This document discusses the nomenclature, properties and reactions of alcohols, phenols, and ethers. It defines each compound group and provides IUPAC names for examples. Alcohols are formed by replacing hydrogen in hydrocarbons with hydroxyl groups. Phenols have hydroxyl groups attached to aromatic systems. Ethers have an alkoxy or aryloxy group in place of hydrogen. The document outlines common preparation methods for each group and describes physical properties like boiling points. It also explains reactions like dehydration, esterification, and oxidation of alcohols.
Importance of amines, classification of amines, Preparation of amines, Physical properties, Chemical properties, Basic nature, tests of amines, Carbylamine test, Hinsberg's test, reactions with nitrous acid, electrophilic reactions, -NH2 group protection, Diazonium salts, Uses, Some important conversions, short questions with answers.
In organic chemistry, a carbonyl group is a functional group composed of a carbon atom double-bonded to an oxygen atom: C=O. It is common to several classes of organic compounds, as part of many larger functional groups.
Importance of amines, classification of amines, Preparation of amines, Physical properties, Chemical properties, Basic nature, tests of amines, Carbylamine test, Hinsberg's test, reactions with nitrous acid, electrophilic reactions, -NH2 group protection, Diazonium salts, Uses, Some important conversions, short questions with answers.
In organic chemistry, a carbonyl group is a functional group composed of a carbon atom double-bonded to an oxygen atom: C=O. It is common to several classes of organic compounds, as part of many larger functional groups.
This presentation will help students to understand the various topics related with halogen compounds in a very short time.it also help teachers during the recapitulation of the chapter content.it will also help students to revise the content in short time especially by those students who r preparing for various competitive exams after class 12th.
Learning Objectives
1. Know that Carboxylic acids contain the functional group -COOH
2. Understand how to draw structural and displayed formulae for Carboxylic Acids
3. 3. Predict physical properties of Carboxylic Acids
Aldehydes and ketones are organic compounds which incorporate a carbonyl functional group, C=O. The carbon atom of this group has two remaining bonds that may be occupied by hydrogen or alkyl or aryl substituents. If at least one of these substituents is hydrogen, the compound is an aldehyde.
Alcohol, phenol, ether are classes of organic compounds which find wide usage in a broad range of industries as well as for domestic purposes. Alcohol is formed when a saturated carbon atom is bonded to a hydroxyl (-OH) group. Phenol is formed when a hydrogen atom in a benzene molecule is replaced by the -OH group.
This presentation will help students to understand the various topics related with halogen compounds in a very short time.it also help teachers during the recapitulation of the chapter content.it will also help students to revise the content in short time especially by those students who r preparing for various competitive exams after class 12th.
Learning Objectives
1. Know that Carboxylic acids contain the functional group -COOH
2. Understand how to draw structural and displayed formulae for Carboxylic Acids
3. 3. Predict physical properties of Carboxylic Acids
Aldehydes and ketones are organic compounds which incorporate a carbonyl functional group, C=O. The carbon atom of this group has two remaining bonds that may be occupied by hydrogen or alkyl or aryl substituents. If at least one of these substituents is hydrogen, the compound is an aldehyde.
Alcohol, phenol, ether are classes of organic compounds which find wide usage in a broad range of industries as well as for domestic purposes. Alcohol is formed when a saturated carbon atom is bonded to a hydroxyl (-OH) group. Phenol is formed when a hydrogen atom in a benzene molecule is replaced by the -OH group.
Richard's entangled aventures in wonderlandRichard Gill
Since the loophole-free Bell experiments of 2020 and the Nobel prizes in physics of 2022, critics of Bell's work have retreated to the fortress of super-determinism. Now, super-determinism is a derogatory word - it just means "determinism". Palmer, Hance and Hossenfelder argue that quantum mechanics and determinism are not incompatible, using a sophisticated mathematical construction based on a subtle thinning of allowed states and measurements in quantum mechanics, such that what is left appears to make Bell's argument fail, without altering the empirical predictions of quantum mechanics. I think however that it is a smoke screen, and the slogan "lost in math" comes to my mind. I will discuss some other recent disproofs of Bell's theorem using the language of causality based on causal graphs. Causal thinking is also central to law and justice. I will mention surprising connections to my work on serial killer nurse cases, in particular the Dutch case of Lucia de Berk and the current UK case of Lucy Letby.
