Reactions Flashcards
1
Q
Base-catalyzed addition of an Alcohol to an Aldehyde or Ketone
A
hemi-acetal or hemi ketal formation
2
Q
A
B
Cyano groups are meta directors. Thus, attack of the sulfonate group would be directed towards the meta position, producing only compound [B].
3
Q
Addition of Cl2 (or any halogen) to an alkene in aqueous solution results in the formation of a _______ or ______ (or halohydrin in general), with the halogen ending up on the cabron atom of the double bond with the _____number of hydrogen atoms.
A
chloro alcohol, chlorohydrin, greatest
4
Q
A
C
SOCl2 converts alcohols into alkyl chlorides and carboxylic acids into acid chlorides.
Na2Cr2O7, H2SO4, H2O (aka the Jones reagent) is an oxidizing agent.
Finally, CH3MgBr is a Grignard reagent, which makes carbon-carbon bonds. Esters are not especially electrophilic, so they require a STRONG reducing agent.
LiAlH4 is a stronger reducing agent than NaBH4, so the best answer choice is C
5
Q
The reagent used in this reaction is called ____
A
Jones reagent - used to oxidize primary alcohols to carboxylic acids
6
Q
A
B
2-butene reacts with osmium tetroxide, OsO4 followed by workup with aqueous Na2SO3 to yield a cis-diol, 2,3-butanediol. Treatment of this vicinal cis-diol with periodic acid, HIO4 oxidizes the diol into aldehyde.
7
Q
A
A
Adding a Grignard reagent to an ester results in a total of TWO additions of the reagent and a final product of a tertiary alcohol.
8
Q
Addition of a Grignard Reagent to an Aldehyde
A
9
Q
Ozonolysis using H2O2 of an internal vs. terminal alkyne
A
10
Q
Describe thin layer chromatography.
What is the plate made out of? How will this effect molecules that are being used in the experiement?
Describe the RF for nonpolar and polar molecules in TLC.
A
In thin layer chromatography (TLC), the chromatography plate is made with polar silica gel.
Polar molecules will be attracted to the plate and travel less, while nonpolar will travel further
more polar = low RF
more nonpolar = high RF
11
Q
Rank the following alkenes in order of DECREASING stability:
a. but-1-ene
b. (E)-but-2-ene
c. (Z)-but-2-ene
A
B > A > C
The greater the number of attached alkyl groups (i.e. the more substituted the carbon atoms on the double bond), the more stable the alkene it will be.
Further, it is more stable to have sterically hindered groups trans to each other, which is why (E)-but-2-ene is more stable than (Z)-but-2-ene.
12
Q
Addition of water to an Aldehyde or Ketone forming a Hydrate
A
13
Q
A
A
Resonance structures should only involve movement of electrons through p-orbitals.
In order for electrons to resonate through a carbon atom, it must be sp2-hybridized (aka have an available p-orbital). This means that any pi-bond, charge, or radical that only has sp3-hybridized carbon atoms adjacent to it will not be resonance stabilized.
14
Q
Addition of a Grignard Reagent to a Ketone
A
15
Q
Addition of a Grignard Reagent to an Epoxide
A
Adds to the less substituted side
16
Q
A conjugate base will be destabilized by ____ _____ groups and will have a decreased acidity
A conjugate base will be stabilized by ____ _____ groups and will have an increased acidity
A
electron donating
electron withdrawing
17
Q
Addition of a Grignard Reagent to an Ester
A
grignard reactions with esters MUST ADD TWICE
18
Q
How to convert from an acyl chloride to an acid chloride?
A
19
Q
A
A
This is an E2 reaction. Sodium ethoxide is a strong base which is reacting with a secondary halide, which means E2 will dominate.
20
Q
Which will reduce an alkyne to a cis alkene? Which will reduce an alkyne to a trans alkene?
A
cis - H2 and Lindlar’s catalyst
Trans - Na or Li with NH3
21
Q
Side-Chain Oxidation of Benzene to form Benzoic Acid
A
22
Q
The strength of base is determined by the availability of the____ ____.
