JEE 2014 :: Advanced 2014 :: Chemistry 2014

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014

 The reactivity of compound Z with different halogens under appropriate conditions is given below

The observed pattern of electrophilic substitution can be explained by

Answer:

Explanation:

Plan  This problem includes concept of effect of steric and electronic effect on reactivity of organic compounds.

 Steric effect of halogens are as follows  Cl₂ < Br₂ < I₂

 The electronic effect of phenolic group directs the approaching electrophile towards ortho and para positions. Tertiary butyl group has large size so it causes steric effect around aromatic nucleus. On the  basis of the above factors the products of the given reactions are as follows

 Hence, orientation in electrophilic substitution reaction is decided by

 (a) The steric effect of the halogen

(b)  the steric effect of the tert-butyl group

(c)  the electronic effect of the phenolic group

(a),(b),and (c) are correct choices.

Hydrogen bonding plays a central role in the following phenomena

Answer:

Explanation:

Plan   This problem can be  solved by using the concept of H-bonding and applications of H-bonding

(a) Ice floats  in water  due to the low density of ice as compared to water which is due to open cage-like structure (formed by intermolecular H-bonding)

(b)  Basic strength of RNH₂ >R₃ N. It is also explained by hydrogen bonding

 Two H- bonds are possible  with water present  in aqueous solution. (stabilize by solvation)

 No H bonding is possible with water present in aqueous solution.(stabilization by solvation is very less)

 (c)  

 More acidic due to the presence of H. (Due to the absence of electron-donating group)

 Less acidic than HCOOH due to presence of CH₃-

(Electron donating group)

(d) Dimerisation of acetic acid in benzene  is due to intermolecular hydrogen bonding

Upon heating with  Cu₂S, the reagent (s)  that  give copper metal is/are

Answer:

Explanation:

(b)   $4CuO$   $\underrightarrow{1100^{0}C}$                      $2Cu_{2}O+O_{2}$

(c)    $Cu_{2}S+2Cu_{2}O$        $\underrightarrow{\triangle}$   $6Cu+SO_{2}$

         $2Cu_{2}O+Cu_{2}S  $      $\underrightarrow{\triangle}$    $6Cu+SO_{2}$

 (d)   $CuSO_{4}$    $\underrightarrow{720^{0}C}$     $CuO+SO_{2}+\frac{1}{2}O_{2}$

                4CuO        $\underrightarrow{1100^{0}C}$   $2Cu_{2}O+O_{2}$

      $2Cu_{2}O+Cu_{2}S $    $\underrightarrow{\triangle}$         $6Cu+SO_{2}$

Reaction is believed to proceed as

$Cu_{2}S\rightleftharpoons 2Cu^{+}+S^{2-}$

$2Cu_{2}O\rightleftharpoons 4Cu^{+}+2O^{2-}$

$S^{2-}+2O^{2-}\rightarrow SO_{2}+6e^{-}$

 $6Cu^{+}+6e^{-}\rightarrow 6Cu:E^{0}_{cell}=0.52$

 Here, copper , sulphide is reduced to copper metal . Solidified copper has bilstered appearance due to evolution of SO₂ and thus obtained copper is known as blister copper.

 Other compounds which give Cu are

(i)  CuO as

$4 CuO$    $\underrightarrow{1100^{0}C}$     $  2Cu_{2}O+O_{2}$

  $2Cu_{2}O+Cu_{2}S$         $\underrightarrow{\triangle}$  $6Cu+SO_{2}$

(ii)   CuSO₄   as

      $CuSO_{4}$               $\underrightarrow{720^{0}C}$  $CuO+SO_{2}+\frac{1}{2}O_{2}$

                4CuO        $\underrightarrow{\triangle}$  $2Cu_{2}O+O_{2}$

  $2Cu_{2}O+Cu_{2}S$     $\underrightarrow{\triangle}$   $6Cu+SO_{2}$

 While CuFeS₂ will not give Cu on heating  . The heating in the presence of O₂  gives Cu₂S  and FeS with the evolution of SO₂      

In  a galvanic cell,  the salt bridge

Answer:

Explanation:

Plan  This problem is based on characteristics of salt bridge

 Functions of salt bridge

(i)  It connects  the two half cells and completes  and complete  the  cell circuit

(ii)  It keeps the solutions of two half-cells and complete the cell circuit but does not participate chemically in the cell reaction.

(iii) It maintains the diffusion of ions from one electrode to another electrode.

(iv)  A cell reaction may also occur in absence  of salt bridge, sometimes, both the electrodes  dip in the same electrolyte  solution  and in such cases we do not require a salt bridge." So, option (c) is incorrect.

 (v) This prevent mixing of two electrolytic solutions hence, option (d) is incorrect choice

Hence, correct choice is (a)

 The pair(s)  of reagents that yield paramagnetic  species is/are

Answer:

Explanation:

Plan: the paramagnetic character of species can be easily explained on the basis of the presence of unpaired electrons. The compounds containing unpaired electron(s)  is/are paramagnetic.

 Reaction of alkali metals with ammonia depends upon the physical state of ammonia whether it is in gaseous state or a liquid state. If ammonia is considered as a gas then the reaction will be

  $(a) Na+NH_{3}\rightarrow NaNH_{3}+\frac{1}{2}H_{2}$

              (excess)

 (NaNH₂+1/2 H₂  are diamaganetic )  If ammonia  is considered as a liquid  then reaction will be

  $M+(x+y)NH_{3}\rightarrow[M(NH_{3})_{x}]^{+}+[e(NH_{3})_{y}]^{-}$

         . ammoniated electron

          . blue colour

           . paramagnetic

            . very strong reducing agent

(b)        $K+O_{2}\rightarrow KO_{2}(K^{+},O_2^-)$

                 (Excess)                      Potassium

                                           superoxide paramagnetic

  (c)   $3Cu+8HNO_{3}\rightarrow3Cu(NO_{3})_{2}+2NO+4H_{2}O$

                                     paramagnetic      paramagnetic     

  (d)

Hence, option (a) , (b)  and (c) are correct choices.

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014