MCQ
$2 - Hexyne$ can be converted into trans- $2 - hexene$ by the action of
- A$H_2-Pd/BaSO_4$
- ✓$Na+Liq.NH_3$
- C$NaBH_4$
- DAll above
✓
Answer
Correct option: B.
$Na+Liq.NH_3$
b
Birch reduction
Birch reduction
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$\text{BDE}$ is affected by $s-$character of the bond and the stability of the radicals formed.
Shorter bonds are typically stronger bonds. $\text{BDEs}$ for some bonds are given below
:$\begin{array}{l} Cl - Cl ( g ) \longrightarrow Cl ^*( g )+ Cl ^*( g ) \Delta H ^{\circ}=58 \ kcal \ mol ^{-1} \end{array}$
$ H _3 C - Cl ( g ) \longrightarrow H _3 C ^*( g )+ Cl ^{\circ}( g ) \Delta H ^{\circ}=85 \ kcal \ mol ^{-1} $
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$(1)$ Correct match of the $C - H$ bonds (shown in bold) in Column $J$ with their $\text{BDE}$ in Column $K$ is
$(A) P - iii, Q - iv, R - ii, S - i$
$(B) P - i, Q - ii, R - iii, S - iv$
$(C) P - iii, Q - ii, R - i, S - iv$
$(D) P - ii, Q - i, R - iv, S - iii$
$(2)$ For the following reaction
$CH _4( g )+ Cl _2( g ) \xrightarrow{\text { light }} CH _3 Cl ( g )+ HCl ( g )$
the correct statement is
$(A)$ Initiation step is exothermic with $\Delta H ^{\circ}=-58 \ kcal \ mol ^{-1}$
$(B)$ Propagation step involving ${ }^{\circ} CH _3$ formation is exothermic with $\Delta H ^{\circ}=-2 \ kcal \ mol ^{-1}$.
$(C)$ Propagation step involving $CH _3 Cl$ formation is endothermic with $\Delta H ^{\circ}=+27 \ kcal \ mol ^{-1}$.
$(D)$ The reaction is exothermic with $\Delta H ^{\circ}=-25 \ kcal \ mol ^{-1}$.
→$\text{BDE}$ is affected by $s-$character of the bond and the stability of the radicals formed.
Shorter bonds are typically stronger bonds. $\text{BDEs}$ for some bonds are given below
:$\begin{array}{l} Cl - Cl ( g ) \longrightarrow Cl ^*( g )+ Cl ^*( g ) \Delta H ^{\circ}=58 \ kcal \ mol ^{-1} \end{array}$
$ H _3 C - Cl ( g ) \longrightarrow H _3 C ^*( g )+ Cl ^{\circ}( g ) \Delta H ^{\circ}=85 \ kcal \ mol ^{-1} $
$ H - Cl ( g ) \quad \longrightarrow H ^*( g ) \quad+ Cl ^*( g ) \Delta H ^{\circ}=103 \ kcal \ mol ^{-1} $
$(1)$ Correct match of the $C - H$ bonds (shown in bold) in Column $J$ with their $\text{BDE}$ in Column $K$ is
| Column $J$ Molecule | Column $K$ $\operatorname{BDE}( kcal mol -1)$ |
| $(P)$ $H - C H \left( CH _3\right)_2$ | ${ (i) } 132$ |
| $(Q)$ $H - CH _2 Ph$ | ${ (ii) } 110$ |
| $(R)$ $H - C H = CH _2$ | ${ (iii) } 95$ |
| $(S)$ $H - C \equiv CH$ | ${ (iv) } 88$ |
$(B) P - i, Q - ii, R - iii, S - iv$
$(C) P - iii, Q - ii, R - i, S - iv$
$(D) P - ii, Q - i, R - iv, S - iii$
$(2)$ For the following reaction
$CH _4( g )+ Cl _2( g ) \xrightarrow{\text { light }} CH _3 Cl ( g )+ HCl ( g )$
the correct statement is
$(A)$ Initiation step is exothermic with $\Delta H ^{\circ}=-58 \ kcal \ mol ^{-1}$
$(B)$ Propagation step involving ${ }^{\circ} CH _3$ formation is exothermic with $\Delta H ^{\circ}=-2 \ kcal \ mol ^{-1}$.
$(C)$ Propagation step involving $CH _3 Cl$ formation is endothermic with $\Delta H ^{\circ}=+27 \ kcal \ mol ^{-1}$.
$(D)$ The reaction is exothermic with $\Delta H ^{\circ}=-25 \ kcal \ mol ^{-1}$.
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