Deep Behavioral Phenotyping in Systems Neuroscience for Functional Atlasing a...Ana Luísa Pinho
Functional Magnetic Resonance Imaging (fMRI) provides means to characterize brain activations in response to behavior. However, cognitive neuroscience has been limited to group-level effects referring to the performance of specific tasks. To obtain the functional profile of elementary cognitive mechanisms, the combination of brain responses to many tasks is required. Yet, to date, both structural atlases and parcellation-based activations do not fully account for cognitive function and still present several limitations. Further, they do not adapt overall to individual characteristics. In this talk, I will give an account of deep-behavioral phenotyping strategies, namely data-driven methods in large task-fMRI datasets, to optimize functional brain-data collection and improve inference of effects-of-interest related to mental processes. Key to this approach is the employment of fast multi-functional paradigms rich on features that can be well parametrized and, consequently, facilitate the creation of psycho-physiological constructs to be modelled with imaging data. Particular emphasis will be given to music stimuli when studying high-order cognitive mechanisms, due to their ecological nature and quality to enable complex behavior compounded by discrete entities. I will also discuss how deep-behavioral phenotyping and individualized models applied to neuroimaging data can better account for the subject-specific organization of domain-general cognitive systems in the human brain. Finally, the accumulation of functional brain signatures brings the possibility to clarify relationships among tasks and create a univocal link between brain systems and mental functions through: (1) the development of ontologies proposing an organization of cognitive processes; and (2) brain-network taxonomies describing functional specialization. To this end, tools to improve commensurability in cognitive science are necessary, such as public repositories, ontology-based platforms and automated meta-analysis tools. I will thus discuss some brain-atlasing resources currently under development, and their applicability in cognitive as well as clinical neuroscience.
Cancer cell metabolism: special Reference to Lactate PathwayAADYARAJPANDEY1
Normal Cell Metabolism:
Cellular respiration describes the series of steps that cells use to break down sugar and other chemicals to get the energy we need to function.
Energy is stored in the bonds of glucose and when glucose is broken down, much of that energy is released.
Cell utilize energy in the form of ATP.
The first step of respiration is called glycolysis. In a series of steps, glycolysis breaks glucose into two smaller molecules - a chemical called pyruvate. A small amount of ATP is formed during this process.
Most healthy cells continue the breakdown in a second process, called the Kreb's cycle. The Kreb's cycle allows cells to “burn” the pyruvates made in glycolysis to get more ATP.
The last step in the breakdown of glucose is called oxidative phosphorylation (Ox-Phos).
It takes place in specialized cell structures called mitochondria. This process produces a large amount of ATP. Importantly, cells need oxygen to complete oxidative phosphorylation.
If a cell completes only glycolysis, only 2 molecules of ATP are made per glucose. However, if the cell completes the entire respiration process (glycolysis - Kreb's - oxidative phosphorylation), about 36 molecules of ATP are created, giving it much more energy to use.
IN CANCER CELL:
Unlike healthy cells that "burn" the entire molecule of sugar to capture a large amount of energy as ATP, cancer cells are wasteful.
Cancer cells only partially break down sugar molecules. They overuse the first step of respiration, glycolysis. They frequently do not complete the second step, oxidative phosphorylation.
This results in only 2 molecules of ATP per each glucose molecule instead of the 36 or so ATPs healthy cells gain. As a result, cancer cells need to use a lot more sugar molecules to get enough energy to survive.
Unlike healthy cells that "burn" the entire molecule of sugar to capture a large amount of energy as ATP, cancer cells are wasteful.
