C5h10o2 (Wiki.answers forced me to make capitalization changes)
methyl butanoate
To make methyl butanoate, you would need methanol (CH3OH) and butanoic acid (C4H8O2) as the reactants. The reaction would be a condensation reaction between an alcohol (methanol) and a carboxylic acid (butanoic acid) catalyzed by an acid catalyst, such as sulfuric acid.
Methyl orange turns red in the presence of calcium hydroxide.
The structural isomers of C5H10O2 ester are pentyl formate, pentyl acetate, and methyl butanoate. These molecules have the same molecular formula but different arrangements of atoms.
The molecular formula for methyl butyrate, also known as methyl butanoate, is C5H10O2.
C5h10o2 (Wiki.answers forced me to make capitalization changes)
Methyl butanoate is not very soluble in water. Due to its nonpolar nature, it tends to dissolve more readily in organic solvents such as ethanol or diethyl ether.
methyl butanoate
To make methyl butanoate, you would need methanol (CH3OH) and butanoic acid (C4H8O2) as the reactants. The reaction would be a condensation reaction between an alcohol (methanol) and a carboxylic acid (butanoic acid) catalyzed by an acid catalyst, such as sulfuric acid.
Methyl orange turns red in the presence of calcium hydroxide.
The structural isomers of C5H10O2 ester are pentyl formate, pentyl acetate, and methyl butanoate. These molecules have the same molecular formula but different arrangements of atoms.
2-methyl-2-butene is the major product (85%), while 2-methyl-1-butene the minor (15%) in an E1 reaction.
Methyl 2-methylbutyrate can be prepared by the esterification of 2-methylbutyric acid with methanol in the presence of a strong acid catalyst, such as sulfuric acid. The reaction is usually carried out under reflux conditions and the ester is then isolated by extraction and distillation.
3-methyl butanoic acid is more acidic than butanoic acid because the presence of the methyl group in 3-methyl butanoic acid increases the electron-withdrawing effect, making the molecule more acidic by stabilizing the conjugate base.
A methyl group can be added to an alkane through a process called alkylation, where a methyl halide, such as methyl iodide, reacts with the alkane in the presence of a strong base, such as sodium hydroxide. This reaction results in the substitution of a hydrogen atom in the alkane with a methyl group, forming a new alkylated compound.
Iodoform is used in the iodoform test to detect the presence of a methyl ketone functional group. When Iodoform reacts with a methyl ketone in the presence of a base and acidic conditions, it forms a yellow precipitate of iodoform, which confirms the presence of the CH3CO- group.