Alcohol Phenol And Ethers
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New answer posted
4 months agoContributor-Level 10
11.82
(i) In aryl alkyl ethers the +R effect of the alkoxy group leads to an increase in the electron density of the benzene ring as they push electrons into the ring making the benzene ring activated towards electrophilic substitution reactions. This could be understood more clearly from the following resonating structures : -
(ii) It could be clearly seen from the above resonating structures that the electron density increases more at the ortho and para positions as compared to the meta positions. Hence, we can conclude that the alkoxy group directs the incoming substituents to ortho and para positions in the benzene ring.
For example
New answer posted
4 months agoContributor-Level 10
11.81
1-propoxypropane reacts with hydrogen iodide to give propan-1-ol and 1-iodopropane as the products.
2. Methoxybenzene reacts with hydrogen iodide to give phenol and iodomethane
Benzyl ethyl ether reacts with hydrogen iodide to give benzyl iodide and ethanol
New answer posted
4 months agoContributor-Level 10
11.80
The preparation of ether by acid dehydration of primary alcohol involves the nucleophilic addition of alcohol molecule to the protonated alcohol molecule as shown below: -
However, under these conditions secondary and tertiary alcohols forms alkenes rather than ethers. The reason for this being that due to stearic hindrance, nucleophilic attack by the alcohol molecule on the protonated alcohol molecule does not take place. Instead protonated 20 and 30 alcohols lose a molecule of water to form stable carbocations. The stable carbocations so formed prefers to lose a proton to form alkenes instead of forming ethers by undergoing
New answer posted
4 months agoContributor-Level 10
11.79
According to the question we have to perform the following conversion: -
The mechanism of the above reaction is as follows : - The mechanism is given below: -
In the first step, the alcohol gets protonated by the acid present to give a protonated alcohol.
In the second step, the nucleophilic attack of another alcohol molecule on the protonated alcohol gives us 1-propoxypropane as the desired product.
New answer posted
4 months agoContributor-Level 10
11.78
Williamson synthesis is basically a SN2 reaction of a primary alkyl halide with an alkoxide ion. The basic mechanism for this reaction is
Now consider this reaction,
This reaction proceeds as conventional Williamson synthesis. But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution reaction, which will result in the formation of alkenes. The reason is alkoxides are better nucleophiles as well as strong bases. Therefore, they react with alkyl halides resulting in an elimination reaction.
New answer posted
4 months agoContributor-Level 10
11.77
During the reaction, an alkoxide ion is formed which is then added to an alkyl halide to form the ether via SN2 mechanism.
In the primary step, the nucleophile is formed (O- ) which will the approach to the alkyl halide and after the transition stage, the substitution takes place.
During the reaction, an alkoxide ion is formed which is then added to an alkyl halide to form the ether via SN2 mechanism.
During the reaction, an alkoxide ion is formed which is then added to an alkyl halide to form the ether via SN2 mechanism.
During the reaction, an alkoxide ion is formed which is then added to an alkyl halide to form the ether via
New answer posted
4 months agoContributor-Level 10
11.76
Ethoxy-2-methylpropane
Chloro-1-methoxyethane 4-Nitroanisole
1-Methoxypropane
Ethoxy-4,4-dimethylcyclohexane Ethoxy benzene
New answer posted
4 months agoContributor-Level 10
11.75
Due to the presence of -OH group, ethanol undergoes intermolecular hydrogen bonding which results in the association of molecules.
Therefore, extra energy is required to break those hydrogen bonds. Whereas methoxymethane does not undergo those hydrogen bonding which implies ethanol has a higher boiling point than that of methoxymethane.
New answer posted
4 months agoContributor-Level 10
11.74
1. Acidified KMnO4 (potassium permanganate)
Potassium permanganate is a strong oxidant and is able to react with many functional groups. Here KMnO4 will readily react with primary carbon (where hydrogen is attached) and transforms that to acid.
2. PCC (Pyridinium Chlorochromate)
This is actually a milder version of chromic acid. This works as a sort of elimination reaction. The formation of aldehyde occurs because of the action chromium (a good leaving group) which will be replaced when the C-H bond is broken.
3. Bromine water
Bromine water is actually an aqueous form of bromine. Here in aqueous solution, phenol ionizes to f
New answer posted
4 months agoContributor-Level 10
11.73
1. The conversion of Propene to propane-2-ol takes place according to markovnikoff rule. The positive part of H2O that is H+ goes to the carbon which has more hydrogen and the negative part that is OH-goes to carbon that has less number of carbons
2. NaOH act as a base in the conversion of benzyl chloride to benzyl On hydrolysis removal of NaCl takes place and OH is inserted in place of Cl.
3. Ethyl magnesium chloride (Grignard reagent) attacks on the carbon of the (The partial positive and negative charge is because of the electronegativity difference). After the formation of the addition product, hydrolysis takes place which furth
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