Cancer cells only partially break down sugar molecules. They overuse the first step of respiration, glycolysis. They frequently do not complete the second step, oxidative phosphorylation.
This results in only 2 molecules of ATP per each glucose molecule instead of the 36 or so ATPs healthy cells gain. As a result, cancer cells need to use a lot more sugar molecules to get enough energy to survive.
introduction to WARBERG PHENOMENA:
WARBURG EFFECT Usually, cancer cells are highly glycolytic (glucose addiction) and take up more glucose than do normal cells from outside.
Otto Heinrich Warburg (; 8 October 1883 – 1 August 1970) In 1931 was awarded the Nobel Prize in Physiology for his "discovery of the nature and mode of action of the respiratory enzyme.
WARNBURG EFFECT : cancer cells under aerobic (well-oxygenated) conditions to metabolize glucose to lactate (aerobic glycolysis) is known as the Warburg effect. Warburg made the observation that tumor slices consume glucose and secrete lactate at a higher rate than normal tissues.
Richard's aventures in two entangled wonderlandsRichard Gill
Since the loophole-free Bell experiments of 2020 and the Nobel prizes in physics of 2022, critics of Bell's work have retreated to the fortress of super-determinism. Now, super-determinism is a derogatory word - it just means "determinism". Palmer, Hance and Hossenfelder argue that quantum mechanics and determinism are not incompatible, using a sophisticated mathematical construction based on a subtle thinning of allowed states and measurements in quantum mechanics, such that what is left appears to make Bell's argument fail, without altering the empirical predictions of quantum mechanics. I think however that it is a smoke screen, and the slogan "lost in math" comes to my mind. I will discuss some other recent disproofs of Bell's theorem using the language of causality based on causal graphs. Causal thinking is also central to law and justice. I will mention surprising connections to my work on serial killer nurse cases, in particular the Dutch case of Lucia de Berk and the current UK case of Lucy Letby.
Earliest Galaxies in the JADES Origins Field: Luminosity Function and Cosmic ...Sérgio Sacani
We characterize the earliest galaxy population in the JADES Origins Field (JOF), the deepest
imaging field observed with JWST. We make use of the ancillary Hubble optical images (5 filters
spanning 0.4−0.9µm) and novel JWST images with 14 filters spanning 0.8−5µm, including 7 mediumband filters, and reaching total exposure times of up to 46 hours per filter. We combine all our data
at > 2.3µm to construct an ultradeep image, reaching as deep as ≈ 31.4 AB mag in the stack and
30.3-31.0 AB mag (5σ, r = 0.1” circular aperture) in individual filters. We measure photometric
redshifts and use robust selection criteria to identify a sample of eight galaxy candidates at redshifts
z = 11.5 − 15. These objects show compact half-light radii of R1/2 ∼ 50 − 200pc, stellar masses of
M⋆ ∼ 107−108M⊙, and star-formation rates of SFR ∼ 0.1−1 M⊙ yr−1
. Our search finds no candidates
at 15 < z < 20, placing upper limits at these redshifts. We develop a forward modeling approach to
infer the properties of the evolving luminosity function without binning in redshift or luminosity that
marginalizes over the photometric redshift uncertainty of our candidate galaxies and incorporates the
impact of non-detections. We find a z = 12 luminosity function in good agreement with prior results,
and that the luminosity function normalization and UV luminosity density decline by a factor of ∼ 2.5
from z = 12 to z = 14. We discuss the possible implications of our results in the context of theoretical
models for evolution of the dark matter halo mass function.
This presentation explores a brief idea about the structural and functional attributes of nucleotides, the structure and function of genetic materials along with the impact of UV rays and pH upon them.
Observation of Io’s Resurfacing via Plume Deposition Using Ground-based Adapt...Sérgio Sacani
Since volcanic activity was first discovered on Io from Voyager images in 1979, changes
on Io’s surface have been monitored from both spacecraft and ground-based telescopes.
