The question is incomplete; the complete question is;
What alkene would give the products below after reaction with O3, followed by reduction with (CH3)2S? Write the condensed structural formula (CH3)2C=O and CH3-CH2-CH=O
Answer:
2-methylpent-2-ene (C6H12)
Explanation:
Ozonolysis is a wonderful method for determining the location of double bonds in an alkene since the oxygenated carbons in the carbonyl compounds formed after ozonolysis are the ones that were initially joined by the double bonds in the original alkene.
Hence if an alkene yields (CH3)2C=O and CH3-CH2-CH=O, the original alkene must be C6H12, that is, 2-methylpent-2-ene
Answer:
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
Explanation:
The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.
The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.
n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.
Answer:
uranium is classifide as actinide a chemical element atomic number 92 and is a solid at room temperature
An occluded front forms when a warm air mass is caught between two cooler air masses.