3 Marks Question

Question 13 Marks

Describe the following:
i. Acetylation
ii. Cannizzaro reaction
iii. Cross aldol condensation
iv. Decarboxylation

Answer

$i.$ Acetylation: Acetylation simply involves the addition of an acetyl group to a compound. An acetyl group is made up of a carbonyl group, or carbon double bonded to oxygen, with a methyl group $(-CH_3)$ on the end. The part of the acetyl group that's attached to the compound is often represented with $'R\ '.$

$ii.$ Cannizzaro reaction: Aldehydes which do not have an alpha$-$hydrogen atom, undergo self oxidation and reduction $($disproportionation$)$ reaction on treatment with concentrated alkali. In this reaction, one molecule of the aldehyde is reduced to alcohol while another is oxidized to carboxylic acid salt.

$iii.$ Cross aldol condensation: When aldol condensation is carried out between two different aldehydes and $/$ or ketones, it is called cross aldol condensation. If both of them contain alpha$-$hydrogen atoms, it gives a mixture of four products. This is illustrated below by aldol reaction of a mixture of ethanal and propanal.

$iv.$ Decarboxylation: Carboxylic acids lose carbon dioxide to form hydrocarbon when their sodium salts are heated with soda lime $( \ce{NaOH}$ and $\ce{CaO}$ in the ratio of $3: 1)$.
This reaction is known as decarboxylation.
$\ce{R-COONa {NaOH \ } R-H+Na_2 CO_3}$

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Question 23 Marks

The $K _{ sp }$ for $\text{AgCl}$ at $298 K$ is $1.0 \times 10^{-10}$. Calculate the electrode potential for $Ag ^{+} / Ag$ electrode immersed in $1.0 \text{M KCl}$ solution. Given $E _{ A g ^{+} / A g }^\theta=0.80 V$

Answer

$AgCl(s) \rightleftharpoons Ag^{+}+Cl^{-}$
$K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right]$
${\left[Cl^{-}\right]=1.0 M}$
${\left[Ag^{+}\right]=\frac{k_{s p}}{\left[Cl^{-}\right]}=\frac{1 \times 10^{-10}}{1}=1 \times 10^{-10} M}$
Now, $Ag ^{+}+ e ^{-} \longrightarrow Ag(s)$
$E=E^\theta-\frac{0.059}{1} \log \frac{1}{\left[Ag^{+}\right]}$
$=0.80-\frac{0.059}{1} \log \frac{1}{10^{-10}}$
$=0.80-0.059 \times 10=0.21 V$

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Question 33 Marks

The conductivity of $2.5 \times 10^{-4} M$ methanoic acid is $5.25 \times 10^{-5} S cm ^{-1}$ and its $\wedge_{ m }^0$ has a value $400 S cm ^2 mol^{-}$ 1. Calculate its molar conductivity and degree of dissociation.

Answer

$\Lambda_m=\frac{1000 \times K }{ M } S \ cm ^2 mol^{-1}$
$\Lambda_m=\frac{1000 \times 5.25 \times 10^{-5}}{2.5 \times 10^{-4}}- S \ cm ^2 mol^{-1}$
$=210 S \ cm ^2 mol^{-1}$
$\wedge_m^0 HCOOH =\lambda^{\circ} HCOO ^{-}+\lambda^{\circ} H ^{+}$
$=(50.5+349.5) S \ cm ^2 mol^{-1}$
$=400 S \ cm ^2 mol^{-1}$
$\alpha=\frac{\Lambda_m}{\Lambda_m^{\circ}}$
$\alpha=\frac{210}{400}$
$=0.525$

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Question 43 Marks

What happens when
i. n-butyl chloride is treated with alcoholic KОН.
ii. bromobenzene is treated with Mg in the presence of dry ether.
iii. chlorobenzene is subjected to hydrolysis.

Answer

i. When n - butyl chloride is treated with alcoholic KOH, the formation of but-1-ene takes place. This reaction is a dehydrohalogenation reaction.
$\underset{\text { n- Butyl chloride }}{ CH _3- CH _2- CH _2}- CH _2- Cl \xrightarrow[\text { (dehydrohalogenation) }]{ KOH (\text { alc } / \Delta} CH _3-\underset{\text { But-1-ene }}{ CH _2- CH }= CH _2+ KCl + H _2 O$
ii. When bromobenzene is treated with Mg in the presence of dry ether, (Grignard reagent) phenylmagnesium bromide is formed.

ii. The chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm form phenol(replacement by hydroxyl group).

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Question 53 Marks

In the button cell widely used in watches and other devices the following reaction takes place:
$Zn(s)+Ag_2 O(s)+H_2 O(l) \rightarrow Zn^{2+}(a q)+2 Ag(s)+2 OH^{-}(a q)$
Determine $\Delta_r G^{(-)}$and $E ^{(-)}$for the reaction
Given $Z n \rightarrow Z n^{2+}+2 e^{-}, E ^0=0.76 V$
Given $Ag \rightarrow Ag ^{+}+2 e^{-}, E ^0=0.344 V$

Answer

Zn is oxidized and $Ag _2 O$ is reduced $($as $Ag ^{+}$ions change to Ag $)$
$E_{\text {cell }}^0=E^0\left[A g_2 O / A g\right](r e d)+E^0\left[Z n / Z n^{2+}\right](o x)$
$=0.344+0.76$
$=1.104 V$
$\Delta_r G^0=-n F E^0 \text { cell }=-2 \times 96500 \times 1.104 J$
$=-2.13 \times 10^5 J$

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Question 63 Marks

Name the reagents which are used in the following conversions:
i. A primary alcohol to an aldehyde
ii. Butan-2-one to butan-2-ol
iii. Phenol to 2, 4, 6-tribromophenol

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Question 73 Marks

Give the structure of the products you would except when each of the following alcohol reacts with (i) Butan-lol (ii) 2- Methylbutan-2-ol
a. $HCl - ZnCl _2$
b. HBr
c. $SOCl _2$

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Question 83 Marks

For the reaction:
$2 A+B \rightarrow A_2 B$
the rate $= k [ A ][ B ]^2$ with $k =2.0 \times 10^{-6} mol^{-2} L^2 s^{-1}$. Calculate the initial rate of the reaction when $[ A ]=0.1 mol$ $L ^{-1},[B]=0.2 mol L ^{-1}$. Calculate the rate of reaction after $[ A ]$ is reduced to $0.06 mol L ^{-1}$.

Answer

The initial rate of the reaction is:
$\text { Rate }=k[A][B]^2$
$=\left(2.0 \times 10^{-6} mol^{-2} L^2 s^{-1}\right)\left(0.1 mol^{-1}\right)\left(0.2 mol^{-1}\right)^2=8.0 \times 10^{-9} mol^{-2} L^2 s^{-1}$
When [A] is reduced from $0.1 mol L ^{-1}$ to $0.06 mol^{-1}$, the concentration of A reacted
$=(0.1-0.06) mol L^{-1}=0.004 mol L^{-1}$
Therefore, concentration of B reacted $=\frac{1}{2} \times 0.04 mol L ^{-1}=0.02 mol L ^{-1}$
Then, concentration of B available, $[B]=(0.2-0.02) mol L^{-1}$
$=0.18 mol L^{-1}$
After $[ A ]$ is reduced to $0.06 mol^{-1}$, the rate of the reaction is given by,
$\text { Rate }=k[A][B]^2$
$=\left(2.0 \times 10^{-6} mol^{-2} L^2 s^{-1}\right)\left(0.06 mol^{-1}\right)(0.18)^2$
$=3.89 \times 10^{-9} mol^{-1} L^{-1} s^{-1}$

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