Here, we present the highest spatial resolution images of Io ever obtained from a groundbased telescope. These images, acquired by the SHARK-VIS instrument on the Large
Binocular Telescope, show evidence of a major resurfacing event on Io’s trailing hemisphere. When compared to the most recent spacecraft images, the SHARK-VIS images
show that a plume deposit from a powerful eruption at Pillan Patera has covered part
of the long-lived Pele plume deposit. Although this type of resurfacing event may be common on Io, few have been detected due to the rarity of spacecraft visits and the previously low spatial resolution available from Earth-based telescopes. The SHARK-VIS instrument ushers in a new era of high resolution imaging of Io’s surface using adaptive
optics at visible wavelengths.
THE IMPORTANCE OF MARTIAN ATMOSPHERE SAMPLE RETURN.Sérgio Sacani
The return of a sample of near-surface atmosphere from Mars would facilitate answers to several first-order science questions surrounding the formation and evolution of the planet. One of the important aspects of terrestrial planet formation in general is the role that primary atmospheres played in influencing the chemistry and structure of the planets and their antecedents. Studies of the martian atmosphere can be used to investigate the role of a primary atmosphere in its history. Atmosphere samples would also inform our understanding of the near-surface chemistry of the planet, and ultimately the prospects for life. High-precision isotopic analyses of constituent gases are needed to address these questions, requiring that the analyses are made on returned samples rather than in situ.
Comparing Evolved Extractive Text Summary Scores of Bidirectional Encoder Rep...University of Maribor
Slides from:
11th International Conference on Electrical, Electronics and Computer Engineering (IcETRAN), Niš, 3-6 June 2024
Track: Artificial Intelligence
https://www.etran.rs/2024/en/home-english/
Comparing Evolved Extractive Text Summary Scores of Bidirectional Encoder Rep...
Alcohols phenols
1.
2. After studying this topic, you will be able to:
•name alcohols, phenols and ethers according to
the IUPAC system of nomenclature
3. If H of hydrocarbon is replaced by:
• One or more hydroxyl (OH) group directly
attached to carbon atom(s), of an aliphatic
system we get an ALCOHOL eg: C2H5OH
• C6H5CH2OH ALCOHOL AROMATIC
• If (–OH) group(s) directly attached to carbon
atom(s) of an aromatic system, we get a
PHENOL (C6H5OH).
• an alkoxy or aryloxy group (R–O/Ar–O), we get
ETHER eg: CH3OCH3 .
4.
5. Structure Name (common) IUPAC Name Type of phenol
Phenol Phenol Monohydric
o- cresol 2-Methylphenol Monohydric
Catechol Benzene-1,2-diol
Dihydric
Resorcinol Benzene-1,3-diol
Dihydric
Hydroquinone Benzene-1,4-diol
Dihydric
6. Structure Name (common) IUPAC Name Type of ethers
CH3OCH3 Dimethyl ether Methoxymethane Symmetrical
CH3CH2OCH2CH3 Diethyl ether Ethoxyethane Symmetrical
CH3CH2CH2OCH3 Methyl n- propyl
ether
Methoxypropane Unsymmetrical
C6H5OCH3 Methyl phenyl
ether (Anisole)
Methoxybenzene
(Anisole)
Unsymmetrical
C6H5OC2H5 Ethyl phenyl ether Ethoxybenzene Unsymmetrical
(i)4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol (ii) 2, 5-Dimethylhexane-1,3-diol
(iii) 3-Bromocyclohexanol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol
7. Preparation of Alcohols
• 1. Hydration of alkenes (Markovnikov addition of
water)
CH3CH=CH2 H2O, H2SO4 CH3CH(OH)CH3
(Markovnikov rule, OH- goes least hydrogenated carbon)
9. • 3.Reduction of aldehydes and ketones
• Aldehyde / ketone on reduction give alcohol
• Any one of the following Reducing agents:
(i)H2 Pt or Pd or Ni (ii)NaBH4 (iii)LiAlH4
• RCHO NaBH4 RCH2OH (PRIMARY ALCOHOL)
• CH3CH2CH2CHO H2 / Pt CH3CH2CH2CH2OH
• RCOR LiAlH4 RCH(OH)R (SECONDARY ALCOHOL)
• CH3CH2COCH3 H2 / Pt CH3CH2 CH CH3
OH
Acetone on reduction with NaBH4 gives? CH3CH(OH)CH3
Ethanal on reduction with LiAlH4 gives? CH3CH2OH
Alcohol ----oxidation -------ald/ketone -------oxidation -----carboxylic acid
10. • 4.Grignard reagent and aldehydes / ketones
• >Cδ+=Oδ- + Rδ-Mgδ+X >C-O- Mg+X H2O >C-OH
R R
HCHO + CH3MgCl H-C-OMgCl H2O H-C-OH
• H CH3 H CH3
• CH3CHO + CH3MgCl
• CH3COCH3 + CH3MgBr ?
