Advertisements
Advertisements
प्रश्न
Why dextro and laevorotatory isomers of Butan-2-ol are difficult to separate by fractional distillation?
उत्तर
Dextro and laevorotatory isomers of Butan-2-ol are enantiomers of each other and both have the same boiling point and hence they cannot be separated by fractional distillation.
APPEARS IN
संबंधित प्रश्न
Give reasons : n-Butyl bromide has higher boiling point than t-butyl bromide.
Arrange the set of compounds in order of increasing boiling points.
Bromomethane, Bromoform, Chloromethane, Dibromomethane.
Arrange the set of compounds in order of increasing boiling points.
1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
For the same alkyl group, an alkyl bromide has a higher boiling point than alkyl fluoride because:
Arrange the following compounds in the increasing order of their densities.
(a)
(b)
(c)
(d)
Arrange the following compounds in increasing order of their boiling points.
(a) \[\begin{array}{cc}
\ce{CH3}\phantom{.................}\\
\backslash\phantom{.............}\\
\ce{CH - CH2Br}\\
/\phantom{.............}\\
\ce{CH3}\phantom{.................}
\end{array}\]
(b) \[\ce{CH3CH2CH2CH2Br}\]
(c) \[\begin{array}{cc}
\phantom{...}\ce{CH3}\\
\phantom{}|\\
\ce{H3C - C - CH3}\\
\phantom{}|\\
\phantom{..}\ce{Br}
\end{array}\]
Reaction of \[\ce{C6H5CH2Br}\] with aqueous sodium hydroxide follows ______.
Assertion: The boiling points of alkyl halides decrease in the order:
\[\ce{RI > RBr > RCl > RF}\]
Reason: The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.
Which out of the following is an intensive property?
Write the structure of the following organic halogen compound.
4-tert-Butyl-3-iodoheptane
