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 The synthesis of alkyl fluorides best accomplished by

Answer:

Explanation:

Alkyl fluorides can be prepared by action of mercurous fluoride or antimony trifluorides  ( inorganic fluorides)  on corresponding alkyl halide. This reaction is known as Swarts reaction. 

                $CH_{3}Br+AgF\rightarrow CH_{3}F+AgBr$

                                                                    methyl fluoride

  But, when action of NaI/acetone takes place on alkyl chloride or bromide, alkyl iodide forms. This reaction is called the ' Finkelstein reaction '.

    $C_{2}H_{5}Cl\xrightarrow[\text{acetone}]{\text{Nacl}}C_{2}H_{5}I+NaCl$

   Free radical fluorination is highly explosive reaction so not preferred for the preparation of fluoride.

Which of the following  compound will exhibit  geometrical isomerism?

Answer:

Explanation:

Alkene in which different groups are attached with the double-bonded carbon atoms, exhibit geometrical isomerism.

It will show geometrical isomerism,

1,1- diphenyl-1- propane being an alkane  (saturated compound) does not show geometrical isomerism.

 Time Saving Technique.

                We do not need the check all options, but it should remind that double-bonded compounds show geometrical isomerism. Thus (d) is eliminated, (i.e, propane). Now, eliminate the terminal alkene and get the correct response.

In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is [ at. mass Ag=108, Br=80]

Answer:

Explanation:

Given,

  weight of organic compound =250  mg

 Weight of AgBr=141 mg

 $\therefore$   According to formula of %  of bromine by Carius method

       %  of Br=   Atomic weight of Br / Molecular  weight of AgBr

                 x    Weight of AgBr/ Weight of organic bromide   x 100

   $\therefore$   % of Br=    $\frac{80}{188}\times\frac{141}{250}\times 100$

                                    =  $\frac{1128000}{47000}=24$ %

Assertion (A)   Nitrogen and oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.

 Reason (R)  The reaction between nitrogen and oxygen requires high temperature.

Answer:

Explanation:

Nitrogen is an inert gas because of the presence of strong bond. That's why although there is 78% N₂ in the atmosphere but nitrogen oxide in not formed under ordinary conditions.

But when temperature is high enough i.e, =2000K, it reacts with oxygen to form nitrogen oxide.

$N_{2}+O_{2} $    $\underrightarrow{-2000K} $  2NO

Thus, Assertion  and reason are true and Reason  is the correct explanation  of the Assertion

The colour of $KMnO_{4}$  is due to 

Answer:

Explanation:

$KMnO_{4}   \rightarrow K^{+}+MnO_{4}^{-}$ 

 $\therefore$ In $MnO_{4}^{-}$  , Mn has +7  oxidation state having no electron in d-orbitals.

 It is considered that higher the oxidation state of metal, greater is the tendency to occur  $L \rightarrow M$  charge transfer because ligand is able to donate the electrons into the vacant d-orbital of metal.

 Since, charge transfer is Laporte as well as spin allowed, therefore, it shows colour.

 Time Saving Technique.   

 There is no need to check all four options. Just find out the oxidation state of metal ion. If oxidation state is highest and ligand present there is of electron-donating nature, gives LMCT , which shows more intense colour.

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• Main 2015 - Chemistry 2015
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