1-CH=CH-CHO is equivalent to -CH----CH+---CHO
consider the -ve charge as a lone pair on O as in -OH
Q1. When we take high concentration of strong base in a non-ionising solvent, bimolecular mechanism are favoured and predominant mechanism is
(a) SN2 (b) E2
(c) AN1Ar (d) E1CB
Q2. In the reaction
2-Bromopropane ------> (Na-Ethoxide/Ethanol)
the major product will be
(a) alcohol
(b) alkene
(c) ether
(d) NOT
Q3. Liq water is in equilibrium with its vapour at a constant temperature in a closed container
H2O(l) <=> H2O(v).
Which of the following statements is incorrect?
(a) PH2O is constant at every instant of time
(b) The const temp must be the BP of water
(c) [H2O(l)]/[H2O(v)] is const
(d) PH2O = atmospheric pressure
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26 Answers
1SORRY IGNORED THIS THREAD AFTER I ANSWERED....
@IITIMCOMING
MINE WERE FOR CORRECT ONES[1]
@ASISH::::::::HOW IS D CORRECT FOR 3RD ONE??[WELL I GUESS I CN IMAGINE Y THEY THINK D IS CORRECT BUT THAT WOULD MAKE QUESTION INCOMPLETE[1]]
11benzene-CH=CH-CHO
it would be -R if -CHO would have been directly connected however here double bonded carbon is NEXT to benzene ring.
-CHO >> -I effect will still be shown
106but then through the CHO group its -R effect should be much more dominant
14) -I +R seems correct, double bond can act as +R
8) x=2 if you want the given product, 3 moles would lead to a different product altogether
1asish ok tell y D is correct.then i will give my opinion[100% my goes wrong a lways[1]] about what i think
1strong base would lead to and attack while non ionaisability will lead to unstability of carbon-cation. so we have option of sn2 and e2 with us. but e2 will win the race as the incomng neucleoplhitcity reduces even in non polar solvents. moreover strong bases favour e2 more than sn2. (most people go for e2 only because of strong bas but it many not only be the reason)
1068. then alkynide ion is formed again naa?. for which another mole of NaNH2 required..
7. given B
217)->A) SO2 is acidic and converts hexavalent chromium to trivalent chromium and colour changes from orange to green.
8)->x= 3 as two moles of HBr is to be removed to form an alkyne........then as it forms a terminal alkyne propyne which has acidic hydrogen and easily will combine with one mole of base.
106any ways it (Q3) was a single option correct one
thx for the others guys
More
Q4. -CH=CH-CHO is (±I,±R) group ... (pick your choice)
Q5. HA (Ka = 10-5, 1 mole/L) is mixed with NaA (1 mol/L) at 298 K ... approx pH is (5/10)
Q6. BOH (Kb = 10-4; 1 mol/L) is mixed with BA (1 mol/L) at 298 K .... approx pH is (5/10)
Q7. A/R
Stmnt 1: SO2 converts acidified K2Cr2O7 green
Stmnt 2: SO2 converts Cr2O72- ion to Cr3+ which is green
(A or B)
(i) x mole NaNH2
Q8. In the reaction 1,2-dibromopropane -----------> CH3-C≡C-C2H5
(ii) C2H5Br
x = ?
3he has asked for the "INCORRECT STATEMENTS" ..... so ur sayin B,C wrt incorrect or correct statements???
1ABOUT 1:::::::I MUST SAY THIS!!!!!!!IF ANY BOOK ASKS SUCH QUESTIONS BURN IT AWAY[1][1][1][1]........ALTHOUGH WE CAN HAVE AN ANSWER BUT IT IS QUITE IDIOTIC FOR AUTHORS TO ASK SUCH QUESTIONS[3][3]
1WELL.....AGAIN I DON FIND ANYTHING TO BE DEBATED IN 3rd one here:::::CLEARLY IT SEEMS B AND C!!!!
11) I'm taking it in a relative factor.
Now , we r taking high concentration of strong base in a non-ionising solvent,
if the reactant participating is primary , then SN2 is surely supported.
if the ractant is secondary , then more preference to E2 , as stronger and bulkier the base more is the elimination product.
if the reactant is tertiary , then its E2 product.
thus, relatively , E2 product is formed.
2) 2-Bromopropane ------> (Na-Ethoxide/Ethanol)
20 halide. thus, stronger and bulkier the base more is the elimination product. thus , alkene is formed.
3) the equilibrium cn be maintained only at the boiling point as the constant temperature.
