Chemistry M.C.Q (1 Marks)

Boiling point of n-pentane $=309 \mathrm{~K} $

isopentane $=301 \mathrm{~K} $

neopentane $ =282.5$

As branching increases molecules attain the shape of a sphere results in smaller area of contact thus weak intermolecular forces between spherical molecules, which are overcome at relatively lower temperature. Leading to decrease in boiling point.

Molar mass of $CH _4=16\,g / mol$

Weight of 2 moles of $CH _4=16 \times 2$

$=32\,g$

It's decarboxylation in presence of sodalime so missing reagent is $\mathrm{CaO} .$ $(\mathrm{NaOH}+\mathrm{CaO})$ sodalime

                                                                 $(A)$

$NH _{4} OH \stackrel{\Delta}{\longrightarrow} NH _{3}+ H _{2} O$

                              $(B)$

$Cu ^{+2}( aq )+4 NH _{3} \rightarrow\left[ Cu \left( NH _{3}\right)_{4}\right]^{+2}($ deep blue $)$

                                              $(C)$

Number of sigma bonds $=10$ 

Number of $\pi$ -bonds $=3$

(less than four 'C)

acc. to EN and Inductive effect.

Thus, pair of electrons is present in sp-hybridised orbital.

$\mathrm{KHSO}_{4} \longrightarrow \mathrm{K}^{\oplus}+\mathrm{H\stackrel{\ominus}{S}O}_{4}$

Due to common ion effect backward reaction will take so the formation $\mathrm{NO}_{2}^{+}$ decrease so nitration process will become slower.

$C{{H}_{3}}-C{{H}_{2}}-C\equiv CH\xrightarrow[liq.\,N{{H}_{3}}]{Na\,N{{H}_{2}}/}$ $C{{H}_{3}}-C{{H}_{2}}-C\equiv C\,Na\xrightarrow[-NaBr]{C{{H}_{3}}C{{H}_{2}}-Br}$ $\underset{3-Hexyne}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C\equiv C-C{{H}_{2}}-C{{H}_{3}}}}\,$

$=s p^{3-2}=s p$

In ethyne or acetylene, $C_{2} H_{2},\left(H-C_{2 \pi}^{=} C-H\right)$

hybridisation of carbon $=s p^{3-2}=s p$

So, higher the branches, lower is the boiling point.

So, the order of boiling point of isomeric pentanes is $n -$ pentane $\,>\,$ iso $-$ pentane $\,>\,$ neo $-$ pentane.

$2C{H_3}C{H_2}C{H_2}Br + 2Na\xrightarrow{{ether}}$ $C{H_3}{(C{H_2})_4}C{H_3} + 2NaBr$

$\begin{array}{*{20}{c}}
  {C{H_2} - C{H_2}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,} \\ 
  {\,Br\,\,\,\,\,\,\,\,Br\,\,\,\,\,\,} 
\end{array}$ $ + Pb \to \mathop {C{H_2} = C{H_2}}\limits_{Ethene}  + \mathop {PbB{r_2}}\limits_{Leadbromide} $

As leaded gasoline burns, lead metal gets deposited in the engine which is removed by adding ethylene dibromide. The lead bromide is volatile and is carried off with the exhaust gases from the engine

The main constituents are saturated straight-chain and branched-chain hydrocarbons which are also called as paraffin, as well as ring-shaped cycloparaffins which are also known as naphthenes. Kerosene is less volatile liquid than gasoline.

$CH _3 Br + CH _3 Cl \stackrel{ Na ( s )}{\longrightarrow} CH _3 CH _3$

(with loss of $NaCl$ and then $NaBr$ )

Option $B$ will give $1$ mono brominated products, as $Br$ can go on $C _1$.

Option $C$ will give $1$ mono brominated products, as $Br$ can go on $C _1 / C _2 / C _3 / C _4$.

Option $D$ will give $2$ mono brominated products, as $Br$ can go on $C _1 / C _2 / C _3 / C _4$.

Therefore,Option $B$ is correct answer for this question.

When $n$-butane is heated in the presence of $AlCl _{3} / HBr$.

It will be converted into Iso-butane.

It is a rearrangement reaction.

The boiling point of $n$-butane is $-1$ degrees celsius.

The boiling point of neopentane is $9.5$ degrees celsius.

The boiling point of isobutane is $-11.7$ degrees celsius.

So $n$-heptane has the highest boiling point.

