Reaction of Aldehydes and ketones
Summary
TLDRThe video script covers the process of adding nucleophiles to carbonyl compounds, specifically focusing on aldehydes and ketones. It explains the formation of a tetrahedral intermediate and discusses reduction reactions using lithium aluminum hydride and sodium borohydride. Additionally, the script delves into the use of Grignard reagents and organolithium compounds to form alcohols from carbonyl compounds, providing detailed examples and explanations of primary, secondary, and tertiary alcohol formations.
Takeaways
- π§ͺ Nucleophiles are attracted to the partially positive carbon in carbonyl groups, leading to the formation of a tetrahedral intermediate.
- π The tetrahedral intermediate is formed when a nucleophile attacks the carbonyl group, but it requires breaking the double bond and redistributing electrons.
- βοΈ No good leaving groups are present in the initial tetrahedral intermediate, so a proton from water or acid is picked up by the oxygen to form an alcohol.
- π οΈ Lithium aluminum hydride (LiAlH4) is a strong reducing agent used to reduce ketones and aldehydes to alcohols, but it requires careful handling to avoid water contamination.
- π§ The 'quenching' process involves adding water at the end of the reaction with LiAlH4 to provide the hydrogen for the alcohol formation.
- π± Sodium borohydride is a milder reducing agent than LiAlH4 and can be used in an alcoholic solvent, with the alcohol providing the hydrogen for the reduction.
- π The reducing power of sodium borohydride is selective, only reducing ketones and aldehydes, unlike LiAlH4 which is a more general reducing agent.
- π The reduction of aldehydes by LiAlH4 results in primary alcohols, while ketones are reduced to secondary alcohols.
- πΏ Grignard reagents or organolithium compounds, which contain a nucleophilic carbon, can also attack carbonyl groups to form tetrahedral intermediates and eventually alcohols.
- π The final product of the reaction with Grignard reagents depends on the type of carbonyl compound: aldehydes form primary alcohols, ketones form tertiary alcohols.
- π The script provides examples of how different reducing agents and nucleophiles interact with carbonyl compounds, resulting in various types of alcohols.
Q & A
What is the role of nucleophiles in reactions involving carbonyl groups?
-Nucleophiles are attracted to the partially positive carbon in carbonyl groups (aldehydes or ketones). They attack the carbonyl carbon, leading to the formation of a tetrahedral intermediate. This process is crucial in various reactions such as reductions and the formation of alcohols.
Why does the carbon in the carbonyl group have a partial positive charge?
-The carbon in the carbonyl group has a partial positive charge due to the electronegativity of the oxygen atom, which pulls electron density away from the carbon, making it more electrophilic and thus more susceptible to nucleophilic attack.
What happens when a nucleophile attacks a carbonyl carbon?
-When a nucleophile attacks a carbonyl carbon, the double bond is broken, and the electrons are given to the oxygen atom, resulting in a tetrahedral intermediate. This intermediate is key in further reaction steps, such as the formation of alcohols.
What is the tetrahedral intermediate and why is it significant?
-The tetrahedral intermediate is a reaction product formed when a nucleophile attacks a carbonyl carbon, leading to a four-coordinate carbon. It is significant because it represents a key step in many carbonyl reactions, and its properties determine the subsequent reaction pathways.
How does the presence of good leaving groups affect the tetrahedral intermediate?
-If the tetrahedral intermediate has good leaving groups, it can undergo further reactions such as elimination to form new products. However, in the context of the script, the intermediate formed does not have good leaving groups, leading to the formation of an alcohol.
What is the purpose of adding water or acid to the tetrahedral intermediate?
-Adding water or acid to the tetrahedral intermediate allows the oxygen to pick up a proton, which is crucial for the formation of alcohols. This step is part of the workup or quenching process in many carbonyl reactions.
What are lithium aluminum hydride and sodium borohydride, and how do they differ in their reducing abilities?
-Lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4) are both reducing agents used in organic chemistry. LiAlH4 is a very strong reducing agent capable of reducing aldehydes, ketones, and even some esters, while NaBH4 is a milder reducing agent typically used for aldehydes and ketones, making it a selective reducer.
Why is lithium aluminum hydride considered a strong reducing agent?
-Lithium aluminum hydride is considered a strong reducing agent because it can donate hydride ions (H-) effectively, reducing various functional groups such as carbonyls to alcohols. Its strong reducing power also requires careful handling to avoid unwanted side reactions.
How does sodium borohydride reduce carbonyl compounds?
-Sodium borohydride reduces carbonyl compounds by donating a hydride ion to the carbonyl carbon, forming a tetrahedral intermediate. The subsequent addition of water or acid leads to the protonation of the oxygen, resulting in the formation of an alcohol.
What is the difference between the reduction of aldehydes and ketones by lithium aluminum hydride?
-Lithium aluminum hydride reduces aldehydes to primary alcohols and ketones to secondary alcohols. This is because the reduction involves the addition of a hydride ion and a proton to the carbonyl carbon, leading to different products depending on the initial carbonyl compound.
How do Grignard reagents and organolithium compounds react with carbonyl compounds?
