JEE 2016 :: Advanced 2016 :: Chemistry 2016

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016

Treatment of compound O with  $KMnO_{4}/H^{+}$ gave P, which on heating with ammonia gave Q. The compound Q on treatment with  $Br_{2}/NaOH$ produced R on strong heating, Q gave S which on further treatment with  ethyl 2-bromopropanoate in the presence of KOH followed by acidification gave a compound T

The compound T is

Answer:

Explanation:

For ' invert sugar ',  the correct statement(s) is (are)

( Given: specific rotations of (+) - sucrose,(+)- maltose, L-(-)-glucose and L-(+)-fructose in aqueous solution are +66°, +140°, -52° and 92° , respectively)

Answer:

Explanation:

If there is inversion of specific rotation from (+) to (-), then invert sugar is formed.

(a) $C_{12}H_{22}O_{11}[(+)maltose140^{0}]+H_{2}O\rightarrow Glucose [D(+) 52^{0}]$

(b) $ C_{12}H_{22}O_{11}[(+)sucrose+66^{0}]+H_{2}O$

                                                 $\rightarrow Glucose [D(+) 52^{0}]+Fructose [L(-)-92^{0}] $

                                                                                                   -40° for 2 moles mixture

                                                                                                    -20° for 1 mole mixture

 There is formation of invert sugar, Thus, correct

(c) Specific rotation of invert sugar is -20° per mole. Thus, correct

(d) $Br_{2} $ water is a weak oxidising agent. 

           It oxidises -CHO to -COOH.

  $- CH_{2} OH$ group is not affected

 $HNO_{3}$ (a strong oxidising agent) oxidises invert sugar to saccharic acid. Thus, incorrect

Mixture (s) showing positive deviation from Raoult's law at 35° C is (are) 

Answer:

Explanation:

When the intermolecular attraction between two components A and B  in the mixture is same as between A and A or B and B, hence it is a  case of ideal solution. When intermolecular attraction between  A and B in a mixture is smaller than the between A and A or B and B, then mixture is more vaporised, bp is lowered. It is a case of positive deviation from Raoult's law. When intermolecular attraction between A and B is higher than that between  A and A or B and B, then mixture is less vaporised bp is increased. It is a case of negative deviation.

(a) Methanol  molecules ( $CH_{3}OH_{}$) are hydrogen-bonded. In a mixture of CCl₄  and $CH_{3}OH_{}$, extent of H-bonding is decreased. Mixture is more vaporised thus, positive deviation from Raoult's law.

(b)  Acetone molecules have higher intermolecular attraction due to dipole-dipole interaction. With CS₂, this interaction is decreased thus, positive deviation.

(c) Mixture of benzene and toluene forms ideal solution

(d) Phenol and aniline have higher interaction due to intermolecular H-bonding. Hence, negative deviation

The nitrogen containing compound produced in the reaction of  $HNO_{3}$ with  $P_{4}O_{10}$

Answer:

Explanation:

$P_{4}O_{10}$  is  a dehydrating agent and converts $HNO_{3}$ into  $N_{2}O_{5}$

$2HNO_{3}\rightarrow N_{2}O_{5}+H_{2}O$

$P_{4}O_{10}+6H_{2}O\rightarrow 4H_{3}PO_{4}$

(a)  $P_{4}+20HNO_{3}\rightarrow4H_{3}PO_{4}+20NO_{2}+4H_{2}O$

 Thus , (a) is incorrect.

(b)  $N_{2}O_{5}$  has no unpaired electron and is thus, diamagnetic , (b) is correct

(c)          

 There is no N-N bond,  thus (c) is incorrect

 (d)  $N_{2}O_{5}+Na\rightarrow NaNO_{3}+NO_{2}$

 $N_{2}O_{5}$ vapours of brownish colour. Thus, (d) is correct.

Extraction of copper from copper pyrite $(CuFeS_{2})$  involves

Answer:

Explanation:

$CuFeS_{2}$ ( copper pyrite) is converted into  copper in the following steps,

Step I: Crushing (grinding) followed by concentration by the froth-floatation process.

Step II: Roasting of ore  in the presence of  $SiO_{2}$ which removes iron as slag $(FeSiO_{3}).$

$2CuFeS_{2}+O_{2}\rightarrow Cu_{2}S+2FeS+SO_{2}$

$2FeS+3O_{2}\rightarrow 2SO_{2}+2FeO$

$FeO+SiO_{2}\rightarrow FeSiO_{3}(slag)$

Step III,

        Self reduction in Bessemer converter

     $2Cu_{2}S+3O_{2}\rightarrow 2Cu_{2}O+2SO_{2}$

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

Copper obtained is blister copper (98% pure)

Step IV,

Refining of blister copper is done by electrolysis

impure copper→ Anode

Pure copper →  Cathode

At anode  :  $Cu\rightarrow Cu^{2+}+2e^{-}$

At cathode: $Cu^{2+}+2e^{-}\rightarrow Cu$

Carbon-reduction method is not used, Thus (d) is incorrect

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016