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Topic 6 – Alkyl halide and carbonyl compounds
Recap 1. Apply your knowledge of curly arrows to work out structures for the products of these
steps.
Organic compounds containing a halogen • Dense liquids and solids which are insoluble in water . • The carbon-‐halogen bond strength decreases in the order C-‐F > C-‐Cl > C-‐Br > C-‐I. • Alkyl fluorides tend to be less reactive than other alkyl halides, mainly due to the higher
strength of the C-‐F bond. • Halogen atoms are electronegative, so C-‐halogen bonds will be polarized like this:
Nucleophiles could therefore attack the electron-‐deficient carbon and substitute the halogen atom directly like this:
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Nucleophilic substitution
2. Draw the organic products in the following reactions:
3. Think about the bulk of the organohalide – compare 3∘ and 1∘. Which of these direct substitutions by do you think will be faster, and why?
A wide range of charged and uncharged nucleophiles may be employed
Nucleophile + H3C I Product + I Product class
HO
RO
N C
R C C
H2N
R3N
H3C I
H3C I
H3C I
H3C I
H3C I
H3C I
H3C OH
H3C OR
H3C C
H3C C
H3C NH2
H3C NR3
N
C R
+ I
+ I
+ I
+ I
+ I
+ I
alcohol
ether
nitrile
alkyne
amine
tetraalkylammoniumsalt
+
+
+
+
+
+
δδ
Br OH
NH2
CN
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Mechanism – SN2 (2nd order) SN2 reaction favoured by a primary alkyl halide as the starting material.
4. Sketch the energy profile for a generic SN2 reaction below. The starting materials and products are shown. Add in a transition state and show the activation energy (Ea) involved in the rate determining step (RDS).
Since both starting materials are involved in the transition state, and since there is only one step, the reaction rate is proportional to the concentrations of both the starting materials.
C Br
H3CH2C
CH3H
CH3O CCH3O
CH2CH3
CH3H
Br
CH2CH3
C
CH3H
H3CO Br~1/2 ~1/2
nucleophile electrophile transition stateleaving group
δδ
C X
H
HH
C X
Me
HH
C X
Me
HMe
C X
Me
MeMe
Nu
Nu
methyl halide2,000,000 (FAST)
primary alkyl halide40,000
secondary alkyl halide500
tertiary alkyl halide<1 (V. SLOW)
Nu
Nu
relative rate:
relative rate:
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Mechanism – SN1 (1st order)
The intermediate in an SN1 reaction is a carbocation.
• The C atom has six electrons and a positive charge. • This is an unstable, highly reactive intermediate. • Not all carbocations have the same relative (un)stability .
SN1 reaction more likely when a tertiary carbocation is the intermediate
C Br
CH3
CH3H3C
OHCHO
CH3
CH3CH3
Br
CH3
C
CH3H3C
planarcarbocationintermediate
leaving group
slow
+
HOC OH
CH3
H3CH3C
fast
H3C CCH3CH3
carbocations have sp2
hybridised carbons (trigonlal planar) with a vacant p-orbital
fast
vacant p-orbital
Attack of nucleophile to either side of the planar carbocation intermediate is equally likely!
HCH3C
H
CH3CH3C
H3CHC
H3C
H3Ca tertiary alkyl
carbocationa secondary alkyl
carbocationa primary alkyl
carbocation
HCH
Ha methyl
carbocation
increasing stabilitymoststable
leaststable
C X
H
HH
C X
H3C
HH
C X
H3C
HH3C
C X
H3C
H3CH3C
methyl halide(V. SLOW) <1
primary alkyl halide1
secondary alkyl halide12
tertiary alkyl halide(FAST) 1,200,000
relative rate:
relative rate:
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5. Fill in the missing reagents.