-Grignard reagents (RMgX) and organolithium compounds (RLi) are nucleophilic and can attack the carbonyl carbon, forming a tetrahedral intermediate. The reaction typically results in the formation of alcohols, with the nature of the alcohol depending on the carbonyl compound (e.g., aldehydes form primary alcohols, ketones form tertiary alcohols).
Outlines
π§ͺ Nucleophiles and Carbonyl Reduction Reactions
This paragraph discusses the fundamental concepts of nucleophilic addition to carbonyl groups, which are electrophilic due to the partial positive charge on the carbon atom. The process involves the formation of a tetrahedral intermediate when a nucleophile attacks the carbonyl carbon. The intermediate is characterized by the presence of a negatively charged oxygen atom, which, in the absence of good leaving groups, will pick up a proton from a water molecule or an acid to form an alcohol. The paragraph also introduces two reducing agents, lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4), which are used to reduce aldehydes and ketones to primary and secondary alcohols, respectively. The mechanism of these reductions is simplified to illustrate the transfer of hydride ions and protons from water to form the alcohol products.
πΏ Grignard and Organolithium Reagents in Carbonyl Reactions
The second paragraph delves into the use of Grignard reagents (RMgX) and organolithium reagents, which are nucleophilic carbons that can attack carbonyl groups. The reaction proceeds through the formation of a tetrahedral intermediate, which, upon protonation, yields alcohols. The type of alcohol produced depends on the type of carbonyl compound being reduced: aldehydes yield primary alcohols, while ketones yield tertiary alcohols. The paragraph provides examples of these reactions, including the reduction of formaldehyde to a primary alcohol using phenylmagnesium chloride and the reduction of a ketone to a tertiary alcohol using methyllithium. The summary emphasizes the role of the counter ion, typically magnesium, in the intermediate and the importance of the proton source in the final step of the reaction.
Mindmap
Keywords
π‘Nucleophile
π‘Carbonyl
π‘Tetrahedral Intermediate
π‘Lithium Aluminum Hydride (LiAlH4)
π‘Sodium Borohydride (NaBH4)
π‘Grignard Reagent
π‘Reduction
π‘Electrophile
π‘Protonation
π‘Secondary Alcohol
π‘Primary Alcohol
π‘Tertiary Alcohol
Highlights
Introduction of nucleophiles to carbonyl compounds, such as aldehydes and ketones, and the formation of a tetrahedral intermediate.
Explanation of the electrophilic nature of the carbonyl carbon and its attraction to nucleophiles.
Mechanism of bond breaking and electron transfer to oxygen in the formation of a tetrahedral intermediate.
Identification of the tetrahedral intermediate as a common step in carbonyl reactions.
Discussion on the lack of good leaving groups in the intermediate, leading to the formation of an alcohol.
Introduction of lithium aluminum hydride as a strong reducing agent for carbonyl compounds.
Procedure for using lithium aluminum hydride, including the importance of excluding water.
Simplified mechanism of lithium aluminum hydride reducing carbonyls to alcohols.
Introduction of sodium borohydride as a selective reducing agent for ketones and aldehydes.
Differences in reactivity between lithium aluminum hydride and sodium borohydride.
Examples of carbonyl reduction to alcohols using both reducing agents.
Explanation of the addition of Grignard reagents or organolithium to carbonyls.
Formation of secondary, primary, and tertiary alcohols depending on the type of carbonyl attacked by Grignard reagents.
Use of phenylmagnesium chloride in the reaction with formaldehyde to produce a primary alcohol.
Demonstration of the reaction of methyllithium with a ketone to yield a tertiary alcohol.
Emphasis on the nucleophilic carbon in Grignard reagents and its role in attacking the carbonyl group.
Summary of the outcomes of attacking aldehydes and ketones with different reagents to form various alcohols.
