The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is ______.
Use: Atomic mass of N (in amu) = 14
Explanation:
Positive carbylamine test is given by primary amine compounds. Hexamethylene diamine is a primary amine compound and can give positive carbylamine test.
So, the monomer $X$ is hoxamethylene diamine.
$
\begin{aligned}
&\begin{aligned}
x \rightarrow \mathrm{H_2N}-\left(\mathrm{CH}_2\right)_6-\mathrm{NH}_2 \quad & \text { Molarmars } & \\
= & 6 \times 12 \mathrm{~g} / \mathrm{mol}+2 \times 14 \mathrm{~g} / \mathrm{mol} +16 \times 1 \mathrm{~g} / \mathrm{mol} & \\
= & 116 \mathrm{~g} / \mathrm{mol} &
\end{aligned}\\
&\text { Moles of } X \text { analyzed using Dumas method }=10 \mathrm{~mol}
\end{aligned}
$According to Dumas method, the nitrogen present in the compound is released as nitrogen gas $\left(\mathrm{N}_2\right)$,
The general chemical equation fol Dumas method can be written as$
\mathrm{C}_x \mathrm{H}_y \mathrm{~N}_z+\left(2 x+\frac{y}{2}\right) \mathrm{CuO}(\mathrm{~s}) \xrightarrow{}
$(organic
Compound)$
x \mathrm{CO}_2(g)+\frac{y}{2} H_2 \mathrm{O}(gold )+\frac{z}{2} \mathrm{~N}_2(g)+\left(2 x+\frac{y}{2}\right) \mathrm{Cu}(s)
$The compound used in Dumas method is hexamethylene diamine. Its formula is $\mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2$$
\text { So, the equation can be written as, }
$$
\mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2+20 \mathrm{CuO} \rightarrow
$$
6 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}+1 \mathrm{~N}_2+20 \mathrm{Cu}
$$
\text { So, } 1 \text { mole of } \mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2 \text { gives. } 1 \text { mole of } \mathrm{N}_2 \text {. }
$Given 10, moles of $\mathrm{C}_{6}\mathrm{H}_{16} \mathrm{~N}_2$ (hexamethylene diamine) gives 10 moles of $\mathrm{N}_2$ gas.Moles of $\mathrm{N}_2$ gas evolved $=10$ mol Mass can be calculated as, \begin{aligned}
&\text { Moles }=\frac{\text { mass }}{\text { molarmass }}\\
&\text { Moloumass of } \mathrm{N}_2=28 \mathrm{~g} / \mathrm{mol}
\end{aligned}mass $=$ moles $\times$ molar mass$
\begin{aligned}
&\begin{aligned}
& =10 \text { mol } x 28 \mathrm{~g} / \mathrm{mol} \\
& =280 \mathrm{~g}
\end{aligned}\\
&\text { Answer: } 280 \%
\end{aligned}
$
Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are
The correct functional group X and the reagent/reaction condition Y in the following scheme are
On complete hydrogenation, natural rubber produces
Among cellulose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest is
Match the chemical substances in Column I with type of polymers/type of bonds in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 $\times$ 4 matrix given in the ORS.
| Column I | Column II | ||
|---|---|---|---|
| (A) | cellulose | (P) | natural polymer |
| (B) | nylon-6, 6 | (Q) | synthetic polymer |
| (C) | protein | (R) | amide linkage |
| (D) | sucrose | (S) | glycoside linkage |
Monomer A of a polymer on ozonolysis yields two moles of HCHO and one mole of CH$_3$COCHO.
(A) Deduce the structure of A.
(B) Write the structure of "all $cis$" forms of polymer of compound A.
Explanation:
(A) The monomer A on ozonolysis yields two molecules of HCHO and a molecule of CH$_3$COCHO. During ozonolysis, the double or triple bonds in the molecule are replaced by carbonyl groups.
Therefore, if we remove the oxygen atoms from 2 molecules of HCHO and a molecule of CH$_3$COCHO and replaced them with double bond, we can deduce the structure of monomer A.
The structure of monomer A we got is the structure of isoprene. Therefore, monomer A is isoprene.
(B) The isoprene is a monomer for natural rubber.
To write the “all $cis$” form of the polymer of monomer A that is isoprene, we need to write the structure of polymer such that the identical groups present across double bonds should lie in the same plane.
The “all $cis$” for of the polymer is shown below.
Final Answer
Hints:
The ozonolysis eliminates the double or triple bond from the molecule and replace it with carbonyl groups. The removal of oxygen atoms and replacing it with double bonds will give the structure of monomer A.