Formaldehyde on reaction with Grignard reagent followed by hydrolysis gives 1o
alcohol
ANY OTHER aldehyde on reaction with Grignard reagent followed by hydrolysis gives
2o alcohol
KETONE on reaction with Grignard reagent followed by hydrolysis gives 3o alcohol
11. • 5. Carboxylic acid on reduction
• LiAlH4 is an expensive reagent, so is not used
commercially
• Commercial method
• RCOOH R’OH, H+ RCOOR’ H2, Pt RCH2OH + R’OH
RCOOH (i) LiAlH4 (ii) H2O RCH2OH
12. Preparation of Phenols
• From Haloalkanes (Dow’s Process)
• From Benzene sulphonic acid
• Oleum conc H2SO4 : H2S2O7
16. Physical Properties
• Boiling point
Increase with increase in number of carbon atoms as (van
der waal forces increase)
decrease with increase of branching in carbon
chain(decrease in van der Waals forces with decrease in
surface area)
CH3CH2CH2CH2OH CH3CH2CHCH3
OH
Alcohols and phenols have higher boiling point than
hydrocarbons, ethers, haloalkanes and haloarenes of
comparable molar masses (alcohols have intermolecular
hydrogen bonding, hydrocarbons van der waal forces,
ethers and haloalkanes have dipole -dipole interaction)
17. Alkane bp oC Alkyl
halide
bp oC alcohol bp oC ether bp oC
propane -42 Methyl
chloride
-24 ethanol 79 Dimethyl
ether
-25
18. Solubility
• alcohols and phenols are soluble in water is
due to their ability to form hydrogen bonds
with water molecules.
• The solubility decreases with increase in size
of alkyl/aryl (hydrophobic) groups.
• Several of the lower molecular mass alcohols
are miscible with water in all proportions.
19. Back Exercise
11.3 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O
and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary
and tertiary alcohols.
11.4 Explain why propanol has higher boiling point than that of the hydrocarbon,
butane?
11.5 Alcohols are comparatively more soluble in water than hydrocarbons of
molecular masses. Explain this fact.
11.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
11.7 Give the structures and IUPAC names of monohydric phenols of molecular
formula, C7H8O.
11.9 Give the equations of reactions for the preparation of phenol from cumene.
11.10 Write chemical reaction for the preparation of phenol from chlorobenzene.
11.11 Write the mechanism of hydration of ethene to yield ethanol.
11.12 You are given benzene, conc. H2SO4 and NaOH. Write the equations for the
preparation of phenol using these reagents.
21. 11.13 Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by
an SN
2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?
(i) CH=CH2 H2SO4/H2O CH(OH)CH3
CH2Br aq NaOH
22. The bond between O–H is broken(H+ goes) when alcohols react
as nucleophiles.
