Draw and name the product obtained in the following acid-base reactions:
1. 2-isopropylpentanoic acid + NaOH
2. 2-isopropylpentanoic acid + NH3
To determine the products of acid-base reactions, we need to identify the reactants and understand how they interact with each other. In acid-base reactions, acids donate protons (H+) while bases accept protons (H+). Let's analyze each reaction:
1. 2-isopropylpentanoic acid + NaOH:
In this reaction, the acid (2-isopropylpentanoic acid) and the base (NaOH) will interact. The NaOH will donate a hydroxide ion (OH-) which will accept a proton (H+) from the acid. The resulting product will be the salt formed by the cation (Na+) from the base and the anion (the remaining portion of the acid after donating its proton).
To name the product, we need to identify the anion. 2-Isopropylpentanoic acid will lose its hydrogen atom from the carboxylic acid group and form a carboxylate ion. So, the product will be named as Sodium 2-isopropylpentanoate.
2. 2-isopropylpentanoic acid + NH3:
In this case, ammonia (NH3) is a weak base and will also accept a proton (H+) from the acid. Similar to the previous reaction, the product will be the remaining portion of the acid after donating a proton, along with NH4+.
To name the product, we can identify the anion as before, which is the remaining portion of the acid after donating its proton. The product in this case will be named as 2-isopropylpentanoic ammonium ion.
Please note that the products may vary depending on reaction conditions and additional factors. The named products provided are based on typical acid-base reactions under standard conditions.