JEE 2014 :: Main 2014 :: Chemistry 2014

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 The major  organic  compound formed by the reaction of 1,1,1- trichloroethane with silver powder is 

Answer:

Explanation:

 The reaction is 

   $2CH_{3}-CCl_{3}$    $\xrightarrow[{\triangle}]{6Ag}$   $CH_{3}-C\equiv C-CH_{3}+6AgCl$

The most  suitable reagent for the conversion of   $R-CH_{2}-OH\rightarrow R-CHO $    is 

Answer:

Explanation:

$R-CH_{2}OH$    $\underrightarrow{PCC}$   $R-CH=O$ 

 Pyridinium chlorochromate is the mild oxidising agent which causes conversion of alcohol to aldehyde stage. While others causes conversion of alcohol  to acid

 On heating an aliphatic primary amine with chloroform and ethanolic  potassium hydroxide , the organic compound formed is

Answer:

Explanation:

This reaction is an example of carbylamine reaction which includes conversion of amine to isocyanide.

          $R-\ddot{N}H_{2}+CHCl_{3}$  $\xrightarrow[{KOH}]{C_{2}H_{5}OH}$   $R-N^{+}\equiv C^{-}:$

                                                                                          Alkyl isocyanide

Note:   This reaction include  rearrangement reaction of nitrene in which migration of alkyl group from carbon to nitrogen takes place.

 Given below  are the half  -cell reactions

$Mn^{2+}+e^{-}\rightarrow Mn;E^{0}=-1.18eV$

$2(Mn^{3+}+e^{-}\rightarrow Mn^{2+});E^{0}=+1.51eV$ the E° for    $3Mn^{2+}\rightarrow Mn+2 Mn^{3+}$ will be

Answer:

Explanation:

 Standard electrode potential  of reaction  [E°]  can be calculated as

   $E_{cell}^{0}=E_{R}-E_{P}$

where,  $E_{R}$   = SRP of reactant

           $E_{P}$   = SRP of product

   If      $E_{cell}^{0}$   = +ve  , then reaction is  spontaneous  otherwise non-spontaneous

$Mn^{3+}$      $\underrightarrow{E_{1}^{0}=1.51V}$        $Mn^{2+}$                

  $Mn^{2+}$       $\underrightarrow{E_{2}^{0}=1.181V}$     Mn

   For   $Mn^{2+}$ disproportionation

             E° =-1.51 V  - 1.18 v   =-2.69 v <  0

 The ratio of masses of oxygen and nitrogen of a  particular gaseous mixture is 1:4. The ratio of the number of  their molecule is 

Answer:

Explanation:

$\frac{n_{O_{2}}}{n_{N_{2}}}=\frac{\frac{(m_{O_{2})}}{(M_{O_{2}})}}{\frac{(m_{N_{2}})}{(M_{N_{2}})}}$

where, 

  $m_{O_{2}}=$   given mass of O₂

 $m_{N_{2}}=$  given mass of N₂

 $M_{O_{2}}=$   molecular mass of O₂

 $M_{N_{2}}=$      molecular mass of N₂

 $n_{O_{2}}=$    number of moles O₂

 $n_{N_{2}}=$   number of moles N₂

            $=\left[\frac{m_{O_{2}}}{m_{N_{2}}}\right]\frac{28}{32}=\frac{1}{4}\times\frac{28}{32}=\frac{7}{32}$

• Main 2014 - Physics 2014
• Main 2014 - Chemistry 2014
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