
What are the numerical values of N and M?
Propane \[\xrightarrow[{h\upsilon }]{{C{l_2}}}\] N (isomeric products) ${C_3}{H_6}C{l_2}\xrightarrow{{fractional - distillation}}$ M (isomeric products)
$A.$ 6, 6
$B.$ 5, 4
$C.$ 4, 4
$D.$ 3, 3
Answer
219.3k+ views
Hint: We will use the concept of substitution reaction of alkanes with a halogen atom in the presence of sunlight. Alkanes, in organic chemistry, is an acyclic saturated hydrocarbon. In other words, an alkane consists of hydrogen and carbon atoms arranged in a tree structure in which all the carbon-carbon bonds are single.
Complete step-by-step answer:
> Alkanes have the general formula ${C_n}{H_{2n + 2}}$. The first member of the alkane family is methane which has the formula $C{H_4}$. In a substitution reaction, one atom is swapped with another atom. These are very useful reactions in the chemical industry because they allow chemists to change one compound into something more useful, building up designer molecules like drugs. Alkanes undergo a substitution reaction with halogens in the presence of light. For instance, in ultraviolet light, methane reacts with halogen molecules such as chlorine and bromine.
For example:
> Methane + bromine $ \to $ methyl bromide + hydrogen bromide
$C{H_4} + B{r_2} \to C{H_3}Br + HBr$
> This reaction is a substitution reaction because one of the hydrogen atoms from the methane is replaced by a bromine atom.
> Now moving onto the question, the four possibilities for two chlorines in ${C_3}{H_6}C{l_2}$ is for the $Cl's$ to be (1,1), (2,2), (1,2) or (1,3). i.e. 1,1-dichloropropane, 2,2-dichloropropane, 1,2-dichloropropane, and 1,3-dichloropropane.
- Therefore, the correct option is $C.$
Note: Alkanes react with halogens such as Cl, I, Br differently. With Cl it reacts vigorously and forms alkyl chloride even reaction can be so vigorous that all hydrogen of alkane can be replaced by chlorine and formation of carbon tetrachloride takes place in methane. In halogens Cl is more reactive than Br and I.
Complete step-by-step answer:
> Alkanes have the general formula ${C_n}{H_{2n + 2}}$. The first member of the alkane family is methane which has the formula $C{H_4}$. In a substitution reaction, one atom is swapped with another atom. These are very useful reactions in the chemical industry because they allow chemists to change one compound into something more useful, building up designer molecules like drugs. Alkanes undergo a substitution reaction with halogens in the presence of light. For instance, in ultraviolet light, methane reacts with halogen molecules such as chlorine and bromine.
For example:
> Methane + bromine $ \to $ methyl bromide + hydrogen bromide
$C{H_4} + B{r_2} \to C{H_3}Br + HBr$
> This reaction is a substitution reaction because one of the hydrogen atoms from the methane is replaced by a bromine atom.
> Now moving onto the question, the four possibilities for two chlorines in ${C_3}{H_6}C{l_2}$ is for the $Cl's$ to be (1,1), (2,2), (1,2) or (1,3). i.e. 1,1-dichloropropane, 2,2-dichloropropane, 1,2-dichloropropane, and 1,3-dichloropropane.
- Therefore, the correct option is $C.$
Note: Alkanes react with halogens such as Cl, I, Br differently. With Cl it reacts vigorously and forms alkyl chloride even reaction can be so vigorous that all hydrogen of alkane can be replaced by chlorine and formation of carbon tetrachloride takes place in methane. In halogens Cl is more reactive than Br and I.
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