Alcohol Phenol And Ethers
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New answer posted
8 months agoContributor-Level 10
11.57
Due to the presence of –OH group, alcohols form hydrogen-bonds with water but hydrocarbons cannot form hydrogen-bonds with water.
Due to inter moleculer hydrogen bonding between Alcohol and water molecular they remain tightly bounded to water molecules and have higher solubility. Whereas in case of hydrocarbon there is no chance of hydrogen bonding.
New answer posted
8 months agoContributor-Level 10
11.56
Here, propanol undergoes intermolecular H-bonding because of the presence of -OH group while butane has no such property
(intermolecular Hydrogen bonding in propanol)
Therefore, extra energy will be required to break those hydrogen bonds which in turn causes higher boiling point for propanol when compared to butane.
New answer posted
8 months agoContributor-Level 10
11.55
The structures of all isomeric alcohols of C5H12O are given below:
Naming is done by the conventional method. The -OH group is attached on the first carbon.
(b) 3-Methylbutan-1-ol
Butane is the longest chain and methyl is the substituent group.
(c) 3-Methylbutan-1-ol
the longest chain is butane and conventional naming method is used.
(d) 2,2-Dimethylpropan-1-ol
Isomer is made by transforming the principal carbon into tertiary type. The longest chain is butane and named accordingly.
(e) Pentan-2-ol
the longest chain is pentane and the numbering is chosen from the minimum position.
(f) 3-Methylbutan-2-ol
Butane is the longest chain,
New answer posted
8 months agoContributor-Level 10
11.54
Butane is the longest chain.
Propane is principle chain and substituents are numbered accordingly.
Hexane is the longest chain. There are three -OH substituents and 2 methyl groups.
-OH of phenol is numbered a 1.
Propane is the principle chain and the alkoxy group is ethyl group.
the longest chain is pentane and substituents are numbered accordingly.
In such cases, cyclo groups are named first, followed by conventional naming methods.
The cyclo group is named first, followed by conventional naming methods.
The longest chain is butane when looked for minimum numbering case.
Cyclo group is named first. Pentane having a double bond is named
New answer posted
8 months agoContributor-Level 10
11.54
1. 2,2,4-Trimethylpentan-3-ol
The naming of the compound usually starts with numbering the carbons in the chain. The lower set of locants are chosen for this purpose, while in this case numbering under this condition is done from the left side. Once the carbons are mentioned, the position of -OH group is numbered and -ol is added as the suffix.
2. 5-Ethylheptane-2,4-diol
Here the longest chain is the straight chain. For such situations, numbering should be such that the functional groups should be denoted by the smallest number. Ethyl group is named at the beginning as it's a side chain.
3. &n
New answer posted
8 months agoContributor-Level 10
11.53
1. When N –propyl methyl ether reacts with HBr, it forms propanol and bromomethane, n- propyl methyl ether will cleave at O . and H+ will attack at O, and Br- will attack CH +
2. When Ethoxybenzene Reacts with HBr, it forms Phenol and bromoethane, Ethoxybenzene cleaves at H+ will attack at O, and Br- will attack C H +
3. When nitrating mixture reacts with ethoxy benzene introduction of nitro group is occurred at para position as it will give the stable product without
4. As HI is a strong nucleophile it will protonate the oxygen, to form a good leaving And I- will attack at C (CH3)3+ to give tert- butyl iodide and ethanol.
New answer posted
8 months agoContributor-Level 10
11.52
Set (ii) is appropriate Because CH3Br is only a nucleophile whereas CH3ONa is nucleophile as well as strong base, so the elimination reaction can occur,
New answer posted
8 months agoContributor-Level 10
11.51
During Williamson synthesis of ethers, an alkyl halide reacts with an alkoxide (ion with –ve charge on the oxygen of alcohol and + ve charge on alkali metal like Na) ion. it is an SN2 reaction. In the reaction, alkyl halides should be least hindered. Hence, an alkyl halide is obtained from ethanol and alkoxide ion from 3-methylpentan-2-ol. The reactions are shown below:
New answer posted
8 months agoContributor-Level 10
11.50
1. Phenol on mixing with chloroform and NaOH at 340K followed by Acidic hydrolysis, salicyl aldehyde is formed. When carbon tetrachloride (CCl4) is used at the place of chloroform salicylic acid is This type of reaction is known as Reimer - Tiemann reaction.
2. The sodium phenoxide reacts with CO2 under pressure 4-7 atm at a 400K temperature to form sodium salicylate, which on acidification yields salicylic acid. This type of reaction is known as Kolbe's
New answer posted
8 months agoContributor-Level 10
11.49.
Resonating structures of o-nitrophenoxide ions that are formed by the loss of a proton from o-nitrophenol are as follows:
Resonating structures of p-nitrophenoxide ions that are formed by the loss of a proton from p- nitrophenol are as follows:
Resonating structures of phenoxide ions that are formed by the loss of a proton from phenol are as follows:
It is clearly evident from the above structures that due to —R-effect of— NO2NO2 group, o-and p-nitrophenoxide ions are more stable than phenoxide ions. Consequently, o- and p- nitrophenols are more acidic than phenols.
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