JEE 2014 :: Advanced 2014 :: Chemistry 2014

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

 Match the following four starting materials (P, Q, R, S)  given in list I with the corresponding reaction scheme. (I, II, III, IV)  provided in List II and select  the correct answer using the code  given below the lists,

Answer:

Explanation:

Plan This problem  can be solved by using  various concepts synthesis of benzene, electrophilic substitution reaction and directive  influence  of various  substituents, including oxidation  and reduction

 $\Rightarrow$     -OH  and -NH₂  are o/p - directing groups.

 $\Rightarrow$          N-acetylation is more favourable than C- acylation

 $\Rightarrow$           N- sulphonation is more favourable than C-sulphonation.

 $\Rightarrow$            NO₂  is a meta- directing group.

 $\Rightarrow$      H₂S-NH₃  reduces only one NO₂ group selectively in the presence of two NO₂ groups.

Using above concepts the correct sequence of reaction can be written as

Different possible thermal decomposition pathways for peroxyesters are shown   below. Match each pathways  from List I with an appropriate  structure from List II and select the correct answer using the code given below the lists.

Answer:

Explanation:

Plan This problerm can be solved by using the stability of the radical obtained after fragmentation of peroxy ester.

 Allylic radicals  are more stable  than alkyl  radical, so when there is a possibility of formation of allyl radical, it will undergo fragmentation through the formation of alkyl radical i.e, fragmentation  produces  stable radical

 On the  basis of stability of radical, fragmentation can be done as 

P → 1,Q→ 3, R → 4, S→2

 Thus, (a) is the correct choice

Match the orbital overlap figures shown in List I with the description given in List-II  and select the correct answer using the code given below the lists.

Answer:

Explanation:

Plan  This problem includes basic concept of bonding. It can be solved by using the concept of molecular orbital theory

Any orbital has two-phase +ve and -ve. In the following  diagram +ve phase  is shown by darkening the lobes and -ve by without  darkening the lobes

 When two same phase overlap  with each other it forms bonding molecular orbital otherwise antibonding

 Axial overlapping leads to the formation of σ  -bond and sideways overlapping leads to the formation of  $\pi$- bond

 On the basis of above two concepts correct matching can be done as shown below

P → 2, Q → 3, R→  1, S→ 4

Hence ,(c)  is the correct option

Match  each coordinates  compound  in List I  with    an appropriate pair of  characteristics from list  II  and select the correct answer using the code given below the lists (en=H₂NCH₂CH₂NH₂. atomic numbers : Ti=22.: Cr=24: Co=27: Pt=78)

Answer:

Explanation:

Plan  This problem is based on concept of VBT  and magnetic properties of coordination compound.

 Draw VBT  for each coordination compound

 If an unpaired electron is present then the coordination compound will be paramagnetic otherwise diamagnetic.

coordination compounds of [MA₄B₂] type show geometrical isomerism.

Molecular orbital electrons  configuration (MOEC) for various coordination compound can be drawn using VBT as

 MOEC for [Cr(NH₃)₄Cl₂]Cl   is

 Number of unpaired elecrtons (n)=3

 Magnetic properties = paramagnetic

 Geometrical isomers of (Cr(NH₃)₄Cl₂]⁺ are

MOEC of [Ti(H₂O)₅ Cl](NO₃)₂   is

   n=1

 Magnetic properties= paramagnetic

  Ionisation  isomers of [Ti(H₂O)₅Cl)(NO₃)₂ are     [Ti(H₂O)₅Cl(NO₃)₂    and [Ti(H₂O)₅(NO₃)]Cl(NO₃)

MOEC   of [Pt(en)(NH₃)Cl]NO₃   is 

n=0

 Magnetic property= diamgnetic

 Ionsiation isomers are [Pt(en)(NH₃)Cl)NO₃ and [Pt(en)(NH₃(NO₃)]Cl

 MOEC of [Co(NH₃)₄ (NO₃)₂]NO₃    is

 n=0

Magnetic property= Diamagnetic  Geometrical isomers are

  Thus, magnetic  property and isomerism in a given coordination compound can be summarised as

 (P)    [Cr(NH₃)₄ Cl₂]Cl  → Paramagnetic and exhibits cis-trans isomerism (3)

 (Q)    [Ti(H₂O)₅Cl](NO₃)₂   →  Paramagnetic and exhibits ionisation isomerism (1)

  (R)   [Pt(en) (NH₃)Cl]NO₃  →  Diamagnetic and exhibits  ionisation isomerism(4)

(S)   [Co(NH₃)₄ (NO₃)₂  ]NO₃   → Diamagnetic and exhibits cis-trans isomerism  (2)

     P → 3, Q→ 1, R→  4, S→ 2

 Hence, (b) is the correct choice

As an aqueous solution of metal ion M₁ reacts separately with reagents  Q and R  in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M₂  always forms tetrahedral complexes with these reagents. An aqueous solution of M₂ on  reaction with  reagent S gives white precipitate  which dissolves in  excess of S. The  reactions are summarized in the scheme given below

Reagent S is 

Answer:

Explanation:

Zn²⁺  on treatment with excess Of KOH produces  [Zn(OH)₄]^2-.

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