Electron-______ substituents ______ basicity while electron-withdrawing substituents ______ basicity.
In general, the basicity of amines follows the trend _______ > ______ > _____> ammonia.
A
electron pair, donating, increase, decrease
basicity of amines = secondary > primary > tertiary > ammonia
23
Q
How many sigma (σ) and how many pi (π) bonds are present in pent-3-en-1-yne?
A. 3 σ and 3 π
B. 10 σ and 3 π
C. 7 σ and 4 π
D. 4 σ and 3 π
E. 5 σ and 5 π
A
B
DRAW THIS OUT
24
Q
A
A
The first step is a standard oxidation using the Jones reagent. This oxidizes the secondary alcohol to a ketone and the primary alcohol to a carboxylic acid (shown below). The resulting β-ketoacid undergoes a decarboxylation reaction in acid and heat. This decarboxylation produces a ketone (answer A) and carbon dioxide.
25
Ozonolysis (Reducing Conditions)
Reducing reagent can also be (CH3)2S
26
27
If there is a 50:50 mixture of two enantiomers, it is known as a ___ mixture. Racemic mixtures have an optical rotation of \_\_\_\_\_
racemic, 0
28
C
The desired product can be synthesized from benzene through chlorination followed by nitration:
The **Cl– substituent is an ortho-para directing group,** while the **NO2– is a meta director**. If we start with nitration followed by chlorination, the meta derivative would be the product.
29
30
T/F? An amine with a lone pair can count as a chiral center due to amine inversion.
False
due to amine inversion, amines with lone pairs cannot count as chiral centers
31
A
choose a reducing agent that is strong enough to reduce the aldehyde but not the acid. NaBH4 achieves this.
**LiAlH4 is a powerful reducing agent that reduces both the aldehyde and the carboxylic acid.** KMnO4 and H2CrO4 are oxidizing agents. H2/Ni is used in hydrogenation (also a reduction) of unsaturated compounds.
32
C
Grignard addition to aldehyde produces a secondary alcohol. In this reaction, the Grignard reagent acts as a nucleophile and attacks the carbonyl carbon, forming a halomagnesium alkoxide. Protonation of the alkoxide forms the alcohol.
33
Free Radical Halogenation with Chlorine
34
Addition of a Grignard Reagent to an Acyl Chloride
35
Addition of a Grignard Reagent to an Amide
36
37
C
38
E
39
T/F? When considering hybridization of a carbon atom, radicals are NOT counted
TRUE
40
D
41
Addition of a 2˚ Amine to an Aldehyde or Ketone
42
What is the trend for the rate of dehydration for alcohols?
the bulkier the alcohol = faster the rate of dehydration will be.
The rate of dehydration of alcohols follows the order: 3° \> 2° \> 1°. Thus, [D], which is a tertiary alcohol, will undergo dehydration the fastest.
43
Allylic/Benzylic Bromination
Br will add at the one carbon away from the double bond
44
The following compounds all have similar molecular weights. Which compound would be isolated FIRST in the process of fractional distillation?
lowest boiling point will boil off first
when molecular weight is equal, the higher degree of **branching** that an alkane has, the lower the degree of intermolecular forces and the lower the boiling point. Therefore, the most highly branched alkane (E) will have the lowest boiling point.
45
Bromination of a double bond will add in syn or anti fashion?
syn
46
The first step of this reaction is an oxidation under strongly acidic conditions. Because the starting material is a primary alcohol, it will be oxidized down to a carboxylic acid. This will then be converted into an acyl chloride. This will react with an alcohol to form an ester, as shown below.
47
Lab Tests Chart
48
Ozonolysis in oxidizing conditions will use _____ as its reagent to form a ____ and an \_\_\_\_\_
H2O2, ketone, COOH
49
D
The acidity of carboxylic acid is determined by the stability of the conjugate base carboxylate.
Aliphatic carboxylic acids are weaker than benzoic acid and derivatives, since the conjugate base of benzoic acid (and derivatives) can be stabilized by resonance to the pi system of the benzene ring. Thus, we can eliminate C.