Breaking of RO-H
1. Acidity of alcohols and phenols
2ROH + 2Na 2RONa + H2
ROH + NaOH no reaction
2C6H5OH + 2Na 2C6H5ONa + H2
C6H5OH + NaOH C6H5ONa + H2O
23. • The acidic character of alcohols is due to the
polar nature of O–H bond.
• An electron-releasing group (–CH3, –C2H5)
increases electron density on oxygen tending
to decrease the polarity of O-H bond.
• Alcohols are, however, weaker acids than
water.
24. The acid strength of alcohols decreases in the following
order:
• Primary> secondary>tertiary
•
• (O is more electronegative than H so attracts shared pair towards
itself thus making the bond polar. When e- density on O is
increased, its tendency to attract electrons is decreased making bond
less polar)
25. Acidity of phenols
• Phenol is a stronger acid than alcohols and water.
• WHY?
• Which is more polar: alcohol or phenol? C3-O C2-O
• Due to high electronegativity of sp2 hybrid carbon attached to –OH in
phenol, electron density decreases on O making O-H bond in phenols more
polar than alcohols (which have sp3 hybrid carbon)
25
26. Acidity of phenols
• Which ion is more stable: phenoxide or
alkoxide?
• When H+ is released from alcohol, it forms alkoxide
ion, which is less stable.
• When H+ is released from phenol, it forms phenoxide
ion, which is stable due to resonance.
26
27. Resonance Recap
• 1. More the number of resonance structures
higher is the stability.
• 2.Structures which involve separation of
charges are less stable so contribute less
towards resonance hybrid.
• 3.Structures in which –ve charge is delocalised
on more electronegative atom are more stable
than those where there is localisation of
charge.
28. Compare resonance structures of phenol and phenoxide ion
Structures II,III and IV of phenol involve separation of charges thus they
contribute less towards resonance hybrid.
Phenoxide ion involves delocalisation of negative charge so is more
stable. The reaction shifts in forward direction, towards more stable
phenoxide ion making phenols more acidic than alcohols.
In Alkoxide ion negative charge
is localised on oxygen so it is
less stable than alcohol
30. Effect of substituent on acidity
• Higher the pKa = - log Ka value weaker is the acid
• Cresols are weaker acids than phenol while nitrophenols are stronger acids than
phenol.
• Nitro group is e- withdrawing group, it withdraws electrons from phenoxide ion
making it more stable thus increasing acidity of phenols.
• Methyl group is e- donating group, it destabilizes phenoxide ion making phenols
less acidic.
32. Esterification
• Alcohols and phenols react with carboxylic acids,
acid chlorides and acid anhydrides to form
esters.
• ROH + R’COOH H+ R’COOR + H2O
CH3.CH2.OH + HOOC.CH2.CH3 H+ CH3.CH2.OOC.CH2.CH3 + H2O
ROH + ( R’CO)2O H+ R’COOR + R’COOH
Water/ carboxylic acid formed in the reactions above is removed as soon as it is
formed so that reaction moves in forward direction. (Le Chatlier’s principle)
32
34. Esterification
ROH + R’COCl pyridine R’COOR + HCl
CH3CH2OH + ClOCCH2CH3 pyridine CH3CH2OOCCH2CH3 + HCl
• Pyridine is a weak base. It is added to neutralize
HCl formed during the reaction so that reaction
moves in forward direction. (Le Chatlier’s
principle)
34
35. Reactions involving breaking of carbon – oxygen (C–O) bond
1. Reaction with hydrogen halides
Alcohols react with hydrogen halides to form alkyl halides.
ROH + HX (anhy ZnCl2) → R–X + H2O
LUCAS REAGENT (conc. HCl and ZnCl2) TEST
Alcohols are soluble in Lucas reagent while their halides are immiscible
and produce turbidity in solution.
Tertiary alcohols, turbidity is produced immediately
Secondary alcohols, turbidity is produced within 5 minutes
Primary alcohols do not produce turbidity at room temperature.
