Question
Explain Friedel$-$Craft’s acylation reaction.
✓
Answer
The reaction in which hydrogen atom of benzene is replaced by an acyl group 
I in the presence of anhydrous $AlCl_3$ is called Friedel$-$Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous $AlCl_3,$ forms acetophenone $(1-$Phenyl ethanone).
Electrophile $: R - C += O$ acylium ion Formation of the electrophile :
I in the presence of anhydrous $AlCl_3$ is called Friedel$-$Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous $AlCl_3,$ forms acetophenone $(1-$Phenyl ethanone).Electrophile $: R - C += O$ acylium ion Formation of the electrophile :
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the order of following reactions whose rate laws are expressed as follows. $C_A$ and $C_B$ are the concentrations of reactants $A$ and $B$ respectively :
(1) $-\frac{d c}{d t}=k C _A^0$
(2) $-\frac{d c}{d t}=k C _A^{3 / 2}$
(3) $-\frac{d c}{d t}=k C _A^{1 / 2} \times C _B^2$
(4) $-\frac{d c}{d t}=k C _A^{5 / 2} \times C _B^0$
(5) $-\frac{d c}{d t}=k C _A^{1 / 3} \times C _B^{2 / 3}$
→(1) $-\frac{d c}{d t}=k C _A^0$
(2) $-\frac{d c}{d t}=k C _A^{3 / 2}$
(3) $-\frac{d c}{d t}=k C _A^{1 / 2} \times C _B^2$
(4) $-\frac{d c}{d t}=k C _A^{5 / 2} \times C _B^0$
(5) $-\frac{d c}{d t}=k C _A^{1 / 3} \times C _B^{2 / 3}$
Explain the construction and working of a dry cell (or Leclanche’s cell).###Write a note on dry cell.
Whatis HDP ? How is it prepared ? Give its properties and uses ?
Predict the sign of $\Delta S$ in the following processes. Give reasons for your answer: ### Explain with reason sign conventions of $\Delta S$ in the following reactions.
(a) $N _2 O _{4( g )} \longrightarrow 2 NO _{2( g )}$
(b) $Fe _2 O _{3( s )}+3 H _{2( g )} \longrightarrow 2 Fe _{( s )}+3 H _2 O _{( g )}$
(c) $N _{2( g )}+3 H _{2( g )} \longrightarrow 2 NH _{3( g )}$
(d) $MgCO _{3( s )} \longrightarrow MgO _{( s )}+ CO _{2( g )}$
(e) $CO _{2( g )} \longrightarrow CO _{2( s )}$
(f) $C l _{2( g )} \longrightarrow 2 Cl _{( g )} \quad$
→(a) $N _2 O _{4( g )} \longrightarrow 2 NO _{2( g )}$
(b) $Fe _2 O _{3( s )}+3 H _{2( g )} \longrightarrow 2 Fe _{( s )}+3 H _2 O _{( g )}$
(c) $N _{2( g )}+3 H _{2( g )} \longrightarrow 2 NH _{3( g )}$
(d) $MgCO _{3( s )} \longrightarrow MgO _{( s )}+ CO _{2( g )}$
(e) $CO _{2( g )} \longrightarrow CO _{2( s )}$
(f) $C l _{2( g )} \longrightarrow 2 Cl _{( g )} \quad$
Calculate $K p, \Delta G^0$ for the reaction,
$\begin{aligned}
& C _{( s )}+ H _2 O _{( g )} \rightleftharpoons CO _{( g )}+ H _{2( g )} \\
& \text { at } 990 K \text { if the equilibrium concentrations are as follows : } \\
& {\left[ H _2 O \right]=1.10 \text { mol dm- }-3,} \\
& {[ CO ]=\left[ H _2\right]=0.2 mol dm ^{-3}} \\
& R =0.08206 L \text { atm K} mol ^{-1} \text {. }
\end{aligned}$
→$\begin{aligned}
& C _{( s )}+ H _2 O _{( g )} \rightleftharpoons CO _{( g )}+ H _{2( g )} \\
& \text { at } 990 K \text { if the equilibrium concentrations are as follows : } \\
& {\left[ H _2 O \right]=1.10 \text { mol dm- }-3,} \\
& {[ CO ]=\left[ H _2\right]=0.2 mol dm ^{-3}} \\
& R =0.08206 L \text { atm K} mol ^{-1} \text {. }
\end{aligned}$
Describe different types of reversible electrodes with examples.
What are isotonic and hypertonic solutions?
Classify the following semiconductors into n or p-type.
(i) B doped with Si
(ii) As doped with Si
(iii) P doped with Si
(vi) Ge doped with In.
(i) B doped with Si
(ii) As doped with Si
(iii) P doped with Si
(vi) Ge doped with In.
Aniline is less basic than ammonia. Explain.
Equilibrium constant of the reaction,
$2Cu^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$
is $1.2 \times 10^6$. What is the standard potential of the cell in which the reaction takes place ?
→$2Cu^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$
is $1.2 \times 10^6$. What is the standard potential of the cell in which the reaction takes place ?