1. Any tertiary alkyl halide is taken and reacted with aq NaOH in i>H2O ii>DMSO.The reactions undergone are
a. SN2,SN2
B. SN1,SN1
c. SN1,SN2
d. SN2,SN1
Multi Ans Correct ---
2. In conversion of EAA to 3-methyl-2-pentanol
a> Et-Br is first introduced to carbanion of EAA
b> Me-Br is first introduced to carbanion of EAA
c> Hydrolysis is carried out in dilute medium
d> Hydrolysis is carried out in conc. medium.
3. Which of the statements is/are correct ??
a> CO has larger bond length than CO2
b> Sulphonic acids evolve CO2 with NaHCO3
c> Sulphonic acids evolve SO2 with NaHCO3
d>Acetophenone gives iodoform test.
39H2O is a polar protic solvent, and DMSO(dimethyl sulfoxide) is a weakly polar aprotic solvent. Polar protic solvents support SN1 and tertiary substrates provide that.
In polar aprotic solvents, complete solvation does not take place. The cation is solvated but the anion remains free. It would support SN2 but tertiary substrate will not give SN2. It would rather give 90% elimination and 10% substitution. So c) is a possible answer for first where substitution is in poor yield in DMSO.
What's EAA?
Triple bonds are shorter than double bonds...so CO will have shorter bond length than CO2.
Acetophenone, having COCH3 group, does give yellow ppt in iodoform test.
Any acid evolves carbon dioxide on treatment with sodium bicarbonate.
So b), d) options for that.
11: b
reason: as it is a tertiary halide hence the carbocation formed will be higly stable moreover there is steric hinderance for SN2 to take place.
However ideally the reaction is with NaOH which is a strong base hence alpha beta elimination would be more probable than SN2
39Bhai ishan in DMSO, the charge density on sulphur is less due to two methyl groups lessening the positive charge. Hence it cannot solvate anion of NaOH, OH- is free. When the base is in such a form, it prefers a simple(bimolecular) mechanism as opposed to a complex(unimolecular) mechanism. SN1 toh hoga hi nahi. SN2 bohot kam hoga aur E2 will be the pathway followed.