6. Draw the products in each case, assuming the reactions proceed by the SN2 mechanism.
Summary SN2 reaction:
Substitution, nucleophilic, second order
The reaction rate is determined by the concentration of both nucleophile and electrophile
Involves the direct, single step substitution of an alkyl halide by a nucleophile with no intermediates
The SN2 reaction proceeds with inversion
The rate of the SN2 reaction of alkyl halides decreases in the order methyl > 1° > 2° >> 3°
Usually only methyl, 1°and 2° alkyl halides react by this pathway - less crowded
Summary SN1 reaction:
Substitution, nucleophilic, first order
The reaction rate is determined by the concentration the electrophile only
Involves the two step substitution of an alkyl halide by a nucleophile via a carbocation intermediate
The SN1 reaction of chiral compounds proceeds with racemisation
The rate of the SN1 reaction of alkyl halides decreases in the order 3° >> 2° > 1° > methyl
Usually only 3° alkyl halides react by this pathway
IOH
C
OCH2CH3
N(CH3)3 I
+ NaI+ NaI
+ NaI
CCH3
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Elimination of HX Dehydrohalogenation – require H and X on adjacent carbons.
7. We’ve seen in the elimination of water from an alcohol reaction that if we lose H+ from a carbocation we can generate a double bond. Look again at the SN1 reaction below. There’s a carbocation in the middle. Can you draw the mechanism for what happens if that carbocation loses an H+ (from the carbon next to where the charge is)?
8. So what does the energy profile look like? The start is exactly the same as the SN1 reaction. We lose a group to give a carbocation. But instead of attack by a nucleophile (which would be a substitution reaction) we lose a proton to give an alkene. Sketch this on the profile below.
HCH
CH2Br
H3C
HC CH2
H3C
1-bromopropane
c. OH heat
+ H2O + Br
HCBr
CH2H
H3Cc. OH heat
2-bromopropane
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9. Draw the E1 reaction mechanism for the following reaction. The H+ does not magically leave on its own. It will be carried away by a mild base, such as water, that might be present in the reaction. Try to include this in the second step of the mechanism.
Why is it important that the base is “mild”. What might happen if you added in something that was stronger, like hydroxide?
Zaitsev’s Rule – where there is a choice, the most substituted alkene forms.
Examples of eliminations
c. OH heat
+ H2O + Br
H3C CHH
HCCl
CH2
H
HC CH CH3H3C
HC CH2H2CH3C + H2O + Br
major product
C CH
H
H
HC C
R
H
H
H
C CR
H
R
H
C CR
R
H
H
C CR
R
R
HC C
R
R
R
R
unsubstituted ethylene
monosubstituted alkene
disubstituted alkene
trisubstituted alkene
tetrasubstituted alkene
Brhot, conc K OH
ethanol+ H2O + K Br
C OHH3C
H3C
H3Chot, conc H2SO4 C CH2
H3C
H3C+ H2O
hot, conc K OH
ethanolCl
+ H2O
+ K Clmajor
trisubstituted C=C bond3 carbon substituents
minordisubstituted C=C bond2 carbon substituents
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10. Fill in the reagents in the following sequence of reactions.
11. Give products for the following reactions involving alcohols and amines.
OHOH
Br
Br
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Carbonyl Compounds – aldehydes and ketones • The double bond is composed of one σ and one π bond. • The electron density of a π bond is above and below the plane of the
double bond. • The C=O double bond is polarised. • Aldehydes have at least one H attached to the carbonyl group,
ketones have two carbon groups attached to the carbonyl group. • Carbon of the carbonyl group is sp2 hybridised.
12. Think about the electronegativity of carbon vs. oxygen and therefore how the carbonyl group is going to be polarised. How, then, might the group react with a nucleophile?
Reactions of carbonyl group – nucleophilic addition Dominated by weak and accessible π-‐bond
• Polarity of C=O directs addition reaction • Reaction conducted in two stages
§ Nucleophile (eg hydride or carbanion) adds to C § Then O– adds to H+
overlap of electrons in sp2-hybridisedorbitals gives a σ-bond
oxygen has two lonepairs of electrons
overlap of electrons in p-orbitalsgive rise to a π-bond
in formaldehyde, overlap of electronsin an sp2-hybridised orbital from C and a 1s orbital from H gives a σ-bond
OCH
HC OH
H
C O
H
Hδδ
C O
H
Hδδ
Nustage 1
C O
H
H
Nu stage 2
H C O
H
H
Nu
H
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13. Let’s draw a nucleophile actually reacting with a carbonyl group. You’ll initially form a compound with a negatively charged oxygen atom.