Benzoic acids with **electron-withdrawing substituents are more acidic than unsubstituted benzoic acid, while benzoic acid with electron-donating substituents are less acidic.**
50
Acid-catalyzed Cleavage of Ethers when one side is 2˚/3˚
(Nucleophile attacks more substituted side via SN1)
51
D
The primary basis for this trend is the hybridization of the carbon in each compound.
The greater the s character of the hybrid orbitals, the lower in energy the electrons of the anions will be, and the more stable is the anion, the stronger the acid will be.
The carbon on ethyne is sp-hybridized, which has 50% s character, the carbon on ethene is sp2 hybridized, which has 33.3% s character, and the carbon on ethane is sp3 hybridized, which has 25% s character.
52
B
The nitro group is strongly deactivating and a meta director, chloro is weakly deactivating and an ortho/para director, and methyl is activating and an ortho/para director.
Starting from benzene, the best method for preparing the desired compound is to do the Friedel-Crafts alkylation to form toluene. Nitration of toluene would mostly form p-nitrotoluene since -CH3 is an ortho/para director. Chlorination of p-nitrotoluene would yield only 2-chloro-4-nitrotoluene
53
This is a classic example of a Williamson Ether Synthesis.
The strongly basic NaH deprotonates the alcohol in the first step to create a nucleophilic alkoxide.
This alkoxide goes on to perform an SN2 reaction with the alkyl halide, resulting in choice C (see sequence below).
54
B
In general, primary and secondary alcohols undergo substitution with PBr3 to yield alkyl bromide.
This reaction does not involve formation of a carbocation and usually occurs without rearrangement.
55
A
56
Treating an alkene with KMnO4 and H3O+ will produce:
ketone and a COOH - oxidative cleavage
57
Oxidative cleavage of an internal vs terminal alkyne, using KMnO4
58
This should be a SN2 reaction, which occurs most rapidly with strong nucleophiles. The ability of the nucleophile to attack the electrophilic carbon (the carbon attached to bromine) depends on the amount of steric hindrance it encounters during a backside attack.
Keep in mind that the relative rate for SN2 reactions in terms of the type of electrophile is as follows: Methyl \> 1o \> 2o \>\>\>\>\>\> 3o (SN2 doesn’t occur with 3o substrates). Therefore, the methyl bromide (answer choice A) will have the least steric hindrance and undergo an SN2 reaction most rapidly.
59
E
Resonance structures should only involve movement of electrons through p-orbitals. Check every structure to be sure that every atom has the same number of hydrogen atoms attached to it.
The parent structure has two hydrogen atoms attached to the nitrogen atom. This means that every single resonance structure should have two hydrogen atoms attached to it.
The structure in answer C has only one hydrogen attached to the nitrogen. This means it is NOT a valid resonance structure (it is actually a tautomer).
60
Hydration of an alkene
Formation of an alcohol
61
E
The more stable the conjugate base, the more acidic the acid.
Phenols are relatively strong acids because the phenolate anion is resonance stabilized. Knowing that the **conjugate base will be DEstabilized by electron DONATING groups**, answer choices B, C, & D can be eliminated.
**Next, electron WITHDRAWING groups will help stabilize the negative charge on the conjugate base**, so answer choice E is the best choice.
Nitro groups are electron withdrawing groups both through resonance and induction.
62
E
The reaction given is like an electrophilic aromatic chlorination. The carbonyl group is a meta director, and chlorine will substitute the hydrogen meta to the carbonyl.
63
64
The first step of this reaction is an acid-catalyzed Fisher esterification. This results in the methyl ester (shown below).
The next step is a Grignard reaction. When a Grignard reacts with an ester, it adds twice, resulting in a tertiary alkoxide. Finally, the alkoxide is protonated in aqueous acid to yield neutral tertiary alcohol A.
65
The first step of this reaction is the nitration of the toluene starting material.
The methyl substituent is **slightly electron donating, so it is activating and ortho/para directing**.
The major product will be the less sterically-hindered para-nitrotoluene.
