I would call it something like 2-bromo propanoic acid
Br Br | |H - C - C - H| |H H
To convert n-propylbromide into iso-propylbromide, you can use a dehydrobromination reaction. By treating n-propylbromide with a strong base such as sodium hydroxide or potassium tert-butoxide, the hydrogen atom at the center carbon atom of the n-propyl chain can be abstracted, leading to the formation of iso-propylbromide.
onvert the ethyl bromide into Grignard's reagent, ethyl magnesium bromide then allow to react with dry ice (Solid carbon dioxide) then acidic hydrolysis produces the propionic acid. CH3-Br + Mg --- anhydrous ether---> CH3-CH2-Mg-Br CH3-CH2-Mg-Br + CO2 ----H+/H2O---> CH3-CH2-COOH
The chemical formula (CH3)3CBr represents a compound with 4 carbon (C) atoms, 9 hydrogen (H) atoms, and 1 bromine (Br) atom. To find the total number of atoms, you simply add them together: 4 (C) + 9 (H) + 1 (Br) = 14 atoms. Thus, there are 14 atoms in (CH3)3CBr.
CH2=CH2-CH-CH2-Br 1-butene, 3- OH, 4-Br | OH Not sure but would be my best bet
1 - bromopropane is the IUPAC name for CH3-CH2-CH2-CH2-Br.
Ethylbromide is the name of the compound CH3-CH2-Br.
The compound Cl-CH2-CH2-CH2-CH=CH-CH2-Br is 1-bromo-6-chloro-2-hexene.
Br Br | |H - C - C - H| |H H
To convert n-propylbromide into iso-propylbromide, you can use a dehydrobromination reaction. By treating n-propylbromide with a strong base such as sodium hydroxide or potassium tert-butoxide, the hydrogen atom at the center carbon atom of the n-propyl chain can be abstracted, leading to the formation of iso-propylbromide.
The reaction between 2-pentene and bromine is an addition reaction. The double bond in 2-pentene reacts with bromine to form a dibromoalkane. This reaction proceeds via an electrophilic addition mechanism.
1-dehydrohalogenation of n-propylbromide which gives the propene. CH3-CH2-CH2-Br + KOH(Alcoholic) -----> CH3-CH=CH2 2-again hydrohalogenation with HBr gives mostly iso-propylbromide,(Markonikov's rule). CH3-CH=CH2 + HBr ------> CH3-CHBr-CH3 3-The reaction of iso-propylbromide with Sodium metal in presence of anhydrous ether (Wurtz reaction) gives the 2,3-dimethylbutane. 2(CH3-CHBr-CH3) + 2Na ---anhydrous ether--->CH3-CHCH3-CHCH3-CH3
onvert the ethyl bromide into Grignard's reagent, ethyl magnesium bromide then allow to react with dry ice (Solid carbon dioxide) then acidic hydrolysis produces the propionic acid. CH3-Br + Mg --- anhydrous ether---> CH3-CH2-Mg-Br CH3-CH2-Mg-Br + CO2 ----H+/H2O---> CH3-CH2-COOH
First, ethyl bromide can be converted to ethyl magnesium bromide. Then, ethyl magnesium bromide can react with carbon dioxide to form propionic acid. Hydrolysis of the resulting compound will yield propionic acid.
1-bromobutane will undergo an SN2 reaction with sodium hydroxide to form 1-butanol and sodium bromide. In this reaction, the hydroxide ion replaces the bromine atom on the carbon chain, resulting in the formation of the alcohol product.
Br-CH2-CH2-C=-C-CH2-CH3
CH3-C(Br)(CH3)-CH3 + H2O = CH3-C(OH)(CH3)-CH3 + HBr