Let’s pretend we treat that intermediate compound with some very mild acid in water, to give a neutral organic compound that we could isolate and purify. Complete the mechanism above.
Notice how in this reaction the nucleophile interacts with the organic compound (containing the C=O) first, making this mechanism different from the electrophilic addition to double bonds that we saw previously.
14. Complete this mechanism where hydride reacts with a ketone, and the intermediate product is worked up to give a neutral organic compound.
This is an example of a nucleophilic addition reaction. But in this case look at what we’ve achieved: the opposite of an oxidation. So what else can we call this reaction, when we use hydride as the nucleophile? Notice how the carbonyl carbon changes from being sp2 hybridised in the starting material to being sp3 hybridised in the product. This is OK, and has happened in lots of other reactions you’ve already seen. The carbon changes its hybridisation as part of the reaction pathway.
• To generate the hydride ion, use LiAlH4 or NaBH4 • The second stage of the reaction is always addition of H+ • Aldehydes give primary alcohols, ketones give secondary alcohols
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15. Give the organic products for the following nucleophilic addition reactions
16. Here are some toe-‐curling mechanism gaffes. What’s wrong in each case?
17. This reaction gives a chiral product. Which enantiomer do we get?
ONu
O
Nu
O
Nu
O
Nu
ONu
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Grignard reaction Formation of a Grignard reagent involves the reaction of an alkyl-‐ or aryl-‐halide and magnesium.
18. The carbon atom in a Grignard reagent is nucleophilic. It’s powerfully nucleophilic. It’s reactive. We have to handle Grignard reagents in the absence of water. Why? Draw the products of the reaction.
So why doesn’t the following reaction work as shown?
This is a very versatile reaction.
• Can use most organohalide compounds to generate Grignard reagent. • Grignard reagent reacts with a carbonyl grooup. • Must employ anhydrous conditions.
Examples:
C nucleophile
use Grignard reagent: RMgX
R Mg X
R Mg X
HC O
H
formaldehyde
nucleophilicaddition
HC O
H
H3CH H
C O
H
H3CH
acid-base reaction
STEP 1 STEP 2
an alkoxideanion
a primary alcohol
methyl magnesium
bromide
MgBr
H3Cδ δ
MgBr
R X + Mg R Mg X X = Cl, Br, IR = alkyl, aromatic
δ δδδ
eg CH3Br + Mg CH3MgBr
I
+ Mg
Mg I
Cl+ Mg
MgCl
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19. A very neat application of Grignard reagents is in the synthesis of carboxylic acids (functional groups that we’ll talk about in the next lecture). What’s the Mystery Reagent below?
20. Some practice. Give the products of the following reactions. In the first reaction, will we see an optical rotation from the reaction mixture?
RC O
H
other aldehydes
RC O
H
H3CH R
C O
H
H3CH
secondary alcohol
methyl magnesium
bromide
MgBr
H3Cδ δ
MgBr
RC O
R
all ketones
RC O
R
H3CH R
C O
R
H3CH
tertiary alcohol
methyl magnesium
bromide
MgBr
H3Cδ δ
MgBr
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21. Draw the products of the following sequence of reactions.
1. In the Grignard reaction to form 3-‐methylhexan-‐3-‐ol there are three possible ketones that could be used as a starting material. Identify the three ketones and the Grignard reagent they need to react with to give the desired product.
22. Supply the missing details for this sequence, which neatly illustrates how we can use carbon-‐based nucleophiles to build up complex molecules quickly.
Cr2O72 / H
conc H2SO4heat
dil H2SO4
1. CH3CH2MgBr 2. H
conc KOHheat
Cl
2 products, use one in next step
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