Step two is the reduction of the nitro group to an amine. The final step is an acetyl protection of the amine.
66
A
The closer the electronegative atom is to the negative charges, the more of a stabilizing effect.
stable = most acidic
67
The best way to assess acidity of the compound is to look at the stability of the ___ \_\_\_\_
conjugate base
assess stability through resonance
68
D
69
Electron donors and withdrawers
70
C
A Bronsted-Lowry base is a proton acceptor. This means that the molecule on the left side of the equation that gains a proton is the base. CH3MgBr becomes CH4, so it has gained a proton going from left to right in the equation. This means it acts as a base. Additionally, the propynylmagnesium bromide gains a proton going from right to left, so it is also a base.
71
EAS of Benzene
72
A
Alkenes react with osmium tetroxide (OsO4) to yield cis-1,2-diols through syn addition.
This reaction can also be accomplished by KMnO4, though the diol formed is prone to cleavage through further oxidation since KMnO4 is a stronger oxidizing
73
Which of the following is associated with an E1 reaction?
A. An antiperiplanar conformation
B. A pentavalent transition state
C. A bimolecular rate law
D. A carbocation intermediate
E. Stereoselectivity in the product
D
DO NOT GET TRICKED. E2 REACTIONS REQUIRE AN ANTIPERIPLANAR CONFORMATION
74
In order to convert an alcohol to an alkyl halide, what can be used as a reagent?
This is commonly done to create what reagent in a popular reaction?
PBr3 is an excellent reagent to use to form an alkyl halide; this is used to create a grignard reagent
75
D
Carboxylic acids are reduced to primary alcohols in the presence of LiAlH4.
Lithium tri-tert-butoxyaluminum hydride is a derivative of lithium aluminum hydride (LiAlH4).
Because it is sterically hindered, **LiAlH(O-t-Bu)3 can reduce carboxylic acid derivatives, such as acyl chloride, into aldehyde only.**
76
Halogenation across an alkene
77
C
Treatment of alkene with HBr in the presence of the peroxide leads to the addition of a bromine atom and a hydrogen atom on the carbon-carbon double bond, via a free-radical addition mechanism forming an anti-Markovnikov product. The peroxide initiates a free radical reaction where the attacking species to the alkene is the bromine atom ●Br. The bromine atom adds to the less substituted carbon atom producing the more stable radical.
78
Intermolecular forces will determine relative melting point.
The stronger the intermolecular forces, the higher the melting point. Hydrogen-bonding is the strongest intermolecular force, so the molecule with the highest capability for H-bonding will have the highest melting point.
79
Addition of a Grignard Reagent to CO2
80
The greater the ___ character of the bond orbital, the greater the interaction between the bonding atoms is, and the \_\_\_\_\_the bond will be.
The percent s character follows the order:
s \> sp \> sp2 \> sp3.
S, shorter
81
How are volatile compounds separated best?
using a method that involves the gas phase - gas liquid chromatography
82
A Brønsted-Lowry acid is a proton donor.
This means that the molecule on the left side of the equation that loses a proton is the acid, which will then become the negatively charged conjugate base on the right side of the equation.
The HCN acts as the acid on the left side of the equation. On the right side of the equation, the proton donor is H3O+.
83
Addition of a Grignard Reagent to a Nitrile
84
An important characteristic of an SN2 reaction is the formation of a pentavalent ______ \_\_\_\_\_\_\_
This is a high-energy transition state where the incoming nucleophile performs a\_\_\_\_\_ attack while the leaving group begins to leave.
This transition state (can/cannot) be isolated and thus is not a \_\_\_\_\_
transition state, backside, cannot, intermediate
85
C
The parent chain is an amide with 6 (hex-) carbons (a hexanamide). There are two methyl substituents attached to the nitrogen (N,N-dimethyl) and a methoxy group attached to carbon 2 (2-methoxy). The IUPAC name is therefore, N,N-dimethyl-2-methoxyhexanamide.
86
In the reaction, the Grignard reacts as a nucleophile the epoxide opens, and the protonation of the oxygen turns it into an alcohol.
