In JEE Main Chemistry, the topic Organic Compounds containing Oxygen is very essential. Under the wide category of General Organic Chemistry, it plays a critical role. From the Organic Compounds containing Oxygen chapter, at least one or two questions are asked, totalling roughly four marks.
As a result, the total weightage of this chapter in the JEE Main exams is up to 1-2 percent. If you thoroughly understand the theoretical section of this chapter, you will have no trouble answering the chapter's questions. Ethers, Alcohols, and the Cannizzaro Reaction are a few of the concepts investigated under the Organic Compounds containing Oxygen heading.
There are four major oxygen containing compounds in organic chemistry: A carboxylic acid is an organic molecule with a carbon (C) atom double linked to an oxygen (O) atom and a single attached hydroxyl group (OH).
Alkene combines with diborane to generate trialkyl boranes, which then react with alkaline H2O2 to form alcohols, resulting in anti-water Markovnikov's addition.
3. By using esters
The OH group has a big influence on the alcohol's characteristics.
The boiling points of the lower members, such as methanol, ethanol, and 1propanol, are greater.
Alcohols with smaller alkyl groups are more water soluble, but as the size of the alkyl group grows larger, the solubility declines.
Substitution occurs when alcohols react with a hydrogen halide, resulting in an alkyl halide and water.
R—OH + HX → R—X + H2O
Alkyl halides are formed when alcohols combine with PX3 and PX5.
3 ROH + PBr3 → 3RBr + H3PO3
(1° or 2° alcohol)
ROH + PCl5 → RCl + POCl3 + HCl
ROH + SOCl2 → R — Cl + SO2 + HCl
The creation of an ester is the third method of replacing hydroxyl hydrogen with an acyl group. Acid chloride, acid anhydride, or carboxylic acid are all frequent sources of acyl groups (RCO). Esters form when they react with alcohol.
Alcohol is a non-acidic substance. A Grignard reagent, which is a strong base, can remove hydrogen from an alcohol's (OH) group.
CH3OH + CH3MgI → CH3OMgI + CH4
With ceric ammonium nitrate solution, an organic molecule containing one oxygen produces a red hue. Any monohydric alcohol, including phenol, can give off a red colour when it contains one oxygen.
Chlorobenzene is heated at 350°C under high pressure with aqueous NaOH in this procedure. The reaction creates C6H5ONa+ (sodium phenoxide), which is converted to phenol by acidification. Dow's method is the name for this reaction.
This process involves melting sodium benzene sulphonate with sodium hydroxide to make sodium phenoxide, which is then acidified to produce phenol.
Phenol is created when a diazonium sulphate solution is steam distilled.
Decarboxylation occurs when phenolic acids are heated with soda lime, resulting in phenols.
Phenol (C6H5OH) is a colourless solid.
The m.p. is 41°C.
The b.p. 182°C.
It becomes coloured on exposure to air.
It is fairly soluble in water.
The ortho and para hydroxy acetophenones are produced by the Fries rearrangement of phenyl acetate with AlCl3. The ortho isomer is removed from the mixture using steam distillation.
When an alkaline phenol solution is forcefully shaken with benzoyl chloride, phenyl benzoate is produced. It's known as the Schotten Baumann response.
Aryl chloride is generated when phenol is treated with PCl5 (poor yield). Triphenyl phosphate is the major product of this process.
When phenol is treated with ammonia at 300°C under high pressure in the presence of anhydrous AlCl3, ZnCl2, or CaCl2, aniline is produced.
When phenols are heated with Zn dust, aromatic hydrocarbons are formed.
When carbon dioxide gas is passed through sodium phenoxide at 400 degrees Celsius and 67 pounds per square inch of air pressure, sodium salicylate is created, which when acidified yields salicylic acid, but some para isomer is also produced.
2ROH + 2Na → 2RO-Na+ + H2
RO-Na+ + R’X → R-O-R’ + NaX
It involves sodium alkoxide reacting with an alkyl halide. The action of Na on alcohol produces sodium alkoxide.
ROH + HOR → R-O-R + H2O (at 140 degrees celsius temperature)
Ethers boil at a substantially lower temperature than the alcohols from which they are formed or similar molecular weight alcohols.
The boiling points are very similar to those of similar molecular complexity substituted alkanes.
Though ethers are neutral compounds, these form oxonium salts with inorganic acids.
This reaction is due to lone pairs of electrons on oxygen of the ether functional group.
R-O-R + HX → RX + R’OH + R’X + H2O
Reactivity of HX : HI > HBr > HCl
Cleavage takes place only under vigorous conditions, i.e., concentrated acids (usually HI or HBr) and at high temperatures.
Action of HI
(a) At room temperature
(b) At 100° C (excess of HI)
Thus, the metameric ethers can be easily identified by the cleavage with HI.
Thus, the reaction of PCl5 or SOCl2 can be used to identify the metameric ethers.
Aldehydes and ketones belong to a class of compounds having general formula CnH2nO and are represented as RCHO and RR'CO, respectively.