$C{H_3}I + 2Na + IC{H_3}\mathop {}\limits_{{\rm{}}} \xrightarrow[Ether]{Dry} C{H_3} - C{H_3} + 2NaI$

${C_2}{H_5}I + 2Na + {C_2}{H_5}I\mathop {}\limits_{{\rm{}}}   \xrightarrow[Ether]{Dry}  \mathop {{C_2}{H_5} - {C_2}{H_5}}\limits_{{\rm{Butane}}}  + 2NaI$

${C_3}{H_7}I + 2Na + {C_3}{H_7}I\mathop {\xrightarrow{{Dry}}}\limits_{Ether} \mathop {{C_3}{H_7} - {C_3}{H_7}}\limits_{{\text{Hexane}}}  + 2NaI$

$1$-chloro propane $( CH _3- CH _2-CH_2(Cl))$ and $2$-chloro propane $( CH _3 - CH ( Cl )- CH _3 \text { ) }$

Knocking has adverse effects. It results in early wear-and-tear of the piston. Carbon is deposited in the inner part of the cylinders [Internal Combustion Engines have cylinders, inside which pistons move]. Also, Efficiency is reduced.

So we want to reduce knocking. We add something which can reduce it. Such substances are called Anti - knockingAgents.

A popular anti-knocking agent was Tetra Ethyl Lead (TEL) $\left( C _2 H _5\right)_4 Pb$.

In presence of alc. $KOH$ dehydrohalogenation occur and alkene is formed.

It contains $98-99\%$ carbon. It is used in making black ink, paints and shoe polishes.

Syn gas is a mixture of $CO , CO _2$ and $H _2$.

Producer gas is a mixture of flammable gases (principally carbon monoxide and hydrogen) and nonflammable gases (mainly nitrogen and carbon dioxide).

Semi-water gas is a mixture of water gas and producer gas made by passing a mixture of air and steam through heated coke.

$2 RX +2 Na \stackrel{\text { Ether }}{\longrightarrow} \underset{\text { alkane }}{ R - R }$

$(ii)$ Kolbe's reaction

$RCOONa + H _2 O \stackrel{\text { Electrolysis }}{\longrightarrow} \underset{\text { alkane }}{ R - R }$

$(iii)$ Ulmann's reaction

$2 C _6 H _5 I +2 Cu \rightarrow \underset{\text { biphenyl } }{ C _6 H _5}- C _6 H _5$

$(iv)$ Frankland reaction

$\underset{\text { alkyl halide }}{2 RX }+ Zn \rightarrow \underset{\text { alkane }}{ R - R }$

$C{H_4} = \frac{4}{{16}}\; \times \;100\; = \;25\% $.

${\rm{Molecular}}\,\,{\rm{mass}} = \frac{{{\rm{Weight}}\,}}{{V\,ml.}} \times {\rm{22400}}$

$ = \frac{{{\rm{11}}}}{{{\rm{5600}}}} \times 22400 = 44$

$\mathop {{C_2}{H_4}}\limits_{{\rm{Ethene}}} + 3{O_2} \to 2C{O_2} + 2{H_2}O$

All hydrocarbons saturated or unsaturated on complete combustion always produce $C{O_2}$ and ${H_2}O$.

$C{H_3} - C{H_3} + 3\frac{1}{2}{O_2} \to 2\,C{O_2} + 3{H_2}O$

$\mathop {C{H_3} - CHB{r_2}}\limits_{{\rm{Gemdihalide}}} $ $\mathop {\mathop {C{H_2}}\limits_{\mathop {|\,\,\,\,\,\,\,}\limits_{Br\,\,\,} } - \mathop {C{H_2}}\limits_{\mathop {|\,\,\,\,\,\,\,\,}\limits_{Br\,\,\,\,} } }\limits_{vic{\rm{ - }}{\rm{dihalide}}} $

$\mathop {{H_3}C - {H_2}C - \mathop {\mathop {{H_2}C - HC - C{H_3}}\limits^{\,\,\,\,|} }\limits^{\,\,\,\,\,\,\,\,\,C{H_3}} }\limits_{{\rm{isopropyl group 2 - methylpentane}}} $

$\mathop {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,C{H_{3\,}}} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {{H_3}C - {H_2}C - HC - C - C{H_3}} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
\end{array}}\limits_{2,2,3 - trimethylpen\tan e} $

$\mathop {\mathop {\mathop {\mathop {\mathop {{H_3}C - {H_2}C - C - C{H_2} - C{H_3}}\limits^{\,\,|} }\limits^{\,\,\,\,\,\,\,\,C{H_3}} }\limits_{\,|} }\limits_{\,\,\,\,\,\,C{H_3}} }\limits_{{\rm{3, 3 - dimethylpentane}}} $

$\mathop {C - C - C - C}\limits_{{\rm{straight chain}}} $ $\mathop {C - \mathop C\limits_{\mathop |\limits_C } - C}\limits_{{\rm{Branched chain}}} $