Grignards tend to react with the l**ess substituted side** of an epoxide but that is not an option in this problem.
There is another important concept with epoxides. In the presence of acid, the nucleophile adds to the more substituted carbon. In basic solution, it adds to the less substituted carbon.
87
What kind of compounds will give the following peaks in an IR spectrum?
3200-3400
650-850
1000-1300
3200-3600
amines = 3200 - 3400
benzene rings = 650 - 850
ethers = 1000 - 1300
alcohols = 3200 - 3600
88
How to name a bicyclo compound?
1) count total number of carbons = parent chain
2) find the number of carbons that are coming off the middle - these will be in brackets, from high \> low
89
2-chlorobutane in ethanol forms but-2-ene as the major product. However, 2-chlorobutane in acetone forms 2-methoxybutane as the major product. Which of the following best explains this observation?
A. Ethanol is a reagent that removes a hydrogen from the alkyl halide and pushes the chloride leaving group off, however acetone cannot remove a hydrogen.
B. Ethanol does not interact with the carbocation intermediate, however acetone interacts with the carbocation intermediate.
C. Ethanol forms a solvent shell around the carbocation to improve its stability.
D. Ethanol is a polar aprotic solvent, while acetone is polar protic.
E. Ethanol forms a network of hydrogen bonding that reduces the nucleophilicity of sodium methoxide.
E
The difference in reactions comes down to the solvent choice. In the first reaction with ethanol (a protic solvent) the ethanol has properties of hydrogen bonding – think of it like this huge spider web of interactions between each solvent molecule. So now our nucleophile is getting stuck in this huge web of hydrogen bonding going on in the solvent, which reduces its nucleophilicity, which means the strength of the nucleophile is decreased. It’s easier to act as a base and to just take a hydrogen off the alkyl halide, which is what it does and why the first reaction goes E2.
In contrast, acetone cannot participate in hydrogen bonding by itself, since it is an aprotic solvent. So our nucleophile is free to act as a nucleophile, and will come in and attack the alkyl halide in a single step SN2 style, giving us 2-methoxybutane.
90
Determining between S1/SN2 or E1/E2 reaction chart
91
Select the correct IUPAC name for the following structure:
A. 2-isopropyl-3-methylpropanal
B. 3-isopropylbutan-4-al
C. 3-methanal-2-methylpentane
D. 2-isopropylbutanal
E. 2-ethyl-3-methylbutanal
E
We must number the chain starting with the aldehyde carbon as carbon 1, but which 4-carbon chain should we use?
If we used the 4-carbon chain that branches down at carbon 2, we end up with an ethyl group coming off carbon 2 and a methyl group coming off carbon 3.
If we use the 4-carbon chain that branches up at carbon 2, we only have an isopropyl substituent at carbon 2.
**Remember when naming alkyl chains we want the longest chain with the most branch points**.
These two chains are tied in length (4 carbons) but the first numbering scheme gives two branch points (or alkyl substituents) while the second scheme gives only one branch point; thus we will use the first scheme. Lastly, we name the alkyl chain (butane) and replace the “-e” with the suffix “-al” (butanal).
92
What is the major product of the following reaction?
A. butanal
B. 2-butanone
C. butane
D. 1,2-butanediol
E. 1-butene
B
Alkynes undergo hydration (additional of water) readily when treated with aqueous solution of Hg2+ in the presence of a strong acid. The reaction follows Markovnikov’s rule. Thus, hydration of 1-butyne results in formation of 2-butanone.
93
Which of the statements best describes the hybridization, geometry and bond angles about the central carbon in the carbocation drawn below?
A. sp3, tetrahedral, & 109.5o
B. sp3, trigonal pyramidal, & 109.5o
C. sp2, tetrahedral, 109.5o
D. sp2, trigonal planar, 120o
E. sp2, trigonal pyramidal, 120o
D
The cation does not count as an orbital!
94
Potassium dichromate in acidic solution is a strong oxidizing agent and is used as a colorimetric indicator for oxidation of organic compounds, since it changes color from orange to green upon reduction.