Transcripts
alright first before we actually do
anything we're adding nucleophiles to
the carbonyl so we're gonna have a
carbonyl whether it's an aldehyde or a
ketone makes really no difference but
what happens is that the carbon of the
carbonyl is partial positive so it
attracts nucleophiles and when the
nucleophile attacks of course you cannot
have a 5 on that carbon so you need to
break the double bond and when you break
a bond you give the electrons to the
most electronegative element in this
case it's going to be the oxygen at the
top and you're going to now get a
nucleophile attached to a tetrahedral
carbon and at the very top I have an
oxygen then right now it's negative this
intermediate right here is the
intermediate that we arrive at anytime
we attack any carbonyl so we're going to
call it the tetrahedral intermediate and
when you get a tetrahedral intermediate
you check to see if you have good
leaving groups and this one doesn't
happen to have any good leaving groups
carbons this are 2 CH 3 s right here are
going to be bad leaving group and the
nucleophile that we just added we're
going to assume that it's also a bad
leaving group so things you don't have
any bad leaving groups in solution then
the oxygen is gonna pick up a proton
from water or from maybe hydronium if
you add a little bit of acid so it picks
up a proton and we end up with an
alcohol so that's what we're what we're
gonna see in this reactions when we
attack the carbonyl of the ketones and
the aldehydes the first reaction we're
gonna do is a reduction with lithium
aluminum hydride and lithium aluminum
hydride has four hydrates and it's a
very strong reducing agent when you
actually use lithium aluminum hydride
you have to keep water out of it so the
first step you just add the
I'm aluminum hydride in an ether solvent
and then at the very end of the reaction
you're gonna do what I like to call the
quenching people like to call workup but
we're going to be adding water and
that's where the hydrogen's on the
oxygen is going to come from so the
hydride is going to be a hydrogen that
has two electrons and a simplified
mechanism of course would be a hydride
attacking the carbonyl forming the
tetrahedral intermediate and if you form
the tetrahedral intermediate in this
particular case you would have two
carbons or maybe a carbon and a hydrogen
and we just added the hydride this is
the hydride from the lithium aluminum
hydride and then nothing can leave so
the next step would be when we add the
water at the very end the oxygen picks
up a proton from the water so this
hydrogen right here on the oxygen came
from the water this hydrogen came from
the lithium aluminum hydride now this is
a very simplified mechanism that doesn't
take into account what the aluminum or
the liam are doing but we also have a
second hydride reagent not as strong as
lithium aluminum hydride but quite
useful nonetheless it's sodium
borohydride we have seen it before in
3331 but now it's a much easier reducing
agent to use I don't need to have two
different steps and I don't need to have
syringes for this reaction sodium
borohydride can be used in an alcohol
and we see exactly the same thing the
hydride is going to be attached to the
carbon and this hydrogen right here came
from the alcohol
so whatever alcohol you're doing it's
gonna up you're using sorry it's going
to provide the hydrogen that goes on the
oxygen so we have two reducing agents
lithium aluminum hydride in
probably strong reducing agent sodium
borohydride is only good for ketones and
aldehydes so in that sense it's a
selectively reducing agent I have some
examples right here just to make sure
that we understand that lithium aluminum
hydride and sodium borohydride react
with carbonyls so I have a double bond
right here on the Left I am NOT going to
be touching the double bond it's going
to remain in place lithium aluminum
hydride can only reduce the ketone this
is a ketone so I reduce a ketone to a
secondary alcohol this is another ketone
I'm using sodium borohydride notice that
I still have an alcohol this one is
ethanol but all I'm gonna be doing is
reducing the carbonyl I'm not gonna
touch anything else in the molecule the
one at the bottom is an aldehyde and
notice that I have an alcohol on the
Left lithium aluminum hydride can only
reduce the carbonyl so it will take that
out of the hide to the primary alcohol
so aldehydes are reduced to primary
alcohols ketones are reduced to
secondary alcohols and we have two
examples of that all right this is one
of my favorite reactions it's the
addition of a green yard or an
organolithium if you have it to a
carbonyl and the green yard is a carbon
that's actually negatively charged and
it's basically a salt of magnesium so I
have two negative charges and I have my
magnesium in in the middle between them
but what's important again is that a
carbon that's attached to the magnesium
that's a nucleophilic carbon it has a
negative charge so it will attack the
carbonyl it will form a tetrahedral
intermediate and this tetrahedral
intermediate actually
shows that for every negative charge
there has to be a counter ion so the
magnesium is going to be next to the
negative oxygen there are no good
leaving groups in my tetrahedral
intermediate so the oxygen picks up a
proton from the acid at the end of the
reaction and I get in this particular
case I get a secondary alcohol anytime
you you attack an aldehyde with a
grignard you're gonna make a secondary
alcohol if you attack formaldehyde
that's gonna produce a primary alcohol
if you attack all the Heights secondary
alcohols and if you attack a ketone
you're gonna get a tertiary alcohol out
of that reaction all right let's do a
couple of problems in here let's attack
formaldehyde formaldehyde is your most
simple aldehyde and you can attack it
with green yards for organolithium so
I'm going to use phenol magnesium
chloride and again the carbon that's
attached to the magnesium this is a
nucleophilic carbon so it's going to
attack the carbon eel form a tetrahedral
intermediate and the tetrahedral
intermediate I have two hydrogen's
attached to it to the carbons and I just
attached a benzene ring to it it doesn't
have any good leaving groups it has two
hydrogen's terrible living groups and a
phenyl group you just came in let's not
kick it out so when you quench the
reaction when you're working it up with
water or maybe hydronium the oxygen is
going to pick up the hydrogen from the
water
and that's gonna give you your in this
case primary alcohol if you take an a a
ketone let's take two fentanyl and let's
use an organolithium for variety so we
are gonna use methyl lithium
so you add methyl lithium to it and then
you go ahead and you add water to the
reaction but the methyl is going to
attack the carbonyl form your
tetrahedral intermediate I just added a
second methyl so now I have two methyls
my oxygen is negative no good leaving
groups carbons are not leaving groups so
when I add the water that's when the
hydrogen picks up the the hydrogen right
there the oxygen picks up the hydrogen
and this one right here is a tertiary
alcohol so whenever I attack a ketone
I'm going to add a third carbon group to
the carbonyl and I get a tertiary
alcohol out of that reaction
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