2. Reaction with SOCl2, PCl3 and PCl5
ROH + PCl3 → RCl +H3PO3
ROH + PCl5 →RCl + POCl3 + HCl
ROH + SOCl2 → RCl + SO2 + HCl
36. 2. Dehydration of alcohols
CH3CHCH2CH3
conc H2SO4 CH3CH=CHCH3 + CH2=CHCH2CH3
Major Minor
Saytzeff or Zaitsev Rule states that the more substituted alkene will be the
major product. (H will be removed from least hydrogenated carbon during
elimination reaction)
OH
37. • Ques: Why does the relative ease of dehydration
of alcohols follow the following order:
Tertiary > Secondary > Primary
• Ans: Mechanism of dehydartion involves
carbocation intermediates. Tertiary carbocations
are more stable and therefore are easier to form
than secondary and primary carbocations thus
tertiary alcohols are the easiest to dehydrate.
38. Oxidation
• Oxidation of alcohols involves the formation of a carbon-oxygen
double bond with cleavage of an O-H and C-H bonds. These are also
known as dehydrogenation reactions as these involve loss of
dihydrogen from an alcohol molecule.
•
• RCH2OH [O] RCHO [O] RCOOH
Primary alcohol on oxidation gives aldehyde same number of C atoms which
on further oxidation gives carboxylic acid with same number of carbon
atoms
RCH(OH )CH3
[O] RCOCH3
[O] RCOOH
Secondary alcohol on oxidation gives ketone same number of C atoms which
on further oxidation gives carboxylic acid with lesser number of carbon
atoms
38
39. Alcohol to aldehyde/ ketone:Weak oxidising agent like PCC (pyridium chloro chromate)
and CrO3
Alcohol to Carboxylic acid: strong oxidising agent like acidified/ alkaline K2Cr2O7 or
acidified/ alkaline KMnO4
CH3CH2OH CrO3 or PCC CH3CHO
CH3CH(OH )CH3 CrO3 CH3COCH3
PCC does not effect double bond
CH2=CHCH2CH2OH PCC CH2=CHCH2CH2CHO
CH3CH2OH acidified KMnO4 CH3COOH
Tertiary alcohols do not undergo oxidation reaction. Under
strong reaction conditions such as strong oxidising agents
(KMnO4) and elevated temperatures, cleavage of various C-C
bonds takes place and a mixture of carboxylic acids
containing lesser number of carbon atoms is formed.
40. Reaction with Cu at 573K
CH3CH2OH Cu/573K CH3CHO
CH3CH(OH )CH3 Cu/573K CH3COCH3
CH3CH(OH )CH3 Cu/573K CH2=CHCH3
CH3 CH3
41. Reactions of phenols
• Electrophilic aromatic substitution
• The –OH group attached to the benzene ring
activates it towards electrophilic substitution.
Also, it directs the incoming group to ortho and
para positions in the ring.
41
44. o-Nitrophenol is steam volatile due to intramolecular
hydrogen bonding while p-nitrophenol is less volatile due to
intermolecular hydrogen bonding which causes the association
of molecules.
O and p nitro phenol can be separated by distillation
45. With conc HNO3, picric acid is produced, but yield is low
Phenol + sulphuric acid o and p hydroxysulphonic acid
Nitration Picric (good yield)
46. Halogenation
1. Halogenation does not require a Lewis acid like FeCl3 or AlCl3
because polarisation of bromine molecule takes place even in the
absence of Lewis acid. It is due to the highly activating
effect of –OH group attached to the benzene ring.
47. Halogenation
In presence of CS2 the product was monobromophenol while
bromine water gives tribromo derivative (a white ppt) Why?
In aq medium, phenoxide ion is formed which further
activates benzene ring while in non polar solvents it is not
formed
55. Some commercially important alcohols
• Methanol (wood spirit)
Poisonous, causes blindness and death
• Used as solvent for paint and varnishes and in
preparing formaldehyde
• Ethanol
