Answer:
Option C. 2,2-dimethyl-1-propanol
Explanation:
The first attached picture, is the initial reactant.
Now, what happens in each step?, at first we have an alkyl bromide reacting with Mg/ether. This converts the reactant in a grignard reagent, which is often used in several ketones/ladehydes reactions, and even some acid and esthers.
Now, in the second step, we see this grignard reagent, reacting with CO₂ and H₃O⁺. At first, the grignard reagent will attach to the oxygen of the CO₂ by the MgBr, and the carbonated chain will attach to the carbonile. Then in acid medium, the OMGBr leaves the molecule, and form a carboxylic acid.
Finally, in the last step with LiAlH₄ and water, this compound just reduces the carboxylic acid to a primary alcohol, therefore, the only choice available is option c, 2,2-dimethyl-1-propanol.
See picture 2 below for the drawing.
Hope this helps
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 formula of the missing product is, 
Explanation :
The given incomplete equation representing a reaction,
When the sodium metal react with the water then it gives a colorless solution of sodium hydroxide and hydrogen gas. In the solution, sodium hydroxide is present in the form of ions i.e, 
and 
ions.
The balanced chemical reaction will be,
Therefore, the formula of the missing product is,
Answer: The molar concentration of acetic acid in the vinegar is 0.539 M.
Explanation:
The formula used is:
where,
and 
are the concentration and volume of base.
and 
are the concentration and volume of an acid.
Given:
Molar concentration of NaOH = 0.1798 M
Volume of NaOH = 30.01 mL
Volume of acetic acid = 10.0 mL
Now putting all the given values in the above formula, we get:
Thu, the molar concentration of acetic acid in the vinegar is 0.539 M.