All of the following alcohols will result in a color change when added to a solution of K2Cr2O7 in H2SO4 EXCEPT for one. Which one is the EXCEPTION?
A. 2-butanol
B. 2-methyl-2-propanol
C. 1-butanol
D. 1-propanol
E. 2-ethyl-1-hexanol
In acidified K2Cr2O7, primary alcohol is oxidized into aldehyde and further oxidized into carboxylic acid, and secondary alcohol is oxidized into ketone. Tertiary alcohols cannot be oxidized. Thus, primary and secondary alcohols will result in a color change, and tertiary alcohols will not. This reaction is called the Jones reaction and is used to distinguish tertiary alcohols from primary and secondary alcohols.
B
95
Which of the following is true of E2 and SN2 reactions?
A. E2's rate law = k[substrate][base] whereas SN2's rate law = k[substrate]
B. E2 and SN2 both favor tertiary substrates
C. E2 and SN2 are both one step reactions
D. E2 and SN2 are both favored in polar protic solvent
E. E2 and SN2 are both NOT favored with a strong base / nucleophile
C
96
Looking at this reaction, there are two concepts that should jump out at you. The first is that it is a strong nucleophile/base with a polar protic solvent. The other main concept is that the bromines attached to the benzene ring are aryl halides and these will not react.
The most important factor in this case is the carbon being examined is primary, meaning that we can’t go SN1 or E1. Additionally, all elimination reactions (E1, E2) are impossible because there are no H’s on the carbon next-door to the carbon bonded to our Br leaving group. Hence, there are no beta H’s to eliminate.
97
Which of the following alkyl radicals is the most stable and which is the least stable?
A. II least stable; III most stable
B. III least stable; I most stable
C. I least stable; II most stable
D. I least stable; III most stable
E. II least stable; I most stable
B
98
An unknown sample was spotted on a silica gel TLC plate and developed in hexane. The Rf value was calculated to be 0.35. Which of the following statements is true if ethyl acetate was used as the developing solvent instead of hexane?
A. The compound would travel further, therefore the Rf value would increase
B. The compound would travel less far, therefore the Rf value would increase
C. The solvent would travel further, therefore the Rf value would decrease
D. The compound would travel the same distance, and the Rf value would be unchanged
E. The solvent and the compound would travel further, and the Rf value would be unchanged
A
Ethyl acetate is more polar than hexane, thus it has greater eluting power. Changing the solvent form hexane to ethyl acetate would cause the compound to travel further increasing its Rf value. In general, the eluting power of solvents follows the order: alkanes \< toluene \< halogenated hydrocarbons \< diethyl ether \< ethyl acetate \< acetone \< alcohol \< acetic acid.
99
Which C–C bond would have an IR absorption that occurs at the highest frequency?
The frequency at which IR vibration occurs is determined by the strength of the bond and the mass of the bonded atom. Since we are comparing the same bonded atoms here (carbon to carbon), the difference in the IR frequency will be due to the difference in the strength of the bond. The stronger the bond, the higher the IR stretching frequency will be. Triple bond is the strongest carbon-carbon bond, and thus is expected to have the highest IR stretching frequency. In general, the IR stretching frequency of C-C bond follows the order:
100
Which of the following is true regarding the two structures shown below?
A. They are identical structures
B. They are diastereomers
C. They are NOT optically active
D. They have the same vapor pressure
E. They are meso compounds
These two compounds are enantiomers, which means they are non-superimposable mirror images and are optically active. Enantiomers have the same physical and chemicals properties, including answer [D]. Diastereomers have different physical and chemical properties.
Definitions:
Enantiomers are molecules that are non-superimposable, mirror images of each other. In a pair of enantiomers, every chiral center on one enantiomer will have the opposite configuration to its corresponding chiral center on the other enantiomer. For example, if one enantiomer had the configuration S, S at its two chiral centers, the other enantiomer would have the configuration R, R.
101
The first reaction in this synthesis is an example of dehydration synthesis because in the process of making an ester, we lose H2O. The second reaction is a reduction by LiAlH4 which is a very strong reducing agent and can reduce esters, carboxylic acids, ketones, and aldehydes whereas NaBH4 can only reduce ketones and aldehydes.
