GC question and OC question

[Troyvdg]

An aqueous sulfuric acid solution is 39.2% H2SO4 by mass and has a specific gravity of 1.25. how many milliliters of this solution are required to make 100mL of a .2M sulfuric acid solution?

Answer: 4 mL

could someone walk me through this? Thanks in advance!

For Ochem.

Q1. which of the following is most easily oxidized?

a. R-COOH
b. R2C=O
c. R-CHO *answer
d. R2CH-OH
e. R3C-OH

the explanation says C because, aldehydes are oxidized to carboxylic acids under mild conditions, but how does that make sense? they require strong oxidizing reagents like Chromic acid, KMnO4, or nitric acid. Shouldn't the answer be D, since secondary alcohols can be oxidized to ketones using weaker oxidizing reagents such as, PCC or swern?

Q2. What is the product of the following reaction?

n-propyl chloride + KOH -->

a. isopropyl chloride
b. isopropyl alcohol
c. n-propyl alcohol
d. propylene *Answer
e. 2-chloro-1-propanol

I also cannot make sense of the explanation for this question. They say the answer is D because a strong base is used. But n-propyl chloride is a primary halide, shouldn't this be SN2 making C the answer instead of B?

---

Members don't see this ad.

 

[aqz]

GChem:

Every time you're given % mass, you want to somehow get to units in grams.

They give you specific gravity, which is the density ratio to water's density, which is 1 g/ml. In other words, H2SO4's density is going to be 1.25 g/ml.

Now you want to get to grams so you can use that % mass. So what you do is assume the solution has 100 ml and multiply 100ml by 1.25 g/ml = 125 grams

Then you use that % mass = 40% or 0.4, multiply it by 125....so 125x0.4 = 50 grams

You also know that H2SO4's molar mass is about 100 grams/mol so then 50 grams * 1mol/100 grams = 0.5 mol

Since you assumed 100 ml at the beginning, you would have 0.5 mol/100 ml or 0.5 mol/0.1 L = 5M concentration.

Then you just use M1V1 = M2V2 to find the answer. So...5M * X = 0.2M * 100 ml ..... X = 4 ml


OChem Q1

This question is poorly worded because they don't give you a standard for what "easy" means. The reason aldehyde is the answer could be one of two reasons:

1) Aldehyde is the only substance out of all of the listed ones to be able to get fully oxidized into a carboxylic acid. There's a ranking of oxidation states from lowest to highest starting from alcohol to aldehyde/ketone to acids and then finally to CO2. Although it might be "easier" to use PCC, it's also "easier" to get to carboxylic acid.

2) Steric hinderance of the alcohols makes it harder to oxidize. Again, no standard = ambiguous answer, so just throw this question out.

OChem Q2

EDIT: Actually Kaplan just does a poor job of being clear once again.

According to Destroyer roadmap #3, KOH acts as a strong base in presence of Alc (Alcohol), but I think it's safe to assume KOH will act as a nucleophile in aqueous solutions. The difference between the two is in alcoholic presence, KOH becomes KO-, a strong base but weak nuc, whereas in aqueous solution KOH becomes K+ OH-, strong base and strong nuc.

 

[412smb412]

2. i have never come across such a question...
3. strong base = E2

 

[ctraNA1]

2. According to DAT Destroyer, aldehydes are easily oxidized to carboxylic acids in air without any reagent at all.