C6H5-NH2-CH3-Cl is a chemical compound with a benzene ring (C6H5), an amino group (NH2), a methyl group (CH3), and a chlorine atom (Cl) attached. It is a specific arrangement of atoms that can have various chemical and physical properties depending on the arrangement and bond types.
Ethanoic acid can be converted into methenamine by reacting it with formaldehyde in the presence of aqueous ammonia. This reaction undergoes a condensation reaction to form methenamine as the final product.
The structure formulas of the four isomers of C3H9N are: Propamine : CH3-CH2-CH2-NH2 : or : CH3-[CH2]2-NH2 Prop-2-amine : CH3-CH-NH2-CH3 EthyMethylamine : CH3-NH-CH2-CH3 TriMethylamine : N(CH3)3
No, substances containing the -NH2 group are basic. The NH2- ion is extremely basic.
I would suggest it being something like (CH3)-NH2 , more or less like NH3 (ammonia) or (CH3)-NH3OH , more or less like NH4OH (ammonium hydroxide) which I do NOT prefer
Valine: HO2CCH(NH2)CH(CH3)2 Tryptophan: C11H12N2O2 Threonine: HO2CCH(NH2)CH(OH)CH3 Phenylalanine: HO2CCH(NH2)CH2C6H5 Methionine: HO2CCH(NH2)CH2CH2SCH3 Lysine: HO2CCH(NH2)(CH2)4NH2 Leucine: HO2CCH(NH2)CH2CH(CH3)2 Isoleucine: HO2CCH(NH2)CH(CH3)CH2CH3
C6H5-NH2-CH3-Cl is a chemical compound with a benzene ring (C6H5), an amino group (NH2), a methyl group (CH3), and a chlorine atom (Cl) attached. It is a specific arrangement of atoms that can have various chemical and physical properties depending on the arrangement and bond types.
It belongs to the amine group or family.
Ethanoic acid can be converted into methenamine by reacting it with formaldehyde in the presence of aqueous ammonia. This reaction undergoes a condensation reaction to form methenamine as the final product.
The structure formulas of the four isomers of C3H9N are: Propamine : CH3-CH2-CH2-NH2 : or : CH3-[CH2]2-NH2 Prop-2-amine : CH3-CH-NH2-CH3 EthyMethylamine : CH3-NH-CH2-CH3 TriMethylamine : N(CH3)3
Reaction: CH3C(=O)CH3 + NH2NHC(=O)NH2 --> (CH3)2C=NNHC(=O)NH2 1) N from NH2 acts as the nucleophile, attacks the carbonyl group and changes into a tetrahedral intermediate. Positive charge on NH2, negative charge on oxygen. (CH3)2(O-C)--(NH2)+NHC(=O)NH2 2) oxygen takes the hydrogen from nitrogen through rearrangement because it is more electronegative. (CH3)2(HOC)--(NH)NHC(=O)NH2 3) Acid protonates the alcohol once more to make it a better leaving group. (CH3)2(H2O+C)--(NH)NHC(=O)NH2 4) Nitrogen from NH2 donates its electrons to generate a double bond, bumping off the water molecule and leaving a positive charge on nitrogen. (CH3)2C=NH+NHC(=O)NH2 5) Water pulls off the hydrogen that was still attached to the nitrogen, thus generating the condensation product between acetone and semicarbazide. (CH3)2C=NNHC(=O)NH2
CH3-C=O-NH2 +HCl----> CH3-C=O-OH+ NH4Cl
Ch3-c(=o)-nh2
The compound CH3-CH2-NH2 is classified as an amine. Specifically, it is a primary amine because the nitrogen atom is bonded to one alkyl group (ethyl group, in this case) and two hydrogen atoms. Amines are characterized by the presence of the amino group (-NH2) and can act as bases and nucleophiles in chemical reactions.
No, substances containing the -NH2 group are basic. The NH2- ion is extremely basic.
I would suggest it being something like (CH3)-NH2 , more or less like NH3 (ammonia) or (CH3)-NH3OH , more or less like NH4OH (ammonium hydroxide) which I do NOT prefer
There are two possibillities:1-aminopropaan CH2(NH2 )CH2CH32-aminopropaan CH3CH(NH2)CH3