Name the following compound: CH3 – CH2 – CH2 – CH2 – CH2 – CH2 – CH2 – CH3
1 answer:
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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:
The empirical formula of a compound is Na2Cr2O7
Explanation:
Empirical formula is defined as the lowest ratio of elements present in any substance. In order to determine the empirical formula for a substance containing these elements we have to follow some steps.
1.In the first step, we have to divide the given mass of an element with its molecular mass to convert the mass to moles.
Similarly, the moles of Cr and O in that compound were calculated and found as 0.396 and 1.3875, respectively.
2. Then the next step is to determine the lowest moles and divide all the moles with that lowest mole. As here the lowest mole is 0.396. So on dividing the moles of Na, Cr and O with 0.396, we will obtain the ratio as 1:1:3.5.
3. As the ratio consists of a fractional number, we have to multiply it with the smallest possible integer to convert it to whole number. In this case, if we multiply it with 2 , we will get the ratio as 2:2:7.
Thus, the empirical formula for the compound can be written as Na2Cr2O7.
Answer:
80 miles/hour
Explanation:
Distance = 400 miles
Time = 5 hours
Speed = ?
Speed= 80mi/hr