A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,
2 answers:
You might be interested in
Answer:
Explanation:
In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).
A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.
Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.
Answer:
<span>Molecular geometry around each carbon atom in a saturated hydrocarbon is Tetrahedral.
Explanation:
</span> In saturated hydrocarbons (-CH₂-) the central atom (carbon) is bonded to either three or two hydrogen atoms and one or two carbon atoms. So, the central atom is having four electron pairs and all pairs are bonding pairs and lacks any lone pair of electron. According to Valence Shell Electron Pair Repulsion (VSEPR) Theory the central atom with four bonding pair electrons and zero lone pair electrons will attain a tetrahedral geometry with bond angles of 109°.
<span>Molecular geometry around each carbon atom in a saturated hydrocarbon is Tetrahedral.
Explanation:
</span> In saturated hydrocarbons (-CH₂-) the central atom (carbon) is bonded to either three or two hydrogen atoms and one or two carbon atoms. So, the central atom is having four electron pairs and all pairs are bonding pairs and lacks any lone pair of electron. According to Valence Shell Electron Pair Repulsion (VSEPR) Theory the central atom with four bonding pair electrons and zero lone pair electrons will attain a tetrahedral geometry with bond angles of 109°.