One of the only ways is to make use of nucleophilic aromatic substitution (SN_Ar). Add a nitro group para to the OH, attack the OH carbon with 1 eq. ammonia, work it up with another eq., convert NH2 to a diazonium salt with NaNO2 + HCl @ 0~4 degrees C, then remove it with H3PO2.
mecanism of sulfanilic asid
It s the structure and bonding of Aniline, Phenol, Benzene and Toluene that causes the differences in the boiling points.
The formula is for benzyl acetate (as an example).
No they don't react together. but phenol is soluble in CHCl3.
Positively charged carbon atom of the aldehyde's carbonyl group attacks p-position of aniline, having a negative charge. Compound 1 forms, with a positive carbon attacking a p-position of another aniline molecule.NH2-C6H5 + RC(O)H -> NH2-C6H4-CHR(OH) (compound 1)NH2-C6H4-CHR(OH) + NH2-C6H5 -> NH2-C6H4-CH(R)-C6H4-NH2 + H2OIn case of formaldehyde, the product is 4,4'-methylenedianiline (MDA).
yeah
React aniline with HCl/NaNO2 (diazotisation) followed by reaction with KOH to give phenol. Nitration of phenol with fuming nitric acid gives picric acid (or trinitrophenol).
In o-aniline phenol Intramolecular hydrogen bonding occurs which is not possible in meta and is responsible for less solubility
In Reimer- Tieman reaction the electrophile is dichlorocarbene
It s the structure and bonding of Aniline, Phenol, Benzene and Toluene that causes the differences in the boiling points.
The formula is for benzyl acetate (as an example).
There is no reaction between phenol and sodium carbonate
This reaction is also called p-Hydroxy methylation of Phenol and phenol is converted into p-hydroxy benzyl alc.
reaction b/w phenol and zinc yield benzene by reduction.
benzanilide
There is no reaction when their molar ratio is around 1...but when phenol is present in excess it gives triphenylphosphate....
No they don't react together. but phenol is soluble in CHCl3.
Positively charged carbon atom of the aldehyde's carbonyl group attacks p-position of aniline, having a negative charge. Compound 1 forms, with a positive carbon attacking a p-position of another aniline molecule.NH2-C6H5 + RC(O)H -> NH2-C6H4-CHR(OH) (compound 1)NH2-C6H4-CHR(OH) + NH2-C6H5 -> NH2-C6H4-CH(R)-C6H4-NH2 + H2OIn case of formaldehyde, the product is 4,4'-methylenedianiline (MDA).