Alcohols, Phenols and Ethers Quiz-5
Dear Readers,
JEE Advanced is a high level exam that checks your concept by putting various types of questions. The questions are in a different format from JEE mains.m, herein the questions are : single correct multiple choice questions, multiple correct multiple choice questions, statement based questions and comprehension type questions..
Q1.
Wine with 24 proof is……by volume of alcohol
Wine with 24 proof is……by volume of alcohol
Solution
(a)
(a)
Q2.
Solution
(c)
(c)
Q3. 
Solution
a
a
Q4.
When ethanol is heated with conc. H2 SO4 at 140°C(413K), the product obtained as:
Solution
(c)
(c)
Q5.
Which of the following is oxetene?
Which of the following is oxetene?
Solution
b
b
Q6. When glycerol is treated with a small amount of HI orPI3, the product obtained is:
Solution
b
b
Q7.
Solution
(c) EDG favours SN1 reactivity with HBr (The positive charge that develops on benzylic C in this SN^1reaction is most effectively delocalised by (OMe) group (+R and –I). The (p-NO2) group withdraws e ̅ density from the ring by –I and –R(p-Cl) group withdraws e ̅s’ from the ring by –I only. More EDG (e.g-OMe)stabilies positive chargeon the benzyl C atom, whereas EWG (e.g,-NO2,-Cl) destabilises the positive charge. Hence, the reactivity order is:I>II>III>IV). EWG retards SN^1 reactivity with HBr. ED effect of (-OMe) (+R,-I) is greater than (Me) group (+I and H.C.). So, reactivity of III>II. In (IV), (-OMe) group is at m-position and exerts only –I effect and so reactivity of (IV) is less than (I) So the reactivity order is : III>II>I>IV
(c) EDG favours SN1 reactivity with HBr (The positive charge that develops on benzylic C in this SN^1reaction is most effectively delocalised by (OMe) group (+R and –I). The (p-NO2) group withdraws e ̅ density from the ring by –I and –R(p-Cl) group withdraws e ̅s’ from the ring by –I only. More EDG (e.g-OMe)stabilies positive chargeon the benzyl C atom, whereas EWG (e.g,-NO2,-Cl) destabilises the positive charge. Hence, the reactivity order is:I>II>III>IV). EWG retards SN^1 reactivity with HBr. ED effect of (-OMe) (+R,-I) is greater than (Me) group (+I and H.C.). So, reactivity of III>II. In (IV), (-OMe) group is at m-position and exerts only –I effect and so reactivity of (IV) is less than (I) So the reactivity order is : III>II>I>IV
Q8.In CH_3 CH_2 OH, the bond that undergoes heterolytic cleavage most readily is:
Solution
(d) Reaction of alcohol with Na indicates that (O-H) bond is most readly broken (2ROH+2Na→2RONa+H2)
(d) Reaction of alcohol with Na indicates that (O-H) bond is most readly broken (2ROH+2Na→2RONa+H2)
Q9.The intestinal antiseptic Salol is correctly represented as:
Solution
(b) Salol is phenyl salicylate
(b) Salol is phenyl salicylate
Q10.
Which is the best method for the conversion of (A) pentan-3-ol to 3- bromopentane (B)?
Which is the best method for the conversion of (A) pentan-3-ol to 3- bromopentane (B)?
-
-
-
- Both (a) and (b)
Solution
(d) Method (c) would give rearranged product also. It would give a mixture of 2-bromo and 3-bromo pentane. In methods (a) and (b), no rearrangement occurs and it gives (B) exclusively. The tosyl group, a good leaving group, is then easily displaced by reaction with Br⊖ in an SN2 reaction
(d) Method (c) would give rearranged product also. It would give a mixture of 2-bromo and 3-bromo pentane. In methods (a) and (b), no rearrangement occurs and it gives (B) exclusively. The tosyl group, a good leaving group, is then easily displaced by reaction with Br⊖ in an SN2 reaction
