Answer:D
Explanation:
The high boiling point of HF is not attributable to the dispersion forces mentioned in the question. In HF, a stronger attraction is in operation, that is hydrogen bonding. This ultimately accounts for the high boiling point and not solely the dispersion model as in F2.
In the modern periodic table the elements are arranged in order of increasing atomic number.
Elements are separate particles that contain the properties of only one type of element (pure substance) and an atom represents that element as the smallest non divisible particle that retains the properties of that element. Compounds can be formed by conjoining different atoms together in different ratios and shapes, so a combination of elements.
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
