Hi, I already answered those questions. I need someone correct me if I\'m wrong
ID: 1044916 • Letter: H
Question
Hi, I already answered those questions. I need someone correct me if I'm wrong and provide me the right answer for those questions that I'm wrong. Please just indicate the answer I don't need explanation.
***PLEASE CHECk ALL 41 QUESTIONS AND DONT ANSWER PARTIAL, I WILL RATE YOU WITH ALL MY 3 ACCOUNTS. THANKS
1. Which of the following carbon substrates will form the most stable carbocation (hint: remember what the substrate can do after the leaving group leaves) CI ci 2. Which of the following carbon substrates will form the most stable carbocation (hint: remember what the substrate can do after the leaving group leaves) CI A. C.Explanation / Answer
1.
(B) After hydride shift it will produce tertiary carbocation
2.
(B) After methide shift it will produce tertiary carbo cation.
3.
(B) SN2 Primary alkyl halides are readily undergo SN2 reaction.
4.
(A) SN2 reaction mechanism is concerted mechanism
5.
(B) Strong nucleophile is needed for SN2 reactions
6.
(B) Polar aprotic
7.
(A) Back side attack takes place to result a product inverted configutration
8.
(B) No . A single product is formed
9.
(B) NO . Because no intermediate is formed no rearrangement takes place.
10.
(B)
11.
(A) SN1 . Because tertiary alkyl halides readily undergo this mechanism
12.
(B) It is a two step mechanism
13.
(B) In general it requires weak nucleophile
14.
(A) Polar protic solvent is preferred to stabilise the intermediate carbocation.
15.
(B) Both sides
16.
(B) There is no chiral cneter.
17.
(B) It already gives stable tertiary carbocation.
18.
(A) Only carbon substrate. Becase the first step involing the cleavage of C - X bond is slow and rate determining. In this step only carbon substrate involves.
19.
(A)
20.
(B)
21. (B)
22. (A)
23 (B)
25. (A) Yes. Rearrangement takes place to give stable tertiary carbocation.
26. (A) Just on the carbon substrate
27. (B) Not good because F- most stable than CL-
28 . (B) No rearragement takes place
29. (A) GOOD
30 (B) NO rearrangement takes place.
31. (B) NOt good it gives mixuture of products.
32. (A) Yes to form the stable tertiary carbocation intermediate
33. (B) NOt good
34.
(A)
35. (A) Good
35. (B)
36. (B)
37. (B) Not good it gives mixture of products
38. (A) Yes. TO form stable tertiary carbocation intermediate
39. C. due to the presence of - OH bond which is respomsible to form intermolecular hydrogen bond.
40. C. 3. All carbonyl compounds ( having - CO - group) show IR line around 1700 cm-1
41. (A)
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