5 Marks Questions

Question 15 Marks

1. What are electrophiles and nucleophiles? Explain with examples.
2. Derive the structure of 3 - Nitrocyclohexene.

Answer

1. Electrophiles: The name electrophiles means electron loving. Electrophiles are electron defficient. They may be positive ions or neutral molecules.
$
Ex: H^{+}, Cl^{+}, Br^{+}, NO_2^{+}, R_3 C^{+}, RN_2^{+}, AlCl_3, BF_3
$
Nucleophiles: The name nucleophiles means 'nucleus loving' and indicates that it attacks the region of low electron density (positive centres) in a subtracts molecule. They are electron rich they may be negative ions or neutral molecules.
$
\text { Ex: } Cl^{-}, Br^{-}, CN^{-}, OH^{-}, RCH_2^{-}, NH_3, RNH_2, H_2 O, ROH \text { etc. }
$
2. A six-membered ring containing a carbon - carbon double bond is called as cyclohexene. Now giving numbers to the carbons:

Attach nitro group to 3rd carbon.
Hence the structure of 3 - Nitrocyclohexene is:

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Question 25 Marks

1. What is the general molecular formula of saturated monohydric alcohols?
2. Write structural formulae for compounds named as-
a. 1-Bromoheptane
b. 5-Bromoheptanoic acid

Answer

1. Monohydric alcohols are the compounds derived from an alkane by replacing one H by - OH group.
Example:
$
\underset{\text { Methane }}{CH_4} \xrightarrow{\text { replacing } H \text { with } OH} \underset{\text { Methanol }}{CH_3-OH}
$
Therefore, the general molecular formula of saturated monohydric alcohols is $C _{ n } H _{2 n +1} OH$.
2. i. Structural formula of 1-Bromoheptane:
$
CH_3-CH_2-CH_2-CH_2-CH_2-CH_2-CH_2 Br
$
ii. Structural formula of 5-Bromoheptanoic acid:
$
CH_3-CH_2-CH(Br)-CH_2-CH_2-CH_2-COOH
$

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Question 35 Marks

At 1127 K and 1 atmosphere pressure, a gaseous mixture of CO and $CO _2$ in equilibrium with solid carbon has 90.55\% CO by mass.
$
C(s)+CO_2(g) \rightleftharpoons 2 CO(g)
$
Calculate $K _{ C }$ for the reaction at the above temperature.

Answer

Let the total mass of the gaseous mixture be 100 g .
Mass of $CO =90.55 g$
And, mass of CO2 $=(100-90.55)=9.45 g$
Now, number of moles of CO,
$
n_{CO}=\frac{90.55}{28}=3.234 mol
$
Number of moles of CO2,
$
n_{CO_2}=\frac{9.45}{44}=0.215 mol
$
Partial pressure of CO ,
$
\begin{aligned}
& P_{CO}=\frac{n_{CO}}{n_{CO}+n_{CO_2}} \times P_{\text {total }} \\
& =\frac{3.234}{3.234+0.215} \times 1 \\
& =0.938 atm
\end{aligned}
$
Partial pressure of CO2,
$
\begin{aligned}
& p_{CO_2}=\frac{n_{CO_2}}{n_{co}+n_{CO_2}} \times p_{\text {total }} \\
& =\frac{0.215}{3.234+0.215} \times 1 \\
& =0.062 atm
\end{aligned}
$
Therefore, $Kp =\frac{\left[ CO ^2\right.}{\left[ CO _2\right]}$
$
\begin{aligned}
& =\frac{(0.938)^2}{0.062} \\
& =14.19
\end{aligned}
$
For the given reaction, $\Delta n =2-1=1$
We know that
$
\begin{aligned}
& K_{p}=K_{C}(RT)^{\Delta n} \\
& \Rightarrow 14.19=K_{c}(0.082 \times 1127) \\
& \Rightarrow K_{C}=\frac{14.19}{0.082 \times 1127} \\
& =0.154 \text { (approximately) }
\end{aligned}
$

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Question 45 Marks

Reaction between $N _2$ and $O _2$ - takes place as follows:
$
2 N_2(g)+O_3(g) \rightleftharpoons 2 N_2 O(g)
$
If a mixture of 0.482 mol of $N _2$ and 0.933 mol of $O _2$ is placed in a 10 L reaction vessel and allowed to form $N _2 O$ at a temperature for which $K_c=2.0 \times 10^{-37}$ determines the composition of the equilibrium mixture.

Answer

Let $x$ moles of $N _2(g)$ take part in the reaction. According to the equation, $x / 2$ moles of $O _2-( g )$ will react to form r moles of $N _2 O ( g )$.
$
2 N_2(g)+O_2(g) \rightleftharpoons 2 N_2 O(g)
$
Initial conc. $( Mol / L )\left[ N _2\right]=\frac{0.482}{10}\left[ O _2\right]=\frac{0.933}{10}$
At equilibrium point: $\frac{0.482-x}{10} \frac{0.933-\frac{x}{2}}{10} \frac{x}{10}$
The value of the equilibrium constant $\left(2.0 \times 10^{-37}\right)$ is extremely small. This means that only small amounts of reactants have reacted. Therefore, x is extremely small and can be omitted as far as the reactants are concerned.
Applying Law of Chemical Equilibrium $K_c=\frac{\left[N_2 O(g)\right]^2}{\left[N_2(g)\right]^2\left[O_2(g)\right]^2}$
$
\begin{aligned}
& 2.0 \times 10^{-37}=\frac{\left(\frac{x}{10}\right)^2}{\left(\frac{0.482}{10}\right)^2 \times\left(\frac{0.933}{10}\right)}=\frac{0.01 x^2}{2.1676 \times 10^{-4}} \\
& x^2=43.352 \times 10^{-40} \text { or } x=6.6 \times 10^{-20}
\end{aligned}
$
As $x$ is extremely small, it can be neglected.
Thus, in the equilibrium mixture
Molar conc. of $N _2=0.0482 mol L ^{-1}$
Molar conc. of $O _2=0.0933 mol L ^{-1}$
Molar conc. of $N _2 O =0.1 \times x=0.1 \times 6.6 \times 10^{-20} mol L^{-1}$ $=6.6 \times 10^{-21} mol L^{-2}$

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Question 55 Marks

Attempt any five of the following:
1. Why does the iodination of benzene is carried out in the presence of nitric acid or iodic acid?
2. Can a catalyst change the position of equilibrium in a reaction?
3. Why is benzene extraordinarily stable though it contains three double bonds?
4. Bring out the following conversion ethane to ethene.
5. Why are alkanes called paraffins?
6. Draw the New man's projection formula of the staggered form of 1,2-dichloroethane.
7. Give the IUPAC name of the lowest molecular weight alkane that contains a quaternary carbon.

Answer

1. The iodination of benzene is usually brought about by refluxing benzene with iodine and conc. $HNO _3$ or $HIO _3$. $HNO _3$ or $HIO _3$ oxidises HI to $I _2$ and prevents the backward reaction to occur.

2. A catalyst speeds up the forward and back reaction to the same extent. Because adding a catalyst doesn't affect the relative rates of the two reactions, a catalyst cannot change the position of equilibrium in a chemical reaction
3. Due to resonance, benzene is extraordinarily stable.

5. Paraffins means little affinity. Alkanes due to strong C-C and $C - H$ bonds are relatively chemically inert.
6. New man’s projection formula of staggered form of 1,2-dichloroethane:

7. IUPAC name of the lowest molecular weight alkane that contains a quaternary carbon is 2,2 -dimethyl propane. Structure:

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