficient than before
How do electron withdrawing groups reduce a nucleophiles ability
nucleophiles- electron acceptor
Nucleophiles need electron density to react with electrophiles; if an EWG is ‘withdrawing’ electrons, this is taking away the source of the nucleophile’s strength
What are examples of some good electron withdrawing groups
aldehydes, ketones, carboxylic acids, =O, NO2, esters (all bonded to carbon e.g C=O)
What are some examples of some electron donating groups
OH, O-, NH2, amine groups NH-R, ethers, alkyl groups also weak donating groups
Does this make a stablized carbocation
secondary substrate that can form stabilised carbocation by conjugation with electron-rich aryl ring
mechanism for making 1,2 diols (cyclic electrons)
mechanism for making 1,2 diols (cyclic electrons)
Curly arrow mechanism from C-D
Describe the base NaOMe
strong small base/ E2 to form most substituted alkene
What is the difference between enantiomers and diastereoisomers
enantiomers are complete opposites of each other/ both stereochemsitry changes
-diastereoisomers is where only one changes its stereochemistry
Provide suitable reagents for the preparation of M from L
Determine the structure of this compound based on the analytical data below
1/ shows you more of what you don’t have (no carbonyl or C triple bonds)
2a/ DBE=8 aromatic. one at end is Br as 3 peaks separated by 2. The 152 and 76 peaks do not contain bromine as they don’t have this pattern in ratio 1:2:1. Peaks are much smaller
2b/ the 264nm means aromatic
2c/ C12 [two bromine rings] cannot be next to each other as too many hydrogens, so propose structure A from knowing it must be para.
4/ 4 hydrogens in the same chemical environment, 4 hydrogen in another same environment. Points to high symmetry
4b/ the two peaks are In the form of a para substitution
3/CNMR- 4 peaks, 12 carbons
What is benzene mass number on a mass spec
78
This is an isomer with the formula CnH2nO2. Identify the isomer
where do aldehydes and ketones fall on this scale
above 190ppm about 200ppm. Anything below this is usually an ester or amide
HNMR- What is the structure of this ester
the answer must be strucutre B shown from the yellow highlighted areas
-on the graph the CH2 quartet group is more deshielded than the singlet (which is from the O-CH3 group). Therefore the CH2 group must be next to the two oxygens instead of the 1 oxygen
Identify the C9H10 isomer
C=O marking is supposed to be C=C
DBE=5 means likely aromatic
CNMR- 7 carbon environments (count the small lines too)
mass spec- has a 78 peak which confirms benzene ring
How does this HNMR spec correlate with the molecule
How does this CNMR spec correlate with the molecule
green and grey are likely to be the most desheilded
-these peaks are not finite
-as general rule if a carbon has more protons around it it is likely to be more right on the scale
these are Jones conditions
note- think about at which oxygen does the reaction take place
-tertiary alc cannot be oxidised so reaction must happen at secondary OH
-dichromate is in acidic solution so is protonated
dont know where the water comes from
and dont know why the last step happens [chromate ester becomes as leaving group]
- know that the product must be a ketone because it is a secondary OH being oxidised
what are the reactants to go from an alcohol to an aldehyde in this reaction
what is the mechanism for this
[alot of similarities to jones reaction]
Ru is electrophile
OH is nucleophillic
what would this form
will take your primary alc to aldehyde only
and secondary alc to ketone
what would this form
what is the mechanism for this reaction
-youve made a bond to an iodine
-therefore you have to break a bond with iodine
-as acetate group is a good leaving group, that bond is broken
what does this make
what is the mechanism for this
what does this form
-radicals are generated
-you need to use an organic solvent that water is miscible in [THF]
what would these make
reduction reactions to make alkane
Both NaBH4 and BH3 are reducing agents
-NaBH4 is a metal hydride and BH3 is a lewis acidic reducing agent
-the BH3 will react with the more electron rich carbonyl species
-the metal hydride will react with the more electropositive carbonyl species
-ketone is more electropositive and amide is more electron rich
what does this make
how to go from a nitrile to an acid
sulfuric acid and water
what are the products of both of these reactions
how are gingnard reagents formed
halogen group with Mg
what is the wittig reaction
-what is the reagent used
-and what product is formed that is the driving force for this reaction because it has a very strong bond
PPH3
-treat halide with PPH3
-red H is deprotonated by any strong base
forming the O=PPH3 is the driving force of the reaction because it is very thermodynamically favourable
what are the products what all of these carbonyl groups are reacted with a gringard [R]
what does this form, and what is the mechanism
how do you make an amide from a
-carboxylic acid
-acyl chloride
reagents for acid to amide
what is the mechanism for this reaction
cyclic acetal reaction/ the acid must be in water
-having water in there makes sure the equilibrium lies a lot to the right
predict the major product of the following reaction
mechanism for this reaction
product of the aldol reaction
product of the aldol reaction