Oxidation can be affected with acidic K2Cr2O7 or with PCC (Pyridine Chlorochromate, mild oxidising agent) in CH2Cl2.
2. Oxidation of Methylbenzenes
3. By Heating a Mixture of the Calcium Salts of Formic Acid and any one of its Homologues
Used for the preparation of aliphatic aromatic ketones or aromatic ketones.
The polar carbonyl group makes aldehydes and ketones polar compounds and hence they have higher boiling points than non-polar compounds of comparable molecular weights.
A specific oxidant for RCHO is Ag(NH3)2+. Ketones do not give this reaction.
Ketones resist mild oxidation, but with strong oxidants at high temperature, they undergo cleavage of C-C bond on either side of the carbonyl group.
CH3COR are readily oxidised by NaOI (NaOH + I2) to iodoform, CHI3, and RCO2Na.
(a) Reduction to alcohols
Reduction to alcohols can be achieved either by hydrogenation in the presence of Pt or Pd or by LiAlH4/NaBH4.
(b) Reduction to Hydrocarbons
The carbon of the carbonyl group is electrophilic and thus initially attacked by a nucleophile.
(a) Addition of Grignard Reagent
(b) Addition of Hydrogen Cyanide
(c) Addition of Sodium Bisulfite
Bisulfite compound formation is confined to aldehydes, methyl ketones, and some cyclic ketones.
(d) Nucleophilic Addition of Derivatives of Ammonia
The carbonyl compounds react with alcohols, R'OH, to yield hemiacetals.
The acid catalysed conversion of ketoximes to N substituted amides is known as Beckmann rearrangement.
The molecules of carboxylic acids are polar and exhibit hydrogen bonding.
The first four members are miscible with water. The higher acids are virtually insoluble. The simplest aromatic acid, benzoic acid, contains too many carbon atoms to show appreciable solubility in water.
The carboxylic acids react with metal to liberate hydrogen and are soluble in NaOH and NaHCO3 solutions.
Carboxylic acid reacts with halogenated compounds and forms acid halide.
Carboxylic acid gets reduced to alcohols.
Hell Volhard Zelinsky Reaction: Aliphatic carboxylic acids react with Br2 or Cl2 in presence of phosphorus (or phosphorus halides) to give haloacids.
Schmidt Reaction: Carboxylic acids react with hydrazoic acid in presence of concentrated H2SO4 to give amine.
Benzoic acid is attacked by the usual electrophilic reagents Chlorine, Bromine, Nitric and sulphuric acids to give m-derivatives.
Acid derivatives are the compounds in which OH group of carboxyl group has been replaced by Cl, COR, NH2, or OR'.
The carboxylate ion is very unreactive towards nucleophilic substitution because it is negatively charged, hence the reaction is essentially irreversible.
2. Reaction with Grignard Reagent
RCOONH4 → RCONH2 + H2O (on heating)
2. Hoffmann’s Bromamide Reaction
Amides (having a primary nitrogen atom) react with bromine in the presence of alkali to form a primary amine having one carbon atom less than the parent amide.
1. Give the structures and IUPAC names of the products expected from the given reaction: catalytic reduction of butanal.
Solution: The catalytic reduction of butanal takes in the presence of a reducing agent. This reaction of butanal will lead to the formation of butanol.
Key Points: Aldehyde is reduced to give alcohols.
2. Write structures of the products of the given reaction.
Solution: The product formed in the above reaction will be 2-Methylbutanol. The structure of this compound is given below.
Key Points: Aldehyde group is reduced to alcohols. The chain with which the functional group is attached is considered as the parent chain.
Ans: The correct answer is option b.
Trick: The given reaction is dependent on the formation of stable carbocation.
2. Predominant product is
Ans: The correct answer is option b.
Trick: OH group is activating in nature. So, NO2 will attack at ortho and para position, but para position is already blocked; therefore, NO2 is only attached at ortho position w.r.t. OH group.
3. The structure of the starting compound P used in the reaction given below is:
Trick: The given reaction is chloroform (CHCl3) reaction.
1. Arrange the sets of compounds in order of their increasing boiling points:
Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol
Ans: Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
2. What will be the product formed in dinitration of 3-methylphenol?
In this article, we have provided important information regarding the chapter Organic Compounds Containing Oxygen such as important concepts, reactions, etc. Students should work on more solved examples and previous year question papers for securing good grades in the JEE Main exams.
1. In the JEE Main examination, how important is the alcohol chapter?
Nearly 1-2 questions arise in the exam from this chapter covering about 8 marks which makes about 2% of the total marks.
2. What are the main points to remember while tackling problems involving halogen-containing organic compounds?
Students should practise writing the mechanism of the reaction for solving the questions from the JEE Main Chemistry Organic Compounds Containing Halogens.
3. Are previous year question papers enough for scoring good marks in JEE Main?
NCERT and previous year JEE Main papers are adequate to score good marks (200+) in JEE Main. Solving previous 10-year IIT-JEE exams gives us a significant edge because 3-4 questions with the same answers are sure to be repeated every year.