102
Which reagent(s) would best accomplish the following transformation?
A. CH3CH2CH2CH3, AlCl3
B. CH3CH2CH2CH2Cl, OH– 2. H2O
C. CH3CH2CH2CH2OH, CH3CH2CH2CH2OK
D. CH3CH2CH2CH2MgBr
E. 1. CH3CH2CH2C(O)Cl, AlCl3 2. Zn(Hg), HCl, reflux
E
Treatment of benzene with butyryl chloride produces a ketone, in a process called Friedel-Crafts acylation.
103
Which of the following would be the easiest to reduce with lithium aluminum hydride?
A. Methyl butanoate
B. Butanal
C. Butanone
D. Butanoic acid
E. Butane
B
## Footnote
Ease of reduction follows the order:
Carboxylic acid \< ester \< ketone \< aldehyde
Thus, butanoic acid would be the hardest to reduce, followed by the ester methyl butanoate, then by the ketone butanone. Butanal, which is an aldehyde, would be easiest to reduce.
Butane is an alkane, and is in the most reduced form already.
104
What is the major product of the following reaction?
A. butanal
B. 2-butanone
C. butane
D. 1,2-butanediol
E. 1-butene
B
Alkynes undergo hydration (additional of water) readily when treated with aqueous solution of Hg2+ in the presence of a strong acid. The reaction follows Markovnikov’s rule. Thus, hydration of 1-butyne results in formation of 2-butanone.
105
What is the product of the following reaction?
106
Which of the following carboxylic acids is prepared from p-bromotoluene by the following reaction sequence:
Free radical chlorination of p-bromotoluene leads to formation of 1-bromo-4-(chloromethyl)benzene. The hydrogens on the benzene ring are not susceptible to free radical halogenation; halogenation of benzene is achieved by electrophilic aromatic substitution. Addition of CN– results in the substitution of the chloro via nucleophilic substitution. Nitrile hydrolysis leads to formation of the carboxylic acid.
107
Electron withdrawing groups make a molecule ___ basic, while electron donating groups make a molecule ____ basic
Molecules [D] and [E] have electron-withdrawing groups (-NO2 on [D] and CN– on [E] are electron-withdrawing) making the electrons on the oxygen less available. This makes these molecules less basic.
Between [A] and [C], CH3 is an electron donating group, so [A] will be more basic than [C].
108
Acetylene structure
109
D
NaNH2 in liquid ammonia would abstract the proton in prop-1-yne forming propynide, which then acts as a nucleophile attacking 2-bromobutane via an SN2 reaction to yield 4-methylhex-2-yne.
110
Difference between E1/E2/S1/SN2 Chart
111
Tollen’s test (silver mirror test) is used to differentiate \_\_\_from \_\_\_\_
Tollen’s reagent gives a ____ result with aldehydes and\_\_\_\_ results with ketones, except α-hydroxy ketones.
A positive test is when the test tube turns a ____ reflective color.
aldehydes, ketones
aldehydes = positive
ketones = negative
silver
112
D
alkyl -C-H vibrates at a lower frequency (2850 – 2900 cm-1)
alkenyl =C-H (3010 – 3090 cm-1)
aromatic Ar-H (~3030 cm-1)
alkynyl ≡C-H has the highest frequency of vibration (~3300 cm-1).
113
C
Alkyl halide, in the presence of a Lewis acid such as AlCl3, will undergo an acid-base reaction forming a carbocation and AlCl4–. The electrophilic carbocation then attacks the benzene forming an alkyl benzene as the final product. The alkyl halide given, upon reacting with AlCl3, will initially form a primary carbocation, which will undergo rearrangement to the more stable tertiary carbocation through a hydride shift. The carbocation then attacks the benzene forming tert-butylbenzene.
114
A
Aldehydes and ketones react with phosphorus ylides to yield alkenes and triphenyl phosphine oxide (a by-product). The reaction is known as the Wittig reaction.
115
