Write the chemical equation when 1-propanol react with
(i) excess of HBr
(ii) H2SO4 at 410 K
(iii ) H2SO4 at 443K
(iv) acidified KMnO4
(i) Well, when 1-propanol meets an excess of HBr, it's like a party that never ends. So here's the equation: 1-propanol + HBr (excess) = an awesome bromination product + lots of laughter + some H2O.
(ii) When you mix 1-propanol with H2SO4 at 410 K, you get a hot and steamy reaction: 1-propanol + H2SO4 (410 K) = funky product + some giggles + H2O.
(iii) Now, if you heat up 1-propanol with H2SO4 at 443 K, things get even more heated: 1-propanol + H2SO4 (443 K) = an explosive reaction + a few fireworks + H2O.
(iv) If you add acidified KMnO4 to 1-propanol, it's like mixing a clown with a magician – you get an amazing transformation: 1-propanol + acidified KMnO4 = a mind-blowing oxidation product + some magical laughter + H2O. Voila!
(i) When 1-propanol reacts with excess HBr, the following chemical reaction occurs:
CH3CH2CH2OH + HBr -> CH3CH2CH2Br + H2O
(ii) When 1-propanol reacts with H2SO4 at 410 K, dehydration occurs, resulting in the formation of propene:
CH3CH2CH2OH -> CH2=CHCH3 + H2O
(iii) When 1-propanol reacts with H2SO4 at 443 K, dehydration occurs, resulting in the formation of a different compound, propionaldehyde:
CH3CH2CH2OH -> CH3CH2CHO + H2O
(iv) When 1-propanol reacts with acidified KMnO4, oxidation occurs, leading to the formation of carboxylic acid, propanoic acid:
CH3CH2CH2OH + acidified KMnO4 -> CH3CH2COOH + MnO2 + H2O
To determine the chemical equations when 1-propanol reacts with different substances, we first need to understand the chemical properties and reactions of 1-propanol.
1-propanol, or C3H7OH, is an alcohol with three carbon atoms in its carbon chain. It can undergo several types of reactions depending on the reagents present. Let's go through each reaction step by step:
(i) When 1-propanol reacts with excess HBr (hydrobromic acid), it undergoes an example of nucleophilic substitution reaction called the Williamson ether synthesis. The reaction proceeds as follows:
C3H7OH + HBr → C3H7Br + H2O
In this reaction, the hydroxyl group (OH) of 1-propanol is replaced by the bromine atom (Br) from HBr, forming 1-bromopropane. Excess HBr ensures the reaction goes to completion.
(ii) When 1-propanol reacts with H2SO4 (sulfuric acid) at 410 K, it undergoes dehydration. This reaction involves the removal of a water molecule (H2O) from 1-propanol, resulting in the formation of propene (C3H6). The chemical equation can be represented as follows:
C3H7OH → C3H6 + H2O
(iii) When 1-propanol reacts with H2SO4 at 443 K, the reaction proceeds similarly to the previous case but at a higher temperature. The increased temperature favors elimination reactions, resulting in the production of propene. Therefore, the reaction equation remains the same:
C3H7OH → C3H6 + H2O
(iv) When 1-propanol reacts with acidified KMnO4 (potassium permanganate), it undergoes oxidation. 1-propanol is oxidized to form a carboxylic acid. The detailed balanced chemical equation is more complex, but it can be summarized as follows:
C3H7OH + 2[KMnO4] + 3[H2SO4] → 3[H2O] + 3[MnSO4] + 3[K2SO4] + 3[H2O] + C3H5(COOH)3
In this reaction, 1-propanol is oxidized to form a carboxylic acid represented by C3H5(COOH)3, which contains three carboxyl groups.
Remember to conduct these reactions under appropriate laboratory conditions and safety precautions.