$ \underset {Coke \,Red \,hot} {2C} \, + \underbrace {O_2 \, + 4N_2}_\text {Air} \, \longrightarrow \, \underbrace {2CO \, + \,4N_2}_\text {Producer $\,$gas} $

$H _2 S _2 O _7 \text { (Oleum) } \rightarrow H _2 SO _4+ SO _3$

$R - H + SO _3 \rightarrow R - SO _2 OH$

${C_2}{H_5} - I + 2Na + I - {C_2}{H_5}     \xrightarrow[Ether]{Dry}   \mathop {}\limits_{{\rm{Ether}}} \mathop {{C_2}{H_5} - {C_2}{H_5} + 2NaI}\limits_{{\rm{Butane}}} $

$C{H_3}MgI + {C_2}{H_5}OH \to C{H_4} + {C_2}{H_5}OMgI$

Alkyl group of Grignard’s reagent is involved in the formation of alkane.

$C{l_2}    \xrightarrow{hv}   C{l^ \bullet } + C{l^ \bullet }$

$C{H_3} - C{H_3} + C{l^ \bullet } \to C{H_3}C{l^ \bullet } + \dot C{H_3}$

$Ca{C_2} + 2{H_2}O \to {C_2}{H_2} + Ca{(OH)_2}$

$\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {C{H_3} - C{H_2} - C - C{H_3}} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
\end{array}$

This reaction is used for the preparation of pure alkanes.

$C{H_2} = C{H_2} + {H_2}  \xrightarrow{Ni}C{H_3} - C{H_3}$

$CH \equiv CH + 2{H_2}  \xrightarrow{Ni}  C{H_3} - C{H_3}$

$C{H_3}COONa$ $ \rightleftharpoons $ $C{H_3}CO{O^ - } + Na + $

Anode :$2C{H_3}CO{O^ - }\xrightarrow{{ - 2{e^ - }}}C{H_3} - C{H_3} + 2C{O_2}$

Cathode : $2N{a^ + } + 2{e^ - } \to 2Na$

$2Na + 2{H_2}O \to 2NaOH + {H_2}$

${1^o}\, \Rightarrow {\rm{ Primary \,6, }}\,\,{{\rm{2}}^{\rm{o}}} \Rightarrow {\rm{Secondary\, 2}}$

${3^o}\, \Rightarrow {\rm{ Tertiary \,2, }}\,\,{{\rm{4}}^{\rm{o}}} \Rightarrow {\rm{Quanternary}}\,1$

$\underbrace {CO + {H_2}}_{{\text{Water}}\,{\text{gas}}} + \mathop {{H_2}}\limits_{{\text{Excess}}} \mathop {\xrightarrow{{{\text{CO}}\,\,{\text{or}}\,{\text{Ni}}}}}\limits_{{\text{heat}}} \mathop {Mixure\,of\,hydrocarbons\,}\limits_{(Petrol)}  + {H_2}O$

$CH _4+ HNO _3 \underset{400^{\circ} C }{\stackrel{\Delta}{\longrightarrow}} \underset{\text { (Nitromethane) }}{ CH _3 NO _2}+ H _2 O$

Open chain alkane product having even number of carbons and symmetrical can  be achieved by Wurtz reaction.

$\therefore$ Mol. formula $= C_5H_{12}$, Structure $\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,C{H_3}} \\ 
  | \\ 
  {C{H_3} - C - C{H_3}} \\ 
  | \\ 
  {\,\,\,\,\,\,\,C{H_3}} 
\end{array}$ neo-pentane

$C{H_3}Cl\xrightarrow[{\operatorname{Re} d.}]{{LiAl{H_4}}}C{H_4}$

$C{H_3}Cl\xrightarrow[\begin{subarray}{l} 
  Dry \\ 
  ether 
\end{subarray} ]{{Mg}}\mathop {\mathop C\limits^{.\,.} }\limits^\Theta  {H_3}\mathop {Mg}\limits^ \oplus  Cl\xrightarrow{{\mathop H\limits^ \oplus  OR}}C{H_4}$

$C{l^ \bullet } + C{H_3}Cl \to {\,^ \bullet }C{H_2}Cl + HCl$

$^ \bullet C{H_2}Cl + C^{12} \to C{H_2}C{l_2} + C{l^ \bullet }$

Bromination of alkanes follows the same mechanism as chlorination. The only  difference is that chlorination produces alkyl chlorides, whereas bromination  forms alkyl bromides 

$C{H_3} - Br\xrightarrow{{Wurtz}}C{H_3} - C{H_3}$