The complete question is
If the compound below is oxidized, the resulting product is ___.
the compound given is Butanal
Select one:
a. methane and propane
b. butanal
c. butanoic acid
d. butane
Answer:
C. Butanoic Acid
Explanation:
When Butanal is oxidized using reagents like KMnO4, Tollens reagent etc.
When Aldehyde is oxidized corresponding Carboxylic Acid is produced
Butanal → Butanoic Acid
Answer:
Multiply 3 by Avogadro's number.
Explanation:
The mole can be defined as the amount of a substance that contains Avogadro’s number of particles, 6.02 x 10²³.
For elementary particles:
Number of particles=
number of moles x 6.02 x10²³
From the question,
Number of moles = 3moles
Number of particles = 3 x 6.02 x10²³
Since the compound has 1.38 time that of oxygen gas at the same conditions of temperature and pressure, we have the relationship:
MW/MWoxygen = 1.38
MW = 44.16
Since there is water formed during the reaction, the formula of the compound must be:
XaHb
where a and b are the coefficients of each element.
If the compound reactions with oxygen forming water and an oxide of the element X, the combustion reaction must be:
XaHb + ((2a + (b/2))/2) O2 = a (XO2) + (b/2)(H2O)
Using dimensional analysis:
10 (1/44.16) (b/2 / 1) (18) = 16.3
Solving for b:
b = 8
The compound now is XaH8. Most probably, the compound is C3H8 since it has a molecular formula of 44 and it reacts with O2 to form water and CO2.
Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called <u>halohydrin formation</u>. This is a <u>markovnikov reaction</u> with <u>anti configuration</u>. Therefore the halogen in this case "Br" and the "OH" must have <u>different configurations</u>. Additionally, in this molecule both carbons have the <u>same substitution</u>, so the "OH" can go in any carbon.
Finally, in the product we will have <u>chiral carbons</u>, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
I hope it helps!
