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chem year 1 > organic paper questions > Flashcards

organic paper questions Flashcards

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1
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4
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What is NaOAc

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5
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What sterics is the Hydrogen on here

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6
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What will the other sterics be when the I switches position

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7
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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

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8
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9
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Which side would the equilibrium between ring flips lie

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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

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10
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Why couldn’t you do a ring flip for this compound

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t-Bu will only sit in equatorial because axial would be far too unfavourable/ high in energy (conformational lock)

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13
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how do you know the reaction is Sn2- it cannot form a stable cation so must be Sn2

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14
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15
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Draw a curly arrow mechanism for the following activation of an alcohol

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no stereochemistry to worry about because its primary

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16
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An alcohol being activated by being turned into a sultanate ester. What are the steps in this reaction

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17
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Sultanate esters are good leaving groups. What happens in this substitution reaction

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18
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Alcohols can also be activated by being converted into acyl chlorides. What reactant is used to do this

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18
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Alcohols can also be activated by being converted into acyl chlorides. What reactant is used to do this

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19
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What is the mechanism of this reaction

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20
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Sn2 inversion where there is stereochemistry

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21
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Why would there be no reaction

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22
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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

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23
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24
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Predict the major product and draw a curly arrow mechanism for the following reaction

A

heat favours elimination over substitution

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25
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elimination

A
  1. NaOMe is a small and not very sterically hindered base so can attack for a more substituted product.
  2. Secondary substrate cannot form a stabilised carbocation intermediate so must be E2
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26
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elimination

A
  1. 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
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27
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28
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  1. acetone- polar aprotic so SN2
  2. Top Br removed because its on sp3 hybridised carbon and other is sp2
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29
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31
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32
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Draw “curly arrow” mechanisms to determine the major products from the following reactions

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33
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How would this propagation step react and form

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forms tertiary radical rather than primary as it is more stable/ transition state lower in energy

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34
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propagation step product

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35
Q

What would each of these termination steps look like

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36
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A-B; alkene + HBr= halogenolakane but under UV light so have to do initiation of the (t-BuO)2

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37
Q

What does an alcohol and H2SO4 make

A

alkene (elimination reaction)

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38
Q

halogenoalkane and CN

A

nitrile

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39
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40
Q

What is the curly arrow mechanism for this (Br2 ring)

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41
Q
A

ring opening Br+, but instead of Br- attacking, water attacks

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42
Q

A bromohydrin alcohol can be treated with a base to get an epoxide. What Is the mechanism for this (deprotonation)

A

this reaction is fast

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43
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44
Q

What is the mechanism for this

A

must form a stabilised tertiary carbocation intermediate

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45
Q

hyboronation has a ** four-membered transition state**; what would the this transition state

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46
Q

What is the mechanism for an alkene + mCPBA to form an epoxide

A

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

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47
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Alkene + mCPBA = epoxide (include transition state)

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48
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49
Q

OsO4 makes 1,2 diol

A

syn addition

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50
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OsO4 makes 1,2 diol

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51
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dihydroxylation= making 1,2 diols

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52
Q

what is the mechanism for this

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53
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A

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

54
Q

Why is CH3- the strongest base

A

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

55
Q

Which would be a better base

A

base- accepts protons
O2- wants to accept protons the most because it is the most unstable, so O2 is the better base

56
Q

Which will have the stronger conj acid

A

weak base= strong conj acid
so the weakest base which is water- conj acid H3O+

57
Q

which are electron withdrawing group =O, NO2, F, Cl,

A

all

58
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How do electron withdrawing groups improve an electrophiles ability

A

electrophile- electron withdrawing
electron-withdrawing effect makes any carbon center even more electron deficient than before

59
Q

How do electron withdrawing groups reduce a nucleophiles ability

A

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

60
Q

What are examples of some good electron withdrawing groups

A

aldehydes, ketones, carboxylic acids, =O, NO2, esters (all bonded to carbon e.g C=O)

61
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What are some examples of some electron donating groups

A

OH, O-, NH2, amine groups NH-R, ethers, alkyl groups also weak donating groups

62
Q

Does this make a stablized carbocation

A

secondary substrate that can form stabilised carbocation by conjugation with electron-rich aryl ring

63
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mechanism for making 1,2 diols (cyclic electrons)

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63
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mechanism for making 1,2 diols (cyclic electrons)

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66
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67
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68
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69
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70
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71
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72
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Curly arrow mechanism from C-D

A
73
Q

Describe the base NaOMe

A

strong small base/ E2 to form most substituted alkene

74
Q

What is the difference between enantiomers and diastereoisomers

A

enantiomers are complete opposites of each other/ both stereochemsitry changes
-diastereoisomers is where only one changes its stereochemistry

75
Q

Provide suitable reagents for the preparation of M from L

A
76
Q

Determine the structure of this compound based on the analytical data below

A

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

77
Q

What is benzene mass number on a mass spec

A

78

78
Q

This is an isomer with the formula CnH2nO2. Identify the isomer

A
79
Q

where do aldehydes and ketones fall on this scale

A

above 190ppm about 200ppm. Anything below this is usually an ester or amide

80
Q

HNMR- What is the structure of this ester

A

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

81
Q

Identify the C9H10 isomer

A

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

82
Q

How does this HNMR spec correlate with the molecule

A
83
Q

How does this CNMR spec correlate with the molecule

A

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

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87
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88
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89
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90
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91
Q

these are Jones conditions
note- think about at which oxygen does the reaction take place

A

-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]

  1. know that the product must be a ketone because it is a secondary OH being oxidised
92
Q

what are the reactants to go from an alcohol to an aldehyde in this reaction

A
93
Q

what is the mechanism for this
[alot of similarities to jones reaction]

A

Ru is electrophile
OH is nucleophillic

94
Q

what would this form

A

will take your primary alc to aldehyde only
and secondary alc to ketone

95
Q

what would this form

A
96
Q

what is the mechanism for this reaction

A

-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

97
Q

what does this make

A
98
Q

what is the mechanism for this

A
99
Q

what does this form

A

-radicals are generated
-you need to use an organic solvent that water is miscible in [THF]

100
Q

what would these make

A

reduction reactions to make alkane

101
Q
A

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

102
Q

what does this make

A
103
Q

how to go from a nitrile to an acid

A

sulfuric acid and water

104
Q

what are the products of both of these reactions

A
105
Q

how are gingnard reagents formed

A

halogen group with Mg

106
Q

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

A

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

107
Q

what are the products what all of these carbonyl groups are reacted with a gringard [R]

A
108
Q

what does this form, and what is the mechanism

A
109
Q
A
110
Q

how do you make an amide from a
-carboxylic acid
-acyl chloride

A
110
Q

reagents for acid to amide

A
111
Q

what is the mechanism for this reaction

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112
Q

cyclic acetal reaction/ the acid must be in water

A

-having water in there makes sure the equilibrium lies a lot to the right

113
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114
Q

predict the major product of the following reaction

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115
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mechanism for this reaction

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116
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127
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128
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129
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130
Q

product of the aldol reaction

A
131
Q

product of the aldol reaction

A

chem year 1 (26 decks)

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